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http://www.zentralblatt-math.org/ioport/en/?q=au:Salva%2C%20N* | History
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Simultaneous determination of unknown coefficients through a phase-change process with temperature-dependent thermal conductivity. (English)
JP J. Heat Mass Transf. 5, No. 1, 11-39 (2011).
Summary: Formulas are obtained for the simultaneous determination of two or three unknown thermal coefficients of a semi-infinite material with temperature-dependent thermal conductivity through a phase-change process with an over-specified condition on the fixed face through a moving boundary problem (inverse Stefan problem) or a free boundary problem (Stefan problem), respectively. We complete and improve the analysis done in [the second author, “The determination of unknown thermal coefficients through phase change process with temperature-dependent thermal conductivity", Int. Commun. Heat Mass Transfer 25, No.~1, 139‒147 (1998), \url{doi:10.1016/S0735-1933(97)00145-0}] and we also study the sensitivity of the solution depending on different thermal parameters, applied to aluminium. | {"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": 0.8550733327865601, "perplexity": 2412.3903785365765}, "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/1368698958430/warc/CC-MAIN-20130516100918-00094-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://tohoku.pure.elsevier.com/en/publications/experimental-evaluation-of-thermal-radiation-effects-on-natural-c | # Experimental evaluation of thermal radiation effects on natural convection with a Rayleigh number of 108–109 by using an interferometer
Takuma Kogawa, Eita Shoji, Junnosuke Okajima, Atsuki Komiya, Shigenao Maruyama
Research output: Contribution to journalArticlepeer-review
5 Citations (Scopus)
## Abstract
In this study, the radiation effects on the temperature field and perturbation of the thermal boundary layer of natural convection for gas were evaluated by an interferometer. The natural convection investigated in this study was that of a cavity, which had differentially heated vertical walls, surrounded by adiabatic walls. The target Rayleigh number was of the order of 108–109. By using the interferometer adopted in a phase-shifting technique, surface and gas radiation effects were evaluated. To compare the experimental results, a numerical calculation was also conducted using a large-eddy simulation coupled with radiative heat transfer. By varying the emissivity of the adiabatic walls while maintaining the emissivity of the isothermal walls, the surface radiation effect on the natural convection was analyzed. To neglect the gas radiation effect, air was used as the working fluid. When the adiabatic walls were black-body surfaces, the temperature difference around the adiabatic walls was higher than that of the reflective surface as a result of the optical path difference. From this result, the high temperature difference due to the surface radiation effect was confirmed. To investigate the gas radiation effect on natural convection, trifluoromethane gas, which has significant radiative absorption, was used as a working fluid. By evaluating the interference fringe by the interferometer, the wave motion of the thermal boundary layer around the heated and adiabatic walls was visualized. Comparing the numerical calculation results (when varying the calculation conditions) revealed that the wave motion and turbulence boundary layer around the heated and adiabatic walls were caused by the gas radiation effect.
Original language English 1239-1249 11 International Journal of Heat and Mass Transfer 132 https://doi.org/10.1016/j.ijheatmasstransfer.2018.11.162 Published - 2019 Apr
## Keywords
• Gas radiation
• Interferometer
• LES
• Natural convection
• REM
• Surface radiation
• Turbulence
## ASJC Scopus subject areas
• Condensed Matter Physics
• Mechanical Engineering
• Fluid Flow and Transfer Processes
## Fingerprint
Dive into the research topics of 'Experimental evaluation of thermal radiation effects on natural convection with a Rayleigh number of 10<sup>8</sup>–10<sup>9</sup> by using an interferometer'. 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": 0.899243950843811, "perplexity": 2124.1976524799506}, "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/1627046154158.4/warc/CC-MAIN-20210801030158-20210801060158-00551.warc.gz"} |
http://playzona.net/Illinois/good-standard-error-mean.html | Address 5048 W 81st Ave, Schererville, IN 46375 (219) 227-8015 http://www.integrous.biz
# good standard error mean Blue Island, Illinois
For data that are non-normal, the standard deviation can be a terrible estimator of scale. So, we should draw another sample and determine how much it deviates from the population mean. For smaller sample sizes, the use of n-1 is generally considered an appropriate correction, allowing the calculation of a sample standard deviation using N-1 as the denominator (Bessel's correction). Sometimes it is too difficult or costs too much money to have lots of samples.
And although I don't work with Power Law distributions, the Pareto distribution can reduce to a single parameter also, with no defined variance. This interval is a crude estimate of the confidence interval within which the population mean is likely to fall. Sep 29, 2014 Joshka Kaufmann · University of Lausanne I agree with Bernardo that distribution of your data is crucial. This often leads to confusion about their interchangeability.
On the other hand, if you assume Poisson distribution, then the mean should be approximately the square of the SD. The standard error of the mean does basically that. The margin of error and the confidence interval are based on a quantitative measure of uncertainty: the standard error. Just as the standard deviation is a measure of the dispersion of values in the sample, the standard error is a measure of the dispersion of values in the sampling distribution.
Next, consider all possible samples of 16 runners from the population of 9,732 runners. Notice, however, that once the sample size is reasonably large, further increases in the sample size have smaller effects on the size of the standard error of the mean. We could subtract the sample mean from the population mean to get an idea of how close the sample mean is to the population mean. (Technically, we don't know the value S is 3.53399, which tells us that the average distance of the data points from the fitted line is about 3.5% body fat.
In that case, the statistic provides no information about the location of the population parameter. The standard error of the mean permits the researcher to construct a confidence interval in which the population mean is likely to fall. Biochemia Medica 2008;18(1):7-13. The Cauchy distribution has, as Bernardo points out, no defined variance.
National Center for Health Statistics does not report an average if the relative standard error exceeds 30%. In fact, we might want to do this many, many times. Oct 1, 2014 Jochen Wilhelm · Justus-Liebig-Universität Gießen Thank you Ronán for your clarification. I'd say that the DS (and the variance) is in fact a statistical property of any distribution, but that it only has a meaning in the case of a normal distribution.
Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. Kind regards, Nicholas Name: Himanshu • Saturday, July 5, 2014 Hi Jim! The standard deviation of those means is then calculated. (Remember that the standard deviation is a measure of how much the data deviate from the mean on average.) The standard deviation I love the practical, intuitiveness of using the natural units of the response variable.
Sampling from a distribution with a large standard deviation The first data set consists of the ages of 9,732 women who completed the 2012 Cherry Blossom run, a 10-mile race held The standard error is a measure of the variability of the sampling distribution. As will be shown, the standard error is the standard deviation of the sampling distribution. M.
What the standard error gives in particular is an indication of the likely accuracy of the sample mean as compared with the population mean. Therefore, the standard error of the estimate is a measure of the dispersion (or variability) in the predicted scores in a regression. Ricky Ramadhian · Lampung University when data , CV >=1, it should be repeated the experiment or how? Conveniently, it tells you how wrong the regression model is on average using the units of the response variable.
The concept of a sampling distribution is key to understanding the standard error. Because of random variation in sampling, the proportion or mean calculated using the sample will usually differ from the true proportion or mean in the entire population. Set the sample size to a small number (e.g. 1) and generate the samples. Each of these averages is a little bit different from the average that would come from measuring every 42cm long redfish (which is not possible anyway).
For example, you have a mean delivery time of 3.80 days with a standard deviation of 1.43 days based on a random sample of 312 delivery times. See unbiased estimation of standard deviation for further discussion. The mean age for the 16 runners in this particular sample is 37.25. Ramadhian:Is this question in regards to the reliability of your data, or is it more about effect sizes (or something else)?
They are quite similar, but are used differently. And that means that the statistic has little accuracy because it is not a good estimate of the population parameter. Standard errors provide simple measures of uncertainty in a value and are often used because: If the standard error of several individual quantities is known then the standard error of some If you have several treatments or different samplings you would like to compare, the overall distribution of your variable might be spread out for example.
Oct 9, 2014 Debashis Chakraborty · Indian Agricultural Research Institute I apreciate Sorensen's comments. The following expressions can be used to calculate the upper and lower 95% confidence limits, where x ¯ {\displaystyle {\bar {x}}} is equal to the sample mean, S E {\displaystyle SE} Another way to find the standard error of the mean is to use an equation that needs only one sample. Taken together with such measures as effect size, p-value and sample size, the effect size can be a very useful tool to the researcher who seeks to understand the reliability 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": 0.9453752040863037, "perplexity": 419.92753132860054}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583509845.17/warc/CC-MAIN-20181015205152-20181015230652-00062.warc.gz"} |
http://en.wikipedia.org/wiki/Simplification | # Simplification
For simplification of fractions, see Equivalent fractions. For other uses, see Simplification (disambiguation).
