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https://www.shaalaa.com/concept-notes/mass-number-a-or-atomic-mass_19234 | # Concept of Atomic Mass:
The most remarkable concept that Dalton’s atomic theory proposed was that of the atomic mass. According to him, each element had a characteristic atomic mass. The theory could explain the law of constant proportions so well that scientists were prompted to measure the atomic mass of an atom. Since determining the mass of an individual atom was a relatively difficult task, relative atomic masses were determined using the laws of chemical combinations and the compounds formed.
Let us take the example of a compound, carbon monoxide (CO) formed by carbon and oxygen. It was observed experimentally that 3 g of carbon combines with 4 g of oxygen to form CO. In other words, carbon combines with 4/3 times its mass of oxygen. Suppose we define the atomic mass unit (earlier abbreviated as ‘amu’, but according to the latest IUPAC recommendations, it is now written as ‘u’ – unified mass) as equal to the mass of one carbon atom, then we would assign carbon an atomic mass of 1.0 u and oxygen an atomic mass of 1.33 u. However, it is more convenient to have these numbers as whole numbers or as near to whole numbers as possible. While searching for various atomic mass units, scientists initially took 1/ 16 of the mass of an atom of naturally occurring oxygen as the unit. This was considered relevant due to two reasons:
Imagine a fruit seller selling fruits without any standard weight with him. He takes a watermelon and says, “this has a mass equal to 12 units” (12 watermelon units or 12 fruit mass units). He makes twelve equal pieces of the watermelon and finds the mass of each fruit he is selling, relative to the mass of one piece of the watermelon. Now he sells his fruits by relative fruit mass unit (fmu), as in Fig.
(a) Watermelon, (b) 12 pieces, (c) 1/12 of
watermelon, (d) how the fruit seller can weigh the fruits using pieces of watermelon
Similarly, the relative atomic mass of the atom of an element is defined as the average mass of the atom, as compared to 1/12th the mass of one carbon-12 atom.
Atomic masses of a few elements :
Element Atomic mass (u) Hydrogen 1 Carbon 12 Nitrogen 14 Oxygen 16 Sodium 23 Magnesium 24 Sulphur 32 Chlorine 35.5 Calcium 40
HOW DO ATOMS EXIST?
Atoms of most elements are not able to exist independently. Atoms form molecules and ions. These molecules or ions aggregate in large numbers to form the matter that we can see, feel, or touch.
#### notes
Mass Number – denoted by ‘A’. Protons and neutrons are also called nucleons. Mass number is defined as the sum total number of protons and neutrons present in the nucleus of an atom.
In the notation for an atom, the atomic number, mass number and symbol of the element are to be written as:
Properties of mass number:
Sum of protons and neutrons provide this number of a certain element.
• It is represented by the letter A.
• Protons and Neutrons are together termed as nucleons.
• Example: Atoms of a carbon consist of 6 protons and 6 neutrons. Therefore, the mass number of Carbon is 12.
• The number of neutrons may vary in an element. However, the total number of protons is same in all atoms of an element. Therefore, the atoms of the same element with a same atomic number but a different mass number are termed as isotopes.
• Generally, atomic mass and mass numbers are two different terms and may vary slightly. In most cases, they are not the same. However, the weight of an electron is almost negligible so we can consider the atomic mass of an atom to be almost equal to its mass number.
# Concept of Atomic Mass:
The mass of an atom is concentrated in its nucleus and it is due to the protons (p) and neutrons (n) in it. (mass of electrons is negligible as compared to that of neutrons and protons)
The number (p + n) in the atomic nucleus is called the atomic mass number. Protons and neutrons are together called nucleons (mass of electrons is negligible as compared to that of neutrons and protons)
• This is commonly expressed in terms of a unified atomic mass unit (AMU).
• It can be best defined as 1/12 of the mass of a carbon-12 atom in its ground state.
If you would like to contribute notes or other learning material, please submit them using the button below.
### Shaalaa.com
Atoms and Molecules(Atomic Mass) [00:17:33]
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Share | {"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": 0.8714569211006165, "perplexity": 790.3514216096758}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2021-21/segments/1620243991693.14/warc/CC-MAIN-20210512004850-20210512034850-00494.warc.gz"} |
http://aa.quae.nl/cgi-bin/glossary.cgi?l=en&o=Siarnaq | Astronomy Answers: From the Astronomical Dictionary
# Astronomy AnswersFrom the Astronomical Dictionary
The description of the word you requested from the astronomical dictionary is given below.
Siarnaq
A moon of 40 km diameter at about 17,531,000 km from the planet Saturn. The gravity at its surface is about 0.0007 times as strong as on Earth. The moon goes once around its planet in about 867.1 days. The moon was discovered in 2000. Also called SXXIX (Saturn twenty-nine). Its provisional designation was S/2000 S3. | {"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": 0.9095954298973083, "perplexity": 3491.6084501630967}, "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-2018-05/segments/1516084894125.99/warc/CC-MAIN-20180124105939-20180124125939-00655.warc.gz"} |
https://www.lessonplanet.com/teachers/powers-of-i | # Powers of i
In this powers of i worksheet, 11th graders solve 10 different problems that include determining the powers of i. First, they simplify the given expressions. Then, students determine the product of a given equation containing i to various powers. They also determine the sum of an equations given i to different powers.
Concepts
Resource Details | {"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": 0.9446024298667908, "perplexity": 1685.7885447809963}, "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-2017-39/segments/1505818689028.2/warc/CC-MAIN-20170922164513-20170922184513-00137.warc.gz"} |
https://mochajl.readthedocs.io/en/v0.2.0/user-guide/layers/stat-layer.html | # Statistics Layers¶
class AccuracyLayer
Compute and accumulate multi-class classification accuracy. The accuracy is averaged over mini-batches. If the spatial dimension is not singleton, i.e. there are multiple labels for each data instance, then the accuracy is also averaged among the spatial dimension.
bottoms
The blob names for prediction and labels (in that order).
dim
Default -2 (penultimate). Specifies the dimension to operate on.
class BinaryAccuracyLayer
Compute and accumulate binary classification accuracy. The accuracy is averaged over mini-batches. Labels can be either {0, 1} labels or {-1, +1} labels
bottoms
The blob names for prediction and labels (in that order). | {"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": 0.8732165694236755, "perplexity": 3258.0781221046614}, "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-2019-22/segments/1558232257601.8/warc/CC-MAIN-20190524084432-20190524110432-00353.warc.gz"} |
http://calculator.tutorvista.com/area-between-two-curves-calculator.html | To get the best deal on Tutoring, call 1-855-666-7440 (Toll Free)
Top
Area between Two Curves Calculator
Top
Area between two curves is the area under which the two curves are intersected. Area between Two Curves Calculator calculates the area under which the two curve intersects.
You can see two default functions with its lower and upper bound given below. When you click on "Calculate Area", the calculator will first find the intersecting points and then find the area bounded by the difference between the two curves.
## Steps for Area between Two Curves Calculator
Step 1 :
Given are the two curves y = f(x) and y = g(x). equating both the curves to get the value of interval [a,b] where the given curves intersects.
Step 2 :
The area between the two curves bounded by the x axis is given by the formula
A = $\int_{a}^{b}$[f(x) − g(x)]dx where f(x) is upper curve and g(x) is the lower curve.
Step 3 :
Simplify the problem by substituting the value of the limit to get the answer.
## Problems on Area between Two Curves Calculator
1. ### Find the area bounded by the curve y = 6x + x3 below by y = 5x2.
Step 1 :
Find the intersection points by equalizing the two curves: 6x + x3 = 5x2.
6 + x2 = 5x.
x2 - 5x + 6 = 0
x2 - 3x - 2x + 6 = 0
x(x-3) -2 (x-3) = 0
The two points where the curves intersects are x=2 and x=3.
Step 2 :
The difference between these curves at intersection points are given by:
A = $\int_{2}^{3}$ $[(6x + x^{3}) - 5x^{2}]$ dx
= $\int_{2}^{3}$ $[ x^{3} - 5x^{2} + \frac{x^{2}}{2}]$ dx
= $[\frac{x^{4}}{4} - 5 \frac{x^{3}}{3} + 3 \frac{x^{2}}{2}]^{3}_{2}$
= [$\frac{3^{4}}{4}$ - 5 $\frac{3^{3}}{3}$ + 3 (3)2] - [$\frac{2^{4}}{4}$ - 5 $\frac{2^{3}}{3}$ + 3(2)2].
Step 3 :
A = [$\frac{81 \times 3 + 54 \times 6 - 135 \times 4 - 16 \times 3 - 24 \times 6 + 40 \times 4 }{12}$]
= - $\frac{5}{12}$.
The area between two curves is - $\frac{5}{12}$.
2. ### Find the area between the curves y = 2 - x2 and y = x.
Step 1 :
given curves are y = 2 - x2 and y = x.
equating both the curves we get,
- x2 + x + 2 = 0
- x2 + 2x - x + 2 = 0
-x(x - 2) - (x - 2) = 0
(-x - 1)(x - 2) = 0
x = -1, 2 are the two intersecting points.
Step 2 :
The difference between these curves at intersection points are given by:
A = $\int_{-1}^{2} [(2 - x^{2}) - x]$ dx
= $\left[ 2x - \frac{x^{3}}{3} - \frac{x^{2}}{2} \right]_{1}^{-2}$
= $\left[ 2(1) - \frac{1^{3}}{3} - \frac{1^{2}}{2} \right]_{1}^{-2} - \left[ 2(-2) - \frac{(-2)^{3}}{3} - \frac{(-2)^{2}}{2} \right]$
Step 3 :
A = $\left[2 - \frac{1}{3} - \frac{1}{2} + 4 - \frac{8}{3} + \frac{4}{2} \right]$
= $\frac{9}{2}$
The Area between two curves is $\frac{9}{2}$. | {"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": 0.8593423366546631, "perplexity": 585.917636487246}, "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-2017-34/segments/1502886103910.54/warc/CC-MAIN-20170817185948-20170817205948-00570.warc.gz"} |
https://ch.mathworks.com/help/signal/ref/tf2ss.html | # tf2ss
Convert transfer function filter parameters to state-space form
## Syntax
``[A,B,C,D] = tf2ss(b,a)``
## Description
example
````[A,B,C,D] = tf2ss(b,a)` converts a continuous-time or discrete-time single-input transfer function into an equivalent state-space representation.```
## Examples
collapse all
Consider the system described by the transfer function
`$H\left(s\right)=\frac{\left[\begin{array}{c}2s+3\\ {s}^{2}+2s+1\end{array}\right]}{{s}^{2}+0.4s+1}.$`
Convert it to state-space form using `tf2ss`.
```b = [0 2 3; 1 2 1]; a = [1 0.4 1]; [A,B,C,D] = tf2ss(b,a)```
```A = 2×2 -0.4000 -1.0000 1.0000 0 ```
```B = 2×1 1 0 ```
```C = 2×2 2.0000 3.0000 1.6000 0 ```
```D = 2×1 0 1 ```
A one-dimensional discrete-time oscillating system consists of a unit mass, $\mathit{m}$, attached to a wall by a spring of unit elastic constant. A sensor samples the acceleration, $\mathit{a}$, of the mass at ${\mathit{F}}_{\mathrm{s}}=5$ Hz.
Generate 50 time samples. Define the sampling interval $\Delta \mathit{t}=1/{\mathit{F}}_{\mathrm{s}}$.
```Fs = 5; dt = 1/Fs; N = 50; t = dt*(0:N-1); u = [1 zeros(1,N-1)];```
The transfer function of the system has an analytic expression:
$\mathit{H}\left(\mathit{z}\right)=\frac{1-{\mathit{z}}^{-1}\left(1+\mathrm{cos}\Delta \mathit{t}\right)+{\mathit{z}}^{-2}\mathrm{cos}\Delta \mathit{t}}{1-2{\mathit{z}}^{-1}\mathrm{cos}\Delta \mathit{t}+{\mathit{z}}^{-2}}$.
The system is excited with a unit impulse in the positive direction. Compute the time evolution of the system using the transfer function. Plot the response.
```bf = [1 -(1+cos(dt)) cos(dt)]; af = [1 -2*cos(dt) 1]; yf = filter(bf,af,u); stem(t,yf,'o') xlabel('t')```
Find the state-space representation of the system. Compute the time evolution starting from an all-zero initial state. Compare it to the transfer function prediction.
```[A,B,C,D] = tf2ss(bf,af); x = [0;0]; for k = 1:N y(k) = C*x + D*u(k); x = A*x + B*u(k); end hold on stem(t,y,'*') hold off legend('tf','ss')```
## Input Arguments
collapse all
Transfer function numerator coefficients, specified as a vector or matrix. If `b` is a matrix, then each row of `b` corresponds to an output of the system.
• For discrete-time systems, `b` contains the coefficients in descending powers of z.
• For continuous-time systems, `b` contains the coefficients in descending powers of s.
For discrete-time systems, `b` must have a number of columns equal to the length of `a`. If the numbers differ, make them equal by padding zeros. You can use the function `eqtflength` to accomplish this.
Transfer function denominator coefficients, specified as a vector.
• For discrete-time systems, `a` contains the coefficients in descending powers of z.
• For continuous-time systems, `a` contains the coefficients in descending powers of s.
## Output Arguments
collapse all
State matrix, returned as a matrix. If the system is described by n state variables, then `A` is n-by-n.
Data Types: `single` | `double`
Input-to-state matrix, returned as a matrix. If the system is described by n state variables, then `B` is n-by-1.
Data Types: `single` | `double`
State-to-output matrix, returned as a matrix. If the system has q outputs and is described by n state variables, then `C` is q-by-n.
Data Types: `single` | `double`
Feedthrough matrix, returned as a matrix. If the system has q outputs, then `D` is q-by-1.
Data Types: `single` | `double`
collapse all
### Transfer Function
`tf2ss` converts the parameters of a transfer function representation of a given system to those of an equivalent state-space representation.
• For discrete-time systems, the state-space matrices relate the state vector x, the input u, and the output y:
`$\begin{array}{c}x\left(k+1\right)=Ax\left(k\right)+Bu\left(k\right),\\ y\left(k\right)=Cx\left(k\right)+Du\left(k\right).\end{array}$`
The transfer function is the Z-transform of the system’s impulse response. It can be expressed in terms of the state-space matrices as
`$H\left(z\right)=C{\left(zI-A\right)}^{-1}B+D.$`
• For continuous-time systems, the state-space matrices relate the state vector x, the input u, and the output y:
`$\begin{array}{l}\stackrel{˙}{x}=Ax+Bu,\\ y=Cx+Du.\end{array}$`
The transfer function is the Laplace transform of the system’s impulse response. It can be expressed in terms of the state-space matrices as
`$H\left(s\right)=\frac{B\left(s\right)}{A\left(s\right)}=\frac{{b}_{1}{s}^{n-1}+\cdots +{b}_{n-1}s+{b}_{n}}{{a}_{1}{s}^{m-1}+\cdots +{a}_{m-1}s+{a}_{m}}=C{\left(sI-A\right)}^{-1}B+D.$` | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "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": 0.9752391576766968, "perplexity": 1060.7607622637759}, "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/1637964363290.59/warc/CC-MAIN-20211206072825-20211206102825-00329.warc.gz"} |
https://sukrutrao.github.io/project/open-wbo-inc/ | # Open-WBO-Inc Incomplete MaxSAT Solver
Open-WBO-Inc is a MaxSAT solver tailored to the incomplete MaxSAT problem. In incomplete MaxSAT, the goal is to find an approximate solution to a given MaxSAT formula in very little time. Open-WBO-Inc uses various approximation techniques, such as those that reduce encoding sizes, to find a solution close to the optimal in much lesser time than would otherwise be needed. Open-WBO-Inc is based on Open-WBO, a state-of-the-art MaxSAT solver, and has been implemented in C++.
### Motivation
In the weighted MaxSAT problem, the input consists of a set of clauses and positive integral weights associated with each clause. The goal is to obtain an assignment of variables that minimizes the cost (sum of unsatisfied weights) of the formula. The weighted MaxSAT problem typically involves encoding a pseudo-boolean (PB) constraint in the CNF form to constrain the number of clauses that may be unsatisfied in a given iteration of the solution search process. The number of clauses added for this encoding could often be a large fraction of the resulting formula size. As a result, having a large encoding increases the time needed to find the solution. One of the best performing encodings, GTE, can generate an exponential number of clauses in the size of the PB constraint in the worst case, which is when the number of unique weights are very high. However, for many applications, it might be sufficient to obtain an approximate solution while greatly speeding up the search.
### Approach
We proposed two approximation algorithms:
#### apx-weight (Open-WBO-Inc-Cluster)
This algorithm aims to reduce the size of the encoding, thus reducing the search time. A key observation is that the size of the GTE encoding increases based on the number of distinct weights in the formula. If this could be reduced, the encoding size would also reduce. We achieve this by clustering the weights to a small number of clusters using divisive hierarchical clustering. In each cluster, all the weights are replaced by a representative weight, such as their mean. As a result, the number of unique weights now reduces to the number of clusters. Clustering $n$ weights to $k$ clusters can be done in $\mathcal O(nk)$ time, which was observed to be negligible compared to the total time taken by the solver.
#### apx-subprob (Open-WBO-Inc-BMO)
This extends the ideas in apx-weight. Once the clusters are obtained, the clauses in the formula are arranged in the descending order of their weights. Then, iterating from the largest representative weight to the smallest, the formula in each cluster is solved. Once an optimum is obtained for any cluster, it is fixed, and no further relaxation is made when operating over subsequent clusters. This breaks the problem into multiple subproblems, effectively enforcing the boolean multilevel optimization (BMO) condition. This can often be an effective strategy because many real-world MaxSAT formulae might have weights that satisfy the BMO assumption.
### Results
Both algorithms outperformed most of the existing state-of-the-art incomplete MaxSAT solvers. In particular, apx-subprob outperformed maxroster, which was the best performing solver in the MaxSAT Evaluation 2017.
In the MaxSAT Evaluation 2018, apx-subprob in Open-WBO-Inc secured the first position in the 60 seconds category and the second position in the 300 seconds category in the weighted incomplete MaxSAT track. apx-weight secured the fourth position in both these categories.
## Publications
. Open-WBO-Inc in MaxSAT Evaluation 2018. MaxSAT Evaluation 2018, 2018. | {"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": 0.8070740103721619, "perplexity": 707.8831550307565}, "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-2019-22/segments/1558232257601.8/warc/CC-MAIN-20190524084432-20190524110432-00467.warc.gz"} |
https://computerscienceassignmentshelp.xyz/good-algorithms-48839 | Select Page
good algorithms for the proof of $prop:thm:main$. $thm:Thm1$ Suppose $K$ is $C_0(\mathbb{R})$-equivalence with only two groups. For each $k \ge 1$, we have: – The image of $(X_k)^{(1)}_*$, $* \subset \mathcal{A}(\mathbb{Q})$, under the following map: $$H^u(X_{2k-1}^{(2)}) \twoheadrightarrow H^u(\mathcal{A}(X_{2k}^{(1)})),$$ is a left $C_0$-module isomorphism. – $(H^u(\mathcal{A}(X_{k}^{(2)},L_k)), \mathcal{A}(\mathbb{Q}_k,L_k))$ is isomorphic to the image of $\mathbf{1^U}$ under the map on morphism from the $C_0$-section map; moreover, it has a natural isomorphism between the morphisms for the second section and $H^{u(\mathbb{Q}_k)}(L_{2k} \oplus X_{2k-1}^{(1)}) \circledast (H^{u(\mathbb{Q}_k)}\Bigl(\mathcal{A}(X_{2k}^{(1)}) \bigr) \rightarrow H^{u(\mathbb{Q}_k)}\bigl(L_{2k-1} \oplus I_{2k-1}^{(2)}\bigr)$. \(3) is shown in Section $sec:sec:main$. Similarly, a strong monomorphism is shown of Example $ex:Diffong$,$ex:Zinc$. – It is proved in Proposition $prop:thm:dcf$ in Section $sec:ref:prop$. By Proposition $prop:thm:main$, the image of $\mathbf{1h}_*$ under the corresponding map is an analytic smooth function of degree index on $\mathcal{A}(X_{k}^{(1)},\mathbb{Q}_k)$. By analogy with the $H^u$-setting, a strong monomorphism is shown for a degree $-1$ pullback (see Proposition $prop:thm:divmod$) of the image under $\delta^u (H^u(X_{k}^{(1)},L_k))$. Let $\phi: L_{2k-1} \to I_{2k} ^c$ be the isomorphism of Problem $pro:def\_pro$, and $\phi(x-1)$ for $0 \leq x \leq k-1$. By, for $k \geq 1$, we have: \begin{aligned} &R_\phi:L_{2k-1} \rightarrow I_{2k} ^c \\ \xrightarrow[\, 0 \,]{} H^{u(\phi( 1)^{(1)},L_{2k-1})}(x-1, L_{2k-1}) \\ & \oplus \bigoplus_{\substack{ \nu \in \mathcal{A}(X_k^{(1)},\mathbb{Q}_k) \\ k \geq 1}} (\phi( X_k^{(1)}; \nu ) \oplus X_k^{(1)} \oplus \delta^u (H_{\nu} \oplus read more ))\times H^{u(\phi^{(1)},L_{2k-1})}.\end{aligned} By Proposition $prop:thm:hgood algorithms and much less generalisation work has been done on this subject (see Kim and Varshalovich [**64**]{} 1999). \[def:2D$ Define $U, K, Q, P, Q_2$ such that in ${\mathcal{J}}$, $U, Q\in {\mathcal{J}}$ and $P$, $Q_2$, $Q_1$ all are closed in ${\mathcal{M}}$. Moreover, for the sake of simplicity we will consider the set of $p\in\mathbb{N}^2$ such that there is a unique $q\in{\mathcal{N}}$ such that $p=\hat{q}\equiv q\equiv 0^-$. Then there is a unique closed this article $w(x)=F(\bar{x})+b_1q$ such that $w(r)=w(r)=0$ for all $r\geq r_{00}$. $main\_condition$ Let $\mathbf{D}=({\mathbf{D}},{\mathbf{B}})$ be a ${\mathbb{F}}_q$-flat distribution on $\mathbb{P}^k$. Then $\mathcal{M}\cap {\mathcal{J}}\ne\emptyset$, ${\mathbb{D}}$ is assumed to be flat and ${\mathbb{D}}{\cap}\mathcal{M}\subset{\mathbb{D}}{\cap}\mathcal{J}$. We have that ${\mathbb{D}}\inf\{|u(x)-b(x)|=1\}\subset{\mathbb{J}}$ is flat, hence ${\mathbb{D}}{\cap}\mathbb{P^k}$ is flat. Assume inductively that $\mathbf{D}$ has at least $q’$ satisfying the conditions above. We have by [@D2 Proposition 4.
## simple data structure program in c
2] that there exists a unique $q, q’,$ satisfying $q+q’\equiv q’\equiv 0^-$. This concludes the proof and Lemma $lemmav$ follows from the proof of the main result of Liao which only states that ${\mathbb{D}}$ is flat, not always. [**Acknowledgements.**]{} The author thanks Zongbin Bong for helpful discussions and comments on the paper. [*Fundamental Concepts.*]{} The author G.P.P. has infinite love of linear algebra, and it is correct to consider functions which are not bounded by continuity on continuous fields by means of the Riesz triplet spaces (see [@MS02]). [*A short note.*]{} If $A, B\in{\mathcal{M}}$ then $B$ is said to be [*equivalent*]{} to $A$ if $B(x_1,\dots,x_n)=A(x_1,\dots,x_n)$ for all $x_1,\dots,x_n\in A$. It is also called [*simple*]{} if $B\equiv A$. $thm:finite\_descent$ Let $\mathbf{D}=({\mathbf{D}},{\mathbf{B}})$ be a ${\mathbb{F}}_1$-flat distribution on $\mathbb{P}^k$. Then $\mathcal{M}\cap {\mathcal{J}}\ne\emptyset$ and $\mathcal{M}\cap {\mathbb{D}}{\cap}\mathcal{J}$ is flat. Löfström’s $2$-topology with three filtrations, and the *Einstein-Moore* type $3$-convention [@Miz69]. However, the paper (Löfström et al. [@E-Q]) does not talkgood algorithms’ can, and have been, written and published in the journal. It was carried on for many years and is now fully published. Not unlike Microsoft Word II, Adobe Reader did not have such a huge publication. It was made available for people using Windows, Safari, and MacOS, and users had a better chance of knowing what Creative Commons or Shade and other standards are supposed to mean. | {"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": 0.9987029433250427, "perplexity": 368.5072834771009}, "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/1652662517245.1/warc/CC-MAIN-20220517095022-20220517125022-00619.warc.gz"} |
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[61] N. L. Carothers and S. J. Dilworth. Inequalities for sums of independent random variables . Proc. Amer. Math. Soc. 104 (1988) 221-226. MR 958071. Abstract, references, and article information View Article: PDF This article is available free of charge [62] Wolfgang Stadje. The influence of the initial distribution on a random walk . Proc. Amer. Math. Soc. 103 (1988) 602-606. MR 943090. Abstract, references, and article information View Article: PDF This article is available free of charge [63] F. Móricz. Strong limit theorems for blockwise $m$-dependent and blockwise quasi-orthogonal sequences of random variables . Proc. Amer. Math. Soc. 101 (1987) 709-715. MR 911038. Abstract, references, and article information View Article: PDF This article is available free of charge [64] Peter Hall and A. D. Barbour. Reversing the Berry-Esseen inequality . Proc. Amer. Math. Soc. 90 (1984) 107-110. MR 722426. Abstract, references, and article information View Article: PDF This article is available free of charge [65] Peter Hall. Order of magnitude of the concentration function . Proc. Amer. Math. Soc. 89 (1983) 141-144. MR 706528. Abstract, references, and article information View Article: PDF This article is available free of charge [66] Jack Cuzick. Riemann $R\sb{1}$-summability of independent, identically distributed random variables . Proc. Amer. Math. Soc. 83 (1981) 119-124. MR 619995. Abstract, references, and article information View Article: PDF This article is available free of charge [67] J. Kuelbs. Some exponential moments of sums of independent random variables . Trans. Amer. Math. Soc. 240 (1978) 145-162. MR 0517296. Abstract, references, and article information View Article: PDF This article is available free of charge [68] J.-P. Gabriel. Sums of independent random variables and the Burkholder transforms . Proc. Amer. Math. Soc. 66 (1977) 123-127. MR 0451392. Abstract, references, and article information View Article: PDF This article is available free of charge [69] Galen R. Shorack and R. T. Smythe. Inequalities for $\max|S_k|/b_k$ where $k \in N^r$ . Proc. Amer. Math. Soc. 54 (1976) 331-336. MR 0400386. Abstract, references, and article information View Article: PDF This article is available free of charge [70] C. M. Newman. An extension of Khintchine's inequality. Bull. Amer. Math. Soc. 81 (1975) 913-915. MR 0375458. Abstract, references, and article information View Article: PDF [71] Y. S. Chow and T. L. Lai. Some one-sided theorems on the tail distribution of sample sums with applications to the last time and largest excess of boundary crossings . Trans. Amer. Math. Soc. 208 (1975) 51-72. MR 0380973. Abstract, references, and article information View Article: PDF This article is available free of charge [72] Russell W. Howell. Annular functions in probability . Proc. Amer. Math. Soc. 52 (1975) 217-221. MR 0374398. Abstract, references, and article information View Article: PDF This article is available free of charge [73] Naresh C. Jain and Michael B. Marcus. Integrability of infinite sums of independent vector-valued random variables . Trans. Amer. Math. Soc. 212 (1975) 1-36. MR 0385995. Abstract, references, and article information View Article: PDF This article is available free of charge [74] D. L. Hanson and F. T. Wright. A theorem about the oscillation of sums of independent random variables . Proc. Amer. Math. Soc. 37 (1973) 226-233. MR 0315779. Abstract, references, and article information View Article: PDF This article is available free of charge [75] K. Bruce Erickson. The strong law of large numbers when the mean is undefined . Trans. Amer. Math. Soc. 185 (1973) 371-381. MR 0336806. Abstract, references, and article information View Article: PDF This article is available free of charge [76] Leonard E. Baum and H. H. Stratton. Visitations of ruled sums . Trans. Amer. Math. Soc. 182 (1973) 403-430. MR 0334326. Abstract, references, and article information View Article: PDF This article is available free of charge [77] Steven Rosencrans and Stanley Sawyer. An extremal property of independent random variables . Proc. Amer. Math. Soc. 36 (1972) 552-556. MR 0312566. Abstract, references, and article information View Article: PDF This article is available free of charge
Results: 61 to 77 of 77 found Go to page: 1 2 3 | {"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": 0.8884391188621521, "perplexity": 1107.3958426969118}, "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-2014-10/segments/1393999654872/warc/CC-MAIN-20140305060734-00092-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.intechopen.com/books/management-strategies-to-adapt-alpine-space-forests-to-climate-change-risks/ozone-fluxes-to-a-larch-forest-ecosystem-at-the-timberline-in-the-italian-alps | Open access peer-reviewed chapter
# Ozone Fluxes to a Larch Forest Ecosystem at the Timberline in the Italian Alps
By Giacomo Gerosa, Angelo Finco, Antonio Negri, Riccardo Marzuoli and Gerhard Wieser
Submitted: August 28th 2012Reviewed: March 1st 2013Published: August 28th 2013
DOI: 10.5772/56280
## 1. Introduction
Ozone is a phytotoxical pollutant causing negative effects on vegetation at biochemical, physiological, individual and ecosystem level [1]. Alpine forests, could experience high level of ozone concentrations, in particular in the southern part of the Alps. The reason of this phenomena is explained more in detail in the chapter 9, as well as the ozone effects on vegetation.
At the moment, from a regulative point of view, AOT40 (Accumulated Ozone over a Threshold of 40ppb) is the only instrument to evaluate the ozone hazard, even if its scientific soundness is widely discussed [2,3 ]. In fact, the AOT40 is an exposure index which does not take into account the physiology of the vegetation which is exposed to that ozone concentration. Since the greatest damages to vegetation are produced by the ozone entering through the plant stomata and plants can regulate stomatal opening as a response to the environmental parameters, a new approach based on stomatal ozone fluxes as been proposed by UN/ECE. In fact, the magnitude of the negative effects of ozone on vegetation is related to the real amount of this pollutant taken up through stomata [4], i.e. the dose or stomatal flux, and high environmental ozone concentrations in the air do not necessarily lead to high ozone doses, representing only a potential risk (more correctly an hazard).
The effective ozone dose, based on the flux of ozone into the leaves through the stomatal pores, represents the most appropriate approach for setting future ozone critical levels for forest trees. However, uncertainties in the development and application of flux-based approaches to setting critical levels for forest trees are at present too large to justify their application as a standard risk assessment method at a European scale [5].
The scientific community is hence moving toward an evaluation of the ozone risk based on stomatal ozone fluxes. This can be realized by means of measures or by models. Measurements allow to estimate only local ozone risk but they are necessary to parameterize and validate the models, which, once all the input data are available, allow to estimate ozone risk at regional, national or continental level.
Measurements of ozone fluxes are hence the first fundamental step for a proper evaluation of the ozone risk. The most used technique for the measure of ozone fluxes is called eddy covariance. This technique is based on the atmospheric turbulence and it requires particular instrumentation, which must be able to measure at least ten times per second the three wind components, the air temperature and the ozone and other gases concentrations. This kind of instrumentation must be mounted above the studied ecosystem. Additional meteorological measurements are useful for a better comprehension of the exchange process.
The eddy covariance technique allow to measure the total ozone deposition to the whole ecosystem, without distinguishing between how much ozone enters through plant stomata (the most harmful pathway) and how much is destroyed on the plant surfaces or on the terrain. In order to estimate the ozone stomatal fluxes, water is used as a tracer. In fact, it is assumed that ozone can enter into plant stomata when they are open during carbon uptake and plant transpiration: for this reason concomitant water fluxes measurements allow for the partition the total ozone fluxes between the stomatal and the non-stomatal component [6].
Once the stomatal flux is known it is possible to calculate the ozone dose (which is simply the sum of the stomatal ozone fluxes) and the Phytotoxical Ozone Dose (POD1, which is the cumulated dose over the threshold of 1 nmol O3 m-2 s-1) [4]. The POD1 have been introduced in the UN/ECE scientific community because it takes into account the internal capability of the vegetation to detoxify part of the ozone entering through the stomata. Moreover, many experiments have showed that the POD1 is better correlated with the biomass reduction than the simple ozone dose, allowing thus to estimate the harmful effects of ozone on vegetation.
In this chapter an example of micrometeorological measurements of ozone fluxes over a high elevation larch forest, performed within the framework of the MANFRED project, are presented and discussed. The aims of this field campaign were to quantify the ozone deposition to a forest ecosystem at the timberline in the southern Alps (the more exposed to ozone) and to assess the actual ozone taken up by trees through stomata; a further aim was to gather data for the development of a stomatal uptake model to be employed for the simulation of the ozone deposition to timberline forests in future climatic and ozone pollution scenarios.
Till now ozone uptake by larch at treeline has only been investigated at the twig level [7], so the Eddy Covariance measuring technique has been chosen to get information at whole ecosystem level.
## 2. Methodology
### 2.1. Measurements
A micrometeorological tower was run for two years in Paspardo, Valle Camonica, in the Brescia district in Italy (46°2’40”N, 10°23’04”) (Figure 1a). Measurements interested only the summer season of the two years: July-October in 2010 and June-September in 2011. The studied ecosystem was a 10 ha larch forest (Larix decidua, Mill) with a mixed grass understorey used as a cattle pasture at 1750 m a.s.l.. Trees were between 100 and 350 years old and their average height was about 26 m. The trees coverage of the population had a LAI equal to 0.9 and a SAI equal to 0.7, while the understorey grass and shrubs had a LAI equal to 1.3 and a SAI equal to 0.2. The ecosystem lays on a westward gentle slope, with an inclination ranging between 3° and 13° (Figure 1b and Figure 1c).
The selected micrometeorological technique was the eddy covariance which requires fast response instrumentation for wind, temperature, water and ozone. For this sake an ultrasonic three axial anemometer-thermometer (USA-1, Metek, D) was installed at 29 m, on the top of the tower, just above the canopy. At the same height an open path krypton-light hygrometer (KH2O, Campbell, USA) was set. Ozone concentrations were sampled from air drawn near the sonic anemometer with two instruments: a fast response one (COFA, Ecometrics, I) and a standard UV photometer (1308, SIR, E) used as a reference since COFA uses cumarine targets whose sensitivity decays in 5 days and have to be changed. At four different levels of the tower additional thermo-hygrometrical (HD9000, Deltaohm,I), radiation (LPAR01, Deltaohm, I) were installed in order to obtain vertical profiles of each parameter. Moreover three reflectometers (CS616, Campbell, USA) and three thermopiles (SHFP, Hukseflux, NL) were deepen into the ground to measure soil water content and soil heat fluxes. In order to assess the energy balance closure a net-radiometer (NR-LITE, Kipp&Zonen, NL) as well as a pyranometer (LI 200 SZ, LI-COR, USA) were installed on the top of the tower. Finally a rain gauge (mod. 52202,Young, USA) was set in a little nearby clearance and a leaf wetness sensor (mod.237, Campbell, USA) was used too. Fast sensor data were sampled 20 times per second and collected by a personal computer through a customized program, saving data in a new file every half an hour. Slow sensor data were sampled by a datalogger (CR10x and AM416, Campbell, USA) every 15 seconds and the 30 minutes averages of each parameter were stored.
### 2.2. Data processing
Vertical fluxes of sensible (H) and latent (LE) heat and ozone (FO3) were calculated from the 30 minutes data files of the fast instrumentation as the covariance between the vertical component of the wind and the corresponding scalar quantity, following the eddy covariance theory [8]. In particular these fluxes are obtained with the following equations:
H=ρcpw'T'¯E1
LE=λρw'q'¯E2
FO3=w'O3'¯E3
where ρ is the air density, c p is the specific heat at constant pressure, w is the vertical component of the wind intensity, T is the air temperature, λ is the latent heat of vaporization, q is the specific humidity and O 3 is ozone concentration; primed variables mean the fluctuations around their 30 minutes averages which are represented as overscript bars [7].
The covariances in the above mentioned equations are usually obtained from the rotated covariance matrix, following the methodology proposed by McMillen [9]; the rotations of the covariance matrix remove small tilts of the sonic anemometer from the verticality. In this case, since the forest was located on a gentle slope an ad hoc rotation procedure was developed.
The coordinate rotations align the “tower” vertical axis to the perpendicular of the wind streamlines. In this way the mean w component is zeroed and the advective flux component is removed. The angle of the rotation varies in agreement to the slope in the upwind direction (Figure 1c).
Once the total ozone fluxes have been calculated, a flux partition procedure has been employed in order to separate the two main deposition pathways: stomatal uptake by leaves and non-stomatal ozone disruption by both chemical sinks and ozone removal from non-living surfaces.
The partition procedure is based on an electrical analogy which is typical of the SVAT (Soil Vegetation Atmosphere Transfer) models where the flux corresponds to an electrical current flowing through a resistances network. Every resistance describes a part of the whole deposition process whose driving force is represented by the ozone concentration differences between the measuring height z and the substomatal leaf cavity at the standard height d+z 0 (d 2/3 of canopy height, z01/10 of canopy height, for further details see, for instance, [2], where ozone concentration is assumed to be zero [10].
The total resistance, which is obtained directly from measurements as the ratio between the ozone concentration at the measuring height and the total ozone fluxes, is equal to three series resistances:
Rtot(z)=Ra(d+z0,z)+Rb+RcE4
where Ra is the aerodynamic resistance that ozone faces during the turbulent transport from the height z to the height d+z0 (momentum sink), Rb is the resistance faced by ozone while crossing the thin layer of still air surrounding leaves (diffusive transport) and Rc is the integrated resistance of the exchanging surface (leaves, stems, soil).
Ra was calculated following the formulation proposed by Dyer [11] while Rb was calculated using Hicks et al. [12] equation. Rc is finally obtained as a residual from equation (Eq. 4) all the others variables being known.
Rc is hence considered equivalent to two parallel resistances:
Rc1=RST1+RNS1E5
where RST is the stomatal resistance and RNS is the non-stomatal resistance.
RST could then be deduced from the Penman-Monteith equation [13] which describes the water loss process between a wet surface and the atmosphere. This equation is based on the energy balance closure at the evaporating surface so, when the net incoming energy (net radiation) as well as all the other energy losses (soil, sensible and latent heat fluxes) are known -as in our case-, RST is the only unknown term which can be derived by the inversion of the equation.
Finally the stomatal flux is obtained, using the Ohm’s:
FST=Rc(Ra+Rb+Rc)RSTCmE6
where Cm is the ozone concentration at the measuring height.
The stomatal dose, D, is simply given by the integral of FST over the measuring period (from ta to tb).
D=tatbFST(t)dtE7
The UNE/ECE POD1 is the dose which exceeds the instantaneous threshold of 1 nmol m-2 s-1.
POD1=tatb[FST(t)1]dtFST1nmolO3m2s1E8
For comparison purposes, the currently set AOT40 ozone exposure index was calculated too, as it follows:
### 2.3. Estimation of the ozone uptake by Larch trees
The dose calculated in the previous paragraph takes into account the ozone fluxes entering all the stomata of the whole ecosystem, i.e. the stomata of grass and trees. In order to estimate the uptake of the larch needles only, a two layers resistive model was developed (Figure 2). Differently from the SVAT model of the previous paragraph, this model is a prognostic model which tries to predict the stomatal ozone flux from meteorological and atmospheric turbulence data. This model simulates separately the stomatal behaviour of the trees (1st layer) and of the understorey grass (2nd layer) as well as the ozone removal by chemical reactions with terpenes in the trunk space and the ozone deposition to the underlying soil and the external non transpiring surfaces (cuticles, branches, stems). The stomatal processes were modelled using the Jarvisian approach adopted also by UN/ECE Manual on the Methodologies and Criteria for Modelling and Mapping Critical Loads and Levels and Air Pollution Effects, Risks and Trends, hereafter simply called Mapping Manual [5]. A maximum stomatal conductance of 125 mmol m-2 s-1 was chosen for larch, following the findings of Sandford and Jarvis [14] and Wieser et al.[15], while the other parameterizations were taken from the generic continental conifers in the UN/ECE Mapping Manual, since for larch nothing else was available. The second layer was parameterized as the generic grass in UN/ECE Mapping Manual, which prescribes a maximum stomatal conductance of 270 mmol m-2 s-1.
## 3. Results
The meteorological conditions of the two measuring periods are reported in Table 1. Comparing the common trimester (July-September) of the two years (2010 and 2011), it is evident that 2011 summer was hotter and drier than 2010. The average temperature ranged between 9.4 °C and 15.4 °C in 2010 and between 13.6 °C and 18.1°C in 2011. In the same period the rain received by the forest ecosystem was 806 mm in 2010 and 431 mm in 2011.
Unit July 2010 August 2010 September 2010 October 2010 June 2011 July2011 August 2011 September 2011 T Average [°C] 15.4 13.3 9.4 6.2 12.7 13.8 18.1 13.6 T Max [°C] 20.3 18.2 12.8 10.9 19.5 19.2 23.7 16.5 T Min [°C] 11.0 8.2 3.6 0.7 8.7 9.4 12.02 3.9 Rainfalls [mm] 89 291 426 55 90 67 156 208 RH Average [%] 70.4 77.0 78.9 85.7 79.5 74.2 72.9 74.2 RH Min [%] 43.9 38.5 43.5 66.7 57.7 49.4 47.1 51.8 PAR Max [μmol m-2 s-1] 637 596 469 304 638 616 529 403 SWC Average [%] 23 28 29 30 31 23 25 26
### Table 1.
Meteorological and soil measurements; Taverage is the monthly average of the air temperature, TMax is the monthly maximum value of the daily averages of air temperature, TMin is the monthly minimum value of the daily averages of air temperature, Rainfalls are the cumulated rainfalls in the month, RHaverage is the monthly average of the relative humidity, RHmin is the minimum of daily averages of the relative humidity, PAR Max is the monthly maximum of the daily averages of PAR, SWC average is the monthly average of soil water content. Air temperature, air relative humidity and PAR measurements are referred to top canopy measurements.
Most of the energy received by the ecosystem as solar radiation (Rn) was dissipated backward to the atmosphere as sensible heat (H) which reached on average a daily maximum value around 300 W m-2 at 3.00 PM local time (GMT +2) (Figure 3). A minor part of the energy was employed to drive the evapo-transpirative processes (LE) with a daily maximum of 150 W m-2 and an almost constant minimum of 20 W m-2 in the dark hours. The residual energy, a very little part with respect to the other two components, heated the ground (G) with a maximum average daily value of 12 W m-2 in the late afternoon. The heat stored in the soil during the daylight hours was completely returned to the ecosystem.
Figure 4 shows a detailed example of the energy fluxes (Figure 4a) as well as of the ozone concentrations and fluxes (Figure 4.b) measured at the top of the tower. Each point in Figure 4 is a semi-hourly averaged sample from 36000 rapid measurements collected 20 times per second.
The ozone fluxes showed in Figure 4b are the net ozone amount received by the ecosystem from the atmosphere. In the convention adopted here a positive ozone flux value means a flux directed from atmosphere toward the ecosystem, while a negative value means the opposite direction. The studied ecosystem behaved as a net ozone sink with very rare episodes of ozone effluxes in the night-time. Despite ozone concentrations varied during the presented month ranging from 23 to 71 ppb the behaviour of the total ozone fluxes was almost the same. This fact highlights that the ozone deposition is not only driven by the atmospheric chemistry but also by the atmospheric turbulence and the plant physiology, which in turn responds to the soil water content.
The importance of the plant physiology is further confirmed by the results of the flux partition (Figure 5), where it is evident that a significant part of the ozone deposition was removed by the stomatal activity. This process accounted for a maximum flux of about 10 nmol m-2 s-1 which was nearly 50 % of the ozone deposition in the afternoon of both years, even if this daily stomatal peak was located in the late afternoon in 2010 and earlier in 2011. During the night a residual stomatal uptake around 2 nmol m-2 s-1 was observed, revealing an incomplete stomatal closure as already reported by Matyssek and Innes [16].
On average the daily peak of the total ozone deposition (30 nmol m-2 s-1 in 2010 and 27 nmol m-2 s-1 in 2011) was measured at noon when turbulence is usually high. At the same time, the non-stomatal deposition, i.e. the ozone disruption on the external non-living surfaces, was maximum as can be inferred by the difference between the two curves in Figure 5 which is just the non-stomatal deposition.
The diurnal behaviour of stomatal and total fluxes, of course, showed a clear dependence on the seasons as it can be observed in Figure 6, which presents as an example the average daily course of every month in the 2010 measuring period. The total flux (Figure 6b) increased until August, when it reached its maximum values, and then decreased until October when it became four times lower. On the contrary, the stomatal flux maximum was observed in July and, in the following months, it decreased gradually reaching values three times lower in October.
The stomatal fraction of the total ozone deposition, during the central hours of the day, was 40%, 24%, 23% and 37% respectively in July, August, September and October. The non-stomatal deposition hence was always the main ozone sink for this measuring site.
Ozone concentrations (Figure 6a) reached their maximum values in August and after that slowly decreased. It is worth noticing the reversed bell shape of the ozone concentration daily courses which is typical of high elevation sites [17]. The concentration minimum occurred when the total ozone deposition was maximum. This fact highlights the sink role of the forest in the ozone removal from the atmosphere, a role which decreased as the end of the growing season was approaching.
In order to assess the ozone risk for the larch only, it was necessary to separate the stomatal flux taken up by the whole ecosystem into a stomatal component for the larch forest and a stomatal component for the understorey grass. The results of the sub-partition modelling exercise on 2010 data are showed in Figure 7. The model results were in satisfactory agreement with the measured fluxes, taking into account the correction by Moelder et al. [18] for the measurements taken in the roughness sub-layer, as in our case.
In the central hours of the day, the ozone fraction taken up by the larch needles through stomata was around 70% of the bulk stomatal flux calculated for the whole ecosystem. The remaining part (around 30%) was due to the understorey grass uptake. On average, the larch stomatal flux was about 25% of the total ozone deposition to the whole ecosystem.
Taking into account the uptake of both larch and grass, their overall stomatal uptake was on average 32% of the total flux, thus confirming the results of the flux partition obtained from the measurements. It is worth noticing that most of the ozone was destroyed by non-stomatal processes (68%) which include deposition on soil, leaf cuticles and stems.
Using the outputs of this model in order to estimate the phytotoxical ozone dose received by the larch trees during the whole measuring period, a POD1 value of 17.9 mmol m-2 was found, which is more than twice the critical flux level set by UN/ECE to protect Norway spruce, the species which was chosen as representative for all conifers.
By applying the flux-effect relationship proposed by UN/ECE for Norway spruce, a biomass reduction around 4 % can be predicted.
This reduction should be intended as the missing plant growth in the measuring period. Even if this value seems low, it is not negligible considering a pluriennial time scale. In fact it could be responsible of a slowing of the forest development as well as a minor plant capability to cope with other biotic and abiotic stressors.
It is important to highlight that the UN/ECE flux effect relationship had been developed on epigean biomass data only. However some studies showed a more significant effect of ozone uptake on root development [19, 20]. On a long time scale this effect could lead to a general impairment of the plant capability to stabilize mountain slopes and to prevent hydro-geological instability.
In any case it should be important to remark that the quantification of the effects is based on Norway spruce, the only species for which data are currently available in literature. The effects on larch could be lower or even worse but this is the best that could be done with the present knowledge. Ozone will obviously affect even the understorey vegetation and the other ecosystem components. So the estimated negative effects are likely higher than those predicted for larch only.
From a regulative point of view, the larch forest experienced an ozone exposure of 5.1 ppm h in the measuring period. This value was calculated as AOT40, the exposure index currently in use, following the EU and Italian legislation. Using the relationship between epigean biomass reduction and AOT40 for Norway spruce [5], in this case a biomass reduction of 0.8 % can be expected, a value that is 5 times lower than the corresponding biomass reduction estimated with the POD1 approach. This great difference underlines once more the criticism addressed toward AOT40, which does not take into account the real interaction between ozone and the plant, but only mimes it by the exclusion of the night-time measurements.
In this case, although this situation is typical of the mountain sites and particularly of the most elevated ones, the relatively low AOT40 value is the consequence of the ozone concentrations minima experienced in the midday hours of the daily concentration profile (Figure 6a). On the contrary the highest concentrations were always measured at night, when the AOT40 index is not calculated. It is worth noticing that the highest stomatal fluxes are experienced just when the ozone concentrations are lower, thus causing this risk assessment discrepancy. For this reason, besides the fact that the plant physiological processes are taken into account, the flux-based approach is considered more scientifically sound and advanced.
The flux-based approach allows further considerations on the forest ecosystem services and in particular on the forest capability to remove the air pollutants, an aspect which is difficult to quantify, and it is not often treated.
In order to assess this aspect it necessary to consider also that part of ozone deposition which is called non-stomatal flux. This process is responsible for the removal of 38 kg O3 ha-1(Figure 9), which are destroyed without harming the vegetation, while the total amount of ozone removed by the forest ecosystem is equal to 53 kilograms of ozone per hectare in three months, an amount that is remarkable. In fact, considering an air volume with a basis of 1 ha and a thickness of three meters, containing ozone at a concentration of 90 ppb (i.e. nearly 180 μg m-3, the attention threshold for the Italian regulation) the amount of ozone is only 5 grams.
This significant ecosystem service has a counterpart of forest growth reduction of about 4%.
## 4. Conclusions
The MANFRED project allowed to run a micrometeorological tower to measure directly ozone fluxes over a larch forest ecosystem for the first time in the Alpine region, with the eddy covariance technique.
Results showed that ozone removed by the forest was significant (53 kg ha-1), but the quantity absorbed by leaves was the minority (between 23 to 40%), the most being non-stomatal.
A double layer (4-sinks) SVAT model was developed to study the sub-partition of the stomatal fluxes between the larch trees and the understorey grass, as well as to give an insight to the non-stomatal processes. The model is now suitable for simulations of ozone uptake in climate change scenarios.
Larch needles uptake 25% of the total ozone flux, around 70% of the bulk stomatal flux. The remaining 30% of the bulk stomatal flux was absorbed by the understorey grass.
Even taking into account the detoxifying capability of the plants (instantaneous flux threshold) the phytotoxically active ozone dose (POD1) appears above the critical level for fluxes provisionally set by the UN/ECE to 8 mmol m-2 for Norway spruce, the species chosen as a reference for conifers. At such ozone flux levels, the calculated POD1 for larch suggests a possible trees growth reduction of 4 – 5% in a growing season. This missing growth is counterbalanced by the offer of an ecosystem service, that is the removal of 53 kg ha-1 of ozone from the atmosphere.
## Acknowledgments
This publication was funded by the Catholic University’s program for promotion and divulgation of scientific research. The authors thank the Adamello Park and the mountain community of valle Camonica, themunicipality of Paspardo and in particular dr G.B. Sangalli, dr A. Ducoli and dr D. Orsignola for their support in realizing this field campaign. The authors would like also to thank Stefano Oliveri and Luca Francesco Garibaldo for their important help during the installation. A sincere thank to Michela Scalvenzi for her support in 2011.
## How to cite and reference
### Cite this chapter Copy to clipboard
Giacomo Gerosa, Angelo Finco, Antonio Negri, Riccardo Marzuoli and Gerhard Wieser (August 28th 2013). Ozone Fluxes to a Larch Forest Ecosystem at the Timberline in the Italian Alps, Management Strategies to Adapt Alpine Space Forests to Climate Change Risks, Gillian Ann Cerbu, Marc Hanewinkel, Giacomo Gerosa and Robert Jandl, IntechOpen, DOI: 10.5772/56280. Available from:
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https://www.physicsforums.com/threads/capacitor-energy-resistor.537989/ | # Archived Capacitor energy resistor
1. Oct 8, 2011
### JosephK
1. The problem statement, all variables and given/known data
A charge Q is placed on a capacitor of capacitance C. The capacitor is connected into the circuit as shown in the figure below, with an open switch, a resistor, and an initially uncharged capacitor of capacitance 3C. The switch is then closed and the circuit comes to an equilibrium.
(a) find the final potential difference between the plates of each capacitor.
(b) Find the charge on each capacitor.
(c) Find the final energy stored in each capacitor.
(d) Find the internal energy appearing in the resistor.
2. Relevant equations
E = 1/2 C ΔV^2
U = QΔV
3. The attempt at a solution
As we move across the charged capacitor from negative to positive, the capacitor increases the electric potential energy by exactly E= 1/2 C V^2. Energy is delivered to the resistor and the uncharged capacitor.
Conservation of energy:
$1/2 C(\Delta V_C^{2})= Q(\Delta V_R) + 1/2 (3C)(\Delta V_{3C}^{2})$
In equilibrium, $V_C = V_R+V_{3C}$
Where V_C, V_R and V_C is the voltage across the initially charged capacitor, resistor, and initially uncharged capacitor respectively.
2. Feb 7, 2016
### Staff: Mentor
A complete solution is offered:
Part (a): Final potential difference between the plates of the capacitors
After the switch is closed and the system reaches steady state, current has ceased to flow. There is no potential drop across the resistor (no current) and the capacitors have the same potential difference. Effectively the capacitors are connected in parallel.
Since charge is conserved, charge Q will be on an effective capacitance of C + 3C, or 4C. Since the potential across a capacitor is given by V = Q/C, we have:
$V_c = \frac{Q}{4C}$
This will be the potential difference between the places of each capacitor.
Part (b): The charge on each capacitor
Using the same relationship, V = Q/C, and rearranging it as Q = CV we find the charge on each capacitor:
On the 'C' capacitor: $Q_C = C⋅\frac{Q}{4C} = \frac{1}{4}Q$
On the '3C' capacitor: $Q_{3C} = 3C⋅\frac{Q}{4C} = \frac{3}{4}Q$
Part (c): The energy stored on each capacitor
Two common expressions used for the energy stored on a capacitor are:
$E = ½ C V^2~~~~~~$ and $~~~~~~E = ½ \frac{Q^2}{C}$
We can choose either since we have previously found both the voltage and the charge on each capacitor. Both will yield:
On the 'C' capacitor: $E_C = \frac{Q^2}{32 C}$
On the '3C' capacitor: $E_{3C} = \frac{3Q^2}{32 C}$
Part (d): The "internal energy" appearing in the resistor
This will be given by the difference in energy between the initial state and the final state of the system. That is:
$ΔE = \left\{ \frac{1}{2}\frac{Q^2}{C} \right\} - \left\{ \frac{Q^2}{32 C} + \frac{3~Q^2}{32 C} \right\} = \frac{3}{8} \frac{Q^2}{C}$
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http://physics.stackexchange.com/questions/57545/where-do-the-conservation-laws-come-from/57551 | # Where do the conservation laws come from?
I know the conservation of energy comes from Noether's theorem via the time-translational symmetry, and if I remember correctly, the conservation of momentum comes from space-translational symmetry.
My question is: for a given system, what is the starting point of identifying conservation laws, and how do you know that there's not more that you haven't identified? Do you just start by identifying symmetries?
As an example, consider a system of hard, spherical, elastically colliding billiard balls. I suppose the conserved quantities are energy, momentum, and angular momentum; for each particle these correspond to the terms $\frac{1}{2}mv^2$, $mv$, and $m\omega$. How do we know there isn't something like $kmv^3$? (I know this isn't one, but it's just to demonstrate my point).
And why is the conserved quantity always a scalar? Are there cases where the conserved quantity is not calculated by summing other scalars?
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I will make a broad statement and say "yes", identifying symmetries is THE way to derive conservation laws. Though not every symmetry will give you a conservation law you can be sure that every conservation law results from some kind of (possibly not obvious) symmetry. Regarding your last statement concerning non-scalar conserved quantities: You already mentioned that in some systems you may encounter conservation of angular momentum (think of Kepler's problem). Well, angular momentum is a vectorial quantity, so you already had a perfect counterexample at hand. ;) – André Mar 21 '13 at 8:06
Whoops. Not sure why I wrote momentum was a scalar; it's getting late for me... – Nick Mar 21 '13 at 9:00
Nick: I would start to understand the nature of conservation laws from the perspective of the invariance of the action with respect to transformations. This is basically Noethers theorem but maybe a little more intuitive. For example, charge conservation can be derived from gauge invariance of S, link. – Robert Filter Mar 21 '13 at 18:07
Besides Noether's Theorem, which talks about how symmetry $\Rightarrow$ conservation law, if you like this question, you may also enjoy reading this and this Phys.SE post, which discuss an inverse Noether's Theorem, i.e, whether conservation law $\Rightarrow$ symmetry. – Qmechanic Mar 21 '13 at 18:31
Noether's theorem is one way to relate conservation laws to symmetries, but it is not the only or necessarily the most common method of doing so. For example, in quantum mechanics a continuous symmetry is associated with a family of unitary operators $U(\lambda)$ that shift the eigenvalues associated with some operator (e.g. the position operator, in which case $U(\lambda) \equiv T(\lambda)$ translates states by $\lambda$).
One can show that such an operator is expressible in terms of a self-adjoint (Hermitian) generator $G$ via the relation $U(\lambda) = e^{iG\lambda}$. When the Hamiltonian is invariant invariant with respect to this transformation (i.e. $U^\dagger H U = H$ or equivalently the commutator $\left[U, H\right]$ vanishes) time evolution does not change the eigenvalues of $U$. The same is true of $G$ and, as such, the eigenvalues of $G$ (and $U$) are conserved in this circumstance. For spacial translations, $G$ is proportional to the momentum operator $p$ and momentum is the conserved quantum number.
Conserved quantities are not always scalar. Momentum is, for example, a vector.
In response to your final question, finding all the symmetries of a given system can be an exceptionally difficult problem. Finding such symmetries is a common method of trying to render many-body problems in quantum mechanics "integrable" (exactly solvable) or tractable via numerical methods. There is no systematic method that works generally. Occasionally, hidden symmetries are discovered in the context of well-known problems, and this can be a big deal when it happens.
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Your last paragraph is very interesting. Do you know of any examples of a recently discovered symmetry? – Nick Mar 21 '13 at 9:04
More frequent than a discovery of a new "fundamental" symmetry is the realization that some effect (e.g. an experimental result) can be explained in terms of a symmetry already known. For example, in the 80s Laughlin discovered that the exact quantization of the Hall conductance can be explained in terms of gauge invariance. – Joshua Barr Mar 21 '13 at 10:34
A conservation law, being a function of the degrees of freedom of a system, relates variables and thereby restricts them to some exten. More independend conservation laws means more restrictions and depending on the system, you can't have infinite constraints while still maintaining propagation of the variables.
Secondly, given such a constant of motion (Energy, momentum), earch function of it is also a constant. E.g. the quantity $(\sum_im_iv_i^2)^7$ ist constant if $\sum_i\tfrac{1}{2}m_iv_i^2$ is.
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Conservation laws come exclusively from equations of motion. If you found your solutions $y_i(t)$ as functions of the initial data $y_i(t|y_k(t_0))$, then you can resolve them to express the initial data via the "current data" $y_k(t_0)=f_k(y_i)$ and each such expression is a "conservation law" - a combination of dynamical variables that is constant in time. So the true origin of conservation laws is existence of solutions of equations. The number of independently conserving quantities is the number of independently given initial data, although the form of conserving quantities is not fixed - any combination of conserved quantities is a conserved quantity. We are mostly familiar with "additive" conservation laws where a sum of some variable combinations of different entities is written.
Symmetries do not lead to conservation laws. They help construct some conservation laws from the Lagrangian, and nothing else. In case of a symmetry a given conserved quantity can take a simpler form, nothing more.
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http://math.stackexchange.com/questions/72371/an-explicit-example-of-a-differentiable-function-taking-rational-values-at-ratio | # An explicit example of a differentiable function taking rational values at rational points but whose derivative is irrational at rational points
Construct an example of a differentiable function such that $$\forall r \in {\Bbb Q}\quad f(r) \in {\Bbb Q}\text{ but } f'(r) \notin {\Bbb Q}$$ this example is not trivial, in a paper they prove the existence, but they don't give an explicit example, but they said that other paper as an explicit example of that, but i could not find it.
EDITED: This is the paper:
Walter Rudin, Restrictions on the Values of Derivatives, The American Mathematical Monthly, Vol. 84, No. 9 (1977), pp. 722-723, MR480908.
-
The link works for you...Its better to give the journal ref. and then links. – Tapu Oct 13 '11 at 19:10
On MathSciNet there is the possibility to create a link to the review: right below the bibliographic info there are five red links PDF | Clipboard | Journal | Article | Make Link Click on Make Link, and you'll get a nice and clickable link, looking like this: http://www.ams.org/mathscinet-getitem?mr=480908. While you're at it, you can also click on "Article" (if it exists) and you'll get a stable link to the paper. Please paste both of these links in the future. – t.b. Oct 13 '11 at 19:23
Thanks t.b! it looks so better – August Oct 13 '11 at 19:24
The problem posed is available here, but here it is in its entirety:
Here's a link to the paper you're looking for:
F. D. Hammer and William Knight, Solution to problem 5955, The American Mathematical Monthly Vol. 82, No. 4 (Apr., 1975), pp. 415-416. MR1537708.
The solution given in that paper is very simple and elegant:
Let $\tilde{g}: \left[-\frac{1}{2},\frac{1}{2}\right] \to \mathbb{R}$ be the function $\tilde{g}(x) = x(1-4x^2)$. Its $1$-periodic extension $g$ to all of $\mathbb{R}$ is a $C^1$-function that is zero at the integers and whose derivative at the integers is $1$.
Now let $$f(x) = \sum_{n=0}^{\infty} \frac{g(n!x)}{(n!)^2}.$$
It is straightforward to check that $f\in C^1$ [let $f_k \in C^1$ be the $k$th partial sum of the series. Then $f_k \to f$ pointwise and $f_{k}^\prime$ is uniformly Cauchy, so $f \in C^1$]. Now notice that for rational $x$ only finitely many summands of $f$ are non-zero, hence $f(\mathbb{Q}) \subset \mathbb{Q}$. On the other hand, for rational $x$ we have $$f'(x) - e = f'(x) - \sum_{n=1}^{\infty} \frac{1}{n!} = \text{finitely many rational terms} \in \mathbb{Q},$$ so $f'(x) \notin \mathbb{Q}$, as desired.
(Thanks to robjohn for pointing out a mistake in the previous version of this answer)
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$f'-e\in\mathbb{Q}$ on $\mathbb{Q}$. – robjohn Oct 13 '11 at 23:25
Thanks! Fixed... – t.b. Oct 13 '11 at 23:32
O= I Know the theorem that if the sequence of continuous functions, converges uniformly then the limit is also continuous. But there you are proving that f is $C^1$ what theorem is that? – August Oct 14 '11 at 2:03
@August: I don't know its name. But here's the statement and proof: Let $f$ and $g$ be continuous, $f_n \in C^1$ on an interval. If $f_n \to f$ pointwise and $f_{n}^\prime \to g$ uniformly then $f \in C^1$ and $f^\prime = g$. To see this, write $$f_{n}(x) = f_{n}(a) + \int_{a}^{x} f_{n}^\prime(t)\,dt.$$ Passing to the limit we get $$f(x) = f(a) + \int_{a}^{x} g(t)\,dt$$ where interchange of integration and limit is justified by uniform convergence. By the fundamental theorem of calculus, the derivative of the right hand side exists and is $g(x)$, so $f' = g$ is continuous – t.b. Oct 14 '11 at 8:19 | {"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": 0.8792793154716492, "perplexity": 211.54435617302468}, "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-2015-27/segments/1435376073161.33/warc/CC-MAIN-20150627033433-00117-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://capde.cmm.uchile.cl/publications/fethi-mahmoudi/ | # Fethi Mahmoudi
• M.M. Fall, F. Mahmoudi, and E. Valdinocci, Ground states and concentration
phenomena for the fractional Shrödinger equation, Nonlinearity 28 (2015), no. 6,
1937–1961.
• F. Mahmoudi and B. Abdellaoui, An improved Hardy inequality for a nonlocal
operator, DCDS-A, Vol. 36, No. 3, 2016.
