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https://tohoku.pure.elsevier.com/en/publications/x-ray-irradiation-effects-on-the-superconductive-properties-of-yb | # X-ray irradiation effects on the superconductive properties of YBa2Cu3Oy and GdBa2Cu3Oz
Research output: Contribution to journalArticlepeer-review
1 Citation (Scopus)
## Abstract
We observed a metastable change in the superconducting properties of YBa2Cu3Oy (YBCO) and GdBa2Cu3Oz (GdBCO) induced by X-ray irradiation. Tc0, which is defined as the temperature below which the electric resistance is zero within the measurement accuracy, was increased by up to 1.9 K in YBCO and GdBCO. The irradiation effect was more pronounced for the YBCO sample with lower Tc0. These observations are consistent with those of light irradiation, and are attributable to hole doping in the CuO2 plane via the generation of electron-hole pairs by the X-ray and subsequent trapping of the electrons at oxygen vacancies. The relaxation of the irradiation effect was observed to be within 100 h from the viewpoint of conductivity and Raman spectra, and is explained as the detrapping of the electrons and recombination with holes, similarly to the case of light irradiation.
Original language English 1465-1470 6 Sensors and Materials 29 10 https://doi.org/10.18494/SAM.2017.1627 Published - 2017
## Keywords
• Photo-doping
• Superconductor | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9092316627502441, "perplexity": 3349.0617814061347}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304287.0/warc/CC-MAIN-20220123141754-20220123171754-00052.warc.gz"} |
http://mathhelpforum.com/differential-geometry/175736-inifinite-product-vector-spaces-print.html | # On inifinite product of vector spaces
• March 24th 2011, 06:43 PM
bkarpuz
On inifinite product of vector spaces
Dear MHF members,
I have the following problem.
Problem. For a family of Banach spaces $\{A_{n}\}_{n\in\mathbb{N}}$, let $A_{\Pi}:=\prod_{k\in\mathbb{N}}A_{k}=\big\{a=(a_{1 },a_{2},\ldots,a_{n},\ldots):\ \|a\|<\infty\big\}$,
where $\|a\|:=\sup_{n\in\mathbb{N}}\|a_{n}\|$.
1. Show that $A_{\oplus}:=\oplus_{k\in\mathbb{N}}A_{k}=\{a=(a_{1 },a_{2},\ldots,a_{n},\ldots):\ \lim_{n\to\infty}\|a_{n}\|=0\}$ is the closure of the algebraic direct sum $A_{\Sigma}$ in the Banach space $A_{\Pi}$.
2. If $\pi$ denotes the quotient map onto the quotient space $A_{\Pi}/A_{\oplus}$, then show that the norm $\|\pi(a)\|=\limsup_{n\to\infty}\|a_{n}\|$.
Thanks for being interested.
bkarpuz
• March 24th 2011, 07:14 PM
Drexel28
Quote:
Originally Posted by bkarpuz
Dear MHF members,
I have the following problem.
Problem. For a family of Banach spaces $\{A_{n}\}_{n\in\mathbb{N}}$, let $A_{\Pi}:=\prod_{k\in\mathbb{N}}A_{k}=\big\{a=(a_{1 },a_{2},\ldots,a_{n},\ldots):\ \|a\|<\infty\big\}$,
where $\|a\|:=\sup_{n\in\mathbb{N}}\|a_{n}\|$.
1. Show that $A_{\oplus}:=\oplus_{k\in\mathbb{N}}A_{k}=\{a=(a_{1 },a_{2},\ldots,a_{n},\ldots):\ \lim_{n\to\infty}\|a_{n}\|=0\}$ is the closure of the algebraic direct sum $A_{\Sigma}:=\sum_{k\in\mathbb{N}}A_{k}$ in the Banach space $A_{\Pi}$.
2. If $\pi$ denotes the quotient map onto the quotient space $A_{\Pi}/A_{\oplus}$, then show that the norm $\|\pi(a)\|=\limsup_{n\to\infty}\|a_{n}\|$.
Thanks for being interested.
bkarpuz
Presumably you mean that you're considering $A_k$ embedded into $A_\Pi$ in the natural way (all coordinates zero, except the $k^{\text{th}}$) and by the 'algebraic direct sum' $A_\Sigma$ you mean $\displaystyle \left\{\sum_{k\in S}A_k:S\subseteq\mathbb{N}\text{ and }\#(S)<\infty\right\}$. If so, let's show that $\overline{A_\Sigma}=A_\oplus$. To see that $A_\oplus\subseteq\overline{A_\Sigma}$ let $(a_n)_{n\in\mathbb{N}}\in A_\oplus$. Then, for any $\varepsilon>0$ we have, since $\|a_n\|_n\to 0$ there exists some $N\in\mathbb{N}$ such that $n\geqslant N\implies \|a_n\|_n<\varepsilon$. Define then $(b_n)_{n\in\mathbb{N}}$ by
$b_n=\begin{cases}a_n & \mbox{if}\quad n\leqslant 0\\ 0 & \mbox{if}\quad n>N\end{cases}$
Evidently then $b_n\in A_\Sigma$ and moreover $\displaystyle \|a_n-b_n\|=\sup_{\substack{n\in\mathbb{N}\\ n>N}}\|a_n\|\leqslant \varepsilon$. Since $\varepsilon$ was arbitrary the conclusion follows (technically you have to adjust the above argument so that $b_n\ne a_n$ but that's easy enough).
Conversely, suppose that $(a_n)_{n\in\mathbb{N}}\notin A_\oplus$. Then, there exists some $\varepsilon>0$ such that for every $N\in\mathbb{N}$ there exists some $N'\geqslant N$ such that $\|a_{N'}\|_{N'}\geqslant \varepsilon$. Evidently this implies that for every $N\in\mathbb{N}$ one has that $\displaystyle \sup_{\substack{n\in\mathbb{N}\\ n\geqslant N}}\|a_n\|_{n}\geqslant \varepsilon$. So, let $(b_n)_{n\in\mathbb{N}}\in A_\Sigma$ be arbitrary. Then, by definition there exists some $N\in\mathbb{N}$ such that for every $n\geqslant N$ one has that $b_n=0$. Consequently
\displaystyle \begin{aligned}\|(a_n)-(b_n)\| &=\sup_{n\in\mathbb{N}}\|a_n-b_n\|_n\\ &\geqslant \sup_{\substack{n\in\mathbb{N}\\ n\geqslant N}}\|a_n-b_n\|_n\\ &= \sup_{\substack{n\in\mathbb{N}\\ n\geqslant N}}\|a_n\|_n\\ &\geqslant \varepsilon\end{aligned}
Thus, $(b_n)\notin B_{\varepsilon}((a_n))$. Since $(b_n)\in A_\Sigma$ was arbitrary it follows that $(a_n)\notin \overline{A_\Sigma}$.
That should give you the basic idea...I might have skipped a minor detail in the case $(a_n)\in A_\Sigma$ but it's easily fixable. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 52, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9993522763252258, "perplexity": 228.0976678463537}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982293692.32/warc/CC-MAIN-20160823195813-00264-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://www.lessonplanet.com/teachers/graphing-a-number-sequence | ## Graphing a Number Sequence
In this number sequence worksheet, students solve and graph 4 different sets of number sequences on the graph at the bottom of the sheet. They graph each number sequence given in the color above each table at the top.
Subjects
Math
3 more...
Resource Types
Worksheets
2 more...
Language
English
#### What Members Say
Lesson Planet helps me search for many lesson plans.
Kathy N., Teacher | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8808350563049316, "perplexity": 4355.674462786434}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719139.8/warc/CC-MAIN-20161020183839-00007-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.mathematik.rwth-aachen.de/cms/mathematik/Forschung/Publikationen/Bibliographie/~ckus/Einzelansicht/?file=721123&lidx=1 | # Stable and convergent discontinuous galerkin methods for hyperbolic and viscous systems of conservation laws
Despite the classical well-posedness theorem for entropy weak solutions of scalar conservation laws, some theoretical and numerical evidence cast doubt on the appropriateness of this solution paradigm for multidimensional hyperbolic systems. It has been conjectured that the more general entropy measure-valued (EMV) solutions ought to be considered as the adequate notion of the solution. Building on previous results, we prove that bounded solutions of a certain class of space-time discontinuous Galerkin (DG) schemes converge to an EMV solution. The novelty in our work is that no streamline-diffusion terms are used for stabilization, in contrary to the main role of such stabilizations in the existing analysis of DG schemes. Our approach conforms to the way DG schemes were originally proposed, and are most often used in practiceIn the case of scalar problems, this result is strengthened to obtain the convergence to the entropy weak solution, via the proof of $L_\infty$-boundedness of the solution as well as its consistency with all entropy inequalities. As a main step in the boundedness proof, we show the coercivity of the shock-capturing operator employing new arguments from polynomial inequalities. For viscous conservation laws, we extend our framework to general convection-diffusion systems, with both nonlinear convection and nonlinear diffusion, such that the entropy stability of the scheme is preserved. Starting from a mixed formulation, we handle the difficulties arising from the nonlinearity of the viscous flux by an additional projection. We prove the entropy stability of the corresponding primal form for different treatments of the viscous flux; thus unifying the existing results in the literature as well as establishing the entropy stability for less-analyzed methods. Our analysis is also valid for the case of degenerate diffusion. Considering quasilinear elliptic problems in scalar settings, we prove that the proposed approach for viscous discretization is asymptotically consistent and adjoint consistent. For the special case of strongly monotone and globally Lipschitz problems, we prove the uniqueness and stability of the numerical solution. For this class of operators, we also provethe optimal convergence to the exact solution with respect to mesh size, in both energy and $L_2$ norms. Such optimal convergence rates for asymptotically (adjoint) consistent schemes have been observed before in numerical experiments. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9348024129867554, "perplexity": 308.48313320658787}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103983398.56/warc/CC-MAIN-20220702010252-20220702040252-00150.warc.gz"} |
http://cp3-origins.dk/a/18416 | The elastic $$I=1/2$$, $$s$$- and $$p$$-wave kaon-pion scattering amplitudes are calculated using a single ensemble of anisotropic lattice QCD gauge field configurations with $$N_{\mathrm{f}} = 2+1$$ flavors of dynamical Wilson-clover fermions at $$m_{\pi} = 230\mathrm{MeV}$$. A large spatial extent of $$L = 3.7\mathrm{fm}$$ enables a good energy resolution while partial wave mixing due to the reduced symmetries of the finite volume is treated explicitly.The $$p$$-wave amplitude is well described by a Breit-Wigner shape with parameters $$m_{K^{*}}/m_{\pi} = 3.808(18)$$ and $$g^{\mathrm{BW}}_{K^{*}K\pi} = 5.33(20)$$ which are insensitive to the inclusion of $$d$$-wave mixing and variation of the $$s$$-wave parametrization. An effective range description of the near-threshold $$s$$-wave amplitude yields $$m_{\pi}a_0 = -0.353(25)$$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9493858218193054, "perplexity": 724.0340003241388}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948119.95/warc/CC-MAIN-20180426070605-20180426090605-00405.warc.gz"} |
https://www.physicsforums.com/threads/studying-for-exam-need-help.52708/ | # Studying for exam, need help
1. Nov 14, 2004
### johnnyICON
Studying for exam, need help!!
Hi can someone help me out. My girlfriend has an exam tomorrow and she got stuck on this question. Her professor decided not to give out any solutions she's not too sure if she is heading in the right direction. Any help would be great, thank you in advance.
Here is the question:
Prove by using the definition of the limit of a sequence that:
$$\lim_{n \to \infty} \frac{n + 1}{n^2} + 3 = 3$$
Last edited: Nov 14, 2004
2. Nov 14, 2004
### mattmns
split up the fraction
$$\frac{n}{n^2} + \frac{1}{n^2}$$
3. Nov 15, 2004
### johnnyICON
How do I get just one term of n?
4. Nov 15, 2004
### matt grime
You must show that given e>0 there is an N such that n>N implies
$$\frac{n+1}{n^2}<e$$
agreed?
Well,
$$\frac{n+1}{n^2}<\frac{n+1}{(n+1)^2} = \frac{1}{n+1}$$
so pick N such that N+1>1/e
5. Nov 15, 2004
### johnnyICON
Second question, if you could get back to me asap, we're at school cramming right now
Let$$a_n = \frac{n^2-1}{2n^2+3}$$ Prove by using the definition of the limit of a sequence that $$\lim_{n \to \infty}a_n = \frac{1}{2}$$
6. Nov 15, 2004
### matt grime
Well, have you simplified a_n -1/2?
every question like this reduces to showing something tends to zero. that thing tends to zero for obvious reasons just like the previous example
7. Nov 15, 2004
### johnnyICON
I just have $$|\frac{n^2-1}{2n^2+3} - \frac{1}{2}|$$ and I don't know how to get it to a single n term.
8. Nov 15, 2004
### johnnyICON
Thanks for your help anyway. We gotta go now.
Similar Discussions: Studying for exam, need help | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8415725827217102, "perplexity": 1513.0905150185574}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189495.77/warc/CC-MAIN-20170322212949-00342-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://wikimili.com/en/Newton's_cradle | Last updated
The Newton's cradle is a device that demonstrates the conservation of momentum and the conservation of energy with swinging spheres. When one sphere at the end is lifted and released, it strikes the stationary spheres, transmitting a force through the stationary spheres that pushes the last sphere upward. The last sphere swings back and strikes the nearly stationary spheres, repeating the effect in the opposite direction. The device is named after 17th-century English scientist Sir Isaac Newton and designed by French scientist Edme Mariotte. It is also known as Newton's pendulum, Newton's balls, Newton's rocker or executive ball clicker (since the device makes a click each time the balls collide, which they do repeatedly in a steady rhythm). [1] [2]
## Operation
When one of the end balls ("the first") is pulled sideways, the attached string makes it follow an upward arc. When it is let go, it strikes the second ball and comes to nearly a dead stop. The ball on the opposite side acquires most of the velocity of the first ball and swings in an arc almost as high as the release height of the first ball. This shows that the last ball receives most of the energy and momentum of the first ball. The impact produces a compression wave that propagates through the intermediate balls. Any efficiently elastic material such as steel does this, as long as the kinetic energy is temporarily stored as potential energy in the compression of the material rather than being lost as heat. There are slight movements in all the balls after the initial strike but the last ball receives most of the initial energy from the impact of the first ball. When two (or three) balls are dropped, the two (or three) balls on the opposite side swing out. Some say that this behavior demonstrates the conservation of momentum and kinetic energy in elastic collisions. However, if the colliding balls behave as described above with the same mass possessing the same velocity before and after the collisions, then any function of mass and velocity is conserved in such an event. [3]
## Physics explanation
Newton's cradle can be modeled fairly accurately with simple mathematical equations with the assumption that the balls always collide in pairs. If one ball strikes four stationary balls that are already touching, these simple equations can not explain the resulting movements in all five balls, which are not due to friction losses. For example, in a real Newton's cradle the fourth has some movement and the first ball has a slight reverse movement. All the animations in this article show idealized action (simple solution) that only occurs if the balls are not touching initially and only collide in pairs.
### Simple solution
The conservation of momentum (mass × velocity) and kinetic energy (1/2 × mass × velocity2) can be used to find the resulting velocities for two colliding perfectly elastic objects. These two equations are used to determine the resulting velocities of the two objects. For the case of two balls constrained to a straight path by the strings in the cradle, the velocities are a single number instead of a 3D vector for 3D space, so the math requires only two equations to solve for two unknowns. When the two objects weigh the same, the solution is simple: the moving object stops relative to the stationary one and the stationary one picks up all the other's initial velocity. This assumes perfectly elastic objects, so there is no need to account for heat and sound energy losses.
Steel does not compress much, but its elasticity is very efficient, so it does not cause much waste heat. The simple effect from two same-weight efficiently elastic colliding objects constrained to a straight path is the basis of the effect seen in the cradle and gives an approximate solution to all its activities.
For a sequence of same-weight elastic objects constrained to a straight path, the effect continues to each successive object. For example, when two balls are dropped to strike three stationary balls in a cradle, there is an unnoticed but crucial small distance between the two dropped balls, and the action is as follows: the first moving ball that strikes the first stationary ball (the second ball striking the third ball) transfers all its velocity to the third ball and stops. The third ball then transfers the velocity to the fourth ball and stops, and then the fourth to the fifth ball. Right behind this sequence is the second moving ball transferring its velocity to the first moving ball that just stopped, and the sequence repeats immediately and imperceptibly behind the first sequence, ejecting the fourth ball right behind the fifth ball with the same small separation that was between the two initial striking balls. If they are simply touching when they strike the third ball, precision requires the more complete solution below.
#### Other examples of this effect
The effect of the last ball ejecting with a velocity nearly equal to the first ball can be seen in sliding a coin on a table into a line of identical coins, as long as the striking coin and its twin targets are in a straight line. The effect can similarly be seen in billiard balls. The effect can also be seen when a sharp and strong pressure wave strikes a dense homogeneous material immersed in a less-dense medium. If the identical atoms, molecules, or larger-scale sub-volumes of the dense homogeneous material are at least partially elastically connected to each other by electrostatic forces, they can act as a sequence of colliding identical elastic balls. The surrounding atoms, molecules, or sub-volumes experiencing the pressure wave act to constrain each other similarly to how the string constrains the cradle's balls to a straight line. For example, lithotripsy shock waves can be sent through the skin and tissue without harm to burst kidney stones. The side of the stones opposite to the incoming pressure wave bursts, not the side receiving the initial strike.
#### When the simple solution applies
For the simple solution to precisely predict the action, no pair in the midst of colliding may touch the third ball, because the presence of the third ball effectively makes the struck ball appear heavier. Applying the two conservation equations to solve the final velocities of three or more balls in a single collision results in many possible solutions, so these two principles are not enough to determine resulting action.
Even when there is a small initial separation, a third ball may become involved in the collision if the initial separation is not large enough. When this occurs, the complete solution method described below must be used.
Small steel balls work well because they remain efficiently elastic with little heat loss under strong strikes and do not compress much (up to about 30 μm in a small Newton's cradle). The small, stiff compressions mean they occur rapidly, less than 200 microseconds, so steel balls are more likely to complete a collision before touching a nearby third ball. Softer elastic balls require a larger separation to maximize the effect from pair-wise collisions.
### More complete solution
A cradle that best follows the simple solution needs to have an initial separation between the balls that measures at least twice the amount that any one ball compresses, but most do not. This section describes the action when the initial separation is not enough and in subsequent collisions that involve more than two balls even when there is an initial separation. This solution simplifies to the simple solution when only two balls touch during a collision. It applies to all perfectly elastic identical balls that have no energy losses due to friction and can be approximated by materials such as steel, glass, plastic, and rubber.
For two balls colliding, only the two equations for conservation of momentum and energy are needed to solve the two unknown resulting velocities. For three or more simultaneously colliding elastic balls, the relative compressibilities of the colliding surfaces are the additional variables that determine the outcome. For example, five balls have four colliding points and scaling (dividing) three of them by the fourth gives the three extra variables needed to solve for all five post-collision velocities.
Newtonian, Lagrangian, Hamiltonian, and stationary action are the different ways of mathematically expressing classical mechanics. They describe the same physics but must be solved by different methods. All enforce the conservation of energy and momentum. Newton's law has been used in research papers. It is applied to each ball and the sum of forces is made equal to zero. So there are five equations, one for each ball—and five unknowns, one for each velocity. If the balls are identical, the absolute compressibility of the surfaces becomes irrelevant, because it can be divided out of both sides of all five equations, producing zero.
Determining the velocities [4] [5] [6] for the case of one ball striking four initially-touching balls is found by modeling the balls as weights with non-traditional springs on their colliding surfaces. Most materials, like steel, that are efficiently elastic approximately follow Hooke's force law for springs, ${\displaystyle F=k\cdot x}$, but because the area of contact for a sphere increases as the force increases, colliding elastic balls follow Hertz's adjustment to Hooke's law, ${\displaystyle F=k\cdot x^{1.5}}$. This and Newton's law for motion (${\displaystyle F=m\cdot a}$) are applied to each ball, giving five simple but interdependent differential equations that are solved numerically. When the fifth ball begins accelerating, it is receiving momentum and energy from the third and fourth balls through the spring action of their compressed surfaces. For identical elastic balls of any type with initially touching balls, the action is the same for the first strike, except the time to complete a collision increases in softer materials. 40% to 50% of the kinetic energy of the initial ball from a single-ball strike is stored in the ball surfaces as potential energy for most of the collision process. Thirteen percent of the initial velocity is imparted to the fourth ball (which can be seen as a 3.3-degree movement if the fifth ball moves out 25 degrees) and there is a slight reverse velocity in the first three balls, the first ball having the largest at −7% of the initial velocity. This separates the balls, but they come back together just before as the fifth ball returns. This is due to the pendulum phenomenon of different small angle disturbances having approximately the same time to return to the center. When balls are "touching" in subsequent collisions is complex, but still determinable by this method, especially if friction losses are included and the pendulum timing is calculated exactly instead of relying on the small angle approximation.
The differential equations with the initial separations are needed if there is less than 10 μm separation when using 100-gram steel balls with an initial 1 m/s strike speed.
The Hertzian differential equations predict that if two balls strike three, the fifth and fourth balls will leave with velocities of 1.14 and 0.80 times the initial velocity. [7] This is 2.03 times more kinetic energy in the fifth ball than the fourth ball, which means the fifth ball would swing twice as high in the vertical direction as the fourth ball. But in a real Newton's cradle, the fourth ball swings out as far as the fifth ball. To explain the difference between theory and experiment, the two striking balls must have at least ≈10 μm separation (given steel, 100 g, and 1 m/s). This shows that in the common case of steel balls, unnoticed separations can be important and must be included in the Hertzian differential equations, or the simple solution gives a more accurate result.
### Effect of pressure waves
The forces in the Hertzian solution above were assumed to propagate in the balls immediately, which is not the case. Sudden changes in the force between the atoms of material build up to form a pressure wave. Pressure waves (sound) in steel travel about 5 cm in 10 microseconds, which is about 10 times faster than the time between the first ball striking and the last ball being ejected. The pressure waves reflect back and forth through all five balls about ten times, although dispersing to less of a wavefront with more reflections. This is fast enough for the Hertzian solution to not require a substantial modification to adjust for the delay in force propagation through the balls. In less-rigid but still very elastic balls such as rubber, the propagation speed is slower, but the duration of collisions is longer, so the Hertzian solution still applies. The error introduced by the limited speed of the force propagation biases the Hertzian solution towards the simple solution because the collisions are not affected as much by the inertia of the balls that are further away.
Identically-shaped balls help the pressure waves converge on the contact point of the last ball: at the initial strike point one pressure wave goes forward to the other balls while another goes backward to reflect off the opposite side of the first ball, and then it follows the first wave, being exactly 1 ball-diameter behind. The two waves meet up at the last contact point because the first wave reflects off the opposite side of the last ball and it meets up at the last contact point with the second wave. Then they reverberate back and forth like this about 10 times until the first ball stops connecting with the second ball. Then the reverberations reflect off the contact point between the second and third balls, but still converge at the last contact point, until the last ball is ejected—but it is less of a wavefront with each reflection.
### Effect of different types of balls
Using different types of material does not change the action as long as the material is efficiently elastic. The size of the spheres does not change the results unless the increased weight exceeds the elastic limit of the material. If the solid balls are too large, energy is being lost as heat, because the elastic limit increases with the radius raised to the power 1.5, but the energy which had to be absorbed and released increases as the cube of the radius. Making the contact surfaces flatter can overcome this to an extent by distributing the compression to a larger amount of material but it can introduce an alignment problem. Steel is better than most materials because it allows the simple solution to apply more often in collisions after the first strike, its elastic range for storing energy remains good despite the higher energy caused by its weight, and the higher weight decreases the effect of air resistance.
## Uses
The most common application is that of a desktop executive toy. Another use is as an educational physics demonstration, as an example of conservation of momentum and conservation of energy. A similar principle, the propagation of waves in solids, was used in the Constantinesco Synchronization gear system for propeller / gun synchronizers on early fighter aircraft.[ further explanation needed ]
## History
Christiaan Huygens used pendulums to study collisions. His work, De Motu Corporum ex Percussione (On the Motion of Bodies by Collision) published posthumously in 1703, contains a version of Newton's first law and discusses the collision of suspended bodies including two bodies of equal mass with the motion of the moving body being transferred to the one at rest.
The principle demonstrated by the device, the law of impacts between bodies, was first demonstrated by the French physicist Abbé Mariotte in the 17th century. [1] [8] Newton acknowledged Mariotte's work, among that of others, in his Principia .
There is much confusion over the origins of the modern Newton's cradle. Marius J. Morin has been credited as being the first to name and make this popular executive toy.[ citation needed ] However, in early 1967, an English actor, Simon Prebble, coined the name "Newton's cradle" (now used generically) for the wooden version manufactured by his company, Scientific Demonstrations Ltd. [9] After some initial resistance from retailers, they were first sold by Harrods of London, thus creating the start of an enduring market for executive toys.[ citation needed ] Later a very successful chrome design for the Carnaby Street store Gear was created by the sculptor and future film director Richard Loncraine.[ citation needed ]
The largest cradle device in the world was designed by MythBusters and consisted of five one-ton concrete and steel rebar-filled buoys suspended from a steel truss. [10] The buoys also had a steel plate inserted in between their two-halves to act as a "contact point" for transferring the energy; this cradle device did not function well because concrete is not elastic so most of the energy was lost to a heat buildup in the concrete. A smaller scale version constructed by them consists of five 15-centimetre (6 in) chrome steel ball bearings, each weighing 15 kilograms (33 lb), and is nearly as efficient as a desktop model.
The cradle device with the largest diameter collision balls on public display was visible for more than a year in Milwaukee, Wisconsin, at the retail store American Science and Surplus (see photo). Each ball was an inflatable exercise ball 66 cm (26 in) in diameter (encased in steel rings), and was supported from the ceiling using extremely strong magnets. It was dismantled in early August 2010 due to maintenance concerns.[ citation needed ]
Newton's cradle has been used more than 20 times in movies, [11] often as a trope on the desk of a lead villain such as Paul Newman's role in The Hudsucker Proxy, Magneto in X-Men, and the Kryptonians in Superman II. It was used to represent the unyielding position of the NFL towards head injuries in Concussion. [12] It has also been used as a relaxing diversion on the desk of lead intelligent/anxious/sensitive characters such as Henry Winkler's role in Night Shift, Dustin Hoffman's role in Straw Dogs, and Gwyneth Paltrow's role in Iron Man 2. It was featured more prominently as a series of clay pots in Rosencrantz and Guildenstern Are Dead , and as a row of 1968 Eero Aarnio bubble chairs with scantily-clad women in them in Gamer. [13] In Storks , Hunter the CEO of Cornerstore has one not with balls, but with little birds. Newton’s Cradle is an item in Nintendo’s Animal Crossing where it is referred to as “executive toy”. [14] In 2017, an episode of the Omnibus podcast, featuring Jeopardy! champion Ken Jennings and musician John Roderick, focused on the history of Newton's Cradle. [15] Newton's cradle is also featured on the desk of Deputy White House Communications Director Sam Seaborn in The West Wing .
Rock band Jefferson Airplane used the cradle on the 1968 album Crown of Creation as a rhythm device to create polyrhythms on an instrumental track.
## Related Research Articles
In physics and engineering, fluid dynamics is a subdiscipline of fluid mechanics that describes the flow of fluids—liquids and gases. It has several subdisciplines, including aerodynamics and hydrodynamics. Fluid dynamics has a wide range of applications, including calculating forces and moments on aircraft, determining the mass flow rate of petroleum through pipelines, predicting weather patterns, understanding nebulae in interstellar space and modelling fission weapon detonation.
In Newtonian mechanics, linear momentum, translational momentum, or simply momentum is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction. If m is an object's mass and v is its velocity, then the object's momentum is
In physics, mathematics, and related fields, a wave is a propagating dynamic disturbance of one or more quantities, sometimes as described by a wave equation. In physical waves, at least two field quantities in the wave medium are involved. Waves can be periodic, in which case those quantities oscillate repeatedly about an equilibrium (resting) value at some frequency. When the entire waveform moves in one direction it is said to be a traveling wave; by contrast, a pair of superimposed periodic waves traveling in opposite directions makes a standing wave. In a standing wave, the amplitude of vibration has nulls at some positions where the wave amplitude appears smaller or even zero.
In physics, a collision is any event in which two or more bodies exert forces on each other in a relatively short time. Although the most common use of the word collision refers to incidents in which two or more objects collide with great force, the scientific use of the term implies nothing about the magnitude of the force.
An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same. In an ideal, perfectly elastic collision, there is no net conversion of kinetic energy into other forms such as heat, noise, or potential energy.
An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved due to the action of internal friction.
In physics, equations of motion are equations that describe the behavior of a physical system in terms of its motion as a function of time. More specifically, the equations of motion describe the behavior of a physical system as a set of mathematical functions in terms of dynamic variables. These variables are usually spatial coordinates and time, but may include momentum components. The most general choice are generalized coordinates which can be any convenient variables characteristic of the physical system. The functions are defined in a Euclidean space in classical mechanics, but are replaced by curved spaces in relativity. If the dynamics of a system is known, the equations are the solutions for the differential equations describing the motion of the dynamics.
In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. Since force is a vector quantity, impulse is also a vector quantity. Impulse applied to an object produces an equivalent vector change in its linear momentum, also in the resultant direction. The SI unit of impulse is the newton second (N⋅s), and the dimensionally equivalent unit of momentum is the kilogram meter per second (kg⋅m/s). The corresponding English engineering unit is the pound-second (lbf⋅s), and in the British Gravitational System, the unit is the slug-foot per second (slug⋅ft/s).
In physics, action is a numerical value describing how a physical system has changed over time. Action is significant because the equations of motion of the system can be derived through the principle of stationary action. In the simple case of a single particle moving with a specified velocity, the action is the momentum of the particle times the distance it moves, added up along its path, or equivalently, twice its kinetic energy times the length of time for which it has that amount of energy, added up over the period of time under consideration. For more complicated systems, all such quantities are added together. More formally, action is a mathematical functional which takes the trajectory, also called path or history, of the system as its argument and has a real number as its result. Generally, the action takes different values for different paths. Action has dimensions of energy × time or momentum × length, and its SI unit is joule-second.
The Kerr metric or Kerr geometry describes the geometry of empty spacetime around a rotating uncharged axially-symmetric black hole with a quasispherical event horizon. The Kerr metric is an exact solution of the Einstein field equations of general relativity; these equations are highly non-linear, which makes exact solutions very difficult to find.
In solid-state physics crystal momentum or quasimomentum is a momentum-like vector associated with electrons in a crystal lattice. It is defined by the associated wave vectors of this lattice, according to
In general relativity, an exact solution is solution of the Einstein field equations whose derivation does not invoke simplifying assumptions, though the starting point for that derivation may be an idealized case like a perfectly spherical shape of matter. Mathematically, finding an exact solution means finding a Lorentzian manifold equipped with tensor fields modeling states of ordinary matter, such as a fluid, or classical non-gravitational fields such as the electromagnetic field.
The coefficient of restitution, is the ratio of the final to initial relative speed between two objects after they collide. It normally ranges from 0 to 1 where 1 would be a perfectly elastic collision. A perfectly inelastic collision has a coefficient of 0, but a 0 value does not have to be perfectly inelastic. It is measured in the Leeb rebound hardness test, expressed as 1000 times the COR, but it is only a valid COR for the test, not as a universal COR for the material being tested.
The physicist Sir Isaac Newton first developed this idea to get rough approximations for the impact depth for projectiles traveling at high velocities.
Soft-body dynamics is a field of computer graphics that focuses on visually realistic physical simulations of the motion and properties of deformable objects. The applications are mostly in video games and films. Unlike in simulation of rigid bodies, the shape of soft bodies can change, meaning that the relative distance of two points on the object is not fixed. While the relative distances of points are not fixed, the body is expected to retain its shape to some degree. The scope of soft body dynamics is quite broad, including simulation of soft organic materials such as muscle, fat, hair and vegetation, as well as other deformable materials such as clothing and fabric. Generally, these methods only provide visually plausible emulations rather than accurate scientific/engineering simulations, though there is some crossover with scientific methods, particularly in the case of finite element simulations. Several physics engines currently provide software for soft-body simulation.
Lamb waves propagate in solid plates or spheres. They are elastic waves whose particle motion lies in the plane that contains the direction of wave propagation and the plane normal. In 1917, the English mathematician Horace Lamb published his classic analysis and description of acoustic waves of this type. Their properties turned out to be quite complex. An infinite medium supports just two wave modes traveling at unique velocities; but plates support two infinite sets of Lamb wave modes, whose velocities depend on the relationship between wavelength and plate thickness.
A Galilean cannon is a device that demonstrates conservation of linear momentum. It comprises a stack of balls, starting with a large, heavy ball at the base of the stack and progresses up to a small, lightweight ball at the top. The basic idea is that this stack of balls can be dropped to the ground and almost all of the kinetic energy in the lower balls will be transferred to the topmost ball - which will rebound to many times the height from which it was dropped. At first sight, the behavior seems highly counter-intuitive, but in fact is precisely what conservation of momentum predicts. The principal difficulty is in keeping the configuration of the balls stable during the initial drop. Early descriptions involve some sort of glue/tape, tube, or net to align the balls.
The index of physics articles is split into multiple pages due to its size.
The physics of a bouncing ball concerns the physical behaviour of bouncing balls, particularly its motion before, during, and after impact against the surface of another body. Several aspects of a bouncing ball's behaviour serve as an introduction to mechanics in high school or undergraduate level physics courses. However, the exact modelling of the behaviour is complex and of interest in sports engineering.
## References
1. "Newton's Cradle". Harvard Natural Sciences Lecture Demonstrations. Harvard University. 27 February 2019.
2. Palermo, Elizabeth (28 August 2013). "How Does Newton's Cradle Work?". Live Science.
3. Gauld, Colin F. (August 2006). "Newton's Cradle in Physics Education". Science & Education. 15 (6): 597–617. Bibcode:2006Sc&Ed..15..597G. doi:10.1007/s11191-005-4785-3. S2CID 121894726.
4. Herrmann, F.; Seitz, M. (1982). "How does the ball-chain work?" (PDF). American Journal of Physics. 50. pp. 977–981. Bibcode:1982AmJPh..50..977H. doi:10.1119/1.12936.
5. Lovett, D. R.; Moulding, K. M.; Anketell-Jones, S. (1988). "Collisions between elastic bodies: Newton's cradle". European Journal of Physics. 9 (4): 323. Bibcode:1988EJPh....9..323L. doi:10.1088/0143-0807/9/4/015.
6. Hutzler, Stefan; Delaney, Gary; Weaire, Denis; MacLeod, Finn (2004). "Rocking Newton's Cradle" (PDF). American Journal of Physics. 72. pp. 1508–1516. Bibcode:2004AmJPh..72.1508H. doi:10.1119/1.1783898.C F Gauld (2006), Newton's cradle in physics education, Science & Education, 15, 597–617
7. Hinch, E.J.; Saint-Jean, S. (1999). "The fragmentation of a line of balls by an impact" (PDF). Proc. R. Soc. Lond. A. 455. pp. 3201–3220.
8. Schulz, Chris (17 January 2012). "How Newton's Cradles Work". HowStuffWorks. Retrieved 27 February 2019.
9. Concussion – Cinemaniac Reviews Archived 11 February 2017 at the Wayback Machine
10. Animal Crossing Official Game Guide by Nintendo Power. Nintendo.
## Literature
• Herrmann, F. (1981). "Simple explanation of a well-known collision experiment". American Journal of Physics. 49 (8): 761. Bibcode:1981AmJPh..49..761H. doi:10.1119/1.12407.
• B. Brogliato: Nonsmooth Mechanics. Models, Dynamics and Control, Springer, 2nd Edition, 1999. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8042682409286499, "perplexity": 773.6463319117121}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057225.38/warc/CC-MAIN-20210921131252-20210921161252-00328.warc.gz"} |
https://ncertmcq.com/rs-aggarwal-class-9-solutions-chapter-14-statistics-ex-14a/ | RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A.
Other Exercises
Question 1.
Solution:
Statistics is a science which deals with the collection, presentation, analysis and interpretations of numerical data.
Question 2.
Solution:
(i) Numerical facts alone constitute data
(ii) Qualitative characteristics like intelligence, poverty etc. which cannot be measured, numerically, don’t form data.
(iii) Data are an aggregate of facts. A single observation does not form data.
(iv) Data collected for a definite purpose may not be suited for another purpose.
(v) Data is different experiments are comparable.
Question 3.
Solution:
(i) Primary data : The data collected by the investigator himself with a definite plan in mind are called primary data.
(ii) Secondary data : The data collected by some one other than the investigator are called secondary data. The primary data is more reliable and relevant.
Question 4.
Solution:
(i) Variate : Any character which is capable of taking several different values is called a variate or variable.
(ii) Class interval : Each group into which the raw data is condensed, is called a class interval
(iii) Class size : The difference between the true upper limit and the true lower limit of a class is called class size.
(iv) Class Mark : $$\frac { upper\quad limit+lower\quad limit }{ 2 }$$ is called a class mark
(v) Class limits : Each class is bounded by two figures which are called class limits which are lower class limit and upper class limit.
(vi) True class limits : In exclusive form, the upper and lower limits of a class are respectively are the true upper limit and true lower limit but in inclusive form, the true lower limit of a class is obtained by subtracting O.S from lower limit of the class and for true limit, adding 0.5 to the upper limit.
(vii) Frequency of a class : The number of times an observation occurs in a class is called its frequency.
(viii) Cumulative frequency of a class : The cumulative frequency corresponding to a class is the sum of all frequencies upto and including that class.
Question 5.
Solution:
The given data can be represent in form of frequency table as given below:
Question 6.
Solution:
The frequency distribution table of the given data is given below :
Question 7.
Solution:
The frequency distribution table of the
Question 8.
Solution:
The frequency table is given below :
Question 9.
Solution:
The frequency table of given data is given below :
Question 10.
Solution:
The frequency distribution table of the given data in given below :
Question 11.
Solution:
The frequency table of the given data:
Question 12.
Solution:
The cumulative frequency of the given table is given below:
Question 13.
Solution:
The given table can be represented in a group frequency table in given below :
Question 14.
Solution:
Frequency table of the given cumulative frequency is given below :
Question 15.
Solution:
A frequency table of the given cumulative frequency table is given below :
Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8734211325645447, "perplexity": 1696.7766476312365}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303709.2/warc/CC-MAIN-20220121192415-20220121222415-00363.warc.gz"} |
https://mathoverflow.net/questions/211746/a-question-on-representation-of-graphs | # A question on representation of graphs
Take a complete graph $K_n$. You want to assign a vectors from $\Bbb F_2^d$ to every edge such that sum of vectors in every simple cycle does not sum to $0$ vector. The question is what is minimum $d$ that is needed. Does this question have any connection to representation of symmetric group over $\Bbb F_2^d$?
Update: Fundamental roadblock here seems to be related to finding a nice way to provide labelling that encompasses every even alternating/reverse alternating cycle which prevents a better than $O(n\log{n})$ solution.
ADDED by David Speyer: I'd like to explain why I find this question interesting and challenging, and wish that people better at extremal combinatorics had turned some attention to it.
If we instead ask that every disjoint union of simple cycles have nonzero sum, the problem is easy (at least to leading order). A winning strategy is to use $d=n \log_2 n$ bits, divided into $n$ blocks of length $\log_2 n$. For edge $ij$, put the binary label of $j$ in the $i$-th block and the binary label of $j$ in the $i$-th block. And it is easy to show you can't do much better. For simplicity, take $n=2m$ even. Look at the $(2m-1) (2m-3) \cdots 5 \cdot 3 \cdot 1$ perfect matchings of the graph. If any two of them get the same sum, then their symmetric difference is a disjoint union of cycles with weight $0$. So we need to have at least $(2m-1) (2m-3) \cdots 5 \cdot 3 \cdot 1$ different vectors, so we need to use at least $\log_2 (2m-1) (2m-3) \cdots 5 \cdot 3 \cdot 1 \approx \frac{n}{2} \log_2 n$ bits. We have found the answer up to a factor of $2$.
Now, out of the $n!$ disjoint unions of cycles, roughly $(e/n) n!$ are a single cycle. One would think that reducing the number of things to avoid by a factor of $n$ shouldn't be able to effect the number of bits very much. Yet it seems incredibly hard to show this! How can we show that the cycles are basically randomly distributed among the disjoint unions of cycles?
Here is one way we could attempt to give a lower bound. Once again, take $n=2m$. Look at the $m!$ matchings from the first $m$ vertices to the second, which we can identify with the group $S_m$. Adding up the edges in a matching gives a coloring of $S_m$ with $2^d$ colors. If we make $S_m$ into a graph by joining $\sigma$ and $\tau$ when $\sigma^{-1} \tau$ is a single cycle, then this must be a proper coloring. So we are trying to find a lower bound for the chromatic number of the Cayley graph of $S_m$ with respect to the cycles. There is literature on the chromatic number of Cayley graphs, but I couldn't find anything that would help. I did find that a random graph on $m!$ vertices where an edge would appear with probability $e/m$ should be expected to have chromatic number $\frac{m! (e/m)}{2 \log m!} \approx \frac{e m!}{2 m^2 \log m}$. If that is the case here, we once again get $d \geq \log_2 m!$ (discarding lower terms). Is there some $S_m$ representation theory which would allow us to actually compute the chromatic number?
It really frustrates me that there is an exponential separation between by upper and lower bounds.
I will give the bounty for a construction which beats $(1-\delta) n \log_2 n$, or a lower bound which beats $(\log n)^{1+\delta}$, for any $\delta>0$.
Finally, I'd like to share a wild musing of mine. Consider two problems in graph theory and optimization. In the first problem, we are given an $n \times n$ matrix of weights $w_{ij}$, and we are trying to find a permuation $\sigma$ of $n$ which minimizes $\sum_i w_{i \sigma(i)}$. The second problem is the same, but we require that $\sigma$ is an $n$-cycle. The first problem is the assignment problem and there are good algorithms for it. The second problem is the traveling salesman problem and it is NP Hard. I am reminded of this problem: A large fraction of permutations are cycles, yet restricting myself to cycles makes thing incredibly harder.
• I do not think the tag representation theory is a proper one here. – Alireza Abdollahi Jul 17 '15 at 7:03
• @AlirezaAbdollahi Actually I do not know. I think this question may have a connection. – Brout Jul 17 '15 at 10:39
• I see. Maybe a the graph-theory tag is needed. – Alireza Abdollahi Jul 17 '15 at 11:37
• Tossed in coding-theory tag. Here is my thinking: Let $V$ be the free vector space on the $\binom{n}{2}$ edges of your graph. Let $K$ be the kernel of the map $V \to \mathbb{F}_2^d$ given by your labels. You want $K$ to be large while not containing the characteristic function of any cycle. This reminds me of the coding theory goal of finding large $K$ that does not contain any vector of small Hamming weight. – David E Speyer Jul 21 '15 at 3:43
• @DavidSpeyer: maybe you could come up with a more evocative title for this question? – Sam Hopkins Jul 27 '15 at 3:11
It's important to clarify what definition of "cycle" you have in mind. In algebraic-graph-theory contexts like this one, the natural definition is that it's a set of edges with even degree at each vertex; these sets form the vectors of a $\mathbb{Z}_2$-vector space, the cycle space of the graph, using the symmetric difference operation as the vector sum.
With this definition, the minimum $d$ you're looking for is just the circuit rank of the graph, i.e. the dimension of the cycle space, $m-n+1$ where $m$ is the number of edges and $n$ is the number of vertices. Your labeling gives a linear map from the cycle space to the label space, this map has a nontrivial kernel unless $d$ is large enough, and a vector in the kernel is a cycle with zero label sum. In the other direction, it is easy to find labelings with dimension equal to the circuit rank that avoid cycles summing to zero. For instance, find a spanning tree of the graph, label each tree edge with the zero vector, and label each non-tree edge with a basis vector (independent of all the other nonzero labels).
This all works regardless of whether the graph is complete. But if you have some other definition of cycles in mind, such as simple cycles, then your labeling problem looks messier.
• No, the cycle basis article is about a different (non-algebraic) invariant of directed graphs. – David Eppstein Jul 17 '15 at 16:00
I have now tried a few different strategies which are all winding up at $d \approx n \log_2 n$. I'll describe the simplest of these. (David Epstein's strategy with the spanning tree gives $d \approx n^2/2$.)
Let $2^{m-1} < n \leq 2^m$, so we can encode the vertices of the graph with $m$ bits. I will show how to take $d=mn$. Think of those $mn$ bits as $n$ blocks of length $m$. For an edge joining vertices $i$ and $j$, put the binary label of $j$ in the $i$-th block and the binary label of $i$ in the $j$-th block. Put zeroes in all other blocks.
Suppose we have a simple cycle. Let this cycle contain the section $(i,j,k)$. Then the $j$-th block will contain the binary labels of $i$ and $k$, which will not cancel, and nothing else.
I have some minor tricks which can shave small numbers of bits off this, but I am waiting to see if someone can come up with something much better first.
My best lower bound is $\approx \log_2 n$. This is horribly far from what I believe the truth to be, but I record it nonetheless. Let $n=2p$ for $p$ prime. For $0 \leq k < p$, let $w_k = \sum_{i=1}^p v(i, p+1+(i \bmod p))$. If $d < \log_2 p$, then two of the $w_k$ are equal by the pigeonhole principle, say $w_{k_1} = w_{k_2}$. But then $w_{k_1} + w_{k_2}$ is the sum over the edges in a $2p$ cycle.
• As a heuristic, there are about $n!$ simple cycles in $K_n$. If we want to get $d$ below $\log_2 n! \approx n \log_2 n$, we need to get the images of the simple cycles in $\mathbb{F}_2^d$ to be abnormally skewed away from $0$. It is easy to think of weights that skew towards zero, but hard to find ways to skew away. – David E Speyer Jul 21 '15 at 4:45
• Apparently so. I can't prove any lower bounds, which is frustrating, but there are upwards of $(2^n)!$ simple cycles which only use edges between the two graphs. It doesn't surprise me that I need something like $2^n$ more coordinates to kill them all. – David E Speyer Jul 21 '15 at 6:44
• I don't understand what ring structure you want to use on $\mathbb{F}_2^d$ for this strategy, nor particularly why you think these quadratic equations will be easier to make nonzero than the linear ones you started with. Assuming that you are using the finite field of order $2^d$, and arguing heuristically, there are $2^{nd}$ possible choices of the $x_i$. Each quadratic vanishes at about $1/2$ these choices, and there are $\approx n!$ quadratics. (continued) – David E Speyer Jul 21 '15 at 12:06
• So I would expect odds of success for an individual choice to be $1/2^{n!}$. So I expect to win if $2^{nd}/2^{n!}>1$, or $d \approx n!$. This is much worse than the $\log_2 n!$ I gave above. Do you have a reason to think you can do better than random? – David E Speyer Jul 21 '15 at 12:06
• Why? (Plus more characters.) – David E Speyer Jul 21 '15 at 12:09
Here is a modest improvement on the upper bound that shows $d=O(n\log\log n)$.
The base of the construction is the following. Order the vertices according to some permutation as $v_1,\ldots,v_n$. Fix $2n$ independent vectors from $\mathbb F_2^d$ and assign two vectors, $\ell_i$ and $r_i$, to each vertex $v_i$. Assign to the edge $v_iv_j$ the vector $r_i+\ell_j$ if $i<j$ and $\ell_i+r_j$ if $i>j$. Notice that this usually gives a nonzero sum for long cycles.
Our strategy will be to take $t\approx \log \frac nk$ permutations (with $t\cdot 2n$ independent vectors) and take the sums on each edge to take care of all cycles longer than $k$. The standard probabilistic approach with $\approx k\log n$ further independent vectors takes care of all cycles whose length is at most $k$. This in total gives $d\approx t\cdot n + k\log n =O(n\log\log n)$.
Claim. $t\approx \log \frac nk$ permutations are enough to take care of all cycles longer than $k$.
Proof of the claim. Divide $n$ into groups of size $k$. Fix an order inside each group, this will be the same in each permutation, which thus practically act on $m=\frac nk$ elements. For every cycle $C=c_1\ldots c_k$ of length $k$, either there are three different groups that contain three consecutive vertices from $C$ (thus $c_{i-1}\in G_1$, $c_{i}\in G_2$, $c_{i+1}\in G_3$), or two different groups, one of which contains the first two of three consecutive vertices and the other group the third one (thus $c_{i-1},c_{i}\in G_1$, $c_{i+1}\in G_2$)
In this latter case, all we need is a permutation where $G_2$ is after $G_1$, then $c_i$ will be between its neighbors and thus the respective $\ell$ and $r$ vectors won't be negated when taking the sum over the edges of $C$. Similarly, in the first case, we need that $G_2$ is between $G_1$ and $G_3$.
Therefore, it is enough to show that for $m=\frac nk$ elements there are $O(\log m)$ permutations such that for any three elements $G_1,G_2,G_3$ there is a permutation in which $G_1<G_2<G_3$. A standard probabilistic argument shows that $O(\log m)$ random permutations work. In more detail, there are $m^3$ ordered triples, each permutation has probability $\frac 16$ to work for each triple, thus if we take $t$ independent permutations, the chance for any ordered tripe that none of the permutations work for it is $(1-\frac 16)^t$ and taking the union bound we need $m^3(1-\frac 16)^t<1$.
• I am not understanding this sentence well " Notice that this gives a nonzero sum for a cycle c1c2…ck if and only if the length of the cycle is even and every eventh vertex comes before (or after) every oddth vertex. Therefore, it has a good chance to work for long cycles" and this phrase well "two groups with an eventh element and a third group with an oddth element". What is eventh, oddth mean? – Brout Aug 6 '15 at 5:29
• In the first for $k=6$, I want that in the order $<$ given by the permutation satisfies $c_1,c_3,c_5<c_2,c_4,c_6$ or $c_1,c_3,c_5>c_2,c_4,c_6$. I hope from this you can also decipher what I meant in the second phrase... – domotorp Aug 6 '15 at 5:33
• "Our strategy will be to take t random permutations (with t⋅2n independent vectors) and take the sums on each edge to take care of all cycles longer than k." t random permutations of what? – Brout Aug 7 '15 at 17:34
• This is what I mean by alternating permutation en.wikipedia.org/wiki/Alternating_permutation (1<3>2<5>4<6>1). – Brout Aug 7 '15 at 17:45
• I have added it. – domotorp Aug 9 '15 at 18:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8951634168624878, "perplexity": 180.34176577956188}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578530527.11/warc/CC-MAIN-20190421100217-20190421121538-00046.warc.gz"} |
https://physics.stackexchange.com/questions/626939/why-is-there-no-blackbody-radiation-in-the-high-frequency-section-of-plancks-cu | # Why is there no blackbody radiation in the high frequency section of Planck's curve?
Upon examining the curve describing blackbody thermal radiation, I noticed that the curves approaches (but never reaches) zero when increasing wavelength, but on the other hand it actually does reach zero at high frequency so I was wondering why?
• Good question. This confused many physicist for years and was at the birth of quantum mechanics. Apr 4 at 1:20
• What makes you think it reaches zero? Apr 4 at 4:14
• Let's state the behaviour carefully. Plotted against frequency, intensity scales as $\nu^3/(e^{\beta h\nu}-1)$, reaching $0$ at $\nu=0$ but not at any finite $\nu>0$. Plotted against wavelength, intensity scales as $\lambda^{-5}/(e^{\beta hc/\lambda}-1)$, which is asymptotic to $\lambda^{-5}e^{-\beta hc/\lambda}$ ($\lambda^{-4}/(\beta hc)$) for small (large) $\lambda>0$, so the $\lambda\to0^+,\,\lambda\to\infty$ one-sided limits are both $0$.
– J.G.
Apr 4 at 20:47
• There is a contradiction in the body of your question. May 11 at 11:05
In the frequency domain, the power spectral density has the form $$f^2$$ as the frequency approaches zero and the form $$f^3e^{-f}$$ as the freuency approaches infinity. In neither case does it reach zero at any finite frequency. You can argue about at which end it approaches zero more rapidly. Plotted on a log frequency axis, the high-frequency end of the spectrum would appear to approach zero very abruptly because of the exponential. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9318242073059082, "perplexity": 572.0091943301684}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363405.77/warc/CC-MAIN-20211207170825-20211207200825-00182.warc.gz"} |
https://www.godelia.org/2020/08/26/degeneration-in-quantum-mechanics/ | # Degeneration in quantum mechanics
Degeneration occurs when two or more independent wave functions have the same eigenvalue.
It is said that $$n$$ functions $$f_ {1}, f_ {2,} \ldots, f_ {n}$$ are linearly independent if the condition
$$\sum_{i}c_{i}f_{i}=0$$
it is satisfied when all the constants $$c_ {i}$$ are equal to zero.
The degree of degeneration of a system is the number of linearly independent functions with the same eigenvalue.
Theorem. Any linear combination of $$n$$ functions of a degenerate level of energy $$E$$ is also a eigenfunction of the Hamiltonian with energy $$E$$.
The proof is straightforward:
$$\hat{H}\psi_{1}=E\psi_{1},\hat{H}\psi_{2}=E\psi_{2},\ldots,\hat{H}\psi_{n}=E\psi_{n}$$
$$\varphi=\sum_{i}c_{i}\psi_{i}$$
$$\hat{H}\varphi=\hat{H}\sum_{i}c_{i}\psi_{i}=E\sum_{i}c_{i}\psi_{i}$$
$$\hat{H}\sum_{i}c_{i}\psi_{i}=\sum_{i}c_{i}\hat{H}\psi_{i}=\sum_{i}c_{i}E\psi_{i}=E\sum_{i}c_{i}\psi_{i}=E\varphi$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.973669171333313, "perplexity": 184.473286445459}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363337.27/warc/CC-MAIN-20211207075308-20211207105308-00600.warc.gz"} |
http://mathhelpforum.com/calculus/47894-finding-distance-given-velocity-cos.html | # Math Help - Finding distance, given velocity with cos.
1. ## Finding distance, given velocity with cos.
What a wonderful website this is. Thank you all in advance for any help or guidance you can give me.
So, okay.
Find the distance traveled in 15 seconds by an object moving at a velocity of v(t) = 20 + 7 cos t feet per second.
Again, thank you for any help. I have no idea what to do with the cosine.
Edit: You know, I put this here because I was told it had to do with integrals? It might fit better in Trigonometry though. Sorry about that!
2. Originally Posted by AppleTini
What a wonderful website this is. Thank you all in advance for any help or guidance you can give me.
So, okay.
Find the distance traveled in 15 seconds by an object moving at a velocity of v(t) = 20 + 7 cos t feet per second.
Again, thank you for any help. I have no idea what to do with the cosine.
$s(t)=\int v(t)\,dt$
Therefore, $s(t)=\int (20+7\cos t)\,dt=20t+7\sin(t){\color{red}+C}$
Is there an initial condition for position?
--Chris
3. Originally Posted by Chris L T521
Is there an initial condition for position?
No, it doesn't seem so.
The problem is that I'm taking this AP Calculus class online [bad idea, for the record], and the online book really is not helpful at all. It hasn't actually talked about integrals or anything, and I certainly don't really remember them from two years ago. So I'm pretty much on my own. Thank you so much, though.
4. It MAY seem logical to assume one means the first 15 seconds.
The integral on [0,15] gives 300 + 7*sin(15) = 304.552
Unfortunately, it may NOT be a logical assumption. If we don't start watching for five seconds, the integral on [5,20] gives 300 + 7(sin(20)-sin(5)) = 313.103
We could play this game for a long time. One needs those initial conditions or some other guidance as to when or where we are clocking the 15 seconds.
In any case, you can estimate the answer well enough.
The constant "20" is, well, constant. It ALWAYS manages 300 in 15 seconds. The tricky part is the cosine, as you said. Interestingly, the cosine is periodic. Every $2\pi$ it has gotten you nowhere. In this way, we can eliminate some of the cosine problem. $15 = 2*(2\pi) + 0.387*(2\pi)$. This means you really have to worry only about the last 2.434 seconds of the trip, since the previous $4\pi$ seconds didn't get us anywhere! Of course, you still have to know where to start or stop. Just any 15 seconds will not due. We need to know which ones to watch.
5. Originally Posted by AppleTini
What a wonderful website this is. Thank you all in advance for any help or guidance you can give me.
So, okay.
Find the distance traveled in 15 seconds by an object moving at a velocity of v(t) = 20 + 7 cos t feet per second.
Again, thank you for any help. I have no idea what to do with the cosine.
Edit: You know, I put this here because I was told it had to do with integrals? It might fit better in Trigonometry though. Sorry about that!
Actually the distance s(t) is given by
$s(t) = \int |v(t)|~dt$
Fortunately in this case you get the same answer as has been given in the previous posts.
-Dan
6. displacement (a vector quantity) = $\int_{t_1}^{t_2} v(t) dt$
distance traveled (a scalar quantity) = $\int_{t_1}^{t_2} |v(t)| dt$
7. Well to find distance you have to integrate the whole v(t) funcn.
integrating dat you will get
s(t) = 20t + 7sint
put t=15
you will get
s= 140 + 7sin15
sin15 =.26 approx.
so you get the ans..
8. Originally Posted by vishalgarg
Well to find distance you have to integrate the whole v(t) funcn.
integrating dat you will get
s(t) = 20t + 7sint
put t=15
you will get
s= 140 + 7sin15
sin15 =.26 approx.
so you get the ans..
How does 20 times 15 equal 140?
-Dan | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8928672671318054, "perplexity": 633.2228722750094}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416400379320.39/warc/CC-MAIN-20141119123259-00047-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://gilkalai.wordpress.com/2012/07/23/a-weak-form-of-borsuk-conjecture/?like=1&source=post_flair&_wpnonce=8149a33672 | ## A Weak Form of Borsuk Conjecture
Problem: Let P be a polytope in $R^d$ with n facets. Is it always true that P can be covered by n sets of smaller diameter?
I also asked this question over mathoverflow, with some background and motivation. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.911348283290863, "perplexity": 419.18337900543605}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823478.54/warc/CC-MAIN-20171019212946-20171019232946-00140.warc.gz"} |
https://www.physicsforums.com/threads/is-it-distribution-function.97674/ | # Homework Help: Is it distribution function ?
1. Oct 31, 2005
### Alexsandro
Could someone help me. I don't able to explain if is FG is a distribution fuction:
Show that if F and G are distribution functions and $0 \leq \lambda \leq 1$ then $\lambda.F + (1 - \lambda).G$ is a distribution function. Is the product F.G a distribution function?
Last edited: Oct 31, 2005
2. Oct 31, 2005
### mathman
In order see if a function is a distribution function, go back to the definition.
F(x) is a d.f. if F-> 1 as x -> inf, F-> 0 as x-> -inf. F(y)>=F(x) for y>x.
It should be easy for you to verify that in both examples you have a distribution function. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9651009440422058, "perplexity": 1439.2407694455424}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267867666.97/warc/CC-MAIN-20180625111632-20180625131632-00182.warc.gz"} |
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You are really close! Change your signal and steering vectors to be complex. Specifically for the steering vectors, these coefficients are meant to act as phase shifts. Using a real sinusoid will introduce a phase shift term in the opposite angle direction, which you don't want. Doing this alone you will see an improvement in your pseudospectrum. In regards ...
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You're almost there; you just need to connect a few dots. Let your $\frac{10}{5 + i 2 \pi f} = G(f)$. Then $g(t) = 10 u(t) e^{-5 t}$. Now we get into that exponent part. Your $h(t) = g(t - t_0)$ is correct, but you're applying it incorrectly. You need to apply it to the part that I've labeled as $g(t)$: $h(t) = g(t - t_0) = 10 u(t - t_0) e^{-5 (t - t_0)}... 0 The principal idea behind my recommended approach is to identify characteristic points in your cycle. You can then make a mapping from time within the cycle (as it is occuring) and the location of that point within your representative cycle. Once you have a set of mapping points, create a best fit function. This will then give you a mapping function of ... 1 You may be able to use a modified transform to compute change in a time stretch parameter (I will use$\zeta$) versus time (t). Your success in being able to do this depends on the cross correlation properties of your pattern with time stretched versions of itself. Let me explain: First notice that the Fourier Transform is a correlation over all possible ... 0 Hello check this source To begin with, let’s remember what the fundamental frequency is and in what tasks it may be needed. The fundamental frequency, which is also referred to as F0, is the vibration frequency of the ligaments when pronouncing voiced sounds. When pronouncing unvoiced sounds, for example, by whispering or uttering hissing and ... 0 Wifi uses OFDM, which uses symbols that can transport many bytes at once. That means you can't have arbitrary long packets, but always need to use the next multiple of a symbol duration. 1 A more general expression states that for$ M \geq N$: $$\sum_{n= N}^{n = M} c = (M-N+1) \cdot c$$ where the derivation simply relies on fact that the epxression has (M-N+1) terms : $$\sum_{n= N}^{n = M} c = \{ c + c + ... + c\} = (M-N+1) \cdot c$$ And when applied for your particular case (with$N = -M$) it becomes:$$\sum_{n= -M}^{n = M} c = (M-(-... 0 For instance, from$-3$to$3$, you have$-3,\,-2\,-1,\,0\,1,\,2,\,3$, hence$2\times 3+1$terms. More generally, the sum from$-M$to$M$is composed of$2M+1$terms: indices with$m$strictly negative (a total of$M$), those which$m$strictly positive (a total of$M$), plus one at zero ($1$). If all terms are the same constant$c$, the total is$(2M+...
0
From what I can glance : The chattering seems to be high-frequency compared to your signal of interest, it should not be hard to filter this chattering noise. You simply need to identify the frequency band of your signal and the frequency band of this noise. Could your perform an FFT to analyze the frequencies of your noise? Then design a filter that will ...
0
Looking at documents like Lecture notes on Distributions, Hasse Carlsson, or Convolution dans l'espace $\mathcal{D}'_+(\mathbb{R})$, convolution of distributions can be defined under some technical conditions. However, when one of the operand has a compact support, as $\delta(t)$ does, the convolution is well-defined. From Wikipedia:Distribution-Convolution:...
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Well, by definition of the $\delta$ distribution, you have: $\int_{-\infty}^{\infty} f(t) \delta(t-T)\, \textrm{d}t = f(T)$ The autocorrelation of a function $g(t)$ can be computed via: $\int_{-\infty}^{\infty} g^{*}(t)g(t + \tau)\, \textrm{d}t$, with $g^*$ as the complex conjugate of $g$. Since $\delta(t)$ is real-valued, this is conjugation can be ...
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http://math.stackexchange.com/questions/62298/what-can-the-writer-assume-in-a-proof | # What can the writer assume in a proof?
When writing a proof, what level of mathematical understanding can I assume my reader has?
For example, can I assume they know all odd integers can be represented by $2q+1$? (Right?)
Or that all even integers can be represented by $2k$?
When do I have the power to draw on a theorem?
Can I always assume that my previous problem was right and cite it to help me with my current proof?
-
You can give the result before the new theorem. Then you either cite where you can find a proof or give a proof yourself. – Jonas Teuwen Sep 6 '11 at 13:55
And if the proof is not too long and not distracting, go with Jonas's second option. – J. M. Sep 6 '11 at 13:59
So I assume they know very little and try to help the reader as much as I can. and ensure that I don't get too off topic? – Dennis Hayden Sep 6 '11 at 14:01
I hope your readers don't "know all odd integers can be represented by 4q+1"... :) – t.b. Sep 6 '11 at 14:04
The question cannot be answered generally. You'll have to know which audience you're targeting, and there is no "standard audience". The assumptions can be very different according to whether you're writing an article for a professional journal, or helping a highschooler on math.SE (or something in between). – Henning Makholm Sep 6 '11 at 14:09
In mathematical writing, as in all writing, you need to think carefully about who your readers will be (and who you want them to be). It is rarely possible to write in a way which will be equally satisfactory to every conceivable person.
In this case, although you don't say so explicitly, it sounds like you are writing up problem sets for a course, so your intended reader is the grader of the problem set and/or the course instructor (perhaps the same person). With such a small audience it is feasible to simply ask them about their preferences, and while you probably don't want to do this before you turn in every single problem set, some questions in the beginning will probably make things go smoothly. In general though, this type of reader is someone who knows the material well -- probably better than you do -- and in most cases broadly knows what you are trying to say even before you say it. However, they are also looking for gaps and mistakes in your arguments. So in this case I would start by erring on the side of including more details / supporting reasoning, while understanding that if you do not express any given idea / argument in the best possible way you are more likely to be understood anyway than with a reader who really doesn't know what you're trying to tell her.
With regard to your specific questions:
can I assume they know all odd integers can be represented by $4q+1$[?]
Gosh, I hope not, since this is not true: e.g. $3 \neq 4q + 1$.
[Added: It seems that the "$4$" was just a typo which has since been corrected to $2$. In this case there may well be something to prove. It is relatively common to define an integer to be odd if it isn't even, and then one has to justify that an odd number is of the form $2k+1$. In fact, in an honors course for future math majors that I am currently teaching this came up in the first week. I defined an integer to be even if it is of the form $2k$ for some integer $k$ and odd if it is of the form $2k+1$, but there was still something to show: every integer is either even or odd and not both. The "not both" is easy, but the first part requires something: a few days later I proved it by mathematical induction, and then later stated it as a special case of the theorem about division with remainder. So no, for my intended audience I did not want to just assume that familiar facts about even and odd numbers are true, although I probably would do so in a course pitched either at a higher or a lower level.]
Or that all even integers can be represented by $2k$?
This is a standard definition of an even number (in fact, I can't think of any other standard definition at the moment). So this may well have come up before in class or in the course text. If not, and you are not working with any other definition of even, then you can just say something like "if $x$ is even -- that is, $x$ is of the form $2k$ for some integer $k$ -- ..." and move on.
Can I always assume that my previous problem was right and cite it to help me with my current proof?
I agree with all of Pete's advice. One additional point, if this is an intro number theory course that focuses on learning how to write proofs: Some people define odd numbers to be those numbers which are not of the form $2k$. In that case, it is not obvious that odd numbers are of the form $2 m+1$, and does require proof (in a more advanced course, this would count as standard knowledge). Other people define odd numbers to be numbers of the form $2m+1$, in that case it is not obvious that every number is either even or odd. (continued) – David Speyer Sep 6 '11 at 14:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.844550609588623, "perplexity": 193.7801757456301}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464051151584.71/warc/CC-MAIN-20160524005231-00110-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://mathmaine.com/2013/04/25/summary-logarithms/ | # Logarithms
Unlike the two most “friendly” arithmetic operations, addition and multiplication, exponentiation is not commutative. You will get a different result if you swap the value in the base with the one in the exponent (unless, of course, they are the same value):
$3^2 \ne 2^3$
The most significant impact of this lack of commutativity arises when you need to solve an equation that involves exponentiation: two different inverse functions are needed, one to undo the exponent (a root), and a different one to undo the base (a logarithm).
Just as there are many versions of the addition function (adding 2, adding 5, adding 7.23, etc.), and many versions of the “root” function (square roots, cube roots, etc.), there are also many versions of the “logarithm” function. Each version has a “base”, which corresponds to the base of its inverse exponential expression.
### Inverse Functions: Logarithms & Exponentials
Logarithms are labelled with a number that corresponds to the base of the exponential that they undo. For example, the symbols to the left of the equal sign below are read “log base b of b to the c”:
$log_b(b^c)=c$
“Log base b” is the inverse function of “b to the…”, so the two cancel each other out, leaving behind the argument of the inside function… “c” in this case. This same concept applies in more complex situations as well:
$log_4(4^{2x-5})=2x-5$
In each case, a function applied to its inverse leaves behind the argument (no matter how simple or complex it may be). It does not matter which inverse function is on the “outside”, if two inverse functions are “next” to each other with no other function or operation “in the way”, they cancel each other out. “Four to the…” is the inverse of “log base 4”, therefore:
$4^{log_4(3y+2)}=3y+2$
Note that if the bases are not the same in both the logarithm and the exponential, then they are not inverse functions… just as adding adding four is not the inverse of subtracting three.
Inverse functions only cancel one another out when they are applied directly to one another, with nothing else “in the way”: when the result of the inner function is not modified in any way before being used as an input to the outer function. Examples of where functions that are inverses of one another do not cancel each other out include:
$5^{(7+log_5(x))}\\*~\\*log_5(6\cdot 5^x)\\*~\\*log_5(2+5^x)$
All of the examples above do something to the result of the inner (inverse) function before evaluating the outer function. In the first example, 7 is added to the result of the logarithm. In the second, the power of 5 is multiplied by 6. In the third, the power of 5 has a 2 added to it. Therefore, none of the above examples can be simplified using inverse function principles.
However, the Laws of Exponents or Logarithms will allow us to simplify the first two examples a bit, as you will see below. The last example above cannot be further simplified.
### Logarithms Produce Exponents
The result of a logarithm is always an exponent… an exponent of the logarithm’s base. So, logarithms answer questions such as “what exponent of 2 produces a result of 16?”:
$log_{2}(16)=4$
Evaluating a logarithm always produces a power of the logarithm’s base.
The two equivalent equations below are often referred to as the definition of a logarithm. The first equation shows that “c” is the result of a logarithm, therefore if used as a power of the logarithm’s base “b“, you will get back to “a“… which was your starting point:
$log_b(a)=c~~~ means~that\\*~\\*a=b^c$
Alternatively, you can take “log base b” of both sides of the second equation, simplify the right side by canceling out inverse functions, and you will end up with the first equation:
$a=b^c\\*~\\*log_b(a)=log_b(b^c)\\*~\\*log_b(a)=c$
### Common and Natural Logarithms
Two bases, 10 and e, are used so frequently with logarithms that they have their own notation:
$log(a)=log_{10}(a)~~~Common~Logarithm\\*~\\*ln(a)=log_e(a)~~~~~Natural~Logarithm$
Since our number system uses ten digits, and each digit in a multi-digit number represents a different power of 10, for example:
$123 = 1\cdot 10^2+2\cdot 10^1+3\cdot 10^0$
it makes sense that we would be interested in powers of 10 more so than other bases. This greater interest has led to a slightly shorter notation for what are called “Common Logarithms”, or “log base 10”, as shown above. If no base is indicated for a logarithm, it is assumed to be base 10, or a Common Logarithm.
Another common base for logarithms is “e“, a number you may not have encountered before. “e” is an irrational number (its decimal representation never terminates or repeats), which explains why it, like “pi”, is usually referred to by name. “e” arises when modeling situations where growth occurs continuously, for example when an initial population of 100 grows continuously at 2%:
$y=100\cdot e^{0.02t}$
If you need to solve such an equation for “t”, the easiest way to do so is by using the inverse function of “e to the…” , which is log base e. This function is referred to as a “Natural Logarithm”, or when using function notation by writing “ln”:
$\frac{y}{100}=e^{0.02t}\\*~\\*ln\left(\frac{y}{100}\right)=ln(e^{0.02t})\\*~\\*ln\left(\frac{y}{100}\right)=0.02t\\*~\\*ln\left(\frac{y}{100}\right)\div 0.02=t$
Without logarithms, we could not get the “t” out of the exponent in the example above to solve for it.
### Laws of Logarithms
Since the result of a logarithm is an exponent, it should come as no surprise that three of the Laws of Logarithms mirror the Laws of Exponents. To derive them, let us work with two real numbers expressed as a power of a base. Any base will do, but since we work with numbers that are powers of 10 the most in our daily lives, base 10 has been used below. This also allows us to use Common Logarithms in the derivation, so that we don’t need to indicate a “base 10” below each logarithm.
Any real number can be represented as a power of ten. For example:
$3=10^{0.4771213...}\\*~\\*12=10^{1.0791812...}$
Let’s use this idea to express a number “A” as a power of ten, as shown in equation (1) below. Take Log base 10 of both sides of this equation, simplify, and the resulting equation (2) is mathematically equivalent to equation (1), but has been solved for “a“, the power of ten:
$(1)~~~~A=10^a\\*~\\*~~~~~~~~~log(A)=log(10^a)\\*~\\*(2)~~~~log(A)=a$
Since two quantities are needed to explain the Laws of Logarithms, we’ll define a second value in the same way:
$(3)~~~~B=10^b\\*~\\*~~~~~~~~~log(B)=log(10^b)\\*~\\*(4)~~~~log(B)=b$
These two definitions can now be used to derive the following “laws”:
1) The logarithm of a product equals the sum of the logarithms of the factors:
$log(A\cdot B)\\*~\\*=log(10^a \cdot 10^b)~~~~~~~~~Substituting~(1)~and~(3)~from~above\\*~\\*=log(10^{a+b})~~~~~~~~~~~~~Laws~of~exponents\\*~\\*=a+b~~~~~~~~~~~~~~~~~~~~Inverse~functions~cancel\\*~\\*=log(A)+log(B)~~~~~Substituting~(2)~and~(4)~from~above\\*~\\*\therefore \mathbf{log(A\cdot B)=log(A)+log(B)}$
As you will hopefully recall from the Laws of Exponents, when like bases are being multiplied we add the exponents. This is exactly the process that this property of logarithms describes. The first line requires us to calculate the Common Logarithm, or power of ten, of the product AB. We know by the laws of exponents that this exponent must be the sum of a (A’s power of ten) and b (B’s power of ten), which is the same thing as the sum of log(A) and log(B).
Note that this identity is used just as often from left-to-right as it is from right-to-left. You are welcome to convert a sum of logarithms (with the same base) into the logarithm of a product, or the logarithm of a product into a sum of logarithms. So, don’t think of this identity as a one-way-street… it is frequently used in either direction.
2) The logarithm of a quotient equals the logarithm of the numerator less the logarithm of the denominator:
$log\left(\dfrac{A}{B}\right)\\*~\\*=log\left(\dfrac{10^a}{10^b}\right)~~~~~~~~~~~~Substituting~(1)~and~(3)\\*~\\*=log(10^{a-b})~~~~~~~~~~~~~Laws~of~exponents\\*~\\*=a-b~~~~~~~~~~~~~~~~~~~~Inverse~functions~cancel\\*~\\*=log(A)-log(B)~~~~~Substituting~(2)~and~(4)\\*~\\*\therefore \mathbf{log\left(\dfrac{A}{B}\right)=log(A)-log(B)}$
Once again, from the Laws of Exponents, when like bases are divided we subtract the exponents. This identity mirrors that law of exponents in the same way as the previous one, and is also used in either direction… so keep an eye out for expressions that look like either side of the last line above. Sometimes you will prefer to have things expressed as the logarithm of a quotient, and other times you will prefer to express things as a difference of logarithms.
3) The logarithm of an exponential equals the exponent times the logarithm of the base:
$log(B^c)\\*~\\*=log([10^b]^c)~~~~Substituting~(3)\\*~\\*=log(10^{bc})~~~~~~Laws~of~exponents\\*~\\*=bc~~~~~~~~~~~~~~~Inverse~functions~cancel\\*~\\*=log(B)\cdot c~~~~~Substituting~(4)\\*~\\*=c\cdot log(B)\\*~\\*\therefore \mathbf{log(B^c)=c\cdot log(B)}$
This is very handy for both a) getting a coefficient that is in front of a logarithm “out of the way” so that an inverse function may be used, or b) getting the expression in an exponent out of the exponent, and down where we can work with it more easily. Once again, this identity is often used in either direction.
4) And lastly, the base change formula. This formula allows the value of a logarithm with any base to be calculated using logarithms with a different base. The formula is usually used to convert logarithms into base 10 or base e, both of which are found on many calculators (as well as in tables of logarithms at the back of textbooks).
$(5)~~~b^c=a\\*~\\*~~~~~~~~log_b(b^c)=log_b(a)~~~~~~~Taking~the~log_b~of~both~sides~of~(5)\\*~\\*(6)~~~c=log_b(a)~~~~~~~~~~~~~~Inverse~functions~cancel~each~other\\*~\\*The~above~is~one~way~to~solve~for~c.~Here~is~another\\*~\\*~~~~~~~~log_x(b^c)=log_x(a)~~~~~~Taking~the~log_x~of~both~sides~of~(5)\\*~\\*~~~~~~~~c\cdot log_x(b)=log_x(a)~~~~~~~Log~of~an~exponential~rule~(3)~above\\*~\\*(7)~~~c=\dfrac{log_x(a)}{log_x(b)}\\*~\\*~~~~~~\therefore \mathbf{log_b(a)=\dfrac{log_x(a)}{log_x(b)}}~~~~~Substituting~(6)~into~(7)$
The result on the last line above is called the base change formula, as it allows a logarithm base “b” to be converted into a ratio of logarithms in any base you choose… labelled “x” above. Note that this derivation uses Law #3 from above.
### Using the Laws of Logarithms
Returning to a couple of examples shown above, let’s use the Laws of Logarithms to change the way they look a bit. The first example was:
$5^{(7+log_5(x))}\\*~\\*=5^7 \cdot 5^{log_5(x)}~~~Laws~of~exponents\\*~\\*=5^7 \cdot x~~~~~~~~~Inverse~functions~cancel$
And the second example was:
$log_5(6\cdot 5^x)\\*~\\*=log_5(6)+log_5(5^x)~~~Laws~of~Logarithms\\*~\\*=log_5(6)+x~~~~~~~~~~~Inverse~functions~cancel\\*~\\*=\dfrac{log(6)}{log(5)}+x~~~~~~~~~~~Base~change~formula$
Be careful! The following do not follow a pattern found in any of the first three laws of logarithms, therefore they cannot be manipulated in any way using those laws. The only actions you could take on them would be to evaluate them (if you know the value of “a” and “b”), or use the base change formula to rewrite in terms of logarithms with a different base:
$log(a+b)\\*~\\*log(a-b)$
### Why Logarithms?
Logarithms have many uses, some of which appear in the media regularly, and some of which are more special-purpose. In addition to their usefulness in mathematics (as the inverse function of an exponential, which allows us to solve for a variable in the exponent), logarithms are often used to make it easier to think about and compare values in situations that involve numbers that span a very large range. By talking about values as exponents of a base, usually base 10, it is easier to work with both tiny numbers and huge ones. For example, consider the following four values. Their decimal representations require a range of five places to the right of the decimal to 10 places to the left of the decimal:
$log(0.0001)=~~~~~~~-4\\*~\\*log(10)=~~~~~~~~~~~~~~~~1\\*~\\*log(10,000)=~~~~~~~~~~4\\*~\\*log(1,000,000,000)=9$
yet by taking the logarithm of each value, and referring to each value by its power of ten, we can use a much smaller number of digits (one in this example), with a much smaller range (negative four to nine in this case), which can make working with quantities that span such a large range much simpler.
Graphing data with a large range
Logarithms are often used when graphing data about something that is growing. By either graphing the logarithm of the data on the vertical axis, or graphing the raw data on “semi-log” graph paper (which has a linearly scaled horizontal axis, and a logarithmic vertical axis), you can easily tell if the data reflects a constant growth rate or not.
A constant growth rate produces a curve that rises faster and faster as you move to the right when graphed on “normal” graph paper, but will produce a graph with a constant slope when plotted using semi-log graph paper. This is very useful when looking at multiple years of sales or earnings data from growing companies, as you can instantly see if the company’s growth is accelerating (the slope of the graph becomes greater), remaining constant (the slope is constant), or slowing (the slope of the graph becomes smaller).
A good example of using a graph on semi-log paper to analyze changes in growth rates can be found in this 2009 swine flu graph from Wikipedia:
The “Total” (blue points above the other curves) shows a very fast and relatively constant initial growth rate through early to mid-May (we can infer this from the fact that the graph is almost linear during this period), followed by a transition to a slower but also relatively constant long term growth rate through the end of June. Notice that each value along the vertical axis is double the previous one, even though the axis tick marks are all the same distance apart. This is a semi-log (base 2) graph, because the distances from the origin on the vertical axis are proportional to the “Log base 2” of the number of cases.
Other quantities in our world and universe for which this approach is useful include:
The Richter scale
The Richter scale is used to measure earthquake strengths. Seismographs measure the amplitude (size or strength) of waves that travel through the earth from the site of an earthquake to the measuring station. The Richter scale is based on the logarithm of the measured wave amplitude, so a Richter Scale measurement is a power of 10. An earthquake of magnitude 3 is ten times more powerful than one of magnitude 2 (a micro-earthquake that is not able to be felt).
The chart below shows the magnitude reported by the U.S. Geological Survey for the more significant earthquakes around the earth over a 24 hour period. The most intense earthquake on that day measured a 6.6 on the Richter scale, while the least intense one measured 2.5. Since these numbers are the result of logarithms, they are exponents – so their difference (4.1) tells us the exponent of the ratio of the original values. The strongest earthquake that day was $10^{4.1}=12,589$ times more powerful than the weakest one.
Decibels
The strength of a signal or sound is usually measured in decibels, which is calculated by dividing the signal’s strength by a standard reference level, taking the logarithm of the result, then multiplying that by 10. Therefore, a decibel measurement is ten times an exponent of ten. To interpret the meaning of a 20 dB measurement, divide it by ten and use that result as a power of ten
$10^{20/10}=10^2=100$
which tells you that the measured signal is one hundred times stronger than a 0 dB signal. A 30 dB signal is
$10^{30/10}=10^3=1,000$
one thousand times stronger than a 0 dB signal.
pH
The quantity of hydrogen ion activity determines how a solution reacts with many elements. A solution with a large quantity of active hydrogen ions is called “acidic”, and a solution with relatively few active hydrogen ions is called “basic”. pH is a measure of how acidic or basic a solution is, and is calculated by taking the logarithm of the reciprocal of the hydrogen ion activity in a solution. Therefore, a higher level of hydrogen ion activity will result in a a lower pH value. A pH value of 7 is considered neutral (neither acidic nor basic), so a solution with a pH of 5 would have
$10^{7-5}=10^2=100$
one hundred times as many active hydrogen ions as a neutral solution. A solution with a pH of 10 would have
$10^{7-10}=10^{-3}=\dfrac{1}{1,000}$
one thousandth as many active hydrogen ions as a neutral solution.
Pitch
The higher pitched of two musical notes that we perceive to be an octave apart has a frequency that is double that of the lower pitch. Therefore the formula for converting a sound frequency to a pitch uses $log_2$ to convert a frequency into a “number of doublings”. The number of doublings (relative to a standard pitch) tells us how many octaves (and fractions of an octave) apart the sounds are. Comparing the pitch of a 3,520 Hz tone with a 440 Hz tone, we can determine that they are
$log_2(3,520/440)\\*~\\*=log_2(8)\\*~\\*=log_2(2^3)\\*~\\*=3$
exactly three octaves apart.
Multiplication, Division, and Powers
If you do not have a calculator handy, but do have a table of logarithms handy (something that used to be true more often than it is today), you can use the table to speed up the process of manually calculating products, quotients, or powers. By converting the relevant numbers into powers of ten, you can then use the laws of exponents to add, subtract, or multiply the exponents to arrive at the exponent of the answer:
$(123,000)( 917,000)\\*~\\*=(1.23)(10^5)(9.17)(10^5)\\*~\\*=(1.23)(9.17)(10^{10})~~~~~~~~Laws~of~Exponents\\*~\\*=10^{log(1.23 \cdot 9.17)} \cdot 10^{10}~~~~~~~~Inverse~functions\\*~\\*=10^{log(1.23)+log(9.17)} \cdot 10^{10}~~~Laws~of~Logarithms\\*~\\*=10^{0.0899051+0.9623693} \cdot 10^{10}\\*~\\*=10^{1.0522744} \cdot 10^{10}$
The last step is to look up exponent of ten in the body of the table of logarithms to convert it back into the answer to the original problem. Since the two numbers I started with had three significant digits each, I rounded the answer to 6 significant digits:
$10^{1.0522744} \cdot 10^{10}\\*~\\*=10^{0.0522744} \cdot 10^{11}\\*~\\*=1.12791\cdot 10^{11}\\*~\\*=112,791,000,000$
A similar process, which also relies on the laws of exponents, can be used to raise a number to a power:
$3.1416^6\\*~\\*=10^{log(3.1416^6)}\\*~\\*=10^{6 \cdot log(3.1416)}~~~Laws~of~Logarithms\\*~\\*=10^{6 \cdot 0.49715}~~~~~~Using~table~of~logarithms\\*~\\*=10^{2.9829}\\*~\\*=10^2 \cdot 10^{0.9829}\\*~\\*=100 \cdot 9.6139~~~Using~table~of~logarithms~in~reverse\\*~\\*=961.39$
While the examples above may seem lengthy, with a little practice the process becomes a faster process than doing the calculation by hand. However… today’s calculators and computers are faster yet.
### Summary
The result of a logarithm is an exponent… an exponent of the logarithm’s base.
Logarithms make it easier to read comparable percentage changes for both very large and very small numbers on the same graph by converting them to powers of a common base.
Logarithms allow us to isolate a variable that is in the exponent of an exponential expression. Roots allow us to solve for a variable that is in the base of an exponential expression.
The laws of logarithms, like the laws of exponents, provide us with ways to manipulate the appearance of certain expressions or equations without affecting the quantities or relationships that they represent. They are another useful tool in our Algebraic toolbox, particularly when working with exponential expressions or equations.
For a slightly different approach to introducing logarithms, with additional useful or interesting information, check out “It’s the law too – the Laws of Logarithms“. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 33, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8559771180152893, "perplexity": 417.5410868885211}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202450.86/warc/CC-MAIN-20190320190324-20190320212035-00018.warc.gz"} |
https://ask.sagemath.org/questions/41219/revisions/ | # Revision history [back]
### Calculations in quotient of a free algebra
I want to define Sweedler's four-dimensional Hopf algebra, which is freely generated by $x,y$ and subject to the relations \begin{align} x^2 = 1, \qquad y^2 = 0, \qquad x\cdot y = - y\cdot x~ , \end{align} but I don't see how to do it.
I have tried the following:
sage: A.<x,y> = FreeAlgebra(QQbar)
sage: I = A*[x*x - 1, y*y, x*y + y*x]*A
sage: H.<x,y> = A.quo(I)
sage: H
Quotient of Free Algebra on 2 generators (x, y) over Algebraic Field by the ideal (-1 + x^2, y^2, x*y + y*x)
But then I get
sage: H.one() == H(x*x)
False
So is this currently possibly using a different method?
Thanks
### Calculations in quotient of a free algebra
I want to define Sweedler's four-dimensional Hopf algebra, which is freely generated by $x,y$ and subject to the relations \begin{align} x^2 = 1, \qquad y^2 = 0, \qquad x\cdot y = - y\cdot x~ , \end{align} but I don't see how to do it.
I have tried the following:
sage: A.<x,y> = FreeAlgebra(QQbar)
sage: I = A*[x*x - 1, y*y, x*y + y*x]*A
sage: H.<x,y> = A.quo(I)
sage: H
Quotient of Free Algebra on 2 generators (x, y) over Algebraic Field by the ideal (-1 + x^2, y^2, x*y + y*x)
But then I get
sage: H.one() == H(x*x)
False
So is this currently possibly using a different method?
Thanks
### Calculations in quotient of a free algebra
I want to define (the algebra part of) Sweedler's four-dimensional Hopf algebra, which is freely generated by $x,y$ and subject to the relations $$x^2 = 1, \qquad y^2 = 0, \qquad x\cdot y = - y\cdot x~ ,$$ but I don't see how to do it.
I have tried the following:
sage: A.<x,y> = FreeAlgebra(QQbar)
sage: I = A*[x*x - 1, y*y, x*y + y*x]*A
sage: H.<x,y> = A.quo(I)
sage: H
Quotient of Free Algebra on 2 generators (x, y) over Algebraic Field by the ideal (-1 + x^2, y^2, x*y + y*x)
But then I get
sage: H.one() == H(x*x)
False
So is this currently possibly using a different method?
Thanks
4 retagged tmonteil 22063 ●25 ●157 ●407 http://wiki.sagemath.o...
### Calculations in quotient of a free algebra
I want to define (the algebra part of) Sweedler's four-dimensional Hopf algebra, which is freely generated by $x,y$ and subject to the relations $$x^2 = 1, \qquad y^2 = 0, \qquad x\cdot y = - y\cdot x~ ,$$ but I don't see how to do it.
I have tried the following:
sage: A.<x,y> = FreeAlgebra(QQbar)
sage: I = A*[x*x - 1, y*y, x*y + y*x]*A
sage: H.<x,y> = A.quo(I)
sage: H
Quotient of Free Algebra on 2 generators (x, y) over Algebraic Field by the ideal (-1 + x^2, y^2, x*y + y*x)
But then I get
sage: H.one() == H(x*x)
False
So is this currently possibly using a different method?
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https://www.arxiv-vanity.com/papers/1608.06198/ | arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.
# Quantum Control Landscapes are Almost Always Trap Free: A Geometric Assessment
Benjamin Russell Department of Chemistry, Princeton University Herschel Rabitz Department of Chemistry, Princeton University Re-Bing Wu Department of Automation, Tsinghua University
April 12, 2020
###### Abstract
A proof is presented that almost all closed, finite dimensional quantum systems have trap free (i.e., free from local optima) landscapes for a large and physically general class of circumstances. This result offers an explanation for why gradient-based methods succeed so frequently in quantum control. The role of singular controls is analyzed using geometric tools in the case of the control of the propagator, and thus in the case of observables as well. Singular controls have been implicated as a source of landscape traps. The conditions under which singular controls can introduce traps, and thus interrupt the progress of a control optimization, are discussed and a geometrical characterization of the issue is presented. It is shown that a control being singular is not sufficient to cause control optimization progress to halt, and sufficient conditions for a trap free landscape are presented. It is further shown that the local surjectivity (full rank) assumption of landscape analysis can be refined to the condition that the end-point map is transverse to each of the level sets of the fidelity function. This mild condition is shown to be sufficient for a quantum system’s landscape to be trap free. The control landscape is shown to be trap free for all but a null set of Hamiltonians using a geometric technique based on the parametric transversality theorem. Numerical evidence confirming this analysis is also presented. This new result is the analogue of the work of Altifini, wherein it was shown that controllability holds for all but a null set of quantum systems in the dipole approximation. These collective results indicate that the availability of adequate control resources remains the most physically relevant issue for achieving high fidelity control performance while also avoiding landscape traps.
## 1 Introduction: Control Landscape Analysis
The study of quantum control landscapes (i.e., specific objective cost that depends on the of quantum time evolution operator as a function of an external field) is a topic of prime interest for assessing the viability of reaching a desired control outcome. As background, prior work has focused on applying differential geometry to several issues in quantum optimal control and quantum mechanics generally [1, 2, 3, 4, 5, 6, 7, 8, 9]. Other recent work [10] has also focused on applying geometric and Lie algebraic methods to controllability in quantum systems. Following similar principles, in this paper geometric methods are brought to bear on the analysis of quantum control landscapes. Specifically, we apply the geometric understanding of the ‘size’ (measure) of the set of singular values of smooth maps (Sard’s theorem [11]) to quantum control landscapes. Further, we apply the parametric transversality theorem, ([12] lemma 1, chapter 2) to reach the main conclusion. In particular, the present work facilitates understanding of the wide prevalence of trap free quantum control landscapes seen in practice, while also accommodating the fact that quantum control landscapes possessing traps appear highly atypical.
There has been success, in both experiment [13, 14, 15] and extensive simulation [16, 17, 18, 19] with the application of gradient based (or other local gradient estimation) algorithms in quantum control. The gradient ascent algorithm is sensitive the critical topology of the function being optimized. In particular, gradient ascent can converge to a local optimum if started from specific initial conditions within the basin of attraction. Consideration of this prospect garnered some controversy. In this regard, some potential issues have also been identified [20, 21] and debated [22] which may affect convergence to a global optima on the control landscape for specific systems. It has been shown that the presence of singular controls [23, 16, 20, 21, 22] can encumber the gradient ascent procedure in some very specific systems, all of which had to be specially constructed to have this unfavorable property. For the definition of singular controls, see sec. (2) and the appendix (A). Particularly the case of constant, or even zero, control fields has received attention [20, 21], partly because of the tractability of analyzing this case. Some examples of non-constant singular controls have been found numerically in [23], but they were all found to be saddles rather than local optima. There is also mounting numerical evidence that situations where singular controls inhibit progress during the gradient ascent procedure are, in several senses, rare [23, 24, 25, 14]. This collected evidence for the common ease of quantum control optimization, and the evident rarity of landscape traps motivated the pursuit of the central result of the present paper.
Finally we note that this work does not address the rate of convergence to the global optima and that this issue is also dependent on the particular search algorithm employed. Some numerical efforts demonstrated that a favorable rate of convergence is typically the case and remains so as the number of levels rises [26, 18, 19].
### 1.1 Summary of The Central Theorem Established in this Paper
Here we first summarize the key finding in this paper, followed by a detailed proof and discussion of the result. In particular we show that only a null set of quantum systems (within the space of all systems with a given number of levels) possess traps caused by singular critical controls. The sense in which this set is null can be understood as follows: if quantum systems are generated at random there is zero probability of finding an example with any singular traps. For a rigorous introduction to the analysis of measures and null sets, see [27].
The crux of the proof rests on a novel application of the parametric transverality theorem from differential geometry [28]. This result has significant implications for quantum control, which can be expressed informally by the following statement:
Consider a parameterized family of time dependent Hamiltonians which depend on finitely many control variables and finitely many real additional parameters . If, for all and (other than those corresponding to the global optima of the fidelity function), it is possible to increase fidelity by applying variations to and respectively, then it follows that the landscape as a function only of , is trap free for almost all fixed values of (i.e., for all but a null set of values).
This informal statement assumes that there is a physically suitable fidelity function quantifying the ‘cost’ of a Hamiltonian driven time evolution. For the above statement to hold, it is also required that the chosen fidelity function does not possess traps of its own, i.e. built into its mathematical structure. This circumstance is the case for many fidelity functions in popular use [29]. An example of a parameterized family of quantum systems, which could be assessed using the above boxed statement is given below:
H[En,λm](t) =(λ0λ1λ1−λ0)+E0sin(t)(0110) (1) +E1cos(t)(0i−i0)+(E2sin(2t)+E3cos(3t))(100−1)
The key implication of the results in this work is that gradient ascent methods will almost always succeed (independent of any initial guess for a control) for optimizing quantum dynamics (i.e., discovering a pulse of maximal fidelity) for almost all quantum systems. This conclusion applies to quantum systems with any finite number of levels. The result has practical significance for finding optimal control fields in simulation as well as in learning control experiments [30] attempting to discover shaped pulses which maximize fidelity in the laboratory. Such experiments, whether guided with a gradient procedure or an other suitable method are essentially a laboratory realization of the analogous simulation, wherein, a control pulse is systematically updated using the algorithmic rule until a high fidelity pulse is found. Henceforth in this work, the distinction between a control function , and vector of control variables will be suppressed unless the distinction is vital.
No attempt is made in this work to address the size of the basin of attraction of local traps in any given quantum control landscape. This interesting and important question is assessed in [16] where it is found that the attractor basin of traps in a few specific examples are small in an appropriate measured sense.
### 1.2 Quantum Systems and the Goals of Control
This paper studies the control of the quantum propagator for finite level quantum systems with Hamiltonians of the form:
^H(t)=^H0+E(t)^Hc (2)
where both the drift and control Hamiltonians are, respectively, traceless and is drawn from a finite, but potentially very high dimensional parameterized space of time dependent control fields. The traceless condition is taken in order that only evolution operators need to be considered rather than , which contains information about a physically redundant overall phase. For a more detailed discussion about the distinction between and with respect to singular controls and the effect of this specialization on the control landscape see [20, 31].
Initially this work studies the maximally controlled system, for which every Hamiltonian matrix element considered as under control; in this circumstance it is reasonably shown that the landscape is trap free. Importantly, incrementally reducing this latter high degree of control back down to the form of (2) is shown to introduce no traps, unless the reduction procedure is engineered to do so by fixing particular matrix elements to specific values within a set shown to be null at each step of the procedure.
In the case of controlling the propagator of a quantum system, the fidelity can be represented by a functional to be optimized:
F1[E]=R{Tr(^G†^UT)} (3)
where is some fixed final time and a target unitary. A phase independent version of this function is given by:
(4)
For several applications of this functional and further discussion see [20, 31]. These functionals both have the form:
F[E]=J(VT[E]) (5)
where is defined as (for example) and is the end-point map, that assigns to a control to the final time propagator which solves the corresponding Schördinger equation at time . The overall conclusion of the paper applies equally to both cost functions which have been shown to both possess only global and local maxima and saddle points [31].
Throughout this paper will be assumed to be large enough for the system to be fixed time controllable, that is to say, for any end-point unitary , there is a control which implements it in time . This property is also referred to as accessibility, it compliments the property of straightforward controllability, which states that: for each end-point unitary , the exists a control which implements at some final time . i.e., there exists at least one control such that for any given . A proof of the existence of such a time , which includes our present case, is given in ([32] theorem 30, Ch. 3).
There are several common objectives in quantum control. These can be broadly classified physically as:
1. Maximizing the expectation of a given (Hermitian) observable at time
2. Controlling an initial quantum state to reach a desired quantum state at time
3. Controlling the quantum propagator so that it is driven to a desired goal . In quantum information sciences applications is typically interpreted as implementing a quantum gate.
Although we will focus on the third of these objectives, the conclusions in this work apply to all of the above tasks as the objective function in the cases (1), (2) can be expressed as a function of the quantum propagator . For a discussion of the correspondence between tasks (2) and (3) from a geometric perspective, see appendix (C.1).
### 1.3 The Gradient Algorithm and the Prospect of Encountering Traps
Employment of the gradient algorithm to determine a control that maximizes fidelity naturally requires the computation of the gradient of the functional . The gradient of the functionals (3), (4) can be computed in closed form [33]. The variation of the end point w.r.t. the full Hamiltonian is given by:
δUT=UT∫T0U†tiδH(t)Utdt (6)
For more complex forms of control coupling, the gradient is not as simple when non-linear coupling to a control field is included, as in the case of control via polarizability [34]. However, this case is essentially the same as the first variation of the end point map with respect to the control in the linear coupling case, except, takes a different form. For further discussion of gradient decent/ascent methods see [35]. Employment of the gradient algorithm is important for understanding the topology of both the set of critical points, and that of the local (if any exist) and global maxima, as exemplified in figure (1).
Here we fix terminology relating to traps in the control landscape of a quantum system.
• A control is called a critical control or a critical point w.r.t. a given , if .
• A control is called a a second order critical point w.r.t. a given if (i.e., a critical control) and the Hessian is negative semi-definite. Such an may or may not be a true local optimum (i.e., a trap) depending on the nature of the higher derivatives with respect to .
• A control is called a trap w.r.t. a given if it is a local, but not global optima of the same .
### 1.4 The Three Assumptions of Landscape Analysis
There are three key assumptions of landscape analysis (see table 2), which are known to be sufficient for a quantum control landscape to be trap free. A clear statement, both in control theoretic and differential geometric terms, can be found in section (5). A key result in much of the recent work in landscape analysis is that these three assumptions imply that the gradient algorithm will converge to a globally optimal control without getting ‘stuck’ in a local optima (i.e., a trap). There is numerical evidence [18, 19] that this hypothesis holds with a wide scope of validity as well as experimental evidence [14, 15, 36, 37].
Earlier important work [10] has shown that the controllability assumption (1) in Table (2) discussed later is almost always satisfied given a pair . That is, the set of Hamiltonians for which controllability fails is a null set. It is however, an open problem to completely characterize the set on which controllability fails. The two most widely applied and useful criteria for controllability are the Lie algebra rank condition (LARC) and the connected graph criterion [10]. A clear example for numerically checking controllability by the Lie algebra rank condition can be found in [38] where an insightful graphical representation of the process is presented. However, only limited work has been performed on the the satisfaction of the local surjectivity assumption (2) in Table (2) and its impact on the performance of gradient algorithms. In this work we present an analogous result to that of [10] for controllability, which applies to the local surjectivity assumption (2). In addition, the final assumption (3), is that there are sufficient control resources (i.e., the control space is sufficient for the end-point map to be globally surjective) to freely explore the landscape. The new result, considering all three assumptions, explains why gradient ascent convergence to a global optimum is so typical in practice despite the fact that some engineered special examples with traps are known.
## 2 Singular Controls and Properties of the End-Point Map
### 2.1 Kinematic and Dynamical Optima
The scenario here concerns discovering a control scheme driving the time evolution operator to a desired goal at time (i.e., ). This process is represented by the following commutative diagram,
is an objective function to maximize, and is a pre-defined space of control fields. Applying the chain rule to yields:
δFδE=dJdVT[E]∘δVT[E]δE (7)
• A control is said to be singular if the set of all for which there exists a corresponding (i.e., a value of ) doesn’t span at the point . This is to say that the Fréchet derivative:
δVT∣∣E:δE↦δUT (8)
is not of maximal rank at the point in the space of controls. The co-dimension of the image of given by is called the co-rank of the control.
There are two types of critical points of : ones for which , and those for which is not of full rank in . Consideration of is referred to as the kinematic control landscape and is referred to as the dynamic control landscape. Figure (2) clarifies the relationship between the kinematical and dynamical landscapes. Ultimately, the goal is to understand if singularities of can introduce critical points of which are not critical points of , i.e. singular controls which introduce new critical points into the landscape of . Understanding which systems have no such singular critical controls will elucidate the circumstances for which the critical point structure of and are the same. For these systems an analysis of the kinematic landscape alone suffices to understand the full control landscape of . This is a desirable goal as it facilitates reaching the conclusion that one only needs to consider the prospect of traps arising directly from a fidelity function, and thus that an appropriate choice of fidelity function is sufficient to result in a trap free landscape [29].
With the remarks above in mind we make the following definitions and observation:
• A Kinematic Critical Point is a control such that .
• A Dynamic Critical Point is a control such that .
It is clear from eqn. (7) that all kinematic critical points are dynamic but that the converse is not true unless is full rank for all . It is important to understand the nature of both types of critical points. Ultimately, the most salient question is: do singular controls introduce local optima into quantum control landscapes and if so, what is the practical ramifications of this both in simulations and laboratory learning control [30] scenarios?
### 2.2 Vt is a smooth map
Here properties of needed in order to apply the parametric transversality theorem are established. Considering controls only drawn from either or any finite dimensional vector space of smooth functions, has the properties of being a smooth function of , a smooth function of and a smooth function of and . We use this fact without giving a proof, however a proof can be given by using the so called ‘convenient calculus’ [39]. The proof is long and involved as well as a direct parallel of many existing proofs, so it is omitted. We further observe, by the smoothness of the matrix exponential and of matrix multiplication, that the end-point map is also smooth on any space of piecewise smooth controls. This fact will be required in ensuing geometric analysis.
## 3 Climbing the Landscape: Transverality to the Level Sets of J is Sufficient to Climb
This section will show the failure of local surjectivity will not necessarily cause a gradient assent to halt. We show that a significantly weaker condition rather than local surjectivity is sufficient to exclude traps, namely: being transverse to the level sets of the fidelity function. We also argue that the set of Hamiltonians for which a search will halt are a null set under some physically reasonable assumptions.
### 3.1 Transversality
Firstly, the concept of transversality is introduced as an abstract property of smooth maps between manifolds. Secondly, the end point map is shown to possess the property of transversality by taking specific instances of the manifolds in the definition of transversality.
The gradient of a smooth function on a smooth manifold is always perpendicular to level sets of this function (if the same Riemannian metric is used throughout). If local surjectivity of fails somewhere on a specific level set of on , it may not matter as far as climbing the landscape when using gradient assent. All that matters is an ability to ascend the landscape, not necessarily to traverse the level set itself. If there does not exist a which causes to vary in a specific direction within the level set of containing , this is not problematic for gradient ascent, but would still indicate the failure of surjectivity. A control may be singular, even up to co-rank of the dimension of the level sets of containing , but as long as there exists a such that has a non-zero component in the direction of the gradient on , then it will not impair the ability to climb the landscape (i.e., increase ) by introducing a small variation of . With this in mind, we state the following definition of a transverse map:
• , Given two smooth manifolds, , a submanifold , and a smooth map (with ), is called transverse to (denoted ):
Image(dϕ∣∣p)⊕Tϕ(p)L=Tϕ(p)M ∀p∈ϕ−1(L) (9)
The concept above is illustrated in figure (3). In this work, only the case that is globally surjective (i.e., an onto function) will be important. This case corresponds to only considering controllable systems. This renders redundant the condition that is a subset of the image of because the image of is the whole of for quantum systems which are fixed time controllable. The results reported in [10] and (Theorem 12, Chap. 6 [32]), can be paraphrased as: for almost all , is globally surjective.
## 4 Applying the Parametric Transversality Theorem to Quantum Control
In this section we will utilize the parametric transversality theorem as the key to facilitate proving the rarity of quantum systems with traps. The proof of this theorem (which follows from Sard’s theorem [11]) is complex and is omitted, but well known within differential geometry.
Here the statement of the parametric transversality theorem is given prior to it’s application to quantum control.
###### Theorem 4.1 (Parametric Transversality Theorem, [12] lemma 1, chapter 2)
Given smooth manifolds, and , a submanifold , and a parameterized family of smooth maps where (parameterized by ), then if defined by: is transverse to (when variations of are considered), then almost all values of have transverse to . Furthermore, the pre-image of is a submanifold of with codimension in equal to the codimension of in .
The following result of differential geometry aids in building intuition about Theorem 4.2:
Given a family of smooth maps parameterized (by drawn from a smooth manifold), for almost all values of s, is transverse of a given submanifold if the family as a whole is transverse to the same .
It is also clarifying to note that if , then a map being transverse to is equivalent to it being locally surjective.
The parametric transversality theorem can be applied to the case of quantum control landscapes for finite level systems. One can set to be a (finite, large dim, manifold of control fields), to be and to be any level set of , the cost functional. In order to analyze the quantum control landscapes, is given by (for some fixed ). The parameter (the set of maps ) can be taken to parameterize values in a set of (potentially time dependent) Hamiltonians. For each time dependent Hamiltonian , there is an end-point mapping . This is clarified in equation (10) below and within the ensuing discussion.
### 4.1 The Central Theorem
In this section we apply the parametric transversality theorem to a large class of quantum control problems. We show that this analysis allows one to conclude that only a null set of quantum systems have singular critical points.
In this section we denote by , a high but finite, dimensional space of control fields taken to consist of piecewise constant (with pieces) functions bounded in magnitude by . We will denote by a basis for this space, and by an orthonormal basis of .
Consider, initially, the scenario of having total control over a system’s Hamiltonian as a function of time (i.e., all matrix elements under control) within the confines of the space , one can then show that successively restricting the degree of control does not generically introduce any singular critical points. The ensuing physical and mathematical argument in no way rests on the assumption that all such Hamiltonians can be created in the laboratory in practice. We initially postulate the existence of such a rich space of control fields and the full degree of coupling permitted by these fields, and then progressively restrict the degree of control while assessing the effect that each restriction has on traps in the landscape. Specifically, we initially study every curve of the form:
iH(t)=∑jfj(t)Bj=∑j,kaj,k gk(t)Bj (10)
as a Hamiltonian where is given by:
gk(t)={1 Tkp≤t≤T(k+1)p,0 else (11)
The case of a Hamiltonian with drift and control , referred to in the single (piecewise constant) field dipole approximation, is included within the above form (10). It is now clear that there exists such that the end-point map is surjective as a map from for some values of . Applying a variation , the end point change becomes:
U†TδUT =∫T0U†tδ(iH(t))Ut dt (12) =∑j,k∫T0δaj,k gk(t) U†tBjUt dt
Assuming that is an orthonormal basis of , the duration of each piecewise constant segment of the control is short enough (i.e. is large enough) and the speed of the curve is not too high (i.e. is small enough), then the set is also a (not necessarily orthonormal) basis of . Here is the piece defined by . To understand exactly what is meant be ‘short enough’, one must understand the singularities of the matrix exponential (C). This implies that any variation can be created by some variation admissible within this space of piecewise constant controls, and thus that the end-point map is locally surjective everywhere on this space of controls.
Adopting the above premises, one can now apply the parametric transversality theorem to conclude that fixing the value of any one of the control the parameters to a given constant will not introduce singular critical points into the landscape for all but a null set of fixed values.
###### Theorem 4.2
For a system with with Hamiltonian of the form (10) and space of piecewise constant controls as above such that the end-point map is locally surjective everywhere in the control space, fixing any single control parameter introduces singular critical points into the new control landscape (a function of the remaining, unfixed variables only) only for a null set of values of .
• Let be the end point map of the Schrödinger equation. This map can be considered to depend on , as these parameters are sufficient to determine , so we will denote the end-point map accordingly. As such, (i.e., one control field for each ) where we have taken to be sufficient for to be surjective as explained above. This is equivalent to saying: , which in turn implies: is transverse to every sub-manifold of . Henceforth, will have left unwritten, assumed to have these values taken so that is locally surjective everywhere in its domain.
We can now decompose the domain of into two parts by selecting one parameter , and considering as a one parameter family of maps of the remaining variables . One can now apply the parametric transversality theorem by considering the values of as , the values of the remaining variables as and as and conclude that only a null set of values cause this new restricted map to fail to be locally surjective.
Given theorem (4.2), one can assume that one value has been fixed to a specific value not in the null set of values which introduce singular critical points. Fixing another , and applying an identical argument, one finds that only a null set , introduce singular critical points. This process can now proceed inductively. Note that these null sets need not be the same set of values. Proceeding by induction, one sees that fixing any number of the free parameters describing the system, does not introduce singular critical points except for a null set of fixed values at each step. It may, however, cause the end-point map to fail to be globally surjective, the property generically preserved by fixing parameters is that of being locally surjective within in image of , without reference to what that image is. This eventual breakdown of global surjectivity corresponds to loss of controllability and thus a shrinking of the reachable set.
We have shown that local surjectivity, which implies transversality to the level sets of any smooth function, only fails for a null set of quantum systems. This technique can be applied to show that transversality of to the level sets of a specific objective function is also generic. Considering a family of systems parameterized by a vector of numbers (for example: the coupling constants of a spin chain system controlled by a magnetic field). Then, by a nearly identical argument, if is transverse to a given level set of fidelity when variations of both and are admissible, then all but a null set of fixed values of leave transverse to the same set when only variations of are considered. We further note that the argument of the central theorem does not rest on the initial appeal to a space of piecewise constant controls. The assumption of a finite dimensional space of controls is physically unrestrictive as the space of controls can be given a very high dimension leaving the space of control fields still including all those which can be physically implemented. One way to achieve this truncation of control space, is to only consider control fields with ‘bounded variation’ (band limited in the terminology of Fourier series). Such control fields possess no frequencies above a certain critical frequency. If this critical frequency is high enough, then any discarded component of the control field would not affect the time evolution differently from noise. An identical argument could be made if one started with any finite dimensional space of controls which can be shown to render the end-point map locally surjective.
The following table 1 clarifies the correspondence between the abstract statement of the parametric transversality theorem (PTT) and its application to quantum control.
### 4.2 Context and Physical Relevance
The result of section (4.1) is a parallel of the central result of Altifini [10]. Altifiti’s key result can be restated as: for almost all (i.e., all but a null set) the map is globally surjective (onto) on . Clearly, this result includes cases with more than one control field as it applies to the case with only a single control field and adding additional control fields cannot destroy controllability. In the same spirit, the conclusion of section 4.1 is that almost all Hamiltonians will not produce singular traps.
It has been discussed in [34] that while many practical control scenarios are described by a Hamiltonian in the dipole approximation, laboratory scenarios can include more complex forms of coupling. As such, the mathematical existence of traps in systems in the dipole approximation, does not necessarily translate into the existence of traps in laboratory experiments. Complex (non-linear) coupling of control fields within the system’s Hamiltonian and coupling to the environment can be present in reality, even if they are only very small. Noise in the control field is also present in practice. If fixed values of complex coupling to external fields are chosen at random, with certainty traps will not be present in the landscape. This implies that, if a well validated model of a quantum system is shown to possess traps, even the least inaccuracy in the model or the presence of any additional types of coupling to the control fields, will very likely remove these traps by effectively perturbing the Hamiltonian matrix elements out of the failing set.
### 4.3 Numerical Confirmation
In order to confirm the conclusions, of the central Theorem 4.2, numerical simulations were run. pairs (i.e., four level systems) were generated at random and optimizations with random initial controls were run using a gradient ascent algorithm. The control fields tested were defined to be piecewise constant with pieces with appropriately bounded magnitude, as per the central result above. Initially, a singular control was generated at random (see appendix 24) by choosing at random (i.e., a procedure specifically seeking a singular control, rather than one designed to avoid them). The bounded magnitude premise of the central theorem was imposed via rejection sampling on . We will denote by the control created from formula 24 with a given value of (with normalized s.t. ). The time evolution simulation is given by:
dUtdt=(a+E(t)b)Ut (13)
Stochastic gradient ascent (over ) was then run to maximize the quantity:
⟨B, U†T∇J∣∣UT⟩ (14)
One readily confirms that this quantity will be maximized for a singular critical point and at no other point. It was found that in all generated cases , no singular critical controls consistent with the magnitude bound premise of Theorem 4.2 were discovered as the algorithm did not converge. As many initial conditions were tested, this is strong numerical confirmation that such controls do not exist.
## 5 Discussion and Conclusions
Theorem 4.2 has been presented describing the structure of typical quantum control landscapes by introducing a novel tool from differential geometry extracted from the parameteric transversality theorem. The technique used to obtain Theorem 4.2 is novel, and potentially has scope well beyond quantum control. We have shown that quantum systems with singular critical controls are rare in the space of all possible quantum control systems, i.e., all systems evolving under the Schrödinger equation with some coupling to external control fields. In order for the transversality result to apply specifically to the dipole approximation with a single field it would be required to show that the restrictions on the maximally controllable Hamiltonian do not leave the remaining Hamiltonian within the null set possessing traps. However, as the set possessing traps is null, there exists a perturbation to any controllable system possessing traps, which removes them. We have further formalized a sufficient condition for systems with singular critical points to possess no second order critical points (B). Attempting to show that this condition almost always holds will be the focus of further work also based on the parametric transversality theorem.
We have refined the surjectivity assumption of quantum control landscape theory to that of transversality. We argued that the end point map being transverse to all level sets of fidelity is sufficient for the dynamical and kinematical landscapes to share the same critical point structure. A novel technique for showing that a very large class of realistic systems possess this property was also presented.
The original and current status of the three assumptions of quantum control landscape analysis are given here, so that the new and original forms and statues of each can be compared.