In propositional logic, simplification[1][2][3] (equivalent to conjunction elimination and also called and elimination) is a valid immediate inference, argument form and rule of inference which makes the inference that, if the conjunction A and B is true, then A is true, and B is true. The rule makes it possible to shorten longer proofs by deriving one of the conjuncts of a conjunction on a line by itself.
An example in English:
It's raining and it's pouring.
Therefore it's raining.
The rule can be expressed in formal language as:
$\frac{P \land Q}{\therefore P}$
or as
$\frac{P \land Q}{\therefore Q}$
where the rule is that whenever instances of "$P \land Q$" appear on lines of a proof, either "$P$" or "$Q$" can be placed on a subsequent line by itself.
## Formal notation
The simplification rule may be written in sequent notation:
$(P \land Q) \vdash P$
or as
$(P \land Q) \vdash Q$
where $\vdash$ is a metalogical symbol meaning that $P$ is a syntactic consequence of $P \land Q$ and $Q$ is also a syntactic consequence of $P \land Q$ in logical system;
and expressed as a truth-functional tautology or theorem of propositional logic:
$(P \land Q) \to P$
and
$(P \land Q) \to Q$
where $P$ and $Q$ are propositions expressed in some formal system.
## References
1. ^ Copi and Cohen[citation needed]
2. ^ Moore and Parker[citation needed]
3. ^ Hurley[citation needed] | {"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": 0.9449661374092102, "perplexity": 357.0603742438485}, "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-2015-11/segments/1424936461332.16/warc/CC-MAIN-20150226074101-00082-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/125971-how-can-i-find-limits-function-equation.html | # Thread: How can i find the limits of this function/equation?
1. ## How can i find the limits of this function/equation?
Hello!
I have no idea where to post this so I'll just put it here.
The function f(x)= 1 - [2x/(1+(x^2))]
Sorry if this is confusing, ive written it all out on paint so please open the attached file
QUESTION:
to display the graph of y=f(x) for -10<x<10, a suitable interval for y, a<x<b must be chosen. Suggest appropriate values for a and b.
I would suppose that you substitute x=-10 and x=10 to find a and b accordingly yet I cant get the right answer
Thank you!
2. For one, you just look at it. Imagine what it does at its extremes.
If x increases without bound, getting larger and larger...
a) What does '1' do? Nothing.
b) What does the rational expression do? It shrinks. This piece disappears.
c) So, we seem to be sneaking up on x = 1 from the bottom.
If x decreases without bound, getting larger and larger in the negative direction...
a) What does '1' do? Nothing.
b) What does the rational expression do? It shrinks. This piece disappears.
c) So, we seem to be sneaking up on x = 1 from the top.
We just defined an horizontal asymptote. x = 1.
What do you suppose is next? | {"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": 0.923706591129303, "perplexity": 1121.6268828482746}, "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-2017-39/segments/1505818685850.32/warc/CC-MAIN-20170919145852-20170919165852-00569.warc.gz"} |
http://mathhelpforum.com/calculus/17081-polar-coordinate-03-a.html | 1. Polar Coordinate 03
Find the area inside r= 3 Cos[x] and outside of r= 1 + Cos[x]
2. Hello, camherokid!
Find the area inside $r \,= \,3\cos x$ and outside $r \:= \:1 + \cos x$
I hope you made a sketch . . .
$r\:=\:3\cos x$ is a circle with center $\left(\frac{3}{2},\,0\right)$ and radius $r = \frac{3}{2}$
$r\:=\:1 + \cos x$ is a cardioid with intercepts: $(2,\,0),\: \left(1,\,\frac{\pi}{2}\right),\:\left(1,\,\frac{3 \pi}{2}\right)$
. . and "dimples in" to the origin from the left.
The polar formula for the area between two curves is: . $A \;=\;\frac{1}{2}\int^{\beta}_{\alpha}\left(r_{_2}^ 2 - r_{_1}^2\right)\,d\theta$
The curves intersect when: . $3\cos x \;=\;1 + \cos x\quad\Rightarrow\quad 2\cos x \:=\:1\quad\Rightarrow\quad \cos x \:=\:\frac{1}{2}$
. . Hence, they intersect at: . $\theta \:=\:\pm\frac{\pi}{3}$
Therefore: . $A \;=\;\frac{1}{2}\int^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\bigg[(3\cos\theta)^2 - (1 + \cos x)^2\bigg]\,d\theta$
3. Originally Posted by camherokid
Find the area inside r= 3 Cos[x] and outside of r= 1 + Cos[x]
We will use the formula:
$A = \int_{ \alpha}^{ \beta} \frac {1}{2} (r_o^2 - r_i^2)~dx$
where $A$ is the area between the curves $r_o$ and $r_i$, $\alpha$ and $\beta$ are the limits of integration (the points of intersection), $r_o$ is the outer curve, and $r_i$ is the inner curve.
First find the points of intersection:
this is where $3 \cos x = 1 + \cos x$
$\Rightarrow \cos x = \frac {1}{2}$
$\Rightarrow x = \frac {\pi}{3}, \frac {5 \pi}{3}$
we want to go from $\frac {5 \pi}{3}$ to $\frac {\pi}{3}$, but we must go from a smaller angle to a bigger angle. changing $\frac {5 \pi}{3}$ to $- \frac {\pi}{3}$ fixes this problem. so our area is given by:
$A = \int_{- \pi / 3}^{ \pi / 3} \frac {1}{2} \left[ (3 \cos x)^2 - (1 + \cos x )^2 \right]~dx$
EDIT: Beaten by Soroban! any way, that's ok. I'm a bit rusty on polar areas, so it's good to have a confirmation that I did the right thing. Soroban, can you check the other posts camherokid put up today and make sure I didn't make any mistakes
4. Originally Posted by camherokid
Find the area inside r= 3 Cos[x] and outside of r= 1 + Cos[x]
the graph
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http://math.stackexchange.com/questions/232402/when-do-we-have-radi-i-for-an-ideal-i-of-a-ring-r | # When do we have $Rad(I)=I$ for an ideal $I$ of a ring $R$?
This is kind of a follow-up question about calculating the radical of an ideal. Since
$Rad(I)$ is the intersection of all the prime ideals of $R$ that contain $I$,
which is a property I learned from this article in wikipedia, we have that $$Rad(I)=I$$ whenever $I$ is a prime ideal. My question is:
Can this be true for some $I$ which is not a prime ideal? [EDIT: And when is this NOT true?] Is there an equivalent easy-to-check conditions for this kind of $I$?
Let $R={\Bbb Z}[x]$, for example. $I=\langle x,2\rangle$ is a prime ideal and thus $Rad(I)=I$. For any ideal $I\unlhd R$, (say $I=\langle x^2+1\rangle$ or $I=\langle x^2+2\rangle$, etc.) the key point is to check $$Rad(I)\subset I$$ since $I\subset Rad(I)$ is always true. But I don't know a quick way to check this relation.
-
By definition, every intersection of a set of prime ideals is semiprime (aka a radical ideal).
Let $\cap P_i=I$, where the $P_i$ are all prime. Then the set of all prime ideals containing $I$ contains the $P_i$. Thus, $\cap\{P\mid P\supseteq I\}\subseteq \cap P_i$. The left hand intersection involves "more" prime ideals, and so the intersection of more ideals should be smaller than just the set of $P_i$.
Thus in total: $$\cap P_i=I\subseteq\cap\{P\mid P\supseteq I\}\subseteq \cap P_i$$
So, there is equality all across.
It is relatively easy to find examples where prime ideals do not intersect to a prime ideal. For example, the intersection of two prime ideals, neither of which contains the other, cannot be prime. (Explain why!)
-
What I learn from your answer is that to check if $Rad(I)=I$, I need to check if $I$ is some intersection of prime ideals. But how can I apply this to check, say, $I=\langle x^2+1\rangle$ in ${\Bbb Z}[x]$? – Jack Nov 7 '12 at 20:59
For the commutative case, $rad(I)=\{x\in R\mid \exists n\in \mathbb{N}, x^n\in I\}$. That might help you do specific commutative examples. Isn't it the case here that $x^2+1$ generates a prime ideal? – rschwieb Nov 7 '12 at 21:00
You can, in fact, have $I=\mathrm{Rad}(I)$ for nonprime $I$. For example, consider the ideal $\langle xy\rangle$ of the ring $\mathbb{Q}[x,y]$. This isn't prime since the generator isn't irreducible, but can easily be seen to be radical.