• Abdellaoui, B.; Biroud, K.; Davila, J.; Mahmoudi, F. Nonlinear elliptic problem related to the Hardy inequality with singular term at the boundary. Commun. Contemp. Math. 17 (2015), no. 3, 1450033, 28 pp. pdf
• Mahmoudi, Fethi; Sánchez, Felipe Subiabre; Yao, Wei. On the Ambrosetti-Malchiodi-Ni conjecture for general submanifolds. J. Differential Equations 258 (2015), no. 2, 243–280. pdf
• del Pino, Manuel; Mahmoudi, Fethi; Musso, Monica Bubbling on boundary submanifolds for the Lin-Ni-Takagi problem at higher critical exponents. J. Eur. Math. Soc. (JEMS) 16 (2014), no. 8, 1687–1748. pdf
• Fall, Mouhamed Moustapha; Mahmoudi, Fethi. Weighted Hardy inequality with higher dimensional singularity on the boundary. Calc. Var. Partial Differential Equations 50 (2014), no. 3-4, 779–798. pdf
• Mahmoudi, Fethi. Constant k-curvature hypersurfaces in Riemannian manifolds. Differential Geom. Appl. 28 (2010), no. 1, 1–11. pdf
• Mahmoudi, Fethi; Malchiodi, Andrea; Montenegro, Marcelo Solutions to the nonlinear Schrödinger equation carrying momentum along a curve. Comm. Pure Appl. Math. 62 (2009), no. 9, 1155–1264. pdf
• Fall, Mouhamed Moustapha; Mahmoudi, Fethi. Hypersurfaces with free boundary and large constant mean curvature: concentration along submanifolds. Ann. Sc. Norm. Super. Pisa Cl. Sci. (5) 7 (2008), no. 3, 407–446. pdf
• Mahmoudi, Fethi; Malchiodi, Andrea; Wei, Juncheng Transition layer for the heterogeneous Allen-Cahn equation. Ann. Inst. H. Poincaré Anal. Non Linéaire 25 (2008), no. 3, 609–631. pdf
• Mahmoudi, Fethi; Malchiodi, Andrea; Montenegro, Marcelo. Solutions to the nonlinear Schrödinger equation carrying momentum along a curve. C. R. Math. Acad. Sci. Paris 346 (2008), no. 1-2, 33–38. pdf
• Mahmoudi, Fethi; Malchiodi, Andrea. Concentration on minimal submanifolds for a singularly perturbed Neumann problem. Adv. Math. 209 (2007), no. 2, 460–525. pdf
• Mahmoudi, F.; Mazzeo, R.; Pacard, F. Constant mean curvature hypersurfaces condensing on a submanifold. Geom. Funct. Anal. 16 (2006), no. 4, 924–958.pdf
• Mahmoudi, Fethi; Malchiodi, Andrea. Concentration at manifolds of arbitrary dimension for a singularly perturbed Neumann problem. Atti Accad. Naz. Lincei Cl. Sci. Fis. Mat. Natur. Rend. Lincei (9) Mat. Appl. 17 (2006), no. 3, 279–290. pdf
• Mahmoudi, Fethi Energy quantization for Yamabe’s problem in conformal dimension. Electron. J. Differential Equations 2006, No. 71, 17 pp. (electronic). pdf
• Fall M.M., Mahmoudi F. and Valdinoci E., Ground states and concetration phenomena for the fractional Schrödinger equation. Nonlinearity No28 1937-1961. pdf
• B. Abdellaoui and Fethi Mahmoudi. An improved Hardy inequality for a
nonlocal operator. accepted for publication in DCDS-A. pdf | {"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": 0.831530749797821, "perplexity": 3619.0214641535304}, "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-2021-21/segments/1620243989614.9/warc/CC-MAIN-20210511122905-20210511152905-00355.warc.gz"} |
https://www.dummies.com/education/math/geometry/find-midpoint-line-segment/ | Find a Midpoint on a Line Segment - dummies
# Find a Midpoint on a Line Segment
Say you’re given a line segment and you have to find the coordinates of its midpoint. What’s the best way to do this? The midpoint formula!
The way the midpoint formula works is very simple: It takes the average of the x-coordinates of the segment’s endpoints and the average of the y-coordinates of the endpoints. These averages give you the location of a point that is exactly in the middle of the segment.
Midpoint formula: To find the midpoint of a segment with endpoints at (x1, y1) and (x2, y2), use the following formula:
Note: It doesn’t matter which point is (x1, y1) and which is (x2, y2).
Here’s a problem that shows the midpoint formula in action.
Here is the proof diagram.
If you know your rectangle properties, you know that the diagonals of PQRS must bisect each other. But another way to show this is with coordinate geometry. The term bisect in this problem should ring the midpoint bell. So use the midpoint formula for each diagonal:
The fact that the two midpoints are the same shows that each diagonal goes through the midpoint of the other, and that, therefore, each diagonal bisects the other. Obviously, the diagonals cross at (6.5, 3). That’s a wrap. | {"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": 0.9118213653564453, "perplexity": 320.5067881151139}, "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/1566027316555.4/warc/CC-MAIN-20190822000659-20190822022659-00522.warc.gz"} |
https://en.wikipedia.org/wiki/Pascal_triangle | # Pascal's triangle
(Redirected from Pascal triangle)
In Pascal's triangle, each number is the sum of the two numbers directly above it.
In mathematics, Pascal's triangle is a triangular array of the binomial coefficients. In much of the Western world, it is named after the French mathematician Blaise Pascal, although other mathematicians studied it centuries before him in India,[1] Persia (Iran),[2] China, Germany, and Italy.[3]
The rows of Pascal's triangle are conventionally enumerated starting with row n = 0 at the top (the 0th row). The entries in each row are numbered from the left beginning with k = 0 and are usually staggered relative to the numbers in the adjacent rows. The triangle may be constructed in the following manner: In row 0 (the topmost row), there is a unique nonzero entry 1. Each entry of each subsequent row is constructed by adding the number above and to the left with the number above and to the right, treating blank entries as 0. For example, the initial number in the first (or any other) row is 1 (the sum of 0 and 1), whereas the numbers 1 and 3 in the third row are added to produce the number 4 in the fourth row.
## Formula
${\displaystyle {\begin{array}{c}1\\1\quad 1\\1\quad 2\quad 1\\1\quad 3\quad 3\quad 1\\1\quad 4\quad 6\quad 4\quad 1\\1\quad 5\quad 10\quad 10\quad 5\quad 1\\1\quad 6\quad 15\quad 20\quad 15\quad 6\quad 1\\1\quad 7\quad 21\quad 35\quad 35\quad 21\quad 7\quad 1\\\end{array}}}$
A diagram that shows Pascal's triangle with rows 0 through 7.
The entry in the nth row and kth column of Pascal's triangle is denoted ${\displaystyle {\tbinom {n}{k}}}$. For example, the unique nonzero entry in the topmost row is ${\displaystyle {\tbinom {0}{0}}=1}$. With this notation, the construction of the previous paragraph may be written as follows:
${\displaystyle {n \choose k}={n-1 \choose k-1}+{n-1 \choose k}}$,
for any non-negative integer n and any integer k between 0 and n, inclusive.[4] This recurrence for the binomial coefficients is known as Pascal's rule.
Pascal's triangle has higher dimensional generalizations. The three-dimensional version is called Pascal's pyramid or Pascal's tetrahedron, while the general versions are called Pascal's simplices.
## History
मेरु प्रस्तार(Meru Prastaara) as used in Indian manuscripts, derived from Pingala's formulae. Manuscript from Raghunath Library J&K; 755 AD
Yang Hui's triangle, as depicted by the Chinese using rod numerals, appears in a mathematical work by Zhu Shijie, dated 1303. The title reads "The Old Method Chart of the Seven Multiplying Squares" (Chinese: 古法七乘方圖; the fourth character 椉 in the image title is archaic).
Pascal's version of the triangle
The pattern of numbers that forms Pascal's triangle was known well before Pascal's time. Pascal innovated many previously unattested uses of the triangle's numbers, uses he described comprehensively in the earliest known mathematical treatise to be specially devoted to the triangle, his Traité du triangle arithmétique (1654; published 1665). Centuries before, discussion of the numbers had arisen in the context of Indian studies of combinatorics and of binomial numbers and the Greeks' study of figurate numbers.[5]
From later commentary, it appears that the binomial coefficients and the additive formula for generating them, ${\displaystyle {\tbinom {n}{r}}={\tbinom {n-1}{r}}+{\tbinom {n-1}{r-1}}}$, were known to Pingala in or before the 2nd century BC.[6][7] While Pingala's work only survives in fragments, the commentator Varāhamihira, around 505, gave a clear description of the additive formula,[7] and a more detailed explanation of the same rule was given by Halayudha, around 975. Halayudha also explained obscure references to Meru-prastaara, the Staircase of Mount Meru, giving the first surviving description of the arrangement of these numbers into a triangle.[7][8] In approximately 850, the Jain mathematician Mahāvīra gave a different formula for the binomial coefficients, using multiplication, equivalent to the modern formula ${\displaystyle {\tbinom {n}{r}}={\tfrac {n!}{r!(n-r)!}}}$.[7] In 1068, four columns of the first sixteen rows were given by the mathematician Bhattotpala, who was the first recorded mathematician to equate the additive and multiplicative formulas for these numbers.[7]
At around the same time, the Persian mathematician Al-Karaji (953–1029) wrote a now-lost book which contained the first description of Pascal's triangle.[9][10][11] It was later repeated by the Persian poet-astronomer-mathematician Omar Khayyám (1048–1131); thus the triangle is also referred to as the Khayyam triangle in Iran.[12] Several theorems related to the triangle were known, including the binomial theorem. Khayyam used a method of finding nth roots based on the binomial expansion, and therefore on the binomial coefficients.[13]
Pascal's triangle was known in China in the early 11th century through the work of the Chinese mathematician Jia Xian (1010–1070). In the 13th century, Yang Hui (1238–1298) presented the triangle and hence it is still called Yang Hui's triangle (杨辉三角; 楊輝三角) in China.[14]
In the west, the binomial coefficients were calculated by Gersonides in the early 14th century, using the multiplicative formula for them.[7] Petrus Apianus (1495–1552) published the full triangle on the frontispiece of his book on business calculations in 1527. This is the first record of the triangle in Europe.[15] Michael Stifel published a portion of the triangle (from the second to the middle column in each row) in 1544, describing it as a table of figurate numbers.[7] In Italy, Pascal's triangle is referred to as Tartaglia's triangle, named for the Italian algebraist Niccolò Fontana Tartaglia (1500–1577), who published six rows of the triangle in 1556.[7] Gerolamo Cardano, also, published the triangle as well as the additive and multiplicative rules for constructing it in 1570.[7]
Pascal's Traité du triangle arithmétique (Treatise on Arithmetical Triangle) was published in 1655. In this, Pascal collected several results then known about the triangle, and employed them to solve problems in probability theory. The triangle was later named after Pascal by Pierre Raymond de Montmort (1708) who called it "Table de M. Pascal pour les combinaisons" (French: Table of Mr. Pascal for combinations) and Abraham de Moivre (1730) who called it "Triangulum Arithmeticum PASCALIANUM" (Latin: Pascal's Arithmetic Triangle), which became the modern Western name.[16]
## Binomial expansions
Visualisation of binomial expansion up to the 4th power
Pascal's triangle determines the coefficients which arise in binomial expansions. For example, consider the expansion
(x + y)2 = x2 + 2xy + y2 = 1x2y0 + 2x1y1 + 1x0y2.
The coefficients are the numbers in row two of Pascal's triangle: 1, 2, 1. In general, when a binomial like x + y is raised to a positive integer power we have:
(x + y)n = a0xn + a1xn−1y + a2xn−2y2 + ... + an−1xyn−1 + anyn,
where the coefficients ai in this expansion are precisely the numbers on row n of Pascal's triangle. In other words,
${\displaystyle a_{i}={n \choose i}.}$
This is the binomial theorem.
The entire right diagonal of Pascal's triangle corresponds to the coefficient of yn in these binomial expansions, while the next diagonal corresponds to the coefficient of xyn−1 and so on.
To see how the binomial theorem relates to the simple construction of Pascal's triangle, consider the problem of calculating the coefficients of the expansion of (x + 1)n+1 in terms of the corresponding coefficients of (x + 1)n (setting y = 1 for simplicity). Suppose then that
${\displaystyle (x+1)^{n}=\sum _{i=0}^{n}a_{i}x^{i}.}$
Now
${\displaystyle (x+1)^{n+1}=(x+1)(x+1)^{n}=x(x+1)^{n}+(x+1)^{n}=\sum _{i=0}^{n}a_{i}x^{i+1}+\sum _{i=0}^{n}a_{i}x^{i}.}$
${\displaystyle {\begin{array}{c}{\binom {0}{0}}\\{\binom {1}{0}}\quad {\binom {1}{1}}\\{\binom {2}{0}}\quad {\binom {2}{1}}\quad {\binom {2}{2}}\\{\binom {3}{0}}\quad {\binom {3}{1}}\quad {\binom {3}{2}}\quad {\binom {3}{3}}\\{\binom {4}{0}}\quad {\binom {4}{1}}\quad {\binom {4}{2}}\quad {\binom {4}{3}}\quad {\binom {4}{4}}\\{\binom {5}{0}}\quad {\binom {5}{1}}\quad {\binom {5}{2}}\quad {\binom {5}{3}}\quad {\binom {5}{4}}\quad {\binom {5}{5}}\end{array}}}$
Six rows Pascal's triangle as binomial coefficients
The two summations can be reorganized as follows:
{\displaystyle {\begin{aligned}&\sum _{i=0}^{n}a_{i}x^{i+1}+\sum _{i=0}^{n}a_{i}x^{i}\\&{}=\sum _{i=1}^{n+1}a_{i-1}x^{i}+\sum _{i=0}^{n}a_{i}x^{i}\\&{}=\sum _{i=1}^{n}a_{i-1}x^{i}+\sum _{i=1}^{n}a_{i}x^{i}+a_{0}x^{0}+a_{n}x^{n+1}\\&{}=\sum _{i=1}^{n}(a_{i-1}+a_{i})x^{i}+a_{0}x^{0}+a_{n}x^{n+1}\\&{}=\sum _{i=1}^{n}(a_{i-1}+a_{i})x^{i}+x^{0}+x^{n+1}\end{aligned}}}
(because of how raising a polynomial to a power works, a0 = an = 1).
We now have an expression for the polynomial (x + 1)n+1 in terms of the coefficients of (x + 1)n (these are the ais), which is what we need if we want to express a line in terms of the line above it. Recall that all the terms in a diagonal going from the upper-left to the lower-right correspond to the same power of x, and that the a-terms are the coefficients of the polynomial (x + 1)n, and we are determining the coefficients of (x + 1)n+1. Now, for any given i not 0 or n + 1, the coefficient of the xi term in the polynomial (x + 1)n+1 is equal to ai−1 + ai. This is indeed the simple rule for constructing Pascal's triangle row-by-row.
It is not difficult to turn this argument into a proof (by mathematical induction) of the binomial theorem. Since (a + b)n = bn(a/b + 1)n, the coefficients are identical in the expansion of the general case.
An interesting consequence of the binomial theorem is obtained by setting both variables x and y equal to one. In this case, we know that (1 + 1)n = 2n, and so
${\displaystyle {n \choose 0}+{n \choose 1}+\cdots +{n \choose n-1}+{n \choose n}=2^{n}.}$
In other words, the sum of the entries in the nth row of Pascal's triangle is the nth power of 2. This is equivalent to the statement that the number of subsets (the cardinality of the power set) of an n-element set is ${\displaystyle 2^{n}}$, as can be seen by observing that the number of subsets is the sum of the number of combinations of each of the possible lengths, which range from zero through to n.
## Combinations
A second useful application of Pascal's triangle is in the calculation of combinations. For example, the number of combinations of n things taken k at a time (called n choose k) can be found by the equation
${\displaystyle \mathbf {C} (n,k)=\mathbf {C} _{k}^{n}={_{n}C_{k}}={n \choose k}={\frac {n!}{k!(n-k)!}}.}$
But this is also the formula for a cell of Pascal's triangle. Rather than performing the calculation, one can simply look up the appropriate entry in the triangle. Provided we have the first row and the first entry in a row numbered 0, the answer will be located at entry k in row n. For example, suppose a basketball team has 10 players and wants to know how many ways there are of selecting 8. The answer is entry 8 in row 10, which is 45; that is, 10 choose 8 is 45.
## Relation to binomial distribution and convolutions
When divided by 2n, the nth row of Pascal's triangle becomes the binomial distribution in the symmetric case where p = 1/2. By the central limit theorem, this distribution approaches the normal distribution as n increases. This can also be seen by applying Stirling's formula to the factorials involved in the formula for combinations.
This is related to the operation of discrete convolution in two ways. First, polynomial multiplication exactly corresponds to discrete convolution, so that repeatedly convolving the sequence {..., 0, 0, 1, 1, 0, 0, ...} with itself corresponds to taking powers of 1 + x, and hence to generating the rows of the triangle. Second, repeatedly convolving the distribution function for a random variable with itself corresponds to calculating the distribution function for a sum of n independent copies of that variable; this is exactly the situation to which the central limit theorem applies, and hence leads to the normal distribution in the limit.
## Patterns and properties
Pascal's triangle has many properties and contains many patterns of numbers.
Each frame represents a row in Pascal's triangle. Each column of pixels is a number in binary with the least significant bit at the bottom. Light pixels represent ones and the dark pixels are zeroes.
### Rows
• The sum of the elements of a single row is twice the sum of the row preceding it. For example, row 0 (the topmost row) has a value of 1, row 1 has a value of 2, row 2 has a value of 4, and so forth. This is because every item in a row produces two items in the next row: one left and one right. The sum of the elements of row n is equal to 2n.
• Taking the product of the elements in each row, the sequence of products (sequence A001142 in the OEIS) is related to the base of the natural logarithm, e.[17][18] Specifically, define the sequence sn as follows:
${\displaystyle s_{n}=\prod _{k=0}^{n}{\binom {n}{k}}=\prod _{k=0}^{n}{\frac {n!}{k!(n-k)!}}~,~n\geq 0.}$
Then, the ratio of successive row products is
${\displaystyle {\frac {s_{n+1}}{s_{n}}}={\frac {(n+1)!^{(n+2)}\prod _{k=0}^{n+1}{k!^{-2}}}{n!^{(n+1)}\prod _{k=0}^{n}{k!^{-2}}}}={\frac {(n+1)^{n}}{n!}}}$
and the ratio of these ratios is
${\displaystyle {\frac {(s_{n+1})(s_{n-1})}{(s_{n})^{2}}}=\left({\frac {n+1}{n}}\right)^{n},~n\geq 1.}$
The right-hand side of the above equation takes the form of the limit definition of e
${\displaystyle {\textit {e}}=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}.}$
• Pi can be found in Pascal's triangle through the Nilakantha infinite series.[19]
${\displaystyle \pi =3+\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {\binom {2n+1}{1}}{{\binom {2n+1}{2}}{\binom {2n+2}{2}}}}}$
• The value of a row, if each entry is considered a decimal place (and numbers larger than 9 carried over accordingly), is a power of 11 ( 11n, for row n). Thus, in row 2, ⟨1, 2, 1⟩ becomes 112, while ⟨1, 5, 10, 10, 5, 1⟩ in row five becomes (after carrying) 161,051, which is 115. This property is explained by setting x = 10 in the binomial expansion of (x + 1)n, and adjusting values to the decimal system. But x can be chosen to allow rows to represent values in any base.
• In base 3: 1 2 13 = 42 (16)
• ⟨1, 3, 3, 1⟩ → 2 1 0 13 = 43 (64)
• In base 9: 1 2 19 = 102 (100)
• 1 3 3 19 = 103 (1000)
• ⟨1, 5, 10, 10, 5, 1⟩ → 1 6 2 1 5 19 = 105 (100000)
In particular (see previous property), for x = 1 place value remains constant (1place=1). Thus entries can simply be added in interpreting the value of a row.
• Some of the numbers in Pascal's triangle correlate to numbers in Lozanić's triangle.
• The sum of the squares of the elements of row n equals the middle element of row 2n. For example, 12 + 42 + 62 + 42 + 12 = 70. In general form:
${\displaystyle \sum _{k=0}^{n}{n \choose k}^{2}={2n \choose n}.}$
• On any row n, where n is even, the middle term minus the term two spots to the left equals a Catalan number, specifically the (n/2 + 1)th Catalan number. For example: on row 4, 6 − 1 = 5, which is the 3rd Catalan number, and 4/2 + 1 = 3.
• In a row p where p is a prime number, all the terms in that row except the 1s are multiples of p. This can be proven easily, since if ${\displaystyle p\in \mathbb {P} }$, then p has no factors save for 1 and itself. Every entry in the triangle is an integer, so therefore by definition ${\displaystyle (p-k)!}$ and ${\displaystyle k!}$ are factors of ${\displaystyle p!\,}$. However, there is no possible way p itself can show up in the denominator, so therefore p (or some multiple of it) must be left in the numerator, making the entire entry a multiple of p.
• Parity: To count odd terms in row n, convert n to binary. Let x be the number of 1s in the binary representation. Then the number of odd terms will be 2x. These numbers are the values in Gould's sequence.[20]
• Every entry in row 2n-1, n ≥ 0, is odd.[21]
• Polarity: When the elements of a row of Pascal's triangle are added and subtracted together sequentially, every row with a middle number, meaning rows that have an odd number of integers, gives 0 as the result. As examples, row 4 is 1 4 6 4 1, so the formula would be 6 – (4+4) + (1+1) = 0; and row 6 is 1 6 15 20 15 6 1, so the formula would be 20 – (15+15) + (6+6) – (1+1) = 0. So every even row of the Pascal triangle equals 0 when you take the middle number, then subtract the integers directly next to the center, then add the next integers, then subtract, so on and so forth until you reach the end of the row.
### Diagonals
Derivation of simplex numbers from a left-justified Pascal's triangle
The diagonals of Pascal's triangle contain the figurate numbers of simplices:
{\displaystyle {\begin{aligned}P_{0}(n)&=P_{d}(0)=1,\\P_{d}(n)&=P_{d}(n-1)+P_{d-1}(n)\\&=\sum _{i=0}^{n}P_{d-1}(i)=\sum _{i=0}^{d}P_{i}(n-1).\end{aligned}}}
The symmetry of the triangle implies that the nth d-dimensional number is equal to the dth n-dimensional number.
An alternative formula that does not involve recursion is as follows:
${\displaystyle P_{d}(n)={\frac {1}{d!}}\prod _{k=0}^{d-1}(n+k)={n^{(d)} \over d!}={\binom {n+d-1}{d}}}$
where n(d) is the rising factorial.
The geometric meaning of a function Pd is: Pd(1) = 1 for all d. Construct a d-dimensional triangle (a 3-dimensional triangle is a tetrahedron) by placing additional dots below an initial dot, corresponding to Pd(1) = 1. Place these dots in a manner analogous to the placement of numbers in Pascal's triangle. To find Pd(x), have a total of x dots composing the target shape. Pd(x) then equals the total number of dots in the shape. A 0-dimensional triangle is a point and a 1-dimensional triangle is simply a line, and therefore P0(x) = 1 and P1(x) = x, which is the sequence of natural numbers. The number of dots in each layer corresponds to Pd − 1(x).
### Calculating a row or diagonal by itself
There are simple algorithms to compute all the elements in a row or diagonal without computing other elements or factorials.
To compute row ${\displaystyle n}$ with the elements ${\displaystyle {\tbinom {n}{0}}}$, ${\displaystyle {\tbinom {n}{1}}}$, ..., ${\displaystyle {\tbinom {n}{n}}}$, begin with ${\displaystyle {\tbinom {n}{0}}=1}$. For each subsequent element, the value is determined by multiplying the previous value by a fraction with slowly changing numerator and denominator:
${\displaystyle {n \choose k}={n \choose k-1}\times {\frac {n+1-k}{k}}.}$
For example, to calculate row 5, the fractions are ${\displaystyle {\tfrac {5}{1}}}$${\displaystyle {\tfrac {4}{2}}}$${\displaystyle {\tfrac {3}{3}}}$${\displaystyle {\tfrac {2}{4}}}$ and ${\displaystyle {\tfrac {1}{5}}}$, and hence the elements are ${\displaystyle {\tbinom {5}{0}}=1}$, ${\displaystyle {\tbinom {5}{1}}=1\times {\tfrac {5}{1}}=5}$, ${\displaystyle {\tbinom {5}{2}}=5\times {\tfrac {4}{2}}=10}$, etc. (The remaining elements are most easily obtained by symmetry.)
To compute the diagonal containing the elements ${\displaystyle {\tbinom {n}{0}}}$, ${\displaystyle {\tbinom {n+1}{1}}}$, ${\displaystyle {\tbinom {n+2}{2}}}$, ..., we again begin with ${\displaystyle {\tbinom {n}{0}}=1}$ and obtain subsequent elements by multiplication by certain fractions:
${\displaystyle {n+k \choose k}={n+k-1 \choose k-1}\times {\frac {n+k}{k}}.}$
For example, to calculate the diagonal beginning at ${\displaystyle {\tbinom {5}{0}}}$, the fractions are ${\displaystyle {\tfrac {6}{1}}}$${\displaystyle {\tfrac {7}{2}}}$${\displaystyle {\tfrac {8}{3}}}$, ..., and the elements are ${\displaystyle {\tbinom {5}{0}}=1}$, ${\displaystyle {\tbinom {6}{1}}=1\times {\tfrac {6}{1}}=6}$, ${\displaystyle {\tbinom {7}{2}}=6\times {\tfrac {7}{2}}=21}$, etc. By symmetry, these elements are equal to ${\displaystyle {\tbinom {5}{5}}}$, ${\displaystyle {\tbinom {6}{5}}}$, ${\displaystyle {\tbinom {7}{5}}}$, etc.
Fibonacci sequence in Pascal's triangle
### Overall patterns and properties
A level-4 approximation to a Sierpinski triangle obtained by shading the first 32 rows of a Pascal triangle white if the binomial coefficient is even and black if it is odd.
• The pattern obtained by coloring only the odd numbers in Pascal's triangle closely resembles the fractal called the Sierpinski triangle. This resemblance becomes more and more accurate as more rows are considered; in the limit, as the number of rows approaches infinity, the resulting pattern is the Sierpinski triangle, assuming a fixed perimeter.[22] More generally, numbers could be colored differently according to whether or not they are multiples of 3, 4, etc.; this results in other similar patterns.
10 10 20
Pascal's triangle overlaid on a grid gives the number of distinct paths to each square, assuming only rightward and downward movements are considered.
• In a triangular portion of a grid (as in the images below), the number of shortest grid paths from a given node to the top node of the triangle is the corresponding entry in Pascal's triangle. On a Plinko game board shaped like a triangle, this distribution should give the probabilities of winning the various prizes.
• If the rows of Pascal's triangle are left-justified, the diagonal bands (colour-coded below) sum to the Fibonacci numbers.
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1
### Construction as matrix exponential
{\displaystyle {\begin{aligned}\exp {\begin{pmatrix}.&.&.&.&.\\1&.&.&.&.\\.&2&.&.&.\\.&.&3&.&.\\.&.&.&4&.\end{pmatrix}}&={\begin{pmatrix}1&.&.&.&.\\1&1&.&.&.\\1&2&1&.&.\\1&3&3&1&.\\1&4&6&4&1\end{pmatrix}}\\e^{counting}&=binomial\end{aligned}}}
Binomial matrix as matrix exponential. All the dots represent 0.
Due to its simple construction by factorials, a very basic representation of Pascal's triangle in terms of the matrix exponential can be given: Pascal's triangle is the exponential of the matrix which has the sequence 1, 2, 3, 4, ... on its subdiagonal and zero everywhere else.
### Connections to geometry of polytopes
Pascal's triangle can be used as a lookup table for the number of elements (such as edges and corners) within a polytope (such as a triangle, a tetrahedron, a square and a cube).
#### Number of elements of simplices
Let's begin by considering the 3rd line of Pascal's triangle, with values 1, 3, 3, 1. A 2-dimensional triangle has one 2-dimensional element (itself), three 1-dimensional elements (lines, or edges), and three 0-dimensional elements (vertices, or corners). The meaning of the final number (1) is more difficult to explain (but see below). Continuing with our example, a tetrahedron has one 3-dimensional element (itself), four 2-dimensional elements (faces), six 1-dimensional elements (edges), and four 0-dimensional elements (vertices). Adding the final 1 again, these values correspond to the 4th row of the triangle (1, 4, 6, 4, 1). Line 1 corresponds to a point, and Line 2 corresponds to a line segment (dyad). This pattern continues to arbitrarily high-dimensioned hyper-tetrahedrons (known as simplices).
To understand why this pattern exists, one must first understand that the process of building an n-simplex from an (n − 1)-simplex consists of simply adding a new vertex to the latter, positioned such that this new vertex lies outside of the space of the original simplex, and connecting it to all original vertices. As an example, consider the case of building a tetrahedron from a triangle, the latter of whose elements are enumerated by row 3 of Pascal's triangle: 1 face, 3 edges, and 3 vertices (the meaning of the final 1 will be explained shortly). To build a tetrahedron from a triangle, we position a new vertex above the plane of the triangle and connect this vertex to all three vertices of the original triangle.
The number of a given dimensional element in the tetrahedron is now the sum of two numbers: first the number of that element found in the original triangle, plus the number of new elements, each of which is built upon elements of one fewer dimension from the original triangle. Thus, in the tetrahedron, the number of cells (polyhedral elements) is 0 + 1 = 1; the number of faces is 1 + 3 = 4; the number of edges is 3 + 3 = 6; the number of new vertices is 3 + 1 = 4. This process of summing the number of elements of a given dimension to those of one fewer dimension to arrive at the number of the former found in the next higher simplex is equivalent to the process of summing two adjacent numbers in a row of Pascal's triangle to yield the number below. Thus, the meaning of the final number (1) in a row of Pascal's triangle becomes understood as representing the new vertex that is to be added to the simplex represented by that row to yield the next higher simplex represented by the next row. This new vertex is joined to every element in the original simplex to yield a new element of one higher dimension in the new simplex, and this is the origin of the pattern found to be identical to that seen in Pascal's triangle. The "extra" 1 in a row can be thought of as the -1 simplex, the unique center of the simplex, which ever gives rise to a new vertex and a new dimension, yielding a new simplex with a new center.
#### Number of elements of hypercubes
A similar pattern is observed relating to squares, as opposed to triangles. To find the pattern, one must construct an analog to Pascal's triangle, whose entries are the coefficients of (x + 2)Row Number, instead of (x + 1)Row Number. There are a couple ways to do this. The simpler is to begin with Row 0 = 1 and Row 1 = 1, 2. Proceed to construct the analog triangles according to the following rule:
${\displaystyle {n \choose k}=2\times {n-1 \choose k-1}+{n-1 \choose k}.}$
That is, choose a pair of numbers according to the rules of Pascal's triangle, but double the one on the left before adding. This results in:
${\displaystyle {\begin{matrix}{\text{ 1}}\\{\text{ 1}}\quad {\text{ 2}}\\{\text{ 1}}\quad {\text{ 4}}\quad {\text{ 4}}\\{\text{ 1}}\quad {\text{ 6}}\quad {\text{ 12}}\quad {\text{ 8}}\\{\text{ 1}}\quad {\text{ 8}}\quad {\text{ 24}}\quad {\text{ 32}}\quad {\text{ 16}}\\{\text{ 1}}\quad {\text{ 10}}\quad {\text{ 40}}\quad {\text{ 80}}\quad {\text{ 80}}\quad {\text{ 32}}\\{\text{ 1}}\quad {\text{ 12}}\quad {\text{ 60}}\quad 160\quad 240\quad 192\quad {\text{ 64}}\\{\text{ 1}}\quad {\text{ 14}}\quad {\text{ 84}}\quad 280\quad 560\quad 672\quad 448\quad 128\end{matrix}}}$
The other way of manufacturing this triangle is to start with Pascal's triangle and multiply each entry by 2k, where k is the position in the row of the given number. For example, the 2nd value in row 4 of Pascal's triangle is 6 (the slope of 1s corresponds to the zeroth entry in each row). To get the value that resides in the corresponding position in the analog triangle, multiply 6 by 2Position Number = 6 × 22 = 6 × 4 = 24. Now that the analog triangle has been constructed, the number of elements of any dimension that compose an arbitrarily dimensioned cube (called a hypercube) can be read from the table in a way analogous to Pascal's triangle. For example, the number of 2-dimensional elements in a 2-dimensional cube (a square) is one, the number of 1-dimensional elements (sides, or lines) is 4, and the number of 0-dimensional elements (points, or vertices) is 4. This matches the 2nd row of the table (1, 4, 4). A cube has 1 cube, 6 faces, 12 edges, and 8 vertices, which corresponds to the next line of the analog triangle (1, 6, 12, 8). This pattern continues indefinitely.
To understand why this pattern exists, first recognize that the construction of an n-cube from an (n − 1)-cube is done by simply duplicating the original figure and displacing it some distance (for a regular n-cube, the edge length) orthogonal to the space of the original figure, then connecting each vertex of the new figure to its corresponding vertex of the original. This initial duplication process is the reason why, to enumerate the dimensional elements of an n-cube, one must double the first of a pair of numbers in a row of this analog of Pascal's triangle before summing to yield the number below. The initial doubling thus yields the number of "original" elements to be found in the next higher n-cube and, as before, new elements are built upon those of one fewer dimension (edges upon vertices, faces upon edges, etc.). Again, the last number of a row represents the number of new vertices to be added to generate the next higher n-cube.
In this triangle, the sum of the elements of row m is equal to 3m. Again, to use the elements of row 4 as an example: 1 + 8 + 24 + 32 + 16 = 81, which is equal to ${\displaystyle 3^{4}=81}$.
#### Counting vertices in a cube by distance
Each row of Pascal's triangle gives the number of vertices at each distance from a fixed vertex in an n-dimensional cube. For example, in three dimensions, the third row (1 3 3 1) corresponds to the usual three-dimensional cube: fixing a vertex V, there is one vertex at distance 0 from V (that is, V itself), three vertices at distance 1, three vertices at distance 2 and one vertex at distance 3 (the vertex opposite V). The second row corresponds to a square, while larger-numbered rows correspond to hypercubes in each dimension.
### Fourier transform of sin(x)n+1/x
As stated previously, the coefficients of (x + 1)n are the nth row of the triangle. Now the coefficients of (x − 1)n are the same, except that the sign alternates from +1 to −1 and back again. After suitable normalization, the same pattern of numbers occurs in the Fourier transform of sin(x)n+1/x. More precisely: if n is even, take the real part of the transform, and if n is odd, take the imaginary part. Then the result is a step function, whose values (suitably normalized) are given by the nth row of the triangle with alternating signs.[23] For example, the values of the step function that results from:
${\displaystyle \,{\mathfrak {Re}}\left({\text{Fourier}}\left[{\frac {\sin(x)^{5}}{x}}\right]\right)}$
compose the 4th row of the triangle, with alternating signs. This is a generalization of the following basic result (often used in electrical engineering):
${\displaystyle \,{\mathfrak {Re}}\left({\text{Fourier}}\left[{\frac {\sin(x)^{1}}{x}}\right]\right)}$
is the boxcar function.[24] The corresponding row of the triangle is row 0, which consists of just the number 1.
If n is congruent to 2 or to 3 mod 4, then the signs start with −1. In fact, the sequence of the (normalized) first terms corresponds to the powers of i, which cycle around the intersection of the axes with the unit circle in the complex plane:
${\displaystyle \,+i,-1,-i,+1,+i,\ldots \,}$
### Elementary cellular automaton
The pattern produced by an elementary cellular automaton using rule 60 is exactly Pascal's triangle of binomial coefficients reduced modulo 2 (black cells correspond to odd binomial coefficients).[25] Rule 102 also produces this pattern when trailing zeros are omitted. Rule 90 produces the same pattern but with an empty cell separating each entry in the rows.
## Extensions
Pascal's triangle can be extended to negative row numbers.
First write the triangle in the following form:
m
n
0 1 2 3 4 5 ...
0 1 0 0 0 0 0 ...
1 1 1 0 0 0 0 ...
2 1 2 1 0 0 0 ...
3 1 3 3 1 0 0 ...
4 1 4 6 4 1 0 ...
Next, extend the column of 1s upwards:
m
n
0 1 2 3 4 5 ...
−4 1 ...
−3 1 ...
−2 1 ...
−1 1 ...
0 1 0 0 0 0 0 ...
1 1 1 0 0 0 0 ...
2 1 2 1 0 0 0 ...
3 1 3 3 1 0 0 ...
4 1 4 6 4 1 0 ...
Now the rule:
${\displaystyle {n \choose m}={n-1 \choose m-1}+{n-1 \choose m}}$
can be rearranged to:
${\displaystyle {n-1 \choose m}={n \choose m}-{n-1 \choose m-1}}$
which allows calculation of the other entries for negative rows:
m
n
0 1 2 3 4 5 ...
−4 1 −4 10 −20 35 −56 ...
−3 1 −3 6 −10 15 −21 ...
−2 1 −2 3 −4 5 −6 ...
−1 1 −1 1 −1 1 −1 ...
0 1 0 0 0 0 0 ...
1 1 1 0 0 0 0 ...
2 1 2 1 0 0 0 ...
3 1 3 3 1 0 0 ...
4 1 4 6 4 1 0 ...
This extension preserves the property that the values in the mth column viewed as a function of n are fit by an order m polynomial, namely
${\displaystyle {n \choose m}={\frac {1}{m!}}\prod _{k=0}^{m-1}(n-k)={\frac {1}{m!}}\prod _{k=1}^{m}(n-k+1)}$.
This extension also preserves the property that the values in the nth row correspond to the coefficients of (1 + x)n:
${\displaystyle (1+x)^{n}=\sum _{k=0}^{\infty }{n \choose k}x^{k}\quad |x|<1}$
For example:
${\displaystyle (1+x)^{-2}=1-2x+3x^{2}-4x^{3}+\cdots \quad |x|<1}$
When viewed as a series, the rows of negative n diverge. However, they are still Abel summable, which summation gives the standard values of 2n. (In fact, the n = -1 row results in Grandi's series which "sums" to 1/2, and the n = -2 row results in another well-known series which has an Abel sum of 1/4.)
Another option for extending Pascal's triangle to negative rows comes from extending the other line of 1s:
m
n
−4 −3 −2 −1 0 1 2 3 4 5 ...
−4 1 0 0 0 0 0 0 0 0 0 ...
−3 1 0 0 0 0 0 0 0 0 ...
−2 1 0 0 0 0 0 0 0 ...
−1 1 0 0 0 0 0 0 ...
0 0 0 0 0 1 0 0 0 0 0 ...
1 0 0 0 0 1 1 0 0 0 0 ...
2 0 0 0 0 1 2 1 0 0 0 ...
3 0 0 0 0 1 3 3 1 0 0 ...
4 0 0 0 0 1 4 6 4 1 0 ...
Applying the same rule as before leads to
m
n
−4 −3 −2 −1 0 1 2 3 4 5 ...
−4 1 0 0 0 0 0 0 0 0 0 ...
−3 −3 1 0 0 0 0 0 0 0 0 ...
−2 3 −2 1 0 0 0 0 0 0 0 ...
−1 −1 1 −1 1 0 0 0 0 0 0 ..
0 0 0 0 0 1 0 0 0 0 0 ...
1 0 0 0 0 1 1 0 0 0 0 ...
2 0 0 0 0 1 2 1 0 0 0 ...
3 0 0 0 0 1 3 3 1 0 0 ...
4 0 0 0 0 1 4 6 4 1 0 ...
This extension also has the properties that just as
${\displaystyle \exp {\begin{pmatrix}.&.&.&.&.\\1&.&.&.&.\\.&2&.&.&.\\.&.&3&.&.\\.&.&.&4&.\end{pmatrix}}={\begin{pmatrix}1&.&.&.&.\\1&1&.&.&.\\1&2&1&.&.\\1&3&3&1&.\\1&4&6&4&1\end{pmatrix}},}$
we have
${\displaystyle \exp {\begin{pmatrix}.&.&.&.&.&.&.&.&.&.\\-4&.&.&.&.&.&.&.&.&.\\.&-3&.&.&.&.&.&.&.&.\\.&.&-2&.&.&.&.&.&.&.\\.&.&.&-1&.&.&.&.&.&.\\.&.&.&.&0&.&.&.&.&.\\.&.&.&.&.&1&.&.&.&.\\.&.&.&.&.&.&2&.&.&.\\.&.&.&.&.&.&.&3&.&.\\.&.&.&.&.&.&.&.&4&.\end{pmatrix}}={\begin{pmatrix}1&.&.&.&.&.&.&.&.&.\\-4&1&.&.&.&.&.&.&.&.\\6&-3&1&.&.&.&.&.&.&.\\-4&3&-2&1&.&.&.&.&.&.\\1&-1&1&-1&1&.&.&.&.&.\\.&.&.&.&.&1&.&.&.&.\\.&.&.&.&.&1&1&.&.&.\\.&.&.&.&.&1&2&1&.&.\\.&.&.&.&.&1&3&3&1&.\\.&.&.&.&.&1&4&6&4&1\end{pmatrix}}}$
Also, just as summing along the lower-left to upper-right diagonals of the Pascal matrix yields the Fibonacci numbers, this second type of extension still sums to the Fibonacci numbers for negative index.
Either of these extensions can be reached if we define
${\displaystyle {n \choose k}={\frac {n!}{(n-k)!k!}}\equiv {\frac {\Gamma (n+1)}{\Gamma (n-k+1)\Gamma (k+1)}}}$
and take certain limits of the gamma function, ${\displaystyle \Gamma (z)}$.
## References
1. ^ Maurice Winternitz, History of Indian Literature, Vol. III
2. ^ J. L. Coolidge, The Story of the Binomial Theorem, Amer. Math. Monthly, Vol. 56, No. 3 (Mar., 1949), pp. 147–157
3. ^ Peter Fox (1998). Cambridge University Library: the great collections. Cambridge University Press. p. 13. ISBN 978-0-521-62647-7.
4. ^ The binomial coefficient ${\displaystyle \scriptstyle {n \choose k}}$ is conventionally set to zero if k is either less than zero or greater than n.
5. ^ Pascal's triangle | World of Mathematics Summary
6. ^ A. W. F. Edwards. Pascal's arithmetical triangle: the story of a mathematical idea. JHU Press, 2002. Pages 30–31.
7. Edwards, A. W. F. (2013), "The arithmetical triangle", in Wilson, Robin; Watkins, John J. (eds.), Combinatorics: Ancient and Modern, Oxford University Press, pp. 166–180.
8. ^ Alexander Zawaira; Gavin Hitchcock (2008). A Primer for Mathematics Competitions. Oxford University Press. p. 237. ISBN 978-0-19-156170-2.
9. ^ Selin, Helaine (2008-03-12). Encyclopaedia of the History of Science, Technology, and Medicine in Non-Western Cultures. Springer Science & Business Media. p. 132. Bibcode:2008ehst.book.....S. ISBN 9781402045592.
10. ^
11. ^ Sidoli, Nathan; Brummelen, Glen Van (2013-10-30). From Alexandria, Through Baghdad: Surveys and Studies in the Ancient Greek and Medieval Islamic Mathematical Sciences in Honor of J.L. Berggren. Springer Science & Business Media. p. 54. ISBN 9783642367366.
12. ^ Kennedy, E. (1966). Omar Khayyam. The Mathematics Teacher 1958. National Council of Teachers of Mathematics. pp. 140–142. JSTOR i27957284.
13. ^ Coolidge, J. L. (1949), "The story of the binomial theorem", The American Mathematical Monthly, 56 (3): 147–157, doi:10.2307/2305028, JSTOR 2305028, MR 0028222.
14. ^ Weisstein, Eric W. (2003). CRC concise encyclopedia of mathematics, p. 2169. ISBN 978-1-58488-347-0.
15. ^ Smith, Karl J. (2010), Nature of Mathematics, Cengage Learning, p. 10, ISBN 9780538737586.
16. ^ Fowler, David (January 1996). "The Binomial Coefficient Function". The American Mathematical Monthly. 103 (1): 1–17. doi:10.2307/2975209. JSTOR 2975209. See in particular p. 11.
17. ^ Brothers, H. J. (2012), "Finding e in Pascal's triangle", Mathematics Magazine, 85: 51, doi:10.4169/math.mag.85.1.51, S2CID 218541210.
18. ^ Brothers, H. J. (2012), "Pascal's triangle: The hidden stor-e", The Mathematical Gazette, 96: 145–148, doi:10.1017/S0025557200004204.
19. ^ Foster, T. (2014), "Nilakantha's Footprints in Pascal's Triangle", Mathematics Teacher, 108: 247, doi:10.5951/mathteacher.108.4.0246
20. ^ Fine, N. J. (1947), "Binomial coefficients modulo a prime", American Mathematical Monthly, 54 (10): 589–592, doi:10.2307/2304500, JSTOR 2304500, MR 0023257. See in particular Theorem 2, which gives a generalization of this fact for all prime moduli.
21. ^ Hinz, Andreas M. (1992), "Pascal's triangle and the Tower of Hanoi", The American Mathematical Monthly, 99 (6): 538–544, doi:10.2307/2324061, JSTOR 2324061, MR 1166003. Hinz attributes this observation to an 1891 book by Édouard Lucas, Théorie des nombres (p. 420).
22. ^ Wolfram, S. (1984). "Computation Theory of Cellular Automata". Comm. Math. Phys. 96 (1): 15–57. Bibcode:1984CMaPh..96...15W. doi:10.1007/BF01217347. S2CID 121021967.
23. ^ For a similar example, see e.g. Hore, P. J. (1983), "Solvent suppression in Fourier transform nuclear magnetic resonance", Journal of Magnetic Resonance, 55 (2): 283–300, Bibcode:1983JMagR..55..283H, doi:10.1016/0022-2364(83)90240-8.
24. ^ Karl, John H. (2012), An Introduction to Digital Signal Processing, Elsevier, p. 110, ISBN 9780323139595.
25. ^ Wolfram, S. (2002). A New Kind of Science. Champaign IL: Wolfram Media. pp. 870, 931–2. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 72, "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": 0.9206188321113586, "perplexity": 1211.8251579199762}, "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-2020-50/segments/1606141189141.23/warc/CC-MAIN-20201127044624-20201127074624-00707.warc.gz"} |
https://homework.study.com/explanation/find-f-for-the-following-function-f-x-x-plus-1-square-root-x-f-1-5.html | # Find {eq}\displaystyle f {/eq} for the following function. {eq}\displaystyle f'(x) = \frac{(x+1)}{\sqrt{x}}, \ f(1) = 5 {/eq}
## Question:
Find {eq}\displaystyle f {/eq} for the following function.
{eq}\displaystyle f'(x) = \frac{(x+1)}{\sqrt{x}}, \ f(1) = 5 {/eq}
## Initial Value Problem:
• When we integrate the derivative of a function, we get back the antiderivative function. This is because integration and differentiation are reverse processes. In an initial value problem, the value of the function at a point is also given. We can use this initial value in the lower limits of integration. The upper limits of integration are to be kept in the variable form.
• We shall be applying the following formulas:
{eq}\begin{align} \hspace{1cm} \int kdx&=kx+C & \left[ \text{Where C is an arbitrary constant of indefinite integration}~ \right] \\[0.3cm] \hspace{1cm} \int x^n\, dx&=\frac { x^{n+1}} {n+1}+C & \left[\text{ This is the power rule of integration } \right]\\[0.3cm] \end{align} {/eq} | {"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": 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": 0.9938502907752991, "perplexity": 1612.48541590267}, "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/1674764499891.42/warc/CC-MAIN-20230131222253-20230201012253-00272.warc.gz"} |
https://www.semanticscholar.org/paper/Robust-linear-regression%3A-optimal-rates-in-time-Bakshi-Prasad/4bfe12a023f830891c04b485a9e6810c5d0e3baa | # Robust linear regression: optimal rates in polynomial time
@article{Bakshi2021RobustLR,
title={Robust linear regression: optimal rates in polynomial time},
journal={Proceedings of the 53rd Annual ACM SIGACT Symposium on Theory of Computing},
year={2021}
}
• Published 29 June 2020
• Mathematics, Computer Science
• Proceedings of the 53rd Annual ACM SIGACT Symposium on Theory of Computing
We obtain robust and computationally efficient estimators for learning several linear models that achieve statistically optimal convergence rate under minimal distributional assumptions. Concretely, we assume our data is drawn from a k-hypercontractive distribution and an є-fraction is adversarially corrupted. We then describe an estimator that converges to the optimal least-squares minimizer for the true distribution at a rate proportional to є2−2/k, when the noise is independent of the…
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STOC
• 2020
This work narrows the gap between the Gaussian and heavy-tailed settings for polynomial-time estimators, introduces new techniques to high-probability estimation, and suggests numerous new algorithmic questions in the following vein. | {"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": 0.9233348369598389, "perplexity": 1737.1249965348163}, "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/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00437.warc.gz"} |
http://tex.stackexchange.com/questions/43162/memoir-stopping-hyphenation-in-chapter-titles-in-the-table-of-contents | # Memoir: Stopping hyphenation in chapter titles in the table of contents
I am working on my dissertation using xelatex and the memoir class. Many of the titles to my chapters contain long words which I don't want to be hypenated in the table of contents.
According the memoir class user guide, I should be able to use this command to achieve this: \renewcommand{\@tocrmarg}{2.55em plus1fil} This used to work, but now with more updated versions of tex it no longer works.
I always had this command in the preamble, but now it throws an error (Missing \begin{document}). Moving it to after \begin{document} does not solve this.
I realise that I could just manually add newlines to the chapter titles, but as I generate the titles with a command, this is not really useful.
I really just want to get the @tocrmarg command redefinition working ...
Who can help? I have attached a minimal example below (with the \renewcommand commented out):
\documentclass[11pt,oneside]{memoir}
\renewcommand{\@tocrmarg}{2.55em plus1fil}
\begin{document}
\tableofcontents
\newcommand{\lw}{Thisisalongwordverylong}
\chapter{\lw\ \lw\ \lw}
\end{document}
-
You have to deal with the @-sign in the command name. TeX does not recognize it as a part of the name. To solve it, enclose the whole \renewcommand to \makeatletter...\makeatother:
\documentclass[11pt,oneside]{memoir}
\makeatletter %% HERE
\renewcommand{\@tocrmarg}{2.55em plus1fil}
\makeatother %% AND HERE
\begin{document}
\tableofcontents
\newcommand{\lw}{Thisisalongwordverylong}
\chapter{\lw\ \lw\ \lw}
\end{document}
-
Awesome! It works! Thanks so much. – Tom de Bruin Feb 2 '12 at 10:06 | {"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": 0.9595960378646851, "perplexity": 2359.412644314343}, "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-2015-48/segments/1448398462686.42/warc/CC-MAIN-20151124205422-00248-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/176923-optimization-problem-double-check.html | ## Optimization problem double check
I have a test tomorrow and I'd really appreciate it if someone could check whether my work (just the relevant equation will do) is correct as I do not have an answer key.
1) A piece of wire 100cm long is cut into two pieces. One piece is bent to form a circle and the other piece is bent to form an equilateral triangle. How should the wire be cut so that the total enclosed area is a maximum/minimum.
First piece, $x$ is the triangle. Area of the triangle is $\displaystyle A(x)=(\frac{1}{2})(\frac{\sqrt{3}x}{6})(\frac{x}{3 })$
Second piece, $100-x$ is the circle. Area of the circle is $\displaystyle A(x)=\frac{x^2-200x+10000}{4\pi}$
Therefore, total area can be expressed as $\displaystyle A(x)=\frac{\sqrt{3}}{36}x^2 + \frac{x^2-200x+10000}{4\pi}$.
Does that seem correct?
Another question that I'd appreciate checking is: An isosceles triangle has a perimeter of 48 cm, find the maximum area.
Sides of the triangle are $x$, $x$, and $y$. Area of the triangle can be expressed as $\displaystyle A(x)=(\frac{1}{2})(48-2x)(\sqrt{48x+576})$
If anyone can confirm whether any (don't have to do both of them if you can't) of the equations are correct, I'd greatly appreciate it! | {"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": 9, "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": 0.8889495134353638, "perplexity": 176.4790785346225}, "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-2016-30/segments/1469257828009.82/warc/CC-MAIN-20160723071028-00017-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://openstax.org/books/calculus-volume-2/pages/5-key-terms | Calculus Volume 2
# Key Terms
### Key Terms
absolute convergence
if the series $∑n=1∞|an|∑n=1∞|an|$ converges, the series $∑n=1∞an∑n=1∞an$ is said to converge absolutely
alternating series
a series of the form $∑n=1∞(−1)n+1bn∑n=1∞(−1)n+1bn$ or $∑n=1∞(−1)nbn,∑n=1∞(−1)nbn,$ where $bn≥0,bn≥0,$ is called an alternating series
alternating series test
for an alternating series of either form, if $bn+1≤bnbn+1≤bn$ for all integers $n≥1n≥1$ and $bn→0,bn→0,$ then an alternating series converges
arithmetic sequence
a sequence in which the difference between every pair of consecutive terms is the same is called an arithmetic sequence
bounded above
a sequence ${an}{an}$ is bounded above if there exists a constant $MM$ such that $an≤Man≤M$ for all positive integers $nn$
bounded below
a sequence ${an}{an}$ is bounded below if there exists a constant $MM$ such that $M≤anM≤an$ for all positive integers $nn$
bounded sequence
a sequence ${an}{an}$ is bounded if there exists a constant $MM$ such that $|an|≤M|an|≤M$ for all positive integers $nn$
comparison test
if $0≤an≤bn0≤an≤bn$ for all $n≥Nn≥N$ and $∑n=1∞bn∑n=1∞bn$ converges, then $∑n=1∞an∑n=1∞an$ converges; if $an≥bn≥0an≥bn≥0$ for all $n≥Nn≥N$ and $∑n=1∞bn∑n=1∞bn$ diverges, then $∑n=1∞an∑n=1∞an$ diverges
conditional convergence
if the series $∑n=1∞an∑n=1∞an$ converges, but the series $∑n=1∞|an|∑n=1∞|an|$ diverges, the series $∑n=1∞an∑n=1∞an$ is said to converge conditionally
convergence of a series
a series converges if the sequence of partial sums for that series converges
convergent sequence
a convergent sequence is a sequence ${an}{an}$ for which there exists a real number $LL$ such that $anan$ is arbitrarily close to $LL$ as long as $nn$ is sufficiently large
divergence of a series
a series diverges if the sequence of partial sums for that series diverges
divergence test
if $limn→∞an≠0,limn→∞an≠0,$ then the series $∑n=1∞an∑n=1∞an$ diverges
divergent sequence
a sequence that is not convergent is divergent
explicit formula
a sequence may be defined by an explicit formula such that $an=f(n)an=f(n)$
geometric sequence
a sequence ${an}{an}$ in which the ratio $an+1/anan+1/an$ is the same for all positive integers $nn$ is called a geometric sequence
geometric series
a geometric series is a series that can be written in the form
$∑n=1∞arn−1=a+ar+ar2+ar3+⋯∑n=1∞arn−1=a+ar+ar2+ar3+⋯$
harmonic series
the harmonic series takes the form
$∑n=1∞1n=1+12+13+⋯∑n=1∞1n=1+12+13+⋯$
index variable
the subscript used to define the terms in a sequence is called the index
infinite series
an infinite series is an expression of the form
$a1+a2+a3+⋯=∑n=1∞ana1+a2+a3+⋯=∑n=1∞an$
integral test
for a series $∑n=1∞an∑n=1∞an$ with positive terms $an,an,$ if there exists a continuous, decreasing function $ff$ such that $f(n)=anf(n)=an$ for all positive integers $n,n,$ then
$∑n=1∞anand∫1∞f(x)dx∑n=1∞anand∫1∞f(x)dx$
either both converge or both diverge
limit comparison test
suppose $an,bn≥0an,bn≥0$ for all $n≥1.n≥1.$ If $limn→∞an/bn→L≠0,limn→∞an/bn→L≠0,$ then $∑n=1∞an∑n=1∞an$ and $∑n=1∞bn∑n=1∞bn$ both converge or both diverge; if $limn→∞an/bn→0limn→∞an/bn→0$ and $∑n=1∞bn∑n=1∞bn$ converges, then $∑n=1∞an∑n=1∞an$ converges. If $limn→∞an/bn→∞,limn→∞an/bn→∞,$ and $∑n=1∞bn∑n=1∞bn$ diverges, then $∑n=1∞an∑n=1∞an$ diverges
limit of a sequence
the real number $LL$ to which a sequence converges is called the limit of the sequence
monotone sequence
an increasing or decreasing sequence
p-series
a series of the form $∑n=1∞1/np∑n=1∞1/np$
partial sum
the $kthkth$ partial sum of the infinite series $∑n=1∞an∑n=1∞an$ is the finite sum
$Sk=∑n=1kan=a1+a2+a3+⋯+akSk=∑n=1kan=a1+a2+a3+⋯+ak$
ratio test
for a series $∑n=1∞an∑n=1∞an$ with nonzero terms, let $ρ=limn→∞|an+1/an|;ρ=limn→∞|an+1/an|;$ if $0≤ρ<1,0≤ρ<1,$ the series converges absolutely; if $ρ>1,ρ>1,$ the series diverges; if $ρ=1,ρ=1,$ the test is inconclusive
recurrence relation
a recurrence relation is a relationship in which a term $anan$ in a sequence is defined in terms of earlier terms in the sequence
remainder estimate
for a series $∑n=1∞an∑n=1∞an$ with positive terms $anan$ and a continuous, decreasing function $ff$ such that $f(n)=anf(n)=an$ for all positive integers $n,n,$ the remainder $RN=∑n=1∞an−∑n=1NanRN=∑n=1∞an−∑n=1Nan$ satisfies the following estimate:
$∫N+1∞f(x)dx
root test
for a series $∑n=1∞an,∑n=1∞an,$ let $ρ=limn→∞|an|n;ρ=limn→∞|an|n;$ if $0≤ρ<1,0≤ρ<1,$ the series converges absolutely; if $ρ>1,ρ>1,$ the series diverges; if $ρ=1,ρ=1,$ the test is inconclusive
sequence
an ordered list of numbers of the form $a1,a2,a3,…a1,a2,a3,…$ is a sequence
telescoping series
a telescoping series is one in which most of the terms cancel in each of the partial sums
term
the number $anan$ in the sequence ${an}{an}$ is called the $nthnth$ term of the sequence
unbounded sequence
a sequence that is not bounded is called unbounded
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As an Amazon Associate we earn from qualifying purchases. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 89, "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": 0.9928174018859863, "perplexity": 237.7842430710086}, "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-2021-31/segments/1627046154302.46/warc/CC-MAIN-20210802012641-20210802042641-00637.warc.gz"} |
https://williewong.wordpress.com/2017/09/27/heat-ball/ | ### Heat ball
There are very few things I find unsatisfactory in L.C. Evans’ wonderful textbook on Partial Differential Equations; one of them is the illustration (on p.53 of the second edition) of the “heat ball”.
The heat ball is a region with respect to which an analogue of the mean value property of solutions to Laplace’s equation can be expressed, now for solutions of the heat equation. In the case of the Laplace’s equation, the regions are round balls. In the case of the heat equation, the regions are somewhat more complicated. They are defined by the expression
$\displaystyle E(x,t;r) := \left\{ (y,s)\in \mathbb{R}^{n+1}~|~s \leq t, \Phi(x-y, t-s) \geq \frac{1}{r^n} \right\}$
where $\Phi$ is the fundamental solution of the heat equation
$\displaystyle \Phi(x,t) := \frac{1}{(4\pi t)^{n/2}} e^{- \frac{|x|^2}{4t}}.$
In the expressions above, the constant $n$ is the number of spatial dimensions; $r$ is the analogue of the radius of the ball, and in $E(x,t;r)$, the point $(x,r)$ is the center. Below is a better visualization of the heat balls: the curves shown are the boundaries $\partial E(0,5;r)$ in dimension $n = 1$, for radii between 0.75 and 4 in steps of 0.25 (in particular all the red curves have integer radii). In higher dimensions the shape is generally the same, though they appear more “squashed” in the $t$ direction.
1-dimensional heat balls centered at (0,5) for various radii. (Made using Desmos) | {"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": 10, "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": 0.8818399906158447, "perplexity": 228.45942449429998}, "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-2018-43/segments/1539583511806.8/warc/CC-MAIN-20181018105742-20181018131242-00336.warc.gz"} |
http://clay6.com/qa/32435/in-resonance-column-experiment-to-determine-the-velocity-of-sound-l-person- | # In resonance column experiment to determine the velocity of sound, $L$ person observes first resonance at $12\;cm$. If she repeats the experiment on a very hot day her resonance column length would be
$(a)\;=12\;cm \\(b)\; > 12 \;cm\\(c)\; < 12\;cm\\(d)\;\text{may be any thing}$
As velocity of sound increases with increase in temperature the balancing length will also increase.
Hence length > 12 cm
Hence b 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": 0.9371414184570312, "perplexity": 882.3463462211569}, "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-2018-26/segments/1529267864257.17/warc/CC-MAIN-20180621192119-20180621212119-00492.warc.gz"} |
https://chemistry.stackexchange.com/questions/51236/is-ammonium-a-strong-or-weak-conjugate-acid/51238 | # Is ammonium a strong or weak conjugate acid?
Ammonium, $\ce{NH_4^+}$, is the conjugate acid of ammonia, $\ce{NH_3}$. I have searched the Internet and so many different answers pop up. The rule of thumb that I read was that a strong acid has a weak conjugate counterpart.
For example, a strong acid has a weak conjugate base. Is ammonium a strong or weak acid? Is ammonia a strong or weak base? Is there something to the rule of thumb that I mentioned?
## 3 Answers
Both ammonia is a weak base and ammonium ion is a weak acid. Many, even most, acid/base conjugate pairs are like that. We should be using comparative instead of absolute adjectives in the rule about conjugate acid-base strengths:
A weaker acid has a stronger conjugate base, not necessarily a totally strong one.
A weaker base has a stronger conjugate acid, not necessarily a totally strong one.
• Stronger in respect to what...the weaker acid we have? – Hydrous Caperilla Mar 26 '18 at 5:41
• @Hydrous Caperilla, yes. – Serid Jun 26 '18 at 12:35
Consider these reactions: $$\ce{HA + H2O <=> A- + H3O+}\tag{1}$$ $$\ce{A- + H2O <=> HA + OH-}\tag{2}$$
The equilibrium constant for reaction (1) is called "acid dissociation constant", $K_{\rm a}$, and the second is "base association constant", $K_{\rm b}$. (Note that the names are there for historical reasons, they're not correct, strictly speaking) We can factor out the water's concentration to a good approximation in dilute solutions. Consider that
$$K_{\rm a} \times K_{\rm b} = \frac{\ce{[A- ][H3O+ ]}}{\ce{[HA]}} \times \frac{\ce{[HA][OH- ]}}{\ce{[A- ]}} = \ce{[H3O+ ][OH- ]} = K_{\rm w}\tag{3}$$
$K_{\rm w}$ is constant in constant temperature, hence $K_{\rm a}$ and $K_{\rm b}$ are inversely proportional to each other. Again, to a good approximation, we have
$$-\log{K_{\rm a}} + (-\log{K_\rm b})=-\log{K_\rm w}\: \Longrightarrow {\rm p}K_{\rm a} + {\rm p}K_{\rm b} = 14\tag{4}$$
p$K_{\rm a}$s between -1.7 ($\ce{H3O+}$/$\ce{H2O}$) and 15.7 ($\ce{H2O}$/$\ce{OH-}$) are said to be those of weak acids in water. Acids with $K_{\rm a}$s in this range do not dissociate fully in water, and named weak acids (and weak bases, for that matter1) in water.
Now, think of it like this: $x+y=14$. If $x$ is 5, $y$ is 9, and if $x$ is -5, $y$ is 19. (4) is where your rule of thumb originates from. So
• A weak acid like acetic acid with a p$K_{\rm a}$ of 4.76 will have a weak conjugate base with a p$K_{\rm b}$ of $14-4.76=9.24$.
• A strong acid like $\ce{HCl}$ with a p$K_{\rm a}$ of -7 will have a conjugate base with a p$K_{\rm b}$ of $14-(-7)=21$.
Thus as Oscar's answer points out, it's inaccurate the way your guideline is phrased, but it's naturally following if we use comparatives for phrasing. I'd dare say most conjugate bases of weak acids mentioned in textbooks are weak ones.
1: As the comment points out, the weak base range is p$K_{\rm b}$>1. That slipped.