It is possible to check that these assumptions imply that the dynamical landscape almost always possess the same critical point structure as the kinematical one. This can be achieved by observing that the composition of two functions, , has the same critical point structure as that of if satisfies both transversality and globally surjectivity. Finally, we note in the current assumptions of Table 2 that assumptions 1 and 2, are almost always satisfied. As a result, it is clear that the primary determining factor for the ease for optimization in quantum control, for the vast majority of Hamiltonians, is the availability of sufficient control resources, assumption (3).
We also note that transversality only requires that the end-point map has rank 1 and the is contained within its range. This is in contrast to the requirement that is full rank, because full means that the rank is equal to the dimension of the group , which is . The condition that local surjectivity be satisfied becomes more demanding on the rank of as the number of level rises, which is in contrast to transversality.
## 6 Acknowledgments
The authors would like to thank Robert Kosut for many helpful discussions and comments during the drafting of this work, and Daniel Burgarth for some corrections. Benjamin Russell acknowledges the support of DOE (grant no. DE-FG02-02ER15344. Herschel Rabitz acknowledges the support of the Army Research Office (Grant No. W911NF-16-1-0014). We also acknowledge partial support from the Templeton foundation.
## Appendix A Singular Controls in the Dipole Approximation and Beyond
Here we give formulas for singular controls in the case of a single control field in the dipole approximation. In order to express all relevant quantities in terms of Lie algebraic objects in we make the following definitions:
a:=−iH0 (15) b:=−iHc
Much of what is presented in this section and throughout is applicable, appropriately modified, to control systems of the analogous form on compact, connected, semi-simple Lie groups other than .
In this notation, the Schrödinger equation for a controlled quantum system (2) reads:
dU(t)dt=(a+E(t)b)U(t) (16)
The authors are only aware of singular control trapping studies in the dipole approximation expressed here other than in a recent work [34].
It is possible to explicitly express in terms of [40];
δVT[w]=U(T)∫T0δE(t)U(t)†bU(t)dt (17)
The right translation of the is given by:
U(T)†δU(T) =∫T0δE(t)U(t)†bU(t)dt (18) =∫T0δE(t)btdt
where we have defined so that it solves the adjoint equation . We note that this form (i.e., the integral of an adjoint orbit in a Lie algebra) is only possible because we have a Lie group of time evolution operators and that form is unique to such a scenario.
If the control , which drove the system along the trajectory , is singular, then there exists, be definition, at least one such that:
⟨BU(T,0),δU(T,0)⟩U(T,0)=0 ∀δE (19)
where is any inner product on the tangent space and . A convenient choice of inner product is given by the unique (up to a constant multiple) bi-invariant Riemannian metric on . This metric is expressed as the right (or left, as the two coincide) translation of the Killing form :
K(X,Y)=\Tr(X†Y) (20)
The condition that a control is singular can now be written, by applying formulas (18, 19), in terms of this inner product:
K(U(T)†δU(T),B)=0⇒ (21) K(∫T0δE(t)btdt,B)=0⇒ ∫T0δE(t)K(bt,B)dt=0
In scenarios where the set of considered is large enough, one can apply the fundamental lemma of the calculus of variations and conclude that:
K(bt,B)=0 ∀t∈[0,T] (22)
The interpretation of this equation is that infinitesimally varying doesn’t in turn vary in the direction for any time during the evolution. From this result, the form of the singular controls can be deduced by differentiating w.r.t .
Noting that has dropped out as , one differentiates again w.r.t to find:
assuming that the denominator is never zero. If it is zero, further differentiation is required which yields another expression containing higher levels of nested commutators. Another formula can also be obtained by applying a symmetry of the Killing form to obtain from (22)
K(b,Bt)=0 ∀t∈[0,T] (25)
where is defined similarly to above. Closely following [23], the form of the singular controls can be similarly determined in the case of controlling the density matrix.
From formula (24), one sees that not all singular controls are constant. One also observes that there is no reason, a priori, to preclude the possibility of encountering singular controls during gradient ascent for systems in the dipole approximation. However, formula (24) also indicates that there are very few singular controls relative to the size of the total control space. This can be deduced from the one-to-one correspondence between , and singular controls. Given that the dimension of is , and the control space typically will have far larger dimension for an level system, this indicates that singular controls are not prevalent in the control space. If the Hamiltonian contains quadratic coupling to a single control field, a similar implicit formula for a singular control can be found [34] by a like procedure.
## Appendix B Second Order Analogue of the Central Theorem 4.2
The parametric transversality theorem can be applied to the Hessian of the fidelity by defining a new map :
QT(E,δE):=(VT(E),dVT∣∣E(δE)) (26)
where is the push-forward of the map . This approach is appropriate for showing that only a null set of systems, which possess singular critical points, possess second order critical points. Another definition is now needed:
• Given two vector spaces define . That is is the set of rank deficient linear maps from to .
It is now possible to state the condition that a given quantum system has no second order critical points in terms of transversality of the map to a specific submanifold of the range of the same map.
If is the submanifold constructed from the union of all the level sets (of J) which contain a singular critical value of , define a submanifold of as:
L=S×Σ (27)
The condition that there are no second order critical points now reads: , which renders it amenable to assessment via application of the parametric transversality theorem. Checking for which systems this condition holds, and similarly for higher order conditions, will form the basis of further work.
## Appendix C Singularities of the Matrix Exponential
The matrix exponential possess singularities at, and only at, any matrix for which has as an eigenvalue for any . In any quantum system with a piecewise constant Hamiltonian (with pieces of duration labeled as , which are not time dependent), the propagator can be expressed as a product of exponentials:
UT=eδtiH(K)⋯eδtiH(1) (28)
The eigenvalues of are where are the eigenvalues of [41] for any square matrix . If the difference of the highest and lowest magnitude eigenvalue of (which equals the magnitude largest eigenvalue of ) is less that , then the map is locally surjective. If all meet this premise, then clearly the overall end-point map is surjective. If a given does not meet the premise, then reducing sufficiently will restore the premise. Specifically, systems such that:
δt<2π∣∣E(n)max−E(n)% min∣∣ (29)
where is the largest eigenval of (and similarly) for all , will have no singularities.
More generally, the Lie theoretic exponential map, , has the property that it is invertible (i.e., a diffeomorphism) in a neighborhood of , i.e., is invertible near to zero (Corollary 3.44, [42]). This indicates that it is free from singularities, and thus that for piecewise constant controls, with sufficiently small pieces, the end-point map is also free from singularities.
### c.1 The Relationship between Singular Trajectories when Controlling the Propagator and when Controlling a Quantum State
It has been noted in [23] that there exist controls for systems of the type (2) which are singular for the control of the the propagator and not for the control a quantum state. Here we discuss the geometric basis of this finding using the construction of complex protective space as the quantum state space of pure states is a homogeneous space of . We also discuss why this construction has no analogue in the case of a mixed state.
In geometric quantum mechanics the space of pure states, up to equivalence of states by normalization and global phase, is a manifold known as a complex projective space [8]. For other applications of this construction in quantum control see [43, 1, 2]. This relationship can be formally expressed as:
CPn−1≅SU(n)/U(n−1) (30)
where the symbol refers to the quotient of into cosets. The special case of the space , is the familiar Bloch sphere for a qubit. This construction is only possible because acts transitively on , see [44] for details of this construction. As unitary evolution preserves the degree of mixedness of a mixed state, does not act transitively on the manifold of all density matrices representing mixed states. Due to this lack of transitive action (i.e., not all states can be transformed into all others via unitary evolution), this space cannot be represented as a homogeneous space of . Because of this circumstance, it is not as straight forward to form an analogous construction for mixed states.
The space where the quantum state resides is a quotient of the space where the time evolution operator resides. Because of this, each direction in the tangent space at some point in corresponds to more than one direction in the tangent space at some point in . As such, the existence of a direction , which does not correspond to any admissible variation , is not sufficient for such a direction to exist in when the time evolution of the state is standardly defined by . Thus, not all controls of the form (24) are singular for the control of a pure state. This construction provides insight into which controls can be singular for the control of a pure state, as it reveals the exact sense of: several directions in which and be steered by variation of corresponding to a given variation of . Thus, it is clear that not every control which is singular for the control of the propagator is singular for the control a pure state. The practical significance of this result is that if the landscape for the control of the propagator is trap free, then the landscape for the control of the state is also trap free. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9243329763412476, "perplexity": 400.1933039213625}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347436466.95/warc/CC-MAIN-20200603210112-20200604000112-00479.warc.gz"} |
https://www.albert.io/ie/general-chemistry/calculating-the-molar-mass-of-an-ideal-gas | Free Version
Moderate
Calculating the Molar Mass of an Ideal Gas
CHEM-01MBY1
If $1.76\text{ g}$ of an ideal gas occupy $1.0\text{ L}$ at standard temperature and pressure (STP), what is the molar mass of the gas?
$\left(R=0.0821\cfrac{L\times atm}{mol\times K}\right)$
A
$4.0\text{ g/mol}$
B
$16\text{ g/mol}$
C
$20\text{ g/mol}$
D
$40\text{ g/mol}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9069265723228455, "perplexity": 781.0509080348277}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170404.1/warc/CC-MAIN-20170219104610-00207-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/coulombs-law-use.485130/ | # Coulomb's law use
• Start date
• #1
4
0
1. The problem statement, all variables given/known data
A particle of charge 6 µC is held fixed while another particle of charge 8 µC is released
from rest at a distance of 1.4 m from the first particle. If the mass of the second particle is
3x10-6 kg, what is its speed when it is very far away from the first particle?
F=ma
F= kQ1Q2 / r2
## The Attempt at a Solution
I pretty much worked through most the problem. In the end, I used coulomb's law in combination with newtons 2nd law to get a= 73469 m/s2. However what's bugging me is the distance that the speed is "very far away". I'm under the assumption that means at a point where the force from the first particle no longer affects the second. However, when I back-tracked using the answer(454m/s) the distance ends up being 1.4( the original distance) with the velocity equations.
Related Introductory Physics Homework Help News on Phys.org
• #2
4
0
Apologies, I searched google and found a similar problem: rocket is launched straight up from the earth's surface at a speed of 1.60×10^4 m/s.
What is its speed when it is very far away from the earth?
One of the helpers suggested conservation of energy, and I applied this to this problem it worked! Sorry!
• #3
gneill
Mentor
20,801
2,778
You do realize that the acceleration is not constant? It decreases as the distance between the particles grows. So your V2 = Vo + 2ad formula is not valid over the trajectory of the second particle.
Why not try a conservation of energy approach?
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1K | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8955975770950317, "perplexity": 1045.9302476360717}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657149819.59/warc/CC-MAIN-20200714083206-20200714113206-00101.warc.gz"} |
https://www.dsprelated.com/freebooks/filters/Elementary_Filter_Sections.html | ## Elementary Filter Sections
This section gives condensed analysis summaries of the four most elementary digital filters: the one-zero, one-pole, two-pole, and two-zero filters. Despite their relative simplicity, they are quite valuable to master in practice. In particular, recall from Chapter 9 that every causal, finite-order, LTI filter (any difference equation of the form Eq.(5.1)) may be factored into a series and/or parallel combinationof such sections. Implementing high-order filters as parallel and/or series combinations of low-order sections offers several advantages, such as numerical robustness and easier/safer control in real time.
### One-Zero
Figure B.1 gives the signal flow graph for the general one-zero filter. The frequency response for the one-zero filter may be found by the following steps:
By factoring out from the frequency response, to balance the exponents of , we can get this closer to polar form as follows:
We now apply the general equations given in Chapter 7 for filter gain and filter phase as a function of frequency:
A plot of and for and various real values of , is given in Fig.B.2. The filter has a zero at in the plane, which is always on the real axis. When a point on the unit circle comes close to the zero of the transfer function the filter gain at that frequency is low. Notice that one real zero can basically make either a highpass ( ) or a lowpass filter ( ). For the phase response calculation using the graphical method, it is necessary to include the pole at .
### One-Pole
Fig.B.3 gives the signal flow graph for the general one-pole filter. The road to the frequency response goes as follows:
The one-pole filter has a transfer function (hence frequency response) which is the reciprocal of that of a one-zero. The analysis is thus quite analogous. The frequency response in polar form is given by
A plot of the frequency response in polar form for and various values of is given in Fig.B.4.
The filter has a pole at , in the plane (and a zero at = 0). Notice that the one-pole exhibits either a lowpass or a highpass frequency response, like the one-zero. The lowpass character occurs when the pole is near the point (dc), which happens when approaches . Conversely, the highpass nature occurs when is positive.
The one-pole filter section can achieve much more drastic differences between the gain at high frequencies and the gain at low frequencies than can the one-zero filter. This difference is achieved in the one-pole by gain boost in the passband rather than attenuation in the stopband; thus it is usually desirable when using a one-pole filter to set to a small value, such as , so that the peak gain is 1 or so. When the peak gain is 1, the filter is unlikely to overflow.B.1
Finally, note that the one-pole filter is stable if and only if .
### Two-Pole
The signal flow graph for the general two-pole filter is given in Fig.B.5. We proceed as usual with the general analysis steps to obtain the following:
The numerator of is a constant, so there are no zeros other than two at the origin of the plane.
The coefficients and are called the denominator coefficients, and they determine the two poles of . Using the quadratic formula, the poles are found to be located at
When the coefficients and are real (as we typically assume), the poles must be either real (when ) or form a complex conjugate pair (when ).
When both poles are real, the two-pole can be analyzed simply as a cascade of two one-pole sections, as in the previous section. That is, one can multiply pointwise two magnitude plots such as Fig.B.4a, and add pointwise two phase plots such as Fig.B.4b.
When the poles are complex, they can be written as
since they must form a complex-conjugate pair when and are real. We may express them in polar form as
where
is the pole radius, or distance from the origin in the -plane. As discussed in Chapter 8, we must have for stability of the two-pole filter. The angles are the poles' respective angles in the plane. The pole angle corresponds to the pole frequency via the relation
where denotes the sampling interval. See Chapter 8 for a discussion and examples of pole-zero plots in the complex -plane.
If is sufficiently large (but less than 1 for stability), the filter exhibits a resonanceB.2 at radian frequency . We may call or the center frequency of the resonator. Note, however, that the resonance frequency is not usually the precise frequency of peak-gain in a two-pole resonator (see Fig.B.9 on page ). The peak of the amplitude response is usually a little different because each pole sits on the other's skirt,'' which is slanted. (See §B.1.5 and §B.6 for an elaboration of this point.)
Using polar form for the (complex) poles, the two-pole transfer function can be expressed as
(B.1)
Comparing this to the transfer function derived from the difference equation, we may identify
The difference equation can thus be rewritten as
(B.2)
Note that coefficient depends only on the pole radius R (which determines damping) and is independent of the resonance frequency, while is a function of both. As a result, we may retune the resonance frequency of the two-pole filter section by modifying only.
The gain at the resonant frequency , is found by substituting into Eq.(B.1) to get
(B.3)
See §B.6 for details on how the resonance gain (and peak gain) can be normalized as the tuning of is varied in real time.
Since the radius of both poles is , we must have for filter stability8.4). The closer is to 1, the higher the gain at the resonant frequency . If , the filter degenerates to the form , which is a nothing but a scale factor. We can say that when the two poles move to the origin of the plane, they are canceled by the two zeros there.
#### ResonatorBandwidth in Terms of Pole Radius
The magnitude of a complex pole determines the damping or bandwidth of the resonator. (Damping may be defined as the reciprocal of the bandwidth.)
As derived in §8.5, when is close to 1, a reasonable definition of 3dB-bandwidth is provided by
(B.4) (B.5)
where is the pole radius, is the bandwidth in Hertz (cycles per second), and is the sampling interval in seconds.
Figure B.6 shows a family of frequency responses for the two-pole resonator obtained by setting and varying . The value of in all cases is , corresponding to . The analytic expressions for amplitude and phase response are
where and .
### Two-Zero
The signal flow graph for the general two-zero filter is given in Fig.B.7, and the derivation of frequency response is as follows:
As discussed in §5.1, the parameters and are called the numerator coefficients, and they determine the two zeros. Using the quadratic formula for finding the roots of a second-order polynomial, we find that the zeros are located at
If the zeros are real [ ], then the two-zero case reduces to two instances of our earlier analysis for the one-zero. Assuming the zeros to be complex, we may express the zeros in polar form as and , where .
Forming a general two-zero transfer function in factored form gives
from which we identify and , so that
is again the difference equation of the general two-zero filter with complex zeros. The frequency , is now viewed as a notch frequency, or antiresonance frequency. The closer R is to 1, the narrower the notch centered at .
The approximate relation between bandwidth and given in Eq.(B.5) for the two-pole resonator now applies to the notch width in the two-zero filter.
Figure B.8 gives some two-zero frequency responses obtained by setting to 1 and varying . The value of , is again . Note that the response is exactly analogous to the two-pole resonator with notches replacing the resonant peaks. Since the plots are on a linear magnitude scale, the two-zero amplitude response appears as the reciprocal of a two-pole response. On a dB scale, the two-zero response is an upside-down two-pole response.
### Complex Resonator
Normally when we need a resonator, we think immediately of the two-pole resonator. However, there is also a complex one-pole resonator having the transfer function
(B.6)
where is the single complex pole, and is a scale factor. In the time domain, the complex one-pole resonator is implemented as
Since is complex, the output is generally complex even when the input is real.
Since the impulse response is the inverse z transform of the transfer function, we can write down the impulse response of the complex one-pole resonator by recognizing Eq.(B.6) as the closed-form sum of an infinite geometric series, yielding
where, as always, denotes the unit step function:
Thus, the impulse response is simply a scale factor times the geometric sequence with the pole as its term ratio''. In general, is a sampled, exponentially decaying sinusoid at radian frequency . By setting somewhere on the unit circle to get
we obtain a complex sinusoidal oscillator at radian frequency rad/sec. If we like, we can extract the real and imaginary parts separately to create both a sine-wave and a cosine-wave output:
These may be called phase-quadrature sinusoids, since their phases differ by 90 degrees. The phase quadrature relationship for two sinusoids means that they can be regarded as the real and imaginary parts of a complex sinusoid.
By allowing to be complex,
we can arbitrarily set both the amplitude and phase of this phase-quadrature oscillator:
The frequency response of the complex one-pole resonator differs from that of the two-pole real resonator in that the resonance occurs only for one positive or negative frequency , but not both. As a result, the resonance frequency is also the frequency where the peak-gain occurs; this is only true in general for the complex one-pole resonator. In particular, the peak gain of a real two-pole filter does not occur exactly at resonance, except when , , or . See §B.6 for more on peak-gain versus resonance-gain (and how to normalize them in practice).
#### Two-PolePartial Fraction Expansion
Note that every real two-pole resonator can be broken up into a sum of two complex one-pole resonators:
(B.7)
where and are constants (generally complex). In this parallel one-pole'' form, it can be seen that the peak gain is no longer equal to the resonance gain, since each one-pole frequency response is tilted'' near resonance by being summed with the skirt'' of the other one-pole resonator, as illustrated in Fig.B.9. This interaction between the positive- and negative-frequency poles is minimized by making the resonance sharper ( ), and by separating the pole frequencies . The greatest separation occurs when the resonance frequency is at one-fourth the sampling rate ( ). However, low-frequency resonances, which are by far the most common in audio work, suffer from significant overlapping of the positive- and negative-frequency poles.
To show Eq.(B.7) is always true, let's solve in general for and given and . Recombining the right-hand side over a common denominator and equating numerators gives
which implies
The solution is easily found to be
where we have assumed im, as necessary to have a resonator in the first place.
Breaking up the two-pole real resonator into a parallel sum of two complex one-pole resonators is a simple example of a partial fraction expansion (PFE) (discussed more fully in §6.8).
Note that the inverse z transform of a sum of one-pole transfer functions can be easily written down by inspection. In particular, the impulse response of the PFE of the two-pole resonator (see Eq.(B.7)) is clearly
Since is real, we must have , as we found above without assuming it. If , then is a real sinusoid created by the sum of two complex sinusoids spinning in opposite directions on the unit circle.
### The BiQuad Section
The term biquad'' is short for bi-quadratic'', and is a common name for a two-pole, two-zero digital filter. The transfer function of the biquad can be defined as
(B.8)
where can be called the overall gain of the biquad. Since both the numerator and denominator of this transfer function are quadratic polynomials in (or ), the transfer function is said to be bi-quadratic'' in (or ).
As derived in §B.1.3, for real second-order polynomials having complex roots, it is often convenient to express the polynomial coefficients in terms of the radius and angle of the positive-frequency pole. For example, denoting the denominator polynomial by , we have
This representation is most often used for the denominator of the biquad, and we think of as the resonance frequency (in radians per sample-- , where is the resonance frequency in Hz), and determines the Q'' of the resonance (see §B.1.3). The numerator is less often represented in this way, but when it is, we may think of the zero-angle as the antiresonance frequency, and the zero-radius affects the depth and width of the antiresonance (or notch).
As discussed on page , a common setting for the zeros when making a resonator is to place one at (dc) and the other at (half the sampling rate), i.e., and in Eq.(B.8) above . This zero placement normalizes the peak gain of the resonator if it is swept using the parameter.
Using the shift theorem for z transforms, the difference equation for the biquad can be written by inspection of the transfer function as
where denotes the input signal sample at time , and is the output signal. This is the form that is typically implemented in software. It is essentially the direct-form I implementation. (To obtain the official direct-form I structure, the overall gain must be not be pulled out separately, resulting in feedforward coefficients instead. See Chapter 9 for more about filter implementation forms.)
### Biquad Software Implementations
In matlab, an efficient biquad section is implemented by calling
outputsignal = filter(B,A,inputsignal);
where
A complete C++ class implementing a biquad filter section is included in the free, open-source Synthesis Tool Kit (STK) [15]. (See the BiQuad STK class.)
Figure B.10 lists an example biquad implementation in the C programming language.
typedef double *pp; // pointer to array of length NTICK typedef word double; // signal and coefficient data type typedef struct _biquadVars { pp output; pp input; word s2; word s1; word gain; word a2; word a1; word b2; word b1; } biquadVars; void biquad(biquadVars *a) { int i; dbl A; word s0; for (i=0; igain * a->input[i]; A -= a->a1 * a->s1; A -= a->a2 * a->s2; s0 = A; A += a->b1 * a->s1; a->output[i] = a->b2 * a->s2 + A; a->s2 = a->s1; a->s1 = s0; } }
Next Section:
Allpass Filter Sections
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A Sum of Sinusoids at the Same Frequency is Another Sinusoid at that Frequency | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9727904796600342, "perplexity": 1150.8633078923408}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780058222.43/warc/CC-MAIN-20210926235727-20210927025727-00663.warc.gz"} |
http://mathhelpforum.com/calculus/183658-definite-integrals-lnx.html | # Thread: definite integrals lnx
1. ## definite integrals lnx
I have an integration whose lower limit has resulted in a negative number. What happens here? So for example;
ln (4/-2)
Does this mean it is equal to zero?
2. ## Re: definite integrals lnx
Originally Posted by Googl
I have an integration whose lower limit has resulted in a negative number. What happens here? So for example;
ln (4/-2)
Does this mean it is equal to zero?
no ... if the log function was the result of integration, you should know that the log's argument involves an absolute value.
3. ## Re: definite integrals lnx
Originally Posted by Googl
I have an integration whose lower limit has resulted in a negative number. What happens here? So for example; ln (4/-2)
Does this mean it is equal to zero?
Why don't you post the actual question?
I for one, have no idea what that question means.
4. ## Re: definite integrals lnx
Originally Posted by skeeter
no ... if the log function was the result of integration, you should know that the log's argument involves an absolute value.
Thanks.
As in ln (4/2) instead of ln (4/-2)
The question is long. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9917668700218201, "perplexity": 2057.344861309575}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719815.3/warc/CC-MAIN-20161020183839-00276-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://math-quiz.co.uk/start_test.php?id=70 | # Test: Angles, Straight Lines - Ambitious
Double click on maths expressions to zoom
Question 1: How many degrees are in $2.5$ right angles?
$112.5°$
$225°$
$235°$
$240°$
Question 2: How many degrees are in $\frac{1}{3}$ of a right angle?
$60°$
$120°$
$90°$
$30°$
Question 3: Find an angle corresponding to $\frac{3}{5}$ revolution?
$72°$
$200°$
$216°$
$270°$
Question 4: Find an angle corresponding to $\frac{5}{18}$ revolution?
$70°$
$60°$
$100°$
$90°$
Question 5: Two angles are complementary. One of them is $73°$. What is the other angle?
$27°$
$283°$
$167°$
$17°$
Please note, you have solved only half of the test. For the complete test get a Must Have account. Get started | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 25, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8435961008071899, "perplexity": 1976.8728084097925}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00560.warc.gz"} |
https://eprints.soton.ac.uk/260941/ | The University of Southampton
University of Southampton Institutional Repository
# Adaptive near minimum error rate training for neural networks with application to multiuser detection in CDMA communication systems
Chen, S., Samingan, A.K. and Hanzo, L. (2005) Adaptive near minimum error rate training for neural networks with application to multiuser detection in CDMA communication systems. Signal Processing, 85 (7), 1435-1448.
Record type: Article
## Abstract
Adaptive training of neural networks is typically done using some stochastic gradient algorithm that tries to minimize the mean square error (MSE). For many applications, such as channel equalization and code-division multiple-access (CDMA) multiuser detection, the goal is to minimize the error probability. For these applications, adopting the MSE criterion may lead to a poor performance. A novel adaptive near minimum error rate algorithm called the least bit error rate (LBER) is developed for training neural networks for these kinds of applications. The proposed method is applied to multiuser detection in CDMA communication systems. Simulation results show that the LBER algorithm has a good convergence speed and a small radial basis function (RBF) network trained by this adaptive algorithm can closely match the performance of the optimal Bayesian multiuser detector. The results also confirm that training the neural network multiuser detector using the least mean square (LMS) algorithm, although converging well in the MSE, can produce a poor error rate performance.
PDF
science.pdf - Other
Published date: July 2005
Organisations: Southampton Wireless Group
## Identifiers
Local EPrints ID: 260941
URI: https://eprints.soton.ac.uk/id/eprint/260941
ISSN: 0165-1684
PURE UUID: 68f0368d-93b0-42c4-af33-9f14b755380f
ORCID for L. Hanzo: orcid.org/0000-0002-2636-5214
## Catalogue record
Date deposited: 06 Jun 2005
## Contributors
Author: S. Chen
Author: A.K. Samingan
Author: L. Hanzo | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.912279486656189, "perplexity": 2348.418512551641}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202588.97/warc/CC-MAIN-20190321234128-20190322020128-00222.warc.gz"} |
http://physics.stackexchange.com/questions/105503/would-gravity-on-the-surface-of-a-planet-which-is-being-consumed-by-a-black-hole | # Would gravity on the surface of a planet which is being consumed by a black hole change?
Assuming that the black hole starts grow in the exact center of the planet and that the general structure of the planet does not degrade as it is eaten from the inside, would the gravity on the surface of the planet be affected by the black hole growing.
My uneducated guess is that as the matter becomes more compacted the gravity on the surface would actually decrease due to the fact that the distance to the mass, with respect to the surface, is increasing.
Also would the evaporation of the black hole actually decrease the total amount of mass in the center of the planet?
-
Assuming that the center of the planet just collapses to a black hole, people on the surface won't notice any difference in gravity at all until the ground starts to fall out from under them (the planet will no longer be stable, as there will be no way to support the ground under your feet, ultimately).
This is due to a result known as Birchoff's theorem that says that one can treat any spherically symmetric mass distribution as if all of its mass were concentrated at it's center.
-
Ah I see now, for every piece of matter that is now farther away from any given point on the surface there is an equal amount of matter that is closer as well. Would the total amount of matter actually decrease due to hawking radiation? – Peter Micheal Lacey-Bordeaux Mar 28 at 17:43
@PeterMichealLacey-Bordeaux: If you're taking about an Earth-sized black hole, you'd have to wait an age of the universe to see anything measurable, but probably. – Jerry Schirmer Mar 28 at 19:11 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9035036563873291, "perplexity": 201.6709848030768}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414119648891.34/warc/CC-MAIN-20141024030048-00172-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/145029-maximum-minimum.html | # Math Help - maximum/minimum
1. ## maximum/minimum
I feel my IQ drops with every second I look at this excercise, that's why I ask you guys:
Given $C\subset \mathbb{R}^3$ with $(x,y,z)$ that satisfy $2x^2+2y^2+z^2=1$ and $x=y^2+z^2$
Find $(x,y,z)\in C$ such that the distance to the origin is maximal/minimal
However, I can't find points that satisfy to both equations of $C$
We can derive
(1) $x\geq 0$
(2) $y^2=1-2x^2-x\geq 0$
(3) $z^2=2x^2+2x-1\geq 0$
From (2) we get $x\in [0,\frac{1}{2}]$
From (3) we get $x\geq \frac{1}{2}$
Only $x= \frac{1}{2}$ seems ok. But it gives $y^2=z^2=0$, ...scheisse
So, I can't find any $x$ that could possibly satisfy the equations, let alone $y,z$
Can someone fix my brains? What's wrong here?
2. This looks like a Lagrange multiplier problem. Have you heard of this method?
For $C$ did you mean to square the $x$ in $x=y^2+z^2$?
3. Yes, I'm familiar with the method. And the excercise itself is not my problem.
My problem is the definition of $C$, with $x=y^2+z^2$, from wich I derive that $C$ is empty
So I think there's a mistake in the excercise. I think indeed it must be $x^2=y^2+z^2$.
But no, I didn't mean to square the $x$, this is exactly the excercise.
4. We can first eliminate $z$
From $x = y^2 + z^2$
$z^2 = x - y^2$
$2x^2 + 2y^2 + z^2 = 1$
$2x^2 + 2y^2 + x - y^2 = 1$
$2x^2 + x + y^2 = 1$
We have
$D^2 = x^2 + y^2 + z^2 = x^2 + x = ( x + \frac{1}{2} )^2 - \frac{1}{4}$
Then we need to find the range of $x$
$2x^2 + x + y^2 = 1
$
$2(x + \frac{1}{4} )^2 + y^2 = 1 + \frac{1}{8} = \frac{9}{8}$
Let $x + \frac{1}{4} = \frac{1}{\sqrt{2}} \frac{3}{2\sqrt{2}} \cos{t}$
$y = \frac{3}{2\sqrt{2}} \sin{t}$
so
$x \leq \frac{3}{4} - \frac{1}{4} = \frac{1}{2}$
Since the quadratic function $x^2 + x$ is increasing for $x \geq 0 > - \frac{1}{2}$ , we have the max. of $D^2$ is $(\frac{1}{2})^2 + (\frac{1}{2}) = \frac{3}{4}$ .
5. Why should C be empty? C consists of points satisfying both $2x^2+ 2y^2+ z^2= 1$, an ellipse, and $x= y^2+ z^2$, a parboloid with axis along the x- axis. Certainly those intersect.
We can write the first equation as $2x^2+ 2y^2+ 2z^2- z^2= 2x^2+ 2(y^2+ z^2)+ z^2$ $= 2x^2+ 2x+ z^2= 2(x^2+ x+ 1/4- 1/4)+ 2z^2= 2(x+ 1/2)^2+ z^2= 1$, so the two surfaces intersect on an ellipse.
6. Yeah, thank you both.
Needed my brain to get fixed, perhaps I wasn't thinking clear.
But the mistake I made was at (3). Somehow I thought the positive root was $x=\frac{1}{2}$ wich is clearly wrong.
getting the positive root of $2x^2+2x-1$ is $x_0 =-\frac{1}{2}+\frac{1}{2}\sqrt{3}$. This gives $x\in [-\frac{1}{2}+\frac{1}{2}\sqrt{3},\infty)$
Hence we need to take the intersection of (1),(2) and we find $x\in [-\frac{1}{2}+\frac{1}{2}\sqrt{3},\frac{1}{2}]$
I guess, no-one really did an effort to see that I made a lame calculation error at (3). Now clearly C is not empty and consists of the points
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https://malegislature.gov/Laws/GeneralLaws/PartI/TitleXXII/Chapter168/Section6 | # General Laws
## Section 6 First meeting of subscribers
Section 6. The first meeting of the subscribers to the agreement of association shall be called by a notice signed either by that subscriber to the agreement who is designated therein for the purpose, or by a majority of the subscribers. Such notice shall state the time, place and purpose of the meeting. Seven days at least before the day appointed for the meeting, a copy of the notice shall be given to each subscriber, or left at his residence or usual place of business, or deposited in the post office, postage prepaid, and addressed to him at his residence or usual place of business. Another copy of said notice and an affidavit by one of the signers that the notice has been duly served shall be recorded with the records of the proposed corporation. If, however, all the incorporators shall, in writing endorsed upon the agreement of association, waive such notice and fix the time and place of the meeting, no notice shall be required. The subscribers to the agreement of association shall hold the franchise until the organization has been completed. At the first meeting, or at any adjournment thereof, the incorporators shall organize by the election by ballot of a temporary clerk, by the adoption of by-laws and by the election, in such manner as the by-laws may determine, of a president, a clerk of the corporation, a treasurer, a board of not less than eleven trustees, and such other officers as the by-laws may prescribe. All the officers so elected shall be sworn to the faithful performance of their duties. The temporary clerk shall make and attest a record of the proceedings until the clerk has been elected and sworn, including a record of the election and qualification of the clerk. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8168588876724243, "perplexity": 3047.9934290382976}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936462009.45/warc/CC-MAIN-20150226074102-00030-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/how-to-draw-the-energy-band-diagram-for-a-semiconductor-oxide-semicondutor-structure.340010/ | # How to draw the energy band diagram for a semiconductor-oxide-semicondutor structure?
1. Sep 24, 2009
### homeless123
1. The problem statement, all variables and given/known data
A schematic diagram of a silicon-oxide-silicon structure is shown. The top and bottom silicon electrodes are uniformly doped. The top silicon electrode is N-type and its dopant concentration is 2E16 cm-3. The bottom silicon electrode is p-type and its dopant concentration is 1E16cm-3. The oxide layer is 30nm thick and free from defects. The bottom electrode is always grounded and a voltage can be applied to the top electrode.
Pls draw the energy diagram when zero bias is applied to the top electrode.
2. Relevant equations
We are taught how to draw the energy diagram for a metal-oxide-semiconductor structure. In this question, we are asked to draw a SOS structure.
3. The attempt at a solution
Since the Fermi level at n-type is higher than that at p-type, the electrons will fly to p-type to make the Fermi levels at the two sides flat. Then the conduction band and valence band will bend upwards at p-type silicon as well as the oxide. But I do not know whether I should draw the two band bending or not at n-type semicondutor. I checked the reference books for MOSFET structure. If the gate is n+ poly-si, they never draw bending at the gate side.
Pls help, really do not know whether need to bend or not at the n-type silicon side. Thanks a lot
Can you offer guidance or do you also need help?
Similar Discussions: How to draw the energy band diagram for a semiconductor-oxide-semicondutor structure? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8420983552932739, "perplexity": 1168.2918940213526}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824824.76/warc/CC-MAIN-20171021171712-20171021191712-00754.warc.gz"} |
https://ece.uwaterloo.ca/~dwharder/NumericalAnalysis/Appendices/linearsystems/ | ## Linear Systems of Equations
What does it actually mean to solve the system Mx = b?
For some students, it's entirely likely that after doing all the math in your linear algebra course, you forgot the original reason why you defined these matrices, and why you're learning Gaussian elimination.
# One Dimension
For the one dimensional case, one solves mx = b for x. In this case, the solution is simple: x = b/m.
# Two Dimensions
Let's write out the matrix vector equaiton Mx = b. Given the matrix
and the vectors x = (x1, x2)T and b = (b1, b2)T, then expanding we get:
It's important here to remember that all the values mi, j and bi are given, a concrete example, is useful.
### Example of a 2×2 System of Linear Equations
What does Mx = b mean when
?
All this is, in disguise, is
4x1 + 2x2 = 3
3x1 + 5x2 = 4
which is equivalent to
4x1 + 2x2 − 3 = 0
3x1 + 5x2 − 4 = 0
However, note that the left hand side describes a plane. In Figure 1, we show f1(x) = 4x1 + 2x2 − 3, and in Figure 2, we show f2(x) = 3x1 + 5x2 − 4.
Figure 1. The function f1(x) = 4x1 + 2x2 − 3.
Figure 2. The function f2(x) = 3x1 + 5x2 − 4.
The first function, f1(x) is zero on the line x1 = 3/4 − 2/4x2, while the second function, f2(x) is zero on the line x1 = 4/3 − 5/3x2. What it means to solve a linear system is that we are finding a value of x = (x1, x2)T such that both f1(x) = 0 and f2(x) = 0.
In this example, there is one unqiue point, and that point may be seen in Figure 3.
Figure 3. Both function f1(x) and f2(x).
These two functions are simultaneously zero at only one point, seen in Figure 3 as the point where the grey (z = 0), red, and blue planes intersect. From the linear algebra you learned, you could solve this system to find that the solution is x = (0.5, 0.5)T.
Thus, solving a system of n linear equations is no more than finding simultaneous roots of n different linear functions. This is why we can use such techniques as when we perform Newton's method in n dimensions: we convert a system of n nonlinear functions into a simplified problem with n linear functions, find the root of the linear functions and show that, under the appropriate conditions, the solution to the linear functions is a good approximation to the nonlinear problem.
# Why are we Solving Systems of Linear Equations?
The most obvious example comes in examples such as trying to solve circuit in Figure 4. Kirchhoff's voltage laws tells us that the sum of the voltages in each of the loops is zero.
Figure 4. A simple circuit.
In the first loop, we get 7 V − 1 (i1 + i2) − 2 i1 = 0, and in the second, we get 2 i1 − 4 i2 = 0. Solving these yeilds the values i1 = 2 A and i2 = 1 A. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9410375952720642, "perplexity": 442.361644511048}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121869.65/warc/CC-MAIN-20170423031201-00411-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://www.mat.univie.ac.at/~schachermayer/pubs/pubabs.php?id=85 | [ Home page ] [ Publications and Preprints ] [ Curriculum vitae ]
Walter Schachermayer
## Is there a predictable criterion for mutual singularity of two probability measures on a filtered space?
W. Schachinger, W. Schachermayer
Theory of Probability and its Applications, Vol. 44 (1999), No. 1, pp. 51-59.
### Abstract:
The theme of providing predictable criteria for absolute continuity and for mutual singularity of two density processes on a filtered probability space is extensively studied, e.g., in the monograph by J. Jacod and A. N. Shiryaev [JS]. While the issue of absolute continuity is settled there in full generality, for the issue of mutual singularity one technical difficulty remained open ([JS], p210): "We do not know whether it is possible to derive a predictable criterion (necessary and sufficient condition) for \$P_T'\perp P_T\$,...". It turns out that to this question raised in [JS] which we also chose as the title of this note, there are two answers: on the negative side we give an easy example, showing that in general the answer is no, even when we use a rather wide interpretation of the concept of "predictable criterion". The difficulty comes from the fact that the density process of a probability measure P with respect to another measure P' may suddenly jump to zero.
On the positive side we can characterize the set, where P' becomes singular with respect to P -- provided this does not happen in a sudden but rather in a continuous way -- as the set where the Hellinger process diverges, which certainly is a "predictable criterion". This theorem extends results in the book of J. Jacod and A. N. Shiryaev [JS].
[JS] --- J. Jacod, A. N. Shiryaev: Limit Theorems for Stochastic Processes. Berlin: Springer 1987.
#### Keywords:
continuity and singularity of probability measures, Hellinger processes, stochastic integrals, stopping times
### Preprints:
[PostScript (123 k)] [PS.gz (47 k)] [PDF (191 k)] [DOI: 10.1137/S0040585X97977367]
Publications marked with have appeared in refereed journals.
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http://mathhelpforum.com/advanced-applied-math/14513-rotation-weighted-pulley.html | # Thread: rotation of a weighted pulley
1. ## rotation of a weighted pulley
A light wheel of radius a has a uniform semicircular rim of mass M, and my rotate freely in a vertical plane about a horizontal axis through its center. A light string passes around the wheel and suspends a mass m.
The system is governed by the equation:
(M+m)a^2 (d^2(x)/dt^2)= a*g*(m -(2Msin(x))/(pi))
where x is the angle between the downward vertical and the diameter through the center of mass of the heavy rim.
Find the equilibrium points and the conditions for their existence and stability.
Let k = m/M. how does the wheel behave when k is large?
-----------------
I have some thoughts on the problem, but I am not sure if they are right:
i think when k is large, d^2(x)/dt^2 tends to g/a, which means the wheel behaves like a simple pendulum??
and one of the equilibrium points is, of course, (k,x) = (0,0). then (0, pi) and (0,2*pi) are equilibrium points as well.
another observation is when k = 2/pi, sin (x) = 1. i think bifurcation occurs here, but i am not sure what would happen when k>(2/pi), becuase it seems like that means sin(x) >1...
2. Originally Posted by totoro
A light wheel of radius a has a uniform semicircular rim of mass M, and my rotate freely in a vertical plane about a horizontal axis through its center. A light string passes around the wheel and suspends a mass m.
The system is governed by the equation:
(M+m)a^2 (d^2(x)/dt^2)= a*g*(m -(2Msin(x))/(pi))
where x is the angle between the downward vertical and the diameter through the center of mass of the heavy rim.
Find the equilibrium points and the conditions for their existence and stability.
Let k = m/M. how does the wheel behave when k is large?
-----------------
I have some thoughts on the problem, but I am not sure if they are right:
i think when k is large, d^2(x)/dt^2 tends to g/a, which means the wheel behaves like a simple pendulum??
and one of the equilibrium points is, of course, (k,x) = (0,0). then (0, pi) and (0,2*pi) are equilibrium points as well.
another observation is when k = 2/pi, sin (x) = 1. i think bifurcation occurs here, but i am not sure what would happen when k>(2/pi), becuase it seems like that means sin(x) >1...
At the equilibrium points all the derivatives of x are zero, so:
m = 2Msin(x)/(pi)
or:
sin(x) = (pi/2) m/M.
Now m and M are both positive, and we want x in the range [0, 2 pi), so
x = arcsin((pi/2) m/M), and x = pi - arcsin((pi/2) m/M).
and there are no real solutions unless (pi/2) (m/M)<=1.
Or in terms of k, for equlibrium points to exist k <= 2/pi, and they are:
x = arcsin((pi/2) k), and x = pi - arcsin((pi/2) k).
RonL
3. thanks!
so given these fixed points, i have yet to determine their stability. with only the equation of the second derivative how can i determine the stability of the fixed points?
also i need to find out where the bifurcation occurs and what type they are. i don't know if there is a point, ie a value of k, such that the two fixed points found would coalesce into 1 (tangent bifurcation)? or is it another type of bifurcation?
4. Originally Posted by totoro
thanks!
so given these fixed points, i have yet to determine their stability. with only the equation of the second derivative how can i determine the stability of the fixed points?
If x0 is one of the equilibrium points we consider a solution which starts
nearby, say x(t)=x0+epsilon(t) x'(0)=0, where epsilon(t) is "small"
(M+m)a^2 (d^2(x)/dt^2) ~= a*g*(m -(2M [sin(x0)-epsilon(t)cos(x0)]))/(pi))
which may be written:
A d^2epsilon(t)/dt^2 ~= K1 + K2 epsilon(t) .............. ...(1),
where:
K1=a*g*(m - 2Msin(x0))/pi
K2 = a*g*(2M)*cos(x0)/pi
A=(M+m)a^2
Then the equilibrium x0 is stable if the solution of (1) is bounded (by some
multiple of epsilon(0)) for all t>0, for sufficently small epsilon(0)>0.
also i need to find out where the bifurcation occurs and what type they are. i don't know if there is a point, ie a value of k, such that the two fixed points found would coalesce into 1 (tangent bifurcation)? or is it another type of bifurcation?
They coalesce when k=1, as then (pi/2) m/M = pi/2 and the two solutions
for the equilibrium x's are equal.
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http://mathhelpforum.com/trigonometry/173459-trig-transformation-graph-easy.html | # Thread: Trig transformation graph - Easy
1. ## Trig transformation graph - Easy
I forgot how to do this.
6 cos(4x).
6 sin(8x)
i know the amplitude is 6. I'm having trouble finding the period.
Can someone tell me how to find what the period is?
2. The period of $\sin(ax)$ is $\dfrac{2\pi}{a}$. Same applies for $\cos(ax)$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9854238033294678, "perplexity": 1409.848074145412}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501172050.87/warc/CC-MAIN-20170219104612-00438-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://paulooliva.blogspot.com/2013/03/goodmans-theorem.html?showComment=1366234997744 | ## Monday, 4 March 2013
### Goodman's theorem
Let HA denote Heyting (intuitionistic) arithmetic and HA$$^\omega$$ be Heyting arithmetic extended to the language of finite types, i.e. with quantifiers for each finite type. AC stands for the axiom of choice. Finally, recall that a formula is arithmetical if it only contains quantifications over numbers.
In [1,2] Goodman proved the following amazing result:
Theorem. If HA$$^\omega$$ + AC proves an arithmetical formula $$A$$ then HA already proves $$A$$.
In proof theory one would simply say that HA$$^\omega$$ + AC is conservative over HA. The fact that HA$$^\omega$$ is conservative over HA was already known and is easy to show. But the conservation of HA$$^\omega$$ + AC over HA is quite a surprising result.
Recall that adding the axiom of choice to classical mathematics leads to all sorts of strange things (e.g. the Banach-Tarski paradox). Goodman's theorem essentially says that this is not the fault of the axiom of choice, but rather a fault of the combination of the axiom of choice with classical logic. Classical logic on its own makes perfect sense. The axiom of choice in an intuitionistic setting is harmless. But when these two are put together all hell breaks loose.
Proof of Goodman's theorem. Goodman's proof involves two major proof-theoretic techniques: forcing and realizability. Realizability is used to eliminate the axiom of choice, since the axiom of choice has a trivial realizer. But then one ends up with a proof of "$$t$$ realizes $$A$$". Forcing is used to recover the truth of $$A$$ given that $$A$$ has a realizer. That is done by choosing the forcing conditions to be approximations of the Skolem functions of the sub-formulas of $$A$$. So, the axiom of choice is replaced by finite approximations of Skolem functions! Let's see a few of the details:
We present here a sketch of Beeson's proof [3] of Goodman's theorem. Beeson's proof has the advantage (over Goodman's proof) that the two techniques of forcing and realizability are clearly separated. The proof consists of the following steps:
(1) Let $$A$$ be an arithmetical formula such that HA$$^\omega$$ + AC $$\vdash A$$.
(2) Let HA$$^\omega_a$$ denote the extension of HA$$^\omega$$ with a new function symbol $$a$$. By the soundness of Kleene realizability relative to $$a$$ we have HA$$^\omega_a \vdash t$$ realizes $$A$$, where $$t$$ is a term of HA$$^\omega_a$$.
(3) By the soundness of the forcing interpretation we have HA$$^\omega \vdash \exists p (p \Vdash t$$ realizes $$A)$$, where the forcing conditions are chosen as in the main lemma (below). Forcing is done as usual, with the crucial difference that in the forcing of the atomic formulas the forcing condition $$p$$ replaces the "generic" function $$a$$, i.e. $$p \Vdash a(n) = m$$ is defined as $$p(n) = m$$, where I'm actually using the equality symbol "=" for partial equality.
(4) By the main lemma HA$$^\omega \vdash \forall p \exists q \leq p (q \Vdash (t$$ realizes $$A) \Rightarrow A)$$.
(5) By (3) and (4) we have HA$$^\omega \vdash \exists q (q \Vdash A)$$.
(6) As for arithmetical formula HA$$^\omega \vdash (q \Vdash A) \Leftrightarrow A$$ we have, HA$$^\omega \vdash A$$.
Steps (2) and (3) are pretty standard realizability and forcing interpretations. And only the soundness theorems of these are used. Step (6) is also easy to check by a simple induction on $$A$$. So, the crucial bit of the proof is the choice of the forcing conditions and the main lemma which we discuss next.
Main lemma. Fix an arithmetical sentence $$A$$. Then there is a set $$C$$ of forcing conditions such that $$\mbox{HA}^\omega \vdash \forall p \exists q \leq p (q \Vdash (t \mbox{ realizes } A) \Rightarrow A).$$
Proof of main lemma. The set of forcing conditions $$C$$ consists of finite functions $$p$$ which are approximations to the Skolem functions of all sub-formulas of $$A$$. More precisely, these are partial functions $$p$$ with finite domain such that for each sub-formula $$B(x, y)$$ of $$A$$ we have $$\exists x B(x, y) \wedge (p_{\exists x B(x, y)}(y) \mbox{ defined }) \Rightarrow B(p_{\exists x B(x, y)}(y), y).$$ Hence, whenever the approximation to the Skolem function $$p_{\exists x B(x, y)}$$ is defined then it produces the right witness. Relative to these (approximation of) Skolem functions, it's easy to show the following:
(i) HA$$^\omega \vdash \forall p \exists q \leq p (q \Vdash (B(y) \Rightarrow \{j_B\}(y) \mbox{ realizes } B ))$$, for some index $$j_B$$ which we can construct (using the Skolem functions).
(ii) HA$$^\omega \vdash \forall p \exists q \leq p (q \Vdash (t \mbox{ realizes } B ) \Rightarrow B)$$.
Remark. Ulrich Kohlenbach [4] has shown the interesting fact that Goodman's theorem does not hold for fragments of HA$$^\omega$$. This means that in order to eliminate AC from the proof of an arithmetical formula $$A$$ we might have to use a more complex induction than in the proof which is allowed to use AC. Thierry Coquand [5] has just published another proof of Goodman's theorem.
[1] Goodman, N., The theory of the Gödel functionals, Journal of Symbolic Logic 41, 574-583 (1976)
[2] Goodman, N. Relativized realizability in intuitionistic arithmetic of all finite types. Journal of Symbolic Logic 43, pp. 23-44 (1978)
[3] Beeson, M., Goodman's theorem and beyond. Pacific J. Math. 84, 1-16 (1979)
[4] Kohlenbach, U., A note on Goodman's theorem. Studia Logica 63, 1-5 (1999)
[5] Coquand, T., About Goodman's theorem. Annals of Pure and Applied Logic 164(4), 437-442 (2013) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9853195548057556, "perplexity": 499.37556449054256}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676590866.65/warc/CC-MAIN-20180719105750-20180719125750-00131.warc.gz"} |
http://math.stackexchange.com/questions/585868/lawvere-theories-an-equivalence | # Lawvere theories: an equivalence.
I'm having trouble understanding Lawvere theories (as defined below).
Definition: A Lawvere Theory is a category $\mathcal{L}$ with finite products and with a distinguished object $A$ such that every object of $\mathcal{L}$ is (isomorphic to) a finite power of $A$; that is, for any $X\in\operatorname{Ob}(\mathcal{L})$, there is an $n\in\mathbb{N}$ with $X\cong A^{n}$. The object $A$ is called the fundamental object of $\mathcal{L}$. An arrow $\omega :A^{n}\to A$ is called an $n$-ary operation (and, in particular, arrows of the type $A^{0}=1\to A$ are called constants).
Here's an exercise in Turi's Category Theory Lecture Notes:
Let $\mathbb{N}^{\text{op}}$ be the opposite category of natural numbers and all functions. Show that Lawvere theories are equivalent to product preserving functors $$\mathbb{N}^{\text{op}}\to\mathbf{C}$$ that are bijective on objects.
The problem: I'm not sure what to do. The $\mathbf{C}$ doesn't help as it's undefined.
My attempt: Let $\mathcal{L}:\mathbb{N}^{\text{op}}\to\mathbf{C}$ be a product preserving functor bijective on objects. Then for any $(n_{i})_{I}\in\mathbb{N}$ for any set $I$, if $\prod_{i\in I}{n_{i}}$ exits, then $\mathcal{L}\left(\prod_{i\in I}{n_{i}}\right) = \prod_{i\in I}{\mathcal{L}(n_{i})}$ and for all $m, n\in\mathbb{N}$, $c\in Ob(\mathbf{C})$,
• $\mathcal{L}(n)=\mathcal{L}(m)\Rightarrow n=m$
• there exists $m_{c}\in\mathbb{N}$ with $\mathcal{L}(m_{c})=c$.
I then set up the commutative diagram(s) for the product with an arbitrary $(f_{i})_{I}$ such that $f_{i}: k\to n_{i}$ in $\mathbb{N}^{\text{op}}$ and took $\mathcal{L}$ of everything.
Then I got stuck.
I want to force this $\mathcal{L}$ to be a Lawvere theory. Once I've done that, I'll take a Lawvere theory and try to go in the other direction.
Thank you :)
Second attempt (based on the comments): The trick is to consider what happens to $1\in Ob(\mathbb{N}^{\text{op}})=\mathbb{N}$. Note that the product in $\mathbb{N}^{\text{op}}$ is the coproduct in $\mathbb{N}$, so is simply addition.
Suppose $L:\mathbb{N}^{\text{op}}\to\mathbb{C}$ is a product preserving functor bijective on objects. Then for all $c\in Ob(\mathbf{C})$, there exists a unique $m_{c}\in Ob(\mathbb{N}^{\text{op}})=\mathbb{N}$ with $L(m_{c})=c$ and for any $m, n\in\mathbb{N}, L(m)=L(n)\Rightarrow m=n$.
Let $L(1)=A$ and note that if a product $\sum_{i\in I}{a_{i}}$ exists in $\mathbb{N}^{\text{op}}$ for $a_{i}\in Ob(\mathbb{N}^{\text{op}})=\mathbb{N}$, it is in $\mathbb{N}$ and so $I$ would be a finite set. Hence finite products must exist in $\mathbf{C}$. Now for $X\in\mathbf{C}$, $X=L(m_{X})$ for some $m_{X}\in\mathbb{N}$ so $X=L(m_{X})=L(\sum_{j=1}^{m_{X}}{1})=\prod_{j=1}^{m_{X}}{L(1)}=A^{m_{X}}$. Thus $X\cong A^{m_{X}}$. Therefore, the pair $\langle L, \mathbf{C}\rangle$ is a Lawvere theory.
Ideas for the converse: Let $L:\mathbb{N}^{\text{op}}\to\mathcal{L}$ such that $L(1)=A$ and . . .
• $L(u)=U\cong V=L(v)$ iff $L(u)=L(v),$
• $Y\cong A^{m}=L(m_{A^{m}})$ implies $L(m_{A^{m}})=Y$, or
• $L(m)=Y$ if (and only if) $Y\cong A^{m}$.
I've explored these ideas and they don't seem fruitful. I would like an explicit proof now please, $\color{red}{\large\text{not just hints}}$. This is really bugging me.
-
The category $\mathcal{C}$ is a Lawvere theory, if you have a bijective-on-objects product-preserving functor $\mathbb{N}^\mathrm{op} \to \mathcal{C}$. – Zhen Lin Nov 29 '13 at 17:29
@Zhen Lin: Thank you. Given the definition above, though, I'm afraid that's little more than an assertion to me at the moment; I want to be able to show it :/ – Alice Nov 29 '13 at 17:34
It says nothing of the sort. It says "equivalent". More precisely there is an equivalence between the category of Lawvere theories and the category of bijective-on-objects product-preserving functors with domain $\mathbb{N}^\mathrm{op}$, if you choose the appropriate notion of morphism on both sides. – Zhen Lin Nov 29 '13 at 17:52
Yes, you can do that. – Zhen Lin Nov 29 '13 at 17:57
The product in $\mathbb{N}^\mathrm{op}$ is the coproduct in $\mathbb{N}$, which is addition. – Zhen Lin Dec 1 '13 at 22:41
Let $\mathbf C$ be a Lawvere theory: i.e. a category with finite products such that there's an object $A \in \mathbf C$ for which for every other $X \in \mathbf C$ there's a $n \in \mathbb N$ such that $A^n\cong X$.
Let's consider $\mathbb N$ the category of natural numbers and all function between them.
Clearly we have a function between the objects of $\mathbb N$ and the objects of $\mathbf C$ defined as $$\mathcal L\colon \mathbb N \to \mathbf C,$$ with $\mathcal L(n) = A^n$ for $n \in \mathbb N \setminus\{0\}$ and $\mathcal L(0)=\bullet$ the terminal object of $\mathbf C$.
Let $f \colon n \to m$ be a morphism in $\mathbb N$, or else a function between the sets $\{0,\dots,n-1\}$ and $\{0,\dots,m-1\}$.
For every $i \in \{0,\dots,m-1\}$ we can consider the family of morphism $\langle \pi_i^j\rangle_{j=1,\dots,n}$ where $\pi^i_j=1_A$ if $i \in f^{-1}(\{j\})$ other wise $\pi^i_j \colon A \to \bullet$ is the unique map in the terminal object $\bullet \in \mathbf C$.
These morphisms give us a morphism $$\pi^i \colon A \to A^{f^{-1}(\{i\})}$$ for universal property of products and then we get the product morphism $$\mathcal L(f) = \prod_{i=1}^n \pi^i \colon A^n=\mathcal L(n) \to A^m=\mathcal L(m).$$
Clearly if $f=1_n$ for a $n$ then for every $i \in n$ we have $1_n^{-1}(\{i\})=\{i\}$ and so the $\pi^i=1_A$ for every $i$ and so $\mathcal L(1_n)=1_{A^n}$.
Doing the calculations you can prove also that $\mathcal L(g \circ f)=\mathcal L(f) \circ \mathcal L(g)$, for every pair $f \colon n \to m$ and $g \colon m \to k$, of course the calculation are a little complicated (I would rather not to write here).
This functor is clearly product preserving, indeed for every $n \in \mathbb N$ we have that $\mathcal L(n)=A^n$ which is the n-fold product, and every projection of $\mathbb N^\text{op}$ i.e. a map $p \colon 1 \to n$ in $\mathbb N$ we have that
$$\mathcal L(p) = \prod_{i=1}^n \pi^i \colon A^n \to A,$$ where every $\pi^i \colon A \to \bullet$ for $i \not \in \text{Im }p(0)$ and $\pi^i = 1_A$ for $i \in \text{Im }p(0)$. So $\mathcal L(p)$ is the $p(0)$-th projection of the product $A^n$, of course now one should verify that these data verify the universal property which involve again some long calculations.
Forgive the lack of some details, but I think that adding them wouldn't make the answer more clear.
Hope this helps.
-
Thank you so much for your patience in writing that out for me, @Giorgio Mossa! I'm confident that I can take it from here. [I first need to check the existence of $\bullet$.] I hope you understand if I accept once I've got the damn thing! – Alice Dec 4 '13 at 11:58
@Shaun sure of course. Btw $\bullet$ the terminal object is also the empty product so it must exists in every category with finite products. :) – Giorgio Mossa Dec 4 '13 at 12:08
Again, thank you. I'm not sure I understand how $\mathcal{L}$ is bijective on objects though. What if there exists an $X$ such that $A^{n}\neq X\cong A^{n}$? [The same question, I suppose, covers how we can overlook $A^{n}\times\bullet\cong A^{n}$.] – Alice Dec 4 '13 at 19:23
Oh, I see; it's a bit of a misnomer. – Alice Dec 4 '13 at 19:30
[I'm following Turi's lecture notes. Maybe the term was covered in additional exercises or some other supplement.] – Alice Dec 4 '13 at 19:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9796623587608337, "perplexity": 171.05873444257853}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644063881.16/warc/CC-MAIN-20150827025423-00244-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://math.libretexts.org/Courses/Misericordia_University/MTH_226%3A_Calculus_III/Chapter_14%3A_Functions_of_Multiple_Variables_and_Partial_Derivatives/3.19%3A_Lagrange_Multipliers | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 3.19: Lagrange Multipliers
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Solving optimization problems for functions of two or more variables can be similar to solving such problems in single-variable calculus. However, techniques for dealing with multiple variables allow us to solve more varied optimization problems for which we need to deal with additional conditions or constraints. In this section, we examine one of the more common and useful methods for solving optimization problems with constraints.
### Lagrange Multipliers
In the previous section, an applied situation was explored involving maximizing a profit function, subject to certain constraints. In that example, the constraints involved a maximum number of golf balls that could be produced and sold in $$1$$ month $$(x),$$ and a maximum number of advertising hours that could be purchased per month $$(y)$$. Suppose these were combined into a single budgetary constraint, such as $$20x+4y≤216$$, that took into account both the cost of producing the golf balls and the number of advertising hours purchased per month. The goal is still to maximize profit, but now there is a different type of constraint on the values of $$x$$ and $$y$$. This constraint and the corresponding profit function
$f(x,y)=48x+96y−x^2−2xy−9y^2 \nonumber$
is an example of an optimization problem, and the function $$f(x,y)$$ is called the objective function. A graph of various level curves of the function $$f(x,y)$$ follows.
In Figure $$\PageIndex{1}$$, the value $$c$$ represents different profit levels (i.e., values of the function $$f$$). As the value of $$c$$ increases, the curve shifts to the right. Since our goal is to maximize profit, we want to choose a curve as far to the right as possible. If there were no restrictions on the number of golf balls the company could produce or the number of units of advertising available, then we could produce as many golf balls as we want, and advertise as much as we want, and there would be not be a maximum profit for the company. Unfortunately, we have a budgetary constraint that is modeled by the inequality $$20x+4y≤216.$$ To see how this constraint interacts with the profit function, Figure $$PageIndex{2}$$ shows the graph of the line $$20x+4y=216$$ superimposed on the previous graph.
As mentioned previously, the maximum profit occurs when the level curve is as far to the right as possible. However, the level of production corresponding to this maximum profit must also satisfy the budgetary constraint, so the point at which this profit occurs must also lie on (or to the left of) the red line in Figure $$\PageIndex{2}$$. Inspection of this graph reveals that this point exists where the line is tangent to the level curve of $$f$$. Trial and error reveals that this profit level seems to be around $$395$$, when $$x$$ and $$y$$ are both just less than $$5$$. We return to the solution of this problem later in this section. From a theoretical standpoint, at the point where the profit curve is tangent to the constraint line, the gradient of both of the functions evaluated at that point must point in the same (or opposite) direction. Recall that the gradient of a function of more than one variable is a vector. If two vectors point in the same (or opposite) directions, then one must be a constant multiple of the other. This idea is the basis of the method of Lagrange multipliers.
Method of Lagrange Multipliers: One Constraint
Theorem $$\PageIndex{1}$$: Let $$f$$ and $$g$$ be functions of two variables with continuous partial derivatives at every point of some open set containing the smooth curve $$g(x,y)=k$$, where $$k$$ is a constant. Suppose that $$f$$, when restricted to points on the curve $$g(x,y)=k$$, has a local extremum at the point $$(x_0,y_0)$$ and that $$\vecs ∇g(x_0,y_0)≠0$$. Then there is a number $$λ$$ called a Lagrange multiplier, for which
$\vecs ∇f(x_0,y_0)=λ\vecs ∇g(x_0,y_0).$
Proof
Assume that a constrained extremum occurs at the point $$(x_0,y_0).$$ Furthermore, we assume that the equation $$g(x,y)=k$$ can be smoothly parameterized as
$$x=x(s) \; \text{and}\; y=y(s)$$
where $$s$$ is an arc length parameter with reference point $$(x_0,y_0)$$ at $$s=0$$. Therefore, the quantity $$z=f(x(s),y(s))$$ has a relative maximum or relative minimum at $$s=0$$, and this implies that $$\dfrac{dz}{ds}=0$$ at that point. From the chain rule,
\begin{align*} \dfrac{dz}{ds} &=\dfrac{∂f}{∂x}⋅\dfrac{∂x}{∂s}+\dfrac{∂f}{∂y}⋅\dfrac{∂y}{∂s} \\[5pt] &=\left(\dfrac{∂f}{∂x}\hat{\mathbf i}+\dfrac{∂f}{∂y}\hat{\mathbf j}\right)⋅\left(\dfrac{∂x}{∂s}\hat{\mathbf i}+\dfrac{∂y}{∂s}\hat{\mathbf j}\right)\\[5pt] &=0, \end{align*}
where the derivatives are all evaluated at $$s=0$$. However, the first factor in the dot product is the gradient of $$f$$, and the second factor is the unit tangent vector $$\vec{\mathbf T}(0)$$ to the constraint curve. Since the point $$(x_0,y_0)$$ corresponds to $$s=0$$, it follows from this equation that
$\vecs ∇f(x_0,y_0)⋅\vecs{\mathbf T}(0)=0, \nonumber$
which implies that the gradient is either the zero vector $$\vecs 0$$ or it is normal to the constraint curve at a constrained relative extremum. However, the constraint curve $$g(x,y)=k$$ is a level curve for the function $$g(x,y)$$ so that if $$\vecs ∇g(x_0,y_0)≠0$$ then $$\vecs ∇g(x_0,y_0)$$ is normal to this curve at $$(x_0,y_0)$$ It follows, then, that there is some scalar $$λ$$ such that
$\vecs ∇f(x_0,y_0)=λ\vecs ∇g(x_0,y_0) \nonumber$
$$\square$$
To apply Theorem $$\PageIndex{1}$$ to an optimization problem similar to that for the golf ball manufacturer, we need a problem-solving strategy.
Problem-Solving Strategy: Steps for Using Lagrange Multipliers
1. Determine the objective function $$f(x,y)$$ and the constraint function $$g(x,y).$$ Does the optimization problem involve maximizing or minimizing the objective function?
2. Set up a system of equations using the following template: \begin{align} \vecs ∇f(x,y) &=λ\vecs ∇g(x,y) \\[5pt] g(x,y)&=k \end{align}.
3. Solve for $$x$$ and $$y$$ to determine the Lagrange points, i.e., points that satisfy the Lagrange multiplier equation.
4. If the objective function is continuous on the constraint and the constraint is a closed curve (like a circle or an ellipse), then the largest of the values of $$f$$ at the solutions found in step $$3$$ maximizes $$f$$, subject to the constraint; the smallest of those values minimizes $$f$$, subject to the constraint.
But in other cases, we need to evaluate the objective functions $$f$$ at points from the constraint on either side of each Lagrange point to determine whether we have obtained a relative maximum or a relative minimum.
Note that it is possible that our objective function will not have a relative maximum or a relative minimum at a given Lagrange point. This can occur in a couple situations, but most often when the Lagrange point is also a critical point of the objective function giving us a saddle point. Most of the time we will still get a relative extremum at a saddle point subject to a constraint, but sometimes we will not. See Figure $$\PageIndex{3}$$ for an example of this case.
Figure $$\PageIndex{3}$$: Graph of $$f(x,y)=x^2-y^3$$ along with the constraint $$(x-1)^2 + y^2 = 1$$. Note that there is no relative extremum at $$(0,0)$$, although this point will satisfy the Lagrange Multiplier equation with $$\lambda=0$$.
Example $$\PageIndex{1}$$: Using Lagrange Multipliers
Use the method of Lagrange multipliers to find the minimum value of $$f(x,y)=x^2+4y^2−2x+8y$$ subject to the constraint $$x+2y=7.$$
Solution
1. The objective function is $$f(x,y)=x^2+4y^2−2x+8y.$$ The constraint function is equal to the left-hand side of the constraint equation when only a constant is on the right-hand side. So here $$g(x,y)=x+2y$$. The problem asks us to solve for the minimum value of $$f$$, subject to the constraint (Figure $$\PageIndex{4}$$).
2. We then must calculate the gradients of both $$f$$ and $$g$$:
$\vecs \nabla f \left( x, y \right) = \left( 2x - 2 \right) \hat{\mathbf{i}} + \left( 8y + 8 \right) \hat{\mathbf{j}} \\ \vecs \nabla g \left( x, y \right) = \hat{\mathbf{i}} + 2 \hat{\mathbf{j}}.$
The equation $$\vecs \nabla f \left( x, y \right) = \lambda \vecs \nabla g \left( x, y \right)$$ becomes
$\left( 2 x - 2 \right) \hat{\mathbf{i}} + \left( 8 y + 8 \right) \hat{\mathbf{j}} = \lambda \left( \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} \right),$
which can be rewritten as
$\left( 2 x - 2 \right) \hat{\mathbf{i}} + \left( 8 y + 8 \right) \hat{\mathbf{j}} = \lambda \hat{\mathbf{i}} + 2 \lambda \hat{\mathbf{j}}.$
Next, we set the coefficients of $$\hat{\mathbf{i}}$$ and $$\hat{\mathbf{j}}$$ equal to each other:
\begin{align} 2 x - 2 &= \lambda \\ 8 y + 8 &= 2 \lambda. \end{align}
The equation $$g \left( x, y \right) = k$$ becomes $$x + 2 y = 7$$. Therefore, the system of equations that needs to be solved is
\begin{align} 2 x - 2 &= \lambda \\ 8 y + 8 &= 2 \lambda \\ x + 2 y &= 7. \end{align}
3. This is a linear system of three equations in three variables. We start by solving the second equation for $$λ$$ and substituting it into the first equation. This gives $$λ=4y+4$$, so substituting this into the first equation gives $2x−2=4y+4.\nonumber$ Solving this equation for $$x$$ gives $$x=2y+3$$. We then substitute this into the third equation: \begin{align*} (2y+3)+2y&=7 \\[5pt]4y&=4 \\[5pt]y&=1. \end{align*} Since $$x=2y+3,$$ this gives $$x=5.$$
4. Next, we evaluate $$f(x,y)=x^2+4y^2−2x+8y$$ at the point $$(5,1)$$, $f(5,1)=5^2+4(1)^2−2(5)+8(1)=27.$To ensure this corresponds to a minimum value on the constraint function, let’s try some other points on the constraint from either side of the point $$(5,1)$$, such as the intercepts of $$g(x,y)=0$$, Which are $$(7,0)$$ and $$(0,3.5)$$.
We get $$f(7,0)=35 \gt 27$$ and $$f(0,3.5)=77 \gt 27$$.
So it appears that $$f$$ has a relative minimum of $$27$$ at $$(5,1)$$, subject to the given constraint.
Exercise $$\PageIndex{1}$$
Use the method of Lagrange multipliers to find the maximum value of
$f(x,y)=9x^2+36xy−4y^2−18x−8y \nonumber$
subject to the constraint $$3x+4y=32.$$
Hint
Use the problem-solving strategy for the method of Lagrange multipliers.
Subject to the given constraint, $$f$$ has a maximum value of $$976$$ at the point $$(8,2)$$.
Let’s now return to the problem posed at the beginning of the section.
Example $$\PageIndex{2}$$: Golf Balls and Lagrange Multipliers
The golf ball manufacturer, Pro-T, has developed a profit model that depends on the number $$x$$ of golf balls sold per month (measured in thousands), and the number of hours per month of advertising y, according to the function
$z=f(x,y)=48x+96y−x^2−2xy−9y^2, \nonumber$
where $$z$$ is measured in thousands of dollars. The budgetary constraint function relating the cost of the production of thousands golf balls and advertising units is given by $$20x+4y=216.$$ Find the values of $$x$$ and $$y$$ that maximize profit, and find the maximum profit.
Solution:
Again, we follow the problem-solving strategy:
1. The objective function is $$f(x,y)=48x+96y−x^2−2xy−9y^2.$$ To determine the constraint function, we divide both sides by $$4$$, which gives $$5x+y=54.$$ The constraint function is equal to the left-hand side, so $$g(x,y)=5x+y.$$ The problem asks us to solve for the maximum value of $$f$$, subject to this constraint.
2. So, we calculate the gradients of both $$f$$ and $$g$$: \begin{align*} \vecs ∇f(x,y)&=(48−2x−2y)\hat{\mathbf i}+(96−2x−18y)\hat{\mathbf j}\\[5pt]\vecs ∇g(x,y)&=5\hat{\mathbf i}+\hat{\mathbf j}. \end{align*} The equation $$\vecs ∇f(x,y)=λ\vecs ∇g(x,y)$$ becomes $(48−2x−2y)\hat{i}+(96−2x−18y)\hat{\mathbf j}=λ(5\hat{\mathbf i}+\hat{\mathbf j}),\nonumber$ which can be rewritten as $(48−2x−2y)\hat{\mathbf i}+(96−2x−18y)\hat{\mathbf j}=λ5\hat{\mathbf i}+λ\hat{\mathbf j}.\nonumber$ We then set the coefficients of $$\hat{\mathbf i}$$ and $$\hat{\mathbf j}$$ equal to each other: \begin{align*} 48−2x−2y&=5λ \\[5pt] 96−2x−18y&=λ. \end{align*} The equation $$g(x,y)=k$$ becomes $$5x+y=54$$. Therefore, the system of equations that needs to be solved is \begin{align*} 48−2x−2y&=5λ \\[5pt] 96−2x−18y&=λ \\[5pt]5x+y&=54. \end{align*}
3. We use the left-hand side of the second equation to replace $$λ$$ in the first equation: \begin{align*} 48−2x−2y&=5(96−2x−18y) \\[5pt]48−2x−2y&=480−10x−90y \\[5pt] 8x&=432−88y \\[5pt] x&=54−11y. \end{align*} Then we substitute this into the third equation: \begin{align*} 5(54−11y)+y&=54\\[5pt] 270−55y+y&=54\\[5pt]216&=54y \\[5pt]y&=4. \end{align*} Since $$x=54−11y,$$ this gives $$x=10.$$
4. We then substitute $$(10,4)$$ into $$f(x,y)=48x+96y−x^2−2xy−9y^2,$$ which gives \begin{align*} f(10,4)&=48(10)+96(4)−(10)^2−2(10)(4)−9(4)^2 \\[5pt] & =480+384−100−80−144=540.\end{align*} Therefore the maximum profit that can be attained, subject to budgetary constraints, is $$540,000$$ with a production level of $$10,000$$ golf balls and $$4$$ hours of advertising bought per month. Let’s check to make sure this truly is a maximum. The endpoints of the line that defines the constraint are $$(10.8,0)$$ and $$(0,54)$$ Let’s evaluate $$f$$ at both of these points: \begin{align*} f(10.8,0)&=48(10.8)+96(0)−10.8^2−2(10.8)(0)−9(0^2) \\[5pt] &=401.76 \\[5pt] f(0,54)&=48(0)+96(54)−0^2−2(0)(54)−9(54^2) \\[5pt] &=−21,060. \end{align*} The second value represents a loss, since no golf balls are produced. Neither of these values exceed $$540$$, so it seems that our extremum is a maximum value of $$f$$, subject to the given constraint.
Exercise $$\PageIndex{2}$$: Optimizing the Cobb-Douglas function
A company has determined that its production level is given by the Cobb-Douglas function $$f(x,y)=2.5x^{0.45}y^{0.55}$$ where $$x$$ represents the total number of labor hours in $$1$$ year and $$y$$ represents the total capital input for the company. Suppose $$1$$ unit of labor costs $$40$$ and $$1$$ unit of capital costs $$50$$. Use the method of Lagrange multipliers to find the maximum value of $$f(x,y)=2.5x^{0.45}y^{0.55}$$ subject to a budgetary constraint of $$500,000$$ per year.
Hint
Use the problem-solving strategy for the method of Lagrange multipliers.
Subject to the given constraint, a maximum production level of $$13890$$ occurs with $$5625$$ labor hours and $$5500$$ of total capital input.
In the case of an objective function with three variables and a single constraint function, it is possible to use the method of Lagrange multipliers to solve an optimization problem as well. An example of an objective function with three variables could be the Cobb-Douglas function in Exercise $$\PageIndex{2}$$: $$f(x,y,z)=x^{0.2}y^{0.4}z^{0.4},$$ where $$x$$ represents the cost of labor, $$y$$ represents capital input, and $$z$$ represents the cost of advertising. The method is the same as for the method with a function of two variables; the equations to be solved are
\begin{align*} \vecs ∇f(x,y,z)&=λ\vecs ∇g(x,y,z) \\[5pt] g(x,y,z)&=k. \end{align*}
Example $$\PageIndex{3}$$: Lagrange Multipliers with a Three-Variable objective function
Maximize the function $$f(x,y,z)=x^2+y^2+z^2$$ subject to the constraint $$x+y+z=1.$$
Solution:
1. The objective function is $$f(x,y,z)=x^2+y^2+z^2.$$ To determine the constraint function, we set it equal to the variable expression on the left-hand side of the constraint equation: $$x+y+z=1$$ which gives the constraint function as $$g(x,y,z)=x+y+z.$$
2. Next, we calculate $$\vecs ∇f(x,y,z)$$ and $$\vecs ∇g(x,y,z):$$ \begin{align*} \vecs ∇f(x,y,z)&=⟨2x,2y,2z⟩ \\[5pt] \vecs ∇g(x,y,z)&=⟨1,1,1⟩. \end{align*} This leads to the equations \begin{align*} ⟨2x,2y,2z⟩&=λ⟨1,1,1⟩ \\[5pt] x+y+z&=1 \end{align*} which can be rewritten in the following form: \begin{align*} 2x&=λ\\[5pt]2y&=λ \\[5pt]2z&=λ \\[5pt]x+y+z&=1. \end{align*}
3. Since each of the first three equations has $$λ$$ on the right-hand side, we know that $$2x=2y=2z$$ and all three variables are equal to each other. Substituting $$y=x$$ and $$z=x$$ into the last equation yields $$3x=1,$$ so $$x=\frac{1}{3}$$ and $$y=\frac{1}{3}$$ and $$z=\frac{1}{3}$$ which corresponds to a critical point on the constraint curve.
4. Then, we evaluate $$f$$ at the point $$\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$$: $f\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)=\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2=\frac{3}{9}=\frac{1}{3}$ Therefore, a possible extremum of the function is $$\frac{1}{3}$$. To verify it is a minimum, choose other points that satisfy the constraint from either side of the point we obtained above and calculate $$f$$ at those points. For example, \begin{align*} f(1,0,0)&=1^2+0^2+0^2=1 \\[5pt] f(0,−2,3)&=0^2++(−2)^2+3^2=13. \end{align*} Both of these values are greater than $$\frac{1}{3}$$, leading us to believe the extremum is a minimum, subject to the given constraint.
Exercise $$\PageIndex{3}$$:
Use the method of Lagrange multipliers to find the minimum value of the function
$f(x,y,z)=x+y+z \nonumber$
subject to the constraint $$x^2+y^2+z^2=1.$$
Hint
Use the problem-solving strategy for the method of Lagrange multipliers with an objective function of three variables.
Evaluating $$f$$ at both points we obtained, gives us, \begin{align*} f\left(\dfrac{\sqrt{3}}{3},\dfrac{\sqrt{3}}{3},\dfrac{\sqrt{3}}{3}\right)&=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{3}}{3}=\sqrt{3} \\ f\left(−\dfrac{\sqrt{3}}{3},−\dfrac{\sqrt{3}}{3},−\dfrac{\sqrt{3}}{3}\right)&=−\dfrac{\sqrt{3}}{3}−\dfrac{\sqrt{3}}{3}−\dfrac{\sqrt{3}}{3}=−\sqrt{3}\end{align*} Since the constraint is continuous, we compare these values and conclude that $$f$$ has a relative minimum of $$−\sqrt{3}$$ at the point $$\left(−\dfrac{\sqrt{3}}{3},−\dfrac{\sqrt{3}}{3},−\dfrac{\sqrt{3}}{3}\right)$$, subject to the given constraint.
### Problems with Two Constraints
The method of Lagrange multipliers can be applied to problems with more than one constraint. In this case the objective function, $$w$$ is a function of three variables:
$w=f(x,y,z)$
and it is subject to two constraints:
$g(x,y,z)=0 \; \text{and} \; h(x,y,z)=0.$
There are two Lagrange multipliers, $$λ_1$$ and $$λ_2$$, and the system of equations becomes
\begin{align*} \vecs ∇f(x_0,y_0,z_0)&=λ_1\vecs ∇g(x_0,y_0,z_0)+λ_2\vecs ∇h(x_0,y_0,z_0) \\[5pt] g(x_0,y_0,z_0)&=0\\[5pt] h(x_0,y_0,z_0)&=0 \end{align*}
Example $$\PageIndex{4}$$: Lagrange Multipliers with Two Constraints
Find the maximum and minimum values of the function
$f(x,y,z)=x^2+y^2+z^2 \nonumber$
subject to the constraints $$z^2=x^2+y^2$$ and $$x+y−z+1=0.$$
Solution:
1. The objective function is $$f(x,y,z)=x^2+y^2+z^2.$$ To determine the constraint functions, we first subtract $$z^2$$ from both sides of the first constraint, which gives $$x^2+y^2−z^2=0$$, so $$g(x,y,z)=x^2+y^2−z^2$$. The second constraint function is $$h(x,y,z)=x+y−z+1.$$
2. We then calculate the gradients of $$f,g,$$ and $$h$$: \begin{align*} \vecs ∇f(x,y,z)&=2x\hat{\mathbf i}+2y\hat{\mathbf j}+2z\hat{\mathbf k} \\[5pt] \vecs ∇g(x,y,z)&=2x\hat{\mathbf i}+2y\hat{\mathbf j}−2z\hat{\mathbf k} \\[5pt] \vecs ∇h(x,y,z)&=\hat{\mathbf i}+\hat{\mathbf j}−\hat{\mathbf k}. \end{align*} The equation $$\vecs ∇f(x,y,z)=λ_1\vecs ∇g(x,y,z)+λ_2\vecs ∇h(x,y,z)$$ becomes $2x\hat{\mathbf i}+2y\hat{\mathbf j}+2z\hat{\mathbf k}=λ_1(2x\hat{\mathbf i}+2y\hat{\mathbf j}−2z\hat{\mathbf k})+λ_2(\hat{\mathbf i}+\hat{\mathbf j}−\hat{\mathbf k}),$ which can be rewritten as $2x\hat{\mathbf i}+2y\hat{\mathbf j}+2z\hat{\mathbf k}=(2λ_1x+λ_2)\hat{\mathbf i}+(2λ_1y+λ_2)\hat{\mathbf j}−(2λ_1z+λ_2)\hat{\mathbf k}.$ Next, we set the coefficients of $$\hat{\mathbf i}$$ and $$\hat{\mathbf j}$$ equal to each other: \begin{align*}2x&=2λ_1x+λ_2 \\[5pt]2y&=2λ_1y+λ_2 \\[5pt]2z&=−2λ_1z−λ_2. \end{align*} The two equations that arise from the constraints are $$z^2=x^2+y^2$$ and $$x+y−z+1=0$$. Combining these equations with the previous three equations gives \begin{align*} 2x&=2λ_1x+λ_2 \\[5pt]2y&=2λ_1y+λ_2 \\[5pt]2z&=−2λ_1z−λ_2 \\[5pt]z^2&=x^2+y^2 \\[5pt]x+y−z+1&=0. \end{align*}
3. The first three equations contain the variable $$λ_2$$. Solving the third equation for $$λ_2$$ and replacing into the first and second equations reduces the number of equations to four: \begin{align*}2x&=2λ_1x−2λ_1z−2z \\[5pt] 2y&=2λ_1y−2λ_1z−2z\\[5pt] z^2&=x^2+y^2\\[5pt] x+y−z+1&=0. \end{align*} Next, we solve the first and second equation for $$λ_1$$. The first equation gives $$λ_1=\dfrac{x+z}{x−z}$$, the second equation gives $$λ_1=\dfrac{y+z}{y−z}$$. We set the right-hand side of each equation equal to each other and cross-multiply: \begin{align*} \dfrac{x+z}{x−z}&=\dfrac{y+z}{y−z} \\[5pt](x+z)(y−z)&=(x−z)(y+z) \\[5pt]xy−xz+yz−z^2&=xy+xz−yz−z^2 \\[5pt]2yz−2xz&=0 \\[5pt]2z(y−x)&=0. \end{align*} Therefore, either $$z=0$$ or $$y=x$$. If $$z=0$$, then the first constraint becomes $$0=x^2+y^2$$. The only real solution to this equation is $$x=0$$ and $$y=0$$, which gives the ordered triple $$(0,0,0)$$. This point does not satisfy the second constraint, so it is not a solution. Next, we consider $$y=x$$, which reduces the number of equations to three: \begin{align*}y &= x \\[5pt] z^2 &= x^2 +y^2 \\[5pt] x + y -z+1&=0. \end{align*} We substitute the first equation into the second and third equations: \begin{align*} z^2 &= x^2 +x^2 \\[5pt] &= x+x-z+1 =0. \end{align*} Then, we solve the second equation for $$z$$, which gives $$z=2x+1$$. We then substitute this into the first equation, \begin{align*} z^2 &= 2x^2 \\[5pt] (2x^2 +1)^2 &= 2x^2 \\[5pt] 4x^2 + 4x +1 &= 2x^2 \\[5pt] 2x^2 +4x +1 &=0, \end{align*} and use the quadratic formula to solve for $$x$$: $x = \dfrac{-4 \pm \sqrt{4^2 -4(2)(1)} }{2(2)} = \dfrac{-4\pm \sqrt{8}}{4} = \dfrac{-4 \pm 2\sqrt{2}}{4} = -1 \pm \dfrac{\sqrt{2}}{2}.$ Recall $$y=x$$, so this solves for $$y$$ as well. Then, $$z=2x+1$$, so $z = 2x +1 =2 \left( -1 \pm \dfrac{\sqrt{2}}{2} \right) +1 = -2 + 1 \pm \sqrt{2} = -1 \pm \sqrt{2} .$ Therefore, there are two ordered triplet solutions: $\left( -1 + \dfrac{\sqrt{2}}{2} , -1 + \dfrac{\sqrt{2}}{2} , -1 + \sqrt{2} \right) \; \text{and} \; \left( -1 -\dfrac{\sqrt{2}}{2} , -1 -\dfrac{\sqrt{2}}{2} , -1 -\sqrt{2} \right).$
4. We substitute $$\left(−1+\dfrac{\sqrt{2}}{2},−1+\dfrac{\sqrt{2}}{2}, −1+\sqrt{2}\right)$$ into $$f(x,y,z)=x^2+y^2+z^2$$, which gives \begin{align*} f\left( -1 + \dfrac{\sqrt{2}}{2}, -1 + \dfrac{\sqrt{2}}{2} , -1 + \sqrt{2} \right) &= \left( -1+\dfrac{\sqrt{2}}{2} \right)^2 + \left( -1 + \dfrac{\sqrt{2}}{2} \right)^2 + (-1+\sqrt{2})^2 \\[5pt] &= \left( 1-\sqrt{2}+\dfrac{1}{2} \right) + \left( 1-\sqrt{2}+\dfrac{1}{2} \right) + (1 -2\sqrt{2} +2) \\[5pt] &= 6-4\sqrt{2}. \end{align*} Then, we substitute $$\left(−1−\dfrac{\sqrt{2}}{2}, -1+\dfrac{\sqrt{2}}{2}, -1+\sqrt{2}\right)$$ into $$f(x,y,z)=x^2+y^2+z^2$$, which gives \begin{align*} f\left(−1−\dfrac{\sqrt{2}}{2}, -1+\dfrac{\sqrt{2}}{2}, -1+\sqrt{2} \right) &= \left( -1-\dfrac{\sqrt{2}}{2} \right)^2 + \left( -1 - \dfrac{\sqrt{2}}{2} \right)^2 + (-1-\sqrt{2})^2 \\[5pt] &= \left( 1+\sqrt{2}+\dfrac{1}{2} \right) + \left( 1+\sqrt{2}+\dfrac{1}{2} \right) + (1 +2\sqrt{2} +2) \\[5pt] &= 6+4\sqrt{2}. \end{align*} $$6+4\sqrt{2}$$ is the maximum value and $$6−4\sqrt{2}$$ is the minimum value of $$f(x,y,z)$$, subject to the given constraints.
Exercise $$\PageIndex{4}$$
Use the method of Lagrange multipliers to find the minimum value of the function
$f(x,y,z)=x^2+y^2+z^2$
subject to the constraints $$2x+y+2z=9$$ and $$5x+5y+7z=29.$$
Hint
Use the problem-solving strategy for the method of Lagrange multipliers with two constraints.
$$f(2,1,2)=9$$ is a relative minimum of $$f$$, subject to the given constraints
## Key Concepts
• An objective function combined with one or more constraints is an example of an optimization problem.
• To solve optimization problems, we apply the method of Lagrange multipliers using a four-step problem-solving strategy.
### Key Equations
• Method of Lagrange multipliers, one constraint
$$\vecs ∇f(x,y)=λ\vecs ∇g(x,y)$$
$$g(x,y)=k$$
• Method of Lagrange multipliers, two constraints
$$\vecs ∇f(x_0,y_0,z_0)=λ_1\vecs ∇g(x_0,y_0,z_0)+λ_2\vecs ∇h(x_0,y_0,z_0)$$
$$g(x_0,y_0,z_0)=0$$
$$h(x_0,y_0,z_0)=0$$
### Glossary
constraint
an inequality or equation involving one or more variables that is used in an optimization problem; the constraint enforces a limit on the possible solutions for the problem
Lagrange multiplier
the constant (or constants) used in the method of Lagrange multipliers; in the case of one constant, it is represented by the variable $$λ$$
method of Lagrange multipliers
a method of solving an optimization problem subject to one or more constraints
objective function
the function that is to be maximized or minimized in an optimization problem
optimization problem
calculation of a maximum or minimum value of a function of several variables, often using Lagrange multipliers | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9912946224212646, "perplexity": 219.68558407451417}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103821173.44/warc/CC-MAIN-20220630122857-20220630152857-00682.warc.gz"} |
https://deepai.org/publication/amr-mul-an-approximate-maximally-redundant-signed-digit-multiplier | DeepAI
# AMR-MUL: An Approximate Maximally Redundant Signed Digit Multiplier
In this paper, we present an energy-efficient, yet high-speed approximate maximally redundant signed digit (MRSD) multiplier (called AMR-MUL) based on a parallel structure. For the reduction stage, we suggest several approximate Full-Adder (FA) reduction cells with average positive and negative errors obtained by simplifying the structure of an exact FA cell. The optimum selection of these cells for each partial product reduction stage provides the lowest possible error, turning this task into a design space exploration problem. We also provide a branch-and-bound design space exploration algorithm to find the optimal assignment of reduction cells based on a predefined constraint (i.e., the width of the approximate part) by the user. The effectiveness of the proposed (Radix-16) multiplier design is assessed under different digit counts and approximate border column. The results show that the energy consumption of the MRSD multiplier is reduced by 7x at the cost of a 1.6
• 1 publication
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### Approximate Reduction of Finite Automata for High-Speed Network Intrusion Detection (Technical Report)
We consider the problem of approximate reduction of non-deterministic au... | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8939505219459534, "perplexity": 3272.615593184207}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446708046.99/warc/CC-MAIN-20221126180719-20221126210719-00612.warc.gz"} |
https://larsson-research.de/publication/larsson-ttns-2019/ | # Computing vibrational eigenstates with tree tensor network states (TTNS) [Editor's Pick]
### Abstract
During my postdoctoral studies in the group of Prof. G. Chan (Caltech), as side project, I worked on the computation of vibrational spectra using tree-tensor network states (TTNS). There, I applied methods developed in the condensed matter community to molecular systems. Compared to established approaches for computing vibrational spectra, such as the multilayer multiconfiguration time-dependent Hartree (ML-MCTDH) method, the new approach is faster and much more robust and the diagrammatic language from condensed matter physics I use gains much more insights. I am currently working on using the TTNS method to compute high-lying vibrational states of the Zundel ion.
Type
Publication
The Journal of Chemical Physics | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8147500157356262, "perplexity": 2239.009793090402}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178366477.52/warc/CC-MAIN-20210303073439-20210303103439-00497.warc.gz"} |
http://en.wikipedia.org/wiki/Contact_geometry | Contact geometry
Contact form redirects here. For a web email form, see Form_(web)#Form-to-email_scripts.
The standard contact structure on R3. Each point in R3 has a plane associated to it by the contact structure, in this case as the kernel of the one-form dzy dx. These planes appear to twist along the y-axis.
In mathematics, contact geometry is the study of a geometric structure on smooth manifolds given by a hyperplane distribution in the tangent bundle and specified by a one-form, both of which satisfy a 'maximum non-degeneracy' condition called 'complete non-integrability'. From the Frobenius theorem, one recognizes the condition as the opposite of the condition that the distribution be determined by a codimension one foliation on the manifold ('complete integrability').
Contact geometry is in many ways an odd-dimensional counterpart of symplectic geometry, which belongs to the even-dimensional world. Both contact and symplectic geometry are motivated by the mathematical formalism of classical mechanics, where one can consider either the even-dimensional phase space of a mechanical system or the odd-dimensional extended phase space that includes the time variable.
Applications
Contact geometry has — as does symplectic geometry — broad applications in physics, e.g. geometrical optics, classical mechanics, thermodynamics, geometric quantization, and applied mathematics such as control theory. Contact geometry also has applications to low-dimensional topology; for example, it has been used by Kronheimer and Mrowka to prove the property P conjecture and by Yakov Eliashberg to derive a topological characterization of Stein manifolds.
Contact forms and structures
Given an n-dimensional smooth manifold M, and a point pM, a contact element of M with contact point p is an (n − 1)-dimensional linear subspace of the tangent space to M at p.[1][2] A contact element can be given by the zeros of a 1-form on the tangent space to M at p. However, if a contact element is given by the zeros of a 1-form ω, then it will also be given by the zeros of λω where λ ≠ 0. Thus, { λω : λ ≠ 0 } all give the same contact element. It follows that the space of all contact elements of M can be identified with a quotient of the cotangent bundle T*M,[1] namely:
$\text{PT}^*M = \text{T}^*M /\! \sim \ \text{ where, for } \omega_i \in \text{T}^*_pM, \ \ \omega_1 \sim \omega_2 \ \iff \ \exists \ \lambda \neq 0 \ : \ \omega_1 = \lambda\omega_2.$
A contact structure on an odd dimensional manifold M, of dimension 2k+1, is a smooth distribution of contact elements, denoted by ξ, which is generic at each point.[1][2] The genericity condition is that ξ is non-integrable.
Assume that we have a smooth distribution of contact elements, ξ, given locally by a differential 1-form α; i.e. a smooth section of the cotangent bundle. The non-integrability condition can be given explicitly as:[1]
$\alpha \wedge (\text{d}\alpha)^k \neq 0 \ \text{where} \ (\text{d}\alpha)^k = \underbrace {\text{d}\alpha \wedge \ldots \wedge \text{d}\alpha}_{k-\text{times}}.$
Notice that if ξ is given by the differential 1-form α, then the same distribution is given locally by β = ƒ⋅α, where ƒ is a non-zero smooth function. If ξ is co-orientable then α is defined globally.
Properties
It follows from the Frobenius theorem on integrability that the contact field ξ is completely nonintegrable. This property of the contact field is roughly the opposite of being a field formed by the tangent planes to a family of nonoverlapping hypersurfaces in M. In particular, you cannot find a piece of a hypersurface tangent to ξ on an open set of M. More precisely, a maximally integrable subbundle has dimension n.
Relation with symplectic structures
A consequence of the definition is that the restriction of the 2-form ω = dα to a hyperplane in ξ is a nondegenerate 2-form. This construction provides any contact manifold M with a natural symplectic bundle of rank one smaller than the dimension of M. Note that a symplectic vector space is always even-dimensional, while contact manifolds need to be odd-dimensional.
The cotangent bundle T*N of any n-dimensional manifold N is itself a manifold (of dimension 2n) and supports naturally an exact symplectic structure ω = dλ. (This 1-form λ is sometimes called the Liouville form). There are several ways to construct an associated contact manifold, one of dimension 2n − 1, one of dimension 2n + 1.
Projectivization
Let M be the projectivization of the cotangent bundle of N: thus M is fiber bundle over a M whose fiber at a point x is the space of lines in T*N, or, equivalently, the space of hyperplanes in TN. The 1-form λ does not descend to a genuine 1-form on M. However, it is homogeneous of degree 1, and so it defines a 1-form with values in the line bundle O(1), which is the dual of the fibrewise tautological line bundle of M. The kernel of this 1-form defines a contact distribution.
Energy surfaces
Suppose that H is a smooth function on T*N, that E is a regular value for H, so that the level set $L=\{(q,p)\in T^*N|H(q,p)=E\}$ is a smooth submanifold of codimension 1. A vector field Y is called an Euler (or Liouville) vector field if it is transverse to L and conformally symplectic, meaning that the Lie derivative of dλ with respect to Y is a multiple of dλ in a neighborhood of L.
Then the restriction of $i_Yd\lambda$ to L is a contact form on L.
This construction originates in Hamiltonian mechanics, where H is a Hamiltonian of a mechanical system with the configuration space N and the phase space T*N, and E is the value of the energy.
The unit cotangent bundle
Choose a Riemannian metric on the manifold N and let H be the associated kinetic energy. Then the level set H =1/2 is the unit cotangent bundle of N, a smooth manifold of dimension 2n-1 fibering over N with fibers being spheres. Then the Liouville form restricted to the unit cotangent bundle is a contact structure. This corresponds to a special case of the second construction, where the flow of the Euler vector field Y corresponds to linear scaling of momenta p's, leaving the q's fixed. The vector field R, defined by the equalities
λ(R) = 1 and dλ(RA) = 0 for all vector fields A,
is called the Reeb vector field, and it generates the geodesic flow of the Riemannian metric. More precisely, using the Riemannian metric, one can identify each point of the cotangent bundle of N with a point of the tangent bundle of N, and then the value of R at that point of the (unit) cotangent bundle is the corresponding (unit) vector parallel to N.
First jet bundle
On the other hand, one can build a contact manifold M of dimension 2n + 1 by considering the first jet bundle of the real valued functions on N. This bundle is isomorphic to T*N×R using the exterior derivative of a function. With coordinates (xt), M has a contact structure
1. α = dt + λ.
Conversely, given any contact manifold M, the product M×R has a natural structure of a symplectic manifold. If α is a contact form on M, then
ω = d(etα)
is a symplectic form on M×R, where t denotes the variable in the R-direction. This new manifold is called the symplectization (sometimes symplectification in the literature) of the contact manifold M.
Examples
As a prime example, consider R3, endowed with coordinates (x,y,z) and the one-form dzy dx. The contact plane ξ at a point (x,y,z) is spanned by the vectors X1 = y and X2 = x + y z.
By replacing the single variables x and y with the multivariables x1, ..., xn, y1, ..., yn, one can generalize this example to any R2n+1. By a theorem of Darboux, every contact structure on a manifold looks locally like this particular contact structure on the (2n + 1)-dimensional vector space.
An important class of contact manifolds is formed by Sasakian manifolds.
Legendrian submanifolds and knots
The most interesting subspaces of a contact manifold are its Legendrian submanifolds. The non-integrability of the contact hyperplane field on a (2n + 1)-dimensional manifold means that no 2n-dimensional submanifold has it as its tangent bundle, even locally. However, it is in general possible to find n-dimensional (embedded or immersed) submanifolds whose tangent spaces lie inside the contact field. Legendrian submanifolds are analogous to Lagrangian submanifolds of symplectic manifolds. There is a precise relation: the lift of a Legendrian submanifold in a symplectization of a contact manifold is a Lagrangian submanifold. The simplest example of Legendrian submanifolds are Legendrian knots inside a contact three-manifold. Inequivalent Legendrian knots may be equivalent as smooth knots.
Legendrian submanifolds are very rigid objects; typically there are infinitely many Legendrian isotopy classes of embeddings which are all smoothly isotopic. Symplectic field theory provides invariants of Legendrian submanifolds called relative contact homology that can sometimes distinguish distinct Legendrian submanifolds that are topologically identical.
Reeb vector field
If α is a contact form for a given contact structure, the Reeb vector field R can be defined as the unique element of the kernel of dα such that α(R) = 1. Its dynamics can be used to study the structure of the contact manifold or even the underlying manifold using techniques of Floer homology such as symplectic field theory and embedded contact homology.
Some historical remarks
The roots of contact geometry appear in work of Christiaan Huygens, Isaac Barrow and Isaac Newton. The theory of contact transformations (i.e. transformations preserving a contact structure) was developed by Sophus Lie, with the dual aims of studying differential equations (e.g. the Legendre transformation or canonical transformation) and describing the 'change of space element', familiar from projective duality.
References
1. ^ a b c d Arnold, V. I. (1989), Mathematical Methods of Classical Mechanics, Springer, pp. 349 − 370, ISBN 0-387-96890-3
2. ^ a b Arnold, V. I. (1989). "Contact Geometry and Wave Propagation". Monographie de L'Enseignement Mathématique. Conférences de l'Union Mathématique Internationale (in English) (Univ. de Genève).
Applications to differential equations
• V. I. Arnold, Geometrical Methods In The Theory Of Ordinary Differential Equations, Springer-Verlag (1988), ISBN 0-387-96649-8
Contact three-manifolds and Legendrian knots
• William Thurston, Three-Dimensional Geometry and Topology. Princeton University Press(1997), ISBN 0-691-08304-5
Information on the history of contact geometry
• Lutz, R. Quelques remarques historiques et prospectives sur la géométrie de contact , Conf. on Diff. Geom. and Top. (Sardinia, 1988) Rend. Fac. Sci. Univ. Cagliari 58 (1988), suppl., 361–393.
• Geiges, H. A Brief History of Contact Geometry and Topology, Expo. Math. 19 (2001), 25–53.
• Arnold, V.I. (trans. E. Primrose), Huygens and Barrow, Newton and Hooke: pioneers in mathematical analysis and catastrophe theory from evolvents to quasicrystals. Birkhauser Verlag, 1990.
• Contact geometry Theme on arxiv.org | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8535981774330139, "perplexity": 574.7860943089769}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657136494.66/warc/CC-MAIN-20140914011216-00182-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://www.quantumstudy.com/a-uniform-cylinder-of-length-l-and-mass-m-having-cross-sectional-area-a-is-suspended-with-its-length-vertical/ | # A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical…
Q: A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half-submerged in a liquid of density ρ at equilibrium position. When the cylinder is given a small downward push and released to starts oscillating vertically with a small amplitude. If the force constant of the spring is k , the frequency of oscillation of the cylinder is
(a) $\large \frac{1}{2\pi} (\frac{k – A \rho g}{M})^{1/2}$
(b) $\large \frac{1}{2\pi} (\frac{k + A \rho g}{M})^{1/2}$
(c) $\large \frac{1}{2\pi} (\frac{k + \rho g L^2}{M})^{1/2}$
(d) $\large \frac{1}{2\pi} (\frac{k + A \rho g}{A \rho g})^{1/2}$
Ans: (b)
Sol: Let cylinder is displaced by an amount x from its mean position .
The net restoring force , $\large F = -(k x + A x \rho g)$
$\large M a = -(k x + A x \rho g)$
$\large a = -\frac{(k + A \rho g)}{M} x$
In S.H.M , a = -ω2 x
Hence , $\large \omega^2 = \frac{(k + A \rho g)}{M}$
$\large \omega = \sqrt{\frac{(k + A \rho g)}{M}}$
Frequency of oscillation $\large f = \frac{\omega}{2\pi}$
$\large f = \frac{1}{2\pi}\sqrt{\frac{(k + A \rho g)}{M}}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9835326075553894, "perplexity": 682.8606552111929}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057447.52/warc/CC-MAIN-20210923195546-20210923225546-00697.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=601074 | # Integrating 2nd order ODE using midpoint rule
by Niles
Tags: integrating, midpoint, order, rule
P: 1,863 Hi I am trying to integrate Newtons equations for my system $$a = \frac{F}{m} = \frac{d^2x}{dt^2}$$ This is only for the first coordinate of the particle. I wish to do it for y and z as well, but let us just work with x for now to make it simple. The force in the x-direction depends on the velocity in the x-direction, vx, and the y- and z-coordinate. In other words $$F=F(v_x, y, z)$$ Now, I wish to solve this equation, and I have currently implemented an Euler method. This is how I iterate $$v_{n+1} = v_n + dt\cdot a(v_{x,n},y_n,z_n) \\ x_{n+1} = x_{n} + dt\cdot v_{n}$$ I now want to improve the error, and use a 2nd order Runge-Kutta method, i.e. the midpoint rule as briefly summarized here: http://www.efunda.com/math/num_ode/num_ode.cfm I am not quite sure how to do this. In the link they say that now I should generally write $$y_{n+1} = y_{n} + dt\cdot f(x_n + dt/2, y_n + k_1/2)$$ where $$k_1 = dt\cdot f(x_n, y_n).$$ This is where my confusion arises: What does $f(x_n + dt/2, y_n + k_1/2)$ correspond to for me? I would really appreciate a hint or two with this. Best, Niles.
P: 26 Here you have it explained: Computational physics page 292, "13.4 More on finite difference methods, Runge-Kutta methods"
P: 1,863 Thanks!
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https://jack.valmadre.net/papers/2012-cvpr-filters/ | Consider a 3D object deforming non-rigidly over time, for example:
The object is observed by a single moving camera. We want to recover the 3D structure from these 2D projection correspondences.
Let’s assume that the camera motion relative to the background can be recovered using rigid structure from motion. With known cameras, each 2D projection of a point defines a 2x3 linear system of equations for its 3D position in that frame.
The 1D nullspace of this system corresponds geometrically to a ray departing from the camera centre.
This means that using only projection constraints, there are infinite solutions for each point in each frame (anywhere along those rays). However, we know intuitively that points should not jump around arbitrarily from frame to frame. Assuming that the points have mass, their motion should be smooth (since they must be accelerated by a finite force). This notion was leveraged by Akhter et al (2008), who required that the trajectory of each point be expressed as a sum of low frequency sinusoids i.e. lie on the subspace of a truncated DCT basis.
Stacking the equations for the observation of a point in all F frames yields a 2Fx3F system.
Introducing a K-dimensional DCT basis for the x, y and z components of the trajectory, the system of equations becomes 2Fx3K.
While this problem is not necessarily underconstrained if 3K ≤ 2F, Park et al (2010) found that significant camera motion was also required to obtain an accurate reconstruction, establishing the notion of reconstructability. The following figure shows reconstruction error versus camera speed for varying basis size K, generated from a synthetic experiment in which human motion capture sequences (of 100 frames) are observed by an orbiting camera.
The key observation here is that faster moving cameras enable more accurate reconstruction. However, you have to choose the basis size correctly or else you will either run out of capacity to represent the points’ motion (to the right) or revert to an ill-posed problem (to the left).
Park et al defined a measure of reconstructability depending on how well the camera trajectory was represented by the basis, since, if the camera centre lies on the basis, its trajectory is a trivial solution. We defined a new measure based on a theoretical bound on the reconstruction error. Our measure does not depend on the camera centre, rather, it incorporates the condition number of the linear system of equations, the ratio of the largest eigenvalue to the smallest.
The implications of this become clearer when we alternatively consider minimising the distance of the trajectory from the DCT subspace, subject to the 2D projections being exactly satisfied.
Since the DCT matrix is rank 3K, the orthogonal projection matrix in the objective will have a nullspace of dimension 3K. When we reduce K, we reduce the size of this nullspace and therefore reduce the likelihood of having a poorly-conditioned system. Consider the more general form
We propose that the conditioning problem can be mostly avoided by choosing
where multiplication by the matrix G is equivalent to convolution with the filter g (note that we define vector convolution to operate independently in x, y, z.) We typically choose simple finite difference filters: some combination of (-1, 1) and (-1, 2, -1). If the filter has support m, then the matrix M has a nullspace of size 3(m-1), since we only compute the convolution with parts of the signal where the two inputs overlap completely (i.e. “valid” mode in Matlab’s conv() function).
Using these “trajectory filters,” we are able to achieve 3D error at the limit of reconstructability without having to choose the basis size K.
The reason that this works becomes evident examining the eigenspectra of the DCT matrix (left) versus the first-difference (middle) and second-difference (right) filters. The filters do not have the many zero eigenvalues which cause the system to become poorly conditioned.
A nice twist here is that the Discrete Cosine Transform diagonalises symmetric convolution in an analogous way to the Fourier transform diagonalising periodic convolution, so the eigenspectra above are actually the DCT transform of each filter. This means we can enforce the filters as a weighting in the DCT domain, although in practice it is usually more efficient to work in the time domain because the systems are sparse. This also makes it possible to reveal the filter which is equivalent to using a particular truncation of the DCT basis.
The final results from our example are shown below. Red points denote the output and black the ground truth.
### Code
I’ve tried to provide sufficient code and data to reproduce the figures in the paper. It’s all Matlab. Never tested with GNU Octave. This code is provided for free use, no warranty whatsoever.
Run setup.m to set the path as required.
The main figures (reconstruction error versus camera speed) can be generated by running solver_experiment.m. Note that averaging these experiments over as many trials as we did probably requires a compute cluster or significant hours. You can reduce the number of trials though, for slightly less-smooth curves.
This experiment depends on Honglak Lee et al’s implementation of their feature-sign search algorithm for the comparison to sparse coding methods. Ensure that l1ls_featuresign.m is found in src/lee-2006/.
I’m also providing the exact 100-frame mocap sequences from the CMU Motion Capture Database which we used. Thanks to Mark Cox for converting them to point clouds. Put this file in data/ before running solver_experiment.m.
The actual reconstruction code is super simple and happens in reconstruct_filters.m and reconstruct_linear.m. The linear systems for projection constraints are constructed in trajectory_projection_equations.m, projection_equation.m and independent_to_full.m.
The qualitative examples from real photos are generated by running real_scene_demo.m. To do this, you’ll need to download the “Real scene” data from Hyun Soo Park’s project page and unzip it to data/RealSceneData/. I’ve also re-distributed two small functions from his code in src/park-2010/. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8326136469841003, "perplexity": 856.9797816535608}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145960.92/warc/CC-MAIN-20200224132646-20200224162646-00167.warc.gz"} |
https://math.stackexchange.com/questions/244718/double-absolute-values | # double absolute values
I am having a little bit of problem with an inequality with nested absolute values:
$$|z^2-1| \ge |z+|1-z^2||$$
I've tried solving it by making three cases, $z\ge1$, $z\le-1$ and $z$ between $1$ and $-1$ and thus getting rid of absolute values for $z^2-1$ and $1-z^1$, and I am only left with 1 absolute value. But solutions at the end are not what they should be based on the graph. Here, $z$ is real, and WolframAlpha gives this solution.
What I am doing wrong?
Note that for any $a$ and $b$, we have $|a|\ge |b|$ iff $a^2 \ge b^2$. Apply this with $a$ the left-hand side, and $b$ the right-hand side of our expression. Thus our inequality is equivalent to $$(z^2-1)^2\ge z^2+2z|1-z^2| +(1-z^2)^2.$$ Since $(z^2-1)^2=(1-z^2)^2$, we are trying to solve the inequality $$z^2+2z|1-z^2| \le 0.\tag{1}$$ Sure killed an awful lot of absolute value signs!
The inequality $(1)$ holds at $z=0$. And it is obvious that it cannot hold for positive $z$. So (remembering that from now on $z$ is negative), we are looking at the inequality $$z+2|1-z^2| \ge 0.$$ The rest is routine. We can divide into two cases, $z\le -1$ and $-1\lt z\lt 0$. It turns out that the inequality holds for all $z \le 0$, except for the numbers in the open interval $(a,b)$, where $a=-\frac{\sqrt{17}+1}{4}$ and $b= -\frac{\sqrt{17}-1}{4}$.
I am presuming that $z$ is real. The problem is that the outer absolute on the right may change sense at other places. Say $z \lt -1$. Then $|z+|1-z^2||=|z+z^2-1|$, but now you are testing whether $z+z^2-1 \gt 0$ which doesn't change sense at those points. So you need to find some secondary cases based on what you get for the prime cases.
$|z^2-1| \ge |z+|1-z^2||$
Case 1: Suppose $z \ge 1$. Then $|z^2 - 1| = z^2 - 1$ and $|1 - z^2| = z^2 - 1$:
$z^2-1 \ge |z+(z^2-1)||$
Also $z + (z^2 - 1) > 0$ so:
$z^2-1 \ge z+(z^2-1)$
$0 \ge z$
This is a contradiction.
Case 2: Suppose $z \le -1$. Then $|z^2 - 1| = z^2 - 1$ and $|1 - z^2| = z^2 - 1$:
$z^2-1 \ge |z+(z^2-1)||$
There is a root of $z^2 + z - 1$, so we must case on that.
Case 2a: Suppose $z \le -\frac{\sqrt5 + 1}{2}$, then:
$z^2-1 \ge z^2+z-1$
Case 2b: Suppose $-\frac{\sqrt5 + 1}{2} \le z \le -1$, then
$z^2-1 \ge -z^2-z+1 \Rightarrow 2z^2 + z \ge 0$. $z \le -\frac{\sqrt5 + 1}{2}$ always satisfies this.
Case 3: Suppose $-1 \le z \le 1$, then
$1-z^2 \ge |z+1-z^2)|$
This has a root at $\frac{1-\sqrt5}{2}$, so we case there,
Case 3a: $\frac{1-\sqrt5}{2} \le z \le 1$
$1-z^2 \ge z+1-z^2)$
$0 \ge z$. $\frac{1-\sqrt5}{2} \le z \le 0$ satisfies this.
Case 3b: $-1 \le z \le \frac{1-\sqrt5}{2}$.
$1-z^2 \ge z^2-z-1$
$2z^2 - z \le 0$. This does not hold for negative $z$, so it is a contradiction.
We conclude that $z \le -\frac{\sqrt5 + 1}{2}$ or $\frac{1-\sqrt5}{2} \le z \le 0$.
The graphing method is definitely easier here. It also may be easier to consider the potential roots first and then use more cases instead of cases-with-subcases, though ultimately those are similar arguments.