More generally, if you take a monomial ideal, i.e. an ideal generated by monomials, in a polynomial ring over some field, its radical will be generated by the "roots" of those monomials. E.g. if $I=\langle x^2y^4,z^3\rangle$, then $\sqrt{I}=\langle xy^2,z\rangle$.
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https://thephysicist.in/which-is-the-most-stable-nuclide/ | # Which is the most stable nuclide?
Which is the most stable nuclide? The popular verdict says $^{56} _{26}Fe$ is the most stable nuclide since it has the highest binding energy per nucleon $(\frac{B}{A}=8.790 \ MeV/A)$. However, this is incorrect. The most stable of all the nuclides is $^{62} _{28}Ni$. Its binding energy per nucleon is $(\frac{B}{A}=8.794 \ MeV/A)$. This post discusses why this incorrect information has prevailed for so long even in respectable academic spheres.
Contents
## How is nuclear stability estimated?
The stability of a nucleus is measured along the same principle of physics that is applied elsewhere; the lesser the energy of a system, the more stable it is. That is, for a system to be stable, has to release some energy during its formation from its constituents. Conversely, this much amount of energy must be supplied to the stable system to destabilize and break it. In the case of a nucleus, it is taken as a system that consists of protons and neutrons. Neither a proton nor a neutron is an elementary particle. A neutron or a proton is considered a hadron, a composite particle made up of two or more quarks. These quarks interact with the neighbouring quarks through nuclear forces (weak and strong). These forces are responsible for the energy of a nucleus. By default, a quark is a matter particle that has a nonzero rest mass. Thus it has some rest mass energy in accordance with the mass-energy equivalence equation: $E_0=mc^2$. This energy, too, accounts for the energy of the nucleus it is a part of. The stability of the nucleus is measured by a parameter called the binding energy.
### The Binding Energy of a Nucleus
A stable nucleus has less energy than the sum of the energies of its individual constituents. This difference is called the binding energy. If a nucleus has a mass M that is made up of $x$ protons (each of mass $m_p$) and $y$ neutrons (each of mass $m_n$), then, for the nucleus to be stable, the following condition must be satisfied. $$M<xm_p+ym_n$$
This tiny difference in mass, $\delta m$ is equivalent to a humongous amount of energy $E= \delta m c^2$. This is the nucleus’s binding energy, which is released when the nucleus is formed out of the constituent particles. On a statistical basis, nuclear physicists also define another parameter, binding energy per nucleon for stability estimates.
#### Binding Energy per Nucleon
Each proton or neutron inside a nucleus is considered as a nucleon. The total number of protons in a nucleus is denoted by $Z$. The total number of neutrons is denoted by $N$. Thus the total number of nucleons is denoted by $A=Z+N$. If the nucleus has a binding energy $B$, the binding energy per nucleon is given by $\frac{B}{A}$, expressed in electronvolts per $A$. The following scatter plot shows the variation of $\frac{B}{A}$ with respect to $A$ for various elements of the periodic table.
Credit: BCcampus Open Publishing
The binding energy per nucleon on a nuclide also shows how tightly bound a nucleus is. The higher the binding energy the tighter it is, thus the more stable and harder to break it apart. Binding energy per nucleon is also called “average binding energy ($\overline{B}$)”.
## How is the most stable nuclide identified?
The B/A vs A curve may look smooth and continuous, it is actually a scatter plot. The mass numbers are distinct. The fluctuations in the value of B/A around the central tendency is partly due to the shell effect. Yet, a lot can be inferred just by studying the central tendency curve which appears to be continuous. The curve appears to be smooth for $A \geq 40$. It attains a maximum around $A=60$ and then begins to fall smoothly for heavier elements. Thus, a reasonable identification of the most stable nuclide can be done by precisely locating the maximum. It may sound easy but it’s very hard and ambiguous to pinpoint this maximum. Through repeated experiments done using the mass spectrometers of various precisions, it has been confirmed that the iron group (chromium-Cr, manganese-Mn, Iron-Fe, Cobalt-Co, and nickel-Ni) is the most stable group of all elements. However, due to the narrow margin between $^{56} _{26} Fe$, $^{58} _{26} Fe$, $^{60} _{28} Ni$, and $^{62} _{28} Ni$, it had been very hard to attribute the maximum to any one of these based on the statistical data. This is where the astrophysical data was used to reach a verdict.
### Stellar Nucleosynthesis
Stellar nucleosynthesis is the process of the formation of heavier nuclei from lighter ones. Heavier nuclei require higher temperatures to be synthesized in the stellar cores. Starting from the fusion of helium-$^4 _2He$ from hydrogen-$^1 _1 H$, the process is exothermic. Thus, a sustainable nuclear chain reaction giving rise to heavier nuclei is feasible but only till the mass number A is near or below 60. Above A=60, nucleosynthesis becomes endothermic and cannot sustain as a chain reaction. Nuclides with A>>60 are formed only when stars collapse and produce shock waves that force smaller nuclides to combine. A thorough study of the end products of the sustained chain reactions in stellar nucleosynthesis has time and again confirmed that $^{56} _{26} Fe$ is the most abundant one of all. Thus, it was taken to be the most stable nuclide in conjunction with the binding energy data. However, with an increase in the sophistication of measurement of A with high-resolution mass spectrometers, this conclusion has been proven to be incorrect.
In his 1995 paper published in the American Journal of Physics, physicist M P Fewell has clearly settled the confusion. He has shown various reasons for the ambiguity that had prevailed earlier and has put forth his thorough justifications to conclude $^{62} _{28} Ni$ as the most stable nucleus of all. Below are a few of his reasonings.
Credit: Hyperphysics
#### The Processes in Stellar Nucleosynthesis
There are three processes in stellar nucleosynthesis: charged particle capture and photodisintegration. In a charged particle capture, an existing nucleus captures particles such as alpha ($^4 _2 He^{++}$), muon ($\mu ^-$) and electron ($e^-$), etc. and transforms into heavier nuclide. In a photodisintegration process, an existing nucleus captures a highly energetic photon and disintegrates into a lighter nuclide. Although both the processes require very high temperatures, photon capture tends to occur more frequently than the charged particle captures because charged particles encounter Coulomb forces in addition to the nuclear forces. That’s why sustained chain reactions in stellar nucleosynthesis end up with an abundance of $^{56} _{26} Fe$, where the equilibrium between both the processes is established. However, when estimated independent of the stellar nucleosynthesis, it is $^{62} _{28} Ni$ wins over $^{56} _{26} Fe$ albeit by a very narrow margin. Since the margin is less than 0.05%, it has been ignored over time.
## Conclusion
A revised and improved investigation of the data yields the conclusive answer that indeed it is $^{62} _{28} Ni$ that should be considered as the most stable nucleus of all.
Credit: Hyperphysics
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error: Alert: Content selection is disabled!! | {"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": 0.928739607334137, "perplexity": 535.9708314215824}, "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-2023-14/segments/1679296943750.71/warc/CC-MAIN-20230322051607-20230322081607-00467.warc.gz"} |
https://brilliant.org/problems/integration-no-no-its/ | # Integration? No no its ...
Calculus Level 4
$\int_{1}^{x}A(x)B(x)dx*\int_{1}^{x}C(x)D(x)dx-\int_{1}^{x}A(x)C(x)dx*\int_{1}^{x}B(x)D(x)dx=f(x)$
If f(x) is a nth degree polynomial and satisfies above equation for all real x, then area bounded by f(x) and the line y=x-1 can be represented as
$\tfrac{a}{b*c}$
Find the value of a-b+c
Assumptions:
n is an even natural number.
a,b may not be numbers.
c is a number.
A, B, C, D are non constant continuous functions of x.
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.9281026721000671, "perplexity": 1732.8631972591888}, "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-2018-17/segments/1524125947033.92/warc/CC-MAIN-20180424174351-20180424194351-00169.warc.gz"} |
https://knowridge.com/2019/01/physicists-measure-weak-force-inside-atoms-for-first-time/ | # Physicists measure ‘weak force’ inside atoms for first time
Researchers have reported the first measurements of the weak interaction between protons and neutrons inside an atom.