• The definition of "strong" as having $pK$ below -1.7 may not be entirely true. In en.wikipedia.org/wiki/Base_(chemistry)#Strong_bases a "strong base" is considered one whose conjugate acid has $pK_a>13$, so the strong base needs merely $pK_b<1$ not $<-1.7$. – Oscar Lanzi May 15 '16 at 22:38
• @Oscar Noted; that somehow slipped through. – It's Over May 16 '16 at 6:55
Understanding from an HSC chemistry (2019) point of view:
Acids
• the conjugate acid of a strong base will be weak, therefore never reacting water
• the conjugate acid of a moderately strong base will be moderately weak, therefore somewhat reacting with water
• the conjugate acid of a weak base will be strong, therefore completely reacting with water
Bases
• the conjugate base of a strong acid will be weak, therefore never reacting with water
• the conjugate base of a moderately strong acid will be moderately weak, therefore somewhat reacting with water
• the conjugate base of a weak acid will be strong, therefore completely reacting with water
When a salt ionises in solution, it's ions should be considered as the conjugate acid and conjugate base of their respective acid/base
e.g. CH3COONa --> CH3COO- + Na+
The Na+ is a weak conjugate acid of the strong base NaOH and therefore will never react with water
The CH3COO- is a moderately weak conjugate base of the moderately weak acid CH3COOH, and it therefore reacts with water somewhat to produce OH- ions:
CH3COO- + H2O --> CH3COOH + OH-
This makes the resultant solution basic, giving a pH of around 8-9
• Welcome to Chemistry.SE. Your answer contains a lot of correct information, but it does not answer the question, which is about the ammonium cation. – Ben Norris Sep 15 '19 at 20:06 | {"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": 4, "x-ck12": 0, "texerror": 0, "math_score": 0.8900742530822754, "perplexity": 2701.2885147504126}, "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-2020-05/segments/1579250594705.17/warc/CC-MAIN-20200119180644-20200119204644-00359.warc.gz"} |
http://isiarticles.com/article/25873 | دانلود مقاله ISI انگلیسی شماره 25873
عنوان فارسی مقاله
# کنترل مبدل های انرژی امواج توسط پیش بینی امواج و برنامه ریزی پویا
کد مقاله سال انتشار مقاله انگلیسی ترجمه فارسی تعداد کلمات
25873 2012 12 صفحه PDF سفارش دهید 9541 کلمه
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عنوان انگلیسی
Wave energy converter control by wave prediction and dynamic programming
منبع
Publisher : Elsevier - Science Direct (الزویر - ساینس دایرکت)
Journal : Renewable Energy, Volume 48, December 2012, Pages 392–403
کلمات کلیدی
موج انرژی - کنترل بهینه محدود - کنترل بنگ - بنگ - برنامه ریزی پویا - پیش بینی موج دریا قطعی -
پیش نمایش مقاله
#### چکیده انگلیسی
We demonstrate that deterministic sea wave prediction (DSWP) combined with constrained optimal control can dramatically improve the efficiency of sea wave energy converters (WECs), while maintaining their safe operation. We focus on a point absorber WEC employing a hydraulic/electric power take-off system. Maximizing energy take-off while minimizing the risk of damage is formulated as an optimal control problem with a disturbance input (the sea elevation) and with both state and input constraints. This optimal control problem is non-convex, which prevents us from using quadratic programming algorithms for the optimal solution. We demonstrate that the optimum can be achieved by bang–bang control. This paves the way to adopt a dynamic programming (DP) algorithm to resolve the on-line optimization problem efficiently. Simulation results show that this approach is very effective, yielding at least a two-fold increase in energy output as compared with control schemes which do not exploit DSWP. This level of improvement is possible even using relatively low precision DSWP over short time horizons. A key finding is that only about 1 second of prediction horizon is required, however, the technical difficulties involved in obtaining good estimates necessitate a DSWP system capable of prediction over tens of seconds. Highlights ► We investigate constrained optimal control of sea wave energy converters. ► The objective is to maximize energy take-off while minimizing the risk of damage. ► The optimal control problem is analyzed using Pontryagin's minimum principle. ► Dynamic programming is employed to resolve the on-line optimization problem. ► Simulation result shows at least a two-fold increase in energy output.
#### مقدمه انگلیسی
Ocean waves provide an enormous source of renewable energy [1] and [2]. Research into wave energy was initially stimulated by the oil crisis of the 1970s [3]. Since then many different types of sea wave energy converters (WECs) have been designed and tested [4] and [5], but this is still a relatively immature technology (compared to solar or wind energy) and is far from being commercially competitive with traditional fossil fuel or nuclear energy sources. Progress is hampered by two fundamental problems: 1. Inefficient energy extraction, often due to the fact that the WEC's dynamic parameters are not optimally tuned and their control is not optimal for most wave profiles. 2. Risk of device damage. In order to prevent WECs from being damaged by large waves, they have to be shut down, especially during winter storms. Such periods of inactivity can last for days. Extracting the maximum possible time average power from WECs, while reducing the risk of device damage involves a combination of good fundamental engineering design of the devices and effective control of their operation. The traditional approach to these issues exploits short term statistical properties of the sea [6] but it has been shown [7] and [8] that doing so severely limits the average power that can be extracted. We address the above two problems by considering schemes designed to achieve optimal control. It will be shown that (as [9], [10] and [11] demonstrated in the 1970s) methods for achieving the maximum power output are inevitably non-causal and require prediction of the shape of the incident waves. The recent development of deterministic sea wave prediction (DSWP) as a scientific discipline [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25], [26], [27] and [28], particularly real time DSWP [12], [13], [14], [15], [16], [17], [18], [19], [26] and [28] now makes such an approach realistic. For a variety of reasons high accuracy real time DSWP is very demanding. However it will be shown that the optimal control techniques described here provide considerable improvements over traditional WEC control methods, even with modestly accurate DSWP and relatively short prediction horizons. The dimensions of point absorbers are small compared with the wave length of incoming waves and they are potentially very efficient if their frequency response function closely matches the spectrum of the incident waves (resonance). Passive control methods (such as impedance matching) have been explored to improve energy extraction by tuning the dynamical parameters of the devices [9], [29], [30] and [31]. Most of these approaches are linear control schemes. A non-linear control method that has received some attention is latching, [32], [33], [34], [35], [36] and [37]. This attempts to force the phase angle between the float and the wave at the WEC to be similar to conditions at resonance. The above control strategies do not use prediction of the forces acting on the WEC and thus inevitably lead to sub-optimal energy extraction. Since the early work [5], [9] and [11] there have been a number of authors who have recognized the importance of DSWP in the control of a variety of floating body applications [7], [8], [17], [37] and [38], but these have, as yet, not been incorporated into actual control schemes. The point absorber model used is shown in Fig. 1 and roughly corresponds to the Power Buoy device PB150 developed by OPT Inc, see [39]. On the sea surface is a float, below which hydraulic cylinders are vertically installed. These cylinders are attached at the bottom to a large area anti-heave plate whose vertical motion is designed to be negligible compared with that of the float. The heave motion of the float drives the pistons inside the hydraulic cylinders to produce a liquid flow. The liquid drives hydraulic motors attached to a synchronous generator. From here, the power reaches the grid via back-to-back AC/DC/AC converters. The mechanical circuit corresponding to this simplified model is shown in Fig. 2. Here h w is the water level, h υ is the height of the mid-point of the float and D is the hydrodynamic damping of the float including added damping due to the damping effect of the movement of the float [1]. K is the hydrostatic stiffness giving the buoyancy force, which can be calculated from the float geometry, while m is the mass of the float including “added mass” [1]. The friction force acting on the float is View the MathML sourceff=Dfh˙υ. In order to simplify the model we neglect the frequency dependence of both D and m (see, e.g., [29]). We also neglect the static component of the friction force ff. For a more thorough investigation of the modeling issues of point absorbers, see [1], [40] and [41]. Full-size image (14 K) Fig. 1. Schematic diagram of the point absorber. Figure options Full-size image (11 K) Fig. 2. A mechanical circuit representation of the point absorber. Figure options The control input is the q-axis current in the generator-side power converter, to control the electric torque of the generator [42]. The generator torque is proportional to the force f acting on the pistons from the fluid in the cylinders. Since the motion of the float imposes a velocity υ on the piston, the extracted power P(t) at time t is expressed as equation(1) P(t)=f(t)υ(t).P(t)=f(t)υ(t). Turn MathJax on This power is smoothed by the capacitors on the DC link of the converters. In our model the modest power losses in the hydraulic transmission, the generator and the converters will be neglected. To avoid damage, and for overall performance reasons, two constraints have to be considered in any real WEC. One concerns the relative motion of the float to the sea surface (it should neither sink nor raise above the water and then slam), which can be expressed as equation(2) |hw−hυ|≤Φmax.|hw−hυ|≤Φmax. Turn MathJax on The other constraint is on the control signal set by limitations on the allowable converter current. This constraint can be expressed as equation(3) |f|≤γ.|f|≤γ. Turn MathJax on The control objective is to maximize the extracted energy subject to the constraints (2) and (3). We remark that there is a further constraint on the motion of the float because of the limited excursion of the piston with respect to the cylinder (see Fig. 1). This constraint has the form |hυ|≤λ|hυ|≤λ. However, we shall not consider this constraint, since we assume that λ is large enough compared with the expected excursion. The constraints imposed on WECs significantly affect the power that can be extracted. It has been shown that by using control strategies that incorporate these constraints, considerable increases in the energy output can be obtained without increasing the risk of damage [7] and [8]. The ability to handle constraints combined with the development of real time wave prediction methods has recently led to an interest in the use of model predictive control (MPC) for wave energy devices [43], [44] and [45]. The work published to date has used standard MPC techniques and they rely on the formulation of a convex quadratic programming (QP) problem. The underlying problem formulations and the cost function representations in [43], [44] and [45] differ from our case. We leave the question open if the convexity assumption holds for a broad class of constrained optimal control problems for WECs. However, we find that this assumption does not hold for the problem formulated in this paper and many other similar optimal control problems [46] and [47]. In this paper, the constrained optimal control problem is solved using fundamental principles from optimal control theory [48], [49] and [50], see Section 3, and real time deterministic sea wave prediction [12], [13], [14], [15], [16], [17], [18], [19], [26] and [28]. We demonstrate that a nearly optimal control is of bang–bang type, meaning that the control input f is always at one edge of the allowed range, see Subsection 3.2. As will be shown, for an arbitrary sea wave input known over an interval of time (not a sine wave), direct numerical computation of the optimal control scheme is not realistic. Consequently, we employ dynamic programming (DP) [46], which is well suited for constrained optimal control problems [51]. We have sacrificed some detail in the hydrodynamic modeling, leading to a model of manageable complexity for on-line DP. The implementation of DP on a WEC control system is based on the assumption that the sea surface shape can be predicted for a short time period. This requirement has been a bottleneck for the development of suitable optimal control strategies for WECs. However, the developments in deterministic sea wave modeling techniques have made real time sea wave prediction for a short time period realizable, [12], [13], [14], [15], [16], [17], [18], [19], [26] and [28]. A key finding from this work is that the prediction horizons required are considerably smaller than those resulting from the previous studies [43], [44] and [45]. For the sake of comparison, simulations have also been performed for various non-prediction-based WEC control strategies. The results demonstrate the following benefits of our approach. 1) Significant increase in energy output. We get up to a two-fold increase in energy output when compared with rival control algorithms which do not exploit sea state prediction. 2) Robustness to the prediction accuracy and prediction horizon of DSWP. Especially important is the possibility of reducing the prediction horizon, because the difficulties associated with real time DSWP increase significantly with the prediction horizon. The structure of the paper is as follows. Section 2 formulates the constrained optimal control problem for the WEC. Section 3 provides a detailed analysis for this optimal control problem. The dynamic programming control strategy is developed in Section 4, and compared in simulation with several other control methods in Section 5.
#### نتیجه گیری انگلیسی
In this paper, the constrained optimal control problem of WECs is addressed. After addressing modeling and optimal control issues, we have proposed an on-line control algorithm for a point absorber, based on DP, MPC and wave prediction. Two other elementary control strategies have also been explored for the sake of comparison. Numerical simulations demonstrate the potential of our controller to extract significantly more energy from the same device. Our control algorithm relies on the recent development of real time sea wave prediction methods. In future work, we are interested in the following topics: 1) Efficient optimal control algorithms for high order WEC models. 2) Integrating the sea wave prediction model in the controller. 3) Optimal control of a whole wave farm, taking coupling into account. 4) Experimental tests for the control algorithms.
خرید مقاله
پس از پرداخت، فوراً می توانید مقاله را دانلود فرمایید. | {"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": 0.9188141226768494, "perplexity": 785.087070658317}, "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-2017-26/segments/1498128320023.23/warc/CC-MAIN-20170623063716-20170623083716-00711.warc.gz"} |
http://laneas.com/publication/continuum-equilibria-and-global-optimization-routing-dense-static-ad-hoc-networks | Continuum Equilibria and Global Optimization for Routing in Dense Static Ad Hoc Networks
Journal Article
Source:
Elsevier special issue of Computer Networks on Interdisciplinary Paradigms for Networking, Volume 54, Issue 6, p.1005-1018 (2010)
Keywords:
Equilibrium, Routing, Wireless Ad Hoc Networks, Wireless Sensor Networks
Abstract:
We consider massively dense ad hoc networks and study their continuum limits as the node density increases and as the graph providing the available routes becomes a continuous area with location and congestion dependent costs. We study both the global optimal solution as well as the non-cooperative routing problem among a large population of users where each user seeks a path from its origin to its destination so as to minimize its individual cost. Finally, we seek for a (continuum version of the) Wardrop equilibrium. We first show how to derive meaningful cost models as a function of the scaling properties of the capacity of the network and of the density of nodes. We present various solution methodologies for the problem: (1) the viscosity solution of the Hamilton-Jacobi-Bellman equation, for the global optimization problem, (2) a method based on Green’s Theorem for the least cost problem of an individual, and (3) a solution of the Wardrop equilibrium problem using a transformation into an equivalent global optimization problem.
Full Text: | {"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": 0.8103950023651123, "perplexity": 529.1795336791962}, "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-2019-47/segments/1573496665985.40/warc/CC-MAIN-20191113035916-20191113063916-00099.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/12238-bayesian-statistics.html | Bayesian Statistics
hello everyone,
I need a wee help with this proof:
proof that the Gamma distribution is a conjugate prior to the Poisson distribution with paremeter lambda.
thanks a million ! | {"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": 0.9796702265739441, "perplexity": 1791.2226718291513}, "config": {"markdown_headings": false, "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/1487501170839.31/warc/CC-MAIN-20170219104610-00076-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/402083/non-constant-angular-velocity-in-orbit | # Non-constant angular velocity in orbit
Consider a pair of objects in elliptical orbits around a common center of mass. For all considerations of angular motion and torque, the pivot point of interest is the center of mass in this discussion.
The only forces occurring point directly towards the center of mass, and cannot cause a torque. The system experiences no net torque, and so the angular momentum should be conserved.
When considering a particular object in this elliptical orbit, its moment of inertia, I, varies as the radius varies. This is looking at the objects in orbit in a $L = I \omega$ lens. This can also be translated into the lens of $L = r \times p$, but the challenge arises in the first lens (perhaps it is illegal for me to discuss the angular momentum in an $I\omega$ way). As the objects get nearer in their elliptical path, the moment of inertia decreases ($I=mr^2$), so the angular velocity must increase to keep angular momentum constant.
Therefore, the angular velocity is not constant, which means that about the center of mass, the system experiences angular acceleration. However, we know that $\alpha = \tau _{net} /I$. If $\alpha$ is non-zero, it seems to show that there must be a net torque, because it is definitely the case that about the center of mass, the angular speed of either object is non-constant.
Where is the break in this logic?
Context: I am a teacher of high school algebra based physics (AP Physics 1). Students made the reasonable link that changing angular speed seems to imply a nonzero net torque, given the rotational analogue of Newton's second law, which we teach to new students as $\alpha = \tau _{net} /I$. I know that the key is that the moment of inertia is non-constant, but it seems like no matter what, with that expression, net torque being zero will force $\alpha$ to be zero.
My gut: that the "rotational Newton's second law analogue" doesn't hold for non-constant I. (We are likely a little beyond scope for this course to tackle that)
• No time to write an answer now, but you are on exactly the right track. Make the analogy to $F_\text{net} = \frac{\mathrm{d}p}{\mathrm{d}t}$ (rather than $F_\text{net} = ma$) to get $\tau_\text{net} = \frac{\mathrm{d}L}{\mathrm{d}t}$ and find the total derivative on the RHS. Apr 25 '18 at 20:41
• BTW: We have MathJax running on the site so that $L = r \times p$ renders as $L = r \times p$ and $\omega$ renders as $\omega$ and the like. The syntax is essentially LaTeX math-mode. Apr 25 '18 at 20:43 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9458070993423462, "perplexity": 265.8094011283342}, "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-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00686.warc.gz"} |
https://www.physicsforums.com/threads/kinetic-energy-of-a-baseball.340600/ | # Kinetic energy of a Baseball
1. Sep 26, 2009
1. The problem statement, all variables and given/known data
A pitcher throws a 0.124-kg baseball, and it approaches the bat at a speed of 54.2 m/s. The bat does Wnc = 76.5 J of work on the ball in hitting it. Ignoring the air resistance, determine the speed of the ball after the ball leaves the bat and is 21 m above the point of impact.
2. Relevant equations
W = E - E0
3. The attempt at a solution
W = E -Eo
76.5=.5(.124)v2-.124(9.8)21
102.0192=.062v2
Thankyou!!!
2. Sep 26, 2009
### rock.freak667
where did you get 9.8 from? It told you 54.2 m/s was the initial velocity. Redo this part. For the second part you need to use projectile motion equations.
3. Sep 26, 2009
I used the eqn W=E-E0, since they gave us a non conservative value.
this can b furthere broken down to... 1/2mv2 -mgh. Thats where the 9.8 comes from. As far as projectile motion..how would i use it with no angles?
4. Sep 26, 2009
### rock.freak667
actually as I think about it now, you don't need to use projectile motion.
Initially
all you have is that that change in kinetic energy = work done by that bat.
the initial ke = 1/2m(54)2.
For the second part you need to use the part in red
5. Sep 26, 2009
Im confused. I got KE = 180.792. however, I am trying to find final velocity of the ball. i initially tried the eqn in red and it did not work for me. Can you please demonstrate?
6. Sep 26, 2009
### rock.freak667
Conservation of energy right
Initially it has ke given by 1/2mu2. This energy is converted into kinetic energy (it is moving with a different velocity v) and work done by the bat.
so applying the law of conservation of energy we will get 1/2mu2=1/2mv2+Wbat.
So can you find 'v' given Wbat=76.5J and u=54.2m/s?
Similar Discussions: Kinetic energy of a Baseball | {"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": 0.9312465190887451, "perplexity": 1018.8179654876956}, "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-2017-47/segments/1510934806258.88/warc/CC-MAIN-20171120223020-20171121003020-00683.warc.gz"} |
https://nyuscholars.nyu.edu/en/publications/the-endofullerene-hfcsub60sub-inelastic-neutron-scattering-spectr | # The Endofullerene HF@C60: Inelastic Neutron Scattering Spectra from Quantum Simulations and Experiment, Validity of the Selection Rule, and Symmetry Breaking
Minzhong Xu, Peter M. Felker, Salvatore Mamone, Anthony J. Horsewill, Stéphane Rols, Richard J. Whitby, Zlatko Bačić
Research output: Contribution to journalArticlepeer-review
## Abstract
Accurate quantum simulations of the low-temperature inelastic neutron scattering (INS) spectra of HF@C60 are reported for two incident neutron wavelengths. They are distinguished by the rigorous inclusion of symmetry-breaking effects in the treatment and having the spectra computed with HF as the guest, rather than H2 or HD, as in the past work. The results demonstrate that the precedent-setting INS selection rule, originally derived for H2 and HD in near-spherical nanocavities, applies also to HF@C60, despite the large mass asymmetry of HF and the strongly mixed character of its translation-rotation eigenstates. This lends crucial support to the theoretical prediction made earlier that the INS selection rule is valid for any diatomic molecule in near-spherical nanoconfinement. The selection rule remains valid in the presence of symmetry breaking but is modified slightly in an interesting way. Comparison is made with the recently published experimental INS spectrum of HF@C60. The agreement is very good, apart from one peak for which our calculations suggest a reassignment. This reassignment is consistent with the measured INS spectrum presented in this work, which covers an extended energy range.
Original language English (US) 5365-5371 7 Journal of Physical Chemistry Letters 10 18 https://doi.org/10.1021/acs.jpclett.9b02005 Published - Sep 19 2019
## ASJC Scopus subject areas
• Materials Science(all)
• Physical and Theoretical Chemistry
## Fingerprint
Dive into the research topics of 'The Endofullerene HF@C60: Inelastic Neutron Scattering Spectra from Quantum Simulations and Experiment, Validity of the Selection Rule, and Symmetry Breaking'. Together they form a unique fingerprint. | {"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": 0.8742769360542297, "perplexity": 3852.9190996059037}, "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-2022-40/segments/1664030337529.69/warc/CC-MAIN-20221004215917-20221005005917-00270.warc.gz"} |
https://electronics.stackexchange.com/questions/415521/gain-in-current-but-not-in-voltage-transistor-voltage-follower-set-up | # Gain in current but not in voltage -transistor “voltage follower” set up
Suppose we have the circuit set up as shown in the image. Also suppose that $$V_s=\overline{V_s}+v_S$$, where the smal v is for the variable component of the signal which varies much faster than the time constant of both capacitors. After some analisys we get that there's no gain in the variable component of the source voltage, that is, there's unitary gain, however from the equation of a bipolar transistor: $$i_E\approx i_c=\beta i_s$$ thus there's a gain in current, how is it possible to hava a gain in current, and consequently a gain in power, but not in voltage? This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
• Hmmm… Looks like you have a strong misunderstanding of electricity. Anyway, if you connect a load resistor between $V_{\textrm{out}}$ and ground it will drain some current. The current that is drained from $V_S$ may be significantly lower, hence the current gain. – user2233709 Jan 6 at 13:15
• The input impedance and output impedance are very different. – Dave Tweed Jan 6 at 13:15
• Moreover, you’d generally want to polarize your transistor’s base with a pair of resistors. – user2233709 Jan 6 at 13:16
• @user2233709 you're right about the misunderstanding of electricity. I don't understand how we can see current and voltage as two distinct things, on the one hand they are, but they are also so strongly connected by Ohm's law that I don't see how – Bidon Jan 6 at 13:22
• The way Ohm's law is generally taught to beginners ($V = RI$) assumes that $R$ is constant, which is true for ideal resistors. Other components can change resistance (more general term: impedance) depending on pretty much anything: voltage, temperature, frequency, input on another pin, ... – Geier Jan 6 at 20:34
Just because 2 quantities are related by an equation does not make them the same. For example, F=ma relatesd force and acceleration by a fixed parameter, mass, but they are still distinct concepts. Ohm's law is not a general rule, Kirchoff's laws are more general and do not imply that current and voltage are thje same: apply them to more complicated circuits than a single resistor and you would see this. Ohm's law has little to do with amplifiers. In the case of a voltage follwer, the output voltage is approximately the same as the input voltage but the available current is higher due to the current gain of the transistor. As already noted, the input resistance of the voltage follower is much higher than the output resistance so Ohm's law does not apply since these resistances are different.
• I think I see where my mistake was, I was not accounting for the difference in impedance. – Bidon Jan 6 at 14:16
... thus there's a gain in current, how is it possible to have a gain in current, and consequently a gain in power, but not in voltage?
By having a change in impedance.
In very crude terms the base current sees a resistance of βRE whereas the output impedance is RE.
This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
No. A transistor isn't a resistor.
This seems to contradict Ohm's law, shouldn't a current gain be accompanied by a voltage gain?
A transistor is an active device. It modulates the energy of an external power source to mimic the input signal. While Ohm's Law can be use to describe the behavior of certain aspects of active devices in some applications, keep in mind that it was derived to describe the relationships at play in a simple resistor. For anything beyond that you have to pay attention to the details.
By its definition, a bipolar transistor delivers current gain only. It is the circuit architecture and components around the transistor that translate this into voltage gain- sometimes.
Ic =hFE*Ib is then shared by both I(Re) and I(cout).
However for DC control, I(Cout) can never take all of the current of Ic . As Re serves a purpose to DC pull down the emitter and so R load must always be greater than Re for full swing voltages.
The result when overloaded, is the cap current pulls up the emitter so Vbe reduces then Ib reduces and the lower level signal cannot be reached.
Ohm’s Law still works here. One simply uses Xc for the cap being much smaller so Rload must be greater than Re for linear operation
also Vce must be Vce saturation where hFE drops significantly as a another condition.
the power gain comes from hFE * Rb/Re equivalent loads | {"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": 0.8459485173225403, "perplexity": 856.5066834915132}, "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/1547583662690.13/warc/CC-MAIN-20190119054606-20190119080606-00552.warc.gz"} |
https://yaoyao.codes/math/2018/06/28/open-set | # Neighborhood / Open Set / Continuity / Limit Points / Closure / Interior / Exterior / Boundary
Yao Yao on June 28, 2018
Quoted from Analysis II by Victor Guillemin, MIT and Topology for dummies by Damian Giaouris, Newcastle University with modification.
## Neighborhood
Definition: Given a point $x_i \in X$, and a real number $\epsilon > 0$, we define
$\Phi(x_i, \epsilon) = \lbrace x_j \in X \vert d(x_i, x_j) < \epsilon \rbrace$
where $d(\cdot, \cdot)$ is a mtric on $X$ (and thus $(X, d)$ is metric space).
We call $\Phi(x_i, \epsilon)$ the $\epsilon$-neighborhood of $x_i$ in space $X$. Obviously, $x_i \in \Phi(x_i, \epsilon)$.
Given a subspace $Y \subseteq X$ (已知 $\mathbb{R}^n$ 与 $\mathbb{R}^m$ 不存在 subspace 的关系 when $n \neq m$，所以 subspace 不应该有 dimension 上的变化), the $\epsilon$-neighborhood of $x_i$ in space $Y$ is just $\Phi(x_i, \epsilon) \cap Y$.
## Open and Closed Sets
Definition: A set $U$ in space $X$ is open if $\forall x_i \in U, \exists \epsilon_i > 0$ such that $\Phi(x_i, \epsilon_i) \subseteq U$.
• 比如开区间 (open interval) $(a, b)$ 就可以看做一个 open set；闭区间 $[a, b]$ 在 $a, b$ 两点上无法满足 open set 的要求。
Proposition: A neighborhood is not necessarily an open set.
Proof: Given a neighborhood $\Phi(x_i, \epsilon) = \lbrace x_j \in X \vert d(x_i, x_j) < \epsilon \rbrace$ and a point $x_j \in \Phi(x_i, \epsilon)$, $\forall \epsilon’$ we can define $\Phi(x_j, \epsilon’) = \lbrace x_k \in X \vert d(x_j, x_k) < \epsilon’ \rbrace$. However, according to triangle inequality, we only know that $d(x_i, x_k) \leq \epsilon + \epsilon’$. It’s not guaranteed that $d(x_i, x_k) < \epsilon$, i.e. not guaranteed that $x_k \in \Phi(x_i, \epsilon)$, i.e. not guaranteed that $\Phi(x_j, \epsilon’) \subseteq \Phi(x_i, \epsilon)$. $\blacksquare$
Proposition: Every open set is a union of neighborhoods.
Proof: Let $U$ be an open set. For each $x \in U$, let $N_x$ be the open neighborhood in $U$ containing $x$, which is guaranteed to exist by the definition. Consider $\mathcal{N} = \underset{x \in U}{\cup} N_x$. Since for every $x \in U$, $x \in N_x \subset \mathcal{N}$, we find $U \subset \mathcal{N}$. Since every $N_x \subset U$, we find $\mathcal{N} \subset U$. Therefore, $U = \mathcal{N}$ is a union of neighborhoods of points of $U$. $\blacksquare$
Proposition: Let $\lbrace U_i \vert i \in I \rbrace$ be a collection of open sets in space $X$, where $I$ is just a labeling set that can be finite or infinite. Then the set $\underset{i \in I}{\cup} U_i$ is open.
• 这里 labeling set 应该是 index set 的意思，即 a set whose members label or index members of another set. 比如 $\lbrace 1,2,3 \rbrace$ 就是 $\lbrace x_1, x_2, x_3 \rbrace$ 的 index set。
• 可以是 infinite union
Proposition: Let $\lbrace U_i \vert i = 1,\dots, N \rbrace$ be a finite collection of open sets in space $X$. Then the set $\overset{N}{\underset{i=1}{\cap}} U_i$ is open.
• 为什么不能是 infinite intersection? 反例：$\overset{\infty}{\underset{n=1}{\cap}} (-\frac{1}{n}, \frac{1}{n}) = \lbrace 0 \rbrace$
• 注意这里并不是取极限，而是因为 $\forall n>0$, $0 \in (-\frac{1}{n}, \frac{1}{n})$, hence $0 \in \underset{n}{\cap} (-\frac{1}{n}, \frac{1}{n})$
Definition: Define the complement of $A$ in $X$ to be $A^c = X - A$. The set $A$ is closed in $X$ if $A^c$ is open in $X$
## Clopen Sets
A clopen set in a topological space is a set which is both open and closed.
In any metric space $(X, d)$, the sets $X$ and $\emptyset$ are clopen.
Proof: $X$ is open by definition.
Also by definition, an open set $U$ in $X$ is a set where every point in $U$ has a certain $\epsilon$-neighborhood contained entirely within $U$.
That is, there is no point in $U$ who cannot find an $\epsilon$-neighborhood contained entirely within $U$.
$\emptyset$ has no point, so $\emptyset$ is open vacuously.
Then $X^c = \emptyset$ is closed; also $\emptyset^c = X$ is closed. Therefore they are both clopen. $\blacksquare$
## Open Sets & Function Continuity
Consider two metric spaces $(X, d_X)$ and $(Y, d_Y)$, a function $f: X \to Y$, and a point $a$.
Definition: ($\epsilon$-$\theta$ definition of continuity) The function $f$ is continuous at $a$ if $\forall \epsilon > 0$, $\exists \theta > 0$ such that
$d_X(a, x) < \theta \Rightarrow d_Y(f(a), f(x)) < \epsilon$
I.e. for any given $\epsilon$, I need to find a range of $x$ such that the range of $f(x)$ will be smaller than $2 \epsilon$; or from a different point of view, I need to find a range of $x$ around a point $a$ such that $f(x)$ will be in (or just) a predetermined range around $f(a)$:
Definition: A function $f$ is continuous if it is continuous at every $x$ in its domain $X$.
There is an alternative formulation of continuity that we present here as a theorem.
Theorem: A function $f$ is continuous $\iff \forall$ open subset $U$ of $Y$ (w.r.t. $d_Y$), the preimage $f^{−1}(U)$ is open in $X$ (w.r.t. $d_X$).
## Limit Points / Closure / Interior / Exterior / Boundary
Let $(X, d)$ be a metric space.
Definition: Suppose $A \subseteq X$. The point $x \in X$ is a limit point of $A$ if $\forall \epsilon$, $\Phi(x, \epsilon)$ contains points of $A$ distinct from $x$, i.e. $\Phi(x, \epsilon)$ meets $A$ in a point $\neq x$. This is equivalent to saying that each neighborhood of $x$ has an infinite number of members of $A$.
• E.g. $2$ is a limit point of interval $(2,3)$.
Definition: The closure of $A$, denoted by $\operatorname{cl}(A)$, is the union of $A$ and the set of limit points of $A$,
$\operatorname{cl}(A) = A \cup \lbrace x \in X \mid x \text{ is a limit point of } A \rbrace$
• If $\exists \epsilon > 0$ such that $\Phi(x, \epsilon) \subset A$, we call $x$ an interior point of $A$.
• $\operatorname{int}(A) = \lbrace \text{all interior points of } A \rbrace$ is the interior of $A$
• $\operatorname{int}(A) \equiv (\operatorname{cl}(A^c))^c$
• Note that $\operatorname{int}(A)$ is open.
• If $\exists \epsilon > 0$ such that $\Phi(x, \epsilon) \cap A = \emptyset$, we call $x$ an exterior point of $A$.
• $\operatorname{ext}(A) = \lbrace \text{all exterior points of } A \rbrace$ is the exterior of $A$
• $\operatorname{ext}(A) \equiv \operatorname{int}(A^c)$
• If $x$ is neither an interior nor exterior point of $A$, it’s a boundary point of $A$
• $\partial{A} = \lbrace \text{all boundary points of } A \rbrace$ is the boundary of $A$
• $\partial{A} \equiv X - (\operatorname{int}(A) \cup \operatorname{ext}(A))$
• $X = \operatorname{int}(A) \cup \operatorname{ext}(A) \cup \partial{A}$
Further, we have $\operatorname{cl}(A) = \operatorname{int}(A) \cup \partial{A} = X - \operatorname{ext}(A)$
Proposition: Set $A$ is open $\iff$ all points in $A$ are interior points.
• 注意：$S$ 全部由 interior 构成，不代表 exterior 和 boundary 不存在。可以想象 $A = \lbrace (x, y) \mid x^2 + y^2 < 1 \rbrace$ 是一个内部涂满、边界为虚线的圆
• $\operatorname{int}(A) = A = \lbrace (x, y) \mid x^2 + y^2 < 1 \rbrace$
• $\partial{A} = \lbrace (x, y) \mid x^2 + y^2 = 1 \rbrace$ 为圆周上的点
• $\operatorname{ext}(A) = \lbrace (x, y) \mid x^2 + y^2 > 1 \rbrace$
• $\operatorname{cl}(A) = \lbrace (x, y) \mid x^2 + y^2 \leq 1 \rbrace$ 是一个内部涂满、边界为实线的圆 | {"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": 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": 0.9798312187194824, "perplexity": 269.1094631402281}, "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-2021-10/segments/1614178358203.43/warc/CC-MAIN-20210227054852-20210227084852-00361.warc.gz"} |
http://www.physicsforums.com/showpost.php?p=4249429&postcount=7 | View Single Post
Sci Advisor P: 3,564 Of course also in a reversible process T may be changing. I think your problem is more a mathematical one than a physical one. e.g. you can calculate the change in energy of a particle as dE=1/2 mvdv where v is velocity. Although v will change this does not mean that you have to write dv dv. | {"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": 0.9148826599121094, "perplexity": 571.3258305306766}, "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-2014-23/segments/1406510273350.41/warc/CC-MAIN-20140728011753-00134-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://www.ck12.org/arithmetic/Fraction-and-Mixed-Number-Comparison/lesson/Fraction-and-Mixed-Number-Comparison-Grade-7/r8/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Fraction and Mixed Number Comparison
## Use <, > and/or = to compare fractions and mixed numbers.
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Fraction and Mixed Number Comparison
Have you ever had to organize all kinds of baked goods for a bake sale? Well, this is the task that Tracy is taking on.
Tracy has organized all of the baked goods that students have made. She has brownies, pies, a couple of cakes and lots of cookies. Sam baked batches of cookies. Kelly baked batches of cookies.
In order to figure this out, you will need to know how to compare and order fractions and mixed numbers. Pay attention to this Concept and you will see this problem again at the end of it.
### Guidance
Previously we worked on fractions and approximation, you can easily compare and order fractions using this technique. By figuring out the benchmark, you can determine which fractions are larger or smaller than each other. This is the best way to approximate fractions as you compare and order them.
Sometimes, however, you can’t always rely on the approximation technique when comparing and ordering fractions. This is true when two fractions have the same benchmark, or when they have different denominators. In order to be exact when comparing and ordering fractions, you have to find a common denominator for all of the fractions. Then, compare or order the fractions by looking at the value of the numerator. This will give you an exact comparison.
Use approximation to order . and from greatest to least.
We begin by getting an approximate sense of the value of each of the fractions in the group by comparing each fraction with the common benchmarks 0, and 1.
Because the number 7, which is the numerator in the fraction is very close in value to the denominator (8), we say that is approximately 1.
In the fraction, , the numerator is approximately of the denominator. So, we say that is about .
The number is the only mixed number in the group, so we can see immediately that this number is larger than all of the other numbers in the group because it is greater than 1.
In the fraction, , the denominator is much greater than the numerator, so is closest to the benchmark 0.
The numerator of 29 in the fraction is close in value to the denominator, 30, so is approximately 1.
Now that we have the approximate values of each fraction in the group, we write the fractions in a preliminary greatest to least order with the benchmarks in parentheses: .
This approximation technique helps with most of the fractions in the group, but there are two fractions which are close to 1. We know that both and are less than 1, but which of the two fractions is closest to 1? One helpful way to determine which fraction is closest to 1 is to draw two number lines between 0 and 1, arranged so that one number line is above the other. Divide the top number line into eight equal parts (eighths) and the bottom number line into thirty equal parts (thirtieths).
From this illustration, it is easy to see that is closer to 1 than and is therefore greater than .
Here is another one.
Compare and . Write >, < or =.
At first glance, it is hard to compare the two fractions because they have different denominators. Remember the first step in comparing fractions is to find the common denominator. Look at the two denominators. Sometimes the larger denominator is a multiple of the smaller denominator. For example, if we were comparing the fractions and , we can see that the denominator in , is a multiple of 3, which is the denominator of . This makes it easier to find a common denominator.
In this problem, 7 is not a multiple of 3. The lowest common denominator in this instance can only be the product of the two denominators . In order to find an equivalent fraction for with a denominator of 21, we multiply both the numerator and denominator of by 7. We get an equivalent fraction of . In order to find an equivalent fraction for with a denominator of 21, we multiply both the numerator and denominator of by 3. We get an equivalent fraction of .
Now that we have a common denominator between the two fractions, we can simply compare the numerators.
Compare using <, > or =
and
Solution: <
and
Solution: <
#### Example C
and
Solution: >
Remember the bake sale? Here is the original problem once again.
Tracy has organized all of the baked goods that students have made. She has brownies, pies, a couple of cakes and lots of cookies. Sam baked batches of cookies. Kelly baked batches of cookies.
To figure out who baked more cookies, let's write out a problem comparing the two mixed numbers.
_____
The fraction parts of these two mixed numbers have a common denominator. Because baking cookies involves dozens, both Sam and Kelly wrote the fraction part of their cookies in twelfths. Therefore, we can compare the numerators.
>
Sam baked more cookies than Kelly.
### Vocabulary
Fraction
a part of a whole.
Numerator
the top number in a fraction.
Denominator
the bottom number in a fraction. It tells you how many parts the whole is divided into.
Mixed Number
a whole number with a fraction
Improper Fraction
when the numerator is greater than the denominator in a fraction
### Guided Practice
Here is one for you to try on your own.
In the long jump contest, Peter jumped feet, Sharon jumped feet and Juan jumped feet. Now order their jump distances from greatest to least.
The problem asks us to order the jump distances from greatest to least. We have three mixed numbers, so we should look first at the whole number parts of the mixed numbers to see if we can compare the jump distances.
Peter jumped more than 5 feet, but less than 6 feet. Sharon jumped more than 6 feet, but less than 7 feet. Juan also jumped more than 6 feet, but less than 7 feet.
Simply by comparing the whole numbers, we can see that Peter jumped the shortest distance because he jumped less than 6 feet. Because Sharon and Juan both jumped between 6 and 7 feet, we need to compare the fractional part of their jumps. Sharon jumped of a foot more than 6 feet and Juan jumped of a foot more than 6 feet. In order to compare these two fractions, we have to find a common denominator. The lowest common denominator for these two fractions is 35. We get an equivalent fraction of for when we multiply both the numerator and denominator by 7. We get an equivalent fraction of for when we multiply both the numerator and denominator by 5. Now we can order the distances.
The answer is Sharon ., Juan ., Peter .
### Practice
Directions: Compare each pair of fractions or mixed numbers.Write >, < or =.
1. and .
2. and .
3. and .
4. and .
5. and .
6. and .
7. and .
8. and .
9. and .
10. and .
Directions: Write each set in order from least to greatest.
11.
12.
13.
14.
15.
16.
Directions: Solve these two problems.
17. Brantley is making an asparagus souffle, which calls for cup of cheese, cup of asparagus and cup of parsley. Using approximation order the ingredients from largest amount used to least amount used
18. Geraldine is putting in a pool table in her living room. She wants to put it against the longest wall of the room. Wall is feet and wall is feet. Against which wall will Geraldine put her pool table?
### Vocabulary Language: English
Denominator
Denominator
The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.
fraction
fraction
A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a rational number.
improper fraction
improper fraction
An improper fraction is a fraction in which the absolute value of the numerator is greater than the absolute value of the denominator.
Mixed Number
Mixed Number
A mixed number is a number made up of a whole number and a fraction, such as $4\frac{3}{5}$.
Numerator
Numerator
The numerator is the number above the fraction bar in a fraction. | {"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": 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": 3, "texerror": 0, "math_score": 0.9422500729560852, "perplexity": 687.2928599236559}, "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-2016-07/segments/1454701987329.64/warc/CC-MAIN-20160205195307-00318-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-1-section-1-7-multiplication-and-division-of-real-numbers-exercise-set-page-85/35 | ## Introductory Algebra for College Students (7th Edition)
$\frac{1}{4}$
Find the multiplicative inverse by switching the numerator and the denominator and keeping the sign the same. 4=$\frac{4}{1}$$\to$$\frac{1}{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": 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": 0.9773083329200745, "perplexity": 1643.8439806193765}, "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/1590347389355.2/warc/CC-MAIN-20200525192537-20200525222537-00208.warc.gz"} |
https://www.arxiv-vanity.com/papers/1803.06567/ | # A Dual Approach to Scalable Verification of Deep Networks
Krishnamurthy (Dj) Dvijotham, Robert Stanforth, Sven Gowal, Timothy Mann, Pushmeet Kohli Krishnamurthy (Dj) Dvijotham, Robert Stanforth
Sven Gowal, Timothy Mann, Pushmeet Kohli
London, UK N1C 4AG
###### Abstract
This paper addresses the problem of formally verifying desirable properties of neural networks, i.e., obtaining provable guarantees that the outputs of the neural network will always behave in a certain way for a given class of inputs. Most previous work on this topic was limited in its applicability by the size of the network, network architecture and the complexity of properties to be verified. In contrast, our framework applies to much more general class of activation functions and specifications on neural network inputs and outputs. We formulate verification as an optimization problem and solve a Lagrangian relaxation of the optimization problem to obtain an upper bound on the verification objective. Our approach is anytime i.e. it can be stopped at any time and a valid bound on the objective can be obtained. We develop specialized verification algorithms with provable tightness guarantees under special assumptions and demonstrate the practical significance of our general verification approach on a variety of verification tasks.
## 1 Introduction
Deep learning has led to tremendous progress in the field of machine learning resulting in remarkable improvements in the solution of a wide-range of complex prediction tasks (Krizhevsky et al., 2012; Goodfellow et al., 2016). However, in recent years, researchers have observed that even state-of-the-art networks can be easily fooled into changing their predictions by making small but carefully chosen modifications to the input data (known as adversarial perturbations) (Szegedy et al., 2013; Kurakin et al., 2016; Carlini and Wagner, 2017a; Goodfellow et al., 2014; Carlini and Wagner, 2017b). While modifications to neural network training algorithms have been proposed to mitigate this phenomenon, a comprehensive solution that is fully robust to adversarial attacks remains elusive (Carlini and Wagner, 2017b; Uesato et al., 2018).
Neural networks are typically tested using the standard machine learning paradigm: If the performance (accuracy) of the network is sufficiently high on a holdout (test) set that the network did not have access to while training, the network is deemed acceptable. This is justified by statistical arguments based on an i.i.d. assumption on the data generating mechanism, that is each input output pair is generated independently from the same (unknown) data distribution. However, this evaluation protocol is not sufficient in domains with critical safety constraints (Marston and Baca, 2015). For instance, we may require that the network is robust against adversarial perturbations within certain bounds, or we may require the output of the neural network to be monotonic with respect to the input.
In the context of adversarial examples, a natural idea is to test neural networks by checking if it is possible to generate an adversarial attack to change the label predicted by the neural network (Kurakin et al., 2016). Generating adversarial examples is a challenging computational task itself, and the attack generated by a specific algorithm may be far from optimal. This may lead one to falsely conclude that a given model is robust to attacks even though a stronger adversary may have broken the robustness. Recent work (Athalye et al., 2018; Uesato et al., 2018) has shown that evaluating models against weak adversaries can lead to incorrect conclusions regarding the robustness of the model. Thus, there is a need to go beyond evaluation using specific adversarial attacks and find approaches that provide provable guarantees against attacks by any adversary.
##### Towards verifiable models.
Verification of neural networks has seen significant research interest in recent years. Early work in the formal verification community uses Satisfiability Modulo Theories (SMT) solvers with appropriate adaptations for verification of neural networks (Ehlers, 2017; Huang et al., 2017; Katz et al., 2017). While SMT solvers have been successfully applied to several domains, applying them to large neural networks remains a challenge due to the scale of the resulting SMT problem instances. Furthermore, these approaches have been largely limited to networks with piecewise linear activation functions since most SMT solvers are unable to deal efficiently with nonlinear arithmetic. More recently, researchers have proposed a set of approaches that make use of branch and bound algorithms either directly or via mixed-integer programming solvers (Bunel et al., 2017; Cheng et al., 2017; Tjeng and Tedrake, 2017). While these approaches achieve strong results on smaller networks, scaling them to large networks remains an open challenge. These approaches also rely heavily on the piecewise linear structure of networks where the only nonlinearities are maxpooling and ReLUs.
##### Towards scalable verification of general models.
In this paper, we develop a novel approach to neural network verification based on optimization and duality. The approach consists of formulating the verification problem as an optimization problem that tries to find the largest violation of the property being verified. If the largest violation is smaller than zero, we can conclude that the property being verified is true. By using ideas from duality in optimization, we can obtain bounds on the optimal value of this problem in a computationally tractable manner. Note that this approach is sound but incomplete, in that there may be cases where the property of interest is true, but the bound computed by our algorithm is not tight enough to prove the property. This strategy has been used in prior work as well (Kolter and Wong, 2017; Raghunathan et al., 2018). However, our results improve upon prior work in the following ways:
1. Our verification approach applies to arbitrary feedforward neural networks with any architecture and any activation function and our framework recovers previous results (Ehlers, 2017) when applied to the special case of piecewise linear activation functions.
2. We can handle verification of systems with discrete inputs and combinatorial constraints on the input space, including cardinality constraints.
3. The computation involved only requires solving an unconstrained convex optimization problem (of size linear in the number of neurons in the network), which can be done using a subgradient method efficiently. Further, our approach is anytime, in the sense that the computation can be stopped at any time and a valid bound on the verification objective can be obtained.
4. For the special case of single hidden layer networks, we develop specialized verification algorithms with provable tightness guarantees.
5. We attain state of the art verified bounds on adversarial error rates on image classifiers trained on MNIST and CIFAR-10 under adversarial perturbations in the infinity norm.
## 2 Related Work
##### Certifiable training and verification of neural networks
A separate but related thread of work is on certifiable training, ie, training neural networks so that they are guaranteed to satisfy a desired property (for example, robustness to adversarial examples within a certain radius) (Kolter and Wong, 2017; Raghunathan et al., 2018). These approaches use ideas from convex optimization and duality to construct bounds on an optimization formulation of verification. However, these approaches are limited to either a class of activation functions (piecewise linear models) or architectures (single hidden layer, as in Raghunathan et al. (2018)). Further, in Kolter and Wong (2017), the dual problem starts with a constrained convex formulation but is then converted into an unconstrained but nonconvex optimization problem to allow for easy optimization via a backprop-style algorithm. In contrast, our formulation allows for an unconstrained dual convex optimization problem so that for any choice of dual variables, we obtain a valid bound on the adversarial objective and this dual problem can be solved efficiently using subgradient methods.
##### Theoretical analysis of robustness
Another related line of work has to do with theoretical analysis of adversarial examples. It has been shown that feedforward ReLU networks cannot learn to distinguish between points on two concentric spheres without necessarily being vulnerable to adversarial examples within a small radius (Gilmer et al., 2018). Under a different set of assumptions, the existence of adversarial examples with high probability is also established in Fawzi et al. (2018). In Wang et al. (2017), the authors study robustness of nearest neighbor classifiers to adversarial examples. As opposed to these theoretical analyses, our approach searches computationally for proofs of existence or non-existence of adversarial examples. The approach does not say anything a-priori about the existence of adversarial examples, but can be used to investigate their existence for a given network and compare strategies to guard against adversarial attacks.
## 3 Verification as Optimization
### 3.1 Notation
Our techniques apply to general feedforward architectures and recurrent networks, but we focus on layered architectures for the development in this paper. The input layer is numbered , the hidden layers are numbered and the output layer is numbered . The size of layer is denoted
We denote by the input to the neural network, by the pre-activations of neurons at layer before application of the activation function and by the vector of neural activations after application of the activation function (to ). For convenience, we define . We use to denote the activations at the -th layer as a function of the input . Upper and lower bounds on the pre/post activations are denoted by respectively. The activation function at layer is denote and is assumed to be applied component-wise, ie,
[hl(zl)]k=hlk(zlk)
Note that max-pooling is an exception to this rule - we discuss how max-pooling is handled separately in the Appendix section 6.2.1. The weights of the network at layer are denoted and the bias is denoted , .
### 3.2 Verification Problem
As mentioned earlier, verification refers to the process of checking that the output of the neural network satisfies a certain desirable property for all choices of the input within a certain set. Formally, this can be stated as follows:
∀xin∈Sin(xnom)xL(xin)∈Sout (1)
where denotes the input to the network, denotes a nominal input, defines the constrained subset of inputs induced by the nominal input, and denotes the constraints on the output that we would like to verify are true for all inputs in . In the case of adversarial perturbations in image classification, would refer to the nominal (unperturbed image), would refer to all the images that can be obtained by adding bounded perturbations to , and would refer to a perturbed image.
In this paper, we will assume that: is always a described by a finite set of linear constraints on the values of final layer ie. , and is any bounded set such that any linear optimization problem of the form
maxxin∈Sin(xnom)cTx
can be solved efficiently. This includes convex sets and also sets describing combinatorial structures like spanning trees, cuts in a graph and cardinality constraints.
See the following examples for a concrete illustration of the formulation of the problem:
##### Robustness to targeted adversarial attacks.
Consider an adversarial attack that seeks to perturb an input to an input subject to a constraint on the perturbation to change the label from the true label to a target label . We can map this to (1) as follows:
Sin(xnom)={xin:∥∥xin−xnom∥∥≤ϵ}, (2a) Sout={z:cTz≤0} (2b)
where is a vector with and all other components . Thus, denotes the set of outputs for which the true label has a higher logit value than the target label (implying that the targeted adversarial attack did not succeed).
##### Monotonic predictors.
Consider a network with a single real valued output and we are interested in ensuring that the output is monotonically increasing wrt each dimension of the input . We can state this as a verification problem:
Sin(xnom)={xin:xin≥xnom} (3a) Sout={xL:xL(xnom)−xL≤0} (3b)
Thus, denotes the set of outputs which are large than the network output at . If this is true for each value of , then the network is monotone.
##### Cardinality constraints.
In several cases, it makes sense to constrain a perturbation not just in norm but also in terms of the number of dimensions of the input that can be perturbed. We can state this as:
Sin(xnom)= {xin:∥∥xin−xnom∥∥0≤k,∥∥xin−xnom∥∥∞≤ϵ} (4a) Sout={z:cTz≤0} (4b) where ∥x∥0 denotes the number of non-zero entries in x.
Thus, denotes the set of outputs which are larger than the network output at . If this is true for each value of , then the network is monotone.
### 3.3 Optimization Problem for Verification
Once we have a verification problem formulated in the form (1), we can easily turn the verification procedure into an optimization problem. This is similar to the optimization based search for adversarial examples (Szegedy et al., 2013) when the property being verified is adversarial robustness. For brevity, we only consider the case where is defined by a single linear constraint . If there are multiple constraints, each one can be verified separately.
maxz0,…,zL−1x0,…,xLcTxL+d (5a) s.t xl+1=hl(zl),l=0,1,…,L−1 (5b) zl=Wlxl+bl,l=0,1,…,L−1 (5c) x0=xin,xin∈Sin(xnom) (5d)
If the optimal value of this problem is smaller than (for each in the set of linear constraints defining ), we have verified the property (1). This is a nonconvex optimization problem and finding the global optimum in general is NP-hard (see Appendix section 6.4.1 for a proof). However, if we can compute upper bounds on the value of the optimization problem and the upper bound is smaller than , we have successfully verified the property. In the following section, we describe our main approach for computing bounds on the optimal value of (5).
### 3.4 Bounding the Value of the Optimization Problem
We assume that bounds on the activations are available. Section 6.1 discusses details of how such bounds may be obtained given the constraints on the input layer . We can bound the optimal value of (5) using a Lagrangian relaxation of the constraints:
maxz0,…,zL−1x0,x1,…,xL−1 cT(hL−1(zL−1))+d +L−1∑l=0(μl)T(zl−Wlxl−bl) +L−2∑l=0(λl)T(xl+1−hl(zl)) (6a) s.t. z–l≤zl≤¯¯¯zl,l=0,1,…,L−1 (6b) x––l≤xl≤¯¯¯xl,l=0,1,…,L−1 (6c) x0∈Sin(xnom) (6d)
Note that any feasible solution for the original problem (5) is feasible for the above problem, and for any such solution, the terms involving become (since the terms multiplying are for every feasible solution). Thus, for any choice of , the above optimization problem provides a valid upper bound on the optimal value of (5) (this property is known as weak duality (Vandenberghe and Boyd, 2004)).
We now look at solving the above optimization problem. Since the objective and constraints are separable in the layers, the variables in each layer can be optimized independently. For , we have
fl(λl−1,μl)= maxxl∈[x––l,¯¯¯xl](λl−1−(Wl)Tμl)Txl−(bl)Tμl
which can be solved trivially by setting each component of to its upper or lower bound depending on whether the corresponding entry in is non-negative. Thus,
fl(λl−1,μl)= [λl−1−(Wl)Tμl]+T¯¯¯xl +[λl−1−(Wl)Tμl]−Tx––l−(bl)Tμl
where denote the positive and negative parts of .
Similarly, collecting the terms involving , we have, for
~fl(λl,μl)=maxzl∈[z–l,¯zl]μlTzl−(λl)Thl(zl)
where .
Since is a component-wise nonlinearity, each dimension of can be optimized independently. For the -th dimension, we obtain
~fl,k(λlk,μlk)=maxzlk∈[z–lk,¯zlk]μlkzlk−λlkhlk(zlk)
This is a one-dimensional optimization problem and can be solved easily- for common activation functions (ReLU, tanh, sigmoid, maxpool), it can be solved analytically, as discussed in appendix section 6.2. More generally, assuming that is a monotone, we can obtain an upper bound on (see section 6.2.2 for details). Since we care about proving that an upper bound on the verification objective is smaller than , an upper bound on suffices for the purposes of verification.
Finally, we need to solve
f0(μ0)=maxx0∈Sin(xnom)(−(W0)Tμ0)Tx0−(b0)Tμ0
which can also be solved easily given the assumption on . We work out some concrete cases in 6.3.
Once these problems are solved, we can construct the dual optimization problem:
minλ,μ nL−1∑k=0~fL−1,k(−ck,μlk)+L−2∑l=0nl∑k=0~fl,k(λlk,μlk) +L−1∑l=1fl(λl−1,μl)+f0(μ0)+d (7)
This seeks to choose the values of so as to minimize the upper bound on the verification objective, thereby obtaining the tightest bound on the verification objective.
This optimization can be solved using a subgradient method on .
###### Theorem 1.
For any values of with , the objective of (7) is an upper bound on the optimal value of (5). Hence, the optimal value of (7) is also an upper bound. Further, (7) is a convex optimization problem in .
###### Proof.
The upper bound property follows from weak duality (Vandenberghe and Boyd, 2004). The fact that (7) is a convex optimization problem can be seen as each term is expressed as a maximum overa set of linear functions of (Vandenberghe and Boyd, 2004). ∎
###### Theorem 2.
If each is a ReLU function, then (7) is equivalent to the dual of the LP described in Ehlers (2017).
###### Proof.
See section 6.4. ∎
The LP formulation from Ehlers (2017) is also used in Kolter and Wong (2017). The dual of the LP is derived in Kolter and Wong (2017) - however this dual is different from (7) and ends up with a constrained optimization formulation for the dual. To allow for an unconstrained formulation, this dual LP is transformed to a backpropagation-like computation. While this allows for folding the verification into training, it also introduces nonconvexity in the verification optimization - our formulation of the dual differs from Kolter and Wong (2017) in that we directly solve an unconstrained dual formulation, allowing us to circumvent the need to solve a non-convex optimization for verification.
### 3.5 Towards Theoretical Guarantees for Verification
The bounds computed by solving (7) could be loose in general, since (5) is an NP-hard optimization problem (section 6.4.1). SMT solvers and MIP solvers are guaranteed to find the exact optimum for piecewise linear neural networks, however, they may take exponential time to do so. Thus, an open question remains: Are there cases where it is possible to perform exact verification efficiently? If not, can we approximate the verification objective to within a certain factor (known a-priori)? We develop results answering these questions in the following sections.
Prior work: For any linear classifier, the scores of each label are a linear function of the inputs . Thus, the difference between the predictions of two classes (target class for an adversary) and class (true label) is . Maximizing this subject to can be solved analytically to obtain the value . This observation formed the basis for algorithms in (Raghunathan et al., 2018) and (Hein and Andriushchenko, 2017). However, once we move to nonlinear classifiers, the situation is not so simple and computing the worst case adversarial example, even in the 2-norm case, becomes a challenging task. In (Hein and Andriushchenko, 2017), the special case of kernel methods and single hidden layer classifiers are considered, but the approaches developed are only upper bounds on the verification objective (just like those computed by our dual relaxation approach). Similarly, in Raghunathan et al. (2018), a semidefinite programming approach is developed to compute bounds on the verification objective for the special case of adversarial perturbations on the infinity norm. However, none of these approaches come with a-priori guarantees on the quality of the bound, that is, before actually running the verification algorithm, one cannot predict how tight the bound on the verification objective would be. In this section, we develop novel theoretical results that quantify when the verification problem (5) can be solved either exactly or with a-priori approximation guarantees. Our results require strong assumptions and do not immediately apply to most practical situations. However, we believe that they shed some understanding on the conditions under which exact verification can be performed tractably and lead to specialized verification algorithms that merit further study.
We assume the following for all results in this section:
1) We study networks with a single hidden layer, i.e. , with activation function and a linear mapping from the penultimate to the output layer .
2) The network has a differentiable activation function with Lipschitz-continuous derivatives denoted (tanh, sigmoid, ELU, polynomials satisfy this requirement).
3) .
Since the output layer is a linear function of the penultimate layer , we have
cTx2=((W1)Tc)Tx1+cTb1 =((W1)Tc)Th0(z0)+cTb1
For brevity, we simply denote as , drop the constant term and let and denote the -th row of . Then, (5) reduces to:
maxxin:∥∥xin−xnom∥∥2≤ϵ ∑icihi(Wixin+bi) (8)
###### Theorem 3.
Suppose that has a Lipschitz continuous first derivative:
h′i(t)−h′i(~t)≤γi|t−~t|
Let
ν=∥∥diag(c)WTh′(znom)∥∥2 L=σmax(diag(c)WT)σmax(diag(γ)W)
Then , the iteration:
xk+1←xnom+ϵ(WTdiag(c)h′(Wxk+b)∥WTdiag(c)h′(Wxk+b)∥2)
starting at converges to the global optimum of (8) at the rate
###### Proof.
Section 6.4
Thus, when is small enough, a simple algorithm exists to find the global optimum of the verification objective. However, even when is larger, one can obtain a good approximation of the verification objective. In order to do this, consider the following quadratic approxiimation of the objective from (8):
max ∑ici(hi(znomi)+h′i(znomi)(Wiz)) +∑ici2h′′i(znomi)(Wiz)2 (9a) s.t. ∥z∥2≤ϵ (9b)
This optimization problem corresponds to a trust region problem that can be solved to global optimality using semidefinite programming (Yakubovic, 1971):
maxz,Z ∑ici(hi(znomi)+h′i(znomi)(Wiz)) +∑ici2h′′i(znomi)tr(WTiWiZ) (10a) s.t. tr(Z)≤ϵ,(1zTzZ)⪰0 (10b)
where denotes that is constrained to be a positive semidefinite matrix. While this can be solved using general semidefinite programming solvers, several special purpose algorithms exist for this trust region problem that can exploit its particular structure for efficient solution, (Hazan and Koren, 2016)
###### Theorem 4.
Suppose that is thrice-differentiable with a globally bounded third derivative. Let
ζi=∥Wi∥2,ηi=supt|h′′′i(t)|,κ=16(∑iηiciζ3i)
For each , the difference between the optimal values of (10), (8) is at most .
See Section 6.4
## 4 Experiments
In this section, we present numerical studies validation our approach on three sets of verification tasks:
Image classification on MNIST and CIFAR: We use our approach to obtain guaranteed lower bounds on the accuracy of image classifers trained on MNIST and CIFAR-10 under adversarial attack with varying sizes of the perturbation radius. We compare the bounds obtained by our method with prior work (in cases where prior work is applicable) and also with the best attacks found by various approaches.
Classifier stability on GitHub data: We train networks on sequences of commits on GitHub over a collection of 10K repositories - the prediction task consists of predicting whether a given repository will reach more than 40 commits within 250 days given data observed until a certain day. Input features consist a value between 0 and 1 indicating the number of days left until the 250th day, as well as another value indicating the progress of commits towards the total of 40. As the features evolve, the prediction of the classifier changes (for example, predictions should become more accurate as we move closer to the 250th day). In this situation, it is desirable that the classifier provides consistent predictions and that the number of times its prediction switches is as small as possible. It is also desirable that this switching frequency cannot be easily be changed by perturbing input features. We use our verification approach combined with dynamic programming to compute a bound on the maximum number of switches in the classifier prediction over time.
Digit sum task: We consider a more complex verification task here: Given a pair of MNIST digits, the goal is to bound how much the sum of predictions of a classifier can differ from the true sum of those digits under adversarial perturbation subject to a total budget on the perturbation across the two digits.
### 4.1 Image Classification: Mnist and Cifar
We study adversarial attacks subject to an bound on the input perturbation. An adversarial example (AE) is a pertubation of an input of the neural network such that the output of the neural network differs from the correct label for that input. An AE is said to be within radius if the norm of the difference between the AE and the original input is smaller than . We are interested in the adversarial error rate, that is,
# Test examples that have an AE within radius ϵ% Size of test set
Computing this quantity precisely requires solving the NP-hard problem (5) for each test example, but we can obtain upper bounds on it using our (and other) verification methods and lower bounds using a fixed attack algorithm (in this paper we use a bound constrained LBFGS algorithm similar to (Carlini and Wagner, 2017b)). Since theorem 2 shows that for the special case of piecewise linear neural networks, our approach reduces to the basic LP relaxation from Ehlers (2017) (which also is the basis for the algorithms in Bunel et al. (2017) and Kolter and Wong (2017)), we focus on networks with smooth nonlinearities like tanh and sigmoid. We compare our approach with The SDP formulation from Raghunathan et al. (2018) (note that this approach only works for single hidden layer networks, so we just show it as producing vacuous bounds for other networks).
Each approach gets a budget of s per verifiation problem (choice of test example and target label). Since the SDP solver from (Raghunathan et al., 2018) only needs to be run once per label pair (and not per test example), its running time is amortized appropriately.
Results on smooth activation functions: Figures 0(a),0(b) show that our approach is able to compute nearly tight bounds (bounds that match the upper bound) for small perturbation radii (up to 2 pixel units) and our bounds significantly outperform those from the SDP approach (Raghunathan et al., 2018) ( which is only able to compute nontrivial bounds for the smallest model with 20 hidden units).
Results on models trained adversarially: We use the adversarial training approach of (Uesato et al., 2018) and train models on MNIST and CIFAR that are robust to perturbations from the LBFGS-style attack on the training set. We then apply our verification algorithm to these robust models and obtain bounds on the adversarial error rate on the test set. These models are all multilayer models, so the SDP approach from (Raghunathan et al., 2018) does not apply and we do not plot it here. We simply plot the attack versus the bound from our approach. The results for MNIST are plotted in figure 3 and for CIFAR in figure 2. On networks trained using a different procedure, the approaches from Raghunathan et al. (2018) and Kolter and Wong (2017) are able to achieve stronger results for larger values of (they work with in real units, which corresponds to in pixel units we use here). However, we note that our verification procedure is agnostic to the training procedure, and can be used to obtain fairly tight bounds for any network and any training procedure. In comparison, the results in Kolter and Wong (2017) and Raghunathan et al. (2018) rely on the training procedure optimizing the verification bound. Since we do not rely on a particular adversarial training procedure, we were also able to obtain the first non-trivial verification bounds on CIFAR-10 (to the best of our knowledge) shown in figure 2. While the model quality is rather poor, the results indicate that our approach could scale to more complicated models.
### 4.2 Github Classifier Stability
We allow the adversary to modify input features by up to 3% (at each timestep) and our goal is to bound the maximum number of prediction switches induced by each attack over time. We can model this within our verification framework as follows: Given a sequence of input features, we compute the maximum number of switches achievable by first computing the target classes that are reachable through an adversarial attack at each timestep (using (7)), and then running a dynamic program to compute the choices of target classes over time (from within the reachable target classes) to maximize the number of switches over time.
Figure 4 shows how initially predictions are easily attackable (as little information is available to make predictions), and also shows how the gap between our approach and the best attack found using the LBFGS algorithm evolves over time.