Here is a solution
$$|z+|1-z^2||\leq |1-z^2| \implies -|1-z^2|\leq z+|1-z^2|\leq |1-z^2|$$
$$\implies -2|1-z^2|\leq z\leq 0 \,.$$
From the above inequality, the solution $z$ should lie in the set $\left\{z\leq 0\right\} \cap \left\{ z\geq-2|1-z^2| \right\}$.
working out $z \geq -2|1-z^2|,$ gives
$$\left\{z \geq -2|1-z^2| \right\} = \left\{z \geq -2(1-z^2) \right\} \cup \left\{z \geq -2(-1+z^2) \right\}$$
$$= ( -.78, 1.28 ) \cup \left\{ (-\infty, -1.28)\cup (-.78,\infty) \right\}$$
$$= (-\infty, -1.28) \cup ( -.78, \infty ) .$$
Thus, the solution is given by
$$\left\{z\leq 0\right\} \cap \left\{ z\geq -2|1-z^2| \right\}$$
$$=\left\{z\leq 0\right\} \cap \left\{ (-\infty, -1.28) \cup ( -.78, \infty ) \right\}$$
$$=\left\{\left\{z\leq 0\right\} \cap (-\infty, -1.28)\right\} \cup \left\{\left\{z\leq 0\right\} \cap ( -.78, \infty )\right\}$$
$$= \left( -\infty,-1.28\right) \cup \left(-.78, 0 \right).$$
Note: I approximated the roots when I was solving the inequalities. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.984227180480957, "perplexity": 163.5538585046446}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00598.warc.gz"} |
https://open.library.ubc.ca/soa/cIRcle/collections/ubctheses/24/items/1.0370937 | # Open Collections
## UBC Theses and Dissertations
### Infrared quantum information Chaurette, Laurent
#### Abstract
Scattering amplitudes in massless gauge field theories have long been known to give rise to infrared divergent effects from the emission of very low energy gauge bosons. The traditional way of dealing with those divergences has been to abandon the idea of measuring amplitudes by only focusing on inclusive cross-sections constructed out of physically equivalent states. An alternative option, found to be consistent with the S-matrix framework, suggested to dress asymptotic states of charged particles by shockwaves of low energy bosons. In this formalism, the clouds of soft bosons, when tuned appropriately, cancel the usual infrared divergences occurring in the standard approach. Recently, the dressing approach has received renewed attention for its connection with newly discovered asymptotic symmetries of massless gauge theories and its potential role in the black hole information paradox. We start by investigating quantum information properties of scattering theory while having only access to a subset of the outgoing state. We give an exact formula for the von Neuman entanglement entropy of an apparatus particle scattered off a set of system particles and show how to obtain late-time expectation values of apparatus observables. We then specify to the case of quantum electrodynamics (QED) and gravity where the unobserved system particles are low energy photons and gravitons. Using the standard inclusive cross-section formalism, we demonstrate that those soft bosons decohere nearly all momentum superpositions of hard particles. Repeating a similar computation using the dressing formalism, we obtain an analogous result: In either framework, outgoing hard momentum states at late times are fully decohered from not having access to the soft bosons. Finally, we make the connection between our results and the framework of asymptotic symmetries of QED and gravity. We give new evidence for the use of the dressed formalism by exhibiting an inconsistency in the scattering of wavepackets in the original inclusive cross-section framework. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8621944785118103, "perplexity": 664.8456103681248}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662619221.81/warc/CC-MAIN-20220526162749-20220526192749-00142.warc.gz"} |
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http://planetmath.org/topologicalentropy | topological entropy
Primary tabs
Synonym:
entropy
Type of Math Object:
Definition
Major Section:
Reference
Mathematics Subject Classification
entropy of ergodic and mixing processes
Very interesting definition. If I grok it correctly, it seems to be saying that ergodic processes will have a low entropy in general, which is surprising and counter-intuitive. The only systems I can think of that would have high entropy would be dissipative systems, where, on iteration, large portions of the space ''X'' are abandoned during iteration, and never visited again. So ... its this entropy in fact a measure of dissipation? Can any other intuitive interpretations can be added?
--linas
Re: entropy of ergodic and mixing processes
Never mind. I've (once again) got the definition exactly upside-down.
Re: entropy of ergodic and mixing processes
Soo .. is there any way to simply erase/retract/edit/modify this line of posts? I clearly misread the defn, (again), so the comment I just made earlier is nonsense. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 23, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9727462530136108, "perplexity": 2331.805006627683}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00338-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://me598.wikidot.com/homework-3-problem-10 | Homework 3 Problem 10
Problem
Give an example of a Lagrangian system that has a degenerate Lagrangian $L:T \mathbb{R}^{2}\rightarrow \mathbb{R}$ but a second-order Lagrangian vector field $Z_{L}$.
Melih's Solution:
Let
(1)
\begin{align} L:T\mathbb{R}^{2}\rightarrow \mathbb{R}:(q_{1},q_{2},\dot{q}_{1},\dot{q}_{2})\mapsto \pounds =\frac{1}{2}\dot{q}_{1}^{2}+\frac{1}{2}q_{1}^{2}q_{2}+\lambda \dot{q}_{2} \end{align}
Using this, we can obtain a coordinate expression for the this Lagrangian as $4\times 4$ matrix. Let $B$ be the the skew-symmetrization of $\frac{\partial ^{2}L}{\partial \dot{q}_{i}\partial q_{j}}$.
(2)
\begin{array} {ll} \Omega_L& =\begin{bmatrix} B & \left[ \frac{\partial ^{2}L}{\partial \dot{q}_{i}\partial \dot{q}_{j}}\right] \\ \left[ -\frac{\partial ^{2}L}{\partial \dot{q}_{i}\partial \dot{q}_{j}}\right] & 0 \end{bmatrix}\\~\\ &= \begin{bmatrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix} \end{array}
This is singular at every point so the Lagrangian is denerate. Also note that the dynamics of that bead moving in the plane has a constraint on the velocity in $2$-direction.
The Legendre transform; $FL:T\mathbb{R}^{2}\rightarrow T^{\ast }\mathbb{R}^{2}$ $:(q_{i},\dot{q}_{i})\mapsto \left( q_{i},\frac{\partial L}{\partial \dot{q}_{i}}\right)$ is found as,
(3)
\begin{align} FL=(q_{1},q_{2},\dot{q}_{1},\lambda ) \end{align}
The action of $L$; $A:T\mathbb{R}^{2}\rightarrow \mathbb{R}:(q_{1},q_{2}, \dot{q}_{1},\dot{q}_{2})\mapsto A=FL(v)\cdot v$ for $v\in T_{q}\mathbb{R}^{2}\Rightarrow$
(4)
\begin{align} A=\left\langle \frac{\partial L}{\partial \dot{q}_{i}},\dot{q}% _{i}\right\rangle =\dot{q}_{1}^{2}+\lambda \dot{q}_{2}. \end{align}
The energy of $L$;
(5)
\begin{align} E:T\mathbb{R}^{2}\rightarrow \mathbb{R}:(q_{1},q_{2},\dot{q}_{1},\dot{q}_{2})\mapsto E=A-L=(\dot{q}_{1}^{2}+\lambda\dot{q}_{2})-(\frac{1}{2}\dot{q}_{1}^{2}+\frac{1}{2}q_{1}^{2}q_{2}+\lambda\dot{q}_{2})=\frac{1}{2}\dot{q}_{1}^{2}-\frac{1}{2}q_{1}^{2}q_{2}\Rightarrow \end{align}
(6)
\begin{align} dE=\dot{q}_{1}d\dot{q}_{1}-q_{1}q_{2}dq_{1}-\frac{1}{2}q_{1}^{2}dq_{2}. \end{align}
Since, $dE=Z_{L}\lrcorner \Omega _{L}+w$, where
(7)
\begin{align} \Omega _{L}=\frac{\partial^{2}L}{\partial \dot{q}_{i}\partial q_{j}}dq_{i}\wedge dq_{j}+\frac{\partial^{2}L}{\partial \dot{q}_{i}\partial \dot{q}_{j}}dq_{i}\wedge d\dot{q}_{j}=dq_{1}\wedge d\dot{q}_{1} \end{align}
and $w$ is the one form exterior force;
(8)
\begin{align} Z_{L} =(\dot{q}_{1},-q_{1}q_{2}) \end{align}
(9)
\begin{align} w =-\frac{1}{2}q_{1}^{2}dq_{2} \end{align}
i.e. there exists a second order vector field, $Z_{L}$, such that;
(10)
\begin{align} Z_{L_{q_{1}}} &=&\dot{q}_{1} \end{align}
(11)
\begin{align} Z_{L_{\dot{q}_{1}}} &=&-q_{1}q_{2} \end{align}
with the constraint; $\dot{q}_{2}=0$.
Discussion
Please check the Lagrangian I suggested.
page revision: 3, last edited: 21 Apr 2007 21:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 10, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999972581863403, "perplexity": 4551.640732846266}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948511435.4/warc/CC-MAIN-20171210235516-20171211015516-00350.warc.gz"} |
http://tex.stackexchange.com/questions/297590/unable-to-plot-both-axes-and-arccos-function-in-pgfplots | # Unable to plot both axes and arccos function in pgfplots
What I want to plot is 3 functions: x=0, y=0, and x=cos^{-1}(x).
The plot and graph of the first two look like this:
\begin{center}
\begin{tikzpicture}
\begin{axis}[domain = -1:1, samples = 500, grid = both]
\addlegendentry{$x=0$}
\addlegendentry{$y=0$}
\end{axis}
\end{tikzpicture}
\end{center}
\begin{center}
\begin{tikzpicture}
\begin{axis}[domain = -1:1, samples = 500, grid = both]
\addlegendentry{$x=0$}
\addlegendentry{$y=0$}
\addlegendentry{$\cos^{-1}(x)$}
\end{axis}
\end{tikzpicture}
\end{center}
Everything gets messed up. Why is this?
-
The horizontal domain gives the impression that nothing has changed but because you have suddenly increased the vertical axis to 150 domain=-1:1 becomes invisible.
Also you are using 500 samples to constant plots. Supply individually instead, here is with 130
\begin{tikzpicture}
\begin{axis}[grid = both,samples=2]
\addlegendentry{$x=0$}
\addlegendentry{$y=0$}
\addplot[draw = blue,samples = 101,domain = -1:1] {acos(x)};
\addlegendentry{$\cos^{-1}(x)$}
\end{axis}
\end{tikzpicture}
-
That makes a lot of sense! It didn't occur to me that this was the issue because in my mind domain was the opposite direction. And thank you for the suggestion about the samples! – Abe Fehr Mar 6 at 13:17
@AbeFehr My pleasure – percusse Mar 6 at 13:54
As it turns out, pgfplots (and the underlying PGF, I think) treat the input to trigonometric functions as degrees, not radians. You can force it to be in radians by the key trig format=rad. This also fixes your y-domain issues:
\documentclass{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.13}
\begin{document}
\begin{tikzpicture}
\begin{axis}[domain = -1:1, samples = 500, grid = both]
\addlegendentry{$x=0$}
\addlegendentry{$y=0$}
\addlegendentry{$\cos^{-1}(x)$} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8496245741844177, "perplexity": 2405.8231202005554}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783397795.31/warc/CC-MAIN-20160624154957-00144-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/fresnel-power-coefficients.434944/ | # Fresnel power Coefficients
1. Oct 4, 2010
### Gogsey
Hi,
I'm doing a question using the Fresnel power coefficients, not the Fresnel amplitude coefficients. We can use matlab for this, but we are given a fixed angle of incidence of 45 deg so its really just one calculation, as opposed to an earlier question where the angle varied between 0 and 90 deg. We are given the refractive indices.
But what are these equations, lol. We're supposed to look them up but I can't find them.
The earlier question talked about the Fresnel amplitude coefficients, and this current question tells us to use the Fresnel power coefficients so I assume they are different.
Thanks a lot
Liam
2. Oct 5, 2010
### hikaru1221
I've never heard of Fresnel power coefficient, but I guess it's similar to reflection coefficient. Then if you can calculate the relation between amplitudes, since power ~ intensity ~ amplitude^2, you can calculate the coefficient, right?
3. Oct 5, 2010
### Gogsey
Hi Thanks,
yeah all we had to do was square the Fresnel amplitude reflection coefficient to get the power reflection coefficient and use 1-R2 to get the Fresnel power transmission coefficient. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9666847586631775, "perplexity": 1128.4269933854312}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860118321.95/warc/CC-MAIN-20160428161518-00177-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/self-studier-w-kleppner-kolenkow-question.404135/ | # Self-studier w/ kleppner & kolenkow question
1. May 18, 2010
### pton265
Hi,
This is my first post. I'm reviewing mechanics out of K&K and have a question about problem 4.5:
"Mass m whirls on a frictionless table, held to a circular motion by a string which passes through a hole in the table. The string is slowly pulled through the hole so that the radius of the circle changes from l1 to l2. Show that the work done in pulling the string equals the increase in kinetic energy of the mass."
I'm assuming the mass starts with uniform circular motion at radius l1, and I analyze in polar coordinates with the center of this circle as the origin. My initial intuition about the motion:
(1) Angular acceleration is non-zero (positive), but there cannot be a $$\widehat{}\theta$$ component of acceleration (the force is always radial) - which is only true if
2$$\dot{}r$$$$\dot{}\theta$$ = -r$$\ddot{}\theta$$.
(2) The only way for the string to do work (increase the magnitude of m's velocity) is if m's trajectory (and, therefore, velocity) has some radial component. That is, the force must at some point have a non-orthogonal component with respect to the trajectory. Since the force is everywhere radial, the $$\widehat{}\theta$$ component of velocity is unchanged, while the radial component of velocity increases (in the negative $$\widehat{}r$$ direction). This statement does not contradict (1), where I state angular acceleration is non-zero (right?!). Physically, all of this corresponds to the mass breaking from uniform circular motion and spiraling inward toward the center of the table (i.e. where the hole is). When it reaches l2, it will NOT be in uniform circular motion because it's velocity must have some radial component (inward).
Now, the only solution (http://hep.ucsb.edu/courses/ph21/problems/p7sol.pdf [Broken]) I've found takes (1) to be true, but (2) to be false. The final velocity in the solution has a magnitude such that the velocity can not possibly have a radial component. In other words, once the mass reaches l2, it pops back into uniform circular motion (albeit, with higher velocity). How can the trajectory take this form (i.e. that of consecutively smaller concentric circles)?? Is the solution wrong? If not, where is my error?
Please bear in mind that the only assumed knowledge at this point is of translational motion, linear momentum, and the Work-energy theorem in one dimension (KK chs.1-4) - not angular momentum, rotational motion, etc.
My apologies if there is already a similar thread - I've searched PF pretty thoroughly.
Thanks very much for any and all help...
L
Last edited by a moderator: May 4, 2017
Can you offer guidance or do you also need help?
Draft saved Draft deleted
Similar Discussions: Self-studier w/ kleppner & kolenkow question | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9383972883224487, "perplexity": 647.587669843904}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886124662.41/warc/CC-MAIN-20170823225412-20170824005412-00255.warc.gz"} |
https://blog.stata.com/tag/optimization/ | ### Archive
Posts Tagged ‘optimization’
## Programming an estimation command in Stata: Using optimize() to estimate Poisson parameters
$$\newcommand{\xb}{{\bf x}} \newcommand{\betab}{\boldsymbol{\beta}}$$I show how to use optimize() in Mata to maximize a Poisson log-likelihood function and to obtain estimators of the variance–covariance of the estimator (VCE) based on independent and identically distributed (IID) observations or on robust methods.
This is the eighteenth post in the series Programming an estimation command in Stata. I recommend that you start at the beginning. See Programming an estimation command in Stata: A map to posted entries for a map to all the posts in this series.
Using optimize()
There are many optional choices that one may make when solving a nonlinear optimization problem, but there are very few that one must make. The optimize*() functions in Mata handle this problem by making a set of default choices for you, requiring that you specify a few things, and allowing you to change any of the default choices.
When I use optimize() to solve a Read more…
Categories: Programming Tags:
## Programming an estimation command in Stata: A review of nonlinear optimization using Mata
$$\newcommand{\betab}{\boldsymbol{\beta}} \newcommand{\xb}{{\bf x}} \newcommand{\yb}{{\bf y}} \newcommand{\gb}{{\bf g}} \newcommand{\Hb}{{\bf H}} \newcommand{\thetab}{\boldsymbol{\theta}} \newcommand{\Xb}{{\bf X}}$$I review the theory behind nonlinear optimization and get more practice in Mata programming by implementing an optimizer in Mata. In real problems, I recommend using the optimize() function or moptimize() function instead of the one I describe here. In subsequent posts, I will discuss optimize() and moptimize(). This post will help you develop your Mata programming skills and will improve your understanding of how optimize() and moptimize() work.
This is the seventeenth post in the series Programming an estimation command in Stata. I recommend that you start at the beginning. See Programming an estimation command in Stata: A map to posted entries for a map to all the posts in this series.
A quick review of nonlinear optimization
We want to maximize a real-valued function $$Q(\thetab)$$, where $$\thetab$$ is a $$p\times 1$$ vector of parameters. Minimization is done by maximizing $$-Q(\thetab)$$. We require that $$Q(\thetab)$$ is twice, continuously differentiable, so that we can use a second-order Taylor series to approximate $$Q(\thetab)$$ in a neighborhood of the point $$\thetab_s$$,
$Q(\thetab) \approx Q(\thetab_s) + \gb_s'(\thetab -\thetab_s) + \frac{1}{2} (\thetab -\thetab_s)’\Hb_s (\thetab -\thetab_s) \tag{1}$
where $$\gb_s$$ is the $$p\times 1$$ vector of first derivatives of $$Q(\thetab)$$ evaluated at $$\thetab_s$$ and $$\Hb_s$$ is the $$p\times p$$ matrix of second derivatives of $$Q(\thetab)$$ evaluated at $$\thetab_s$$, known as the Hessian matrix. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8295153975486755, "perplexity": 771.6493558824662}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668534.60/warc/CC-MAIN-20191114182304-20191114210304-00082.warc.gz"} |
http://wikis.controltheorypro.com/MATLAB_tf | MATLAB tf
MATLAB tf In order to prevent spam, users must register before they can edit or create articles.
## 1 Introduction to MATLAB's tf command
Transfer functions are easily to created in MATLAB using the tf command. (The tf command requires the.) This command accepts two vectors - one for the numerator and one for the denominator.
## 2 Basic Usage for MATLAB's tf command
The basic usage of MATLAB's tf command is
>> sys = tf(num, den);
where:
num is a vector representing the coefficients of the numerator and
den is a vector representing the coefficients of the denominator.
For example, a transfer function like this
$LaTeX: H\left( s \right) = \frac{s+2}{s^2+3+10}$
becomes
>> num = [1 2];
>> den = [1 3 10];
Notice that MATLAB determines the order from the number of elements in the vectors. If the numerator had been $LaTeX: s^2+2$ then
>> num = [1 0 2];
would be the correct vector. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8868213295936584, "perplexity": 2650.321076144421}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218191396.90/warc/CC-MAIN-20170322212951-00259-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://mattbaker.blog/2015/03/20/a-p-adic-proof-that-pi-is-transcendental/ | # A p-adic proof that pi is transcendental
Ferdinand von Lindemann
In my last blog post, I discussed a simple proof of the fact that pi is irrational. That pi is in fact transcendental was first proved in 1882 by Ferdinand von Lindemann, who showed that if $\alpha$ is a nonzero complex number and $e^\alpha$ is algebraic, then $\alpha$ must be transcendental. Since $e^{i \pi} = -1$ is algebraic, this suffices to establish the transcendence of $\pi$ (and setting $\alpha = 1$ it shows that $e$ is transcendental as well). Karl Weierstrass proved an important generalization of Lindemann’s theorem in 1885.
The proof by Lindemann that pi is transcendental is one of the crowning achievements of 19th century mathematics. In this post, I would like to explain a remarkable 20th century proof of the Lindemann-Weierstrass theorem due to Bezivin and Robba [Annals of Mathematics Vol. 129, No. 1 (Jan. 1989), pp. 151-160], which uses p-adic analysis in a key way. Their original argument was made substantially more elementary by Beukers in this paper; we refer the reader to [American Mathematical Monthly Vol. 97 Issue 3 (Mar. 1990), pp. 193-197] for a lovely exposition of the resulting proof, which rivals any of the usual approaches in its simplicity. But I’d like to focus here on the original Bezivin-Robba proof, which deserves to be much better known than it is. In the concluding remarks, we will briefly discuss a 21st century theorem of Bost and Chambert-Loir that situates the Bezivin-Robba approach within a much broader mathematical framework.
An equivalent assertion
Let $\overline{{\mathbb Q}}$ be the subfield of ${\mathbb C}$ consisting of all complex numbers which are algebraic (over ${\mathbb Q}$). The Lindemann-Weierstrass theorem is the following statement:
(L-W) Let $\alpha_1,\ldots,\alpha_m \in \overline{{\mathbb Q}}$ be distinct algebraic numbers. Then $e^{\alpha_1},\ldots,e^{\alpha_m}$ are linearly independent over $\overline{{\mathbb Q}}$.
A relatively simple argument shows that (L-W) is equivalent to a rather different-looking assertion about formal power series which are represented by rational functions.
It will be convenient to work with power series expansions around infinity rather than zero. Recall that a function $f : {\mathbb C} \to {\mathbb C}$ is analytic at $\infty$ if the function $g(w) = f(1/w)$ is analytic at $w=0$. If $g(w)=b_0 + b_1 w + b_2 w^2 + \cdots$ is the power series expansion for $g(w)$ around $0$, we call
$f(z) = b_0 + b_1 \frac{1}{z} + b_2 \frac{1}{z^2} + \cdots$
the power series expansion for $f(z)$ around $z=\infty$. We will be particularly interested in functions $f(z)$ for which $b_0 = 0$ (i.e., which vanish at infinity).
We say that a formal power series $v(x) \in {\mathbb C}[[\frac{1}{x}]]$ is analytic at $\infty$ if the power series $u(w) := v(1/w)$ has a nonzero radius of convergence around $w=0$. And by abuse of terminology, we say that $v(x)$ is a rational function if there are polynomials $P(x),Q(x)$ with $Q(x)$ not identically zero such that the power series expansion of $f(x)=\frac{P(x)}{Q(x)}$ around $x=\infty$ is equal to $v(x)$. A rational function vanishes at infinity if and only if ${\rm deg}(P) < {\rm deg}(Q)$.
Let $K$ be a field, and let ${\mathcal F}_K := \frac{1}{x}K[[\frac{1}{x}]]$ be the ring of formal power series over $K$ in $\frac{1}{x}$ which vanish at infinity. Let ${\mathcal D} : {\mathcal F}_{\mathbb C} \to {\mathcal F}_{\mathbb C}$ be the differential operator ${\mathcal D}(v) = v + v'$. We will show that (L-W) is equivalent to the following statement:
(B-R) If $v \in {\mathcal F}_{\mathbb Q}$ is analytic at infinity and ${\mathcal D}(v)$ is a rational function, then $v$ is also a rational function.
Note that the conclusion of (B-R) can fail for functions with an essential singularity at infinity; for example, ${\mathcal D}(e^{-x}) = 0$ but $e^{-x}$ is not a rational function.
Proof of the equivalence
The proof that (L-W) and (B-R) are equivalent is based on properties of the Laplace transform. Define the formal Laplace transform ${\mathcal L} : {\mathbb C}[[z]] \to {\mathcal F}_{\mathbb C}$ by
${\mathcal L}(\sum_{n=0}^\infty a_n z^n) = \sum_{n=0}^\infty \frac{n! a_n}{x^{n+1}}.$
(This is just the extension of the usual Laplace transform to the setting of formal power series.) The map ${\mathcal L} : {\mathbb C}[[z]] \to {\mathcal F}_{\mathbb C}$ is clearly a bijection.
We will make use of the following standard facts from complex analysis:
(L1) $f(z) \in {\mathbb C}[[z]]$ defines an entire function of exponential growth (i.e. $|f(z)| \leq C_1 e^{C_2 |z|}$ for some $C_1, C_2$) if and only if ${\mathcal L}(f)$ is analytic at infinity.
(L2) $f(z) \in {\mathbb C}[[z]]$ is the power series expansion around $z=0$ of an exponential polynomial $p_1(z)e^{a_1 z} + \cdots + p_n(z)e^{a_n z}$ if and only if ${\mathcal L}(f)$ is a rational function. This gives a bijection between exponential polynomials and rational functions vanishing at infinity.
The proof of (L2), which is based on the partial fractions decomposition of rational functions and the fact that ${\mathcal L}(e^{az}) = \frac{1}{1-ax}$, shows that $p_i(z) \in \overline{{\mathbb Q}}[z]$ and $a_i \in \overline{{\mathbb Q}}$ for all $i$ if and only if ${\mathcal L}(f) \in \overline{{\mathbb Q}}(x).$
We will also need the following lemma, whose proof we leave as an exercise:
Lemma: Define $\delta : {\mathbb C}[[z]] \to {\mathbb C}[[z]]$ by $\delta(f(z)) = (z-1)f(z)$, and let ${\mathcal D} : {\mathcal F}_{\mathbb C} \to {\mathcal F}_{\mathbb C}$ be as above. Then $\delta$ and ${\mathcal D}$ are bijections, and ${\mathcal D}({\mathcal L}(f))= {\mathcal L}(\delta(f)).$
To see that (L-W) implies (B-R), suppose $v \in {\mathcal F}_{\mathbb Q}$ is analytic at infinity and ${\mathcal D}(v)$ is a rational function. By (L2), there is an exponential polynomial $f(z)= \sum p_i(z) e^{\alpha_i z}$ with the $\alpha_i$ distinct algebraic numbers and $p_i(z) \in \overline{{\mathbb Q}}[z]$ such that ${\mathcal L}(f) = {\mathcal D}(v)$. The function $g(z) := \frac{f(z)}{z-1}$ satisfies $\delta(g(z)) = f(z)$, so by the Lemma we have ${\mathcal L}(g)=v$. As $v$ is analytic at infinity, we know by (L1) that $g$ is entire, and hence $f(1)=0$. By (L-W), we must have $p_i(1)=0$, i.e. $(z-1) \mid p_i(z)$, for all $i$. Thus $g(z)$ is also an exponential polynomial, which implies by (L2) that $v(x)$ is a rational function.
To see that (B-R) implies (L-W), assume for the sake of contradiction that $f(1)=0$, where $f(z) := \sum_{i=1}^m \beta_i e^{\alpha_i z}$, the $\alpha_i$ are distinct and algebraic, and the $\beta_i$ are algebraic and nonzero. Replacing $f(z)$ by the product of its Galois conjugates $\sum_{i=1}^m \sigma(\beta_i) e^{\sigma(\alpha_i) z}$, we may assume without loss of generality that the power series expansion of $f(z)$ lies in ${\mathbb Q}[[z]]$. (This is a standard reduction which appears in many proofs of (L-W).) The Laplace transform of $f(z)$ is
${\mathcal L}(f) = \sum \frac{\beta_i}{1-\alpha_i x},$
which has only simple poles. Moreover, since the $\alpha_i$ are distinct and $f(1)=0$ we must have $m \geq 2$ and some $\alpha_i$ is non-zero; thus ${\mathcal L}(f)$ has at least one simple pole. On the other hand, since $f(1)=0$, the function $g(z) := \frac{f(z)}{z-1}$ is entire and of exponential growth, so by (L1) $v := {\mathcal L}(g) \in {\mathcal F}_{\mathbb C}$ is analytic at infinity. The Lemma tells us that ${\mathcal L}(f) = {\mathcal D}(v)$, so ${\mathcal D}(v)$ has only simple poles. However, it is easy to see that if $u$ is a rational function then ${\mathcal D}(u) = u + u'$ can never have a simple pole. Thus $v$ is not a rational function, contradicting (B-R).
Rationality of formal power series
In order to prove (B-R), we need to show that if $v \in {\mathcal F}_{\mathbb Q}$ is analytic at infinity and ${\mathcal D}(v)$ is a rational function, then $v$ is also a rational function. For this, we need some kind of robust criterion for determining whether a formal power series with coefficients in ${\mathbb Q}$ represents a rational function. There is a long history of such results culminating in what one might call the Borel-Polya-Dwork-Bertrandias criterion, which will turn out to be exactly what we need. We interrupt our regularly scheduled proof to give a brief history of these developments.
Borel
Around 1894, Emile Borel noticed that if $f(z)=\sum_{n=0}^\infty a_n z^n$ is a power series with integer coefficients defining an analytic function on a closed disc of radius $R > 1$ in ${\mathbb C}$, then $f(z)$ must in fact be a polynomial. This is a simple consequence of Cauchy’s integral formula, which shows that if $|f| \leq M$ on the disc then $|a_n| < \frac{M}{2\pi R^{n+1}}$. Since the $a_n$ are assumed to be integers, the inequality implies that $a_n = 0$ for all sufficiently large $n$.
Borel extended this argument to show:
Theorem (Borel): If $f(z)=\sum_{n=0}^\infty a_n z^n$ is the power series expansion around $z=0$ of a meromorphic function on a closed disc of radius $R > 1$ in ${\mathbb C}$, and the coefficients $a_n$ are all integers, then $f(z)$ is a rational function.
The proof is based on the following well-known characterization of rational functions, whose proof we omit (see Lemma 9 in this blog post by Terry Tao):
Lemma (Kronecker): Let ${\mathbf a} = \{ a_n \}_{n \geq 0}$ be a sequence of complex numbers. Then the following are equivalent:
(R1) $f(z) =\sum_{n=0}^\infty a_n z^n$ represents a rational function.
(R2) The Kronecker-Hankel determinant
$K_N({\mathbf a}) = \begin{vmatrix} a_0 & a_1 & a_2 & \dots & a_N \\ a_1 & a_2 & a_3 & \dots & a_{N+1} \\ \hdotsfor{5} \\ a_N & a_{N+1} & a_{N+2} & \dots & a_{2N} \end{vmatrix}$
is zero for $N$ sufficiently large.
The idea behind the proof of the more general result of Borel is to use the above Cauchy estimate (applied to the product of $f(z)$ with some polynomial), together with standard facts about determinants, to show that if $f$ is meromorphic on a closed disc of radius $R > 1$ then $K_N({\mathbf a}) \to 0$ as $N \to \infty$. If the $a_n$ are all integers, this forces $K_N({\mathbf a}) = 0$ for $N$ sufficiently large.
Polya
Around 1916, George Polya realized that the proof of Borel’s theorem via Kronecker-Hankel determinants could be generalized by replacing the radius of convergence with the transfinite diameter of the region of convergence.
The transfinite diameter is a measure of the size of a set which generalizes the radius of a disc. It has many uses in complex analysis and potential theory (as well as in number theory). The diameter of a bounded set $A$ in some metric space $X$ is the maximum distance between two points of $A$, and one can generalize this to the $N^{\rm th}$ diameter $\delta_N(A)$, which by definition is the supremum over all $N$-tuples $(z_1,\ldots,z_N) \in A^N$ of the geometric mean of the pairwise distances between the $z_i$:
$\delta_N(A) = \sup_{z_1,\ldots,z_N \in A} \left( \prod_{i \neq j} |z_i - z_j| \right)^{\frac{1}{n(n-1)}}.$
It turns out that $\{ \delta_N \}_{N \geq 2}$ forms a monotonically decreasing sequence and thus one can define the transfinite diameter
$\delta_\infty(A) := \lim_{N \to \infty} \delta_N(A).$
The transfinite diameter of a disc in any algebraically closed normed field (e.g. ${\mathbb C}$) is its radius, and the transfinite diameter of a real line segment is one-quarter of its length.
It will be convenient for the statement of Polya’s theorem, and for our application to the Lindemann-Weierstrass theorem, to work with $g(z) = \frac{1}{z} f(\frac{1}{z})$ instead of $f(z)$ in Borel’s theorem, and to study the transfinite diameter of the complement of the region of convergence.
Theorem (Polya): If $g(z)=\sum_{n=0}^\infty \frac{a_n}{z^{n+1}}$ is a power series with integer coefficients which can be continued to a meromorphic function on the complement of a bounded set $A \subset {\mathbb C}$ containing $0$ with $\delta_\infty(A) < 1$, then $g(z)$ is a rational function.
The condition $\delta_\infty(A) < 1$ in Polya’s theorem is sharp: the series $g(z) = \sum_{n=0}^\infty \binom{2n}{n} z^n$ has integer coefficients and can be extended to the analytic function $\sqrt{1 - \frac{4}{z}}$ on the complement of the real segment $[0,4]$, which has transfinite diameter equal to 1. However, $\sqrt{1 - \frac{4}{z}}$ is not a rational function.
Dwork
Bernard Dwork noticed around 1960 that Borel’s theorem has a $p$-adic analogue, and this observation is a key ingredient in Dwork’s famous proof of Weil’s conjecture that the zeta function of an algebraic variety over a finite field is a rational function. Dwork realized, in fact, that one could deduce both Borel’s theorem and its $p$-adic analogue from the following global result. (For the statement, we let ${\mathbb C}_v$ denote the completion of an algebraic closure of the $v$-adic completion of ${\mathbb Q}$. For $v = \infty$ this is just ${\mathbb C}$; for $v$ corresponding to a prime number $p$ it is a p-adic analogue of the complex numbers.)
Theorem (Dwork): Suppose $f(z)=\sum_{n=0}^\infty a_n z^n$ is a power series with rational coefficients. Let $S$ be a finite set of places of ${\mathbb Q}$, containing the infinite place, such that:
(D1) For $p \not\in S$, $|a_n|_p\leq 1$ for all $n \geq 0$ (i.e., $a_n$ is a $p$-adic integer).
(D2) For $v \in S$, $f(z)$ extends to a meromorphic function on a disc $D_v$ of radius $R_v$ in ${\mathbb C}_v$ and $\prod_{v \in S} R_v > 1$.
Then $f(z)$ is a rational function.
The proof of Dwork’s theorem in the special case where $f$ is analytic (rather than just meromorphic) in each $D_v$ is not difficult. In this case, for $v \in S$ corresponding to a prime number $p$, the $p$-adic convergence of $f$ on $D_p$ means that $|a_n| R_p^n \to 0$ as $n \to \infty$. This implies that there is a constant $M_p$ such that $|a_n|_p \leq \frac{M_p}{R_p^{n+1}}$ for all $n$. And as above, the Cauchy estimate implies that $|a_n|_\infty \leq \frac{M_\infty}{R_\infty^{n+1}}$ for some constant $M_\infty$. Thus (setting $M = \prod_{v \in S} M_v$ and $R = \prod_{v \in S} R_v$)
$\prod_{v \in S} |a_n|_v \leq \frac{M}{R^{n+1}} \to 0$
as $n \to \infty$. On the other hand, the product formula shows that if $a_n \neq 0$ then
$\prod_{v \in S} |a_n|_v \geq \prod_{{\rm all \;} v} |a_n|_v = 1.$
It follows that $a_n = 0$ for $n$ sufficiently large, and $f$ is a polynomial.
Bertrandias
The transfinite diameter makes sense in any metric space, and in particular we can define it for subsets of the “p-adic complex numbers” ${\mathbb C}_p$. Bertrandias put several of the above ingredients together and proved the following common generalization of the theorems of Borel, Polya, and Dwork around 1963.
Theorem (Bertrandias): Let $g(z)=\sum_{n=0}^\infty \frac{a_n}{z^{n+1}}$ with $a_n \in {\mathbb Q}$ for all $n \geq 0$. Let $S$ be a finite set of places of ${\mathbb Q}$, containing the infinite place, such that:
(B1) For $p \not\in S$, $|a_n|_p \leq 1$ for all $n \geq 0$ (i.e., $a_n$ is a $p$-adic integer).
(B2) For $v \in S$, $g(z)$ extends to a meromorphic function on the complement of a bounded set $K_v \subset {\mathbb C}_v$ (which is assumed to be a finite union of discs if $v$ is non-Archimedean) and $\prod_{v \in S} \delta_\infty(K_v) < 1$.
Then $g(z)$ is a rational function.
The proof is based on Kronecker-Hankel determinants and the product formula, like the proof of Dwork’s theorem above. For simplicity we have assumed that the $a_n$ lie in ${\mathbf Q}$, but the statement and proof of Bertrandias’s theorem generalize easily to any number field $K$. We will only use the special case of the theorem of Bertrandias in which each extension of $g(z)$ is assumed to be analytic.
The proof of assertion (B-R)
We are finally ready to explain Bezivin and Robba’s proof of assertion (B-R), which as we have seen implies the Lindemann-Weierstrass theorem (and hence the transcendence of $\pi$). Perhaps the most interesting aspect of the proof is that it is the p-adic places which will be used to verify the hypotheses of Bertrandias’s theorem.
Let $\omega(x)= v(x) + v'(x)$, which by assumption is a rational function, and let
$\omega(x) = \sum_{i,j} \frac{c_{ij}}{(x - \gamma_i)^j}$
be the partial fraction expansion for $\omega$, where $\gamma_1,\ldots,\gamma_m$ are distinct algebraic numbers. Using the formal inverse $(I + \frac{d}{dx})^{-1} = \sum_{k \geq 0} (-1)^k \frac{d^k}{dx^k}$ for ${\mathcal D}$, one verifies easily that $v$ has the following explicit partial fraction expansion:
(*) $v(x) = \sum_{i,j} c_{ij} \sum_{k \geq 0} \binom{k+j-1}{j-1} \frac{k!}{(x - \gamma_i)^{k+j}}.$
Let $S_1$ be a finite set of places of ${\mathbb Q}$ containing the Archimedean place such that for $p \not\in S_1$, all of the nonzero $c_{i,j}$ and $\gamma_i$ have p-adic absolute value 1, and such that $|\gamma_i - \gamma_j|_p = 1$ for all $i \neq j$. The explicit formula (*) shows that for $p \not\in S_1$ the coefficients $a_n$ of $v(x) = \sum_{n \geq 0} \frac{a_n}{x^{n+1}}$ are $p$-adic integers. Thus $v(x)$ satisfies hypothesis (B1) for any set of places $S$ containing $S_1$.
For $v \in S_1$, formula (*) shows that the series defining $v(x)$ converges outside a disc $K_v \subset {\mathbb C}_v$ of some positive radius $R_v$.
For $p \not\in S_1$, formula (*) shows that the series defining $v(x)$ converges in the complement of a set $K_p \subset {\mathbb C}_p$ which is a union of discs $D_i$ centered at the various $\gamma_i$. Since the series $\sum_{k=0}^\infty k! x^k$ has p-adic radius of convergence equal to $p^{\frac{1}{p-1}}$, we can take the radii of the discs $D_i$ to be $p^{-\frac{1}{p-1}}$. By our assumptions on $S_1$, the discs $D_1,\ldots D_m$ are distinct, and it is a simple exercise using the non-Archimedean triangle inequality to prove that
$\delta_\infty \left( \bigcup_{i=1}^m D_i \right) = p^{-\frac{1}{m(p-1)}}.$
Since the series $\sum_{{\rm primes \;} p} \frac{1}{p \log p}$ diverges, the infinite product $\prod_{p \not\in S_1} \delta_\infty(K_p)$ diverges to zero. Thus there exists a set of places $S$ containing $S_1$ such that
$\prod_{v \in S} \delta_\infty(K_v) < 1.$
For this choice of $S$, $v(x)$ satisfies both (B1) and (B2) and thus $v(x)$ is a rational function. Q.E.D.
Concluding remarks
1. My formulation of (B-R), and the accompanying exposition of the proof that (L-W) and (B-R) are equivalent, differs a bit from Bezivin and Robba’s. They work with power series $u(x) \in {\mathbb C}[[x]]$ and the differential operator ${\mathcal D}'(u) = x^2 u' + (x-1) u$ instead, which amounts to the same thing via the transformation $v(x) = \frac{1}{x} u(\frac{1}{x})$. (I thank Xander Flood for helping me with the details of how to translate smoothly between the two settings.)
2. In their paper, Bezivin and Robba generalize assertion (B-R) to an arbitrary linear differential operator ${\mathcal D}$ with polynomial coefficients for which $\infty$ is a totally irregular singular point. In the special case ${\mathcal D}(v) = v' + v$, the proof is significantly simpler than the general case because one has an explicit inverse operator. In the general case, one needs to use techniques from the theory of $p$-adic differential equations to establish the properties (B1) and (B2).
3. The converses of the theorems of Borel, Polya, Dwork, and Bertrandias are clearly true as well, so these results give a precise characterization of rational functions among formal power series of a certain type.
4. A proof of the theorem of Bertrandias appears in Chapter 5 of Amice’s unfortunately out-of-print book Les Nombres p-adiques.
5. For a deeper understanding of p-adic transfinite diameters, it is very useful to work with Berkovich spaces. See for example my book with Robert Rumely, in which we prove (as in the classical case) that the transfinite diameter of a compact set $K \subset {\mathbf A}^1_{\rm Berk}$ coincides with its capacity, defined in terms of a probability measure of minimum energy supported on $K$.
6. The theorem of Bost and Chambert-Loir mentioned in the introduction is a generalization of the theorem of Bertrandias giving a criterion for a formal meromorphic function on an algebraic curve to be the germ of a rational function. The proof uses Arakelov geometry. Bost and Chambert-Loir view their theorem, and its proof, as an arithmetic counterpart of the following theorem from algebraic geometry:
Theorem (Hartshorne): Let $X$ be a complex projective surface and $H$ an ample effective divisor on $X$. Then any formal meromorphic function along $H$ is the restriction of a rational function on $X$.
For more details and background related to the theorem of Bost and Chambert-Loir, see http://www.math.u-psud.fr/~chambert/publications/pdf/toronto2008.pdf | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 293, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9879916310310364, "perplexity": 158.83278290727222}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424945.18/warc/CC-MAIN-20170725002242-20170725022242-00438.warc.gz"} |
http://www.eoht.info/page/Negative%20entropy | In animate thermodynamics, negative entropy is a mathematical synonym for order, in an entropic sense. The term comes from Austrian physicist Erwin Schrödinger's famous 1944 booklet What is Life?, wherein he tried to explain the second law to a lay audience, stating that negative entropy is the amount of order that an organism "sucks from its environment" as its lives or "avoids decay to thermodynamical equilibrium or of maximum entropy". [1]
Mathematics
The idea of the verbal expression 'negative entropy' being synonymous to 'order' stems from a combination of the following three expressions:
$\frac{1}{X} = X^{-1} \,\!$ Rule for inverse functions $\log(a^b) = b \log(a) \,\!$ Rule for logarithms $S = k \log W \!$ Entropy expression from statistical mechanics.
In short, Schrodinger equates the multiplicity W of the Boltzmann entropy equation with disorder, pure and simple, which he reasons applied to all systems; then equates the inverse of multiplicity with order, as in:
$W^{-1} = Order \,\!$
then carries the negative sign over to the left side of the statistical entropy expression, using the rule for logarithms, to argue that negative S equals order.
Derivation
In his 1944 book What is Life?, Schrödinger reasoned that it is not energy that living beings feed on that keeps them at bay from decay but “negative entropy”. In rephrasing this statement, he says “the essential thing in metabolism is that the organism succeeds in freeing itself from all the entropy it cannot help producing while alive.” In making these ball-park statements, Schrödinger calls on the statistical concept of order and disorder, connections that were revealed, as he says, by the investigations of Boltzmann and Gibbs in statistical physics. On this basis, he situates the following definition:
where k is the Boltzmann constant and D, he says, is a “quantitative measure of the atomistic disorder of the body in question”. Here, to note, Schrodinger fails to mention that this expression is generally valid only for ideal gases. In any event, Schrödinger reasons that this statistical expression applies to living organisms. Moreover, to make is verbal argument mathematical, he states that “if D is a measure of disorder, its reciprocal, 1/D, can be regarded as a direct measure of order.” In addition, “since the logarithm of 1/D is just the minus of the logarithm of D, we can write can write Boltzmann’s equation thus:
or
Hence, as Schrödinger states:
“The awkward expression negative entropy can be replaced by a better one: entropy, taken with the negative sign [ – entropy], is itself a measure of order.”
Thus, he concludes “the device by which an organism maintains itself stationary at a fairly high level of orderliness”, a state he equates with a low level of entropy, consists in “sucking orderliness from its environment”.
Negentropy
In 1953, through the guise of information theory, Schrödinger's negative entropy usage was shortened into the term "negentropy" by French physicist Léon Brillouin. [2]
Difficulties
After his lecture, wherein he discussed negative entropy, Schrodinger famously had to add a note to Chapter 6, where explains that:
“My remarks on negative entropy have met with doubt and opposition from physicist colleagues.”
He goes on to explain that had he been lecturing to them, he would have turned the discussion to free energy, but judged the concept too intricate for the lay audience. In a 1946 review of Schrödinger’s What is Life?, author H.J. Muller stated, supposedly, that biologists have, in the previous decades, commonly defined negative entropy as "potential energy". [3] Muller, it seems, is referring here to Schrodinger's note on free energy. When concept of negative entropy is taken literally and measured in actual organisms, the concept looses its meaning, and a move to free energy discussions prevails. [4] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8068132400512695, "perplexity": 1988.2447456649224}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303385.49/warc/CC-MAIN-20220121131830-20220121161830-00002.warc.gz"} |
https://stats.stackexchange.com/questions/424702/does-order-of-events-matter-in-bayesian-update | # Does order of events matter in Bayesian update?
I'm wondering whether the order of events can lead to different Bayesian update.
For example, consider a coin-tossing problem with unknown $$p$$, the probability of Head. Initially, $$p$$ is known to follow some beta distribution: $$p\sim Beta(a_0,b_0).$$ Suppose that we have a sequence of observations that do not have to be an outcome of coin-tossing. For example, the first observation is "$$\mathbb E[p]>\frac{1}{2}$$" while the second observation is "Head".
If I want to update $$p$$ using Baye's rule, it will be a lot easier if I can process the second event first and then the first event later as Beta is a conjugate prior of binomial experiments.
However, if I have to update $$p$$ in the order of the events (first observation first, and then the second one later), the process requires a bit more of computation.
So, my question is that does the order of events matter in Bayesian updating? If not, what can be a theoretical background that justifies it?
• How can you observe that "$\mathbb E[p]>\frac{1}{2}$"? Nov 28, 2019 at 1:37
AFAIK, you cannot say that $$p > \frac{1}{2}$$ or even $$\mathbb E[p]>\frac{1}{2}$$ is an "observation" or "event", but rather a constraint on your model parameter(s). The term "observation" is usually reserved for specific realizations of a random variable (i.e. draws from a distribution). In your model, one could plausibly observe $$p$$ (numbers between 0 and 1) or the outcomes of $$Bernoulli(p)$$ (either 0 or 1). There is no way to observe $$\mathbb E[p]>\frac{1}{2}$$, this information lies outside your probabilistic model.
As a rough rule (some caveats apply) you should be able to simulate observations from your probabilistic model, given model parameters. How would you simulate a model where $$\mathbb E[p]>\frac{1}{2}$$ is a possible observation?
If you start with the $$p\sim Beta(a_0,b_0); a_0, b_0 \in \mathbb{R}^+$$ model and then learn that $$\mathbb E[p]>\frac{1}{2}$$, it means your initial model was incorrect and you should change your model to reflect the constraint ($$\mathbb E[p]>\frac{1}{2}$$ implies $$\frac{a_0}{a_0 + b_0} > \frac{1}{2}$$, so some combinations of $$a_0$$ and $$b_0$$ are ruled out). Adding a constraint cannot be AFAIK directly handled in the language of Bayesian updating.
Hope that helps.
• Thank you for your comment, @Martin Modrak. I got your point. I modified the question. What happens if the first obvservation is $\mathbb E[p]>\frac{1}{2}?$ Within the model with the Beta conjugate, there are infinitely many possibilities that can draw $\mathbb E[p]>\frac{1}{2}$. In that case, are the two events interchangable? Sep 4, 2019 at 2:55
• @Andeanlll I don't think $\mathbb E[p]>\frac{1}{2}$ can be treated as an observation either. I tried to expand my answer on that. Sep 4, 2019 at 5:19
In Bayesian inference, terms like "observation" and "event" are just conveniences; there is no fundamental importance to them, so don't get hung up on them.
In particular, there is no physical causality or time's arrow -- no "events". Whether you can carry out some calculations in more than one order depends solely on the form of the model. If, algebraically, the results are the same assuming different orders of assignments to some variables (i.e., "observations"), then, terrific, you can do whatever is convenient. If not, well, so what?
About the representation of p > 1/2, you could represent that as a likelihood function which is just a step at 1/2. That is, it is zero to the left of 1/2, and any positive constant to the right. Note that ordinary "observations" yield likelihood functions which vary smoothly, but the smoothness is not a requirement.
In order for this to be the case, the random variables must be exchangeable.
Your example is a little different since $$p>1/2$$ isn't an event. An event should be in the support of the likelihood. In this case, events are only constituted by binomial random variables, or sums thereof. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.909518301486969, "perplexity": 378.88779250514403}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663012542.85/warc/CC-MAIN-20220528031224-20220528061224-00149.warc.gz"} |
https://www.physicsforums.com/threads/shm-energies.211961/ | # SHM energies
1. Jan 30, 2008
in SHM
average K.E when done w.r.t to time is equal to average potential energy calculated w.r.t to time.
but today in class when my sir asked me to to prove average K.e = average P.E
I just tried integrating it w.r.t to displacement and then divided it with A ,
I thought this gives us av . energies in 1/4 vibration and as all the four parts are identical
average will remain same , but i found that av . K.E=2*av. P.E
I then done this again w.r.t to time and got the answer
but still I am confused , about how can that come i.e K=2*P.E
Can you offer guidance or do you also need help?
Draft saved Draft deleted
Similar Discussions: SHM energies | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9540147185325623, "perplexity": 2835.106504049171}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189474.87/warc/CC-MAIN-20170322212949-00112-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/conditional-probability.626763/ | # Conditional Probability
1. Aug 9, 2012
1. I am going over some past Probability exam papers and cannot solve this question. Any help or advice would be much appreciated!
2. David eats cereal for lunch 60% of days. If he had ice cream for breakfast, then the probability that he eats cereal for lunch is only 0.25. If he didn't eat ice cream for breakfast then he would not eat cereal for lunch with probability 0.3.
(a) Are the events that David ate cereal for lunch and ice cream for breakfast independent?
(b) Show that P(David has ice cream for breakfast)=2/9
3. I (think I) have been able to work out part (a): Let the event cereal for lunch be A and ice cream for breakfast be B. The events A and B are independant if P(A$\cap$B)= P(a)* P(B). So P(A given B)*P(B)= P(A)* P(B) which gives us P(A given B)= P(A) and then when we sub in the values given in the question we get 0.25= 0.6 which is not true and thus proves they are not independent. Is this correct?
For part b I have been using the partition theorem to try to show that P(B)=2/9. So P(B)= P(B given A)*P(A)+ P(B given Ac)*P(ac) which gives me P(B)*(0.75)=. And this is as far as I can get because I cannot work out how to find P(B$\cap$Ac)?? More than likely I have gone completely wrong from the very beginning. Any help would be very much appreciated please!
Thank you:)
2. Aug 9, 2012
### Ray Vickson
I prefer to use a notation where the meaning of the symbols is apparent at once, so let I = {ice cream for breakfast} and C = {cereal for lunch}. You are given P(C) = 0.6 = 3/5, P(C|I) = 0.25 = 1/4 and P(Cc|Ic) = 0.3 = 3/10. Thus, P(C|Ic) = 7/10. Since P(C|I) and P(C|Ic) are different, C and I are dependent, as you said.
P(C) = P(C & I) + P(C & Ic) = P(C|I)P(I) + P(C|Ic)P(Ic), and P(Ic) = 1-P(I). You can solve for P(I).
RGV
3. Aug 9, 2012
Thank you! You have been very helpful
Similar Discussions: Conditional Probability | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8259921073913574, "perplexity": 1280.9442413143338}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948512584.10/warc/CC-MAIN-20171211071340-20171211091340-00453.warc.gz"} |
https://brilliant.org/discussions/thread/how-is-that-possible/ | ×
# How is that possible??
This discussion has been deleted!
Note by Atul Shivam
1 year, 4 months ago
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I guess that you cannot let $$S=1+2+4+8+16+\ldots \infty$$ because the sum of the geometric progression does not converge (approaches infinity). Also, you cannot subtract equation $$(2)$$ with $$(1)$$ because both $$S$$ and $$2S$$ do not exist. · 1 year, 4 months ago
I think you mean to say ,we can't find sum of geometric progression which converges to infinity!!! Is it so ??? · 1 year, 4 months ago
Yes sorry for not being clear. · 1 year, 4 months ago
Got a reason. If S is infinity. Then 2*infinity is also infinity. This means there will be nothing like 2S for infinity.Also you can't apply operations like addition or subtraction etc to infinity. If you will say that assuming n terms where n tends to infinity and apply operations. Then the result will be something which is likely to be. · 1 year, 4 months ago
Before writing the statement "S is equal to.." you must prove the convergence of the series. If the series converges, then only it has a limiting value, which you may call S. If you assume that S exists before proving its existence, then anomalous results as the above may pop up. Here the series does not converge (rather it diverges) simply because its the sequence of its partial sums is not bounded. · 1 year, 2 months ago
It looks like to be on the same pattern as we derive formula for infinite sum of G.P. But there was |r| < 1. We can't use this pattern in such case where |x| > 1. Because the terms are increasing at a large rate and uncertainty of one last infinitely large term may result in false value. · 1 year, 4 months ago
I don't think so as we can add or subtract anything to infinity and it will not affect anything · 1 year, 4 months ago
For |x|> 1 G.P. diverges. So, sum can't be found. · 1 year, 4 months ago
What do you want to say, Can you please elaborate · 1 year, 4 months ago
Well, i just want to say that infinity doesn't follow elementary operations. I need some time to give a proper explanation. I am very sorry that i couldn't explain it. · 1 year, 4 months ago
Oh!!! Is it so · 1 year, 4 months ago
Uncertainty of one term may result in this false value. Because if you get to same approach by using sigma instead of value. You will get the desired result. Explain me if you know the exact reason. · 1 year, 4 months ago
Because infinite does not follow general Algebraic rule of operations(+,-,x,÷). · 1 year, 4 months ago
Hey are you still in ISM dhanbad ??? · 1 year, 4 months ago
Let us take an another geometric progression $$T_n= 1,\frac {1}{2},\frac {1}{4},\frac {1}{8},\frac {1}{16}......$$ and let $$S$$ be it's sum
I.e $$S= (1+\frac {1}{2}+\frac {1}{4}+\frac {1}{8}+\frac {1}{16}......)----(1)$$ $$(2S =2+1+\frac {1}{2}+\frac {1}{4}+\frac {1}{8}+\frac {1}{16}......)----(2)$$ Now $$2S-2= (1+\frac {1}{2}+\frac {1}{4}+\frac {1}{8}+\frac {1}{16}......)$$
$$2S=4$$ and hence $$S= 2$$ which is absolutely correct as $$S$$ has actual value of $$+2$$. Why it seems to be correct and not the above one
sorry for my poor English I am not very much fluent $$:-)$$ · 1 year, 4 months ago
That's what I said earlier your procedure is correct when S is finite but same thing is not applicable when S is infinite because the value of $$\displaystyle \infty - \infty$$ cannot be determined. I am in 1st year of college(my age is 18). · 1 year, 4 months ago
I was just asking because I am in the same city:) · 1 year, 4 months ago | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.965385913848877, "perplexity": 722.8639018945985}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171670.52/warc/CC-MAIN-20170219104611-00619-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://istopdeath.com/find-the-volume-pyramid-567/ | # Find the Volume pyramid (5)(6)(7)
The volume of a pyramid is equal to .
Substitute the values of the length , the width , and the height into the formula to find the volume of the pyramid.
Combine and .
Cancel the common factor of .
Factor out of .
Cancel the common factor.
Rewrite the expression.
Multiply.
Multiply by .
Multiply by .
Find the Volume pyramid (5)(6)(7) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8725098967552185, "perplexity": 1869.0144207539165}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00252.warc.gz"} |
https://dsp.stackexchange.com/questions/14862/measure-the-snr-of-a-signal-in-the-frequency-domain | # Measure the SNR of a signal in the frequency domain?
How do you measure the SNR of a sinewave in the frequency domain? (Assuming no filtering.)
Suppose: $$x(n) = s(n) + n(n)$$
Create the signal (nCycles of a sinewave) in MATLAB...:
sampleRate = 1024;
f0 = sampleRate/8;
nCycles = 11;
time = 0:1/sampleRate:nCycles/f0;
signal = sin(2*pi*f0*time);
Create the noise and scale it by the desired SNR...
SNR = 10;
noise = randn(size(signal));
% scale the noise to obtain the desired SNR
noise = noise / norm(noise) * norm(signal) / 10^(SNR/20);
x = signal + noise;
Calculate the resulting SNR (in the time domain):
actualSNR = 20*log10(norm(signal)/norm(x - signal));
disp(['SNR = ',num2str(actualSNR),' dB'])
Plot the noisy signal in time and frequency
subplot(2,1,1)
% plot the signal in time
plot(time,signal)
hold on; grid on
plot(time,x,'r')
% plot the noisy signal in frequency
NFFT = 4096;
X = fftshift(fft(x,NFFT));
X = X/max(abs(X));
f = sampleRate/2*linspace(-1,1,NFFT);
subplot(2,1,2)
plot(f,20*log10(abs(X)))
grid on;
• Specifying a SNR without mention of a bandwidth has little meaning. I.e., for a pure tone, the narrower the bandwidth, the higher the SNR. That's one of the main reasons to use a filter - increase the SNR. – rickhg12hs Mar 8 '14 at 20:45
• I disagree with you. Since I didn't mention any filters, it's quite clear what the bandwidth of interest is. – Seth Mar 10 '14 at 15:43
• Well, maybe in your signal world. Why someone wouldn't use a simple filter to increase the SNR is at least questionable. Nice OP edit. – rickhg12hs Mar 11 '14 at 1:50
It can be a little tricky. If your sine wave happens to fall at an FFT bin center, things are a little easier. If your sine wave happens to not fall at a bin center, you have to consider spectral leakage. You'll also need to consider how your window function affects your signal.
A simple technique to estimate the signal power would be to sum the squared magnitudes of the FFT bins with most of the signal's energy (i.e. the largest bins). To get the noise energy, sum the squared magnitude of all of the other bins. This will get you close.
If you know the SNR per sample in the time domain (SNRt) the SNR in the frequency domain (SNRf) will be
$$SNR_f = SNR_t \sqrt{N}$$
where N is the number of non zero elements in the basis vector in the Fourier transform. The fastest Fourier transforms (FFT with $n=2^{integer}$) has most zeros and thus lowest SNRf. The highest SNRf comes with $N$ as a prime number with no zeros in the basis vectors, which is also thus the most arithmetically intense choice of sample size. Quality costs. Also pay attention to the fact that the number of zeros in the basis vectors in the FFT depends on the frequency leading to SNRf being frequency dependent. Lowest SNRf is at the middle frequency where every second element of the cosine part is zero.
Now tell me why people still use sample sets of 2^integer when it is the choice of lowest quality? Today we have the means to do the extra computations of a zero free Fourier transform. It would led to better image quality, better sound and higher data rates.
• How would you know the SNR per sample in the time domain? Can you elaborate on "The highest SNRf comes with N as a prime number with no zeros in the basis vectors"? – Seth Nov 10 '14 at 4:46
• SNR in time domain is either given, measured, assumed or calculated. I was specifically thinking about the cosine transformation or the even FFT when talking about zero free basis. Check for example the basis vector link which uses samle size 2^4=16 and having half of the components zero. Change that to 17 and get no zeros link – David Jonsson Nov 12 '14 at 12:51
• The intent of the question was to determine the SNR given no information about the time domain. Doesn't the FFT become the DFT if N is not a power of 2? – Seth Nov 12 '14 at 22:06
• Sure, the inverse is so similar, just a change of sign in the exponent, that the relation holds. So for the inverse Fourier transform, with signal to noise ratio in a specific frequency SNRif, the SNR in the resulting time domain would be SNRit = √N SNRif – David Jonsson Nov 23 '14 at 21:03
• Addition. With a properly scaled transformation there is no change in variance or standard deviation in a Fourier transform, but there is an increase in the expected value for the DC signal. It grows as $$µ_f = \sqrt N µ_t$$ this is not dependent on the number of zeros in the basis vectors. – David Jonsson Dec 18 '14 at 16:09
Here is a better answer (than the one I gave before) based on properly scaled Fourier transformations where basis vectors are normalized to length 1.
Noise, a normal distribution of a random variable, has the following variance, based on https://en.wikipedia.org/wiki/Variance#Basic_properties for variance of a linear combination with no correlation, applies to both the Fourier transformation and it's inverse
$${Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X_{i})$$
The same variance for all Xi means $${Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X) = \left(\sum _{{i=1}}^{{N}}a_{i}^{2}\right)\operatorname {Var}(X)$$
Example 1
The first row DC basis vector in a even Fourier transform
$$\left(\frac{1}{\sqrt N } , \frac{1}{\sqrt N } , \frac{1}{\sqrt N } , ... \right)$$ gives a variance of
$$Var(Transform) = N \left( \frac{1}{\sqrt N } \right)^2 Var(X) = Var(X)$$
Example 2
A basis vector in an even fast Fourier transform with a wavelength of 4 samples, a repeating series of
$$\left( \sqrt{\frac{2}{N}} , 0 , -\sqrt{\frac{2}{N}} , 0 , ... \right)$$
This is the basis vector with most zero elements, every second. The variance becomes
$$Var(Transform) = \frac{N}{2} \left( \sqrt{\frac{2}{N}} \right)^2 Var(X) = Var(X)$$
So variance, and thus standard deviation, is independent of basis vectors and zero elements. But what happens to the signal, the expected value? Maybe someone else can show that?
Another solution can be to model the noise using the non-sine frequencies. This relies on having a reasonable parametric model whose parameters can be estimated from the non-sine frequencies. E.g. if you know the noise is white or pink, this is fairly straightforward. Once you've got the parameters estimated, it's also easy to estimate how much noise there is at the sine frequency, and sum up all noise contributions. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8891549706459045, "perplexity": 770.1228017392398}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250614086.44/warc/CC-MAIN-20200123221108-20200124010108-00087.warc.gz"} |
http://cvgmt.sns.it/paper/2911/ | # On the isoperimetric properties of planar N-clusters
created by caroccia on 22 Jan 2016
[BibTeX]
Phd Thesis
Inserted: 22 jan 2016
Last Updated: 22 jan 2016
Year: 2015
Notes:
This Thesis aims to highlight some isoperimetric questions involving the, so-called, $N$-clusters. We first briefly recall the theoretical framework we are adopting. This is done in Chapter one. In chapter two we focus on the standard isoperimetric problem for planar N-cluster for large values of $N$ and we provide an equidistribution energy-type results under some suitable assumption. The third Chapter is devoted to a stability results of the hexagonal honeycomb tiling. Finally in the fourth Chapter we consider a generalization of the Cheeger constant, defined as a minimization of a suitable energy among the class of the $N$-clusters. We show how this problem is related to the optimal partition problem for the first Dirichlet eigenvalue of the Laplacian introduced by Caffarelli and Fang-Hua Lin in 2007. We conclude, in Chapter five, with some remarks and some possible future direction of investigation. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9040785431861877, "perplexity": 505.86881344745393}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812880.33/warc/CC-MAIN-20180220050606-20180220070606-00091.warc.gz"} |
http://clay6.com/qa/24338/cracking-of-rubber-is-due-to | Browse Questions
Cracking of rubber is due to
$\begin{array}{1 1}(a)\;Smog\\(b)\;Acid\;rain\\(c)\;Green\;house\;effect\\(d)\;High\;temperature\end{array}$
Cracking of rubber is due to smog
Hence (a) is the correct answer. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9912134408950806, "perplexity": 4122.938193670156}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280872.69/warc/CC-MAIN-20170116095120-00453-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://fusionenergy.lanl.gov/Documents/MTF/Why_MTF/Why-MTF-Comments.html | Why Magnetized Target Fusion Offers
A Low-Cost Development Path
for Fusion Energy
Richard E. Siemon
Irvin R. Lindemuth
Kurt F. Schoenberg
Los Alamos National Laboratory
Los Alamos, New Mexico
Submitted to Comments on Plasma Physics and Controlled Fusion, November 12, 1997
I. Introduction
Reasonably priced energy supplies have become an expectation of the developed world and a necessary ingredient for development of Third World countries. The problem of providing large supplies of low-cost energy is a long-term, complex one that requires sustained R&D efforts, in spite of the shadow cast on long-term R&D by the federal deficit problem. The role of fusion energy as a power source was thoroughly reviewed and strongly endorsed in 1995 by the President’s Committee of Advisors on Science and Technology Fusion Review Panel chaired by John Holdren. He argued [Holdren 95]:
The options available for meeting the world’s demand for energy in 2050 and beyond are those already in use – fossil fuels, biomass energy, nuclear fission, hydropower, geothermal energy, wind energy, and solar energy – plus, potentially, nuclear fusion.
In these circumstances, it should be obvious that there is great merit in the pursuit of diversity in energy options for the next century. There are not so many possibilities altogether. The greater the number of these that can be brought to the point of commercialization, the greater will be the chance that overall energy needs can be met without encountering excessive costs from or unmanageable burdens upon any one source.
In the past decade the critical issue for fusion has shifted from one of scientific feasibility to one of commercial viability. The specific problem is that all fusion technologies currently being pursued involve extremely costly facilities for the required steps of further development. In the present international fiscal environment, it is imperative to find a more cost effective development path for fusion energy.
The conventional regime of Magnetic Fusion Energy (MFE), with plasma density n ~ 1014 cm-3 and magnetic field provided by superconducting magnets, has been relatively well explored [Sheffield 96]. Tokamaks are the major devices studied in MFE, and tokamak research has tremendously advanced our understanding of plasma physics. The International Tokamak Experimental Reactor (ITER) design illustrates the technology and cost for an ignited plasma demonstration in the MFE regime. The estimated \$10-billion price for ITER calls into question whether fusion can ever be developed based on tokamak-like technology. Factors of a few, or maybe ten at most, in any parameter such as size, neutron wall loading, and so forth are about all that one can credibly seek in optimizing a tokamak system. Certainly research seeking to reduce the ITER-like system size by factors of a few is extremely important and needs to be pursued. But we strongly suspect that the necessary breakthrough, which would allow fusion to be developed in a more timely and affordable manner, will involve a qualitatively different and significant departure from the MFE tokamak regime and technology.
Another approach to fusion, Inertial Confinement Fusion (ICF), represents a good alternative to MFE in that the regime of density and pressure is completely different, the physics issues are quite distinct, and the technology required has fairly little in common with a tokamak-like system [Lindl 95]. Thus, the issues that are likely to emerge as limitations for one approach are unlikely to apply to the other. Unfortunately, the cost of developing ICF is also high. The price of the National Ignition Facility (NIF), which will demonstrate ICF ignition, is over \$1 billion. The anticipated cost of developing efficient inertial fusion drivers such as heavy ion beams is also high [Bangerter 97]. For the development of fusion energy, something less expensive would obviously be desirable.
A Lower Cost Alternative—Magnetized Target Fusion
To find a lower cost approach, we start by noting that the cost of development is directly linked to the system size, which in the case of MFE is mostly dictated by the maximum magnetic field strength obtainable with superconducting magnets. The critical constraint with ICF is the costly high-power drivers needed to achieve the extreme conditions of density and pressure.
We also note that countless examples can be found in the magnetic fusion literature showing that fusion reactions can be created in smaller-sized systems if one admits larger magnetic field, higher plasma density, and pulsed operation as with imploding liners [Sherwood 81, Lindemuth 83, Robson 76, Vekshtein 90, Ryutov 96, Gross 76]. In this paper we will review the basic reason for that tendency, and examine some of the consequences. We will conclude that the most interesting regime of density is n ~1020 cm-3, which is high compared with MFE, but low compared with ICF. This density regime at 10 keV temperature corresponds to megabars of pressure (millions of atmospheres), which is intrinsically pulsed in nature.
We define the intermediate density regime to be Magnetized Target Fusion (MTF). The name is chosen based on two general characteristics that we assume for MTF: 1) as with ICF, PdV work heats the fuel by compressing it inside an imploding wall, or "pusher" in the parlance of ICF, and 2) magnetic field is embedded in the fuel to insulate it from the pusher.
Although numerous variations in approach can be envisioned, we have in mind the magnetically-driven imploding liner method for MTF. In the liner approach:
• fuel with an embedded magnetic field would be preheated and positioned inside a volume of centimeter dimensions, which is surrounded by a thin metal shell (or liner) that will act as the pusher,
• a current introduced on the outer surface of the liner would cause it to implode by self-pinching magnetic forces at a velocity of approximately 106 cm/sec,
• the liner would be made thick enough that the pinching current does not vaporize it, and therefore the liner would be a flux-conserving metal shell during the implosion,
• at peak compression a significant fraction of the liner kinetic energy would be converted to thermal energy of the fuel, and
• the dwell time of the liner at peak compression and the final fuel density and temperature would be designed to give significant fusion energy generation.
The liner velocity required is termed hypervelocity because the kinetic energy density exceeds the heat of vaporization for liner materials. The technology for precision implosions creating millions of atmospheres of pressure is a challenge in its own right. In the 1970s when a number of MTF-related efforts were underway, most of the effort was directed towards developing this demanding technology, and very few integrated tests with a preheated plasma were ever done. In what must be viewed as a serendipitous coincidence, the Department of Energy's Office of Defense Programs (DP) in the last decade has significantly advanced the technology of imploding liners with the same parameters of implosion velocity and kinetic energy as those needed for investigating fusion reactions in the MTF regime. The purpose of the Defense Program work is to study and understand hydrodynamics in the megabar pressure regime and has no connection with nuclear fusion. However, the existence of DP expertise and facilities offers an important near term advantage for resuming MTF research.
The magnetic field to insulate fuel from its surroundings is the essential ingredient of MTF. In fact, the benefit of a magnetic field in a fusion target was recognized in the 40’s by Fermi at Los Alamos and at approximately the same time by Sakharov in the former Soviet Union. We will derive below the advantages in terms of reduced energy and power that must be delivered to the fusion fuel. The advantages of MTF can also be expressed in terms of requirements on driver technology. By preheating MTF fuel to between 100 and 500 eV, the volume compression needed to reach 10 keV temperature is 100-1000. The volume compression ratios for ICF are typically 30,000 to 60,000, which requires a much more precise implosion system. The characteristic implosion velocity for MTF is 0.3-3.0 cm per microsecond, which is 10 to 100 times smaller than for ICF. The peak pressure for MTF is 1-10 megabars, and for ICF, 100s of gigabars. These impressive differences justify careful examination of ways to introduce a magnetic field.
II. The Technical Case for Magnetized Target Fusion
A. Lawson Condition for Pulse Duration and Energy Confinement Time
In a pulsed system, as opposed to steady-state, the pulse duration, t burn, is an important new variable. The pulse duration determines the amount of fuel that reacts or "burns," given the reaction cross section, leading to an nt burn requirement in a similar way that nt E is determined from power balance in a steady-state system. For deuterium (DT) fuel the thermonuclear reaction rate per unit volume is
R = nDnT <s DT v> = 1/4 n2 <s DT v> (1)
where nD=nT is the deuterium and tritium density, n =nD+nT is the total ion or electron density, and <s DT v> is the averaged product of cross section and relative velocity for a Maxwellian velocity distribution. At 10 keV, <s DT v> @ 10-16 cm3/sec. The total density decreases at the rate 2R as the fuel is consumed, and the frequency of fusion reactions per ion for either deuterium or tritium ions is given by 2R/n:
(dnD/dt)/nD = (dn/dt)/n = ½ n <s DT v> (2)
Assuming for simplicity that DT fuel is held at constant temperature so that <s DT v> is constant in time while it burns, Eqn. 2 can be integrated to give
n/n0 = 1/(1+ n0t burn <s DT v>/2) (3)
where t burn is the burn time. Equation 3 can be recast in terms of f, the fractional burnup of fuel, as:
f /(1-f) = n0t burn <s DT v>/2 (4)
where f º 1-n/n0. For complete burnup, the gain would be Gmax= 300 at 10 keV. This is simply the ratio of energy for a 14.1 MeV neutron and 3.5 MeV alpha divided by the
60 keV of thermal energy for a DT ion pair with electrons.
Figure 1. Fusion energy output relative to plasma energy vs. the product of density and burn time.
As a function of burn time, the gain plotted in Fig. 1 is Gmax times the fractional burnup. We can define a Lawson condition using Fig. 1. With nt burn ~ 3x1014 cm-3 sec the gain relative to thermal energy is around five, enough to allow for net gain with realistic efficiencies. The net gain relative to initially stored electrical energy is the gain of Fig. 1 times the efficiency of heating fuel to 10 keV temperature. For example, if 50% of the stored electrical energy is converted to liner kinetic energy [Gerwin 78], and 50% of the liner kinetic energy is converted to thermal plasma energy at peak compression, then the net gain would be 1/4 of the gain plotted in Fig. 1.
A plasma heated to 10 keV will cool by numerous mechanisms. The total power losses per unit volume are conventionally written as 3nT/t E, where t E is the global energy confinement time. In deriving Fig. 1 we ignored losses, which is equivalent to assuming
t E>>t burn. To obtain the minimum possible system size for the purpose of low–cost development, we would require t E ~ t burn . That is, if t E were much less than t burn the fuel would cool before it burned. On the other hand if t E were much larger than t burn , the plasma should be made smaller to equalize the two, which requires less energy, assuming the energy confinement time increases with system size. For approximate estimates, the relevant energy confinement time and the burn time should both satisfy a Lawson-like nt , which we will take for the purposes of demonstrating feasibility to be the same as ITER, and approximately an energy breakeven condition according to Figure 1:
Lawson requirement: nt ~ nt E ~ nt burn ~ 3x1014 cm-3 sec
This nt E corresponds to 1.5% burnup fraction in a pulsed system.
B. Pressure of High-Density Fuel Dictates Pulsed Technology
The first requirement for containing fuel is equilibrium or pressure balance to prevent the fuel from expanding during the required burn time. There are a continuum of possibilities ranging from ICF with zero magnetic field where pressure is supported by the inertia of surrounding low-temperature fuel, to full magnetic confinement where plasma pressure is less than or equal to the confining magnetic pressure. In the MTF regime we consider the possibility where plasma pressure is larger than or equal to the magnetic field pressure, because the main role of magnetic field is insulation and not confinement.
Broadly speaking, the relevant technology changes as the density increases. We assume Ti ~ Te ~ 10 keV. At densities from 1014 cm-3 up to about 1016 cm-3 plasma pressure can be contained by superconducting magnets, where the higher density corresponds to magnetic confinement with b = 1. Plasma b º 2nkT/(B2/8p ), where B is the magnetic field. At pressure or density too high for superconductors, pulsed magnets can be used up to pressures that fracture known materials. Strength limitations set an upper limit on the density at about 1018 cm-3. This density corresponds to magnetic field of about 1 MG if magnetic pressure confines the plasma. To date, the largest magnetic fields reported are pulsed fields of about 20 MG, which can be obtained by imploding liners [Pavlovskii 96]. If 20 MG were used for plasma confinement, the corresponding maximum density is around 1021 cm-3. Above that density, plasma pressure must be held by the inertia of material walls, although magnetic field can be utilized for its insulating properties. For ICF the density of the ignited hot spot is expected to be about 1025 cm-3, which corresponds to a pressure of 200 Gbar. We see that the technology for fusion changes radically as one moves from MFE density to ICF density.
C. Fusion Fuel Diffuses Before Burning
Another basic point useful to recall for the following discussion is that s DT, the cross section for fusion, is much smaller than s C, the cross section for Coulomb scattering, almost independent of density. By definition the frequency of collisions is given by the product of cross section and flux. The rate of fusion reactions is given by the right–hand side of Eqn. 3:
Frequency of fusion reactions = ½ n <s DT v> (5)
The effective fusion cross section can be taken as <s DT v>/vi, ~ 1 barn (10-24 cm2) at
10 keV where vi is the ion thermal speed. Similarly the Coulomb collision frequency can be written as a product of the Coulomb cross section and particle flux, n multiplied by vi:
Ion-ion Coulomb collision frequency = n ii = 1/ t ii = n vis C (6)
Thus at 10 keV and 1014 cm-3 s C ~ 7000 barns. This Coulomb collision frequency, or reciprocal of the ion-ion collision time, is extensively discussed in the standard textbooks. Because of the accumulated effects of small-angle scattering, the frequency of Coulomb collisions is proportional to lnL , a factor that depends weakly upon temperature and density. The Coulomb logarithm is often taken as a constant about equal to 20, but even for rough estimates we will calculate lnL when it arises, because the range of density we will consider (1014–1026 cm-3) corresponds to lnL changing by more than a factor of 3.
At a temperature of 10 keV, the cross section or frequency for Coulomb scattering is larger than the cross section or frequency of fusion reactions by a factor of 2000-6000 for density between 1026 cm-3 and 1014 cm-3 respectively. Therefore, the number of collisions (N) that occur during a burn time is calculated to be:
N = t burn / t ii = 2 f vis C / <s DT v> (7)
For nt E = 3x1014 cm-3 sec, the burn time is between 60 and 180 ion-ion collision times as density varies from 1026 cm-3 to 1014 cm-3.
In summary, we conclude that, independent of the fuel density over a wide range of density, collisional diffusive processes are unavoidable when fusion fuel is assembled for a time long enough to produce energy gain.
D. The Nature of Energy Diffusion
Even if fuel is held in pressure balance for the necessary burn time, it has been historically difficult to achieve the required global energy confinement time. Much of MFE fusion research has been devoted to understanding the many modes of plasma motion that transport energy in addition to classical collisional processes. With ICF, there is less uncertainty about loss processes, because the absence of a magnetic field simplifies the transport physics. In that case electron thermal conduction is the dominant loss process. In the ICF approach parameters are chosen so that even electron thermal conduction is consistent with the Lawson condition. One could say that ICF is the "worst case" for thermal losses when compared with any type of magnetic configuration.
Classical diffusion. We review now the lower bound on energy confinement represented by classical diffusion. In MFE fusion literature, the global energy confinement time is usually expressed in terms of thermal diffusivity:
t E ~ a2/c , (8)
where a is the characteristic dimension across which heat diffuses and c is the thermal diffusivity. The value of c (same as thermal conductivity divided by density) is derived by calculating the energy flux in the presence of a temperature gradient.
Thermal diffusion can also be viewed as a random walk of particles. After each collision, a particle moves one step at random either up or down the temperature gradient. Heat conduction is the diffusion of cold particles up the gradient and hot particles down the gradient with no net flux of particles. The essential feature of the random walk is that after N collisions, there is a binomial distribution for particle location, and it has a width proportional to N1/2. If the step size is l , then the standard deviation of the distribution of particle locations after N collisions is a given by
a = N1/2 l . (9)
If the collision time is t , the number of collisions is N = t / t , so we can also write Eqn. 9 as
t = (a/l )2 t (10)
Eqn. 9 indicates that if N collisions are needed before heat dissipates, then the fuel must have a characteristic size greater than a. Equivalently, Eqn. 10 gives the time to dissipate heat (energy confinement time) in terms of the number of steps across the characteristic size, (a/l ), and the time per step or collision time.
Classical diffusion without a magnetic field. To apply the random-walk argument to electron thermal conduction, we equate the step size to a plasma mean free path l . Electrons have a larger thermal speed and a shorter collision time, such that the mean free path l is the same for either ions or electrons:
l = 1 / ns C = vi t ii = ve t ee (11)
where vi,e is the ion, or electron, thermal speed. Electrons collide more frequently by a factor of (mi/me)1/2 , or about 60 for a DT mixture. Therefore, if we consider high density where ions make about 60 collisions, then electrons make about 3600 collisions during the fusion burn time. The size of a plasma with burn time long enough to allow 3600 electron-electron collisions is
a = (3600)1/2 l . (12)
For ICF, where the ignition hot spot density is about 1025 cm-3, the mean free path is
0.7 microns; this simple estimate of Eqn. 12 for hot spot radius is 42 microns. More detailed calculations [Lindl 95] give about the same value.
Classical diffusion with a magnetic field. To apply the random-walk argument to magnetized plasma is more difficult, because the step size depends upon complicated particle orbits in the magnetic field. However, for poloidal-field dominated configurations like the Reversed Field Pinch, the spheromak, and the Field-Reversed Configuration (FRC), and for tokamaks, detailed studies give the simple prescription that the step size can be taken as the ion gyro radius calculated in the poloidal magnetic field [Boozer 83]. (In a torus the toroidal direction is the long way around the torus, and the poloidal direction is the short way around.) In the direction perpendicular to a magnetic field, the classical ion heat conduction dominates because the ions have a larger gyro radius. Therefore, we can estimate that the minimum required size of a fusion system to diffuse heat slowly enough to meet Lawson, say 180 ion-ion collision times, is
a = (180)1/2 ri, (13)
where ri is the ion gyro radius in the poloidal magnetic field. The tokamak banana-regime formulas for neoclassical transport theory give about 20 ri instead of the approximate estimate of 13 ri given by Eqn. 13. Because of anomalous transport, the design radius of ITER is about 5 times larger than the neoclassical limit (i.e. aiter @ 100 ri).
E. Characteristic Step Sizes Decrease as Density Increases
Comparing Eqns. 8 and 10, we see that c has the form of a step size squared times a collision frequency. For classical transport,
Electron thermal conduction: c e ~ l 2n ee. (14)
Ion cross field transport: c i ~ ri 2n ii. (15)
The mean free path (l ), which depends on temperature and density, is plotted in Fig. 2 for 10 keV temperature. The gyro radius (ri), which depends mainly on density, is also plotted in Fig. 2, assuming constant poloidal beta (b i), where b i is the ratio of ion pressure to poloidal field pressure (b i = 8p nkTi/Bp2). The density dependence can be seen by writing the gyro radius as:
ri = vi/w ci = (c/w pi)b i1/2 (16)
where w ci is the ion cyclotron frequency in the poloidal magnetic field, c is the speed of light, and w pi is the ion plasma frequency,
w pi = (4p ne2/mi)1/2 (cgs units). (17)
Poloidal beta in tokamaks and the above mentioned configurations is observed not to differ much from unity.
In the spirit of a survey of minimum system size for fusion, Fig. 2 gives useful guidance. The dimensions of a system without magnetic insulation become unacceptably large at low density. The classical limit for the size of a magnetized plasma is seen to be quite small as density increases. If the anomaly factor assumed in the ITER design, and observed with tokamaks having density in the vicinity of 1014 cm-3, were to apply at higher density, then Lawson should be possible at 1020 cm-3 in a tokamak with a minor radius of 2.8 mm! This dramatic reduction in size at higher density provides much of the motivation for MTF.
Figure 2. Plots of characteristic step sizes and poloidal magnetic field strength assuming poloidal beta = 1 vs. fuel density for a plasma with 10 keV temperature.
Speculation on anomalous transport. Anomalous transport mechanisms are still a subject of unfinished research. Clearly, all possibilities cannot be anticipated, but the following can be noted. Generally the form of c is a product of characteristic lengths times a frequency. The characteristic lengths in a plasma normally identified are l , l D,
c/w pi, c/w pe, ri, and re. As already noted, c/w pi and ri are only different by a factor of order unity, and therefore the gyro radius in Fig. 2 is also approximately the same as c/w pi. The gyro radius re (and thus c/w pe) is smaller than the gyro radius ri by a factor of (mi/me)1/2. The Debeye length l D has the same density dependence as the electron gyro radius. Therefore the variation of all the usual characteristic lengths with density is correctly inferred from Fig. 2, and a reasonable conjecture is that the tendency towards smaller size at higher density is true for anomalous transport as well as for classical transport.
III. Plasma Energy Reduced at High Density
To quantify the variation of diffusion step sizes with density in terms that come closer to economic value, we show in Fig. 3 the thermal energy contained by a plasma with characteristic dimension of a. Three different configurations are included in Fig. 3: ICF-relevant unmagnetized fuel, tokamaks, and a generic MTF plasma taken to be a compact torus (CT). We assume that when density is varied for a given configuration, size is adjusted to be the minimum necessary to provide nt E = 3x1014 cm-3 sec at 10 keV temperature. Specific assumptions for each configuration are summarized in the table following Fig. 3.
A. ICF Energy Requirements
For ICF we see a very strong dependence of energy upon density, and thus the importance of compressing to high density. By compressing to density of approximately 1025 cm-3, the energy in the hot spot according to Fig. 3 is approximately 30 kJ, which is similar to the value anticipated in the design of NIF [Lindl 95]. Achieving such a high density requires an implosion velocity of about 30-40 cm per microsecond and a radial convergence of between 30 and 40. The NIF laser design, with 1.8 MJ and 500 TW, has enough energy and power to produce these conditions even with the inefficiency of indirect drive. However, if the hot-spot density were to be reduced, the energy requirements would be considerably increased as shown in Fig. 3, and the power requirements would also be increased to achieve the same nt E. Thus, the ICF approach utilizes very high density to achieve fusion with minimum energy, but the driver requirements are extremely demanding and expensive.
B. Tokamak Energy Requirements
Tokamaks are included in Fig. 3 for academic interest, even though high-density operation of a tokamak-like configuration is not being considered. The poloidal magnetic field required at any given density is plotted in Fig. 2. For the assumed value of safety factor (q) and aspect ratio, the toroidal field required would be approximately a factor of ten higher than the poloidal field. Thus, the magnetic energy would be 100 times as large as the plasma thermal energy plotted in Fig. 2. The cost of a tokamak is well known to be strongly tied to the cost of the magnets.
The important aspect of the tokamak is that much more is known about transport than for any other configuration. A useful summary of tokamak transport formulas can be found in the textbook by Kadomtsev [Kadomtsev 92]. We plot both the classical limit for confinement (neoclassical in the banana, transition, and Pfirsch-Schluter regimes as density increases) and some empirically based models for anomalous transport. The anomalous transport curves show the anticipated tendency that system size becomes small at increasing density. One concludes from these plots that if the technology were available to operate tokamaks at higher density, the size and cost could be reduced.
C. MTF Energy Requirements.
For MTF compression by a liner, there are many possible magnetic configurations. To make estimates for Fig. 3, we have chosen a compact toroid (CT) plasma as generic for any magnetic configuration. Specifically the CT curves in Fig. 3 are calculated assuming the plasma is an FRC, which has ideally only poloidal magnetic field [Tuszewski 88]. Similar values apply to a spheromak. In that case a toroidal field comparable in magnitude to the poloidal field of the FRC would be required [Jarboe 94]. CTs require more energy than a tokamak at a given density because CTs need more volume to achieve the same effective radius or insulating distance. A prolate FRC, as is commonly studied in experiments, has an effective radius equal to the distance from the field null to the outer edge, which is approximately 0.3 of the small radius of the prolate spheroid. Thus the FRC estimate for energy may be conservatively high in Fig. 3, although modeling of wall-plasma interactions tend to show spatial profiles that resemble an FRC-like profile (Siemon 97).
Figure 3. Energy requirements vs. fuel density for various configurations and transport assumptions assuming nt E = 3x1014 cm-3 sec, T = 10 keV, and poloidal b = 1.
Configuration Transport Comments ICF Electron thermal conduction Spherical plasma with size given by Eqn. 12. Density of ~1025 cm-3 corresponds to NIF. Tokamak Neoclassical, anomalous neo-Alcator, and anomalous ITER-89P Aspect ratio (2.9), poloidal beta (1.0), and safety factor q (3.0) are held constant at ITER-like values. Compact Torus (CT) Classical or Bohm Geometry of a prolate FRC assumed for illustration with length to diameter ratio of 3.
The amount of energy required for fusion conditions depends upon the global energy confinement time. Fig. 3 indicates that compressed plasma energy between about 30 kJ and 10 MJ is required in the MTF regime (density of 1020 cm-3), if plasma transport is between classical and Bohm. For the larger Bohm requirement of 10 MJ, the required liner kinetic energy would be tens of MJs, a few times the final plasma energy. One striking difference between the MFE and MTF regimes of density is that Bohm is an acceptable possibility at MTF density, while as shown in Fig. 3, Bohm is totally unacceptable at 1014 cm-3.
The curve labeled Bohm deserves additional comment. In the early days of fusion research Bohm was introduced as an empirical diffusivity [Spitzer 62] equivalent to the following:
c BOHM = c i (w cit ii)/16, (18)
where w cit ii = l /ri is the magnetization parameter. The factor of 16 has no theoretical basis. It is interesting to note that apart from the factor of 16, c BOHM is the geometric mean or logarithmic average of c i and c e given in Eqns. 14 and 15. Thus Bohm can be thought of as intermediate between classical magnetized and unmagnetized confinement. Kadomtsev describes how there are situations where macroscopic convection can lead to energy transport with a global Bohm confinement time [Kadomtsev 92]. Studies of wall-confined MTF-type plasma by Vekshtein show how classical confinement can lead to a Bohm-like scaling [Vekshtein 90]. Even more interesting is that experimental data from a number of carefully studied magnetic configurations, including Reversed Field Pinches, spheromaks, and FRCs, is generally as good as Bohm or better.
Global energy confinement time can be worse than Bohm when other non-diffusive processes dominate. Examples are radiation because of impurities, or plasma flow out of the system at a speed comparable to the thermal speed. Radiation by impurities is always a concern and places an upper limit on the allowed impurity concentration. Plasma flow cannot be ruled out in general, but the conjecture here is that target plasma configurations can be found for which a pressure equilibrium exists between the metal liner boundary and the fuel, and thus flow is reduced to nothing worse than convective motions. Close proximity of a conducting boundary should provide a stabilizing influence on magneto-hydrodynamic modes, especially since magnetic fields do not penetrate a conducting boundary on the short time scale of interest for MTF. Spheromaks and FRCs are two examples of CTs for which there are data to support this conjecture. We conclude Bohm represents a reasonable, even conservative, expectation for achievable global energy confinement based on previous experimental results, assuming impurities can be avoided by careful experimental technique.
IV. The Size and Cost of Ignition Facilities
Only a rough connection can be made between cost and plasma energy plotted in Fig. 3. For each of the configurations, however, one would expect that the indicated reduction of energy as density increases would result in a reduction of costs for the required facility to create the ignition-grade plasma. Even an approximate connection is adequate for present purposes, given the many decades of system size plotted in Fig. 3. Note that the left-hand scale varies by 12 orders of magnitude. We list in Table 1 costs for recently designed ignition-class facilities in each of the regimes of MFE, ICF, and MTF.
In the case of MTF we base the cost for an ignition facility upon the ATLAS pulsed-power facility, recently designed and under construction by Defense Programs at Los Alamos [Trainor 97]. ATLAS should be able to deliver 5-10 MJ to an imploding liner, which makes it suitable for a considerable range of possible MTF experiments. Although the primary mission of ATLAS is not MTF, a reasonable number of additional experiments to test MTF are consistent with current plans for the facility. For the purpose of estimating MTF ignition-grade facility costs, we assume that 1) the 35-MJ of stored energy in ATLAS is enough to implode a liner-plasma configuration to ignition (see Fig. 3), and 2) the additional cost for the plasma target preparation is small compared with the \$50-million cost of the ATLAS facility. The purpose of Table I is to compare facility costs needed for a fusion energy development program. The fact that ATLAS is being built for other reasons is simply a fortunate circumstance. The research effort expended to date on MTF has been minuscule compared with the other two approaches to fusion, and so the cost of achieving ignition conditions is obviously much less certain. However, the advantage appears so large that the accuracy of the estimate is not very important.
Table 1. Approximate Cost of Ignition Facilities
Concept Plasma Thermal Energy Facility Cost MFE/ITER 1 GJ \$10 billion ICF/NIF 30 kJ \$1 billion MTF/ATLAS ~ 10 MJ ~\$50 million
V. Near Term Prospects for MTF Research
A. Typical MTF Parameters
The main points of this paper, which are contained in Fig. 3 and Table 1, argue for starting a new thrust in fusion energy research. In this section we discuss some aspects of how to begin that effort. Our concept for a liner-driven plasma implosion suggests approximate values for initial and final plasma parameters as given in Table 2.
Table 2. Representative Conditions for an Adiabatic Implosion
Parameter Desired Final Conditions Required Initial Plasma if Kv=100 Required Initial Plasma if Kv=1000 Temperature 10 keV 460 eV 100 eV Density 1020 cm-3 1018 cm-3 1017 cm-3 B Field 10 MG 100 kG 100 kG Liner inner radius 5 mm 5 cm 5 cm
To illustrate the required initial target-plasma conditions, we assume adiabatic compression (pVg =const) with a volumetric compression Kv = 100 or 1000, corresponding to cylindrical, or spherical, radial compression of 10 respectively. The adiabatic approximation is justified according to time-dependent calculations taking thermal and radiation losses into account [Lindemuth 83], and the parameter space for MTF is found to be quite large, assuming an implosion velocity on the order of 106 cm/sec.
B. Target Plasma Possibilities.
Among the many possible magnetic configurations that would be possible for the target plasma, the ones currently receiving attention in our awareness are: 1) the MAGO-type of accelerated diffuse-z-pinch plasma [Lindemuth, 96], 2) an expanded high-density-fiber z pinch inside a conducting boundary [Wysocki 97], and 3) compact toroids [Ryutov 96]. An approach that uses energy from a high-power e-beam driver to form a magnetized plasma has also been reported [Chang 78].
Extensive research on compact toroids, the spheromak and Field-Reversed Configuration, began in about 1980. The review articles by Tuszewski and Jarboe have hundreds of references [Tuszewski 88, and Jarboe 94]. By definition, a CT is a self-contained magnetized plasma that can be moved from one spatial location to another. Thus, CTs are an obvious candidate for inserting a plasma target into an imploding metal liner. Unfortunately, most fusion-related liner research ended about the same time that CT work began, so most of the information gained from CT research was not available to the early liner researchers. A few experiments studying the implosion of an FRC-type of CT were done in Russia [Kurtmullaev 82]. Most CT research was done at much lower density than is needed for MTF. The RACE experiments at LLNL are a notable exception [Hammer 91]. There is no obvious problem in forming CTs at higher density, and experiments to move in that direction would be desirable.
The MAGO and expanded fiber z pinch are diffuse z-pinch magnetic configurations. The outstanding attraction of these approaches is that the technology for plasma formation is reasonably compatible with liner implosion technology, and is less complicated than for CTs. For MAGO at least, plasma density and temperature appear suitable for proceeding with MTF implosion experiments [Lindemuth 95]. More refined measurements are still needed to characterize global energy confinement in both the MAGO and expanded fiber plasmas. The diffuse z pinch has well known limitations with regard to stability, and containment of energetic particle orbits. However, simulations show [Sheehey 89] that an unstable plasma inside a conducting boundary can evolve to a stable state (known as a Kadomtsev-stable profile). In such a state, the energy confinement may be adequate on the time scale of an MTF implosion. The fact that most alpha particles generated near peak compression would be lost is not a major consideration for the batch-burn approach we have assumed for MTF.
1. Liner Technology and Facilities are Available.
The advances in liner technology of the past few years are impressive [Chernyshev 97]. More than enough liner velocity and implosion symmetry has been demonstrated compared with the detailed requirements for an MTF liner system discussed elsewhere [Lindemuth 96, Siemon 97, Ryutov 96, and Schoenberg 98]. A quasi-spherical implosion of unmagnetized plasma has also been reported [Degnan 96].
A number of existing facilities supported by DOE’s Defense Programs and DOD would be suitable for a variety of MTF experiments. These include the Z capacitor bank at Sandia National Laboratory, the Shiva Star capacitor bank at Phillips Air Force Laboratory, the Pegasus capacitor bank at Los Alamos National Laboratory, the Ranchero explosively-driven electrical generators at Los Alamos, and the ATLAS capacitor bank under construction at Los Alamos.
These facilities and expertise allow significant leveraging of research dollars, which gives additional incentive for MTF research.
D. Major Technical Issues.
MTF can be conceptually separated into three inter-related aspects: target plasma formation and confinement properties, liner-driver implosion, and target-plasma compression. The major technical issues are:
Issues of Target Plasma Formation and Confinement Properties
• plasma parameters on the proper adiabat for heating to ignition
• suitable magnetic topology for magnetohydrodynamic stability and adequate thermal insulation
• plasma-wall interactions leading to high-Z impurities and concomitant plasma radiation losses
Liner-Driver Implosion Issues
• symmetric implosions of a liner at approximately 106 cm/s (a velocity well within the range of what has been demonstrated in Defense Program experiments).
• development of liner implosion configurations that match target-plasma requirements for a conducting boundary throughout the implosion
• convergence ratios of roughly 10 in a stable quasi-spherical geometry
Target Plasma Compression Issues
• technical compatibility between plasma formation and liner-implosion technologies
• accelerated mixing of wall and plasma material during the implosion, resulting for example from Rayleigh-Taylor instabilities in the liner
• plasma thermal transport during the implosion
• diagnostic methods under conditions of energetic implosions
We recommend a multi-institutional MTF research program to address these important experimental and theoretical questions. In addition, studies are needed on how MTF would best be utilized for electricity generation or other applications. Qualitatively the intrinsically pulsed nature of MTF makes it similar to ICF in its potential application. Early studies of an electrical power plant based on liner technology [Krakowski 78] indicated the basic feasibility of a pulsed liner-driven system, and identified numerous technology issues that must be solved.
An intriguing more recent study of power generation using MHD conversion of fusion energy [Logan 93] indicates that MTF is well matched to the requirements of an MHD conversion system. The energy from 14-MeV neutrons would be used to vaporize and heat a lithium-containing blanket to 1 or 2 eV. Then MHD conversion gives higher efficiency and a greatly reduced balance of plant cost leading to considerably less expensive electricity compared with conventional MFE reactor concepts.
VI. Conclusions
We briefly reviewed some very elementary features of all the standard fusion approaches. The main assumptions were that the fusion fuel is deuterium and tritium with a 10 keV Maxwellian velocity distribution. We emphasized the variation of quantities with fuel density and observed that the system size becomes small, and energy requirements are much reduced, when fuel density is made considerably larger than in conventional MFE systems. This general conclusion, which has been noted by many researchers in the past, warrants renewed attention today as the fusion program restructures itself within today’s budget limitations.
The reasons for embarking on an MTF research effort at the present time are several:
• The cost of development for fusion has become a major consideration in recent years, and MTF appears to offer advantages compared with MFE and ICF.
• The pulsed power facilities of Defense Programs, both DOE and DOD, are remarkably well matched to what is needed to investigate MTF.
• In the twenty years since MTF-like concepts were last seriously pursued in the United States, the theoretical understanding and experimental methods of plasma science as well as the technology of high-energy liner implosions have advanced significantly.
The interesting regime we call Magnetized Target Fusion occurs at fuel density of about 1020 cm-3. The MTF regime may be an optimum in the sense of using the maximum possible magnetic field for insulation of the fuel, and thus the smallest possible system size without going to the extreme density of ICF. This new thrust in fusion research has the potential to achieve the lowest possible development cost.
We believe that the arguments presented here are robust in nature and give a valid basis for recommending a new research thrust in magnetic fusion energy. Given the global importance of long-term energy R&D, adding MTF as a new complementary element to MFE and ICF in the portfolio of fusion approaches seems well justified.
Acknowledgements
We appreciate receiving encouragement to examine MTF from John Browne, Steve Younger, and Al Sattelberger. We also thank our colleagues Carl Ekdahl, Bob Reinovsky, and others in the pulsed power community at Los Alamos for providing expert advice on liner technology.
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Wysocki 97. F. J. Wysocki, R. E. Chrien, G. Idzorek, H. Oona, D. O. Whiteson, R. C. Kirkpatrick, I. R. Lindemuth, and P. T. Sheehey, "Progress With Developing A Target For Magnetized Target Fusion", Proc. of 11th IEEE International Pulsed Power Conference (Baltimore, MD, June 29- July 2, 1997, to be published). | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8292484879493713, "perplexity": 2523.663350932596}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583657151.48/warc/CC-MAIN-20190116093643-20190116115643-00314.warc.gz"} |
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1. CJM 2015 (vol 67 pp. 759)
Carey, Alan L; Gayral, Victor; Phillips, John; Rennie, Adam; Sukochev, Fedor
Spectral Flow for Nonunital Spectral Triples We prove two results about nonunital index theory left open in a previous paper. The first is that the spectral triple arising from an action of the reals on a $C^*$-algebra with invariant trace satisfies the hypotheses of the nonunital local index formula. The second result concerns the meaning of spectral flow in the nonunital case. For the special case of paths arising from the odd index pairing for smooth spectral triples in the nonunital setting we are able to connect with earlier approaches to the analytic definition of spectral flow. Keywords:spectral triple, spectral flow, local index theoremCategory:46H30
2. CJM 2007 (vol 59 pp. 3)
Biller, Harald
Holomorphic Generation of Continuous Inverse Algebras We study complex commutative Banach algebras (and, more generally, continuous inverse algebras) in which the holomorphic functions of a fixed $n$-tuple of elements are dense. In particular, we characterize the compact subsets of~$\C^n$ which appear as joint spectra of such $n$-tuples. The characterization is compared with several established notions of holomorphic convexity by means of approximation conditions. Keywords:holomorphic functional calculus, commutative continuous inverse algebra, holomorphic convexity, Stein compacta, meromorphic convexity, holomorphic approximationCategories:46H30, 32A38, 32E30, 41A20, 46J15
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https://www.arxiv-vanity.com/papers/1710.07232/ | April 4, 2021
Vertical Integration from the Large Hilbert Space
Theodore Erler111, Sebastian Konopka222
Institute of Physics of the ASCR, v.v.i.
Na Slovance 2, 182 21 Prague 8, Czech Republic
Abstract
We develop an alternative description of the procedure of vertical integration based on the observation that amplitudes can be written in BRST exact form in the large Hilbert space. We relate this approach to the description of vertical integration given by Sen and Witten.
## 1 Introduction and Summary
Computing a superstring scattering amplitude requires inserting a configuration of picture changing operators (PCOs) for each Riemann surface contributing to the amplitude. A choice of PCOs roughly corresponds to a section of a fiber bundle: The base of the fibre bundle consists of the moduli space of Riemann surfaces with the relevant genus, spin structure, and number of punctures, and the fiber at each point consists of copies of the Riemann surface with the corresponding value of the moduli; this parameterizes the possible ways of inserting PCOs on that surface. The worldsheet path integral defines a measure on this fiber bundle which can be pulled back to any submanifold; in particular, pulling the measure back onto a section of the fiber bundle and integrating defines a superstring amplitude with a prescribed configuration of PCOs on each Riemann surface. A significant complication with this procedure, however, is the appearance of spurious singularities in the superstring measure [1]. One can try to look for a global section which avoids spurious singularities everywhere, but this may be inconvenient in practice, or there may be an obstruction to the existence of such a global section.333It is known that the supermoduli space of super-Riemann surfaces cannot be holomorphically projected down to the ordinary (bosonic) moduli space of Riemann surfaces [2]. To our knowledge, the implications of this fact from the point of view of PCOs has not been worked out, but one possibility is that the PCO positions cannot be chosen globally as holomorphic functions of the moduli.
A remedy proposed by Sen [3], and later made more explicit by Sen and Witten [4], is to divide the moduli space into regions so that on each region we can choose a local section which avoids spurious poles. Simply adding the contributions from the local sections together, however, does not define a gauge invariant amplitude. To correct for this, at the interface between the different regions of the moduli space we must integrate “along the fiber” to connect local sections—that is, one must deform one choice of PCOs into another while keeping the moduli fixed. This is called vertical integration. The amplitude is then defined by a closed integration cycle in the fiber bundle composed of local sections connected by “vertical segments.” Importantly, the nature of the superstring measure implies that spurious singularities can be rendered harmless on the vertical segments. Therefore we can obtain gauge invariant amplitudes free from unphysical divergences in the measure.
In this paper we investigate a different, more algebraic, approach to this problem, motivated by recent studies of superstring field theories [5, 6]. Consider an -point amplitude expressed as an -fold bra state:
⟨Ap|: H⊗n→C, (1.1)
where is a CFT vector space containing BRST invariant physical states. The superscript indicates that the amplitude contains picture changing operators inserted in some way on the constituent Riemann surfaces. Gauge invariance of the amplitude is equivalent to the statement that this bra state is annihilated by a sum of BRST operators acting on each external state:
⟨Ap|(Q⊗I⊗n−1+...+I⊗n−1⊗Q)=0. (1.2)
Here denotes the BRST operator and is the identity operator on . It is well-known that the cohomology of is trivial in the “large Hilbert space” introduced by Friedan, Martinec, and Shenker [7], that is, the CFT state space obtained by bosonizing the ghosts into the system and allowing for states which depend on the zero mode of the ghost. This implies that the amplitude can be expressed in the form
⟨Ap|=⟨αp|(Q⊗I⊗n−1+...+I⊗n−1⊗Q). (1.3)
We will call the -fold bra state a gauge amplitude, following the terminology of [6]. The gauge amplitude lives in the large Hilbert space. However, the physical amplitude must reside in the “small Hilbert space” where the zero mode of the ghost is absent. This requires that the amplitude satisfies
⟨Ap|(η⊗I⊗n−1+...+I⊗n−1⊗η)=0, (1.4)
where denotes the zero mode of the eta ghost. This is consistent with (1.3) provided that the object
⟨αp|(η⊗I⊗n−1+...+I⊗n−1⊗η) (1.5)
is annihilated by the BRST operator. Since carries picture , it is natural to interpret this object as an amplitude containing one fewer PCO insertion:
⟨αp|(η⊗I⊗n−1+...+I⊗n−1⊗η)=⟨Ap−1|. (1.6)
We can now apply this procedure again, expressing as the BRST variation of a gauge amplitude , and apply once again to arrive at an amplitude containing two fewer PCO insertions. Continuing this process unfolds a hierarchy of amplitudes and gauge amplitudes:
⟨Ap| =⟨αp|(Q⊗I⊗n−1+...+I⊗n−1⊗Q) ⟨Ap−1| =⟨αp|(η⊗I⊗n−1+...+I⊗n−1⊗η) ⟨Ap−1| =⟨αp−1|(Q⊗I⊗n−1+...+I⊗n−1⊗Q) ⋮ ⟨A1| =⟨α2|(η⊗I⊗n−1+...+I⊗n−1⊗η) ⟨A1| =⟨α1|(Q⊗I⊗n−1+...+I⊗n−1⊗Q) ⟨A0| =⟨α1|(η⊗I⊗n−1+...+I⊗n−1⊗η), (1.7)
where at the end we obtain an amplitude containing no PCO insertions at all.444The number of PCO insertions in is determined by the requirement that the amplitude is nonzero acting on NS states at picture and Ramond states at picture . This means that the amplitudes with fewer than PCO insertions will need to act on states with nonstandard picture to obtain a nonzero result. Generally, such amplitudes will encounter divergences from spurious singularities. We discuss such amplitudes formally, as intermediate objects used to obtain the final amplitude which is gauge invariant and free from spurious singularity. The structure here is reminiscent of descent equations which appear in analysis of anomalies in gauge theories [8]. This leads the following procedure for deriving gauge invariant amplitudes. First we start with the amplitude , and insert the operator at some point on each constituent Riemann surface. This defines the gauge amplitude . We then take the BRST variation to arrive at the amplitude containing one PCO. We then insert another on the Riemann surfaces of to derive , and continue in this way until we arrive at the amplitude containing PCO insertions. The crucial point is that the insertions of do not need to vary continuously with the moduli to ensure gauge invariance. Gauge invariance is automatic since the final amplitude is expressed in BRST exact form. We may therefore allow the insertions to “jump” across spurious poles discontinuously as a function of the moduli to avoid unphysical divergences.
The primary goal of this work is to show that the above algebraic procedure gives a viable alternative to defining a consistent measure on the moduli space for superstring amplitudes. The approach has some advantages. The computation of vertical corrections is arguably simpler and more flexible, and certain essential properties, such as gauge invariance of the amplitude and independence from various choices, are evident from the nature of the construction. A second goal of our work will be understanding the relationship between the algebraic approach and the conceptually rather different idea of vertical integration. Our motivation is to form a link between Sen’s discussion of superstring field theories [9] and other techniques which have been independently developed based on the large Hilbert space [5, 6, 10]. As investigations continue into quantum effects in superstring field theories [11, 12], and in the geometrical formulation based on super-Riemann surfaces [13], it may be useful to have an understanding of the relationship between these approaches.
Vertical integration is a general idea which can be implemented in many ways. To give ourselves a concrete objective, we will focus on the connection to the vertical integration procedure as implemented by Sen and Witten [4]. In the spirit of that work, we discuss only on-shell amplitudes and ignore the fact that the moduli space of Riemann surfaces is noncompact. The boundary of moduli space is associated with the infrared physics of superstring perturbation theory, about which there has been extensive discussion in recent years. One way of dealing with infrared divergences is to extend amplitudes off-shell using the formalism of string field theory. Suffice it to say that our discussion can be easily adapted in this context, which provided part of the motivation for this work. For simplicity we discuss PCOs in the holomorphic sector only, as would be relevant for the heterotic string. For type II strings we have a similar story also in the antiholomorphic sector.
This paper is organized as follows. In section 2 we review the definition of superstring measure in the PCO formalism. In section 3 we describe the algebraic construction of superstring amplitudes, deriving a set of recursive equations for the vertical corrections at the interface between local sections needed to ensure gauge invariance. We give examples and prove that on-shell amplitudes are independent of the choice of vertical corrections derived by this procedure. In section 4 we discuss the construction of Sen and Witten. To give a clear formalization of their procedure, we employ an analogue of differential forms on the lattice, called difference forms. The Sen-Witten vertical corrections are defined by “integration” of a discretized measure—characterized by difference forms—over a collection of links in a -dimensional cubical lattice, where is the number of PCOs in the amplitude. The sites of the lattice correspond to combinations of PCOs taken from adjoining local sections, and the collection of links which define the “integration cycle” are called lattice chains. We describe how vertical corrections of this form may be constructed from the algebraic point of view. The algebraic construction introduces a collection of auxiliary amplitudes containing PCOs whose vertical corrections are characterized by lattice chains inside lower dimensional lattices of respective dimension . The algebraic construction functions by extending lattice chains from lower dimensional into higher dimensional lattices in such a way as to be consistent with gauge invariance and so that the chains of higher dimensional lattices project down to the chains of lower dimensional lattices. We conclude with some examples.
## 2 Superstring Measure
In this section we review the superstring measure in the PCO formalism [3]. The purpose is to fix a convenient notation for our calculations and to simplify some signs.
Given a Riemann surface with genus and punctures, we can remove disks around each puncture and cut what remains into components with the topology of a sphere with three holes. We cover the disks with holomorphic local coordinates with . The origin of these coordinates corresponds to the location of the punctures on the Riemann surface. On the spheres we introduce holomorphic coordinates . The Riemann surface can be reconstructed by gluing the boundaries of these components with holomorphic transition functions:
zi =fij(zj) (2.1) zi =fa(wa). (2.2)
The transition functions exist between coordinates which are identified by the gluing, and encode all information about the moduli of the Riemann surface. Since we work with the heterotic string, we are interested in the moduli space of Riemann surfaces with spin structure in the leftmoving sector. This is a -fold covering of the bosonic moduli space which comes in two disconnected components, representing the even and odd spin structures. We use to denote one of these disconnected components. That is, is the moduli space of genus Riemann surfaces with punctures together with either an even or odd spin structure, and denotes a point in this moduli space.
Given transition functions , we may define an -fold bra state called a surface state
⟨Σ|:H⊗n→C. (2.3)
The surface state is defined so that the quantity
⟨Σ|Φ1⊗...⊗Φn (2.4)
represents a correlation function on a Riemann surface assembled with the transition functions , with the vertex operators corresponding to the states inserted at the punctures in the respective coordinates . If the vertex operators are conformally invariant, the correlation function only depends on the transition functions through the moduli of the Riemann surface they represent. For generic vertex operators, the surface state will depend more nontrivially on the choice of transition functions. The surface state is BRST invariant:
⟨Σ|Q=0. (2.5)
Since this will not cause confusion, we use a shorthand notation where represents a sum of BRST operators acting on each state:
Q → Q⊗I⊗n−1+...+I⊗n−1⊗Q. (2.6)
In particular represents a correlation function with a contour integral of the BRST current surrounding all punctures. If we deform the contour inside the surface and shrink to a point, this gives zero. Also important for our discussion is the fact that is well-defined in the small Hilbert space. This implies that it is annihilated by the zero mode of the eta ghost:
⟨Σ|η=0, (2.7)
where denotes a sum of eta zero modes acting on each state.
Let us first describe the measure without PCOs, as would be relevant for computing amplitudes in bosonic string theory. We fix a choice of surface state for each point in the moduli space, and write simply as , leaving the dependence on moduli implicit. To define differential forms that can be integrated over the moduli space, we need to insert the appropriate -ghosts inside correlation functions. Using the idea of the Schiffer variation, following [14], we may express the -ghost insertions as contour integrals surrounding the punctures. Around the th puncture we have a -ghost insertion of the form
b(v(a)μ)=∮dwa2πiv(a)μ(m,wa)b(wa)+∮d¯wa2πi¯v(a)μ(m,¯wa)¯b(¯wa), (2.8)
where the contours are oriented counterclockwise respectively in the and coordinates on the Riemann surface. The contour integrals are weighted by functions called Schiffer vector fields. The lower index corresponds to coordinates on the moduli space, with . We introduce the operator
Tμ ≡(∮dw12πiv(1)μ(m,w1)T(w1))⊗I⊗n−1+...+I⊗n−1⊗(∮dwn2πiv(n)μ(m,wn)T(wn)) (2.9) +(∮d¯w12πi¯v(1)μ(m,¯w1)¯T(¯w1))⊗I⊗n−1+...+I⊗n−1⊗(∮d¯wn2πi¯v(n)μ(m,¯wn)¯T(¯wn)).
The Schiffer vector fields are defined so that the following equation holds:
∂∂mμ⟨Σ|=−⟨Σ|Tμ. (2.10)
Additional properties are
[Q,bμ] =Tμ (2.11) [Tμ,bν] =∂∂mμbν−∂∂mνbμ, (2.12)
where is defined as in (2.9) with the energy momentum tensor replaced by the -ghost, and represents a graded commutator with respect to Grassmann parity. Let be coordinate 1-forms on the moduli space, and introduce operator-valued 1-forms:
T ≡dmμTμ, (2.13) b ≡dmμbμ. (2.14)
To simplify signs, we assume that the coordinate 1-forms are uniformly Grassmann odd objects, so they anticommute through each other and also though Grassmann odd worldsheet operators. In this convention, the operator is Grassmann odd, and is Grassmann even. The identities (2.10)-(2.12) imply
d⟨Σ| =−⟨Σ|T (2.15) [Q,b] =−T (2.16) db =12[T,b], (2.17)
where
d=dmμ∂∂mμ (2.18)
is the exterior derivative on the moduli space. The measure for scattering amplitudes can then be expressed
⟨Ω|=⟨Σ|eb. (2.19)
This is a differential form of inhomogeneous degree. In particular, the operator is defined by the series expansion
eb=I⊗n+b+12!b2+...+1(6g+2n−6)!b6g+2n−6. (2.20)
The series terminates since there are only independent 1-forms . The last term is a top degree form, and this is the part of the measure that should be integrated over the moduli space to obtain the amplitude. Using the identities (2.15)-(2.17), it is straightforward to show that
⟨Ω|Q=−d⟨Ω|. (2.21)
Assuming we can ignore contributions from the boundaries of moduli space, this implies that BRST trivial states decouple from scattering amplitudes.