The detection of the elusive force verifies a prediction of the Standard Model, the most widely accepted model explaining the behavior of three of the four known fundamental forces in the universe.
“This observation determines the most important component of the weak interaction between the neutron and the proton—and also between the neutron and all other nuclei,” says lead author W. Michael Snow, a professor in the Indiana University-Bloomington College of Arts and Sciences’ physics department and the director of the university’s Center for Spacetime Symmetries.
Snow is also a cospokesperson on the NDPGamma Experiment at Oak Ridge National Laboratory, where researchers conducted the experiments.
“The result deepens our understanding of one of the four fundamental forces of nature,” he adds.
These four forces are the strong force, electromagnetism, the weak force, and gravity. Protons and neutrons are made of smaller particles called quarks that the strong force binds together. The weak force exists in the distance inside and between protons and neutrons.
The goal of the experiment was to isolate and measure one component of this weak interaction.
Inside the atom
To detect the weak interaction inside protons and neutrons, the experiment’s leaders used a device called NPDGamma at Oak Ridge National Laboratory that controls the spin direction of cold neutrons the laboratory’s Spallation Neutron Source generates.
After the angular momentum, or spin, of these neutrons lined up, the team smashed them into protons in a liquid hydrogen target to produce gamma rays.
“The goal of the experiment was to isolate and measure one component of this weak interaction, which manifested as gamma rays that could be counted and verified with high statistical accuracy,” says coauthor David Bowman, team leader for neutron physics at Oak Ridge. “You have to detect a lot of gammas to see this tiny effect.”
Any “lopsidedness” in the direction of the resulting rays can only come from the weak force between the protons and neutrons.
By counting more gamma ray emissions opposite to the neutron spin than along the neutron spin, the researchers observed the influence of the weak interaction. The small size of this lopsidedness, about 30 parts per billion, is the smallest gamma asymmetry ever measured.
Researchers conducted the experiments to detect the weak force over nearly 20 years, with Snow playing a role in the work since the beginning.
“I’ve been involved in the experiment since the original proposal almost two decades ago,” says Snow, whose work on the project has spanned two major phases, including an initial phase that took place at Los Alamos National Laboratory.
What’s next?
Next, Snow is eager to delve deeper into new questions the recently reported study prompted, including exploring the connection between the weak force between the neutrons and protons and the strong force between the quarks inside them.
As part of this effort, researchers plan to search for the effect of the weak interaction on slow neutron spin rotation in liquid helium.
“There is a theory for the weak force between the quarks inside the proton and neutron, but the way that the strong force between the quarks translates into the force between the proton and the neutron is not fully understood,” says Snow. “That’s still an unsolved problem.”
He compared the measurement of the weak force in relation with the strong force as a kind of tracer, similar to a tracer in biology that reveals a process of interest in a system without disturbing it.
“The weak interaction allows us to reveal some unique features of the dynamics of the quarks within the nucleus of an atom,” Snow adds.
The NPDGamma result also helps enable a new search for possible violations of time reversal symmetry. This experiment, called the Neutron OPtics Time Reversal EXperiment, NOPTREX, will address the mystery of why there is more matter than antimatter in the universe. Snow is the cospokesperson for NOPTREX.
The paper appears in the journal Physical Review Letters.
Source: Indiana University. | {"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": 0.916985273361206, "perplexity": 826.1748441458456}, "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/1550247481249.5/warc/CC-MAIN-20190216230700-20190217012700-00358.warc.gz"} |
http://math.stackexchange.com/questions/109563/uniqueness-of-compact-topology-for-a-group/109569 | # Uniqueness of compact topology for a group
Suppose $G$ is a compact $T_2$ group. Can there be other compact $T_2$ topologies on $G$ which also turn $G$ into a topological group? ($T_2$ refers to the Hausdorff separation axiom)
-
The topology of a compact Hausdorff space is maximal compact and minimal Hausdorff; that is, no finer topology is compact, and no coarser topology is compact. So if you have another compact Hausdorff topology, then it is neither finer nor coarser to the original one. – Mariano Suárez-Alvarez Feb 15 '12 at 9:00
It may be worth stating that if you pick a topology once and for all and ask about uniqueness of smooth structures (if it has any at all!), then the answer is yes. – Jason DeVito Feb 15 '12 at 13:49
Take the circle group $G=S^1=\mathbb R/\mathbb Z$. Any non-continuous automorphism of $\mathbb R$ which fixes pointwise the subgroup $\mathbb Z$ passes to the quotient and gives an automorphism $f$ of the abstract group $G$, which is not continuous. Now define a topology on $G$ so that a set $U$ is open iff $f(U)$ is open in the usual topology. This new topology is of course Hausdorff and compact, but it is different to the usual topology.
Interesting, the resulting topological group is isomorphic (as topological group) to the original topological group, still the topology on the set $G$ is different. But I wonder more whether I could somehow prevent the implicit use of the axiom of choice. One idea would be to prescribe the Borel $\sigma$-algebra of a second-countable space and only allow topologies whose open sets belong to that $\sigma$-algebra. – Thomas Klimpel Feb 15 '12 at 15:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9127157330513, "perplexity": 204.19104120994754}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645378542.93/warc/CC-MAIN-20150827031618-00004-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://research.nsu.ru/ru/publications/search-for-rare-decays-of-z-and-higgs-bosons-to-j-%CF%88-and-a-photon- | # Search for rare decays of Z and Higgs bosons to J / ψ and a photon in proton-proton collisions at √s = 13 TeV
Результат исследования: Научные публикации в периодических изданияхстатьярецензирование
8 Цитирования (Scopus)
## Аннотация
A search is presented for decays of Z and Higgs bosons to a J / ψ meson and a photon, with the subsequent decay of the J / ψ to μ+μ-. The analysis uses data from proton-proton collisions with an integrated luminosity of 35.9fb-1 at s=13TeV collected with the CMS detector at the LHC. The observed limit on the Z → J / ψγ decay branching fraction, assuming that the J / ψ meson is produced unpolarized, is 1.4 × 10 - 6 at 95% confidence level, which corresponds to a rate higher than expected in the standard model by a factor of 15. For extreme-polarization scenarios, the observed limit changes from - 13.6 to + 8.6 % with respect to the unpolarized scenario. The observed upper limit on the branching fraction for H → J / ψγ where the J / ψ meson is assumed to be transversely polarized is 7.6 × 10 - 4, a factor of 260 larger than the standard model prediction. The results for the Higgs boson are combined with previous data from proton-proton collisions at s=8TeV to produce an observed upper limit on the branching fraction for H → J / ψγ that is a factor of 220 larger than the standard model value.
Язык оригинала английский 94 94 27 European Physical Journal C 79 2 https://doi.org/10.1140/epjc/s10052-019-6562-5 Опубликовано - 30 янв 2019
## Fingerprint
Подробные сведения о темах исследования «Search for rare decays of Z and Higgs bosons to J / ψ and a photon in proton-proton collisions at √s = 13 TeV». Вместе они формируют уникальный семантический отпечаток (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": 0.998865008354187, "perplexity": 2495.112834550992}, "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-2021-43/segments/1634323585121.30/warc/CC-MAIN-20211017052025-20211017082025-00564.warc.gz"} |
http://smurf.mimuw.edu.pl/node/1766 | ## Classical propositional logic
A formula of propositional logic is in the conjunctive normal form (CNF) if it is a conjunction of (possibly many) disjunctions of (possibly many) propositinal variables and negated propositional variables. Eg., $$(p_1\lor\lnot p_2\lor p_3)\land(p_2\lor p_4\lor \lnot p_5)\land(\lnot p_1\land p_2)$$ is a CNF formula.
A formula of propositional logic is in $$k$$-CNF if it is in CNF and each disjunction has at most $$k$$ disjuncts (variables or their negations). The formula given above is in 3-CNF, byt not in 2-CNF.
A formula of propositional logic is in the disjunctive normal form (DNF) if it is a disjunction of (possibly many) conjunctions of (possibly many) propositinal variables and negated propositional variables.
Exercise 1
Show that for each propositional formula $$\varphi$$ there exists a propositional formula $$\psi$$ in DNF, equivalent to $$\varphi$$, i.e., $$\varphi\leftrightarrow\psi$$ is a tautology.
Exercise 2
Show that for each propositional formula $$\varphi$$ there exists a propositional formula $$\psi$$ in CNF, equivalent to $$\varphi$$, i.e., $$\varphi\leftrightarrow\psi$$ is a tautology.