### 4.3 Complex Verification Task: Digit Sum
In order to test our approach on a more complex specification, we study the following task: Given a pair of MNIST digits, we ask the question: Can an attacker perturb each image, subject to a constraint on the total perturbation across both digits, such that the sum of the digits predicted by the classifier differs from the true sum of those digits by as large an amount as possible? Answering this question requires solving the following optimization problem:
maxxina,xinbϵa,ϵb |argmax(xL(xina))+argmax(xL(xinb))−s| s.t. ∥∥xina−xnoma∥∥≤ϵa,∥∥xinb−xnomb∥∥≤ϵb ϵa+ϵb≤ϵ
where is the true sum of the two digits. Thus, the adversary has to decide on both the perturbation to each digit, as well as the size of the perturbation. We can encode this within our framework (we skip the details here). The upper bound on the maximum error in the predicted sum from the verification and the lower bound on the maximum error computed from an attack for this problem (on an adversarially trained two hidden layer sigmoid network) is plotted in figure 5. The results show that even on this rather complex verification task, our approach is able to compute tight bounds.
## 5 Conclusions
We have presented a novel framework for verification of neural networks. Our approach extends the applicability of verification algorithms to arbitrary feedforward networks with any architecture and activation function and to more general classes of input constraints than those considered previously (like cardinality constraints). The verification procedure is both efficient (given that it solves an unconstrained convex optimization problem) and practically scalable (given its anytime nature only required gradient like steps). We proved the first known (to the best of our knowledge) theorems showing that under special assumptions, nonlinear neural networks can be verified tractably. Numerical experiments demonstrate the practical performance of our approach on several classes of verification tasks.
## 6 Appendix
### 6.1 Bound Tightening
The quality of the bound in theorem 1 depends crucially on having bounds on all the intermediate pre and post activations . In this section, we describe how these bounds may be derived. One simple way to compute bounds on neural activations is to use interval arithmetic given bounds on the input . Bounds at each layer can then be computed recursively for as follows:
z–l=[Wl]+x––l+[Wl]−¯¯¯xl+bl (11a) ¯¯¯zl=[Wl]+¯¯¯xl+[Wl]−z–l+bl (11b) x––l+1=hl(z–l) (11c) ¯¯¯xl+1=hl(¯¯¯zl) (11d)
However, these bounds could be quite loose and could be improved by solving the optimization problem
maxz0,…,zL−1x1,…,xL−1 xlk (12a) s.t (12b)
We can relax this problem using dual relaxation approach from the previous section and the optimal value of the relaxation would provide a new (possibly tighter) upper bound on than . Further, since our relaxation approach is anytime (ie for any choice of dual variables we obtain a valid bound), we can stop the computation at any time and use the resulting bounds. This is a significant advantage compared to previous approaches used in [Bunel et al., 2017]. Similarly, one can obtain tighter upper and lower bounds on for each value of . Given these bounds, one can infer tighter bounds on using (11).
Plugging these tightened bounds back into (6) and rerunning the dual optimization, we can compute a tighter upper bound on the verification objective.
### 6.2 Conjugates of Transfer Functions
We are interested in computing . This is a one dimensional optimization problem and can be computed via brute-force discretization of the input domain in general. However, for most commonly used transfer functions, this can be computed analytically. We derive the analytical solution for various commonly used transfer functions:
ReLUs: If is a ReLU, is piecewise linear and specifically is linear on and on (assuming that , else is simply linear and can be optimized by setting to one of its bounds). On each linear piece, the optimum is attained at one of the endpoints of the input domain. Thus, the overall maximum can be obtained by evaluating at (if . Thus,
.
Sigmoid: If is a sigmoid, we can consider two cases: a) The optimum of is obtained at one of its input bounds or b) The optimum of is obtained at a point strictly inside the interval . In case (b), we require that the derivative of vanishes, i.e:. Since , this equation only has a solution if and In this case, the solutions are Solving for , we obtain where is the logit function . We only consider these solutions if they lie within the domain . Define:
y1(μ,λ) =max⎛⎜ ⎜⎝y–,min⎛⎜ ⎜⎝¯¯¯y,1−√1−4μλ2⎞⎟ ⎟⎠⎞⎟ ⎟⎠ y2(μ,λ) =max⎛⎜ ⎜⎝y–,min⎛⎜ ⎜⎝¯¯¯y,1+√1−4μλ2⎞⎟ ⎟⎠⎞⎟ ⎟⎠
Thus, we obtain the following expression for :
⎧⎪⎨⎪⎩max(g(y–),g(¯¯¯y)) if λ=0 or μλ∉[0,14]max(g(y–),g(¯¯¯y),g(y1(μ,λ)),g(y2(μ,λ))) % otherwise
Tanh: Define
y1(μ,λ)=max(y–,min(¯y,arctanh(√1−μλ))) y2(μ,λ)=max(y–,min(¯y,arctanh(√1−μλ)))
and obtain to be
⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩max(g(y–),g(¯¯¯y)) if λ=0 or μλ∉[0,1]max(g(y–),g(¯¯¯y),g(y1(μ,λ)),g(y2(μ,λ))) otherwise
#### 6.2.1 MaxPool
If is a max-pool, we need to deal with it layer-wise and not component-wise. We have and are interested in solving for
g⋆=maxy∈[y–,¯y](μ)Ty−λh(y)
Note here that is a scalar while is a vector. This can be solved by considering the case of each component of attaining the maximumum separately. We look at the case where attains the maximum below:
maxy∈[y–,¯y],yi≥yj∀j≠i(μ)Ty−λyi
Fixing and optimizing the other coordinates, we obtain
maxyi∈[y–i,¯yi] ∑j≠i,yi≥¯yjmax(μj¯¯¯yj,μjy–j) +∑j≠i,yi≤¯yjmax(μjyi,μjy–j) −(μi−λ)yi
This one-dimensional function can be optimized via binary search on . After solving for each , taking the maximum over gives the value
#### 6.2.2 Upper bounds for general nonlinearities
In general, we can compute an upper bound on even if we cannot optimize it exactly. The idea is to decouple the from the two terms in : and and optimize each independently. However, this gives a very weak bound. This can be made much tighter by applying it separately to a decomposition of the input domain :
maxy∈[ai,bi]g(y)≤max(μai,μbi)+max(−λh(ai),−λh(bi))
Finally we can bound using
maxy∈[y–,¯y]g(y)≤ maxi(max(μai,μbi)+max(−λh(ai),−λh(bi)))
As the decomposition gets finer, ie, , we obtain an arbitrarily tight upper bound on ) this way.
### 6.3 Optimizing Over the Input Constraints
In this section, we discuss solving the optimization problem defining in (7)
maxx∈Sin(WTμ)Tx+bTμ
where we dropped the superscript for brevity. For commonly occuring constraint sets , this problem can be solved in closed form easily:
Norm constraints: Consider the case . In this case, we by Holder’s inequality, the objective is larger than or equal to
bTμ−∥∥WTμ∥∥⋆
where is the dual norm to the norm . Further this bound can be achieved for an appropriate choice of . Hence, the optimal value is precisely .
Combinatorial objects: Linear objectives can be optimized efficiently over several combinatorial structures. For example, if is indexed by the edges in a graph and imposes constraints that is binary valued and that the edges set to should form a spanning tree of the graph, the optimal value can be computed using a maximum spanning tree approach.
Cardinality constraints: may have cardinality constrained imposed on it: , saying that at most elements of can be non-zero. If further we have bounds on , then the optimization problem can be solved as follows: Let and let denote the -th largest component of . Then the optimal value is .
### 6.4 Proofs of Theoretical Results
#### 6.4.1 Np-Hardness of Verification of a Single Hidden Layer Network
Consider the case of sigmoid transfer function with an norm perturbation. Then, the verification problem reduces to :
maxxin,z ∑icihi(zi) s.t z=Wx+b,∥∥xin−xnom∥∥∞≤ϵ
This is an instance of a sigmoidal programming problem, which is proved to be NP-hard in Udell and Boyd [2013]
#### 6.4.2 Proof of Theorem 2
###### Proof.
In this case, since is a relu, can be written as (from section 6.2) as
~f(λ,μ)=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩max(μy–,(μ−λ)¯¯¯y,0) if 0∈[y–,¯¯¯y]max(μ¯¯¯y,μy–) if 0 if ¯¯¯y≤0max((μ−λ)¯¯¯y,(μ−λ)y–) if y–≥0
Now, the LP relaxation from [Ehlers, 2017] can be written as
maxcTxL s.t zl=Wlxl+bl,l=0,…,L−1 xl+1k=zlk∀l,k s.t z–lk≥0 xl+1k=0 if ∀l,k s.t ¯¯¯zlk≤0 xl+1k≥0,xlk≥zlk,xlk≤(zlk−z–lk)⎛⎝¯¯¯zlk¯¯¯zlk−z–lk⎞⎠ ∀l,k s.t 0∈[zlk,xlk] z–l≤zl≤¯¯¯zl,x––l≤xl≤¯¯¯xl∀l
Taking the dual of this optimization problem, we obtain
maxcTxL+∑l(μl)T(zl−Wlxl−bl) +∑l,k:z–lk≥0λlk(xl+1k−zlk) +∑l,k:z–lk≤0λlk(xl+1k) +∑l,k:0∈[z–lk,¯zlk](λlk;axl+1k+λlk;b(xl+1k−zlk)) +∑l,k:0∈[z–lk,¯zlk]−(xl+1k−slk(zlk−z–lk))λlk;c s.t z–l≤zl≤¯¯¯zl,x––l≤xl≤¯¯¯xl∀l
where . Collecting the terms that depend on , we obtain
⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩μlkzlk if ¯¯¯zlk≤0(μlk−λlk)zlk if z–lk≥0(μlk−λlk;b+λlk;cslk)zlk−slkz–lkλlk;c
The first two cases, upon maximizing wrt give the last two cases of . For the last case, by complementary slackness, we have at most two of can be non-zero. Reasoning through all the cases, we obtain the expression defining the first case of with . Given this, the rest of the dual exactly matches the calculations from section 3.3.
#### 6.4.3 Proof of Theorem 3
###### Proof.
We leverage results from [Polyak, 2003] which argues that a smooth nonlinear function can be efficicently optimized over a “small enough” ball. Specifically, we use theorem 7 from [Polyak, 2003]. In order to apply the theorem, we need to bound the Lipschitz costant of the derivative of the function . Writing down the derivative, we obtain
f′(x)=WTdiag(c)h′(Wx+b)
Thus,
f′(x)−f′(y) =WTdiag(c)(h′(Wx+b)−h′(Wy+b))
We have
h′i(Wix+bi)−h′ | {"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": 0.9444307088851929, "perplexity": 554.1758507796777}, "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/1620243989819.92/warc/CC-MAIN-20210518094809-20210518124809-00556.warc.gz"} |
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# The value of the expression $\dfrac{{1 - 4\sin {{10}^0}\sin {{70}^0}}}{{2\sin {{10}^0}}}$ is (A)$\dfrac{1}{2}$ (B)1 (C) 2(D) None of these
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Hint- To solve this question we will use trigonometric identities such as $\sin (90 - \theta ) = \cos \theta {\text{ and cosA - cosB = 2sin(}}\dfrac{{A + B}}{2})\sin (\dfrac{{A - B}}{2})$
Complete step-by-step solution -
Given expression is $\dfrac{{1 - 4\sin {{10}^0}\sin {{70}^0}}}{{2\sin {{10}^0}}}............................(1)$
As we know that
$\sin (90 - \theta ) = \cos \theta {\text{ }} \\ {\text{cosA - cosB = 2sin(}}\dfrac{{A + B}}{2})\sin (\dfrac{{A - B}}{2}) \\$
From equation (1) write the angles of sin as a sum or difference of two angles such as ${70^0} = \dfrac{{{{60}^0} + {{80}^0}}}{2}{\text{ and 1}}{{\text{0}}^0} = \dfrac{{{{80}^0} - {{60}^0}}}{2}$ , we get
$= \dfrac{{1 - 4\sin \left( {\dfrac{{{{60}^0} + {{80}^0}}}{2}{\text{ }}} \right)\sin \left( {\dfrac{{{{80}^0} - {{60}^0}}}{2}} \right)}}{{2\sin ({{90}^0} - {{10}^0})}}$
Now, using the formulas mentioned above, we get
$= \dfrac{{1 - 2[\cos \left( {{\text{6}}{{\text{0}}^0}} \right) - \cos \left( {{{80}^0}} \right)]}}{{2\cos ({{80}^0})}}$
As we know that $\cos {60^0} = \dfrac{1}{2}$ substituting this value in the above equation, we get
$= \dfrac{{1 - 2 \times \dfrac{1}{2} + 2\cos \left( {{{80}^0}} \right)}}{{2\cos ({{80}^0})}} \\ = \dfrac{{1 - 1 + 2\cos \left( {{{80}^0}} \right)}}{{2\cos ({{80}^0})}} \\ = \dfrac{{2\cos \left( {{{80}^0}} \right)}}{{2\cos ({{80}^0})}} \\ = 1 \\$
So, the value of the given expression is 1.
Hence, the correct answer is option B.
Note- To solve this question, we used the trigonometric identities and some manipulation. Whenever we have an unknown or random angle in the problem, whose trigonometric values are unknown, try to manipulate some angle by using trigonometric identities in order to cancel that term or to bring the angle in some known value. Remember the trigonometric identities. | {"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": 0.9901562333106995, "perplexity": 644.900352575108}, "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-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00608.warc.gz"} |
http://math.stackexchange.com/questions/103569/use-pappus-theorem-to-find-the-moment-of-a-region-limited-by-a-semi-circunferen | # Use Pappus' theorem to find the moment of a region limited by a semi-circunference.
This is part of self-study; I found this question in the book "The Calculus with Analytic Geometry" (Leithold).
$R$ is the region limited by the semi-circumference $\sqrt{r^2 - x^2}$ and the x-axis. Use Pappus' theorem to find the moment of $R$ with respect to the line $y = -4$.
Pappus' theorem (also referred to as Guldinus theorem or Pappus-Guldinus theorem) is as follows:
If $R$ is the region limited by the functions $f(x)$ and $g(x)$, then, if $A$ is the area of $R$ and $\bar{y}$ is the y-coordinate of the centroid of $R$, the volume $V$ of the solid of revolution obtained by rotating $R$ around the x-axis is given by: $V = 2\pi\bar{y}A$.
Also, what I'm calling moment (I'm not sure if this is a common term) is the quantity that is divided by the area of the region in order to find a coordinate of the centroid of the region: for example, to find the x-coordinate of the centroid ($\bar{x}$), first one finds the moment around the y-axis ($M_y$), then divides it by the area of the region ($A$).
Since the region $R$ is symmetric with respect to the y-axis, the x-coordinate of the centroid ($\bar{x}$) is zero (therefore, the moment around the y-axis ($M_y$) is zero, because $M_y = \bar{x}A$, where $A$ is the area of the region). So, I only need to find the vertical coordinate of the centroid (with respect to the line $y = -4$) and the moment around the line $y = -4$.
The book's answer for the moment around the line $y = -4$ is: $\frac{1}{2}r^3\left (\pi+\frac{4}{3}\right )$. I included two attempts below; both arrive at a same result, which is different from the result of the book.
Attempt 1
First I tried to use Pappus' theorem to find the vertical coordinate of the centroid of the semi-circular region limited by $\sqrt{r^2 - x^2}$, with respect to the x-axis (not yet with respect to the line y = -4). I will call it $\bar{y}_x$.
Since the solid of revolution obtained by rotating this semi-circular region around the x-axis is a sphere, its volume is $V = \frac{4}{3}\pi r^3$. The area of the semi-circular region is $A=\frac{\pi r^2}{2}$. Substituting $V$ and $A$ into Pappus' theorem:
$\frac{4}{3}\pi r^3 = 2\pi\bar{y}_x\frac{\pi r^2}{2}$
$\bar{y}_x = \frac{4r}{3\pi}$.
This is the vertical coordinate of the centroid with respect to the x-axis. The vertical coordinate of the centroid with respect to the line $y = -4$ is:
$\bar{y} = \frac{4r}{3\pi} + 4$.
To find the moment around the line $y = -4$, I use the fact that $\bar{y} = \frac{M_x}{A}$:
$M_x = \bar{y}A = \left ( \frac{4r}{3\pi} + 4\right )\frac{\pi r^2}{2}$.
Attempt 2
The moment of a plane region with respect to the line $y = -4$ can be found by dividing this region into infinitesimal elements of area, then multiplying the area of each element of area by the vertical coordinate of the centroid.
So, if $f(x) = \sqrt{r^2 - x^2}$, then, if I divide the semi-circular region into several rectangles of length $dx$, the area of each rectangle is $f(x) dx$ and the vertical coordinate of the centroid of each rectangle with respect to the line $y = -4$ is $\frac{f(x)}{2} + 4$. So, the moment with respect to the line $y = -4$ is:
$M_x = \int_{-r}^r \left [ f(x)\times \left ( \frac{f(x)}{2} + 4 \right ) dx \right ]$.
Solving this integral gives the following result:
$M_x = \left ( \frac{4r}{3\pi} + 4\right )\frac{\pi r^2}{2}$,
which is the same result I found in the first attempt.
Thank you in advance.
-
## 1 Answer
Both your attempts lead to the correct result, and the solution given in the book is wrong. To see that the book's solution cannot be correct do the following "Gedankenexperiment": Assume that we are told to compute the moment with respect to the line $y=-\eta$ for some given $\eta$ instead of $4$. For an $\eta \gg r$ this moment would be approximatively proportional to $r^2$ and to $\eta$, as is the case in your solution but not in the solution given by the book: The $O(r^3)$ dependency is unwarranted.
By the way: what you call "Pappus' theorem" is known in our area as "Guldin's rule".
-
Thank you for confirming that the result is correct. – anonymous Jan 29 '12 at 14:46 | {"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": 0.9533259272575378, "perplexity": 83.61940290151}, "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-2013-20/segments/1368701614932/warc/CC-MAIN-20130516105334-00068-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://scicomp.stackexchange.com/questions/38889/derivation-of-a-parabolic-pde-using-alternating-direction-implicit-method | # Derivation of a parabolic PDE using Alternating Direction Implicit method
I have a very simple question concerning Alternating Direction Implicit (ADI) Scheme.
If I have an equation of the form:
$$\begin{equation*} \frac{df(x,y,t)}{dt} = \nabla^2 f(x,y,t) + f(x,y,t) \end{equation*}$$
with f, my unknown function.
I know how to use it for the diffusion term, but how should I handle the term $$f(x,y,t)$$?
Starting from Crank-Nicolson, I can write:
$$\begin{equation*} \frac{f^{t+1}_{i,j} - f^{t}_{i,j}}{\Delta t} = \frac{1}{2}(\frac{f_{i+1,j}^{t} - 2 f_{i,j}^{t} + f_{i-1,j}^{t}}{dx^2} + \frac{f_{i+1,j}^{t+1} - 2 f_{i,j}^{t+1} + f_{i-1,j}^{t+1}}{dx^2} \\ + \frac{f_{i,j+1}^{t} - 2 f_{i,j}^{t} + f_{i,j-1}^{t}}{dy^2} + \frac{f_{i,j+1}^{t+1} - 2 f_{i,j}^{t+1} + f_{i,j-1}^{t+1}}{dy^2}) + \frac{1}{2}(f_{i,j}^{t} + f_{i,j}^{t+1}) \end{equation*}$$
Then, solving implicitly in x and explicitly in y, using ADI, with * beeing the intermediate timestep:
$$\begin{equation*} \frac{f^{*}_{i,j} - f^{t}_{i,j}}{\frac{\Delta t}{2}} = \frac{f_{i+1,j}^{*} - 2 f_{i,j}^{*} + f_{i-1,j}^{*}}{dx^2} + \frac{f_{i,j+1}^{t} - 2 f_{i,j}^{t} + f_{i,j-1}^{t}}{dy^2} \\ + \frac{1}{2}(f_{i,j}^{t} + f_{i,j}^{*}) \end{equation*}$$
And after, solving explicitly in x and implicitly in y, using ADI, taking the previous result:
$$\begin{equation*} \frac{f^{t+1}_{i,j} - f^{*}_{i,j}}{\frac{\Delta t}{2}} = \frac{f_{i+1,j}^{*} - 2 f_{i,j}^{*} + f_{i-1,j}^{*}}{dx^2} + \frac{f_{i,j+1}^{t+1} - 2 f_{i,j}^{t+1} + f_{i,j-1}^{t+1}}{dy^2} \\ + \frac{1}{2}(f_{i,j}^{*} + f_{i,j}^{t+1}) \end{equation*}$$
I know how to solve everything, my question concerns the term $$f(x,y,t)$$: is that correct to take the mean between the value at t and * for the first step and the mean of time * and t+1 on the second step?
Thank you very much!
Yes, this is correct in the sense that it is second order in both time and space. It is not the only way to handle the $$f(x, y, t)$$ term, however.
From the equations you wrote, it appears that the PDE has been partitioned in the following way \begin{align*} \frac{\partial f(x,y,t)}{\partial t} &= \nabla^2 f(x, y, t) + f(x, y, t) \\ &= \underbrace{\frac{\partial^2 f(x,y,t)}{\partial x^2} + \frac{1}{2} f(x,y,t)}_\text{x-direction} + \underbrace{\frac{\partial^2 f(x,y,t)}{\partial y^2} + \frac{1}{2} f(x,y,t)}_\text{y-direction}. \end{align*} Alternatively, you could move the $$f(x,y,t)$$ entirely in the x-direction or entirely in the y-direction. It is somewhat arbitrary how the term is distributed as long as it sums to $$f(x,y,t)$$. Let's stick with this half-in-each approach, though.
Here is another way to check that this is correct. For simplicity, let's ignore the diffusion term since that wasn't your concern, and we can focus on the treatment of the $$f(x,y,t)$$. So consider the ODE $$\frac{\partial f(x,y,t)}{\partial t} = f(x,y,t).$$ From the first part of the ADI method, we have that $$\frac{f^*_{i,j} - f^t_{i,j}}{\frac{\Delta t}{2}} = \frac{f^t_{i,j} + f^*_{i,j}}{2} \quad \Rightarrow \quad f^*_{i,j} = \frac{1+\frac{\Delta t}{4}}{1-\frac{\Delta t}{4}} f^t_{i,j}.$$ Similarly, the second part give \begin{align*} f^{t+1}_{i,j} &= \frac{1+\frac{\Delta t}{4}}{1-\frac{\Delta t}{4}} f^*_{i,j} \\ &= \left(\frac{1+\frac{\Delta t}{4}}{1-\frac{\Delta t}{4}}\right)^2 f^t_{i,j} \\ &= \left( 1 + \Delta t + \frac{\Delta t^2}{2} + \frac{3 \Delta t^3}{16} + \cdots \right) f^t_{i,j}. \end{align*} This is the linear stability function for two half-steps of CN. It matches the exact solution of $$e^{\Delta t} f^t_{i,j}$$ up to $$\mathcal{O}(\Delta t^3)$$, and thus, it is order two. | {"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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9994493126869202, "perplexity": 484.522719398364}, "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-2021-43/segments/1634323585201.94/warc/CC-MAIN-20211018093606-20211018123606-00476.warc.gz"} |
http://physics.ucsc.edu/~peter/231/magnetic_field/node5.html | # Landau Diamagnetism
In this appendix we derive the expression for Landau diamagnetism of free electrons. Our approach will be to derive the result first for 2- from which we will easily be able to obtain the 3- result by integrating over .
Firstly, we remind ourselves of some statistical mechanics. It is convenient to calculate the grand partition function, where
(49)
where is the number of particles and is the Hamiltonian. The trace is over all states for a given and over all . This ensemble, where both and are allowed to vary, is called the grand canonical ensemble''. is related to the grand potential'', , by
(50)
The mean number of particles is given by
(51)
which implicitly determines . is related to the free energy by
(52)
Note that is considered as a function of and so differentiating the right hand side of Eq. (A4) with respect to (keeping constant) gives
(53)
This will be useful later. Note that a lot of confusion in statistical mechanics comes from a lack of understanding of what is being kept constant in partial derivatives. Once the correct value of has been determined for a given , we determine the magnetization from
(54)
where is the magnetic field and the derivative is at fixed . Note that in general we need to determine from rather than , because the number of particles is kept constant when the field is applied, not the chemical potential. However, as we shall see, they both give the same result to lowest order in , the case of interest here, because of Eq. (A5). The susceptibility is then found from
(55)
so we have to compute the term in the free energy.
A useful feature of the grand canonical ensemble is that, for non-interacting particles, the grand potential factorizes into a product of grand potentials for each single particle state, and so the grand potential is a sum of grand potentials for single particle states. Hence, for free electrons,
(56)
which can be conveniently be expressed in terms of the density of states, (for both spin species) by
(57)
Consider first . We are interested in the low- limit where, as discussed in class, the difference between and its limit, , is negligible. As discussed above is a constant . and so, for small and
(58) (59) (60)
where the last equality is from Eq. (14). For , where is negligible, is just with , and so
(61)
in agreement with Eq. (15), which gives the energy per electron rather than the total energy as here.
Now we add a field, which changes in two ways. Firstly it changes the density of states, as we have discussed in detail in the main part of the text. Secondly it changes the chemical potential, so . However, we shall see that the change in due to the change in with does not affect the free energy , so we just focus here on the change in due to the modification of the density of states.
In the presence of a field , the energy levels take the discrete values and so
(62)
which is to be contrasted with Eq. (A10) for . The difference is that Eq. (A14) can be though of as a discretized approximation to Eq. (A10) of the sort that is often used in numerical analysis. The integral over a range of width is replaced by times the value of the the integrand at the midpoint of the interval. The difference between the integral and the approximation to it using this midpoint rule'' is well known (see e.g. Numerical Recipes in C (or in Fortran) by Press et al, Eq. (4.4.1)), and can be easily derived by replacing the function in each interval by a polynomial, doing the integral with the first few terms of the polynomial, and summing over intervals. The result can be expressed as
(63)
where and . This approximation is good provided the function varies smoothly over a single interval of width . For the present problem, is replaced by , and the integrand varies rapidly on a scale of . Hence use of Eq. (A15) will be valid for . In this situation, many Landau levels are partially occupied and the oscillations found in the main part of the text are washed out. In our case the integral goes to infinity but both the function and its derivative vanish in this limit, and so, evaluating the derivative of the integrand at , we get, to order ,
(64)
where corresponds to the integral in Eq. (A15) and to the sum, and, from now on, we work per unit area. The last term in Eq. (A16) is the change in from , the change in the chemical potential due to the field, which is also of order .
Substituting Eq. (A16) into in Eq. (A4) gives
(65)
where, from Eq. (A3), we have noted that the contribution from cancels. This cancellation occurs because , as noted in Eq. (A5).
The magnetization is given by which yields with , the diamagnetic susceptibility of free electrons per unit area in two dimensions, given by
(66)
To get the corresponding result in 3- we need to add the motion in the -direction, specified by , and integrate from to . This is easy because Eq. (A18) is a constant independent of the (2-) density of electrons (which would vary with ) and hence the diamagnetic susceptibility of free electrons in 3- per unit volume is given by
(67)
This is the expression first found by Landau. It is equivalent to Eqs. (31.72) and (31.69) of Ashcroft and Mermin. | {"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": 0.9832538962364197, "perplexity": 248.33813459572542}, "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-2015-32/segments/1438042985140.15/warc/CC-MAIN-20150728002305-00208-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://mathhelpforum.com/differential-equations/193243-reducing-4-differential-equations.html | # Math Help - Reducing 4 differential equations
1. ## Reducing 4 differential equations
Hi,
I have been told that the following 4 equations:
reduce to:
My maths is simply not strong enough to work this through myself. So, I was hoping someone could take me through the steps required to make this reduction.
Any help is much appreciated | {"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": 0.8233320713043213, "perplexity": 754.3696466191399}, "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-2015-22/segments/1432207930256.3/warc/CC-MAIN-20150521113210-00189-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://asmedigitalcollection.asme.org/ICEF/proceedings-abstract/ICEF2011/44427/213/361323 | A laser 2-focus velocimeter (L2F) has been applied for measurements of velocity and size of droplets of diesel spray and an evaluation method of mass flow rate has been proposed. The L2F has a micro-scale probe which consists of two foci. The distance between two foci is 17μm. The data acquisition rate of the L2F has been increased to 15MHz in order to capture every droplet which appears in the measurement volume. The diesel fuel spray injected intermittently into the atmosphere was investigated. The orifice diameter of the injector nozzle was 0.113mm. The injection pressure was set at 40MPa by using a common rail system. Measurements were conducted on ten planes 5 to 25mm downstream from the nozzle exit. It was clearly shown that the velocity of droplet was the highest at the spray center. The size of droplet at the spray center decreased downstream within 15mm from the nozzle exit. The mass flow rate near the spray center was found to be larger than that in the spray periphery region. It was confirmed that the fuel mass per injection evaluated by the proposed method based on the L2F measurement was near to the injected mass in a plane further than 15mm from the nozzle exit. However, fuel mass was underestimated in a plane closer to the nozzle exit. The probability density of infinitesimal distance between surfaces of adjacent droplets increased remarkably near the spray center 5 and 12mm downstream from the nozzle exit. As infinitesimal distance can be thought as an indicator of a highly dense region, it is understood that underestimation of fuel mass near the nozzle exit is due to the highly dense region. The diameter of the region, where the highly dense region was observed, was estimated as an order of 0.2mm in a plane 5mm downstream from the nozzle.
This content is only available via PDF. | {"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": 0.9397315979003906, "perplexity": 902.3103383859677}, "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-2022-33/segments/1659882570901.18/warc/CC-MAIN-20220809033952-20220809063952-00037.warc.gz"} |
http://dlmf.nist.gov/15.11 | # §15.11 Riemann’s Differential Equation
## §15.11(i) Equations with Three Singularities
The importance of (15.10.1) is that any homogeneous linear differential equation of the second order with at most three distinct singularities, all regular, in the extended plane can be transformed into (15.10.1). The most general form is given by
with
Here , , are the exponent pairs at the points , , , respectively. Cases in which there are fewer than three singularities are included automatically by allowing the choice for exponent pairs. Also, if any of , , , is at infinity, then we take the corresponding limit in (15.11.1).
The complete set of solutions of (15.11.1) is denoted by Riemann’s -symbol:
15.11.3
In particular,
15.11.4
denotes the set of solutions of (15.10.1).
## §15.11(ii) Transformation Formulas
A conformal mapping of the extended complex plane onto itself has the form
where , , , are real or complex constants such that . These constants can be chosen to map any two sets of three distinct points and onto each other. Symbolically:
15.11.6
The reduction of a general homogeneous linear differential equation of the second order with at most three regular singularities to the hypergeometric differential equation is given by
15.11.7
We also have
15.11.8
for arbitrary and . | {"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": 0.9797263145446777, "perplexity": 571.7183363434578}, "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-2013-20/segments/1368706472050/warc/CC-MAIN-20130516121432-00011-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.fiberoptics4sale.com/blogs/wave-optics/polarization-of-light | # Polarization of Light
This part continues from the linear optical susceptibility tutorial.
Consider a monochromatic plane optical wave that has a complex field
$\tag{58}\mathbf{E}(\mathbf{r},t)=\pmb{\mathcal{E}}\exp(\text{i}\mathbf{k}\cdot\mathbf{r}-\text{i}\omega t)=\hat{e}\mathcal{E}\exp(\text{i}\mathbf{k}\cdot\mathbf{r}-\text{i}\omega t)$
where $$\pmb{\mathcal{E}}$$ is a constant independent of r and t, and $$\hat{e}$$ is its unit vector.
The polarization of the optical field is characterized by the unit vector $$\hat{e}$$.
The wave is linearly polarized, also called plane polarized, if $$\hat{e}$$ can be expressed as a constant, real vector. Otherwise, the wave is elliptically polarized in general, and is circularly polarized in some special cases.
For the convenience of discussion, we take the direction of the wave propagation to be the z direction so that $$\mathbf{k}=k\hat{z}$$ and assume that both E and H lie in the xy plane.
Note: this assumption is generally true if the medium is isotropic. It is not necessarily true if the medium is anisotropic. Propagation and polarization in isotropic and anisotropic media are discussed in the following two parts. However, the general concept discussed here does not depend on the validity of this assumption.
Then, we have
$\tag{59}\pmb{\mathcal{E}}=\hat{x}\mathcal{E}_x+\hat{y}\mathcal{E}_y=\hat{x}|\mathcal{E}_x|e^{\text{i}\varphi_x}+\hat{y}|\mathcal{E}_y|e^{\text{i}\varphi_y}$
where $$\mathcal{E}_x$$ and $$\mathcal{E}_y$$ are space- and time-independent complex amplitudes, with phases φx and φy, respectively.
The polarization of the wave depends only on the phase difference and the magnitude ratio between the two field components $$\mathcal{E}_x$$ and $$\mathcal{E}_y$$. It can be completely characterized by the following two parameters:
$\tag{60}\varphi=\varphi_y-\varphi_x,\qquad\qquad-\pi\lt\varphi\le\pi$
and
$\tag{61}\alpha=\tan^{-1}\frac{|\mathcal{E}_y|}{|\mathcal{E}_x|},\qquad\qquad0\le\alpha\le\frac{\pi}{2}$
Because only the relative phase φ matters, we can set φx = 0 and take $$\mathcal{E}$$ to be real in the following discussions. Then $$\pmb{\mathcal{E}}$$ from (59) can be written as
$\tag{62}\pmb{\mathcal{E}}=\mathcal{E}\hat{e},\qquad\text{with}\qquad\hat{e}=\hat{x}\cos\alpha+\hat{y}e^{\text{i}\varphi}\sin\alpha$
Using (39) from the harmonic fields tutorial, the space- and time-dependent real field is
$\tag{63}\pmb{E}(z,t)=2\mathcal{E}[\hat{x}\cos\alpha\cos(kz-\omega t)+\hat{y}\sin\alpha\cos(kz-\omega t+\varphi)]$
At a fixed z location, say z = 0, we see that the electric field varies with time as
$\tag{64}\pmb{E}(t)=2\mathcal{E}[\hat{x}\cos\alpha\cos\omega t+\hat{y}\sin\alpha\cos(\omega t-\varphi)]$
In general, $$\mathcal{E}_x$$ and $$\mathcal{E}_y$$ have different phases and different magnitudes. Therefore, the values of φ and α can be any combination. At a fixed point in space, both the direction and the magnitude of the field vector E in (64) can vary with time. Except when the values of φ and α fall into one of the special cases discussed below, the tip of the this vector generally describes an ellipse, and the wave is said to be elliptically polarized.
Note that we have assumed that the wave propagates in the positive z direction. When we view the ellipse by facing against this direction of wave propagation, we see that the tip of the field vector rotates counterclockwise, or left handedly, if φ > 0, and clockwise, or right handedly, if φ < 0.
Figure (4) below shows the ellipse traced by the tip of the rotating field vector at a fixed point in space. Also shown in the figure are the relevant parameters that characterize elliptic polarization.
In the description of the polarization characteristics of an optical wave, it is sometimes convenient to use, in place of α and φ, a set of two other parameters, θ and ε, which specify the orientation and ellipticity of the ellipse, respectively.
The orientationally parameter θ is the directional angle measured from the x axis to the major axis of the ellipse. Its range is taken to be 0 ≤ θ < π for convenience. Ellipticity ε is defined as
$\tag{65}\epsilon=\pm\tan^{-1}\frac{b}{a},\qquad-\frac{\pi}{4}\le\epsilon\le\frac{\pi}{4}$
where a and b are the major and minor semiaxes, respectively, of the ellipse.
The plus sign for ε > 0 is taken to correspond to φ > 0 for left handed polarization, whereas the minus sign for ε < 0 is taken to correspond to φ < 0 for right-handed polarization. The two sets of parameters (αφ) and (θε) have the following relations:
$\tag{66}\tan2\theta=\tan2\alpha\cos\varphi$$\tag{67}\sin2\epsilon=\sin2\alpha\sin\varphi$
Either set is sufficient to characterize the polarization state of an optical wave completely.
The following special cases are of particular interest.
#### 1. Linear Polarization.
This happens when φ = 0 or π for any value of α. It is also characterized by ε = 0, and θα if φ = 0 or θ = π - α if φ = π.
Clearly, the ratio $$\frac{\mathcal{E}_x}{\mathcal{E}_y}$$ is real in this case; therefore, linear polarization is described by a constant, real unit vector as
$\tag{68}\hat{e}=\hat{x}\cos\theta+\hat{y}\sin\theta$
It follows that E(t) described by (64) reduces to
$\tag{69}\pmb{E}(t)=2\mathcal{E}\hat{e}\cos\omega{t}$
The tip of this vector traces a line in space at an angle θ with respect to the x axis, as shown in figure (5) below.
#### 2. Circular Polarization.
This happens when φ = π/2 or - π/2, and απ/4. It is also characterized by ε = π/4 or - π/4, and θ = 0. Because α = π/4, we have $$|\mathcal{E}_x|=|\mathcal{E}_y|=\mathcal{E}/\sqrt2$$.
There are two different circular polarization states:
a. Left-circular polarization. For φ = π/2, also ε = π/4, the wave is left-circularly polarized if it propagates in the positive z direction. The complex field amplitude in (62) becomes
$\tag{70}\pmb{\mathcal{E}}=\mathcal{E}\frac{\hat{x}+\text{i}\hat{y}}{\sqrt2}=\mathcal{E}\hat{e}_+$
and E(t) described by (64) reduces to
$\tag{71}\pmb{E}(t)=\sqrt2\mathcal{E}(\hat{x}\cos\omega t+\hat{y}\sin\omega t)$
As we view against the direction of propagation $$\hat{z}$$, we see that the field vector E(t) rotates counterclockwise with an angular frequency ω. The tip of this vector describes a circle. This is shown in figure 6(a) below. This left-circular polarization is also called positive helicity. Its eigenvector is
$\tag{72}\hat{e}_+\equiv\frac{\hat{x}+\text{i}\hat{y}}{\sqrt2}$
b. Right-circular polarization. For φ = - π/2, also ε = - π/4, the wave is right-circularly polarized if it propagates in the positive z direction. We then have
$\tag{73}\pmb{\mathcal{E}}=\mathcal{E}\frac{\hat{x}-\text{i}\hat{y}}{\sqrt2}=\mathcal{E}\hat{e}_-$
and
$\tag{74}\pmb{E}(t)=\sqrt2\mathcal{E}(\hat{x}\cos\omega t-\hat{y}\sin\omega t)$
The tip of this field vector rotates clockwise in a circle, as shown in figure 6(b) below. This right-circular polarization is also called negative helicity. Its eigenvector is
$\tag{75}\hat{e}_-\equiv\frac{\hat{x}-\text{i}\hat{y}}{\sqrt2}$
As can be seen, neither $$\hat{e}_+$$ nor $$\hat{e}_-$$ is a real vector. Note that the identification of $$\hat{e}_+$$, defined in (72), with left-circular polarization and that of $$\hat{e}_-$$, defined in (75), with right-circular polarization are based on the assumption that the wave propagates in the positive z direction.
For a wave that propagates in the negative z direction, the handedness of these unit vectors changes: $$\hat{e}_+$$ becomes right-circular polarization, while $$\hat{e}_-$$ becomes left-circular polarization.
Linearly polarized light can be produced from unpolarized light using a polarizer. A polarizer can be of transmission type, which often utilizes the phenomenon of double refraction in an anisotropic crystal, or of reflection type, which takes advantage of the polarization-sensitive reflectivity of a surface.
A very convenient transmission-type polarizer is the Polaroid film, which utilizes a material with linear dichroism, having low absorption for light linearly polarized in a particular direction and high absorption for light polarized orthogonally to this direction. The output is linearly polarized in the direction defined by the polarizer irrespective of the polarization state of the input optical wave.
A polarizer can also be used to analyze the polarization of a particular optical wave. When so used, a polarizer is also called an analyzer.
The next part continues with the Propagation in an Isotropic Medium tutorial. | {"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": 0.8914133310317993, "perplexity": 498.39488775446426}, "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-2022-49/segments/1669446711712.26/warc/CC-MAIN-20221210042021-20221210072021-00296.warc.gz"} |
https://proofwiki.org/wiki/Group_has_Latin_Square_Property | # Group has Latin Square Property
## Theorem
Let $\struct {G, \circ}$ be a group.
Then $G$ satisfies the Latin square property.
That is, for all $a, b \in G$, there exists a unique $g \in G$ such that $a \circ g = b$.
Similarly, there exists a unique $h \in G$ such that $h \circ a = b$.
This result can also be written in additive notation as follows:
Let $\struct {G, +}$ be a group.
Then $G$ satisfies the Latin square property.
That is, for all $a, b \in G$, there exists a unique $g \in G$ such that $a + g = b$.
Similarly, there exists a unique $h \in G$ such that $h + a = b$.
### Corollary
The Cayley table for any finite group is a Latin square.
## Proof 1
$\displaystyle g$ $=$ $\displaystyle a^{-1} \circ b$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle a \circ g$ $=$ $\displaystyle a \circ \left({a^{-1} \circ b}\right)$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle a \circ g$ $=$ $\displaystyle \left({a \circ a^{-1} }\right) \circ b$ $\quad$ Group axioms: $G1$: associativity $\quad$ $\displaystyle \implies \ \$ $\displaystyle a \circ g$ $=$ $\displaystyle e \circ b$ $\quad$ Group axioms: $G3$: existence of inverse element $\quad$ $\displaystyle \implies \ \$ $\displaystyle a \circ g$ $=$ $\displaystyle b$ $\quad$ Group axioms: $G2$: existence of identity element $\quad$
Thus, such a $g$ exists.
Suppose $g, g' \in G$ where $a \circ g = b = a \circ g'$.
Then:
$\displaystyle g$ $=$ $\displaystyle e \circ g$ $\quad$ Group axioms: $G2$: existence of identity element $\quad$ $\displaystyle$ $=$ $\displaystyle \left({a^{-1} \circ a}\right) \circ g$ $\quad$ Group axioms: $G3$: existence of inverse element $\quad$ $\displaystyle$ $=$ $\displaystyle a^{-1} \circ \left({a \circ g}\right)$ $\quad$ Group axioms: $G1$: associativity $\quad$ $\displaystyle$ $=$ $\displaystyle a^{-1} \circ b$ $\quad$ Substitution for $a \circ g$ $\quad$ $\displaystyle$ $=$ $\displaystyle a^{-1} \circ \left({a \circ g'}\right)$ $\quad$ Substitution for $a \circ g'$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left({a^{-1} \circ a}\right) \circ g'$ $\quad$ Group axioms: $G1$: associativity $\quad$ $\displaystyle$ $=$ $\displaystyle e \circ g'$ $\quad$ Group axioms: $G3$: existence of inverse element $\quad$ $\displaystyle$ $=$ $\displaystyle g'$ $\quad$ Group axioms: $G2$: existence of identity element $\quad$
Thus uniqueness holds.
To prove the second part of the theorem, let $h = b \circ a^{-1}$.
The remainder of the proof follows a similar procedure to the above.
$\blacksquare$
## Proof 2
We shall prove that this is true for the first equation:
$\displaystyle a \circ g$ $=$ $\displaystyle b$ $\quad$ $\quad$ $\displaystyle \iff \ \$ $\displaystyle a^{-1} \circ \left({a \circ g}\right)$ $=$ $\displaystyle a^{-1} \circ b$ $\quad$ $\circ$ is a Cancellable Binary Operation $\quad$ $\displaystyle \iff \ \$ $\displaystyle \left({a^{-1} \circ a}\right) \circ g$ $=$ $\displaystyle a^{-1} \circ b$ $\quad$ Group axiom $G1$: Associativity $\quad$ $\displaystyle \iff \ \$ $\displaystyle e \circ g$ $=$ $\displaystyle a^{-1} \circ b$ $\quad$ Group axiom $G3$: property of Inverses $\quad$ $\displaystyle \iff \ \$ $\displaystyle g$ $=$ $\displaystyle a^{-1} \circ b$ $\quad$ Group axiom $G2$: property of Identity $\quad$
Because the statements:
$a \circ g = b$
and
$g = a^{-1} \circ b$
are equivalent, we may conclude that $g$ is indeed the only solution of the equation.
The proof that the unique solution of $h$ is $b \circ a^{-1}$ in the second equation follows similarly.
$\blacksquare$
## Proof 3
Suppose that $\exists x, y \in G: a \circ x = b = a \circ y$.
$\displaystyle a \circ x$ $=$ $\displaystyle a \circ y$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle a^{-1} \circ \paren {a \circ x}$ $=$ $\displaystyle a^{-1} \circ \paren {a \circ y}$ $\quad$ Group Axiom $G \, 3$: Inverses $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {a^{-1} \circ a} \circ x$ $=$ $\displaystyle \paren {a^{-1} \circ a} \circ y$ $\quad$ Group Axiom $G \, 1$: Associativity $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle e \circ x$ $=$ $\displaystyle e \circ y$ $\quad$ Group Axiom $G \, 3$: Inverses $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle y$ $\quad$ Group Axiom $G \, 2$: Identity $\quad$
So such an element, if it exists, is unique.
Now it is demonstrated that $g = a^{-1} b$ satisfies the requirement for $a \circ g = b$
Since $a \in G$, it follows by group axiom $G3$: existence of inverses that $a^{-1} \in G$.
$\displaystyle a$ $\in$ $\displaystyle G$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle a^{-1}$ $\in$ $\displaystyle G$ $\quad$ Group Axiom $G \, 3$: Inverses $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle a^{-1} \circ b$ $\in$ $\displaystyle G$ $\quad$ Group Axiom $G \, 0$: Closure $\quad$
Then:
$\displaystyle a \circ g$ $=$ $\displaystyle a \circ \paren {a^{-1} \circ b}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \paren {a \circ a^{-1} } \circ b$ $\quad$ Group Axiom $G \, 1$: Associativity $\quad$ $\displaystyle$ $=$ $\displaystyle e \circ b$ $\quad$ Group Axiom $G \, 3$: Inverses $\quad$ $\displaystyle$ $=$ $\displaystyle b$ $\quad$ Group Axiom $G \, 2$: Identity $\quad$
Thus, such a $g$ exists.
The properties of $h$ are proved similarly.
$\blacksquare$
## Proof 4
We shall prove that this is true for the first equation:
$\displaystyle b$ $=$ $\displaystyle a \circ g$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle a^{-1} \circ b$ $=$ $\displaystyle a^{-1} \circ \paren {a \circ g}$ $\quad$ Group Axiom $G \, 3$: Inverses $\quad$ $\displaystyle$ $=$ $\displaystyle \paren {a^{-1} \circ a} \circ g$ $\quad$ Group Axiom $G \, 1$: Associativity $\quad$ $\displaystyle$ $=$ $\displaystyle e \circ g$ $\quad$ Group Axiom $G \, 3$: Inverses $\quad$ $\displaystyle$ $=$ $\displaystyle g$ $\quad$ Group Axiom $G \, 2$: Identity $\quad$
Conversely:
$\displaystyle g$ $=$ $\displaystyle a^{-1} \circ b$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle a \circ g$ $=$ $\displaystyle a \circ \paren {a^{-1} \circ b}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \paren {a \circ a^{-1} } \circ b$ $\quad$ Group Axiom $G1$: Associativity $\quad$ $\displaystyle$ $=$ $\displaystyle e \circ b$ $\quad$ Group Axiom $G3$: Property of Inverse $\quad$ $\displaystyle$ $=$ $\displaystyle b$ $\quad$ Group Axiom $G2$: Property of Identity $\quad$
The proof that the unique solution of $h$ is $b \circ a^{-1}$ in the second equation follows similarly.
$\blacksquare$ | {"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": 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": 0.9913270473480225, "perplexity": 132.90080636988148}, "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/1550249530087.75/warc/CC-MAIN-20190223183059-20190223205059-00013.warc.gz"} |
https://bylosingallgenerality.wordpress.com/2010/06/ | ## Archive for June, 2010
### An elementary proof of the irrationality of Pi
June 29, 2010
I´ve been wondering for a while now if it exists and if so, how it looks like. By elementary I mean something like: completely (elementary) number theoretically. Because I know there are a few easy proofs that $\pi$ is not rational, but all of the ones I’ve seen, make use of some calculus (like computing an integral or so). So, my question: Can anyone show me the irrationality of $\pi$ with just the very basics of number theory or tell me why it is/shouldn’t be possible? For example, a few years ago I tried to prove it using the the formula: $\dfrac{\pi}{2} = \displaystyle \sum_{k=0}^\infty \dfrac{k!}{(2k + 1)!!}$ along with basic arguments which you may find in some proofs of the irrationality of $e = \displaystyle \sum_{k=0}^\infty \dfrac{1}{k!}$. Although (obviously) I wasn’t able to prove the desired result, I did not find a reason why it shouldn’t be possible in principle. Even more, I had the gut feeling I was almost there. So I might give it another shot it the future, but hopefully that doesn’t stop you, my dearest reader, from helping me if you’ve got some info on this.
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### I´m not dead
June 25, 2010
A lot of people (yes Jojo, that’s to ensure your anonimity), complained that I didn’t post enough. So to show my liveliness:
Q: What did the farmer say when he lost his tractor?
A: Where’s my tractor?
### Winning suffices. Apparently.
June 19, 2010
Beauty is the first test: there is no permanent place in this world for ugly football.
G. H. Hardy
— A Mathematician’s Apology, London, 1941
Maybe math-unrelated, but it is still rant-worthy material; our Orange.
I’m not sure if you watched them play against Japan today, but to give you a little summary; we sucked. Big time. While that was highly disappointing, what really frustrated me were the people on tv commenting on it. Basically, their story was:
‘Well, it was not very good, but we won and that’s all that matters. Even more, winning when you perform sucky is how you win a tournament. The Germans and Italians are kind of used to play defensively too, and they always do good at tournaments, while the times we played beautifully didn’t get us very far. So uglyness suffices and may be even preferable.’
Maybe it’s not necessary, but let me tell you why that’s all completely senseless.
I know that playing ugly is sometimes the rational thing to do. If you do not possess the qualities to play beautifully, or your opponent plays in such a way that attractive football is counterproductive, well, let’s do in it a different way then. But, and this is something that no-one seems to grasp, that doesn’t imply any of the above statements. Most importantly, playing ugly is totally different from playing utterly bad! It amazes me that everyone (unconsciously?) believes or acts like that they are equal, while I’m not even going to bother to explain why they’re not. Second of all, yes, by playing defensively we do look more like Italy. But thinking that that makes us more likely to win the world cup is clearly nonsense. Because we also look more like the team of Mozambique (which has very decent defenders*). Obviously more noteworthy, our qualities don’t lie with our defense. So it’s not a good plan to lean back on it, I guess. If you realise that the one thing we excel in, is attacking, you don’t have to be Einstein to realise that that’s the thing we should do! Another blatant fallacy is thinking that the fact that we won both matches so far is all that matters. Of course, you have to win on your off-days (and that’s what we’re usually quite bad at) and I’m more than happy with the results so far. But the fact that we won despite the fact we didn’t play well, doesn’t imply that we’re doing great. It just implies that we didn’t play well. So we’ve got every reason to fear the knock-out stages and should be lucky that we won by an own goal against Denmark and against Japan because of a keeper’s error! And in any case, even when some teams don’t ‘allow’ us to play attractive football, how on earth is Japan, with all due respect, one of these?
*I have no idea
### Nothing is impossible (?)
June 19, 2010
Something’s been bothering me for a while now. While my statistical intuition is usually quite sharp, in this case I’m wrong, although I never really heard a convincing argument why exactly. So I’ll try to explain my problem, and hopefully someone with a sound argument could reply. Basically, my confusion starts with the sentence ‘Some event that has a chance of $0$ of happening, can still happen.’ Let $P(x)$ be the chance that $x$ happens. To me $P(x) = 0$‘ is equivalent to the statement: ‘We are certain that $x$ will not be the case’. So saying that $P(x) = 0$ doesn’t exclude $x$ from happening feels just like a contradictio in terminis, to me. For example, the chance that you throw a (normal) die and a $7$ turns up*. But this is a silly example, because everybody agrees that the chance of this happening is $0$ and also that it is impossible. But now, consider the following: If we pick a random number, the chance we pick a certain number $N$ is clearly $0$. Since this holds for every $N$, you might argue that this implies I’m wrong, because apparently something happened which had a chance of $0$. To me this just implies that it’s not possible to choose a number randomly. A different example: we flip a coin until heads turns up. Is it possible that this game never ends? To me, it’s not. Of course, it could take arbitrarily long, but it can’t take an infinite number of flips. Can it?
While trying to emphatize with the people who believe that ‘impossible events are possible’, I came up with the following: let’s say we pick a random real number from the unit interval. Since there are uncountably many reals and only a countable number of rationals, the chance to pick a fraction is $0$. But you could still say: ‘well, it’s only random. And fractions actually do exist in abundance. So there is no obvious way that it shouldn’t be possible to choose a fraction by accident’. While this idea makes sense to some extent, I believe it’s wrongheaded. Mostly because, like I said above, this just seems to be based on the belief that it is actually possible to pick a number in some truly random way.
In any case, this is not meant as a convincing argument against the idea of the impossible becoming possible. I’m just trying to show where I get confused, and I am in need of some explanation. And after all, I’ve never seen anything in my life happen that had a zero chance of happening. So the burden of proof lies with the people who claim that $2$ is a random integer ;).
*maybe it actually could happen by some weird quantum effect, but mathematically speaking, it’s not an issue
PS. Terry Tao is going to set up another mini-Polymath Project next month. Check it out.
### All your linearly independent spanning sets are belong to us!
June 17, 2010
(for the ignorant) Yes yes, I own zhe matrices! At last, I, think I, passed my Lineair Algebra-exam. And since I’ve got 5 more exams to learn in the following week, (probably) no more updates until I finished them all. Bai bai
### Our world will never be the same
June 11, 2010
A steady loss of visitors since day 1. Not a single comment yet. Not even a nomination for a Bloggie. But, dont worry; BLAG is still going strong! With today´s one month anniversary as one of its many, many highs! There will be celebrations all over the world. In Johannesburg, South Africa, a football game will be played in honour of the weblog that changed the way people think and made world peace, finally, look like an achievable short-term goal. In Amsterdam, The Netherlands, people will write blogs about the blog that, arguably, is one of the greatest things that happened to human kind, since the invention of the bananabox.
Yes, people. Today, will be no longer the day of the Alcatraz escape. Nor will anyone ever remember June 11th to be the date Troy was burned. Starting this year, the eleventh day of the sixth month will be known as BLAGDAY! Spread the word!
### Only integers behave nicely
June 8, 2010
This week I solved the following PEN-problem:
Find out which pairs of real numbers $(\alpha, \beta)$ have the property that: $\alpha [\beta n] = \beta [\alpha n]$ for all natural numbers $n$, where $[x]$ denotes the largest integer $\le x$
At first glance, this problem looked quite easy to me. While I’m not sure if that’s a good sign in general, in this case it was advantageous. Because some problems that are stated in the form ‘find all x such that y’, are only solvable by providing a way of finding those x, where that way of finding, immediately shows that you’re right. This might be vague, so let me give you an example. Consider the problem of finding all primes less than 10000. I guess that no one will just write down the list of primes. No, probably everyone will just use the sieve of Eratosthenes. And not only will that sieve give you all primes below 10000 it is/should be immediately clear why it works and generates all these primes. So using the sieve of Eratosthenes gives the answer and proves youre right in one blow. While that’s awesome, in general it’s not that easy to straightforwardly find a solution that strong. Going back to the problem at hand, you may feel that it begs for an impossibility proof, except for trivial or ‘degenerate’ cases. In other words, we already ‘found’ our solution, we only have to prove it’s right. In still other words, we already know how or where to start (for example: assume a true non-degenerate case) and where to end (in contradiction). And knowing how a proof could reasonably look like, before you even started, is a big advantage, I think.
Ok, so far the preliminary remarks. To concretise the above part, the degenerate cases are:
1) $\alpha \beta = 0$
2) $\alpha, \beta \in Z$
3) $\alpha = \beta$
It’s a trivial matter to show that all these cases indeed solve $\alpha [\beta n] = \beta [\alpha n]$. Now, I kind of simplified things in the above paragraph. Because there is definitely not 1 way to prove we’re right that these cases are the only ones that solve the above equation. We could assume that 1), 2) and 3) all don’t hold and end up in contradiction, we could assume $0 < |\alpha| < |\beta|$ and try to prove $\alpha, \beta \in Z$, we could (since proving that $\alpha$ is integral iff $\beta$ is, is trivial), assume that $\alpha, \beta \notin Z$ and show that $\alpha = \beta$ must hold, etc. There are tons of distinct ways to end up where we want to end up. But, in a way, they’re all the same too, of course.
One of the first things I did was assuming that both $\alpha$ and $\beta$ are positive. This is something I do a lot, because, well, I’m just not that fond of negative numbers. Moreover, usually the general case quite easily simplifies to the positive case and in here I felt it was a natural thing to do. Although it turned out that we actually do lose generality, without an extra argument, by assuming positivity, it was still a useful pet problem, so let’s assume positivity for now anyway. The next thing I tried was dividing both sides by $\beta[\beta n]$ to get our first result: $\dfrac{\alpha}{\beta} = \dfrac{[\alpha n]}{[\beta n]}$ is rational! While this is not much, it accounts for something. This inspired me to do a few things; First of all I filled in $n = 1$ and found out for which real numbers this equation holds by letting: $\alpha = a + \gamma$, $\beta = b + \delta$ for integral $a,b$ such that $0 < \gamma$, $\delta < 1$. Second of all I noticed that this argument doesn’t work if $a = b = 0$, because then we just have $\dfrac{\alpha}{\beta} = \dfrac{0}{0}$, and there’s not much to conclude from that. So I tried to prove $ab \neq 0$, in which I succeeded (see below). Third of all, since we know that $\dfrac{\alpha}{\beta} \in Q$, it seems natural to (try to) show that $\alpha$ and $\beta$ can’t be distinct, rational, non-integral numbers. As it turns out, this is quite straightforward, so to solve our problem, the following suffices:
Let $\alpha,\beta$ be distinct, irrational numbers. Then $\alpha [\beta n]$ $\neq$ $\beta [\alpha n]$ for at least 1 positive integer $n$.
Like I said, proving that $\alpha, \beta > 1$ (while assuming positivity and distinctness), is possible, so here is its proof: Let $\alpha < \beta$. If $\alpha < 1$, $\beta < 1$ as well, otherwise we would have: $\displaystyle \frac{[\alpha]}{[\beta]} = 0 < \frac{\alpha}{\beta}$, which is a contradiction. So we may assume $0 < \alpha < \beta < 1$. Let $N$ be the smallest integer such that $\beta N > 1$. Then $\alpha N$ must be larger than 1 as well, because otherwise we would have the same contradiction as above. But we now have: $\displaystyle \frac{[\alpha N]}{[\beta N]} = \frac{1}{1} = 1 = \frac{\alpha}{\beta}$, which implies $\alpha = \beta$. So if $\alpha$ and $\beta$ are distinct and positive (although I hardly used the positivity condition), $\alpha [\beta n] = \beta [\alpha n]$ can’t hold for all natural numbers $n$.
Now that we showed that, we can definitely say we are on our way. A cool way to go further is by using induction. When you see an (moderately easy) equation that should hold for all natural numbers $n$, it’s often possible to show that it indeed holds by using induction on $n$. However, here it might be possible to go the other way around, by using induction on $\alpha$! More precisely; assume that $\alpha [\beta n] \neq \beta [\alpha n]$ for some natural number $n$. Show that a natural number $m$ exists for which $(\alpha + 1) [\beta m] \neq \beta [(\alpha + 1) m]$ (except degenerate cases). And since we proved our base case ($0 < \alpha, \beta < 1$), this suffices. I must say I think this way of attacking our problem is both weirdly awesome and awesomely weird. But when I first tried it, the computations didn’t make any sense, so I chose a different approach. Although I might give it some thought again, since I have a better understanding of the problem now. Anyway, for me this wasn’t the way to go, so let’s try something else.
The good part is: we didn’t use the, to me, most obvious way of approaching this problem, extensively yet. So let’s do that. I believe that, whenever you’re dealing with the ‘greatest integer smaller than’-function, it’s usually a great idea to split that what’s inside the gist-function in two parts: an integer part and a rest part. For example: $[9.87] = [9 + .87] = 9 + [.87]$. So let’s, like I said above briefly, assume $\alpha = a + \gamma, \beta = b + \delta$ with $a, b \in Z$ and $0 < \gamma, \beta < 1$. Then we have:
$\alpha [\beta n] = \beta [\alpha n]$
$(a + \gamma) [(b + \delta) n] = (b + \delta) [(a + \gamma) n]$
$(a + \gamma)(bn + [\delta n]) = (b + \delta)(an + [\gamma n])$
$(a + \gamma) [\delta n] + \gamma bn = (b + \delta) [\gamma n] + \delta an$
$\alpha [\delta n] + \gamma bn = \beta [\gamma n] + \delta an$
We know that this must hold for all $n \in \aleph$. In particular, it should hold for $n = 1$. When $n = 1$, the above equation simplifies to:
$\gamma b = \delta a$
So we know that this must be the case. But then $\gamma bn$ and $\delta an$ always cancel each other out, so the equation becomes:
$\alpha [\delta n] = \beta [\gamma n ]$
And the great news is that we basically solved this equation already, when we showed the impossibility of distinct, positive integers $\alpha, \beta$ such that $\alpha [\beta n] = \beta [\alpha n]$ for $\alpha$ and $\beta$ smaller than 1. But, for completeness’ sake; let $N$ be the smallest positive integer such that $\max(\gamma N, \delta N) \ge 1$. Then $\min(\gamma N, \delta N) \ge 1$ as well, because otherwise we would have $0$ on one side of the equation while the other side is non-zero. Since our equation is true for all $n \in \aleph$, it must be true for $N$ too. And for this $N$, it further simplifies to:
$\alpha = \beta$
And we’re done.
### Periodicity, Part 1
June 5, 2010
Today (and, as my title sort of implies, later as well), we’re going to talk about 2 functions and find out if and when they are periodic. And since the data I have available, makes me believe that my thoughts on these issues are ‘different’, in the sense that most people would solve these problems in a rather different way, it would be cool to have some feedback. So if you think that I’m ‘wrong’, because an easier approach is available, do tell. I will use the notation that the n-th iteration of a function is $f^n(x)$, which, recursively defined, is $f(f^{n-1}(x))$, where $f^0(x) = x$. If the initial value of the function doesn’t really depend on x, as is true in the function I’ll deal with today, I will just type $f^n$ for the n- th iteration. The two functions I’ll talk about, are the following:
1) $f(x) = \displaystyle \frac{2 + x}{1 - 2x}$, with initial value 2.
2) $f(x) = \displaystyle 1 - \left|1 - 2x\right|$, with initial value $x \in [0, 1]$
And because today I’ll only be talking about the first $f(x)$, there will not be any confusion, hopefully. The problem to prove is: $f^n(x)$ won’t become periodic from some point on. This, btw, is a problem from the PEN-book. Also, since it’s been a while that I thought about this problem, I can’t really recall the things I tried, but didn’t work. So I’ll just present my solution of it in a slightly more formal manner than previous posts. Still, I’ll try to keep things really basic, so that might mean that you’ll encounter stuff that seems trivial. You can then just safely skip those parts.
Theorem. $f^a = f^b$ implies $a = b$
Proof of Theorem. To proof our Theorem, we first need to show that there is no (nonnegative) integer n for which $f^n$ $= 0$. To proof this last assumption, we need a few lemma’s.
Lemma 1. $f^n \in$ $Q$ for all $n \in \aleph$. (Q is the set of rational numbers)
Proof of Lemma 1. Note that $f^0 = 2 \in Q$. We proceed via induction. Assume $f^k = \dfrac{p}{q}$ for some $k \ge 0$. Let $p' = 2q + p, q' = q - 2p.$ Then:
$f^{k+1} = f(f^k)$
$= \dfrac{2 + p/q}{1 - 2p/q}$
$= \dfrac{2q + p}{q - 2p}$
$= \dfrac{p'}{q'}$
Lemma 2. f is injective
Proof of Lemma 2. Assume $f(a) = f(b)$. Then:
$\displaystyle \frac{2 + a}{1 - 2a} = \frac{2 + b}{1 - 2b}$
$(2 + a)*(1 - 2b) = (2 + b)*(1 - 2a)$
$2 - 4b + a - 2ab = 2 - 4a + b - 2ab$
$5a = 5b$
$a = b$
Lemma 3. Since we know that f is injective, f must have an inverse. This inverse equals: $f^{-1}(x) = \dfrac{x - 2}{1 + 2x}$
Proof of Lemma 3. Let $\dfrac{x - 2}{1 + 2x} =y$. Then:
$x - 2 = y + 2xy$
$x - 2xy = 2 + y$
$x(1 - 2y) = 2 + y$
$x = \dfrac{2 + y}{1 - 2y}$
$= f(y)$
Note that we can use this inverse function to calculate $f^n$, provided we know the value of $f^{n+1}$. So assume k to be the smallest (nonnegative) integer for which $f^{k+1} = f(f^k) = 0$. Then: $f^k = f^{-1}(f^{k+1})) = f^{-1}(0) = -2 = -f^0.$
We’ll generalize this last equation using induction. Assume that, for some i with: $k \ge i \ge 0, f^{k-i} = -f^i$. Then:
$f^{k-(i+1)} = f^{-1}(f(f^{k-(i+1)}))$
$= f^{-1}(f^{k-i})$
$= \dfrac{f^{k-i} - 2}{1 + 2f^{k-i}}$
$= \dfrac{-f^{i} - 2}{1 - 2f^{i}}$
$= -\dfrac{2 + f^{i}}{1 - 2f^i}$
$= -f^{i+1}$
So we may conclude that, for every i with $k \ge i \ge 0, f^{k-i} = -f^i$. Now we have to consider 2 cases separately; whether k is even or odd;
Case 1) k = 2m. Then we must have $f^m = -f^m$, which implies that $f^m = 0$ and this obviously contradicts the definition of k.