The measure , however, can only compute superstring scattering amplitudes between states of nonstandard picture. Such amplitudes will typically suffer from unphysical divergences due to spurious singularities. Therefore, it is useful to generalize the measure to accommodate correlation functions containing additional operator insertions, in particular PCOs. One concrete way to do this is as follows.555In the description of [3], PCOs are inserted in the coordinates representing the Riemann surface with the disks around the punctures removed. In this approach, the Schiffer vector fields must be chosen to vanish at the location of the PCOs in order to ensure that deformations of the moduli are independent from deformations of the PCO positions in the coordinates . This is equivalent to the approach we take, but expressed in a different coordinate system on the Riemann surface. Suppose we have a correlation function including operators , in addition to the vertex operators representing the external states. We remove a disk from the Riemann surface containing the location of all operators , but no vertex operators. We fix a coordinate system on this disk denoted with , so that each operator has a corresponding position in this coordinate system. For short we write the complete set of operator insertions as
Op=O1(y1)...Op(yp), (2.22)
where the upper index indicates the number of operator insertions. We build the remaining part of the Riemann surface by removing disks around the punctures, covered by coordinates with and corresponding to the location of the punctures. Including the disk , the surface now has holes; we cut what remains into components with the topology of a sphere with three holes, and introduce coordinates on these components. The Riemann surface may be reconstructed by specifying transition functions between coordinates identified by gluing:
zi =fij(zj) (2.23) zi =fa(wa) (2.24) zi =f(y). (2.25)
Note that at this level the coordinate is on the same footing as the coordinates , but the coordinate will play a distinct role in defining the measure. From the transition functions we define a surface state acting on copies of :
⟨Σ′|:H⊗n+1→C. (2.26)
We use the prime to indicate that acts on states, including a state represented by the coordinate . Assuming that the first copy of represents operators inserted in the coordinate , we may then represent correlation functions containing operators through the -fold bra state
⟨Σ′|(Op|0⟩)⊗I⊗n. (2.27)
Suppose that for every point in the moduli space we chose transition functions building a Riemann surface with moduli . From this we can define a surface state for every ; we write simply as , leaving the dependence on implicit. We assume that the transition functions have been defined so that the coordinate covers all parts of each Riemann surface where we care to insert . Note that the moduli space carries information about the location of the punctures represented by the coordinates , but does not carry information about the coordinate . We introduce a collection of Schiffer vector fields and defined so that the analogue of (2.15)-(2.17) hold:
d⟨Σ′| =−⟨Σ|T′ (2.28) [Q,b′] =−T′ (2.29) db′ =12[T′,b′], (2.30)
where
b′≡dmμb′μ (2.31)
and
b′μ ≡(∮dy2πivμ(m,y)b(y))⊗I⊗n+I⊗(∮dw12πiv(1)μ(m,w1)b(w1))⊗I⊗n−1+...+I⊗n⊗(∮dwn2πiv(n)μ(m,wn)b(wn))
With this we define the measure with operator insertions as
(2.33)
We label the measure according to the operator insertions it contains.
It is useful to think of the measure as a differential form on a fiber bundle . The base of consists of the moduli space of genus Riemann surfaces with punctures together with an even or odd spin structure, and are coordinates on the base. The fiber at the point consists of copies of the Riemann surface with the corresponding value of the moduli, and are coordinates on the fiber. We introduce coordinate 1-forms on the fiber and define the exterior derivative on :
d=dmμ∂∂mμ+dyi∂∂yi+d¯yi∂∂¯yi. (2.34)
We assume that are uniformly Grassmann odd objects which anticommute with each other, the s, and Grassmann odd worldsheet operators. Using the identities (2.28)-(2.30), it is straightforward to show that the generalization of (2.21) in the presence of operator insertions takes the form
(2.35)
where now includes differentiation along the fiber directions. On the right hand side, represents a sum of operator insertions
(Q−d)Op (2.36) −(dO1(y1))...Op(yp)−...−(−1)O1+...+Op−1O1(z1)...(dOp(yp)),
and, for example
dO1(y1)=dy1∂O1(y1)+d¯y1¯∂O1(y1). (2.37)
Also important in our discussion is the property
(−1)Op⟨Ω,Op|η=−⟨Ω,ηOp|, (2.38)
where represents a sum of operator insertions
ηOp=(ηO1(y1))...Op(yp)+...+(−1)O1+...+Op−1O1(y1)...(ηOp(yp)). (2.39)
This is nonzero only if contains some operators in the large Hilbert space.
The measure which is relevant for computing superstring scattering amplitudes in the PCO formalism is
⟨Ω,Xp|, (2.40)
where refers to a collection of operator insertions of the form
Xp≡[X(y1)−dξ(y1)] ... [X(yp)−dξ(yp)], (2.41)
and is a picture changing operator. If the number of insertions is chosen appropriately, we obtain nonvanishing correlation functions with Neveu-Schwarz and Ramond external states at the standard pictures and . The measure is defined in the small Hilbert space:
⟨Ω,Xp|η=0. (2.42)
Furthermore, since
X(y)−dξ(y)=(Q−d)ξ(y), (2.43)
we have the property
⟨Ω,Xp|Q=−d⟨Ω,Xp|. (2.44)
The second term in (2.35) drops out since squares to zero. Therefore, the superstring measure produces a total derivative on the fiber bundle when acting on BRST trivial states.
Naively, we can define a gauge invariant amplitude by integrating the pullback of the superstring measure on a global section of . The difficulty, however, is in finding a global section of which avoids spurious singularities in the measure. However, it is always possible to find sections of which avoid spurious singularities locally. We can then attempt to define the amplitude by summing contributions from local sections on disjoint regions of moduli space which avoid spurious poles. Generally there will be discontinuities in the choice of PCOs between disjoint regions, and the amplitude will require additional contributions—the “vertical corrections”—to cancel boundary terms between different regions when the amplitude contains BRST trivial states. The vertical corrections can be seen to arise from integrating the superstring measure “along the fiber” at junctions between different regions of the moduli space so as to join local sections into a closed integration cycle in . This is vertical integration. Next we describe the algebraic approach to the PCO formalism, where the origin of vertical corrections is somewhat different.
## 3 Algebraic Approach
In the algebraic approach outlined in the introduction, PCOs are derived by repeatedly inserting in the measure followed by application of the BRST operator. If the location of is not a continuous function of the moduli, (2.35) implies that the BRST operator produces boundary terms from the integration over moduli space at the locus of discontinuities. These boundary terms are the vertical corrections.
We assume that the moduli space is decomposed into regions where the location of varies continuously as a function of the moduli:
M=∪αMα. (3.1)
The contribution to the amplitude from will turn out to be the pullback of the superstring measure on a local section of defined on . To connect with the discussion of [4], we assume that the regions form closed polyhedra which are glued along their faces in such a way as to define a dual triangulation of . However, it should be clear that the general procedure applies regardless of the choice of decomposition of the moduli space. By definition, all faces of codimension in a dual triangulation appear at the junction between distinct polyhedra. We will write for the codimension face at the junction of distinct polyhedra , so we have
codimension 0: Mα codimension 1: Mαβ=Mα∩Mβ, α,β distinct codimension 2: Mαβγ=Mα∩Mβ∩Mγ α,β,γ distinct ⋮ ⋮ . (3.2)
See figure 3.1. If the intersection of the polyhedra is empty, we assume that is the empty set. The faces and are equal as sets, but it is useful to consider them as having opposite orientations as integration cycles in the moduli space. More generally, we assume that
∫M...αi...αj...=−∫M...αj...αi.... (3.3)
In this sense, is totally antisymmetric in the indices . In particular, is the empty set if any two indices are equal. Fixing an orientation on the moduli space induces an orientation on the polyhedra, and the orientation of the higher codimension faces will be determined by
∫∂Mα0...αk=−∑β∫Mα0...αkβ. (3.4)
In this setup we can formulate a useful version of Stokes’ theorem. Suppose on each codimension face we have a differential form which is antisymmetric in the indices . Stokes’ theorem implies
1(k+1)!∑α0...αk∫Mα0...αkdωα0...αk=1(k+2)!∑α0...αk+1∫Mα0...αk+1(δω)α0...αk+1. (3.5)
We introduce an operation , which acts on an object with antisymmetric indices to produce an object with antisymmetric indices . It is defined as
(δω)α0...αk+1=k+1∑n=0(−1)nωα0...ˆαn...αk+1, (3.6)
where the hat over the index indicates omission. The operation is nilpotent,
δ2=0, (3.7)
and is related to the Čech coboundary operator.
### 3.1 The Construction
We propose to express the amplitude in the form666Since refers to only one connected component of the moduli space of Riemann surfaces with spin structure, technically only gives the contribution to the total amplitude coming from either the even or the odd spin structures. Since the location of spurious poles depends on the spin structure, in general we must adjust the choice of dual triangulation, local sections, and vertical corrections separately for the even and odd spin structures. The complete amplitude is then given by adding these contributions.
⟨Ap|=∑α∫Mα⟨Ω,Xpα|+12!∑αβ∫Mαβ⟨Ω,Xpαβ|+13!∑αβγ∫Mαβγ⟨Ω,Xpαβγ|+... . (3.8)
The first term is the contribution to the amplitude from the pullback of the superstring measure (2.40) onto local sections of on each polyhedron. The operator insertions in the first term are given by
Xpα=[X(y1α(m))−dξ(y1α(m))] ... [X(ypα(m))−dξ(ypα(m))], (3.9)
where the points parameterize the location of the PCOs as a function of , and characterize the local section of . The remaining terms in the amplitude are the vertical corrections, and can be arranged hierarchically according to the codimension of the faces in the dual triangulation. The vertical corrections are defined by integrating a measure over the face of the dual triangulation, where denotes a collection of operator insertions whose positions are prescribed functions of . The insertions are defined to be antisymmetric in the indices , and for even (odd) codimension the insertions are Grassmann even (odd). Generally, will be expressed through combinations of and , and the goal of the present discussion is to determine what form the insertions take.
The central condition characterizing the vertical corrections is that they lead to a gauge invariant amplitude. From (2.35) we know that
(−1)k⟨Ω,Xpα0...αk|Q=−d⟨Ω,Xpα0...αk|−⟨Ω,(Q−d)Xpα0...αk∣∣. (3.10)
Using Stokes’ theorem (3.5), gauge invariance implies that the operator insertions satisfy
(Q−d)Xpα0...αk−(δXp)α0...αk=0(. (3.11)
The operator acts on the insertions corresponding to the faces of one fewer codimension:
(δXp)α0...αk=k∑n=0(−1)nXpα0...ˆαn...αk. (3.12)
All terms in (3.11) are evaluated at a common point .
To solve (3.11), we propose that the physical amplitude can be expressed as the BRST variation of a gauge amplitude:
⟨Ap|=⟨αp|Q. (3.13)
The gauge amplitude is expressed in a form analogous to (3.8):
⟨αp|=∑α∫Mα⟨Ω,Ξpα|−12!∑αβ∫Mαβ⟨Ω,Ξpαβ|+13!∑αβγ∫Mαβγ⟨Ω,Ξpαβγ|−... . (3.14)
For convenience, we take the signs in this series to alternate. On each face of the dual triangulation we have a measure defined by a collection of operator insertions . The insertions are antisymmetric in the indices , and for even (odd) codimension they are Grassmann odd (even). Typically, the insertions depend on the zero mode of the ghost. Taking the BRST variation of the gauge amplitude gives a formula for the insertions :
Xpα0...αk=(Q−d)Ξpα0...αk+(δΞp)α0...αk(. (3.15)
The operator acts on the insertions corresponding to the faces of one fewer codimension,
(δΞp)α0...αk=k∑n=0(−1)nΞpα0...ˆαn...αk, (3.16)
and all terms in (3.15) are evaluated at a common point on . Note that, schematically, gauge invariance requires that is annihilated by , and this follows from (3.15) because
(Q−d−δ)(Q−d+δ)=(Q−d)2−δ2=0. (3.17)
Since the physical amplitude is defined in the small Hilbert space, we know that the insertions must be independent of the zero mode:
ηXpα0...αk=0. (3.18)
From (3.15), we therefore learn that satisfies
(Q−d)ηΞpα0...αk−(δηΞp)α0...αk=0. (3.19)
Interestingly, this implies that the operator insertions given by define a gauge invariant amplitude. Since carries picture , it is natural to interpret as defining an amplitude with one fewer PCO insertion:
ηΞpα0...αk=Xp−1α0...αk. (3.20)
Thus we have the relation
⟨αp|η=⟨Ap−1|, (3.21)
where is defined by insertions . We can apply this procedure again, relating to the amplitude containing two fewer PCO insertions, and continue all the way down until we have the amplitude where PCOs are absent.
This leads to the following procedure for deriving gauge invariant amplitudes. The “insertions” defining an amplitude without PCOs can be trivially written
X0α=1, X0α0...αk=0 (k≥1). (3.22)
The second equation says that there are no vertical corrections in the absence of PCOs. Since is independent of the zero mode, it can be expressed in -exact form:
X0α0...αk=ηΞ1α0...αk. (3.23)
The expression for is not unique, but let us assume that we have made some choice. We can then plug into (3.15) to derive an expression for the insertions defining the amplitude with a single PCO. By construction, will be independent of the zero mode and can be expressed in -exact form:
X1α0...αk=ηΞ2α0...αk. (3.24)
Substituting into (3.15) gives the insertions defining the amplitude with two PCOs. Continuing this process for steps we arrive at the insertions , as desired.
The solution generated by this procedure is not unique. For most purposes it does not matter how the solution is chosen as long as the PCO insertions in the final amplitude avoid spurious poles. As we will demonstrate later, Sen and Witten give a class of solutions for the vertical corrections which can be generated by this procedure, but not the most general solution.
### 3.2 Examples
Let us give some examples to see what the vertical corrections look like. Consider first an amplitude containing one PCO. We must find a set of insertions satisfying
X0α0...αk=ηΞ1α0...αk. (3.25)
We can choose for example
Ξ1α=ξ(y1α(m)), Ξ1α0...αk=0 (k≥1), (3.26)
where gives the location of a insertion on the Riemann surface as a function of in each polyhedron. We may determine the insertions by substituting into (3.15):
X1α =(Q−d)Ξ1α X1αβ =(Q−d)Ξ1αβ+Ξ1β−Ξ1α X1αβγ =(Q−d)Ξ1αβγ+Ξ1βγ−Ξ1αγ+Ξ1αβ (3.27) ⋮ .
This gives
X1α =X(y1α)−dξ(y1α) X1αβ =ξ(y1β)−ξ(y1α) X1αβγ =0 (3.28) ⋮ .
The vertical corrections on the faces of codimension 2 and higher vanish. Here and in later equations we will not explicitly indicate the dependence of the fiber coordinates on the moduli, unless needed for clarity.
As expected, is the pullback of the superstring measure (2.40) onto a local section of defined by . The insertions have a simple interpretation in terms of vertical integration. Let us make a brief detour to spell out what this means in the current setup. Let denote the local section of defined on each face of the dual triangulation. Let denote submanifolds of —the “vertical segments”—which, with a suitable orientation, connect the local sections to form a closed integration cycle in . We assume that the orientation of is antisymmetric in the indices, and postulate that the projection from down to the moduli space maps the vertical segments down to the faces of the dual triangulation. This implies that the vertical segments can be parameterized by coordinates on together with coordinates tangent to the fiber. The basic idea is to express the amplitude as
⟨Ap|=∑α∫Mα⟨Ω,Xp|+12!∑α,β∫Mαβ⟨Ω,Xp|+13!∑α,β,γ∫Mαβγ⟨Ω,Xp|+... , (3.29)
where in each term we take the pullback of the superstring measure (2.40) on the corresponding submanifold of . If we integrate out the fiber coordinates on the vertical segments, this gives an expression for the amplitude as postulated in (3.8). We can work this out fairly easily in the case where there is only one PCO. Let us choose a coordinate system on corresponding to coordinates on together with an additional coordinate parameterizing the fiber direction. The submanifold is defined by specifying the fiber coordinate as a function of and . Since must join the local sections and , we require that
y1(m,t)|t=1=y1β(m), y1(m,t)|t=0=y1α(m). (3.30)
We then find
∫Mαβ⟨Ω,X1| =∫Mαβ∫t ⟨Ω,X(y1(m,t))−dξ(y1(m,t))| (3.31) =∫Mαβ∫10dtddt⟨Ω,ξ(y1(m,t))| =∫Mαβ⟨Ω,ξ(y1β(m))−ξ(y1α(m))| =∫Mαβ⟨Ω,X1αβ|.
The only part of the measure with the 1-form is a total derivative with respect to , and integrating out the fiber coordinate gives (3.2). Note that, in this case, the vertical correction only depends on the boundary of , not on how is chosen in the interior. This is a special occurrence since we are dealing with only one PCO. With more PCOs, the part of the measure proportional to is not a total derivative, and generally the vertical corrections will depend on the choice of vertical segments. This ambiguity corresponds in the algebraic formalism to the different possible ways of expressing an amplitude in exact form.
Let us continue to the case of two PCOs. We must find a set of insertions satisfying
X1α0...αk=ηΞ2α0...αk. (3.32)
We can find a solution by multiplying by an insertion of :
Ξ2α =ξ(y2α(m))[X(y1α(m))−dξ(y1α(m))] Ξ2αβ =ξ(y2αβ(m))[ξ(y1β(m))−ξ(y1α(m))] Ξ2αβγ =0 (3.33) ⋮ .
Here gives the location of a new insertion on the codimension 0 faces as a function of , and gives the location of a | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.984817385673523, "perplexity": 530.8420530114815}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153814.37/warc/CC-MAIN-20210729011903-20210729041903-00167.warc.gz"} |
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In each of Exercises 40 through 46 following, a linear space V is given and a mapping T : V→V is defined as indicated. In each case determine whether T is a linear mapping. If T is linear, determine the kernel (or null space) and range, and compute the dimension of each of these subspaces wherever they are finite-dimensional.
40. V is the (real) linear space of all real polynomials p on R. If p E V, then T(p) is defined by setting
T(p)(x) = p(x+1), x E R
41. V is the linear space of all real functions f defined and differentiable on the open interval (0,1). If f E V, then T(f) is defined by setting
T(f)(x) = xf'(x), x E (0,1)
42. V is the linear space of all real functions f defined and continuous on the closed interval [0, 2π]. If f E V, then T(f) is defined by setting
T(f)(x) = 0∫2πf(t)sin(x-t)dt, x E [0, 2π]
43. V is the linear space of all real functions f defined and continuous on the closed interval [0, 2π]. If f E V, then T(f) is defined by setting
T(f)(x) = 0∫2πf(t)cos(x-t)dt, x E [0, 2π]
---
(See attached file for full problem description with accurate equations) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.876526951789856, "perplexity": 987.5841346799222}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764495012.84/warc/CC-MAIN-20230127195946-20230127225946-00038.warc.gz"} |
https://ccrma.stanford.edu/~bilbao/master/node206.html | Next: The Interpolated Rectilinear Scheme Up: Finite Difference Schemes for Previous: Finite Difference Schemes for
## The Rectilinear Scheme
The finite difference scheme corresponding to a rectilinear mesh is obtained by applying centered differences to the wave equation, over a rectangular grid with indices and (which refer to points with spatial coordinates and ). The difference scheme, given originally as (4.53) is
(A.14)
and the amplification polynomial equation is of the form (A.5), with
for . From (A.7), we thus have
and we have
Condition (A.8) is thus satisfied, and condition (A.9) gives the bound
for stability
which implies that the amplification factor for such values of . Because , this bound is the same as the bound for passivity of the associated mesh scheme, given in (4.63). The amplification factors, however, are distinct at all spatial frequencies only for . If , then the factors are degenerate for , and for and we are then in the situation discussed in §A.1.2 where linear growth of the solution may occur. This is an important special case, because it corresponds to the standard finite difference scheme for the rectilinear waveguide mesh (i.e., the realization without self-loops). The waveguide mesh implementation does not allow such growth at these frequencies.
As far as assessing the computational requirements of the finite difference scheme, first consider the case . Five adds are required at each grid point in order to update. Given that , we can write the computational and add densities for the scheme as
for
For , however, scheme (A.13) simplifies to
(A.15)
which may be operated on alternating grids, i.e., need only be calculated for even (or odd). The computational and add densities, for are then
for
where we note that the reduced scheme (A.14) requires only four adds for updating at a given grid point; in addition, the multiplies by may be accomplished, in a fixed-point implementation, by simple bit-shifting operations. The increased efficiency of this scheme must be weighed against the danger of instability, and the fact that because grid density is reduced, the scheme is now applicable over a smaller range of spatial frequencies. The numerical phase velocities of the schemes, at the stability limit, and away from it, at , are plotted in Figure A.1. It is interesting to note that away from the stability limit, the numerical dispersion is somewhat less directionally-dependent; this important factor may be useful from the point of view of frequency-warping techniques [157] which may be used to reduce numerical dispersion effects for schemes which are relatively directionally-independent. This idea has been discussed in the waveguide mesh context (where self-loops will be present) in [175].
Next: The Interpolated Rectilinear Scheme Up: Finite Difference Schemes for Previous: Finite Difference Schemes for
Stefan Bilbao 2002-01-22 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9848318696022034, "perplexity": 850.0237699127686}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860116173.76/warc/CC-MAIN-20160428161516-00098-ip-10-239-7-51.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/43985/how-to-cross-reference-a-section-and-item-in-a-list-in-single-document?answertab=oldest | How to cross reference a section and item in a list in single document?
I want to cross reference sections as well as some items in the list in my document. for ex: \label helps us to cross reference the sections. If we renew the command to provide the cross reference for any item in an item, then its not available for section reference. But how can we do both in a single document?
-
There is no difference in using \label-\ref when it comes to sectioning commands or enumerated lists. Here is an example showing how to properly use it:
\documentclass{article}
\begin{document}
\section{A section}\label{sec:label}
Here is some text.
\begin{enumerate}
\item An item
\item Another item \label{enum:label}
\end{enumerate}
Reference to section~\ref{sec:label} and item~\ref{enum:label}.
\end{document}
For more advanced (even automated) enumerated list labelling/referencing, you can use the label and ref options to enumerate provided by the enumitem package.
-
how about custom lists like starlist(*) or bulletlist? – volatNumbers Feb 9 '12 at 18:33
@volatNumbers: For "custom lists" you would need to include a minimal working example (MWE), since there may be many variables at play. Note that referencing requires a counter. If you have no counter, for the reference, you have to do something different. – Werner Feb 9 '12 at 18:38
@Werner: It would be possible to refer to the page where the label is placed, p.~\pageref{customlist:label} (with \phantomsection before the \label{customlist:label}, if hyperref is used) - depends on the wishes of the OP (you named it: MWE!). – Stephen Feb 9 '12 at 18:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8433188199996948, "perplexity": 1909.4471347640274}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042981576.7/warc/CC-MAIN-20150728002301-00041-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://space.stackexchange.com/questions/31314/what-are-common-guidance-strategies-for-a-finite-orbital-maneuver | # What are common guidance strategies for a finite orbital maneuver?
Suppose we have our spacecraft in a circular orbit around the Moon and we want to transfer it to an elliptic orbit with a medium-thrust (10-100N) retrograde burn. Based on the current state estimate, we want to determine a propellant-optimal steering strategy to achieve the transfer. The thrusters operate at full throttle when they're activated, and thrust levels decline with propellant consumption (the propulsion system operates in blow-down mode).
It seems to me that it is best to consider such a problem as a two-point boundary value problem, with the initial and final state vector given by some set of non-singular orbital elements (such as the equinoctial elements), and the dynamics given by some accordingly modified form of the Gauss planetary equations. Are there any standard and/or proven methods to solve this problem in real-time, aboard the spacecraft? Or are there other common guidance strategies that are better recommended?
The requirement of solving this particular guidance problem in real-time on-board the spacecraft is really exigent. As far as I know the usual thing to do is solve the guidance problem on Earth and then uplink it to the spacecraft. Maybe the spacecraft control, understanding control as the ability of the spacecraft to follow the desired path in the presence of disturbances, can be autonomous by some sort of feedback law, MPC or event control. But maybe not because the dynamics are so slow and it may be more simple to manage control from Earth.
Anyway when facing such problems I barely always find direct methods much easier and intuitive to employ rather than the indirect TBVP's. The key idea of direct methods is first discretize and then optimize whereas indirect ones do in opposite order. A good book is "Spacecraft Trajectory Optimization" written by Bruce A.Conway.
• Thanks for your response. In Falck et al. (2014) (arc.aiaa.org/doi/10.2514/6.2014-3714), two closed-loop guidance algorithms for low-thrust vehicles are described and compared. Can't these be used to solve the guidance problem in real-time? Or do I understand their working incorrectly? – woeterb Oct 15 '18 at 9:29
• Taking a glance at this paper, what I see is that it can be implemented in real-time because it employs sub-optimal laws with analytical expressions. If you can find a form to solve your problem analytically, or at most requiring an optimizer to solve a LP or QP with a moderate number of parameters, then ok. – Julio Oct 15 '18 at 12:43 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.830216109752655, "perplexity": 675.7332970744455}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735963.64/warc/CC-MAIN-20200805153603-20200805183603-00190.warc.gz"} |
https://www.physicsforums.com/threads/quick-question-on-the-dirac-delta-function.219072/ | # Quick Question on the Dirac Delta Function
1. Mar 1, 2008
### G01
The Dirac delta function, $$\delta (x)$$ has the property that:
(1) $$\int_{-\infty}^{+\infty} f(x) \delta (x) dx = f(0)$$
Will this same effect happen for the following bounds on the integral:
(2) $$\int_{0}^{+\infty} f(x) \delta (x) dx = f(0)$$
My intuition tells me that it should, but the fact that the peak of the delta function lies on one of the bounds makes me think I should double check my reasoning. So, does anyone know if (2) above is correct?
Thanks for any advice you can offer.
Last edited: Mar 1, 2008
2. Mar 1, 2008
### Vid
According to mathematica integral from 0 to infinity and the integral from negative infinity to 0 are equal to f(0)/2.
3. Mar 1, 2008
### jostpuur
I think I've seen sometimes integration limits like this
$$\lim_{\epsilon\to 0^+}\int\limits_{0-\epsilon}^{\infty}$$
to avoid this problem, which suggests that there probably is not very simple answer to the question. For example
$$\lim_{n\to\infty} \int\limits_0^{\infty} f(x) \sqrt{\frac{n}{\pi}}e^{-nx^2} dx = \frac{1}{2}f(0)$$
but suppose you modify the kernel like this
$$\sqrt{\frac{n}{\pi}} e^{-n(x-x_0(n))^2}$$
where $x_0:\mathbb{N}\to\mathbb{R}$ is some sequence such that $x_0(n)\to 0$ as $n\to\infty$. That still behaves as a delta function if the zero is contained in an open integration interval, but if the zero is on the boundary of the integration interval, the answer could be anything, depending on the chosen $x_0$.
4. Mar 1, 2008
### tiny-tim
$$\delta (x)$$ isn't really a function.
It only has meaning inside $$\int_{-\infty}^{+\infty} f(x) \delta (x) dx$$.
When it is on its own as part of a calculation, that is because it will be put inside such an integral in the next step of the calculation.
This is because it is a symbol which is defined by the property $$\int_{-\infty}^{+\infty} f(x) \delta (x) dx = f(0)$$
Inside $$\int_{0}^{+\infty} f(x) \delta (x) dx$$ , it's a symbol without any meaning.
(Though of course, you can write $$\int_{-\infty}^{+\infty} \theta (x) f(x) \delta (x) dx$$, which looks pretty similar.)
5. Mar 3, 2008
### Rainbow Child
The answer comes by joining the posts of jostpuur and tiny-tim.
The second integral is
$$I=\int_{0}^{+\infty} f(x) \delta (x) dx = \int_{-\infty}^{+\infty} \theta (x) f(x) \delta (x) dx$$
where $\theta(x)$ is the step function. Thus $I$ equals
$$I=\theta(0)\,f(0)[/itex] and the question now is what's the definition of the step function. The one which is most commonly used is that $\theta(0)=\frac{1}{2}$, thus [tex]I=\frac{1}{2}\,f(0)$$
6. Mar 3, 2008
### jostpuur
This doesn't fully make sense. If f is a function, then
$$\int\limits_0^{\infty} f(x) dx = \int\limits_{-\infty}^{\infty} \theta(x)f(x) dx$$
is true with arbitrary value $\theta(0)$. Why should the $\theta(0)=1/2$ be chosen when there is $f(x)\delta(x)$ in the integrand instead?
I think the original question is of similar nature as the question about what 0/0 is supposed to be. As I showed in my first response, if you try to approach this integral with some limits, you can get different results with different ways of taking the limit. Thus, we say that the limits cannot be used to give proper meaning for the this integral.
7. Mar 3, 2008
Because when the integrand contains a Dirac delta, it is not a function, and so the condition "if f is a function" does not apply.
Well, it's more like asking what half of infinity is. But, yeah, to make Dirac deltas at all consistent (let alone rigorous), you need to enforce a definition of them as a limit of some sequence of well-defined functions. As you demonstrated, which exact sequence you choose will affect the answers you get if you ask questions like the one in this thread. That said, most people who are that concerned about consistency forgo the Dirac delta entirely and instead use distributions. The rest of us just live with the fact that certain operations (again, like the OP) aren't defined.
8. Mar 3, 2008
### jostpuur
I'll be more explicit!
Consider the sequence of following functions
$$\delta_{n,1}(x) = \sqrt{\frac{n}{\pi}} \exp\big(-n(x-n^{-1/4})^2\big) = \sqrt{\frac{n}{\pi}} \exp\big(-(\sqrt{n}x -n^{1/4})^2\big)$$
It is easy to believe, that the distributions, represented by these functions, converge towards the delta distribution, because this is just the usual Gaussian peak representation, but displaced slightly. Let a and b be such that $a\leq 0 < b$. Let us calculate the next integral.
$$\lim_{n\to\infty}\int\limits_a^b f(x)\delta_{n,1}(x) dx = \lim_{n\to\infty} \sqrt{\frac{n}{\pi}} \int\limits_a^b f(x)\exp\big(-(\sqrt{n}x-n^{1/4})^2\big) dx = \lim_{n\to\infty} \frac{1}{\sqrt{\pi}} \int\limits_{\sqrt{n}a-n^{1/4}}^{\sqrt{n}b-n^{1/4}} f\big(\frac{y}{\sqrt{n}} + n^{-1/4}\big) e^{-y^2} dy$$
$$= \frac{1}{\sqrt{\pi}} \int\limits_{-\infty}^{\infty} f(0) e^{-y^2} dy = f(0)$$
However, notice that this is indeed true even with a=0.
If we instead chose the sequence
$$\delta_{n,2}(x) = \sqrt{\frac{n}{\pi}} e^{-nx^2}$$
we would have
$$\lim_{n\to\infty} \int\limits_a^b f(x)\delta_{n,2}(x) dx = f(0),\quad\quad a<0<b$$
and
$$\lim_{n\to\infty} \int\limits_a^b f(x)\delta_{n,2}(x) dx = \frac{1}{2}f(0),\quad\quad a=0<b.$$
This means, that the distributions, represented by functions $\delta_{n,1}$ and $\delta_{n,2}$, converge towards the same delta distribution. This is so, because if you want to calculate
$$\int\limits_a^b f(x)\delta(x) dx,\quad\quad a<0<b$$
you can choose either of these sequences, and they give the same result. But when you calculate
$$\int\limits_0^b f(x)\delta(x) dx,$$
choosing different representations for the delta distribution, you get different numbers out of the integral. Thus, the delta distribution does not contain enough information to calculate this integral.
Last edited: Mar 3, 2008
9. Mar 3, 2008
### jostpuur
I was not applying any function assumption to any calculation with distributions. I merely noted, that with integrals of functions, the value of $\theta(0)$ does not matter, and then asked that why should some specific value for it be chosen when we integrate distributions.
Not precisely true. Distributions can be defined without sequences of functions, but the sequences of functions still remain as an important way to handle distributions.
Yes. This was my point. I explained it even more explicitly in my last post #8.
10. Mar 4, 2008
### tiny-tim
The endearing quality of distributions
Because that's how distributions are defined, and what they were created for!
It is distributions' most endearing (and, in my opinion, only endearing ) quality.
11. Mar 4, 2008
### arildno
Not at all.
Let F be some function space over R, I some open interval in R, and define the functional D as follows:
$$D(f,I)=f(0), 0\in{I},f\in{F},D(f,I)=0,0\notin{I},f\in{F}$$
This is the delta functional, and it can be shown to be linear, i.e, a distribution.
This is how you rigorously define the Dirac Delta "function".
12. Mar 4, 2008
### Rainbow Child
But that's the whole point! $\delta(x)$ is a distribution thus it does not behaves like ordinary functions. Your point of view is that the integral does not exist, because it can take multiple values, but I say that it can be defined if you choose the value of $\theta(0)$ yielding to
$$I=\int\limits_0^{\infty} f(x)\,\delta(x) dx = \theta(0)\,f(x)$$
13. Mar 4, 2008
### jostpuur
When you choose the value for theta(0), in effect you are choosing the value of the entire integral. The number, that is supposed to come out of the integral, is not something that should be chosen. It should come out from some calculation. Right now the calculations don't give a unique number, and I don't feel like choosing some unique number by force would be a very good way to deal with this.
Similar Discussions: Quick Question on the Dirac Delta Function | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9586115479469299, "perplexity": 461.92220268549374}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128319912.4/warc/CC-MAIN-20170622220117-20170623000117-00398.warc.gz"} |
https://docs.plasmapy.org/en/latest/formulary/relativity.html | # Relativistic functions (plasmapy.formulary.relativity)¶
Functionality for calculating relativistic quantities ($$v \to c$$).
## Functions¶
Lorentz_factor(V) Return the Lorentz factor. relativistic_energy(m, v) Calculate the relativistic energy (in Joules) of an object of mass m and velocity v. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9715425372123718, "perplexity": 4345.748839635387}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400201826.20/warc/CC-MAIN-20200921143722-20200921173722-00277.warc.gz"} |
https://www.physicsforums.com/threads/proof-of-godels-1st-theorem-missing-o-consistency-requirement-whats-wrong.647039/ | Proof of Godel's 1st Theorem missing ω-consistency requirement. What's wrong?
1. Oct 25, 2012
andrewkirk
Below is a proof of one of the key steps in Godel's first incompleteness theorem. It appears to prove the theorem. However, it doesn't assume that T⋃Q is ω-consistent, which I have read is necessary for the proof to work. The alternative is to use Rosser's Trick to avoid needing to assume ω-consistency. But the proof doesn't do that either.
This leads me to believe that my proof must have an invalid step in it, that requires ω-consistency to validate it. But I cannot see where that would be required. That's probably because my grasp of the concept of ω-consistency is very new and very tenuous.
I think my omission is probably in the part of the proof I've laid out in detail below. But it's possible that it lies somewhere else, like in the Representability Theorem or the couple of steps at the end.
If anybody can help me identify where I have gone wrong, I would appreciate it.
Strong Undecidability of Q
For a collection S of sentences in formal language L, define:
Cn(S) = {#(θ) : Sentence(θ) ⋀ (S⊢ θ)}
where #(F) denotes the Godel number of formula F and Sentence(ψ) means that ψ is a well-formed sentence in L.
The strong undecidability of Q theorem states that for any L-theory T, if T⋃Q is consistent in L, then T is undecidable, by which we mean that Cn(T) is not recursive.
Here Q denotes Robinson Arithmetic.
Proof
Assume T⋃Q is consistent in L and Cn(T⋃Q) is recursive (meaning that the relation that defines it as a subset of ω is recursive, aka μ-recursive).
Because of the recursivity of Cn(T⋃Q), the Representability Theorem tells us that the relation it defines must be representable in Q, which means there must exist a formula BewT⋃Q in wff(LN) with one free variable, say c, such that, for any θ in wff(LN):
A. (#(θ)∈Cn(T⋃Q)) ⇒ (Q⊢BewT⋃Q[c:=#(θ)] and
B. (#(θ)∉Cn(T⋃Q)) ⇒ (Q⊢¬BewT⋃Q[c:=#(θ)]).
Bew is short for the German word beweisbar, meaning provable.
Note that representability of Cn(T⋃Q) is a bigger requirement than it might at first seem, as any wff must have finite length and, since both sets Cn(T⋃Q) and ω - Cn(T⋃Q) are infinite, this precludes BewT⋃Q from being a simple infinite list of all numbers in Cn(T⋃Q). BewT⋃Q doesn't have to be recursive, but it must be concise.
Now by the Diagonal Lemma, applied to the wff (¬BewT⋃Q), we know there exists a sentence G in wff(LN) such that:
1. Q ⊢ (G↔¬BewT⋃Q[c:=#(G)])
This appears to be a theorem saying that G is true in T iff it is not provable in T, which immediately arouses suspicion. Let us examine this formally:
2. T ⊢G [Hypothesis]
3. #(G)∈Cn(T) [from previous line, by definition of Cn(T)]
4. #(G)∈Cn(T⋃Q) [as Cn(T)⊂Cn(T⋃Q)]
5. Q ⊢BewT⋃Q[c:=#(G)]) [from A. above]
6. Q ⊢¬G [from lines 1 and 5, via Modus Ponens]
7. T⋃Q ⊢¬G [as Q⊂T⋃Q]
8. T⋃Q ⊢¬G [from line 2, as Q⊂T⋃Q]
Hence T⋃Q⊢(G→(G⋀¬G)), so if T⋃Q is consistent, we must have:
9. T⋃Q⊬G
However it then follows that:
10. #(G)∉Cn(T⋃Q) [from previous line, by definition of Cn(T⋃Q)]
11. Q ⊢¬BewT⋃Q[c:=#(G)] [from B. above]
12. Q ⊢G [by lines 1 and 11, via Modus Ponens]
13. T⋃Q ⊢G [from previous line, as Q⊂T⋃Q]
So we have (T⋃Q⊢G)⋀¬(T⋃Q⊢G), which is a contradiction outside T.
Hence we must conclude that one of the assumptions we have made is false. If we insist on retaining the assumption of consistency then the only other available assumption is the one that Cn(T⋃Q) is recursive, so we must reject that assumption.
A bit more argument, which I've omitted here, shows that if Cn(T⋃Q) is not recursive then neither is Cn(T).
That means that T is undecidable and if we assume T is axiomatisable then it follows that T is not complete.
2. Oct 26, 2012
Preno
Maybe it would be easier if you just put your questions concerning the proof of Godel's theorem in a single thread?
Anyway, Rosser's trick is used in the proof of the Representability Theorem. Without Rosser's trick, you would get the Representability Theorem in the following form:
#(θ)∈Cn(T⋃Q)) ⇒ (Q⊢BewT⋃Q[c:=#(θ)]
#(θ)∉Cn(T⋃Q)) ⇒ ¬ (Q⊢BewT⋃Q[c:=#(θ)]
rather than in the form that you use:
#(θ)∈Cn(T⋃Q)) ⇒ (Q⊢BewT⋃Q[c:=#(θ)]
#(θ)∉Cn(T⋃Q)) ⇒ (Q⊢¬BewT⋃Q[c:=#(θ)]
3. Oct 27, 2012
andrewkirk
Thanks Preno. I thought it might be in the Representability Theorem. I'll go and work through the version of the proof of that that I have, now, to see if I can find where either omega-consistency is used or Rosser's Trick is built in.
If I have more questions I'll just start a thread called "Questions about Godel's Incompleteness Theorem" and put them there. I'd rename this thread to that but the OP is already locked. They lock posts quickly around here. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9604184627532959, "perplexity": 1488.2434374576858}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267160754.91/warc/CC-MAIN-20180924205029-20180924225429-00099.warc.gz"} |
http://www.gnu.org/software/auctex/manual/auctex/Environments.html | [ < ] [ > ] [ << ] [ Up ] [ >> ] [Top] [Contents] [Index] [ ? ]
## 2.4 Inserting Environment Templates
A large apparatus is available that supports insertions of environments, that is ‘\begin{}’ — ‘\end{}’ pairs.
AUCTeX is aware of most of the actual environments available in a specific document. This is achieved by examining your ‘\documentclass’ command, and consulting a precompiled list of environments available in a large number of styles.
You insert an environment with C-c C-e, and select an environment type. Depending on the environment, AUCTeX may ask more questions about the optional parts of the selected environment type. With C-u C-c C-e you will change the current environment.
Command: LaTeX-environment arg
(C-c C-e) AUCTeX will prompt you for an environment to insert. At this prompt, you may press <TAB> or <SPC> to complete a partially written name, and/or to get a list of available environments. After selection of a specific environment AUCTeX may prompt you for further specifications.
If the optional argument arg is not-nil (i.e. you have given a prefix argument), the current environment is modified and no new environment is inserted.
As a default selection, AUCTeX will suggest the environment last inserted or, as the first choice the value of the variable LaTeX-default-environment.
User Option: LaTeX-default-environment
Default environment to insert when invoking ‘LaTeX-environment’ first time.
If the document is empty, or the cursor is placed at the top of the document, AUCTeX will default to insert a ‘document’ environment.
Most of these are described further in the following sections, and you may easily specify more. See Customizing Environments.
You can close the current environment with C-c ], but we suggest that you use C-c C-e to insert complete environments instead.
Command: LaTeX-close-environment
(C-c ]) Insert an ‘\end’ that matches the current environment.
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http://physics.stackexchange.com/questions/46928/constructing-the-space-of-quantum-states | # Constructing the space of quantum states
I want to learn how to construct spaces of quantum states of systems.
As an exercize, I tried to build the space of states and to find hamiltonian spectrum of the quantum system whose Hamiltonian is the Hamiltonian of the harmonic oscillator with the quadratic term: $\hat{H}=\hat{H}_{0}+\hat{H}_{1}$, where
$\hat{H}_{0}=\hbar\omega\left(\hat{a}^{\dagger}\hat{a}+1/2\right)$, $\hat{H}_{1}=\dot{\imath}\gamma\left(\hat{a}^{\dagger}\right)^{2}-\dot{\imath}\gamma\left(\hat{a}\right)^{2}$; $\hat{a}$, $\hat{a}^{\dagger}$-ladder operators, $\gamma$-real parameter
For this purpose, we should define a complete set of commuting observables (CSCO).
As for the harmonic oscillator, we can define a "number" operator $N=\hat{a}^{\dagger}\hat{a}$.
We can prove the following statement:
Let be $a$ and $a^{\dagger}$ Hermitian conjugated operators and $\left[a,a^{\dagger}\right]=1$. Define operator $N=aa^{\dagger}$. Then we can prove that $\left[N,a^{p}\right]=-pa^{p}, \left[N,a^{\dagger p}\right]=pa^{\dagger p}$ and that the only algebraic functions of $a$ and $a^{\dagger}$, which commute with $N$, are the functions of $N$. (For example, see Messiah, Quantum Mechanics, exercises after chapter $12$)
Using this statement, we conclude(am i right?) that operator $N$ forms a CSCO. So, sequence of eigenvectors of operator $N$ forms the basis of the space of states. So, I've come to the conclusion that the space of states of the described system is the same as the space of states of harmonic oscillator.
But operators $a$, $a^{\dagger}$ can always be determined (as i think), so this arguments will be valid, so I've come to the conclusion, that spaces of states of all systems will be the same. After that I realized that I am mistaken.
Would you be so kind to explain where is a mistake in the arguments above? And can you give some references/articles/books where i can read some additional information about constructing spaces of states for different systems?
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http://lists.gnu.org/archive/html/help-emacs-windows/2005-11/msg00042.html | help-emacs-windows
[Top][All Lists]
## Re: [h-e-w] Unix utilities for Emacs on MS Windows
From: Lennart Borgman Subject: Re: [h-e-w] Unix utilities for Emacs on MS Windows Date: Sat, 26 Nov 2005 11:12:07 +0100 User-agent: Mozilla Thunderbird 1.0.7 (Windows/20050923)
```Ismael Valladolid Torres wrote:
```
```Not much feedback to give. Windows is a dangerous place to live in and
in order to avoid viruses and spyware I usually prefer being a non
form of a zipped or tarred archive I can simply uncompress it and add
the resulting bin to my personal path. If I needed to run an installer,
before I would need to contact my IT department in order to become an
```
I tried to write the installer in such a way that you do not need admin priv.
```
```
```Moreover I am simply used to UNIX way of life where all the
configuration goes to plain text files under /etc or under your home in
form of dotfiles. Usually I am not aware of how much garbage an
installer puts in my registry, so better simply uncompress things and
run them being in full control of their configuration.
```
That makes me remember someone wanted a list of all changes to the registry. I forgot to make such a list (which will be very short).
```
```
```My way: Getting a compiled emacs archive, uncompressing it to C:\emacs,
adding C:\emacs\bin to my path, setting HOME to C:\cygwin\home\ismaeval2
and putting there my .emacs, adding a shortcut to gnuclienw.exe to my
SendTo folder, and that's it!
```
You need to start gnuserv too yourself I guess when you do it this way. For me as a typical command line user it is also important to make it possible to start editing a new file from the command line so I add that also in the installer. (And I add a shortcut to the SendTo folder too;-)
```
Why do you add Emacs to your path? Is that needed outside of Emacs?
```
```Hope this makes sense. :)
```
Yes, it surely does. When I wrote EmacsW32 I tried to target both new and old users, but that is not very easy ;-)
```
``` | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.93787682056427, "perplexity": 2207.9062059052476}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207929096.44/warc/CC-MAIN-20150521113209-00309-ip-10-180-206-219.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/110557/about-the-bloch-conjecture-on-entire-curves | About the Bloch conjecture on entire curves
The Bloch conjecture states the following:
Bloch's conjecture. Let $X$ be a compact complex Kähler variety such that the irregularity $q = h^0(X,\Omega^1_X)$ is larger than the dimension $n = \dim X$. Then, every entire curve drawn in $X$ is analytically degenerate.
Here $X$ may be singular and $\Omega^1_X$ can be defined in any reasonable way (direct image of the $\Omega^1_{\widetilde X}$ of a desingularization $\widetilde X$ or direct image of $\Omega^1_U$ where $U$ is the set of regular points in the normalization of $X$). By an entire curve I mean a non constant holomorphic map $f\colon\mathbb C\to X$, and analytically degenerate means that there exists a closed analytic subset $Y\subsetneq X$ such that $f(\mathbb C)\subset Y$.
This conjecture has been proven, thanks to the works of Ochiai, by Kawamata and, independently, by Wong.
A standard Albanese map argument permits to reduce the conjecture to the following statement:
Let $A$ be an abelian variety and $f\colon\mathbb C\to A$ an entire curve. Then, the Zariski closure $\overline{f(\mathbb C)}$ is a translate of a subtorus.
In particular a subvariety of an abelian variety does not have any entire curve (Brody hyperbolicity) if and only if it does not contain any translate of a subtorus. Thus, in a simple abelian variety every subvariety is hyperbolic. More generally, if a subvariety of an abelian variety is not a translate of a subtorus, then every entire curve in $X$ is analytically degenerate.
Question 1. Is there any geometric characterization or sufficient condition in order to insure that the Albanese variety of a projective algebraic manifold is simple?
Question 2. Is there any geometric characterization or sufficient condition, other than having big irregularity, in order to insure that the image of a projective algebraic manifold via the Albanese map is a proper subvariety?
Notice that by the universal property of the Albanese map, if the image of a projective algebraic manifold via the Albanese map is a proper subvariety then this image is necessarily not a translate of a subtorus.
N.B. I changed the last part of my post. Now, Question 2 is no more as in the previously, not reedited post. In particular the comment of ulrich refers to my previous Question 2.
-
What is an analytically degenerate curve? – aglearner Oct 24 '12 at 16:42
I define this at line 7 from above. – diverietti Oct 24 '12 at 16:42
A subvariety $X$ of abelian variety $A$ is of general type unless there is an abelian subvariety $B$ of $A$, translation by which preserves $X$; this is proved in Ueno's Springer Lecture Notes on classification theory, but I do not have the precise reference right now. If the abelian variety is not simple, then using products you can construct subvarieties with arbitrarily large irregularity but not of general type. – ulrich Oct 24 '12 at 18:43
By "Abelian Variety", do you mean a complete, irreducible and reduced group scheme of finite type over $\mathbb{C}$? – Filippo Alberto Edoardo Oct 25 '12 at 8:37
Ahahahahahahahahahahahah. Exactly! – diverietti Oct 25 '12 at 8:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9361647367477417, "perplexity": 283.07993609480064}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414119646180.24/warc/CC-MAIN-20141024030046-00258-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/133638/how-does-this-equation-to-find-the-radius-from-3-points-actually-work | # How does this equation to find the radius from 3 points actually work?
I had searched online and found an equation that solves the radius of a circle from 3 points that are located on the circumference of that specific circle. Where I had found this formula did not state its derivation or anything of the likes, however it is to find the radius. With the 3 points, you should form a triangle. Let us call this Triangle $ABC$.
Using the distance formula, we can calculate the distance between $AB$, $BC$, and $AC$. To simplify things, let us call these distances A, B, and C respectively. We also need the area of the triangle. After finding an altitude of the specified triangle, we can use the area of a triangle equation to solve for it. Let us call the area of this triangle $K$.
This is the formula:
$r = \dfrac{ABC}{4K}$
Which is essentially saying
$radius = \dfrac{\text{Product~of~the~triangle~side~lengths}}{\text{The~area~of~the~triangle~multiplied~by~4}}$
This above pseudo equation was just to clarify the formula, I understand everyone here are exceptional mathematicians, and as I mature, I also hope to become one as well.
This, to my astonishment, finds the correct radius, however I am unable to comprehend how this method works. I hope one of the kind people on Math StackExchange are willing to help me out on understanding this formula and its derivation. Many thanks.
-
the radius the same as a distance from a vertex to the 'circumcentre' of the triangle and simplifies to the Law of Sines. en.wikipedia.org/wiki/Law_of_sines#Relation_to_the_circumcircle – Ronald Apr 18 '12 at 21:58
@Ronald yes, and normally, to find the circumcentre I would find the perpendicular bisector of 2 sides, and the POI of those 2 lines would be the circumcentre. However, in this method I can't understand how it works to find the radius. – Backslash Apr 18 '12 at 22:01
@Ronald if you could explain it in a little more depth I would be very grateful. – Backslash Apr 18 '12 at 22:06
Let $a$ be the angle opposite to side $A$. First show that if $R$ is the radius of the circle, then $\frac{A}{\sin a} = 2R$. This isn't hard (just drop a perpendicular from $O$ to $A$, and use the definition of $\sin$ on the similar triangles).
Then, the area of the triangle $ABC$, $K = \frac{1}{2}BC \sin a$, and so $\frac{A}{\sin a} = \frac{A}{\frac{2K}{BC}} = \frac{ABC}{2K} = 2R$, and so $\frac{ABC}{4K} = R$.
I have only one question, however. Why is there a $\text sina$ after the area of a triangle equation? Is that supposed to be there? – Backslash Apr 19 '12 at 3:18
That is the general formula for the area of a triangle in terms of the side lengths and angles. If we have a right triangle, then $\sin a = 1$ and this reduces to base x height. en.wikipedia.org/wiki/Triangle#Computing_the_area_of_a_triangle – Michael Biro Apr 19 '12 at 4:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9002147316932678, "perplexity": 183.6363178538954}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375097710.59/warc/CC-MAIN-20150627031817-00199-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://stats.stackexchange.com/questions/370147/the-universal-approximation-theorem-vs-the-no-free-lunch-theorem-whats-the-ca | # The Universal Approximation Theorem vs. The No Free Lunch Theorem: What's the caveat?
The universal approximation theorem:
A neural network with 3 layers and suitably chosen activation functions can any approximate continuous function on compact subsets of $$R^n$$.
The no free lunch theorem:
If a learning algorithm performs well on some data sets, it will perform poorly on some other data sets.
I sense a contradiction here: the first theorem implies that NNets are the "one learning approach to rule them all", while second says that such a learning approach doesn't exist.
I'm pretty certain NFLT holds, so there must be a caveat, but I can't put my finger on it?
What is the caveat in the universal approximation theorem so that NFLT holds?
• The first theorem does not imply that NNets are the "one learning approach to rule them all". There are other functions that can approximate any continuous function on compact subsets of $R^n$, see the Stone-Weierstrass theorem for example, and then there's the issue of generalizability, which the UAT does not address. – jbowman Oct 4 '18 at 17:08
• Stone-Weierstrass theorem - means polynomial linear regression also satisfies UAT – seanv507 Oct 4 '18 at 17:22
• While what you're saying is correct in terms of traininggeneralization, I don't think it correctly addresses the question. The NFL theorem has nothing to do with generalization. Rather it says that if you sample functions from a uniform distribution of the input/output space, then the overall expected cost of any algorithm will be the same at iteration step $m$. – Alex R. Oct 4 '18 at 17:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8995171189308167, "perplexity": 436.46732334541}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247486480.6/warc/CC-MAIN-20190218114622-20190218140622-00051.warc.gz"} |
https://learn.careers360.com/ncert/question-the-slant-height-of-a-frustum-of-a-cone-is-4-cm-and-the-perimeters-circumference-of-its-circular-ends-are-18-cm-and-6-cm-find-the-curved-surface-area-of-the-frustum/ | # 2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
We are given the perimeter of upper and lower ends thus we can find r1 and r2.
$2\pi r_1\ =\ 18$
$r_1\ =\ \frac{9}{\pi}\ cm$
And,
$2\pi r_2\ =\ 6$
$r_2\ =\ \frac{3}{\pi}\ cm$
Thus curved surface area of the frustum is given by : $=\ \pi \left ( r_1\ +\ r_2 \right )l$
$=\ \pi \left ( \frac{9}{\pi}\ +\ \frac{3}{\pi} \right )4$
$=\ 48\ cm^2$
Exams
Articles
Questions | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.956436276435852, "perplexity": 541.526318994339}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347425148.64/warc/CC-MAIN-20200602130925-20200602160925-00527.warc.gz"} |
https://lazyprogrammer.me/deep-learning-tutorial-part-13-logistic/ | # Deep Learning Tutorial part 1/3: Logistic Regression
April 22, 2015
This is part 1/3 of a series on deep learning and deep belief networks. I’ve wanted to do this for a long time because learning about neural networks introduces a lot of useful topics and algorithms that are useful in machine learning in general.
Unfortunately, while the material I’ve read focusing on logistic regression and the multiple layer perceptron (building blocks of the deep belief network) are great and accessible to a wide audience, I’ve found most of the material I’ve encountered about deep learning are highly technical and hard to follow.
So, I’ve decided to create this series in order to teach the most practical aspects of deep learning and neural networks – enough so that you can implement one yourself, but not so much that you’ll get bogged down by all the theory.
Part 1 will focus on logistic regression. Part 2 will focus on the multilayer perceptron (a.k.a. artificial neural network) and backpropagation. Part 3 will focus on restricted Boltzmann machines and deep networks. Each is designed to be a stepping stone to the next.
The topic of this post (logistic regression) is covered in-depth in my online course, Deep Learning Prerequisites: Logistic Regression in Python. We derive all the equations step-by-step, and fully implement all the code in Python and Numpy. To solidify the concepts, we apply the method to real world datasets, including an e-commerce dataset and facial expression recognition.
Let us begin.
Logistic Regression doesn’t do Regression
Despite its name, Logistic Regression is actually a classification algorithm.
This means the output gives us a label, not a real number.
HOWEVER: the methods you read about in this series can be applied to both regression and classification. Just the equations for the outputs and the error function differ. I will note these differences where appropriate, but the tutorials will focus on classification.
Diagram of how Logistic Regression works
I’ve included a few pictures here so you get used to looking at how we visualize a neural network.
Here’s one where X (input) is 3-dimensional and Y (output) is 2-dimensional.
Here’s one where the weights use the symbol theta and the summation operation and sigmoid function are shown explicitly.
Here’s one where the weights use the variable “w” and the bias is explicitly shown as “b”. Here the sigmoid function uses the Greek letter “phi”, but more often you see the letter “sigma”.
A Little Math
So what do these diagrams mean about how we calculate the output from a set of inputs?
Notice first that we can have more than one output Y.
For K classes/labels, as in the digit recognition problem, we would have K outputs, and Y(k) = 1 if the label is the kth digit, otherwise it is 0.
The only exception is the 2-class case. In this situation, we only need 1 output because Y = 1 is the first class and Y = 0 is the second class.
We’ll focus on this scenario first.
The equation in its compact form is this:
The inside part is the dot product of the weights and the input:
As in linear regression we assume there is an x0 and that it is 1.
The “sigma” part is the sigmoid function:
If we graph the sigmoid, it looks like this:
There are 2 things we can tell from the above equation:
1) For logistic regression to work, the classes must be linearly separable. This is because the dot product between “w” and “x” is a line/plane.
(i.e. ax + by + c = 0)
w0 + w1x1 + w2x2 + … = 0 is the plane (more correctly, hyperplane) here.
So here is a situation where logistic regression would work well:
Here is a situation where it wouldn’t work well.
But we will cover that more in parts 2 and 3.
2) The sigmoid means the output Y is between 0 and 1.
So if w*x = 0, we land right on the hyperplane, and Y = 0.5.
If w*x > 0, we get Y > 0.5, and vice versa for w*x < 0.
As w*x approaches infinity, Y approaches 1, and vice versa.
Probabilistic Interpretation
Because Y is between 0 and 1, we can interpret it as a probability.
This makes more sense if you consider the following:
If we fall right on the barrier/plane between the two classes – our probability of being in either class is 0.5.
If we are further away from that barrier, the probability of being in either class increases.
We usually denote Y as P(Y=1|X) and P(Y=0|X).
Note that while we use some probabilistic concepts here, the way in which we use them is different than for say, a Bayesian classifier.
Also note that P(Y=0|X) = 1 – P(Y=1|X).
Maximizing the Likelihood
We have seen squared error used as an error function before, as with linear regression.
In fact, if we were doing regression, we could use the same thing here.
For classification, we take a different approach.
You may have seen this error function before:
t is the target and y is the output of the network/model. (This introduces some ambiguity because we usually write p(y=1 | x) as the output and y as the target).
This is called the cross-entropy error.
Where does this come from?
Let us go back to first principles.
Instead of minimizing error, we maximize likelihood. This seems like a logical place to start – maximizing the probability that our model parameters are correct.
Consider N IID (independent and identically distributed) training samples and corresponding labels (we’ll call them “t” here).
$$L = \prod_{i=1}^{N} y_i ^{t_i} (1 – y_i)^{1 – t_i}$$
The likelihood of the model given the entire dataset can be represented by this equation. We can use the product rule because each sample is independent.
(Sidenote 1: This is the same thing we do when we want to say, find the maximum likelihood estimate for the mean. We calculate the joint probability aka. likelihood P(data|mean) and find the “argmax” mean that gives us the highest likelihood, hence the term – “maximum likelihood”)
(Sidenote 2: This is the same likelihood you see when we do Bayesian inference – posterior ~ likelihood x prior or P(param | data) ~ P(data | param) P(param))
(Sidenote 3: If you wanted to do regression, you would simply not have a sigmoid at the end, and you would use the squared error. The exponential of the squared error is a Gaussian, because in regression we often assume the error is Gaussian distributed. By making these 2 changes, we would just be doing linear regression.)
Recall y = P(y=1|x).
The target t can be 1 or 0.
When t is 1, only the left part of the product matters (the right side evaluates to 1). All the y’s here are the probability that the output of the network is 1. Given that the target is 1, we want to maximize this probability.
When t is 0, only the right part of the product matters. Recall that 1-y is the probability that the output of the network is 0. So when t = 0, we want to maximize this probability.
Since each sample is independent, we can get the joint probability by multiplying all the individual probabilities together.
2 key points:
1) There is no analytic solution, we must use iterative methods. In this tutorial we will cover gradient descent, but there are others (such as conjugate gradient, and L-BFGS).
The added advantage of learning gradient descent now is that it is also used to train neural networks.
2) As is usual with these ML problems, we will work with the log likelihood instead of the likelihood. Just try taking the derivative of both, and you will see why.
If you take the log of the above expression, notice you’ll get the same error function we started with!
We take the negative because we want something to minimize. We call this the “error” or “cost” function.
Maximizing the likelihood is equivalent to minimizing the negative likelihood.
It is also equivalent to minimizing the negative log-likelihood. This is because log() is a monotonically increasing function.
How do we actually minimize the negative log-likelihood if we can’t simply set the derivative = 0 and solve for the weights?
This is where gradient descent enters the picture.
Note that gradient descent is just a numerical method – it can be applied whenever you want to solve for the minima of a function, not just for machine learning.
Here is a picture of what we’re trying to do:
We start at some random weight, w = random().
Then we update the weight by going in the direction of the derivative of the error function (slope), which we have previously stated is the negative log-likelihood.
With squared error it is easy to see that the error function is quadratic, and so we are descending down a parabola in that case. The minimum is global.
With log-likelihood the extremum is also global.
It may help to plot the function E(y,t) = tlog(y) + (1-t)log(1-y) to see why.
The equation for updating the weights is:
Here j indexes the dimension, so j = 1…D.
t indexes the iteration number (not to be confused with the other t, which was the target).
“Eta” is called the “learning rate”. This hyperparameter determines how far along the error surface we travel on each iteration. Bigger values mean we go further, which means our weights might converge to the final solution faster, but it also means we may “overshoot” that solution.
Since w is a vector, we can usually speed up our code by doing vector operations (i.e. in MATLAB or Python). In this case, we can use this equation:
The full training algorithm is:
for i = 1…number of epochs:
error = negative log-likelihood ( -L(Y|X,w) )
w = w – learning rate * error gradient
The number of epochs is yet another hyperparameter. There are many ways to determine when to stop the gradient descent process.
Some other methods you may want to look into:
• Stopping when the gradient is small enough
• Stopping when the training error is no longer decreasing or approaching 0
• Stopping when the error on a held out test set starts to increase (overfitting)
We call things like learning rate and epochs “hyperparameters”. These are parameters that are not part of the model itself, but can still be optimized, perhaps via cross-validation.
Biological Inspiration
In computational neuroscience, a logistic regression unit is sometimes referred to as a “neuron”. How are the two related?
Here is a diagram of a typical neuron.
Some notable components:
• Dendrites: These are the “inputs” into the neuron – they take electrical signals from other neurons’ axons.
• Cell body / Nucleus: This part of the neuron “sums up” all the inputs and propagates this summed signal to the axon.
• Axon: This is the “output” of the neuron. It sends the signal from this neuron to other neurons’ dendrites.
So dendrites are our logistic unit’s X, and axons are the Y.
The brain is essentially a network of neurons, or rather, a neural network. An artificial neuron network, which is the topic discussed in Part 2 of this tutorial, is a network of connected logistic regression units.
Another notable feature of neurons is the behavior of the “action potential”.
Observe a typical amplitude/potential (voltage) vs. time signal:
Notice how the potential rises gradually and then spikes. We call this the “all-or-nothing” principle. If the sum of the inputs to the neuron is high enough, a spike is generated. Otherwise, the voltage stays relatively low.
This is reflected in the logistic units’ binary output. The output if a sigmoid is interpreted as P(Y=1|X) – the probability of being “on”, or in other words, the probability that a spike is generated.
Inhibitory vs. Excitatory neurons:
It is well-known that the signal a neuron sends can either “excite” or “inhibit” the receiving neuron. These are reflected in the logistic model by the weights. A positive weight is excitatory. A negative weight is inhibitory.
Researchers have tried to create models with “spiking” neurons, however, it has been difficult to get them to actually learn anything.
The topic of this post (logistic regression) is covered in-depth in my online course, Deep Learning Prerequisites: Logistic Regression in Python. We derive all the equations step-by-step, and fully implement all the code in Python and Numpy. To solidify the concepts, we apply the method to real world datasets, including an e-commerce dataset and facial expression recognition.
#ann #dbn #deep belief networks #gradient descent #mlp #Multilayer Perceptron #neural networks #rbm #restricted Boltzmann machines
### Deep Learning and Artificial Intelligence Newsletter
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# Funky Mathematical
Physics Concepts
The Anti-Textbook*
A Work In Progress. See physics.ucsd.edu/~emichels for the latest versions of the Funky Series.
By Eric L. Michelsen
T
ijx
v
x
T
ijy
v
y
T
ijz
v
z
+
dR
I
real
Imaginary
I
I
R
i
-i
I study mathematics to learn how to think.
I study physics to have something to think about.
Perhaps the greatest irony of all is not that the square root of two is
irrational, but that Pythagoras himself was irrational.
* Physical, conceptual, geometric, and pictorial physics that didnt fit in your textbook.
Please cite as: Michelsen, Eric L., Funky Mathematical Physics Concepts, physics.ucsd.edu/~emichels, 8/1/2012.
2006 values from NIST. For more physical constants, see http://physics.nist.gov/cuu/Constants/ .
Speed of light in vacuum c = 299 792 458 m s
1
(exact)
Boltzmann constant k = 1.380 6504(24) x 10
23
J K
1
Stefan-Boltzmann constant = 5.670 400(40) x 10
8
W m
2
K
4
Relative standard uncertainty 7.0 x 10
6
A
, L = 6.022 141 79(30) x 10
23
mol
1
Relative standard uncertainty 5.0 x 10
8
Molar gas constant R = 8.314 472(15) J mol
-1
K
-1
Electron mass m
e
= 9.109 382 15(45) x 10
31
kg
Proton mass m
p
= 1.672 621 637(83) x 10
27
kg
Proton/electron mass ratio m
p
/m
e
= 1836.152 672 47(80)
Elementary charge e = 1.602 176 487(40) x 10
19
C
Electron g-factor g
e
= 2.002 319 304 3622(15)
Proton g-factor g
p
= 5.585 694 713(46)
Neutron g-factor g
N
= 3.826 085 45(90)
Muon mass m
## = 1.883 531 30(11) x 10
28
kg
Inverse fine structure constant
1
= 137.035 999 679(94)
Planck constant h = 6.626 068 96(33) x 10
34
J s
Planck constant over 2 = 1.054 571 628(53) x 10
34
J s
0
= 0.529 177 208 59(36) x 10
10
m
Bohr magneton
B
= 927.400 915(23) x 10
26
J T
1
Reviews
... most excellent tensor paper.... I feel I have come to a deep and abiding understanding of
relativistic tensors.... The best explanation of tensors seen anywhere! -- physics graduate student
physics.ucsd.edu/~emichels Funky Mathematical Physics Concepts emichels at physics.ucsd.edu
Contents
Introduction..................................................................................................................................... 7
Why Funky? .................................................................................................................................. 7
How to Use This Document ........................................................................................................... 7
Why Physicists and Mathematicians Dislike Each Other ............................................................. 7
Thank You..................................................................................................................................... 7
Scope............................................................................................................................................. 7
Notation......................................................................................................................................... 8
Random Topics .............................................................................................................................. 11
Whats Hyperbolic About Hyperbolic Sine?................................................................................. 11
Basic Calculus You May Not Know............................................................................................. 12
The Product Rule ......................................................................................................................... 14
Integration By Pictures............................................................................................................. 14
Theoretical Importance of IBP.................................................................................................. 15
Delta Function Surprise................................................................................................................ 16
Spherical Harmonics Are Not Harmonics ..................................................................................... 18
The Binomial Theorem for Negative and Fractional Exponents..................................................... 19
When Does a Divergent Series Converge?.................................................................................... 20
Algebra Family Tree .................................................................................................................... 21
Convoluted Thinking ................................................................................................................... 22
Vectors ........................................................................................................................................... 23
Small Changes to Vectors ............................................................................................................ 23
Why (r, , |) Are Not the Components of a Vector ....................................................................... 23
Laplacians Place......................................................................................................................... 24
Vector Dot Grad Vector ............................................................................................................... 32
Greens Functions.......................................................................................................................... 34
Complex Analytic Functions.......................................................................................................... 46
Residues ...................................................................................................................................... 47
Contour Integrals ......................................................................................................................... 48
Evaluating Integrals ..................................................................................................................... 48
Choosing the Right Path: Which Contour?................................................................................ 50
Evaluating Infinite Sums.............................................................................................................. 56
Multi-valued Functions ................................................................................................................ 58
Conceptual Linear Algebra ........................................................................................................... 60
Matrix Multiplication ............................................................................................................... 60
Determinants ............................................................................................................................... 61
Cramers Rule.......................................................................................................................... 62
Area and Volume as a Determinant........................................................................................... 63
The Jacobian Determinant and Change of Variables.................................................................. 64
Expansion by Cofactors............................................................................................................ 66
Proof That the Determinant Is Unique....................................................................................... 68
Getting Determined.................................................................................................................. 69
Getting to Home Basis ............................................................................................................. 70
Contraction of Matrices............................................................................................................ 73
Trace of a Product of Matrices.................................................................................................. 73
Linear Algebra Briefs................................................................................................................... 74
Probability, Statistics, and Data Analysis...................................................................................... 75
Probability and Random Variables ............................................................................................... 75
Precise Statement of the Question Is Critical............................................................................. 76
physics.ucsd.edu/~emichels Funky Mathematical Physics Concepts emichels at physics.ucsd.edu
How to Lie With Statistics ........................................................................................................... 77
Choosing Wisely: An Informative Puzzle..................................................................................... 77
Multiple Events............................................................................................................................ 78
Combining Probabilities........................................................................................................... 79
To B, or To Not B? .................................................................................................................. 81
Continuous Random Variables and Distributions.......................................................................... 82
Population and Samples ........................................................................................................... 83
Variance .................................................................................................................................. 83
Standard Deviation................................................................................................................... 84
New Random Variables From Old Ones ................................................................................... 84
Some Distributions Have Infinite Variance, or Infinite Average.................................................... 86
Samples and Parameter Estimation............................................................................................... 87
Combining Estimates of Varying Uncertainty........................................................................... 90
Statistically Speaking: What Is The Significance of This?............................................................. 90
Predictive Power: Another Way to Be Significant, but Not Important........................................ 93
Bias and the Hood (Unbiased vs. Maximum-Likelihood Estimators)............................................ 94
Correlation and Dependence ........................................................................................................ 96
Data Fitting (Curve Fitting).......................................................................................................... 97
Goodness of Fit ........................................................................................................................ 99
Fitting To Histograms ................................................................................................................ 103
Guidance Counselor: Practical Considerations for Computer Code to Fit Data ............................ 107
Numerical Analysis...................................................................................................................... 110
Round-Off Error, And How to Reduce It .................................................................................... 110
How To Extend Precision In Sums Without Using Higher Precision Variables ........................ 111
Numerical Integration............................................................................................................. 112
Sequences of Real Numbers ................................................................................................... 112
Root Finding.............................................................................................................................. 112
Simple Iteration Equation....................................................................................................... 112
Newton-Raphson Iteration...................................................................................................... 114
Pseudo-Random Numbers.......................................................................................................... 116
Generating Gaussian Random Numbers.................................................................................. 117
Generating Poisson Random Numbers.................................................................................... 118
Generating Weirder Random Numbers ................................................................................... 119
Exact Polynomial Fits ................................................................................................................ 119
Twos Complement Arithmetic .................................................................................................. 121
How Many Digits Do I Get, 6 or 9?............................................................................................ 122
How many digits do I need? ................................................................................................... 123
How Far Can I Go? ................................................................................................................ 123
Software Engineering................................................................................................................. 124
Object Oriented Programming................................................................................................ 124
The Best of Times, the Worst of Times ...................................................................................... 125
Cache Withdrawal: Matrix Multiplication............................................................................... 130
Cache Summary..................................................................................................................... 132
IEEE Floating Point Formats And Concepts ............................................................................... 132
Precision in Decimal Representation....................................................................................... 140
Underflow.............................................................................................................................. 141
Fourier Transforms and Digital Signal Processing..................................................................... 147
Model of Digitization ............................................................................................................. 148
Complex Sequences and Complex Fourier Transform............................................................. 148
Sampling................................................................................................................................ 148
Basis Functions and Orthogonality ......................................................................................... 150
Real Sequences ...................................................................................................................... 151
Normalization and Parsevals Theorem................................................................................... 152
Continuous and Discrete, Finite and Infinite ........................................................................... 153
physics.ucsd.edu/~emichels Funky Mathematical Physics Concepts emichels at physics.ucsd.edu
White Noise and Correlation .................................................................................................. 153
Why Oversampling Does Not Improve Signal-to-Noise Ratio................................................. 153
Filters TBS?? ......................................................................................................................... 154
What Happens to a Sine Wave Deferred? ................................................................................... 154
Nonuniform Sampling and Arbitrary Basis Functions ................................................................. 156
Two Dimensional Fourier Transforms ........................................................................................ 158
Note on Continuous Fourier Series and Uniform Convergence.................................................... 158
Tensors, Without the Tension...................................................................................................... 160
Approach ............................................................................................................................... 160
Two Physical Examples ............................................................................................................. 160
Magnetic Susceptibility.......................................................................................................... 160
Mechanical Strain .................................................................................................................. 164
When Is a Matrix Not a Tensor?............................................................................................. 166
Heading In the Right Direction............................................................................................... 166
Some Definitions and Review.................................................................................................... 166
Vector Space Summary.......................................................................................................... 167
When Vectors Collide ............................................................................................................ 168
Tensors vs. Symbols........................................................................................................ 169
Notational Nightmare............................................................................................................. 169
Tensors? What Good Are They?................................................................................................ 169
A Short, Complicated Definition ............................................................................................ 169
Building a Tensor ...................................................................................................................... 170
Tensors in Action....................................................................................................................... 171
Tensor Fields ......................................................................................................................... 172
Dot Products and Cross Products as Tensors........................................................................... 172
The Danger of Matrices.......................................................................................................... 174
Reading Tensor Component Equations ................................................................................... 174
Adding, Subtracting, Differentiating Tensors .......................................................................... 175
Higher Rank Tensors ................................................................................................................. 175
Tensors In General ................................................................................................................. 177
Change of Basis: Transformations.............................................................................................. 177
Matrix View of Basis Transformation..................................................................................... 179
Non-Orthonormal Systems: Contravariance and Covariance ....................................................... 179
What Goes Up Can Go Down: Duality of Contravariant and Covariant Vectors....................... 182
The Real Summation Convention ........................................................................................... 183
Transformation of Covariant Indexes...................................................................................... 183
Indefinite Metrics: Relativity...................................................................................................... 183
Is a Transformation Matrix a Tensor?......................................................................................... 184
How About the Pauli Vector?..................................................................................................... 184
Cartesian Tensors....................................................................................................................... 185
The Real Reason Why the Kronecker Delta Is Symmetric........................................................... 186
Tensor Appendices .................................................................................................................... 186
Pythagorean Relation for 1-forms........................................................................................... 186
Geometric Construction Of The Sum Of Two 1-Forms: .......................................................... 187
Fully Anti-symmetric Symbols Expanded............................................................................ 188
Metric? We Dont Need No Stinking Metric!............................................................................. 189
References: ................................................................................................................................ 191
Differential Geometry.................................................................................................................. 192
Manifolds .................................................................................................................................. 192
Coordinate Bases ................................................................................................................... 192
Covariant Derivatives ................................................................................................................ 194
Christoffel Symbols ................................................................................................................... 196
Visualization of n-Forms............................................................................................................ 197
Review of Wedge Products and Exterior Derivative.................................................................... 197
physics.ucsd.edu/~emichels Funky Mathematical Physics Concepts emichels at physics.ucsd.edu
1-D........................................................................................................................................ 197
2-D........................................................................................................................................ 197
3-D........................................................................................................................................ 198
Math Tricks ................................................................................................................................. 199
Math Tricks That Come Up A Lot .............................................................................................. 199
The Gaussian Integral............................................................................................................. 199
Math Tricks That Are Fun and Interesting .................................................................................. 199
Phasors...................................................................................................................................... 200
Future Funky Mathematical Physics Topics................................................................................ 200
Appendices................................................................................................................................... 201
References................................................................................................................................. 201
Glossary .................................................................................................................................... 201
Formulas.................................................................................................................................... 205
Index......................................................................................................................................... 205
a cos a
sin a
1
u
n
i
t
tan a
cot a
sec a
c
s
c
a
O
A
B
C
D
a
cos a
From OAD: sin = opp / hyp
sin
2
+ cos
2
= 1
From OAB: tan = opp / adj
tan
2
+ 1 = sec
2
(and with OAD) tan = sin / cos
sec = hyp / adj = 1 / cos
From OAC: cot = adj / opp
cot
2
+ 1 = csc
2
(and with OAD) cot = cos / sin
csc = hyp / opp = 1 / sin
physics.ucsd.edu/~emichels Funky Mathematical Physics Concepts emichels at physics.ucsd.edu
Introduction
Why Funky?
The purpose of the Funky series of documents is to help develop an accurate physical, conceptual,
geometric, and pictorial understanding of important physics topics. We focus on areas that dont seem to
be covered well in most texts. The Funky series attempts to clarify those neglected concepts, and others
that seem likely to be challenging and unexpected (funky?). The Funky documents are intended for serious
students of physics; they are not popularizations or oversimplifications.
Physics includes math, and were not shy about it, but we also dont hide behind it.
Without a conceptual understanding, math is gibberish.
http://physics.ucsd.edu/~emichels for the latest versions of the Funky Series, and for contact information.
Were looking for feedback, so please let us know what you think.
How to Use This Document
This work is not a text book.
There are plenty of those, and they cover most of the topics quite well. This work is meant to be used
with a standard text, to help emphasize those things that are most confusing for new students. When
standard presentations dont make sense, come here.
You should read all of this introduction to familiarize yourself with the notation and contents. After
that, this work is meant to be read in the order that most suits you. Each section stands largely alone,
though the sections are ordered logically. Simpler material generally appears before more advanced topics.
You may read it from beginning to end, or skip around to whatever topic is most interesting. The Shorts
chapter is a diverse set of very short topics, meant for quick reading.
If you dont understand something, read it again once, then keep reading.
Dont get stuck on one thing. Often, the following discussion will clarify things.
The index is not yet developed, so go to the web page on the front cover, and text-search in this
document.
Why Physicists and Mathematicians Dislike Each Other
Physics goals and mathematics goals are antithetical. Physics seeks to ascribe meaning to mathematics
that describe the world, to understand it, physically. Mathematics seeks to strip the equations of all
physical meaning, and view them in purely abstract terms. These divergent goals set up a natural conflict
between the two camps. Each goal has its merits: the value of physics is (or should be) self-evident; the
value of mathematical abstraction, separate from any single application, is generality: the results can be
applied to a wide range of applications.
Thank You
I owe a big thank you to many professors at both SDSU and UCSD, for their generosity even when I
wasnt a real student: Dr. Herbert Shore, Dr. Peter Salamon, Dr. Arlette Baljon , Dr. Andrew Cooksy, Dr.
George Fuller, Dr. Tom ONeil, Dr. Terry Hwa, and others.
Scope
What This Text Covers
This text covers some of the unusual or challenging concepts in graduate mathematical physics. It is
also very suitable for upper-division undergraduate level, as well. We expect that you are taking or have
taken such a course, and have a good text book. Funky Mathematical Physics Concepts supplements those
other sources.
physics.ucsd.edu/~emichels Funky Mathematical Physics Concepts emichels at physics.ucsd.edu
What This Text Doesnt Cover
This text is not a mathematical physics course in itself, nor a review of such a course. We do not cover
all basic mathematical concepts; only those that are very important, unusual, or especially challenging
(funky?).
This text assumes you understand basic integral and differential calculus, and partial differential
equations. Further, it assumes you have a mathematical physics text for the bulk of your studies, and are
using Funky Mathematical Physics Concepts to supplement it.
Notation
Sometimes the variables are inadvertently not written in italics, but I hope the meanings are clear.
?? refers to places that need more work.
TBS To be supplied (one hopes) in the future.
Interesting points that you may skip are asides, shown in smaller font and narrowed margins. Notes to
myself may also be included as asides.
Common misconceptions are sometimes written in dark red dashed-line boxes.
Formulas: We write the integral over the entire domain as a subscript , for any number of
dimensions:
3
1-D: 3-D: dx d x
} }
Evaluation between limits: we use the notation [function]
a
b
to denote the evaluation of the function
between a and b, i.e.,
[f(x)]
a
b
f(b) f(a). For example,
0
1
3x
2
dx = [x
3
]
0
1
= 1
3
- 0
3
= 1
We write the probability of an event as Pr(event).
Column vectors: Since it takes a lot of room to write column vectors, but it is often important to
distinguish between column and row vectors, I sometimes save vertical space by using the fact that a
column vector is the transpose of a row vector:
( ) , , ,
T
a
b
a b c d
c
d
| |
|
|
=
|
|
\ .
For Greek letters, pronunciations, and use, see Funky Quantum Concepts. Other math symbols:
Symbol Definition
for all
- there exists
such that
iff if and only if
proportional to. E.g., a b means a is proportional to b
perpendicular to
therefore
~ of the order of (sometimes used imprecisely as approximately equals)
is defined as; identically equal to (i.e., equal in all cases)
physics.ucsd.edu/~emichels Funky Mathematical Physics Concepts emichels at physics.ucsd.edu
implies
tensor product, aka outer product
direct sum
In mostly older texts, German type (font: Fraktur) is used to provide still more variable names:
Lat
in
Germ
an
Capital
Germa
n
Lowercase Notes
A
A a
Distinguish capital from U, V
B
B b
C
C c
Distinguish capital from E, G
D
D d
Distinguish capital from O, Q
E
E e
Distinguish capital from C, G
F
F f
G
G g
Distinguish capital from C, E
H
H h
I
I i
Capital almost identical to J
J
J j
Capital almost identical to I
K
K k
L
L l
M
M m
Distinguish capital from W
N
N n
O
O o
Distinguish capital from D, Q
P
P p
Q
Q q
Distinguish capital from D, O
R
R r
Distinguish lowercase from x
S
S s
Distinguish capital from C, G, E
T
T t
Distinguish capital from I
U
U u
Distinguish capital from A, V
V
V v
Distinguish capital from A, U
W
W w
Distinguish capital from M
physics.ucsd.edu/~emichels Funky Mathematical Physics Concepts emichels at physics.ucsd.edu
X
X x
Distinguish lowercase from r
Y
Y y
Z
Z z
physics.ucsd.edu/~emichels Funky Mathematical Physics Concepts emichels at physics.ucsd.edu
Random Topics
x
sinh a
area = a/2
y
y
=
x
x
2
y
2
= 1
cos a
sin a
x
2
+ y
2
= 1
x
y
area = a/2
1
u
n
i
t
cosh a
1 unit
a
From where do the hyperbolic trigonometric functions get their names? By analogy with the circular
functions. We usually think of the argument of circular functions as an angle, a. But in a unit circle, the
area covered by the angle a is a / 2 (above left):
2
( 1)
2 2
a a
area r r t
t
= = =
Instead of the unit circle, x
2
+ y
2
= 1, we can consider the area bounded by the x-axis, the ray from the
origin, and the unit hyperbola, x
2
y
2
= 1 (above right). Then the x and y coordinates on the curve are
called the hyperbolic cosine and hyperbolic sine, respectively. Notice that the hyperbola equation implies
the well-known hyperbolic identity:
2 2 2 2
cosh , sinh , 1 cosh sinh 1 x a y a x y = = = =
Proving that the area bounded by the x-axis, ray, and hyperbola satisfies the standard definition of the
hyperbolic functions requires evaluating an elementary, but tedious, integral: (?? is the following right?)
2
1
2 2
1
2
2
2 2 2
3
1 1 1 1
1
: 1
2 2
1 2 1
For the integral, let sec , tan sec sec 1 tan
sin
1 sec 1 tan sec tan sec
cos
x
x
x x x x
a
area xy y dx Use y x
a x x x dx
x dx d y
x dx d d d
u u u u u u
u
u u u u u u u u
u
= = =
=
= = = =
= = =
}
}
} } } }
Try integrating by parts:
2
2 3
1
1 1
tan sec tan sec , sec
tan sec sec tan sec
x x
x
U dV d dU d V
d UV V dU d
u u u u u u u
u u u u u u u
= = = =
= =
} } }
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This is too hard, so we try reverting to fundamental functions sin( ) and cos( ):
( )
3 2
2
2
3 2 2
1 1
1 1
2 2
1
1
1
1
sin cos sin cos , cos
2
sin sin sin
2 2 2 cos cos : sec tan
cos cos cos
sec ln sec tan ln 1 ln 1 ln1
x x
x x
x
x
x
U dV d dU d V
d UV V dU d Use xy
xy d xy xy x x xy x x
u u u u u u u
u u u
u u u u u u
u u u
u u u u
= = = =
= = = =
| |
= = + = + = +
|
\ .
} } }
}
2 2
2
ln 1 ln 1
1
a
a xy xy x x x x
e x x
= + + = +
= +
Solve for x in terms of a, by squaring both sides:
( )
( )
2 2 2 2 2
2
2 1 1 2 1 1 2 1
1 2
2 cosh
2
a
a a
e
a a
a a
a a
e x x x x x x x xe
e xe
e e
e e x x a
= + + = + =
+ =
+
+ = =
The definition for sinh follows immediately from
( )
2 2 2 2 2
2
2
2 2 2 2
cosh sinh 1 1
2 2
sinh 1 1
2 4 4 4 2
a a
a a a a a a a a
x y y x
e e
e e e e e e e e
a y
= = =
| |
+ + + +
= = = = = |
|
\ .
Basic Calculus You May Not Know
Amazingly, many calculus courses never provide a precise definition of a limit, despite the fact that
both of the fundamental concepts of calculus, derivatives and integrals, are defined as limits! So here we
go:
Basic calculus relies on 4 major concepts:
1. Functions
2. Limits
3. Derivatives
4. Integrals
1. Functions: Briefly, (in real analysis) a function takes one or more real values as inputs, and
produces one or more real values as outputs. The inputs to a function are called the arguments. The
simplest case is a real-valued function of a real-valued argument e.g., f(x) = sin x. Mathematicians would
write (f : R
1
R
1
), read f is a map (or function) from the real numbers to the real numbers. A function
which produces more than one output may be considered a vector-valued function.
2. Limits: Definition of limit (for a real-valued function of a single argument, f : R
1
R
1
):
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L is the limit of f(x) as x approaches a, iff for every > 0, there exists a (> 0) such that |f(x) L| <
whenever 0 < |x a| < . In symbols:
lim ( ) iff 0, such that ( ) whenever 0
x a
L f x f x L x a c o c o
## = > - < < <
This says that the value of the function at a doesnt matter; in fact, most often the function is not
defined at a. However, the behavior of the function near a is important. If you can make the function
arbitrarily close to some number, L, by restricting the functions argument to a small neighborhood around
a, then L is the limit of f as x approaches a.
Surprisingly, this definition also applies to complex functions of complex variables, where the absolute
value is the usual complex magnitude.
Example: Show that
2
1
2 2
lim 4
1 x
x
x
=
.
Solution: We prove the existence of given any by computing the necessary from . Note that for
2
2 2
1, 2( 1)
1
x
x x
x
= = +
( )
2
2 2
4 0 1
1
x
whenever x
x
c o
## < < <
So we solve for x in terms of . Since we dont care what the function is at x = 1, we can use the
simplified form, 2(x + 1). When x = 1, this is 4, so we suspect the limit = 4. Proof:
2( 1) 4 2 ( 1) 2 1 1 1
2 2 2
x x x or x
c c c
c c + < + < < < < +
So by setting = /2, we construct the required for any given . Hence, for every , there exists a
satisfying the definition of a limit.
3. Derivatives: Only now that we have defined a limit, can we define a derivative:
0
( ) ( )
'( ) lim
x
f x x f x
f x
x A
+ A
A
4. Integrals: A simplified definition of an integral is an infinite sum of areas under a function
divided into equal subintervals:
( )
1
( ) lim (simplified definition)
N
b
a N
i
x
b a i
f x dx f b a
N N
=
A
| |
|
\ .
}
For practical physics, this definition would be fine. For mathematical preciseness, the actual definition
of an integral is the limit over any possible set of subintervals, so long as the maximum of the subinterval
size goes to zero. This is called the norm of the subdivision, written as ||x
i
||:
( )
0
1
( ) lim (precise definition)
i
N
b
i i
a x
i
f x dx f x x
A
=
A
}
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(Left) Simplified definition of an integral as the limit of a sum of equally spaced samples.
(Right) Precise definition requires convergence for arbitrary, but small, subdivisions.
Why do mathematicians require this more precise definition? Its to avoid bizarre functions, such as:
f(x) is 1 if x is rational, and zero if irrational. This means f(x) toggles wildly between 1 and 0 an infinite
number of times over any interval. However, with the simplified definition of an integral, the following is
well defined:
3.14
0 0
( ) 3.14 ( ) 0 (with simplified definition of integral) f x dx but f x dx
t
= =
} }
But properly, and with the precise definition of an integral, both integrals are undefined and do not
exist.
The Product Rule
Given functions U(x) and V(x), the product rule (aka the Leibniz rule) says that for differentials,
( ) d UV U dV V dU = +
This leads to integration by parts, which is mostly known as an integration tool, but it is also an
important theoretical (analytic) tool, and the essence of Legendre transformations.
Integration By Pictures
We assume you are familiar with integration by parts (IBP) as a tool for performing indefinite
integrals, usually written as:
( ) '( ) ( ) ( ) ( ) '( )
dV dU
U dV UV V dU which really means
U x V x dx U x V x V x U x dx
=
=
} }
} }
This comes directly from the product rule above: ( ) U dV d UV V dU = , and integrate both sides.
Note that x is the integration variable (not U or V), and x is also the parameter to the functions U(x) and
V(x).
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U(x)
V(x)
U(a) U(b)
V(a)
V(b)
U(b)V(b)
U(a)V(a)
V dU
U dV
U(x)
V(x)
U(a) U(b)
V
max
V(a) = V(b) = 0
V dU
1
U dV
1
2
V(x)
U(a) = 0
U(b),
V(b) = 0
V(a)
V dU= U dV
integration
path
Three cases if integration by parts: (left) U(x) and V(x) increasing. (Middle) V(x) decreasing
to 0. (Right) V(x) progressing from zero, to finite, and back to zero.
The diagram above illustrates IBP in three cases. The left is the simplest case where U(x) and V(x) are
monotonically increasing functions of x (note that x is not an axis, U and V are the axes, but x is the
integration parameter). IBP says
| | | | ( ) ( ) ( ) ( ) ( ) ( )
b
x a
boundary term
b b
x x
b
a x a a
U x V x U b V b U a V dU V dV U V d a U
= =
=
=
= =
} } }
The LHS (left hand side) of the equation is the red shaded area; the term in brackets on the right is the
big rectangle minus the white rectangle; the last term is the blue shaded area. The left diagram illustrates
IBP visually as areas. The term in brackets is called the boundary term (or surface term), because in
some applications, it represents the part of the integral corresponding to the boundary (or surface) of the
region of integration.
The middle diagram illustrates another common case: that in which the surface term UV is zero. In
this case, UV = 0 at x = a and x = b, because U(a) = 0 and V(b) = 0. The shaded area is the integral, but the
path of integration means that dU > 0, but dV < 0. Therefore V dU > 0, but U dV < 0.
The right diagram shows the case where one of U(x) or V(x) starts and ends at 0. For illustration, we
chose V(a) = V(b) = 0. Then the surface term is zero, and we have:
| | ( ) ( ) 0
b
b
x a
x
b
x a a
U x V x U d V d V U
=
= =
= =
} }
For V(x) to start and end at zero, V(x) must grow with x to some maximum, V
max
, and then decrease
back to 0. For simplicity, we assume U(x) is always increasing. The V dU integral is the blue striped area
below the curve; the U dV integral is the area to the left of the curves. We break the dV integral into two
parts: path 1, leading up to V
max
, and path 2, going back down from V
max
to zero. The integral from 0 to
V
max
(path 1) is the red striped area; the integral from V
max
back down to 0 (path 2) is the negative of the
entire (blue + red) striped area. Then the blue shaded region is the difference: (1) the (red) area to the left
of path 1 (where dV is positive, because V(x) is increasing), minus (2) the (blue + red) area to the left of
path 2, because dV is negative when V(x) is decreasing:
max max max
max
0
0
1 2 2
0 0
1
2
1
V
V V V
path path path
path
V V
V V
path pat
x a
h
b
U dV U dV U dV U
V d
U
U
dV dV
= =
+
=
= =
= + =
=
}
} } } } }
Theoretical Importance of IBP
Besides being an integration tool, an important theoretical consequence of IBP is that the variable of
integration is changed, from dV to dU. Many times, one differential is unknown, but the other is known:
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Under an integral, integration by parts allows one to exchange a derivative
that cannot be directly evaluated, even in principle, in favor of one that can.
The classic example of this is deriving the Euler-Lagrange equations of motion from the principle of
stationary action. The action of a dynamic system is defined by
( ( ), ( )) S L q t q t dt
}
`
where the lagrangian is a given function of the trajectory q(t). Stationary action means that the action
does not change (to first order) for small changes in the trajectory. I.e., given a small variation in the
trajectory, q(t):
0 ( , )
L L d
S L q q q q dt S q q dt Use q q
q q dt
L L d
q q dt
q q dt
o o o o o o o
o o
( c c
= = + + = + =
(
c c
( c c
= +
(
c c
} }
}
` ` ` `
`
`
The quantity in brackets involves both q(t) and its time derivative, q-dot. We are free to vary q(t)
arbitrarily, but that fully determines q-dot. We cannot vary both q and q-dot separately. We also know
that q(t) = 0 at its endpoints, but q-dot is unconstrained at its endpoints. Therefore, it would be simpler if
the quantity in brackets was written entirely in terms of q(t), and not in terms of q-dot. IBP allows us to
eliminate the time derivative of q(t) in favor of the time derivative of L/q-dot. Since L(q, q-dot) is
given, we can easily determine L/q-dot. Therefore, this is a good trade. Integrating the 2
nd
term in
brackets by parts gives:
0
'
, . ,
( )
U
t f
t
V
L d L d
Let U dU dt dV q dt V q
q dt q dt
L
d
L d
q
dt
t UV V dU
q
q t
q
o o
o o
=
=
| | c c
= = = =
|
c c
\ .
c
c
( c
= =
(
c
} }
`
` `
' V U
d L
d
d q
t q
t o
| | c
|
c
\ .
}
`
The boundary term is zero because q(t) is zero at both endpoints. The variation in action S is now:
0 ( )
L d L
S q dt q t
q dt q
o o o
( c c
= =
(
c c
}
`
The only way S = 0 can be satisfied for any q(t) is if the quantity in brackets is identically 0. Thus
IBP has lead us to an important theoretical conclusion: the Euler-Lagrange equation of motion.
This fundamental result has nothing to do with evaluating a specific difficult integral. IBP: its not just
for doing hard integrals any more.
Delta Function Surprise
Rarely, one needs to consider the 3D -function in coordinates other than rectangular. The 3D -
function is written
3
(r r). For example, in 3D Greens functions, whose definition depends on a
3
-
function, it may be convenient to use cylindrical or spherical coordinates. In these cases, there are some
unexpected consequences [Wyl p280]. This section assumes you understand the basic principle of a 1D
and 3D -function. (See the introduction to the delta function in Funky Quantum Concepts.)
Recall the defining property of
3
(r - r):
3 3 3 3
( ') 1 ' ( " ") ( ') ( ) ( ') d for all d f f o o
= =
} }
r r r r r r r r r .
The above definition is coordinate free, i.e. it makes no reference to any choice of coordinates, and is
true in every coordinate system. As with Greens functions, it is often helpful to think of the -function as a
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function of r, which is zero everywhere except for an impulse located at r. As we will see, this means that
it is properly a function of r and r separately, and should be written as
3
(r, r) (like Greens functions
are).
Rectangular coordinates: In rectangular coordinates, however, we now show that we can simply
break up
3
(x, y, z) into 3 components. By writing (r r) in rectangular coordinates, and using the
defining integral above, we get:
3
3
' ( ', ', ') ( ', ', ') 1
( ', ', ') ( ') ( ') ( ') .
x x y y z z dx dy dz x x y y z z
x x y y z z x x y y z z
o
o o o o
=
=
} } }
r r
In rectangular coordinates, the above shows that we do have translation invariance, so we can simply write:
3
( , , ) ( ) ( ) ( ) x y z x y z o o o o = .
In other coordinates, we do not have translation invariance. Recall the 3D infinitesimal volume
element in 4 different systems: coordinate-free, rectangular, cylindrical, and spherical coordinates:
3 2
sin d dx dy dz r dr d dz r dr d d | u u | = = = r
The presence of r and imply that when writing the 3D -function in non-rectangular coordinates, we must
include a pre-factor to maintain the defining integral = 1. We now show this explicitly.
Cylindrical coordinates: In cylindrical coordinates, for r > 0, we have (using the imprecise notation
of [Wyl p280]):
2
3
0 0
3
' ( ', ', ')
( ', ', ') 1
1
( ', ', ') ( ') ( ') ( '), ' 0
'
r r z z
dr d dz r r r z z
r r z z r r z z r
r
t
| |
| o | |
o | | o o | | o
=
=
= >
} } }
r r
Note the 1/r pre-factor on the RHS. This may seem unexpected, because the pre-factor depends on
the location of
3
( ) in space (hence, no radial translation invariance). The rectangular coordinate version of
3
( ) has no such pre-factor. Properly speaking,
3
( ) isnt a function of r r; it is a function of r and r
separately.
In non-rectangular coordinates,
3
( ) does not have translation invariance, and includes a pre-factor
which depends on the position of
3
( ) in space, i.e. depends on r.
At r = 0, the pre-factor blows up, so we need a different pre-factor. Wed like the defining integral to
be 1, regardless of |, since all values of | are equivalent at the origin. This means we must drop the (|
|), and replace the pre-factor to cancel the constant we get when we integrate out |:
2
3
0 0
3
0
( ', ', ') 1, ' 0
1
( ', ', ') ( ) ( '), ' 0,
2
assuming that ( ) 1.
dr d dz r r r z z r
r r z z r z z r
r
dr r
t
| o | |
o | | o o
t
o
= =
= =
=
} } }
}
This last assumption is somewhat unusual, because the -function is usually thought of as symmetric about
0, where the above radial integral would only be . The assumption implies a right-sided -function,
whose entire non-zero part is located at 0
+
. Furthermore, notice the factor of 1/r in (r 0, z z). This
factor blows up at r = 0, and has no effect when r 0. Nonetheless, it is needed because the volume
element r dr d| dz goes to zero as r 0, and the 1/r in (r 0, z z) compensates for that.
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Spherical coordinates: In spherical coordinates, we have similar considerations. First, away from the
origin, r > 0:
2
2 3
0 0 0
3
2
sin ( ', ', ') 1
1
( ', ', ') ( ') ( ') ( '), ' 0 . [Wyl 8.9.2 p280]
' sin '
dr d d r r r
r r r r r
r
t t
u | u o u u | |
o u u | | o o u u o | |
u
=
= >
} } }
Again, the pre-factor depends on the position in space, and properly speaking,
3
( ) is a function of r, r, ,
and separately, not simply a function of r r and . At the origin, wed like the defining integral to
be 1, regardless of | or . So we drop the (| |) ( ), and replace the pre-factor to cancel the
constant we get when we integrate out | and :
2
2 3
0 0 0
3
2
0
sin ( 0, ', ') 1, ' 0
1
( 0, ', ') ( ), ' 0,
4
assuming that ( ) 1.
dr d d r r r
r r r
r
dr r
t t
u | u o u u | |
o u u | | o
t
o
= =
= =
=
} } }
}
Again, this definition uses the modified (r), whose entire non-zero part is located at 0
+
. And similar to the
cylindrical case, this includes the 1/r
2
factor to preserve the integral at r = 0.
2D angular coordinates: For 2D angular coordinates and |, we have:
2
2
0 0
2
sin ( ', ') 1, ' 0
1
( ', ') ( ') ( '), ' 0.
sin '
d d
t t
u | u o u u | | u
o u u | | o u u o | | u
u
= >
= >
} }
Once again, we have a special case when = 0: we must have the defining integral be 1 for any value of |.
Hence, we again compensate for the 2 from the | integral:
2
2
0 0
2
sin ( ', ') 1, ' 0
1
( 0, ') ( ), ' 0.
2 sin
d d
t t
u | u o u u | | u
o u | | o u u
t u
= =
= =
} }
Similar to the cylindrical and spherical cases, this includes a 1/(sin ) factor to preserve the integral at
= 0.
Spherical Harmonics Are Not Harmonics
See Funky Electromagnetic Concepts for a full discussion of harmonics, Laplaces equation, and its
solutions in 1, 2, and 3 dimensions. Here is a brief overview.
Spherical harmonics are the angular parts of solid harmonics, but we will show that they are not truly
harmonics. A harmonic is a function which satisfies Laplaces equation:
2
( ) 0 V u = r , with r typically in 2 or 3 dimensions.
Solid harmonics are 3D harmonics: they solve Laplaces equation in 3 dimensions. For example, one
form of solid harmonics separates into a product of 3 functions in spherical coordinates:
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( )
( )
( )
( )
( )
1
1
( , , ) ( ) ( ) ( ) (cos ) sin cos
( ) is the radial part
( ) (cos ) is the polar angle part, the associated Legendre functions
( ) sin cos is the azimuthal part
l l m
l l l l l
l l
l l
m
l
l l
r R r P Q A r B r P C m D m
where R r Ar B r
P P
Q C m D m
u | u | u | |
u u
| | |
+
+
u = = + +
= +
=
= +
The spherical harmonics are just the angular (, |) parts of these solid harmonics. But notice that the
angular part alone does not satisfy the 2D Laplace equation (i.e., on a sphere of fixed radius):
2
2 2
2 2 2 2 2
2
2 2 2 2
1 1 1
sin , but for fixed :
sin sin
1 1
sin
sin sin
r r
r r r r r
r r
u
u u u u |
u
u u u u |
c c c c c | | | |
V = + +
| |
c c c c c \ . \ .
c c c | |
= +
|
c c c \ .
But direct substitution of spherical harmonics into the above Laplace operator shows that the result is
not 0 (we let r = 1). We proceed in small steps:
2
2
2
( ) sin cos ( ) ( ) Q C m D m Q m Q | | | | |
|
c
= + =
c
For integer m, the associated Legendre functions, P
l
m
(cos ), satisfy, for given l and m:
( )
2
2 2
1 1
sin (cos ) (cos )
sin
m m
l l
l l
P m P
r r
u u u
u u u
| | + c c | |
= +
|
|
|
c c
\ .
\ .
Combining these 2 results (r = 1):
( ) ( )
( )
( )
( )
2
2
2 2
2 2
1 1
( ) ( ) sin ( ) ( )
sin sin
1 (cos ) ( ) (cos ) ( )
1 (cos ) ( )
m m
l l
m
l
P Q P Q
l l m P Q m P Q
l l P Q
u | u u |
u u u u |
u u u u
u u
(
c c c | |
V = +
(
|
c c c \ . (
= + +
= +
Hence, the spherical harmonics are not solutions of Laplaces equation,
i.e. they are not harmonics.
The Binomial Theorem for Negative and Fractional Exponents
You may be familiar with the binomial theorem for positive integer exponents, but it is very useful to
know that the binomial theorem also works for negative and fractional exponents. We can use this fact to
easily find series expansions for things like ( )
1/ 2 1
and 1 1
1
x x
x
+ = +
.
First, lets review the simple case of positive integer exponents:
( )
( ) ( )( )
0 1 1 2 2 3 3 0
1 1 2 !
...
1 1 2 1 2 3 !
n n n n n n
n n n n n n n
a b a b a b a b a b a b
n
+ = + + + +
[For completeness, we note that we can write the general form of the m
th
term:
( )
!
, integer 0; integer, 0
! !
th n m m
n
m term a b n m m n
n m m
= > s s
.]
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But were much more interested in the iterative procedure (recursion relation) for finding the (m + 1)
th
term from the m
th
term, because we use that to generate a power series expansion. The process is this:
1. The first term (m = 0) is always a
n
b
0
= a
n
, with an implicit coefficient C
0
= 1.
2. To find C
m+1
, multiply C
m
by the power of a in the m
th
term, (n m)
3. divide it by (m + 1), [the number of the new term were finding]:
1
( )
1
m m
n m
C C
m
+
=
+
4. lower the power of a by 1 (to n m), and
5. raise the power of b by 1 to (m + 1).
This procedure is valid for all n, even negative and fractional n. A simple way to remember this is:
For any real n, we generate the (m + 1)th term from the mth term by differentiating with respect to
a, and integrating with respect to b.
The general expansion, for any n, is then:
( ) ( ) 1 2 ...( 1)
, real; integer 0
!
th n m m
n n n n m
m term a b n m
m
+
= >
Notice that for integer n > 0, there are n+1 terms. For fractional or negative n, we get an infinite series.
Example 1: Find the Taylor series expansion of
1
1 x
. Since the Taylor series is unique, any method
we use to find a power series expansion will give us the Taylor series. So we can use the binomial
theorem, and apply the rules above, with a = 1, b = (x):
( ) ( )
( )
( )
( )( )
( )
( )( )( )
( )
1 1 2 3 1 2 3 4
2
1 1 2 1 2 3
1
1 1 1 1 1 ...
1 1 1 2 1 2 3
1 ... ...
m
x x x x
x
x x x
= + = + + + +
= + + + + +
Notice that all the fractions, all the powers of 1, and all the minus signs cancel.
Example 2: Find the Taylor series expansion of ( )
1/ 2
1 1 x x + = + . The first term is a
1/2
= 1
1/2
:
( )
( ) ( ) ( )
( )
( )
( )( ) ( )
1/ 2 1/ 2 1/ 2 1 3/ 2 2 5/ 2 3
1 2 3
1 1 1 1 1 1 1 3 1
1 1 1 1 1 ...
2 1 2 2 1 2 2 2 2 1 2 3
2 3 !!
1 1 3
1 ... 1
2 8 48
2 !
!! 2 4 ... 2 1
m m
m
x x x x
m
x x x x
m
where p p p p or
+
| | | || |
+ = + + + +
| | |
\ . \ .\ .
= + + +
When Does a Divergent Series Converge?
Consider the infinite series
2
1 ... ...
n
x x x + + + + +
When is it convergent? Apparently, when |x| < 1. What is the value of the series when x = 2 ?
Undefined! you say. But there is a very important sense in which the series converges for x = 2, and its
value is 1! How so?
Recall the Taylor expansion (you can use the binomial theorem, see above):
( )
1 2
1
1 1 ... ...
1
n
x x x x
x
= = + + + + +
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It is exactly the original infinite series above. So the series sums to 1/(1 x). This is defined for all x
= 1. And its value for x = 2 is 1.
Why is this important? There are cases in physics when we use perturbation theory to find an
expansion of a number in an infinite series. Sometimes, the series appears to diverge. But by finding the
analytic expression corresponding to the series, we can evaluate the analytic expression at values of x that
make the series diverge. In many cases, the analytic expression provides an important and meaningful
answer to a perturbation problem. This happens in quantum mechanics, and quantum field theory.
This is an example of analytic continuation. A Taylor series is a special case of a Laurent series, and
any function with a Laurent expansion is analytic. If we know the Laurent series (or if we know the values
of an analytic function and all its derivatives at any one point), then we know the function everywhere,
even for complex values of x. The original series is analytic around x = 0, therefore it is analytic
everywhere it converges (everywhere it is defined). The process of extending a function which is defined
in some small region to be defined in a much larger (even complex) region, is called analytic continuation
(see Complex Analysis, discussed elsewhere in this document).
TBS: show that the sum of the integers 1 + 2 + 3 + ... = 1/12. ??
Algebra Family Tree
Doo
Properties Examples
gro
up
Finite or infinite set of elements and
operator (), with closure, associativity, identity
element and inverses. Possibly commutative:
ab = c w/ a, b, c group elements
rotations of a square by n 90
o
continuous rotations of an object
ring Set of elements and 2 binary operators
(+ and *), with:
commutative group under +
left and right distributivity:
a(b + c) = ab + ac, (a + b)c = ac + bc
usually multiplicative associativity
integers mod m
polynomials p(x) mod m(x)
inte
gral
domain,
or
domain
A ring, with:
commutative multiplication
multiplicative identity (but no inverses)
no zero divisors ( cancellation is valid):
ab = 0 only if a = 0 or b = 0
integers
polynomials, even abstract
polynomials, with abstract variable x, and
coefficients from a field
fiel
d
rings with multiplicative inverses (&
identity)
commutative group (excluding 0) under
multiplication.
distributivity, multiplicative inverses
Allows solving simultaneous linear equations.
Field can be finite or infinite
integers with arithmetic modulo 3 (or
any prime)
real numbers
complex numbers
vect
or space
field of scalars
group of vectors under +.
Allows solving simultaneous vector
equations for unknown scalars or vectors.
Finite or infinite dimensional.
physical vectors
real or complex functions of space:
f(x, y, z)
kets (and bras)
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Hil
bert
space
vector space over field of complex
numbers with:
a conjugate-bilinear inner product,
<av|bw> = (a*)b<v|w>,
<v|w> = <w|v>*
a, b scalars, and v, w vectors
Mathematicians require it to be infinite
dimensional; physicists dont.
real or complex functions of space:
f(x, y, z)
quantum mechanical wave functions
Convoluted Thinking
Convolution arises in many physics, engineering, statistics, and other mathematical areas.
f(t)
t t
g(t)
Two functions, f(t) and g(t).
increasing
t
g(t
1
-)
f()
t
1
(f *g)(t
1
)
g(t
2
-)
f()
t
2
(f *g)(t
2
)
g(t
0
-)
f()
t
0
< 0
(f *g)(t
0
)
(Left) (f *g)(t
0
), t
0
< 0. (Middle) (f *g)(t
1
), t
1
> 0. (Right) (f*g)(t
2
), t
2
> t
1
.
The convolution is the magenta shaded area.
Given two functions, f(t) and g(t), the convolution of f(t) and g(t) is a function of a time-displacement,
defined by (see diagram above):
( ) * ( ) ( ) ( ) the integral covers some domain of interest f g t d f g t where t t t
A A
}
When t < 0, the two functions are backing into each other (above left). When t > 0, the two
functions are backing away from each other (above middle and right).
Of course, we dont require functions of time. Convolution is useful with a variety of independent
variables. E.g., for functions of space, f(x) and g(x), f *g(x) is a function of spatial displacement, x.
Notice that convolution is
(1) commutative: * * f g g f =
(2) linear in each of the two functions:
( ) ( )
( )
* * *
* * *
f kg k f g kf g and
f g h f g f h
= =
+ = +
The verb to convolve means to form the convolution of. We convolve f and g to form the
convolution f *g.
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Vectors
Small Changes to Vectors
Projection of a Small Change to a Vector Onto the Vector
r
dr
dr
dr d|r|
r'
r
' ' r ~ r r r r
r r'
' r r
r |r|
(Left) A small change to a vector, and its projection onto the vector.
(Right) Approximate magnitude of the difference between a big and small vector.
It is sometimes useful (in orbital mechanics, for example) to relate the change in a vector to the change
in the vectors magnitude. The diagram above (left) leads to a somewhat unexpected result:
(multiplying both sides by and using ) d dr or r r
d r dr
= =
=
r r r r
r r
And since this is true for any small change, it is also true for any rate of change (just divide by dt):
r r = r r` `
Vector Difference Approximation
It is sometimes useful to approximate the magnitude of a large vector minus a small one. (In
electromagnetics, for example, this is used to compute the far-field from a small charge or current
distribution.) The diagram above (right) shows that:
' ' , ' ~ >> r r r r r r r
Why (r, , |) Are Not the Components of a Vector
(r, , |) are parameters of a vector, but not components. That is, the parameters (r, , |) uniquely
define the vector, but they are not components, because you cant add them. This is important in much
physics, e.g. involving magnetic dipoles (ref Jac problem on mag dipole field). Components of a vector
are defined as coefficients of basis vectors. For example, the components v = (x, y, z) can multiply the
basis vectors to construct v:
x y z = + + v x y z
There is no similar equation we can write to construct v from its spherical components (r, , |).
Position vectors are displacements from the origin, and there are no
## , , r defined at the origin.
Put another way, you can always add the components of two vectors to get the vector sum:
( ) ( ) ( )
( , , ) rectangular components. Let a b c Then a x b y c z = + = + + + + + w v w x y z
We cant do this in spherical coordinates:
( ) ( , , ) spherical components. , ,
w w w v w v w v w
Let r Then r r u | u u | | = + = + + + w v w
However, at a point off the origin, the basis vectors
## , , r are well defined, and can be used as a basis
for general vectors. [In differential geometry, vectors referenced to a point in space are called tangent
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vectors, because they are tangent to the space, in a higher dimensional sense. See Differential Geometry
elsewhere in this document.]
Laplacians Place
What is the physical meaning of the Laplacian operator? And how can I remember the Laplacian
operator in any coordinates? These questions are related because understanding the physical meaning
allows you to quickly derive in your head the Laplacian operator in any of the common coordinates.
Lets take a step-by-step look at the action of the Laplacian, first in 1D, then on a 3D differential
volume element, with physical examples at each step. After rectangular, we go to spherical coordinates,
because they illustrate all the principles involved. Finally, we apply the concepts to cylindrical coordinates,
as well. We follow this outline:
1. Overview of the Laplacian operator
2. 1D examples of heat flow
3. 3D heat flow in rectangular coordinates
4. Examples of physical scalar fields [temperature, pressure, electric potential (2 ways)]
5. 3D differential volume elements in other coordinates
6. Description of the physical meaning of Laplacian operator terms, such as
2 2 2 2
, , , ,
T T T T
T r r r r
r r r r r r
c c c c c c | | | |
V
| |
c c c c c c \ . \ .
Overview of Laplacian operator: Let the Laplacian act on a scalar field T(r), a physical function of
space, e.g. temperature. Usually, the Laplacian represents the net outflow per unit volume of some physical
quantity: something/volume, e.g., something/m
3
. The Laplacian operator itself involves spatial second-
derivatives, and so carries units of inverse area, say m
2
.
1D Example: Heat Flow: Consider a temperature gradient along a line. It could be a perpendicular
wire through the wall of a refrigerator (below left). It is a 1D system, i.e. only the gradient along the wire
matters.
Refrigerator Room
t
e
m
p
e
r
a
t
u
r
e
x
metal
wire
Refrigerator
Warmer
Room
t
e
m
p
e
r
a
t
u
r
e
x
current
carrying wire
heat flow
heat flow
Let the left and right sides of the wire be in thermal equilibrium with the refrigerator and room, at 2 C
and 27 C, respectively. The wire is passive, and can neither generate nor dissipate heat; it can only conduct
it. Let the 1D thermal conductivity be k = 100 mW-cm/C. Consider the part of the wire inside the insulated
wall, 4 cm thick. How much heat (power, J/s or W) flows through the wire?
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( )
25
100 mW-cm/C 625 mW
4
dT C
P k
dx cm
= = =
There is no heat generated or dissipated in the wire, so the heat that flows into the right side of any
segment of the wire (differential or finite) must later flow out the left side. Thus, the heat flow must be
constant along the wire. Since heat flow is proportional to dT/dx, dT/dx must be constant, and the
temperature profile is linear. In other words, (1) since no heat is created or lost in the wire, heat-in = heat-
out; (2) but heat flow ~ dT/dx; so (3) the change in the temperature gradient is zero:
2
2
0
d dT d T
dx dx
dx
| |
= =
|
\ .