Exercise 3
Show that for each propositional formula $$\varphi$$ in CNF there exists a propositional formula $$\psi$$ in 3-CNF such that $$\psi$$ is satisfiable if and only if $$\varphi$$ is satisfiable.
Exercise 4
Give a polynomial time algorithm for the following decision problem:
Input: Propositional formula $$\varphi$$ in DNF.
Question: Is $$\varphi$$ satisfiable?
Exercise 5
Give a polynomial time algorithm for the following decision problem:
Input: Propositional formula $$\varphi$$ in 2-CNF.
Question: Is $$\varphi$$ satisfiable?
Exercise 6
We consider formulas built using connectives of conjunction and disjunction, only.
For such a formula $$\varphi$$ let $$\hat{\varphi}$$ denote its dualisation, i.e., the formula obtained from $$\varphi$$ by replacing every occurrence of $$\wedge$$ by $$\vee$$ and every occurrence of $$\vee$$ by $$\wedge$$.
* Prove that $$\varphi$$ is a tautology if and only if $$\lnot\hat{\varphi}$$ is a tautology.
* Prove that $$\varphi\leftrightarrow\psi$$ is a tautology if and only if $$\hat{\varphi}\leftrightarrow\hat{\psi}$$ is a tautology.
* Propose a method to define dualisation for formulas contaning addtionally logical constants $$\bot$$ and $$\top$$, such that the above equivalences remain valid.
Exercise 7
Prove that for any function $$f:\{0,1\}^k\to\{0,1\}$$ there exists a formula $$\varphi$$ using only the connectives $$\to$$ i $$\bot$$ and variables from the set $$\{p_1,\ldots, p_k\}$$ with the property that for any valuation $$\varrho$$ the following equality holds:
$$[[\varphi]]_\varrho = f(\varrho(p_1),\ldots, \varrho(p_k))$$.
(In other words, the formula $$\varphi$$ defines the function $$f$$.)
Exercise 8
Consider an infinite set of lads, each of which has a finite number of fiancees. Moreover, for each $$k\in N$$, any $$k$$ lads has at least $$k$$ fiancees. Demonstrate that it is possible to marry each lad with one of his fiancees, without commiting bigamy.
Exercise 9
Let $$k$$ be a fixed natural number. Prove, using the compactness theorem, that if every finite subgraph of an infinite graph $$G=\langle V,E\rangle$$ is $$k$$-collorable, then $$G$$ itself is $$k$$-collorable, too.
Exercise 10
For the formula $$\gamma =\ r\leftrightarrow (p_1\lor p_2)$$ the following equivalence holds: $$\varrho\models\gamma$$ if and ony if $$\varrho(r)=\max(\varrho(p_1),\varrho(p_2))$$.
Investigate whether there exists a set of formulas $$\Gamma$$ such that $$\varrho\models\Gamma$$ if and only if $$\varrho(r)=\max_{n\in\mathbb{N}}(\varrho(p_n))$$.
Exercise 11
Is the sequent $$\{p,q\to p,\lnot q\}\vdash\{p,q\}$$ provable in the Gentzen system for propositional logic?
Exercise 12
Decide if the following sequents are provable in the Gentzen system for propositional logic?
* $$(p\to q) \lor (q\to p)$$
* $$(p\to ( q \to p)) \to p$$
Exercise 13
In the Gentzen system, the sequent $$\Gamma,p\vdash\Delta,p$$ is an axiom, where $$p$$ is a propositional variable. Prove that every sequent of the form $$\Gamma,\varphi\vdash\Delta,\varphi$$ is provable in the Gentzen system. What can you assert about the size of this proof?
## Three-valued propositional logics
Exercise 14
A logic $$L$$ is called monotonic, if the conditions $$\Delta\models\varphi$$ and $$\Gamma\supseteq\Delta$$ imply $$\Gamma\models\varphi.$$
For a three-valued Sobociński logic we define that $$\Delta\models\varphi$$, iff for every valuation of propositional variables into $$\{0,\frac12,1\}$$, if the values of all sentences in $$\Delta$$ are 1, then the value of $$\varphi$$ is 1, too.
Is this logic monotonic?
Exercise 15
Answer the same question as in Exercise 14 for the logics of Heyting-Kleene-Łukasiewicz and Bochvar, as well as for the logic of lazy (short) Pascal evaluation.
Exercise 16
What is the computational complexity of the following decision problem:
Given: A formula $$\varphi$$ of propositional logic.
Question: Is there a valuation $$\varrho$$ of propositional variables into $$\{0,\frac12,1\}$$ such that in the three-valued logic of Bochvar $$[[\varphi]]_\varrho=1$$?
## Intuitionistic propositional logic
Exercise 16
Prove the following formula in the Natural Deduction system for the intuitionistic logic
* $$p \to \neg \neg p$$
* $$\neg (p \lor q) \to\neg p \land\neg q$$
* $$\neg p \land\neg q \to\neg (p \lor q)$$
* $$\neg p \lor\neg q \to\neg (p \land q)$$
Exercise 17
Prove that the following formulas are not tautologies of the intuitionistic logic. Use Kripke models:
* $$((p\to q) \to p) \to p$$
* $$\neg (p \land q) \to (\neg p \lor\neg q)$$
Exercise 18
Prove inexpressibility of connectives in the intuitionistic logic:
* $$\lor$$ can not be expressed using only $$\land$$, $$\to$$ and $$\bot$$
* $$\land$$ can not be expressed using only $$\lor$$, $$\to$$ and $$\bot$$
* $$\to$$ can not be expressed using only $$\lor$$, $$\land$$ and $$\bot$$ | {"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": 0.9825029373168945, "perplexity": 419.7238774266443}, "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/1566027316021.66/warc/CC-MAIN-20190821131745-20190821153745-00371.warc.gz"} |
http://www.frontierlattices.ch/indepth/elbm | ## Entropic Lattice Boltzmann Method (ELBM)
##### Lattice Boltzmann methods provide access to high Reynolds numbers through keeping low the Mach number and the Knudsen number as the two independent parameters of the simulation
Lattice Boltzmann methods(LBM) were introduced in the late 80’s – early 90’s as a new approach to CFD, and begun to find wide acceptance during the past decade. In LBM, one does not attempt a direct discretization of the governing fluid dynamics equations for mass, momentum and energy; instead, a kinetic equation of the Boltzmann type for a controlled number of discrete velocities is solved numerically on a regular grid. The entropic LBM is an advancement of LBM which satisfies the Second Law of thermodynamics (entropy of the system always increases).
The simplest Entropic lattice Boltzmann equation for the incompressible flow simulations can be understood with this example. Let ${v}_i$, $i=1,\dots,Q$ be a set of discrete velocities representing links of a regular -dimensional lattice, and $f_i(x,t)$ be the populations of the velocities at the node $x$ at the discrete time $t$. Using the notion of the entropy $H(f)=\sum_i f_i\ln(f_i/W_i)$ (weights $W_i$ depend on the choice of the lattice), the equilibrium $f_i^{\rm eq}(\rho,u)$ (analog of the Maxwell distribution) is derived as the minimizer of $H$ under fixed local (nodal) density $\rho=\sum_i f_i$ and momentum $\rho u=\sum_i f_i v_i$. The entropic lattice Bhatnagar-Gross-Krook equation (ELBGK) describes the dynamics of the populations due to the free streaming of particles along the direction of the lattice links and the local relaxation to the equilibrium at the nodes:
$f_i(x+v_i,t+1)-f_i(x,t)=\alpha\beta(f_i^{\rm eq}-f_i)$
where $\beta$ is a parameter relates to the kinematic viscosity while function $\alpha$ maintains the entropy balance in the relaxation step at every grid node and is found as the non-trivial root of the equation (termed the entropy estimate)
$$H(f+\alpha(f^{\rm eq}-f))=H(f)$$
Entropy estimate tells us that the entropy value should stay constant at the vanishing viscosity ($\beta=1$). This condition defines $\alpha$ as the maximal step of the over-relaxation without violating the Second Law ($H$-function decrease in the relaxation). Entropy estimate results in a confinement of the populations within the entropy contour during the relaxation, and leads to the unconditional stability of ELBGK. Observe that the entire nonlinearity (collision) in the ELBGK equation is on the right hand side, and is completely local in space, while the propagation in space (left hand side) is linear and exact. Furthermore, if the simulation is fully resolved, the entropy estimate leads to $\alpha=2$, and the ELBGK equation becomes its predecessor, the LBGK equation
$f_i(x+v_i,t+1)-f_i(x,t)=2\beta(f_i^{\rm eq}-f_i)$
With this stunningly simple formulation, the ELBM overcomes the stability problems of regular lattice Boltzmann method, while still retaining its locality, efficiency and flexibility.