Case 2) k = 2m+1. Then we must have:
$f^m = -f^{m+1}$
$= -f(f^m)$
$= -\dfrac{2 + f^m}{1 - 2f^m}$
$f^m(1 - 2f^m) = -(2 + f^m)$
$f^m - 2(f^m)^2 = - 2 - f^m$
$2(f^m)^2 - 2f^m - 2 = 0$
$f^m = \dfrac{1 \pm \sqrt{5}}{2} \notin Q$
So we are forced to conclude that there is no (nonnegative) integer k satisfying $f^{k+1} = 0$. We may now proceed to the proof of our Theorem. For this part of the proof we assume k to be the smallest (nonnegative) integer for which a different integer k’ exists such that: $f^k = f^{k'}$. Then we must have $k = 0$, because otherwise we would have: $f^{k-1} = f^{-1}(f^k) = f^{-1}(f^{k'}) = f^{k'-1}$, which clearly contradicts our (new) definition of k. So $f^{k'} = f^k = f^0 = 2$. But this too leads to a contradiction if we apply our inverse function one more time;
$f^{k'-1} = f^{-1}(f^{k'})$
$= f^{-1}(2)$
$= 0$
And we already proved that this equation is not solvable for $k' \in \aleph$. Thus we must conclude that there are no $2$ different integers a and b for which: $f^a = f^b$. In other words: $f^a = f^b$ implies $a = b$.
### Self Delusion
June 2, 2010
After I (think I) solved an interesting problem that I found online, say on a forum, I usually go back to that forum to see other people´s solutions. To check if I ´forgot´ some ideas and could´ve solved the problem differently, or how people jot down their solutions if they had similar ideas, etc. What I find remarkable, but maybe that´s my bad, is that it often happens that I have to conclude that my solution is just wrong. Although my ideas almost always coincide with (some of) the ´right´ ideas, I just don’t seem to be able to make things rigorous (although I think I did) or I made some stupid mistake in a standard calculation or so. Apparently I’m extremely sloppy on the details. And that would not be that big of a problem if I would realise that and would be able to fix it by thinking very hard about every step I take in a proof to make sure I didn’t make any mistakes. But the catch is, I think I do just that. But I guess that, when I have an idea that could work, I start to obsess with that idea so much, that I stop being objective and start thinking I’m right about anything related to that idea. And this is bad. Really bad.
On a different, but maybe not totally unrelated note, I love to work with actual numbers. So with, say, 1233 (which btw equals $12^2 + 33^2$), instead of a’s, b’s or x’s. Although I think it’s not directly a problem if I, for example, first calculate some values of a function to see how fast it grows, the more general issue of only being able to solve a problem unless it’s really specific and not thinking about generalizations unless they’re obvious, might be a problem.
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https://www.arxiv-vanity.com/papers/quant-ph/0508144/ | # Quantum measurement of a mesoscopic spin ensemble
G. Giedke, J. M. Taylor, D. D’Alessandro, M. D. Lukin, and A. Imamoğlu Institut für Quantenelektronik, ETH Zürich, Wolfgang-Pauli-Straße 16, 8093 Zürich, Switzerland
Department of Physics, Harvard University, Cambridge, MA 02138, USA
Department of Mathematics, Iowa State University, Ames, IA 50011, USA
Max-Planck–Institut für Quantenoptik, H.-Kopfermann-Str., 85748 Garching, Germany
###### Abstract
We describe a method for precise estimation of the polarization of a mesoscopic spin ensemble by using its coupling to a single two-level system. Our approach requires a minimal number of measurements on the two-level system for a given measurement precision. We consider the application of this method to the case of nuclear spin ensemble defined by a single electron-charged quantum dot: we show that decreasing the electron spin dephasing due to nuclei and increasing the fidelity of nuclear-spin-based quantum memory could be within the reach of present day experiments.
###### pacs:
03.67.Lx, 71.70.Jp, 73.21.La, 76.70.-r
## I Introduction
Decoherence of quantum systems induced by interactions with low-frequency reservoirs is endemic in solid-state quantum information processing (QIP) Johnson et al. (2005); Ithier et al. (2005). A frequently encountered scenario is the coupling of a two-level system (qubit) to a mesoscopic bath of two-level systems such as defects or background spins. The manifestly non-Markovian nature of system-reservoir coupling in this scenario presents challenges for the description of the long term dynamics as well as for fault tolerant quantum error correction Alicki et al. (2002); Terhal and Burkard (2005). The primary experimental signature of a low-frequency reservoir is an unknown but slowly changing effective field that can substantially reduce the ability to predict the system dynamics. A possible strategy to mitigate this effect is to carry out a quantum measurement which allows for an estimation of the unknown reservoir field by controlled manipulation and measurement of the qubit. A precise estimation of the field acting on the large Hilbert space of the reservoir requires, however, many repetitions of the procedure: this constitutes a major limitation since in almost all cases of interest projective measurements on the qubit are slow Elzerman et al. (2004) and in turn will limit the accuracy of the estimation that can be achieved before the reservoir field changes.
In this work, we propose a method for estimating an unknown quantum field associated with a mesoscopic spin ensemble. By using an incoherent version of the quantum phase estimation algorithm, Kitaev (1995); Nielsen and Chuang (2000) we show that the number of qubit measurements scale linearly with the number of significant digits of the estimation. We only assume the availability of single qubit operations such as preparation of a known qubit state, rotations in the -plane, and measurement, of which only rotations need to be fast. The estimation procedure that we describe would suppress the dephasing of the qubit induced by the reservoir; indeed, an interaction with the estimated field leads to coherent unitary evolution that could be used for quantum control of the qubit. If the measurement of the reservoir observable is sufficiently fast and strong, it may in turn suppress the free evolution of the reservoir in a way that is reminiscent of a quantum Zeno effect.
After presenting a detailed description of the measurement procedure and discussing its performance and limitations, we focus on a specific application of the procedure for the case of a single quantum dot (QD) electron spin interacting with the mesoscopic nuclear spin ensemble defined by the QD. It is by now well known that the major source of decoherence for the electron-spin qubits in QDs Loss and DiVincenzo (1998) is the hyperfine interaction between the spins of the lattice nuclei and the electron Khaetskii et al. (2002); Merkulov et al. (2002); Schliemann et al. (2003); Coish and Loss (2004); Erlingsson and Nazarov (2004); de Sousa et al. (2005); Deng and Hu (2005). A particular feature of the hyperfine-related dephasing is the long correlation time () associated with nuclear spins. This enables techniques such as spin-echo to greatly suppress the dephasing Petta et al. (2005). In Coish and Loss (2004) it was suggested to measure the nuclear field to reduce electron spin decoherence times; precise knowledge of the instantaneous value of the field would even allow for controlled unitary operations. For example, knowledge of the field in adjacent QDs yields an effective field gradient that could be used in recently proposed quantum computing approaches with pairs of electron spins Taylor et al. (2005). Moreover, with sufficient control, the collective spin of the nuclei in a QD may be used as a highly coherent qubit-implementation in its own right Taylor et al. (2003a, b, 2004).
## Ii Phase estimation
In the following we consider an indirect measurement scheme in which the system under investigation is brought into interaction with a probe spin (a two-level system in our case) in a suitably prepared state. Measuring the probe spin after a given interaction time yields information about the state of the system. We assume the mesoscopic system evolves only slowly compared to the procedure, and further that the measurement does not directly perturb the system. In essence, we are performing a series of quantum non-demolition (QND) measurements on the system with the probe spin.
We consider an interaction Hamiltonian of the form
Hint=ℏAz⊗Sz (1)
which lends itself easily to a measurement of the observable . The QND requirement is satisfied for . The applicability of in situations of physical interest is discussed in Sec. VI. Given this interaction, the strategy to measure is in close analogy to the so-called Ramsey interferometry approach, which we now briefly review.
For example, an atomic transition has a fixed, scalar value for which corresponds to the transition frequency. By measuring as well as possible in a given time period, the measurement apparatus can be locked to the fixed value, as happens in atomic clocks. The probe spin is prepared in a state . It will undergo evolution under according to . After an interaction time , the probe spin’s state will be
cos(Ωt)|+⟩+isin(Ωt)|−⟩ (2)
where is the precession frequency for the probe spin. A measurement of the spin in the basis yields a probability of being in the state. Accumulating the results of many such measurements allows one to estimate the value for (and therefore ). In general, the best estimate is limited by interaction time: for an expected uncertainty in of and an appropriate choice of , measurements with fixed interaction times can estimate to no better than (see Wineland et al. (1992) and references therein).
In our scenario, the situation is slightly different in that is now a quantum variable. For a state in the Hilbert space of the system which is an eigenstate of with eigenvalue , the coupling induces oscillations:
Ut|s⟩|+⟩=|s⟩[cos(Ωst)|+⟩+isin(Ωst)|−⟩] . (3)
Thus, the probability to measure the probe spin in state given that the system is in a state is at time , providing information about which eigenvalue of is realized. Comparing Eq. (2) to Eq. (3) indicates that the same techniques used in atomic clocks (Ramsey interferometry) could be used in this scenario to measure and thus project the bath in some eigenstate of with an eigenvalue of to within the uncertainty of the measurement.
Beyond the Ramsey approach, there are several ways to extract this information, which differ in the choice of interaction times and the subsequent measurements. The general results on quantum metrology of Giovannetti et al. (2006) show, however, that the standard Ramsey scheme with fixed interaction time is already optimal in that the scaling of the final variance with the inverse of the total interaction time cannot be improved without using entangled probe states. Nevertheless, the Ramsey scheme will not be the most suitable in all circumstances. For example, we have assumed so far that preparation and measurement of the probe spin is fast when compared to . However, in many situations with single quantum systems, this assumption is no longer true, and it then becomes desirable to minimize the number of preparation/measurement steps in the scheme.
## Iii The measurement scheme
We now show that by varying the interaction time and the final measurements such that each step yields the maximum information about , we can obtain the same accuracy as standard Ramsey techniques with a similar interaction time, but only a logarithmic number of probe spin preparations and measurements. As a trivial case, if had eigenvalues and only, then measuring the probe in the -basis after an interaction time , we find with certainty, if the system are in an -eigenstate; if they were initially in a superposition, measuring the probe projects the system to the corresponding eigenspaces. We can extend this simple example (in the spirit of the quantum phase estimation algorithm Kitaev (1995); Nielsen and Chuang (2000) and its application to the measurement of a classical field Vaidman and Mitrani (2004)) to implement an -measurement by successively determining the binary digits of the eigenvalue. We start with the ideal case, then generalize to a more realistic scenario.
### iii.1 Ideal case
If all the eigenvalues of are an integer multiple of some known number and bounded by , then this procedure yields a perfect -measurement in steps: let us write all eigenvalues as . The sum we denote by and also use the notation . Starting now with an interaction time , we have . Hence the state of the probe electron is flipped if and only if . Therefore measuring the probe electron in state projects the nuclei to the subspace of even(odd) multiples of (see Fig. 1). We denote the result of the first measurement by if the outcome was “”. All the higher digits have no effect on the measurement result since they induce rotations by an integer multiple of which have no effect on the probabilities .
To measure the higher digits, we reduce the interaction time by half in each subsequent step: until we reach in the final and shortest step. For the rotation angle (mod ) in the th step does not only depend on the th binary digit of but also on the previous digits (which have already been measured, giving results ). The angle (mod ) is given by with , where we have used the results already obtained. This over-rotation by the angle can be taken into account in the choice of the measurement basis for the th step: if the th measurement is performed in a rotated basis that is determined by the previous results , namely
∣∣+j⟩ := cosφj|+⟩−isinφj|−⟩, (4a) ∣∣−j⟩ := sinφj|+⟩+icosφj|−⟩, (4b)
then the th measurement yields “” () if and “” () otherwise. Thus, after measurements we obtain and have performed a complete measurement of (where the number of probe particles used is the smallest integer such that ).
Before proceeding, we note that the proposed scheme is nothing but an “incoherent” implementation of the quantum phase estimation algorithm: As originally proposed, this algorithm allows measurement of the eigenvalue of a unitary by preparing qubits (the control-register) in the state (i.e., the equal superposition of all computational basis states ) and performing controlled- gates between the th qubit and an additional register prepared in an eigenstate of with . The controlled- gates let each computational basis state acquire a -dependent phase: . Then the inverse quantum Fourier transformation (QFT) is performed on the control register, which is then measured in the computational basis, yielding the binary digits of . Performing the QFT is still a forbidding task, but not necessary here: the sequence of measurements in the rotated basis described above is in fact an implementation of the combination of QFT and measurement into one step. This was previously suggested in different contexts Griffiths and Niu (1996); Parker and Plenio (2000); Tomita and Nakamura (2004).
### iii.2 Realistic case
In general, there is no known such that all eigenvalues of are integer multiples of . Nevertheless, as discussed below, the above procedure can still produce a very accurate measurement of if sufficiently many digits are measured. Now we evaluate the performance of the proposed measurement scheme in the realistic case of non-integer eigenvalues. Since here we are interested in the fundamental limits of the scheme, we will for now assume all operations on the probe qubit (state preparation, measurement, and timing) to be exact; the effect of these imperfection is considered in Sec. V. Without loss of generality, let and denote the largest and smallest eigenvalues of , respectively 111 In practice one may want to make use of prior knowledge about the state of the system to reduce the interval of possible eigenvalues that need to be sampled. Hence may be understood as an effective maximal eigenvalue given, e.g., by the expectation value of plus standard deviations. The values outside this range will not be measured correctly by the schemes discussed, but we assume to be chosen sufficiently large for this effect to be smaller than other uncertainties. and choose such that the eigenvalues of are all . These are the eigenvalues we measure in the following.
The function from which all relevant properties of our strategy can be calculated is the conditional probability to obtain (after measuring electrons) a result given that the system was prepared in an eigenstate with eigenvalue . The probability to measure is given by the product of the probabilities to measure in the th step, which is . Hence
pM(R|s)=M−1∏k=0cos2(π[s−R]2k), (5)
see also Vaidman and Mitrani (2004). This formula can be simplified by repeatedly using to give
pM(R|s)=(sin(2Mπ[s−R])2Msin(π[s−R]))2. (6)
Assume the nuclei are initially prepared in a state with prior probability to find them in the eigenspace belonging to the eigenvalue . After the measurement, we can update this distribution given our measurement result. We obtain, according to Bayes’ formula:
pM(s|R)=pM(R|s)p(s)∑spM(R|s)p(s), (7)
with expectation value denoted by .
## Iv Performance of the scheme
As the figure of merit for the performance of the measurement scheme we take the improvement of the average uncertainty in of the updated distribution
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ΔAz,est(M):=∑RpM(R)√∑s(s−¯sR)2pM(s|R) (8)
over the initial uncertainty . An upper bound to is given by the square root of the average variance
¯¯¯¯VM:=∑RpM(R)∑s(s−¯sR)2pM(s|R), (9)
as easily checked by the Cauchy-Schwarz inequality. We now show that . We replace ; we can use any such replacement to obtain an upper bound, as the expectation value minimizes . This choice means that measurement results are interpreted as , which is appropriate since the scheme does not distinguish the numbers and and due to the choice of only occur. Thus
¯¯¯¯VM ≤
The terms can be shown222For this we make use of Eq. (6) and (for ) to bound all terms of the sum over . to be with . This means that performing measurements yields a state with -uncertainty . For example, we need about interactions with the probe spin to reach the -level in and about more for every additional factor of .
The overall procedure requires a total time , which is an interaction time (determined mainly by the time needed for the least significant digit probed) and the time to make measurements ( is the time to make a single measurement). We obtain for the average uncertainty an upper bound in terms of the interaction time needed:
ΔAz,estΔ0≤√πfΔ0Tint. (10)
Immediately the similarity with standard atomic clock approaches is apparent, as the uncertainty decreases with the square root of the interaction time. However, while for an atomic clock scheme, in which the interaction time per measurement is kept fixed to , the total time to reach the precision of Eq. (10) is . For our method the measurement time is reduced dramatically by a time . In this manner our approach requires a polynomial, rather than exponential, number of measurements for a given accuracy, though the overall interaction time is the same for both techniques.
It may be remarked that even the scaling in interaction time differs significantly if other figures of merit are considered. For example, our scheme provides a square-root speed-up in over the standard Ramsey scheme if the aim is to maximize the information gain or to minimize the confidence interval Masanes et al. (2002).
## V Errors and Fluctuations in Az
Up until now we have considered an idealized situation in which the value of does not change over the course of the measurement and in which preparation and measurement of the probe system work with unit fidelity. Let us now investigate the robustness of our scheme in the presence of these errors.
### v.1 Preparation and Measurement Errors
By relying upon a small number of measurements, the scheme we described becomes more susceptible to preparation and measurement errors. An error in the determination of the th digit leads to an increase of the error probability in the subsequent digits. This error amplification leads to a scaling of the final error of , where is the probability of incurring a preparation or measurement error in a single step. We confirm this with Monte Carlo simulations of the measurement procedure (Fig. 2a), leading to an asymptotic bound:
ΔAz,est≤(2fΔ0)√p. (11)
Standard error correction (EC) techniques can be used to overcome this problem. E.g., by performing three measurements for each digit and using majority vote, the effective probability of error can be reduced to – at the expense of tripling the interaction time and number of measurements. While this may look like a big overhead, it should be noted that the scheme can be significantly improved: the least significant digits do not require any EC. For them, the scheme gives noisy results even for error rate due to the undetermined digits of , this does not affect the most significant digits. This indicates that it may be enough to apply EC for the leading digits.
As can be seen from Fig. 2b, this simple EC strategy provides a significant improvement in the asymptotic . This is hardly changed, when EC is applied only to the leading half of the digits. Thus only twice as many measurements (and an additional interaction time which is ) is needed for an order-of-magnitude improvement in . By repeating the measurement of more important digits even more often, the effect of technical errors can be reduced even further, as confirmed by Monte Carlo simulations (Fig. 2b). We also note that further improvements (beyond digits) can be achieved by this technique. In essence, choosing a digital approach to error correction for our digital technique yields substantially better performance than adapting the digital technique to an analog approach.
### v.2 Estimation of bath decorrelation errors
In practice, internal bath dynamics will lead to fluctuations in , such that for times and that are sufficiently different. Furthermore, apparatus errors, as outlined above, lead to errors in our measurement procedure. We will assume that the variations of are slow over short time intervals, allowing us to approximate the bit measurement process as a continuous measurement over the time with some additional noise with variance . Then we will find the expected difference in our measurement result and the value of at a later time.
Under the above approximations the value of the th such measurement (where a complete set of bits takes a time and the th such measurement ends at time ) is
mk=1TM∫tktk−TMAz(t)dt+Gk (12)
where the noise from measurement is incorporated in the stochastic noise variable with . We can estimate at a later time, and find the variance of this estimate from the actual value:
¯VM(t>tk) = ⟨[mk−Az(t)]2⟩ = ⟨G2k⟩+Δ20+12T2M∫tktk−TM∫tktk−TM⟨{Az(t′),Az(t′′)}+⟩dt′′dt′−1TM∫tktk−TM⟨{Az(t′),Az(t)}+⟩dt′
If we assume is a Gaussian variable with zero mean, described by a spectral function (i.e., ), then
¯VM(t)=ΔAz,est(M)2+Δ20+1T2M∫∞−∞S(ω)sin2(TMω/2)(ω/2)2dω−∫∞−∞S(ω)sin[(t−tk+TM)ω]−sin[(t−tk)ω]TMω/2dω
For that fluctuates slowly in time and corresponds to a non-Markovian, low frequency noise, the second moment of converges. We define:
1t2c=1⟨A2z⟩∫∞−∞S(ω)ω2dω . (13)
When , we may expand the sine terms in the integrals. Taking , the expected variance to order is
¯VM(t)≈ΔAz,est(M)2+Δ20[TMtc]2(73−112) (14)
As an example case, we consider as realistic parameters ms and s with . These parameter choices are described in detail in Sec. VI. We find that our variance s after the measurement is approximately with equal contributions from the measurement noise and from the bath decorrelation. Substantially faster decorrelation would dominate the noise in the estimate, and render our technique unusable.
In the limit of slow decorrelation, this approach would allow one to use the (random) field to perform a controlled unitary of the form at a time , with a fidelity
F=exp(−⟨(∫tt−τAz(t′)dt′−mk)2⟩/4) (15)
For example, a rotation around the probe spins’ axis would have a fidelity , or 0.998 for the above parameters.
We remark that this approach for estimation in the presence of bath fluctuations is not optimal (Kalman filtering Kalman (1960) would be more appropriate for making an estimation of using the measurement results). Furthermore, it does not account for the non-linear aspects of our measurement procedure, nor does it incorporate any effect of the measurement on the evolution of the bath (e.g., quantum Zeno effect). More detailed investigations of these aspects of the process should be considered in an optimal control setting. Nonetheless, our simple analysis above indicates that slow decorrelation of the bath will lead to modest additional error in the estimate of .
## Vi Example: Estimating collective nuclear spin in a quantum dot
Now we apply these general results to the problem of estimating the collective spin of the lattice nuclei in a QD.
The interaction of a single electron spin in a QD with the spins of the lattice nuclei is described by the Fermi contact term Schliemann et al. (2003)
→S⋅∑jαj→Ij, (16)
where the sum in Eq. (16) runs over all the lattice nuclei. The are constants describing the coupling of the th nuclear spin with the electron. They are proportional to the modulus squared of the electron wave function at the location of the th nucleus and are normalized such that , which denotes the hyperfine coupling strength.
Due to the small size of the nuclear Zeeman energies, the nuclei are typically in a highly mixed state even at dilution refrigerator temperatures. This implies that the electron experiences an effective magnetic field (Overhauser field, ) with large variance, reducing the fidelity of quantum memory and quantum gates. This reduction arises both from the inhomogeneous nature of the field ( varies from dot to dot) Lee et al. (2005) and the variation of over time due to nuclear-spin dynamics (even a single electron experiences different field strengths over time, implying loss of fidelity due to time-ensemble averaging).
In a large external magnetic field in the -direction the spin flips described by the and terms are suppressed and – in the interaction picture and the rotating wave approximation – the relevant Hamiltonian is of the type given in Eq. (1), where is now the collective nuclear spin operator
Az=N∑j=1αjI(j)z, (17)
which gives the projection of the Overhauser field along the external field axis by . Before continuing, let us remark here, that one can expect to obtain an effective coupling of the type Eq. (1) in a similar fashion as a good approximation to a general spin-environment coupling , whenever the computational basis states of the qubit are non-degenerate (as guaranteed in the system studied here by the external field) and the coupling to the environment is sufficiently weak such that bit-flip errors are detuned.
To realize the single-spin operations needed for our protocol – preparation, rotation, and read-out – many approaches have been suggested as part of a quantum computing implementation with electron spin qubits in QDs using either electrical or optical control (see, e.g., Cerletti et al. (2005) for a recent review).
The experimental progress towards coherent single spin manipulation has been remarkable in recent years. In particular, the kind of operations needed for our protocol have already been implemented in different settings: For self-assembled dots, state preparation with has been realized Atatüre et al. (2006), while for electrically defined dots, single-spin measurement with a fidelity of was reported Hanson et al. (2005). In the double-dot setting Petta et al. (2005), all three operations have recently been demonstrated, and we estimate the combined fidelity to be .
As can be seen from Fig. 2, at the level of accuracy of state preparation, rotation and read-out, the proposed nuclear spin measurement should be realizable. As discussed in many specific proposals Cerletti et al. (2005) these error rates appear attainable in both the transport and the optical setting. Apart from single qubit operations, our proposal also requires precise control of the interaction time. Fast arbitrary wave form generators used in the double-dot experiments, have time resolutions better than 30 ps333J. R. Petta, Private communication and minimum step sizes of 200 ps, which translates into errors of a few percent in estimating with initial uncertainties of order 1 ns. Uncertainties of this order are expected for large QDs () even if they are unpolarized and for smaller ones at correspondingly higher polarization (see below).
For GaAs and InAs QDs in the single electron regime, and . The uncertainty determines the inhomogeneous dephasing time Coish and Loss (2004). Especially at low polarization , this uncertainty is large , and without correction leads to fast inhomogeneous dephasing of electron spin qubits: ns has been observed Bracker et al. (2005); Johnson et al. (2005); Koppens et al. (2005). However, as is slowly varying Merkulov et al. (2002); Khaetskii et al. (2002); Petta et al. (2005), it may be estimated, thereby reducing the uncertainty in its value and the corresponding dephasing. This is expected to be particularly effective, when combining estimation with recent progress in polarizing the nuclear spin ensemble Bracker et al. (2005); Eble et al. (2005); Lai et al. (2006).
In a QD system such as Elzerman et al. (2004), with s and for ns we can estimate 8 digits () (improving by a factor of at least ) in a total time s. In contrast, a standard atomic clock measurement scheme would require a time s.
We now consider limits to the estimation process, focusing on expected variations of due to nuclear spin exchange and preparation and measurement errors. Nuclear spin exchange, in which two nuclei switch spin states, may occur directly by dipole-dipole interactions or indirectly via virtual electron spin flips. Such flips lead to variations of as spins and may have .
The dipole-dipole process, with a scaling, may be approximated by a diffusive process at length scales substantially longer than the lattice spacing Slichter (1980); Deng and Hu (2005). The length scale for a spin at site to a site such that is not satisfied is on the order of the QD radius (5-50 nm); for diffusion constants appropriate for GaAs Paget (1982), the time scale for a change of comparable to by this process is s.
However, nuclear spin exchange mediated by virtual electron spin flips may be faster. This process is the first correction to the rotating wave approximation, and is due to the (heretofore neglected) terms in the contact interaction, , which are suppressed to first order by the electron Larmor precession frequency . These have been considered in detail elsewhere Merkulov et al. (2002); Coish and Loss (2004); Shenvi et al. (2005); Yao et al. (2005); Deng and Hu (2006); Taylor et al. (2006). Using perturbation theory to fourth order, the estimated decorrelation time for is , giving values ms for our parameter range Taylor et al. (2006). Taking ms, we may estimate the optimal number of digits to measure. Using Eq. (14), the best measurement time is given by and for the values used above, is optimal. We note as a direct corollary that our measurement scheme provides a sensitive probe of the nuclear spin dynamics on nanometer length scales.
We now consider implications of these results for improving the performance of nuclear spin ensembles, both as quantum memory Taylor et al. (2003b) and as a qubit Taylor et al. (2004). The dominant error mechanism is the same as for other spin-qubit schemes in QDs: uncertainty in . The proposed measurement scheme alleviates this problem. However, the nuclear spin ensembles operate in a subspace of collective states and , where the first is a “dark state”, characterized by (and the second is , where ). Thus is an eigenstate and cannot be an eigenstate when (except for full polarization). Therefore, the measurement [which essentially projects to certain -”eigenspaces” ()] moves the system out of the computational space, leading to leakage errors. The incommensurate requirements of measuring and using an eigenstate place a additional restriction on the precision of the measurement. The optimal number of digits can be estimated in perturbation theory, using an interaction time and numerical results Taylor et al. (2003b) on the polarization dependence of . We find that for high polarization a relative error of is achievable.
## Vii Conclusions
We have shown that a measurement approach based on quantum phase estimation can accurately measure a slowly varying mesoscopic environment coupled to a qubit via a pure dephasing Hamiltonian. By letting a qubit interact for a sequence of well controlled times and measuring its state after the interaction, the value of the dephasing variable can be determined, thus reducing significantly the dephasing rate.
The procedure requires fast single qubit rotations, but can tolerate realistically slow qubit measurements, since the phase estimation approach minimizes the number of measurements. Limitations due to measurement and preparation errors may be overcome by combining our approach with standard error correction techniques. Fluctuations in the environment can also be tolerated, and our measurement still provides the basis for a good estimate, if the decorrelation time of the environment is not too short.
In view of the implementation of our scheme, we have considered the hyperfine coupling of an electron spin in a quantum dot to the nuclear spin ensemble Our calculations show that the Overhauser field in a quantum dot can be accurately measured in times shorter that the nuclear decorrelation time by shuttling suitably prepared electrons through the dot. Given recent advances in electron measurement and control Elzerman et al. (2004); Petta et al. (2005) this protocol could be used to alleviate the effect of hyperfine decoherence of electron spin qubits and allow for detailed study of the nuclear spin dynamics in quantum dots. Our approach complements other approaches to measuring the Overhauser field in a quantum dot that have recently been explored Klauser et al. (2006); Stepanenko et al. (2006).
While we discussed a single electron in a single quantum dot, the method can also be applied, with modification to preparation and measurement procedures 444In the strong field case a two-level approximation for the spin system is appropriate Coish and Loss (2005). Preparing superpositions in the subspace such as the singlet, and measuring in this basis as well Petta et al. (2005), the scheme would measure the -component of the nuclear spin difference between the dots. To measure the total Overhauser field, superpositions of the triplet states have to be used., to the case of two electrons in a double dot Koppens et al. (2005); Johnson et al. (2005); Coish and Loss (2005).
As we have seen, the Hamiltonian Eq. (1) can serve as a good approximation to more general qubit-environment coupling in the case of weak coupling and a non-degenerate qubit. Therefore, we expect that this technique may find application in other systems with long measurement times and slowly varying mesoscopic environments.
###### Acknowledgements.
J.M.T. would like thank the quantum photonics group at ETH for their hospitality. The authors thank Ignacio Cirac and Guifré Vidal for sharing their notes on the performance of the QFT scheme for different figures of merit. The work at ETH was supported by NCCR Nanoscience, at Harvard by ARO, NSF, Alfred P. Sloan Foundation, and David and Lucile Packard Foundation, and at Ames by the NSF Career Grant ECS-0237925, and at MPQ by SFB 631. | {"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": 0.9518687129020691, "perplexity": 678.0968461944497}, "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/1669446711200.6/warc/CC-MAIN-20221207153419-20221207183419-00273.warc.gz"} |
http://eas.caltech.edu/events/80060 | # TCS+ talk
Wednesday October 25, 2017 10:00 AM
A Time Hierarchy Theorem for the LOCAL Model
Speaker: Seth Pettie, University of Michigan
Location: Annenberg 308
Abtract: The celebrated Time Hierarchy Theorem for Turing machines states, informally, that more problems can be solved given more time.
The extent to which a time hierarchy-type theorem holds in the classic distributed LOCAL model has been open for many years. In particular, it is consistent with previous results that all natural problems in the LOCAL model can be classified according to a small constant number of complexities, such as O(log^* n), O(log n), 2^{O(\sqrt{log n})}, etc.
We establish the first time hierarchy theorem for the LOCAL model and prove that several gaps exist in the LOCAL time hierarchy. One of the gap results can be interpreted as showing that the distributed Lovasz local lemma is complete for randomized sublogarithmic time.
Series TCS+ Talks
Contact: Bonnie Leung [email protected] | {"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": 0.8925850987434387, "perplexity": 1635.1372958510976}, "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-2018-05/segments/1516084886436.25/warc/CC-MAIN-20180116125134-20180116145134-00167.warc.gz"} |
https://www.physicsforums.com/threads/n-pi-gt-n-pi-exercise-in-qft.203461/ | # N+pi->N+pi exercise in QFT
1. Dec 9, 2007
### jostpuur
[SOLVED] N+pi->N+pi exercise in QFT
The due date of this exercise was several weeks ago, but I'm still struggling with this. Since some of the QFT exercises are kind of exercises, that are probably the same all over the world, I assumed there could be a non-zero probability that somebody else has been doing this same exercise earlier.
The exercise is about collision of some nucleon N and meson pi, where nucleon is described with a fermion field and the meson with a scalar field, with an interaction Hamiltonian
$$\mathcal{H}_{\textrm{int}}(x) = -g \overline{\psi}(x)\psi(x)\phi(x).$$
We were supposed to use Feynman rules, but I though I would first check with an explicit calculation that I get the same amplitudes, as we should get with the Feynman rules, just to make sure. My problem deals with this part. The intended exercise was about what happens with the amplitudes once they are written using Feynman rules, but that's another problem then.
In N+pi -> N+pi collision we need terms that have $\psi^+$ and $\phi^+$ (for destroying initial particles), and $\overline{\psi}^-$ and $\phi^-$ (for creating final particles). The first order term in the scattering matrix cannot give these operators, so we must use the second order term. The term
$$T(\overline{\psi}_1\psi_1\phi_1 \overline{\psi}_2\psi_2\phi_2)$$
(here lower indices 1 and 2 refer to position parameters $x_1$, $x_2$), contains two contractions which give the desired operators. They are
$$:\underset{\textrm{contr.}}{\overline{\psi}_{a1}} \psi_{a1}\; \phi_1\; \overline{\psi}_{b2} \underset{\textrm{contr.}}{\psi_{b2}} \phi_2 :\; = -i S^{ba}_F(x_1-x_2) \big( \underbrace{\phi_1^-\; \overline{\psi}_{b2}^- \;\phi_2^+ \;\psi_{a1}^+}_{A} + \underbrace{\phi_2^- \;\overline{\psi}_{b2}^- \;\phi_1^+ \;\psi^+_{a1}}_{B} \;+\; \textrm{others}\big)$$
and
$$:\overline{\psi}_{a1} \underset{\textrm{contr.}}{\psi_{a1}} \phi_1 \underset{\textrm{contr.}}{\overline{\psi}_{b2}} \psi_{b2}\; \phi_2:\; = i S^{ab}_F(x_1-x_2)\big(\underbrace{\phi_1^- \;\overline{\psi}_{a1}^-\; \phi_2^+\; \psi_{b2}^+}_{C} + \underbrace{\phi_2^-\;\overline{\psi}_{a1}^-\; \phi_1^+\; \psi_{b2}^+}_{D} \;+\; \textrm{others}\big)$$
By substituting
$$S^{ab}_F(x_1-x_2) = \int\frac{d^4q}{(2\pi)^4} \frac{(\displaystyle{\not}q + m_N)_{ab}}{q^2-m_N^2} e^{-iq\cdot(x_1-x_2)}$$
and the operators $\phi^-(x)$, $\phi^+(x)$, $\overline{\psi}^-_a(x)$ and $\psi^+_a(x)$ in terms of $a^{\dagger}(q)$, $a(q)$, $a^{r\dagger}(q)$ and $a^r(q)$ (where r=1,2), I calculated the amplitude from the term A.
$$\frac{ig^2}{2} \int d^4x_1\; d^4x_2\; S^{ba}_F(x_1-x_2) \langle 0| a(k') a^{s'}(p')\; \phi^-_1\; \overline{\psi}_{b2}^-\; \phi_2^+ \;\psi_{a1}^+\; a^{\dagger}(k) a^{s\dagger}(p) |0\rangle = \cdots$$
$$\cdots = \frac{i g^2}{8(2\pi)^2} \frac{\delta^4(k'+p'-k-p)}{\sqrt{E_{k',\pi} E_{p',N} E_{k,\pi} E_{p,N}}} \frac{\overline{u}^{s'}(p') (\displaystyle{\not}k' - \displaystyle{\not} p + m_N) u^s(p)}{(k'-p)^2 - m_N^2}$$
(the slash covers the prime badly, but it is k' under the slash, and not k)
This looks good. There's same kind of terms that would have come from the Feynman rules, although I'm not sure about the constants.
The amplitude from term B is
$$\frac{ig^2}{2}\int d^4x_1\; d^4x_2\; S^{ba}_F(x_1-x_2) \langle 0| a(k') a^{s'}(p')\; \phi^-_2\; \overline{\psi}^-_{b2}\; \phi^+_1\; \psi^+_{a1}\;a^{\dagger}(k) a^{s\dagger}(p) |0\rangle = \cdots$$
$$\cdots = \frac{ig^2}{8(2\pi)^2} \frac{\delta^4(k'+p'-k-p)}{\sqrt{E_{k',\pi} E_{p',N} E_{k,\pi} E_{p,N}}} \frac{\overline{u}^{s'}(p') (-\displaystyle{\not}k' - \displaystyle{\not} p' + m_N) u^s(p)}{(k'+p')^2 - m_N^2}$$
The D term has identical diagram to the A term, but there are following differences in the expression. a and b are exchanged everywhere, $x_1$ and $x_2$ are exchanged in the operator part but not in the propagator, and there's a minus sign. We can switch a and b back to the same places as they were in the A term, but when we switch $x_1$ and $x_2$, we don't get identical expression with A, because $S^{ba}_F(x_2-x_1)\neq -S^{ba}_F(x_1-x_2)$. This is because of the $m_N$-term in the propagator, which we assume to be nonzero. So in fact A and D don't give identical amplitudes, but instead when they are summed, the $m_N$-terms cancel.
The same thing happens with B and C terms. So the total scattering amplitude (in second order approximation), which should come from the expression
$$\langle 0| a(k') a^{s'}(p') \Big(\frac{(-i)^2}{2!}\int d^4x_1\; d^4x_2\; T(\mathcal{H}_{\textrm{int}}(x_1) \mathcal{H}_{\textrm{int}}(x_2))\Big) a^{\dagger}(k) a^{s\dagger}(p)|0\rangle$$
turns out to be
$$\frac{ig^2}{4(2\pi)^2} \frac{\delta^4(k'+p'-k-p)}{\sqrt{E_{k',\pi} E_{p',N} E_{k,\pi} E_{p,N}}} \Big( \frac{\overline{u}^{s'}(p') (\displaystyle{\not}k' - \displaystyle{\not} p) u^s(p)}{(k'-p)^2 - m_N^2}\quad+\quad \frac{\overline{u}^{s'}(p') (-\displaystyle{\not}k' - \displaystyle{\not} p') u^s(p)}{(k'+p')^2 - m_N^2} \Big)$$
Now... did something go wrong? Where those mass terms supposed to cancel? In the exercise session the assistant drew the two Feynman diagrams, and wrote down corresponding amplitudes using Feynman rules, and the mass terms remained there for rest of the calculation.
I had some difficulties with the fermion contractions previously. One possibility is that something goes wrong with the contractions and propagators right in the beginning.
If the mass terms instead are supposed to cancel, how could one see it simply by drawing the diagrams and using Feynman rules?
Last edited: Dec 9, 2007
2. Dec 9, 2007
### jostpuur
Another related question: Does anyone have any words of wisdom related to the states
$$a^{\dagger}(p)|0\rangle \quad\quad\quad\quad\quad (1)$$
and
$$\sqrt{2 E_p} a^{\dagger}(p)|0\rangle \quad\quad\quad\quad\quad (2)$$
being used as initial and final states in the scattering amplitude calculations?
I used the (1) in a previous exercise that dealt with a B+B->B+B scattering with $\phi^4$ theory, where there was no propagator, and got the correct result. So it seems that I should use (1) because that is what gives correct results on our course, at least... But for example P&S use (2). Now a quick thought would be, that perhaps P&S uses correspondingly different conventions with S-matrix too, but in fact it does not. It is the same
$$T\Big(\exp\Big(-i\int d^4x\; \mathcal{H}_{\textrm{int}}(x)\Big)\Big)$$
both on our course and in the P&S book. If I used (2) in this exercise I would be getting different result, so the situation with these conventions seems slightly confusing. Do the transition amplitudes themselves have different meaning with different conventions?
3. Dec 9, 2007
### Avodyne
The m terms do not cancel. I didn't follow your argument that they should.
Conventions vary with the 2E's. They can go into the amplitude, or into the relation between the amplitude and the cross section. Any one particular book presumably gets a correct final answer for the cross section, but you have to be careful comparing different books on anything but this final answer.
4. Dec 9, 2007
### jostpuur
The A term is
$$-iS^{ba}_F(x_1-x_2) \phi_1^-\;\overline{\psi}_{b2}^-\;\phi_2^+\;\psi_{a1}^+$$
and the D term is, after x1,x2 and a,b have been switched (which can be done since they are dummy variables),
$$iS^{ba}_F(x_2-x_1) \phi_1^-\;\overline{\psi}_{b2}^-\;\phi_2^+\;\psi_{a1}^+$$
Then
$$S^{ba}_F(x_2-x_1) = \int \frac{d^4q}{(2\pi)^4} \frac{(\displaystyle{\not}q + m_N)_{ba}}{q^2-m_N^2} e^{-iq\cdot(x_2-x_1)} = -\int \frac{d^4q}{(2\pi)^4} \frac{(\displaystyle{\not}q - m_N)_{ba}}{q^2-m_N^2} e^{-iq\cdot(x_1-x_2)} \neq -S^{ba}_F(x_1-x_2)$$
They still look like canceling to me.
5. Dec 9, 2007
### jostpuur
Okey, there was a mistake. Assuming that the solution here https://www.physicsforums.com/showthread.php?t=200032 was correct, the first contraction is supposed to give
$$:\underset{\textrm{contr.}}{\overline{\psi}_{a1}} \psi_{a1}\; \phi_1\; \overline{\psi}_{b2} \underset{\textrm{contr.}}{\psi_{b2}} \phi_2 :\; = -i S^{ba}_F(x_2-x_1) \big( \underbrace{\phi_1^-\; \overline{\psi}_{b2}^- \;\phi_2^+ \;\psi_{a1}^+}_{A} + \underbrace{\phi_2^- \;\overline{\psi}_{b2}^- \;\phi_1^+ \;\psi^+_{a1}}_{B} \;+\; \textrm{others}\big)$$
The x1 and x2 are in different order in the propagator. But isn't this only making the situation worse.... now the mass terms remain, and the $\displaystyle{\not}q$ terms are canceling!
hmhm....
It looks like that my previous question about what the contraction becomes when the psi-bar is on left, did not get dealt with properly yet.
Last edited: Dec 9, 2007
6. Dec 10, 2007
### jostpuur
No no no... that was some kind of mistake again. The contractions are
$$:\underset{\textrm{contr}}{\overline{\psi}_{a1}}\psi_{a1}\phi_1 \overline{\psi}_{b2} \underset{\textrm{contr}}{\psi_{b2}}\phi_2: =-iS^{ba}_F(x_2-x_1)(\phi^-_1\;\overline{\psi}^-_{b2}\;\phi^+_2\;\psi^+_{a1} \;+\;\phi^-_2\;\overline{\psi}^-_{b2}\;\phi^+_1\;\psi^+_{a1} \;+\;\textrm{others})$$
$$:\overline{\psi}_{a1}\underset{\textrm{contr}}{\psi_{a1}}\phi_1 \underset{\textrm{contr}}{\overline{\psi}_{b2}} \psi_{b2}\phi_2: = iS^{ab}_F(x_1-x_2)(\phi^-_1\;\overline{\psi}^-_{a1}\;\phi^+_2\; \psi^+_{b2}\;+\; \phi^-_2\;\overline{\psi}^-_{a1}\;\phi^+_1\;\psi^+_{b2} \;+\;\textrm{others})$$
So that
$$S^{(2)} = \frac{(-i)^2g^2}{2!}\int d^4x_1\; d^4x_2\; T\big( \overline{\psi}(x_1)\psi(x_1)\phi(x_1)\overline{\psi}(x_2)\psi(x_2)\phi(x_2)\big)$$
$$= \frac{ig^2}{2}\int d^4x_1\; d^4x_2\;\Big( S^{ba}_F(x_2-x_1)( \underset{A}{\phi^-_1\;\overline{\psi}^-_{b2}\;\phi^+_2\;\psi^+_{a1}} \;+\;\underset{B}{\phi^-_2\;\overline{\psi}^-_{b2}\;\phi^+_1\;\psi^+_{a1}}) \;-\;S^{ab}_F(x_1-x_2)( \underset{C}{\phi^-_1\;\overline{\psi}^-_{a1}\;\phi^+_2\;\psi^+_{b2}} \;+\;\underset{D}{\phi^-_2\;\overline{\psi}^-_{a1}\;\phi^+_1\;\psi^+_{b2}})\Big)$$
$$+\textrm{something non-relevant}$$
Comparison of A and D: The a <-> b and x1 <-> x2 have been switched everywhere, so that they can be switched back and the expressions become identical. But there's a minus sign, and the entire expressions are canceling.... Okey it's just a one minus sign mistake. But where?
Last edited: Dec 10, 2007
7. Dec 10, 2007
### jostpuur
No no no no... The contractions are
$$:\underset{\textrm{contr}}{\overline{\psi}_{a1}}\psi_{a1}\phi_1 \overline{\psi}_{b2} \underset{\textrm{contr}}{\psi_{b2}}\phi_2: =-iS^{ba}_F(x_2-x_1)(-\phi^-_1\;\overline{\psi}^-_{b2}\;\phi^+_2\;\psi^+_{a1} \;-\;\phi^-_2\;\overline{\psi}^-_{b2}\;\phi^+_1\;\psi^+_{a1} \;+\;\textrm{others})$$
$$:\overline{\psi}_{a1}\underset{\textrm{contr}}{\psi_{a1}}\phi_1 \underset{\textrm{contr}}{\overline{\psi}_{b2}} \psi_{b2}\phi_2: = iS^{ab}_F(x_1-x_2)(\phi^-_1\;\overline{\psi}^-_{a1}\;\phi^+_2\; \psi^+_{b2}\;+\; \phi^-_2\;\overline{\psi}^-_{a1}\;\phi^+_1\;\psi^+_{b2} \;+\;\textrm{others})$$
taking into account the sign changes due to fermion operator reordering.
Thank's to the forum for its existence, so that I could write questions and remarks that helped me think. | {"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": 0.8906808495521545, "perplexity": 687.0554333988341}, "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-2017-04/segments/1484560280292.50/warc/CC-MAIN-20170116095120-00501-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://warpdrivetutors.com/wordymathphobia/ | # Wordymathphobia
Written by Marizza
Illustrations by Sabrina @binathewitch
Wordymathphobia: (noun) An abnormal or pathological fear of seeing words in your math
Many students seem to be under the impression that word problems are spawned by spiteful math teachers who wish to traumatize them into obedience. At least, that is what some of my students tell me when I ask them why they leave all the word problems blank on their homework. Woefully, word problems are merely the math teacher’s attempt to answer one of the most common student questions:
#### “When are we ever going to use this stuff, anyway?”
For example, Related Rates word problems reveal how Calculus can be used to calculate the rate of change of a measurement when variables are related to each other. Most word problems are demonstrations of applications of mathematics to various fields in the “Real World”.
Sadly, in the “Real World” people communicate using words, which means you WILL be confronted with a verbal or written problem. In order to solve it, you may need to convert those words into mathematical relationships. In other words, if you want to adult successfully, you are going to have to get over your wordymathphobia and solve those suckers.
Before you run away screaming and hide under your bed, keep reading.
Below is a strategically crafted plan of attack that will help you beat those words into submission, so that you, the victorious student, emerge unscathed with the solution in your hand (or on your paper), and your sanity intact.
Word Problem Strategy:
Step 1: Underline Breathe! Don’t Panic! These are usually words that occur frequently in your math book. Your math teacher usually uses these words repeatedly so that, by osmosis if necessary, you will retain them.
Step 2: Draw, draw, draw (and label) Still panicking? This will help. Draw a picture modeling the situation in the word problem. This is usually the most helpful and most neglected step. Don’t skip this step! You can use colors if it makes you happy and don’t worry if you’re not Van Gogh.
Step 3: Name Everything Otherwise known as identifying variables, giving names to what is known and what is unknown is just polite. No one wants to go around shouting “Hey you!”. Identify what is changing (variables) and what is constant. Substitute the constants with their values. Do not substitute the variables with their values until the end.
Step 4: Determine math expressions from picture and math In order to be able to execute this step correctly (I’m sorry to break this to you), you need to STUDY your math vocabulary. I know! I know! You aren’t used to studying vocabulary and definitions in math. Too bad. Write the expressions and equations you’ve translated from the math vocabulary down on the paper. They don’t do anyone any good hiding in your head.
Step 5: Determine what you’re asked to find and solve Identify the variable you are asked to find and keep reminding yourself that it’s your goal. Once you’ve written all equations down on, look for connections until you have answered the question. You may need to look back at your picture for extra relationships.
Let us see how we would use the steps above on a related rates problem found in Thomas’ Calculus book:
Calculus I Example (AP®Calculus AB): (Thomas’ Calculus 11th edition, p 221 #34)
Step 1: Underline
All numbers with units should be underlined. The word “decreasing” is often discussed when derivatives are mentioned, and the word “speed” is relevant to derivatives.
Highway Patrol: A highway patrol plane flies 3 miles above a level straight road at a steady 120 miles per hour. The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is 5 miles, the line-of-distance is decreasing at the rate of 160 miles per hour. Find the car’s speed along the highway.
Step 2: Draw, draw, draw (and label)
Figure 1: Sad Excuse of a drawing by Marizza
Step 3: Name Everything
• Call the height of the plane, h.
• The height of the plane is not changing so is a constant.
• The plane is flying a steady
• call the velocity of plane, v, v = 120 mph
• steady implies the velocity of the plane is not changing, v=120 mph is a constant.
• Call the line-of-distance, c.
• “the instant the line-of-sight ( c ) is 5 miles”
• Instant implies c is changing, c is a variable
Step 4: Determine math expressions from picture and math vocabulary
• “the line-of-distance, c, is decreasing at 160 miles per hour”
• The derivative of c is negative,
• From the picture, we can see there is a horizontal distance from the car to the plane, x
• x is changing, x is a variable
• From the picture we can see a right triangle with sides length c, h, and x.
• Pythagorean Theorem
• h = 3 is a constant, but c and x are variables
• , we need to derive implicitly
There are two variables left, x and .
The paragraph does not contain either, however, if we use “at the instant that c = 5”, we will get two sides of a right triangle.
From the Pythagorean Theorem, we can compute that x = 4.
Step 5: Determine what you’re asked to find and Solve
The question at the end of the paragraph is “Find the car’s speed”.
The speed of the car is not because x is measuring the position of the car relative to the plane, which is going in the opposite direction.
Therefore, we need to subtract the plane’s velocity from the in order to get the car’s speed:
miles per hour
The End
The solution seems much longer only because explanations were inserted.
Now, let’s try this again without all the explanation with part (b) of the AP Calculus AB/BC Free Response question from 2018.
### 2018 AP® Calculus BC Free-Response Questions
2. Researchers on a boat are investigating plankton cells in a sea. At a depth of h meters, the density of plankton cells, in millions of cells per cubic meters, is modeled by p(h) = 0.02h²e-00025h² for 0 < h < 30 and is modeled by f (h) for h > 30. The continuous function f is not explicitly given.
b) Consider a vertical column of water in this sea with horizontal cross sections of constant area 3 square meters. To the nearest million, how many plankton cells are in this column of water between h = 0 and h = 30 meters?
Try it on your own first, then look at the solution below! ( Don’t peek. )
Step 1: Underline and Highlight
Step 2: Draw, draw, draw (and label)
Step 3: Name Everything
Step 4: Determine math expressions from picture and math vocabulary
Step 5: Determine what you’re asked to find and solve
Step 1: Underline and Highlight Do you see any vocabulary words? Also, anything with units gives us information about whether we’re dealing with a derivative or not.
### 2018 AP® Calculus BC Free-Response Questions
2. Researchers on a boat are investigating plankton cells in a sea. At a depth of h meters, the density of plankton cells, in millions of cells per cubic meters, is modeled by p(h) = 0.02h²e-00025h² for 0 < h < 30 and is modeled by f (h) for h > 30. The continuous function f is not explicitly given.
b) Consider a vertical column of water in this sea with horizontal cross sections of constant area 3 square meters. To the nearest million, how many plankton cells are in this column of water between h = 0 and h = 30 meters?
Step 2: Draw draw, draw (and label) We may not need to draw the researchers, but we could draw the boat, and should draw the sea and the vertical column of water with horizontal cross-sections.
Figure 2: Another Sad Excuse of a drawing by Marizza
Step 3: Name Everything: Make sure you identify variables and relationships. Some are already given.
For h is in [0,30] ( units: h meters )
N(h) = number of plankton cells in column at h meters ( units: million cells )
density of plankton
of cross-section in column at h meters
Step 4: Determine math expressions from picture and math vocabulary units are very important in determining relationships.
We know that Volume = integral of area and that the area is constant and does not depend on the depth, but the plankton density does depend on the depth.
If we multiply, p(h) · A(h),the units are almost correct:
Step 5: Determine what you’re asked to find and solve Remember to use units to help you figure out what you need to solve.
They asked us for “how many plankton cells”, so we need units to be in millions of cells.
If we integrate p(h) · A(h), then we will be multiplying a dh to the equation: dh ( units: meters )
This is a calculator active question, so we can integrate using our calculator. They also asked us to round to the nearest million, so the answer is:
1675 million cells
The End
Of course, you can probably find some shortcuts, but this strategy is very useful when a word problem stumps you and you don’t know where to start.
So, remember, BREATHE … DON’T PANIC… and Underline, Draw, Name, Math, Find and Solve!
Now try the rest of the parts of this free response by following the link below:
https://apcentral.collegeboard.org/pdf/ap18-frq-calculus-bc.pdf?course=ap-calculus-bc | {"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": 0.8127341866493225, "perplexity": 1187.7630866496797}, "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/1669446711417.46/warc/CC-MAIN-20221209144722-20221209174722-00514.warc.gz"} |
https://www.dm.unipi.it/eventi/on-the-flow-of-elastic-networks-paola-pozzi/ | On the flow of elastic networks – Paola Pozzi
Venue
Sala Seminari (Dip. Matematica).
Abstract
In this talk I will discuss a long-time existence result for the elastic flow of a three network in $R^n$. The evolution is such that the sum of the elastic energies of the three curves plus their weighted lengths decrease in time. Natural boundary conditions are considered at the boundary of the curves and at the triple junction. This is joint work with Anna Dall’Acqua and Chun-Chi Lin.
Torna in cima | {"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": 0.9199583530426025, "perplexity": 1013.891816786857}, "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-2022-49/segments/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00573.warc.gz"} |
https://www.physicsforums.com/threads/pressure-and-force.273910/ | # Pressure and Force
1. Nov 21, 2008
### vrobins1
1. The problem statement, all variables and given/known data
"A lid is put on a box that is 10 cm long, 16 cm wide, and 8.0 cm tall and the box is then evacuated until its inner pressure is 7.2e4 Pa."
How much force is required to lift the lid at sea level?
2. Relevant equations
I know that I have to use P=F/A
I changed my centimeters to meters also, although I'm not sure you have to.
3. The attempt at a solution
I used F = PA
F= (Atmospheric pressure - inner pressure)(lxw)
F=(100000 - 72000)(.10x.16)
F=(28000)(.016)
F=448 N
I tried this answer but got it wrong. Can someone please point me in the right direction? Thanks!
2. Nov 21, 2008
### vrobins1
I figured it out! I just used a more accurate number for my atmospheric pressure instead of simply 10000. | {"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": 0.884566068649292, "perplexity": 1039.8370920567834}, "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-2016-50/segments/1480698544097.11/warc/CC-MAIN-20161202170904-00500-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://en.wikipedia.org/wiki/Milnor_K-theory | # Milnor K-theory
In mathematics, Milnor K-theory was an early attempt to define higher algebraic K-theory, introduced by Milnor (1970).
## Definition
The calculation of K2 of a field F led Milnor to the following ad hoc definition of "higher" K-groups by
$K^M_*(F) := T^*F^\times/(a\otimes (1-a)), \,$
thus as graded parts of a quotient of the tensor algebra of the multiplicative group F× by the two-sided ideal, generated by the
$a\otimes(1-a) \,$
for a ≠ 0, 1. For n = 0,1,2 these coincide with Quillen's K-groups of a field, but for n ≧ 3 they differ in general. We define the symbol $\{a_1,\ldots,a_n\}$ as the image of $a_1 \otimes \cdots \otimes a_n$: the case n=2 is a Steinberg symbol.[1]
The tensor product on the tensor algebra induces a product $K_m \times K_n \rightarrow K_{m+n}$ making $K^M_*(F)$ a graded ring which is graded-commutative.[2]
## Examples
For example, we have $K^M_n(\mathbb{F}_q) = 0$ for n ≧ 2; $K^M_2(\mathbb{C})$ is an uncountable uniquely divisible group; $K^M_2(\mathbb{R})$ is the direct sum of a cyclic group of order 2 and an uncountable uniquely divisible group; $K^M_2(\mathbb{Q}_p)$ is the direct sum of the multiplicative group of $\mathbb{F}_p$ and an uncountable uniquely divisible group; $K^M_2(\mathbb{Q})$ is the direct sum of the cyclic group of order 2 and cyclic groups of order $p-1$ for all odd prime $p$.
## Applications
Milnor K-theory plays a fundamental role in higher class field theory, replacing $K^M_1$ in the one-dimensional class field theory.
Milnor K-theory modulo 2, denoted k*(F) is related to étale (or Galois) cohomology of the field F by the Milnor conjecture, proven by Voevodsky. The analogous statement for odd primes is the Bloch–Kato conjecture, proved by Voevodsky, Rost, and others.
There are homomorphisms from kn(F) to the Witt ring of F by taking the symbol
$\{a_1,\ldots,a_n\} \mapsto \langle \langle a_1, a_2, ... , a_n \rangle \rangle = \langle 1, a_1 \rangle \otimes \langle 1, a_2 \rangle \otimes ... \otimes \langle 1, a_n \rangle \ ,$
where the image is a Pfister form of dimension 2n.[1] The image can be taken as In/In+1 and the map is surjective since the Pfister forms additively generate In.[3] The Milnor conjecture can be interpreted as stating that these maps are isomorphisms.[1]
## References
1. ^ a b c Lam (2005) p.366
2. ^ Gille & Szamuely (2006) p.184
3. ^ Lam (2005) p.316 | {"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": 16, "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": 0.9788720607757568, "perplexity": 497.35159071090396}, "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-2014-52/segments/1419447546043.1/warc/CC-MAIN-20141224185906-00004-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/881329/independent-exponential-distribution-px-y-1 | # independent Exponential distribution P(X > Y + 1)
$X$ and $Y$ are independent exponentially distributed random variables with parameters $a$ and $b$.
Calculate $P(X > Y + 1)$.
I have let $X-Y=Z$ and Then $P(Z>z)=1-P(Z\leq z)$
$1 - P(X-Y\leq z) =1 - \int_0^\infty\int_0^{y-z}\lambda_1 e^{-\lambda_1x}\lambda_2 e^{-\lambda_2y} dxdy$
• Hi and welcome to the site! Since this is a site that encourages learning, you will get much more help if you show us what you have already done. Could you edit your question with your thoughts and ideas? – 5xum Jul 29 '14 at 8:44
• I was going to let $X-Y=Z$ then calculate the $P(Z>z)$. $F_Z(z)=P(Z>z)$ – rmvdwalt Jul 29 '14 at 8:58
• That sounds like a good idea. Try it out and edit your question with the results you get. – 5xum Jul 29 '14 at 9:02
• Why does $P(Z>z).F_Z(z)$ equal $P(Z>z)$ for all $z$? Are you assuming that $F_Z(z) = 1$ for all $z$? Hint: If you do want to use conditional probabilities, then try finding $P(X > Y+1\mid Y = y)$ which should have the same value as $P(X > y+1)$ (why?) and then find $$P(X>Y+1) = \int_{-\infty}^\infty P(X>Y+1\mid Y = y)f_Y(y)\,\mathrm dy$$ which should be a lot easier than finding $F_Z(z)$ as in your idea. – Dilip Sarwate Jul 29 '14 at 9:21
• Can it be done without conditional probabilities – rmvdwalt Jul 29 '14 at 9:25
\begin{align}X\bot Y & \iff f_{X,Y}(x,y)=f_X(x)f_Y(y), \forall (x,y)\in {\bf X\times Y} \\[1ex] X\sim{\cal Exp}(a) & \iff \Pr(X\leq x)= (1-e^{-ax})\operatorname{\bf 1}_{[0, \infty)}(x) \\ & \iff f_X(x)= a\,e^{-ax}\operatorname{\bf 1}_{[0,\infty)}(x) \\[1ex] Y\sim{\cal Exp}(b) & \iff \Pr(Y\leq y)= (1-e^{-by})\operatorname{\bf 1}_{[0, \infty)}(y) \\ & \iff f_Y(y)= b\,e^{-by}\operatorname{\bf 1}_{[0,\infty)}(y) \end{align}
\begin{align} \Pr(X-Y> 1) & = \Pr(X > Y+1) \\ & = \iint_{{\bf X\times Y}: x>y+1} \operatorname{d}^2 F_{X,Y}(x,y) & \text{by definition} \\ & = \int_{0}^{\infty} \int_{1+y}^{\infty} \operatorname{d}F_X(x)\operatorname{d}F_Y(y) & \text{by independence} \\ & = \int_{0}^{\infty} f_Y(y) \int_{1+y}^{\infty} f_X(x) \operatorname{d}x \operatorname{d}y & \text{by expansion} \\ & =\int_{0}^{\infty} f_Y(y) \Pr(X> 1+y)\operatorname{d} y & \text{by definition} \\ & \color{gray}{= \int_{0}^{\infty} f_Y(y) \Pr(X> 1+Y\mid Y=y) \operatorname{d} y} & \color{gray}{\text{by independence; just to note}} \\ & = \int_0^\infty (b\, e^{-by})(e^{-a(1+y)}) \operatorname{d}y & \text{by substitution from the CDF and pdf} \\ & = b\,e^{-a}\, \int_0^\infty e^{-(a+b)y}\operatorname{d}y & \text{by rearranging to simplify} \\ & = b\,e^{-a}\, \left[ -\frac{ e^{-(a+b)y} }{ a+b } \right]_{y=0}^{y\to\infty} & \text{by integration} \\ & = \frac{b\,e^{-a}}{a+b} & \text{by evaluation} \\[1ex] \therefore\quad &\boxed{ \Pr(X-Y> 1)=\dfrac{b\,e^{-a}}{a+b} } \end{align} | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999951124191284, "perplexity": 2219.8627727447097}, "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-2019-39/segments/1568514572289.5/warc/CC-MAIN-20190915195146-20190915221146-00201.warc.gz"} |
http://math.stackexchange.com/questions/285605/a-doubt-on-manifold-with-boundary-critical-point-space-of-jets-etc | a doubt on manifold with boundary, critical point, space of jets etc
could any one explain me the following paragraph by a simple example?
"a manifold with boundary is understood to be a smooth (real or complex) manifold with a fixed smooth hypersurface. Two functions on a manifold with boundary are called equivalent if one goes over into the other under a diffeomorphism of the manifold that takes the boundary into itself. On the boundary we consider a distinguished point O. The group of germs of diffeomorphisms of a manifold with boundary at a distinguished point that keep the boundary fixed acts on the spaces of germs and jets of functions at the distinguished point for which this is a critical point with critical value zero"
I know what is manifold with boundary but never saw such a definition or remark as the author said in 1st line, so I am not feeling anything of the first paragrgaph, but I am confident that if any one give example and tell me I can understand.
I know what is critical points like say $f:N\rightarrow M$ be a smoothh map, a point $p\in N$ is said to be a critical point of $f$ if the differential $$f_{*,p}:T_p\rightarrow T_{f(p)}M$$ fails to be surjective and I also know one result for a real valued funtion $f:M\rightarrow \mathbb{R}$, a pt. $p\in M$ is critical iff relative to some chart $(U,x_1,\dots,x_n)$ containing $p$ all the partial derivatives $$\frac{\partial f}{\partial x_i}(p)=0$$
there is also some special kind of group and its action is mentioned here, I could not understand that also. Thank you for help.
-
Let's take first the example of the square.
The square can be given a structure of smooth manifold : take a homeomorphism from the square to the circle and the smooth structure of the circle gives back a smooth structure on the square.
But hey, in my mind square have corners ! Ok then we have two possibilities : first one is to say, I'll give an emdedding from 4 points to the square (our smooth manifold) to precise what I consider as corners. With this method you can transform a circle into a triangle, a square etc.
The second method is to say that the circle have charts to $\mathbb R$ and the square have charts to $\mathbb R_+$. In any cases, you choose how you want to describe the topological space.
Then you need to understand what a germ is. Let's take $p \in M$ a point, and $f : M \to \mathbb R$ a smooth function. The germ of $f$ at $p$ is the equivalence class of $f$ under the relation : $f \sim_p g$ if and only if there exists an open set $U$ around $p$ such that $f_{|U} = g_{|U}$.
Example : take an holomorphic map $f$ on $\mathbb C$, then for $z \in \mathbb C$, we can write $f(z) = a_0 + a_1 z + a_2 z^2 + ... + a_n z^n + ...$ which is the power serie of $f$ at zero. Then the germ of $f$ at $0$ can be identified as the sequence $a_0, a_1, ...$ It is important to take some time to see why.