(At the edges of the wall, the 1D approximation breaks down, and the inevitable nonlinearity of the
temperature profile in the x direction is offset by heat flow out the sides of the wire.)
Now consider a current carrying wire which generates heat all along its length from its resistance
(diagram above, right). The heat that flows into the wire from the room is added to the heat generated in
the wire, and the sum of the two flows into the refrigerator. The heat generated in a length dx of wire is
2 2
resistance per unit length, and
gen
P I dx where I const = =
In steady state, the net outflow of heat from a segment of wire must equal the heat generated in that
segment. In an infinitesimal segment of length dx, we have heat-out = heat-in + heat-generated:
2
2
2
2 2
2
out in gen
a a dx
a dx a
dT dT
P P P I dx
dx dx
dT dT
I dx
dx dx
d dT d T
dx I dx I
dx dx
dx
+
+
= + = +
=
| |
= =
|
\ .
The negative sign means that when the temperature gradient is positive (increasing to the right), the
heat flow is negative (to the left), i.e. the heat flow is opposite the gradient. Many physical systems have a
similar negative sign. Thus the 2
nd
derivative of the temperature is the negative of heat outflow (net inflow)
from a segment, per unit length of the segment. Longer segments have more net outflow (generate more
heat).
3D Rectangular Volume Element
Now consider a 3D bulk resistive material, carrying some current. The current generates heat in each
volume element of material. Consider the heat flow in the x direction, with this volume element:
dx
x
y
z
Outflow surface area
is the same as inflow
flow
The temperature gradient normal to the y-z face drives a heat flow per unit area, in W/m
2
. For a net
flow to the right, the temperature gradient must be increasing in magnitude (becoming more negative) as
we move to the right. The change in gradient is proportional to dx, and the heat outflow flow is
proportional to the area, and the change in gradient:
2
2
out in
out in
P P d dT d T
P P k dx dy dz k
dx dx dx dy dz
dx
| |
= =
|
\ .
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Thus the net heat outflow per unit volume, due to the x contribution, goes like the 2
nd
derivative of T.
Clearly, a similar argument applies to the y and z directions, each also contributing net heat outflow per unit
volume. Therefore, the total heat outflow per unit volume from all 3 directions is simply the sum of the
heat flows in each direction:
2 2 2
2 2 2
out in
P P T T T
k
dx dy dz
x y z
| |
c c c
= + + |
|
c c c
\ .
We see that in all cases, the
net outflow of flux per unit volume = change in (flux per unit area), per unit distance
We will use this fact to derive the Laplacian operator in spherical and cylindrical coordinates.
General Laplacian
We now generalize. For the Laplacian to mean anything, it must act on a scalar field whose gradient
drives a flow of some physical thing.
Example 1: My favorite is T(r) = temperature. Then VT(r) drives heat (energy) flow, heat per unit
time, per unit area:
/
( )
heat t
k T where k thermal conductivity
area
heat flowvector
= V
q r
q
Then ~
r
T
q radial component of heat flow
r
c
=
c
Example 2: T(r) = pressure of an incompressible viscous fluid (e.g. honey). Then VT(r) drives fluid
mass (or volume) flow, mass per unit time, per unit area:
/
( )
mass t
k T where k some viscous friction coefficient
area
mass flow density vector
= V
j r
j
Then ~
r
T
j radial component of mass flow
r
c
=
c
Example 3: T(r) = electric potential in a resistive material. Then VT(r) drives charge flow, charge
per unit time, per unit area:
charge /
( )
t
T where electrical conductivity
area
current density vector
o o = V
j r
j
Then ~
r
T
j radial component of current density
r
c
=
c
Example 4: Here we abstract a little more, to add meaning to the common equations of
electromagnetics. Let T(r) = electric potential in a vacuum. Then VT(r) measures the energy per unit
distance, per unit area, required to push a fixed charge density through a surface, by a distance of dn,
normal to the surface:
energy/distance
( ) electric charge volume density T where
area
V r
Then T/r ~ net energy per unit radius, per unit area, to push charges of density out the same
distance through both surfaces.
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In the first 3 examples, we use the word flow to mean the flow in time of some physical quantity, per
unit area. In the last example, the flow is energy expenditure per unit distance, per unit area. The
requirement of per unit area is essential, as we soon show.
Laplacian In Spherical Coordinates
To understand the Laplacian operator terms in other coordinates, we need to take into account two
effects:
1. The outflow surface area may be different than the inflow surface area
2. The derivatives with respect to angles ( or |) need to be converted to rate-of-change per unit
distance.
Well see how these two effects come into play as we develop the spherical terms of the Laplacian
operator. The cylindrical terms are simplifications of the spherical terms.
Spherical radial contribution: We first consider the radial contribution to the spherical Laplacian
operator, from this volume element:
dr
x
y
z
|
## Outflow surface area
is differentially
larger than inflow
flow
d
s
i
n
d
|
d
d = sin d| d
The differential volume element has thickness dr, which can be made arbitrarily small compared to the
lengths of the sides. The inner surface of the element has area r
2
dO. The outer surface has infinitesimally
more area. Thus the radial contribution includes both the surface-area effect, but not the converting-
derivatives effect.
The increased area of the outflow surface means that for the same flux-density (flow) on inner and
outer surfaces, there would be a net outflow of flux, since flux = (flux-density)(area). Therefore, we must
take the derivative of the flux itself, not the flux density, and then convert the result back to per-unit-
volume. We do this in 3 steps:
( )( )
( )
( )
( )
( )
( )
( ) ( )
2
2
2 2
2 2
- ~
1 1
flux area flux density r d
r
d flux
r d
dr r r
d flux
outflow
r d r
volume area dr r r r r
r d r
c | |
= O
|
c \ .
c c | |
= O
|
c c
\ .
c c c c | | | |
= = O =
| |
c c c c
\ . \ . O
The constant dO factor from the area cancels when converting to flux, and back to flux-density. In
other words, we can think of the fluxes as per-steradian.
We summarize the stages of the spherical radial Laplacian operator as follows:
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2 2
2
2
2
2
1
( ) ( )
( )( )
change in radial flux per unit length, per unit solid-angle; positive is increasing flux
1
r
T r T
r r
r
T
r
area flow per unit area
r T
r d
r T
r r
r
c c
V =
c c
c
=
c
c
= =
c O
c c
=
c c
c
r r
2
2
2
per unit area
change in radial flux per unit length, per unit area
net outflow of flux per unit volume
1
r T
r r
r T
r r
r
c
=
c c
=
c c
c c
unit length, per unit area
Following the steps in the example of heat flow, let T(r) = temperature. Then
2
2
2
2
2
radial heat flow per unit area, W/m
Watts
change in radial heat flux per unit length, per unit solid-angle
1
net outflow of heat flux per unit volume
T
r
r T
r
r T
r r
r T
r r
r
c
=
c
c
=
c
c c
=
c c
c c
=
c c
Spherical azimuthal contribution: The spherical | contribution to the Laplacian has no area-change,
but does require converting derivatives. Consider the volume element:
d|
Outflow surface area
is identical to inflow
x
y
z
|
flow
The inflow and outflow surface areas are the same, and therefore area-change contributes nothing to
the derivatives.
However, we must convert the derivatives with respect to | into rates-of-change with respect to
distance, because physically, the flow is driven by a derivative with respect to distance. In the spherical |
case, the effective radius for the arc-length along the flow is r sin , because we must project the position
vector into the plane of rotation. Thus, (/|) is the rate-of-change per (r sin ) meters. Therefore,
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1
rate-of-change-per-meter
sin r u |
c
=
c
Performing the two derivative conversions, we get
2
1 1
( ) ( )
sin sin
1
azimuthal flux per unit area
sin
1
change in (azimuthal flux per unit area) per radian
sin
1 1
change in (azimuthal flux per unit area) per unit distance
sin sin
net azimu
T T
r r
T
r
T
r
T
r r
|
u | u |
u |
| u |
u | u |
c c
V =
c c
c
=
c
c c
=
c c
c c
=
c c
=
r r
azimuthal flux
per unit area
change in (azimuthal flux
change in (azimuthal flux per
unit area) per unit distance
thal outflow of flux per unit volume
1 1
sin sin
T
r r u | u |
c c
c c
2
2 2 2
1
sin
T
r u |
c
=
c
Notice that the r
2
sin
2
in the denominator is not a physical area; it comes from two derivative
conversions.
Spherical polar angle contribution:
d
Outflow surface area
is differentially
larger than inflow
x
y
z
|
flow
The volume element is like a wedge of an orange: it gets wider (in the northern hemisphere) as
increases. Therefore the outflow area is differentially larger than the inflow area (in the northern
hemisphere). In particular, ( ) sin area r dr u = , but we only need to keep the dependence, because the
factors of r cancel, just like dO did in the spherical radial contribution. So we have
sin area u
In addition, we must convert the / to a rate-of-change with distance. Thus the spherical polar angle
contribution has both area-change and derivative-conversion.
Following the steps of converting to flux, taking the derivative, then converting back to flux-density,
we get
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( )
2
1 1 1
( ) sin ( )
sin
1
( )( ) 1
1
1 1
## sin change in -flux per
T T
r r
T
r
area flux per unit area
T
r dr
T
r
T
r r
u
u
u u u
u
u
u u
u
u u
u u
u u
u u
c c
V =
c c
c
=
c
c
= =
c
c c
=
c c
c c
=
c c
r r
( )
-flux per
unit area
-flux, per
change
unit radius , per unit distance
1 1 1
## sin change in ( -flux per unit area), per unit distance
sin
net outflow of flux per unit volume
1 1 1
sin
sin
T
r r
T
r r
u
u
u u
u u u
u
u u u
c c
=
c c
=
c c
c c
in ( -flux per
## change in ( -flux per unit
area), per unit distance
1
sin
sin
T
r
u
u
u
u
u u
u
c c
=
c c
Notice that the r
2
in the denominator is not a physical area; it comes from two derivative conversions.
Cylindrical Coordinates
The cylindrical terms are simplifications of the spherical terms.
dr
x
y
z
|
surface area is
differentially larger
than inflow
flow
r
d|
flow
| and z outflow
surface areas are
identical to
inflow
dz
flow
Cylindrical radial contribution: The picture of the cylindrical radial contribution is essentially the
same as the spherical, but the height of the slab is exactly constant. We still face the issues of varying
inflow and outflow surface areas, and converting derivatives to rate of change per unit distance. The
change in area is due only to the arc length r d|, with the z (height) fixed. Thus we write the radial result
directly:
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2
1
( ) ( ) (Cylindrical Coordinates)
( )( )
1
change i
r
T r T
r r r
T
r
flow per unit area area
r T
r d dz
r T
r r
r T
r r r
|
c c
V =
c c
c
=
c
c
= =
c
c c
=
c c
c c
=
c c
r r
per unit area
net outflow of flux per unit volume
1
r T
r r r
=
c c
c c
r
Cylindrical azimuthal contribution: Like the spherical case, the inflow and outflow surfaces have
identical areas. Therefore, the | contribution is similar to the spherical case, except there is no sin factor;
r contributes directly to the arc-length and rate-of-change per unit distance:
( )
2
1 1
( ) ( )
1
azimuthal flux per unit area
1
change in azimuthal flux per unit area per radian
1 1
change in (azimuthal flux per unit area) per unit distance
net azimuthal outflow of flux per unit vo
T T
r r
T
r
T
r
T
r r
|
| |
|
| |
| |
c c
V =
c c
c
=
c
c c
=
c c
c c
=
c c
=
r r
2
2 2
change in (azimuthal flux per
unit area) per unit distance
lume
1 1 1
azimuthal flow
per unit area
change in azimuthal
T T
r r
r
| |
|
c c c
=
c c
c
Cylindrical z contribution: This is identical to the rectangular case: the inflow and outflow areas are
the same, and the derivative is already per unit distance, ergo: (add cylindrical volume element picture??)
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2
vertical flux
per unit area
change in (vertical flux per
unit area) pe
( ) ( )
vertical flux per unit area
change in (vertical flux per unit area) per unit distance
net outflow of flux per unit volume
z
T T
z z
T
z
T
z z
T
z z
c c
V =
c c
c
=
c
c c
=
c c
=
c c
c c
r r
2
2
r unit distance
T
z
c
=
c
In electromagnetic propagation, and elsewhere, one encounters the dot-product of a vector field with
the gradient operator, acting on a vector field. What is this v V operator? Here, v(r) is a given vector
field. The simple view is that v(r) V is just a notational shorthand for
( )
( ) ,
( )
x y z
x y z x y z
v v v
x y z
because v v v v v v
x y z x y z
| | c c c
V + +
|
c c c
\ .
| | | | c c c c c c
V = + + + + = + +
| |
c c c c c c
\ . \ .
v r
v r x y z x y z
by the usual rules for a dot-product in rectangular coordinates.
There is a deeper meaning, though, which is an important bridge to the topics of tensors and
differential geometry.
We can view the v V operator as simply the dot-product of the vector field v(r) with the gradient
of a vector field.
You may think of the gradient operator as acting on a scalar field, to produce a vector field. But the
gradient operator can also act on a vector field, to produce a tensor field. Heres how it works: You are
probably familiar with derivatives of a vector field:
( , , ) be a vector field. Then is a vector field.
Writing spatial vectors as column vectors, ,
Similarly, are
y
x z
x
x
y
y
z
z
A
A A
Let x y z
x x x x
A
x
A
A
A A and
x x
A
A
x
and
y z
c | | c c c
= + +
|
c c c c
\ .
c | |
|
c
| | |
c | c |
= =
| |
c c
| |
\ .
c
|
|
c
\ .
c c
c c
A
A x y z
A
A A
also vector fields.
By the rule for total derivatives, for a small displacement (dx, dy, dz),
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x x
y
x
x
x
y y
z
x
y y
z z z
z
z
y
A
z
A
dx A
A
x
x
dA
A A
dx dx
x x
A
A A
y y
A A dy
dA dy
y y
A A A
y
x
x
dz
y
dA
A dz
z y
d
x y z
| | | |
| |
| |
| |
| |
| c c c
= + + = = + | |
|
c c c
| |
|
\ . | |
| |
| |
\ .
c
c | |
|
c
c
|
c c
|
|
c c
|
c c | |
|
c c
c c
c c
c c
c c
c
\
| c
|
c
c \ .
c
c
c
c
c
.
c
\ .
A A A
A
x
y
z
A
z
A
dz
y
A
z
dy
c | |
|
c
|
c
|
|
c
|
|
|
|
|
| c
|
c \
|
+
.
|
|
This says that the vector dA is a linear combination of 3 column vectors A/x, A/y, and A/z,
weighted respectively by the displacements dx, dy, and dz. The 3 x 3 matrix above is the gradient of the
vector field A(r). It is the natural extension of the gradient (of a scalar field) to a vector field. It is a rank-2
tensor, which means that given a vector (dx, dy, dz), it produces a vector (dA) which is a linear combination
of 3 (column) vectors (VA), each weighted by the components of the given vector (dx, dy, dz).
Note that VA and VA are very different: the former is a rank-2 tensor field, the latter is a scalar field.
This concept extends further to derivatives of rank-2 tensors, which are rank-3 tensors: 3 x 3 x 3 cubes
of numbers, producing a linear combination of 3 x 3 arrays, weighted by the components of a given vector
(dx, dy, dz). And so on.
Note that in other coordinates (e.g., cylindrical or spherical), VA is not given by the derivative of its
components with respect to the 3 coordinates. The components interact, because the basis vectors also
change through space. That leads to the subject of differential geometry, discussed elsewhere in this
document.
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Greens Functions
Greens functions are a method of solving inhomogeneous linear differential equations (or other linear
operator equations):
{ } { } ( ) ( ), f x s x where is a linear operator = L L ,
We use them when other methods are hard, or to make a useful approximation (the Born
approximation). Sometimes, the Greens function itself can be given physical meaning, as in Quantum
Field Theory. Greens functions can generate particular (i.e. inhomogeneous) solutions, and solutions
matching boundary conditions. They dont generate homogeneous solutions (i.e., where the right hand side
is zero). We explore Greens functions through the following steps:
1. Extremely brief review of the -function.
2. The tired, but inevitable, electromagnetic example.
3. Linear differential equations of one variable (1-dimensional), with sources.
4. Delta function expansions.
5. Greens functions of two variables (but 1 dimension).
6. When you can collapse a Greens function to one variable (portable Greens functions:
translational invariance)
7. Dealing with boundary conditions: at least 5 (6??) kinds of BC
8. Green-like methods: the Born approximation
You will find no references to Greens Theorem or self-adjoint until we get to non-homogeneous
boundary conditions, because those topics are unnecessary and confusing before then. We will see that:
The biggest hurdle in understanding Greens functions is the boundary conditions.
Dirac Delta Function
Recall that the Dirac -function is an impulse, an infinitely narrow, tall spike function, defined as
( ) 0, 0, ( ) 1, 0 (the area under the d-function is 1)
a
a
x for x and x dx a o o
= = = >
}
.
The linearity of integration implies the delta function can be offset, and weighted, so that
( ) 0
b a
b a
w x b dx w a o
+
= >
}
Since the -function is infinitely narrow, it can pick out a single value from a function:
( ) ( ) ( ) 0
b a
b a
x b f x dx f b a o
+
= >
}
[It also implies (0) o , but we dont focus on that here.]
(See Funky Quantum Concepts for more on the delta function).
The Tired, But Inevitable, Electromagnetic Example
You probably have seen Poissons equation relating the electrostatic potential at a point to a charge
distribution creating the potential (in gaussian units):
(1)
2
( ) 4 ( ) electrostatic potential, charge density where | t | V = r r
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We solved this by noting three things: (1a) electrostatic potential, |, obeys superposition: the
potential due to multiple charges is the sum of the potentials of the individual charges; (1b) the potential is
proportional to the source charge; and (2) the potential due to a point charge is
1
( ) (point charge at origin) q
r
| = r
The properties (1a) and (1b) above, taken together, define a linear relationship:
1 1 2 2
1 2 1 2
( ) ( ), ( ) ( )
( ) ( ) ( ) ( ) ( )
total
Given and
Then a a
| |
| | |
+ = +
r r r r
r r r r r
To solve Eq (1), we break up the source charge distribution into an infinite number of little point
charges spread out over space, each of charge d
3
r. The solution for | is the sum of potential from all the
point charges, and the infinite sum is an integral, so we find | as
3
1
( ) ( ') '
'
d r | =
}
r r
r r
Note that the charge distribution for a point charge is a -function: infinite charge density, but finite
total charge. [We have also implicitly used the fact that the potential is translationally invariant, and
depends only on the distance from the source. We will remove this restriction later.]
But all of this followed from simple mathematical properties of Eq (1) that have nothing to do with
electromagnetics. All we used to solve for | was that the left-hand side is a linear operator on | (so
superposition applies), and we have a known solution when the right-hand side is a delta function:
2 2
" " int
" " '
1
( ) ( ) ( ')
'
linear linear
unknown given shource given po
operator operator
function function shource at
known
solution
and | t o V = 4 V =
r
r r r r
r r
_ _
_
The solution for a given is a sum of delta-function solutions. Now we generalize all this to arbitrary
(for now, 1D) linear operator equations by letting r x, | f, V
2
L, s, and call the known -
function solution G(x):
{ } { } Given ( ) ( ) ( ) ( ), then ( ) ( ') ' ( ') f x s x and G x x f x s x dx G x x o = = =
}
L L
assuming, as above, that our linear operator, and any boundary conditions, are translationally invariant.
A Fresh, New Signal Processing Example
If this example doesnt make sense to you, just skip it. Signal processing folk have long used a
Greens function concept, but with different words. A time-invariant linear system (TILS) produces an
output which is a linear operation on its input:
{ } { } ( ) ( ) is a linear operation taking input to output o t i t where =
In this case, we arent given {}, and we dont solve for it. However, we are given a measurement
(or computation) of the systems impulse response, called h(t) (not to be confused with a homogeneous
solution to anything). If you poke the system with a very short spike (i.e., if you feed an impulse into the
system), it responds with h(t).
{ } ( ) ( ) ( ) is the system's impulse response h t t where h t o =
Note that the impulse response is spread out over time, and usually of (theoretically) infinite duration.
h(t) fully characterizes the system, because we can approximate any input function as a series of impulses,
and sum up all the responses. Therefore, we find the output for any input, i(t), with:
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( ) ( ') ( ') ' o t i t h t t dt
=
}
h(t) acts like a Greens function, giving the system response at time t to a delta function at t = 0.
Linear differential equations of one variable, with sources
We wish to solve for f(x), given s(x):
{ } { }
2 2
2 2
2 2
( ) ( ), is a linear operator
( ) is called the "source," or forcing function
. ., ( ) ( ) ( ) ( )
f x s x where
s x
d d
E g f x f x f x s x
dx dx
e e
=
| |
+ + = |
|
\ .
L L
We ignore boundary conditions for now (to be dealt with later). The differential equations often have
3D space as their domain. Note that we are not differentiating s(x), which will be important when we get to
the delta-function expansion of s(x).
Greens functions solve the above equation by first solving a related equation: if we can find a function
(i.e., a Greens function) such that
{ }
2
2
2
( ) ( ), ( ) is the Dirac delta function
. ., ( ) ( )
G x x where x
d
E g G x x
dx
o o
e o
=
| |
+ = |
|
\ .
L
then we can use that Greens function to solve our original equation.
This might seem weird, because (0) , but it just means that Greens functions often have
discontinuities in them or their derivatives. For example, suppose G(x) is a step function:
( ) 0, 0
( ) ( )
1, 0
G x x d
G x x
x dx
Then o
= <
=
`
= >
)
.
Now suppose our source isnt centered at the origin, i.e., ( ) ( ) s x x a o = . If { } L is translation
invariant [along with any boundary conditions], then G( ) can still solve the equation by translation:
{ } ( ) ( ) ( ), ( ) ( ) is a solution. f x s x x a f x G x a o = = = L
If s(x) is a weighted sum of delta functions at different places, then because { } L is linear, the
solution is immediate; we just add up the solutions from all the -functions:
{ } ( ) ( ) ( ) ( ) ( )
i i i i
i i
f x s x w x x f x wG x x o = = =
L
Usually the source s(x) is continuous. Then we can use -functions as a basis to expand s(x) as an
infinite sum of delta functions (described in a moment). The summation goes over to an integral, and a
solution is
{ }
{ }
'
( ') '
1
( ) ( ) ( )
( ) ( ) ' ( ') ( ') ( ) ' ( ') ( ')
i
i
x x
w s x dx
i i
i
f x s x w x x
f x s x dx s x x x and f x dx s x G x x
o
o
=
= =
= = =
} }
L
L
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We can show directly that f(x) is a solution of the original equation by plugging it in, and noting that
{ } L acts in the x domain, and goes through (i.e., commutes with) any operation in x:
{ }
{ } { }
( ) ' ( ') ( ')
' ( ') ( ') moving inside the integral
' ( ') ( ') ( ) ( ) ( ). .
f x dx s x G x x
dx s x G x x
dx s x x x s x picks out the value of s x QED o o
=
`
)
=
= =
}
}
}
L L
L L
We now digress for a moment to understand the -function expansion.
Delta Function Expansion
As in the EM example, it is frequently quite useful to expand a given function s(x) as a sum of -
functions:
1
( ) ( ), are the weights of the basis delta functions
N
i i i
i
s x w x x where w o
=
~
[This same expansion is used to characterize the impulse-response of linear systems.]
x
Approximating a function
with delta functions
s(x)
N = 8
x
s(x)
x
x
i
w
i
= area
s(x
i
)x
N = 16
On the left, we approximate s(x) first with N = 8 -functions (green), then with N = 16 -functions
(red). As we double N, the weight of each -function is roughly cut in half, but there are twice as many of
them. Hence, the integral of the -function approximation remains about the same. Of course, the
approximation gets better as N increases. As usual, we let the number of -functions go to infinity: N .
On the right above, we show how to choose the weight of each -function: its weight is such that its
integral approximates the integral of the given function, s(x), over the interval covered by the -function.
In the limit of N , the approximation becomes arbitrarily good.
In what sense is the -function series an approximation to s(x)? Certainly, if we need the derivative
s'(x), the delta-function expansion is terrible. However, if we want the integral of s(x), or any integral
operator, such as an inner product or a convolution, then the delta-function series is a good approximation:
1
( ) *( ) ( ) ( ' ) ( ) ,
( ) ( ) ( )
N
i i i i
i
For s x dx or f x s x dx or f x x s x dx
then s x w x x where w s x x o
=
~ = A
} } }
As N , we expand s(x) in an infinite sum (an integral) of -functions:
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'
'
( ') '
( ) ( ) ( ) ' ( ') ( ')
i
i
x x
x dx
w s x dx
i i
i
s x w x x s x dx s x x x o o
= =
}
,
which if you think about it, follows directly from the definition of (x).
[Aside: Delta-functions are a continuous set of orthonormal basis functions, much like sinusoids from
quantum mechanics and Fourier transforms. They satisfy all the usual orthonormal conditions for a continuous
basis, i.e. they are orthogonal and normalized:
( ) ( ) ( ) dx x a x b a b o o o
=
}
]
Note that in the final solution of the prior section, we integrate s(x):
( ) ' ( ') ( ') f x dx s x G x x =
}
and integrating s(x) is what makes the -function expansion of s(x) valid.
Introduction to Boundary Conditions
We now incorporate a simple boundary condition. Consider a 2D problem in the plane:
{ } ( , ) ( , ) inside the boundary
( ) 0, where the boundary is given.
f x y s x y
f boundary
=
=
L
We define the vector r (x, y), and recall that
( ) ( ) ( ), ( ') ( ') ( ') x y so that x x y y o o o o o o = r r r
[Some references use the notation
(2)
(r) for a 2D -function.]
x
Domain
of f(x, y)
f(boundary) = 0
boundary
y
x
y
(r)
(r r')
Boundary condition does
NOT translate with r
Boundary condition
remains fixed
(Left) The domain of interest, and its boundary. (Right) A solution meeting the BC for the
source at (0, 0) does not translate to another point and still meet the BC.
The boundary condition removes the translation invariance of the problem. The delta-function
response of L{G(r)} translates, but the boundary condition does not. I.e., a solution of
{ } { } ( ) ( ), ( ) 0 ( ') ( ')
( ') 0
G and G boundary G
BUT does NOT G boundary
o o = = =
=
r r r r r r
r
L L
With boundary conditions, for each source point r', we need a different Greens function!
The Greens function for a source point r', call it G
r
(r), must satisfy both:
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{ }
' '
( ) ( ') ( ) 0 G and G boundary o = =
r r
r r r L .
We can think of this as a Greens function of two arguments, r and r', but really, r is the argument, and
r' is a parameter. In other words, we have a family of Greens functions, G
r
(r), labeled by the location of
the delta-function, r'.
Example: Returning to a 1D example in r: Find the Greens function for the equation
2
2
( ) ( ), on the interval [0,1], (0) (1) 0.
d
f r s r subject to f f
dr
= = =
Solution: The Greens function equation replaces the source s(r) with (r r'):
2
'
2
( ) ( ')
r
d
G r r r
dr
o =
Note that G
r
(r) satisfies the homogeneous equation on either side of r:
2
'
2
( ') 0
r
d
G r r
dr
= =
The full Greens function simply matches two homogeneous solutions, one to the left of r, and another
to the right of r, such that the discontinuity at r creates the required -function there. First we find the
homogeneous solutions:
2
2
( ) 0 Integrate both sides:
( ) C is an integration constant. Integrate again:
( ) , are arbitrary constants
d
h r
dr
d
h r C where
dr
h r Cr D where C D
=
=
= +
There are now 2 cases: (left) r < r', and (right) r > r'. Each solution requires its own set of
integration constants.
'
'
'
'
: ' ( )
Only the left boundary condition applies to ' : (0) 0 0
: ' ( )
Only the right boundary condition applies to ' : (1) 0 0,
r
r
r
r
Left case r r G r Cr D
r r G D
Right case r r G r Er F
r r G E F F E
< = +
< = =
> = +
> = + = =
So far, we have
: ( ') : ( ') Left case G r r Cr Right case G r r Er E < = > =
The integration constants C and E are as-yet unknown. Now we must match the two solutions at r = r',
and introduce a delta function there. The -function must come from the highest derivative in L{ }, in this
case the 2
nd
derivative, because if G(r) had a delta function, then the 2
nd
derivative G(r) would have the
derivative of a -function, which cannot be canceled by any other term in L{ }. Since the derivative of a
step (discontinuity) is a -function, G(r) must have a discontinuity, so that G(r) has a -function. And
finally, if G(r) has a discontinuity, then G(r) has a cusp (aka kink or sharp point).
We can find G(r) to satisfy all this by matching G(r) and G(r) of the left and right Greens functions,
at the point where they meet, r = r:
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' '
: ( ') : ( ')
There must be a unit step in the derivative across ' :
1
r r
d d
Left G r r C Right G r r E
dr dr
r r
C E
< = > =
=
+ =
So we eliminate E in favor of C. Also, G(r) must be continuous (or else G(r) would have a -
function), which means
' '
( ' ) ( ' ) ' ( 1) ' 1, ' 1
r r
G r r G r r Cr C r C C r
+
= = = = + =
yielding the final Greens function for the given differential equation:
( ) ( )
' '
( ') ' 1 , ( ') ' ' ' 1
r r
G r r r r G r r r r r r r < = > = =
Heres a plot of these Greens functions for different values of r':
r
r' = 0.3
G
r'
(r)
0
0.5
-0.5
r
0
0.5
-0.5
r
0
0.5
-0.5
0 1
r' = 0.5 r' = 0.8
G
r'
(r) G
r'
(r)
0 1 0 1
To find the solution f(x), we need to integrate over r'; therefore, it is convenient to write the Greens
function as a true function of two variables:
{ }
'
( ; ') ( ) ( ; ' ( '), ( ; ') 0
r
G r r G r G r r r r and G boundary r o = = L ,
where the ; between r and r' emphasizes that G(r ; r') is a function of r, parameterized by r'. I.e., we
can still think of G(r; r') as a family of functions of r, where each family member is labeled by r, and each
family member satisfies the homogeneous boundary condition.
It is important here that the boundary condition is zero, so that any sum of Greens functions still
satisfies the boundary condition.
Our particular solution to the original equation, which now satisfies the homogeneous boundary
condition, is
( ) ( )
1 1
0 0
( ; '), ' ( ; '), '
( ) ' ( ') ( ; ') ' ( ') ' 1 ' ( ') ' 1
which satisfies ( ) 0
r
r
G r r r r G r r r r
f r dr s r G r r dr s r r r dr s r r r
f boundary
> <
= = +
=
} } }
Summary: To solve { }
'
( ) ( ')
x
G x x x o = L , we break G(x) into left- and right- sides of x. Each side
satisfies the homogeneous equation, { }
'
( ) 0
x
G x = L , with arbitrary constants. We use the matching
conditions to achieve the -function at x, which generates a set of simultaneous equations for the unknown
constants in the homogeneous solutions. We solve for the constants, yielding the left-of-x and right-of-x
pieces of the complete Greens function, G
x
(x).
Aside: It is amusing to notice that we use solutions to the homogeneous equation to construct the Greens
function. We then use the Greens function to construct the particular solution to the given equation. So we are
ultimately constructing a particular solution from a homogeneous solution. Thats not like anything we learned in
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When Can You Collapse a Greens Function to One Variable?
Portable Greens Functions: When we first introduced the Greens function, we ignored boundary
conditions, and our Greens function was a function of one variable, r. If our source wasnt at the origin,
we just shifted our Greens function, and it was a function of just (r r). Then we saw that with (certain)
boundary conditions, shifting doesnt work, and the Greens function is a function of two variables, r and
r. In general, then, under what conditions can we write a Greens function in the simpler form, as a
function of just (r r)?
When both the differential operator and the boundary conditions are translation-invariant,
the Greens function is also translation-invariant.
We can say its portable. This is fairly common: differential operators are translation-invariant (i.e.,
they do not explicitly depend on position), and BCs at infinity are translation-invariant. For example, in
E&M it is common to have equations such as
2
( ) ( ), ( ) 0 with boundary condition | | V = = r r
Because both the operator V
2
and the boundary conditions are translation invariant, we dont need to
introduce r' explicitly as a parameter in G(r). As we did when introducing Greens functions, we can take
the origin as the location of the delta-function to find G(r), and use translation invariance to move around
the delta function:
{ }
'
( ; ') ( ) ( ') ( ') ( ')
( ) 0
r
G r r G r G r r and G r r r r
with BC G
o = =
=
L
Non-homogeneous Boundary Conditions
So far, weve dealt with homogeneous boundary conditions by requiring
'
( ) ( ; ')
r
G r G r r to be zero
on the boundary. There are different kinds of boundary conditions, and different ways of dealing with each
kind.
Note that in general, constraint conditions dont have to be specified at the boundary of anything. They are
really just constraints or conditions. For example, one constraint is often that the solution be a normalized
function, which is not a statement about any boundaries. But in most physical problems, at least one condition
does occur at a boundary, so we defer to this, and limit ourselves to boundary conditions.
Boundary Conditions Specifying Only Values of the Solution
In one common case, we are given a general (inhomogeneous) boundary condition, m(r) along the
boundary of the region of interest. Our problem is now to find the complete solution c(r) such that
{ } ( ) ( ), ( ) ( ) c r s r and c boundary m boundary = = L
One approach to find c(r) is from elementary differential equations: we find a particular solution f(x) to
the given equation, that doesnt necessarily meet the boundary conditions. Then we add a linear
combination of homogeneous solutions to achieve the boundary conditions, while preserving the solution of
the non-homogeneous equation. Therefore, we (1) first solve for f(r), as above, such that
{ }
{ }
( ) ( ), ( ) 0, using a Green's function satisfying
( ; ') ( ') ( ; ') 0
f r s r and f boundary
G r r r r and G boundary r o
= =
= =
L
L
(2) We then find homogeneous solutions h
i
(r) which are non-zero on the boundary, using ordinary
methods (see any differential equations text):
{ } ( ) 0, ( ) 0
i i
h r and h boundary = = L
Recall that in finding the Greens function, we already had to find homogeneous solutions, since every
Greens function is a homogeneous solution everywhere except at the -function position, r'.
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(3) Finally, we add a linear combination of homogeneous solutions to the particular solution to yield a
complete solution which satisfies both the differential equation and the boundary conditions:
{ }
{ } { }
{ }
1 1 2 2 1 1 2 2
1 1 2 2
1 1 2 2
( ) ( ) ... ( ), ( ) ( ) ... 0 by superposition
( ) ( ) ( ) ( ) ... ,
( ) ( ) ( ) ( ) ...
( ) ( ) ( ( )
A h r A h r m r A h r A h r
c r f r A h r A h r Therefore
c r f r A h r A h r
f r s r and c boundary m boundary
+ + = + + =
= + + +
= + + +
= = =
Continuing Example: In our 1D example above, we have:
{ } ( ) ( )
2
' '
2
' '
( ') ' 1 , ( ') ' 1 ,
: (0) (1) 0 (0) (1) 0, ( )
r r
r r
and G r r r r G r r r r
r
satisfying BC G G f f s r
c
= < = > =
c
= = = =
We now add boundary conditions to the original problem. We must satisfy c(0) = 2, and c(1) = 3, in
addition to the original problem. Our linearly independent homogeneous solutions are:
1 1 0 0
( ) ( ) (a constant) h r A r h r A = =
To satisfy the BC, we need
1 0 0
1 0 1
(0) (0) 2 2
(1) (1) 3 1
h h A
h h A
+ = =
+ = =
and our complete solution is
1
0
( ) ' ( ') ( ; ') 2 c r dr s r G r r r
(
= + +
(
}
Boundary Conditions Specifying a Value and a Derivative
Another common kind of boundary conditions specifies a value and a derivative for our complete
solution. For example, in 1D:
(0) 1 '(0) 5 c and c = =
But recall that our Greens function does not have any particular derivative at zero. When we find the
particular solution, f(x), we have no idea what its derivative at zero, f '(0), will be. And in particular,
different source functions, s(r), will produce different f(r), with different values of f '(0). This is bad. In the
previous case of BC, f(r) was zero at the boundaries for any s(r). What we need with our new BC is f(0) =
0 and f '(0) = 0 for any s(r). We can easily achieve this by using a different Greens function! We
subjected our first Greens function to the boundary conditions G(0; r) = 0 and G(1; r) = 0 specifically to
give the same BC to f(r), so we could add our homogeneous solutions independently of s(r). Therefore, we
now choose our Greens function BC to be:
{ } (0; ') 0 '(0; ') 0, ( ; ') ( ') G r and G r with G r r r r o = = =
We can see by inspection that this leads to a new Greens function:
( ; ') 0 ', ( ; ') ' ' G r r r r and G r r r r r r = < = >
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r
r' = 0.3
G(r ; r')
0
0.5
r
0
0.5
r
0
0.5
0 1 0 1 0 1
r' = 0.5 r' = 0.8
G(r ; r') G(r ; r')
The 2
nd
derivative of G(r; r) is everywhere 0, and the first derivative changes from 0 to 1 at r.
Therefore, our new particular solution f(r) also satisfies:
1
0
( ) ' ( ') ( ; ') (0) 0, '(0) 0, ( ) f r dr s r G r r and f f s r = = =
}
We now construct the complete solution using our homogeneous solutions to meet the BC:
1 1 0 0
1 0 0
1 0 1
1
0
( ) ( ) (a constant)
(0) (0) 1 1
'(0) '(0) 5 5.
( ) ' ( ') ( ; ') 5 1
h r A r h r A
h h A
h h A Then
c r dr s r G r r r
= =
+ = =
+ = =
(
= + +
(
}
In general, the Greens function depends not only on the particular operator, but also on the kind
of boundary conditions specified.
Boundary Conditions Specifying Ratios of Derivatives and Values
Another kind of boundary conditions specifies a ratio of the solution to its derivative, or equivalently,
specifies a linear combination of the solution and its derivative be zero. This is equivalent to a
homogeneous boundary condition:
'(0)
, (0) 0 '(0) (0) 0
(0)
c
or equivalently if c c c
c
o o = = =
This BC arises, for example, in some quantum mechanics problems where the normalization of the
wave-function is not yet known; the ratio cancels any normalization factor, so the solution can proceed
without knowing the ultimate normalization. Note that this is only a single BC. If our differential operator
is 2
nd
order, there is one more degree of freedom that can be used to achieve normalization, or some other
condition. (This BC is sometimes given as c'(0) c(0) = 0, but this simply multiplies both sides by a
constant, and fundamentally changes nothing.)
Also, this condition is homogeneous: a linear combination of functions which satisfy the BC also
satisfies the BC. This is most easily seen from the form given above, right:
( ) ( )
'(0) (0) 0, '(0) (0) 0,
( ) ( ) ( ) '(0) (0) 0
'(0) (0) '(0) (0) '(0) (0)
If d d and e e
then c r Ad r Be r satisfies c c
because c c A d d B e e
o o
o
o o o
= =
= + =
= +
Therefore, if we choose a Greens function which satisfies the given BC, our particular solution f(r)
will also satisfy the BC. There is no need to add any homogeneous solutions.
Continuing Example: In our 1D example above, with L = d
2
/dr
2
, we now specify BC:
'(0) 2 (0) 0 c c =
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Since our Greens functions for this operator are always two connected line segments (because their 2
nd
derivatives are zero), we have
' : ( ; ') , 0 (0) 0
' : ( ; ')
0 : 2 0
r r G r r Cr D D so that c
r r G r r Er F
BC at C D
< = + = =
> = +
=
With this BC, we have an unused degree of freedom, so we choose D = 1, implying C = 2. We must
find E and F so that G(r; r) is continuous, and G(r; r) has a unit step at r. The latter condition requires
that E = 3, and then continuity requires
' ' 2 ' 1 3 ' , ' 1.
' : ( ; ') 2 1 ' : ( ; ') 3 ' 1
Cr D Er F r r F F r So
r r G r r r and r r G r r r r
+ = + + = + = +
< = + > = +
r
r' = 0.3
G(r ; r')
1
0
1
1.6
2.5
4.0
r
r' = 0.5
G(r ; r')
1
0
1
r
r' = 0.8
G(r ; r')
1
0
1
2.5
4.0
2.5
4.0
and our complete solution is just
1
0
( ) ( ) ' ( ') ( ; ') c r f r dr s r G r r = =
}
Boundary Conditions Specifying Only Derivatives (Neumann BC)
Another common kind of BC specifies derivatives at points of the solution. For example, we might
have
'(0) 0 '(1) 1 c and c = =
Then, analogous to the BC specifying two values for c( ), we choose a Greens function which has
zeros for its derivatives at 0 and 1:
( 0 ; ') 0 ( 1; ') 0
d d
G r r and G r r
dr dr
= = = =
Then the sum (or integral) of any number of such Greens functions also satisfies the zero BC:
1
0
( ) ' ( ') ( ; ') '(0) 0 '(1) 0 f r dr s r G r r satisfies f and f = = =
}
We can now form the complete solution, by adding homogeneous solutions that satisfy the given BC:
1 1 2 2 1 1 2 2
1 1 2 2
( ) ( ) '( ) '( ) '(0) '(0) 0
'(1) '(1) 1
c r f r A h r A h r where A h A h
and A h A h
= + + + =
+ =
Example: We cannot use our previous example where L{ } = d
2
/dr
2
, because there is no solution to
2
2
( ; ') ( ') ( 0 ; ') ( 1; ') 0
d d d
G r r r r with G r r G r r
dr dr
dr
o = = = = =
This is because the homogenous solutions are straight line segments; therefore, any solution with a
zero derivative at any point must be a flat line. So we choose another operator as our example:
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3D Boundary Conditions: Yet Another Method
TBS: Using Greens theorem.
Green-Like Methods: The Born Approximation
In the Born approximation, and similar problems, we have our unknown function, now called (x), on
both sides of the equation:
(1) { } ( ) ( ) x x = L
The theory of Greens functions still works, so that
( ) ( ') ( ; ') ' x x G x x dx =
}
,
but this doesnt solve the equation, because we still have on both sides of the equation. We could try
rearranging Eq (1):
{ }
{ } { } { }
( ) ( ) 0 which is the same as
( ) 0, ( ) ( ) ( )
x x
x with x x x
=
=
L
L' L' L
But recall that Greens functions require a source function, s(x) on the right-hand side. The method of
Greens functions cant solve homogeneous equations, because it yields
{ } ( ) ( ) 0 ( ) ( ') ( ; ') ' 0 ' 0 x s x x s x G x x dx dx = = = = =
} }
L
which is a solution, but not very useful. So Greens functions dont work when (x) appears on both
sides. However, under the right conditions, we can make a useful approximation. If we have an
approximate solution,
{ }
0 0
( ) ( ) x x ~ L
we can use
0
(x) as the source term, and use the method of Greens functions, to get a better
approximation to (x):
{ }
{ }
1 0 1 0
( ) ( ) ( ) ( ') ( ; ') '
( ; ') is the Green's function for , . . ( ; ') ( ')
x x x x G x x dx
where G x x i e G x x x x
o
= =
=
}
L
L L
1
(x) is called the first Born approximation of (x). Of course, this process can be repeated to
arbitrarily high accuracy:
2 1 1
( ) ( ') ( ; ') ' . . . ( ) ( ') ( ; ') '
n n
x x G x x dx x x G x x dx
+
= =
} }
TBS: a real QM example.
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Complex Analytic Functions
For a review of complex numbers and arithmetic, see Funky Quantum Concepts.
Notation: In this chapter, z, w are always complex variables; x, y, r, are always real variables.
Other variables are defined as used.
A complex function of a complex variable f(z) is analytic over some domain if it has an infinite
number of continuous derivatives in that domain. It turns out, if f(z) is once differentiable on a domain,
then it is infinitely differentiable, and therefore analytic on that domain.
A necessary condition for analyticity of f(z) = u(x, y) + iv(x, y) near z
0
is that the Cauchy-Riemann
equations hold, to wit:
,
f f u v u v u v u v v u
i i i i i and
x y x x y y y y x y x y
| | c c c c c c c c c c c c
= + = + = + = =
|
c c c c c c c c c c c c
\ .
A sufficient condition for analyticity of f(z) = u(x, y) + iv(x, y) near z
0
is that the Cauchy-Riemann
equations hold, and the first partial derivatives of f exist and are continuous in a neighborhood of z
0
. Note
that if the first derivative of a complex function is continuous, then all derivatives are continuous, and the
function is analytic. This condition implies
2
1
2 2
0
0 " "
( ) is countour independent if ( ) is single-valued
z
z
u v
u v level lines are perpendicular
f z dz f z
V = V =
V V =
}
Note that a function can be analytic in some regions, but not others. Singular points, or singularities,
are not in the domain of analyticity of the function, but border the domain [Det def 4.5.2 p156]. E.g., \z is
singular at 0, because it is not differentiable, but it is continuous at 0. Poles are singularities near which the
function is unbounded (infinite), but can be made finite by multiplication by (z z
0
)
k
for some finite k [Det
p165]. This implies f(z) can be written as:
1 1 1
0 1 0 1 0 0 1 0
( ) ( ) ( ) ... ( ) ( ) ...
k k
k k
f z a z z a z z a z z a a z z
+
= + + + + + +
The value k is called the order of the pole. All poles are singularities. Some singularities are like
poles of infinite order, because the function is unbounded near the singularity, but it is not a pole because
it cannot be made finite by multiplication by any (z z
0
)
k
, for example e
1/z
. Such a singularity is called an
essential singularity.
A Laurent series expansion of a function is similar to a Taylor series expansion, but the sum runs from
to +, instead of from 1 to . In both cases, an expansion is about some point, z
0
:
( )
( )
( )
0
( )
0
0 0
1
0
1
0
( )
Taylor series: ( ) ( )
!
1 ( )
Laurent series: ( ) ,
2
n
n
n n
n
n
n n
k
around z
n
f z
f z f z b z z where b
n
f z
f z a z z where a dz
i
z z
t
+
=
= + =
= =
}
[Det thm 4.6.1 p163] Analytic functions have Taylor series expansions about every point in the
domain. Taylor series can be thought of as special cases of Laurent series. But analytic functions also have
Laurent expansions about isolated singular points, i.e. the expansion point is not even in the domain of
analyticity! The Laurent series is valid in some annulus around the singularity, but not across branch cuts.
Note that in general, the a
k
and b
k
could be complex, but in practice, they are often real.
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Properties of analytic functions:
1. If it is differentiable once, it is infinitely differentiable.
2. The Taylor and Laurent expansions are unique. This means you may use any of several methods
to find them for a given function.
3. If you know a function and all its derivatives at any point, then you know the function everywhere
in its domain of analyticity. This follows from the fact that every analytic function has a Laurent
power series expansion. It implies that the value throughout a region is completely determined by
its values at a boundary.
4. An analytic function cannot have a local extremum of absolute value. (Why not??)
Residues
Mostly, we use complex contour integrals to evaluate difficult real integrals, and to sum infinite series.
To evaluate contour integrals, we need to evaluate residues. Here, we introduce residues. The residue of a
complex function at a complex point z
0
is the a
1
coefficient of the Laurent expansion about the point z
0
.
Residues of singular points are the only ones that interest us. (In fact, residues of branch points are not
defined [Sea sec 13.1].)
Common ways to evaluate residues
1. The residue of a removable singularity is zero. This is because the function is bounded near the
singularity, and thus a
1
must be zero (or else the function would blow up at z
0
):
1 0 1 1
0
1
For 0, as , 0 a z z a a
z z
= =
2. The residue of a simple pole at z
0
(i.e., a pole of order 1) is
( )
0
1 0
lim ( )
z z
a z z f z
= .
3. Extending the previous method: the residue of a pole at z
0
of order k is
( )
( )
0
1
1 0
1
1
lim ( )
1 !
k
k
k
z z
d
a z z f z
k
dz
,
which follows by substitution of the Laurent series for f(z), and direct differentiation. Noting that
poles of order m imply that a
k
= 0 for k < m, we get:
( ) ( )
( )
1 1 1
0 1 0 1 0 0 1 0
1 1 1
0 1 0 1 0 0 0 1 0
1
1 1
0 1 0 0 0 1 0
1
( ) ( ) ( ) ... ( ) ( ) ...
( ) ( ) ( ) ... ( ) ( ) ( ) ...
1 ! !
( ) 1 ! ( ) ( ) ( )
1! 2!
k k
k k
k k k k
k k
k
k
k k k
k
f z a z z a z z a z z a a z z
z z f z a a z z a z z a z z a z z
k d k
z z f z k a z z a z z a z z
dz
+
+
= + + + + + +
= + + + + + +
+
= + + +
( ) ( )
( )
( )
0
0
1
0 1
1
1
1 0
1
...
lim ( ) 1 !
1
lim ( )
1 !
k
k
k
z z
k
k
k
z z
d
z z f z k a
dz
d
a z z f z
k
dz
=
=
4. If f(z) can be written as
( )
( )
( )
P z
f z
Q z
= , where P is continuous at z
0
, and Q(z
0
) = 0 (and is
continuous at z
0
), then f(z) has a simple pole at z
0
, and
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( )
( ) ( )
( )
0
0
0 0 0
0 0
0 0 0 0
0
0 0
0 0 0
0 0 0
( ) ( )
Res ( ) . ? , ( ) '( ).
'( )
( )
( ) ( )
Res ( ) lim ( ) lim
'( ) '( )
z z
z
z z z z z z
P z P z
f z Why Near z Q z z z Q z Then
d Q z
Q z
dz
P z P z
f z z z f z z z
z z Q z Q z
=
=
= ~
= = =
5. Find the Laurent series, and hence its coefficient of (z z
0
)
1
. This is sometimes easy if f(z) is
given in terms of functions with well-known power series expansions. See the sum of series
example later.
We will include real-life examples of most of these as we go.
Contour Integrals
Contour integration is an invaluable tool for evaluating both real and complex-valued integrals.
Contour integrals are used all over advanced physics, and we could not do physics as we know it today
without them. Contour integrals are mostly useful for evaluating difficult ordinary (real-valued) integrals,
and sums of series. In many cases, a function is analytic except at a set of distinct points. In this case, a
contour integral may enclose, or pass near, some points of non-analyticity, i.e. singular points. It is these
singular points that allow us to evaluate the integral.
You often let the radius of the contour integral go to for some part of the contour:
real
imaginary
C
R
R
.Any arc where
1
1
lim ( ) ~ , 0
R
f z
z
c
c
+
>
has an integral of 0 over the arc.
Beware that this is often stated incorrectly as any function which goes to zero faster than 1/|z| has
a contour integral of 0. The problem is that it has to have an exponent < 1; it is not sufficient to
be simply smaller than 1/|z|. E.g.
1 1
1 z z
<
+
, but the contour integral still diverges.
Jordans lemma: ??.
Evaluating Integrals
Surprisingly, we can use complex contour integrals to evaluate difficult real integrals. The main point
is to find a contour which (a) includes some known (possibly complex) multiple of the desired (real)
integral, (b) includes other segments whose values are zero, and (c) includes a known set of poles whose
residues can be found. Then you simply plug into the residue theorem:
( ) 2 Res ( ), are the finite set of isolated singularities
n
n
C z
n residues
f z dz i f z where z t =
}
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We can see this by considering the contour integral around the unit circle for each term in the Laurent
series expanded about 0. First, consider the z
0
term (the constant term). We seek the value of
O
dz
}
. dz is
a small complex number, representable as a vector in the complex plane. The diagram below (left) shows
the geometric meaning of dz. Below (right) shows the geometric approximation to the desired integral.
real
Imaginary
u
du
unit
circle
dz = e
i(u+/4)
du
dz
1
dz
2
dz
i
dz
N
(Left) Geometric description of dz.
(Right) Approximation of
O
dz
}
as a sum of 32 small complex terms (vectors).
We see that all the tiny dz elements add up to zero: the vectors add head-to-tail, and circle back to the
starting point. The sum vector (displacement from start) is zero. This is true for any large number of dz, so
we have
0
O
dz =
}
Next, consider the z
1
term,
1
O
dz
z
| |
|
\ .
}
, and a change of integration variable to :
2 2
0 0
1
, : 2
i i i i
O
Let z e dz ie d dz e ie d id i
z
t t
u u u u
u u u t
| |
= = = = =
|
\ .
} } }
The change of variable maps the complex contour and z into an ordinary integral of a real variable.
Geometrically, as z goes positively (counter-clockwise) around the unit circle (below left), z
1
goes
around the unit circle in the negative (clockwise) direction (below middle). Its complex angle, arg(1/z) =
, where z = e
i
. As z goes around the unit circle, dz has infinitesimal magnitude c = d, and argument +
t/4. Hence, the product of (1/z) dz always has argument of + + t/4 = t/4; it is always purely
imaginary.
real
Imaginary
A
B
C
D
E
Path of z = e
i
around unit
circle
real
Imaginary
A
B
C
D
E
Path of z = e
i
around unit circle
real
Imaginary
A
B
C
D
E
Path of dz = ie
i
around unit circle
Paths of z, 1/z, and dz in the complex plane
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The magnitude of (1/z) dz = du; thus the integral around the circle is 2ti. Multiplying the integrand by
some constant, a
1
(the residue), just multiplies the integral by that constant. And any contour integral that
encloses the pole 1/z and no other singularity has the same value. Hence, for any contour around the origin
( )
1
1
1
1 1 1
2
2
O
O
a z dz
a z dz i a a
i
t
t
= =
}
}
Now consider the other terms of the Laurent expansion of f(z). We already showed that the a
0
z
0
term,
which on integration gives the product a
0
dz, rotates uniformly about all directions, in the positive (counter-
clockwise) sense, and sums to zero. Hence the a
0
term contributes nothing to the contour integral.
The a
1
z
1
dz product rotates uniformly twice around all directions in the positive sense, and of course,
still sums to zero. Higher powers of z simply rotate more times, but always an integer number of times
around the circle, and hence always sum to zero.
Similarly, a
2
z
2
, and all more negative powers, rotate uniformly about all directions, but in the
negative (clockwise) sense. Hence, all these terms contribute nothing to the contour integral.
So in the end:
The only term of the Laurent expansion about 0 that contributes to the contour integral is the
residue term, a1 z1.
The simplest contour integral: Evaluate
2
0
1
1
I dx
x
=
+
}
.
We know from elementary calculus (let x = tan u) that I = / 2. We can find this easily from the
residue theorem, using the following contour:
real
imaginary
C
I
C
R
i
-i
R
C
I
C denotes a contour, and I denotes the integral over that contour. We let the radius of the arc go to
infinity, and we see that the closed contour integral I
C
= I + I + I
R
. But I
R
= 0, because f(R ) < 1/R
2
.
Then I = I
C
/ 2. f(z) has poles at i. The contour encloses one pole at i. Its residue is
( )
2
1 1 1 1
Res ( ) . 2 Res ( ) 2
2 2 2
1
2 2
C n
z i n
z i
C
f i I i f z i
d z i i
z
dz
I
I
t t t
t
=
=
= = = = = =
+
= =
Note that when evaluating a real integral with complex functions and contour integrals, the is always
cancel, and you get a real result, as you must. Its a good check to make sure this happens.
Choosing the Right Path: Which Contour?
The path of integration is fraught with perils. How will I know which path to choose? There is no
universal answer. Often, many paths lead to the same truth. Still, many paths lead nowhere. All we can do
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is use wisdom as our guide, and take one step in a new direction. If we end up where we started, we are
grateful for what we learned, and we start anew.
We here examine several useful and general, but oft neglected, methods of contour integration. We
use a some sample problems to illustrate these tools. This section assumes a familiarity with contour
integration, and its use in evaluating definite integrals, including the residue theorem.
Example: Evaluate
2
2
sin x
I dx
x
=
}
The integrand is everywhere nonnegative, and somewhere positive, and it is in the positive direction,
so I must be positive. We observe that the given integrand has no poles. It has only a removable
singularity at x = 0. If we are to use contour integrals, we must somehow create a pole (or a few), to use the
residue theorem. Simple poles (i.e. 1
st
-order) are sometimes best, because then we can also use the
indented contour theorem.
real
Imaginary
I
R
= 0
I
r
real
Imaginary
I
R
= 0
I
r
Contours for the two exponential integrals: (left) positive (counter-clockwise) exp(2z);
(right) negative (clockwise) exp(2z)
To use a contour integral (which, a priori, may or may not be a good idea), we must do two things: (1)
create a pole; and (2) close the contour. The same method does both: expand the sin( ) in terms of
exponentials:
( )
( )
2
2 2 2
2 2 2 2 2 2
sin 1 2
4
2
iz iz
i z i z e e
x e e
I dx dz dz dz dz
x z z z i z
(
= = = +
(
(
} } } } }
All three integrals have poles at z = 0. If we indent the contour underneath the origin, then since the
function is bounded near there, the limit as r 0 leaves the original integral unchanged (above left). The
first integral must be closed in the upper half-plane, to keep the exponential small. The second integral can
be closed in either half-plane, since it ~ 1/z
2
. The third integral must be closed in the lower half-plane,
again to keep the exponential small (above right). Note that all three contours must use an indentation that
preserves the value of the original integral. An easy way to insure this is to use the same indentation on all
three.
Now the third integral encloses no poles, so is zero. The 2
nd
integral, by inspection of its Laurent
series, has a residue of zero, so is also zero. Only the first integral contributes. By expanding the
exponential in a Taylor series, and dividing by z
2
, we find its residue is 2i. Using the residue theorem, we
have:
( )
2
2
sin 1
2 2
4
x
I dx i i
x
t t
= = = (
}
Example: Evaluate
2
0
cos( ) cos( ) ax bx
I dx
x
=
}
[B&C p?? Q1]
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This innocent looking problem has a number of funky aspects:
- The integrand is two terms. Separately, each term diverges. Together, they converge.
- The integrand is even, so if we choose a contour that includes the whole real line, the contour
integral includes twice the integral we seek (twice I).
- The integrand has no poles. How can we use any residue theorems if there are no poles?
Amazingly, we can create a useful pole.
- A typical contour includes an arc at infinity, but cos(z) is ill-behaved for z far off the real-axis.
How can we tame it?
- We will see that this integral leads to the indented contour theorem, which can only be applied to
simple poles, i.e., first order poles (unlike the residue theorem, which applies to all poles).
Each of these funky features is important, and each arises in practical real-world integrals. Let us
consider each funkiness in turn.
1. The integrand is two terms. Separately, each term diverges. Together, they converge.
Near zero, cos(x) 1. Therefore, the zero endpoint of either term of the integral looks like
2 2
0 0
0
cos 1 1
~
anywhere
anywhere anywhere ax
dx dx
x
x x
= +
} }
Thus each term, separately, diverges. However, the difference is finite. We see this by power series
expanding cos(x):
( )
( ) ( )
2 4 2 2 2 2
4
2 2 2 2
2 2
2
2 2
2
0
cos( ) 1 ... cos( ) cos( )
2! 4! 2 2
cos( ) cos( )
2 2 2
cos( ) cos( )
~ which is to say, is finite.
2
anywhere
x x a x b x
x ax bx O x and
ax bx a b b a
O x O x
x
ax bx b a
dx
x
= + = + +
= + + = +
}
2. The integrand is even, so if we choose a contour that includes the whole real line, the contour
integral includes twice the integral we seek (twice I).
Perhaps the most common integration contour (below left) covers the real line, and an infinitely distant
arc from + back to . When our real integral (I in this case) is only from 0 to , the contour integral
includes more than we want on the real axis. If our integrand is even, the contour integral includes twice
the integral we seek (twice I). This may seem trivial, but the point to notice is that when integrating from
to 0, dx is still positive (below middle).
real
R
x
imaginary
f(x) even
dx > 0
(Left) A common contour.
(Right) An even function has integral over the real-line twice that of 0 to infinity.
Note that if the integrand is odd (below left), choosing this contour cancels out the original (real)
integral from our contour integral, and the contour is of no use. Or if the integrand has no even/odd
symmetry (below middle), then this contour tells us nothing about our desired integral. In these cases, a
different contour may work, for example, one which only includes the positive real axis (below right).
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real
R
imaginary
x
f(x) asymmetric
x
f(x) odd
dx > 0
(Left) An odd function has zero integral over the real line. (Middle) An asymmetric function
has unknown integral over the real line. (Right) A contour containing only the desired real
integral.
3. The integrand has no poles. How can we use any residue theorems if there are no poles?
Amazingly, we can create a useful pole.
This is the funkiest aspect of this problem, but illustrates a standard tool. We are given a real-valued
integral with no poles. Contour integration is usually useless without a pole, and a residue, to help us
evaluate the contour integral. Our integrand contains cos(x), and that is related to exp(ix). We could try
replacing cosines with exponentials,
( ) ( ) exp exp
cos (does no good)
2
iz iz
z
+
=
but this only rearranges the algebra; fundamentally, it buys us nothing. The trick here is to notice that
we can often add a made-up imaginary term to our original integrand, perform a contour integration, and
then simply take the real part of our result:
( )
{ }
( ) , ( ) ( ). Re ( )
b b
a a
Given I g x dx let f z g z ih z Then I f z dz = + =
} }
For this trick to work, ih(z) must have no real-valued contribution over the contour we choose, so it
doesnt mess up the integral we seek. Often, we satisfy this requirement by choosing ih(z) to be purely
imaginary on the real axis, and having zero contribution elsewhere on the contour. Given an integrand
containing cos(x), as in our example, a natural choice for ih(z) is i sin(z), because then we can write the new
integrand as a simple exponential:
cos( ) ( ) cos( ) sin( ) exp( ) x f z z i z iz = + =
In our example, the corresponding substitution yields
2 2
0 0
cos cos exp( ) exp( )
Re
ax bx iax ibx
I dx I dx
x x
= =
`
)
} }
Examining this substitution more closely, we find a wonderful consequence: this substitution
introduced a pole! Recall that
3
2
sin 1
sin ... ...
3! 3!
z i z z
z z i
z
z
| |
= + = +
|
\ .
We now have a simple pole at z = 0, with residue i.
By choosing to add an imaginary term to the integrand, we now have a pole that we can work with
to evaluate a contour integral!
Its like magic. In our example integral, our residue is:
( )
2
sin sin
... ,
i az i bz a b
i and residue i a b
z
z
| |
= + =
|
\ .
Note that if our original integrand contained sin(x) instead of cos(x), we would have made a similar
substitution, but taken the imaginary part of the result:
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( )
{ }
sin( ) , cos( ) sin( ). Im ( )
b b
a a
Given I x dx let f z z i z Then I f z dz = + =
} }
4. A typical contour includes an arc at infinity, but cos(z) is ill-behaved for z far off the real-
axis. How can we tame it?
This is related to the previous funkiness. Were used to thinking of cos(x) as a nice, bounded, well-
behaved function, but this is only true when x is real.
When integrating cos(z) over a contour, we must remember that cos(z) blows up rapidly off the
real axis.
In fact, cos(z) ~ exp(Im{z}), so it blows up extremely quickly off the real axis. If were going to
evaluate a contour integral with cos(z) in it, we must cancel its divergence off the real axis. There is only
one function which can exactly cancel the divergence of cos(z), and that is i sin(z). The plus sign cancels
the divergence above the real axis; the minus sign cancels it below. There is nothing that cancels it
everywhere. We show this cancellation simply:
( ) ( ) cos sin exp( ) exp exp( ) exp( )
exp( ) exp( ) exp( ) exp( ) exp( )
Let z x iy
z i z iz i x iy ix y and
ix y ix y y
+
+ = = + =
= =
For z above the real axis, this shrinks rapidly. Recall that in the previous step, we added i sin(x) to our
integrand to give us a pole to work with. We see now that we also need the same additional term to tame
the divergence of cos(z) off the real axis. For the contour weve chosen, no other term will work.
5. We will see that this integral leads to the indented contour theorem, which can only be
applied to simple poles, i.e., first order poles (unlike the residue theorem, which applies to all
poles).
Were now at the final step. We have a pole at z = 0, but it is right on our contour, not inside it. If the
pole were inside the contour, we would use the residue theorem to evaluate the contour integral, and from
there, wed find the integral on the real axis, cut it in half, and take the real part. That is the integral we
seek.
But the pole is not inside the contour; it is on the contour. The indented contour theorem allows us to
work with poles on the contour. We explain the theorem geometrically in the next section, but state it
briefly here:
Indented contour theorem: For a simple pole, the integral of an arc of tiny radius around the pole,
of angle , equals (i)(residue). See diagram below.
real
imaginary
real
imaginary
0,
( ) ( )( )
arc
As
f z dz i residue
=
}
arc
(Left) A tiny arc around a simple pole. (Right) A magnified view; we let 0.
Note that if we encircle the pole completely, = 2t, and we have the special case of the residue
theorem for a simple pole:
( ) ( ) 2 f z dz i residue t =
}
However, the residue theorem is true for all poles, not just simple ones (see The Residue Theorem
earlier).
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Putting it all together: We now solve the original integral using all of the above methods. First, we
add i sin(z) to the integrand, which is equivalent to replacing cos(z) with exp(iz):
{ }
2 2
0 0
2
0
cos cos exp( ) exp( )
Re
exp( ) exp( )
, Re
ax bx iax ibx
I dx I dx
x x
iax ibx
Define J dx so I J
x
= =
`
)
=
} }
}
We choose the contour shown below left, with R , and 0.
real
imaginary
R
real
imaginary
R
C
2
C
R
C
There are no poles enclosed, so the contour integral is zero. The contour includes twice the desired
integral, so
(1)
2
exp( ) exp( )
( ) ( ) ( ) 2 ( ) 0
R
C C
iaz ibz
Define f z then f z dz f z dz J f z dz
z
= + + =
} } }
For C
R
, |f(z)| < 1/R
2
, so as R , the integral goes to 0. For C
## , the residue is i(a b), and the arc is t
radians in the negative direction, so the indented contour theorem says:
( ) ( ) ( )
0
lim ( )
C
f z dz i i a b a b
t t
= =
}
Plugging into (1), we finally get
( ) { } ( ) 2 0 Re
2
J a b I J b a
t
t + = = =
In this example, the contour integral J happened to be real, so taking I = Re{J} is trivial, but in general,
theres no reason why J must be real. It could well be complex, and we would need to take the real part of
it.
To illustrate this and more, we evaluate the integral again, now with the alternate contour shown above
right. Again, there are no poles enclosed, so the contour integral is zero. Again, the integral over C
R
= 0.
We then have:
( ) ( )
R
C
f z dz f z dz =
} }
( ) ( ) ( )
2
0
( ) ( ) 0
lim ( ) / 2
2
C C
C
f z dz J f z dz
And f z dz i i a b a b
t
t
+ + + =
= =
} }
}
The integral over C
2
is down the imaginary axis:
( ) ( ) ( ) ( )
2 2
0
2 2
0 ,
exp exp exp exp
( )
C C
Let z x iy iy iy then dz i dy
iaz ibz ay by
f z dz dz i dy
z y
= + = + = =
= =
} } }
We dont know what this integral is, but we dont care! In fact, it is divergent, but we see that it is
purely imaginary, so will contribute only to the imaginary part of J. But we seek I = Re{J}, and therefore
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{ }
0
lim Re is well-defined. I J
=
So we ignore the divergent imaginary contribution from C
2
. We then have
( ) ( ) { } ( ) 0 Re
2 2
i something J a b I J b a
t t
+ + = = =
as before.
Evaluating Infinite Sums
The simplest infinite sum in the world is
2
1
1
n
S
n
=
=
## . The general method for using contour integrals
is to find an countably infinite set of residues whose values are the terms of the sum, and whose contour
integral can be evaluated by other means. Then
1
2 Res ( ) 2
2
C
C n
n
I
I i f z iS S
i
t t
t
=
= = =
The hard part is finding the function f(z) that has the right residues. Such a function must first have
poles at all the integers, and then also have residues at those poles equal to the terms of the series.
Consider the complex function cot(z). Clearly, this has poles at all real integer z, due to the sin(z)
function in the denominator of cot(z). Hence,
( )
( )
( )
( )
( )
cos cos
, Res cot Res 1
sin cos
n n
n n
n n
z z
For z n z
z z
t t
t t t
t t
(
( = = = =
(
(
Thus t cot(tz) can be used to generate lots of infinite sums, by simply multiplying it by a continuous
function of z that is the terms of the infinite series. For example, for the sum above,
2
1
1
n
S
n
=
=
, we simply
define
( )
2 2
1 1
( ) cot , and its residues are Res ( ) , 0
n
f z z f z n
z n
t t = = = .
[In general, to find
1
( )
n
s n
, define
( ) ( ) ( ) cot , and its residues are Res ( ) ( )
z n
f z s z z f z s n t t
=
= = (
.
However, now you may have to deal with the residues for n s 0.]
Continuing our example, now we need the residue at n = 0. Since cot(z) has a simple pole at zero,
cot(z)/z
2
has a 3
rd
order pole at zero. We optimistically try tedious brute force for an m
th
order pole with m
= 3, only to find that it fails:
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2 2
3
2 2 2 2
0 0 0
2
2 2
0 0 0
2
cot 1 cot 1
Res lim lim cot
2! 2!
1
sin 2
cos sin
2
lim cot csc lim lim
2 2 2
sin sin
:
l
2
z z z
z z z
z d z d
z z z
z dz z dz
z z
d d z z z d
z z z
dz dz dz
z z
U VdU UdV
Use d
V V
t t t t
t t
t t
t t t t t t
t t t
t t
t
=
( (
= =
( (
(
(
(
(
= = =
(
(
(
=
=
( )
( )
( ) ( ) ( )
2
4
0
3
0
0
1
sin cos 2 sin 2 2 sin cos
2
im
sin
1
sin cos 2 sin 2 2 cos
2
lim
2 sin
' ' :
1
cos cos2 sin 2 sin 2 1 cos 2 2 cos sin 2
2
lim
2
z
z
z
z z z z z z
z
z z z z z
z
Use L hopital s rule
z z z z z z z z
t t t t t t t t t
t
t t t t t t t t
t
t
t t t t t t t t t t t t t t t
t
| |
|
\ .
| |
|
\ .
=
|
+
=
( ) ( )
2
2
2 2
2
0
2 sin
3 sin cos
1
cos cos 2 1 sin 2 sin 2 1 2 sin 2 sin
2
lim
2
3 sin cos
z
z
z z
z z z z z z z
z z
t t
t t t
t t t t t t t t t t
t
t t t
|
|
\ .
| |
+
|
\ .
=
At this point, we give up on brute force, because we see from the denominator that well have to use
LHopitals rule twice more to eliminate the zero there, and the derivatives wil l get untenably complicated.
But in 2 lines, we can find the a
1
term of the Laurent series from the series expansions of sin and cos.
The z
1
coefficient of cot(z) becomes the z
-1
coefficient of f(z) = cot(z)/z
2
:
( ) ( ) ( )
2 2
2 2 2
3 2
2
2
0
cos 1 / 2 ... 1 1 / 2 1 1 1
cot 1 / 2 1 / 6 1 / 3
sin 3 / 6 ... 1 / 6
1 cot
cot Res
3 3 z
z z z z
z z z z
z z z z z z z z
z z
z
z z
t t t
t t
t =
+ | | | | | |
= ~ = ~ + ~ =
| | |
+ \ . \ . \ .
~ =
Now we take a contour integral over a circle centered at the origin: (no good, because cot(tz) blows up
every integer ! ??)
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real
imaginary
I
C
As R , I
C
0. Hence
2
0
0 0
2 2 2 2
1 1 1 1
1 1 1 1
0 2 2 0,
2 6
C
n n n n
K
I i K K
n n n n
t
t
= = = =
| |
= = + + + = = = |
|
\ .
Multi-valued Functions
Many functions are multi-valued (despite the apparent oxymoron), i.e. for a single point in the
domain, the function can have multiple values. An obvious example is a square-root function: given a
complex number, there are two complex square roots of it. Thus, the square root function is two-valued.
Another example is arc-tangent: given any complex number, there are an infinite number of complex
numbers whose tangent is the given complex number.
[picture??]
We refer now to nice functions, which are locally (i.e., within any small finite region) analytic, but
multi-valued. If youre not careful, such multi-valuedness can violate the assumptions of analyticity, by
introducing discontinuities in the function. Without analyticity, all our developments break down: no
contour integrals, no sums of series. But, you can avoid such a breakdown, and preserve the tools weve
developed, by treating multi-valued functions in a slightly special way to insure continuity, and therefore
analyticity.
A regular function, or region, is analytic and single valued. (You can get a regular function from a
multi-valued one by choosing a Riemann sheet. More below.)
A branch point is a point in the domain of a function f(z) with this property: when you traverse a
closed path around the branch point, following continuous values of f(z), f(z) has a different value at the end
point of the path than at the beginning point, even though the beginning and end point are the same point in
the domain. Example TBS: square root around the origin. Sometimes branch points are also singularities.
A branch cut is an arbitrary (possibly curved) path connecting branch points, or running from a
branch point to infinity (connecting the branch point to infinity). If you now evaluate integrals of
contours that never cross the branch cuts, you insure that the function remains continuous (and thus
analytic) over the domain of the integral.
When the contour of integration is entirely in the domain of analyticity of the integrand,
ordinary contour integration, and the residue theorem, are valid.
This solves the problem of integrating across discontinuities. Branch cuts are like fences in the domain
of the function: your contour integral cant cross them. Note that youre free to choose your branch cuts
wherever you like, so long as the function remains continuous when you dont cross the branch cuts.
Connecting branch points is one way to insure this.
A Riemann sheet is the complex plane plus a choice of branch cuts, and a choice of branch. This
defines a domain on which a function is regular.
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A Riemann surface is a continuous joining of Riemann sheets, gluing the edges together. This looks
like sheets layered on top of each other, and each sheet represents one of the multiple values a multi-
valued analytic function may have. TBS: consider ( ) ( ) z a z b .
real
imaginary
branch cut
real
imaginary
branch cuts
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Conceptual Linear Algebra
Instead of lots of summation signs, we describe linear algebra concepts, visualizations, and ways to
think about linear operations as algebraic operations. This allows fast understanding of linear algebra
methods that is extremely helpful in almost all areas of physics. Tensors rely heavily on linear algebra
methods, so this section is a good warm-up for tensors. Matrices and linear algebra are also critical for
quantum mechanics.
Caution In this section, vector means a column or row of numbers. In other sections, vector
has a more general meaning.
In this section, we use bold capitals for matrices (A), and bold lower-case for vectors (a).
Matrix Multiplication
It is often helpful to view a matrix as a horizontal concatenation of column-vectors. You can think of
it as a row-vector, where each element of the row-vector is itself a column vector.
or
(
(
(
(
= =
(
(
(
(
d
A a b c A e
f
Equally valid, you can think of a matrix as a vertical concatenation of row-vectors, like a column-
vector where each element is itself a row-vector.
Matrix multiplication is defined to be the operation of linear transformation, e.g., from one set of
coordinates to another. The following properties follow from the standard definition of matrix
multiplication:
Matrix times a vector: A matrix B times a column vector v, is a weighted sum of the columns of B:
11 11
21 21
13 13
23 23
3
12 12
22 22
32 32 3 1 33 31 3
y y z
z
x
x
B v B
B v B
B B
B B
B v v B
B B
B B
B v B
B v B
(
( (
(
(
(
(
( (
= + +
(
( (
(
(
( (
(
(
(
(
Bv
We can visualize this by laying the vector on its side above the columns of the matrix, multiplying
each matrix-column by the vector component, and summing the resulting vectors:
13 13
23 23 13
33 33 23
3
12 12
22 22 12
32 32 2
11 11
21 21 11
31 31 21
31
2
2 3 3
z
z
z
y
y y
x
x
x
v
B B
B v v B B
B B B
B
v
B B
B v B B
B v
v
B v B
B v B B
B B B B B
B B
(
(
(
( ( (
(
(
( ( (
(
= = = + + + +
(
( ( (
(
(
( ( (
(
(
(
(
(
(
(
(
(
Bv
The columns of B are the vectors which are weighted by each of the input vector components, v
j
.
Another important way of conceptualizing a matrix times a vector: the resultant vector is a column of
dot-products. The i
th
element of the result is the dot-product of the given vector, v, with the i
th
row of B.
Writing B as a column of row-vectors:
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1 1 1
2 2 2
3 3 3
( (
( (
( (
( (
= = =
( (
( (
( (
( (
r r r v
B r Bv r v r v
r r r v
-
-
-
This view derives from the one above, where we lay the vector on its side above the matrix, but now
consider the effect on each row separately: it is exactly that of a dot-product.
In linear algebra, even if the matrices are complex, we do not conjugate the left vector in these dot-
products. If they need conjugation, the application must conjugate them separately from the matrix
multiplication, i.e. during the construction of the matrix.
We use this dot-product concept later when we consider a change of basis.
Matrix times a matrix: Multiplying a matrix B times another matrix C is defined as multiplying each
column of C by the matrix B. Therefore, by definition, matrix multiplication distributes to the right across
the columns:
, Let then
( ( (
( ( (
= =
( ( (
( ( (
C x y z BC B x y z Bx By Bz
[Matrix multiplication also distributes to the left across the rows, but we dont use that much.]
Determinants
This section assumes youve seen matrices and determinants, but probably didnt understand the
reasons why they work.
The determinant operation on a matrix produces a scalar. It is the only operation (up to a constant
factor) which is (1) linear in each row and each column of the matrix; and (2) antisymmetric under
exchange of any two rows or any two columns.
The above two rules, linearity and antisymmetry, allow determinants to help solve simultaneous linear
equations, as we show later under Cramers Rule. In more detail:
1. The determinant is linear in each column-vector (and row-vector). This means that multiplying
any column (or row) by a scalar multiplies the determinant by that scalar. E.g.,
det det ; det det det k k and = + = + a b c a b c a d b c a b c d b c
2. The determinant is anti-symmetric with respect to any two column-vectors (or row-vectors). This
means swapping any two columns (or rows) of the matrix negates its determinant.
The above properties of determinants imply some others:
3. Expansion by minors/cofactors (see below), whose derivation proves the determinant operator is
unique (up to a constant factor).
4. The determinant of a matrix with any two columns equal (or proportional) is zero. (From anti-
symmetry, swap the two equal columns, the determinant must negate, but its negative now equals
itself. Hence, the determinant must be zero.)
det det det 0 = = b b c b b c b b c
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5. det det det = A B AB . This is crucially important. It also fixes the overall constant factor of
the determinant, so that the determinant (with this property) is a completely unique operator.
6. Adding a multiple of any column (row) to any other column (row) does not change the
determinant:
det det det det det k k k + = + = + a b b c a b c b b c a b c b b c det = a b c
7. det|A + B| det|A| + det|B|. The determinant operator is not distributive over matrix addition.
8. det|kA| = k
n
det|A|.
The ij-th minor, M
ij
, of an nn matrix (A A
ab
) is the product A
ij
times the determinant of the (n
1)(n1) matrix formed by crossing out the i-th row and j-th column:
i
th
row
j
th
column
11 1
11 1, 1
1,1 1, 1
1
det
. . .
' . . '
. . . . .
. . . .
. . . .
. . . .
. . . . .
' . . '
. . .
n
n
ij ij ij
n n n
nn
n
A A
A A
M A A
A A
A A
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
A cofactor is just a minor with a plus or minus sign affixed:
| | ( 1) ( 1) det without row and column
i j i j th th
ij ij ij
C M A i j
+ +
(
= =
A
Cramers Rule
Its amazing how many textbooks describe Cramers rule, and how few explain or derive it. I spent
years looking for this, and finally found it in [Arf ch 3]. Cramers rule is a turnkey method for solving
simultaneous linear equations. It is horribly inefficient, and virtually worthless above 3 3, however, it
does have important theoretical implications. Cramers rule solves for n equations in n unknowns:
, is a coefficient matrix,
is a vector of unknowns,
is a vector of constants,
i
i
Given where
x
b
= Ax b A
x
b
To solve for the i
th
unknown x
i
, we replace the i
th
column of A with the constant vector b, take the
determinant, and divide by the determinant of A. Mathematically:
1 2
1 1 1
Let is the column of . We can solve for as
det ... ...
is the column of
det
th
n i i
i i n
th
i i
where i x
x where i
+
= (
=
A a a a a A
a a b a a
a A
A
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This seems pretty bizarre, and one has to ask, why does this work? Its quite simple, if we recall the
properties of determinants. Lets solve for x
1
, noting that all other unknowns can be solved analogously.
Start by simply multiplying x
1
by det|A|:
1 1 1 2
det det ...
n
x x = A a a a from linearity of det[ ]
1 1 2 2 2
adding a multiple of any column to
det ...
another doesn't change the determinant
n
x x = + a a a a
1 1 2 2 2
det ... ...
n n n
x x x = + + a a a a a ditto (n 2) more times
2 2
det ... det ...
n n
= = Ax a a b a a rewriting the first column
2
1
det ...
det
n
x =
b a a
A
Area and Volume as a Determinant
(a,0)
(c,d)
c
d
(a,b)
(c,d)
c
d
b
a
a c
b
d
Determining areas of regions defined by vectors is crucial to geometric physics in many areas. It is the
essence of the Jacobian matrix used in variable transformations of multiple integrals. What is the area of
the parallelogram defined by two vectors? This is the archetypal area for generalized (oblique, non-
normal) coordinates. We will proceed in a series of steps, gradually becoming more general.
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First, consider that the first vector is horizontal (above left). The area is simply base height: A = ad.
We can obviously write this as a determinant of the matrix of column vectors, though it is as-yet contrived:
det (0)
0
a c
d
= = =
For a general parallelogram (above right), we can take the big rectangle and subtract the smaller
rectangles and triangles, by brute force:
1 1
( )( ) 2 2 2
2 2
A a c b d bc cd ab ab
| | | |
= + + =
| |
\ . \ .
ad cb cd + + + 2bc cd ab det
a c
b d
= =
This is simple enough in 2-D, but is incomprehensibly complicated in higher dimensions. We can
achieve the same result more generally, in a way that allows for extension to higher dimensions by
induction. Start again with the diagram above left, where the first vector is horizontal. We can rotate that
to arrive at any arbitrary pair of vectors, thus removing the horizontal restriction:
( )
the rotation matrix. Then the rotated vectors are
0
det det det
0 0
a c
Let and
d
a c a c
d d
( (
=
( (
| | ( ( (
= =
|
( ( (
\ .
R R R
R R R R det det
0 0
a c a c
d d
=
The final equality is because rotation matrices are orthogonal, with det = 1. Thus the determinant of
arbitrary vectors defining arbitrary parallelograms equals the determinant of the vectors spanning the
parallelogram rotated to have one side horizontal, which equals the area of the parallelogram.
What about the sign? If we reverse the two vectors, the area comes out negative! Thats ok, because
in differential geometry, 2-D areas are signed: positive if we travel counter-clockwise from the first vector
to the 2
nd
, and negative if we travel clockwise. The above areas are positive.
In 3-D, the signed volume of the parallelepiped defined by 3 vectors a, b, and c, is the determinant of
the matrix formed by the vectors as columns (positive if abc form a right-handed set, negative if abc are a
left-handed set). We show this with rotation matrices, similar to the 2-D case: First, assume that the
parallelogram defined by bc lies in the x-y plane (b
z
= c
z
= 0). Then the volume is simply (area of the base)
height:
( ) ( ) ( ) det det
0 0
x x x
z y y y
z
a b c
V area of base height a a b c
a
| |
= = =
|
\ .
b c
where the last equality is from expansion by cofactors along the bottom row. But now, as before, we
can rotate such a parallelepiped in 3 dimensions to get any arbitrary parallelepiped. As before, the rotation
matrix is orthogonal (det = 1), and does not change the determinant of the matrix of column vectors.
This procedure generalizes to arbitrary dimensions: the signed hyper-volume of a parallelepiped
defined by n vectors in n-D space is the determinant of the matrix of column vectors. The sign is positive if
the 3-D submanifold spanned by each contiguous subset of 3 vectors (v
1
v
2
v
3
, v
2
v
3
v
4
, v
3
v
4
v
5
, ...) is right-
handed, and negated for each subset of 3 vectors that is left-handed.
The Jacobian Determinant and Change of Variables
How do we change multiple variables in a multiple integral? Given
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| |
( , , ) and the change of variables to , , :
( , , ), ( , , ), ( , , ). The simplistic
( , , ) ( , , ), ( , , ), ( , , ) ( !)
f a b c da db dc u v w
a a u v w b b u v w c c u v w
f a b c da db dc f a u v w b u v w c u v w du dv dw wrong
= = =
}}}
}}} }}}
fails, because the volume du dv dw associated with each point of f() is different than the volume da
db dc in the original integral.
da
du
dv
dw
db
dc
dv
dw
da
db
dc
du
Example of new-coordinate volume element (du dv dw), and its corresponding old-coordinate
volume element (da db dc). The new volume element is a rectangular parallelepiped. The old-
coordinate parallelepiped has sides straight to first order in the original integration variables.
In the diagram above, we see that the volume (du dv dw) is smaller than the old-coordinate volume
(da db dc). Note that volume is a relative measure of volume in coordinate space; it has nothing to do
with a metric on the space, and distance need not even be defined.
There is a concept of relative volume in any space, even if there is no definition of distance.
Relative volume is defined as products of coordinate differentials.
The integrand is constant (to first order in the integration variables) over the whole volume element.
Without some correction, the weighting of f() throughout the new-coordinate domain is different than
the original integral, and so the integrated sum (i.e., the integral) is different. We correct this by putting in
the original-coordinate differential volume (da db dc) as a function of the new differential coordinates, du,
dv, dw. Of course, this function varies throughout the domain, so we can write
| |
( ) ( )
( , , ) ( , , ), ( , , ), ( , , ) ( , , )
( , , )
f a b c da db dc f a u v w b u v w c u v w V u v w du dv dw
where V u v w takes du dv dw da db dc
}}} }}}
To find V(), consider how the a-b-c space vector daa is created from the new u-v-w space. It has
contributions from displacements in all 3 new dimensions, u, v, and w:
. ,
a a a
da du dv dw Similarly
u v w
b b b
db du dv dw
u v w
c c c
dc du dv dw
u v w
c c c | |
= + +
|
c c c
\ .
c c c | |
= + +
|
c c c
\ .
c c c | |
= + +
|
c c c
\ .
a a
b b
c c
The volume defined by the 3 vectors , , du dv and dw u v w maps to the volume spanned by the
corresponding 3 vectors in the original a-b-c space. The a-b-c space volume is given by the determinant of
the components of the vectors da, db, and dc (written as rows below, to match equations above):
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( ) det det
a a a a a a
du dv dw
u v w u v w
b b b b b b
volume du dv dw du dv dw
u v w u v w
c c c c c c
du dv dw
u v w u v w
c c c c c c
c c c c c c
c c c c c c
= =
c c c c c c
c c c c c c
c c c c c c
where the last equality follows from linearity of the determinant. Note that all the partial derivatives
are functions of u, v, and w. Hence,
( , , ) det ( , , ) the Jacobian ,
( , , ) ( , , ), ( , , ), ( , , ) ( , , )
a a a
u v w
b b b
V u v w J u v w and
u v w
c c c
u v w
f a b c da db dc f a u v w b u v w c u v w J u v w du dv dw
c c c
c c c
c c c
= (
c c c
c c c
c c c
(
}}} }}}
QED.
Expansion by Cofactors
Let us construct the determinant operator from its two defining properties: linearity, and antisymmetry.
First, well define a linear operator, then well make it antisymmetric. [This section is optional, though
instructive.]
We first construct an operator which is linear in the first column. For the determinant to be linear in
the first column, it must be a sum of terms each containing exactly one factor from the first column:
( ) ( ) ( )
11
21
11 21 1
12 1
2
1
2 2
2
. det . . . . . . . . .
n
n
n
n nn n
A
A
A A A
A
A A
A A
Let Then
A A
(
(
(
= = + + +
(
(
A A
.
.
. .
.
.
To be linear in the first column, the parentheses above must have no factors from the first column (else
they would be quadratic in some components). Now to also be linear in the 2
nd
column, all of the
parentheses above must be linear in all the remaining columns. Therefore, to fill in the parentheses we
need a linear operator on columns 2...n. But that is the same kind of operator we set out to make: a linear
operator on columns 1..n. Recursion is clearly called for, therefore the parentheses should be filled in with
more determinants:
( ) ( ) ( )
11 1 21 2 1
det det det det ( )
n n
A A A so far = + + + A M M M
We now note that the determinant is linear both in the columns, and in the rows. This means that det
M
1
must not have any factors from the first row or the first column of A. Hence, M
1
must be the submatrix
of A with the first row and first column stricken out.
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1
st
row
1
st
column
2
nd
row
1
st
column
11 1
11 12 1
21 22 2
21 2
31 32 3
1 2
1 2
. . .
. .
. .
. . .
. . . .
, . . , .
1 2
. . . . .
. . . . .
. .
. .
n
n
n
n
n
n n nn
n n nn
A A
A A A
A A A
A A
A
A A A etc
ij
A A A
A A A
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
M M
Similarly, M
2
must be the submatrix of A with the 2
nd
row and first column stricken out. And so on,
through M
n
, which must be the submatrix of A with the n
th
row and first column stricken out. We now
have an operator that is linear in all the rows and columns of A.
So far, this operator is not unique. We could multiply each term in the operator by a constant, and still
preserve linearity in all rows and columns:
( ) ( ) ( )
1 11 1 2 21 2 1
det det det det
n n n
k A k A k A = + + + A M M M
We choose these constants to provide the 2
nd
property of determinants: antisymmetry. The determinant
is antisymmetric on interchange of any two rows. We start by considering swapping the first two rows:
Define A (A with A
1*
A
2*
).
11 12 1
11 12
1
1
21 2
1 1 2
21 2
. . . .
. .
. . . . . . . .
,
. . . . . . . . . .
. .
' '
.
.
.
.
.
.
.
.
n
n
n
n
n nn n n nn
A A A
A A A
A A
etc
ij ij
A A A
A
A
A
A
A
A
( (
( (
( (
( (
( (
( (
( (
( (
A M
swap swapped
Recall that M
1
strikes out the first row, and M
2
strikes out the 2
nd
row, so swapping row 1 with row 2
replaces the first two terms of the determinant:
( ) ( ) ( ) ( )
1 11 1 2 21 2 1 21 2 1 2 11
det det det ... det det ' ' det .. ' . k A k A k A k A = + + = + + A M A M M M
But M
1
= M
2
, and M
2
= M
1
. So we have:
( ) ( )
1 21 2 2 11 1
det det det ' ... k A k A = + + A M M
This last form is the same as det A, but with k
1
and k
2
swapped. To make our determinant
antisymmetric, we must choose constants k
1
and k
2
such that terms 1 and 2 are antisymmetric on
interchange of rows 1 and 2. This simply means that k
1
= k
2
. So far, the determinant is unique only up to
an arbitrary factor, so we choose the simplest such constants: k
1
= 1, k
2
= 1.
For M
3
through M
n
, swapping the first two rows of A swaps the first two rows of M
3
through M
n
:
31
41 42 4
21 22 2
1
11 12 1
2
3
. . . . ,
. .
.
.
.
.
.
'
.
.
n
n n n
n
n
n
A A A
A etc
A A A
A A A
A A A
(
(
(
(
(
(
(
M
swapped
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Since M
3
through M
n
appear inside determinant operators, and such operators are defined to be
antisymmetric on interchange of rows, terms 3 through n also change sign on swapping the first two rows
of A. Thus, all the terms 1 through n change sign on swapping rows 1 and 2, and det A = det A.
We are almost done. We have now a unique determinant operator, with k
1
= 1, k
2
= 1. We must
determine k
3
through k
n
. So consider swapping rows 1 and 3 of A, which must also negate our determinant:
11 12 1
1
21 2 21 2
31 3
3 1 12
2 2
1
1 1 1
1 2
3
. . .
. . . . .
,
. . . . . . . . .
. .
.
.
.
.
. . . .
. . " " . .
n n
n n
n
n
n
n
n
n n nn
A A A
A
A A A A A
etc
A A A A A
A A
A A A A
( (
( (
( (
( (
( (
( (
( (
A M
swap swapped
Again, M
4
through M
n
have rows 1 & 3 swapped, and thus terms 4 through n are negated by their
determinant operators. Also, M
2
(formed by striking out row 2 of A) has its rows 1 & 2 swapped, and is
also thus negated.
The terms remaining to be accounted for are ( ) ( )
11 1 3 31 3
det and det A k A M M . The new M
1
is the
same as the old M
3
, but with its first two rows swapped. Similarly, the new M
3
is the same as the old M
1
,
but with its first two rows swapped. Hence, both terms 1 and 3 are negated by their determinant operators,
so we must choose k
3
= 1 to preserve that negation.
Finally, proceeding in this way, we can consider swapping rows 1 & 4, etc. We find that the odd
numbered ks are all 1, and the even numbered ks are all 1.
We could also have started from the beginning by linearizing with column 2, and then we find that the
k are opposite to those for column 1: this time for odd numbered rows, k
odd
= 1, and for even numbered
rows, k
even
= +1. The ks simply alternate sign. This leads to the final form of cofactor expansion about any
column c:
( ) ( ) ( )
1 2
1 1 2 2
det ( 1) det ( 1) det ( 1) det
c c n c
c c nc n
A A A
+ + +
= + + + A M M M
Note that
We can perform a cofactor expansion down any column,
or across any row, to compute the determinant of a matrix.
We usually choose an expansion order which includes as many zeros as possible, to minimize the
computations needed.
Proof That the Determinant Is Unique
If we compute the determinant of a matrix two ways, from two different cofactor expansions, do we
get the same result? Yes. We here prove the determinant is unique by showing that in a cofactor
expansion, every possible combination of elements from the rows and columns appears exactly once. This
is true no matter what row or column we expand on. Thus all expansions include the same terms, but just
written in a different order.
Also, this complete expansion of all combinations of elements is a useful property of the cofactor
expansion which has many applications beyond determinants. For example, by performing a cofactor
expansion without the alternating signs (in other word, an expansion in minors), we can fully symmetrize a
set of functions (such as boson wave functions).
The proof: lets count the number of terms in a cofactor expansion of a determinant for an nn matrix.
We do this by mathematical induction. For the first level of expansion, we choose a row or column, and
construct n terms, where each term includes a cofactor (a sub-determinant of an n1 n1 matrix). Thus,
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the number of terms in an nn determinant is n times the number of terms in an n1 n1 determinant. Or,
turned around,
( )( ) # ( 1 1) 1 # terms in n n n terms in n n + + = +
There is one term in a 11 determinant, 2 terms in a 22, 6 terms in a 33, and thus n! terms in an nn
determinant. Each term is unique within the expansion: by construction, no term appears twice as we work
our way through the cofactor expansion.
Lets compare this to the number of terms possible which are linear in every row and column: we have
n choices for the first factor, n1 choices for the second factor, and so on down to 1 choice for the last
factor. That is, there are n! ways to construct terms linear in all the rows and columns. That is exactly the
number of terms in the cofactor expansion, which means every cofactor expansion is a sum of all possible
terms which are linear in the rows and columns. This proves that the determinant is unique up to a sign.
To prove the sign of the cofactor expansion is also unique, we can consider one specific term in the
sum. Consider the term which is the product of the main diagonal elements. This term is always positive,
since TBS ??
Getting Determined
You may have noticed that computing a determinant by cofactor expansion is computationally
infeasible for n > ~15. There are n! terms of n factors each, requiring O(n n!) operations. For n = 15, this
is ~10
13
operations, which would take about a day on a few GHz computer. For n = 20, it would take years.
Is there a better way? Fortunately, yes. It can be done in O(n
3
) operations, so one can easily compute
the determinant for n = 1000 or more. We do this by using the fact that adding a multiple of any row to
another row does not change the determinant (which follows from anti-symmetry and linearity).
Performing such row operations, we can convert the matrix to upper-right-triangular form, i.e., all the
elements of A below the main diagonal are zero.
11 12 1, 1 1
11 12 1
22 2, 1 2
21 22 2
1, 1 1,
1 2
' ' '
0 ' ' '
'
0 0 ' '
0 0 0 '
n n
n
n n
n
n n n n
n n nn
nn
A A A A
A A A
A A A
A A A
A A
A A A
A
(
(
(
(
(
(
( = =
(
(
(
(
(
A A
.
.
.
.
. . . .
. . .
.
.
By construction, det|A| = det|A|. Using the method of cofactors on A, we expand down the first
column of A and first column of every submatrix in the expansion. E.g.,
11
22
33
44
' x x x
0 ' x x
'
0 0 ' x
0 0 0 '
A
A
A
A
(
(
(
=
(
(
A
Only the first term in each expansion survives, because all the others are zero. Hence, det|A| is the
product of its diagonal elements:
1
det det ' ' ' are the diagonal elements of '
n
ii ii
i
A where A
=
= = A A A
[
Lets look at the row operations needed to achieve upper-right-triangular form. We multiply the first
row by (A
21
/ A
11
) and subtract it from the 2
nd
row. This makes the first element of the 2
nd
row zero (below
left):
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11 12 13 1 11 12 13 1 11 12 13 1
22 23 24 22 23 24 22 23 24
31 32 33 34 32 33 34 33 34
41 42 43 44 42 43 44 43 44
0 0 0
0 0 0
0 0 0
n n n
A A A A A A A A A A A A
B B B B B B B B B
A A A A B B B C C
A A A A B B B C C
( ( (
( ( (
( ( (
( ( (
( ( (
A
Perform this operation for rows 3 through n, and we have made the first column below row 1 all zero
(above middle). Similarly, we can zero the 2
nd
column below row 2 by multiplying the (new) 2
nd
row by
(B
32
/ B
22
) and subtracting it from the 3
rd
row. Perform this again on the 4
th
row, and we have the first two
columns of the upper-right-triangular form (above right). Iterating for the first (n 1) columns, we
complete the upper-right-triangular form. The determinant is now the product of the diagonal elements.
About how many operations did that take? There are n(n 1)/2 row-operations needed, or O(n
2
).
Each row-operation takes from 1 to n multiplies (average n/2), and 1 to n additions (average n/2), summing
to O(n) operations. Total operations is then of order
( ) ( ) ( )
2 3
~ O n O n O n
TBS: Proof that det|AB| = det|A| det|B|
Getting to Home Basis
We often wish to change the basis in which we express vectors and matrix operators, e.g. in quantum
mechanics. We use a transformation matrix to transform the components of the vectors from the old basis
to the new basis. Note that:
We are not transforming the vectors; we are transforming the components of the vector from one
basis to another. The vector itself is unchanged.
There are two ways to visualize the transformation. In the first method, we write the decomposition of
a vector into components in matrix form. We use the visualization from above that a matrix times a vector
is a weighted sum of the columns of the matrix:
y y
y y
z
x z
z
x
z
x
x
v v
v
v
v v
(
(
(
(
= = + +
(
(
(
(
v e e e e e e
This is a vector equation which is true in any basis. In the x-y-z basis, it looks like this:
1 0 0 1 0 0
0 1 0 0 , 1 , 0
0 0 1 0 0 1
x x
y y
x y z
z z
v v
v v where
v v
( (
( ( ( (
( (
( ( ( (
= = = = =
( (
( ( ( (
( (
( ( ( (
v e e e
If we wish to convert to the e
1
, e
2
, e
3
basis, we simply write e
x
, e
y
, e
z
in the 1-2-3 basis:
(in the1-2-3 basis) : , ,
x x
y y
x y z
z z
a d g v v a d g
b e h v v where b e h
c f i v v c f i
( (
( ( ( (
( (
( ( ( (
= = = = =
( (
( ( ( (
( (
( ( ( (
v e e e
Thus:
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The columns of the transformation matrix are the old basis vectors written in the new basis. This
is true even for non-ortho-normal bases.
Now let us look at the same transformation matrix, from the viewpoint of its rows. For this, we must
restrict ourselves to ortho-normal bases. This is usually not much of a restriction.
Recall that the component of a vector v in the direction of a basis vector e
i
is given by:
( ) ( ) ( )
i
i x x y y z z
v = = + + e v v e v e e v e e v e
But this is a vector equation, valid in any basis. So i above could also be 1, 2, or 3 for the new basis:
( ) ( ) ( )
1 2 3
1 2 3 1 1 2 2 3 3
, , v v v = = = = + + e v e v e v v e v e e v e e v e
Recall from the section above on matrix multiplication that multiplying a matrix by a vector is
equivalent to making a set of dot-products, one from each row, with the vector:
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
1 1
1 1 1 1 1 1
2 2
2 2 2 2 2 2
3 3
3 3 3 3 3 3
x
x y z
y
x y z
z
x y z
v v v
v or v v
v v v
( (
( ( (
( ( (
( (
( ( (
( ( (
( (
( ( (
( ( (
( (
= = = =
( ( (
( ( (
( (
( (
( ( (
( (
( (
( ( (
( (
( (
e e v e e e e v
e v e v e e e e v
e e v e e e e v
(
(
Thus:
The rows of the transformation matrix are the new basis vectors written in the old basis. This is
only true for ortho-normal bases.
There is a beguiling symmetry in the above two boxed statements about the columns and rows of the
transformation matrix.
For complex vectors, we must use the dot-product defined with the conjugate of the row basis vector,
i.e. the rows of the transformation matrix are the hermitian adjoints of the new basis vectors written in the
old basis:
1
1 1
2
2 2
3
3 3
v
v
v
(
(
( (
(
(
( (
(
(
( (
(
= =
(
( (
(
(
( (
(
(
( (
(
(
e e v
e v e v
e e v
A special case of basis changing comes up often in quantum mechanics: we wish to change to the basis
of eigenvectors of a given operator. In this basis, the basis vectors (which are also eigenvectors) always
have the form of a single 1 component, and the rest 0. E.g.,
1 2 3
1 0 0
0 1 0
0 0 1
( ( (
( ( (
= = =
( ( (
( ( (
e e e
The matrix operator A, in this basis (its own eigenbasis), is diagonal, because:
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1 1 1
1
2 2 2 2
3
3 3 3
=
(
(
= =
`
(
(
=
)
Ae e
Ae e A
Ae e
Finding the unitary (i.e., unit magnitude) transformation from a given basis to the eigenbasis of an
operator is called diagonalizing the matrix. We saw above that the transformation matrix from one basis
to another is just the hermitian adjoint of the new basis vectors written in the old basis. We call this matrix
U:
1
1 1 1
2
2 2 2
3
3 3 3
v
v
v
( (
(
( (
( (
(
( (
( (
(
( (
( (
= = =
(
( (
( (
(
( (
( (
(
( (
( (
( (
e e v e
e v e v U e
e e v e
U transforms vectors, but how do we transform the operator matrix A itself? The simplest way to see
this is to note that we can perform the operation A in any basis by transforming the vector back to the
original basis, using A in the original basis, and then transforming the result to the new basis:
( ) ( ) ( ) ( )
1
1 1 1
new old old new
new new old old old new old new new old
= =
= = = =
v Uv v U v
A v U A v U A U v UA U v A UA U
where we used the fact that matrix multiplication is associative. Thus:
The unitary transformation that diagonalizes a (complex) self-adjoint matrix is the matrix of
normalized eigen-row-vectors.
We can see this another way by starting with:
1
1 2 3 1 2 3 1 1 2 2 3 3
are the otho-normal eigenvectors
are the eigenvalues
i
i
where
( ( (
( ( (
= = =
( ( (
( ( (
AU A e e e Ae Ae Ae e e e
e
Recall the eigenvectors (of self-adjoint matrices) are orthogonal, so we can now pre-multiply by the
hermitian conjugate of the eigenvector matrix:
( ) ( )
1 1
1
2 1 2 3 2 1 1 2 2 3 3
3 3
1 1 1 2 1 2
( (
( (
( (
( (
= = ( (
( (
( (
( (
( (
=
e e
UAU e A e e e e e e e
e e
e e e e
( )
1 2 1
e e ( )
2 2 2
e e
( )
1
2
3
3 3 3
0 0
0 0
0 0
(
(
(
(
(
=
(
(
(
(
(
e e
where the final equality is because each element of the result is the inner product of two eigenvectors,
weighted by an eigenvalue. The only non-zero inner products are between the same eigenvectors
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(orthogonality), so only diagonal elements are non-zero. Since the eigenvectors are normalized, their inner
product is 1, leaving only the weight (i.e., the eigenvalue) as the result.
Warning Many books write the diagonalization as U
1
AU, instead of the correct UAU
1
. This is
confusing and pointless, and these very books then change their notation when they have
to transform a vector, because nearly all books agree that vectors transform with U, and
not U
1
.
Contraction of Matrices
You dont see a dot-product of matrices defined very often, but the concept comes up in physics, even
if they dont call it a dot-product. We see such products in QM density matrices, and in tensor operations
on vectors. We use it below in the Trace section for traces of products.
For two matrices of the same size, we define the contraction of two matrices as the sum of the
products of the corresponding elements (much like the dot-product of two vectors). The contraction is a
scalar. Picture the contraction as overlaying one matrix on top of the other, multiplying the stacked
numbers (elements), and adding all the products:
A
ij
B
ij
}
sum A:B
We use a colon to convey that the summation is over 2 dimensions (rows and columns) of A and B
(whereas the single-dot dot product of vectors sums over the 1 dimensional list of vector components):
, 1
11 11 12 12 13 13
21 21 22 22 23 23 11 11 12 12 13 13 21 21 22 22 23 23 31 31 32 32 33 33
31 31 32 32 33 33
For example, for 33 matrices:
n
ij ij
i j
a b
a b a b a b
a b a b a b a b a b a b a b a b a b a b a b a b
a b a b a b
=
+ +
= + + + = + + + + + + + +
+ + +
A: B
A: B
which is a single number.
If the matrices are complex, we do not conjugate the left matrix (such conjugation is often done in
defining the dot-product of complex vectors).
Trace of a Product of Matrices
The trace of a matrix is defined as the sum of the diagonal elements:
( ) ( )
12 13
21 23
31
11 22 33
1
32
11
22
33
Tr . . : , Tr
n
jj
j
a a
a a
a a
a E g
a
a a a
a
a
=
| |
|
= = + +
|
|
\ .
A A A
The trace of a product of matrices comes up often, e.g. in quantum field theory. We first show that
Tr(AB) = Tr(BA):
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( )
12 13
21 22 23 22 23 21 22 23
31
11 12 13 11 11 12 13
11 22
* *
1 21
32 3 3 3
1 1* *
32 33 3 3 1
1
1 32 3
. Tr ...
as the row of , and as the column of
,
nn
th th
r c
b b
a a
Let c c c
Define
a b b a a a
a a a b b a a
a
a r b c
c
a
a b o
a b a a a
r
a
b
b
= = + + +
| || | | |
| | |
=
| | |
| |
\ .\ . \ .
C AB AB
A B
( ) ( ) ( )
( ) ( ) ( )
11 12 1
11
3
12 13
12
21 22 23 22 21 22 23
21 22 23
1
3
1 13 11 12 13
21 23
31 32 33 31 33 31 32 33
2 *
2
2 2 *2
. . .
. . .
. . .
. . .
T T T
T T T
b
a a a b
a a a b b a a a
b b
a a a b b a a
a a c a or
b
b
a
a
| |
|
|
|
|
|
\ .
| |
| || | | |
|
| | |
=
|
| | |
| | | |
\ .\ . \ .
\ .
B B B
B B B
and so on.
The diagonal elements of the product C are the sums of the overlays of the rows of A on the columns
of B. But this is the same as the overlays of the rows of A on the rows of B
T
. Then we sum the overlays,
i.e., we overlay A onto B
T
, and sum all the products of all the overlaid elements:
Tr( )
T
= AB A: B
Now consider Tr(BA) = B : A
T
. But visually, B : A
T
overlays the same pairs of elements as A : B
T
, but
in the transposed order. When we sum over all the products of the pairs, we get the same sum either way:
( ) ( ) Tr Tr
T T
because = = AB BA A: B B: A
This leads to the important cyclic property for the trace of the product of several matrices:
( ) ( ) ( ) ( ) ( ) ( )
Tr ... Tr ... Tr ... Tr ... because = = AB C CAB AB C C AB
and matrix multiplication is associative. By simple induction, any cyclic rotation of the matrices
leaves the trace unchanged.
Linear Algebra Briefs
The determinant equals the product of the eigenvalues:
1
det are the eigenvalues of
n
i i
i
where
=
= A A
[
This is because the eigenvalues are unchanged through a similarity transformation. If we diagonalize
the matrix, the main diagonal consists of the eigenvalues, and the determinant of a diagonal matrix is the
product of the diagonal elements (by cofactor expansion).
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Probability, Statistics, and Data Analysis
I think probability and statistics are among the most conceptually difficult topics in mathematical
physics. We start with a brief overview of the basics, but overall, we assume you are familiar with simple
probabilities, and gaussian distributions.
Probability and Random Variables
We assume you have an basic idea of probability, and since we seek understanding over mathematical
purity, we give here intuitive definitions. A random variable, say X, is a quantity that you can observe (or
measure), multiple times (at least in theory), and is not completely predictable. Each observation of a
random variable may give a different value. Random variables may be discrete (the roll of a die), or
continuous (the angle of a game spinner after you spin it). A uniform random variable has all its values
equally likely. Thus the roll of a die is a uniform discrete random variable. The angle of a game spinner is
a uniform continuous random variable. But in general, the values of a random variable are not necessarily
equally likely. For example, a gaussian (aka normal) random variable is more likely to be near the
mean.
Given a large sample of observations of any physical quantity X, there will be some structure to the
values X assumes. For discrete random variables, each possible value will appear (close to) some fixed
fraction of the time in any large sample. The fraction of a large sample that a given value appears is that
values probability. For a 6-sided die, the probability of rolling 1 is 1/6, i.e. Pr(1) = 1/6. Because
probability is a fraction of a total, it is always between 0 and 1 inclusive:
0 Pr( ) 1 anything s s
Note that one can imagine systems of chance specifically constructed to not provide consistency between
samples, at least not on realistic time scales. By definition, then, observations of such a system do not constitute a
random variable in the sense of our definition.
Strictly speaking, a statistic is a number that summarizes in some way a set of random values. Many
people use the word informally, though, to mean the raw data from which we compute true statistics.
Conditional Probability
Probability, in general, is a combination of physics and knowledge: the
physics of the system in question, and what you know about its state.
Conditional probability specifically addresses probability when the state of the system is partly
known. A priori probability generally implies less knowledge of state (a priori means in the
beginning or beforehand). But there is no true, fundamental distinction, because all probabilities are in
some way dependent on both physics and knowledge.
Suppose you have 1 bag with 2 white and 2 black balls. You draw 2 balls without replacement. What
is the chance the 2
nd
ball will be white? A priori, its obviously . However, suppose the first ball is
known white. Now Pr(2
nd
ball is white) = 1/3. So we say the conditional probability that the 2
nd
ball will
be white, given that the first ball is white, is 1/3. In symbols,
Pr(2 | ) 1/ 3 nd ball white first ball white =
Another example of how conditional probability of an event can be different than the a priori
probability of that event: I have a bag of white and a bag of black balls. I give you a bag at random. What
is the chance the 2
nd
ball will be white? A priori, its . After seeing the 1
st
ball is white, now Pr(2
nd
ball is
white) = 1. In this case,
Pr(2 | ) 1 nd ball white first ball white =
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Precise Statement of the Question Is Critical
Many arguments arise about probability because the questions are imprecise, each combatant has a
different interpretation of the question, but neither realizes the other is arguing a different question.
Consider this:
You deal 4 cards from a shuffled deck. I tell you 3 of them are aces. What is the probability that the
4
th
card is also an ace?
The question is ambiguous, and could reasonably be interpreted two ways, but the two interpretations
have quite different answers. It is very important to know exactly how I have discovered 3 of them are
aces.
Case 1: I look at the 4 cards and say At least 3 of these cards are aces. There are 193 ways that 4
cards can hold at least 3 aces, and only 1 of those ways has 4 aces. Therefore, the chance of the 4
th
card
being an ace is 1/193.
Case 2: I look at only 3 of the 4 cards and say, These 3 cards are aces. There are 49 unseen cards,
all equally likely to be the 4
th
card. Only one of them is an ace. Therefore, the chance of the 4
th
card being
an ace is 1/49.
It may help to show that we can calculate the 1/49 chance from the 193 hands that have at least 3 aces:
Of the 192 that have exactly 3 aces, we expect that 1/4 of them = 48 will show aces as their first 3 cards
(because the non-ace 1/4 chance of being last) . Additionally, the one hand of 4 aces will always show aces
as its first 3 cards. Hence, of the 193 hands with at least 3 aces, 49 show aces as their first 3 cards, of
which exactly 1 will be the 4-ace hand. Hence, its conditional probability, given that the first 3 cards are
aces, is 1/49.
Lets Make a Deal
This is an example of a problem that confuses many people (including me), and how to properly
analyze it. We hope this example illustrates some general methods of analysis that you can use to navigate
more general confusing questions. In particular, the methods used here apply to renormalizing entangled
quantum states when a measurement of one value is made.
Your in the Big Deal on the game show Lets Make a Deal. There are 3 doors. Hidden behind two of
them are goats; behind the other is the Big Prize. You choose door #1. Monty Hall, the MC, knows whats
behind each door. He opens door #2, and shows you a goat. Now he asks, do you want to stick with your
choice, or switch to door #3? Should you switch?
Without loss of generality (WLOG), we assume you choose door #1 (and of course, it doesnt matter
which door you choose). We make a chart of mutually exclusive events, and their probabilities:
Bgg shows door #2
1/6
gBg shows door #3
1/3
ggB shows door #2
1/3
After you choose, Monty shows you that door #2 is a goat. So from the population of possibilities, we
strike out those that are no longer possible, and renormalize the remaining probabilities:
Bgg shows door #2 1/6
1/3
gBg shows door #3 1/3
ggB shows door #2 1/3
2/3
Another way to think of this: Monty showing you door #2 is equivalent to saying, The big prize is
either the door you picked, or its door #3. Since your chance of having picked right is unaffected by
Monty telling you this, Pr(#3) = 2/3. Monty uses his knowledge to pick a door with a goat. That gives you
information, which improves your ability to guess right on your second guess.
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You can also see it this way: Theres a 1/3 chance you picked right the first time. Then youll switch,
and lose. But theres a 2/3 chance you picked wrong the first time. Then youll switch, and win. So you
win twice as often as you lose, much better odds than 1/3 of winning.
Lets take a more extreme example: suppose there are 100 doors, and you pick #1. Now Monty tells
you, The big prize is either the door you picked, or its door #57. Should you switch? Of course. The
chance you guessed right is tiny. But Monty knows for sure.
How to Lie With Statistics
In 2007, on the front page of newspapers, was a story about a big study of sexual behavior in America.