With the proper choice of the lattice, the LBGK (or resolved ELBGK) recovers the Navier-Stokes equation with the kinematic viscosity, $\nu=c_{\rm s}^2\left(\frac{1}{2\beta}-\frac{1}{2}\right)$, where $c_{\rm s}$ is the lattice speed of sound (a constant depending on the choice of the lattice), so that the relaxation parameter $\beta$ can be matched to the desired value. Note that the kinematic viscosity in the LBM formulation is independent of the time step which is one of the major findings of the method enabling it to reach low viscosity leading to high Reynolds numbers flow regimes.
While the LBGK cannot reach this limit due to disruptive numerical instabilities at the sub-grid scale, the ELBGK is unconditionally stable by respecting the Second Law of thermodynamics. | {"extraction_info": {"found_math": true, "script_math_tex": 22, "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": 22, "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": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9340842962265015, "perplexity": 610.4199747339414}, "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-2014-41/segments/1410657137190.70/warc/CC-MAIN-20140914011217-00329-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://exoplanetmusings.wordpress.com/2013/03/25/764/ | # The Rossiter-McLaughlin Effect
A retrograde hot Jupiter (Credit: ESO)
It is a convenient fact that stars spin in space – a result of angular momentum imparted upon them from the collapse of the progenitor clouds from which they form. This fact, while easily taken for granted, can permit a great deal of information to be gained about surrounding planets whose orbits are fortuitously oriented to cause them to transit.
For the case of a rotating star, one hemisphere of the star will be approaching the observer, and the opposite hemisphere will be receding. The approaching hemisphere is therefore blueshifted, and the receding hemisphere is redshifted. Because the stellar spectrum will accordingly be blueshifted and redshifted, the spectral lines of a rotating star appear widened or broadened. This spectral line broadening is directly proportional to the star’s rotational velocity, and the angle between the line of sight and the stellar rotation axis. Much like with Doppler spectroscopic detection of planets, the constraints you can set to the rotational velocity of a star are only a lower limit because the rotation axis is unknown. Indeed, a rapidly rotating star viewed pole on (i = 0o), there will be no observed spectral line broadening because all of the surface rotation is perpendicular to the line of sight. Because of this inclination degeneracy, the rotational velocity of the star is represented in terms of v sin i.
The redshifting of the receding hemisphere is typically balanced by the blueshifting of the approaching hemisphere, and so there is no net change in the colour of the star. However the transit of a planet will cause an imbalance in this hemispheric Doppler shifting. For a planet in an equatorial, prograde transiting orbit the planet will first cover part of the approaching hemisphere. The Doppler shifting of the rotating stellar surface is no longer balanced and the stellar spectrum has a net redshift. This anomaly disappears as the planet moves to the centre of the stellar disc, as the redshifted and blueshifted hemispheres are both equally apparent again. Then as the planet moves over the receding hemisphere, there is a net blueshift in the spectrum as some of the redshifted light is occulted.
This radial velocity anomaly is known as the Rossiter-McLaughlin effect. It can be used to measure the projected stellar spin-orbit alignment angle (the angle between the spin axis of the star, and the orbital axis of the star+planet orbit). It can be thought of as the projected obliquity of the star relative to the planet orbit, or the inclination of the planetary orbit relative to the stellar equator. For an equatorial orbit (spin-orbit alignment), λ = 0°, and for a polar orbit (spin-orbit misalignment), λ = 90°.
The Rossiter-McLaughlin Effect
As we can see in the diagram above, the RM effect will have a different RV shape depending on the orientation of the planet’s orbit axis relative to the spin axis of the star. If the orbit is nearly polar and the planet transits across only one of the approaching or receding hemispheres, then the RM effect will be entirely blueshifted (if the planet crosses the receding hemisphere as is shown in the third example above) or redshifted (if the planet crosses the approaching hemisphere).
The amplitude of the RM effect, KR, relative to the amplitude of the star’s reflex velocity due to the influence of the planet over the course of the orbit, KO, can be estimated by
$\displaystyle \frac{K_R}{K_O} \approx 0.3\left(\frac{M}{M_J}\right)^{-1/3}\left(\frac{P}{3\text{ days}}\right)^{1/3}\left(\frac{v \sin i}{5\text{ km s}^{-1}}\right)$
Very few stars have their spin axes perfectly perpendicular to our line of sight. If the star is viewed pole-on, the RM effect amplitude will be zero. It’s worth remembering that it isn’t the absolute stellar rotation velocity that determines the Doppler shifting (and thus the RM effect amplitude), but the apparent rotational velocity, v sin i.
Another degeneracy affecting the RM effect is that impact parameters close to zero have the planet occupying both hemispheres more equally in time. This causes the RM effect to be much more symmetric than an off-centre transit with a high impact parameter. For the case of b = 0, the RM effect will be symmetric regardless of the value for λ, and only the RM effect amplitude will have changed. Since the RM effect amplitude is also a function of v sin i, some ambiguity is present in low- impact parameter systems over the value of λ.
The overwhelming majority of planets which have had RM effects measured are hot Jupiters, whose short orbital periods and high mass cause high orbital RV amplitudes, leaving the RM effect amplitude to be small in comparison. Interestingly, however, the magnitude of the RM effect amplitude is not dependent on the planet’s mass or semi-major axis, but on only the stellar v sin i and Rp. A planet of a given radius will have the same RM effect amplitude whether it is close to the star, producing a high orbital RV amplitude, or far from the star, with a much smaller orbital RV amplitude.
Accordingly, transiting planets in long-period orbits may have RM effect amplitudes several times their orbital RV amplitude. The advantages of this become quite apparent when one takes into consideration the timescales involved. Without the Rossiter-McLaughlin effect, a Doppler spectroscopic detection of an Earth-analogue around a solar-type star (with v sin i = 5 km s-1) would require at least a year to secure a full orbit, and furthermore, would require a Doppler precision on the order of 10 cm s-1. With noise issues in the RV data, it may require several orbits to build up enough data to confirm the planet. However, the Rossiter-McLaughlin effect for the same planet would be on the order of ~30 cm s-1 and would last less than a day. The far shorter time needed to observe the RM effect and the far greater amplitude make this a promising tool to validate extrasolar planet candidates whose existence are known from transits alone, but whose mass may take a considerable amount of effort to determine. Instead of wasting months trying to validate a planet candidate and run the risk that the candidate is a false-positive, a single night can conclusively demonstrate that the object is an orbiting planet.
Rossiter-McLaughlin Effect for a transiting Earth analogue
Because the rotational velocity of a star is correlated with its age and mass, which itself is correlated to the spectral type of the star, the true rotational velocity of a star is loosely correlated to the spectral type. Comparing a measured v sin i to the expected rotational velocity can give you a loose estimate for the inclination of the stellar rotation axis to the line of sight. This, combined with λ, can permit you to estimate the true spin-orbit alignment angle, typically denoted with φ (though variations on this definitely exist).
Constraining the spin-orbit alignment angle of extrasolar planetary systems provides clues to their dynamical histories. It is still not clear how hot Jupiter systems form. Giant planets may tidally interact with the planet-forming disk to migrate inward toward the star, or they may be gravitationally scattered inward by close encounters with other planets. Either scenario has testable predictions in the context of the observed spin-orbit alignment of a planetary system.
Peaceful migration through disk-interaction is expected to result in well-aligned planetary systems. Like in our Solar System, planets that form in a protoplanetary disk coplanar with the stellar equator will maintain that orientation, and you’ll end up with planets orbiting close to the stellar equatorial plane. However if hot Jupiters are the result of the scattering of planets through close encounters with each other, then the migration behaviour of hot Jupiters will have been much more chaotic in the early history of the planetary system and the spin-orbit alignment angle may have a wide range of values.
The first few hot Jupiters to have their RM effects measured were all fairly well aligned, consistent with calm migration as was generally believed to be the most likely explanation of their origin. Later work with planets from the SuperWASP survey uncovered several misaligned planets, including even retrograde hot Jupiters. Further work since has found planets across a wide range of values for λ.