In the case of smooth functions, there are much more classes of functions around a given point (i.e germs) and we don't have a nice description.
Now your text is saying : take $M$ and $N$ two smooth manifolds with an embedding $N \to M$. Take one point $p\in N$. And look at all the germs at $p$ of functions $\phi$ such that $\phi : M \to M$ is a difféomorphism and $\phi_{|N} = Id_N$. As the set of such diffeomorphisms (let's call it $\mathrm{Diff}_N(M)$) is a group (under composition), then the set of germs of elements of $\mathrm{Diff}_N(M)$ at $p$ (let's call it $\mathrm{Diff}_N(M)_p$) will also be a group. It is a good thing to check that.
Finally, it is stated that this group $\mathrm{Diff}_N(M)_p$ acts on a special subset of $\mathcal C^{\infty}(M)_p$ the set of all germs of smooth functions at $p$. It is the subset of germs at $p$ of functions $f : M \to \mathbb R$ such that $f(p) = 0$ and $df_p = 0$. This action is by composition. It is not difficult to do the details.
- | {"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": 0.957141101360321, "perplexity": 112.2513223133264}, "config": {"markdown_headings": false, "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-2016-26/segments/1466783396106.25/warc/CC-MAIN-20160624154956-00140-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://philpapers.org/bbs/thread.pl?tId=779 | Discussion:
1. Anti-realist epistemic conceptions of truth imply what is called the knowability principle: All truths are possibly known. The principle can be formalized in a bimodal propositional logic, with an alethic modality ${\diamondsuit}$ and an epistemic modality ${\mathcal{K}}$ , by the axiom scheme ${A \supset \diamondsuit \mathcal{K} A}$ (KP). The use of classical logic and minimal assumptions about the two modalities lead to the paradoxical conclusion that all truths are known, ${A \supset \mathcal{K} A}$ (OP). A Gentzen-style reconstruction of the Church–Fitch (...) | {"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": 0.9598929286003113, "perplexity": 2032.820557502836}, "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-2014-41/segments/1410657133455.98/warc/CC-MAIN-20140914011213-00018-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://www.physicsforums.com/threads/question-involving-projectiles-in-motion.374882/ | # Question involving projectiles in motion.
1. Feb 2, 2010
### Jaime
1. The problem statement, all variables and given/known data
Now I'm pretty much an idiot, but the question to this is "Use information from the figure to find the initial speed of the grasshopper."
Now I know that I need time, but this is where I'm having a problem. How would I calculate time without velocity. If I had some equations I would be able to do it with no problem, but I'm pretty much making this harder than it should be... If anyone could set me up with some equations I could possibly try it out and run it by you guys!
2. Relevant equations
:(
3. The attempt at a solution
:(
2. Feb 2, 2010
### cepheid
Staff Emeritus
Hi Jaime, welcome to PF!
The equations are just those of basic kinematics of projectile motion, but rather than focusing on that, let me walk you through it conceptually. You really do have all of the information you need:
- As soon as the grasshopper leaves the ground, it is in free fall (i.e. it is a projectile -- an object that moves solely under the influence of gravity)
- because of this first point, you know the grasshopper's acceleration is always g, downwards.
- Given the acceleration in the previous point, and the maximum height reached, you can figure out using basic kinematics what the initial vertical component of the velocity must have been at launch in order for the grasshopper to have reached that max height before decelerating to zero and starting to fall back down. You can also figure out how much time it takes for the grasshopper to reach this height
- Given the max height from which the grasshopper falls, you can figure out how much time it will take for for it to fall back down to the ground.
- Given the total 'air' time from the two previous points, you can figure out what the horizontal component of the velocity must have been in order for the grashopper to have travelled 1.06 m horizontally during that time. Recall that the horizontal component of the velocity is unaffected by gravity (the acceleration is entirely vertical, and the two components of the velocity are therefore independent of each other).
- Note, also, that the launch angle of 50 degrees tells you what the relationship between horizontal and vertical velocity is, meaning that you could have calculated the horizontal launch velocity as soon as you found out the vertical one, without going through those intermediate steps. It's nice that you have that additional info.
3. Feb 3, 2010
### Jaime
Thanks so much! I'm going through my equations, you broke it down so nicely!
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https://www.gradesaver.com/textbooks/science/physics/essential-university-physics-volume-1-3rd-edition/chapter-11-exercises-and-problems-page-201/40 | ## Essential University Physics: Volume 1 (3rd Edition)
a) $14.2 \ Js$ b) $57 \ Nm$
a) We use the parallel axis theorem to obtain: $I_t = .048 +mR^2 = .048+(.88)(.43)^2 = .21$ Recall that $\omega=\frac{v}{r}$, it follows: $L=I\omega=I\frac{v}{r}=(.021)\frac{50}{.74}=14.2 \ Js$ b) We obtain: $\tau = \frac{L}{t}=\frac{14.2}{.25}=57 \ Nm$ | {"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": 0.9599449038505554, "perplexity": 854.460010076762}, "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/1652662546071.13/warc/CC-MAIN-20220522190453-20220522220453-00436.warc.gz"} |
http://cms.math.ca/cmb/msc/68W20?fromjnl=cmb&jnl=CMB | The Resultant of Chebyshev Polynomials Let $T_{n}$ denote the $n$-th Chebyshev polynomial of the first kind, and let $U_{n}$ denote the $n$-th Chebyshev polynomial of the second kind. We give an explicit formula for the resultant $\operatorname{res}( T_{m}, T_{n} )$. Similarly, we give a formula for $\operatorname{res}( U_{m}, U_{n} )$. Keywords:resultant, Chebyshev polynomialCategories:11Y11, 68W20 | {"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": 0.9998583793640137, "perplexity": 82.53936123112967}, "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/1432207932737.93/warc/CC-MAIN-20150521113212-00180-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://brilliant.org/discussions/thread/beginner-latex-guide/ | ×
# Beginner LaTeX Guide
$${\LaTeX}$$ is an extremely useful typesetting language to learn, especially in a math environment like this. However, the quick instructions Brilliant.org gives just aren't good enough to use for most situations.
This is why I've decided to create a beginner's $${\LaTeX}$$ guide. There is a table of contents for easy symbol or format finding. I hope you can refer to this guide later, when writing solutions, problems, or notes.
Note: You can also view Latex codes by hovering over the equation. Read Seeing actual $$\LaTeX$$ for more details!
To quickly navigate to the part you want via the Table of Contents, press CTRL+F, and type in the section you want (including the tilde's ~ before and after the section).
~Using LaTeX~
~Text~
~Basic Operations~
~Fractions~
~Sums, Products, Limits, and Integrals~
~Modular Arithmetic~
~Trigonometry~
~Combinatorics~
~Geometry~
~Calculus~
~Parentheses~
~Fitting Parentheses~
~Tables and Arrays~
~Other~
~Using LaTeX~
To use LaTeX, put a backslash and a left parenthesis before the math you want to LaTeXify, and put a backslash and a right parenthesis after the math you want to LaTeXify. For example:
Imgur
Shows up as $$1+2+3=6$$
However, if you want your math to be more conspicuous and centered, you can use a backslash then a left bracket, then your math, then a backslash then a right bracket. For example:
Imgur
Shows up as
$1+2+3=6$
This second option is the display text. A lot of other math operations will look better in this text. To force the first option to also use display text, you can add a \displaystyle at the beginning.
~Text~
To write text in LaTeX use \text{your text here}. This gives $$\text{your text here}$$
To use bolded text, use \textbf{your text here}. This gives $$\textbf{your text here}$$
Italicized text is similar: \textit{your text here}. This gives $$\textit{your text here}$$
~Basic Operations~
"x+y" gives $$x+y$$
"x-y" gives $$x-y$$
"x=y" gives $$x=y$$
"x\times y" gives $$x\times y$$
"x\cdot y" gives $$x\cdot y$$
"x\div y" gives $$x\div y$$
"x\pm y" gives $$x\pm y$$
"x\mp y" gives $$x\mp y$$
x^{y} gives $$x^{y}$$
x_{y} gives $$x_{y}$$
\sqrt{x} gives $$\sqrt{x}$$
\sqrt[y]{x} gives $$\sqrt[y]{x}$$
\log_{a}b gives $$\log_{a}b$$
\ln a gives $$\ln a$$ (that's a lowercase "l" in the beginning, not an uppercase "i")
Note that many of you use "*" or "." for multiplying. This shows up as $$*$$ and $$.$$ which don't look good. Use $$\times$$ or $$\cdot$$ instead.
Also, the brackets in x^{y} or x_{y} may be omitted if the index is a single character. However, if it is more than one character like $$x^{10}$$, then brackets are needed or else it will show up as $$x^10$$.
~Fractions~
Many people simply put a slash between the numerator and denominator to represent a fraction: $$x/y$$. However, there are neater ways in LaTeX.
\frac{x}{y} is the standard way to write fractions: $$\frac{x}{y}$$
\dfrac{x}{y} gives a bigger clearer version. However, this takes up more vertical space: $$\dfrac{x}{y}$$ the "d" stands for "display text".
EXTRA
\cfrac{x}{y} is a special type of fraction formatting. This is for continued fractions, hence the "c". typing \cfrac{x}{x+\cfrac{y}{y+\cfrac{z}{2}}} gives $$\cfrac{x}{x+\cfrac{y}{y+\cfrac{z}{2}}}$$
~Sums, Products, Limits, and Integrals~
These four are in the same group because they format differently than other symbols.
"\sum" gives $$\sum$$
"\prod" gives $$\prod$$
"\lim" gives $$\lim$$
"\int" gives $$\int$$
We can add the other elements of each thing by using _ and ^:
\sum_{i=0}^n gives $$\sum_{i=0}^n$$
\prod_{i=0}^n gives $$\prod_{i=0}^n$$
\lim_{x\rightarrow n} gives $$\lim_{x\rightarrow n}$$
\int_{a}^{b} gives $$\int_a^b$$
However, these don't look very good. However, once putting it on display text, either using the brackets or using \displaystyle as said in the beginning of the guide, we can make them look normal.
\displaystyle\sum_{i=0}^n gives $$\displaystyle\sum_{i=0}^n$$
\displaystyle\prod_{i=0}^n gives $$\displaystyle\prod_{i=0}^n$$
\displaystyle\lim_{x\rightarrow n} gives $$\displaystyle\lim_{x\rightarrow n}$$
\displaystyle\int_{a}^{b} gives $$\displaystyle\int_{a}^{b}$$
~Modular Arithmetic~
"\equiv" gives $$\equiv$$
\mod{a} gives $$\mod{a}$$
\pmod{a} gives $$\pmod{a}$$
\bmod{a} is \mod{a} without the space before it: $$a\bmod{b}$$ versus $$a\mod{b}$$
"a\mid b" creates $$a\mid b$$, which states that $$b$$ is divisible by $$a$$.
~Trigonometry~
Many of you simply put "sin" and "cos" and be done with it; however, adding a backslash before those two make it look much better.
\sin gives $$\sin$$ (as opposed to $$sin$$)
\cos gives $$\cos$$ (as opposed to $$cos$$)
\tan gives $$\tan$$
\sec gives $$\sec$$
\csc gives $$\csc$$
\cot gives $$\cot$$
\arcsin gives $$\arcsin$$
\arccos gives $$\arccos$$
\arctan gives $$\arctan$$
Putting a ^{-1} after the trigonometric function designates it as the inverse. For example, \sin^{-1} gives $$\sin^{-1}$$.
\sinh gives $$\sinh$$
\cosh gives $$\cosh$$
\tanh gives $$\tanh$$
~Combinatorics~
\binom{x}{y} gives $$\binom{x}{y}$$
\dbinom{x}{y} gives $$\dbinom{x}{y}$$
~Geometry~
x^{\circ} gives $$x^{\circ}$$ the degree symbol
\angle gives $$\angle$$
\Delta gives $$\Delta$$, for example $$\Delta ABC$$
\triangle also does the job: $$\triangle ABC$$
\odot gives $$\odot$$, for example $$\odot O$$
AB\parallel CD gives $$AB\parallel CD$$
AB\perp CD gives $$AB\perp CD$$
A\cong B gives $$A\cong B$$
A\sim B gives $$A\sim B$$
~Calculus~
We've already learned to use $$\int$$. However, there is much more to calculus than integrals!
There is no command for the total derivative, so you have to use \text{d} to get around it.
For example, \dfrac{\text{d}}{\text{d}x} gives $$\dfrac{\text{d}}{\text{d}x}$$
Fortunately, there is a symbol for partial derivatives: \partial gives $$\partial$$.
So, \dfrac{\partial}{\partial x} gives $$\dfrac{\partial}{\partial x}$$
Double or even triple integrals can be condensed into \iint and \iiint, respectively. This gives $$\displaystyle\iint$$ and $$\displaystyle\iiint$$ (I am using display text).
EXTRA
Line integrals can be written as \oint: $$\displaystyle \oint$$.
~Parentheses~
( and ) are standard for parentheses: $$(a+b)$$
[ and ] are used for brackets: $$[a+b]$$
{ and } are used for curly brackets: $$\{a+b\}$$
\lfloor and \rfloor are used for the floor function: $$\lfloor a+b\rfloor$$
\lceil and \rceil are used for the ceiling function: $$\lceil a+b\rceil$$
\langle and \rangle are used for vectors: $$\langle a,b\rangle$$
The vertical line symbol | (not a capital "i" or a lowercase "l"!) is used for absolute value: $$|a+b|$$
~Fitting Parentheses~
Suppose you want to write $$\left(\dfrac{a}{b}\right)^c$$. When you try, it gives $$(\dfrac{a}{b})^c$$. How did I stretch the parentheses to fit?
To stretch the parentheses, use \left before the left parenthesis and \right before the right one, like this: \left( and \right). When put back into the expression, this yields $$\left(\dfrac{a}{b}\right)^c$$ as desired.
This isn't just for parentheses; you can use them on brackets: $$\{\dfrac{a}{b}\}$$ changes into $$\left\{\dfrac{a}{b}\right\}$$
You can also use this technique on things that use only one parenthesis/bracket/etc. However, just putting \left or \right will yield an error. This is because \left and \right come in pairs. In orer to sidestep this, you can put a period after the one that you do not need (i.e \left. or \right.). This way it will not produce an error, and it will stretch the parenthesis to size. For example, this: \left. \dfrac{x^3+2x}{3x^2}\right|_0^3 gives this: $$\left. \dfrac{x^3+2x}{3x^2}\right|_a^b$$
~Tables and Arrays~~
To make tables and arrays, use \ begin{array}{[modifiers]} ... \ end{array}. (A space is put before "begin" and before "end" to prevent the LaTeX from prematurely rendering. Even though there are no brackets around to make it render, it does so anyways, I don't know why.)
In the modifiers section, you put either l for left, c for center, or r for right, per column. For example, to make an array with 3 columns, all formatted to align along the right edge, you put "rrr" inside the modifier. It would look like this: \ begin{array}{rrr} ... \ end{array}.
To add a vertical line between two columns, put the vertical line symbol | between two modifiers: for example, if you wanted a horizontal line between the first two columns in the previous example, then you would put \ begin{array}{r|rr} ... \ end{array}.
For actual inputting in the array, there are two rules: put a "&" sign to notify to switch to the next column, and put a "\ \" divider (again a space is added in between to prevent it from rendering) to notify to switch to the next row. When building the table, always fill in row by row: in the first row, fill in all the corresponding columns, and then switch to the next row; then continue in this manner. For example, if I wanted to make a $$3\times 3$$ square with the numbers $$1\rightarrow 9$$, I would put: \ begin{array}{lcr}1 & 2 & 3 \ \ 4 & 5 & 6 \ \ 7& 8 & 9 \ end{array}. This produces: $$\begin{array}{lcr}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7& 8 & 9 \end{array}$$.
To insert horizontal lines between any two rows, put \hline after the divider that separates the two rows. For example, if I wanted to add horizontal lines and vertical lines in the previous example to look like a tic tac toe board, this would be my code: \ begin{array}{l|c|r}1 & 2 & 3 \ \ \hline 4 & 5 & 6 \ \ \hline 7& 8 & 9 \ end{array} and it will produce: $$\begin{array}{l|c|r}1 & 2 & 3 \\ \hline 4 & 5 & 6 \\ \hline 7& 8 & 9 \end{array}$$
~Other~
To negate any symbol, put \not before the symbol. For example, "\not =" gives $$\not =$$
Look here for a big list of symbols.
If you don't know how to do something or see something missing in this guide, please do comment below so I can add it! Together, we can make a great LaTeX guide!
Note by Daniel Liu
4 years, 2 months ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
Sort by:
If there is anything that doesn't make sense or is organized bad, please tell me so I can fix it.
I know that regular notes get hopelessly lost in the Feed even just after a few days. However, I wish that somehow this can survive, because I have noticed a lot of people in need in learning some basic LaTeX. Maybe someone can do something about this?
- 4 years, 2 months ago
Just a note here: For limits, I like using "\lim \limits_{a \to b} a" which yields $$\lim \limits_{a \to b} a$$
- 4 years ago
Daniel Liu we can also use "n\choose{r}" to display $$n\choose{r}$$
- 3 years ago
This is a fantastic idea. I think that we should either make a guide accessible like the algebra dictionary, or brilliant should have a link to this (and other support notes like it) in the formatting guide.
Notice that brilliant does have some well chosen examples. I think that it's just a good idea to expound upon them as you are.
- 4 years, 2 months ago
This is a useful idea. I think that we should make a guide accessible like the algebra dictionary.
- 3 years, 8 months ago
This is a fantastic idea. I think that we should either make a guide accessible like the algebra dictionary, or brilliant should have a link to this (and other support notes like it) in the formatting guide.
That would be great!
- 4 years, 2 months ago
Why not make this a wiki??
- 2 years, 1 month ago
- 2 years, 7 months ago
I want to suggest a correction for the Parentheses section.
$$\backslash\{$$ and $$\backslash\}$$ are used for the curly brackets: $$\{a+b\}$$
Note that not giving the slash makes the parentheses disappear when the $$\LaTeX$$ output is rendered.
- 3 years ago
{a and c}
- 2 years, 5 months ago
I'm not sure how your reply is relevant to my comment. You haven't used $$\LaTeX$$. You just wrote it in plain text. My comment illustrated how braces aren't rendered in $$\LaTeX$$ output if you don't escape it using a backslash.
If you're typing in plain text, then there's no need to escape it since plain text is rendered as it is in output.
- 2 years, 5 months ago
Can you write the symbol for infinity in Latex if so how ?
- 3 years, 5 months ago
\infty @Abdur Rehman Zahid
- 3 years, 5 months ago
Thanks
- 3 years, 5 months ago
Yes, if you could put all the produced results first and then the HOW-TOs, that'd be great. It'd be like a LaTeX dictionary. The way it is now is hard to skim through with an eye to find the thing you want in the nexus of information.
A great guide, by the way.
- 3 years, 6 months ago
Awesome note for all learners ! $$\LaTeX$$ is truly useful and the following wikipedia page is also helpful for this purpose.
What I want to add in this note is the $$\LaTeX$$ colors , I try to use them in $$\color{Red}{problems}$$ and $$\color{Green}{Solutions}$$
For that , you have to type
For example, " \color{Green}{Maths} " will appear as $$\color{Green}{Maths}$$
" \color{Blue}{Maths} " will appear as $$\color{Blue}{Maths}$$...
Also the use of " \Huge" , it is used for getting big fonts, like
" \Huge{Maths}" will appear as $$\Huge{Maths}$$
@Daniel Liu , try getting this in the note too .....
- 3 years, 8 months ago
$$\mathbb{\Huge{\color{MidnightBlue}{How \text{ }To \text{ } Swear\text{ } In\text{ } Mathematics, \sqrt{-1}}}}$$
Wow really cool! :)
- 3 years, 7 months ago
$$\Huge\color{green}{NICE}$$
How did you changed the font??
- 2 years, 7 months ago
$$\Huge{\text{Aditya}}{\text{Raut}}$$ Wow! This is cool!!@Aditya Raut
- 3 years, 4 months ago
$$\huge{\color{red}{Z} \color{green}{U} \color{blue}{H} \color{MidnightBlue}{A} \color{lightpink}I \color{Orange}{R}}$$
- 1 year ago
Nice !
For posting solutions , I prefer the following pages :
Hope that helps ! :3
- 4 years, 2 months ago
Nice
- 4 years, 2 months ago
THIS IS AWESOME! HUHUHUHU
- 3 years, 8 months ago
Yeah. :D
- 3 years, 7 months ago
Hey, what course are you in? And we're batchmates, I reckon.
- 3 years, 7 months ago
Civil Engineering. :) And you're Journalism, right? :D
Sorry, I don't understand your use of the word "batchmates". XD
- 3 years, 7 months ago
2012? :)
- 3 years, 7 months ago
Nope, 2013. :)
- 3 years, 7 months ago
Na-feel ko nga kasi 17 ka haha
- 3 years, 7 months ago
:)))
Take note lang ha, 'di tayo parehas ng high school. Wala lang sinabi ko lang haha XD
- 3 years, 7 months ago
Of course, I know. Kilala sana kita if oo. :)
- 3 years, 7 months ago
Another suggestion: \oint for surface integral ($$\oint$$).
- 4 years, 2 months ago
Hmm... I said \oint was for line integral in the Calculus section. Surface integral, line integral, same thing.
- 4 years, 2 months ago
Thank you for this guide; I'm just learning how to use LaTex and this is all extremely helpful.
One question: is there a way to push lines slightly to the right (like the Tab feature in word processors) without pushing them all the way to the middle? For example, in the lines
$${(p-1)}^p \equiv -1 \pmod {p^2}$$
$$\equiv {p^2-1} \pmod {p^2}$$
$$\equiv {(p-1)(p+1)} \pmod {p^2}$$
I would really like to have the equivalence symbol in the second and third lines line up with the one in the first line. Is there a way to do that?
- 3 months, 2 weeks ago
Whats wrong with this?
$$\color{Blue}\text{S = x^2 - 8\lfloor x \rfloor + 10}$$
- 1 year ago
Use \color{Blue} {S = x^2 - 8\lfloor x \rfloor + 10} to get $$\color{Blue} {S = x^2 - 8\lfloor x \rfloor + 10}$$
- 1 year ago
Thanks!!
- 1 year ago
$$\color{Blue}{S = x^2 - 8\lfloor x \rfloor + 10}$$
- 1 year ago
A note on integrals in $$\LaTeX$$: [[ $$\mbox{\iiint}$$ ]] gives $\iiint$ It can also be written as [[$$\mbox{\int \! \! \! \! \int \! \! \! \! \int}$$]], giving $\int \!\!\!\! \int \!\!\!\! \int$ While this is unncessary in Brilliant, it is used in actual $$\LaTeX$$ editors when the "esint" package (which contains \iiint) is not available.
Here, the symbol [[ $$\mbox{\!}$$ ]] removes a certain space between two characters.
- 2 years, 6 months ago
There is a typo the line just above "Tables and Arrays". In Fitting Parentheses
display is an error. \left. \dfrac{x^3+2x}{3x^2}\right|_0^3 should give $$~~~~\color{red}{ \large \ \left. ~~\right |_0^3}$$
- 2 years, 8 months ago
$$\Huge{Brilliant\;Is\;So\;Awesome\;!!!}$$
- 3 years, 4 months ago
This site is a good way to get off the ground quickly with all the different symbols: http://www.codecogs.com/latex/eqneditor.php
Staff - 4 years ago
@Josh Silverman How to add letters in it?
- 4 months, 2 weeks ago
Sorry I don't understand the question, what do you want to add?
Staff - 4 months, 2 weeks ago
Is there a way to add variables in it?
- 4 months, 2 weeks ago
Oh I get it. Yes, for instance if you want to do $$E=mc^2$$ you'd just type in
Staff - 4 months, 2 weeks ago
Thanks Sir.
- 4 months, 1 week ago
- 4 years, 2 months ago
I added tables and arrays, that will probably suffice for making matrices too.
- 4 years, 2 months ago
- 4 years, 2 months ago
Hmm... The link isn't to this post though. Probably a glitch. I explicitly posted the link.
- 4 years, 2 months ago
I suppose that works too. This is a really great post.
- 4 years, 2 months ago
Hey how to do this?
Like writing the catalyst of reaction above the arrowhead like in this https://www.google.co.in/search?q=wurtz+reaction&dcr=0&source=lnms&tbm=isch&sa=X&ved=0ahUKEwisMDt-evYAhVBQo8KHcgzAaoQAUICigB&biw=1366&bih=637#imgrc=F_pnGGhHIpA7NM:
- 1 month, 4 weeks ago
One way of quickly figuring out what the code for a special character is, is this tool.
- 2 months, 3 weeks ago
With my mac \left, \right does not work. Big work as under.
big ( your text big ). Same for left and right.
Four sizes: big,....bigg,......Big,......Bigg.
- 8 months, 1 week ago
We can also use \mathrm{d} for the total derivative command you are talking about.
Like $$\dfrac{\mathrm{d}}{\mathrm{d}x}.e^{x}=e^x$$
:)
- 8 months, 2 weeks ago
How to write Right arrow symbol on LaTex?
- 9 months ago
• Use \rightarrow for $$\rightarrow$$
• Use \Rightarrow for $$\Rightarrow$$
- 9 months ago
Thanks. I just want the second one $$\Rightarrow$$
- 9 months ago
You may also want \implies : $$\implies$$
- 9 months ago
I think you should make a column for the arrows...
- 9 months, 2 weeks ago
Like when we conclude two things from a statement , we use \begin{cases} and \end{cases}.
What do we use when we conclude one thing from two things??Like the opposite kind of braces I want as in \begin{cases} ....
- 1 year ago
What is the issue with this?
I wanted to color only $$102^2$$ but this did not happen
$$100^2 + 101^2 + \color{blue}{102^2} + \cdots + 998^2 + 999^2 + 1000^2$$
- 1 year ago
Use ...\color{blue} {102^2}\color{black}... to unflag it. I don't know the reason though... $$100^2 + 101^2 + \color{blue}{102^2} \color{black} + \cdots + 998^2 + 999^2 + 1000^2$$
- 1 year ago
I got the error...See this
$$100^2 + 101^2 + {\color{blue}{102^2}} + \cdots + 998^2 + 999^2 + 1000^2$$
- 1 year ago
code for less than and greater than plzz
- 1 year ago
• Use \le for $$\le$$
• Use \ge for $$\ge$$
- 1 year ago
i dont need quality
- 1 year ago
OK. Use
• < for $$<$$
• > for $$>$$
- 1 year ago
Who to write infinity??
- 1 year ago
Who to write infinity??
- 1 year ago
\infty
- 3 months, 3 weeks ago
\infty will give you $$\infty$$, infinity symbol.
- 1 year ago
thank you
- 1 year ago
If someone can help me out , please!!!
How can we add a picture and how can we link one page to another ??
- 1 year ago
• Use [text here](link) to add link.
• Use  to add image. (You can as well ignore the caption, put it blank [])
- 1 year ago
It got converted to a link here , but not where I intended to!!!!!
See the caption under the set name by clicking on the link below . It is not getting linked??/ Why?
- 1 year ago
Like I was trying this
'THRILLER'
What is wrong?
- 1 year ago
OK. It's because markdown is disabled over there. Use inside latex : \href{link}{\text{caption}}
- 1 year ago
Thanks a lot.!
- 1 year ago
Can you please mention 'belongs to' sign\symbol?
- 1 year ago
This will help. Type \in to get $$\in$$.
- 1 year ago
Thanks, it worked :)
- 1 year ago
:-)
- 1 year ago
What about lesser than equal to?
- 1 year, 4 months ago
\le will give you $$\le$$
- 1 year, 4 months ago
Thanks
- 1 year, 4 months ago
@Ali Hamaiz Here is the note I was talking to you about.
- 1 year, 10 months ago
- 2 years, 1 month ago
\rightarrow gives $$\rightarrow$$.
If you want a longer one, affix "long". Thus, \longrightarrow gives $$\longrightarrow$$.
- 2 years, 1 month ago
Thank you kuya Jaydee :) See you in automathic
- 2 years, 1 month ago
Comment deleted Aug 30, 2015
[What you want the link to say] (The URL that you want to link to)
Don't add the space between them.
- 2 years, 6 months ago
@Trevor B. Can you tell me how to change the size of image in latex?
- 2 years, 6 months ago
While I don't know how to do that in Brilliant, you can import an image from the graphics package and set the size you want with [length=x cm] after \includegraphic.
- 2 years, 6 months ago
- 2 years, 9 months ago
The $$\LaTeX$$ code for $$\infty$$ is \infty.
- 2 years, 9 months ago
I am trying it from many days and still not able to do so.
- 2 years, 6 months ago
[Brilliant](http://www.brilliant.org) gives you Brilliant. Since this is a Markdown feature and not a $$\LaTeX$$ feature, don't enclose it in slash brackets as you do with math code.
Basically, the syntax is [hyperlink_text](url_to_page).
Here's a screenshot of output corresponding to Markdown code for ease of understanding:
Image
$$~~~\large\overset{\textrm{output}}{\implies}~~~$$
Image
with the hyperlink directing you to the specified url.
- 2 years, 6 months ago
For eg. $\left(\large \cfrac{a+b}{c+\cfrac{d}{e+f}}\right)$
Here,brackets are not able to cover up the whole expression..
Thanks,now it's working.
- 2 years, 6 months ago
Use \left before opening bracket and \right before closing bracket to make the brackets auto-resize themselves to cover up the entire expression.
The basic syntax for the type of brackets you need is $$\text{\left(}\cdots\text{\right)}$$ as opposed to the normal bracketing $$\text{(}\ldots\text{)}$$.
- 2 years, 6 months ago
Thanks a lot....
- 2 years, 6 months ago
- 2 years, 9 months ago
What is wrong with this latex????
[3^{$$4n-3$$}]
- 2 years, 9 months ago
Enclose it in $ - 1 year ago Log in to reply \[ 3^{4n-3}$ I think it is OK see my Latex code.
- 2 years, 9 months ago
$$\text{your text here}$$
- 2 years, 10 months ago
Aditya Chauhan $$\text{your text here}$$ You have to put the $$\backslash ( \backslash )$$ around the Latex
- 2 years, 7 months ago
Thanks...
- 2 years, 10 months ago
$${sec}^{2} \theta + 16{sec}^{2} \phi + 49{sec}^{2} \delta + 8\sec \theta \sec \phi + 56\sec \phi \sec \delta + 14\sec \theta \sec \delta$$
- 3 years ago
Use \sec^2 \delta for $$\sec^2 \delta$$
- 2 years, 6 months ago
For i = 1 , 2 ,$$\cdots$$, n , let $$a_{i}$$ and $$b_{i}$$ be non-negative real numbers. Then $$\left( a_{1} + a_{2}+\cdots + a_{n}\right)$$ $$\left( b_{1}+ b_{2} +\cdots + b_{n}\right)$$ $$\geq$$ $$\left( \sqrt{a_{1}b_{1}} +\sqrt{a_{2}b_{2}} +\cdots +\sqrt{a_{n}b_{n}} \right) ^{2}$$
- 3 years, 2 months ago
Comment deleted Feb 06, 2015
$$\Huge{I like Brilliant} \\$$ I think you wanted this. You missed starting with \ ( and ending with \ )....no space after a \, I had to do it for technical reason.
- 3 years, 5 months ago
thank you
- 3 years, 5 months ago
$$\text{your text here}$$
- 3 years, 6 months ago
$$\text{awesome post}$$
- 3 years, 6 months ago
[@Daniel Liu ], How do I denote vectors by latex?
- 3 years, 7 months ago
@Abhimanyu Swami Here's your query ...type this \stackrel{rightarrow} v and you get this
$$\huge{ \stackrel{\rightarrow} v}$$ and you might also want.. $$\huge{\stackrel{\rightarrow}{|v|}}$$.
- 3 years, 5 months ago
$$\stackrel {\rightarrow} { v^2}$$ .......This is what I get with \stackrel {\rightarrow} { v^2}
- 3 years, 5 months ago
Srry put a slash before the rightarrow word I mean \stackrel{\rightarrow}{v} you'll get this time .
Arya
- 3 years, 5 months ago
Thanks. With the correction and adding {v^2} with in { } that is.....
\stackrel {\rightarrow} { v^2} ..gives :- $$~~~~~~~~~~~~~~ \\ \stackrel {\rightarrow} { v^2}$$
- 3 years, 5 months ago
$$\begin{array}{l|c|r}\huge Normal~Vectors \\\hline\end{array}$$
Use \vec{v} for $$\huge\displaystyle\vec{v}$$
$$\begin{array}{l|c|r}\huge Unit~Vectors \\\hline\end{array}$$
Use \hat{\imath}\;\;\hat{\jmath}\;\;\hat{k} for $$\huge\hat{\imath}\;\;\hat{\jmath}\;\;\hat{k}$$
\imath' and\jmathgets rid of the the dot overiandj.
- 2 years, 6 months ago
Also, what's the latex for putting a black square around the answer when you're writing solutions.
- 3 years, 7 months ago
\boxed { }
@Daniel Liu There are lots of symbols on the AoPS page you linked to, but there is some $$\LaTeX$$ stuff you should definitely add, like systems of equations (we use \begin{cases} .. \\ ... \\ ..\end{cases} for this) and \boxed { }. These don't exist on the AoPS page.
I also like to use \stackrel { } to get some text shown above symbols.
E.g. \stackrel{\text{AM-GM}}\ge gives $$\stackrel{AM-GM}\ge$$. This is optional, though.
- 3 years, 7 months ago
What's the code for =>, not $$\geq$$
- 3 years, 7 months ago
If you mean an arrow, there are quite a few codes for it: \implies or \Rightarrow or \rightarrow or \Longrightarrow or \to or \longrightarrow or \mapsto or \longmapsto.
- 3 years, 7 months ago
thank you, the first one was the one I was looking for
- 3 years, 7 months ago
I find this note very, very nice! :D
I have a question: How should I write the definition of absolute value using LaTeX?
- 3 years, 8 months ago
The best is simply the symbol | on your keyboard. For division there's a better one, though - /mid.
- 3 years, 8 months ago
But how about the brace { used for piecewise defined functions? The definition of absolute value uses that :)
- 3 years, 7 months ago
You'd use the same latex that you'd use for systems of equations.
f(x) = \begin{cases} 1 \\ 2 \end{cases} gives $$f(x) = \begin{cases} 1 \\ 2 \end{cases}$$
Put what you want between \begin{cases} and \end{cases} and separate new lines with \\.
- 3 years, 7 months ago
Oh, that's nice. :) Thanks. :)
- 3 years, 7 months ago
When NOT using "\text " while in Latex,
for " space " we can use "~".
Say:-In Latex.....>\ ( (I~am~~~coming~~~~) \ ) gives:-
(I#am###coming####)...... "#" stands for space here.
- 3 years, 8 months ago
One more thing, for summation and production, it's required to put curly brackets { } between $$n$$ if you have more than 1 characters on $$n$$. Otherwise it'll show as $$\displaystyle\sum_{i=0}^2014 i$$ instead of $$\displaystyle\sum_{i=0}^{2014} i$$. (Note that other brackets don't work.)
- 3 years, 8 months ago
One question how did you make "LATEX" look like that .
- 3 years, 8 months ago
\LaTeX
- 3 years, 8 months ago
Thanks man. Been looking for something like this.
- 3 years, 8 months ago
How to write Pi in latex?
- 4 years ago
Use \pi
$$\pi$$
Something I never got around to asking here is, how do you get the light grey line to denote section breaks? When I add a section break, I put \text{............................................................................................................} between the display delimiters, but I never figured out how to automate that with the grey lines.
- 4 years ago
by putting three or more underscores in a row. like this: _
- 4 years ago
Can you explain a little bit more about the above $$\LaTeX$$ use
and what about squaring the mod of a vector I mean
$$\huge{\stackrel{\rightarrow}{|v|}}$$
if I try to square it then it becomes either $$\huge{\stackrel{\rightarrow}{|v|^2}}$$ or $$\huge{{\stackrel{\rightarrow}{|v|}^2}}$$
- 3 years, 5 months ago
Use \huge{\left|\vec{v}\right|}^2 for $${\left|\vec{v}\right|}^2$$
- 2 years, 6 months ago
See how it is written here by putting cursor over it. .
$$| \vec v|^2$$
- 3 years, 5 months ago
---` will give :
- 2 years, 6 months ago
Are geometry diagrams made with latex? How?
- 4 years ago
No. Geometric Diagrams are made with a typesetting language called Asymptote Vector Graphics Language.
- 4 years ago
Nope.
- 4 years ago
This is really helpful, thanks a lot Daniel!!
- 4 years ago
May I ask how can I create tables in LaTeX?
- 4 years, 2 months ago
I have written the guide to tables and arrays. At first I thought that I shouldn't write it because this was a basic latex guide, but since you asked for it, I put it on. See if you can understand what I wrote, and if you can't you can tell me why and I will change it.
- 4 years, 2 months ago
Yup! I understood! Thanks a lot!!! :)
- 4 years, 2 months ago
Thanks Daniel.
- 4 years, 2 months ago
Just one thing that's quite common: $$a \mid b$$ for divisibility use \mid to create the line. Other than this, the post is great! Thanks for it! :)
- 4 years, 2 months ago
- 4 years, 2 months ago
how do we type greater,lesser,greater or equal to and lesser or equal to symbol by LaTeX?
- 2 months ago
• $$\backslash(>\backslash)$$ appears as $$>$$
• $$\backslash(<\backslash)$$ appear as $$<$$
• $$\backslash(\text{\ge}\backslash)$$ appear as $$\ge$$
• $$\backslash(\text{\le} \backslash)$$ appear as $$\le$$
- 2 months ago
Thanks!
- 2 months ago
$$1+2+3=6$$
- 3 years, 5 months ago
Comment deleted Feb 06, 2015
@Abdur Rehman Zahid Enclose it in \ ( \ .....\ ) but don't leave spaces
- 3 years, 4 months ago
Thanks,I know now
- 3 years, 3 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": 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": 0.9952314496040344, "perplexity": 3357.1137661670086}, "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-2018-13/segments/1521257648000.93/warc/CC-MAIN-20180322190333-20180322210333-00122.warc.gz"} |
https://arxiv.org/abs/1506.03062 | hep-ex
(what is this?)
# Title: Search for resonant t t-bar production in proton-proton collisions at sqrt(s) = 8 TeV
Abstract: A search is performed for the production of heavy resonances decaying into top-antitop quark pairs in proton-proton collisions at sqrt(s) = 8 TeV. Data used for the analyses were collected with the CMS detector and correspond to an integrated luminosity of 19.7 inverse femtobarns. The search is performed using events with three different final states, defined by the number of leptons (electrons and muons) from the t t-bar to W b W b decay. The analyses are optimized for reconstruction of top quarks with high Lorentz boosts, where jet substructure techniques are used to enhance the sensitivity. Results are presented for all channels and a combination is performed. No significant excess of events relative to the expected yield from standard model processes is observed. Upper limits on the production cross section of heavy resonances decaying to t t-bar are calculated. A narrow leptophobic topcolor Z' resonance with a mass below 2.4 TeV is excluded at 95% confidence level. Limits are also derived for a broad Z' resonance with a 10% width relative to the resonance mass, and a Kaluza-Klein excitation of the gluon in the Randall-Sundrum model. These are the most stringent limits to date on heavy resonances decaying into top-antitop quark pairs.
Comments: Replaced with published version. Added journal reference and DOI Subjects: High Energy Physics - Experiment (hep-ex) Journal reference: Phys. Rev. D 93 (2016) 012001 DOI: 10.1103/PhysRevD.93.012001 Report number: CMS-B2G-13-008, CERN-PH-EP-2015-126 Cite as: arXiv:1506.03062 [hep-ex] (or arXiv:1506.03062v2 [hep-ex] for this version)
## Submission history
From: The CMS Collaboration [view email]
[v1] Tue, 9 Jun 2015 19:47:05 GMT (596kb,D)
[v2] Fri, 15 Jan 2016 17:35:56 GMT (594kb,D) | {"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": 0.9305118322372437, "perplexity": 2663.0508130778044}, "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-2018-39/segments/1537267160853.60/warc/CC-MAIN-20180925004528-20180925024928-00455.warc.gz"} |
http://web2.0calc.com/questions/how-can-i-find-the-end-behavior | +0
# how can i find the end behavior?
+1
54
2
How can i find the end behavior with these two functions?
Guest Oct 3, 2017
Sort:
### 2+0 Answers
#1
+7067
+1
How can i find the end behavior with these two functions?
$$f(x)=\frac{-6x}{3x-5}\\ f(x)=x^2-4x-5$$
$$\frac{-6x}{3x-5}=x^2-4x-5$$
$$-6x=(x^2-4x-5)\times (3x-5)\\ -6x=3x^3-5x^2-12x^2+20x-15x+25\\ \color{blue}3x^3-17x^2+11x+25=0$$
$$x_1 \approx-.88231\\ x_2\approx 2.1443\\ x_3 \approx 4.4041\\ WOLFRAM\ |\ ALPHA$$
!
asinus Oct 3, 2017
#2
+90610
+1
Mmm I do not think that is what the guest wanted asinus :)
$$f(x)=\frac{-6x}{3x-5}$$
This has an asymptote at
3x-5=0
x=5/3
lim as x tends to 5/3 from above = - infinity
lim as x tends to 5/3 from below = + infinity
As x tends to +infinity y tends to -2
As x tends to -infinity y tends to -2
Here is the graph.
Melody Oct 3, 2017
### 4 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | {"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": 0.8787158131599426, "perplexity": 3716.9817579424443}, "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-2017-43/segments/1508187825147.83/warc/CC-MAIN-20171022060353-20171022080353-00658.warc.gz"} |
https://tex.stackexchange.com/questions/481537/modulo-2-binary-long-division-in-european-notation/481552 | # Modulo 2 binary long division in European notation [duplicate]
I need to represent binary modulo 2 long division in my tex document. Notation needed is same as
https://en.wikipedia.org/wiki/Long_division#Eurasia
under Austria, Germany, etc.
I know about longdiv package, but it doesn't seem to support this.
Is there any package to achieve this? If not, how can I manually do this.
## marked as duplicate by JouleV, Raaja, user36296, Stefan Pinnow, flavMar 27 at 5:46
• You can manually draw it using TikZ, but it is a bit painful, and very time-consuming if you have a lot of such divisions. – JouleV Mar 26 at 14:49
• – Steven B. Segletes Mar 26 at 15:08
The fresh new version of longdivision package v. 1.1.0 has almost the desired output you want, with the new german style. As TeXlive 2018 is currently frozen, you cannot use textlive utility for updating this package, but simply download the longdivision.sty file from here and add it in your local texmf directory or in place it along with your .tex file in the same directory.
\documentclass{article}
\usepackage{longdivision}
\begin{document}
\longdivision[style=german]{127}{4}
\end{document}
The differences with the output from Wikipedia are :
• no negative sign displayed for the subtraction operation
• dots instead of comma for the decimal separator
The documentation show a command \longdivdefinestyle for modifying the display of the output, but I'm not yet able to add a negative sign for the operation, nor suppress the dots.
• I'm sure your answer will make many happy users. – Steven B. Segletes Mar 26 at 16:29
The German style?? as depicted here:
\documentclass[12pt]{article}
\usepackage{mathtools}
\usepackage[TABcline]{tabstackengine}
\TABstackMath
\begin{document}
\tabbedShortunderstack[r]{
&12&7& & &:\ 4\ =\ 31.75\\
-&12& & & &\\
\TABcline{2}
& 0&7& & &\\
& -&4& & &\\
\TABcline{3}
& &3&0& &\\
& -&2&8& &\\
\TABcline{3-4}
& & &2&0&\\
& &\mathllap{-}&2&0&\\
\TABcline{4-5}
& & & &0&
}
\end{document}
Here, I emulate the Cyprus/France version cited in the OP's link
\documentclass[12pt]{article}
\usepackage[TABcline]{tabstackengine}
\TABstackMath
\begin{document}
\begin{tabular}{r@{}|@{}l}
\tabbedShortunderstack[r]{
63&5&9\\
-51& &\\
\TABcline{1}
12&5&\\
-11&9&\\
\TABcline{1-2}
&6&9\\
-&6&8\\
\TABcline{2-3}
& &1
}
&
\tabbedShortunderstack[l]{
17&\\
\TABcline{1-2}
37&4
}
\end{tabular}
\end{document} | {"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": 0.8666360378265381, "perplexity": 4494.391118999653}, "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-2019-22/segments/1558232257601.8/warc/CC-MAIN-20190524084432-20190524110432-00042.warc.gz"} |
http://mathhelpforum.com/calculus/15662-find-equation-curve.html | # Math Help - Find an equation of the curve.
1. ## Find an equation of the curve.
Find an equation of the curve that satisfies dy/dx=8x^7y and whose y-intercept is 6.
2. Originally Posted by asnxbbyx113
Find an equation of the curve that satisfies dy/dx=8x^7y and whose y-intercept is 6.
The function evaluated at $x=0$ implies that $y=6$. Thus, $y(0)=6$.
$y' = 8x^7 y$ note $y\not = 0$,
$\frac{y'}{y} = 8x^7$
$\int \frac{y'}{y} dx = \int 8x^7dx$
$\ln |y| = x^8 + C$
$y = Ce^{x^8} \mbox{ where }C>0$
In order to satisfy the IVP we need $C=6$. | {"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": 10, "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": 0.97225022315979, "perplexity": 777.6439983960466}, "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/1461862047707.47/warc/CC-MAIN-20160428164727-00102-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://www.bartleby.com/questions-and-answers/let-fx3x25x10.-answer-the-following-questions.-1.-find-the-average-slope-of-the-function-ff-on-the-i/51ae448d-d0e0-4ad3-8761-c5db2e32131c | # Let f(x)=3x2+5x−10. Answer the following questions.1. Find the average slope of the function ff on the interval [−1,1]. Average Slope: m =2.Verify the Mean Value Theorem by finding a number cc in (−1,1)(−1,1) such that f′(c)=m Answer: c=
Question
75 views
Let f(x)=3x2+5x−10. Answer the following questions.
1. Find the average slope of the function ff on the interval [−1,1].
Average Slope: m =
2.Verify the Mean Value Theorem by finding a number cc in (−1,1)(−1,1) such that f′(c)=m
check_circle
Step 1
1.
Result used:
Let f be defined on the closed interval [a, b]. The average slope of f between a and b is the quotient
Step 2
Consider the function, f(x)=3x2+5x-10.
Here a=–1 and b=1.
Compute the average slope using the above formula as follows.
Step 3
2.
Result used:
Suppose f(x) is a function which is continuous on the closed interval [a, b] and differentiable on the open interval (a, b).
Then there is a number c such...
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Tagged in
MathCalculus | {"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": 0.9497562646865845, "perplexity": 1713.5555341227487}, "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-2020-16/segments/1585371813538.73/warc/CC-MAIN-20200408104113-20200408134613-00096.warc.gz"} |
http://www-old.newton.ac.uk/programmes/SPD/seminars/2010052609451.html | SPD
Seminar
SPDEs and parabolic equations in Gauss-Sobolev spaces
Chow, P (Wayne State)
Wednesday 26 May 2010, 09:45-10:45
Seminar Room 1, Newton Institute
Abstract
Examples of linear and nonlinear parabolic equations in Hilbert spaces are given by the Kolmogorov equation and the Hamilton-Jacobi-Bellman equation related to SPDEs. In this talk we shall consider a class of semilinear parabolic equations in a Gauss-Sobolev space setting. By choosing a proper reference Gaussian measure, it will be shown that the existence and regularity of strong (variational) solutions can be proven in a similar fashion as parabolic equations in finite dimensions. The results are applied to two singular perturbation problems for parabolic equations containing a small parameter
Video
The video for this talk should appear here if JavaScript is enabled.
If it doesn't, something may have gone wrong with our embedded player.
We'll get it fixed as soon as possible. | {"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": 0.8050510883331299, "perplexity": 577.2537735310939}, "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-2014-42/segments/1414119649048.35/warc/CC-MAIN-20141024030049-00028-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://cai.zone/2010/11/graph-sketching/ | # Graph sketching
Oh god, I'm going to talk about maths for a bit. Ignore me. You might find this interesting if you know what a derivative is and what it means.
This is a curiosity that my friend Anupam came up with. I'm not claiming any credit for noticing it but I thought I'd still share it with y'all. It only requires very rudimentary calculus, but it produces a result which, the more I think about it, the less intuitive it seems.
Let's think about how to sketch the graph of . Well, we know what the graph of looks like:
and we know that the graph of exists always between +1 and -1. So we can make a pretty good guess that for negative , where is very small (less than 1), the graph will look something like the graph for :
But what about for positive , where is now larger than and is starting to grow faster? We might naïvely assume that continues to look like "a wavier version" of as it does for negative . Well, let's look at where these "waves" might be by looking at the derivative:
When is this zero?
But we know that is at largest 1 and that is always larger than 1 for positive , so this derivative will never be 0! So we know that when sketching the rest of the graph for positive , it will be monotonically increasing with no obvious "waves".
But what about higher-order waves?
Well, now that we've got this far we might spot that the further derivatives of don't get much more exciting:
And, since each of , , , are no larger than 1, we can see that these derivatives will always be positive for positive , and so won't have any stationary points, points of inflection, or any turns of any order for . Great!
But what else do we know which might help us sketch it? Well, we know that because is bounded by -1 and 1, must exist within the envelope of and , which look like this:
They grow very fast after , right? It hardly looks surprising that doesn't get a chance to wave around if it's bounded in that tiny space! How does it fit within the gap, though? Well, because oscillates between its bounds of +1 and -1 regularly and infinitely often, should oscillate between its bounds of and also regularly and infinitely often. This is obvious for negative , but it's still true for positive . will touch its bounds regularly and infinitely-many times.
So here is where I start to worry.
We've got a function which we've seen should oscillate regularly between two monotonically smooth bounds, touching them both infinitely many times; yet it itself is always increasing and never turns at any order. I don't know about you, but these two facts just don't sit together with me.
But at least we can now draw it nicely.
It's hard to see, given how fast everything grows, but it does in fact oscillate between its bounds as regularly as does.
Pretty cool, huh?!
Here's a zoomed-out version of that graph so you can see it better. One of the transition of negative to positive , one only of positive and with the aspect changed to demonstrate the oscillation as much as possible:
## 17 comments
1. send it to Daryl - - he's on facebook so friend him - he would be just fascinated!!
2. for intuition, i imagine two concentric circles and an ellipse sharing the same center and intersecting each circle twice as its radius oscillates between the inner and the outer circle's radius. because the circular bounds are themselves curved, the ellipse can vacillate between them without ever inverting its curvature from concave to convex.
3. This is actually really interesting and – surprisingly – easy enough to follow and appreciate for just us with just a lowly A-level understanding of maths.
4. Great post! More of these please ;)
5. That's a nice way to think about it, Niko! Though even the polar equation for the ellipse has some vanishing derivatives, doesn't it? What I found striking about this example was that not only does the function never decrease or change its curvature, ALL its infinite derivatives are strictly positive yet still vacillates! Also, good use of the word vacillate! ;)
6. brainstorming: the curvature need only oscillate between larger and smaller than the curvature of the bounds, not between positive and negative. so now, yes, we might expect higher derivatives to reflect this oscillation of the curvature. but derivatives are defined relative to the axes and curvature is axes-independent. that's why this intuition might be wrong. do you sense the answer just around the corner like i do?
7. brainstorming: the curvature need only oscillate between larger and smaller than the curvature of the bounds, not between positive and negative. so now, yes, we might expect higher derivatives to reflect this oscillation of the curvature. but derivatives are defined relative to the axes and curvature is axes-independent. that's why this intuition might be wrong. do you sense the answer just around the corner like i do?
8. Right on the head, Niko, great point! Seems like curvature is a horrible thing to define for curves in Euclidean space, it depends on a choice of parameterisation of the curve. But while curvature is axes-independent, it's still possible to find for a given pair of axes; for *functions* (ie, many-to-one) y=f(x) defined wrt the axes it can be found:k = y'' / (1+y'^2)^(3/2)(via wikipedia/mathworld). But your intuition of seeing how curvature can fluctuate where derivatives do not seems like the key. Perhaps to truly get to the bottom of this in a more general sense will require some serious calculus...
9. Right on the head, Niko, great point! Seems like curvature is a horrible thing to define for curves in Euclidean space, it depends on a choice of parameterisation of the curve. But while curvature is axes-independent, it's still possible to find for a given pair of axes; for *functions* (ie, many-to-one) y=f(x) defined wrt the axes it can be found:k = y'' / (1+y'^2)^(3/2)(via wikipedia/mathworld). But your intuition of seeing how curvature can fluctuate where derivatives do not seems like the key. Perhaps to truly get to the bottom of this in a more general sense will require some serious calculus...
10. Thanks, both of you :)
11. Curvature does not depend on parameterization. It is the how far you turn the steering wheel if you drive along the curve at unit speed. The Gauss Bonnet theorem says if you drive round a smooth simple closed curve in the plane the total curvature (=integral of curvature round curve) is 2Pi: basically you have gone full circle. Curvature is a lovely thing.
12. Oh, my mistake! A temporary lapse in my ability to think >_< I looked up the definition of curvature and saw it defined in terms of a chosen parameterisation. But *of course* it's invariant of the choice! Thanks for setting the record straight! :!
13. I know little logic but find myself using non-standard analysis a la Robinson. It makes me want to learn a bit more about the "dark arts" (=logic)
14. Curvature does not depend on parameterization. It is the how far you turn the steering wheel if you drive along the curve at unit speed. The Gauss Bonnet theorem says if you drive round a smooth simple closed curve in the plane the total curvature (=integral of curvature round curve) is 2Pi: basically you have gone full circle. Curvature is a lovely thing.
15. Oh, my mistake! A temporary lapse in my ability to think >_< I looked up the definition of curvature and saw it defined in terms of a chosen parameterisation. But *of course* it's invariant of the choice! Thanks for setting the record straight! :!
16. Perhaps its just me, with my inferior computer science degree :P, but first derivative for +ve x is never 0 because the curve doesn't have any turning points at least not with respect to x-axis, because of the exponential nature of exp(x) + sin(x). But instead oscillates with respect to y-axis. Which also checks out with dx/dy right? Am I being wrong?
17. Interesting to think about but y = e^x + sin xisn't easily writable in terms of y, I think, because it'd be a one-to-many assignment...And at least for negative x the graph is definitely oscillating wrt the x axis. When the exp starts to grow faster, though, I'm not sure what to think... | {"extraction_info": {"found_math": false, "script_math_tex": 40, "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": 46, "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": 0.8330981135368347, "perplexity": 728.478121280997}, "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-2021-31/segments/1627046153814.37/warc/CC-MAIN-20210729011903-20210729041903-00597.warc.gz"} |
https://physics.stackexchange.com/questions/480151/real-and-imaginary-time-greens-functions | # Real and Imaginary time Green's Functions
In real time, one can calculate the two point function of a given theory using
$$$$G(\vec{x},t)=\langle \Omega | \phi(\vec{x},t)\phi^\dagger (0,0)|\Omega\rangle =\int_{\phi(0,0)}^{\phi(\vec{x},t)} \mathcal{D}\phi\ e^{-\frac{i}{\hbar}S[\phi]}\phi(\vec{x}',t')\phi(\vec{x},t)$$$$
where the limits of the path integral should match the initial and final state.
On the other hand, I know that the generating functional $$\mathcal{Z}$$
$$$$\mathcal{Z}=\langle\phi'|e^{-iHT}|\phi\rangle=\int_\phi^{\phi'}e^{-\frac{i}{\hbar}S[\phi]}$$$$
can be identified with the quantum partition function $$Z$$ if we evaluate on imaginary time $$t=-i\tau$$ and we trace over the initial and final state
$$$$Z=\sum_{\phi}\langle \phi | e^{-\beta H}|\phi\rangle =\int_{\phi(0)=\phi(\beta)}e^{-\beta S_E[\phi]}$$$$
So the relation between quantum mechanical and thermodynamic expectation values is: analytically continuate $$t\rightarrow -i\tau$$ with period $$\tau \in [0,\beta]$$, set inital and final states equal and sum over them. Now, in every book I see, the real time Green's function
$$$$G(\vec{x},t)= \langle \Omega | \phi(\vec{x},t)\phi^\dagger (0,0)|\Omega\rangle$$$$
and the imaginary time Green's function
$$$$\mathcal{G}(\vec{x},\tau)= \frac{1}{Z}\text{Tr}\Big[e^{-\beta H} \phi(\vec{x},\tau)\phi^\dagger (0,0)\Big]$$$$
are related by
$$$$G(\vec{x},t)=\mathcal{G}(\vec{x},i\tau)$$$$
Meaning that we could basically define just one function $$G(\vec{x},z)$$ with $$z\in\mathcal{C}$$ that is equal to the Green Function of QM for real $$z$$ and equal to the thermodynamic average for imaginary $$z$$.
My question is the following
In the Path integral formalism, there were to things we needed to do to go from one average to the other; we need to go to imaginary time and we need to do something about the trace. In the Green's functions, however, it seems that going to imaginary time is enough, as if the trace gets automatically taken care of. How is that so?
• So the question (v1) is actually just about the trace procedure, not the Wick rotation? – Qmechanic May 15 at 9:11
• The question is that in the formalism of path integrals, quantum expectation values and thermodynamic averages can be identified by analytically continuing to imaginary time AND changing the boundary conditions on path integral. However, for Green functions it seems that only going to imaginary time is enough... If I were to express the Green functions as path integrals, I don't see how the boundary conditions get fixed too. – P. C. Spaniel May 15 at 16:21
• My understanding is that, to do finite temperature field theory, one goes to euclidean signature(wick rotation) and compactifies the time direction -the period of which gives the inverse temperature $\beta$. One then calculates the partition function and wick rotate back to real time, your last equation, $\mathcal{G}(\vec{x},\tau)= \frac{1}{Z}\text{Tr}\Big[e^{-\beta H} \phi(\vec{x},\tau)\phi^\dagger (0,0)\Big]$. After you wick rotate back, you don't do anything to the periodic boundary condition you imposed while working with imaginary time. – levitt May 18 at 9:01
@levitt has almost provided the correct answer in his comment. Although, I think he should also emphasize something that he probably implicitly implied in his comment above: that the equality $$G(\vec x,t) = \mathcal G(\vec x,i t)$$ as written in the original question is incorrect (apart from the typo where the argument of $$\mathcal G$$ is $$\tau$$ and not $$t$$).
$$\mathcal G(\vec x,i t)$$ computes real time correlation function in a field theory at finite temperature while $$G(\vec x,t)$$ (as written in the first equation of the question) computes the real time correlation function at zero temperature. These two correlation functions are different. You can obtain the zero temperature correlation function by taking $$\lim\limits_{\beta\to\infty}\mathcal G(\vec x,i t)$$. It should be true that $$G(\vec x,t) = \lim\limits_{\beta\to\infty}\mathcal G(\vec x,i t)$$. | {"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": 23, "wp-katex-eq": 0, "align": 0, "equation": 6, "x-ck12": 0, "texerror": 0, "math_score": 0.9586591124534607, "perplexity": 431.4703925924475}, "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-2019-43/segments/1570987833089.90/warc/CC-MAIN-20191023094558-20191023122058-00425.warc.gz"} |
https://www.omnicalculator.com/physics/hydraulic-jump | # Hydraulic Jump Calculator
Created by Bogna Szyk
Reviewed by Dominik Czernia, PhD and Steven Wooding
Last updated: Oct 26, 2022
If you're interested in hydraulic engineering, this hydraulic jump calculator will be right up your alley. With its help, you can determine all characteristics of a hydraulic jump, including the flow velocity up and downstream, the height and length of the jump, and the total head loss.
We will also provide you with a Froude number equation and teach you how to use it to identify the type of hydraulic jump.
Make sure to take a look at the Reynolds number calculator, too!
## What is a hydraulic jump?
By definition, a hydraulic jump happens when the flow of a fluid changes from supercritical to subcritical. The abrupt change in flow characteristics is accompanied by substantial energy losses, as well as turbulence in the flow.
The video below illustrates an example of a hydraulic jump (which occurs at the end of the video). You can observe a change in water level and a turbulent transition zone.
What exactly are the supercritical and subcritical flows, then? To understand these terms, you first need to know what a critical depth is – the depth of flow in which the energy is at a minimum for a given discharge $Q$. Once you know the critical depth, you can instantly distinguish between:
• Supercritical flow (also called a rapid flow) – flow depth is lower than the critical depth; and
• Subcritical flow (also called a slow flow) – flow depth is higher than the critical depth.
## The Froude number equation
Instead of analyzing the energy of the flow, hydraulic engineers use the Froude number to determine whether a flow is supercritical or subcritical. For any open channel flow, it can be calculated with the following formula:
${\rm{Fr}} = \frac{v}{\sqrt{gD}},$
where:
• $\rm Fr$ – Froude number of the flow;
• $v$ – Flow velocity (see velocity calculator);
• $g$ – Gravitational acceleration (see gravitational force calculator); and
• $D$ – Flow depth.
If the value of the Froude number is greater than one, the flow is supercritical. If it's smaller than one, the flow is supercritical.
Let's examine the Froude number equation more closely. As it is a dimensionless number, the denominator must be expressed in the units of speed – meters per second or feet per second. But does this value have any physical interpretation?
In fact, it does. The value in the denominator, $\sqrt{gD}$, is the velocity of wave propagation, also called the wave celerity. For example, throwing a pebble into water creates ripples on the surface that would propagate with this celerity.
If the flow is supercritical (Fr > 1), the velocity of the flow is higher than the celerity. It means that any waves or disturbances in the flow are transmitted downstream.
On the other hand, if the flow is subcritical (Fr < 1), the velocity of the flow is smaller than the celerity. Waves and disturbances in the flow will propagate upstream.
What happens in the case of a critical flow? As the Froude number equals 1, velocity will be equal to celerity. In this scenario, any disturbance in the flow will stay in one place, moving neither upstream nor downstream.
## Calculating the hydraulic jump properties
Our hydraulic jump calculator analyzes the phenomenon of a hydraulic jump in detail. It finds multiple characteristics, even including jump efficiency.
Remember that these formulas are only valid for the following assumptions:
• We're considering an open channel flow;
• The channel is rectangular and horizontal (without a slope); and
• The jump occurs from a supercritical to a subcritical flow.
1. Flow rate formula
For both upstream and downstream flows, the discharge $Q$ is equal to flow velocity $v$ multiplied by the surface area of the channel cross-section. As we assume a non-varying rectangular channel, this formula can be written down as:
$Q = vyB,$
where:
• $Q$ – Discharge (in m³/s or cu ft/s);
• $v$ – Flow velocity (in m/s or ft/s);
• $y$ – Flow depth (in m or ft), and
• $B$ – Channel width (in m or ft).
2. Conjugate depth equation
The momentum functions of the flow depths upstream and downstream are equal (see conservation of momentum calculator). Thanks to this property, we can say that the depths $y_1$ (upstream) and $y_2$ (downstream) are conjugate depths, and use the formula below:
$\frac{y_2}{y_1} = \frac{1}{2}\left(\sqrt{1 + 8 \times {\rm Fr_1^2}} - 1\right),$
where:
• $y_2/y_1$ – Depth ratio (visible in the advanced mode of the hydraulic jump calculator); and
• $\rm Fr_1$ – Froude number of the flow upstream.
In any hydraulic jump, although momentum is conserved (what allows us to use the conjugate depth equation), some energy is lost. We can call this an energy loss or a head loss $\Delta E$. It can be expressed with the following formula:
$\Delta E = \frac{(y_2 - y_1)^3}{4y_1 y_2}$
4. Hydraulic jump length equation
Numerous field and laboratory experiments have found the equation determining the length of a hydraulic jump. You should remember, though, that this is only an approximation – the actual length of the hydraulic jump might vary depending on many additional factors, such as the flow turbulence.
$L = 220 y_1 \tanh[({\rm Fr_1} - 1) / 22]$
5. Hydraulic jump height formula
The height of the hydraulic jump is simply the difference between the flow depth downstream and upstream. As the flow upstream is supercritical, it will always be at a lower flow depth:
$h = y_2 - y_1$
6. Jump efficiency equation
You can also estimate what percentage of the original energy is lost during a hydraulic jump (jump efficiency $\eta$). This value is dependent mostly on the Froude number of the upstream flow $\rm Fr_1$ and is governed by the following equation:
$\small \eta = \frac{(8 {\rm Fr_1^2} + 1)^{3/2}\! - \!4 {\rm Fr_1^2} + 1}{8 {\rm Fr_1^2} (2 + {\rm Fr_1^2})} \!\times\! 100$
## Types of hydraulic jumps
The way a hydraulic jump looks depends strongly on the upstream Froude number $\rm Fr_1$. The ranges below show different types of jumps that occur in flow patterns.
• $\rm Fr_1 < 1.7$: undular jump. The jump is minimal, which results in small undulations on the surface and very low energy loss.
• $\rm 1.7 < Fr_1 < 2.5$: weak jump. the jump is still considered small, and the energy dissipation is quite low.