The headline point was that on average, heterosexual men have 7 partners in their lives, and women have
only 4.
Innumeracy, a book about math and statistics, uses this exact point from a previous study of sexual
behavior, and noted that one can easily prove that the average number of heterosexual partners of men and
women must be exactly the same (if there are equal numbers of men and women in the population. The US
has equal numbers of men and women to better than 1%).
The only explanation for the survey results is that most people are lying. Men lie on the high side,
women lie on the low side. The article goes on to quote all kinds of statistics and facts, oblivious to the
fact that most people were lying in the study. So how much can you believe anything the subjects said?
Even more amazing to me is that the scientists doing the study seem equally oblivious to the
mathematical impossibility of their results. Perhaps some graduate student got a PhD out of this study, too.
The proof: every heterosexual encounter involves a man and a woman. If the partners are new to each
other, then it counts as a new partner for both the man and the woman. The average number of partners for
men is the total number of new partners for all men divided by the number of men considered. But this is
exactly equal to the total number of new partners for all women divided by the number of women
considered. QED.
An insightful friend noted, Maybe to some women, some guys arent worth counting.
Choosing Wisely: An Informative Puzzle
Heres a puzzle which illuminates the physical meaning of the
n
k
| |
|
\ .
binomial forms. Try it yourself
( )
!
! !
n
n
n choose k
k k n k
| |
|
\ .
is the number of ways of choosing k items from n distinct items, without replacement, where the order
of choosing doesnt matter. In other words,
n
k
| |
|
\ .
is the number of combinations of k items taken from n
distinct items.
The puzzle: Show in words, without algebra, that
1
1
n n n
k k k
+ | | | | | |
= +
| | |
\ . \ . \ .
.
Some purists may complain that the demonstration below lacks rigor (not true), or that the algebraic
demonstration is shorter. However, though the algebraic proof is straightforward, it is dull and
uninformative. Physicists may like the demonstration here because it uses the physical meaning of the
mathematics to reach an iron-clad conclusion.
The solution: The LHS is the number of ways of choosing k items from n + 1 items. Now there are
two distinct subsets of those ways: those ways that include the (n + 1)
th
item, and those that dont. In the
first subset, after choosing the (n + 1)
th
item, we must choose k 1 more items from the remaining n, and
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there are
1
n
k
| |
|
\ .
ways to do this. In the second subset, we must choose all k items from the first n, and
there are
n
k
| |
|
\ .
ways to do this. Since this covers all the possible ways to choose k items from n + 1 items, it
must be that
1
1
n n n
k k k
+ | | | | | |
= +
| | |
\ . \ . \ .
. QED.
Multiple Events
First we summarize the rules for computing the probability of combinations of events from their
individual probabilities, then we justify them:
Pr(A and B) = Pr(A)Pr(B), A and B independent
Pr(A or B) = Pr(A) + Pr(B), A and B mutually exclusive
Pr(not A) = 1 Pr(A)
Pr(A or B) = Pr(A) + Pr(B) Pr(A)Pr(B), A and B independent
For independent events A and B, Pr(A and B) = Pr(A)Pr(B). This follows from the definition of
probability as a fraction. If A and B are independent (have nothing to do with each other), then Pr(A) is
the fraction of trials with event A. Then of the fraction of those with event A, the fraction that also has B is
Pr(B). Therefore, the fraction of the total trials with both A and B is
Pr(A and B) = Pr(A)Pr(B)
For mutually exclusive events, Pr(A or B) = Pr(A) + Pr(B). This also follows from the definition of
probability as a fraction. The fraction of trials with event A Pr(A); fraction with event B Pr(B). If no
trial can contain both A and B, then the fraction with either is simply the sum:
fraction
with A
fraction with B Total
trials
- - -
-
fraction with A or B - -
-
Pr(not A) = 1 Pr(A). Since Pr(A) is the fraction of trials with event A, and all trials must either have
event A or not,
Pr(A) + Pr(not A) = 1
Notice that A and (not A) are mutually exclusive events (a trial cant both have A and not have A), so
By Pr(A or B) we mean Pr(A or B or both). For independent events, you might think that Pr(A or B) =
Pr(A) + Pr(B), but this is not so. A simple example shows that it cant be: suppose Pr(A) = Pr(B) = 0.7.
Then Pr(A) + Pr(B) = 1.4, which cant be the probability of anything. The reason for the failure of simple
addition of probabilities is that doing so counts the probability of (A and B) twice:
fraction
with A
fraction with A and
B
fraction
with B
Total
trials
- -
- -
fraction with A or
B -
- - - -
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Note that Pr(A or B) is equivalent to Pr(A and maybe B) OR Pr(B and maybe A). But Pr(A and maybe
B) includes the probability of both A and B, as does the 2
nd
term, hence it is counted twice. So subtracting
the probability of (A and B) makes it counted only once:
Pr(A or B) = Pr(A) + Pr(B) Pr(A)Pr(B), A and B independent
A more complete statement, which breaks down (A or B) into mutually exclusive events is:
Pr(A or B) = Pr(A and not B) + Pr(not A and B) + Pr(A and B)
Since the right hand side are now mutually exclusive events, their probabilities add:
Pr(A or B) = Pr(A)[1 Pr(B)] + Pr(B)[1 Pr(A)] + Pr(A)Pr(B)
= Pr(A) + Pr(B) 2Pr(A)Pr(B) + Pr(A)Pr(B)
= Pr(A) + Pr(B) Pr(A)Pr(B) .
TBS: Example of rolling 2 dice.
Combining Probabilities
Here is a more in-depth view of multiple events, with several examples. This section should be called
Probability Calculus, but most people associate calculus with something hard, and I didnt want to
scare them off. In fact, calculus simply means a method of calculation.
Probabilities describe binary events: an event either happens, or it doesnt.
Therefore, we can use some of the methods of Boolean algebra in probability.
Boolean algebra is the mathematics of expressions and variables that can have one of only two values:
usually taken to be true and false. We will use only a few simple, intuitive aspects of Boolean algebra
here.
An event is something that can either happen, or not (its binary!). We define the probability of an
event as the fraction of time, out of many (possibly hypothetical) trials, that the given event happens. For
example, the probability of getting a heads from a toss of a fair coin is 0.5, which we might write as
Pr(heads) = 0.5 = 1/2. Probability is a fraction of a whole, and so lies in [0, 1].
We now consider two random events. Two events have one of 3 relationships: independent, mutually
exclusive, or conditional (aka conditionally dependent). We will soon see that the first two are special
cases of the conditional relationship. We now consider each relationship, in turn.
Independent: For now, we define independent events as events that have nothing to do with each
other, and no effect on each other. For example, consider two events: tossing a heads, and rolling a 1 on a
6-sided die. Then Pr(heads) = 1/2, and Pr(rolling 1) = 1/6. The events are independent, since the coin
cannot influence the die, and the die cannot influence the coin. We define one trial as two actions: a toss
and a roll. Since probabilities are fractions, of all trials, will have heads, and 1/6 of those will roll a 1.
Therefore, 1/12 of all trials will contain both a heads and a 1. We see that probabilities of independent
events multiply. We write:
Pr(A and B) = Pr(A)Pr(B) (independent events)
In fact, this is the precise definition of independence: if the probability of two events both occurring is
the product of the individual probabilities, then the events are independent.
[Aside: This definition extends to PDFs: if the joint PDF of two random variables is the product of their
individual PDFs, then the random variables are independent.]
Geometric diagrams are very helpful in understanding the probability calculus. We can picture the
probabilities of A, B, and (A and B) as areas. The sample space or population is the set of all possible
outcomes of trials. We draw that as a rectangle. Each point in the rectangle represents one possible
outcome. Therefore, the probability of an outcome being within a region of the population is proportional
to the area of the region.
(Below, (a)) An event A either happens, or it doesnt. Therefore,
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Pr(A) + Pr(~A) = 1. (always)
B
A
A
and B
0 1
1
Pr(B)
Pr(A)
~A A
0 1
Pr(A)
B
A
A
and B
sample space,
aka population
B
A
(b) independent (c) conditional (d) mutually exclusive
(a)
(Left) An event either happens, or it doesnt. (Middle) The (continuous) sample space is the
square. Events A and B are independent. (Right) A and B are dependent.
(Above, (b)) Pr(A) is the same whether B occurs or not, shown by the fraction of B covered by A is
the same as the fraction of the sample space covered by A. Therefore, A and B are independent.
(Above, (c)) The probability of (A or B) is the red, blue, and magenta areas. Geometrical, then we see
Pr(A or B) = Pr(A) + Pr(B) Pr(A and B) (always)
This is always true, regardless of any dependence between A and B.
Conditionally dependent: From the diagram, when A and B are conditionally dependent, we see
that the Pr(B) depends on whether A happens or not. Pr(B given that A occurred) is written as Pr(B | A),
and read as probability of B given A. From the ratio of the magenta area to the red, we see
Pr(B | A) = Pr(B and A)/Pr(A) (always)
Mutually exclusive: Two events are mutually exclusive when they cannot both happen (diagram
above, (d)). Thus,
Pr(A and B) = 0, and Pr(A or B) = Pr(A) + Pr(B) (mutually exclusive)
Note that Pr(A or B) follows the rule from above, which always applies.
We see that independent events are an extreme case of conditional events: independent events satisfy
Pr(B | A) = Pr(B) (independent)
since the occurrence of A has no effect on B. Also, mutually exclusive events satisfy
Pr(B | A) = 0 (mutually exclusive)
Summary of Probability Calculus
Always
Pr(~A) = 1 Pr(A) Pr(entire sample space) = 1 (diagram
above, (a))
Pr(A or B) = Pr(A) + Pr(B) Pr(A and B) Subtract off any double-count of A and
B (diagram above, (c))
A & B independent All from diagram above, (b)
Pr(A and B) = Pr(A)Pr(B) Precise defn of independent
Pr(A or B) = Pr(A) + Pr(B) Pr(A)Pr(B) Using the and and or rules above
Pr(B | A) = Pr(B) special case of conditional probability
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A & B mutually exclusive All from diagram above, (d)
Pr(A and B) = 0 Defn of mutually exclusive
Pr(A or B) = Pr(A) + Pr(B) Nothing to double-count; special case of
Pr(A or B) from above
Pr(B | A) = Pr(A | B) = 0 Cant both happen
Conditional probabilities All from diagram above, (c)
Pr(B | A) = Pr(B and A) / Pr(A) fraction of A that is also B.
Pr(B and A) = Pr(B | A)Pr(A) = Pr(A | B)Pr(B) Bayes Rule: Shows relationship between
Pr(B | A) and Pr(A | B)
Pr(A or B) = Pr(A) + Pr(B) Pr(A and B) Same as Always rule above
Note that the and rules are often simpler than the or rules.
To B, or To Not B?
Sometimes its easier to compute Pr(~A) than Pr(A). Then we can find Pr(A) from Pr(A) = 1 Pr(~A).
Example: What is the probability of rolling 4 or more with two dice?
The population has 36 possibilities. To compute this directly, we must use
4 5 11 12
33
3 4 5 6 5 4 3 2 1 33 Pr( 4)
36
way to ways to ways to ways to
roll roll roll roll
+ + + + + + + + = > =
Thats a lot of addition. Its much easier to note that:
2 3
3 33
Pr( 4) 1 2 3 Pr( 4) , Pr(
36 36
4) 1 Pr( 4)
ways to ways to
roll roll
and < = + = < = > = < =
In particular, the and rules are often simpler than the or rule. Therefore, when asked for the
probability of this or that, it is sometimes simpler to convert to its complementary and statement,
compute the and probability, and subtract it from 1 to find the or probability.
Example: From a standard 52-card deck, draw a single card. What is the chance it is a spade or a
face-card (or both)? Note that these events are independent.
To compute directly, we use the or rule:
Pr( ) 1/ 4, Pr( ) 3/13,
1 3 1 3 13 12 3 22
Pr( )
4 13 4 13 52 52
= =
+
= + = =
It may be simpler to compute the probability of drawing neither a spade nor a face-card, and
subtracting from 1:
Pr(~ ) 3/ 4, Pr(~ ) 10 /13,
3 10 30 22
Pr( ) 1 Pr(~ ~ ) 1 1
4 13 52 52
= =
= = = =
The benefit of converting to the simpler and rule increases with more or terms.
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Example: Remove the 12 face cards from a standard 52-card deck, leaving 40 number cards (aces
are 1). Draw a single card. What is the chance it is a spade (S), low (L) (4 or less), or odd (O)? Note that
these 3 events are independent.
To compute directly, we can count up the number of ways the conditions can be met, and divide by the
population of 40 cards. There are 10 spades, 16 low cards, and 20 odd numbers. But we cant just sum
those numbers, because we would double (and triple) count many of the cards.
Instead, we can extend the or the rules to 3 conditions, shown below.
S L
O
Venn diagram for Spade, Low, and Odd.
Without proof, we state that the direct computation from a 3-term or rule is this:
Pr( ) 1/ 4, Pr( ) 4/10, Pr( ) 1/ 2
Pr( ) Pr( ) Pr( ) Pr( )
Pr( ) Pr( ) Pr( ) Pr( ) Pr( ) Pr( ) Pr( ) Pr( ) Pr( )
1 4 1 1 4 1 1 4 1 1 4 1
4 10 2 4 10 4 2 10 2 4 10 2
10 16 20 4 5 8 2 31
40 40
S L O
S or L or O S L O
S L S O L O S L O
= = =
= + +
+
| | | | | | | |
= + + +
| | | |
\ . \ . \ . \ .
+ + +
= =
It is far easier to compute the chance that it is neither spade, nor low, nor odd:
Pr(~ ) 3/ 4, Pr(~ ) 6/10, Pr(~ ) 1/ 2
Pr( ) 1 Pr(~ ~ ~ ) 1 Pr(~ ) Pr(~ ) Pr(~ )
3 6 1 9 31
1 1
4 10 2 40 40
S L O
S or L or O S and L and O S L O
= = =
= =
= = =
You may have noticed that converting S or L or O into ~(~S and ~L and ~O) is an example of De
Morgans theorem from Boolean algebra.
Continuous Random Variables and Distributions
Probability is a little more complicated for continuous random variables. A continuous population is a
set of random values than can take on values in a continuous interval of real numbers; for example, if I spin
a board-game spinner, the little arrow can point in any direction: 0 < 2.
u = 0
u =
u
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Board game spinner
Further, all angles are equally likely. By inspection, we see that the probability of being in the first
quadrant is , i.e. Pr(0 < t/2) = . Similarly, the probability of being in any interval d is:
( )
1
Pr in any interval
2
d d u u u
t
=
If I ask, what is the chance that it will land at exactly = ? the probability goes to zero, because the
interval d goes to zero. In this simple example, the probability of being in any interval d is the same as
being in any other interval of the same size. In general, however, some systems have a probability per unit
interval that varies with the value of the random variable (call it X) (I wish I had a simple, everyday
example of this??). So
( ) Pr in an infinitesimal interval around pdf( ) X dx x x dx = , where
pdf(x) the probability distribution function.
pdf(x) has units of 1/x.
By summing mutually exclusive probabilities, the probability of X in any finite interval [a, b] is:
Pr( ) pdf( )
b
a
a X b dx x s s =
}
Since any random variable X must have some real value, the total probability of X being between
and
+ must be 1:
( ) Pr pdf( ) 1 X dx x
< < = =
}
The probability distribution function of a random variable tells you everything there is to know
Population and Samples
A population is a (often infinite) set of all possible values that a random variable may take on, along
with their probabilities. A sample is a finite set of values of a random variable, where those values come
from the population of all possible values. The same value may be repeated in a sample. We often use
samples to estimate the characteristics of a much larger population.
A trial or instance is one value of a random variable.
There is enormous confusion over the binomial (and similar) distributions, because each instance of a
binomial random variable comes from many attempts at an event, where each attempt is labeled either
success or failure. Superficially, an attempt looks like a trial, and many sources confuse the terms.
In the binomial distribution, n attempts go into making a single trial (or instance) of a binomial random
variable.
Variance
The variance of a population is a measure of the spread of any distribution, i.e. it is some measure of
how widely spread out values of a random variable are likely to be [there are other measures of spread,
too].
The variance of a population or sample is the most important parameter in statistics.
Variance is always positive, and is defined as the average squared-difference between the random
values and their average value:
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( ) ( )
2
var is an operator which takes the average X X X where
X X
The units of variance are the square of the units of X. If I multiply a set of random numbers by a
constant k, then I multiply the variance by k
2
:
( ) ( )
( )
2
var var is any set of random numbers
var takes the variance
kX k X where X =
Any function, including variance, with the above property is homogeneous-of-order-2 (2
nd
order
homogeneous??). We will return later to methods of estimating the variance of a population.
Standard Deviation
The standard deviation of a population is another measure of the spread of any distribution, closely
related to the variance. Standard deviation is always positive, and is defined as the square root of the
variance, which equals the root-mean-square (RMS) of the deviations from the average.
( ) ( ) ( )
2
dev var is an operator which takes the average X X X X where
X X
=
The units of standard deviation are the units of X. If I multiply a set of random numbers by a constant
k, then I multiply the standard deviation by the same constant k:
( ) ( )
( )
dev dev is any set of random numbers
dev takes the standard deviation
kX k X where X =
TBS: Why we care about standard deviation, even for non-normal populations. Bounds on percentage
of population contained for any population. Stronger bounds for unimodal populations.
Normal (aka Gaussian) Distribution
From mathworld.wolfram.com/NormalDistribution.html : While statisticians and mathematicians
uniformly use the term normal distribution for this distribution, physicists sometimes call it a Gaussian
distribution and, because of its curved flaring shape, social scientists refer to it as the bell curve.
A gaussian distribution is one of a family of distributions defined as a population with
2
1
2
population average
1
pdf( )
population standard deviation 2
x
x e where
o
o t o
| |
|
\ .
=
[picture??].
and are parameters: can be any real value, and > 0 and real. This illustrates a common feature
named distributions: they are usually a family of distributions, parameterized by one or more parameters.
The gaussian distribution is a 2-parameter distribution: and .
New Random Variables From Old Ones
Given two random variables X and Y, we can construct new random variables as functions of x and y
(trial values of X and Y). One common such new random variable is simply the sum:
,
i i i
Define Z X Y which means trial i z x y + +
X
(x) and pdf
Y
(y) (all we can know about X and Y), what is pdf
Z
(z)? Given a
particular value x of X, we see that
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( ) ( ) : Pr( ) Pr Given x Z within dz of z Y within dz of z x =
But x is a value of a random variable, so the total Pr(Z within dz of z) is the sum (integral) over all x:
( ) ( )
( ) ( ) ( )
( )
( )
Pr( ) pdf ( ) Pr ,
Pr pdf
Pr( ) pdf ( ) pdf
pdf ( ) pdf ( ) pdf
X
Y
X Y
Z X Y
Z within dz of z dx x Y within dz of z x but
Y within dz of z x z x dz so
Z within dz of z dz dx x z x
z dx x z x
=
=
=
=
}
}
}
This integral way of combining two functions, pdf
X
(x) and pdf
Y
(y) with a parameter z is called the
convolution of pdf
X
and pdf
Y
, which is a function of a parameter z.
x y x
Convolution of pdf
X
with pdf
Y
at z = 8
pdf
Y
(y) pdf
X
(x)
z = 8
The convolution evaluated at z is the area under the product pdf
X
(x)pdf
Y
(z x).
From the above, we can easily deduce the pdf
Z
(z) if Z X Y = X + (Y). First, we find pdf
(Y)
(y), and
then use the convolution rule. Note that
( )
( )
pdf ( ) pdf ( )
pdf ( ) pdf ( ) pdf ( ) pdf ( ) pdf ( )
Y Y
Z X Y X Y
y y
z dx x z x dx x x z
=
= =
} }
Since we are integrating from to +, we can shift x with no effect:
pdf ( ) pdf ( ) pdf ( )
Z X Y
x x z z dx x z x
+ = +
}
which is the standard form for the correlation function of two functions, pdf
X
(x) and pdf
Y
(y).
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x
pdf
X
(x)
y
pdf
Y
(y)
x
Correlation of pdf
X
with pdf
Y
at z = 2
z = 2
The correlation function evaluated at z is the area under the product pdf
X
(x + z)pdf
Y
(x).
The pdf of the sum of two random variables is the convolution of the pdfs of those random
variables.
The pdf of the difference of two random variables is the correlation function of the pdfs of those
random variables.
Note that the convolution of a gaussian distribution with a different gaussian is another gaussian.
Therefore, the sum of a gaussian random variable with any other gaussian random variable is gaussian.
Some Distributions Have Infinite Variance, or Infinite Average
In principle, the only requirement on a PDF is that it be normalized:
PDF( ) 1 x dx
=
}
Given that, it is possible that the variance is infinite (or properly, undefined). For example, consider:
3
2 2
0 0
PDF( ) 1
PDF( ) 1, PDF( )
0 0
x x x
x x x dx but x x dx
x
o
= >
= = = =
`
= <
)
} }
The above distribution is normalized, and has finite average, but infinite deviation. Even worse,
2
2 2
0 0
PDF( ) 1
PDF( ) , PDF( )
0 0
x x x
x x x dx and x x dx
x
o
= >
= = = =
`
= <
)
} }
This distribution is normalized, but has both infinite average and infinite deviation.
Are such distributions physically meaningful? Sometimes. The Lorentzian (aka BreitWigner)
distribution is common in physics, or at least, a good approximation to physical phenomena. It has infinite
average and deviation. Its normal and parameterized forms are:
( ) ( )
0
2 2
0
0
1 1 1 1
( ) ( ; , )
1
1 /
location of peak, half-width at half-maximum
L x L x x
x
x x
where x
t t
= =
+
+
This is approximately the energy distribution of particles created in high-energy collisions. Its CDF
is:
0
1 1
CDF ( ) arctan
2
Lorentzian
x x
x
t
| |
= +
|
\ .
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Samples and Parameter Estimation
In statistics, an efficient estimator the most efficient estimator [ref??]. There is none better (i.e.,
none with smaller variance). You can prove mathematically that the average and variance of a sample are
the most efficient estimators (least variance) of the population average and variance. It is impossible to do
any better, so its not worth looking for better ways. The most efficient estimators are least squares
estimators, which means that over many samples, they minimize the sum-squared error from the true value.
We discuss least-squares vs. maximum-likelihood estimators later.
Note, however, than given a set of measurements, some of them may not actually measure the
population of interest (i.e., they may be noise). If you can identify those bad measurements from a sample,
you should remove them before estimating any parameter. Usually, in real experiments, there is always
some unremovable corruption of the desired signal, and this contributes to the uncertainty in the
measurement.
The sample average is defined as:
1
1
n
i
i
x x
n
=
and is the least variance estimate of the average <X> of any population. It is unbiased, which means
the average of many estimates approaches the true population average:
average, over the given parameter if not obvious
many samples over what
x X where = -
Note that the definition of unbiased is not that the estimator approaches the true value for large
samples; it is that the average of the estimator approaches the true value over many samples, even small
samples.
The sample variance and standard deviation are defined as:
( )
2
2
1
2
1
is the sample average, as above :
1
n
i i
i
s x x where x x x
n
s s
=
The sample variance is an efficient and unbiased estimate of Var(X), which means no other estimate of
Var(X) is better. Note that s
2
is unbiased, but s is biased, because the square root of the average is not equal
to the average of the square root. I.e.,
( )
2 2
many samples
s Dev X because s s = =
This exemplifies the importance of properly defining bias:
( ) ( ) lim
many samples
N
s Dev X even though s Dev X
= =
Sometimes you see variance defined with 1/n, and sometimes with 1/(n 1). Why? The population
variance is defined as the mean-squared deviation from the population average. For a finite population, we
find the population variance using 1/N, where N is the number of values in the whole population:
( )
2
1
is the # of values in the entire population
1
( ) is the value of the population
exact population average
N
th
i i
i
N
Var X X where X i
N
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In contrast, the sample variance is the variance of a sample taken from a population. The population
average is usually unknown. We can only estimate <x>. Then to make s
2
unbiased, one can show
that it must use 1/(n 1), where n is the sample size, not population size. (Show this??).
This is actually a special case of curve fitting, where we fit a constant, <x>, to the population. This is a
single parameter, and so removes 1 degree of freedom from our fit errors. Hence, the mean-squared fit
error has 1 degree of freedom less than the sample size. (Show this with algebra.)
For a sample from a population when the average is exactly known, we use n as the weighting for s
2
:
( )
2
2
1
1
n
i
i
s x
n
=
=
## , which is just the above equation with X
i
x
i
, N n.
Notice that infinite populations with unknown can only have samples, and thus always use n1. But
as n , it doesnt matter, so we can compute the population variance either way:
1 1
1 1
( ) lim lim
1
n n
i i
n n
i i
Var X x x
n n
= =
=
, because n = n 1, when n = .
Central Limit Theorem For Continuous And Discrete Populations
The central limit theorem is important because it allows us to estimate some properties of a population
given only sample of the population, with no a priori information. Given a population, we can take a
sample of it, and compute its average. If we take many samples, each will (likely) produce a different
average. Hence, the average of a sample is a new random variable, created from the original.
The central limit theorem says that for any population, as the sample size grows, the sample
average becomes a gaussian random variable, with average equal to the population average, and
variance equal to the population variance divided by n.
2
2
Given a random variable , with mean and variance , then
lim , sample average
X
X
n
X
x Gaussian where x
n
o
o
| |
e
|
|
\ .
Note that the central limit theorem applies only to multiple samples from a single population (though
there are some variations that can be applied to multiple populations). [It is possible to construct large
sums of multiple populations whose averages are not gaussian, e.g. in communication theory, inter-symbol
interference (ISI). But we will not go further into that.]
How does the Central Limit Theorem apply to a discrete population? If a population is discrete,
then any sample average is also discrete. But the gaussian distribution is continuous. So how can the
sample average approach a gaussian for large sample size N? Though the sample average is discrete, the
density of allowed values increases with N. If you simply plot the discrete values as points, those points
approach the gaussian curve. For very large N, the points are so close, they look continuous.
TBS: Why binomial (discreet), Poisson (discreet), and chi-squared (continuous) distributions approach
gaussian for large n (or v).
Uncertainty of Average
The sample average <x> gives us an estimate of the population average . The sample average, when
taken as a set of values of many samples, is itself a random variable. The Central Limit Theorem (CLT)
says that if we know the population standard deviation , the sample average <x> will have standard
deviation
x
n
o
o
< >
= (proof below)
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In statistics,
<x>
is called the standard error of the mean. In experiments,
<x>
is the 1-sigma
uncertainty in our estimate of the average . However, most often, we know neither nor , and must
estimate both from our sample, using <x> and s. For large samples, we use simply s, and then
for "large" samples, i.e "large"
x
s
n
n
o
< >
~
For small samples, we must still use s as our estimate of the population deviation, since we have
nothing else. But instead of assuming that
<x>
is gaussian, we use the exact distribution, which is a little
wider, called a T-distribution [W&M ??], which is complicated to write explicitly. It has a parameter, t,
similar to the gaussian , which measures its spread:
sample average, sample standard deviation
x x
t where x s
s
= =
We then use t, and t-tables, to establish confidence intervals.
Uncertainty of Uncertainty: How Big Is Infinity?
Sometimes, we need to know the uncertainty in our estimate of the population variance (or standard
deviation). So lets look more closely at the uncertainty in our estimate s
2
of the population variance
2
.
The random variable
( )
2
2
1 n s
o
## has chi-squared distribution with n 1 degrees of freedom [W&M Thm
6.16 p201]. So
( )
( )
( )
2
2 2 4
2 2 2
1
2
2 2
2
2 1 ,
1 1 1
2
2 1
1 1
n
s Var s n
n n n
Dev s n
n n
o o o
_
o
o
| |
e = =
|
|
\ .
= =
However, usually were more interested in the uncertainty of the standard deviation estimate, rather
than the variance. For that, we use the fact that s is function of s
2
: s (s
2
)
1/2
. For moderate or bigger
sample sizes, and confidence ranges up to 95% or so, we can use the approximate formula for the deviation
of a function of a random variable (see Functions of Random Variables, elsewhere)
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
1/ 2 1/ 2
2 2 2 2
( ) ' for small Dev(X)
1 1 1 2 1 1
( )
2 2 1
2 1 2 1
Y f X Dev Y f X Dev X
s s Dev s Dev s s
n
n n
o o o
o
= ~
~ = = ~
This allows us to explain the rule of thumb: n > 30 is statistical infinity.
This rule is most often used in estimating the standard error of the mean
<x>
(see above), given by
x
s
n n
o
o
< >
= ~ . We dont know the population deviation, , so we approximate it with s . For small
samples, this isnt so good. Then the uncertainty
<x>
needs to include both the true sampling uncertainty in
<x> and the uncertainty in s. To be confident that our <x> is within our claim, we need to expand our
confidence limits, to allow for the chance that s happens to be low. The Student T-distribution exactly
handles this correction to our confidence limits on <x> for all sample sizes. However, roughly, an upper
bound on would be, say, the 95% limit on s, which is about
( )
95%
1
2 2
2 1
s
s s s
n
o o ~ + ~ +
This might seem circular, because we still have (which we dont know) on the right hand side.
However, its effect is now reduced by the fraction multiplying it. So the uncertainty in is also reduced
by this factor, and we can neglect it. Thus to first order, we have
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( )
95%
1 2
2 2 1
1
2 1
s
s s s s s
n
n
o
| |
~ + ~ + = +
|
|
\ .
So long as \[2/(n 1)] is small compared to 1, we can ignore it. Plug into our formula for
<x>
:
95%
,95%
2
1
1
x x
s s
n n n n
o
o o
< > < >
| |
= ~ = +
|
|
\ .
When n = 30, \[2/(n 1)] = 0.26. That seems like a lot to me, but n = 30 is the generally agreed upon
bound for good confidence that s .
Combining Estimates of Varying Uncertainty
Weight measurements by 1/
2
: When taking data, our measurements often have varying uncertainty:
some samples are better than others. We can still find an average, but what is the best average, and what
is
<x>
of a set of samples, each with its own uncertainty,
i
? We find this from the rule that variances
add (if the random variables are uncorrelated).
In general, if you have a set of estimates of a parameter, but each estimate has a different uncertainty,
how do you combine the estimates for the most reliable estimate of the parameter? Clearly, estimates with
smaller variance should be given more weight than estimates with larger variance. But exactly how much?
In general, you weight each estimate by 1/
2
. For example, to estimate the average of a population from
several samples of different uncertainty (different variance):
2
1
2
2
1
are the estimates of the average
1
are the variances of those estimates
n est
i
est
i
i i
n
i
i i
x
x
x where
o
o
o
=
=
~
This is just a weighted average, where the denominator is the sum of the weights. There are whole
books written on parameter estimation, so it can be a big topic.
Functions of Random Variables
It follows from the definition of probability that the average value of any function of a random variable
is
( ) ( ) pdf ( )
X
f X dx f x x
=
}
We can apply this to our definitions of population average and population variance:
( )
2
pdf ( ) ( ) pdf ( )
X X
X X dx x x and Var X dx x X x
= =
} }
Statistically Speaking: What Is The Significance of This?
Before we compute any uncertainties, we should understand what they mean. Statistical significance
interprets uncertainties. It is one of the most misunderstood, and yet most important, concepts in science.
It underlies virtually all experimental and simulation results. Beliefs (correct and incorrect) about statistical
significance drive experiment, research, funding, and policy.
Understanding statistical significance is a prerequisite to understanding science.
This cannot be overstated, and yet many (if not most) scientists and engineers receive no formal
training in statistics. The following few pages describe statistical significance, surprisingly using almost no
math.
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Overview of Statistical Significance
The term statistically significant has a precise meaning which is, unfortunately,
different than the common meaning of the word significant.
Many experiments compare quantitative measures of two populations, e.g. the IQs of ferrets vs.
gophers. In any real experiment, the two measures will almost certainly differ. How should we interpret
this difference?
We can use statistics to tell us the meaning of the difference. A difference which is not statistically
significant in some particular experiment may, in fact, be quite important. But we can only determine its
importance if we do another experiment with finer resolution, enough to satisfy our subjective judgment of
importance. For this section, I use the word importance to mean a subjective assessment of a measured
result.
The statement We could not measure a difference is very different from There is no important
difference. Statistical significance is a quantitative comparison of the magnitude of an effect and the
resolution of the statistics used to measure it.
This section requires an understanding of probability and uncertainty.
statistical significance is and is not. We then give more specific statements and examples.
Statistical significance is many things:
Statistical significance is a measure of an experiments ability to resolve its own measured result.
It is not a measure of the importance of a result.
Statistical significance is closely related to uncertainty.
Statistical significance is a quantitative statement of the probability that a result is real, and not a
measurement error or the random result of sampling that just happened to turn out that way (by chance).
Statistically significant means measurable by this experiment. Not statistically significant
means that we cannot fully trust the result from this experiment alone; the experiment was too crude to
have confidence in its own result.
Statistical significance is a one-way street: if a result is statistically significant, it is real. However, it
may or may not be important. In contrast, if a result is not statistically significant, then we dont know if
its real or not. However, we will see that even a not significant result can sometimes provide meaningful
and useful information.
If the difference between two results in an experiment is not statistically significant,
that difference may still be very real and important.
Details of Statistical Significance
A meaningful measurement must contain two parts: the magnitude of the result, and the confidence
limits on it, both of which are quantitative statements. When we say, the average IQ of ferrets in our
experiment is 102 5 points, we mean that there is a 95% chance that the actual average IQ is between 97
and 107. We could also say that our 95% confidence limits are 97 to 107. Or, we could say that our 95%
uncertainty is 5 points. The confidence limits are sometimes called error bars, because on a graph,
confidence limits are conventionally drawn as little bars above and below the measured values.
Suppose we test gophers and find that their average IQ is 107 4 points. Can we say on average,
gophers have higher IQs than ferrets? In other words, is the difference we measured significant, or did it
happen just by chance? To assess this, we compute the difference, and its uncertainty (recall that
( )
2 2
107 102 4 5 5 6 (gophers ferrets) IQ A = + =
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This says that the difference lies within our uncertainty, so we are not 95% confident that gophers have
higher IQs. Therefore, we still dont know if either population has higher IQs than the other. Our
experiment was not precise enough to measure a difference. This does not mean that there is no difference.
However, we can say that there is a 95% chance that the difference is between 1 and 11 (5 6). A given
experiment measuring a difference can produce one of two results of statistical significance: (1) the
difference is statistically significant; or (2) it is not. In this case, the difference is not (statistically)
significant at the 95% level.
In addition, confidence limits yield one of three results of importance: (1) confirm that a difference
is important; or (2) not important, or (3) be inconclusive. But the judgment of how much is important is
outside the scope of the experiment. For example, we may know from prior research that a 10 point
average IQ difference makes a population a better source for training pilots, enough better to be
important. Note that this is a subjective statement, and its precise meaning is outside our scope here.
Five of the six combinations of significance and importance are possible, as shown by the following
examples.
Example 1, not significant, and inconclusive importance: With the given numbers, IQ = 5 6, the
importance of our result is inconclusive, because we dont know if the average IQ difference is more or
less than 10 points.
Example 2, not significant, but definitely not important: Suppose that prior research showed
(somehow) that a difference needed to be 20 points to be important. Then our experiment shows that the
difference is not important, because the difference is very unlikely to be as large as 20 points. In this case,
even though the results are not statistically significant, they are very valuable; they tell us something
meaningful and worthwhile, namely, the difference between the average IQs of ferrets and gophers is not
important for using them as a source for pilots. The experimental result is valuable, even though not
significant, because it establishes an upper bound on the difference.
Example 3, significant, but inconclusive importance: Suppose again that a difference of 10 points is
important, but our measurements are: ferrets average 100 3 points, and gophers average 107 2 points.
Then the difference is:
( )
2 2
107 100 2 3 7 4 (gophers ferrets) IQ A = + =
These results are statistically significant: there is better than a 95% chance that the average IQs of
ferrets and gophers are different. However, the importance of the result is still inconclusive, because we
dont know if the difference is more or less than 10 points.
Example 4, significant and important: Suppose again that a difference of 10 points is important, but
we measure that ferrets average 102 3 points, and gophers average 117 2 points. Then the difference is:
( )
2 2
117 102 2 3 15 4 (gophers ferrets) IQ A = + =
Now the difference is both statistically significant, and important, because there is a 95% chance that
the difference is > 10 points. We are better off choosing gophers to go to pilot school.
Example 5, significant, but not important: Suppose our measurements resulted in
5 4 IQ A =
Then the difference is significant, but not important, because we are confident that the difference < 10.
This result established an upper bound on the difference. In other words, our experiment was precise
enough that if the difference were important (i.e., big enough to matter), then wed have measured it.
Finally, note that we cannot have a result that is not significant, but important. Suppose our result was:
11 12 IQ A =
The difference is unmeasurably small, and possibly zero, so we certainly cannot say the difference is
important. In particular, we cant say the difference is greater than anything.
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Thus we see that stating there is a statistically significant difference is (by itself) not saying much,
because the difference could be tiny, and physically unimportant.
We have used here the common confidence limit fraction of 95%, often taken to be ~2. The next
most common fraction is 68%, or ~1. Another common fraction is 99%, taken to be ~3. More precise
gaussian fractions are 95.45% and 99.73%, but the digits after the decimal point are usually meaningless
(i.e., not statistically significant!) Note that we cannot round 99.73% to the nearest integer, because that
would be 100%, which is meaningless in this context. Because of the different confidence fractions in use,
you should always state your fractions explicitly. You can state your confidence fraction once, at the
beginning, or along with your uncertainty, e.g. 10 2 (1).
Caveat: We are assuming random errors, which are defined as those that average out with larger
sample sizes. Systematic errors do not average out, and result from biases in our measurements. For
example, suppose the IQ test was prepared mostly by gophers, using gopher cultural symbols and
metaphors unfamiliar to most ferrets. Then gophers of equal intelligence will score higher IQs because the
test is not fair. This bias changes the meaning of all our results, possibly drastically.
Ideally, when stating a difference, one should put a lower bound on it that is physically important, and
give the probability (confidence) that the difference is important. E.g. We are 95% confident the
difference is at least 10 points (assuming that 10 points on this scale matters).
Examples
Here are some examples of meaningful and not-so-meaningful statements:
Meaningless Statements
(appearing frequently in print)
Meaningful Statements, possibly
subjective
(not appearing enough)
The difference in IQ between groups A
and B is not statistically significant.
because the difference is small?)
We measured an average IQ difference
of 5 points. (With what confidence?)
Group A has a statistically significantly
higher IQ than group B.
(How much higher? Is it important?)
Our data show there is a 99% likelihood
that the IQ difference between groups A and B
is less than 1 point.
resolution to tell if there was an important
difference in IQ.
Our data show there is a 95% likelihood
that the IQ difference between groups A and B
is greater than 10 points.
Statistical significance summary: Statistical significance is a quantitative statement about an
experiments ability to resolve its own result. Importance is a subjective assessment of a measurement
that may be guided by other experiments, and/or gut feel. Statistical significance says nothing about
whether the measured result is important or not.
Predictive Power: Another Way to Be Significant, but Not Important
Suppose that we have measured IQs of millions of ferrets and gophers over decades. Suppose their
population IQs are gaussian, and given by (note the switch to 1 uncertainties):
:101 20 :103 20 (1 ) ferrets gophers o
The average difference is small, but because we have millions of measurements, the uncertainty in the
average is even smaller, and we have a statistically significant difference between the two groups.
Suppose we have only one slot open in pilot school, but two applicants: a ferret and a gopher. Who
should get the slot? We havent measured these two individuals, but we might say, Gophers have
significantly higher IQs than ferrets, so well accept the gopher. Is this valid?
To quantitatively assess the validity of this reasoning, let us suppose (simplistically) that pilot students
with an IQ of 95 or better are 20% more likely to succeed than those with IQ < 95. From the given
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statistics, 65.5% of gophers have IQs > 95, vs. 61.8% of ferrets. The relative probabilities of success are
then:
: 0.382 0.618(1.2) 1.12 : 0.345 0.655(1.2) 1.13 ferrets gophers + = + =
So a random gopher is less than 0.7% more likely to succeed than a random ferret. This is pretty
unimportant. In other words, species (between ferrets and gophers) is not a good predictor of success. Its
so bad that many, many other facts will be better predictors of success. Height, eyesight, years of
schooling, and sports ability are probably all better predictors. The key point is this:
Differences in average, between two populations, which are much smaller than the deviations
within the populations, are poor predictors of individual outcomes.
Bias and the Hood (Unbiased vs. Maximum-Likelihood Estimators)
In experiments, we frequently have to estimate parameters from data. There is a very important
difference between unbiased and maximum likelihood estimates, even though sometimes they are the
same. Sadly, two of the most popular experimental statistics books completely destroy these concepts, and
their distinction.
[Both books try to derive unbiased estimates using the principle of maximum likelihood, which is
impossible since the two concepts are very different. The incorrect argument goes through the exercise of
deriving the formula for sample variance from the principle of maximum likelihood, and (of course) gets the
wrong answer! Hand waving is then applied to wiggle out of the mistake.]
Everything in this section applies to arbitrary distributions, not just gaussian. We follow these steps:
1. Terse definitions, which wont be entirely clear at first.
2. Example of estimating the variance of a population (things still fuzzy).
3. Silly example of the need for maximum-likelihood in repeated trials.
4. Real-world physics examples of different situations leading to different choices between unbiased
and maximum-likelihood.
Terse definitions: In short:
An unbiased statistic is one whose average is exactly right: the average of many samples of an
unbiased statistic is closer to the right answer than one sample is.
In the limit of an infinite number of estimates, the average of an unbiased statistic is exactly the
population parameter.
A maximum likelihood statistic is one which is most likely to have produced the given the data. Note
that if it is biased, then the average of many maximum likelihood estimates does not get you closer to right
answer. In other words, given a fixed set of data, maximum-likelihood estimates have some merit, but
biased ones cant be combined well with other sets of data (perhaps future data, not yet taken). This
concept should become more clear below.
Which is better, an unbiased estimate or a maximum-likelihood estimate? It depends on what you
goals are.
Example of population variance: Given a sample of values from a population, an unbiased estimate
of the population variance is
( )
2
2 1
(unbiased estimate)
1
n
i
i
x x
n
o
=
~
If we take several samples of the population, compute an unbiased estimate of the variance for each
sample, and average those estimates, well get a better estimate of the population variance. Generally,
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unbiased estimators are those that minimize the sum-squared-error from the true value (principle of least-
squares).
However, suppose we only get one shot at estimating the population variance? Suppose Monty Hall
says Ill give you a zillion dollars if you can estimate the variance (to within some tolerance)? What
estimate should we give him? Since we only get one chance, we dont care about the average of many
estimates being accurate. We want to give Mr. Hall the variance estimate that is most likely to be right.
One can show that the most likely estimate is given by using n in the denominator, instead of (n 1):
( )
2
2 1
(maximum-likelihood estimate)
n
i
i
x x
n
o
=
~
This is the estimate most likely to win the prize. Perhaps more realistically, if you need to choose how
long to fire a retro-rocket to land a spacecraft on the moon, do you choose (a) the burn time that, averaged
over many spacecraft, reaches the moon, or (b) the burn time that is most likely to land your one-and-only
craft on the moon?
In the case of variance, the maximum-likelihood estimate is smaller than the unbiased estimate by a
factor of (n 1)/n. If we were to make many maximum-likelihood estimates, each one would be small by
the same factor. The average would then also be small by that factor. No amount of averaging would ever
fix this error. Our average estimate of the population variance would not get better with more estimates.
You might conclude that maximum-likelihood estimates are only good for situations where you get a
single trial. However, we now show that maximum-likelihood estimates can be useful even when there are
many trials of a statistical process.
A silly example: You are a medieval peasant barely keeping your family fed. Every morning, the
benevolent king goes to the castle tower overlooking the public square, and tosses out a gold coin to the
crowd. Whoever catches it, keeps it.
Being better educated than most medieval peasants, each day you record how far the coin goes, and
generate a pdf (probability distribution function) for the distance from the tower. It looks like this:
distance
Gold Coin Toss Distance
pdf
most likely
average
The most-likely distance is notably different than the average distance. Given this information, where
do you stand each day? Answer: At the most-likely distance, because that maximizes your payoff not only
for one trial, but across many trials over a long time. The best estimator is in the eye of the beholder: as a
peasant, you dont care much for least squares, but you do care about most money.
Note that the previous example of landing a spacecraft is the same as the gold coin question: even if
you launch many spacecraft, for each one you would give the burn most-likely to land the craft. The
average of many failed landings has no value.
Real physics examples: Example 1: Suppose you need to generate a beam of ions, all moving at
very close to the same speed. You generate your ions in a plasma, with a Maxwellian thermal speed
distribution (roughly the same shape as the gold coin toss pdf). Then you send the ions through a velocity
selector to pick out only those very close to a single speed. You can tune your velocity selector to pick any
speed. Now ions are not cheap, so you want your velocity selector to get the most ions from the speed
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distribution that it can. That speed is the most-likely speed, not the average speed. So here again, we see
that most-likely has a valid use even in repeated trials of random processes.
Example 2: Suppose you are tracing out the orbit of the moon around the earth by measuring the
distance between the two. Any given days measurement has limited ability to trace out an entire orbit, so
you must make many measurements over several years. You have to fit a model of the moons orbit to this
large set of measurements. Youd like your fit to get better as you collect more data. Therefore, you
choose to make unbiased estimates of the distance, so that on-average, over time, your estimate of the orbit
gets better and better. If instead you chose each days maximum-likelihood estimator, youd be off of the
average (in the same direction) every day, and no amount of averaging would ever fix that.
Wrap up: When you have symmetric, unimodal distributions (symmetric around a single maximum),
then the unbiased and maximum-likelihood estimates are identical. This is true, for example, for the
average of a gaussian distribution. For asymmetric or multi-modal distributions, the unbiased and
maximum-likelihood estimates are different, and have different properties. In general, unbiased estimates
are the most efficient estimators, which means they have the smallest variance of all possible estimators.
Unbiased estimators are also least-squares estimators, which means they minimize the sum-squared error
from the true value. This property follows from being unbiased, since the average of a population is the
least-squares estimate of all its values.
Correlation and Dependence
To take a sample of a random variable X, we get a value of X
i
for each sample point i, i = 1 ... n.
Sometimes when we take a sample, for each sample point we get not one, but two, random variables, X
i
and
Y
i
. The two random variables X
i
and Y
i
may or may not be related to each other. We define the joint
probability distribution function of X and Y such that
Pr( ) pdf ( , )
XY
x X x dx and y Y y dy x y < < + < < + =
This is just a 2-dimensional version of a typical pdf. Since X and Y are random variables, we could
look at either of them and find its individual pdf: pdf
X
(x), and pdf
Y
(y). If X and Y have nothing to do with
each other (i.e., X and Y are independent), then a fundamental axiom of probability says that the
probability of finding x < X < x + dx and y < Y < y + dy is the product of the two pdfs:
pdf ( , ) pdf ( ) pdf ( )
XY X Y
x y x y if X and Y are independent =
The above equation is the definition of statistical independence:
Two random variables are independent if and only if their joint distribution
function is the product of the individual distribution functions.
A very different concept is the correlation. Correlation is a measure of how linearly related two
random variables are. It turns out that we can define correlation mathematically by the correlation
coefficient:
( ) ( )
( )
( ) ( ) ( )
,
, ,
X Y X Y
X X Y Y
Cov X Y
where Cov X Y X X Y Y
o o o o
=
For a discrete variable,
( ) ( )
1
, #
N
i i
x y
i
x x y y
where N elements in population
N
o o
=
= =
If = 0, then X and Y are uncorrelated. If = 0, then X and Y are correlated. Note that
( ) ( )
1
0 ( , ) 0, or for a discrete variable 0
population
i i
i
Cov X Y x x y y
=
= = =
,
where Cov(X, Y) is the covariance of X and Y.
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Two random variables are uncorrelated if and only
if their covariance, defined above, is zero.
Being independent is a stronger statement that uncorrelated. Random variables which are independent
are necessarily uncorrelated (proof??). But variables which are uncorrelated can be highly dependent. For
example, suppose we have a random variable X, which is uniformly distributed over [1, 1]. Now define a
new random variable Y such that Y = X
2
. Clearly, Y is dependent on X, but Y is uncorrelated with X. Y and
X are dependent because given either, we know a lot about the other. They are uncorrelated because for
every Y value, there is one positive and one negative value of X. So for every value of ( ) ( ) X X Y Y ,
there is its negative, as well. The average is therefore 0; hence, Cov(X, Y) = 0.
A crucial point is:
Variances add for uncorrelated variables, even if they are dependent.
This is easy to show. Given that X and Y are uncorrelated,
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
2 2
2 2 2
2 2
Var X Y X Y X Y X X Y Y
X X X X Y Y Y Y X X X X Y Y
( ( + = + + = +
= + + = +
( )
2
( ) ( )
Y Y
Var X Var Y
+
= +
All we needed to prove that variances add is that Cov(X, Y) = 0.
Data Fitting (Curve Fitting)
Suppose we have an ideal process, with an ideal curve mapping an independent variable x to a
dependent variable y. Now we take a set of measurements of this process, that is, we measure a set of data
pairs (x
i
, y
i
), below left:
x
y(x)
Ideal curve,
with non-ideal data
x
Data,
with straight line guess
y(x)
Suppose further we dont know the ideal curve, but we have to guess it. Typically, we make a guess of
the general form of the curve from theoretical or empirical information, but we leave the exact parameters
of the curve free. For example, we may guess that the form of the curve is a straight line (above right)
y mx b = + ,
but we leave the slope and intercept (m and b) of the curve unknown. (We might guess another form,
with other, possibly more parameters.) Then we fit our curve to the data, which means we compute the
values of m and b which best fit the data. Best means that the values of m and b minimize some
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measure of error, called the figure of merit, compared to all other values of m and b. For data with
constant uncertainty, the most common figure of merit is the sum-squared error:
( ) ( ) ( )
2 2
2
1 1 1
sum-squared-error
( ) is our fitting function
n n n
i i i i i
i i i
SSE error measurement curve measurement f x
where f x
= = =
= =
In our example of fitting to a straight line, for given values of m and b, we have:
( )
2
2
1 1
( )
n n
i i i
i i
SSE error y mx b
= =
= +
Curve fitting is the process of finding the values of all our unknown parameters
such that (for constant uncertainty) they minimize the sum-squared error from our data.
The (measurement curve) is often written as (O C) for (observed computed).
We discuss data with varying uncertainty later, but in that more general case, we adjust parameters to
minimize the
2
parameter.
Multiple Linear Regression and Polynomial Fitting
Fitting a polynomial to data is actually a simple example of multiple linear regression (see also the
Numerical Analysis section for exact polynomial fits). A simple example of linear regression is: you
measure some function y of an independent variable x, i.e. you measure y(x) for some set of x = {x
i
}. You
have a model for y(x) which is a linear combination of basis functions:
1 1 2 2
1
( ) ( ) ( ) ... ( ) ( )
k
k k m m
m
y x b f x b f x b f x b f x
=
= + + + =
You use multiple linear regression to find the coefficients b
i
of the basis functions f
i
which compose
the measured function, y(x). Note that:
Linear regression is not fitting data to a straight line.
Fitting data to a line is called fitting data to a line (seriously).
The funky part is understanding what are the random variables or predictors to which we perform
the regression. Most intermediate statistics texts cover multiple linear regression, e,g, [W&M p353], but
we remind you of some basic concepts here:
1. Multiple linear regression predicts the value of some random variable y
i
from k (possibly
correlated) predictors, x
mi
, m = 1, 2, ... k. The predictors may or may not be random variables. In
the example above, the predictors are x
mi
= f
m
(x
i
).
2. Its linear prediction, so our prediction model is that y is a linear combination of the xs, i.e., for
each i:
0 1 1 2 2
1
...
k
i i i k ki m mi
m
y b b x b x b x b x
=
= + + + + =
3. Multiple linear regression determines the unknown regression coefficients b
0
, b
1
, ... b
k
from n
samples of the y and x
m
, by solving the following k + 1 linear equations in k + 1 unknowns [W&M
p355]:
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0 1 1 2 2
1 1 1 1
0 1 1 2 2
1 1 1 1 1
( ) ...
1, 2,... :
...
n n n n
i i k ki i
i i i i
n n n n n
mi mi i mi i k mi ki mi i
i i i i i
b n b x b x b x y
And for each m k
b x b x x b x x b x x x y
= = = =
= = = = =
+ + + + =
=
+ + + + =
Again, all the y
i
and x
mi
are given; we solve for b
m
. Numerically, we slap the sums on the left into a
matrix, and the constants on the right into a vector, and call a function (e.g., gaussj( ) in Numerical Recipes)
to solve for the unknowns.
Polynomials are just a special case of multiple linear regression [W&M p357], where we are predicting
y
i
from powers of x
i
. As such, we let ( )
m
mi i
x x = , and proceed with standard multiple linear regression:
2
0 1 2
1 1 1 1
1 2
0 1 2
1 1 1 1 1
( ) ...
1, 2,... :
...
n n n n
k
i i k i i
i i i i
n n n n n
m m m m k m
i i i k i i i
i i i i i
b n b x b x b x y
And for each m k
b x b x b x b x x y
= = = =
+ + +
= = = = =
+ + + + =
=
+ + + + =
Goodness of Fit
Chi-Squared Distribution
You dont really need to understand the
2
distribution to understand the
2
parameter, but we start here
Notation: ( ) X D x e means X is a random variable with probability distribution function (pdf) = D(x).
Chi-squared (
2
) distributions are a family of distributions characterized by 1-parameter, called
(Greek nu). (Contrast with the gaussian distribution, which has two parameters, the mean, , and standard
deviation, .) So we say chi-squared is a 1-parameter distribution. is almost always an integer. The
simplest case is = 1: if we define a new random variable X from a gaussian random variable , as
2
, ( 0, 1), . . 0, . 1 X where gaussian i e avg std deviation _ _ o = e = = = =
then X has a
2
1
distribution. I.e.,
2
=1
(x) is the probability distribution function (pdf) of the square of a
gaussian.
For general ,
2
(x) is the pdf of the sum of the squares of gaussian random variables:
2
1
, ( 0, 1), . . 0, . 1
i i
i
X where gaussian i e avg std deviation
v
_ _ o
=
= e = = = =
Thus, the random variable X above has a
2
## distribution. [picture??] Chi-squared random variables
are always 0, since they are the sums of squares of gaussian random variables. Since the gaussian
distribution is continuous, the chi-squared distributions are also continuous.
From the definition, we can also see that the sum of two chi-squared random variables is another chi-
squared random variable:
2 2 2
,
n m n m
Let A B then A B _ _ _
+
e e + e
By the central limit theorem, this means that for large , chi-squared itself approaches gaussian.
We can show that
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( )
( ) ( )
2 2
1 1
2 2 2
1, var 2
, var 2 dev 2
v v v
_ _
_ v _ v _ v
= =
= = =
Chi-Squared Parameter
As seen above,
2
is a continuous probability distribution. However, there is also a goodness-of-fit test
which computes a parameter also called chi-squared. This parameter is from a distribution that is often
close to a
2
distribution, but be careful to distinguish between the parameter
2
and the distribution
2
.
The chi-squared parameter is not required to be from a chi-squared distribution, though it often is. All
the chi-squared parameter really requires is that the variances of our errors add, which is to say that our
errors are uncorrelated (not necessarily independent).
The 2 parameter is valid for any distribution of uncorrelated errors.
The 2 parameter has a 2 distribution only if the errors are gaussian.
However, for large , the
2
distribution approaches gaussian, as does the sum of many values of any
distribution. Therefore:
The 2 distribution is a reasonable approximation to the distribution of any 2 parameter with >~
20, even if the errors are not gaussian.
If we know the standard deviation of our measurement error, then the set of {error divided by } has
standard-deviation = 1:
2
dev( ) standard deviation of random variable ,
var( ) variance of random variable ,
dev 1 var 1
X
X
Define X X also written
X X also written
error error
o
o
o o
| | | |
= =
| |
\ . \ .
As a special case, but not required for a
2
parameter, if our errors are gaussian,
2
2
1
(0,1)
error error
gaussian _
o o
| |
e e
|
\ .
Often, the uncertainties vary from measurement to measurement. In that case, we are fitting a curve to
data triples: (x
i
, y
i
,
i
). Still, the error divided by uncertainty for any single measurement is unit deviation:
dev 1, var 1,
i i
i i
error error
and for all i
o o
| | | |
= =
| |
\ . \ .
If we have n measurements, with uncorrelated errors, then because variances add:
2
2
1 1
var . :
n n
i i
n
i i
i i
error error
n For gaussian errors _
o o
= =
| |
| |
= e |
|
|
\ .
\ .
Returning to our ideal process from above, with a curve mapping an independent variable x to a
dependent variable y, we now take a set of measurements with known errors
i
:
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x
y(x)
Then our parameter
2
is defined as
2 2
2 2 2
1 1
,
n n
i i i
n
i i
i i
error measurement curve
If gaussian errors _ _ _
o o
= =
| | | |
(
= e
| |
\ . \ .
If n is large, this sum will be close to the average, and (for zero-mean errors),
2
2
1 1
n n
i i
i i
i i
error error
Var n _
o o
= =
| |
| |
= = = |
|
|
\ .
\ .
Now suppose we have fit a curve to our data, i.e., we guessed a form, and found the parameters which
minimize the
2
parameter. If our fit is good, then our curve is close to the real dependence curve for y as
a function of x, and our errors will be purely random (no systematic error). We now compute the
2
parameter for our fit, as if our fit-curve were the ideal curve above:
2 2
2
1 1
n n
i i i
i i
i i
error measurement fit
_
o o
= =
| | | |
=
| |
\ . \ .
If our fit is good, the number
2
will likely be close to n. (We will soon modify the distribution of the
2
parameter, but for now, it illustrates our principle.)
If our fit is bad, there will be significant systematic fit error in addition to our random error, and our
2
parameter will be much larger than n. Summarizing,
If 2 is close to n, then our errors are no worse than our measurement
errors, and the fit is good. If 2 is much larger than n, then our
errors are worse than our measurement errors, so our fit must be bad.
Degrees of freedom: So far we have ignored something called degrees of freedom. Consider again
the hypothetical fit to a straight line. We are free to choose our parameters m and b to define our fit-line.
But in a set of n data points, we could (if we wanted) choose our m and b to exactly go through two of the
data points:
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x
y(x)
This guarantees that two of our fit errors are zero. If n is large, it wont significantly affect the other
2
being the sum of n squared-errors, it is approximately the sum of (n 2) squared-
errors, because our fit procedure guarantees that 2 of the errors are zero. In this case,
( )
2
var 2 n _ ~ . A
rigorous analysis shows that for the best fit line (which probably doesnt go through any of the data points),
and gaussian measurement errors, then
( )
2
var 2 n _ = , exactly. This concept generalizes quite far:
- even if we dont fit 2 points exactly to the line,
- even if our fit-curve is not a line,
- even if we have more than 2 fit parameters,
the effect is to reduce the
2
parameter to be a sum of less than n squared-errors. The effective number
of squared-errors in the sum is called the degrees of freedom (dof):
( ) # dof n fit parameters
Thus the statistics of our
2
parameter are really
( ) ( )
2 2
, dev 2 dof dof _ _ = =
Reduced Chi-Squared Parameter
Since it is awkward for everyone to know n, the number of points in our fit, we simply divide our chi-
squared parameter by dof, to get the reduced chi-squared parameter. Then it has these statistics:
( )
( ) ( )
2
2
2
1
2 2
2 2
1
dev
2
2
1, dev
n
i i
i
i
measurement fit
reduced
dof dof
dof
dof
reduced reduced
dof dof dof dof dof
_
_
o
_ _
_ _
=
| |
=
|
\ .
= = = = = =
If reduced
2
is close to 1, the fit is good. If reduced
2
is much larger than 1, the fit is bad. By
much larger we mean several deviations away from 1, and the deviation gets smaller with larger dof
(larger n).
Of course, our confidence in
2
or reduced-
2
depends on how many data points went into computing
it, and our confidence in our measurement errors,
i
. Remarkably, one reference on
2
[which I dont
remember] says that our estimates of measurement errors,
i
, should come from a sample of at least five!
That seems to me to be a very small number to have much confidence in .
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Fitting To Histograms
Data analysis often requires fitting a function to binned data, that is, fitting a predicted probability
distribution to a histogram of measured values. While such fitting is very commonly done, it is much less
commonly understood. There are important subtleties often overlooked. This section assumes you are
familiar with the binomial distribution, the
2
goodness of fit parameter (described earlier), and some
basic statistics.
The general method for fitting a model to a histogram of data is this:
- Start with n data points (measurements), and a parameterized model for the PDF of those data.
- Bin the data into a histogram.
- Find the model parameters which best fit the data histogram
For example, a gaussian distribution is a 2-parameter model; the parameters are the average, , and
standard deviation, . If we believe our data should follow a gaussian distribution, and we want to know
the and of that distribution, we might bin the data into a histogram, and fit the gaussian PDF to it:
model PDF
measurement
x
x
i
predicted bin count, model
i
fit error
measured bin count, c
i
Sample histogram with a 2-parameter model PDF ( and ). The fit model is gaussian in this
example, but could be any pdf with any parameters.
We must define best fit. Usually, we use the
2
(chi-squared) goodness of fit parameter as the
figure of merit (FOM). The smaller
2
, the better the fit. Fitting to a histogram is a special case of general
2
fitting. Therefore, we need to know two things for each bin: (1) the predicted (model) count, and (2) the
uncertainty in the measured count. We find these things in the next section.
(This gaussian fit is a simplified example. In reality, if we think the distribution is gaussian, we would
compute the sample average and standard deviation directly, using the standard formulas. More on this
later. In general, the model is more complicated, and there is no simple formula to compute the parameters.
For now, we use this as an example because it is a familiar model to many.)
Chi-squared For Histograms
We now develop the
2
figure of merit for fitting to a histogram. A sample is a set of n measurements
(data points). In principle, we could take many samples of data. For each sample, there is one histogram,
i.e., there is an infinite population of samples, each with its own histogram. But we have only one sample.
The question is, how well does our one histogram represent the population of samples, and therefore, the
population of data measurements.
To develop the
2
figure of merit for the fit, we must understand the statistics of a single histogram bin,
from the population of all histograms that we might have produced from different samples. The key point
is this: given a sample of n data points, and a particular histogram bin numbered i, each data point in the
sample is either in the bin (with probability p
i
), or its not (with probability (1 - p
i
) ). Therefore, the count
in the i
th
histogram bin is binomially distributed, with some probability p
i
, and n trials. (See standard
references on the binomial distribution if this is not clear.) Furthermore, this is true of every histogram bin:
The number of counts in each histogram bin is a binomial random variable.
Each bin has its own probability, pi, but all bins share the same number of trials, n.
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Recall that a binomial distribution is a discrete distribution, i.e. it gives the probability of finding
values of a whole-number random variable; in this case, it gives the probability for finding a given number
of counts in a given histogram bin. The binomial distribution has two parameters:
p is the probability of a given data point being in the bin
n is the number of data points in the sample, and therefore the number of trials in the
binomial distribution.
Recall that the binomial distribution has average, c-bar, and variance,
2
given by:
2
, (1 ) (binomial distribution) c np np p o = =
For a large number of histogram bins, N
bins
, the probability of being in a given bin is of order p ~
1/N
bins
, which is small. Therefore, we approximate
2
(1) ( 1 1)
bins
np c N p o ~ = >> <<
We find c-bar for a bin from the pdf model: typically, we assume the bins are narrow, and the
probability of being in a bin is just
( ) Pr being in bin pdf ( )
i X i i
i p x x ~ A
Then the model average (expected) count is Pr(being in bin) times the number of data points, n:
2
pdf ( ) (narrow bins)
bin center, bin width
pdf ( ) model pdf at bin center
1 1
For example, for a gaussian histogram: pdf ( , ; ) exp
2 2
i X i i
i i
X i
X
model n x x
where x x
x
x
x
o
o o t
~ A
A
| |
| |
= |
|
|
\ .
\ .
However, one can use any more sophisticated method to properly integrate the pdf to find e for each
bin.
We now know the two things we need for evaluating a general
2
goodness-of-fit parameter: for each
histogram bin, we know (1) the model average count, model
i
, and (2) the variance of the measured count,
which is also approximately model
i
. We now compute
2
for the model pdf (given a set of model
parameters) in the usual way:
( )
2
2
1
the measured count in the bin
the model average count in the bin
bins
N
i i th
i
i
i
th
i
c model
where c i
model
model i
_
=
If your model predicts a count of zero (model
i
= 0 for some i), then
2
below.
Reducing the Effect of Noise
To find the best-fit parameters, we take our given sample histogram, and try different values of the
pdf(x) parameters (in this example, and ) to find the combination which produces the minimum
2
.
Notice that the low count bins carry more weight than the higher count bins:
2
weights the terms by
1/model
i
. This reveals the first common misunderstanding:
A fit to a histogram is driven by the tails, not by the central peak. This is usually bad.
Tails are often the worst part of the model (theory), and often the most contaminated (percentage-wise)
by noise: background levels, crosstalk, etc. Three methods help reduce these problems:
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- limiting the weight of low-count bins
- truncating the histogram
- rebinning
Limiting the weight: The tails of the model distribution are often less than 1, and approach zero.
This gives them extremely high weights compared to other bins. Since the model is probably inadequate at
these low bin counts (due to noise, etc.), one can limit the denominator in the
2
sum to at least 1; this also
avoids division-by-zero:
( )
2
2
1
1
1
bins
N
i i
i i
i
i
i
model if model
c model
where d
d otherwise
_
=
>
This is an ad-hoc approach, and the minimum weight can be anything; it doesnt have to be 1. Notice,
though, that this modified
2
value is still a monotonic function of the model parameters, which is critical
for stable parameter fits (it avoids local minima, see Practical Considerations below).
Truncating the histogram: Alternatively, we can truncate the histogram on the left and right sides to
those bins with a reasonable number of counts, substantially above the noise (below left). [Bev p110]
recommends a minim bin count of 10, based on a desire for gaussian errors. I dont think that matters
much. In truth, the minimum count completely depends on the noise level.
model PDF
x
model PDF
x
s
x
f
x
1
x
7
x
2
x
3
x
4
x
5
x
6
1.2
3.9
10.8
3 8
3
Avoiding noisy tails by (left) truncating the histogram, or (right) rebinning.
Truncation requires renormalizing: we normalize the model within the truncated limits to the data
count within those same limits:
pdf ( ) , are the start and final bins to include
pdf ( )
f f f
i norm X i i i
i s i s i s
f
i
i s
norm f
X i i
i s
model n x x c where s f
c
n
x x
= = =
=
=
= A =
=
A
You might think that we should use the model, not the data histogram, to choose our truncation limits.
After all, why should we let sampling noise affect our choice of bins? This approach fails miserably,
however, because our bin choices change as we vary our parameters in the hunt for the optimum
2
.
Changing which bins are included in the FOM causes unphysical steps in
2
as we vary our parameters,
making many local minima. This makes the fit unstable, and generally unusable. For stability: truncate
your histogram based on the data, and keep it fixed during the parameter search.
Rebinning: Alternatively, bins dont have to be of uniform width [Bev p175], so combining adjacent
bins into a single, wider bin with higher count can help improve signal-to-noise ratio (SNR) in that bin
(above right). Note that when rebinning, we evaluate the theoretical count as the sum of the original
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(narrow) bin theoretical counts. In the example of the diagram above right, the theoretical and measured
counts for the new (wider) bin 1 are
1 1
1.2 3.9 10.8 15.9 3 3 8 14 a c mo nd del = + + = = + + =
Other Histogram Fit Considerations
Slightly correlated bin counts: Bin counts are binomially distributed (a measurement is either in a
bin, or its not). However, there is a small negative correlation between any two bins, because the fact that
a measurement lies in one bin means it doesnt lie in any other bin. Recall that the
2
parameter relies on
uncorrelated errors between bins, so a histogram slightly violates that assumption. With a moderate
number of bins (> ~15 ??), this is usually negligible.
Overestimating the low count model: If there are a lot of low-count bins in your histogram, you may
find that the fit tends to overestimate the low-count bins, and underestimate the high-count bins (diagram
below). When properly normalized, the sum of overestimates and underestimates must be zero: the sum of
bin counts equals the sum of the model predicted counts.
model PDF
x
underestimated
overestimated
2
is artificially reduced by overestimating low-count bins, and underestimating high-count
bins.
But since low-count bins weigh more than high-count bins, and since an overestimated model reduces
2
(the model value model
i
appears in the denominator of each
2
term), the overall
2
is reduced if low-
count bins are overestimated, and high-count bins are underestimated.
This effect can only happen if your model has the freedom to bend in the way necessary: i.e., it can
be a little high in the low-count regions, and simultaneously a little low in the high-count regions. Most
realistic models have this freedom. If the model is reasonably good, this effect can cause reduced-
2
to be
consistently less than 1 (which should be impossible).
I dont know of a simple fix for this. It helps to limit the weight of low-count bins to (say) 1, as
described above. However once again, the best approach is to minimize the number of low-count bins in
Noise not zero mean: for counting experiments, such as those that fill in histograms with data, all bin
counts are zero or positive. Any noise will add positive counts, and therefore noise cannot be zero-mean.
If you know the pdf of the noise, then you can put it in the model, and everything should work out fine.
However, if you have a lot of un-modeled noise, you should see that your reduced-
2
is significantly greater
than 1, indicating a poor fit. Some people have tried playing with the denominator in the
2
sum to try to
get more accurate fit parameters in the presence of noise, but there is little theoretical justification for
this, and it usually amounts to ad-hoc tweaking to get the answers you want.
Non-
2
figure of merit: One does not have to use
2
as the fit figure of merit. If the model is not very
good, or if there are problems as mentioned above, other FOMs might work better. The most common
alternative is probably least-squares, which means minimizing the sum-squared-error:
( )
2
1
(sum-squared-error)
bins
N
i i
i
SSE c model
=
This is like
2
where the denominator in each term in the sum is always 1.
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Guidance Counselor: Practical Considerations for Computer Code to Fit
Data
Computer code for finding the best-fit parameters is usually divided into two pieces, one piece you
buy, and one piece you have to write yourself:
- You buy a generic optimization algorithm, which varies parameters without knowledge of what
they mean, looking for the minimum figure-of-merit (FOM). For each trial set of parameters, it
calls your FOM function to compute the FOM as a function of the current trial parameters.
- You write the FOM function which computes the FOM as a function of the given parameters.
Generic optimizers usually minimize the figure-of-merit, consistent with the FOM being a cost or
error that we want reduced. If instead, you want to maximize a FOM, return its negative to the
minimizer.
Generic optimization algorithms are available off-the-shelf, e.g. [Numerical Recipes]. However, they
are sometimes simplistic, and in the real world, often fail with arithmetic faults (overflow, underflow,
domain-error, etc). The fault (no pun intended) lies not in their algorithm, but in their failure to tell you
what you need to do to avoid such failures:
Your job is to write a bullet-proof figure-of-merit function.
This is harder than it sounds, but quite do-able with proper care. A bullet-proof FOM function requires
only two things:
- Proper validation of all parameters.
- A properly bad FOM for invalid parameters (a guiding error).
Guiding errors are similar to penalty functions, but they operate outside the valid parameter space,
rather than inside it.
A simple example: Suppose you wish to numerically find the minimum of the figure-of-merit function
below left. Suppose the physics is such that only p > 1 is sensible.
p
1
2
3
1 2 3
1
( ) f p p
p
= +
4
p
f(p)
1
2
3
1 2 3 4
p
f(p)
1
2
3
1 2 3 4
valid p
(Left and middle) Bad figure-of-merit (FOM) functions. (Right) A bullet-proof FOM.
Your optimization-search algorithm will try various values of p, evaluating f(p) at each step, looking
for the minimum. You might write your FOM function like this:
fom(p) = 1./p + sqrt(p)
But the search function knows nothing of p, or which values of p are valid. It may well try p = 1.
Then your function crashes with a domain-error in the sqrt( ) function. You fix it with (above middle):
float fom(p)
if(p < 0.) return 4.
return 1./p + sqrt(p)
Since you know 4 is much greater than the true minimum, you hope this will fix the problem. You run
the code again, and now it crashes with divide-by-zero error, because the optimizer tried p = 0. Easy fix:
float fom(p)
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if(p <= 0.) return 4.
return 1./p + sqrt(p)
Now the optimizer crashes with an overflow error, p < (max_float). The big flat region to the left
confuses the optimizer. It searches negatively for a value of p that makes the FOM increase, but it never
finds one, and gets an overflow trying. Your flat value for p 0 is no good. It needs to grow upward to the
left to provide guidance to the optimizer:
float fom(p)
if(p <= 0.) return 4. + fabs(p - 1) // fabs() = absolute value
return 1./p + sqrt(p)
Now the optimizer says the minimum is 4 at p = 10
6
. It found the local minimum just to the left of
zero. Your function is still ill-behaved. Since only p > 1 is sensible, you make yet another fix (above
right):
float fom(p)
if(p <= 1.) return 4.
return 1./p + sqrt(p)
Finally, the optimizer returns the minimum FOM of 1.89 at p = 1.59. After 5 tries, you have made
A bullet-proof FOM has only one minimum, which it monotonically approaches from both sides,
even with invalid parameters, and it never crashes on any parameter set.
In this example, the FOM is naturally bullet-proof from the right. However, if it werent, the absolute
value of (p 1) on the error return value provides a V-shape which guides the optimizer into the valid
range from either side. Such guiding errors are analogous to so-called penalty functions, but better,
because they take effect only for invalid parameter choices, thus leaving the valid parameter space
completely free for full optimization.
Multi-parameter FOMs: Most fit models use several parameters, p
i
, and the optimizer searches over
all of them iteratively to find a minimum. Your FOM function must be bullet-proof over all parameters: it
must check each parameter for validity, and must return a large (guaranteed unoptimal) result for invalid
inputs. It must also slope the function toward valid values, i.e. provide a restoring force to the invalid
parameters toward the valid region. Typically, with multiple parameters p
i
, one uses
1
_ _ #
N
i i i i
i
guiding bad FOM big p valid where valid a valid value for p
=
= +
This guides the minimization search when any parameter is outside its valid range.
p
g(p)
1 2 3 4
guiding
error
guiding
error
valid p
Guiding errors lead naturally to a valid solution, and are better than traditional penalty
functions.
A final note:
The big # for invalid parameters may need to be much bigger than you think.
In my thesis research, I used reduced
2
as my FOM, and the true minimum FOM is near 1. I started
with 1,000,000 as my big #, but it wasnt big enough! I was fitting to histograms with nearly a thousand
counts in several bins. When the trial model bin count was small, the error was about 1,000, and the sum-
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squared-error over several bins was > 1,000,000. This caused the optimizer to settle on an invalid set of
parameter values as the minimum! I had to raise big # to 10
9
.
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Numerical Analysis
Round-Off Error, And How to Reduce It
Floating point numbers are stored in a form of scientific notation, with a mantissa and exponent. E.g.,
1.23 10
45
has mantissa m = 1.23 and exponent e = 45
Computer floating point stores only a finite number of digits. float (aka single-precision) stores at
least 6 digits; double stores at least 15 digits. Well work out some examples in 6-digit decimal scientific
notation; actual floating point numbers are stored in a binary form, but the concepts remain the same. (See
IEEE Floating Point in this document.)
Precision loss due to summation: Adding floating point number with different exponents results in
round-off error:
1.234 56 10
2
1.234 56 10
2
+ 6.111 11 10
0
+ 0.061 111 1 10
2
= 1.295 67 10
2
where 0.000 001 1 of the result is lost,
because the computer can only stored 6 digits. (Similar round-off error occurs if the exponent of the
result is bigger than both of the addend exponents.) When adding many numbers of similar magnitude (as
is common in statistical calculations), the round-off error can be quite significant:
float sum = 1.23456789; // Demonstrate precision loss in sums
printf("%.9f\n", sum); // show # significant digits
for(i = 2; i < 10000; i++)
sum += 1.23456789;
printf("Sum of 10,000 = %.9f\n", sum);
1.234567881 8 significant digits
Sum of 10,000 = 12343.28 only 4 significant digits
You lose about 1 digit of accuracy for each power
of 10 in n, the number of terms summed.
I.e.
10
- log digit loss n ~
When summing numbers of different magnitudes, you get a better answer by adding the small numbers
first, and the larger ones later. This minimizes the round-off error on each addition.
E.g., consider summing 1/n for 1,000,000 integers. We do it in both single- and double-precision, so
you can see the error:
float sum = 0.;
double dsum = 0.;
// sum the inverses of the first 1 million integers, in order
for(i = 1; i <= 1000000; i++)
sum += 1./i, dsum += 1./i;
printf("sum: %f\ndsum: %f. Relative error = %.2f %%\n",
sum, dsum, (dsum-sum)/dsum);
sum: 14.357358
dsum: 14.392727. Relative error = 0.002457
This was summed in the worst possible order: largest to smallest, and (in single-precision) we lose
about 5 digits of accuracy, leaving only 3 digits. Now sum in reverse (smallest to largest):
float sumb = 0.;
double dsumb = 0.;
for(i = 1000000; i >= 1; i--)
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sumb += 1./i, dsumb += 1./i;
printf(" sumb: %f\ndsumb: %f. Relative error = %.6f\n",
sumb, dsumb, (dsumb-sumb)/dsumb);
sumb: 14.392652
dsumb: 14.392727. Relative error = 0.000005
The single-precision sum is now good to 5 digits, losing only 1 or 2.
[In my research, I needed to fit a polynomial to 6000 data points, which involves many sums of 6000 terms,
and then solving linear equations. I needed 13 digits of accuracy, which easily fits in double-precision (double,
15-17 decimal digits). However, the precision loss due to summing was over 3 digits, and my results failed.
Simply changing the sums to long double, then converting the sums back to double, and doing all other
calculations in double solved the problem. The dominant loss was in the sums, not in solving the equations.]
Summing from smallest to largest is very important for evaluating polynomials,
which are widely used for transcendental functions. Suppose we have a 5
th
order polynomial, f(t):
2 3 4 5
0 1 2 3 4 5
( ) f t a a x a x a x a x a x = + + + + +
which might suggest a computer implementation as :
f = a0 + a1*t + a2*t*t + a3*t*t*t + a4*t*t*t*t + a5*t*t*t*t*t
Typically, the terms get progressively smaller with higher order. Then the above sequence is in the
worst order: biggest to smallest. (It also takes 15 multiplies.) It is more accurate (and faster) to evaluate
the polynomial as:
f = ((((a5*t + a4)*t + a3)*t + a2)*t + a1)*t + a0
This form adds small terms of comparable size first, progressing to larger ones, and requires only 5
multiplies.
How To Extend Precision In Sums Without Using Higher Precision Variables
(Handy for statistical calculations): You can avoid round-off error in sums without using higher
precision variables with a simple trick. For example, lets sum an array of n numbers:
sum = 0.;
for(i = 0; i < n; i++) sum += a[i];
This suffers from precision loss, as described above. The trick is to actually measure the round-off
error of each addition, and save that error for the next iteration:
sum = 0.;
error = 0.; // the carry-in from the last add
for(i = 0; i < n; i++)
{ newsum = sum + (a[i] + error); // include the lost part of prev add
diff = newsum - sum; // what was really added
error = (a[i] + error) - diff; // the round-off error
sum = newsum;
}
The error variable is always small compared to the sum, because it is the round-off error. Keeping
track of it effectively doubles the number of accurate digits in the sum, until it is lost in the final addition.
Even then, error still tells you how far off your sum is. For all practical purposes, this eliminates any
precision loss due to sums. Lets try summing the inverses of integers again, in the bad order, but with
this trick:
float newsum, diff, sum = 0., error = 0.;
for(i = 1; i <= 1000000; i++)
{ newsum = sum + (1./i + error);
diff = newsum - sum; // what was really added
error = (1./i + error) - diff; // the round-off error
sum = newsum;
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}
printf(" sum: %f\ndsumb: %f. Relative error = %.6f, error = %g\n",
sum, dsumb, (dsumb-sum)/dsumb, error);
sum: 14.392727
dsumb: 14.392727. Relative error = -0.000000, error = -1.75335e-07
As claimed, the sum is essentially perfect.
Numerical Integration
The above method of sums is extremely valuable in numerical integration. Typically, for accurate
numerical integration, one must carefully choose an integration step size: the increment by which you
change the variable of integration. E.g., in time-step integration, it is the time step-size. If you make the
step size too big, accuracy suffers because the rectangles (or other approximations) under the curve dont
follow the curve well. If you make the step size too small, accuracy suffers because youre adding tiny
increments to large numbers, and the round-off error is large. You must thread the needle of step-size,
getting it just right for best accuracy. This fact is independent of the integration interpolation method:
By virtually eliminating round-off error in the sums (using the method above), you eliminate the
lower-bound on step size. You can then choose a small step-size, and be confident your answer is right. It
might take more computer time, but integrating 5 times slower and getting the right answer is vastly better
than integrating 5 times faster and getting the wrong answer.
Sequences of Real Numbers
Suppose we want to generate the sequence 2.01, 2.02, ... 2.99, 3.00. A simple approach is this:
real s;
for(s = 2.01; s <= 3.; s += 0.01) ...
The problem with this is round-off error: 0.01 is inexact in binary (has round-off error). This error
accumulates 100 times in the above loop, making the last value 100 times more wrong than the first. In
fact, the loop might run 101 times instead of 100. The fix is to use integers where possible, because they
are exact:
real s;
int i;
for(i = 201; i <= 300; i++) s = i/100.;
When the increment is itself a variable, note that multiplying a real by an integer incurs only a single
round-off error:
real s, base, incr;
int i;
for(i = 1; i <= max; i++) s = base + i*incr;
Hence, every number in the sequence has only one round-off error.
Root Finding
In general, a root of a function f(x) is a value of x for which f(x) = 0. It is often not possible to find the
roots analytically, and it must be done numerically. [TBS: binary search]
Simple Iteration Equation
Some forms of f( ) make root finding easy and fast; if you can rewrite the equation in this form:
( ) 0 ( ) f x x g x = =
then you may be able to iterate, using each value of g( ) as the new estimate of the root, r.
This is the simplest method of root finding, and generally the slowest to converge.
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It may be suitable if you have only a few thousand solutions to compute, but may be too slow for
millions of calculations.
You start with a guess that is close to the root, call it r
0
. Then
1 0 2 1 1
( ), ( ), ... ( )
n n
r g r r g r r g r
+
= = =
If g( ) has the right property (specifically, |g(x)| < 1 near the root) this sequence will converge to the
solution. We describe this necessary property through some examples. Suppose we wish to solve
/ 2 0 x x = numerically. First, we re-arrange it to isolate x on the left side:
2
x
x = (below left).
x
x/2
1
0.5
x
4x
2
1
0.5
1 1
y=x y=x
Two iteration equations for the same problem. The left converges; the right fails.
From the graph, we might guess r
0
0.2. Then we would find,
1 2 1 3 4 5
6 7
0.2 / 2 0.2236, / 2 0.2364, 0.2431, 0.2465, 0.2483,
0.2491, 0.2496
r r r r r r
R r
= = = = = = =
= =
We see that the iterations approach the exact answer of 0.25. But we could have re-arranged the
equation differently:
2
2 , 4 x x x x = = (above right). Starting with the same guess x = 0.2, we get this
sequence:
1 2 1 3 4
0.2 / 2 0.16, / 2 0.1024, 0.0419, 0..0070 r r r r r = = = = = =
But they are not converging on the nearby root; the sequence diverges away from it. So whats the
difference? Look at a graph of whats happening, magnified around the equality:
x
x/2
0.25
4x
2
y=x
0.25
0.2
r
0
r
1
r
2
x
0.25
y=x
0.25
0.2
r
0
r
1
r
2
When the curve is flatter than y = x (above left), then trial roots that are too small get bigger, and trial
roots that are too big get smaller. So iteration approaches the root. When the curve is steeper than y = x
(above right), trial roots that are too small get even smaller, too big get even bigger; the opposite of what
we want. So for positive slope curves, the condition for convergence is
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0
1, is the exact root
i
y
in the region r r r r where r
r
A
< <
A
Consider another case, where the curve has negative slope. Suppose we wish to solve
1
cos 0 x x
= ,
(x in radians). We re-write it as
1
cos x x
= . On the other hand, we could take the cosine of both sides and
get an equivalent equation: cos x x = . Which will converge? Again look at the graphs:
x
cos x
0.739
cos
-1
x
y=x
0.739
r
0
r
1
r
2
x
y=x
r
0
r
1
r
2
r
3
0.739
0.739
r
3
So long as the magnitude of the slope < 1, the iterations converge. When the magnitude of the slope >
1, they diverge. We can now generalize to all curves of any slope:
The general condition for convergence is
0
1, is the exact root
i
y
in the region r r r r where r
r
A
< <
A
The flatter the curve, the faster the convergence.
Given this, we could have easily predicted that the converging form of our iteration equation is
cos x x = , because the slope of cos x is always < 1, and cos
1
x is always > 1. Note, however, that if the
derivative is > 1/2, then the binary search will be faster than iteration.
Newton-Raphson Iteration
The above method of variable iteration is kind of blind, in that it doesnt use any property of the
given functions to advantage. Newton-Raphson iteration is a method of finding roots that uses the
derivative of the given function to provide more reliable and faster convergence. Newton-Raphson uses the
original form of the equation: ( ) / 2 0 f x x x = = . The idea is to use the derivative of the function to
approximate its slope to the root (below left). We start with the same guess, r
0
= 0.2.
x
x/2 x
0.25
0.0
x
4x
2
x
0
tangent
0.1 0.2
f
0.25 0.1 0.2
tangent
f
x
x
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0 0
1/ 2 1/ 2 3/ 2
1/ 2
1/ 2 1/ 2 1/ 2
( )
'( ) (Note '( ) 0)
'( )
/ 2 4 2 4 1
'( ) 1
4 / 4 1 4 1 4
i
i
i i i i i
i i i
f r f
f r r f r
x f r
r r r r r
f x x r
r r r
A
~ A ~ <
A
= A = =
Heres a sample computer program fragment, and its output:
// Newton-Raphson iteration
r = 0.2;
for(i = 1; i < 10; i++)
{ r -= (2.*r - 4.*r*sqrt(r)) / (1. - 4.*sqrt(r));
printf("r%d %.16f\n", i, r);
}
r1 0.2535322165454392
r2 0.2500122171752588
r3 0.2500000001492484
r4 0.2500000000000000
In 4 iterations, we get essentially the exact answer, to double precision accuracy of 16 digits. This is
much faster than the variable isolation method above. In fact, it illustrates a property of some iterative
Quadratic convergence is when the fractional error (aka relative error) gets squared on each
iteration, which doubles the number of significant digits on each iteration.
You can see this clearly above, where r
1
has 2 accurate digits, r
2
has 4, r
3
has 9, and r
4
has at least 16
(maybe more). Derivation of quadratic convergence??
Also, Newton-Raphson does not have the restriction on the slope of any function, as does variable
isolation. We can use it just as well on the reverse formula (previous diagram, right):
2
2
( ) 4
( ) 4 , '( ) 8 1,
'( ) 8 1
i
i
f r x x
f x x x f x x r
f r x
= = A = =
## , with these computer results:
r1 0.2666666666666667
r2 0.2509803921568627
r3 0.2500038147554742
r4 0.2500000000582077
r5 0.2500000000000000
This converges essentially just as fast, and clearly shows quadratic convergence.
If you are an old geek like me, you may remember the iterative method of finding square roots on an
old 4-function calculator: to find a: divide a by r, then average the result with r. Repeat as needed:
1
/
2
n n
n
a r r
r
+
+
=
You may now recognize that as Newton-Raphson iteration:
2
2
1
( ) 0, '( ) 2 ,
( ) 1
'( ) 2 2 2 2
n n
n n n n n n
n n n
f r r a f r r
r a r f r a a
r r r r r r r
f r r r r
+
= = =
| |
= + A = = = + = +
|
\ .
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If you are truly a geek, you tried the averaging method for cube roots:
2
1
/
2
n n
n
a r r
r
+
+
= . While you
found that it converged, it was very slow; cube-root(16) with r
0
= 2 gives only 2 digits after 10 iterations.
Now you know that the proper Newton-Raphson iteration for cube roots is:
3
3 2
1
2 2 2
1
( ) 0, '( ) 3 , 2
3 3 3 3
n n
n n n n
n n n
r a r a a
f r r a f r r r r r r
r r r
+
| |
= = = = = + = +
|
|
\ .
which gives a full 17 digits in 5 iterations for r
0
= 2, and shows (of course) quadratic convergence:
r1 2.6666666666666665
r2 2.5277777777777777
r3 2.5198669868999541
r4 2.5198421000355395
r5 2.5198420997897464
It is possible for Newton-Raphson to cycle endlessly, if the initial estimate of the root is too far off,
and the function has an inflection point between two successive iterations:
x
f(x)
0
tangent
tangent
Failure of Newton-Raphson iteration.
It is fairly easy to detect this failure in code, and pull in the root estimate before iterating again.
Pseudo-Random Numbers
We use the term random number to mean pseudo-random number, for brevity. Uniformly
distributed random numbers are equally likely to be anywhere in a range, typically (0, 1).
Uniformly distributed random numbers are the starting point for many other statistical
applications.
Computers can easily generate uniformly distributed random numbers, with the linear congruential
method described in [Numerical Recipes in C, 2nd ed., p284??] [New info 10/2010: 3
rd
ed. Describes better
LFSR-based generators.]. E.g., the best such generator (known at publication) is
// Uniform random value, 0 < v < 1, i.e. on (0,1) exclusive.
// Numerical Recipes in C, 2nd ed., p284
static uint32 seed=1; // starting point
vflt rand_uniform(void)
{
do seed = 1664525L*seed + 1013904223L; // period 2^32-1
while(seed == 0);
rand_calls++; // count calls for repetition check
return seed / 4294967296.;
} // rand_uniform()
Many algorithms which use such random numbers fail on 0 or 1, so this generator never returns
them.
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After a long simulation with a large number of calls, its a good idea to check rand_calls to be sure
its < ~400,000,000 = 10% period. This insures the numbers are essentially random, and not predictable.
Arbitrary distribution random numbers: To generate any distribution from a uniform random
number,
1
1
cdf ( ) is the random variable of the desired distribution
cdf inverse of the desired cumulative distribution function of
is a uniform random number on (0,1)
R
R
R U where R
R
U
=
=
To see why, recall that the cumulative distribution function gives the probability of a random
variable being less than or equal to its argument:
cdf ( ) Pr( ) pdf ( ) is a random variable
a
X X
a X a dx x where X
s =
}
x
pdf(x)
1
2
-0.5 0.5
x
cdf(x)
1
-0.5 0.5
u
1
0.5
-0.5
cdf
-1
(x)
Steps to generating the probability distribution function (pdf) on the left.
Also, the pdf of a function F = f(u) of a random variable is (see Probability and Statistics elsewhere in
this document):
pdf ( )
pdf ( ) , '( ) ( )
'( )
X
F
x
x where f x is the derivative of f x
f x
=
Then
1
1
1
1
1
cdf ( ). Using pdf ( ) 1on [0, 1]
pdf ( ) 1
pdf ( ) using ( ) ( ) ,
cdf ( )
( )
pdf ( ) as desired.
R U
U
Q
R
R
R
Let Q U u
u d d
r g u g u and u r
d
du du
d
u
cdf r
du
dr
r
=
| |
= = =
|
\ .
| |
|
\ .
=
?? Need a simple picture.
Generating Gaussian Random Numbers
The inverse cdf method is a problem for gaussian random numbers, because there is no closed-form
expression for the cdf of a gaussian:
2
/ 2
1
cdf( ) ( )
2
a
x
a dx e gaussian
t
=
}
But [Knu] describes a clever way based on polar coordinates to use two uniform random numbers to
generate a gaussian. He gives the details, but the final result is this:
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( ) ( )
( )
2ln cos is uniformon 0, 2
is uniformon 0,1
gaussian u where
u
u u t =
/* Gaussian random value, 0 mean, unit variance. From Knuth, "The Art of
Computer Programming, Vol. 2: Seminumerical Algorithms," 2nd Ed., p. 117.
It is exactly normal if rand_uniform() is uniform. */
PUBLIC double rand_gauss(void)
{ double theta = (2.*M_PI) * rand_uniform();
return sqrt( -2. * log(rand_uniform()) ) * cos(theta);
} // rand_gauss()
Generating Poisson Random Numbers
Poisson random numbers are integers; we say the Poisson distribution is discrete:
n
pdf(n)
0.25
0
u
0.50
0.75
1.00
1 2 3 4 5
n
cdf(n)
0.25
0
0.50
0.75
1.00
1 2 3 4 5
0
1
2
3
4
5
.25 .50 .75 1.00
cdf
-1
(u)
0
Example of generating the (discrete) Poisson distribution.
We can still use the inverse-cdf method to generate them, but in an iterative way. The code starts with
a helper function, poisson( ), that compute the probability of exactly n events in a Poisson distribution with
an average of avg events:
// ---------------------------------------------------------------------------
PUBLIC vflt poisson( // Pr(exactly n events in interval)
vflt avg, // average events in interval
int n) // n to compute Pr() of
{ vflt factorial;
int i;
if(n <= 20) factorial = fact[n];
else
{ factorial = fact[20];
for(i = 21; i <= n; i++) factorial *= i;
}
return exp(-avg) * pow(avg, n) / factorial;
} // poisson()
/*----------------------------------------------------------------------------
Generates a Poisson randum value (an integer), which must be <= 200.
Prefix 'irand_...' emphasizes the discreteness of the Poisson distribution.
----------------------------------------------------------------------------*/
PUBLIC int irand_poisson( // Poisson random integer <= 200
double avg) // avg # "events"
{
int i;
double cpr; // uniform probability
// Use inverse-cdf(uniform) for Poisson distribution, where
// inverse-cdf() consists of flat, discontinuous steps
cpr = rand_uniform();
for(i = 0; i <= 200; i++) // safety limit of 200
{ cpr -= poisson(avg, i);
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if(cpr <= 0) break;
}
return i; // 201 indicates an error
} // irand_poisson()
Other example random number generators: TBS.
Generating Weirder Random Numbers
Sometimes you need to generate more complex distributions, such as a combination of a gaussian with
a uniform background of noise. This is a raised gaussian:
pdf(x)
0
x
uniform pdf
gaussian pdf
Construction of a raised gaussian random variable from a uniform and a gaussian
Since this distribution has a uniform component, it is only meaningful if its limited to some finite
width. To generate distributions like this, you can compose two different distributions, and use the
principle:
The PDF of a random choice of two random variables is the weighted sum of the individual PDFs.
For example, the PDF for an RV (random variable) which is taken from X 20% of the time, and Y the
remaining 80% of the time is:
pdf( ) 0.2pdf ( ) 0.8pdf ( )
X Y
z z z = +
In this example, the two component distributions are uniform and gaussian. Suppose the uniform part
of the pdf has amplitude 0.1 over the interval (0, 2). Then it accounts for 0.2 of all the random values. The
remainder are gaussian, which we assume to be mean of 1.0, and = 1. Then the random value can be
generated from 3 more fundamental random values:
// Raised Gaussian random value: gaussian part: mean=1, sigma=1
// Uniform part (20% chance): interval (0, 2)
if(rand_uniform() <= 0.2)
random_variable = rand_uniform()*2.0;
else
random_variable = rand_gauss() + 1.0; // mean = 1, sigma = 1
Exact Polynomial Fits
Its sometimes handy to make an exact fit of a quadratic, cubic, or quartic polynomial to 3, 4, or 5 data
points, respectively.
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x
-1 0 1
3 points, 2
nd
order
x
-1 0 1 2
4 points, 3
rd
order
x
-2 -1 0 1 2
5 points, 4
th
order
The quadratic case illustrates the principle simply. We seek a quadratic function
2
2 1 0
( ) y x a x a x a = + +
which exactly fits 3 equally spaced points, at x = 1, x = 0, and x = 1, with value y
1
, y
0
, and y
1
,
respectively (shown above). So long as your actual data are equally spaced, you can simply scale and
offset to the x values 1, 0, and 1. We can directly solve for the coefficients a
2
, a
1
, and a
0
:
( ) ( )
2
2 1 0 1
2 1 0 1
2
2 1 0 0 0 0
2
2 1 0 1
2 1 0 1
2 1 1 0 1 1 1 0 0
( 1) ( 1)
(0) (0)
(1) (1)
/ 2 , / 2,
a a a y
a a a y
a a a y a y
a a a y
a a a y
a y y y a y y a y
+ + =
+ =
+ + = =
` `
+ + =
+ + = )
)
= + = =
Similar formulas for the 3
rd
and 4
th
order fits yield this code:
// ---------------------------------------------------------------------------
// fit3rd() computes 3rd order fit coefficients. 4 mult/div, 8 adds
PUBLIC void fit3rd(
double ym1, double y0, double y1, double y2)
{
a0 = y0;
a2 = (ym1 + y1)/2. - y0;
a3 = (2.*ym1 + y2 - 3.*y0)/6. - a2;
a1 = y1 - y0 - a2 - a3;
} // fit3rd()
// ---------------------------------------------------------------------------
// fit4th() computes 4th order fit coefficients. 6 mult/div, 13 add
PUBLIC void fit4th(
double ym2, double ym1, double y0, double y1, double y2)
{
b0 = y0;
b4 = (y2 + ym2 - 4*(ym1 + y1) + 6*y0)/24.;
b2 = (ym1 + y1)/2. - y0 - b4;
b3 = (y2 - ym2 - 2.*(y1 - ym1))/12.;
b1 = (y1 - ym1)/2. - b3;
} // fit4th()
TBS: Alternative 3
rd
order (4 point) symmetric fit, with x {-3, -1, 1, 3}.
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Twos Complement Arithmetic
Twos complement is a way of representing negative numbers in binary. It is universally used for
integers, and rarely used for floating point. This section assumes the reader is familiar with positive binary
numbers and simple binary arithmetic.
0110
Most Significant Bit
(MSB)
Least Significant Bit
(LSB)
2
3
2
2
2
1
2
0
Twos complement uses the most significant bit (MSB) of an integer as a sign bit: zero means the
number is > 0; 1 means the number is negative. Twos complement represents non-negative numbers as
ordinary binary, with the sign bit = 0. Negative numbers have the sign bit = 1, but are stored in a special
way: for a b-bit word, a negative number n (n < 0) is stored as if it were unsigned with a value of 2
b
+ n.
This is shown below, using a 4-bit word as a simple example:
bits unsigned signed
0000 0 0
0001 1 1
0010 2 2
sign 0011 3 3
bit (MSB) 0100 4 4
0101 5 5
0110 6 6
0111 7 7
1000 8 -8
1001 9 -7
1010 10 -6
1011 11 -5
1100 12 -4
1101 13 -3
1110 14 -2
1111 15 -1
With twos complement, a 4-bit word can store integers from 8 to +7. E.g., 1 is stored as 16 1 =
15. This rule is usually defined as follows (which completely obscures the purpose):
0, 0 Let n a n a = < > Example: n = 4, a = 4
complement it (change all 0s to 1s and 1s to 0s). 1011
Lets see how twos complement works in practice. There are 4 possible addition cases:
(1) Adding two positive numbers: so long as the result doesnt overflow, we simply add normally (in
binary).
(2) Adding two negative numbers: Recall that when adding unsigned integers, if we overflow our 4
bits, the carries out of the MSB are simply discarded. This means that the result of adding a + c is
actually (a + c) mod 16. Now, let n and m be negative numbers in twos complement, so their bit patterns
are 16 + n, and 16 + m. If we add their bit patterns as unsigned integers, we get
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( ) ( ) ( ) ( ) 16 16 32 mod16 16 , 0 n m n m n m n m ( + + + = + + = + + + <
which is the 2s complement representation of (n + m) < 0.
E.g., 2 1110 16 + (2)
+ 3 + 1101 + 16 + (3)
5 1011 16 + (5)
So with twos complement, adding negative numbers uses the same algorithm as adding unsigned
integers! Thats why we use twos complement.
(3) Adding a negative and a positive number, with positive result:
( ) ( ) 16 16 mod16 , 0 n a n a n a n a ( + + = + + = + + >
E.g., 2 1110 16 + (2)
+ 5 0101 + 5
3 0011 3
(4) Adding a negative and a positive number, with negative result:
( ) ( ) 16 16 , 0 n a n a n a + + = + + + <
E.g., 6 1010 16 + (6)
+ 3 0011 + 3
3 1101 16 + (3)
In all cases,
With twos complement arithmetic, adding signed integers uses the same algorithm as adding unsigned
integers! Thats why we use twos complement.
The computer hardware need not know which numbers are signed, and which are unsigned: it adds the
same way no matter what.
It works the same with subtraction: subtracting twos complement numbers is the same as subtracting
unsigned numbers. It even works multiplying to the same word size:
( ) ( )
( ) ( ) ( )
: 16 16 mod16 16 , 0, 0, 0
: 16 16 256 16 mod16 , 0, 0, 0
n a a na na n a na
n m n m nm nm n m nm
( + + = + = + < > <
( + + = + + + = < < >
In reality, word sizes are usually 32 (or maybe 16) bits. Then in general, we store b-bit negative
numbers (n < 0) as 2
b
+ n. E.g., for 16 bits, (n < 0) 65536 + n.
How Many Digits Do I Get, 6 or 9?
How many decimal digits of accuracy do I get with a binary floating point number? You often see a
range: 6 to 9 digits. Huh? We jump ahead, and assume here that you understand binary floating point (see
below for explanation).
Wobble, but dont fall down: The idea of number of digits of accuracy is somewhat flawed. Six
digits of accuracy near 100,000 is ~10 times worse than 6 digits of accuracy near 999,999. The smallest
increment is 1 in the least-significant digit. One in 100,000 is accuracy of 10
-5
; 1 in 999,999 is almost 10
-6
,
or 10 times more accurate.
Aside: The wobble of a floating point number is the ratio of the lowest accuracy to the highest accuracy for a
fixed number of digits. It is always equal to the base in which the floating point number is expressed, which is 10
in this example. The wobble of binary floating point is 2. The wobble of hexadecimal floating point (mostly
obsolete now) is 16.
We assume IEEE-754 compliant numbers (see later section). To insure, say, 6 decimal digits of
accuracy, the worst-case binary accuracy must exceed the best-case decimal accuracy. For IEEE single-
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precision, there are 23 fraction bits (and one implied-1 bit), so the worst case accuracy is 2
-23
= 1.2 10
-7
.
The best 6-digit accuracy is 10
-6
; the best 7 digit accuracy is 10
-7
. Thus we see that single-precision
guarantees 6 decimal digits, but almost gets 7, i.e. most of the time, it actually achieves 7 digits. The table
in the next section summarizes 4 common floating point formats.
How many digits do I need?
Often, we need to convert a binary number to decimal, write it to a file, and then read it back in,
converting it back to binary. An important question is, how many decimal digits do we need to write to
insure that we get back exactly the same binary floating point number we started with? In other words,
how many binary digits do I get with a given number of decimal digits? (This is essentially the reverse of
the preceding section.) We choose our number of decimal digits to insure full binary accuracy (assuming
our conversion software is good, which is not always the case).
Our worst-case decimal accuracy has to exceed our best-case binary accuracy. For single precision,
the best accuracy is 2
24
= 6.0 10
8
. The worst case accuracy of 9 decimal digits is 10
8
, so we need 9
decimal digits to fully represents IEEE single precision. Heres a table of precisions for 4 common
formats:
Format
Fract
ion bits
Minimum decimal
digits accuracy
Decimal digits for
exact replication
Decima
l digits
range
IEEE single 23 2
23
= 1.2 10
7
=> 6 2
24
= 6.0 10
8
=>
9
6 9
IEEE double 52 2
52
= 2.2 10
16
=>
15
2
53
= 1.1 10
16
=>
17
15 17
x86 long double 63 2
63
= 1.1 10
19
=>
18
2
64
= 5.4 10
20
=>
21
18 21
SPARC
REAL*16
112 2
112
= 1.9 10
34
=>
33
2
113
= 9.6 10
35
=> 36
33 36
These number of digits agree exactly with the quoted ranges in the IEEE Floating Point section, and
the ULP table in the underflow section. In C, then, to insure exact binary accuracy when writing, and then
reading, in decimal, for double precision, use
sprintf(dec, "%.17g", x);
How Far Can I Go?
A natural question is: What is the range, in decimal, of numbers that can be represented by the IEEE
formats? The answer is dominated by the number of bits in the binary exponent. This table shows it:
Range and Precision of Storage Formats
Format
Signific
ant Bits
Smallest Normal
Number
Largest
Number
Decim
al Digits
IEEE single 24 1.175... 10
38
3.402... 10
+38
6-9
IEEE double 53 2.225... 10
308
1.797...
10
+308
15-17
x86 long double
64 3.362... 10
4932
1.189...
10
+4932
18-21
SPARC REAL*16
113 3.362... 10
4932
1.189...
10
+4932
33-36
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Software Engineering
Software Engineering is much more than computer programming: it is the art and science of designing
and implementing programs efficiently, over the long term, across multiple developers. Software
engineering maximizes productivity and fun, and minimizes annoyance and roadblocks.
Engineers first design, then implement, systems that are useful, fun, and efficient.
Hackers just write code. Software engineering includes:
- Documentation: lots of it in the code as comments.
- Documentation: design documents that give an overview and conceptual view that is infeasible to
- Coding guidelines: for consistency among developers. Efficiency can only be achieved by
cooperation among the developers, including a consistent coding style that allows others to
quickly understand the code. E.g., physics.ucsd.edu/~emichels/Coding%20Guidelines.pdf.
- Clean code: it is easy to read and follow.
- Maintainable code: it functions in a straightforward and comprehensible way, so that it can be
changed easily and still work.
Notice that all of the above are subjective assessments. Thats the nature of all engineering:
Engineering is lots of tradeoffs, with subjective approximations of the costs and benefits.
Dont get me wrong: sometimes I hack out code. The judgment comes in knowing when to hack and
when to design.
Fun quotes:
Whenever possible, ignore the coding standards currently in use by thousands of developers in your
projects target language and environment.
- Roedy Green, How To Write Unmaintainable Code, www.strauss.za.com/sla/code_std.html
Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as
cleverly as possible, you are, by definition, not smart enough to debug it. - Brian W. Kernighan
Coding guidelines make everyones life easier, even yours. - Eric L. Michelsen
Object Oriented Programming
This is a much used and abused term, with no definitive definition. The goal of Object Oriented
Programming (OOP) is to allow reusable code that is clean and maintainable. The best definition Ive seen
of OOP is that it uses a language and approach with these properties:
- User defined data types, called classes, which allow (1) a single object (data entity) to have
multiple data items, and (2) provide user-defined methods (functions and operators) for
manipulating objects of that class.
- Information hiding: a class can define a public interface which hides the implementation details
from the code which uses the class.
- Overloading: the same named function or operator can be invoked on multiple data types,
including both built-in and user-defined types. The language chooses which of the same-named
functions to invoke based on the data types of its arguments.
- Inheritance: new data types can be created by extending existing data types. The derived class
inherits all the data and methods of the base class, but can add data, and override (overload) any
methods it chooses with its own, more specialized versions.
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- Polymorphism: this is more than overloading. Polymorphism allows derived-class objects to be
handled by (often older) code which only knows about the base class (i.e., which does not even
know of the existence of the derived class.) Even though the application code knows nothing of
the derived class, the data object itself insures calling proper specialized methods for itself.
In C++, polymorphism is implemented with virtual functions.
OOP does not have to be a new paradigm. It is usually more effective to make it an
improvement on the good software engineering practices you already use.
The Best of Times, the Worst of Times
We give here some ways to speed up common computations, using matrices as examples. The
principles are applicable to almost any computation performed over a large amount of data.
For the vast majority of programs, execution time is so short that it doesnt matter how efficient it
is; clarity and simplicity are more important than speed.
In rare cases, time is a concern. For some simple examples, we show how to easily cut your execution
times to 1/3 of original. We also show that things are not always so simple as they seem. This section
assumes knowledge of computer programming with simple classes (the beginning of object oriented
programming).
This topic is potentially huge, so we can only touch on some basics. The main point here is:
Computer memory management is the key to fast performance.
We proceed along these lines:
- We start with a simple C++ class for matrix addition. We give run times for this implementation
(the worst of times).
- A simple improvement greatly improves execution times (the best of times).
- We try another expected improvement, but things are not as expected.
- We describe the general operation of memory cache (pronounced cash) in simple terms.
- Moving on to matrix multiplication, we find that our previous tricks dont work well.
- However, due to the cache, adding more operations greatly improves the execution times.
The basic concept in improving matrix addition is to avoid C++s hidden copy operations. However:
Computer memory access is tricky, so things arent always what youd expect. Nonetheless, we
can be efficient, even without details of the computer hardware.
The tricks are due to computer hardware called RAM cache, whose general principles we describe
later, but whose details are beyond our scope.
First, here is a simple C++ class for matrix creation, destruction, and addition. (For simplicity, our
sample code has no error checking; real code, of course, does. In this case, we literally dont want reality
to interfere with science.) The class data for a matrix are the number of rows, the number of columns, and
a pointer to the matrix elements (data block).
typedef double T; // matrix elements are double precision
class ILmatrix // 2D matrix
{ public:
int nr, nc; // # rows & columns
T *db; // pointer to data
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ILmatrix(int r, int c); // create matrix of given size
ILmatrix(const ILmatrix &b); // copy constructor
~ILmatrix(); // destructor
T * operator [](int r) const {return db + r*nc;}; // subscripting
ILmatrix & operator =(const ILmatrix& b); // assignment
ILmatrix operator +(const ILmatrix& b) const; // matrix add
};
The matrix elements are indexed starting from 0, i.e. the top-left corner of matrix a is referenced as
a[0][0]. Following the data are the minimum set of methods (procedures) for matrix addition. Internally,
the pointer db points to the matrix elements (data block). The subscripting operator finds a linear array
element as (row)(#columns) + column. Here is the code to create, copy, and destroy matrices:
// create matrix of given size (constructor)
ILmatrix::ILmatrix(int r, int c) : nr(r), nc(c) // set nr & nc here
{
db = new T[nr*nc]; // allocate data block
} // ILmatrix(r, c)
// copy a matrix (copy constructor)
ILmatrix::ILmatrix(const ILmatrix & b)
{ int r,c;
nr = b.nr, nc = b.nc; // matrix dimensions
if(b.db)
{ db = new T[nr*nc]; // allocate data block
for(r = 0; r < nr; r++) // copy the data
for(c = 0; c < nc; c++)
(*this)[r][c] = b[r][c];
}
} // copy constructor
// destructor
ILmatrix::~ILmatrix()
{ if(db) {delete[] db;} // free existing data
nr = nc = 0, db = 0; // mark it empty
}
// assignment operator
ILmatrix & ILmatrix::operator =(const ILmatrix& b)
{ int r, c;
for(r = 0; r < nr; r++) // copy the data
for(c = 0; c < nc; c++)
(*this)[r][c] = b[r][c];
return *this;
} // operator =()
The good stuff: With the tedious preliminaries done, we now implement the simplest matrix addition
method. It adds two matrices element by element, and returns the result as a new matrix:
ILmatrix ILmatrix::operator +(const ILmatrix& b) const
{
int r, c;
ILmatrix result(nr, nc);
for (r=0; r < nr; r++)
for (c=0; c < nc; c++)
result[r][c] = (*this)[r][c] + b[r][c];
return result; // invokes copy constructor!
} // operator +()
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How long does this simple code take? To test it, we standardize on 300 300 and 400 x 400 matrix
sizes, each on two different computers: computer 1 is a c. 2000 Compaq Workstation W6000 with a 1.7
GHz Xeon. Computer 2 is a Gateway Solo 200 ARC laptop with a 2.4 GHz CPU. We time 100 matrix
int n = 300; // matrix dimension
ILmatrix a(n,n), b(n,n), d(n,n);
d = a + b; // prime memory caches
for(i = 0; i < 100; i++)
d = a + b;
With modern operating systems, you may have to run your code several times before the execution
times stabilize.
[This may be due to internal operations of allocating memory, and flushing data to disk.] We find that,
con computer 1, it takes ~1.36 0.10 s to execute 100 simple matrix additions (see table at end of this
section). Wow, that seems like a long time. Each addition is 90,000 floating point adds; 100 additions is 9
million operations. Our 2.4 GHz machine should execute 2.4 additions per ns. Wheres all the time going?
C++ has a major flaw. Though it was pretty easy to create our matrix class:
C++ copies your data twice in a simple class operation on two values.
So besides our actual matrix addition, C++ copies the result twice before it reaches the matrix d. The
first copy happens at the return result statement in our matrix addition function. Since the variable
result will be destroyed (go out of scope) when the function returns, C++ must copy it to a temporary
variable in the main program. Notice that the C++ language has no way to tell the addition function that
the result is headed for the matrix d. So the addition function has no choice but to copy it into a
temporary matrix, created by the compiler and hidden from programmers. The second copy is when the
temporary matrix is assigned to the matrix d. Each copy operation copies 90,000 8-byte double-precision
numbers, ~720k bytes. Thats a lot of copying.
of writing our own loops to copy data, we can call the library function memcpy( ), which is specifically
optimized for copying blocks of data. Our copy constructor is now:
ILmatrix::ILmatrix(const ILmatrix & b)
{ int r,c;
nr = b.nr, nc = b.nc; // matrix dimensions
if(b.db)
{ db = new T[nr*nc]; // allocate data block
memcpy(db, b.db, sizeof(T)*nr*nc); // copy the data
}
} // copy constructor
Similarly for the assignment operator. This code takes 0.98 0.10 s, 28 % better than the old code.
Not bad for such a simple change, but still bad: we still have two needless copies going on.
For the next improvement, we note that C++ can pass two matrix operands to an operator function, but
not three. Therefore, if we do one copy ourselves, we can then perform the addition in place, and avoid
the second copy. For example:
// Faster code to implement d = a + b:
d = a; // the one and only copy operation
d += b; // += adds b to the current value of d
The expression in parentheses copies a to d, and evaluates as the matrix d, which we can then act
on with the += operator. We can simplify this main code to a single line as:
(d = a) += b;
To implement this code, we need to add a += operator function to our class:
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ILmatrix & ILmatrix::operator +=(const ILmatrix & b)
{
int r, c;
for (r = 0; r < nr; r++)
for (c = 0; c < b.nc; c++)
(*this)[r][c] += b[r][c];
return *this; // returns by reference, NO copy!
}
This code runs in 0.45 0.02 s, or 1/3 the original time! The price, though, is somewhat uglier code.