Hot Jupiter Alignments
As spin-orbit alignment angles came in over the years it became apparent that misaligned planets are preferentially around hotter stars. Specifically above stellar effective temperatures of Teff > 6250 K, the distribution of λ seems to be much more random. This is clearly a clue to understanding the dynamical histories of hot Jupiters, but it’s not currently clear what it means.
Not only will measuring λ for extrasolar planet systems help us understand their formation histories, but it can also prove to be a powerful tool for confirming small, long-period planets with relatively little effort. It can require fairly high quality RV data, but it is well worth the effort to obtain these values, especially for smaller planets to see how the dynamical histories of low-mass, short-period planets compare to those of hot Jupiters. Do these multi-planet systems of low-mass planets form in a similar way to hot Jupiters? So far the evidence appears to point to “no,” but more data is needed to understand this question. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8906633853912354, "perplexity": 993.9407842441948}, "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/1529267864039.24/warc/CC-MAIN-20180621055646-20180621075646-00304.warc.gz"} |
https://eleijonmarck.dev/an-introduction-to-gradient-descent-w-linear-regression/ | # An Introduction to Gradient Descent w. Linear Regression
Gradient descent is one of those “greatest hits” algorithms that can offer a new perspective for solving problems. Unfortunately, it’s rarely taught in undergraduate computer science programs. In this post I’ll give an introduction to the gradient descent algorithm, and walk through an example that demonstrates how gradient descent can be used to solve machine learning problems such as linear regression.
Gradient descent is widely used in Machine Learning and Deep Learning
import pandas as pd
import numpy as np
import altair as alt
data = pd.read_csv('demo.txt')
scatter = alt.Chart(data).mark_circle().encode(
x='x:Q',
y='y:Q'
).properties(
title='data'
)
scatter
# Our goal is to align a line to this dataset
• why would we want to do that?
• we can use this to infer properties of the dataset
• we can use it to predict future behaviour (extrapolate)
from sklearn import linear_model
from sklearn import model_selection
model = linear_model.LinearRegression()
X_train, X_test, y_train, y_test = model_selection.train_test_split(data['x'], data['y'], test_size=0.2, random_state=0)
model.fit(X_train.values.reshape(-1, 1), y_train)
#For retrieving the slope:
print("""
Model intercept (position of the line) \n{:.2f}
Model coefficients (slope of the line) \n{:.2f}
Model score (how close are we to fit a line to the data) \n{:.2f}
""".format(
model.intercept_,
model.coef_[0],
model.score(X_test.values.reshape(-1, 1), y_test)))
This gives us the intercept $b$ 6.69 (where the line should be centered around) and the coefficient of $m$ 1.35. Which given our modeled loss function is a score of 0.27
def make_line_using(m, b):
# y = m * x + b
x = np.arange(100)
y = m * x + b
df = pd.DataFrame(np.matrix([x,y]).T, columns=['x','y'])
line = alt.Chart(df).mark_line().encode(
x='x:Q',
y='y:Q',
tooltip=[alt.Tooltip('y', title='b * x + m')]
).interactive()
return (scatter + (line))
m_guess = model.coef_[0]
b_guess = model.intercept_
make_line_using(m_guess, b_guess)
# Now let's try to implement this ourselves!
### Naive approach to guess until we get a good fit
guessing the beta parameters linear equation
$y = m \times \mathbf{x} + b$
def plot_on_top_of_data(m, b):
# y = B_2 * x + B_1
x = np.arange(100)
y = b * x + m
df = pd.DataFrame(np.matrix([x,y]).T, columns=['x','y'])
line = alt.Chart(df).mark_line().encode(
x='x:Q',
y='y:Q'
)
return (scatter + (line))
Our guess
$y = 2 \times \mathbf{x} + 1$
m_guess = 2
b_guess = 1
plot_on_top_of_data(m_guess, b_guess)
#### Not the most sufficient algorithm but might work.
hmmmmmm ¯\(ツ)/¯ let's think of another approach.
Can we we somehow see if we have a good guess?
# Let's improve our guessing strategy using Gradient Descent
Gradient descent is an optimization algorithm used to minimize some function (loss function) by iteratively moving in the direction of steepest descent as defined by the negative of the gradient. In machine learning, we use gradient descent to update the parameters of our model. Parameters refer to coefficients in Linear Regression and weights in neural networks.
Starting at the top of the mountain, we take our first step downhill in the direction specified by the negative gradient. We continue this process iteratively until we get to the bottom of our graph, or to a point where we can no longer move downhill–a local minimum.
math.stackexchange - Partial derivative in gradient descent
ml-cheatsheet
## Let's introduce the loss function (or cost/error)
A Loss Functions tells us “how good” our model is at making predictions for a given set of parameters. The loss function has its own curve and its own gradients. The slope of this curve tells us how to update our parameters to make the model more accurate.
${\displaystyle \operatorname {MSE} ={\frac {1}{n}}\sum {i=1}^{n}(Y{i}-{\hat {Y_{i}}})^{2}.}$
Given ${\displaystyle n}$ predictions generated to ${\hat{Y}}$, and ${\displaystyle Y}$ is the vector of observed values of the variable being predicted.
Our example with $\hat{Y} = mx_i + b$
Now let’s run gradient descent using our new loss function. There are two parameters in our lost function we can control: m (weight) and b (bias).
Since we need to consider the impact each one has on the final prediction, we need to use partial derivatives. We calculate the partial derivatives of the loss function with respect to each parameter and store the results in a gradient.
Given the loss function:
$$f(m,b) = \frac{1}{N} \sum_{i=1}^{n} (y_i - (mx_i + b))^2$$
The gradient can be calculated as:
$$f'(m,b) = \begin{bmatrix} \frac{df}{dm}\ \frac{df}{db}\ \end{bmatrix} = \begin{bmatrix} \frac{1}{N} \sum -2x_i(y_i - (mx_i + b)) \ \frac{1}{N} \sum -2(y_i - (mx_i + b)) \ \end{bmatrix}$$
def step_gradient(m: int, b: int, points: np.ndarray, learning_rate: float) -> list:
"""
this calculates the gradient step of a **linear function**
WILL NOT WORK for multiple dimensional data,
since the derivates will be on matricies instead
"""
N = float(len(points))
for i in range(0, len(points)):
x = points[i, 0]
y = points[i, 1]
b_gradient += -(2/N) * (y - ((m * x) + b))
m_gradient += -(2/N) * x * (y - ((m * x) + b))
b = b - (learning_rate * b_gradient)
m = m - (learning_rate * m_gradient)
return [b, m]
new_b, new_m = step_gradient(m_guess, b_guess, data.values, learning_rate)
new_b, new_m
(6.6874785109966535, 1.3465666635682905)
# Let's run through it through the whole dataset
def gradient_descent_runner(points, starting_m, starting_b, learning_rate, num_iterations):
m = starting_m
b = starting_b
for i in range(num_iterations):
m, b = step_gradient(m, b, np.array(points), learning_rate)
return [m, b]
number_iterations = 10000
data,
m_guess,
b_guess,
learning_rate,
number_iterations
)
print("Starting gradient descent at guess_m = {0}, guess_b = {1}".format(
m_guess,
b_guess
))
print("Last gradient descent at guess_m = {0}, guess_b = {1}".format(
m,
b
))
Starting gradient descent at guess_m = 1.3450919020620442, guess_b = 6.687439682550092
Last gradient descent at guess_m = 1.4510680203998683, guess_b = 1.4510195909326549
def compute_error_for_line(m, b, points):
total_error = 0
# sum (y_i - y_hat_i) ^ 2
for i in range(0, len(points)):
x = points[i, 0]
y = points[i, 1]
total_error += (y - (m * x + b)) ** 2
# 1 / n
mse = total_error / float(len(points))
return mse
# lets see how bad our guess was
compute_error_for_line(m_guess, b_guess, data.values)
838.9099083602013
print("Starting gradient descent at \n guess_m = {0}, guess_b = {1}, error {2}".format(
m_guess,
b_guess,
compute_error_for_line(m_guess, b_guess, data.values)
))
print("Last gradient descent at \n guess_m = {0}, guess_b = {1}, error {2}".format(
m,
b,
compute_error_for_line(m, b, data.values)
))
Starting gradient descent at
guess_m = 2, guess_b = 1, error 838.9099083602013
guess_m = 1.4510680203998683, guess_b = 1.4510195909326549, error 111.87217648730648
# How does gradient descent work?
In it's most general form:
Gradient descent is based on the observation that if the multi-variable function ${\displaystyle F(\mathbf {x} )}$ is defined and differentiable in a neighborhood of a point ${\displaystyle \mathbf {a} }$ , then ${\displaystyle F(\mathbf {x} )}$ decreases fastest if one goes from ${\displaystyle \mathbf {a} }$ in the direction of the negative gradient of ${\displaystyle F}$ at ${\displaystyle,-\nabla F(\mathbf {a} )}$. It follows that, if
${\displaystyle \mathbf {a} {n+1}=\mathbf {a} {n}-\gamma \nabla F(\mathbf {a} _{n})}$
for ${\displaystyle \gamma \in \mathbb {R} {+}}$ small enough, then ${\displaystyle F(\mathbf {a{n}} )\geq F(\mathbf {a_{n+1}} )}$. In other words, the term ${\displaystyle \gamma \nabla F(\mathbf {a} )}$ is subtracted from ${\displaystyle \mathbf {a} }$ because we want to move against the gradient, toward the minimum.