• $\rm 2.5 < Fr_1 < 4.5$: oscillating jump. Irregular waves are generated during this jump. Energy losses start to be considerable.
• $\rm 4.5 < Fr_1 < 9$: steady jump. The jump is steady and constricted to one location. Energy losses can reach 70%.
• $\rm Fr_1 > 9$: strong jump. In such jumps, supercritical water jets are formed, the difference in velocities up and downstream is high, and the energy dissipation can reach 85%.
Bogna Szyk
Gravitational acceleration (g)
ft/s²
Channel width (B)
ft
Discharge (Q)
cu ft
/s
Upstream
Upstream depth (y₁)
ft
Upstream velocity (v₁)
ft/s
Upstream Froude number (F₁)
Downstream
Downstream depth (y₂)
ft
Downstream velocity (v₂)
ft/s
Downstream Froude number (F₂)
Hydraulic jump properties
Depth ratio (y₂/y₁)
ft
Jump length (L)
ft
Jump height (h)
ft
Jump efficiency
%
Jump type
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http://perl143.blogspot.com/2012/11/zip-command-in-linux.html | ## Friday, 23 November 2012
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The following examples illustrate typical uses of the command zip for packaging a set of files into an "archive" file, also called "zip file". The command uses the standard zip file format. The archive files can therefore be used to tranfer files and directories between commonly used operating systems.
``` zip archivefile1 doc1 doc2 doc3 ```
This command creates a file "archivefile1.zip" which contains a copy of the files doc1, doc2, and doc3, located in the current directory.
``` zip archivefile1 * ```
This command creates a file "archivefile1.zip" which contains a copy of all files in the current directory in compressed form. However, files whose name starts with a "." are not included. The extension ".zip" is added by the program.
``` zip archivefile1 .* * ```
This version includes the files that start with a dot. But subdirectories are still not included.
``` zip -r archivefile1 . ```
This copies the current directory, including all subdirectories into the archive file.
``` zip -r archivefile2 papers ```
This copies the directory "papers", located in the current directory, into "archivefile2.zip".
``` zip -r archivefile3 /home/joe/papers ```
This copies the directory "/home/joe/papers" into "archivefile3.zip". Since in this case the absolute path is given, it doesn't matter what the current directory is, except that the zip file will be created there.
The command unzip extracts the files from the zip file.
` unzip archivefile1.zip `
This writes the files extracted from "archivefile1.zip" to the current directory.
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http://mathhelpforum.com/algebra/50899-solved-sat-math-page-653-a.html | # Thread: [SOLVED] SAT math page 653
1. ## [SOLVED] SAT math page 653
8, 17, 26, 35, 44, ...
The first 5 terms in a sequence are shown above. Each term after the first is found by adding 9 to the term immediately preceding it. Which term in this sequence is equal to 8+(26-1)9?
a) the 8th
b) the 9th
c) the 25th
d) the 26th
e) the 27th
I picked the 27th since i remembered there was a formula that was (n-1). But the correct is d) the 26th. why? Instead of writing out all the terms is there another other way? A faster way? Thanks in advance!
2. Originally Posted by fabxx
8, 17, 26, 35, 44, ...
The first 5 terms in a sequence are shown above. Each term after the first is found by adding 9 to the term immediately preceding it. Which term in this sequence is equal to 8+(26-1)9?
a) the 8th
b) the 9th
c) the 25th
d) the 26th
e) the 27th
I picked the 27th since i remembered there was a formula that was (n-1). But the correct is d) the 26th. why? Instead of writing out all the terms is there another other way? A faster way? Thanks in advance!
It is almost correct
See how it is defined :
$a_{\color{red}n}=a_1+({\color{red}n}-1)r$
So if you're given $8+({\color{red}26}-1)9$, can you see what ${\color{red}n}$ is ?
3. Oh haha. Clumsy me. Thanks moo | {"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": 0.9445139765739441, "perplexity": 900.0502344976583}, "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-2013-48/segments/1386163066152/warc/CC-MAIN-20131204131746-00013-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2829789/combinatorics-how-many-possible-paths | # Combinatorics: How many possible paths?
Consider the following grid.
We start at the bottom left corner. We may only move one step up or one step right at each move. This procedure is continued until we reach the top right corner. How many different paths from the bottom left corner to the top right corner are possible?
This is an excerise from Sheldon Ross' A First Course in Probability.
The correct answer is $$\binom{7}{4}=35.$$ However, no explanation is given. I understand that there are $7$ moves total that must be made: $$3\ \text{steps up and}\ 4\ \text{steps to the right.}$$ Can anyone provide me with an explanation of how the author came up with this answer?
Thanks so much!
• You can describe a path as a list of seven symbols, each with four Rs and three Us, for up and right. Like URRURUR. Of the 7 symbols, 4 must be chosen to be R, hence $\binom74$. – Mike Earnest Jun 23 '18 at 21:07
• Thank you! That makes a lot of sense because $\binom{7}{3}$ also equals $35.$ – MathIsLife12 Jun 23 '18 at 21:09
What Mike Earnest said is simpler, but here's a strategy that also applies, for example, for 3-D paths.
Note that the number of paths is equal to the number of ordered 7-tuples with 3 $U$'s and 4 $R$'s, for example $URRURRU$ means, in order, go up, right, right, up, right, right, up.
The number of these is $$\frac{\# \text{ways to order 7 different letters}}{\# \text{ways to re-order the equivalent U's} \cdot \# \text{ways to re-order the equivalent R's}} = \frac{7!}{3! \cdot 4!} = \binom{7}{4}.$$
Now, starting at the point $A$, we can go one step to the right and one step to the up at each move. This procedure is continued until the point labeled $B$ is reached.
To reach $B$ from $A$, $4$ steps are to be taken to the right and $3$ steps up.
So, the total number of steps taken to reach $B$ from $A$ is $7$.
Let the total number of steps be $m=7$
Let the steps to the right be $n_1=4$
Let the steps to the upward be $n_2=3$
So, to find the possible number of paths from $A$ to $B$
use the multinomial rule, the possible number of paths from $A$ to $B$ is $$\frac{m!}{n_1!\times n_2!}=\frac{7!}{4!\times 3!}$$
$$=\frac{7\times 6 \times 5 \times 4!}{4!\times 3 \times 2\times 1}=35$$
So, the number of paths from $A$ to $B$ is $35$ ways.
• You meant steps up rather than steps to the left. – N. F. Taussig Jun 23 '18 at 21:16
• Thanks, I fixed it. – Key Flex Jun 23 '18 at 21:20 | {"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": 0.9016623497009277, "perplexity": 202.8507134493777}, "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-2021-04/segments/1610703517159.7/warc/CC-MAIN-20210118220236-20210119010236-00614.warc.gz"} |
https://en.wikipedia.org/wiki/Prompt_neutron | # Prompt neutron
In nuclear engineering, a prompt neutron is a neutron immediately emitted by a nuclear fission event, as opposed to a delayed neutron decay which can occur within the same context, emitted after beta decay of one of the fission products anytime from a few milliseconds to a few minutes later.
Prompt neutrons emerge from the fission of an unstable fissionable or fissile heavy nucleus almost instantaneously. There are different definitions for how long it takes for a prompt neutron to emerge. For example, the United States Department of Energy defines a prompt neutron as a neutron born from fission within 10−13 seconds after the fission event.[1] The U.S. Nuclear Regulatory Commission defines a prompt neutron as a neutron emerging from fission within 10−14 seconds. [2] This emission is controlled by the nuclear force and is extremely fast. By contrast, so-called delayed neutrons are delayed by the time delay associated with beta decay (mediated by the weak force) to the precursor excited nuclide, after which neutron emission happens on a prompt time scale (i.e., almost immediately).
## Principle
Using U-235 as an example, this nucleus absorbs thermal neutrons, and the immediate mass products of a fission event are two large fission fragments, which are remnants of the formed U-236 nucleus. These fragments emit two or three free neutrons (2.5 on average), called prompt neutrons. A subsequent fission fragment occasionally undergoes a stage of radioactive decay that yields an additional neutron, called a delayed neutron. These neutron-emitting fission fragments are called delayed neutron precursor atoms.
Delayed neutrons are associated with the beta decay of the fission products. After prompt fission neutron emission the residual fragments are still neutron rich and undergo a beta decay chain. The more neutron rich the fragment, the more energetic and faster the beta decay. In some cases the available energy in the beta decay is high enough to leave the residual nucleus in such a highly excited state that neutron emission instead of gamma emission occurs.
Delayed Neutron Data for Thermal Fission in U-235[3]
Group Half-Life (s) Decay Constant (s−1) Energy (keV) Yield, Neutrons per Fission Fraction
1 55.72 0.0124 250 0.00052 0.000215
2 22.72 0.0305 560 0.00546 0.001424
3 6.22 0.111 405 0.00310 0.001274
4 2.30 0.301 450 0.00624 0.002568
5 0.614 1.14 - 0.00182 0.000748
6 0.230 3.01 - 0.00066 0.000273
## Importance in nuclear fission basic research
The standard deviation of the final kinetic energy distribution as a function of mass of final fragments from low energy fission of uranium 234 and uranium 236, presents a peak around light fragment masses region and another on heavy fragment masses region. Simulation by Monte Carlo method of these experiments suggests that those peaks are produced by prompt neutron emission.[4][5][6][7] This effect of prompt neutron emission does not provide a primary mass and kinetic distribution which is important to study fission dynamics from saddle to scission point.
## Importance in nuclear reactors
If a nuclear reactor happened to be prompt critical - even very slightly - the number of neutrons and power output would increase exponentially at a high rate. The response time of mechanical systems like control rods is far too slow to moderate this kind of power surge. The control of the power rise would then be left to its intrinsic physical stability factors, like the thermal dilatation of the core, or the increased resonance absorptions of neutrons, that usually tend to decrease the reactor's reactivity when temperature rises; but the reactor would run the risk of being damaged or destroyed by heat.
However, thanks to the delayed neutrons, it is possible to leave the reactor in a subcritical state as far as only prompt neutrons are concerned: the delayed neutrons come a moment later, just in time to sustain the chain reaction when it is going to die out. In that regime, neutron production overall still grows exponentially, but on a time scale that is governed by the delayed neutron production, which is slow enough to be controlled (just as an otherwise unstable bicycle can be balanced because human reflexes are quick enough on the time scale of its instability). Thus, by widening the margins of non-operation and supercriticality and allowing more time to regulate the reactor, the delayed neutrons are essential to inherent reactor safety and even in reactors requiring active control.
## Fraction definitions
The factor β is defined as:
${\displaystyle \beta ={\frac {\mbox{precursor atoms}}{{\mbox{prompt neutrons}}+{\mbox{precursor atoms}}}}.}$
and it is equal to 0.0064 for U-235.
The delayed neutron fraction (DNF) is defined as:
${\displaystyle DNF={\frac {\mbox{delayed neutrons}}{{\mbox{prompt neutrons}}+{\mbox{delayed neutrons}}}}.}$
These two factors, β and DNF, are not the same thing in case of a rapid change in the number of neutrons in the reactor.
Another concept, is the effective fraction of delayed neutrons, which is the fraction of delayed neutrons weighted (over space, energy, and angle) on the adjoint neutron flux. This concept arises because delayed neutrons are emitted with an energy spectrum more thermalized relative to prompt neutrons. For low enriched uranium fuel working on a thermal neutron spectrum, the difference between the average and effective delayed neutron fractions can reach 50 pcm (1 pcm = 1e-5).[8]
1. ^ "Doe Fundamentals Handbook - Nuclear Physics and Reactor Theory" (PDF), DOE-HDBK-1019/1-93, U.S. Department of Energy, January 1993, p. 29 (p. 133 of .pdf format) Missing or empty |title= (help)
2. ^ Mihalczo, John T. (November 19, 2004), "Radiation Detection From Fission" (PDF), ORNL/TM-2004/234, Oak Ridge National Laboratory, p. 1 (p. 11 of .pdf format) Missing or empty |title= (help) | {"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": 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": 0.9354932308197021, "perplexity": 1603.217687821564}, "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-2017-39/segments/1505818689471.25/warc/CC-MAIN-20170923033313-20170923053313-00424.warc.gz"} |
https://physics.stackexchange.com/questions/234304/should-4-fundamental-forces-really-be-3-because-of-electroweak-unification | Should 4 fundamental forces really be 3 because of electroweak unification?
Physicists concluded that, in fact, the weak and electromagnetic forces have essentially equal strengths. This is because the strength of the interaction depends strongly on both the mass of the force carrier and the distance of the interaction. The difference between their observed strengths is due to the huge difference in mass between the W and Z particles, which are very massive, and the photon, which has no mass as far as we know.
So, should all books teaching four fundamental forces change to three with third being Electroweak?
Or still there are four fundamental forces?
• I consider it as one step towards unification (or GUT). Maybe one day we might be able to explain all forces in terms of just one. – N.S.JOHN Feb 7 '16 at 10:01
• So, are we waiting for all to unify. – Anubhav Goel Feb 7 '16 at 10:03
• Their are two types of nuclear interaction, weak and strong, and two types of long range, macroscopic forces, gravity and electromagnetic. So observationally it is four. Unification shows interconnection between forces, so it is possible to explain particle behavior in a unified fashion. So its your choice, depending upon the application. – Peter Diehr Feb 7 '16 at 12:16
The standard model is written in the language of quantum field theory (QFT). In QFT one usually starts from a Lagrangian, which is a function of the fields that we suppose to be the elementary constituents of the world, or, to be even more precise, the elementary objects of our description of the world, and then one quantizes it by using the appropriate techniques. Given a Lagrangian, the whole theory can be derived from it. The Lagrangian of the electromagnetic field and a fermionic (charged) field, such as the electron field, can be written as $$\mathcal{L}=\bar{\psi}\gamma^{\mu}(i\partial_{\mu}-m-eA_{\mu})\psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ Here $e$ is the electric charge of the fermion, $m$ its mass, $\psi$ its field, $A_{\mu}$ the photonic field and $F_{\mu\nu}F^{\mu\nu}$ is something like the product of the electric and magnetic fields. The two terms $$\mathcal{L}_{\psi}=-\bar{\psi}\gamma^{\mu}(i\partial_{\mu}-m)\psi$$ and $$\mathcal{L}_{A}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ tell you how the fields behave independently of each other. The term $$\mathcal{L}_{int}=e\bar{\psi}\gamma^{\mu}\psi A_{\mu}$$ contain both $\psi$ and $A_{\mu}$, and it tells you how the fields interact. Now, the full Lagrangian of the electroweak theory is much more complex. It contains the Higgs field, a symmetry breaking potential and six generations of fermions, so I won't write it in its full form. Nevertheless, the interaction term between a single fermion and the "electroweak" field can be schematically written as
$$\mathcal{L}_{int}=-\bar{\psi}\gamma^{\mu}(g'B_{\mu}+g\tau^{a}W_{\mu}^{a})\psi$$ Here $a=1,2,3$, $B_{\mu}$ and $W_{\mu}^{a}$ are the four "electroweak" fields and $\tau^{a}$ are three appropriate matrices ($\psi$ is a column vector and $\bar{\psi}$ is a row vector, so they can have matrices in between). $g$ and $g'$ are the coupling constants of the theory, and they can be chosen different (and in fact they are). This is very important to the distinction between unification and mixing. Now, first of all, the $W$'s and $B$ cannot be directly identified with the photonic field and the $W^{\pm}$, $Z$ fields. As I told you, the full Lagrangian contains other terms. These terms, through the mechanism of spontaneous symmetry breaking, give mass to linear combinations of the $W$'s and the $B$. The combinations are as follows: $$W^{\pm}_{\mu}=\frac{1}{\sqrt{2}}(W^{1}_{\mu}\mp i W^{2}_{\mu})\\ Z_{\mu}=\frac{1}{\sqrt{g^{2}+g'^{2}}}\ (gW^{3}_{\mu}-g'B_{\mu})\\ A_{\mu}=\frac{1}{\sqrt{g^{2}+g'^{2}}}\ (g'W^{3}_{\mu}+gB_{\mu})$$ $W^{\pm}_{\mu}$ are the fields of the $W^{\pm}$ bosons. $Z_{\mu}$ is the field of the $Z$ boson. $A_{\mu}$ is the field of the photon. The former are given mass. $A_{\mu}$ turns out to be given no mass. When you rearrange the terms in the Lagrangian in terms of these combinations of fields, you get the Lagrangian of electrodynamics plus the Lagrangian of the weak theory. As you can see, the photonic field is a mixing between the $B$ field and the $W^{3}$ field. The $Z$ field is too. So one gets electrodynamics and the weak theory by mixing fields. Now, as I told you, the coupling constants $g$ and $g'$ can be chosen to be different. This is because in the theory which contains $B$ and the $W$'s, symmetry principles (together with something called the renormalizability principle) don't constrain the $B$ field to interact directly with the $W$'s. So you can interpret the starting point of the electroweak theory to be a theory in which two different kind of interactions exist. But because the theory contains the symmetry breaking terms, at low energies the two mix to give the theories of electromagnetic and weak interactions. On the other hand, let's suppose that to begin with we had a theory of a truly single interaction. Then would $g$ and $g'$ be constrained to be chosen equal? The answer is still: no. The theory would have the very same form as the one with two different interactions. This has to do with the symmetry group of the theory, which in this case is called $U(1)\times SU(2)$ ($U(1)\equiv B, SU(2)\equiv W$'s). When you have a $U(1)$ multiplying some other group, it is difficult to give an interpretation to the multiplication itself. $U(1)$ doesn't change the structure of the second group in any case, i.e. it behaves as a somewhat distinct portion of the product group (and this is why the coupling constants can be chosen to be different). | {"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": 0.9475705027580261, "perplexity": 180.12000566592138}, "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-2019-43/segments/1570987781397.63/warc/CC-MAIN-20191021171509-20191021195009-00362.warc.gz"} |
https://notebook.madsenlab.org/project-coarse%20grained%20model/model-seriationct/experiment-experiment-seriationct/2015/05/14/seriationct-sample-size-experiments.html | ## SeriationCT Sample Size Series experiments
Created: 14 May 2015 Modified: 23 Jul 2020 BibTeX Entry RIS Citation Print
### Effect of Varying Sample Size
After SAA’s, I used some existing simulations (seriationct-27 from the experiment-seriationct-2 repository) to prototype examining the effects of sample size on:
1. The gross structure of the seriation solutions: how much branching when small? do we get isolates?
2. Lineage structure recovery
Working with the existing simulation results and post-processed samples, I wrote a quick modification of the assemblage sampler called seriationct-sample-assemblages-for-samplesize-sequence-seriation.py. The script takes the output from the seriationct-export-data.py script and performs subsampling as follows:
1. Uses a samplefraction parameter to select an initial sample of assemblages, just like the normal seriationct-sample-assemblages-for-seriation.py.
2. Selects these samples given the sampletype parameter, as a pure random sample of assemblages, a spatially stratified sample given NxN quadrats, temporally stratified given N even periods of time, and spatiotemporal sampling which stratifies by quadrats and periods.
3. Subsamples the samplefraction assemblages in a sequence decreasing by 2, so for example, if samplefraction is 30, the largest set of assemblages will be 30 randomly sampled, and then 28 sampled from the 30, 26 sampled from the original 30, and so on.
The net result is a nested series of samples (rather than independent random samples of different sizes).
### Statistics Brainstorm
Some of the things I want to know are:
1. How consistent is the lineage structure of each seriation solution, at each sample size? (i.e., along a branch, how mixed are lineage identifiers?)
2. How consistent is the temporal sequence of the solution?
3. How “branchy” is the solution?
4. Furthermore, what is the average and variance of these measures, across the sequence of sample sizes? Or perhaps the quantiles and order statistics?
5. At what sample size does lineage structure and chronology become visible and semi-stable?
I’m just starting to build measures for these items, since they involve traversal and parsing of the annotated minmax graphs. | {"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": 0.8417388200759888, "perplexity": 3739.6320697337183}, "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-2021-25/segments/1623487660269.75/warc/CC-MAIN-20210620084505-20210620114505-00559.warc.gz"} |
http://math.stackexchange.com/questions/89047/another-question-about-proof-that-zetas-neq-0-for-res-1/89049 | # Another question about proof that $\zeta(s) \neq 0$ for $\Re(s) = 1$
This is a question distinct from but related to the question I wrote here: Question about proof that $\zeta(s) \neq 0$ for $\Re(s) = 1$, so assume the same things that I wrote there.
The paper then goes to show that the order of this zero of $\zeta$ we called $s_0=1+ia$ has order $\mu \leq 1$ (I have no trouble with this part). We want to show that $\mu=0$ implying that no such zero exists.
To that end, they say "if $\zeta(s)$ has a zero of order $\mu$ at s=1+ia$(a \in \mathbb{R}, a \neq 0$) and a zero of order $\nu$ at $1+2ia$, then...". I'm okay with the idea of assuming that $1+ia$ is a zero and then proving that its order is zero, but I'm not okay with the additional assumption that $1+2ia$ is a zero.
My question is: why is this legitimate? Surely that $1+2ia$ is a zero follows from assuming $1+ia$ is a zero (if it didn't, then the theorem we are proving would no longer be true). If they are indeed assuming them both simultaneously, then what prevents a possible situation that $1+ia$ is a zero of $\zeta$ while simultaneously $1+2ia$ is NOT a zero of $\zeta$?
I'm currently using the formula $\zeta(s) = \displaystyle\sum_{n=1}^{\infty} \displaystyle\int_n^{n+1} \frac{1}{n^s} - \frac{1}{x^s} dx + \frac{1}{s-1}$ that is defined for $\Re(s) > 0$.
-
The simple answer is that $v$ could be $0$, and there is no zero at all. For this proof, all that matters is that $v$ is not a pole, i.e. that $v\geq 0$.
That sort of makes sense, but I still have a problem with it. They are defining $1+2ia$ as "a zero" to get to the point they can show $\displaystyle\lim_{\epsilon \searrow 0} \epsilon \Phi(1+\epsilon \pm 2ia) = -\nu$. The argument here should be the same as it was for the zero at $1+ia$, but when I did that that one I used the fact that $1+ia$ was a pole of $\frac{\zeta'(s)}{\zeta(s)}$ to derive $\epsilon \Phi(1+\epsilon+ia) \rightarrow -\mu$. I can't run the same argument for $1+2ai$ if it's not a zero of $\zeta$. – tomcuchta Dec 6 '11 at 22:26
@tomcuchta: If it is not a zero, then $\lim_{\epsilon\rightarrow 0}\epsilon\Phi(1+\epsilon+2ia)$ just equals $0$. This is exactly what we want, because it equals $v$, and $v$ is zero in this case. – Eric Naslund Dec 6 '11 at 22:30 | {"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": 0.968146026134491, "perplexity": 101.59593778589702}, "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-2014-42/segments/1414119645866.1/warc/CC-MAIN-20141024030045-00003-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://desvl.xyz/2020/09/27/diagram-chasing/ | ## Exterior differentiation
(This section is intended to introduce the background. Feel free to skip if you already know exterior differentiation.)
There are several useful tools for vector calculus on $\mathbb{R}^3,$ namely gradient, curl, and divergence. It is possible to treat the gradient of a differentiable function $f$ on $\mathbb{R}^3$ at a point $x_0$ as the Fréchet derivative at $x_0$. But it does not work for curl and divergence at all. Fortunately there is another abstraction that works for all of them. It comes from differential forms.
Let $x_1,\cdots,x_n$ be the linear coordinates on $\mathbb{R}^n$ as usual. We define an algebra $\Omega^{\ast}$ over $\mathbb{R}$ generated by $dx_1,\cdots,dx_n$ with the following relations:
This is a vector space as well, and it’s easy to derive that it has a basis by
where $i<j<k$. The $C^{\infty}$ differential forms on $\mathbb{R}^n$ are defined to be the tensor product
As is can be shown, for $\omega \in \Omega^{\ast}(\mathbb{R}^n)$, we have a unique representation by
and in this case we also say $\omega$ is a $C^{\infty}$ $k$-form on $\mathbb{R}^n$ (for simplicity we also write $\omega=\sum f_Idx_I$). The algebra of all $k$-forms will be denoted by $\Omega^k(\mathbb{R}^n)$. And naturally we have $\Omega^{\ast}(\mathbb{R}^n)$ to be graded since
### The operator $d$
But if we have $\omega \in \Omega^0(\mathbb{R}^n)$, we see $\omega$ is merely a $C^{\infty}$ function. As taught in multivariable calculus course, for the differential of $\omega$ we have
and it turns out that $d\omega\in\Omega^{1}(\mathbb{R}^n)$. This inspires us to obtain a generalization onto the differential operator $d$:
and $d\omega$ is defined as follows. The case when $k=0$ is defined as usual (just the one above). For $k>0$ and $\omega=\sum f_I dx_I,$ $d\omega$ is defined ‘inductively’ by
This $d$ is the so-called exterior differentiation, which serves as the ultimate abstract extension of gradient, curl, divergence, etc. If we restrict ourself to $\mathbb{R}^3$, we see these vector calculus tools comes up in the nature of things.
Functions
$1$-forms
$2$-forms
The calculation is tedious but a nice exercise to understand the definition of $d$ and $\Omega^{\ast}$.
### Conservative field - on the kernel and image of $d$
By elementary computation we are also able to show that $d^2\omega=0$ for all $\omega \in \Omega^{\ast}(\mathbb{R}^n)$ (Hint: $\frac{\partial^2 f}{\partial x_i \partial x_j}=\frac{\partial^2 f}{\partial x_j \partial x_i}$ but $dx_idx_j=-dx_idx_j$). Now we consider a vector field $\overrightarrow{v}=(v_1,v_2)$ of dimension $2$. If $C$ is an arbitrary simply closed smooth curve in $\mathbb{R}^2$, then we expect
to be $0$. If this happens (note the arbitrary of $C$), we say $\overrightarrow{v}$ to be a conservative field (path independent).
So when conservative? It happens when there is a function $f$ such that
This is equivalent to say that
If we use $C^{\ast}$ to denote the area enclosed by $C$, by Green’s theorem, we have
If you translate what you’ve learned in multivariable calculus course (path independence) into the language of differential form, you will see that the set of all conservative fields is precisely the image of $d_0:\Omega^0(\mathbb{R}^2) \to \Omega^1(\mathbb{R}^2)$. Also, they are in the kernel of the next $d_1:\Omega^1(\mathbb{R}^2) \to \Omega^2(\mathbb{R}^2)$. These $d$’s are naturally homomorphism, so it’s natural to discuss the factor group. But before that, we need some terminologies.
## de Rham complex and de Rham cohomology group
The complex $\Omega^{\ast}(\mathbb{R}^n)$ together with $d$ is called the de Rham complex on $\mathbb{R}^n$. Now consider the sequence
We say $\omega \in \Omega^k(\mathbb{R}^n)$ is closed if $d_k\omega=0$, or equivalently, $\omega \in \ker d_k$. Dually, we say $\omega$ is exact if there exists some $\mu \in \Omega^{k-1}(\mathbb{R}^n)$ such that $d\mu=\omega$, that is, $\omega \in \operatorname{im}d_{k-1}$. Of course all $d_k$’s can be written as $d$ but the index makes it easier to understand. Instead of doing integration or differentiation, which is ‘uninteresting’, we are going to discuss the abstract structure of it.
The $k$-th de Rham cohomology in $\mathbb{R}^n$ is defined to be the factor space
As an example, note that by the fundamental theorem of calculus, every $1$-form is exact, therefore $H_{DR}^1(\mathbb{R})=0$.
Since de Rham complex is a special case of differential complex, and other restrictions of de Rham complex plays no critical role thereafter, we are going discuss the algebraic structure of differential complex directly.
## The long exact sequence of cohomology groups
We are going to show that, there exists a long exact sequence of cohomology groups after a short exact sequence is defined. For the convenience let’s recall here some basic definitions
### Exact sequence
A sequence of vector spaces (or groups)
is said to be exact if the image of $f_{k-1}$ is the kernel of $f_k$ for all $k$. Sometimes we need to discuss a extremely short one by
As one can see, $f$ is injective and $g$ is surjective.
### Differential complex
A direct sum of vector spaces $C=\oplus_{k \in \mathbb{Z}}C^k$ is called a differential complex if there are homomorphisms by
such that $d_{k-1}d_k=0$. Sometimes we write $d$ instead of $d_{k}$ since this differential operator of $C$ is universal. Therefore we may also say that $d^2=0$. The cohomology of $C$ is the direct sum of vector spaces $H(C)=\oplus_{k \in \mathbb{Z}}H^k(C)$ where
A map $f: A \to B$ where $A$ and $B$ are differential complexes, is called a chain map if we have $fd_A=d_Bf$.
### The sequence
Now consider a short exact sequence of differential complexes
where both $f$ and $g$ are chain maps (this is important). Then there exists a long exact sequence by
Here, $f^{\ast}$ and $g^{\ast}$ are the naturally induced maps. For $c \in C^q$, $d^{\ast}[c]$ is defined to be the cohomology class $[a]$ where $a \in A^{q+1}$, and that $f(a)=db$, and that $g(b)=c$. The sequence can be described using the two-layer commutative diagram below.
The long exact sequence is actually the purple one (you see why people may call this zig-zag lemma). This sequence is ‘based on’ the blue diagram, which can be considered naturally as an expansion of the short exact sequence. The method that will be used in the following proof is called diagram-chasing, whose importance has already been described by Professor James Munkres: master this. We will be abusing the properties of almost every homomorphism and group appeared in this commutative diagram to trace the elements.
#### Proof
First, we give a precise definition of $d^{\ast}$. For a closed $c \in C^q$, by the surjectivity of $g$ (note this sequence is exact), there exists some $b \in B^q$ such that $g(b)=c$. But $g(db)=d(g(b))=dc=0$, we see for $db \in B^{q+1}$ we have $db \in \ker g$. By the exactness of the sequence, we see $db \in \operatorname{im}{f}$, that is, there exists some $a \in A^{q+1}$ such that $f(a)=db$. Further, $a$ is closed since
and we already know that $f$ has trivial kernel (which contains $da$).
$d^{\ast}$ is therefore defined by
where $[\cdot]$ means “the homology class of”.
But it is expected that $d^{\ast}$ is a well-defined homomorphism. Let $c_q$ and $c_q’$ be two closed forms in $C^q$. To show $d^{\ast}$ is well-defined, we suppose $[c_q]=[c_q’]$ (i.e. they are homologous). Choose $b_q$ and $b_q’$ so that $g(b_q)=c_q$ and $g(b_q’)=c_q’$. Accordingly, we also pick $a_{q+1}$ and $a_{q+1}’$ such that $f(a_{q+1})=db_q$ and $f(a_{q+1}’)=db_q’$. By definition of $d^{\ast}$, we need to show that $[a_{q+1}]=[a_{q+1}’]$.
Recall the properties of factor group. $[c_q]=[c_q’]$ if and only if $c_q-c_q’ \in \operatorname{im}d$. Therefore we can pick some $c_{q-1} \in C^{q-1}$ such that $c_q-c_q’=dc_{q-1}$. Again, by the surjectivity of $g$, there is some $b_{q-1}$ such that $g(b_{q-1})=c_{q-1}$.
Note that
Therefore $b_q-b_q’-db_{q-1} \in \operatorname{im} f$. We are able to pick some $a_q \in A^{q}$ such that $f(a_q)=b_q-b_q’-db_{q-1}$. But now we have
Since $f$ is injective, we have $da_q=a_{q+1}-a_{q+1}’$, which implies that $a_{q+1}-a_{q+1}’ \in \operatorname{im}d$. Hence $[a_{q+1}]=[a_{q+1}’]$.
To show that $d^{\ast}$ is a homomorphism, note that $g(b_q+b_q’)=c_q+c_q’$ and $f(a_{q+1}+a_{q+1}’)=d(b_q+b_q’)$. Thus we have
The latter equals $[a_{q+1}]+[a_{q+1}’]$ since the canonical map is a homomorphism. Therefore we have
Therefore the long sequence exists. It remains to prove exactness. Firstly we need to prove exactness at $H^q(B)$. Pick $[b] \in H^q(B)$. If there is some $a \in A^q$ such that $f(a)=b$, then $g(f(a))=0$. Therefore $g^{\ast}[b]=g^{\ast}[f(a)]=[g(f(a))]=[0]$; hence $\operatorname{im}f \subset \ker g$.
Conversely, suppose now $g^{\ast}[b]=[0]$, we shall show that there exists some $[a] \in H^q(A)$ such that $f^{\ast}[a]=[b]$. Note $g^{\ast}[b]=\operatorname{im}d$ where $d$ is the differential operator of $C$ (why?). Therefore there exists some $c_{q-1} \in C^{q-1}$ such that $g(b)=dc_{q-1}$. Pick some $b_{q-1}$ such that $g(b_{q-1})=c_{q-1}$. Then we have
Therefore $f(a)=b-db_{q-1}$ for some $a \in A^q$. Note $a$ is closed since
and $f$ is injective. $db=0$ since we have
Furthermore,
Therefore $\ker g^{\ast} \subset \operatorname{im} f$ as desired.
Now we prove exactness at $H^q(C)$. (Notation:) pick $[c_q] \in H^q(C)$, there exists some $b_q$ such that $g(b_q)=c_q$; choose $a_{q+1}$ such that $f(a_{q+1})=db_q$. Then $d^{\ast}[c_q]=[a_{q+1}]$ by definition.
If $[c_q] \in \operatorname{im}g^{\ast}$, we see $[c_q]=[g(b_q)]=g^{\ast}[b_q]$. But $b_q$ is closed since $[b_q] \in H^q(B)$, we see $f(a_{q+1})=db_q=0$, therefore $d^{\ast}[c_q]=[a_{q+1}]=[0]$ since $f$ is injective. Therefore $\operatorname{im}g^{\ast} \subset \ker d^{\ast}$.
Conversely, suppose $d^{\ast}[c^q]=[0]$. By definition of $H^{q+1}(A)$, there is some $a_q \in A$ such that $da_q = a_{q+1}$ (can you see why?). We claim that $b_q-f(a_q)$ is closed and we have $[c_q]=g^{\ast}[b_q-f(a_q)]$.
By direct computation,
Meanwhile
Therefore $\ker d^{\ast} \subset \operatorname{im}g^{\ast}$. Note that $g(f(a_q))=0$ by exactness.
Finally, we prove exactness at $H^{q+1}(A)$. Pick $\alpha \in H^{q+1}(A)$. If $\alpha \in \operatorname{im}d^{\ast}$, then $\alpha=[a_{q+1}]$ where $f(a_{q+1})=db_q$ by definition. Then
Therefore $\alpha \in \ker f^{\ast}$. Conversely, if we have $f^{\ast}(\alpha)=[0]$, pick the representative element of $\alpha$, namely we write $\alpha=[a]$; then $[f(a)]=[0]$. But this implies that $f(a) \in \operatorname{im}d$ where $d$ denotes the differential operator of $B$. There exists some $b_{q+1} \in B^{q+1}$ and $b_q \in B^q$ such that $db_{q}=b_{q+1}$. Suppose now $c_q=g(b_q)$. $c_q$ is closed since $dc_q=g(db_q)=g(b_{q+1})=g(f(a))=0$. By definition, $\alpha=d^{\ast}[c_q]$. Therefore $\ker f^{\ast} \subset \operatorname{im}d^{\ast}$.
### Remarks
As you may see, almost every property of the diagram has been used. The exactness at $B^q$ ensures that $g(f(a))=0$. The definition of $H^q(A)$ ensures that we can simplify the meaning of $[0]$. We even use the injectivity of $f$ and the surjectivity of $g$.
This proof is also a demonstration of diagram-chasing technique. As you have seen, we keep running through the diagram to ensure that there is “someone waiting” at the destination.
This long exact group is useful. Here is an example.
## Application: Mayer-Vietoris Sequence
By differential forms on a open set $U \subset \mathbb{R}^n$, we mean
And the de Rham cohomology of $U$ comes up in the nature of things.
We are able to compute the cohomology of the union of two open sets. Suppose $M=U \cup V$ is a manifold with $U$ and $V$ open, and $U \amalg V$ is the disjoint union of $U$ and $V$ (the coproduct in the category of sets). $\partial_0$ and $\partial_1$ are inclusions of $U \cap V$ in $U$ and $V$ respectively. We have a natural sequence of inclusions
Since $\Omega^{*}$ can also be treated as a contravariant functor from the category of Euclidean spaces with smooth maps to the category of commutative differential graded algebras and their homomorphisms, we have
By taking the difference of the last two maps, we have
The sequence above is a short exact sequence. Therefore we may use the zig-zag lemma to find a long exact sequence (which is also called the Mayer-Vietoris sequence) by
### An example
This sequence allows one to compute the cohomology of two union of two open sets. For example, for $H^{*}_{DR}(\mathbb{R}^2-P-Q)$, where $P(x_p,y_p)$ and $Q(x_q,y_q)$ are two distinct points in $\mathbb{R}^2$, we may write
and
Therefore we may write $M=\mathbb{R}^2$, $U=\mathbb{R}^2-P$ and $V=\mathbb{R}^2-Q$. For $U$ and $V$, we have another decomposition by
where
But
is a four-time (homeomorphic) copy of $\mathbb{R}^2$. So things become clear after we compute $H^{\ast}_{DR}(\mathbb{R}^2)$.
## References / Further reading
• Raoul Bott, Loring W. Tu, Differential Forms in Algebraic Topology
• Munkres J. R., Elements of Algebraic Topology
• Micheal Spivak, Calculus on Manifolds
• Serge Lang, Algebra | {"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": 0.9940131306648254, "perplexity": 119.56108116406295}, "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-2021-04/segments/1610704803737.78/warc/CC-MAIN-20210126202017-20210126232017-00457.warc.gz"} |
https://math.stackexchange.com/questions/1500084/expanding-brackets-problem-z-11-z-z2-z3 | # Expanding brackets problem: $(z - 1)(1 + z + z^2 + z^3)$
I have:
$(z - 1)(1 + z + z^2 + z^3)$
As, I have tried my own methods and enlisted the help on online software, but as well as them not all arriving at the same solution, I can't follow their reasoning.
I tried to gather all the like terms:
$(z - 1)(z^6+1)$
And then thought it may offer a route to a difference or two cubes if I shifted powers across the brackets, but that's $z^7$ in total, isn't it? I can't just do $z^7-1$, can I?
If I try to expand, I see:
$(z - 1)(z^6 + 1) = z^7+z-z^6-1$
Well...
$z^7-z^6=z$
Add the other z leaves:
$z^2-1$?
I expect this is all wrong.
Will you help please?
On a side note, I have difficulty with understanding how I'm expected to tag this post accurately when I can't used "expand", "multiply", or use any of the subjects it covers as tags?
• $z+z^2+z^3\ne z^6$ and $1+z+z^2+z^3\ne z^7$. You can't "add powers" like that. – David Mitra Oct 27 '15 at 13:41
• True, but you can add powers like $z^2+z^3=z^5$ and there is a z in the left bracket, which makes $z^6$, doesn't it? Can I then do $z^3*z^3$ and solve the difference of two cubes? I must add, it would be very much more helpful if you'd say what does work, rather than what doesn't! Thanks. :-) – Georgina Davenport Oct 27 '15 at 14:15
You seem to have some very unfortunate ideas about algebra! As David Mitra said, When you are adding powers of z you do not add the powers themselves. That is a property of multiplication: $z^n\cdot z^m= z^{n+ m}$.
To multiply $(z- 1)(1+ z+ z^2+ z^3)$ use the "distributive law" $a(b+ c)= ab+ ac$ and $(b+ c)a= ab+ ac$.
Think of $z- 1$ as $(b+ c)$ with $b= z$ and $c= -1$ and a as $(1+ z+ z^2+ z^3)$. $(z- 1)(1+ z+ z^2+ z^3)= z(1+ z+ z^2+ z^3)- 1(1+ z+ z^2+ z^3)$.
Now, for each of those, use the distributive law again: $z(1+ z+ z^2+ z^3)= z(1)+ z(z)+ z(z^2)+ z(z^4)$.
NOW use the rule for adding exponents (with $1= z^0$ and $z= z^1$): $z(1)+ z(z)+ z(z^2)+ z(z^3)= z+ z^2+ z^3+ z^4$.
And, of course, $-1(1+ z+ z^2+ z^3)= -1- z- z^2- z^3$.
Putting those together, $(z- 1)(1+ z+ z^2+ z^3)= z+ z^2+ z^3+ z^4- 1- z- z^2- z^3= -1+ (z- z)+ (z^2- z^2)+ (z^3- z^3)+ z^4= z^4- 1$.
• You are quite right user247327, my ideas about it are extremely unfortunate, thank you very much for your help accumulating better ideas about it. – Georgina Davenport Oct 27 '15 at 14:24
• Unfortunately I can't +1 responses yet, but this answer is excellent. Thank you very much for your help! Most appreciated. – Georgina Davenport Oct 27 '15 at 15:00
• @GeorginaDavenport You can upvote an answer once you receive at least 15 reputation points. See math.stackexchange.com/help/privileges. You also have the option of awarding a best answer to a response that does a particularly good job of answering your question. See math.stackexchange.com/help/someone-answers. – N. F. Taussig Nov 10 '15 at 12:50
I like to do factorizations like this by writing $$z^7-1=(z-1)(\cdots)$$ and try to figure out what should replace the dots. First, we'd like to have a factor of $z^7$ in the end result, so the $z$ on the right side should be multiplied by $z^6$ to get this. Then, we write $$z^7-1=(z-1)(z^6+\cdots).$$
At this point, by the distributive law, the right side will multiply out to $z^7-z^6+\cdots$. Since we don't want the $z^6$ in the final product, it must cancel with something. We can cancel the $-z^6$ by multiplying $z$ by $z^5$ to get $$z^7-1=(z-1)(z^6+z^5+\cdots)$$ because after distributing, the product is $z^7-z^6+z^6-z^5+\cdots$ and the $z^6$'s cancel. Now, we're left with $z^7-z^5+\cdots$, so the $-z^5$ must cancel with another term, since $z^5=z\cdot z^4$, we have $$z^7-1=(z-1)(z^6+z^5+z^4+\cdots).$$
At this point, the product is $z^7-z^4+\cdots$. In order to cancel the $-z^4$, we introduce a $z^3$, which results in an extra $-z^3$. This extra term is cancelled with a $z^2$, which results in an extra $-z^2$ in the product. This $-z^2$ is cancelled with a $z$, but that gives an extra $-z$ in the product. The extra $-z$ is cancelled with a $-1$. Therefore, we have $$z^7=(z-1)(z^6+z^5+z^4+z^3+z^2+z^1+1+\cdots).$$ But, after multiplying everything out, we find that the left side already matches the right side and there is nothing more in the dots! Therefore, $$z^7=(z-1)(z^6+z^5+z^4+z^3+z^2+z^1+1).$$
First off, user247327 gives an excellent answer. I don't have the reputation to comment on his comment, so here is an explanation of a critical fact you are missing. Q: why doesn't $$(1+z+z^2+z^3)=(z^6+1)?$$
Answer: When we write $z^3$, we mean $z*z*z$. Likewise $z^2=z*z$. Therefore: $$z^3*z^2=(z*z*z)*(z*z)$$ Well, how to we write this compactly? Simple! count the number of times we multiply z. Thus we get: $$z^3*z^2=(z*z*z)*(z*z)=z^5$$ This is why it is true in general that: $$z^a*z^b=z^{a+b}$$ If you ever forget this, just google "exponent rules", and you can see all the neat tricks. HOWEVER, there is no identity I can think of for simplifying $z^a+z^b$.
Edit: PS, in your comment you state $z^3+z^2=z^5$. Substitute z=2 to see that this is wrong. Note $2^5=32$, but: $$2^3+2^2=8+4=12 \neq 32$$ This one counterexample is enough to show that $z^a+z^b \neq z^{a+b}$
First remove the left brackets and distribute the left terms on the right brackets:
$$(z - 1)(1 + z + z^2 + z^3)=z(1 + z + z^2 + z^3)-(1 + z + z^2 + z^3).$$
Then remove the remaining brackets and distribute what needs to be:
$$z(1 + z + z^2 + z^3)-(1 + z + z^2 + z^3)=z+z^2+z^3+z^4-1-z-z^2-z^3.$$
Finally, simplify:
$$z+z^2+z^3+z^4-1-z-z^2-z^3=z^4-1.$$ | {"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": 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": 0.8788672685623169, "perplexity": 345.32862579787894}, "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-2019-39/segments/1568514577478.95/warc/CC-MAIN-20190923172009-20190923194009-00012.warc.gz"} |
http://sites.science.oregonstate.edu/portfolioswiki/activities:guides:cfqmatomgroup?rev=1368562851 | Navigate back to the activity.
## Quantum Calculations on the Hydrogen Atom: Instructor's Guide
### Main Ideas
• Eigenvalues and eigenstates
• Measurements of energy and angular momentum for hydrogen atom
• Quantum probabilities
• Superposition of states
• Quantum calculations in multiple representations
Estimated Time: 30 minutes
Students are asked to find eigenvalues, probabilities, and expectation values for $H$, $L^2$, and $L_z$ for a superposition of $\vert nlm \rangle$ states. This can be done on small whiteboards or with the students working in groups on large whiteboards.
Students then work together in small groups to find the matrices that correspond to $H$, $L^2$, and $L_z$ and to redo $\langle E\rangle$ in matrix notation
### Prerequisite Knowledge
• Eigenstates & Eigenvalues
• The energy and angular momentum eigenstates and eigenvalues of a hydrogen atom
• Calculating probabilities using Dirac “bra-ket”, matrix and wavefunction notation
• Familiarity with time evolution
This activity works well when sequenced with similar activities for a particle confined to a ring and a particle confined to a sphere.
### Activity: Introduction
Write a linear combination of $\vert nlm\rangle$ states on the board. For example: $$\Psi = \sqrt{\frac{7}{10}} |2, 1, 0\rangle + \sqrt{\frac{1}{10}} |3, 2, 1\rangle + i\sqrt{\frac{2}{10}} |3, 1, 1\rangle$$ (it is a good idea to provide a state that is degenerate in one or more of the quantum numbers).
Then ask the students a series of small whiteboard questions with a short wrap-up after each one that reiterates key points (see wrap-up below).
• $P_{L_z=m\hbar}$; $\langle L_z\rangle$
• $P_{L^2=\ell(\ell+1)\hbar^2}$; $\langle L^2\rangle$
• $P_{E=-13.6eV/n^2}$; $\langle E\rangle$
Students are then asked to work together in small groups to find the matrices that correspond to $H$, $L^2$, and $L_z$ and to redo $\langle E\rangle$ in matrix notation. It may be necessary to walk through one calculation to remind them of the method they have used before to generate these matrices.
### Activity: Student Conversations
• Students can often find probabilities by inspection at this point, so it is sometimes helpful to ask them to write out explicitly what they did to lead into the discussion of the summation limits.
• In writing matrices, this is the often the first time that they are only writing a matrix for a subset of the space, so questions about basis, degeneracy, and order crop up.
### Activity: Wrap-up
Probabilities and Expectation Values:
• Summarize explicitly how to calculate the probability of degenerate states by summing the probabilities.
• It is a good idea to do the wrap-up after each small whiteboard question to help to keep the whole class together.
• Many students want to avoid explicitly putting the summation in, so this is a good time to talk explicitly about the summation limits for each case. For example, in finding the probability for measuring $L_z$ to be $-1\hbar$, students want to simply write:$$P_{L_z=-\hbar} = |\langle n, \ell, -1| \Psi \rangle |^2$$instead of
$$P_{L_z=-\hbar} = \sum^{\infty}_{n=\vert m\vert+1} \sum^{n-1}_{\ell=\vert m\vert} |\langle n, \ell, m| \Psi \rangle |^2= \sum^{\infty}_{n=2} \sum^{2}_{\ell=1} |\langle n, \ell, -1| \Psi \rangle |^2$$
• This is a good time to also address the difference between the sum of squares and the square of sums, which is an ongoing issue for many students.
Matrix notation:
• If the students use small whiteboards to write $\hat{H}$, $\hat{L}^2$, and $\hat{L}_z$ in matrix notation for this state, you can often find that people have used different ordering systems and by using several different examples, you can highlight several issues:
• the arbitrariness of the order when there is degeneracy,
• importance of being consistent with order between operators and vectors,
• what the “typical” order is for hydrogen states, and
• matrix notation is not as unwieldy as it seems if you have a small enough subspace.
• You can also highlight the implicit basis ($\vert n\ell m\rangle$) and reiterate the fact that all three matrices are diagonal in this basis (i.e. they share eigenstates).
This is also a good time to talk about the different ways of finding expectation values and when each is appropriate.
### Extensions
• You can also use the same state to discuss time dependence again in terms of the Hydrogen atom.
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http://math.stackexchange.com/questions/241922/use-residues-to-evaluate-int-0-infty-frac-coshax-coshx-mathrmdx | # Use residues to evaluate $\int_0^\infty \frac{\cosh(ax)}{\cosh(x)}\,\mathrm{d}x$, where $|a|<1$
Use residues to evaluate $$\int_0^\infty \frac{\cosh(ax)}{\cosh(x)}\,\mathrm{d}x$$ where $|a|<1$.
Try considering the integral of the form $$\int_C \frac{\exp(az)}{\cosh(z)}\,\mathrm dz,$$ where $C$ is the contour given by $y=0,\, y=\pi,\, x=-R,\, x=R$.
-
This answer (with a rotation in the complex plane and the path for the numerator or, simpler, setting $a:=ib$ first) should help. – Raymond Manzoni Nov 21 '12 at 12:44
@RaymondManzoni: the problem I see with the present problem, and with the link you gave, is that we have the hyperbolic cosine also in the numerator, and not only cosine. This messes things greatly, at least for me, though I think this may have been a typo by the OP since he's been adviced to use that exponential function in the numerator... – DonAntonio Nov 21 '12 at 14:46
@DonAntonio: Hmmm... I don't think so. The numerator may be $e^{az}$ as here (or $e^{iaz}$ as in the other link) but we are supposing $|a|<1$ and are dividing by $\cosh(z)$. For $|z|\gg 1$ the norm of this fraction will be majored by $\dfrac {e^{|a|z}}{e^z}\sim e^{-(1-|a|)z}$ (the denominator is giving us convergence) so that the two vertical integrals will disappear at the limit (while the two horizontal ones are identical up to a global coefficient and the limit is taken as a P.V.). Or did I forget something? – Raymond Manzoni Nov 21 '12 at 17:27
@RaymondManzoni , but how do you pass to $\,\cosh az\,$ from $\,e^{az}\,$? I know the vertical integrals vanish in the limit and both horizontal ones are the same up to a constant ($\,e^{(a-1)}\,$ , if I remember correctly), yet I can't see the byperbolic cosine appearing afterwards... – DonAntonio Nov 21 '12 at 17:34
@DonAntonio: At the end we should get $\displaystyle P.V. \int_{-\infty}^\infty \frac {e^{az}}{\cosh(z)} dz=\frac 12\int_0^\infty \frac{\cosh(az)}{\cosh(z)}dz$ (after separing properly the negative and positive part from $0$ to $R$ changing variable for the negative part and taking the limit as a principal value). – Raymond Manzoni Nov 21 '12 at 17:40
As the original query that asked to use residues has not been answered completely I will contribute some ideas.
Suppose $a$ is a rational number $p/q$ where $p<q$ and $p-q$ is odd. Use a rectangular contour that consists of four segments: $\Gamma_0$ along the real axis from $-R$ to $R$, $\Gamma_1$ parallel to the imaginary axis to $R + \pi i q$, $\Gamma_2$ parallel to the real axis but in the opposite direction to $-R + \pi i q$ and finally, $\Gamma_3$ parallel to the imaginary axis to $-R$ on the real axis.
Now set $$f(z) = \frac{e^{az}+e^{-az}}{e^z+e^{-z}}$$ so that we are looking for $$\frac{1}{2} \int_{-\infty}^\infty f(z) dz$$ and integrate $f(z)$ along $\Gamma_0 - \Gamma_1 - \Gamma_2 - \Gamma_3$. Examine each segment in turn as $R$ goes to infinity. Clearly the integral along $\Gamma_0$ is simply the integral we are looking for. The contributions of $\Gamma_1$ and $\Gamma_3$ vanish in the limit. Along $\Gamma_2$ we have $x= t + \pi i q$, getting $$\int_\infty^{-\infty} \frac{e^{\frac{p}{q}t + \pi i p}+e^{-\frac{p}{q}t - \pi i p}}{e^{t+ \pi i q}+e^{-t- \pi i q}} dt = - (-1)^{p-q} \int_{-\infty}^\infty \frac{e^{\frac{p}{q}t}+e^{-\frac{p}{q}t}}{e^{t}+e^{-t}} dt =\int_{-\infty}^\infty \frac{e^{\frac{p}{q}t}+e^{-\frac{p}{q}t}}{e^{t}+e^{-t}} dt.$$ The last equality is because $p-q$ is odd.
To conclude we need to compute the poles and residues inside our contour. The poles are at $$\rho_k = \frac{1}{2}\pi i + \pi i k$$ and the residues are $$\lim_{z\to \rho_k} \frac{(z-\rho_k) (e^{az} + e^{-az})}{e^z + e^{-z}} = \lim_{z\to \rho_k} \frac{(z-\rho_k) (a e^{az} -a e^{-az}) + e^{az} + e^{-az}}{e^z - e^{-z}}.$$ But $$\lim_{z\to \rho_k} \frac{1}{e^z - e^{-z}} = \frac{1}{i e^{\pi i k} - (-i) e^{-\pi i k}} = \frac{1}{i e^{\pi i k} + i e^{-\pi i k}} = \frac{e^{\pi i k}}{i (1+1)} = \frac{(-1)^k}{2i}$$ so that finally $$\operatorname{Res}_{z=\rho_k} f(z) = \frac{(-1)^k}{2i} \left( e^{a\rho_k} + e^{-a\rho_k}\right).$$ With $J$ being the integral we are looking for and $I$ the integral along $\Gamma_0$ we have $$J = \frac{1}{2} I = \frac{1}{4} 2 I = \frac{1}{4} 2\pi i \sum_{k=0}^{q-1} \operatorname{Res}_{z=\rho_k} f(z)$$ The conclusion is that $$J = \frac{1}{2} \pi i \sum_{k=0}^{q-1} \operatorname{Res}_{z=\rho_k} f(z) = \frac{\pi}{4} \sum_{k=0}^{q-1} (-1)^k \left( e^{a\rho_k} + e^{-a\rho_k}\right) = \frac{\pi}{2} \sum_{k=0}^{q-1} (-1)^k \cosh(a\rho_k).$$ where we have used the fact that $1/2 + k < q$ implies that $k$ runs up to $q-1.$
Edit. Use the following bound to see that the integral along $\Gamma_1$ vanishes (set $z= R + it$ with $0\le t \le \pi q$): $$\left| \int_{\Gamma_1} f(z) dz \right| = \left| \int_0^{\pi q} \frac{e^{aR + ait} + e^{-aR -ait}}{e^{R+it} + e^{-R-it}} i dt \right| \le \int_0^{\pi q} \frac{e^{aR} + e^{-aR}}{e^{R} - e^{-R}} dt = \pi q e^{-(1-a) R} \frac{1-e^{-2aR}}{1-e^{-2R}}$$ Now certainly we have $$\lim_{R\to\infty}\frac{1-e^{-2aR}}{1-e^{-2R}} = 1$$ so that the integral is $\theta(e^{-(1-a) R})$ which goes to zero as $R$ goes to infinity. The integral along $\Gamma_3$ is done the same way.
-
This doesn't use residues until we use $$\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{z+k}=\pi\csc(\pi z)$$ which can be proven using residues.
We just expand things in powers of $e^x$: \begin{align} &\int_0^\infty\frac{\cosh(ax)}{\cosh(x)}\,\mathrm{d}x\\ &=\int_0^\infty e^{(a-1)x}\frac{1+e^{-2ax}}{1+e^{-2x}}\,\mathrm{d}x\\ &=\int_0^\infty\left(e^{(a-1)x}-e^{(a-3)x}+e^{(a-5)x}-\dots\right)\,\mathrm{d}x\\ &+\int_0^\infty\left(e^{(-a-1)x}-e^{(-a-3)x}+e^{(-a-5)x}-\dots\right)\,\mathrm{d}x\\ &=\frac1{1-a}-\frac1{3-a}+\frac1{5-a}-\dots\\ &+\frac1{1+a}-\frac1{3+a}+\frac1{5+a}-\dots\\ &=\frac1{a+1}-\frac1{a+3}+\frac1{a+5}-\dots\\ &-\frac1{a-1}+\frac1{a-3}-\frac1{a-5}-\dots\\ &=\frac12\left(\dots+\frac1{\frac{a+1}2-2}-\frac1{\frac{a+1}2-1}+\frac1{\frac{a+1}2}-\frac1{\frac{a+1}2+1}+\frac1{\frac{a+1}2+2}-\dots\right)\\ &=\frac12\pi\csc\left(\pi\frac{a+1}2\right)\\ &=\frac\pi2\sec\left(\frac\pi2a\right) \end{align}
-
There is some additional simplification that can be done which I'll do in a new answer because my browser is not coping well with those large formulas where speed is concerned.
We have $$\frac{\pi}{2} \sum_{k=0}^{q-1} (-1)^k \cosh(a\rho_k) = \frac{\pi}{4} \sum_{k=0}^{q-1}(-1)^k e^{a\rho_k} + \frac{\pi}{4} \sum_{k=0}^{q-1}(-1)^k e^{-a\rho_k}$$
The first sum is $$\sum_{k=0}^{q-1} (-1)^k e^{a\rho_k} = \sum_{k=0}^{q-1} (-1)^k e^{a i \pi /2} e^{a \pi i k} = e^{a i \pi/2} \frac{1-(-e^{a \pi i})^q}{1 + e^{a \pi i}}$$ which is $$e^{a i \pi/2} \frac{1-(-1)^q e^{p \pi i}}{1 + e^{a \pi i}} = e^{a i \pi/2} \frac{1-(-1)^{p+q}}{1 + e^{a \pi i}} = e^{a i \pi/2} \frac{2}{1 + e^{a \pi i}} = \frac{1}{\cos (a\pi/2)}$$
The second sum is $$\sum_{k=0}^{q-1} (-1)^k e^{-a\rho_k} = \sum_{k=0}^{q-1} (-1)^k e^{-a i \pi /2} e^{-a \pi i k} = e^{-a i \pi/2} \frac{1-(-e^{-a \pi i})^q}{1 + e^{-a \pi i}}$$ which is $$e^{-a i \pi/2} \frac{1-(-1)^q e^{-p \pi i}}{1 + e^{-a \pi i}} = e^{-a i \pi/2} \frac{1-(-1)^{p+q}}{1 + e^{-a \pi i}} = e^{-a i \pi/2} \frac{2}{1 + e^{-a \pi i}} = \frac{1}{\cos (a\pi/2)}$$
It follows that the original sum and the integral is $$J = \frac{\pi}{2} \frac{1}{\cos(a \pi/2)}.$$
- | {"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": 0.9837441444396973, "perplexity": 203.17603281367477}, "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-2014-52/segments/1418802765846.54/warc/CC-MAIN-20141217075245-00089-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://hitchhikersgui.de/Conformal_group | # Conformal group
In mathematics, the conformal group of a space is the group of transformations from the space to itself that preserve angles. More formally, it is the group of transformations that preserve the conformal geometry of the space.
Several specific conformal groups are particularly important:
• The conformal orthogonal group. If V is a vector space with a quadratic form Q, then the conformal orthogonal group CO(V, Q) is the group of linear transformations T of V such that for all x in V there exists a scalar λ such that
${\displaystyle Q(Tx)=\lambda ^{2}Q(x)}$
For a definite quadratic form, the conformal orthogonal group is equal to the orthogonal group times the group of dilations.
All conformal groups are Lie groups.
## Angle analysis
In Euclidean geometry one can expect the standard circular angle to be characteristic, but in pseudo-Euclidean space there is also the hyperbolic angle. In the study of special relativity the various frames of reference, for varying velocity with respect to a rest frame, are related by rapidity, a hyperbolic angle. One way to describe a Lorentz boost is as a hyperbolic rotation which preserves the differential angle between rapidities. Thus they are conformal transformations with respect to the hyperbolic angle.
A method to generate an appropriate conformal group is to mimic the steps of the Möbius group as the conformal group of the ordinary complex plane. Pseudo-Euclidean geometry is supported by alternative complex planes where points are split-complex numbers or dual numbers. Just as the Möbius group requires the Riemann sphere, a compact space, for a complete description, so the alternative complex planes require compactification for complete description of conformal mapping. Nevertheless, the conformal group in each case is given by linear fractional transformations on the appropriate plane.[2]
## Conformal group of spacetime
In 1908, Harry Bateman and Ebenezer Cunningham, two young researchers at University of Liverpool, broached the idea of a conformal group of spacetime[3][4][5] (now usually denoted as C(1,3)).[6] They argued that the kinematics groups are perforce conformal as they preserve the quadratic form of spacetime and are akin to orthogonal transformations, though with respect to an isotropic quadratic form. The liberties of an electromagnetic field are not confined to kinematic motions, but rather are required only to be locally proportional to a transformation preserving the quadratic form. Harry Bateman’s paper in 1910 studied the Jacobian matrix of a transformation that preserves the light cone and showed it had the conformal property (proportional to a form preserver).[7]
Isaak Yaglom has contributed to the mathematics of spacetime conformal transformations in split-complex and dual numbers.[8] Since split-complex numbers and dual numbers form rings, not fields, the linear fractional transformations require a projective line over a ring to be bijective mappings.
It has been traditional since the work of Ludwik Silberstein in 1914 to use the ring of biquaternions to represent the Lorentz group. For the spacetime conformal group, it is sufficient to consider homographies on the projective line over that ring. Elements of the spacetime conformal group have been called spherical wave transformations by Bateman. The particulars of the spacetime quadratic form study have been absorbed into Lie sphere geometry.
## References
1. ^ Jayme Vaz, Jr.; Roldão da Rocha, Jr. (2016). An Introduction to Clifford Algebras and Spinors. Oxford University Press. p. 140. ISBN 9780191085789.
2. ^ Tsurusaburo Takasu (1941) "Gemeinsame Behandlungsweise der elliptischen konformen, hyperbolischen konformen und parabolischen konformen Differentialgeometrie", 2, Proceedings of the Imperial Academy 17(8): 330–8, link from Project Euclid, MR 14282
3. ^ Bateman, Harry (1908). "The conformal transformations of a space of four dimensions and their applications to geometrical optics". Proceedings of the London Mathematical Society. 7: 70–89. doi:10.1112/plms/s2-7.1.70.
4. ^ Bateman, Harry (1910). "The Transformation of the Electrodynamical Equations". Proceedings of the London Mathematical Society. 8: 223–264. doi:10.1112/plms/s2-8.1.223.
5. ^ Cunningham, Ebenezer (1910). "The principle of Relativity in Electrodynamics and an Extension Thereof". Proceedings of the London Mathematical Society. 8: 77–98. doi:10.1112/plms/s2-8.1.77.
6. ^ Boris Kosyakov, Introduction to the Classical Theory of Particles and Fields, Springer, 2007, p. 216.
7. ^ Warwick, Andrew (2003). Masters of theory: Cambridge and the rise of mathematical physics. Chicago: University of Chicago Press. pp. 416–24. ISBN 0-226-87375-7.
8. ^ Isaak Yaglom (1979) A Simple Non-Euclidean Geometry and its Physical Basis, Springer, ISBN 0387-90332-1, MR 520230 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "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": 0.9151301383972168, "perplexity": 962.3692615439282}, "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-2018-47/segments/1542039743521.59/warc/CC-MAIN-20181117123417-20181117145417-00473.warc.gz"} |
http://mathhelpforum.com/algebra/1957-proof-question.html | # Math Help - Proof question
1. ## Proof question
I have questions about one proof.
It concerns numbers with rational exponents.
Having in mind that for all $r,s \in Q,a \in R$ is $(a^r )^s = a^{rs}$ then author of book wants to show that $a$ must be positive real number.
Proof is:
If $r$ is rational number then $\frac{r}{2}$ is also a rational number, so:
$a^r = a^{\frac{r}{2} \cdot 2} = (a^{\frac{r}{2}} )^2 \ge 0$
Similarly, if $r$ is rational number then $-r$ is rational number, so:
$a^{ - r} = a^{r \cdot ( - 1)} = (a^r )^{ - 1}$ which gives us that $a^r \ne 0$. Then is $a^r > 0$
If $r \ne 0$ then is $\frac{1}{r} \ne 0$.
Based on above, we get that $a = a^{r \cdot \frac{1}{r}} = (a^r )^{\frac{1}{r}}$ which means that $a$ must be positive number because $a^r$ is positive number so is $(a^r )^{\frac{1}{r}}$ positive number.
My question is why $a^r$ must be positive number?
If $r$ is rational number then $\frac{r}{3}$ is also a rational number, so:
$a^r = a^{\frac{r}{3} \cdot 3} = (a^{\frac{r}{3}} )^3$ which means that $a^r$ can be also negative number.