Perhaps we can do even better. Instead of using operator functions, which are limited to only two
matrix arguments, we can write our own addition function, with any arguments we want. The main code is
now:
mat_add(d, a, b); // add a + b, putting result in d
Requiring the new function mat_add( ):
// matrix addition to new matrix: d = a + b
ILmatrix & mat_add(ILmatrix & d, const ILmatrix & a, const ILmatrix & b)
{
int r, c;
for (r = 0; r < d.nr; r++)
for (c = 0; c < d.nc; c++)
d[r][c] = a[r][c] + b[r][c];
return d; // returned by reference, NO copy constructor
This runs in 0.49 0.02 s, slightly worse than the one-copy version. Its also even uglier than the
previous version. How can this be?
Memory access, including data copying, is dominated by the effects of a complex piece of
hardware called memory cache.
There are hundreds of different variations of cache designs, and even if you know the exact design,
you can rarely predict its exact effect on real code. We will describe cache shortly, but even then, there is
no feasible way to know exactly why the zero-copy code is slower than one-copy. This result also held true
for the 400 400 matrix on computer 1, and the 300 300 matrix on computer 2, but not the 400 400
matrix on computer 2. All we can do is try a few likely cases, and go with the general trend. More on this
later.
Beware Leaving out a single character from your code can produce code that works, but runs over 2
times slower than it should. For example, in the function definition of mat_add, if we leave out
the & before argument a:
ILmatrix & mat_add(ILmatrix & d, const ILmatrix a, const ILmatrix & b)
then the compiler passes a to the function by copying it! This completely defeats our goal of zero
copy. [Guess how I found this out.]
Also notice that the memcpy( ) optimization doesnt apply to this last method, since it has no copies
at all.
Below is a summary of matrix addition. The best code choice was a single copy, with in-place
addition. It is medium ugly. While there was a small discrepancy with this on computer 2, 400 400, its
not worth the required additional ugliness.
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Computer 1
times
(ms, ~
100 ms)
Computer 2
times
(ms, ~
100 ms)
Algorithm 300 300 400 400 300 300 400 400
d = a + b, loop
copy
1360 100 % 5900 100
%
1130 100 % 2180 100
%
d = a + b,
memcpy( )
985 = 72 % 4960 = 84
%
950 = 84 % 1793 = 82
%
(d = a) += b 445 = 33 % 3850 = 65
%
330 = 29 % 791 = 36 %
mat_add(d, a, b) 490 = 36 % 4400 = 75
%
371 = 33 % 721 = 33 %
Run times for matrix addition with various algorithms.
Uncertainties are very rough 1. Best performing algorithms are highlighted
Cache Value
In the old days, computations were slower than memory accesses. Therefore, we optimized by
increasing memory use, and decreasing computations. Today, things are exactly reversed:
Modern CPUs (c. 2009) can compute about 50 times faster than they can access main memory.
Therefore, the biggest factor in overall speed is efficient use of memory.
To help reduce the speed degradation of slow memory, computers use a memory cache: a small
memory that is very fast. A typical main memory is 1 Gb, while a typical cache is 1 Mb, or 1000x smaller.
The CPU can access cache memory as fast as it can compute, so cache is ~50x faster than main memory.
The cache is invisible to program function, but is critical to program speed. The programmer usually does
not have access to details about the cache, but she can use general cache knowledge to greatly reduce run
time.
RAM
0
1
2
:
N-1
0
5
10
:
N-5
matrix
A
matrix
B
sequential
CPU data path
big, slow
RAM
small, fast
RAM cache
sequential
(Left) Computer memory (RAM) is a linear array of bytes. (Middle) For convenience, we
draw it as a 2D array, of arbitrary width. We show sample matrix storage. (Right) A very fast
memory cache keeps a copy of recently used memory locations, so they can be quickly used again.
The cache does two things (diagram above):
1. Cache remembers recently used memory values, so that if the CPU requests any of them again, the
cache provides the value instantly, and the slow main memory access does not happen.
2. Cache looks ahead to fetch memory values immediately following the one just used, before the
CPU might request it. If the CPU in fact later requests the next sequential memory location, the
cache provides the value instantly, having already fetched it from slow main memory.
The cache is small, and eventually fills up. Then, when the CPU requests new data, the cache must
discard old data, and replace it with the new. Therefore, if the program jumps around memory a lot, the
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benefits of the cache are reduced. If a program works repeatedly over a small region of memory (say, a
few hundred k bytes), the benefits of cache increase. Typically, cache can follow four separate regions of
memory concurrently. This means you can interleave accesses to four different regions of memory, and
still retain the benefits of cache. Therefore, we have three simple rules for efficient memory use:
For efficient memory use: (1) access memory sequentially, or at most in small steps, (2) reuse
values as much as possible in the shortest time, and (3) access few memory regions concurrently,
preferably no more than four.
There is huge variety in computer memory designs, so these rules are general, and behavior varies
from machine to machine, sometimes greatly. Our data below demonstrate this.
We can now understand some of our timing data given above. We see that the one-copy algorithm
unexpectedly takes less time than the zero-copy algorithm. The one-copy algorithm accesses only two
memory regions at a time: first matrix a and d for the copy, then matrix b and d for the add. The
zero-copy algorithm accesses three regions at a time: a, b, and d. This is probably reducing cache
efficiency. Recall that the CPU is also fetching instructions (the program) concurrently with the data,
which is at least a fourth region. Exact program layout in memory is virtually impossible to know. Also,
the cache on this old computer may not support 4-region concurrent access. The newer machine, computer
2, probably has a better cache, and the one- and zero-copy algorithms perform very similarly.
Heres a new question for matrix addition: the code given earlier loops over rows in the outer loop, and
columns in the inner loop. What if we reversed them, and looped over columns on the outside, and rows on
the inside? The result is 65% longer run time, on both machines. Heres why: the matrices are stored by
rows, i.e. each row is consecutive memory locations. Looping over columns on the inside accesses
memory sequentially, taking advantage of cache look-ahead. When reversed, the program jumps from row
to row on the inside, giving up any benefit from look-ahead. The cost is quite substantial. This concept
works on almost every machine.
Caution FORTRAN stores arrays in the opposite order from C and C++. In FORTRAN, the first index
is cycled most rapidly, so you should code with the outer loop on the second index, and the
inner loop on the first index. E.g.,
DO C = 1, N
DO R = 1, N
A(R, C) = blah blah ...
ENDDO
ENDDO
Scaling behavior: Matrix addition is an O(N
2
) operation, so increasing from 300 300 to 400 400
increase the computations by a factor of 1.8. On the older computer 1, the runtime penalty is much larger,
between 4.5x and 9x slower. On the newer computer 2, the difference is much closer, between 1.8x and
2.2x slower. This is likely due to cache size. A 300 300 double precision matrix takes 720 k bytes, or
under a MB. A 400 400 matrix takes 1280 k bytes, just over one MB. It could be that on computer 1,
with the smaller matrix, a whole matrix or two fits in cache, but with the large matrix, cache is overflowed,
and more (slow) main memory accesses are needed. The newer computer probably has bigger caches, and
may fit both sized matrices fully in cache.
Cache Withdrawal: Matrix Multiplication
We now show that the above tricks dont work well for large-matrix multiplication, but a different
trick cuts multiplication run time dramatically. To start, we use a simple matrix multiply in the main code:
d = a * b;
The straightforward matrix multiply operator is this:
// matrix multiply to temporary
ILmatrix ILmatrix::operator *(const ILmatrix & b) const
{
int r, c, k;
ILmatrix result(nr, b.nc); // temporary for result
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T sum;
for(r = 0; r < nr; r++)
{ for(c = 0; c < b.nc; c++)
{ sum = 0.;
for(k = 0; k < nc; k++) sum += (*this)[r][k] * b[k][c];
result[r][c] = sum;
}
} return result; // invokes copy constructor!
} // operator *()
While matrix addition is an O(N
2
) operation, matrix multiplication is an O(N
3
) operation. Multiplying
two 300 x 300 matrices is about 54,000,000 floating point operations, which is much slower than addition.
Timing the simple multiply routine, similarly to timing matrix addition, but with only 5 multiplies, we find
it takes 7.8 0.1 s on computer 1.
First we try the tricks we already know to improve and avoid data copies: we started already with
memcpy( ). We compare the two-copy, one-copy, and zero-copy algorithms as with addition, but this time,
5 of the 6 trials show no measurable difference. Matrix multiply is so slow that the copy times are
insignificant. The one exception is the one-copy algorithm on computer 2, which shows a significant
reduction of ~35%. This is almost certainly due to some quirk of memory layout and the cache, but we
cant identify it precisely. However, if we have to choose from these 3 algorithms, we choose the one-copy
(which coincidentally agrees with the matrix addition favorite). And certainly, we drop the ugly 3-
argument mat_mult( ) function, which gives no benefit.
Now well improve our matrix multiply greatly, by adding more work to be done. The extra work will
result in more efficient memory use, that pays off handsomely in reduced runtime. Notice that in matrix
multiplication, for each element of the results, we access a row of the first matrix a, and a column of the
second matrix b. But we learned from matrix addition that accessing a column is much slower than
accessing a row. And in matrix multiplication, we have to access the same column N times. Extra bad. If
only we could access both matrices by rows!
Well, we can. We first make a temporary copy of matrix b, and transpose it. Now the columns of b
become the rows of b
T
. We perform the multiply as rows of a with rows of b
T
copy time is insignificant for multiplication, so the cost of one copy and one transpose (similar to a copy) is
negligible. But the benefit of cache look-ahead is large. The transpose method reduces runtime by 30% to
50%.
Further thought reveals that we only need one column of b at a time. We can use it N times, and
discard it. Then move on to the next column of b. This reduces memory usage, because we only need
extra storage for one column of b, not for the whole transpose of b. It costs us nothing in operations, and
reduces memory. That can only help our cache performance. In fact, it cuts runtime by about another
factor of two, to about one third of the original runtime, on both machines. (It does require us to loop over
columns of b on the outer loop, and rows of a on the inner loop, but thats no burden.)
Note that optimizations that at first were insignificant, say reducing runtime by 10%, may become
significant after the runtime is cut by a factor of 3. That original 10% is now 30%, and may be worth
doing.
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Computer 1
times
(ms, ~
100 ms)
Computer 2
times
(ms, ~
100 ms)
Algorithm 300 300 400 400 300 300 400 400
d = a * b 7760 100 % 18,260
100 %
5348 100 % 16,300
100 %
(d = a) *= b 7890 = 102 % 18,210 =
100 %
3485 = 65 % 11,000 = 67
%
mat_mult(d,
a, b)
7720 = 99 % 18,170 =
100 %
5227 = 98 % 16,200 = 99
%
d = a * b,
transpose b
4580 = 59 % 12,700 = 70
%
2900 = 54 % 7800 = 48%
(d = a) *= b,
transpose b
4930 = 64 % 12,630 = 69
%
4250 = 79 % 11,000 = 67
%
d = a * b,
copy b column
2710 = 35 % 7875 = 43
%
3100 = 58 % 8000 = 49
%
(d = a) *= b,
copy b column
2945 = 38 % 7835 = 43
%
2100 = 39 % 5400 = 33
%
Run times for matrix multiplication with various algorithms.
Uncertainties are very rough 1. Best performing algorithms are highlighted
Cache Summary
In the end, exact performance is nearly impossible to predict. However, general knowledge of cache,
and following the three rules for efficient cache use (given above), will greatly improve your runtimes.
Conflicts in memory between pieces of data and instruction cannot be precisely controlled.
Sometimes even tiny changes in code will cross a threshold of cache, and cause huge changes in
performance.
IEEE Floating Point Formats And Concepts
Much of this section is taken from http://docs.sun.com/source/806-3568/ncg_math.html , an excellent
article introducing IEEE floating point. However, many clarifications are made here.
What Is IEEE Arithmetic?
In brief, IEEE 754 specifies exactly how floating point operations are to occur, and to what precision.
It does not specify how the floating point numbers are stored in memory. Each computer makes its own
choice for how to store floating point numbers. We give some popular formats later.
In particular, IEEE 754 specifies a binary floating point standard, with:
- Two basic floating-point formats: single and double.
- The IEEE single format has a significand (aka mantissa) precision of 24 bits, and is 32 bits
overall. The IEEE double format has a significand precision of 53 bits, and is 64 bits overall.
- Two classes of extended floating-point formats: single extended and double extended. The
standard specifies only the minimum precision and size. For example, an IEEE double extended
format must have a significand precision of at least 64 bits and occupy at least 79 bits overall.
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- Accuracy requirements on floating-point operations: add, subtract, multiply, divide, square root,
remainder, round numbers in floating-point format to integer values, convert between different
floating-point formats, convert between floating-point and integer formats, and compare. The
remainder and compare operations must be exact. Other operations must minimally modify the
exact result according to prescribed rounding modes.
- Accuracy requirements for conversions between decimal strings and binary floating-point
numbers. Within specified ranges, these conversions must be exact, if possible, or minimally
modify such exact results according to prescribed rounding modes. Outside the specified ranges,
these conversions must meet a specified tolerance that depends on the rounding mode.
- Five types of floating-point exceptions, and the conditions for the occurrence of these exceptions.
The five exceptions are invalid operation, division by zero, overflow, underflow, and inexact.
- Four rounding directions: toward the nearest representable value, with "even" values preferred
whenever there are two nearest representable values; toward negative infinity (down); toward
positive infinity (up); and toward 0 (chop).
- Rounding precision; for example, if a system delivers results in double extended format, the user
should be able to specify that such results be rounded to either single or double precision.
The IEEE standard also recommends support for user handling of exceptions.
IEEE 754 floating-point arithmetic offers users great control over computation. It simplifies the task
of writing numerically sophisticated, portable programs not only by imposing rigorous requirements, but
also by allowing implementations to provide refinements and enhancements to the standard.
Storage Formats
The IEEE floating-point formats define the fields that compose a floating-point number, the bits in
those fields, and their arithmetic interpretation, but not how those formats are stored in memory.
A storage format specifies how a number is stored in memory.
Each computer defines its own storage formats, though they are obviously all related.
High level languages have different names for floating point data types, which usually correspond to
the IEEE formats as shown:
IEEE Formats and Language Types
IEEE Precision C, C++ Fortran
single float REAL or REAL*4
double double DOUBLE PRECISION or REAL*8
double
extended
long
double
double
extended
REAL*16 [e.g., SPARC]. Note that in many
implementations, REAL*16 is different than long double
IEEE 754 specifies exactly the single and double floating-point formats, and it defines ways to extend
each of these two basic formats. The long double and REAL*16 types shown above are two double
extended formats compliant with the IEEE standard.
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f
127 126 112 111 0
1 15 112 LSB
Double-Extended (SPARC) (33-36 decimal digits)
f
63 62 52 51 0
1 11 52 LSB
Double (15-17 decimal digits)
e f
31 30 23 22 0
1 8 23 LSB
Single (6-9 decimal digits)
s f
95 80 79 78 64 63 62 0
16 1 15 1 63 LSB
j unused
Double-Extended (long double) (x86) (18-21 decimal digits)
s
s
e
e
s e
The following sections describe each of the floating-point storage formats on SPARC and x86
platforms.
When a Bias Is a Good Thing
IEEE floating point uses biased exponents, where the actual exponent is the unsigned value of the e
field minus a constant, called a bias:
exponent = e bias
The bias makes the e field an unsigned integer, and smallest numbers have the smallest e field (as
well as the smallest exponent). This format allows (1) floating point numbers sort in the same order as if
their bit patterns were integers; and (2) true floating point zero is naturally represented by an all-zero bit
pattern. These might seem insignificant, but they are quite useful, and so biased exponents are nearly
universal.
Single Format
The IEEE single format consists of three fields: a 23-bit fraction, f; an 8-bit biased exponent, e; and a
1-bit sign, s. These fields are stored contiguously in one 32-bit word, as shown above.
The table below shows the three constituent fields s, e, and f, and the value represented in single-
format:
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Single-Format Fields Value
1 e 254 (1)
s
2
127
1.f (normal numbers)
e = 0; f 0 (at least one bit in f is nonzero)
(1)
s
2
126
0.f (denormalized
numbers)
e = 0; f = 0 (all bits in f are zero) (1)
s
0.0 (signed zero)
s = 0/1; e = 255; f = 0 (all bits in f are zero) +/ (infinity)
s = either; e = 255; f 0 (at least one bit in f is
nonzero)
NaN (Not-a-Number)
Notice that when 1 e 254, the value is formed by inserting the binary radix point to the left of the
fraction's most significant bit, and inserting an implicit 1-bit to the left of the binary point, thus representing
a whole number plus fraction, called the significand, where 1 significand < 2. The implicit bits value is
not explicitly given in the single-format bit pattern, but is implied by the biased exponent field.
A denormalized number (aka subnormal number) is one which is too small to be represented by an
exponent in the range 1 e 254. The difference between a normal number and a denormalized number
is that the bit to left of the binary point of a normal number is 1, but that of a denormalized number is 0.
The 23-bit fraction combined with the implicit leading significand bit provides 24 bits of precision in
single-format normal numbers.
Examples of important bit patterns in the single-storage format are shown below. The maximum
positive normal number is the largest finite number representable in IEEE single format. The minimum
positive denormalized number is the smallest positive number representable in IEEE single format. The
minimum positive normal number is often referred to as the underflow threshold. (The decimal values are
rounded to the number of figures shown.)
Important Bit Patterns in IEEE Single Format
Common Name Bit Pattern (Hex) Approximate Value
+0 0000 0000 0.0
0 8000 0000 0.0
1 3f80 0000 1.0
2 4000 0000 2.0
maximum normal number 7f7f ffff 3.40282347e+38
minimum positive normal number 0080 0000 1.17549435e38
maximum subnormal number 007f ffff 1.17549421e38
minimum positive subnormal number 0000 0001 1.40129846e45
+ 7f80 0000 + (positive infinity)
ff80 0000 (negative infinity)
Not-a-Number (NaN) 7fc0 0000 (e.g.) NaN
A NaN (Not a Number) can be represented with many bit patterns that satisfy the definition of a NaN;
the value of the NaN above is just one example.
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Double Format
The IEEE double format is the obvious extension of the single format, and also consists of three fields:
a 52-bit fraction, f; an 11-bit biased exponent, e; and a 1-bit sign, s. These fields are stored in two
consecutive 32-bit words. In the SPARC architecture, the higher address 32-bit word contains the least
significant 32 bits of the fraction, while in the x86 architecture the lower address 32-bit word contains the
least significant 32 bits of the fraction.
The table below shows the three constituent fields s, e, and f, and the value represented in double-
format:
Double-Format Fields Value
1 e 2046 (1)
s
2
1023
x 1.f (normal numbers)
e = 0; f 0 (at least one bit in f is nonzero)
(1)
s
2
1022
x 0.f (denormalized
numbers)
e = 0; f = 0 (all bits in f are zero) (1)
s
0.0 (signed zero)
s = 0/1; e = 2047; f = 0 (all bits in f are zero) +/ (infinity)
s = either; e = 2047; f 0 (at least one bit in f is 1) NaN (Not-a-Number)
This is the obvious analog of the single format, and retains the implied 1-bit in the significand. The
52-bit fraction combined with the implicit leading significand bit provides 53 bits of precision in double-
format normal numbers.
Below, the 2
nd
column has two hexadecimal numbers. For the SPARC architecture, the left one is the
lower addressed 32-bit word; for the x86 architecture, the left one is the higher addressed word. The
decimal values are rounded to the number of figures shown.
Important Bit Patterns in IEEE Double Format
Common Name Bit Pattern (Hex) Approximate Value
+ 0 00000000 00000000 0.0
0 80000000 00000000 0.0
1 3ff00000 00000000 1.0
2 40000000 00000000 2.0
max normal number 7fefffff ffffffff
1.797 693 134 862
3157e+308
min positive normal number 00100000 00000000
2.225 073 858 507
2014e308
max denormalized number 000fffff ffffffff
2.225 073 858 507
2009e308
min positive denormalized
number
00000000 00000001
4.940 656 458 412
4654e324
+ 7ff00000 00000000 + (positive infinity)
fff00000 00000000 (negative infinity)
Not-a-Number 7ff80000 00000000 NaN
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(e.g.)
A NaN (Not a Number) can be represented with many bit patterns that satisfy the definition of a NaN;
the value of the NaN above is just one example.
Double-Extended Format (SPARC)
The SPARC floating-point quadruple-precision format conforms to the IEEE definition of double-
extended format. The quadruple-precision format occupies four 32-bit words and consists of three fields: a
112-bit fraction, f; a 15-bit biased exponent, e; and a 1-bit sign, s. These fields are stored contiguously.
The lowest addressed word has the sign, exponent, and the 16 most significant bits of the fraction. The
highest addressed 32-bit word contains the least significant 32-bits of the fraction.
Below shows the three constituent fields and the value represented in quadruple-precision format.
Double-Extended Fields (SPARC) Value
1 e 32766 (1)
s
x 2
16383
1.f (normal numbers)
e = 0, f 0 (at least one bit in f is nonzero)
(1)
s
x 2
16382
0.f (denormalized
numbers)
e = 0, f = 0 (all bits in f are zero) (1)
s
x 0.0 (signed zero)
s = 0/1, e = 32767, f = 0 (all bits in f are zero) +/ (infinity)
s = either, e = 32767, f 0 (at least one bit in f is 1) NaN (Not-a-Number)
In the hex digits below, the left-most number is the lowest addressed 32-bit word.
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Important Bit Patterns in IEEE Double-Extended Format (SPARC)
Name Bit Pattern (SPARC, hex) Approximate Value
+0
00000000 00000000 00000000
00000000
0.0
0
80000000 00000000 00000000
00000000
0.0
1
3fff0000 00000000 00000000
00000000
1.0
2
40000000 00000000 00000000
00000000
2.0
max
normal
7ffeffff ffffffff ffffffff
ffffffff
1.189 731 495 357 231 765 085
759 326 628 0070 e+4932
min
normal
00010000 00000000 00000000
00000000
3.362 103 143 112 093 506 262
677 817 321 7526 e4932
max
subnormal
0000ffff ffffffff ffffffff
ffffffff
3.362 103 143 112 093 506 262
677 817 321 7520 e4932
min pos
subnormal
00000000 00000000 00000000
00000001
6.475 175 119 438 025 110 924
438 958 227 6466 e4966
+
7fff0000 00000000 00000000
00000000
+
ffff0000 00000000 00000000
00000000
Not-a-
Number
7fff8000 00000000 00000000
00000000 (e.g.)
NaN
Double-Extended Format (x86)
The important difference in the x86 long-double format is the lack of an implicit leading 1-bit in the
significand. Instead, the 1-bit is explicit, and always present in normalized numbers. This clearly violates
the spirit of the IEEE standard. However, big companies carry a lot of clout with standards bodies, so Intel
claims this double-extended format conforms to the IEEE definition of double-extended formats, because
IEEE 754 does not specify how (or if) the leading 1-bit is stored. X86 long-double consists of four fields: a
63-bit fraction, f; a 1-bit explicit leading significand bit, j; a 15-bit biased exponent, e; and a 1-bit sign, s
In the x86 architectures, these fields are stored contiguously in ten successively addressed 8-bit bytes.
However, the UNIX System V Application Binary Interface Intel 386 Processor Supplement (Intel ABI)
requires that double-extended parameters and results occupy three consecutive 32-bit words in the stack,
with the most significant 16 bits of the highest addressed word being unused, as shown below.
Double-Extended (long double) Format (x86)
s e f
95 80 79 78 64 63 62 0
1 15 1 63 LSB
j unused
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The lowest addressed 32-bit word contains the least significant 32 bits of the fraction, f[31:0], with bit
0 being the least significant bit of the entire fraction. Though the upper 16 bits of the highest addressed 32-
bit word are unused by x86, they are essential for conformity to the Intel ABI, as indicated above.
Below shows the four constituent fields and the value represented by the bit pattern. x = dont care.
Double-Extended Fields (x86) Value
j = 0, 1 <= e <= 32766 Unsupported
j = 1, 1 <= e <= 32766 (1)
s
x 2
e16383
x 1.f (normal numbers)
j = 0, e = 0; f 0 (at least one bit in f is
nonzero)
(1)
s
x 2
16382
x 0.f (denormalized numbers)
j = 1, e = 0
(1)
s
x 2
16382
x 1.f (pseudo-denormal
numbers)
j = 0, e = 0, f = 0 (all bits in f are zero) (1)
s
x 0.0 (signed zero)
j = 1; s = 0/1; e = 32767; f = 0 (all bits in f are
zero)
+/ (infinity)
j = 1; s = x; e = 32767; f = .1xxx...xx QNaN (quiet NaNs)
j = 1; s = x; e = 32767; f = .0xxx...xx 0 (at
least one of the x in f is 1)
SNaN (signaling NaNs)
Notice that bit patterns in x86 double-extended format do not have an implicit leading significand bit.
The leading significand bit is given explicitly as a separate field, j. However, when e 0, any bit pattern
with j = 0 is unsupported: such a bit pattern as an operand in floating-point operations provokes an invalid
operation exception.
The union of the fields j and f in the double extended format is called the significand. The
significand is formed by inserting the binary radix point between the leading bit, j, and the fraction's most
significant bit.
In the x86 double-extended format, a bit pattern whose leading significand bit j is 0 and whose biased
exponent field e is also 0 represents a denormalized number, whereas a bit pattern whose leading
significand bit j is 1 and whose biased exponent field e is nonzero represents a normal number. Because
the leading significand bit is represented explicitly rather than being inferred from the exponent, this format
also admits bit patterns whose biased exponent is 0, like the subnormal numbers, but whose leading
significand bit is 1. Each such bit pattern actually represents the same value as the corresponding bit
pattern whose biased exponent field is 1, i.e., a normal number, so these bit patterns are called pseudo-
denormals. Pseudo-denormals are merely an artifact of the x86 double-extended storage format; they are
implicitly converted to the corresponding normal numbers when they appear as operands, and they are
never generated as results.
Below are some important bit patterns in the double-extended storage format. The 2
nd
column has
three hex numbers. The first number is the 16 least significant bits of the highest addressed 32-bit word
(recall that the upper 16 bits of this 32-bit word are unused), and the right one is the lowest addressed 32-bit
word.
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Important Bit Patterns in Double-Extended (x86) Format and their
Values
Common Name Bit Pattern (x86)
Approximate
Value
+0
0000 00000000
00000000
0.0
0
8000 00000000
00000000
0.0
1
3fff 80000000
00000000
1.0
2
4000 80000000
00000000
2.0
max normal
7ffe ffffffff
ffffffff
1.189 731 495 357 231
765 05 e+4932
min positive normal
0001 80000000
00000000
3.362 103 143 112 093
506 26 e4932
max subnormal
0000 7fffffff
ffffffff
3.362 103 143 112 093
506 08 e4932
min positive subnormal
0000 00000000
00000001
3.645 199 531 882 474
602 53 e4951
+
7fff 80000000
00000000
+
ffff 80000000
00000000
quiet NaN with greatest fraction
7fff ffffffff
ffffffff
QNaN
quiet NaN with least fraction
7fff c0000000
00000000
QNaN
signaling NaN with greatest
fraction
7fff bfffffff
ffffffff
SNaN
signaling NaN with least
fraction
7fff 80000000
00000001
SNaN
A NaN (Not a Number) can be represented by any of the bit patterns that satisfy the definition of NaN.
The most significant bit of the fraction field determines whether a NaN is quiet (bit = 1) or signaling
(bit = 0).
Precision in Decimal Representation
This section covers the precisions of the IEEE single and double formats, and the double-extended
formats on SPARC and x86. See the earlier section on How Many Digits Do I Get? for more information.
The IEEE standard specifies the set of numerical values representable in a binary format. Each format
has some number of bits of precision (e.g., single has 24 bits). But the decimal numbers of roughly the
same precision do not match exactly the binary numbers, as you can see on the number line:
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10
n
10
n+1
2
m
2
m+1
2
m+2
decimal
binary
10
n+2
Comparison of a Set of Numbers Defined by Decimal and Binary Representation
Because the decimal numbers are different than the binary numbers, estimating the number of
significant decimal digits corresponding to b significant binary bits requires some definition.
Reformulate the problem in terms of converting floating-point numbers between binary and decimal.
You might convert from decimal to binary and back to decimal, or from binary to decimal and back to
binary. It is important to notice that because the sets of numbers are different, conversions are in general
inexact. If done correctly, converting a number from one set to a number in the other set results in
choosing one of the two neighboring numbers from the second set (which one depends on rounding).
All binary numbers can be represented exactly in decimal, but usually this requires unreasonably many
digits to do so. What really matters is how many decimal digits are needed, to insure no loss in converting
from binary to decimal and back to binary.
Most decimal numbers cannot be represented exactly in binary (because decimal fractions include a
factor of 5, which requires infinitely repeating binary digits). For example, run the following Fortran
program:
REAL Y, Z
Y = 838861.2
Z = 1.3
WRITE(*,40) Y
40 FORMAT("y: ",1PE18.11)
WRITE(*,50) Z
50 FORMAT("z: ",1PE18.11)
The output should resemble:
y: 8.38861187500E+05
z: 1.29999995232E+00
The difference between the value 8.388612 10
5
assigned to y and the value printed out is 0.0125,
which is seven decimal orders of magnitude smaller than y. So the accuracy of representing y in IEEE
single format is about 6 to 7 significant digits, or y has about 6 significant digits.
Similarly, the difference between the value 1.3 assigned to z and the value printed out is
0.00000004768, which is eight decimal orders of magnitude smaller than z. The accuracy of representing z
in IEEE single format is about 7 to 8 significant digits, or z has about 7 significant digits.
See Appendix F of http://docs.sun.com/source/806-3568/ncg_references.html for references on base
conversion. They say that particularly good references are Coonen's thesis and Sterbenz's book.
Underflow
Underflow occurs, roughly speaking, when the result of an arithmetic operation is so small that it
cannot be stored in its intended destination format without suffering a rounding error that is larger than
usual; in other words, when the result is smaller than the smallest normal number.
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Underflow Thresholds in Each Precision
single
smallest normal
number
largest subnormal number
1.175 494 35e38
1.175 494 21e38
double
smallest normal
number
largest subnormal number
2.225 073 858 507 201 4e308
2.225 073 858 507 200 9e308
double
extended (x86)
smallest normal
number
largest subnormal number
3.362 103 143 112 093 506 26e4932
3.362 103 143 112 093 505 90e4932
double
extended
(SPARC)
smallest normal
number
largest subnormal number
3.362 103 143 112 093 506 262 677 817 321 752 6e
4932
3.362 103 143 112 093 506 262 677 817 321 752 0e
4932
The positive subnormal numbers are those numbers between the smallest normal number and zero.
Subtracting two (positive) tiny numbers that are near the smallest normal number might produce a
subnormal number. Or, dividing the smallest positive normal number by two produces a subnormal result.
The presence of subnormal numbers provides greater precision to floating-point calculations that
involve small numbers, although the subnormal numbers themselves have fewer bits of precision than
normal numbers. Gradual underflow produces subnormal numbers (rather than returning the answer
zero) when the mathematically correct result has magnitude less than the smallest positive normal number.
There are several other ways to deal with such underflow. One way, common in the past, was to flush
those results to zero. This method is known as Store 0 and was the default on most mainframes before the
The mathematicians and computer designers who drafted IEEE Standard 754 considered several
alternatives, while balancing the desire for a mathematically robust solution with the need to create a
standard that could be implemented efficiently.
How Does IEEE Arithmetic Treat Underflow?
IEEE Standard 754 requires gradual underflow. This method requires defining two representations for
stored values, normal and subnormal.
Recall that the IEEE value for a normal floating-point number is: (1)
s
2
ebias
1.f
where s is the sign bit, e is the biased exponent, and f is the fraction. Only s, e, and f need to be stored
to fully specify the number. Because the leading bit of the significand is 1 for normal numbers, it need not
be stored (but may be).
The smallest positive normal number that can be stored, then, has the negative exponent of greatest
magnitude and a fraction of all zeros. Even smaller numbers can be accommodated by considering the
leading bit to be zero rather than one. In the double-precision format, this effectively extends the minimum
exponent from 10
308
to 10
324
, because the fraction part is 52 bits long (roughly 16 decimal digits.) These
are the subnormal numbers; returning a subnormal number (rather than flushing an underflowed result to
Clearly, the smaller a subnormal number, the fewer nonzero bits in its fraction; computations
producing subnormal results do not enjoy the same bounds on relative roundoff error as computations on
normal operands. However, the key fact is:
Gradual underflow implies that underflowed results never suffer a loss of accuracy any greater than
that which results from ordinary roundoff error.
Addition, subtraction, comparison, and remainder are always exact when the result is very small.
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Recall that the IEEE value for a subnormal floating-point number is: (1)
s
2
bias + 1
0.f
where s is the sign bit, the biased exponent e is zero, and f is the fraction. Note that the implicit power-
of-two bias is one greater than the bias in the normal format, and the leading bit of the fraction is zero.
Gradual underflow allows you to extend the lower range of representable numbers. It is not smallness
that renders a value questionable, but its associated error. Algorithms exploiting subnormal numbers have
smaller error bounds than other systems. The next section provides some mathematical justification for
The purpose of subnormal numbers is not to avoid underflow/overflow entirely, as some other
arithmetic models do. Rather, subnormal numbers eliminate underflow as a cause for concern for a variety
of computations (typically, multiply followed by add). For a more detailed discussion, see Underflow and
the Reliability of Numerical Software by James Demmel, and Combatting the Effects of Underflow and
Overflow in Determining Real Roots of Polynomials by S. Linnainmaa.
The presence of subnormal numbers in the arithmetic means that untrapped underflow (which implies
loss of accuracy) cannot occur on addition or subtraction. If x and y are within a factor of two, then x y is
error-free. This is critical to a number of algorithms that effectively increase the working precision at
critical places in algorithms.
In addition, gradual underflow means that errors due to underflow are no worse than usual roundoff
error. This is a much stronger statement than can be made about any other method of handling underflow,
and this fact is one of the best justifications for gradual underflow.
Most of the time, floating-point results are rounded:
computed result = true result + roundoff
How large can the roundoff be? One convenient measure of its size is called a unit in the last place,
abbreviated ulp. The least significant bit of the fraction of a floating-point number is its last place. The
value represented by this bit (e.g., the absolute difference between the two numbers whose representations
are identical except for this bit) is a unit in the last place of that number. If the true result is rounded to the
nearest representable number, then clearly the roundoff error is no larger than half a unit in the last place of
the computed result. In other words, in IEEE arithmetic with rounding mode to nearest,
0 |roundoff | 1/2 ulp
of the computed result.
Note that an ulp is a relative quantity. An ulp of a very large number is itself very large, while an ulp
of a tiny number is itself tiny. This relationship can be made explicit by expressing an ulp as a function:
ulp(x) denotes a unit in the last place of the floating-point number x.
Moreover, an ulp of a floating-point number depends on the floating point precision. For example, this
shows the values of ulp(1) in each of the four floating-point formats described above:
ulp(1) in Four Different Precisions
single ulp(1) = 2
23
~ 1.192093e07
double ulp(1) = 2
52
~ 2.220446e16
double extended (x86) ulp(1) = 2
63
~ 1.084202e19
112
~ 1.925930e34
Recall that only a finite set of numbers can be exactly represented in any computer arithmetic. As the
magnitudes of numbers get smaller and approach zero, the gap between neighboring representable numbers
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narrows. Conversely, as the magnitude of numbers gets larger, the gap between neighboring representable
numbers widens.
For example, imagine you are using a binary arithmetic that has only 3 bits of precision. Then,
between any two powers of 2, there are 2
3
= 8 representable numbers, as shown here:
The number line shows how the gap between numbers doubles from one exponent to the next.
In the IEEE single format, the difference in magnitude between the two smallest positive subnormal
numbers is approximately 10
45
, whereas the difference in magnitude between the two largest finite
numbers is approximately 10
31
!
Below, nextafter(x, +) denotes the next representable number after x as you move towards +.
Gaps Between Representable Single-Format Floating-Point Numbers
x nextafter(x, +) Gap
0.0 1.4012985e45 1.4012985e45
1.1754944e-38 1.1754945e38 1.4012985e45
1.0 1.0000001 1.1920929e07
2.0 2.0000002 2.3841858e07
16.000000 16.000002 1.9073486e06
128.00000 128.00002 1.5258789e05
1.0000000e+20 1.0000001e+20 8.7960930e+12
9.9999997e+37 1.0000001e+38 1.0141205e+31
Any conventional set of representable floating-point numbers has the property that the worst effect of
one inexact result is to introduce an error no worse than the distance to one of the representable neighbors
of the computed result. When subnormal numbers are added to the representable set and gradual underflow
is implemented, the worst effect of one inexact or underflowed result is to introduce an error no greater than
the distance to one of the representable neighbors of the computed result.
In particular, in the region between zero and the smallest normal number, the distance between any
two neighboring numbers equals the distance between zero and the smallest subnormal number.
Subnormal numbers eliminate the possibility of introducing a roundoff error that is greater than the distance
to the nearest representable number.
Because roundoff error is less than the distance to any of the representable neighbors of the true result,
many important properties of a robust arithmetic environment hold, including these:
- x y <=> x - y 0
- (x y) + y x, to within a rounding error in the larger of x and y
- 1/(1/x) x, when x is a normalized number, implying 1/x 0
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An old-fashioned underflow scheme is Store 0, which flushes underflow results to zero. Store 0
violates the first and second properties whenever x y underflows. Also, Store 0 violates the third property
whenever 1/x underflows.
Let represent the smallest positive normalized number, which is also known as the underflow
threshold. Then the error properties of gradual underflow and Store 0 can be compared in terms of .
gradual underflow: |error| < ulp in
Store 0: |error|
Even in single precision, the round-off error is millions of times worse with Store 0 than gradual
underflow.
Two Examples of Gradual Underflow Versus Store 0
The following are two well-known mathematical examples. The first example computes an inner
product.
sum = 0;
for (i = 0; i < n; i++)
{ sum = sum + a[i] * y[i];
}
With gradual underflow, the result is as accurate as roundoff allows. In Store 0, a small but nonzero
sum could be delivered that looks plausible but is wrong in nearly every digit. To avoid these sorts of
problems, clever programmers must scale their calculations, which is only possible if they can anticipate
The second example, deriving a complex quotient, is not amenable to scaling:
( ) ( ) ( ) ( )
( )
/ /
, / 1,
/
p r s q i q r s p
p iq
a ib assuming r s
r is s r r s
+ +
+
+ = s =
+ +
It can be shown that, despite roundoff, (1) the computed complex result differs from the exact result by
no more than what would have been the exact result if p + iq and r + is each had been perturbed by no more
than a few ulps, and (2) this error analysis holds in the face of underflows, except that when both a and b
underflow, the error is bounded by a few ulps of |a + ib|. Neither conclusion is true when underflows are
flushed to zero.
This algorithm for computing a complex quotient is robust, and amenable to error analysis, in the
presence of gradual underflow. A similarly robust, easily analyzed, and efficient algorithm for computing
the complex quotient in the face of Store 0 does not exist. In Store 0, the burden of worrying about low-
level, complicated details shifts from the implementer of the floating-point environment to its users.
The class of problems that succeed in the presence of gradual underflow, but fail with Store 0, is larger
than the fans of Store 0 may realize. Many frequently used numerical techniques fall in this class:
- Linear equation solving
- Polynomial equation solving
- Numerical integration
- Convergence acceleration
- Complex division
Does Underflow Matter?
In the absence of gradual underflow, user programs need to be sensitive to the implicit inaccuracy
threshold. For example, in single precision, if underflow occurs in some parts of a calculation, and Store 0
is used to replace underflowed results with 0, then accuracy can be guaranteed only to around 10
31
, not
10
38
, the usual lower range for single-precision exponents. This means that programmers need to
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implement their own method of detecting when they are approaching this inaccuracy threshold, or else
abandon the quest for a robust, stable implementation of their algorithm.
Some algorithms can be scaled so that computations don't take place in the constricted area near zero.
However, scaling the algorithm and detecting the inaccuracy threshold can be difficult and time-consuming
for each numerical program.
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Fourier Transforms and Digital Signal Processing
We assume the reader is familiar with basic sampling and Fourier Transform principles. In particular,
you must be familiar with decomposing a function into an orthonormal basis of functions. We describe
important (often overlooked) properties of Discrete Fourier Transforms. We start with the most general
(and simplest) case, then proceed through more specialized cases. Topics:
- Complex sequences, and complex Fourier Transform (its actually easier to start with the complex
case, and specialize to real numbers later)
- Sampling and the Model of Digitization
- Even number of points vs. odd number of points
- Basis Functions and Orthogonality
- Real sequences: even and odd # points
- Normalization and Parsevals Theorem
- Continuous vs. discrete time and frequency; finite vs. infinite time and frequency
- Non-uniformly spaced samples
This section assumes you are familiar with complex arithmetic and exponentials. Understanding
phasors is very helpful, but not essential (see Funky Electromagnetic Concepts for a discussion of phasors).
Brief Definitions:
Fourier Series represents a periodic continuous function as an infinite sum of sinusoids at discrete
frequencies:
1
2
1 1 1 0
are complex (phasors),
( ) ,
1/ (in cycle/s or Hz), 2 (in rad/s)
k
i kf t
k
k
where S t time
s t S e
f period f
t
e t
=
=
f
1
= 1/period, the lowest nonzero frequency, is called the fundamental frequency. f
0
= 0, always.
Fourier Transform (FT)
represents a continuous function as an integral of sinusoids over continuous frequencies:
2
1
( ) (2 ) ( ) , ( ) is complex
2
i ft i t
s t S f e df S e d where S
t e
t e e
t
= =
} }
We do not discuss this here. The function s(t) is not periodic, so there is no fundamental frequency.
S() is a phasor-valued function of angular frequency.
Discrete Fourier Transform (DFT)
represents a finite sequence of numbers as a finite sum of sinusoids:
( )
1
2 /
1 1 1 0
are complex (phasors), 0, ... 1 the sample index,
,
1/ (in cycle/s), 2 (in rad/s)
n
k
i k n j
j k
k
where S j n
s S e
f period f
t
e t
=
=
=
= =
The sequence s
j
may be thought of as either periodic, or undefined outside the sampling interval. As in
the Fourier Series, the fundamental frequency is 1/period, or equivalently 1/(sampled interval), and f
0
= 0,
always.
[Since a DFT essentially treats the input as periodic, it might be better called a Discrete Fourier Series (rather
than Transform), but Discrete Fourier Transform is completely standard.]
Fast Fourier Transform (FFT)
an algorithm for implementing special cases of DFT.
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Inverse Discrete Fourier Transform (IDFT)
gives the sequence of numbers s
j
from the DFT components.
The general digital Fourier Transform is a Discrete Fourier Transform (DFT).
An FFT is an algorithm for special cases of DFT.
Model of Digitization
All realistic systems which digitize analog signals must comprise at least the following components:
0 1 2 3 4 5 6 7 8 9 ...
analog
signal
Anti-alias
Low Pass
Filter (LPF)
Analog to
Digital
Converter
sample
clock, f
samp
filtered analog
signal
digital
samples, s
j
Minimum components of a Digital Signal Processing system, with uniformly spaced samples
Complex Sequences and Complex Fourier Transform
Its actually easier to start with the complex case, and specialize to real numbers later. Given a
sequence of n complex numbers s
j
, we can write the sequence as a sum of sinusoids, i.e. complex
exponentials:
Inverse Discrete Fourier Transform:
( )
1
2 /
0
, 0, ... 1is the sample index
the frequency of the component, in cycle/sample
the complex frequency component (phasor)
n
i k n j
j k
k
th
k
s S e where j n
k
k
n
S
t
=
= =
Note that there are n original complex numbers, and n complex frequency components, so no
information is lost. The transform is exact, unique, and reversible. (In other words, this is not a fit.)
The above equation forces all normalization conventions. We use the simple scheme wherein a
function equals the sum of its components (with no factors of 2 or anything else).
Often, the index j is a measure of time or distance, and the sequence comprises samples of a signal
taken at equal intervals. Without loss of generality, we will refer to j as a measure of time, but it could be
anything. Note that the equation above actually defines the Inverse Discrete Fourier Transform (IDFT),
because it gives the original sequence from the Fourier components. [Mathematicians often reverse the
definitions of DFT and IDFT, by putting a minus sign in the exponent of the IDFT equation above.
Engineers and physicists usually use the convention given here.]
Sampling
Each number in the sequence is called a sample, because such sequences are often generated by
sampling a continuous signal s(t). For n samples, there are n frequency components, S
k
, each at normalized
frequency k/n (defined soon):
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j
complex samples
n = 10
s
j
0 1 2 3 4 5 6 7 8 9
k
Complex Frequency
Components
S
k
0 1 2 3 4 5 6 7 8 9
k = 9
k = 1
k = 2
k = 0
basis
frequencies
signal period, aka sample interval
fundamental
frequency
For simplicity in this diagram, the samples, sinusoids, and component amplitudes
are shown as real, but in general, they are all complex valued.
Note that there are a full n sample times in the sample interval (aka signal period), not (n 1).
The above representation is used by many DFT functions in computer libraries.
Also, there is no need for any other frequencies, because k = 10 has exactly the same values at all the
sample points as k = 0. If the samples are from a continuous signal that had a frequency component at
k = 10, then that component will be aliased down to k = 0, and added to the actual k = 0 component. It is
forever lost, and cannot be recovered from the samples, nor distinguished from the k = 0 (DC) component.
The same aliasing occurs for any two frequencies k and k + n.
The above definition is the only correct meaning for aliasing.
Many (most?) people misuse this word to mean other things (e.g., harmonics or sidebands).
To avoid a dependence on n, we usually label the frequencies as fractions. For n samples, there are n
frequencies, measured in units of cycles/sample, and running from f = 0 to f = (1 1/n) cycles/sample. The
n normalized frequencies are
{ }
1 2 3 1
, 0,1, ... 1, 0, , , ,...
k k
k n
f k n that is f
n n n n n
= = =
`
)
There is no f = 1, just as there is no k = n, because f = 1 is an alias of f = 0. The Fourier components
are written as S(f), a function of f, so we re-label the above diagram with normalized frequencies:
j
complex samples
n = 10
s
j
0 1 2 3 4 5 6 7 8 9
f
Complex Frequency
Components
S
k
0 .1 .2 .3 .4 .5 .6 .7 .8 .9
f =. 9
f = .1
f = .2
f = 0
basis
frequencies
sampled interval
fundamental
frequency
Normalized frequencies are equivalent to measuring time in units of the sample time,
and frequencies in cycles/sample.
For theoretical analysis, it is often more convenient to have the frequency range be 0.5 < f 0.5,
instead of 0 f < 1. Since any frequency f is equivalent to (an alias of) f 1, we can simply move the
frequencies in the range 0.5 < f < 1 down to 0.5 < f < 0:
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f
Complex Frequency
Components
S(f )
-.4 -.3 -.2 -.1 0 .1 .2 .3 .4 .5
f
Complex Frequency
Components
S(f )
0 .1 .2 .3 .4 .5 .6 .7 .8 .9
n = 10
For an even number of samples (and frequencies, diagram above), the resulting frequency set is
necessarily asymmetric, because there is no f = 0.5, but there is an f = +0.5. For an odd number of points
(below), the frequency set is symmetric, and there is neither f = 0.5 nor f = +0.5:
f
Complex Frequency
Components
S(f )
-.4 -.2 0 .2 .4
f
Complex Frequency
Components
S(f )
-.4 -.2 0 .2 .4 .6 .8
n = 5
Some references say that sampling a signal is like
setting it to zero everywhere except the sample times. It is not.
This is a common misconception. It well refuted by [Openheim and Schafer, and dozens of other
signal processing experts]. It is easy to show that that claim is not true, in several ways. One simple way is
this: For a band-limited signal, I can reconstruct the signal between the sample times from just the samples
alone. That makes no sense if sampling amounted to zeroing the signal between samples, and then
transforming.
Basis Functions and Orthogonality
The basis functions of the DFT are the discrete-time exponentials, which are equivalent to sines and
cosines:
( ) 2 /
even: / 2 1, ... 1, 0,1... / 2
( ) , 0,1, ... 1, 0,1, ... 1
: int( / 2), ... 1, 0,1, ... int( / 2)
i k n j
k
n n n
b j e j n k n or
n odd n n
t
+
= = =
Note that:
The DFT and FT are simply decompositions of functions into basis functions,
just like in ordinary quantum mechanics. The transform equations are just the inner products of
the given functions with the basis functions.
The basis functions are orthogonal, normalized (in our convention) such that
k m km
b b n o = . Proof:
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( ) ( ) ( )( ) ( )( )
( )( )
( )( )
1 1 1 1
2 / 2 / 2 / 2 / *
0 0 0 0
1
0
0
1
2 /
0
2 /
( ) ( )
,
1
, . Then:
1
1
1
n n n n
j
i k n j i m n j i n m k j i n m k
k m k m
j j j j
n
j
k m
j
n n
i n m k j
j
n
i n m k
k m
i
b b b j b j e e e e
For k m we have b b e n
r
For k m use r where r e
r
e
b b
e
t t t t
t
t
= = = =
(
= = =
(
= = =
(
= = =
( )( )
( )( )
( )( ) ( )( )
2
2 / 2 / 2 /
1 1 1
0
1 1
i m k
k m km
n m k i n m k i n m k
e
b b n
e e
t
t t t
o
= = = =
( ( (
The orthogonality condition allows us to immediately write the DFT from the definition of the IDFT
above:
Discrete Fourier Transform:
( )
1
2 /
0
1
, the complex frequency component
the normalized frequency of the component
n
i k n j
k j k
j
S s e where S
n
k
kth
n
t
=
=
=
Note that there are 2n independent real numbers in the complex sequence s
j
, and there are also 2n
independent real numbers in the complex spectrum S
k
, as there must be (same number of degrees of
freedom).
Real Sequences
An important special case of sequence s
j
is a real-valued sequence, which is a special case of a
complex-valued sequence. For
( )
/ 2
2 /
/ 2
n
i k n j
j k
k n
s S e
t
~
~
=
to be real, the S
k
must occur in complex conjugate
pairs, i.e., the spectrum S
k
must be conjugate symmetric:
*
(for real, and / 2)
k k j
S S s k n
= <
This implies that S
0
is always real, which is also clear since S
0
is just the average of the real sequence.
Note that there is no k = n/2.
There are n independent real numbers in the real sequence s
j
. We now consider two subcases: n is
even, and n is odd.
For n even,
( )
/ 2
2 /
/ 2 1
n
i k n j
j k
k n
s S e n even
t
= +
=
,
and we use the asymmetric frequency range 0.5 < f 0.5, which corresponds to n/2 < k n/2 (below
left). For an even number of points, since there are no conjugates to k = 0 or k = n/2, we must have that S
0
and S
n/2
are real. All other S
k
are conjugate symmetric.
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-.44 -.33 -.22 -.11 0 .11 .22 .33 .44
-.4 -.3 -.2 -.1 0 .1 .2 .3 .4 .5
f
S
0
& S
5
real
conjugate symmetric
f
n = 10,
s
j
real
Complex Frequency
Components
S
0
real
conjugate symmetric
n = 9,
s
j
real
S(f) S(f)
Complex Frequency
Components
(Left) Sequence with even number of samples, n = 10. (Right) Sequence with odd number, n
= 9.
Therefore, in the spectrum, there are (n/2 1) independent complex frequency components, plus two
real components, totaling n independent real numbers in the spectrum, matching the n independent real
numbers in the sequence s
j
. In terms of sine and cosine components (rather than the complex components),
there are (n/2 + 1) independent cosine components, and (n/2 1) independent sine components.
For an odd number of points (above right),
( )
( )
( ) 1 / 2
2 /
1 / 2
,
n
i k n j
j k
k n
s S e n odd
t
=
=
,
there is no k = n/2 component, and again there is no conjugate to k = 0. Therefore, we must have that
S
0
is real. All other S
k
are conjugate symmetric. Therefore, in the spectrum, there are (n 1)/2 independent
complex frequency components, plus one real component (S
0
), totaling n independent real numbers in the
spectrum, matching the n independent real numbers in the sequence s
j
. In terms of sine and cosine
components (rather than the complex components), there are (n + 1)/2 independent cosine components, and
(n 1)/2 independent sine components.
Normalization and Parsevals Theorem
When the original sequence represents something akin to samples of voltage over time, we can speak
of energy in the signal. The energy of the signal is the sum of the energies of each sample:
2 2
1 1
2
0 0
, "conductance", choosen to be1.
j j j
n n
j j
j j
E Gs s where G
E E s
= =
= = =
= =
The energies of the sinusoidal components in the DFT add as well, because the sinusoids are
orthogonal (show why??):
1
2
0
n
k
k
E S
Parseval Theorem equates the energy of the original sequence to the energy of the sinusoidal
components, by providing the constant of proportionality. First, we evaluate the energy of a single
sinusoid:
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( )
1
2
2 2 2 /
0
1 1 1 1
2 2
2
0 0 0 0
n
i k n j
k k k
j
n n n n
k k k j
k k k j
E S e n S
E E n S n S s
t
=
= = = =
= =
= = =
Besides our normalization choice above, there are several other choices in common use. In general,
between the DFT, IDFT, and Parsevals Theorem, you can choose a normalization for one, which then
fixes the normalization for the other two. For example, some people choose to make the DFT and IDFT
more symmetric by defining:
( )
( )
1
2 /
1 1
0
2
2
1
0 0
2 /
0
1
:
1
:
n
i k n j
j k
n n
k
k j
n
k j
i k n j
k j
j
IDFT s S e
n
S s
DFT S s e
n
t
t
= =
=
`
(alternate normalizations)
Continuous and Discrete, Finite and Infinite
TBS: Finite length implies discrete frequencies; infinite length implies continuous frequencies.
Discrete time implies finite frequencies; continuous time implies infinite frequencies. Finite length is
equivalent to periodic.
White Noise and Correlation
White noise has, on average, all frequency components equal (named by incorrect analogy with white
light); samples of white noise are uncorrelated. Non-white noise has unequal frequency components (on
average); samples of non-white noise are necessarily correlated. (Show this??).
Why Oversampling Does Not Improve Signal-to-Noise Ratio
Sometimes it might seem that if I oversample a signal (i.e., sample above the Nyquist rate), the noise
power stays constant (= noise variance is constant), but I have more samples of the signal which I can
average. Therefore, by oversampling, I should be able to improve my SNR by averaging out more noise,
but keeping all the signal.
This reasoning is wrong, of course, because it implies that by sampling arbitrarily fast, I can filter out
arbitrarily large amounts of noise, and ultimately recover anything from almost nothing. So whats wrong
with this reasoning? Lets take an example.
Suppose I sample a signal at 100 samples/sec, with white noise. Then my Nyquist frequency is 50 Hz,
and I must use a 50 Hz Low Pass Filter (LPF) for anti-aliasing before sampling. This LPF leaves me with
50 Hz worth of noise power (= variance).
Now suppose I double the sampling frequency to 200 samples/sec. To maintain white noise, I must
open my anti-alias filter cutoff to the new Nyquist frequency, 100 Hz. This doubles my noise power. Now
I have twice as many samples of the signal, with twice as much noise power. I can run a LPF to reduce the
noise (say, averaging adjacent samples). At best, I cut the noise by half, reducing it back to its 100
sample/sec value, and reducing my sample rate by 2. Hence, Im right back where I was when I just
sampled at 100 samples/sec in the first place.
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f
samp
= 100 samples/sec
50 Hz =
Nyquist frequency
A
m
p
l
i
t
u
d
e
Discrete white
noise spectrum
f
samp
= 200 samples/sec
50 Hz
A
m
p
l
i
t
u
d
e
100 Hz =
Nyquist
frequency
f
samp
= 200 samples/sec
50 Hz
A
m
p
l
i
t
u
d
e
Discrete correlated
noise spectrum
Discrete white
noise spectrum
100 Hz =
Nyquist
frequency
But wait! Why open my anti-alias LPF? Lets try keeping the LPF at 50 Hz, and sampling at 200
samples/sec. But then, my noise occupies only of the sampling bandwidth: it occupies only 50 Hz of the
100 Hz Nyquist band. Hence, the noise is not white, which means adjacent noise samples are correlated!
Hence, when I average adjacent samples, the noise variance does not decrease by a factor of 2. The factor
of 2 gain only occurs with uncorrelated noise. In the end, oversampling buys me nothing.
Filters TBS??
FIR vs. IIR. Because the data set can be any size, and arbitrarily large:
The transfer function of an FIR or IIR is continuous.
Consider some filter. We must carefully distinguish between the filter in general, which can be applied
to any data set (with any n), and the filter as applied to one particular data set. Any given data set has only
discrete frequencies; if we apply the filter to the data set, the data sets frequencies will be multiplied by the
filters transfer function at those frequencies. But we can apply any size data set to the filter, with
frequency components, f = k/n, anywhere in the Nyquist interval. For every data set, the filter has a transfer
function at all its frequencies. Therefore, the filter in general has a continuous transfer function.
H(f )
f
0.5
H(f )
f
0.5
H(f )
f
0.5
Data sets with different n sample the transfer function H(f) at different points. H(f), in
general, is a continuous curve, defined at all points in the Nyquist interval f e [0, 1) or (-0.5,
+0.5].
What Happens to a Sine Wave Deferred?
... Maybe it just sags, like a heavy load. Or does it explode? [Sincere apologies to Langston
Hughes.] You many ask, The DFT has only a discrete set of frequencies. Can I use a DFT to estimate the
frequency of an unknown signal? What happens if I sample a sinusoid of a frequency in between the
allowed DFT frequencies? What is its spectrum? Good questions. We now demonstrate.
We choose n = 40 samples, which means the allowed frequencies are k(1/n), k = 19, ... 0, ... 20,
measured in cycles per sample (or equivalently, in units of the sampling rate, f
samp
). The frequency spacing
is 1/n = 0.025 cycle/sample. No other frequencies exist in the DFT (we say no others are allowed). First,
we show an allowed-frequency sinusoid of f = 10/n = 0.25 cycle/sample (k = 10). Since the signal is real,
the spectrum is conjugate symmetric (S
k
= S
k
*
); therefore, I show only the positive frequencies, and double
their magnitudes:
2
cos , cycle/sample
j
k k
s j f
n n
t | |
= =
|
\ .
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(Left) A sampled sinusoid of f = 0.25, n = 40. (Right) As expected, its magnitude spectrum
(DFT) has exactly one component at f = 0.25.
[Aside: Notice that when the sample points are connected by straight lines, the sinusoid doesnt look
sinusoidal, but recall that connecting with straight lines is not the proper way to interpolate between samples.]
Now we go off-frequency by half the frequency spacing: f = 10.5/n = 0.2625 cycle/sample, halfway
between two allowed frequencies:
(Left) A sampled sinusoid of f = 0.2625, n = 40. (Right) Its magnitude spectrum (DFT) has
components everywhere, but is peaked around f = 0.2625.
Not too surprisingly, the components are peaked at the allowed frequencies closest to the sinusoid
frequency, but there are also components at all other frequencies. This is the artifact of sampling a pure
sinusoid of a non-allowed frequency for a finite time.
Finally, instead of being half-way between allowed frequencies, suppose were only 0.2 of the way, f =
10.2/n = 0.255 cycle/sample:
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(Left) A sampled sinusoid of f = 0.255, n = 40. (Right) Its magnitude spectrum (DFT) has
components everywhere, is asymetric, and peaked at f = 0.25.
These examples show that a DFT, with its fixed frequencies, can give only a rough estimate of an
unknown sinusoids frequency. The estimate gets worse if the unknown signal is not exactly a sinusoid,
because that means it has an even smaller spectral peak, with more components spread around the
spectrum.
Other methods exist for estimating the frequency of an unknown signal, even one that is non-uniformly
sampled in time. If the signal is fairly sinusoidal, one can correlate with a sinusoidal basis frequency, and
numerically search for the frequency with maximum correlation. This avoids the discrete-frequency
limitation of a DFT. Other methods usually require many periods of data, e.g. [Leahy, Ap J, 1983, vol 266,
p160].
Nonuniform Sampling and Arbitrary Basis Functions
So far, we have used a signal sampled uniformly in time. We now show that one can find a Fourier
transform of a signal with any set of n samples, uniform or not. This has many applications: some
experiments (such as lunar laser ranging) cannot sample the signal uniformly for practical, economic, or
political reasons. Magnetic Resonance Imaging (MRI) often uses non-uniform sampling to reduce imaging
time, which can be an hour or more for a patient.
We write the required transform as a set of simultaneous equations, with t
j
as the arbitrary sample
times, and keeping (for now) the uniformly spaced frequencies:
( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( )
1
0 0
0
1
1 1
0
1
1 1
0
0 0 0 1 0 1 0
1 0 1 1 1 1 1
1 0 1
( ) exp 2 /
( ) exp 2 /
...
( ) exp 2 /
( ) exp 2 exp 2 ... exp 2
( ) exp 2 exp 2 ... exp 2
: : : :
( ) exp 2 exp
n
k
k
n
k
k
n
n k n
k
n
n
n n
s t S i k n t
s t S i k n t OR
s t S i k n t
s t f t f t f t
s t f t f t f t
s t f t
t
t
t
t t t
t t t
t
=
=
=
(
(
(
=
(
(
( ) ( )
0
1
1 1 1 1 1
:
2 ... exp 2
n n n n
S
S
S f t f t t t
( (
( (
( (
( (
( (
(
How can we find the required coefficients, S
k
?
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The exponential functions are no longer orthogonal over the sample times;
they are only orthogonal over uniformly spaced samples.
Nonetheless, we have n unknowns (S
0
, ... S
n1
), and n equations. So long as the basis functions are
linearly independent over the sample times, we can (in principle) solve for the needed coefficients, S
k
. We
have now greatly expanded our ability to decompose arbitrary samples into basis functions:
We can decompose a signal over any set of sample times into any set of linearly independent
(not necessarily orthogonal) basis functions.
Note that Parsevals theorem does not apply to the coefficients, since the basis functions (evaluated at
the non-uniform sample points) are no longer orthogonal. Also, S
0
is no longer the average of the signal
values, since the sinusoids may have nonzero average over the sample points.
There is one more subtlety: what is the fundamental frequency f
0
? Equivalently, what is the signal
period? The two are related, because f
0
= 1/period. There is no unique answer to this. However, since a
finite signal transforms as if it is periodic, the period cannot be (t
n1
t
0
), since the first and last samples
would then have to be identical. The period must be longer than that. A convenient choice is to simply
mimic what happens when the samples are uniform. In that case,
( )
1 0 0
, 1/
1
n
n
period t t f period
n
= =
This choice for period reproduces the traditional DFT when the samples are uniform, and is usually
adequate for non-uniform samples, as well.
Example: DFT of a real, non-uniformly sampled sequence: We can set up the matrix equation to
be solved by recalling the frequency layout for even and odd n, and applying the above. We set t
0
= 0, and
define n/2 as floor(n/2). For illustration of the last two columns, we take n odd:
( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( ) ( )
/ 2
0 0 0
0
/ 2
1 1 1
0
/ 2
1 1 1
0
0
1 1 1
1
1
( ) cos sin 2 /
( ) cos sin
:
( ) cos sin
1.0 1.0 0.0 ... 1.0 0.0
( )
1.0 cos sin ... cos / 2 sin
( )
:
( )
n
k
k
n
k
k
n
n k n n
k
n
s t S k t k t where n
s t S k t k t OR
s t S kt k t
s t
t t n t
s t
s t
e e e t
e e
e
e e e
=
=
=
( = +
= + (
( = +
(
(
(
=
(
(
( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
0
1
1
1 1 1 1 1
/ 2
: : : : : :
1.0 cos sin ... cos / 2 sin / 2
n n n n n
S
n t
S
S t t n t n t
e
e e e e
(
(
(
(
(
(
(
(
(
(
(
This gives us the sine and cosine components separately. For n even, the highest frequency component
is k = n/2, or = 2k/n = 2(1/2) = rad/sample, and the final column of sin() is not present.
Note that this is not a fit; it is an exact, reversible transformation. The matrix is the set of all the basis
functions (across each row), evaluated at the sample points (down each column). The matrix has no
summations in it, and depends on the sample points, but not on the sample values.
Example: basis functions as powers of x: In the continuous world, a Taylor series is a
decomposition of a function into powers of (x a), which are a set of linearly independent (but not
orthogonal) basis functions. Despite this lack of orthogonality, Taylor devised a clever way to evaluate the
basis-function coefficients without solving simultaneous equations.
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Example: sampled standard basis functions: We could choose a standard (continuous)
mathematical basis set, such as Bessel functions, J
n
(t). For n sample points, t
1
, ... t
n
, the Bessel functions
are linearly independent, and we can solve for the coefficients, A
k
. We need a scale factor for the time
(equivalent to 2k/n in the Fourier transform). For example, we might use = the (n 1)
th
zero of J
n1
(t).
Then:
1
0 0
0 1
1
1 1
0 1
1
1 1
0 1
( )
( )
...
( )
n
k k
k n
n
k k
k n
n
n k k n
k n
s t A J t
t
s t A J t
t
s t A J t
t
o
o
o
=
| |
=
|
\ .
| |
=
|
\ .
| |
=
|
\ .
We have n equations and n unknowns, A
0
, ... A
n1
, so we can solve for the A
k
.
Old fashioned FFT implementations required you to have N = a power of 2 number of samples (64,
1024, etc.). Modern FFT implementations are general to any number of samples, and use the prime
decomposition of N to provide the fastest and most accurate DFT known. The worst case is when N is
prime, and no FFT optimization is possible: the DFT is evaluated directly from the defining summations.
But with modern computers, this is so fast that we dont care.
In the old days, if people had a non-power-of-2 number of data points, they used to pad their data,
typically (and horribly) by just adding zeros to the end until they reached a power of 2. This introduced
artifacts into the spectrum, which often obscured or destroyed the very information they sought [Ham p??].
With a modern FFT implementation, there is no need for it, anyway.
If for some reason, you absolutely must constrain N to some preferred values, it is much better to throw
away some data points than to add fake ones.
Two Dimensional Fourier Transforms
One dimensional Fourier transforms often have time or space as the independent variable. Two
dimensional transforms almost always have space, say (x, y), as the independent variables. The most
common 2D transform is of pictures.
In the continuous world of light, lenses can physically project a Fourier transform of an image based on
optics, with no computations. This allows for filtering the image with opaque masks, and re-transforming back to
the original-but-filtered image, all at the speed of light with no computer. But digitized images store the image as
pixels, each with some light intensity. These are computationally processed by computer.
Basis Functions
TBS. Not sines and cosines, or products of sines and cosines. Products of complex exponentials.
Wave fronts at various angles, discrete k
x
and k
y
.
Note on Continuous Fourier Series and Uniform Convergence
The continuous Fourier Series is defined for a periodic signal s(t) over a continuous range of times,
t e [0, T):
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0
2
0
0
( ) , is the frequency of the component
is the complex frequency component
i k th
k
k
k
s t S e where k k
S
t e
e
=
=
Note that the time interval is continuous, but the frequency components are discrete. In general,
periodic signals lead to discrete frequency components.
The continuous Fourier Series is not always uniformly convergent.
Therefore, the order of integrations and summations cannot always be interchanged.
Non-uniform convergence is illustrated by the famous Gibbs phenomenon: when we transform a
square wave to the frequency domain (aka Fourier space), and retain only a finite number of frequency
components to then transform back to the time domain, the square wave comes back with overshoot:
wiggles that are large near the discontinuities:
t
overshoot
t
original square
wave
reconstructed
wave
Gibbs phenomenon: (Left) After losing high frequencies, reconstructed square wave has
overshoot and wiggles. (Right) Retaining more frequencies reduces wiggle time, but not
amplitude.
As we include more and more frequency components, the wiggles get narrower, but do not get lower in
amplitude. This means that there are always some time points for which the inverse transform does not
converge to the original square wave. Such wiggles are commonly observed in many electronic systems,
which must necessarily filter out high frequency components above some cut-off frequency.
However:
Continuous signals have Fourier Series that converge uniformly. This applies to most physical
phenomena, so interchanging integration and summation is valid [F&W p217+].
This is true even if the derivative of the signal is discontinuous.
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Tensors, Without the Tension
Approach
Well present tensors as follows:
1. Two physical examples: magnetic susceptibility, and deformable solids
2. A non-example: when is a matrix not a tensor?
3. Forward looking definitions (dont get stuck on these)
4. Review of vector spaces and notation (dont get stuck on this, either)
5. A short, but at first unhelpful, definition (really, really dont get stuck on this)
6. A discussion which clarifies the above definition
7. Examples, including dot-products and cross-products as tensors
8. Higher rank tensors
9. Change of basis
10. Non-orthonormal systems: contravariance and covariance
11. Indefinite metrics of Special and General Relativity
12. Mixed basis linear functions (transformation matrices, the Pauli vector)
Tensors are all about vectors. They let you do things with vectors you never thought possible. We
define tensors in terms of what they do (their linearity properties), and then show that linearity implies the
transformation properties. This gets most directly to the true importance of tensors. [Most references
define tensors in terms of transformations, but then fail to point out the all-important linearity properties.]
We also take a geometric approach, treating vectors and tensors as geometric objects that exist
independently of their representation in any basis. Inevitably, though, there is a fair amount of unavoidable
algebra.
Later, we introduce contravariance and covariance in terms of non-orthonormal coordinates, but first
with a familiar positive-definite metric from classical mechanics. This makes for a more intuitive
understanding of contra- and co-variance, before applying the concept to the more bizarre indefinite metrics
of special and general relativity.
There is deliberate repetition of several points, because it usually takes me more than once to grok
something. So I repeat:
If you dont understand something, read it again once, then keep reading. Dont get stuck on one
thing. Often, the following discussion will clarify an ambiguity.
Two Physical Examples
We start with two physical examples: magnetic susceptibility, and deformation of a solid. We start
with matrix notation, because we assume it is familiar to you. Later we will see that matrix notation is not
ideal for tensor algebra.
Magnetic Susceptibility
We assume you are familiar with susceptibility of magnetic materials: when placed in an H-field,
magnetizable (susceptible) materials acquire a magnetization, which adds to the resulting B-field. In
simple cases, the susceptibility is a scalar, and
is the magnetization,
is the susceptibility, and
is the applied magnetic field
where _
_
= M H M
H
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The susceptibility in this simple case is the same in any direction; i.e., the material is isotropic.
However, there exist materials which are more magnetizable in some directions than others. E.g.,
imagine a cubic lattice of axially-symmetric molecules which are more magnetizable along the molecular
axis than perpendicular to it:
x
y
z
H
M
H M
H M
xx
= 2
yy
= 1
zz
= 1
x
y
z
x
y
z
more magnetizable
l
e
s
s
m
a
g
n
e
t
i
z
a
b
l
e
Magnetization, M, as a function of external field, H, for a material with a tensor-valued
susceptibility, .
In each direction, the magnetization is proportional to the applied field, but is larger in the x-direction
than y or z. In this example, for an arbitrary H-field, we have
( ) ( )
2 0 0
, , 2 , , 0 1 0
0 0 1
x y z x y z
ij
M M M H H H or
_
| |
|
= = = =
|
|
\ .
M M H H
_
Note that in general, M is not parallel to H (below, dropping the z axis for now):
H
M= (2H
x
, H
y
)
x
y
M need not be parallel to H for a material with a tensor-valued .
But M is a linear function of H, which means:
( ) ( ) ( )
1 2 1 2
k k + = + M H H M H M H
.
This linearity is reflected in the fact that matrix multiplication is linear:
( ) ( ) ( ) ( )
1 2 1 2 1 2 1 2
2 0 0 2 0 0 2 0 0
0 1 0 0 1 0 0 1 0
0 0 1 0 0 1 0 0 1
k k k k
| | | | | |
| | |
+ = + = + = +
| | |
| | |
\ . \ . \ .
M H H H H H H M H M H
The matrix notation might seem like overkill, since is diagonal, but it is only diagonal in this basis of
x, y, and z. Well see in a moment what happens when we change basis. First, let us understand what the
matrix
ij
really means. Recall the visualization of pre-multiplying a vector by a matrix: a matrix times a
column vector H, is a weighted sum of the columns of :
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xx x xx
yx x y
xz xz
yz z
xy xy
yy y y yy
zy zy
yz
zz z zz
x
zx zx
H
H H
H
H H
_ _
_ _
_ _
_ _
_ _
_
_ _
_
_ _
_
_
( (
(
(
(
( (
(
( (
= + +
(
(
(
(
(
( (
(
( (
H
We can think of the matrix as a set of 3 column vectors: the first is the magnetization vector for H =
e
x
; the 2
nd
column is M for H = e
y
; the 3
rd
column is M for H = e
z
. Since magnetization is linear in H, the
magnetization for any H can be written as the weighted sum of the magnetizations for each of the basis
vectors:
( ) ( ) ( ) ( ) , , are the unit vectors in , ,
x x y y z z x y z
H H H where x y z = + + M H M e M e M e e e e
This is just the matrix multiplication above: = M H . (Were writing all indexes as subscripts for
now; later on well see that M, , and H should be indexed as M
i
,
i
j
, and H
i
.)
Now lets change bases from e
x
, e
y
, e
z
, to some e
1
, e
2
, e
3
, defined below. We use a simple
transformation, but the 1-2-3 basis is not orthonormal:
e
y
x
y
z
e
z
e
x
e
2
1
2
3
e
3
e
1
ae
1
be
2
e
x
e
y
ce
1
de
2
old basis new basis
x
y
z
Transformation to a non-orthogonal, non-normal basis. e
1
and e
2
are in the x-y plane, but are
neither orthogonal nor normal. For simplicity, we choose e
3
= e
z
. Here, b and c are negative.
To find the transformation equations to the new basis, we first write the old basis vectors in the new
basis. Weve chosen for simplicity a transformation in the x-y plane, with the z-axis unchanged:
1 2 1 2 3 x y z
a b c d = + = + = e e e e e e e e
Now write a vector, v, in the old basis, and substitute out the old basis vectors for the new basis. We
see that the new components are a linear combination of the old components:
( ) ( )
( ) ( )
1 2 1 2 3
1 2 3 1 1 2 2 3 3
1 2 3
, ,
y
x x y y z z x y z
x
x y x y z
x y x y z
v v v v a b v c d v
av cv bv dv v v v v
v av cv v bv dv v v
= + + = + + + +
= + + + + = + +
= + = + =
v e e e e e e e e
e
e
e e e e e e
_ _
Recall that matrix multiplication is defined to be the operation of linear transformation, so we can
write this basis transformation in matrix form:
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1
2
3
0 0 0
0
0
1 1 0
0
0
x
x
z
z
y y
y
z
x
a v a
b v
c c
d v v d
v
b
v v
v v
( | | ( ( ( (
| ( ( ( ( (
= = + +
| ( ( ( ( (
|
( ( ( ( (
\ .
e e e
The columns of the transformation matrix are the old basis vectors written in the new basis.
This is illustrated explicitly on the right hand side, which is just
x x y y z z
v v v + + e e e .
Finally, we look at how the susceptibility matrix
ij
transforms to the new basis. We saw above that
the columns of are the M vectors for H = each of the basis vectors. So right away, we must transform
each column of with the transformation matrix above, to convert it to the new basis. Since matrix
multiplication AB is distributive across the columns of B, we can write the transformation of all 3 columns
in a single expression by pre-multiplying with the above transformation matrix:
0 0 2 0 0 2 0
1 0 0 0 1 0 2 0
0 0 1 0 0 1 0 0 1 0 0 1
new
a c a c a c
Step of in newbasis b d b d b d
| | | | | | | |
| | | |
= = =
| | | |
| | | |
\ . \ . \ . \ .
But were not done. This first step expressed the column vectors in the new basis, but the columns of
the RHS (right hand side) are still the Ms for basis vectors e
x
, e
y
, e
z
. Instead, we need the columns of
new
to be the M vectors for e
1
, e
2
, e
3
. Please dont get bogged down yet in the details, but we do this
transformation similarly to how we transformed the column vectors. We transform the contributions to M
due to e
x
, e
y
, e
z
to that due to e
1
by writing e
1
in terms of e
x
, e
y
, e
z
:
( ) ( ) ( )
1 1 x y x y
e f e f = + = = = + = e e e M H e M H e M H e
Similarly,
( ) ( ) ( )
( ) ( )
2 2
3 3
x y x y
z z
g h g h = + = = = + =
= = = =
e e e M H e M H e M H e
e e M H e M H e
Essentially, we need to transform among the columns, i.e. transform the rows of . These two
transformations (once of the columns, and once of the rows) is the essence of a rank-2 tensor:
A tensor matrix (rank-2 tensor) has columns that are vectors, and simultaneously, its rows are also
vectors. Therefore, transforming to a new basis requires two transformations:
once for the rows, and once for the columns (in either order).
[Aside: The details (which you can skip at first): We just showed that we transform using the inverse of our
previous transformation. The reason for the inverse is related to the up/down indexes mentioned earlier; please be
patient. In matrix notation, we write the row transformation as post-multiplying by the transpose of the needed
transformation:
0 2 0 0 0 0 2 0 0 0
0 0 1 0 0 0 0 1 0 0
0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1
T
new
a c e f a c e g
Final b d g h b d f h
| | | | | | | || || |
| | | | | |
= =
| | | | | |
| | | | | |
\ . \ . \ . \ .\ .\ .
]
[Another aside: A direction-dependent susceptibility requires to be promoted from a scalar to a rank-2
tensor (skipping any rank-1 tensor). This is necessary because a rank-0 tensor (a scalar) and a rank-2 tensor can
both act on a vector (H) to produce a vector (M). There is no sense to a rank-1 (vector) susceptibility, because
there is no simple way a rank-1 tensor (a vector) can act on another vector H to produce an output vector M.
More on this later.]
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Mechanical Strain
A second example of a tensor is the mechanical strain tensor. When I push on a deformable material,
it deforms. A simple model is just a spring, with Hookes law:
1
applied
x F
k
A = +
We write the formula with a plus sign, because (unlike freshman physics spring questions) we are
interested in how a body deforms when we apply a force to it. For an isotropic material, we can push in
any direction, and the deformation is parallel to the force. This makes the above equation a vector
equation:
1
the strain constant s where s
k
A = = x F
Strain is defined as the displacement of a given point under force. [Stress is the force per unit area
applied to a body. Stress produces strain.] In an isotropic material, the stress constant is a simple scalar.
Note that if we transform to another basis for our vectors, the stress constant is unchanged. Thats the
definition of a scalar:
A scalar is a number that is the same in any coordinate system. A scalar is a rank-0 tensor.
The scalar is unchanged even in a non-ortho-normal coordinate system.
But what if our material is a bunch of microscopic blobs connected by stiff rods, like atoms in a
crystal?
x
F
x
F
(Left) A constrained deformation crystal structure. (Middle) The deformation vector is not
parallel to the force. (Right) More extreme geometries lead to a larger angle between the force
and displacement.
The diagram shows a 2D example: pushing in the x-direction results in both x and y displacements.
The same principle could result in a 3D x, with some component into the page. For small deformations,
the deformation is linear with the force: pushing twice as hard results in twice the displacement. Pushing
with the sum of two (not necessarily parallel) forces results in the sum of the individual displacements. But
the displacement is not proportional to the force (because the displacement is not parallel to it). In fact,
each component of force results in a deformation vector. Mathematically:
xz x xy xy xx xx
y
x
x yx yx
z
z
z yz yz
zz
yy yy y
z zz z zx y y x z
s s s
F s s F
s s
s
F s s
s s
s s F
F
F s s
s s
( | |
( ( (
| (
( ( (
A = + + = =
| (
( ( (
| (
( ( (
\ .
x sF
s
_
Much like the anisotropy of the magnetization in the previous example, the anisotropy of the strain
requires us to use a rank-2 tensor to describe it. The linearity of the strain with force allows us to write the
strain tensor as a matrix. Linearity also guarantees that we can change to another basis using a method
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similar to that shown above for the susceptibility tensor. Specifically, we must transform both the columns
and the rows of the strain tensor s. Furthermore, the linearity of deformation with force also insures that we
can use non-orthonormal bases, just as well as orthonormal ones.
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When Is a Matrix Not a Tensor?
I would say that most matrices are not tensors. A matrix is a tensor when its rows and columns are
both vectors. This implies that there is a vector space, basis vectors, and the possibility of changing basis.
As a counter example, consider the following graduate physics problem:
Two pencils, an eraser, and a ruler cost \$2.20. Four pencils, two erasers, and a ruler cost \$3.45. Four
pencils, an eraser, and two rulers cost \$3.85. How much does each item cost?
We can write this as simultaneous equations, and as shorthand in matrix notation:
2 220
2 1 1 220
4 2 345 4 2 1 345
4 1 1 385
4 2 385
p e r
p
p e r or e
r
p e r
+ + =
| | ( (
| ( (
+ + = =
| ( (
|
( (
+ + = \ .
It is possible to use a matrix for this problem because the problem takes linear combinations of the
costs of 3 items. Matrix multiplication is defined as the process of linear combinations, which is the same
process as linear transformations. However, the above matrix is not a tensor, because there are no vectors
of school supplies, no bases, and no linear combinations of (say) part eraser and part pencil. Therefore, the
matrix has no well-defined transformation properties. Hence, it is a lowly matrix, but no tensor.
However, later (in We Dont Need No Stinking Metric) well see that under the right conditions, we
can form a vector space out of seemingly unrelated quantities.
An ordinary vector associates a number with each direction of space:
x y z
v v v = + + v x y z
The vector v associates the number v
x
with the x-direction; it associates the number v
y
with the y-
direction, and the number v
z
with the z-direction.
The above tensor examples illustrate the basic nature of a rank-2 tensor: it associates a vector with
each direction of space:
xx xy xz
yx yy yz
zx zy zz
T T T
T T T
T T T
(
( (
(
( (
= + +
(
( (
(
( (
T x y z
Some Definitions and Review
These definitions will make more sense as we go along. Dont get stuck on these:
ordinary vector = contravariant vector = contravector = (
1
0
) tensor
1-form = covariant vector = covector = (
0
1
) tensor. (Yes, there are 4 different ways to say the same thing.)
covariant the same. E.g., General Relativity says that the mathematical form of the laws of physics
are covariant (i.e., the same) with respect to arbitrary coordinate transformations.
This is a completely different meaning of covariant than the one above.
rank The number of indexes of a tensor; T
ij
is a rank-2 tensor; R
i
jkl
is a rank-4 tensor. Rank is
unrelated to the dimension of the vector space in which the tensor operates.
MVE mathematical vector element. Think of it as a vector for now.
Caution: a rank (
0
1
) tensor is a 1-form, but a rank (
0
2
) tensor is not always a 2-form. [Dont worry
about it, but just for completeness, a 2-form (or any n-form) has to be fully anti-symmetric in all pairs of
vector arguments.]
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Notation:
(a, b, c) is a row vector; (a, b, c)
T
is a column vector (the transpose of a row vector).
To satisfy our pathetic word processor, we write (
m
n
), even though the m is supposed to be directly
above the n.
T is a tensor, without reference to any basis or representation.
T
ij
is the matrix of components of T, contravariant in both indexes, with an understood basis.
T(v, w) is the result of T acting on v and w.
or v v
,
are two equivalent ways to denote a vector, without reference to any basis or
representation. Note that a vector is a rank-1 tensor.
~ or a a are two equivalent ways to denote a covariant vector (aka 1-form), without reference to
any basis or representation
a
i
the components of the covecter (1-form) a, in an understood basis.
Vector Space Summary
Briefly, a vector space comprises a field of scalars, a group of vectors, and the operation of scalar
multiplication of vectors (details below). Quantum mechanical vector spaces have two additional
characteristics: they define a dot-product between two vectors, and they define linear operators which act
on vectors to produce other vectors.
Before understanding tensors, it is very helpful, if not downright necessary, to understand vector
spaces. Funky Quantum Concepts has a more complete description of vector spaces. Here is a very brief
summary: a vector space comprises a field of scalars, a group of vectors, and the operation of scalar
multiplication of vectors. The scalars can be any mathematical field, but are usually the real numbers, or
the complex numbers (e.g., quantum mechanics). For a given vector space, the vectors are a class of
things, which can be one of many possibilities (physical vectors, matrices, kets, bras, tensors, ...). In
particular, the vectors are not necessarily lists of scalars, nor need they have anything to do with physical
space. Vector spaces have the following properties, which allow solving simultaneous linear equations
both for unknown scalars, and unknown vectors:
Scalars Mathematical Vectors
Scalars form a commutative group
(closure, unique identity, inverses) under
operation +.
Vectors form a commutative group
(closure, unique identity, inverses)
under operation +.
Scalars, excluding 0, form a
commutative group under operation ( ).
Distributive property of ( ) over +.
Scalar multiplication of vector produces another vector.
Distributive property of scalar multiplication over both scalar + and vector +.
With just the scalars, you can solve ordinary scalar linear equations such as:
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
...
...
: : :
...
n n
n n
n n nn n n
a x a x a x c
a x a x a x c
written in matrix form as
a x a x a x c
+ + =
+ + =
=
`
+ + =
)
ax c
All the usual methods of linear algebra work to solve the above equations: Cramers rule, Gaussian
elimination, etc. With the whole vector space, you can solve simultaneous linear vector equations for
unknown vectors, such as
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11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
...
...
: : :
...
n n
n n
n n nn n n
a a a
a a a
written in matrix form as
a a a
+ + =
+ + =
=
`
+ + =
)
v v v w
v v v w
av w
v v v w
where a is again a matrix of scalars. The same methods of linear algebra work just as well to solve
vector equations as scalar equations.
Vector spaces may also have these properties:
Dot-product produces a scalar from two vectors.
Linear operators act on vectors to produce other
vectors.
The key points of mathematical vectors are (1) we can form linear combinations of them to make
other vectors, and (2) any vector can be written as a linear combination of basis vectors:
v = (v
1
, v
2
, v
3
) = v
1
e
1
+ v
2
e
2
+ v
3
e
3
where e
1
, e
2
, e
3
are basis vectors, and
v
1
, v
2
, v
3
are the components of v in the e
1
, e
2
, e
3
basis.
Note that v
1
, v
2
, v
3
are numbers, while e
1
, e
2
, e
3
are vectors. There is a (kind of bogus) reason why
basis vectors are written with subscripts, and vector components with superscripts, but well get to that
later.
The dimension of a vector space, N, is the number of basis vectors needed to construct every vector in
the space.
Do not confuse the dimension of physical space (typically 1D, 2D, 3D, or (in relativity) 4D), with
the dimension of the mathematical objects used to work a problem.
For example, a 33 matrix is an element of the vector space of 33 matrices. This is a 9-D vector
space, because there are 9 basis matrices needed to construct an arbitrary matrix.
Given a basis, components are equivalent to the vector. Components alone (without a basis) are
insufficient to be a vector.
[Aside: Note that for position vectors defined by r = (r, , |), r, , and | are not the components of
a vector. The tip off is that with two vectors, you can always add their components to get another
vector. Clearly, ( )
1 2 1 2 1 2 1 2
, , r r u u | | + = + + + r r , so (r, , |) cannot be the components of a vector.
This failure to add is due to r being a displacement vector from the origin, where there is no consistent
basis: e.g., what is e
r
at the origin? At points off the origin, there is a consistent basis: e
r
, e
, and e
|
are
well-defined.]
When Vectors Collide
There now arises a collision of terminology: to a physicist, vector usually means a physical vector in
3- or 4-space, but to a mathematician, vector means an element of a mathematical vector-space. These
are two different meanings, but they share a common aspect: linearity (i.e., we can form linear
combinations of vectors to make other vectors, and any vector can be written as a linear combination of
basis vectors). Because of that linearity, we can have general rank-n tensors whose components are
arbitrary elements of a mathematical vector-space. To make the terminology confusion worse, an (
m
n
)
tensor whose components are simple numbers is itself a vector-element of the vector-space of (
m
n
)
tensors.
Mathematical vector-elements of a vector space are much more general than physical vectors (e.g.
force, or velocity), though physical vectors and tensors are elements of mathematical vector spaces. To be
clear, well use MVE to refer to a mathematical vector-element of a vector space, and vector to mean a
normal physics vector (3-vector or 4-vector). Recall that MVEs are usually written as a set of components
in some basis, just like vectors are. In the beginning, we choose all the input MVEs to be vectors.
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If youre unclear about what an MVE is, just think of it as a physical vector for now, like force.
Tensors vs. Symbols
There are lots of tensors: metric tensors, electromagnetic tensors, Riemann tensors, etc. There are also
symbols: Levi-Civita symbols, Christoffel symbols, etc. Whats the difference? Symbols arent
tensors. Symbols look like tensors, in that they have components indexed by multiple indices, they are
referred to basis vectors, and are summed with tensors. But they are defined to have specific components,
which may depend on the basis, and therefore symbols dont change basis (transform) the way tensors do.
Hence, symbols are not geometric entities, with a meaning in a manifold, independent of coordinates. For
example, the Levi-Civita symbol is defined to have specific constant components in all bases. It doesnt
follow the usual change-of-basis rules. Therefore, it cannot be a tensor.
Notational Nightmare
If you come from a differential geometry background, you may wonder about some insanely confusing
notation. It is a fact that dx and dx are two different things:
( , , ) is a ,
( ) is a -
d dx dy dz but
x x
=
= V
x vector
d r 1 form
We dont use the second notation (or exterior derivatives) in this chapter, but we might in the
Differential Geometry chapter.
Tensors? What Good Are They?
A Short, Complicated Definition
It is very difficult to give a short definition of a tensor that is useful to anyone who doesnt already
know what a tensor is. Nonetheless, youve got to start somewhere, so well give a short definition, to
point in the right direction, but it may not make complete sense at first (dont get hung up on this, skip if
needed):
A tensor is an operator on one or more mathematical vector elements (MVEs), linear in each
operand, which produces another mathematical vector element.
The key point is this (which we describe in more detail in a moment):
Linearity in all the operands is the essence of a tensor.
I should add that the basis vectors for all the MVEs must be the same (or tensor products of the same)
for an operator to qualify as a tensor. But thats too much to put in a short definition. We clarify this
point later.
Note that a scalar (i.e., a coordinate-system-invariant number, but for now, just a number) satisfies the
definition of a mathematical vector element.
Many definitions of tensors dwell on the transformation properties of tensors. This is mathematically
valid, but such definitions give no insight into the use of tensors, or why we like them. Note that to satisfy
the transformation properties, all the input vectors and output tensors must be expressed in the same basis
(or tensor products of that basis with itself).
Some coordinate systems require distinguishing between contravariant and covariant components of
tensors; superscripts denote contravariant components; subscripts denote covariant components. However,
orthonormal positive definite systems, such as the familiar Cartesian, spherical, and cylindrical systems, do
not require such a distinction. So for now, lets ignore the distinction, even though the following notation
properly represents both contravariant and covariant components. Thus, in the following text, contravariant
components are written with superscripts, and covariant components are written with subscripts, but we
dont care right now. Just think of them all as components in an arbitrary coordinate system.
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Building a Tensor
Oversimplified, a tensor operates on vectors to produce a scalar or a vector. Lets construct a tensor
which accepts (operates on) two 3-vectors to produce a scalar. (Well see later that this is a rank-2 tensor.)
Let the tensor T act on vectors a and b to produce a scalar, s; in other words, this tensor is a scalar function
of two vectors:
s = T(a, b)
Call the first vector a = (a
1
, a
2
, a
3
) in some basis, and the second vector b = (b
1
, b
2
, b
3
) (in the same
basis). A tensor, by definition, must be linear in both a and b; if we double a, we double the result, if we
triple b, we triple the result, etc. Also,
T(a + c, b) = T(a, b) + T(c, b), and T(a, b + d) = T(a, b) + T(a, d).
So the result must involve at least the product of a component of a with a component of b. Lets say
the tensor takes a
2
b
1
as that product, and additionally multiplies it by a constant, T
21
. Then we have built a
tensor acting on a and b, and it is linear in both:
2 1 2 1
21
( , ) . : ( , ) 7 T a b Example a b = = T a b T a b
But, if we add to this some other weighted product of some other pair of components, the result is still
a tensor: it is still linear in both a and b:
1 3 2 1 1 3 2 1
13 21
( , ) . : ( , ) 4 7 T a b T a b Example a b a b = + = + T a b T a b
In fact, we can extend this to the weighted sum of all combinations of components, one each from a
and b. Such a sum is still linear in both a and b:
3 3
1 1
2 6 4
( , ) : 7 5 1
6 0 8
i j
ij ij
i j
T a b Example T
= =
(
(
= =
(
(
T a b
Further, nothing else can be added to this that is linear in a and b.
A tensor is the most general linear function of a and b that exists, i.e. any linear function of a and b
can be written as a 33 matrix.
(Well see that the rank of a tensor is equal to the number of its indices; T is a rank-2 tensor.) The T
ij
are the components of the tensor (in the basis of the vectors a and b.) At this point, we consider the
components of T, a, and b all as just numbers.
Why does a tensor have a separate weight for each combination of components, one from each input
mathematical vector element (MVE)? Couldnt we just weight each input MVE as a whole? No, because
that would restrict tensors to only some linear functions of the inputs.
Any linear function of the input vectors can be represented as a tensor.
Note that tensors, just like vectors, can be written as components in some basis. And just like vectors,
we can transform the components from one basis to another. Such a transformation does not change the
tensor itself (nor does it change a vector); it simply changes how we represent the tensor (or vector). More
on transformations later.
Tensors dont have to produce scalar results!
Some tensors accept one or more vectors, and produce a vector for a result. Or they produce some
rank-r tensor for a result. In general, a rank-n tensor accepts m vectors as inputs, and produces a rank n
m tensor as a result. Since any tensor is an element of a mathematical vector space, tensors can be written
as linear combinations of other (same rank & type) tensors. So even when a tensor produces another (lower
rank) tensor as an output, the tensor is still a linear function of all its input vectors. Its just a tensor-valued
function, instead of a scalar-valued function. For example, the force on a charge: a B-field operates on a
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vector, qv, to produce a vector, f. Thus, we can think of the B-field as a rank-2 tensor which acts on a
vector to produce a vector; its a vector-valued function of one vector.
Also, in general, tensors arent limited to taking just vectors as inputs. Some tensors take rank-2
tensors as inputs. For example, the quadrupole moment tensor operates on the 2
nd
derivative matrix of the
potential (the rank-2 Hessian tensor) to produce the (scalar) work stored in the quadrupole of charges.
And a density matrix in quantum mechanics is a rank-2 tensor that acts on an operator matrix (rank-2
tensor) to produce the ensemble average of that operator.
Tensors in Action
Lets consider how rank-0, rank-1, and rank-2 tensors operate on a single vector. Recall that in
tensor-talk, a scalar is an invariant number, i.e. it is the same number in any coordinate system.
Rank-0: A rank-0 tensor is a scalar, i.e. a coordinate-system-independent number. Multiplying a
vector by a rank-0 tensor (a scalar), produces a new vector. Each component of the vector contributes to
the corresponding component of the result, and each component is weighted equally by the scalar, a:
x y z x y z
v v v v av av av av = + + = + + i j k i j k
, ,
Rank-1: A rank-1 tensor a operates on (contracts with) a vector to produce a scalar. Each component
of the input vector contributes a number to the result, but each component is weighted separately by the
corresponding component of the tensor a:
3
1
( )
x y z i
x y z i
i
a v a v a v a v
=
= + + =
a v
Note that a vector is itself a rank-1 tensor. Above, instead of considering a acting on v, we can
equivalently consider that v acts on a: a(v) = v(a). Both a and v are of equal standing.
Rank-2: Filling one slot of a rank-2 tensor with a vector produces a new vector. Each component of
the input vector contributes a vector to the result, and each input vector component weights a different
vector.
z
x
y
z
x
y
z
x
y
column 1
column 2
column 3
(a) (b) (c)
(a) A hypothetical rank-2 tensor with an x-vector (red), a y-vector (green), and a z-vector
(blue). (b) The tensor acting on the vector (1, 1, 1) producing a vector (heavy black). Each
component (column) vector of the tensor is weighted by 1, and summed. (c) The tensor acting on
the vector (0, 2, 0.5), producing a vector (heavy black). The x-vector is weighted by 0, and so
does not contribute; the y-vector is weighted by 2, so contributes double; the z-vector is weighted
by 0.5, so contributes half.
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3 3 3
1 1 1
( _, )
x x
z z
y z y
z z
z z
x x
y y
y y y y
y y
x x x
x x
y i j
j
x y z x j y j z j
x y z j j j
j j j
x
z
y
x x
z z
x x z
z
y y z
z
B v
v v v B v
B B
B v v B
B B
B v B
B v B B B
B v B
B
B v B
B B
v
B v
= = =
( (
( (
= = = + +
( (
( (
|
( (
( (
(
(
(
(
(
(
(
(
| | | | |
= + + = + +
| | |
\ . \ . \ .
B v
B B B i j k
The columns of B are the vectors which are weighted by each of the input vector components, v
j
; or
equivalently, the columns of B are the vector weights for each of the input vector components
Example of a simple rank-2 tensor: the moment-of-inertia tensor, I
ij
. Every blob of matter has
one. We know from mechanics that if you rotate an arbitrary blob around an arbitrary axis, the angular
momentum vector of the blob does not in general line up with the axis of rotation. So what is the angular
momentum vector of the blob? It is a vector-valued linear function of the angular velocity vector, i.e. given
the angular velocity vector, you can operate on it with the moment-of-inertia tensor, to get the angular
momentum vector. Therefore, by the definition of a tensor as a linear operation on a vector, the
relationship between angular momentum vector and angular velocity vector can be given as a tensor; it is
the moment-of-inertia tensor. It takes as an input the angular velocity vector, and produces as output the
angular momentum vector, therefore it is a rank-2 tensor:
( , _) , ( , ) 2KE = = = I L I L
[Since I is constant in the blob frame, it rotates in the lab frame. Thus, in the lab frame, the above
equations are valid only at a single instant in time. In effect, I is a function of time, I(t).]
[?? This may be a bad example, since I is only a Cartesian tensor [L&L3, p ??], which is not a real tensor.
Real tensors cant have finite displacements on a curved manifold, but blobs of matter have finite size. If you
want to get the kinetic energy, you have to use the metric to compute L. Is there a simple example of a real
rank-2 tensor??]
Note that some rank-2 tensors operate on two vectors to produce a scalar, and some (like I) can either
act on one vector to produce a vector, or act on two vectors to produce a scalar (twice the kinetic energy).
More of that, and higher rank tensors, later.
Tensor Fields
A vector is a single mathematical object, but it is quite common to define a field of vectors. A field in
this sense is a function of space. A vector field defines a vector for each point in a space. For example,
the electric field is a vector-valued function of space: at each point in space, there is an electric field vector.
Similarly, a tensor is a single mathematical object, but it is quite common to define a field of tensors.
At each point in space, there is a tensor. The metric tensor field is a tensor-valued function of space: at
each point, there is a metric tensor. Almost universally, the word field is omitted when calling out tensor
fields: when you say metric tensor, everyone is expected to know it is a tensor field. When you say
moment of inertia tensor, everyone is expected to know it is a single tensor (not a field).
Dot Products and Cross Products as Tensors
Symmetric tensors are associated with elementary dot-products, and anti-symmetric tensors are
associated with elementary cross-products.
A dot product is a linear operation on two vectors: AB = BA, which produces a scalar. Because the
dot-product is a linear function of two vectors, it can be written as a tensor. (Recall that any linear function
of vectors can be written as a tensor.) Since it takes two rank-1 tensors, and produces a rank-0 tensor, the
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dot-product is a rank-2 tensor. Therefore, we can achieve the same result as a dot-product with a rank-2
symmetric tensor that accepts two vectors and produces a scalar; call this tensor g:
g(A, B) = g(B, A)
g is called the metric tensor: it produces the dot-product (aka scalar product) of two vectors. Quite
often, the metric tensor varies as a function of the generalized coordinates of the system; then it is a metric
tensor field. It happens that the dot-product is symmetric: AB = BA.; therefore, g is symmetric. If we
write the components of g as a matrix, the matrix will be symmetric, i.e. it will equal its own transpose.
(Do I need to expand on this??)
On the other hand, a cross product is an anti-symmetric linear operation on two vectors, which
produces another vector: A B = B A. Therefore, we can associate one vector, say B, with a rank-2
anti-symmetric tensor, that accepts one vector and produces another vector:
B( _, A) = B(A, _ )
For example, the Lorentz force law: F = v B. We can write B as a (
1
1
) tensor:
0
( _, ) 0
0
y z
x
z y
z y
x y z i j y x z
j z x z x
z x y
x y z y x
y x
B v B v
v B B
v v v B v B B v B v B v
B B B B B
v B v B v
(
(
(
(
(
(
(
= = = = = + (
(
(
(
(
(
(
i j k
F = v B B v
We see again how a rank-2 tensor, B, contributes a vector for each component of v:
B
i
x
e
i
= B
z
j + B
y
k (the first column of B) is weighted by v
x
.
B
i
y
e
i
= B
z
i B
x
k (the 2
nd
column of B) is weighted by v
y
.
B
i
z
e
i
= B
y
i + B
x
j (the 3
rd
column of B) is weighted by v
z
.
z
x
y
B
i
x
=-B
z
j+B
y
k
B
x
, B
y
, B
z
> 0
z
x
y
B
i
y
=B
z
i-B
x
k
z
x
y
B
i
z
=-B
y
i+B
x
j
A rank-2 tensor acting on a vector to produce their cross-product.
TBS: We can also think of the cross product as a fully anti-symmetric rank-3 tensor, which acts on 2
vectors to produce a vector (their cross product). This is the anti-symmetric symbol c
ijk
(not a tensor).
Note that both the dot-product and cross-product are linear on both of their operands. For example:
( ) ( ) ( )
( ) ( ) ( )
o o
| q | q
+ = +
+ = +
A C B A B C B
A B D A B A D
Linearity in all the operands is the essence of a tensor.
Note also that a rank of a tensor contracts with (is summed over) a rank of one of its operands to
eliminate both of them: one rank of the B-field tensor contracts with one input vector, leaving one
surviving rank of the B-field tensor, which is the vector result. Similarly, one rank of the metric tensor, g,
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contracts with the first operand vector; another rank of g contracts with the second operand vector, leaving
a rank-0 (scalar) result.
The Danger of Matrices
There are some dangers to thinking of tensors as matrices: (1) it doesnt work for rank 3 or higher
tensors, and (2) non-commutation of matrix multiplication is harder to follow than the more-explicit
summation convention. Nonetheless, the matrix conventions are these:
- contravariant components and basis covectors (up indexes) column vector. E.g.,
1
1
2
2
3
3
, basis 1-forms:
v
v
v
(
(
(
(
=
(
(
(
(
e
v e
e
- covariant components and basis contravectors (down indexes) row vector
( ) ( )
1 2 3 1 2 3
, , , basis vectors: , , w w w = w e e e
Matrix rows and columns are indicated by spacing of the indexes, and are independent of their
upness or downness. The first matrix index is always the row; the second, the column:
,
r c rc
c r rc
T T T T where r rowindex c column index = =
Tensor equations can be written as equations with tensors as operators (written in bold):
KE = I(, )
Or, they can be written in component form:
(1) KE = I
ij
i
j
Well be using lots of tensor equations written in component form, so it is important to know how to
read them. Note that some standard notations almost require component form: In GR, the Ricci tensor is
R
v
, and the Ricci scalar is R:
1
2
G R Rg
v v v
=
In component equations, tensor indexes are written explicitly. There are two kinds of tensor indexes:
dummy (aka summation) indexes, and free indexes. Dummy indexes appear exactly twice in any term.
Free indexes appear only once in each term, and the same free indexes must appear in each term (except for
scalar terms). In the above equation, both and are free indexes, and there are no dummy indexes. In eq.
(1) above, i and j are both dummy indexes and there are no free indexes.
Dummy indexes appear exactly twice in any term are used for implied summation, e.g.
3 3
1 1
1 1
2 2
i j i j
ij ij
i j
KE I KE I e e e e
= =
= =
Free indexes are a shorthand for writing several equations at once. Each free index takes on all
possible values for it. Thus,
, , (3 equations)
i i i x x x y y y z z z
C A B C A B C A B C A B = + = + = + = +
and
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00 00 00 01 01 01 02 02 02 03 03 03
10 10 10 11 11 11 12 12 12 13 13 13
20 20 20 21 21 21 22 22 22 23 23 23
30 30
1
2
1 1 1 1
, , ,
2 2 2 2
1 1 1 1
, , ,
2 2 2 2
1 1 1 1
, , ,
2 2 2 2
G R Rg
G R Rg G R Rg G R Rg G R Rg
G R Rg G R Rg G R Rg G R Rg
G R Rg G R Rg G R Rg G R Rg
G R
v v v
=
= = = =
= = = =
= = = =
=
30 31 31 31 32 32 32 33 33 33
1 1 1 1
, , ,
2 2 2 2
Rg G R Rg G R Rg G R Rg = = =
(16 equations).
It is common to have both dummy and free indexes in the same equation. Thus the GR statement of
conservation of energy and momentum uses as a dummy index, and as a free index:
3 3 3 3
0 1 2 3
0 0 0 0
0 0, 0, 0, 0 T T T T T
v
= = = =
V = V = V = V = V =
(4 equations). Notice that scalars apply to all values of free indexes, and dont need indexes of their
own. However, any free indexes must match on all tensor terms. It is nonsense to write something like:
ij i j
A B C = + (nonsense)
However, it is reasonable to have
E.g., angular momentum:
ij i j ij i j j i
A B C M r p r p = =
Since tensors are linear operations, you can add or subtract any two tensors that take the same type
arguments and produce the same type result. Just add the tensor components individually.
. . , , 1, ...
ij ij ij
E g S T U i j N = + = + = S T U
You can also scalar multiply a tensor. Since these properties of tensors are the defining requirements
for a vector space, all the tensors of given rank and index types compose a vector space, and every tensor is
an MVE in its space.
This implies that a tensor field can be differentiated (or integrated), and in particular, it has a gradient.
Higher Rank Tensors
When considering higher rank tensors, it may be helpful to recall that multi-dimensional matrices can
be thought of as lower-dimensional matrices with each element itself a vector or matrix. For example, a 3 x
3 matrix can be thought of as a column vector of 3 row-vectors. Matrix multiplication works out the
same whether you consider the 3 x 3 matrix as a 2-D matrix of numbers, or a 1-D column vector of row
vectors:
( ) ( )
( ) ( )
( , , )
( , , ) ( , , ) ( , , ) ( , , )
( , , )
a b c
x y z d e f ax dy gz bx ey hz cx fy iz
g h i
a b c
x y z d e f x a b c y d e f z g h i ax dy gz bx ey hz cx fy iz
g h i
or
= + + + + + +
= + + = + + + + + +
(
(
(
(
(
(
(
(
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Using this same idea, we can compare the gradient of a scalar field, which is a (
0
1
) tensor field (a 1-
form), with the gradient of a rank-2 (say (
0
2
)) tensor field, which is a (
0
3
) tensor field. First, the gradient of
a scalar field is a (
0
1
) tensor field with 3 components, where each component is a number-valued function:
1 2 3
1 2 3
1 2 3 1 2 3
, , , are basis (co)vectors
( , , ), , ,
f f f
f
x y z
f f f
can be written as D D D where D D D
x y z
c c c
V = = + +
c c c
c c c
= = =
c c c
D
D
.
The gradient operates on an infinitesimal displacement vector to produce the change in the function
when you move through the given displacement: ( )
f f f
df d dx dy dz
x y z
c c c
= = + +
c c c
D r .
Now let R be a (
0
2
) tensor field, and T be its gradient. T is a (
0
3
) tensor field, but can be thought of as a
(
0
1
) tensor field where each component is itself a (
0
2
) tensor.
z
x
y
z
x
y
z
x
y
x-tensor y-tensor z-tensor
A rank-3 tensor considered as a set of 3 rank-2 tensors: an x-tensor, a y-tensor, and a z-tensor.
The gradient operates on an infinitesimal displacement vector to produce the change in the (
0
2
) tensor
field when you move through the given displacement.
11 12 13
21 22 23
31
1 11 12 13
2
1 12 13
21 22 23
3 32 33
1 22 23
31 32 3
1 2
32 33 3 1
3
T T T
y y y
T T T
y y y
T T
T T T T T
x x x
T T T
x x x
T T T
T
z z z
T T T
z z z
T T T
z z
T
y y z x x x
x y z
y
(
c c c
= V = + +
c c c
| | (
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
|
|
|
=
|
|
|
\ .
R R R
T R
.
T
ijx
v
x
T
ijy
v
y
T
ijz
v
z
+
dR
, ,
( ) ( )
k k
ijk ij ijk
k x y z
d T v dR T v
=
= = =
R T v
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Note that if R had been a (
2
0
) (fully contravariant) tensor, then its gradient would be a (
2
1
) mixed
tensor. Taking the gradient of any field simply adds a covariant index, which can then be contracted with a
displacement vector to find the change in the tensor field when moving through the given displacement.
The contraction considerations of the previous section still apply: a rank of an tensor operator contracts
with a rank of one of its inputs to eliminate both. In other words, each rank of input tensors eliminates one
rank of the tensor operator. The rank of the result is the number of surviving ranks from the tensor
operator:
( )
( )
( ) ( ) ( )
( ) ( ) ( )
rank tensor rank inputs rank result
or rank result rank tensor rank inputs
= +
=
Tensors of Mathematical Vector Elements: The operation of a tensor on vectors involves
multiplying components (one from the tensor, and one from each input vector), and then summing. E.g.,
1 1
11
( , ) ... ...
i j
ij
T a b T a b = + + + T a b
Similar to the above example, the T
ij
components could themselves be a vector of a mathematical
vector space (i.e., could be MVEs), while the a
i
and b
j
components are scalars of that vector space. In the
example above, we could say that each of the T
ij;x
, T
ij;y
, and T
ij;z
is a rank-2 tensor (an MVE in the space of
rank-2 tensors), and the components of v are scalars in that space (in this case, real numbers).
Tensors In General
In complete generality then, a tensor T is a linear operation on one or more MVEs:
T(a, b, ...).
Linearity implies that T can be written as a numeric weight for each combination of components, one
component from each input MVE. Thus, the linear operation performed by T is equivalent to a weighted
sum of all combinations of components of the input MVEs. (Since T and the a, b, ... are simple objects,
not functions, there is no concept of derivative or integral operations. Derivatives and integrals are linear
operations on functions, but not linear functions of MVEs.)
Given the components of the inputs a, b, ..., and the components of T, we can contract T with (operate
with T on) the inputs to produce a MVE result. Note that all input MVEs have to have the same basis.
Also, T may have units, so the output units are arbitrary. Note that in generalized coordinates, different
components of a tensor may have different units (much like the vector parameters r and have different
units).
Change of Basis: Transformations
Since tensors are linear operations on MVEs, we can represent a tensor by components. If we know a
tensors operations on all combinations of basis vectors, we have fully defined the tensor. Consider a rank-
2 tensor T acting on two vectors, a and b. We expand T, a, and b into components, using the linearity of
the tensor:
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1 2 3 1 2 3
1 1 2 1 3 1
1 2 2 2 3 2
1 3 2 3 3 3
1 2 3
3 3
1 1
( , ) ( , )
( , ) ( , ) ( , )
( , ) ( , ) ( , )
( , ) ( , ) ( , )
( , ), , ,
( , ) ( ,
ij i j
i j
i
i j
a a a b b b
a b a b a b
a b a b a b
a b a b a b
Define T where
then a b
= =
= + + + +
= + +
+ + +
+ + +
= = = =
=
T a b T i j k i j k
T i i T j i T k i
T i j T j j T k j
T i k T j k T k k
T e e e i e j e k
T a b T e
3 3
1 1
)
i j
j ij
i j
T a b
= =
=
e
The tensors values on all combinations of input basis vectors are the components of the tensor (in the
basis of the input vectors.)
Now lets transform T to another basis. To change from one basis to another, we need to know how to
find the new basis vectors from the old ones, or equivalently, how to transform components in the old basis
to components in the new basis. We write the new basis with primes, and the old basis without primes.
Because vector spaces demand linearity, any change of basis can be written as a linear transformation
of the basis vectors or components, so we can write (eq. #s from Talman):
( ) ( )
1
1 1
1
' [Tal 2.4.5]
' [Tal 2.4.8]
N
k k
i i k i k
k
N
i i
i k k
k k
k
v v v
=
=
= A = A
= A = A
e e e
,
where the last form uses the summation convention. There is a very important difference between
equations 2.4.5 and 2.4.8. The first is a set of 3 vector equations, expressing each of the new basis vectors
in the old basis
Aside: Lets look more closely at the difference between equations 2.4.5 and 2.4.8. The first is a set of 3
vector equations, expressing each of the new basis vectors in the old basis. Basis vectors are vectors, and hence
can themselves be expressed in any basis:
1 2 3 1 2 3
1 1 1 1 2 1 3 1 1 2 3
1 2 3 1 2 3
2 2 1 2 2 2 3 2 1 2 3
1 2 3 1 2 3
3 3 1 3 2 3 3 3 1 2 3
' '
' '
' '
a a a
or more simply b b b
c c c
= A + A + A = + +
= A + A + A = + +
`
= A + A + A = + +
)
e e e e e e e e
e e e e e e e e
e e e e e e e e
where the as are the components of e
1
in the old basis, the bs are the components of e
2
in the old basis,
and the cs are the components of e
3
in the old basis.
In contrast, equation 2.4.8 is a set of 3 number equations, relating the components of a single vector, taking
its old components into the new basis. In other words, in the first equation, we are taking new basis vectors and
expressing them in the old basis (new old). In the second equation, we are taking old components and
converting them to the new basis (old new). The two equations go in opposite directions: the first takes new to
old, the second takes old to new. So it is natural that the two equations use inverse matrices to achieve those
conversions. However, because of the inverse matrices in these equations, vector components are said to
transform contrary (oppositely) to basis vectors, so they are called contravariant vectors.
I think it is misleading to say that contravariant vectors transform oppositely to basis vectors. In fact, that
is impossible. Basis vectors are contravectors, and transform like any other contravector. A vector of (1, 0, 0) (in
some basis) is a basis vector. It may also happen to be the value of some physical vector. In both cases, the
expression of the vector (1, 0, 0) (old basis) in the new-basis is the same.
Now we can use 2.4.5 to evaluate the components of T in the primed basis:
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1 1 1 1
' ( ' , ' ) ( , ) ( , )
N N N N
k l k l k l
ij i j i k j l i j k l i j kl
k l k l
T T T T T
= = = =
= = A A = A A = A A
e e e e e e
Notice that there is one use of the transformation matrix A for each index of T to be transformed.
Matrix View of Basis Transformation
The concept of tensors seems clumsy at first, but its a very fundamental concept. Once you get used
to it, tensors are essentially simple things (though it took me 3 years to understand how simple they
are). The rules for transformations are pretty direct. Transforming a rank-n tensor requires using the
transformation matrix n times. A vector is rank-1, and transforms by a simple matrix multiply, or in tensor
terms, by a summation over indices. Here, since we must distinguish row basis from column basis, we put
the primes on the indices, to indicate which index is in the new basis, and which is in the old basis.
0' 0' 0 0'1 0' 2 0' 3 0
1' 1' 0 1'1 1' 2 1' 3 1
' '
2' 2' 0 2'1 2' 2 2' 3 2
3' 3' 0 3'1 3' 2 3' 3 3
a a
a a
a a
a a
a a
v
v
( | | (
A A A A
| ( (
| A A A A ( (
= = A
| ( (
A A A A
| ( (
| ( (
A A A A
\ .
a' = a
Notice that when you sum over (contract over) two indices, they disappear, and youre left with the
unsummed index. So above when we sum over old-basis indices, were left with a new-basis vector.
Rank-2 example: The electromagnetic field tensor F is rank-2, and transforms using the
transformation matrix twice, by two summations over indices, transforming both stress-energy indices.
This is clumsy to write in matrix terms, because you have to use the transpose of the transformation matrix
to transform the rows; this transposition has no physical significance. In the rank-2 (or higher) case, the
tensor notation is both simpler, and more physically meaningful:
0' 0' 0'1' 0' 2' 0' 3' 0' 0 0'1 0' 2 0' 3 00 01 02 03
1' 0' 1'1' 1' 2' 1' 3' 1' 0 1'1 1' 2 1' 3 10 11
2' 0' 2'1' 2' 2' 2' 3' 2' 0 2'1 2' 2 2' 3
3' 0' 3'1' 3' 2' 3' 3' 3' 0 3'1 3' 2 3' 3
T
F F F F F F F F
F F F F F F F
F F F F
F F F F
( | |
A A A A
| (
| A A A A (
=
| (
A A A A
| (
| (
A A A A
\ .
F' = F
0' 0 1' 0 2' 0 3' 0
12 13 0'1 1'1 2'1 3'1
20 21 22 23 0' 2 1' 2 2' 2 3' 2
30 31 32 33 0' 3 1' 3 2' 3 3' 3
' ' ' '
F
F F F F
F F F F
F F
v v v
v v
( | |
A A A A
| (
| A A A A (
| (
A A A A
| (
| (
A A A A
\ .
= A A
In general, you have to transform every index of a tensor, each index requiring one use of the
transformation matrix.
Non-Orthonormal Systems: Contravariance and Covariance
Many systems cannot be represented with orthonormal coordinates, e.g. the (surface of a) sphere.
Dealing with non-orthonormality requires a more sophisticated view of tensors, and introduces the
concepts of contravariance and covariance.
Consider the following problem from classical mechanics: a pendulum is suspended from a pivot
point which slides horizontally on a spring. The generalized coordinates are (a, ).
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u
a
(a,u)
(a+da,u)
(a,u +du)
(a+da,u+du)
constant
constant a
dr = da a+du u
dr
To compute kinetic energy, we need to compute |v|
2
, conveniently done in some orthogonal
coordinates, say x and y. We start by converting the generalized coordinates to the orthonormal x-y
coordinates, to compute the length of a physical displacement from the changes in generalized coordinates:
2 2 2 2 2 2 2 2 2 2
2 2 2
sin , cos
cos , sin
2 cos cos sin
2 cos
x a l dx da l d
y l dy l d
ds dx dy da l da d l d l d
da l da d l d
u u u
u u u
u u u u u u
u u u
= + = +
= =
= + = + + +
= + +
We have just computed the metric tensor field, which is a function of position in the (a, )
configuration space. We can write the metric tensor field components by inspection:
1 2
2 2
2 2 2 2
2
1 1
,
1 cos
2 cos
cos
i j
ij ij
i j
Let x a x
l
ds g dx dx da l da d l d g
l l
u
u
u u u
u
= =
= =
| |
= = + + =
|
|
\ .
Then |v|
2
= ds
2
/dt
2
. A key point here is that the same metric tensor computes a physical displacement
from generalized coordinate displacements, or a physical velocity from generalized coordinate velocities,
or a physical acceleration from generalized coordinate accelerations, etc., because time is the same for any
generalized coordinate system (no Relativity here!). Note that we symmetrize the cross-terms of the
metric, g
ij
= g
ji
, which is necessary to insure that g(v, w) = g(w, v).
Now consider the scalar product of two vectors. The same metric tensor (field) helps compute the
scalar product (dot-product) of any two (infinitesimal) vectors, from their generalized coordinates:
( , )
i j
ij
d d d d g dv dw = = v w g v w
Since the metric tensor takes two input vectors, is linear in both, and produces a scalar result, it is a
rank-2 tensor. Also, since g(v, w) = g(w, v), g is a symmetric tensor.
Now, lets define a scalar field as a function of the generalized coordinates; say, the potential energy:
2
cos
2
k
U a mg u =
It is quite useful to know the gradient of the potential energy:
( )
a
U U U U
U dU d da d
a a
u
u
u u
c c c c
= V = + = = +
c c c c
D D r
The gradient takes an infinitesimal displacement vector dr = (da, du), and produces a differential in the
value of potential energy, dU (a scalar). Further, dU is a linear function of the displacement vector. Hence,
by definition, the gradient at each point in a- space is a rank-1 tensor, i.e. the gradient is a tensor field.
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Do we need to use the metric (computed earlier) to make the gradient operate on dr? No! The
gradient operates directly on dr, without the need for any assistance by a metric. So the gradient is a
rank-1 tensor that can directly contract with a vector to produce a scalar. This is markedly different from
the dot-product case above, where the first vector (a rank-1 tensor) could not contract directly with an input
vector to produce a scalar. So clearly,
There are two kinds of rank-1 tensors: those (like the gradient) that can contract directly with an
input vector, and those that need the metric to help them operate on an input vector.
Those tensors that can operate directly on a vector are called covariant tensors, and those that need
help are called contravariant, for reasons we will show soon. To indicate that D is covariant, we write its
components with subscripts, instead of superscripts. Its basis vectors are covariant vectors, related to e
1
, e
2
,
and e
3
:
, are covariant basis vectors
i a r
i a
D D D where
u u
u
= = + D
In general, covariant tensors result from differentiation operators on other (scalar or) tensor fields:
gradient, covariant derivative, exterior derivative, Lie derivative, etc.
Note that just as we can say that D acts on dr, we can say that dr is a rank-1 tensor that acts on D to
produce dU:
( ) ( )
i
i
i
U U U
d d dx da d
a x
u
u
c c c
= = = +
c c c
D r r D
The contractions are the same with either acting on the other, so the definitions are symmetric.
Interestingly, when we compute small oscillations of a system of particles, we need both the potential matrix,
which is the gradient of the gradient of the potential field, and the mass matrix, which really gives us kinetic
energy. The potential matrix is fully covariant, and we need no metric to compute it. The kinetic energy matrix
requires us to compute absolute magnitudes of |v|
2
, and so requires us to compute the metric.
We know that a vector, which is a rank-1 tensor, can be visualized as an arrow. How do we visualize
this covariant tensor, in a way that reveals how it operates on a vector (an arrow)? We use a set of equally
spaced parallel planes. Let D be a covariant tensor (aka 1-form):
D(v
1
+ v
2
) = D(v
1
) + D(v
2
)
+ + + + + +
+ + + + +
D(v
1
), D(v
2
) > 0
D(v
3
) < 0
v
3
v
1
v
2
Visualization of a covariant vector (1-form) as oriented parallel planes.
The 1-form is a linear operator on vectors (see text).
The value of D on a vector, D(v), is the number of planes pierced by the vector when laid on the
parallel planes. Clearly, D(v) depends on the magnitude and direction of v. It is also a linear function of v:
the sum of planes pierced by two different vectors equals the number of planes pierced by their vector sum.
There is an orientation to the planes. One side is negative, and the other positive. Vectors crossing in
the negative to the positive direction pierce a positive number of planes. Vectors crossing in the positive
to negative direction pierce a negative number of planes.
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Note also we could redraw the two axes arbitrarily oblique (non-orthogonal), and rescale the axes
arbitrarily, but keeping the intercept values of the planes with the axes unchanged (thus stretching the
arrows and planes). The number of planes pierced would be the same, so the two diagrams above are
equivalent. Hence, this geometric construction of the operation of a covector on a contravector is
completely general, and even applies to vector spaces which have no metric (aka non-metric spaces). All
you need for the construction is a set of arbitrary basis vectors (not necessarily orthonormal), and the values
D(e
i
) on each, and you can draw the parallel planes that illustrate the covector.
The direction of D, analogous to the direction of a vector, is normal to (perpendicular to) the planes
used to graphically represent D.
What Goes Up Can Go Down: Duality of Contravariant and Covariant Vectors
Recall the dot-product is given by
( , )
i j
ij
d d d d g dv dw = = v w g v w
If I fill only one slot of g with v, and leave the 2
nd
slot empty, then g(v, _ ) is a linear function of one
vector, and can be directly contracted with that vector; hence g(v, _ ) is a rank-1 covariant vector. For any
given contravariant vector v
i
, I can define this dual covariant vector, g(v, _ ), which has N components
Ill call v
i
.
( , _)
k
i ik
v g v = = g v
So long as I have a metric, the contravariant and covariant forms of v contain equivalent
information, and are thus two ways of expressing the same vector (geometric object).
The covariant representation can contract directly with a contravariant vector, and the contravariant
representation can contract directly with a covariant vector, to produce the dot-product of the two vectors.
Therefore, we can use the metric tensor to lower the components of a contravariant vector into their
covariant equivalents.
Note that the metric tensor itself has been written with two covariant (lower) indexes, because it
contracts directly with two contravariant vectors to produce their scalar-product.
Why do I need two forms of the same vector? Consider the vector force:
i i
m or F ma = = F a (naturally contravariant)
Since position x
i
is naturally contravariant, so is its derivative v
i
, and 2
nd
derivative, a
i
. Therefore,
force is naturally contravariant. But force is also the gradient of potential energy:
i
i
U or F U
x
c
= V =
c
F (naturally covariant)
Oops! Now force is naturally covariant! But its the same force as above. So which is more natural
for force? Neither. Use whichever one you need. Nurture supersedes nature.
The inverse of the metric tensor matrix is the contravariant metric tensor, g
ij
. It contracts directly with
two covariant vectors to produce their scalar product. Hence, we can use g
ij
to raise the index of a
covariant vector to get its contravariant components.
( , _)
i ik ik i
k kj j
v g v g g g = = = g v
Notice that raising and lowering works on the metric tensor itself. Note that in general, even for
symmetric tensors, T
i
j
T
j
i
, and T
i
j
T
i
j
.
For rank-2 or higher tensors, each index is separately of the contravariant or covariant type. Each
index may be raised or lowered separately from the others. Each lowering requires a contraction with the
fully covariant metric tensor; each raising requires a contraction with the fully contravariant metric tensor.
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In Euclidean space with orthonormal coordinates, the metric tensor is the identity matrix. Hence, the
covariant and contravariant components of any vector are identical. This is why there is no distinction
made in elementary treatments of vector mathematics; displacements, gradients, everything, are simply
called vectors.
The space of covectors is a vector space, i.e. it satisfies the properties of a vector space. However, it is
called dual to the vector space of contravectors, because covectors operate on contravectors to produce
scalar invariants. A thing is dual to another thing if the dual can act on the original thing to produce a
scalar, and vice versa. E.g., in QM, bras are dual to kets. Vectors in the dual space are covectors.
Just like basis contravectors, basis covectors always have components (in their own basis)
1 2 3
(1, 0, 0...), (0,1, 0...), (0, 0,1...), . etc = = =
and we can write an arbitrary covector as
1 2 3
1 2 3
... f f f = + + + f
.
TBS: construction and units of a dual covector from its contravector.
The Real Summation Convention
The summation convention says repeated indexes in an arithmetic expression are implicitly summed
(contracted). We now understand that only a contravariant/covariant pair can be meaningfully summed.
Two covariant or two contravariant indexes require contracting with the metric tensor to be meaningful.
Hence, the real Einstein summation convention is that any two matching indexes, one up (contravariant)
and one down (covariant), are implicitly summed (contracted). Two matching contravariant or covariant
indexes are meaningless, and not allowed.
Now we can see why basis contravectors are written e
1
, e
2
, ... (with subscripts), and basis covectors are
written e
1
, e
2
, ... (with superscripts). It is purely a trick to comply with the real summation convention
that requires summations be performed over one up index and one down index. Then we can write a
vector as a linear combination of the basis vectors, using the summation convention:
1 2 3 1 2 3
1 2 3 1 2 3
i i
i i
v v v v a a a a = + + = = + + = v e e e e a
Note well that there is nothing covariant about e
i
, even though it has a subscript; there is nothing
i
, even though it has a superscript. Its just a notational trick.
Transformation of Covariant Indexes
It turns out that the components of a covariant vector transform with the same matrix as used to
express the new (primed) basis vectors in the old basis:
f
k
= f
j
j
k
[Tal 2.4.11]
Again, somewhat bogusly, eq. 2.4.11 is said to transform covariantly with (the same as) the basis
vectors, so f
i
is called a covariant vector.
For a rank-2 tensor such as T
ij
, each index of T
ij
transforms like the basis vectors (i.e., covariantly
with the basis vectors). Hence, each index of T
ij
is said to be a covariant index. Since both indexes are
covariant, T
ij
is sometimes called fully covariant.
Indefinite Metrics: Relativity
In short, a covariant index of a tensor is one which can be contracted with (summed over) a
contravariant index of an input MVE to produce a meaningful resultant MVE.
In relativity, the metric tensor has some negative signs. The scalar-product is a frame-invariant
interval. No problem. All the math, raising, and lowering, works just the same. In special relativity, the
metric ends up simply putting minus signs where you need them to get SR intervals. The covariant form of
a vector has the minus signs pre-loaded, so it contracts directly with a contravariant vector to produce a
scalar.
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Lets use the sign convention where
## = diag(1, 1, 1, 1). When considering the dual 1-forms for
Minkowski space, the only unusual aspect is that the 1-form for time increases in the opposite direction as
the vector for time. For the space components, the dual 1-forms increase in the same direction as the
vectors. This means that
1, 1, 1, 1
t x y z
t x y z
= = = = e e e e
as it should for the Minkowski metric.
Is a Transformation Matrix a Tensor?
Sort of. When applied to a vector, it converts components from the old basis to the new basis. It
is clearly a linear function of its argument. However, a tensor usually has all its inputs and outputs in the
same basis (or tensor products of that basis). But a transformation matrix is specifically constructed to take
inputs in one basis, and produce outputs in a different basis. Essentially, the columns are indexed by the
old basis, and the rows are indexed by the new basis. It basically works like a tensor, but the
transformation rule is that to transform the columns, you use a transformation matrix for the old basis; to
transform the rows, you use the transformation matrix for the new basis.
Consider a vector
1 2 3
1 2 3
v v v = + + v e e e
This is a vector equation, and despite its appearance, it is true in any basis, not just the (e
1
, e
2
, e
3
) basis.
If we write e
1
, e
2
, e
3
as vectors in some new (e
x
, e
y
, e
z
) basis, the vector equation above still holds:
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
1 2 3
1 1 1 1
2 2 2 2
3 3 3 3
1 2 3
1 2 3
1 2 3
1 1 1 2 2 2 3 3 3
x y z
x y z
x y z
x y z
x y z
x y z
x y z x y z x y z
x y z x y z x y z
v v v
v v v
= + +
= + +
= + +
= + +
( ( (
= + + + + + + + +
e e e
e e e e e e e
e e e e e e e
e e e e e e e
v e e e
e e e e e e e e e e e e e e e e e e
_ _ _
The vector v is just a weighted sum of basis vectors, and therefore the columns of the transformation
matrix are the old basis vectors expressed in the new basis. E.g., to transform the components of a vector
from the (e
1
, e
2
, e
3
) to the (e
x
, e
y
, e
z
) basis, the transformation matrix is
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
1 2 3
1 2 3
1 2 3 1 2 3
1 2 3
1 2 3
x x x
x x x
y y y
y y y
z z z
z z z
(
(
(
(
(
=
(
(
(
(
e e e
e e e e e e
e e e e e e e e e
e e e e e e
e e e
You can see directly that the first column is e
1
written in the x-y-z basis; the 2
nd
column is e
2
in the x-y-
z basis; and the 3
rd
column is e
3
in the x-y-z basis.
In quantum mechanics, the Pauli vector is a vector of three 2x2 matrices: the Pauli matrices. Each 2x2
complex-valued matrix corresponds to a spin-1/2 operator in some x, y, or z direction. It is a 3
rd
rank object
in the tensor product space of R
3
C
2
C
2
, i.e. xyz spinor spinor. The xyz rank is clearly in a
different basis than the complex spinor ranks, since xyz is a completely different vector space than spin-1/2
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spinor space. However, it is a linear operator on various objects, so each rank transforms according to the
transformation matrix for its basis.
0 1 0 1 0
, ,
1 0 0 0 1
x y z
i
i
o
| | ( ( (
=
|
( ( (
|
\ .
,
Its interesting to note that the term tensor product produces, in general, an object of mixed bases, and
often, mixed vector spaces. Nonetheless, the term tensor seems to be used most often for mathematical
objects whose ranks are all in the same basis.
Cartesian Tensors
Cartesian tensors arent quite tensors, because they dont transform into non-Cartesian coordinates
properly. (Note that despite their name, Cartesian tensors are not a special kind of tensor; they arent really
tensors. Theyre tensor wanna-bes.) Cartesian tensors have two failings that prevent them from being true
tensors: they dont distinguish between contravariant and covariant components, and they treat finite
displacements in space as vectors. In non-orthogonal coordinates, you must distinguish contravariant and
covariant components. In non-Cartesian coordinates, only infinitesimal displacements are vectors. Details:
Recall that in Cartesian coordinates, there is no distinction between contravariant and covariant
components of a tensor. This allows a certain sloppiness that one can only get away with if one sticks to
Cartesian coordinates. This means that Cartesian tensors only transform reliably by rotations from one
set of Cartesian coordinates to a new, rotated set of Cartesian coordinates. Since both the new and old
bases are Cartesian, there is no need to distinguish contravariant and covariant components in either basis,
and the transformation (to a rotated coordinate system) works.
For example, the moment of inertia tensor is a Cartesian tensor. There is no problem in its first use,
to compute the angular momentum of a blob of mass given its angular velocity:
( , _)
x x
z z
y z y
z z
z z z
x x x x
x
x x
y y
i i j
z
z z
x
y x y
x x
z z
x x
y y y y y
y y
z
y
j
z
y
I I
I I
L I I
I
I I
L
L I
L I
I
I I I
I
I
I
I
e
e
e
e e
e
e
= =
( ( (
( ( (
= = + +
( ( (
( (
(
(
(
(
(
(
(
(
(
(
(
(
I L
But notice that if I accepts a contravariant vector, then Is components for that input vector must be
covariant. However, I produces a contravariant output, so its output components are contravariant. So far,
so good.
But now we want to find the kinetic energy. Well,
2
1 1 1
( , _)
2 2 2
Ie
| |
= =
|
\ .
L I . But we have a
dot-product of two contravariant vectors. To evaluate that dot-product, in a general coordinate system, we
have to use the metric:
1 1 1
2 2 2
i j i j k i j i
j i j ik j
KE I I g I e e e e e e = = =
However, in Cartesian coordinates, the metric matrix is the identity matrix, the contravariant
components equal the covariant components, and the final not-equals above becomes an equals.
Hence, we neglect the distinction between contravariant components and covariant components, and
incorrectly sum the components of I on the components of e, even though both are contravariant in the
2
nd
sum.
In general coordinates, the direct sum for the dot-product doesnt work, and you must use the metric
tensor for the final dot-product.
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Example of failure of finite displacements: TBS: The electric quadrupole tensor acts on two copies of
the finite displacement vector to produce the electric potential at that displacement. Even in something as
simple as polar coordinates, this method fails.
The Real Reason Why the Kronecker Delta Is Symmetric
TBS: Because it a mixed tensor,
## . Symmetry can only be assessed by comparing interchange of two
indices of the same up- or down-ness (contravariance or covariance). We can lower, say , in
with
the metric:
g g
o| o | o|
o o = =
The result the metric, which is always symmetric. Hence,
## is a symmetric tensor, but not because its
matrix looks symmetric. In general, a mixed rank-2 symmetric tensor does not have a symmetric matrix
representation. Only when both indices are up or both down is its matrix symmetric.
The Kronecker delta is a special case that does not generalize.
Things are not always what they seem.
Tensor Appendices
Pythagorean Relation for 1-forms
Demonstration that 1-forms satisfy the Pythagorean relation for magnitude:
0 dx + 1 dy
|a~| = 1
1 dx + 1 dy
|a~| = 2
2 dx + 1 dy
|a~| = 5
unit
vector
a dx + b dy
|a~| = (a
2
+b
2
)
1/b
1/a
unit
vector
a~
a~
a~
a~
Examples of 3 1-forms, and a generic 1-form.
Here, dx is the x basis 1-form, and dy is the y basis 1-form.
From the diagram above, a max-crossing vector (perpendicular to the planes of a~) has (x, y)
components (1/b, 1/a). Dividing by its magnitude, we get a unit vector:
( )
2 2
1 1
max crossing unit vector . ( ) 1, 1
1 1
b a
Note that x and y
b a
+
= = =
+
x y
d x d y u
The magnitude of a 1-form is the scalar resulting from the 1-forms action on a max-crossing unit
vector:
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( )
( ) ( )
2 2 2 2
2 2
2 2
2 2 2 2 2 2
1 1
( )
1 1 1 1 1 1
a b
a x b y
a b a b
b a b a
a b
a b
ab
b a b a b a
| | | |
+ + +
| |
+ +
\ . \ .
= = = = = = +
+
+ + +
d d x y
a a u
Heres another demonstration that 1-forms satisfy the Pythagorean relation for magnitude. The
magnitude of a 1-form is the inverse of the plane spacing:
O
A
B
X
( )( )
2 2
2 2
2 2
2 2
1 1
1 1
1 1
1 1 1
1 1
OX BO
OXA ~ BOA
OA BA
BO OA
b a
OX
BA
b a
b a
ab a b
OX b a
b a
|
=
= =
+
+
= = = + = +
a
Geometric Construction Of The Sum Of Two 1-Forms:
a~(x) = 2
b~(x) = 1
(a~ + b~)(x) = 3
a~(v
a
) = 1
b~(v
a
) = 0
(a~ + b~)(v
a
) = 1
a~(v
b
) = 0
b~(v
b
) = 1
(a~ + b~)(v
b
) = 1
Example of a~ + b~ Construction of a~ + b~
a~
b~
x
v
b
v
a
a~ + b~
O
step 4
step 5
To construct the sum of two 1-forms, a~ + b~:
1. Choose an origin at the intersection of a plane of a~ and a plane of b~.
2. Draw vector v
a
from the origin along the planes of b~, so b~(v
a
) = 0, and of length such that
a~(v
a
) = 1. [This is the dual vector of a~.]
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3. Similarly, draw v
b
from the origin along the planes of a~, so a~(v
b
) = 0, and b~(v
b
) = 1. [This is
the dual vector of b~.]
4. Draw a plane through the heads of v
a
and v
b
(black above). This defines the orientation of
(a~ + b~).
5. Draw a parallel plane through the common point (the origin). This defines the spacing of planes
of (a~ + b~).
6. Draw all other planes parallel, and with the same spacing. This is the geometric representation of
(a~ + b~).
Now we can easily draw the test vector x, such that a~(x) = 2, and b~(x) = 1.
Fully Anti-symmetric Symbols Expanded
Everyone hears about them, but few ever see them. They are quite sparse: the 3-D fully anti-
symmetric symbol has 6 nonzero values out of 27; the 4-D one has 24 nonzero values out of 256.
3-D, from the 6 permutations, ijk: 123+, 132-, 312+, 321-, 231+, 213-
1 2 3
0 0 0 0 0 1 0 1 0
0 0 1 , 0 0 0 , 1 0 0
0 1 0 1 0 0 0 0 0
ijk
k k k
c
= = =
( ( (
( ( (
=
( ( (
( ( (
4-D, from the 24 permutations, :
0123+ 0132- 0312+ 0321- 0231+ 0213-
1023- 1032+ 1302- 1320+ 1230- 1203+
2013+ 2031- 2301+ 2310- 2130+ 2103-
3012- 3021+ 3201- 3210+ 3120- 3102+
0 1 2 3
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0
, , ,
0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0
0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
1 ,
0 0 0 1 0 0 0 0
0 0 1 0 0 0 0 0
o|o
| | | |
c
o
o
= = = =
( ( ( (
( ( ( (
=
( ( ( (
( ( ( ( =
( ( ( (
( (
( (
( (
=
( (
( (
0 0 0 1 0 0 1 0
0 0 0 0 0 0 0 0
, ,
0 0 0 0 1 0 0 0
1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0
0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0
2 , , ,
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
0 0 1 0
3
0 1 0 0
0 0 0
o
o
( (
( (
( (
( (
( (
( ( ( (
( ( ( (
( ( ( (
=
( ( ( (
( ( ( (
=
0 0 1 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0
, , ,
1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
( ( ( (
( ( ( (
( ( ( (
( ( ( (
( ( ( (
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Metric? We Dont Need No Stinking Metric!
Examples of Useful, Non-metric Spaces
Non-metric spaces are everywhere. A non-metric space has no concept of distance between arbitrary
points, or even between arbitrary nearby points (points with infinitesimal coordinate differences).
However:
Non-metric spaces have no concept of distance,
but many still have a well-defined concept of area, in the sense of an integral.
For example, consider a plot of velocity (of a particle in 1D) vs. time (below, left).
velocity
time
pressure
volume
displacement work
momentum
position
action
A
B
Some useful non-metric spaces: (left) velocity vs. time; (middle) pressure vs. volume;
(right) momentum vs. position. In each case, there is no distance, but there is area.
The area under the velocity curve is the total displacement covered. The area under the P-V curve is
the work done by an expanding fluid. The area under the momentum-position curve (p-q) is the action of
the motion in classical mechanics. Though the points in each of these plots exist on 2D manifolds, the two
coordinates are incomparable (they have different units). It is meaningless to ask what is the distance
between two arbitrary points on the plane. For example, points A and B on the v-t curve differ in both
velocity and time, so how could we define a distance between them (how can we add m/s and seconds)?
In the above cases, we have one coordinate value as a function of the other, e.g. velocity as a function
of time. We now consider another case: rather than consider the function as one of the coordinates in a
manifold, we consider the manifold as comprising only the independent variables. Then, the function is
defined on that manifold. As usual, keeping track of the units of all the quantities will help in
understanding both the physical and mathematical principles.
For example, the speed of light in air is a function of 3 independent variables: temperature, pressure,
and humidity. At 633 nm, the effects amount to speed changes of about +1 ppm per kelvin, 0.4 ppm per
mm-Hg pressure, and +0.01 ppm per 1% change in relative humidity (RH) (see http://patapsco.nist.gov/
s(T, P, H) = s
0
+ aT bP + cH
where a 300 (m/s)/k, b 120 (m/s)/mm-Hg, and c 3 (m/s)/% are positive constants, and the
function s is the speed of light at the given conditions, in m/s. Our manifold is the set of TPH triples, and s
is a function on that manifold. We can consider the TPH triple as a (contravariant, column) vector: (T, P,
H)
T
. These vectors constitute a 3D vector space over the field of reals. s() is a real function on that vector
space.
Note that the 3 components of a vector each have different units: the temperature is measured in
kelvins (K), the pressure in mm-Hg, and the relative humidity in %. Note also that there is no metric on (T,
P, H) space (which is bigger, 1 K or 1 mm-Hg?). However, the gradient of s is still well defined:
s s s
s a b c
T P H
c c c
V = + + = +
c c c
dT dP dH dT dP dH
What are the units of the gradient? As with the vectors, each component has different units: the first is
in (m/s) per kelvin; the second in (m/s) per mm-Hg; the third in (m/s) per %. The gradient has different
units than the vectors, and is not a part of the original vector space. The gradient, Vs, operates on a vector
(T, P, H)
T
to give the change in speed from one set of conditions, say (T
0
, P
0
, H
0
) to conditions incremented
by the vector (T
0
+ T, P
0
+ P, H
0
+ H).
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One often thinks of the gradient as having a second property: it specifies the direction of steepest
increase of the function, s. But:
Without a metric, steepest is not defined.
Which is steeper, moving one unit in the temperature direction, or one unit in the humidity direction?
In desperation, we might ignore our units of measure, and choose the Euclidean metric (thus equating one
unit of temperature with one unit of pressure and one unit of humidity); then the gradient produces a
direction of steepest increase. However, with no justification for such a choice of metric, the result is
probably meaningless.
What about basis vectors? The obvious choice is, including units, (1 K, 0 mm-Hg, 0 %)
T
, (0 K, 1 mm-
Hg, 0 %)
T
, and (0 K, 0 mm-Hg, 1 %)
T
, or omitting units: (1, 0, 0), (0, 1, 0), and (0, 0, 1). Note that these
are not unit vectors, because there is no such thing as a unit vector, because there is no metric by which
to measure one unit. Also, if I ascribe units to the basis vectors, then the components of an arbitrary
vector in that basis are dimensionless.
Now lets change the basis: suppose now I measure temperature in some unit equal to K (almost the
Rankine scale). Now all my temperature measurements double, i.e. T
new
= 2 T
old
. In other words, ( K,
0, 0)
T
is a different basis than (1 K, 0, 0)
T
. As expected for a covariant component, the temperature
T
is cut in half if the basis vector halves. So when the half-size gradient
component operates on the double-size temperature vector component, the product remains invariant, i.e.,
the speed of light is a function of temperature, not of the units in which you measure temperature.
The above basis change was a simple change of scale of one component in isolation. The other
common basis change is a rotation of the axes, mixing the old basis vectors.
Can we rotate axes when the units are different for each component? Surprisingly, we can.
H
T
P
H
T
P
e
1
e
2
e
3
We simply define new basis vectors as linear combinations of old ones, which is all that a rotation
does. For example, suppose we measured the speed of light on 3 different days, and the environmental
conditions were different on those 3 days. We choose those measurements as our basis, say e
1
= (300 K,
750 mm-Hg, 20%), e
2
= (290 K, 760 mm-Hg, 30 %), and e
3
= (290 K, 770 mm-Hg, 10 %). These basis
vectors are not orthogonal, but are (of course) linearly independent. Suppose I want to know the speed of
light at (296 K, 752 mm-Hg, 18 %). I decompose this into my new basis and get (0.6, 0.6, 0.2). I
compute the speed of light function in the new basis, and then compute its gradient, to get
1 2 3
1 2 3
d d d + + e e e . I then operate on the vector with the gradient to find the change in speed: s = Vs(0.6,
0.6, 0.2) = 0.6 d
1
+ 0.6 d
2
0.2 d
3
.
We could extend this to a more complex function, and then the gradient is not constant. For example,
a more accurate equation for the speed of light is
( )
( )
2
0
( , , ) 273 160
P
s T P H c f gH T
T
= + +
where f 7.86 10
4
and g 1.5 10
11
are constants. Now the gradient is a function of position (in
TPH space), and there is still no metric.
Comment on the metric: In desperation, you might define a metric, i.e. the length of a vector, to be s,
the change in the speed of light due to the environmental changes defined by that vector. However, such a
metric is in general non-Euclidean (not a Pythagorean relationship), indefinite (non-zero vectors can have
zero or negative lengths), and still doesnt define a meaningful dot product. Our more-accurate equation
for the speed of light provides examples of these failures.
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References:
[Knu] Knuth, Donald, The Art of Computer Programming, Vol. 2: Seminumerical Algorithms,
2nd Ed., p. 117.
[Mic] Michelsen, Eric L., Funky Quantum Concepts, unpublished.
http://physics.ucsd.edu/~emichels/FunkyQuantumConcepts.pdf .
[Sch] Schutz, Bernard, A First Course in General Relativity, Cambridge University Press,
1990.
[Sch2] Schutz, Bernard, Geometrical Methods of Mathematical Physics, Cambridge University
Press, 1980.
[Tal] Talman, Richard, Geometric Mechanics, John Wiley and Sons, 2000.
physics.ucsd.edu/~emichels Funky Mathematical Physics Concepts emichels at physics.ucsd.edu
Differential Geometry
Manifolds
A manifold is a space: a set of points with coordinate labels. We are free to choose coordinates
many ways, but a manifold must be able to have coordinates that are real numbers. We are familiar with
metric manifolds, where there is a concept of distance. However, there are many useful manifolds which
have no metric, e.g. phase space (see We Dont Need No Stinking Metric above).
Even when a space is non-metric, it still has concepts of locality and continuity.
Such locality and continuity are defined in terms of the coordinates, which are real numbers. It may
also have a volume, e.g. the oft-mentioned phase-space volume. It may seem odd that theres no
definition of distance, but there is one of volume. Volume in this case is simply defined in terms of
the coordinates, dV = dx
1
dx
2
dx
3
..., and has no absolute meaning.
Coordinate Bases
Coordinate bases are basis vectors derived from coordinates on the manifold. They are extremely
useful, and built directly on basic multivariate calculus. Coordinate bases can be defined a few different
ways. Perhaps the simplest comes from considering a small displacement vector on a manifold. We use
2D polar coordinates in (r, ) as our example. A coordinate basis can be defined as the basis in which the
components of an infinitesimal displacement vector are just the differentials of the coordinates:
e
u
e
r
e
r
e
u
e
r
e
u
e
u
e
r
e
r
e
u
p
p + dp
dp = (dr, d)
(Left) Coordinate bases: the components of the displacement vector are the differentials of the
coordinates. (Right) Coordinate basis vectors around the manifold.
Note that e
(the basis vector) far from the origin must be bigger than near, because a small change in
angle, d, causes a bigger displacement vector far from the origin than near. The advantage of a coordinate
basis is that it makes dot products, such as a gradient dotted into a displacement, appear in the simplest
possible form:
( ) ( , ), ( , ) ,
f f f f
Given f r df f r d dr d dr d
r r
u u u u
u u
c c c c | |
= V = + = +
|
c c c c
\ .
p
The last equality is assured from elementary multivariate calculus.
The basis vectors are defined by differentials, but are themselves finite vectors. Any physical vector,
finite or infinitesimal, can be expressed in the coordinate basis, e.g., velocity, which is finite.
Vectors as derivatives: There is a huge confusion about writing basis vectors as derivatives.
From our study of tensors (earlier), we know that a vector can be considered an operators on a 1-form,
which produces a scalar. We now describe how vector fields can be considered operators on scalar
functions, which produce scalar fields. I dont like this view, since it is fairly arbitrary, confuses the much
more consistent tensor view, and is easily replaced with tensor notation.
We will see that in fact, the derivative basis vectors are operators which create 1-forms (dual-basis
components), not traditional basis vectors. The vector basis is then implicitly defined as the dual of the
dual-basis, which is always the coordinate basis. In detail:
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We know from the Tensors chapter that the gradient of a scalar field is a 1-form with partial
derivatives as its components. For example:
1 2 3 1 2 3
( , , ) , , , , , are basis 1-forms
f f f f f f
f x y z where
x y z x y z
| | c c c c c c
V = = + +
|
c c c c c c
\ .
Many texts define vectors in terms of their action on scalar functions (aka scalar fields), e.g. [Wald
p15]. Given a point (x, y, z), and a function f(x, y, z), the definition of a vector v amounts to
( ) ( ) , , , , (a scalar field)
x y z x y z
f f f
v v v such that f x y z f v v v
x y z
c c c
V = + + (
c c c
v v v
Roughly, the action of v on f produces a scaled directional derivative of f: Given some small
displacement dt, as a fraction of |v| and in the direction of v, v tells you how much f will change when
moving from (x, y, z) to (x + v
x
dt, y + v
y
dt, z + v
z
dt):
| | | |
df
df f dt or f
dt
= = v v
If t is time, and v is a velocity, then v[f] is the time rate of change of f. While this notation is compact,
Id rather write it simply as the dot-product of v and Vf, which is more explicit, and consistent with tensors:
df
df f dt or f
dt
= V = V v v
The definition of v above requires an auxiliary function f, which is messy. We remove f by redefining
v as an operator:
(an operator)
x y z
v v v
x y x
| | c c c
+ +
|
c c c
\ .
v
Given this form, it looks like /x, /y, and /z are some kind of basis vectors. Indeed, standard
terminology is to refer to /x, /y, and /z as the coordinate basis for vectors, but they are really
operators for creating 1-forms! Then
| | ( )
, ,
(a scalar field)
x y z i
i
i x y z
f f f
f v v v v f
x y z
=
c c c
= + + = V
c c c
v
The vector v contracts directly with the 1-form Vf (without need of any metric), hence v is a vector
implicitly defined in the basis dual to the 1-form Vf.
Note that if v = v(x, y, z) is a vector field, then
( ) , , ( , , ) ( , , ) (a scalar field) f x y z x y z f x y z V (
v v
These derivative operators can be drawn as basis vectors in the usual manner, as arrows on the
manifold. They are just the coordinate basis vectors shown earlier. For example, consider polar
coordinates (r, ):
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e
u
e
r
e
r
e
u
e
r
e
u
e
u
e
r
Examples of coordinate basis vectors around the manifold.
e
r
happens to be unit magnitude everywhere, but e
is not.
The manifold in this case is simply the flat plane,
2
. The r-coordinate basis vectors are all the same
size, but have different directions at different places. The coordinate basis vectors get larger with r, and
also vary in direction around the manifold.
Covariant Derivatives
Notation: Due to word-processor limitations, the following two notations are equivalent:
( ) ( ), h r
,
,
h r .
This description is similar to one in [Sch].
We start with the familiar concepts of derivatives, and see how that evolves into the covariant
derivative. Given a real-valued function of one variable, f(x), we want to know how f varies with x near a
value, a. The answer is the derivative of f(x), where
df = f '(a) dx and therefore f(a + dx) f(a) + df = f(a) + f '(a) dx
Extending to two variables, g(x, y), wed like to know how g varies in the 2-D neighborhood around a
point (a, b), given a displacement vector dr = (dx, dy). We can compute its gradient:
( ) and therefore , ( , ) ( )
g g
g g a dx b dy g a b g dr
x y
c c
V = + + + ~ + V
c c
dx dy
,
The gradient is also called a directional derivative, because the rate at which g changes depends on the
direction in which you move away from the point (a, b).
The gradient extends to a vector valued function (a vector field) h(x, y) = h
x
(x, y)i + h
y
(x, y)j:
( )
x y x y
x x x x
y y y y
h h
h
x y
h h h h h h
and
x x x y y y
h h h h
dx
x y x y
h h
dh h dr dx dy dx dy
x y
h h h h
dy
x y x y
c c
V = +
c c
c c c c c c
= + = +
c c c c c c
( ( ( (
c c c c
( ( ( (
( ( (
c c c c
c c
( ( ( = V = + = = +
c c ( ( (
c c c c
( ( (
( ( (
c c c c
dx dy
i j i j
, ,
,
, ,
, ,
, ,
,
(
(
(
(
(
We see that the columns of Vh are vectors which are weighted by dx and dy, and then summed to
produce a vector result. Therefore, Vh is linear in the displacement vector dr = (dx, dy). This linearity
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insures that it transforms like a duck . . . I mean, like a tensor. Thus Vh is a rank-2 (
1
1
) tensor: it takes a
single vector input, and produces a vector result.
So far, all this has been in rectangular coordinates. Now we must consider what happens in curvilinear
coordinates, such as polar. Note that were still in a simple, flat space. (Well get to curved spaces later).
Our goal is still to find the change in the vector value of h( ), given an infinitesimal vector change of
position, dx = (dx
1
, dx
2
). We use the same approach as above, where a vector valued function comprises
two (or n) real-valued component functions:
1 2 1 1 2 2 1 2
1 2
( , ) ( , ) ( , ) h x x h x x h x x = + e e
,
, ,
. However, in this
general case, the basis vectors are themselves functions of position (previously the basis vectors were
constant everywhere). So h( ) is really
1 2 1 1 2 1 2 2 1 2 1 2
1 2
( , ) ( , ) ( , ) ( , ) ( , ) h x x h x x x x h x x x x = + e e
,
, ,
Hence, partial derivatives of the component functions alone are no longer sufficient to define the
change in the vector value of h( ); we must also account for the change in the basis vectors.
h(x
1
, x
2
)
h(x
1
+dx
1
, x
2
+dx
2
)
e
1
(x
1
, x
2
)
e
2
(x
1
, x
2
)
e
2
(x
1
+dx
1
, x
2
+dx
2
)
e
1
(x
1
+dx
1
, x
2
+dx
2
)
Components constant,
but vector changes
Vector constant, but
components change
e
1
e
2
dx = (dx
1
, dx
2
)
dx = (dx
1
, dx
2
)
Note that a component of the derivative is distinctly not the same as the derivative of the component
(see diagram above). Therefore, the ith component of the derivative depends on all the components of the
vector field.
We compute partial derivatives of the vector field h(x
1
, x
2
) using the product rule:
1 2
1 2 1 1 2 1 2 2 1 2 1 2
1 2
1 1 1 1 1
1 2 1 2
1 1
1
( , ) ( , ) ( , ) ( , )
( , ) ( , )
n j
j j
j
j
h h h
x x h x x x x h x x
x x x x x
h
x x h x x
x x
=
c c c c c
= + + +
c c c c c
c | |
c
= +
|
|
c c
\ .
e e
e e
e
e
,
, ,
, ,
,
,
This is a vector equation: all terms are vectors, each with components in all n basis directions. This is
equivalent to n numerical component equations. Note that (ch/cx
1
) has components in both (or all n)
directions. Of course, we can write similar equations for the components of the derivative in any basis
direction, e
k
:
1 2
1 2 1 1 2 1 2 2 1 2 1 2
1 2
1 2 1 2
1
( , ) ( , ) ( , ) ( , )
( , ) ( , )
k k k k k
n j
j j
j
k k
j
h h h
x x h x x x x h x x
x x x x x
h
x x h x x
x x
=
c c c c c
= + + +
c c c c c
c | |
c
= +
|
|
c c
\ .
e e
e e
e
e
,
, ,
, ,
,
,
Because we must frequently work with components and component equations, rather than whole
vector equations, let us now consider only the ith component of the above:
1 2
1
( , )
i i
n i
j j
k k k
j
h h
h x x
x x x
=
c | | | | c c
= +
| |
c c c
\ . \ .
e
, ,
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The first term moves out of the summation because each of the first terms in the summation of eq. (1)
are vectors, and each points exactly in the e
j
direction. Only the j = i term contributes to the ith component;
the purely e
j
directed vector contributes nothing to the ith component when j = i.
Recall that these equations are true for any arbitrary coordinate system; we have made no assumptions
about unit length or orthogonal basis vectors. Note that
( )
the th (covariant) component of
k
k
h
h k h
x
c
= V = V
c
,
, ,
Since Vh is a rank-2 tensor, the kth covariant component of Vh is the kth column of Vh:
1 1
1 2
2 2
1 2
h h
x x
h
h h
x x
(
| | | |
c c
(
| |
(
c c
\ . \ .
(
V =
(
( | | | |
c c
( | |
c c
( \ . \ .
, ,
,
, ,
Since the change in h( ) is linear with small changes in position,
1 2
( ), ( , ) dh h dx where dx dx dx = V =
, ,
, ,
Going back to Equations (1) and (2), we can now write the full covariant derivative of h( ) in 3 ways:
vector, verbose component, and compact component:
( )
( )
( )
1 2 1 2
1 1 1
1 2
1
( , ) ( , )
( , )
,
n n n
j j j i
k jk i
k k k
k
j j i
i
n i
i
j j
k
k k
j
i
i
i
j j j i i i
k jk jk jk i
k k k
h h
h h h x x h x x
x x x
h
h h x x
x x
h
h h where
x x x
= = =
=
c
c c
V V = + = + I
c c c
c | | c
V = +
|
c c
\ .
c c | | c
V = + I I I =
|
c c c
\ .
e
e
e
e e
e
, , ,
, ,
,
,
,
, ,
,
,
Aside: Some mathematicians complain that you cant define the Christoffel symbols as derivatives of basis
vectors, because you cant compare vectors from two different points of a manifold without already having the
Christoffel symbols (aka the connection). Physicists, including Schutz [Sch], say that physics defines how to
compare vectors at different points of a manifold, and thus you can calculate the Christoffel symbols. In the end,
it doesnt really matter. Either way, by physics or by fiat, the Christoffel symbols are, in fact, the derivatives of
the basis vectors.
Christoffel Symbols
Christoffel symbols are the covariant derivatives of the basis vector fields. TBS.
Derivatives of e
r
r e
r
r + dr e
r
de
r
= 0
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Visualization of n-Forms
TBS: 1-forms as oriented planes
2-forms (in 3 or more space) as oriented parallelograms
3-forms (in 3 or more space) as oriented parallelepipeds
4-forms (in 4-space): how are they oriented??
Review of Wedge Products and Exterior Derivative
This is a quick insert that needs proper work. ??
1-D
I dont know of any meaning for a wedge-product in 1-D, or even a vector. Also, the 1-D exterior
derivative is a degenerate case, because the exterior of a line segment is just the 2 endpoints, and all
functions are scalar functions. In all higher dimensions, the exterior or boundary of a region is a closed
path/ surface/ volume/ hyper-volume. In 1-D the boundary of a line segment cannot be closed. So instead
of integrating around a closed exterior (aka boundary), we simply take the difference in the function value
at the endpoints, divided by a differential displacement. This is simply the ordinary derivative of a
function, f (x).
2-D
The exterior derivative of a scalar function f(x, y) follows the 1-D case, and is similarly degenerate,
where the exterior is simply the two endpoints of a differential displacement. Since the domain is a 2-D
space, the displacements are vectors, and there are 2 derivatives, one for displacements in x, and one for
displacements in y. Hence the exterior derivative is just the one-form gradient of the function:
( , ) " "
f f
x y
c c
= = +
c c
df dx dy
In 2-D, the wedge product . dx dy
is a two-form, which accepts two vectors to produce the signed
area of the parallelogram defined by them. A signed area can be + or -; a counter-clockwise direction is
positive, and clockwise is negative.
v
w
+
v
w
-
( , ) signed area defined by ( , ) ( , )
( ) ( ) ( ) ( )
( ) ( )
det det
( ) ( )
x x
y y
v w v w w v
v w v w
v w v w
v w
v w
. = = .
=
= =
dx dy dx dy
dx dy dy dx
dx dx
dy dy
, , , , , ,
, , , ,
, ,
, ,
The exterior derivative of a 1-form is the ratio of the closed path integral of the 1-form to the area of
the parallelogram of two vectors, for infinitesimal vectors. This is very similar to the definition of curl,
only applied to a 1-form instead of a vector field.
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1
3
dx
dy
dx
dy 2
4
x
(r+dy)
x
(r)
y
(r)
y
(r+dx)
Path integrals from 2
Consider the horizontal and vertical contributions to the path integral separately:
1 3
2 4
( , ) ( ) ( ) ( , ) ( , )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
x
x
x x
y
y y
x y x, y x, y r x y dr dx dy
dr dr r dx r dy dx dy dx
y
dr dr r dx dy r dy dx dy
x
e e
e
e e
e
e e
= + = =
c
+ = + =
c
c
+ = + =
c
} }
} }
y
dx dy
, ,
, ,
, ,
The horizontal (segments 1 & 3) integrals are linear in dx, because that is the length of the path. They
are linear in dy, because dy is proportional to the difference in
x
. Hence, the contribution is linear in both
dx and dy, and therefore proportional to the area (dx)(dy).
A similar argument holds for the vertical contribution, segments 2 & 4. Therefore, the path integral
varies proportionately to the area enclosed by two orthogonal vectors.
It is easy to show this is true for any two vectors, and any shaped area bounded by an infinitesimal
path. For example, when you butt up two rectangles, the path integral around the combined boundary
equals the sum of the individual path integrals, because the contributions from the common segment cancel
from each rectangle, and hence omitting them does not change the path integral. The area integrals clearly
3-D
In 3-D, the wedge product
( , , ) signed volume defined by ( , , ) ( , , ), .
( ) ( ) ( )
det ( ) ( ) ( ) det
( ) ( ) ( )
x x x
y y y
z z z
u v w u v w u w v etc
u v w
u v w
u v w u v w
u v w
u v w
. . = = . .
= =
dx dy dz dx dy dz
dx dx dx
dy dy dy
dz dz dz
, , , , , , , , ,
, , ,
, , ,
, , ,
is a 3-form which can either:
1. accept 2 vectors to produce an oriented area; it doesnt have a sign, it has a direction. Analogous
to the cross-product. Or,
2. accept 3 vectors to produce a signed volume.
The exterior derivative of a scalar or 1-form field is essentially the same as in the 2-D case, except that
now the areas defined by vectors are oriented instead of simply signed. In this case, the exterior is a
closed surface; the interior is a volume.
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Math Tricks
Here are some math tricks that either come up a lot and are worth knowing about, or are just fun and
interesting.
Math Tricks That Come Up A Lot
The Gaussian Integral
2
ax
e dx
}
You can look this up anywhere, but here goes: well evaluate the basic integral,
2
x
e dx
}
, and throw in the a at the end by a simple change of variable. First, we square the integral,
then rewrite the second factor calling the dummy integration variable y instead of x:
( )
2 2
2 2 2
2
x y
x x y
dx e dx e dy e dx dy e
+
| | | | | |
= =
| | |
\ . \ . \ .
} } } } }
This is just a double integral over the entire x-y plane, so we can switch to polar coordinates. Note that
the exponential integrand is constant at constant r, so we can replace the differential area dx dy with 2tr dr:
x
d(area) = 2r dr
y
r
dr
( )
2 2
2
2
2 2 2
2 2 2
0
0
2
2
,
x y
r
r
x x ax
Let r x y dx dy e dr r e
e
dx e dx e and dx e
a
t
t t
t
t t
+
= + =
(
= =
(
| |
= = =
|
\ .
} } }
} } }
Math Tricks That Are Fun and Interesting
sin
dx
x
}
Continuous Infinite Crossings
The following function has an infinite number of zero crossings near the origin, but is everywhere
continuous (even at x = 0). That seems bizarre to me. Recall the definition:
f(x) is continuous at a iff lim ( ) ( )
x a
f x f a
=
Then let
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0
1
sin , 0
( )
0, 0
lim ( ) 0 (0) (0)
x
x x
x
f x
x
f x f f is continuous
| |
=
|
\ . =
= =
Picture
Phasors
Phasors are complex numbers that represent sinusoids. The phasor defines the magnitude and phase of
the sinusoid, but not its frequency. See Funky Electromagnetic Concepts for a full description.
Future Funky Mathematical Physics Topics
1. Finish theoretical importance of IBP
2. Finish Legendre transformations
3. Sturm-Liouville
4. Pseudo-tensors (ref. Jackson).
5. Tensor densities
6. f(z) =
-
dx exp(x^2)/xz has no poles, but has a branch cut. Where is the branch cut, and what
is the change in f(z) across it?
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Appendices
References
[A&S] Abramowitz and Stegun, ??
[Chu] Churchill, Ruel V., Brown, James W., and Verhey, Roger F., Complex Variables and
Applications, 1974, McGraw-Hill. ISBN 0-07-010855-2.
[Det] Dettman, John W., Applied Complex Variables, 1965, Dover. ISBN 0-486-64670-X.
[F&W] Fetter, Alexander L. and John Dirk Walecka, Theoretical Mechanics for Particles and
Continua, McGraw-Hill Companies, February 1, 1980. ISBN-13: 978-0070206588.
[Jac] Jackson, Classical Electrodynamics, 3
rd
ed.
[M&T] Marion & Thornton, 4th ed.
[One] ONeill, Barrett, Elementary Differential Geometry, 2
nd
ISBN 0-12-526745-2.
[Sch] Schutz, Bernard F., A First Course in General Relativity, Cambridge University Press
(January 31, 1985), ISBN 0521277035.
[Sch2] Schutz, Bernard F., Geometrical Methods of Mathematical Physics, Cambridge
University Press ??, ISBN
[Sea] Sean, Seans Applied Math Book, 1/24/2004.
http://www.its.caltech.edu/~sean/book.html.
[Tal] Talman, Richard, Geometric Mechanics, Wiley-Interscience; 1
st
edition (October 4,
1999), ISBN 0471157384
[Tay] Taylor, Angus E., General Theory of Functions and Integration, 1985, Dover. ISBN 0-
486-64988-1.
[W&M] Walpole, Ronald E. and Raymond H. Myers, Probability and Statistics for Engineers and
Scientists, 3
rd
edition, 1985, Macmillan Publishing Company, ISBN 0-02-424170-9.
Glossary
Definitions of common mathematical physics terms. Special mathematical definitions are noted by
(math). These are technical mathematical terms that you shouldnt have to know, but will make reading
math books a lot easier because they are very common. These definitions try to be conceptual and
accurate, but comprehensible to normal people (including physicists, but not mathematicians).
1-1 A mapping from a set A to a set B is 1-1 if every value of B under the map has only one
value of A that maps to it. In other words, given the value of B under the map, we can
uniquely find the value of A which maps to it. However, see 1-1 correspondence. See
also injection.
1-1 correspondence A mapping, between two sets A and B, is a 1-1 correspondence if it uniquely
associates each value of A with a value of B, and each value of B with a value of A.
Synonym: bijection.
accumulation point syn. for limit point.
1
In inner products, hard to define simply (see text). Crudely, the adjoint of an operator is
the operator which preserves the inner product of two vectors as <v|(O|w>) =
(O
|v>)
## |w>. When viewing matrices as operators, the adjoint of a matrix is the
hermitian conjugate. This has nothing to do with matrix adjoints (below).
2
In matrices, the transpose of the cofactor matrix is called the adjoint of a matrix. This has
nothing to do with linear operator adjoints (above).
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analytic A function is analytic in some domain iff it has continuous derivatives to all orders, i.e. is
infinitely differentiable. For complex functions of complex variables, if a function has a
continuous first derivative in some region, then it has continuous derivatives to all orders,
and is therefore analytic.
analytic geometry the use of coordinate systems along with algebra and calculus to study
geometry. Aka coordinate geometry
bijection Both an injection and a surjection, i.e. 1-1 and onto. A mapping between sets A
and B is a bijection iff it uniquely associates a value of A with every value of B.
Synonym: 1-1 correspondence.
BLUE In statistics, Best Linear Unbiased Estimator.
branch point A branch point is a point in the domain of a complex function f(z), z also complex, with
this property: when you traverse a closed path around the branch point, following
continuous values of f(z), f(z) has a different value at the end point of the path than at the
beginning point, even though the beginning and end point are the same point in the
domain. Example TBS: square root around the origin.
boundary point (math) see limit point.
C or the set of complex numbers.
closed (math) contains any limit points. For finite regions, a closed region includes its
boundary. Note that in math talk, a set can be both open and closed! The surface of a
sphere is open (every point has a neighborhood in the surface), and closed (no excluded
limit points; in fact, no limit points).
cofactor The ij-th minor of an nn matrix is the determinant of the (n1)(n1) matrix formed by
crossing out the i-th row and j-th column. A cofactor is just a minor with a plus or minus
sign affixed, according to whether (i, j) is an even or odd number of steps away from
(1,1): ( 1)
i j
ij ij
C M
+
=
compact (math) for our purposes, closed and bounded [Tay thm 2-6I p66]. A compact region may
comprise multiple (infinite number??) disjoint closed and bounded regions.
congruence a set of 1D non-intersecting curves that cover every point of a manifold. Equivalently, a
foliation of a manifold with 1D curves. Compare to foliation.
contrapositive The contrapositive of the statement If A then B is If not B then not A. The
contrapositive is equivalent to the statement: if the statement is true (or false), the
contrapositive is true (or false). If the contrapositive is true (or false), the statement is
true (or false).
convergent approaches a definite limit
converse The converse of the statement If A then B is If B then A. In general, if a statement is
true, its converse may be either true or false. The converse is the contrapositive of the
inverse, and hence the converse and inverse are equivalent statements.
connected There exists a continuous path between any two points in the set (region). See also:
simply connected. [One p178].
coordinate geometry the use of coordinate systems along with algebra and calculus to study
geometry. Aka analytic geometry
diffeomorphism a C
## inverse, from one manifold onto another. Onto implies the
mapping covers the whole range manifold. Two diffeomorphic manifolds are
topologically identical, but may have different geometries.
divergent not convergent: a sequence is divergent iff it is not convergent.
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domain of a function: the set of numbers (usually real or complex) on which the function is
defined.
entire A complex function is entire iff it is analytic over the entire complex plane. An entire
function is also called an integral function.
essential singularity a pole of infinite order, i.e. a singularity around which the function is
unbounded, and cannot be made finite by multiplication by any power of (z z
0
) [Det
p165].
factor a number (or more general object) that is multiplied with others. E.g., in (a + b)(x +y),
there are two factors: (a + b), and (x +y).
finite a non-zero number. In other words, not zero, and not infinity.
foliation a set of non-intersecting submanifolds that cover every point of a manifold. E.g., 3D real
space can be foliated into 2D sheets stacked on top of each other, or 1D curves packed
around each other. Compare to congruence.
holomorphic syn. for analytic. Other synonyms are regular, and differentiable. Also, a holomorphic
map is just an analytic function.
homomorphic something from abstract categories that should not be confused with homeomorphism.
homeomorphism a continuous (1-1) map, with a continuous inverse, from one manifold onto another.
Onto implies the mapping covers the whole range manifold. A homeomorphism that
preserves distance is an isometry.
identify to establish a 1-1 and onto relationship. If we identify two mathematical things, they are
essentially the same thing.
iff if, and only if,
injection A mapping from a set A to a set B is an injection if it is 1-1, that is, if given a value of B
in the mapping, we can uniquely find the value of A which maps to it. Note that every
value of A is included by the definition of mapping [CRC 30
th
]. The mapping does not
have to cover all the elements of B.
integral function Syn. for entire function: a function that is analytic over the entire complex plane.
inverse The inverse of the statement If A then B is If not A then not B. In general, if a
statement is true, its inverse may be either true or false. The inverse is the contrapositive
of the converse, and hence the converse and inverse are equivalent statements.
invertible A map (or function) from a set A to a set B is invertible iff for every value in B, there
exists a unique value in A which maps to it. In other words, a map is invertible iff it is a
bijection.
isolated singularity a singularity at a point, which has a surrounding neighborhood of analyticity
[Det p165].
isometry a homeomorphism that preserves distance, i.e. a continuous, invertible (1-1) map from
one manifold onto another that preserves distance (onto in the mathematical sense).
isomorphic same structure. A widely used general term, with no single precise definition.
limit point of a domain is a boundary of a region of the domain: for example, the open interval (0, 1)
on the number line and the closed interval [0, 1] both have limit points of 0 and 1. In this
case, the open interval excludes its limit points; the closed interval includes them
(definition of closed). Some definitions define all points in the domain as also limit
points. Formally, a point p is a limit point of domain D iff every open subset containing
p also contains a point in D other than p.
mapping syn. function. A mapping from a set A to a set B defines a value of B for every value
of A [CRC 30
th
].
physics.ucsd.edu/~emichels Funky Mathematical Physics Concepts emichels at physics.ucsd.edu
meromorphic A function is meromorphic on a domain iff it is analytic except at a set of isolated poles
of finite order (i.e., non-essential poles). Note that some branch points are not poles
(such as \z at zero), so a function including such a branch point is not meromorphic.
minor The ij-th minor of an nn matrix is the determinant of the (n1)(n1) matrix formed by
the set of natural numbers (positive integers)
oblique non-orthogonal and not parallel
one-to-one see 1-1.
onto covering every possible value. A mapping from a set A onto the set B covers every
possible value of B, i.e. the mapping is a surjection.
open (math) An region is open iff every point in the region has a finite neighborhood of points
around it that are also all in the region. In other words, every point is an interior point.
Note that open is not not closed; a region can be both open and closed.
pole a singularity near which a function is unbounded, but which becomes finite by
multiplication by (z z
0
)
k
for some finite k [Det p165]. The value k is called the order of
the pole.
positive definite a matrix or operator which is > 0 for all non-zero operands. It may be 0 when acting on a
zero operand, such as the zero vector. This implies that all eigenvalues > 0.
positive semidefinite a matrix or operator which is 0 for all non-zero operands. It may be 0 when
acting on a non-zero operands. This implies that all eigenvalues 0.
PT perturbation theory
Q or the set of rational numbers. Q
+
the set of positive rationals.
R or the set of real numbers.
removable singularity an isolated singularity that can be made analytic by simply defining a value for
the function at that point. For example, f(x) = sin(x)/x has a singularity at x = 0. You can
remove it by defining f(0) = 1. Then f is everywhere analytic. [Det p165]
residue The residue of a complex function at a complex point z
0
is the a
1
coefficient of the
Laurent expansion about the point z
0
.
simply connected There are no holes in the set (region), not even point holes. I.e., you can shrink any
closed curve in the region down to a point, the curve staying always within the region
(including at the point).
singularity of a function: a point on a boundary (i.e. a limit point) of the domain of analyticity, but
where the function is not analytic. [Det def 4.5.2 p156]. Note that the function may be
defined at the singularity, but it is not analytic there. E.g., \z is continuous at 0, but not
differentiable.
smooth for most authors, smooth means infinitely differentiable, i.e. C
## . For some authors,
though, smooth means at least one continuous derivative, i.e. C
1
, with first derivative
continuous. This latter definition looks smooth to our eye (no kinks, or sharp points).
surjection A mapping from a set A onto the set B, i.e. that covers every possible value of B. Note
that every value of A is included by the definition of mapping [CRC 30
th
], however
multiple values of A may map to the same value of B.
term a number (or more general object) that is added to others. E.g., in ax + by + cz, there
are three terms: ax, by, and cz.
uniform convergence a series of functions f
n
(z) is uniformly convergent in an open (or partly open)
region iff its error after the N
th
function can be made arbitrarily small with a single
physics.ucsd.edu/~emichels Funky Mathematical Physics Concepts emichels at physics.ucsd.edu
value of N (dependent only on ) for every point in the region. I.e. given , a single N
works for all points z in the region [Chu p156].
voila French for see there!
WLOG or WOLOG without loss of generality
Z or the set of integers. Z
+
or the set of positive integers (natural numbers).
Formulas
completing the square:
2 2
2
(x-shift / 2 )
2 4
b b
ax bx a x b a
a a
| |
+ = + =
|
\ .
Integrals
2 2 2
2 3
3 2
0
1 1
2
2
ax ax ar
dx e dx x e dr r e
a
a a
t t
= = =
} } }
Statistical distributions
2 2
2 2
: 2
exponential :
avg
avg
v
_ v o v
t o t
= =
= =
error function [A&S]:
2
0
2
( )
x
t
erf x e dt
t
}
gaussian included probability between z and +z:
( )
( )
2
2
/ 2 2 2
/ 2
0
1
pdf ( ) / 2 , 2
2
2
2 / 2
2
z z
u
gaussian gaussian
z z
z
t
p z u du e du Let u t du dt
e dt erf z
t
t
+ +
+
= =
= =
} }
}
Special Functions
( ) ( )
1
0
( ) 1 ! ( ) ( ) 1 ( 1) (1/ 2)
a x
n n a dx x e a a a t
I = I I = I I =
}
Index
The index is not yet developed, so go to the web page on the front cover, and text-search in this
document. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9387505650520325, "perplexity": 2363.9613915609525}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670729.90/warc/CC-MAIN-20191121023525-20191121051525-00419.warc.gz"} |
https://www.examcopilot.com/subjects/radio-navigation/dme/d-m-e-hold-entries | # DME Hold Entries RNAV
An unusual question for Radio Navigation, as holds and hold entries are covered in Air Law.
Sector one is a parallel join and sector three is a direct join. These are the only two possible as you are flying either directly along or back along the hold track when the DMEDME —Distance Measuring Equipment arc is completed.
Sector two, an offset join, cannot be achieved in this fashion.
If you have not studied these sector joins yet, do not worry. The answer is sectors 1 and 3, and that answer only. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8579484224319458, "perplexity": 2503.726041113305}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487586239.2/warc/CC-MAIN-20210612162957-20210612192957-00477.warc.gz"} |
https://arxiv-export-lb.library.cornell.edu/abs/2108.02453v1 | math.AP
(what is this?)
# Title: Optimal decay of compressible Navier-Stokes equations with or without potential force
Abstract: In this paper, we investigate the optimal decay rate for the higher order spatial derivative of global solution to the compressible Navier-Stokes (CNS) equations with or without potential force in three-dimensional whole space. First of all, it has been shown in \cite{guo2012} that the $N$-th order spatial derivative of global small solution of the CNS equations without potential force tends to zero with the $L^2-$rate $(1+t)^{-(s+N-1)}$ when the initial perturbation around the constant equilibrium state belongs to $H^N(\mathbb{R}^3)\cap \dot H^{-s}(\mathbb{R}^3)(N \ge 3 \text{~and~} s\in [0, \frac32))$. Thus, our first result improves this decay rate to $(1+t)^{-(s+N)}$. Secondly, we establish the optimal decay rate for the global small solution of the CNS equations with potential force as time tends to infinity. These decay rates for the solution itself and its spatial derivatives are really optimal since the upper bounds of decay rates coincide with the lower ones.
Subjects: Analysis of PDEs (math.AP) MSC classes: 35Q30, 35Q35, 35B40 Cite as: arXiv:2108.02453 [math.AP] (or arXiv:2108.02453v1 [math.AP] for this version)
## Submission history
From: Minling Li [view email]
[v1] Thu, 5 Aug 2021 08:40:19 GMT (34kb)
Link back to: arXiv, form interface, contact. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9363115429878235, "perplexity": 1041.940522259886}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00614.warc.gz"} |
https://www.physicsforums.com/threads/can-this-4-bit-adder-be-further-simplified.639533/ | # Can this 4 bit adder be further simplified?
1. Sep 28, 2012
### Psinter
Hello, I'm new to logic circuits and I was making a 4 bit binary adder and I thought I could maybe simplify it. However, I can't find anything simpler than what I already got. What I wanted to know is: is my circuit already in its most simple form or am I making something wrong in my attempts to simplificate it (meaning it can still be simplificated)?
The circuit works as follows:
We have the 1st 4 bit number: ABCD
We have the 2nd 4 bit number: EFGH
The total sum with one bit at a time would be: (D + H) and (C + G) and (B + F) and (A + E). In other words:
ABCD
EFGH
---------
XXXX
where, (D + H) is the only one that doesn't receive any carry because it's the initial point.
So I need 4 outputs with their respective carries as illustrated:
As you can see the pattern of circuits start at the second sum. Meaning, (C + G), (B + F), and (A + E) are the same circuits. I tried to simplify 1 of them so I could simplify the rest of the adder. However, I can't find any way to simplify it any further. When I look at the Karnaugh Map I made out of the Truth Table of the circuit that repeats I don't know what to do from there because I don't have any mapping groups there.
Truth Table: $$((C \oplus G) \oplus CARRY) \wedge CARRY$$
\begin{array}{|c|c|c|c|}
\hline
C & G & CARRY & F1 \\ \hline
0 & 0 & 0 & 0 \\ \hline
0 & 0 & 1 & 1 \\ \hline
0 & 1 & 0 & 0 \\ \hline
0 & 1 & 1 & 0 \\ \hline
1 & 0 & 0 & 0 \\ \hline
1 & 0 & 1 & 0 \\ \hline
1 & 1 & 0 & 0 \\ \hline
1 & 1 & 1 & 1 \\
\hline
\end{array}
Karnaugh Map: $$((C \oplus G) \oplus CARRY) \wedge CARRY$$
\begin{array}{|c|c|c|c|c|c|}
\hline
& G \wedge CARRY & 00 & 01 & 11 & 10 \\ \hline
C & & & & & \\ \hline
0 & & & 1 & & \\ \hline
1 & & & & 1 & \\
\hline
\end{array}
Does all this mean that I cannot simplify my adder any further?
2. Oct 4, 2012
### Psinter
This is wrong, never mind, this adder is wrongly designed. This is not an adder. It says that 0 + 0 = 1 with carry 1 if the previous sum has a carry of 1. We know that this is not true so this is wrong.
3. Oct 7, 2012
### Enthalpy
I confirm it's wrong, because each pair of bits must enter an AND gate. The XOR doesn't suffice.
If you like circuitry, you can give a look at carry propagation and anticipation, both for adders and counters.
4. Oct 7, 2012
### skeptic2
How do you want to simplify it - by reducing the number of ICs? One way to simplify it would be to replace all the ICs with a PROM. Use the address lines as inputs and the data lines as outputs. For every combination of inputs (all addresses) program the correct output. You could even have one of the output lines designated as a carry out and one of the inputs as carry in. That way they could be cascaded.
5. Oct 10, 2012
### Enthalpy
http://www.physics.wisc.edu/undergrads/courses/fall2012/623/ds/74LS283.pdf
http://en.wikipedia.org/wiki/Adder_(electronics)
http://www.google.de/search?q="4+bi...DsbZsgbDpYHoCQ&ved=0CB4QsAQ&biw=1200&bih=1728
Beware that what was a "simple" design with TTL gates isn't "simple" with present-day CMOS, and a "simple" breadboard isn't a "simple" chip design - a simple breadbord would buy a 4-bit full adder chip...
Especially, CMOS can make good XOR gates with 3 inputs, as well as good majority logic gates (= at least 2 inputs of 3 are logical 1).
http://book.huihoo.com/design-of-vlsi-systems/ch06/ch06.html
Interesting : the LS283 computed the carry over 4 bits directly to save time, despite it takes more gates. Again, many-inputs-Nand fit LS very well, and heavy output loading was a small penalty ; different story with CMOS.
Carry lookahead saves time by splitting the 1-bit adder carry signal into a "generate carry" and "propagate carry" (if one carry is input to this bit). These two signals can enter the lookahead unit which, for with words, is a tree.
Depicted here as a binary tree:
http://net.pku.edu.cn/~course/cs101/2007/resource/Intro2Algorithm/book6/chap29.htm
but normally it would be a radix-four tree:
http://pages.cs.wisc.edu/~jsong/CS352/Readings/CLAs.pdf [Broken]
Carry lookahead exist for counters as well but are easier because a counter passes through FC and FE before attaining FF, giving more time.
Last edited by a moderator: May 6, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9800403714179993, "perplexity": 696.7682274338797}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267155817.29/warc/CC-MAIN-20180919024323-20180919044323-00217.warc.gz"} |
https://math.stackexchange.com/questions/81583/how-do-i-prove-that-xp-xa-is-irreducible-in-a-field-with-p-elements-when/81637 | # How do I prove that $x^p-x+a$ is irreducible in a field with $p$ elements when $a\neq 0$?
Let $p$ be a prime. How do I prove that $x^p-x+a$ is irreducible in a field with $p$ elements when $a\neq 0$?
Right now I'm able to prove that it has no roots and that it is separable, but I have not a clue as to how to prove it is irreducible. Any ideas?
• When reading recently an article about the Artin-Schreier theorem, some properties of the so-called Artin extensions were used, and, if no mistakes occur here, those are intimately related to the polynomials of the form $x^p-x+a$. Is there indeed any error that occur? and is there any reference to know more in this direction? Thanks in advance. – awllower Nov 13 '11 at 14:05
• @awllower: This question may get you started? – Jyrki Lahtonen Nov 13 '11 at 16:39
• @JyrkiLahtonen: Thanks for the question. I really appreciate this. – awllower Nov 13 '11 at 16:46
• The original version of the question (which I've now edited) omitted the requirement that $p$ be a prime. This requirement was necessary: The polynomial $x^4-x+a$ over $\mathbb{F}_4$ is reducible for every $a \in \mathbb{F}_4$ ! – darij grinberg Sep 18 '16 at 3:20
• This is also a particular case of math.stackexchange.com/questions/136164 . – darij grinberg Sep 18 '16 at 3:34
Greg Martin and zyx have given you IMHO very good answers, but they rely on a few basic facts from Galois theory and/or group actions. Here is a more elementary but also a longer approach.
Because we are in a field with $p$ elements, we know that $p$ is the characteristic of our field. Hence, the polynomial $g(x)=x^p-x$ has the property $$g(x_1+x_2)=g(x_1)+g(x_2)$$ whenever $x_1$ and $x_2$ are two elements of an extension field of $\mathbb{F}_p$. By little Fermat we know that $g(k)=k^p-k=0$ for all $k\in \Bbb{F}_p$. Therefore, if $r$ is one of the roots of $f(x)=x^p-x+a$, then $$f(r+k)=g(r+k)+a=g(r)+g(k)+a=f(r)+g(k)=0,$$ so all the elements $r+k$ with $k \in \Bbb{F}_p$ are roots of $f(x)$, and as there are $p$ of them, they must be all the roots. It sounds like you have already shown that $r$ cannot be an element of $\Bbb{F}_p$.
Now assume that $f(x)=f_1(x)f_2(x)$, where both factors $f_1(x),f_2(x)\in \Bbb{F}_p[x]$. From the above consideration we can deduce that $$f_1(x)=\prod_{k\in S}(x-(r+k)),$$ where $S$ is some subset of the field $\Bbb{F}_p$. Write $\ell=|S|=\deg f_1(x)$. Expanding the product we see that $$f_1(x)=x^\ell-x^{\ell-1}\sum_{k\in S}(r+k)+\text{lower degree terms}.$$ This polynomial was assumed to have coefficients in the field $\Bbb{F}_p$. From the above expansion we read that the coefficient of degree $\ell-1$ is $|S|\cdot r+\sum_{k\in S}k$. This is an element of $\Bbb{F}_p$, if and only if the term $|S|\cdot r\in\Bbb{F}_p$. Because $r\notin \Bbb{F}_p$, this can only happen if $|S|\cdot1_{\Bbb{F}_p}=0_{\Bbb{F}_p}$. In other words $f_1(x)$ must be either of degree zero or of degree $p$.
• Well done, it is a good proof. – awllower Nov 13 '11 at 14:01
• I love your proof. One thing is bothering me, though. And It's probably really obvious. How do you know that if $|S|\cdot r\in F_p$, then $|S|$ must be a multiple of $p$ in order for $|S|\cdot r$ to be in $F_p$? I know it has to do with the fact that $r\notin F_p$, and I have some intuition for it, but I don't know how to prove it. – MathTeacher Nov 14 '11 at 4:21
• @MathMastersStudent: If $|S|$ is not a multiple of $p$, then $|S|\cdot 1$ is an invertible element of $F_p$. So if $|S|\cdot r= b$ with $b\in F_p$, then $r=b|S|^{-1}$ would be in the prime field $F_p$ as well contradicting known facts. – Jyrki Lahtonen Nov 14 '11 at 7:20
• Great answer as usual, Jyrki: +1. Just to show you how pathologically nit-picking some guys are, I would write $|S|\cdot 1_{F_p}=0_{F_p}$ rather than $|S|=0_{F_p}$, since $S$ is an integer and the integers are not included in a finite field... – Georges Elencwajg Jul 24 '13 at 8:05
• This argument also applies to any field of characteritic $p$. – Lao-tzu Dec 31 '14 at 12:41
$x \to x^p$ is an automorphism sending $r$ to $r-a$ for any root $r$ of the polynomial. This operation is cyclic of order $p$, so that one can get from any root to any other by applying the automorphism several times. The Galois group thus acts transitively on the roots, which is equivalent to irreducibility.
• Impressively elegant, zyx: +1 – Georges Elencwajg Jul 24 '13 at 8:12
$f(x)$ is separable since its derivative is $f'(x) = -1 \ne 0$.
Suppose $\theta$ is a root of $f(x) = x^p - x + a$. Using the Frobenius automorphism, we have: \begin{align} f(\theta + 1) &= (\theta + 1)^p - (\theta + 1) + a\\ &= \theta^p + 1^p - \theta - 1 + a\\ &= \theta^p - \theta + a\\ &= f(\theta) = 0 \end{align}
Thus, by induction, if $\theta$ is a root of $f(x)$, then $\theta + j$ is also a root for all $j \in \mathbb F_p$.
By above, if $f(x)$ were to have a root in $\mathbb F_p$, then $0$ would a be a root too, but this contradicts $a \ne 0$. Thus, $f(x)$ has no roots in $\mathbb F_p$. (This can also be shown using Fermat's little theorem.)
Suppose $\theta$ is a root of $f(x)$ in some extension of $\mathbb F_p$. We know that $\theta + j$ is also a root for all $j \in \mathbb F_p$. Since $f(x)$ is of degree $p$, these are all of the roots of $f(x)$.
Clearly, $\mathbb F_p(\theta) = \mathbb F_p(\theta + j)$ for all $j \in \mathbb F_p$. Thus, all $\{\theta + j\}$ have the same degree over $\mathbb F_p$. Since $f(x)$ is separable, it follows that $f(x)$ must be the product of all minimal polynomials of $\{\theta + j\}$. Suppose the minimal polynomials have degree $m$. We have $p = km$ for some $k$. Since $p$ is prime, either $m = 1$; hence $\theta \in \mathbb F_p$, a contradiction. Or $k = 1$; hence $f(x)$ is irreducible because it's the minimal polynomial.
• Ah, I already accepted it. I want to accept it another time but it will get as unaccepted :P Thank You for this wonderful proof :) :) :) – user87543 Aug 3 '13 at 10:03
• A good one! ${}$ – Jyrki Lahtonen Aug 3 '13 at 10:06
• Happy to help! I have it in my notes. I don't actually remember if I came up with it myself or found it somewhere. – Ayman Hourieh Aug 3 '13 at 10:11
• What ever may be the source, Your Answer is good :) – user87543 Aug 3 '13 at 10:16
I think the following idea works. Let $f(x) = x^p-x+a$. They key observation is that $f(x+1)=f(x)$ in the field of $p$ elements. Now factor $f(x) = g_1(x) \cdots g_k(x)$ as a product of irreducibles. Sending $x$ to $x+1$ must therefore permute the factors $\{ g_1(x), \dots, g_k(x) \}$. But sending $x$ to $x+1$ $p$ times in a row comes back to the original polynomial, so this permutation of the $k$ factors has order dividing $p$. It follows that either every $g_j(x)$ is fixed by sending $x$ to $x+1$ - which I think is a property that no nonconstant polynomial of degree less than $p$ can have, but that needs proof - or else there are $k=p$ factors, which can only happen in the case $a=0$.
• This probably needs the separability of $f$, because if some of the $g_1,g_2,\ldots,g_k$ could be equal, the concept of a permutation and its order would be somewhat muddled. – darij grinberg Sep 18 '16 at 3:25
$x^p-x+a$ divides $x^{p^p}-x$. If $f$ is an irreducible divisor of $x^p-x+a$ of degree $d$ then $\mathbf{Z}_p[x]/f$ will be a subfield of the field with $p^p$ elements so $p^p = (p^d)^e$ and so $d=1$ or $e=1$. since $x^p-x+a$ has no roots $e=p.$
One more proof, similar to Greg Martin's: Suppose $\alpha$ is a root of $f(x)=x^p-x+a$ in some splitting field; then \begin{equation*} (\alpha+1)^p - (\alpha+1) + a = \alpha^p + 1 - \alpha - 1 + a = \alpha^p - \alpha + a = 0, \end{equation*} so that $\alpha+1$ is also a root. It follows that the roots of $f$ are $\alpha+i$ for $0\le i < p$. If $f$ factors in $\mathbb{F}[x]$, say $f = gh$, then the sum of the roots of $g$ is $k\alpha + r$ where $\deg g = k$ and $k, r\in\mathbb{F}_P$. Since $g\in \mathbb{F}_p[x]$ it follows that $\alpha\in \mathbb{F}_p$. But that implies that $f$ splits in $\mathbb{F}_p$, which is not the case (for example, neither $0$ nor $1$ is a root). Thus $f$ is irreducible.
• This is a very helpful and simple answer. Can I ask why it is not the case why f splits in Fp? – P-S.D Apr 1 '17 at 23:20
• Also, by f splits in Fp, do you mean that Fp is the splitting field of f? Why would Fp being the splitting field of f imply that 0 or 1 is a root? – P-S.D Apr 1 '17 at 23:26
• @P-S.D If $\alpha\in\mathbb{F}_p$, then the $p$ elements $\alpha+i$, $0\le i<p$, are all in $\mathbb{F}_p$ and are all roots of $f$. But $f$ has degree $p$, so it clearly splits in $\mathbb{F}_p$ and thus each of the $p$ elements of that field, including $0$ and $1$, is a root of $f$. – rogerl Apr 2 '17 at 14:53
The supposition of Greg Martin is truth, if a polinomyal $f$ with $deg(f)=n$ satisfies the property, then $n\ge p$, by contradiction argument, just write the expansion with the Newton's formula and analyse the coeficient of $x^{n-1}$ term, you get $\binom n 1a_{n}+a_{n-1}=a_{n-1}$, if $n\lt p$, this equation is an absurd. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9597019553184509, "perplexity": 125.76826470674644}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998844.16/warc/CC-MAIN-20190618223541-20190619005541-00025.warc.gz"} |
https://www.physicsforums.com/threads/solve-diff-eq-using-power-series.908293/ | # Homework Help: Solve Diff. Eq. using power series
Tags:
1. Mar 19, 2017
### JamesonS
1. The problem statement, all variables and given/known data
(1-x)y^{"}+y = 0
I am here but do not understand how to combine the two summations:
Mod note: Fixed LaTeX in following equation.
$$(1-x)\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^{\infty}a_nx^n = 0$$
Last edited by a moderator: Apr 1, 2017
2. Mar 19, 2017
### Ray Vickson
Here is your equation using PF-compatible TeX:
$$(1-x)\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} x^n+\sum_{n=0}^{\infty}a_nx^n = 0$$
Just replace the "\ begin {equation} ... \ end {equation} " by " ... " (no spaces between the initial and final \$ signs). Also: write "\infty", not "\infinity".
As for your question: write out the first 3 or 4 terms, to see what you get. That will give you insight into what you should do next.
Last edited by a moderator: Apr 1, 2017
3. Mar 19, 2017
### JamesonS
Thanks for the response and Latex help. Writing out the first few terms of each sum:
$$(1-x)[2a_2+6a_3x+12a_4x^2+...]+[a_0+a_1x+a_2x^2+...]$$
I am not sure what to do with the (1-x) term outside the first sum...
Last edited: Mar 19, 2017
4. Mar 19, 2017
### Ray Vickson
What is preventing you from "distributing out" the product? That is, $(1-x) P(x) = P(x) - x P(x).$
Last edited: Mar 20, 2017
5. Mar 31, 2017
### MidgetDwarf
Been a while since I did DE, but doesn't the OP have to be aware of the singular point in this problem? Hence he has to use the method of Frobenius?
6. Apr 1, 2017
### LCKurtz
But the singular point isn't at $x=0$.
7. Apr 2, 2017
### MidgetDwarf
Thanks for the correction. It's been a while. I remembered that there is no singular point if we take the Taylor expansion about x=0? Correct?
8. Apr 2, 2017
### epenguin
Practically after your first equation, or at any later stage, just multiply it out. You have shown that you know how to write a sum of different powers of x in terms of xn by changing the subscript appropriately.
9. Apr 2, 2017
### vela
Staff Emeritus
That's backwards. If you expand about a regular point, then the solution can be written as a Taylor series. If there's a singular point, then you can use the method of Frobenius and end up with a Laurent series.
10. Apr 2, 2017
### MidgetDwarf
Thanks Vela!
It has been a while. I may pop open a differentials to bring the memory back. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9047558307647705, "perplexity": 1586.9298928307878}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867559.54/warc/CC-MAIN-20180526151207-20180526171207-00117.warc.gz"} |
https://www.physicsoverflow.org/36282/question-about-notation-writing-moduli-space-string-theory | # Question about notation used in writing the moduli space in string theory
+ 2 like - 0 dislike
118 views
In physics papers, particularly those by Aspinwall, or textbooks, I encounter things like
$$\mathcal{M} \simeq O(\Gamma_{4,20})\setminus O(4,20)/((O(4)\times O(20))$$
For instance, this is from https://arxiv.org/abs/hep-th/9707014. Am I correct in understanding that the numerator is $O(\Gamma_{4,20})$ folllowed by a set theoretic subtraction of $O(4,20)$?
If I wanted to compute the dimension of $\mathcal{M}$, I know I'd have to subtract the dimension of the numerator from the dimension of the denominator (which is just $(4\times 3/2) + (20 \times 19/2)$). What is the dimension of the numerator?
This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user leastaction
asked Jun 6, 2016
That rather looks like a double coset to me, i.e. you're quotienting a left action of $\mathrm{O}(\Gamma)$ and a right action of $\mathrm{O}(4)\times\mathrm{O}(20)$ out of $\mathrm{O}(4,20)$
This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user ACuriousMind
Ah, thank you! So does that mean that if one writes $G_1\setminus H / G_2$, then the dimension will be $dim(H) - dim(G_1) - dim(G_2)$?
This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user leastaction
The group acting from the left is a discrete group (dimension 0) consisting of transformations of a lattice, similar to a matrix group with integer entries. As for the dimension, it will be at least that, but may be higher if the actions are not faithful/effective.
This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user doetoe
Thank you @doetoe!
This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user leastaction
It is useful to remember that the dimension of $\mathcal{M} \simeq O(\Gamma_{p,q})\setminus O(p,q)/((O(p)\times O(q))$ is just $pq$.
This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user user40085
## 2 Answers
+ 2 like - 0 dislike
This is not really an answer (the answer is ACuriousMind's comment: this is a double coset space), but it may help to consider the construction of the moduli space of elliptic curves, as this can be done in the same way but is very easy.
Every complex elliptic curve is obtained as $\Bbb C$ modulo a lattice. Scaling the lattice by a complex number gives an isomorphic curve, so you can scale in such a way that the lattice is generated by 1 and by $\tau\in\Bbb H$, the complex upper half plane (this is similar to gauge fixing). Not all $\tau$ give different lattices: two sets of generators give the same lattice if they are related by an element of $GL_2(\Bbb Z)$ (there is some residual gauge freedom). Since we work with oriented bases, we can restrict to $SL_2(\Bbb Z)$. It is not hard to see that the action of $SL_2(\Bbb Z)$ on a basis $1,\tau$ corresponds to an action on $\Bbb H$ by Möbius transformations
$$\begin{pmatrix}a & b\\ c & d\end{pmatrix}\tau = \frac{a\tau + b}{c\tau + d}$$
This gives us the moduli space as a quotient
$$\mathcal M \cong SL_2(\Bbb Z)\backslash \Bbb H$$
In general, if you have a space (possibly with some extra structure like a Riemannian metric) on which some group of automorphisms acts transitively (i.e. every point can be mapped to every point) by some mapping of the whole space onto itself, then this space can be written as the quotient of this group by the stabilizer (i.e. the subgroup fixing a given point) of any point. This is the orbit-stabilizer theorem.
In our example, the complex upper half plane $\Bbb H$ has an obvious complex structure as a subset of $\Bbb C$, and its group of holomorphic automorphisms is $SL_2(\Bbb R)$ acting by Möbius transformations, except that $+I$ and $-I$ do the same thing, and the automorphisms are really $SL_2(\Bbb R)/\langle-I\rangle = PSL_2(\Bbb R)$. The stabilizer of the point $i$ is $SO_2(\Bbb R)\subset PSL_2(\Bbb R)$, so that
$$\mathcal M \cong SL_2(\Bbb Z)\backslash PSL_2(\Bbb R)/SO_2(\Bbb R)$$
This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user doetoe
answered Jun 6, 2016 by (125 points)
Thank you @doetoe for a comprehensive answer!
This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user leastaction
+ 2 like - 3 dislike
There are two things going on. One is modulo that is the forwards slash / and the other is set-minus $\setminus$ the backwards slash. The $$G_4(20) = \frac{O(4,20)}{O(4)\times O(20)}$$ is the Grassmanian space defined by $4$-planes. the group $O(\Gamma_{4,20})$ is an orthogonal group over the unimodular transformations, a bit like saying $O(n,\mathbb Z)$, where Aspinwall introduces this on the two-torus earlier in the paper. The moduli space is this orthogonal group "minus" these Grassmanians.
This post imported from StackExchange Physics at 2016-06-08 08:49 (UTC), posted by SE-user Lawrence B. Crowell
answered Jun 6, 2016 by (590 points)
This is wrong. There is not set-minus but one left quotient and one right quotient, as explained in the comments to the question and in doetoe answer.
## Your answer
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https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_25&curid=18141&oldid=151749 | # 2020 AMC 12A Problems/Problem 25
## Problem
The number , where and are relatively prime positive integers, has the property that the sum of all real numbers satisfying is , where denotes the greatest integer less than or equal to and denotes the fractional part of . What is ?
## Solution 1
Let be the unique solution in this range. Note that is also a solution as long as , hence all our solutions are for some . This sum must be between and , which gives and . Plugging this back in gives .
## Solution 2
First note that when while . Thus we only need to look at positive solutions ( doesn't affect the sum of the solutions). Next, we breakdown down for each interval , where is a positive integer. Assume , then . This means that when , . Setting this equal to gives We're looking at the solution with the positive , which is . Note that if is the greatest such that has a solution, the sum of all these solutions is slightly over , which is when , just under . Checking this gives ~ktong
## Video Solution 1 (Geometry)
This video shows how things like The Pythagorean Theorem and The Law of Sines work together to solve this seemingly algebraic problem: https://www.youtube.com/watch?v=6IJ7Jxa98zw&feature=youtu.be
## Video Solution 3 (by Art of Problem Solving)
Created by Richard Rusczyk
## Remarks
### Graph
Let
We make the following table of values:
We graph by branches:
~MRENTHUSIASM (Graph by Desmos: https://www.desmos.com/calculator/ouvaiqjdzj)
### Extensions
Visit the Discussion Page for the underlying arguments and additional questions.
~MRENTHUSIASM | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9867926836013794, "perplexity": 621.1007132794549}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989018.90/warc/CC-MAIN-20210509213453-20210510003453-00619.warc.gz"} |
https://math.dartmouth.edu/~ahb/thesis_html/node134.html | Next: Convergence with number of Up: Dissipation in Deforming Chaotic Previous: Appendix F: Cross correlations
# Appendix G: Numerical evaluation of wavefunction boundary integrals
Boundary methods are a central component of this thesis. Closed integrals of a function over the boundary coordinate are ubiquitous. Generally and a square matrix of integrals
(G.1)
is required, where the indices label multiple functions. For evaluation of the quantum band profile (Chapters 2 and 3), the local density of states (Chapter 6), the tension and area-norm matrices (Chapter 5), and the Vergini matrix and its derivative (Chapter 6), and are basis functions or eigenstates which oscillate about zero on the length scale , the quantum (de Broglie) free-space wavelength. The deformation function boundary integrals (Chapter 4) do not involve any quantum scale, but are also evaluated using the method below. I will present only the case where boundary integrals over become line integrals over ; the generalization to higher is simple.
My tool for evaluation of an integral on a closed curve is the discretization
(G.2)
where is the range of , that is, the length of the line integral (billiard perimeter). The points are spread uniformly (equidistant in ) along the closed curve. Because no point is special, no special quadrature [161] weights arise near any endpoints: all weights are equal. More sophisticated and accurate approximations exist for closed line integral evaluation [58], however this is sufficient for my needs and is very simple to code. Its errors will be discussed and tested below.
A single integral (G.2) requires function evaluations of . Naively one might guess that filling a matrix using (G.1) requires evaluations. However, the correct way to compute (G.1) requires only such evaluations: First fill the rectangular matrices and , from which follows
(G.3)
This matrix multiplication does require operations, but being simple adds and multiplications (and using optimized library code e.g. BLAS), it is very fast and does not affect the scaling. If you like, the matrix multiply `performs' the integration over . In the case where and are the same function, only evaluations are required. Note that if a general weighting function is required in the integrand (G.1), it can easily be incorporated into or , or equivalently be included as a diagonal matrix inserted between and in (G.3).
Subsections
Next: Convergence with number of Up: Dissipation in Deforming Chaotic Previous: Appendix F: Cross correlations
Alex Barnett 2001-10-03 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9679244756698608, "perplexity": 1282.3157587384399}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583509845.17/warc/CC-MAIN-20181015205152-20181015230652-00199.warc.gz"} |
https://forums.powerarchiver.com/topic/3278/crc-error-when-out-of-space/5?lang=en-US | # CRC error when out of space
• When I try to extract a multi-part RAR file (i.e. r01, r02) to a drive that runs out of space during extraction, PowerArchiver reports a CRC error in the part that it is extracting from when the drive runs out of space… freeing up space and re-attempting extraction works fine with no CRC errors.
This has been going on for awhile because I’ve had it happen before with earlier versions. Is it a known problem?
Thanks.
• So what’s the problem? PowerArchiver reports an error because there is no enough space and so it is not possible to restore the original file there.
• But it doesn’t report an error saying that there’s no space left. It says there’s a CRC error in the file it’s extracting when space runs out when that’s not the real problem.
• it is general error that says many things, including crc and lack of space, right? :-).
we will see if we can be more specific, thanks.
• No, the error expliticly mentions a CRC error and only a CRC error.
The exact text is as follows, in a dialog with a red X icon:
“There is a CRC error on file:<file being=”" extracted="" from="" multi-part="" rar="">
Volume:<part being="" extracted="" when="" out="" of="" space="">
Continue?"
…with Yes/No buttons.
If I then try and extract the same file to a drive that has enough space, it succeeds without any problems.</part></file>
• @mbg:
No, the error expliticly mentions a CRC error and only a CRC error.
The exact text is as follows, in a dialog with a red X icon:
“There is a CRC error on file:<file being=”" extracted="" from="" multi-part="" rar="">
Volume:<part being="" extracted="" when="" out="" of="" space="">
Continue?"
…with Yes/No buttons.
If I then try and extract the same file to a drive that has enough space, it succeeds without any problems.</part></file>
great, we will see if we can improve it… thanks!
• This problem has been bugging me for a long time now. CRC error implies that the archive is corrupt and not “not enough space”. To get rid of this problem just check if there’s enough space on the drive for all the files in the archive…and warn user if there’s not enough space. Since every type of archive doesn’t contain extracted file sizes you can do a “free space” check right before poping up the error message and decide on what message to display. That’s VERY easy to code… :cool:
• it was wrong error shown, it will be fixed, i think it is in internal versions.
• This post is deleted!
• please check with official beta 1, and let us know if this has been fixed… thank you! | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8153259754180908, "perplexity": 2360.348164522575}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735916.91/warc/CC-MAIN-20200805065524-20200805095524-00324.warc.gz"} |
https://mathematica.stackexchange.com/questions/113883/huge-difference-after-changing-a-fraction-to-decimal | # Huge difference after changing a fraction to decimal
I have a limit to calculate.
With[{p = 1/2, q = 0.1},
Limit[a^q Integrate[Sin[x]/x^p, {x, a, Infinity}], a -> Infinity]]
gives correct result 0. But change that $\frac{1}{2}$ to $0.5$,
With[{p = 0.5, q = 0.1},
Limit[a^q Integrate[Sin[x]/x^p, {x, a, Infinity}], a -> Infinity]]
gives ComplexInfinity which very far away from the correct answer. Why does machine precision and infinite precision have this huge difference without any warning?
FYI, for any decimal p, using decimal gives incorrect result, while the true value is 0.
• The difference seems to lie in the integration; when p == 1/2 the integral is expressed as a function of FresnelS; in the numerical case of p == 0.5, instead, the integral is expressed with HypergeometricPFQ. Unfortunately I don't know enough about special functions to go any further than that, but others on this site do, so hopefully they may be able to weigh in on this. – MarcoB Apr 28 '16 at 18:59
• Seems like a possible bug to me. This integral looks convergent to me for any p>q regardless, machine or infinite precision. – LLlAMnYP Apr 28 '16 at 19:32
• @LLlAMnYP Mathematica gives a result for the integral. Then claims the limit diverges for decimal input. No bug there. – Daniel Lichtblau Apr 28 '16 at 20:08
• @DanielLichtblau the integral should have a convergent, though not necessarily 0 result for all p>0. The limit as a whole, though I have not taken the time to rigorously prove this, should reduce to zero for a -> Infinity and p>q. If there was a case, that only exactly 1/2 works, while otherwise some important condition is not met, I would agree with you, but IMO here the result should be zero for a range of ps. – LLlAMnYP Apr 28 '16 at 20:36
• Possibly shows a bug in Series handling of HypergeometricPFQ at infinity, when input to PFQ is approximate. – Daniel Lichtblau Apr 28 '16 at 21:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8349422812461853, "perplexity": 983.422675113494}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195529175.83/warc/CC-MAIN-20190723085031-20190723111031-00207.warc.gz"} |
https://www.physicsforums.com/threads/strength-of-magnetic-fields.356238/ | Strength of magnetic fields
• Start date
• #1
7
0
I've been given the question to compute the strength in the van allen belts when given the strength of Earth's magnetic field at the surface. Then, I am supposed to calculate the gyroradius of a 50 MeV proton from the strengths I come up with. I want to know if I am going about this correctly.
Is it ok to use say that B = B (at Earth's surface) * (1r (Earth)/ r (belt radius in terms of Earth radius) ^3
For example, if B at surface is 4 *10^-5 T and inner belt radius = 1.5 (earth radii) then I will get the equation:
B = 4 *10^-5(1/1.5)^3 = 1.185 * 10^-5 T
As for the gyroradius, the formula is r = (mv/qB)--can I go ahead and assume the velocity of the proton is equal to the speed of light. The notes that I am using to carry out this problem does not specify any way of getting the velocity of a particle moving through the belts.
Please let me know if this looks ok! Any help would be appreciated. Thanks!!
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https://www.physicsforums.com/threads/spinning-pencil-think-top-urgent-help-wanted.158488/ | # Spinning Pencil (think top) Urgent help wanted
1. Feb 27, 2007
### ^_^physicist
1. The problem statement, all variables and given/known data
A pencil is set spinning in an upright position. How fast must the spin be for the pencil to remain in the upright position? Assume that the pencil is a uniform cylinder of length a and diameter b. Find the value of the spin in revolutions per second for a=20cm and b=1cm. Express results in revolutions per second
2. Relevant equations
Assuming that this is a top problem: $$S^2*(I_s)^2 > (4mglI)$$
is the only relavent equation I can think of. Where $$I_s$$ = moment of inertia about the symmetry axis, and I= moment about the axes normal to the symmetry axis.
3. The attempt at a solution
Taking the above equation for the necessary speed for "sleeping" to occur (that is maintaining a the vertical), I mainpulate the equation into
S > [(4*m*g*l*I)/(I_s)^2)]^(1/2). Now noting that this is a cylinder I come up with I_s=m*(a^2)/2 and I= m*(a^2/4+b^2/12); however, when I plug these values into the express I am not getting what the back of the book is getting, in fact if I plug in numbers I am off by a signifigant amount (3 orders of magnatute). So, from what I can tell I am making a mistake somewhere on figuring out the I and I_s values.
The back of the book gives the equation as
S > [ (128*g*a)/(b^4)*(a^2/3+b^2/16) ]^(1/2) = (when values are inserted as indicated in the question) 2910 RPS.
Any ideas on how to figure out I and I_s
Last edited: Feb 27, 2007
Can you offer guidance or do you also need help?
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https://physics.stackexchange.com/questions/123263/hamiltonian-with-dirac-delta-function | # Hamiltonian with Dirac Delta function
I've to compute this expression
$$\hat{H} ~=~\frac{1}{4}g_2\int d^3R\int d^3r\ \bar{\Psi}(\vec{R}+\frac{\vec{r}}{2})\bar{\Psi}(\vec{R}-\frac{\vec{r}}{2})$$$$\times \left[ \delta(\vec{r})\nabla_{\vec{r}}^2 +\nabla_{\vec{r}}^2\delta(\vec{r}) \right]\Psi(\vec{R}+\frac{\vec{r}}{2}) \Psi(\vec{R}-\frac{\vec{r}}{2}), \tag{15}$$
where $$\bar{\Psi}$$ is the conjugate of $$\Psi$$.
Using Dirac delta properties, Can I say that
$$\left[ \delta(\vec{r})\nabla_{\vec{r}}^2 +\nabla_{\vec{r}}^2\delta(\vec{r}) \right] = 2 \delta(\vec{r})\nabla_{\vec{r}}^2~?$$
If not, how can I calculate this integral?
I should obtain $$\hat{H} = \frac{1}{4}g_2\int d^3R\ \bar{\Psi}(\vec{R})\left[ \nabla^2(\bar{\Psi}(\vec{R})\ \Psi(\vec{R}))\right]\Psi(\vec{R}). \tag{16}$$ A method should be expanding $$\Phi = V^{-1/2} \sum_\alpha a_\alpha e^{i\textbf{k}_\alpha\cdot\textbf{r}}$$, but I don't have any idea what doing!
This integrals (15) comes from the paper Phys. Rev. A 67 053612 and authors say they do integration for part and then over $$\textbf{r}$$.
Does anyone have ideas how to calculate this integral?
/// Update ///
I tried to calculate the integral using yours suggestions. I'm near the solution! At the last there is an extra term and an extra $$1/2$$. In the followed images, the conjugate is $$\phi^*$$ and I indicate $$\phi_+ = \Psi(\vec{R}+\frac{\vec{r}}{2})$$ and $$\phi_- = \Psi(\vec{R}-\frac{\vec{r}}{2})$$
The first term is
and the second is
Summing,
$$\int d^3\vec{R}\int d^3\vec{r}\nabla^2_{\vec{r}}\left( \phi_+^* \phi^*_-\phi_-\phi_+ \right)\delta(\vec{r})$$
and then, at the last $$\int d^3\vec{R}\ \frac{1}{2}(\phi^*\phi(\phi\nabla^2\phi^*+\phi^*\nabla^2\phi) - \phi^2|\nabla\phi^*|^2-\phi^{*2}|\nabla\phi|^2)$$ and completing it by adding and subtracting $$2\phi\phi^*\nabla\phi^*\cdot\nabla\phi$$
$$\int d^3\vec{R}\ \frac{1}{2}\phi^*\left( \nabla^2(\phi^*\phi)\right)\phi - \int d^3\vec{R}\ \frac{1}{2} (\nabla(\phi\phi^*))^2$$ I did all calculations by hand and then i checked them with mathematica.
is the term $$\int d^3\vec{R}\ (\nabla(\phi\phi^*))^2 = 0$$ for any reasons? I hope yes.
Why is there the constant $$1/2$$?
• – ACuriousMind Jul 5 '14 at 17:51
• @ACuriousMind thanks. The curiosity is that the result of the integral depends on the laplacian of product between the conjugate of $\Psi$ and itself. And, integrating by parts, i don't understand why the conjugate is inside the laplacian – apt45 Jul 5 '14 at 18:01
Hints to the sought-for formula (16) for $\hat{H}$:
1. Use integration by parts in ${\bf r}$-space to remove derivatives from the Dirac delta distributions, cf. comment by user ACuriousMind.
2. Work on the problem from both ends (15) and (16). Use Leibniz rule $$\tag{*}\nabla^2 (fg)~=~ g\nabla^2 f + f \nabla^2 g+ 2 \nabla f\cdot\nabla g,$$ so that $\nabla$ only acts on single objects everywhere. Let's call the last term in eq. (*) for a 'cross-term'.
3. Change the derivative $\nabla_{\bf r}~$ to $~\pm\frac{1}{2}\nabla_{\bf R}$.
4. Perform the $\bf r$-integration.
5. For cross-terms that act on $\Psi\Psi$ or $\bar{\Psi}\bar{\Psi}$, integrate by parts in ${\bf R}$-space, so that there are only cross-terms that act on $\bar{\Psi}\Psi$.
6. Compare!
• Thanks Qmechanic. I've just edited my question with few calculations. I'm near the solution... can you help me? – apt45 Jul 5 '14 at 23:01
• I see that you edited your answer thank you! Are my calculation corrects? When you say "For cross-terms that act on $\Psi\Psi$ or $\bar{\Psi}\bar{\Psi}$, integrate by parts in R-space, so that there are only cross-terms that act on $\bar{\Psi}\Psi$ does it mean that $\int \nabla\Psi \cdot \nabla\Psi = 0$ under assumptions that $\Psi(\infty) \rightarrow 0$ ? – apt45 Jul 6 '14 at 9:05
• Concerning your last sentence: No, only total divergence terms are omitted. – Qmechanic Jul 6 '14 at 11:09
• I solved using the fact that $\int d^3\vec{R} \phi\nabla^2\phi = -\int d^3\vec{R} \nabla\phi\cdot\nabla\phi$ thank you!! – apt45 Jul 6 '14 at 11:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9353086352348328, "perplexity": 563.0291633686679}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027318952.90/warc/CC-MAIN-20190823172507-20190823194507-00519.warc.gz"} |
https://wiki.wina.be/examens/index.php?title=Advanced_Statistical_Mechanics&diff=17701&oldid=17699 | # Advanced Statistical Mechanics: verschil tussen versies
Ga naar: navigatie, zoeken
## Examenvragen
#### 1 Sept 2017
Theory:
1) Consider a discrete Markov process in continuous time. Write down the master equation. Determine sufficient conditions for the existence of a stationary probability distribution. Derive the form of the stationary distribution function under these conditions. Is a uniform PDF possible?
2) Consider the Landau-Ginzburg Hamiltionian in dimension d and for an n-component order parameter m. Discuss the critical behavior (t<0 and t>0) of the (ferromagnetic) order parameter and of the specific heat in mean-field theory. Consider subsequently small fluctuations of the order parameter components and treat them in a Gaussian approximation (take n=2). Calculate and discuss the fluctuation corrections to the specific heat and derive the Gaussian approximation for the critical exponent of the specific heat. Interpret your result.
Exercices:
1) Consider diffusion in one dimension in a finite region (-a<x<a) with impenetrable and fully reflecting endpoints x=+/-a. Use seperation of variables to find the PDF solutions p(x,t) of the diffusion equation for the PDF on (-a,a), respecting these boundary conditions, as well as the initial condition ${\displaystyle p(x,0)=\delta (x)}$ and normalized to unity on (-a,a).
Hints: -The boundary conditions imply that the current on the endpoints vanish at all times. -On the interval (-a,a) the Dirac delta can be represented as ${\displaystyle \delta (x)={\frac {1}{2a}}+{\frac {1}{a}}\sum _{n=1}^{\infty }\cos \left({\frac {n\pi x}{a}}\right)}$
2) Real-space RG. Consider the Migdal-Kadanoff (bond-moving) transformation of the square Ising lattice (with nearest-neighbour coupling K) with rescaling factor b (b=integrer\geq 2). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9821851849555969, "perplexity": 1186.837820421824}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178366477.52/warc/CC-MAIN-20210303073439-20210303103439-00068.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/120145-finding-polar-decomposition-print.html | # finding polar decomposition!
• December 12th 2009, 07:25 PM
mancillaj3
finding polar decomposition!
I need help finding the polar
decomposition
of
$
\left( \begin{array}{cc} 11 & -5 \\ -2 & 10\end{array} \right)
$
• December 14th 2009, 12:51 PM
Opalg
If $A = \begin{bmatrix}11 & -5 \\ -2 & 10\end{bmatrix}$ then the polar decomposition of A is a factorisation $A = UR$, where U is unitary and R is positive. To find R use the fact that $R^2 = A^*A = \begin{bmatrix}125 & -75 \\ -75 & 125\end{bmatrix}$. So R is the positive square root of that matrix, which you can compute by diagonalising it. You should find that the eigenvalues of $R^2$ are 50 and 200, with corresponding normalised eigenvectors $\frac1{\sqrt2}\begin{bmatrix}1 \\ 1\end{bmatrix}$ and $\frac1{\sqrt2}\begin{bmatrix}1 \\ -1\end{bmatrix}$. So $R^2 = PDP^{-1}$, where $P = P^{-1} = \frac1{\sqrt2}\begin{bmatrix}1 &1 \\ 1 & -1\end{bmatrix}$ and $D = \begin{bmatrix}50 & 0 \\ 0 & 200\end{bmatrix}$. The square root is given by $R = PEP^{-1}$, where $E = D^{1/2} = \begin{bmatrix}5\sqrt2 & 0 \\ 0 & 10\sqrt2\end{bmatrix}$.
Having found R, you can then get U as $U = AR^{-1}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.994670569896698, "perplexity": 270.97829995398246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246658116.80/warc/CC-MAIN-20150417045738-00054-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-do-you-find-the-critical-numbers-for-g-t-abs-3t-4-to-determine-the-maximum-a | Calculus
Topics
How do you find the critical numbers for g(t)=abs(3t-4) to determine the maximum and minimum?
Nov 4, 2016
See below.
Explanation:
$g \left(t\right) = \left\mid 3 t - 4 \right\mid = \left\{\begin{matrix}3 t - 4 & \text{if" & t >= 4/3 \\ -3t+4 & "if} & t < \frac{4}{3}\end{matrix}\right.$
$g ' \left(t\right) = \left\{\begin{matrix}3 & \text{if" & t >= 4/3 \\ -3 & "if} & t < \frac{4}{3}\end{matrix}\right.$
$g '$ is never $0$ and is undefined (fals to exist) at $x = \frac{4}{3}$
The only critical number is $\frac{4}{3}$.
We see that $g$ is decreasing left of $\frac{4}{3}$ and increasing on the right.
So $g \left(\frac{4}{3}\right) = 0$ is a local minimum.
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http://www.mun.ca/math/graduate/grad-thesis/ | # Thesis Help
Obtaining a PDF/A-1b thesis with Latex, XeLaTeX and pdfLaTeX
Our institution requires that the final version of the thesis is submitted as a PDF/A-1b ISO-19005-1:2005 document. If your document has a good deal of equations and you like to have also some fancy stuff in the document, like bookmarks and so on, then the task of obtaining the such a document may not be straightforward.
There is a general pdfLaTeX package named pdfx that is supposed to do this job as one of its options. However, there are many instances were the package fails to produce a PDF/A-1b file. Here an alternative to 'pdfx' that is somehow more robust is introduced. The package package is called pdfathesis. It can be used with pdfLaTeX, XeLaTeX, LaTeX and some other LaTeX processors. In addition to the regular files generated when processing the document, the package creates two additional files to be post-processed by the latest version of Ghostscript in order to obatain the PDF/A-1b file.
If you do not know how to install it, please read the documentation included.
If you have already formated your thesis, just include the package near the begining of yor preamble with the command
\usepackage{pdfathesis}
and remove all references to the packages 'hyperref' or 'bookmarks'. If you are lucky enough, after runing LaTeX or XeLaTeX of pdfLaTeX and post-processing with Ghostscript you will have a PDF/A-1b document ready. It may also happen that the fancy stuff that you want in your document is not compatible with the PDF/A specification and, sadly, you might need to get rid of it if you want to graduate.
The package 'pdfathesis' also can be called as a LaTeX class and in such a case it will do the basic formating of your document. Enclosed in the package, there is a self-explanatory template called pdfathesis_template.tex that may help you to start this job. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9512251019477844, "perplexity": 1223.9328303800312}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783399385.17/warc/CC-MAIN-20160624154959-00068-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/272258/how-to-solve-pde-using-techniques-of-separation-variables-in-this-question-solv | # How to Solve PDE using techniques of Separation Variables in this question [SOLVED]
Hi guys my name is Maxwell. This is my first time I asking question in this forum. I hope someone can help me this problem out :
Question says :
$U_{xx} + U_{yy} = U$. Solve this PDE product solution using separation of variables.
What I do is
$X''(x)Y(y) + X(x)Y''(y)=X(x)Y(y)$
$X''(x)Y(y)=X(x)[Y(y)-Y''(y)]$
$\frac{X''(x)}{X(x)}$ = $\frac{Y(y)-Y''(y)}{Y(y)}$= k, where k is a constant
Then I made into 3 cases where $k>0$, $k<0$ and $k=0$
I already got the answer for $k<0$ and $k=0$ which my teacher say correct but for $k>0$ my teacher say wrong because he said for $k>0$ case, we need to divide into another 3 sub cases.
My $k>0 [Let k=p^2 ]$, I got my answer
$X(x)=Ae^{-px}+Be^{px}$
$Y(y)=Ce^{-\sqrt{1-p^2}y}$+$De^{\sqrt{1-p^2}y}$
For this part could somone please solve it for me.Please dont say tips and hints. I need some work shown from you so that I can understand better. Please guys I really need help from you. This my first time in this forum. If someone could solve it, i will be really appreciate it. Thanks in advance
-
Look at this. – JohnD Jan 7 '13 at 17:10
The three cases are $1-p^2 > 0$, $=0$, $< 0$.
-
Why?? Could you show the final step please for the 3 sub cases. Please Im not good at PDE and I'm quite weak.I live in Malaysia and I dont have proper teacher to guide me. If you could show then I can analyse myself and try by myself later – maxwell Jan 7 '13 at 17:17
My X(x) is correct just the Y(y) only it is wrong.So help me to proceed the Y(y) only please – maxwell Jan 7 '13 at 17:24
This should be just like what you did for the $X$. Exponentials in one case, sine and cosine in another, $1$ and $y$ in the third. – Robert Israel Jan 7 '13 at 17:27
I really cant get it Sir. Really sorry Sir I'm totally blur about this. I still unable to get as what you said Sir. – maxwell Jan 7 '13 at 17:31
Is it like this now; for $1-p^2>0$, $Y(y)=Ce^{-\sqrt{1-p^2}y}$+$De^{\sqrt{1-p^2}y}$ and for $1-p^2<0$, $Y(y)=Ccos{\sqrt{1-p^2}y}$+$Dsin{\sqrt{1-p^2}y}$. Could you recheck my answer whether it is right or not Sir?? So that I can show to my teacher. – maxwell Jan 7 '13 at 17:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8645902872085571, "perplexity": 804.8098920288033}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997901076.42/warc/CC-MAIN-20140722025821-00158-ip-10-33-131-23.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/44749-applicatios-integration-volumes-please-help-print.html | • July 28th 2008, 09:44 PM
eawolbert
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
• July 28th 2008, 10:05 PM
Jhevon
Quote:
Originally Posted by eawolbert
$V = \pi \int_{-1}^1 \bigg[ (6 - x^4)^2 - 5^2 \bigg]~dx$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.898021936416626, "perplexity": 567.473457377281}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982939917.96/warc/CC-MAIN-20160823200859-00233-ip-10-153-172-175.ec2.internal.warc.gz"} |
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