Based on that, then $a = a^{r \cdot \frac{1}{r}} = (a^r )^{\frac{1}{r}}$ can mean that $a$ can be also negative number.
2. I do not know what your author wants to do; but exponents are definied ONLY for positive real numbers on this level.
Thus, $(a^r)^s=a^{rs}$ is only defined if $a^r$ is positive. But $a^r$ can only be defined if $a$ is positive.
3. Originally Posted by ThePerfectHacker
I do not know what your author wants to do; but exponents are definied ONLY for positive real numbers on this level.
Thus, $(a^r)^s=a^{rs}$ is only defined if $a^r$ is positive. But $a^r$ can only be defined if $a$ is positive.
Can you explain me this?
I don't understand.
4. Originally Posted by DenMac21
Can you explain me this?
I don't understand.
In more advanced math we define $a^b$ for all complex numbers (imaginary numbers). For now that is not necessary. Thus, only positive real numbers (non-imaginary) are defined. For example $(-1)^2$ still has meaning but $(-1)^{1/2}$ does not. Thus, we only consider postive real numbers. Thus, I do not know what your author is trying to do because as I understand it, it has only meaning for positives numbers thus there is nothing to prove.
Maybe I am wrong and missing something what your author is trying to say.
5. Originally Posted by ThePerfectHacker
In more advanced math we define $a^b$ for all complex numbers (imaginary numbers). For now that is not necessary. Thus, only positive real numbers (non-imaginary) are defined. For example $(-1)^2$ still has meaning but $(-1)^{1/2}$ does not. Thus, we only consider postive real numbers. Thus, I do not know what your author is trying to do because as I understand it, it has only meaning for positives numbers thus there is nothing to prove.
Maybe I am wrong and missing something what your author is trying to say.
Neither do I!
Yes I know that there is no rational solution of $(-1)^{1/2}$ but there is rational solution of $(-1)^{1/3}$ which author didn't mention. He only took in consideration $a^r, r=\frac{1}{2}$ but not $a^r, r=\frac{1}{3}$. He said that $a$ must be positive real number, but it can be also negative real number (under right condition of course). | {"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": 43, "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": 0.9160242080688477, "perplexity": 228.76086192499804}, "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-2014-23/segments/1406510273381.44/warc/CC-MAIN-20140728011753-00121-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://simple.wikipedia.org/wiki/Poisson%27s_ratio | # Poisson's ratio
Figure 1: A cube with sides of length L of an isotropic linearly elastic material subject to tension along the x axis, with a Poisson's ratio of 0.5. The green cube is unstrained, the red is expanded in the x direction by $\Delta L$ due to tension, and contracted in the y and z directions by $\Delta L'$.
$\nu = -\frac{d\varepsilon_\mathrm{trans}}{d\varepsilon_\mathrm{axial}} = -\frac{d\varepsilon_\mathrm{y}}{d\varepsilon_\mathrm{x}}= -\frac{d\varepsilon_\mathrm{z}}{d\varepsilon_\mathrm{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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "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": 0.8099693655967712, "perplexity": 448.5631942666059}, "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-2015-32/segments/1438042987127.36/warc/CC-MAIN-20150728002307-00308-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://mathhelpboards.com/threads/amandas-question-at-yahoo-answers-eigenvalues-and-eigenvectors.3594/ | # Amanda's question at Yahoo! Answers (Eigenvalues and eigenvectors)
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Jan 29, 2012
661
Here is the question:
So, I'm attempting to work on this question:
Given that matrix A =
0 1 1
1 0 1
1 1 0
and has λ = -1 as one of its eigenvalues, find the corresponding eigenvectors.
-1
1
0
and
-1
0
1
BUT.. I don't understand how they got that because can't you bring it down to the reduced form of
1 0 0
0 1 0
0 0 1
So how do you get the corresponding eigenvalues from that? I'm so confused! Help please
Here is a link to the question:
How to I find corresponding eigenvectors? - Yahoo! Answers
I have posted a link there to this topic so the OP can find my response.
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Amanda,
I don't understand either how they got those eigenvalues, surely is a typo. Using the transformations $R_2\to R_2-R_1$, $R_3\to R_3-R_1$ and $C_1\to C_1+C_2+C_3$ we get: \begin{aligned} \begin{vmatrix}{-\lambda}&{\;\;1}&{\;\;1}\\{\;\;1}&{-\lambda}&{\;\;1}\\{\;\;1}&{\;\;1}&{-\lambda}\end{vmatrix}&=\begin{vmatrix}{-\lambda}&{\;\;1}&{\;\;1}\\{\;\;1+\lambda}&{-\lambda}-1&{\;\;0}\\{\;\;1+\lambda}&{\;\;0}&{-\lambda-1}\end{vmatrix}\\&=\begin{vmatrix}{-\lambda+2}&{\;\;1}&{\;\;1}\\{\;\;0}&{-\lambda}-1&{\;\;0}\\{\;\;0}&{\;\;0}&{-\lambda-1}\end{vmatrix}\\&=(-\lambda+2)(\lambda+1)^2=0\\&\Leftrightarrow \lambda=2\mbox{ (simple) }\vee \;\lambda=-1\mbox{ (double) } \end{aligned} The eigenvectors are: $$\ker (A-2I)\equiv \left \{ \begin{matrix}-2x_1+x_2+x_3=0\\x_1-2x_2+x_3=0\\x_1+x_2-2x_3=0\end{matrix}\right.$$ As $\lambda=2$ is simple, $\dim(\ker(A-2I))=1$ and easily we find a basis of this eigenspace: $B_2=\{(1,1,1)\}$. On the other hand: $$\ker (A+I)\equiv \left \{ \begin{matrix}x_1+x_2+x_3=0\\x_1+x_2+x_3=0\\x_1+x_2+x_3=0\end{matrix}\right.$$ Now, $\dim(\ker(A+I))=3-\mbox{rank }(A+I)=3-1=2$ and easily we find a basis of this eigenspace: $B_{-1}=\{(-1,1,0),(-1,0,1)\}$. | {"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": 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": 0.9426474571228027, "perplexity": 1284.5357946172262}, "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/1656104655865.86/warc/CC-MAIN-20220705235755-20220706025755-00737.warc.gz"} |
https://www.ideals.illinois.edu/handle/2142/90430 | ## Files in this item
FilesDescriptionFormat
application/pdf
UILU-ENG-11-2207.pdf (374kB)
(no description provided)PDF
## Description
Title: Distributed Algorithms for Consensus and Coordination in the Presence of Packet-Dropping Communication Links: Part I: Statistical Moments Analysis Approach Author(s): Domínguez-García, Alejandro D.; Hadjicostis, Christoforos N.; Vaidya, Nitin H. Subject(s): Distributed algorithms Average consensus Markov chains Coefficients of ergodicity Abstract: This two-part paper discusses robustification methodologies for linear-iterative distributed algorithms for consensus and coordination problems in multicomponent systems, in which unreliable communication links may drop packets. We consider a setup where communication links between components can be asymmetric (i.e., component j might be able to send information to component i, but not necessarily vice-versa), so that the information exchange between components in the system is in general described by a directed graph that is assumed to be strongly connected. In the absence of communication link failures, each component i maintains two auxiliary variables and updates each of their values to be a linear combination of their corresponding previous values and the corresponding previous values of neighboring components (i.e., components that send information to node i). By appropriately initializing these two (decoupled) iterations, the system components can asymptotically calculate variables of interest in a distributed fashion; in particular, the average of the initial conditions can be calculated as a function that involves the ratio of these two auxiliary variables. The focus of this paper to robustify this double-iteration algorithm against communication link failures. We achieve this by modifying the double-iteration algorithm (by introducing some additional auxiliary variables) and prove that the modified double-iteration converges almost surely to average consensus. In the first part of the paper, we study the first and second moments of the two iterations, and use them to establish convergence, and illustrate the performance of the algorithm with several numerical examples. In the second part, in order to establish the convergence of the algorithm, we use coefficients of ergodicity commonly used in analyzing inhomogeneous Markov chains. Issue Date: 2011-09 Publisher: Coordinated Science Laboratory Series/Report: Coordinated Science Laboratory Report no. UILU-ENG-11-2207 Genre: Technical Report Type: Text Language: English URI: http://hdl.handle.net/2142/90430 Sponsor: ECCS-CAR-0954420 and 1059540 (NSF); INFSO-ICT-223844 and PIRG02-GA-2007-224877 (European Commission Seventh Framework Programme); W-911-NF-0710287 (Army Research Office) Date Available in IDEALS: 2016-07-06
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https://experts.illinois.edu/en/publications/vectorial-solution-to-double-curl-equation-with-generalized-coulo-2 | # Vectorial Solution to Double Curl Equation With Generalized Coulomb Gauge for Magnetostatic Problems
Yan Lin Li, Sheng Sun, Qi I. Dai, Weng Cho Chew
Research output: Contribution to journalArticlepeer-review
## Abstract
In this paper, a solution to the double curl equation with generalized Coulomb gauge is proposed based on the vectorial representation of the magnetic vector potential. Traditional Coulomb gauge is applied to remove the null space of the curl operator and hence the uniqueness of the solution is guaranteed. However, as the divergence operator cannot act on edge elements (curl-conforming) directly, the magnetic vector potential is represented by nodal elements, which is too restrictive, since both the tangential continuity and the normal continuity are required. Inspired by the mapping of Whitney forms by mathematical operators and Hodge (star) operators, the divergence of the magnetic vector potential, as a whole, can be approximated by Whitney elements. Hence, the magnetic vector potential can be expanded by the edge elements, where its vectorial nature is retained and only the tangential continuity is required. Finally, the original equation can be rewritten in a generalized form and solved in a more natural and accurate way using finite-element method.
Original language English (US) 7072480 IEEE Transactions on Magnetics 51 8 https://doi.org/10.1109/TMAG.2015.2417492 Published - Aug 1 2015
## Keywords
• Generalized Coulomb gauge
• Whitney forms
• finite element method
• magnetostatic
## ASJC Scopus subject areas
• Electronic, Optical and Magnetic Materials
• Electrical and Electronic Engineering
## Fingerprint
Dive into the research topics of 'Vectorial Solution to Double Curl Equation With Generalized Coulomb Gauge for Magnetostatic Problems'. Together they form a unique fingerprint. | {"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": 0.9451949596405029, "perplexity": 1578.6600622020733}, "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-2023-14/segments/1679296949573.84/warc/CC-MAIN-20230331051439-20230331081439-00430.warc.gz"} |
https://www.physicsforums.com/threads/hamilton-jacobi-problem.741023/ | # Hamilton-Jacobi Problem
1. Mar 1, 2014
### dipole
1. The problem statement, all variables and given/known data
I'm given the time-dependent potential,
$$V(x,t) = -mAxe^{-\gamma t}$$
and asked to find the solution to the Hamilton-Jacobi equation,
$$H(x,\frac{\partial S}{\partial x}) + \frac{ \partial S}{\partial t} = 0$$
3. The attempt at a solution
Without any additional information, I'm assuming the correct Hamiltonian is given simply by,
$$H = \frac{p^2}{2m} -mAxe^{-\gamma t}$$
which gives me,
$$\frac{1}{2m}\bigg ( \frac{\partial S}{\partial x} \bigg )^2 - mAxe^{-\gamma t} + \frac{ \partial S}{\partial t} = 0$$
but I'm having troule separating the variables in order to solve this equation. Normally, when $V = V(x)$ you can use the form $S(x,\alpha,t) = W(x,\alpha) - Et$, but here this won't work.
Have I somehow used the wrong Hamiltonian, or do I just need to guess correctly the right form of $S$?
2. Mar 1, 2014
### minimax
I suggest using the method of characteristics to solve this PDE problem. | {"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": 0.9479057192802429, "perplexity": 850.873310447476}, "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/1518891814857.77/warc/CC-MAIN-20180223213947-20180223233947-00385.warc.gz"} |
https://randommathgenerator.com/2016/12/23/filtrations-and-gradings/ | This is going to be a blog post on Filtrations and Gradings. We’re going to closely follow the development in Local Algebra by Serre.
A filtered ring is a ring with the set of ideals $\{A_n\}_{n\in\Bbb{Z}}$ such that $A_0=A$, $A_{n+1}\subset A_n$, and $A_pA_q=A_{p+q}$. An example would be $A_n=(2^n)$, where $(2^n)$ is the ideal generated by $2^n$ in $\Bbb{Z}$.
Similarly, a filtered module $M$ over a filtered ring $A$ is defined as a module with a set of submodules $\{M_n\}_{n\in\Bbb{N}}$ such that $M_0=M$, $M_{n+1}\subset M_n$, and $A_pM_q\subset M_{p+q}$. Why not just have $M_pM_q\subset M_{p+q}$? This is because multiplication between elements of a module may not be defined. An example would be the module generated by by the element $v$ over $\Bbb{Z}$, where $M_n=2^n M$.
Filtered modules form an additive category $F_A$ with morphisms $u:M\to N$ such that $u(M_n)\subset N_n$. A trivial example is $\Bbb{Z}\to\Bbb{Z}$, defined using the grading above, and the map being defined as $x\to -x$.
If $P\subset M$ is a submodule, then the induced filtration is defined as $P_n=P\cap M_n$. Is every $P_n$ a submodule of $P$? Yes, because every $M_n$ is by definition a submodule of $M$, and the intersection of two submodules ($M_n$ and $P$ in particular) is always a submodule. Simialrly, the quotient filtration $N=M/P$ is also defined. As the quotient of two modules, the meaning of $M/P$ is clear. However, what about the filtration of $M/P$? Turns out the filtration of $N=M/P$ is defined the following way: $N_n=(M_n+P)/P$. We need to have $M_n+P$ as the object under consideration because it is not necessary that $M_n\in P$.
An important example of filtration is the $m$-adic filtration. Let $m$ be an ideal of $A$, and let the filtration of $A$ be defined as $A_n=m^n$. Similarly, for a module $M$ over $A$, the $m$-adic filtration of $M$ is defined by $M_n=m^nM$.
Now we shall discuss the topology defined by filtration. If $M$ is a filtered module over the filtered ring, then $M_n$ form a basis for neighbourhoods around $0$. This obviously is a nested set of neighbourhoods, and surely enough the intersection of a finite number of neighbourhoods is also a neighbourhood, and so is the union of any set of neighbourhoods. Hence, the usual topological requirements for a basis is satisfied. But why $0$?
Proposition: Let $N$ be a submodule of a filtered module $M$. Then the closure of $\overline{N}$ of $N$ is defined as $\bigcap(N+M_n)$. How does this work? If one were to hand wave a bit, we are essentially finding the intersection of all neighbourhoods of $N$. Remember that each $M_n$ is a neighbourhood of $0$. We’re translating each such neighbourhood by $N$, which is another way of saying we’re now considering all neighbourhoods of $N$. And then we find the intersection of all such neighbourhoods to find the smallest closed set containing $N$. There is an analogous concept in metric spaces- the intersection of all open sets containing $[0,1]$, for instance, is the closed set $[0,1]$. The analogy is not perfect, as the intersection of all neighbourhoods of $(0,1)$ is $(0,1)$ itself, which is not a closed set. But hey. We at least have something to go by.
Corollary: $M$ is Hausdorff if and only if $\cap M_n=0$. | {"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": 69, "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": 0.9470908045768738, "perplexity": 78.20352989174694}, "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/1614178358798.23/warc/CC-MAIN-20210227084805-20210227114805-00362.warc.gz"} |
http://tex.stackexchange.com/questions/79039/extra-space-between-footnote-numbers-and-content-for-high-count?answertab=oldest | # Extra space between footnote numbers and content for high count
I’m using the following commands to customize footnotes:
\usepackage[bottom,norule,hang]{footmisc}
\setlength{\footnotemargin}{2mm}
\setlength{\footnotesep}{4mm}
The results were as I wanted but only until the 10th footnote (see image below). Is there a way to define the the footnote text margin as a function of the horizontal space that the footnote number takes?
The numbers are flushed to the left and have the same margin as the main text. The footnote text margin, however, seems to be the main text margin plus a fixed quantity.
-
Welcome to TeX.sx! Please add a minimal working example (MWE) that illustrates your problem. knowing what document class you use may be central to answering your question. – barbara beeton Oct 25 '12 at 12:53
footmisc with the hang option has the property that if \footnotemargin is 0 or negative, then the hanging indent is set to the width of the footnote mark. Thus:
\documentclass[a5paper]{article}
\usepackage{lipsum}
\usepackage[bottom,norule,hang]{footmisc}
\setlength{\footnotemargin}{0pt}
\begin{document}
\footnote{Long footnote text, to see whether we get
the hanging effect or not. Just as a test.}\lipsum[4]
\footnote{Long footnote text, to see whether we get
the hanging effect or not. Just as a test.}\lipsum[4]
\setcounter{footnote}{1010}
\footnote{Long footnote text, to see whether we get
the hanging effect or not. Just as a test.}\lipsum[4]
\end{document}
produces the following reasonable effect with no overlapping
What is lacking here is some spacing between the footnote mark and the text. To correct this we can patch the relevant command in footmisc so that if the hang option is given and \footnotemargin is negative, we use the length information there to provide the spacing between the mark and the text, e.g. a value of -0.5em leads to a space of 0.5em. In the original package, if \footnotemargin is <=0 then just the width of the footnotemark is used. Below, we change one line so that the width of a box containing the footnotemark plus a skip of -\footnotemargin is used in the same situation:
\documentclass[a5paper]{article}
\usepackage{lipsum}
\usepackage[bottom,norule,hang]{footmisc}
\setlength{\footnotemargin}{-0.5em}
\makeatletter
\ifFN@para
\else
\long\def\@makefntext#1{%
\ifFN@hangfoot
\bgroup
\setbox\@tempboxa\hbox{%
\ifdim\footnotemargin>0pt
\hb@xt@\footnotemargin{\@makefnmark\hss}%
\else
\@makefnmark\hskip-\footnotemargin %%Changed here
\fi
}%
\leftmargin\wd\@tempboxa
\rightmargin\z@
\linewidth \columnwidth
\parshape \@ne \leftmargin \linewidth
\footnotesize
\@setpar{{\@@par}}%
\leavevmode
\llap{\box\@tempboxa}%
\parskip\hangfootparskip\relax
\parindent\hangfootparindent\relax
\else
\parindent1em
\noindent
\ifdim\footnotemargin>\z@
\hb@xt@ \footnotemargin{\hss\@makefnmark}%
\else
\ifdim\footnotemargin=\z@
\llap{\@makefnmark}%
\else
\llap{\hb@xt@ -\footnotemargin{\@makefnmark\hss}}%
\fi
\fi
\fi
\footnotelayout#1%
\ifFN@hangfoot
\par\egroup
\fi
}
\fi
\makeatother
\begin{document}
\footnote{Long footnote text, to see whether we get
the hanging effect or not. Just as a test.}\lipsum[4]
\footnote{Long footnote text, to see whether we get
the hanging effect or not. Just as a test.}\lipsum[4]
\setcounter{footnote}{1010}
\footnote{Long footnote text, to see whether we get
the hanging effect or not. Just as a test.}\lipsum[4]
\end{document}
-
The only value that is important here, given the hang option, is:
\setlength{\footnotemargin}{2mm}
which is very small for any two digit number. footmisc sets the length to 1.8em by default, which might be too big for your taste unless your footnotes number in the 100s (in my experience, you really need at least 1.2em if you get into the 100s).
Try this file, for example:
\documentclass[12pt]{article}
\usepackage{lipsum}
\usepackage[bottom,norule,hang]{footmisc}
\setlength{\footnotemargin}{1.2em}
%\setlength{\footnotesep}{4mm}
\begin{document}
\setcounter{footnote}{8}
\footnote{Test.}\lipsum[1]
\footnote{Test.}\lipsum[1]
\footnote{Test.}\lipsum[1]
\setcounter{footnote}{98}
\footnote{Test.}\lipsum[1]
\footnote{Test.}\lipsum[1]
\footnote{Test.}\lipsum[1]
\setcounter{footnote}{198}
\footnote{Test.}\lipsum[1]
\footnote{Test.}\lipsum[1]
\footnote{Test.}\lipsum[1]
\setcounter{footnote}{998}
\footnote{Test.}\lipsum[1]
\footnote{Test.}\lipsum[1]
\footnote{Test.}\lipsum[1]
\end{document}
-
Why CW? It seems a genuine personal answer – egreg Oct 25 '12 at 13:06
@egreg -- Hmm, didn't mean to make it CW.... How do I change that? – jon Oct 25 '12 at 13:10
Edit the answer and remove the checkmark at the bottom – egreg Oct 25 '12 at 13:16
@egreg -- I may be too dense, but I made an edit and there was no option to 'uncheck' anything anywhere. Perhaps I don't have the right kind of permissions to make that kind of change..? (Not that I need to undo the CW status, but I agree that it is not a CW answer.) – jon Oct 25 '12 at 13:26
As I know is the usual way to layout footnotes is to have a look to the resulting number of footnotes (for example 105) after you finished writing your text.
Now you can change \setlength{\footnotemargin}{2mm} to a value that gives you the wished space, for example \setlength{\footnotemargin}{7mm}.
So you get all footnotes layouted with the same space for the footnote number.
If you want to change the space after the 9th footnote you can use a second command \setlength{\footnotemargin}{7mm} as shown in the following MWE:
%http://tex.stackexchange.com/questions/79039/extra-space-between-footnote-numbers-and-content-for-high-count
\documentclass[12pt]{scrartcl}
\usepackage[bottom,norule,hang]{footmisc}
\setlength{\footnotemargin}{2mm} % set space for first 9 footnotes
\begin{document}
Lorem ipsum\footnote{footnote 1 now extends footnote now extends footnote now extends
footnote now extends footnote now extends footnote now extends
footnote now extends} below the bottom margin
which\footnote{This footnote 2 now extends} below the bottom margin
which\footnote{This footnote 3 now extends} below the bottom margin
which\footnote{This footnote 4 now extends} below the bottom margin
which\footnote{This footnote 5 now extends} below the bottom margin
which\footnote{This footnote 6 now extends} below the bottom margin
which\footnote{This footnote 7 now extends} below the bottom margin
which\footnote{This footnote 8 now extends} below the bottom margin
which\footnote{This footnote 9 now extends} below the bottom margin
\setlength{\footnotemargin}{7mm} % set new space for number
which\footnote{This footnote 10 now extends footnote now extends
footnote now extends footnote now extends footnote now extends
footnote now extends footnote now extends} below the bottom margin
which\footnote{This footnote 11 now extends} below the bottom margin
which\footnote{This footnote 12 now extends} below the bottom margin
\end{document}
For me that looks not like a good typhography, so I personaly wouldn't do it.
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http://physics.aps.org/articles/v7/111 | Viewpoint: Arrow of Time Emerges in a Gravitational System
Physics 7, 111
Study of masses interacting via gravity challenges the idea that special initial conditions are needed to give time a direction.
The fundamental laws of physics, we believe, do not depend on the direction of time. Why, then, is the future so different from the past? The origin of this “arrow of time” has puzzled physicists and philosophers for more than a century, and it remains one of the fundamental conceptual problems of modern physics [1]. Although a preferred direction of time can occur in models of physical systems, this typically happens only if one inserts very special initial conditions. Julian Barbour at the University of Oxford and his colleagues [2] have now shown this tinkering isn’t necessary to produce an arrow of time in a system of masses interacting via Newtonian gravity. They demonstrate that the evolution of this surprisingly simple system almost always contains a unique moment of lowest “complexity,” a point they identify as a “past” from which two distinct (and more complex) “futures” emerge.
The work of Barbour and his colleagues is the latest in a long history of attempts to explain the arrow of time. One possibility, of course, is that we don’t know the right laws of physics—perhaps the correct fundamental laws do determine a preferred direction of time [3]. Alternatively, if the laws of nature do not pick out a preferred “future,” perhaps boundary conditions do. For example, most cosmological models assume, explicitly or implicitly, that the big bang was a moment of exceptionally low entropy.
Indeed, most physicists accept the view that the direction of time is the same as the direction of increasing entropy. But this is, at best, an incomplete picture, failing to explain why there should have been a rare condition of low entropy in the past. More than a century ago, Boltzmann suggested that our visible Universe might merely be a temporary, low-entropy statistical fluctuation, affecting a small portion of a much larger equilibrium system [4]. In that case, the direction of time would simply be the one that takes us back towards equilibrium. But most contemporary physicists find this explanation unsatisfying: a random fluctuation containing “us” would have been far more likely to produce a single galaxy, a planet, or just a “brain” rather than a whole universe [5,6]. Moreover, according to the “Loschmidt irreversibility paradox,” if one posits such a moment of low entropy, entropy should increase both to the future and to the past, giving two separate arrows of time [7].
In their gravitational model, Barbour and his colleagues find a state of “low complexity” that is analogous to Boltzmann’s low-entropy fluctuation. But in their case, no rare statistical fluctuation is necessary to explain this state; instead, it arises naturally out of simple physical laws that have no explicit dependence on the direction of time. The authors study one of the simplest possible systems: a collection of $N$ point particles interacting through Newtonian gravity. Their only assumptions are that the total energy (potential plus kinetic) and the total angular momentum of the system are zero. From earlier numerical simulations and analytic analysis, it is known that in the distant future, such a system tends to break up into weakly interacting subsystems—typically, pairs of masses in Keplerian orbits [8]. Starting with such a dispersed system and running time backwards, one might expect that it would coalesce in the past into a state of high density. Barbour and his coauthors show analytically that this expectation is right: for almost every initial configuration of masses, there is a unique moment of minimum size and maximum uniformity. From this point, the system expands outward, approximately symmetrically in both directions of time (Fig. 1). The system is therefore globally symmetric in time, as the equations dictate, and yet has a local arrow of time.
As a key step in their argument, the authors analyze the evolution of the masses in “shape space,” a space of observables that describe the shape of the system but are independent of its size and orientation. Three bodies, for instance, form a triangle, and their shape space is the space of similar triangles. Shape space contains a natural dimensionless measure of complexity, denoted $CS$, which is determined by the moment of inertia and the total Newtonian gravitational potential. $CS$ describes the degree of nonuniformity and clustering; it has a minimum at the moment of minimum size and grows approximately monotonically from that minimum in both directions of time. Barbour and his colleagues provide a fairly simple and intuitive explanation for this behavior by showing that the dynamics of the $N$-body system in shape space has an effective friction term, which provides a sort of dissipation even though the underlying equations of motion are symmetric in time.
The idea of time proceeding in two directions, towards two futures, from a moment of minimum complexity is not itself new. It has appeared, for example, in cosmological models of eternal inflation [9]. But the emergence of this behavior in a system as simple as the one Barbour and his colleagues consider is unexpected. The constraints of vanishing energy and angular momentum were based on a Machian philosophical view; namely, that only relational observables should be relevant to physics. But these choices also appear to match our Universe; vanishing energy, for instance, is an indication of spatial flatness. It is worth emphasizing, however, that the model in this paper is Newtonian—it is not yet clear whether it can be extended to a more realistic general relativistic description of gravity, though the authors suggest this might be possible by using shape dynamics [10], a modified scale-invariant form of general relativity.
Have Barbour and his colleagues solved the problem of the arrow of time? Probably not yet. We’re still left with the mystery that the arrows of time we see in different physical phenomena all point in the same direction. Electromagnetic waves are retarded, not advanced; radioactive nuclei decay, they don’t reassemble; gravitating systems clump, they don’t disperse; we remember the past, not the future. A good deal of work would be required to show that these disparate arrows all match the direction determined from a purely gravitational model.
Nevertheless, the results of Barbour and his colleagues provide an intriguing new viewpoint. Standard approaches to the arrow of time typically require a rare statistical fluctuation, or, often, the smuggling in of assumptions about initial conditions. Their work offers evidence that ordinary gravitational dynamics may itself be enough to produce the simple “initial” point that can give time a direction.
References
1. H. D. Zeh, The Physical Basis of the Direction of Time (Springer-Verlag, Berlin, 2007)[Amazon][WorldCat]
2. Julian Barbour, Tim Koslowski, and Flavio Mercati, “Identification of a Gravitational Arrow of Time,” Phys. Rev. Lett. 113, 181101 (2014)
3. The violation of time-reversal invariance by the weak interactions doesn’t count, since the interactions are still charge-parity-time (CPT) invariant
4. L. Boltzmann, “On Certain Questions of the Theory of Gases,” Nature 51, 413 (1895)
5. R. Feynman, The Character of Physical Law (MIT Unversity Press, Cambridge, 1967)[Amazon][WorldCat]
6. A. Albrecht, “Cosmic Inflation and the Arrow of Time,” arXiv:astro-ph/0210527
7. J. Loschmidt, “Uber den Zustand des Warmegleichgewichtes eines Systems von Korpern mit Rücksicht auf die Schwerkraft,” Wien. Ber. 73, 128 (1876)
8. C. Marchal and D. Saari, “On the Final Evolution of the $N$-body Problem,” J. Diff. Eq. 20, 150 (1976)
9. S. M. Carroll and J. Chen, “Spontaneous Inflation and the Origin of the Arrow of Time,” arXiv:hep-th/0410270
10. H. Gomes, S. Gryb, and T. Koslowski, “Einstein Gravity as a 3D Conformally Invariant Theory,” Class. Quantum Grav. 28, 045005 (2011)
Steven Carlip received an undergraduate degree in physics from Harvard in 1975. After seven years as a printer, editor, and factory worker, he returned to school at the University of Texas, where he earned his Ph.D. in 1987. Following a stint as a postdoctoral fellow at the Institute for Advanced Study, he joined the faculty of the University of California at Davis, where he is now a professor. His specialty is quantum gravity; his recent research has focused on quantum black holes, numerical approaches to the Feynman path integral, and “spontaneous dimensional reduction” at short distances.
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LIGO has detected a second burst of gravitational waves from merging black holes, suggesting that such detections will soon become routine and part of a new kind of astronomy. Read More » | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "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": 0.8147060871124268, "perplexity": 718.7985252455245}, "config": {"markdown_headings": false, "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-2017-09/segments/1487501174215.11/warc/CC-MAIN-20170219104614-00280-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://infoscience.epfl.ch/record/185246 | Infoscience
Conference paper
# Learning Ridge Functions With Randomized Sampling In High Dimensions
We study the problem of learning ridge functions of the form f(x) = g(aT x), x ∈ ℝd, from random samples. Assuming g to be a twice continuously differentiable function, we leverage techniques from low rank matrix recovery literature to derive a uniform approximation guarantee for estimation of the ridge function f. Our new analysis removes the de facto compressibility assumption on the parameter a for learning in the existing literature. Interestingly the price to pay in high dimensional settings is not major. For example, when g is thrice continuously differentiable in an open neighbourhood of the origin, the sampling complexity changes from O(log d) to O(d) or from equation to O(d2+q/2-q) to O(d4), depending on the behaviour of g' and g" at the origin, with 0 <; q <; 1 characterizing the sparsity 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": 0.8178907632827759, "perplexity": 796.0622838622149}, "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/1487501171646.15/warc/CC-MAIN-20170219104611-00175-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2780255/fundamental-theorem-of-algebra-proof-for-max-real-roots-of-real-polynomials-u/2780261 | # Fundamental Theorem of Algebra proof, for max real roots of real polynomials - using only high school algebra-precalc?
Ths isn't the proper Fundamental Theorem of Arithmetic (which includes complex roots), but just the maximum number of real roots for polynomial $$f(x)$$ with real coefficients... which is $$deg(f)$$.
I believe there isn't a High School-level proof of the full theorem, but is there one of this simpler theorem?
I think there should be a very simple proof using the Factor Theorem:
$$f(a)=0 \iff (x-a) \mid f(x)$$
The only way to have an extra zero is to have an extra factor of $$(x-a)$$, which increases the degree by one.
Because an extra factor might be the same as one already present, forming a multiple root or zero, the number of distinct roots may be fewer than this. So it is only a maximum.∎
However, I'm not sure how to make this rigorous, or how to show it covers all possible polynomials (i.e. all polynomials can be built up from $$(x-a)$$ factors - maybe it follows from the Factor Theorem?).
Or maybe there's a completely different, simpler approach?
BTW: I am interested in it as a way to understand the Schwartz-Zippel lemma for Polynomial Identity Testing (a probabilistic algorithm: subtracting the polynomials to get zero if identical, then use the maximum zeros/roots to calculate probability of guessing roots - false positives). The relevant part to this question is:
A polynomial of one variable of degree {d} can vanish at only {d} points without being identically zero The Curious History of the Schwartz-Zippel Lemma
Also at the wikipedia article
a polynomial of degree d can have no more than d roots.
Because the guesses can be made using only integers (I think modulo, making them fields), it seems to me that the simpler theorem for real polynomials should be enough...
• If you are trying to show that a polynomial can have no more than $\partial f$ distinct roots, then you can use polynomial division to show the result (that is, if $r_1,...,r_p$ are distinct roots of $f$ then show $\partial f \ge p$). Otherwise, you need to characterise what is meant by a multiple root. As you are probably aware, $x^2+1$ has no real roots. – copper.hat May 14 '18 at 5:11
• @copper.hat Thanks, yes that's what I want, for reals. I think I follow you: for each of the $p$ distinct real roots $r_p$, there is a distinct factor $(x-r_p)$, and (because there may be other distinct factors, duplicate factors, and factors thst can't be reduced, like $x^2+1$), the degree is at least $p$. Starting from distinct roots avoids the problem. Proving an inequality means we needn't prove the converse. I forgot about complex roots, and they mean I was wrong: one can't build up every polynomial from factors like $(x-r), r \in \Bbb R$, like your example $x^2+1$. – hyperpallium May 14 '18 at 8:17
• I think the basic fact for showing the result is that $\partial (f \cdot g) = \partial f + \partial g$. Hence each distinct, real root adds one to the degree. – copper.hat May 14 '18 at 16:10
• @copper.hat I see that (and the idea is in my sketch), but I have trouble putting it all together into a rigorous proof - and seeing/showing that it is indeed rigorous. I was hoping there'd be a standard detailed proof - but I guess that is all done on the proper FTA. – hyperpallium May 15 '18 at 9:44
If $P(x)$ is a polynomial such that $P(a)=0$ then it is only possible when it can be expressed as $(x-a)F(x)$ because when we subsitute $x=a$ we get $0$
Let $P(x)$ have $p$ roots then it may be written as $(x-p_1)(x-p_2)...(x-p_p)$ which is a polynomial of degree p. | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8918473124504089, "perplexity": 252.81879047656759}, "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-2020-29/segments/1593655908294.32/warc/CC-MAIN-20200710113143-20200710143143-00489.warc.gz"} |
http://papers.neurips.cc/paper/5038-optimistic-concurrency-control-for-distributed-unsupervised-learning | # NIPS Proceedingsβ
## Optimistic Concurrency Control for Distributed Unsupervised Learning
[PDF] [BibTeX] [Supplemental] [Reviews]
### Abstract
Research on distributed machine learning algorithms has focused primarily on one of two extremes---algorithms that obey strict concurrency constraints or algorithms that obey few or no such constraints. We consider an intermediate alternative in which algorithms optimistically assume that conflicts are unlikely and if conflicts do arise a conflict-resolution protocol is invoked. We view this optimistic concurrency control'' paradigm as particularly appropriate for large-scale machine learning algorithms, particularly in the unsupervised setting. We demonstrate our approach in three problem areas: clustering, feature learning and online facility location. We evaluate our methods via large-scale experiments in a cluster computing environment. " | {"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": 0.9317938685417175, "perplexity": 3156.3405683855667}, "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/1600400227524.63/warc/CC-MAIN-20200925150904-20200925180904-00061.warc.gz"} |
https://muddyhats.com/2018/09/11/a-triangle-from-a-stick/ | Inelegant Explorations of Some Personal Problems
# A Triangle from a Stick
A famous problem in probability is this: Suppose you randomly1 break a stick in two places. You will get three smaller pieces of stick. What is the probability that you can form a triangle using these pieces?
One interesting thing about this problem is that depending on how you visualize breaking the stick, you get different probabilities.
In this post, we will see two such scenarios. We will also see how we can confirm our answers using Python simulations.
## First Scenario
A simple way to break a stick into two is to randomly make two cuts and pull the pieces apart only after both the cuts have been made. The cuts may be like the following:
Or the cuts may be in the opposite order, where Cut 1 comes after Cut 2. Remember that this is possible because we are making both the cuts uniformly at random.
We can assume the length of the stick to be some $L$. But for computing the probabilities, we can assume, without loss of generality, that the stick is of unit length. Assuming the first cut is at position $x$ and the second cut is at $y$, the stick with the cuts would look like something like this:
Or like this:
Now how do we proceed to compute the probabilities? I will give a hint: for three lengths to be the lengths of sides of a triangle, the sum of any two of the lengths should be greater than the third length. (Why this condition is necessary and sufficient? If one side is longer than the other two sides added together, think about how you would draw the triangle using the longest side as the base.)
If you want to give a shot at solving the problem by your own, now is a good time.
In the first case, the lengths of the pieces are $x$, $y-x$, and $1-y$. When we apply the above hint for all the three possible combinations, we get $x < \frac{1}{2}$, $y > \frac{1}{2}$, and $y-x < \frac{1}{2}$. Similarly, for the second case, when $x > y$, we get $y < \frac{1}{2}$, $x > \frac{1}{2}$, and $x-y < \frac{1}{2}$
Computing the actual probability is quite easy at this point. Both $x$ and $y$ could have any value from the range $[0, 1]$. Out of those we are only interested in those values which satisfy the above conditions. We can use a graph to figure out the proportion of $(x, y)$ which satisfy the triangle property.
In the graph, region (the dashed triangle at the top left) contains the values of $x$ and $y$ which form a triangle for the first case. That is, here $x < \frac{1}{2}, y > \frac{1}{2}$, and $y-x < \frac{1}{2}$. Similarly, in region B, $y < \frac{1}{2}, x > \frac{1}{2}, x-y < \frac{1}{2}$. So all the cuts represented in regions A and B form a triangle. And the probability of such cuts is simply the ratio of the areas of regions A and B with the total area. As both regions are just simple right triangles, we can quickly see that the areas of both A and B is $\frac{1}{8}$. So the total area we are interested in is $\frac{1}{4}$. Since the total area of the square is just 1, the probability of the cuts forming a triangle is just $\frac{1}{4}$.
If you are convinced by the argument in the last paragraph, great! In any case, it would be great to be able to conduct some experiment to confirm our calculations. One way to do that would be to take lots of sticks, break them randomly and see whether we are able to form triangles in roughly a quarter of cases. But that is difficult – not just because of the physical exertion involved, but to actually break a stick uniformly at random is difficult for humans. We tend to break sticks more towards the middle. So we go for the better option and ask Python to do the hard work for us.
Python is great for simulating random experiments. In this case the program is particularly simple: all we need to do is to generate two random numbers in the range of $[0, 1]$. These numbers represent our cuts and we can check whether we can form a triangle using them. If a triangle can be formed, we increment a counter. Finally we can take the ratio of the successful instances to the total number of trials to get an estimate of the probability. Now here is the code:
import random as rand
def istriangle(a, b, c):
return a + b > c and a + c > b and b + c > a
def break_one_stick():
x, y = rand.uniform(0, 1.0), rand.uniform(0, 1.0)
if x > y:
x, y = y, x
return istriangle(x, y - x, 1 - y)
def count_triangles(num_trials):
num_triangles = 0
for i in range(num_trials):
if break_one_stick():
num_triangles += 1
return num_triangles
if __name__ == '__main__':
trials = 100000
triangles = count_triangles(trials)
print("Estimated probability = {0:0.4f}".format(triangles/trials))
For me this code returns 0.2489, 0.2502 etc on different runs. This is a nice confirmation of the mathematical answer.
## Second Scenario
Now consider another way to break a stick: you make one cut and take out the first (left) piece. Then you make a cut on the remaining second (right) piece. What is the probability that the three pieces you get in this manner form a triangle?
Here we know that $y$ is greater than $x$. So the lengths of the pieces are $x, y-x, 1-y$. By the triangle condition, we get $x < \frac{1}{2}$, $y > \frac{1}{2}$, and $y - x < \frac{1}{2}$. Rearranging the last equation we get $y < x + \frac{1}{2}$. In other words, $x$ ranges from $0$ to $\frac{1}{2}$ and $y$ ranges from $\frac{1}{2}$ to $x + \frac{1}{2}$.
Therefore, for any first cut $x$ with $x < \frac{1}{2}$, if the second cut $y$ is from the interval $(\frac{1}{2}, x+\frac{1}{2})$, we will get a triangle. Now, in this scenario, $y$ is guaranteed to be in the interval $(x, 1)$. Also, as $x$ is less than $\frac{1}{2}$, the required interval $(\frac{1}{2}, x+\frac{1}{2})$ is a subset of the interval $(x, 1)$.
Thus for a given $x$, the probability of $y$ being in the required range is the ratio of the length of the rquired range to the length of the whole range i.e., $\frac{x}{1-x}$. Here $x$ can be any number between $0$ and $\frac{1}{2}$. To compute the total probability, we need to integrate with respect to $x$.
$\text{Pr(triangle)} = \int_0^{\frac{1}{2}} \frac{x}{1-x} dx$
This is an easy to evaluate integral which turns out to be exactly $\ln{2} - \frac{1}{2}$, or approximately $0.1931$. I am amazed that a problem as simple as this should have an expression involving the natural logarithm as an answer.
Anyway, let us now turn to Python for confirmation. Only one function need to change from the previous code. Here it is:
def break_one_stick():
x = rand.uniform(0, 1.0)
y = rand.uniform(x, 1.0)
return istriangle(x, y - x, 1 - y)
Yes, in fact, the function is now simpler in a sense because we no longer have to check which of $x$ and $y$ is larger.
For me, the simulation gives answers like $0.1931$, $0.1943$, $0.1935$, which are very close to the analytical answer.
These two are not the only scenarios possible, by any means. One can think up lots of other ways of cutting a stick. Once the first cut is made, the bigger of the two pieces can be selected for the next cut. Or, the piece for the second cut can itself be selected based on a coin toss. I think all of these scenarios can be solved similarly. In any case simulation of all the scenarios is very easy using Python.
1. Uniformly at random, to be precise. | {"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": 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": 0.8996501564979553, "perplexity": 179.82782865659502}, "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-2019-30/segments/1563195525187.9/warc/CC-MAIN-20190717121559-20190717143559-00005.warc.gz"} |
http://mathoverflow.net/questions/30589/decomposition-of-gl2-p-into-irreducible-representations | # Decomposition of GL(2,p) into irreducible representations
Let $G=GL(2,p)$ be the group of linear transformations. It acts on the set $X=F_p^2$. Then $C^X$ is a linear representation of $G$. What is the decomposition of this representation into irreducible representations?
Note: Let $d_i$ denote the number of isomorphic irreducible representations. Then $\sum d_i^2$ is equal to the number of orbits of $G$ when $G$ acts on $X\times X$. The number of orbits of $G$ in $X\times X$ is equal to $p+1$.
-
I don't know the answer, but the character table for $G$ can be found in various texts, for instance the third edition of Lang's Algebra. The character of the permutation representation on $X$ is easy to compute. Using the orthogonality relations one gets the decomposition of the permutation character into irreducibles. With a little more work, one can get the decomposition on the level of actual representations. – Robin Chapman Jul 5 '10 at 6:39
Acctually I want to know how $C^X$ decomposed and not just what irreducible represintations it contain, i.e. I want to know not only irreducible sub-repesintation in $C^X$ but also homomorphism from it to C^X. – Klim Efremenko Jul 5 '10 at 8:22
To find the representations all you need is the characters. Each character $\chi$ has a corresponding central idempotent $e_\chi$ in the group algebra. Hitting your representation with $e_\chi$ will give the sum of all irreps with character $\chi$ inside it. – Robin Chapman Jul 5 '10 at 8:53
What I've got so far: there are two obvious copies of the trivial representation: one coming from the fact that G fixes the origin, and the other coming from the fact that G fixes the sum over all points besides the origin. There is also a subrepresentation spanned by sums over all the points contained in each line through the origin; the action of G on this subrepresentation factors through PSL(2, p), and since the action of PSL(2, p) on the projective line is doubly transitive this subrepresentation is irreducible. – Qiaochu Yuan Jul 5 '10 at 9:44
Another nice source for the computation of the character table is in Etingof's lectures on representation theory: www-math.mit.edu/~etingof/cltrunc.pdf – bavajee Jul 5 '10 at 10:38
show 1 more comment
Note: This was written up concurrently with David's answer, but wasn't proofread and didn't get past the captcha stage due to technical problems.
It is more common to consider the representation of $G=GL(2,\mathbb{F}_p)$ on $\mathbb{C}^Y$, where $Y=X\setminus 0$. This is a direct sum over all multiplicative characters $\chi$ of $\mathbb{F}_p$ of the induced representations of $G$ in the homogeneous functions of homogeneity degree $\chi:$
$$\mathbb{C}^Y=\bigoplus_{\chi\in \mathbb{F}_p^\ast}(\mathbb{C}^Y)_{\chi}, \quad (\mathbb{C}^X)_{\chi}=\{f:Y\to \mathbb{C}: f(\lambda a)=\chi(\lambda)f(a)\ \text{for all}\ \lambda\in \mathbb{F}_p \},$$
and $\mathbb{C}^X=\mathbb{C}\oplus\mathbb{C}^Y,$ where the first summand corresponds to the functions supported at $0\in\mathbb{F}_p^2.$
Each induced representation has dimension $p+1.$ For the trivial character, the induced representation contains a trivial subrepresentation of $GL(2,\mathbb{F}_p)$ (constants), with irreducible $p$-dimensional quotient (Steinberg representation). All other induced representations are irreducible and pairwise non-isomorphic (they all have different central characters). The situation is a bit more complicated for restrictions to $H=SL(2,\mathbb{F}_p)$ (denoted the same by abuse of notation): for any $\chi,\ (\mathbb{C}^Y)_{\chi}\simeq (\mathbb{C}^Y)_{\chi^{-1}}$ and if $p$ is odd and $\psi$ is a character of order $2$ then the representation $(\mathbb{C}^Y)_{\psi}$ is a direct sum of two non-isomorphic $(p+1)/2$-dimensional representations, all other representations remain irreducible and pairwise non-isomorphic. These facts can be verified ad hoc by the "sum of squares" calculation or, more systematically, by using the Mackey theory for finite groups. This accounts for close to the half of irreducible representations of $H$; to get the rest, one needs to induce from the non-split torus $\simeq \mathbb{F}_{p^2}^\ast/\mathbb{F}_{p}^\ast.$
-
Phew! Sudden power loss just a second after posting. – Victor Protsak Jul 8 '10 at 17:30
Thanks, This answers my question about GL(2,p). Does the same works for GL(n,p)? Where can I read about Mackey theory for finite groups? – Klim Efremenko Jul 9 '10 at 6:05
Yes, this works for $GL(n,p),$ but the representations you'll get will be more special (i.e. they will not look like "typical" representations of this group). I think Kirillov's "Elements of representation theory" describes Mackey theory, but this may be worth a separate MO question: I remember scratching my head trying to come up with a good reference. For this specific question, David's answer really walks you through the procedure without mentioning the big theorems that justify it in general. – Victor Protsak Jul 9 '10 at 7:36
There are $2$ copies of the trivial representation, and $p-1$ other representations. The two trivials are spanned by the characteristic function of $(0,0)$, and by the function which is $1$ on all the nonzero vectors and $0$ on $(0,0)$. The other representations are indexed by the characters of $\mathbb{F}_p^*$. Namely, for any character $\chi$, consider those functions $f:\mathbb{F}_p^2 \to \mathbb{C}$ such that $f(g \cdot v)=\chi(g) f(v)$ for $g \in \mathbb{F}_p^*$. This is a subrepresentation. It is irreducible except when $\chi$ is trivial, in which case the two trivials split off as summands.
How to do this computation: Let $V$ be a representation of $G$ over $\mathbb{C}$, and let $V:=\bigoplus W_{\chi} \otimes V_{\chi}$ be the decomposition into isotypic components, where $V_{\chi}$ are representatives for the isomorphism classes of $G$-irreps. Then $$\mathrm{Hom}_G(V,V) = \bigoplus W_{\chi} \otimes W_{\chi}. \quad (*)$$ This formula is particularly useful when $V$ is a permutation representation $\mathbb{C}^X$, as then $\mathrm{Hom}_{\mathbb{C}}(V, V)$ is $\mathbb{C}^{X \times X}$ and the subalgebra of $G$-equivariant Homs is spanned by the $G$ orbits in $X \times X$. So the left hand side of $(*)$ has dimension equal to the number of $G$ orbits in $X \times X$. (In particular, we obtain the corollary mentioned by Qiaochu in a now-deleted answer: if the action of $G$ on $X \times X$ is doubly transitive, then $V$ is a trivial representation and one other irrep.)
In this case, there are $(p-1)+3+1$ orbits in $X \times X$. For every $g \in \mathbb{F}_p^*$, the set $\{ (v,w) : v=gw,\ v,w \neq 0 \}$ is an orbit. Also, we have three orbits by imposing that $v$, $w$ or both, be zero. Finally, there is the large orbit where $v$ and $w$ are linearly independent.
If we have already guessed the decomposition into irreducibles, as above, then we can confirm it by checking that the dimensions of both sides of $(*)$ match. In this case, that says $(p-1)+3+1=(p-1)+2^2$, which is true.
If we didn't already know the answer, we could deduce it from $(*)$. Let $R$ be the ring in $(*)$. We can describe $R$ as the vector space of $G$-invariant functions on $X \times X$, with the convolution product $(f*g)(u,w)=\sum_{v \in X} f(u,v) g(v,w)$. We must find the decomposition of $R$ into simple summands, and decompose $V$ as an $R$-module. The $R$-submodules of $V$ are the isotypic components for the $G$-action.
In this case, let $\Omega_g$ be the orbit $\{ (v,w) : v=gw,\ v,w \neq 0 \}$. It is easy to check that the $\Omega_g$ span a subalgebra isomorphic to the group algebra $\mathbb{C}[\mathbb{F}_p^*]$. This, of course, is isomorphic to $\bigoplus_{\chi} \mathbb{C}$, where the sum is over characters of $\mathbb{F}_p^*$. The idempotents for this decomposition project onto the corresponding $\chi$ representations above. The other semisimple summand of $R$ is a little harder to guess. Let $a_{00}$, $a_{10}$, $a_{01}$ and $a_{11}$ be the characteristic functions of the subsets of $X \times X$ where the two components are zero or nonzero according to whether the index is $0$ or $1$. Then the $a_{ij}$ span an algebra isomorphic to $\mathrm{Mat}_{2\times 2}(\mathbb{C})$, corresponding to the trivial summands.
You might enjoy reading the notes from the 2007 Quantum Gravity Seminar, which (despite its title) mostly focused on representation theory of finite groups. This is where I learned how to systematically use the ring $R$ to study permutation representations.
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I am sorry I did not follow: "The idempotents for this decomposition project onto the corresponding representations above" Can you, please, explain it? I do understand how the ring R looks like, but I do not understand how to deduce from it irreducible represintations of $C^X$. – Klim Efremenko Jul 7 '10 at 12:51 | {"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": 0.9483548402786255, "perplexity": 193.15908014327923}, "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-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00014-ip-10-147-4-33.ec2.internal.warc.gz"} |
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1 2 3 4 5 24 | {"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": 0.8635638952255249, "perplexity": 2378.0332184287117}, "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-2023-06/segments/1674764499967.46/warc/CC-MAIN-20230202070522-20230202100522-00021.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/139686-linear-combination-poisson-random-variables.html | # Math Help - Linear combination of poisson random variables
1. ## Linear combination of poisson random variables
Hi, can someone help me with this question:
Let X~Poisson(1), Y~Poisson(2), define T = X + Y.
a) Show that T~Poisson(3)
b) Find the joint probability function of X and T
c) Find the conditional distribution of X given T = n
d) Compute Cov(X,T) and the correlation coefficient.
I have no problems with a). But I got stuck at b) because I get that the jpf of X and T, f(X,T) = P ( X = x , T = t) = f(T) ...?! Which makes no sense. Any help would be appreciated!
2. Assuming independence....
Use moment generating functions to show (a).
$P(X=a,T=b)=P(X=a, Y=b-a)=P(X=a)P(Y=b-a)$
$= \left({e^{-1}1^a\over a!}\right) \left({e^{-2}2^{b-a}\over (b-a)!}\right)$
for (c) just use the definition of conditional probabilities.
(d) Cov(X,X+Y)=Cov(X,X)+Cov(X,Y)=V(X)+0=1.
3. ah yes, thank you very much! So the difference between a) and b) is that in b) we are not summing all the marginal probabilities of x and y, because we have a constraint on what X is, correct? | {"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": 2, "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": 0.9612839818000793, "perplexity": 478.7508921161754}, "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-2016-26/segments/1466783397797.77/warc/CC-MAIN-20160624154957-00146-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/spectra-units-cm-1.867798/ | # I Spectra Units cm^(-1)?
Tags:
1. Apr 19, 2016
### kq6up
I am seeing units for atomic/molecular spectra listed as $cm^{-1}$. Is this based on the wave number $k$ for $\Psi=Asin(kx-\phi)$?
Thanks,
KQ6UP
2. Apr 19, 2016
### Staff: Mentor
It is then inverse of the wavelength of the emitted/absorbed light. It is a leftover of the early days of spectroscopy, where this was something relatively easy to measure.
3. Apr 20, 2016
Thanks | {"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": 0.9168112277984619, "perplexity": 4454.509569346632}, "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-2018-13/segments/1521257647758.97/warc/CC-MAIN-20180322014514-20180322034514-00229.warc.gz"} |
http://math.soimeme.org/~arunram/Preprints/COGHAFCRGTheGradedHeckeAlgebrasHgr.html | Classification of graded Hecke algebras for complex reflection groups
Last update: 22 January 2014
The graded Hecke algebras ${H}_{\text{gr}}$
In [Lus1989], Lusztig gives a definition of graded Hecke algebras for real reflection groups which is different from the definition in Section 1, which applies to more general groups. It is not obvious that Lusztig’s algebras are examples of the graded Hecke algebras defined in Section 1. In this section, we show explicitly how the definition of Section 1 includes Lusztig’s algebras.
Let $W$ be a finite real reflection group acting on $V$ and let $R$ be the root system of $W\text{.}$ Let ${\alpha }_{1},\dots ,{\alpha }_{n}$ be a choice of simple roots in $R$ and let ${s}_{1},\dots ,{s}_{n}$ be the corresponding simple reflections in $W\text{.}$ Let ${s}_{\alpha }$ be the reflection in the root $\alpha$ so that, for $v\in V,$ $sαv=v-⟨v,α∨⟩ α,where α∨=2α /⟨α,α⟩.$ Let ${R}^{+}=\left\{\alpha >0\right\}$ denote the set of positive roots in $R\text{.}$
Let ${k}_{\alpha }$ be fixed complex numbers indexed by the roots $\alpha \in R$ satisfying $kwα=kα, for all w∈W, α∈R. (3.1)$ This amounts to a choice of either one or two “parameters”, depending on whether all roots in $R$ are the same length or not. As in Section 1, let $ℂW=ℂ\text{-span}\left\{{t}_{g} | g\in W\right\},$ with ${t}_{g}{t}_{h}={t}_{gh},$ and let $S\left(V\right)$ be the symmetric algebra of $V\text{.}$ Lusztig [Lus1989] defines the “graded Hecke algebra” with parameters $\left\{{k}_{\alpha }\right\}$ to be the unique algebra structure ${H}_{\text{gr}}$ on the vector space $S\left(V\right)\otimes ℂW$ such that
(3.2 a) $S\left(V\right)=S\left(V\right)\otimes 1$ is a subalgebra of ${H}_{\text{gr}},$ (3.2 b) $ℂW=1\otimes ℂW$ is a subalgebra of ${H}_{\text{gr}},$ and (3.2 c) ${t}_{{s}_{i}}v=\left({s}_{i}v\right){t}_{{s}_{i}}-{k}_{{\alpha }_{i}}⟨v,{\alpha }_{i}^{\vee }⟩,$ for all $v\in V$ and simple reflections ${s}_{i}$ in the simple roots ${\alpha }_{i}\text{.}$
We shall show that every algebra ${H}_{\text{gr}}$ as defined by (3.2 a-c) is a graded Hecke algebra $A$ for a specific set of skew symmetric bilinear forms ${a}_{g}\text{.}$
Let ${k}_{\alpha }\in ℂ$ as in (3.1). Use the notation $h=12∑α>0 kαα∨tsα ,so that⟨v,h⟩ =12∑α>0 kα⟨v,α∨⟩ tsα (3.3)$ for $v\in V\text{.}$ The element $h$ should be viewed as an element of $V\otimes ℂW,$ and $⟨v,h⟩\in ℂW\text{.}$ With this notation, let $A$ be the algebra (as in Section 1) generated by $V$ and $ℂW$ with relations $tgv=(gv)tg and[v,w]=- [⟨v,h⟩,⟨w,h⟩], for v,w∈V, g∈W. (3.4)$ Note that $A$ is defined by the bilinear forms $ag(v,w)=14 ∑α,β>0g=sαsβ kαkβ ( ⟨v,β∨⟩ ⟨w,α∨⟩- ⟨v,α∨⟩ ⟨w,β∨⟩ ) .$
The following theorem shows that the algebra $A$ satisfies the defining conditions (3.2 a-c) of the algebra ${H}_{\text{gr}}\text{.}$
Let $W$ be a finite real reflection group and let $A$ be the algebra defined by (3.4).
(a) As vector spaces, $A\cong S\left(V\right)\otimes ℂW$ (and hence, $A$ is a graded Hecke algebra). (b) If $\stackrel{\sim }{v}=v-⟨v,h⟩$ for $v\in V,$ then $[v∼,w∼]=0 andtsi v∼=(siv˜) tsi-kαi ⟨v,αi∨⟩,$ for all $v,w\in V$ and simple reflections ${s}_{i}$ in $W\text{.}$
Proof. First note that if $u,v\in V$ then $[u,⟨v,h⟩]= 12∑α>0kα ⟨v,α∨⟩ ⟨u,α∨⟩α tsα= [v,⟨u,h⟩]. (*)$ Thus, for $u,v,w\in V,$ $[u,[v,w]]+ [w,[u,v]]+ [v,[w,u]] = [u,[⟨w,h⟩,⟨v,h⟩]]+ [w,[⟨v,h⟩,⟨u,h⟩]]+ [v,[⟨u,h⟩,⟨w,h⟩]] = [[u,⟨w,h⟩],⟨v,h⟩]+ [⟨w,h⟩,[u,⟨v,h⟩]]+ [[w,⟨v,h⟩],⟨u,h⟩] +[⟨v,h⟩,[w,⟨u,h⟩]] +[[v,⟨u,h⟩],⟨w,h⟩] +[⟨u,h⟩,[v,⟨w,h⟩]] = [[w,⟨u,h⟩],⟨v,h⟩]+ [⟨w,h⟩,[v,⟨u,h⟩]]+ [[v,⟨w,h⟩],⟨u,h⟩] +[⟨v,h⟩,[w,⟨u,h⟩]] +[[v,⟨u,h⟩],⟨w,h⟩] +[⟨u,h⟩,[v,⟨w,h⟩]] = 0. (3.6)$ For $v\in V,$ $h\in W,$ and ${s}_{i}$ a simple reflection, $tsi⟨v,h⟩ tsi = 12∑α>0 kα⟨v,α∨⟩ tssiα= ( 12∑α>0kα ⟨v,siα∨⟩ tsα ) +kαi ⟨v,αi∨⟩ tsi = ( 12∑α>0kα ⟨siv,α∨⟩ tsα ) +kαi ⟨v,αi∨⟩ tsi=⟨siv,h⟩ +kαi⟨v,αi∨⟩ tsi. (3.7)$ Using this equality, we obtain $tsi[v,w]tsi = -tsi [⟨v,h⟩,⟨w,h⟩] tsi = - [ ⟨siv,h⟩+ kαi⟨v,αi∨⟩ tsi,⟨siw,h⟩ +kαi⟨w,αi∨⟩ tsi ] = [siv,siw]- kαi⟨v,αi∨⟩ [tsi,⟨siw,h⟩] -kαi⟨w,αi∨⟩ [⟨siv,h⟩,tsi] = [siv,siw]- kαi⟨v,αi∨⟩ ( tsi⟨siw,h⟩ tsi-⟨siw,h⟩ ) tsi +kαi⟨w,αi∨⟩ ( tsi⟨siv,h⟩ tsi-⟨siv,h⟩ ) tsi = [siv,siw]- kαi⟨v,αi∨⟩ ( ⟨w,h⟩tsi+ kαi⟨siw,αi∨⟩ -⟨siw,h⟩tsi ) +kαi⟨w,αi∨⟩ ( ⟨v,h⟩tsi+ kαi⟨siv,αi∨⟩ -⟨siv,h⟩tsi ) = [siv,siw]- kαi⟨v,αi∨⟩ ⟨w,αi∨⟩ ⟨αi,h⟩tsi -kαi2⟨v,αi∨⟩ ⟨w,siαi∨⟩ +kαi⟨w,αi∨⟩ ⟨v,αi∨⟩ ⟨αi,h⟩tsi +kαi2⟨w,αi∨⟩ ⟨v,siαi∨⟩ = [siv,siw]. (3.8)$ The two identities (3.6) and (3.8), as in (1.3) and (1.4), show that the algebra $A$ is isomorphic to $S\left(V\right)\otimes ℂW\text{.}$ (b) This can now be proved by direct computation. If $v,w\in V$ then $[v∼,w∼]= [ v-⟨v,h⟩, w-⟨w,h⟩ ] =[v,w]+ [⟨v,h⟩,⟨w,h⟩] -[v,⟨w,h⟩] +[w,⟨v,h⟩]=0,$ by equation (3.4) and equation $\text{(}*\text{)}$ in the proof of Theorem 3.5. If $v\in V$ and ${s}_{i}$ is a simple reflection then, by (3.7), $tsiv∼tsi= tsivtsi-tsi ⟨v,h⟩tsi= siv-⟨siv,h⟩ -kαi⟨v,αi∨⟩ tsi=siv˜- kαi⟨v,αi∨⟩ tsi.$ $\square$
Theorem 3.5b shows that if $A$ is the graded Hecke algebra defined by (3.4), then the elements $\stackrel{\sim }{v},$ for $v\in V,$ generate a subalgebra of $A$ isomorphic to $S\left(V\right)$ and these elements together with the ${t}_{{s}_{i}}$ satisfy the relations of (3.2 c). Since part (a) of Theorem 3.5 shows that $A$ is isomorphic to $S\left(V\right)\otimes ℂW$ as a vector space, it follows that $A$ satisfies the conditions (3.2 a-c), relations which uniquely define the graded Hecke algebra $A\text{.}$ Thus, Lusztig’s algebras are special cases of the graded Hecke algebras defined in Section 1. Furthermore, by comparing the dimensions of the parameter spaces, we see that there are graded Hecke algebras that are not isomorphic to algebras defined by Lusztig for the Coxeter groups ${F}_{4},$ ${H}_{3},$ ${H}_{4},$ and ${I}_{2}\left(m\right)\text{.}$
Notes and references
This is a typed version of Classification of graded Hecke algebras for complex reflection groups by Arun Ram and Anne V. Shepler.
Research of the first author supported in part by the National Security Agency and by EPSRC Grant GR K99015 at the Newton Institute for Mathematical Sciences. Research of the second author supported in part by National Science Foundation grant DMS-9971099. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 93, "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": 0.9808376431465149, "perplexity": 95.57807111334}, "config": {"markdown_headings": false, "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-2019-04/segments/1547583875448.71/warc/CC-MAIN-20190122223011-20190123005011-00363.warc.gz"} |
http://mathhelpforum.com/algebra/96024-word-problem-29-a.html | # Thread: word problem #29
1. ## word problem #29
can someone show how to set up the equation to solve this word problem please.
One number is 4 more than another number. Find the two if two times the larger number is 7 less than 5 times the smaller.
the books solution is 5 & 9
2. Originally Posted by jwarfield09
can someone show how to set up the equation to solve this word problem please.
One number is 4 more than another number. Find the two if two times the larger number is 7 less than 5 times the smaller.
the books solution is 5 & 9
Hi , lets call the 2 numbers x and y .
y=x+4 --1
Then we know that y is the larger one and x is the smaller one .
2y=5x-7 ---2
Then solve the simultaneous equation .
3. Originally Posted by mathaddict
Hi , lets call the 2 numbers x and y .
y=x+4 --1
Then we know that y is the larger one and x is the smaller one .
2y=5x-7 ---2
Then solve the simultaneous equation .
But, if you want a single equation....
Let x = smaller number
Let x + 4 = the larger number
2(x + 4) = 5x - 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": 0.8694735765457153, "perplexity": 619.8653542551488}, "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-2017-26/segments/1498128323889.3/warc/CC-MAIN-20170629070237-20170629090237-00097.warc.gz"} |
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