Here we have defined and the algorith works with contraints:
learning rate ${\gamma}$, for small ${\displaystyle \gamma \in \mathbb {R} _{+}}$
function $F(\mathbf{x})$, if differentiable; then ${\displaystyle F(\mathbf {a{n}} )\geq F(\mathbf {a{n+1}} )}$
Leading to ($\leadsto$) a monotonic sequence $F(\mathbf {x} {0})\geq F(\mathbf {x} {1})\geq F(\mathbf {x} _{2})\geq \cdots,$
# Coming back to the real world
Scikit learn provides you two approaches to linear regression:
If you can decompose your loss function into additive terms, then stochastic approach is known to behave better (thus SGD) and if you can spare enough memory - OLS method is faster and easier (thus first solution).
1) LinearRegression object uses Ordinary Least Squares solver from scipy, as LR is one of two classifiers which have closed form solution. Despite the ML course - you can actually learn this model by just inverting and multiplicating some matrices.
2) SGDRegressor which is an implementation of stochastic gradient descent, very generic one where you can choose your penalty terms. To obtain linear regression you choose loss to be L2 and penalty also to none (linear regression) or L2 (Ridge regression)
The "gradient descent" is a major part of most learning algorithm. What you will see most often is a improved version of the algorithm Stochastic Gradient Descent.
source - linear regression scitkit | {"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": 0.8059509992599487, "perplexity": 1944.2581319452638}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669795.59/warc/CC-MAIN-20191118131311-20191118155311-00429.warc.gz"} |
http://arxiv-export-lb.library.cornell.edu/abs/1907.12702 | math.NA
(what is this?)
# Title: On an optimal quadrature formula for approximation of Fourier integrals in the space $L_2^{(1)}$
Abstract: This paper deals with the construction of an optimal quadrature formula for the approximation of Fourier integrals in the Sobolev space $L_2^{(1)}[a,b]$ of non-periodic, complex valued functions which are square integrable with first order derivative. Here the quadrature sum consists of linear combination of the given function values in a uniform grid. The difference between the integral and the quadrature sum is estimated by the norm of the error functional. The optimal quadrature formula is obtained by minimizing the norm of the error functional with respect to coefficients. Analytic formulas for optimal coefficients can also be obtained using discrete analogue of the differential operator $d^2/d x^2$. In addition, the convergence order of the optimal quadrature formula is studied. It is proved that the obtained formula is exact for all linear polynomials. Thus, it is shown that the convergence order of the optimal quadrature formula for functions of the space $C^2[a,b]$ is $O(h^2)$. Moreover, several numerical results are presented and the obtained optimal quadrature formula is applied to reconstruct the X-ray Computed Tomography image by approximating Fourier transforms.
Comments: 27 pages, 6 figures Subjects: Numerical Analysis (math.NA) MSC classes: 41A05, 41A15 Cite as: arXiv:1907.12702 [math.NA] (or arXiv:1907.12702v1 [math.NA] for this version)
## Submission history
From: Abdullo Hayotov R [view email]
[v1] Tue, 30 Jul 2019 01:55:27 GMT (2548kb)
Link back to: arXiv, form interface, contact. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9532462954521179, "perplexity": 400.39352070627854}, "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/1600400234232.50/warc/CC-MAIN-20200926040104-20200926070104-00380.warc.gz"} |
https://www.physicsforums.com/threads/spherical-rolling.422499/ | # Spherical rolling
1. Aug 16, 2010
### hmoein
hi , every one!
I have a problem with a sphere rolling on a fixed sphere. My problem is to find relationship between coordinate of center of sphere (X,Y,Z) and orientations (alpha, beta, gamma) or Euler angles of sphere. as we know a sphere has 6 DOF in space (3 coordiantes and 3 rotation) when a sphere rolling on surface we expect that it have 3 dof beacuse of relation beween coordinate and rotation.
for example when a circle roll on a surface the x coordinate of its center is:
X=R*teta (R = radius of circle) and it has one DOF.
Like the circle rolling i want to find the relations for sphere.
thanks
hossein
2. Aug 16, 2010
### Ben Niehoff
Unfortunately, the constraint for a sphere rolling on a 2-dimensional surface cannot be integrated; it is "non-holonomic". Consider, as a simple case, a sphere rolling on a flat plane without slipping.
By rolling the sphere around a closed path, back to its starting point, you can imagine that in general the sphere will not end up in exactly the same orientation as it started; it will be rotated about the normal axis. Therefore, there is not a 1-to-1 correspondence between locations on the plane and orientations of the sphere.
You can construct differential relations, though; however, they will be more difficult to use.
3. Aug 17, 2010
thanks | {"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": 0.8157047629356384, "perplexity": 751.9760476182648}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676590794.69/warc/CC-MAIN-20180719090301-20180719110301-00102.warc.gz"} |
https://www.physicsforums.com/threads/hamiltonian-in-landau-gauge.848816/ | # Homework Help: Hamiltonian in Landau gauge
Tags:
1. Dec 18, 2015
### shinobi20
1. The problem statement, all variables and given/known data
Define n=(x + iy)/(2)½L and ñ=(x - iy)/(2)½L.
Also, ∂n = L(∂x - i ∂y)/(2)½ and ∂ñ = L(∂x + i ∂y)/(2)½.
with ∂n=∂/∂n, ∂x=∂/∂x, ∂y=∂/∂y, and L being the magnetic length.
a=(1/2)ñ+∂n and a=(1/2)n -∂ñ
a and a are the lowering and raising operators of quantum mechanics.
Show that H=ħωc(aa + ½)
2. Relevant equations
L=ħc/eB, ωc=eB/mc (cyclotron frequency), e for the charge of the electron
H = Px2/2m + ( Py2 + eBx/c )2/2m
3. The attempt at a solution
I have tried to find x,y,∂x,∂y in terms of n,ñ,∂n,∂ñ. But I ended up getting only some if the right terms to come out but not all, is my first step wrong? Any suggestions?
2. Dec 19, 2015
### blue_leaf77
Should the exponent "2" of $P_y$ be there?
3. Dec 19, 2015
### shinobi20
Sorry, it was a typo. Do you have any suggestions?
4. Dec 19, 2015
### blue_leaf77
You should post your initial attempt before we can discuss further. In particular, how the old variables look like in terms of the new ones?
5. Dec 19, 2015
### shinobi20
This is what I've done so far. My problem is that everything is there except for the ½. I wrote ∂ for ∂n and ∂(bar) for ∂ñ.
#### Attached Files:
• ###### IMG20151220132914.jpg
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6. Dec 20, 2015
### blue_leaf77
According to this link https://en.wikipedia.org/wiki/Landau_quantization, the Gauge you should be using is the symmetric gauge and hence the original Hamiltonian should be different than that you are using. For instance, in Landau gauge, the operator ${y}$ is not present.
7. Dec 20, 2015
### shinobi20
Why can't I show it using the Landau gauge? The choice is just for simplification of computation right?
8. Dec 20, 2015
### blue_leaf77
$x$ and $y$ appear symmetrically in the gauge transformation, but they do not in the original Hamiltonian.
9. Dec 20, 2015
### shinobi20
Oh I see that, then I'll try it again using the symmetric gauge. 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": 0.9143356084823608, "perplexity": 3290.9098596867198}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676588972.37/warc/CC-MAIN-20180715203335-20180715223335-00368.warc.gz"} |
https://mathhelpboards.com/threads/fast-fourier-transform-and-its-inverse.6550/ | [SOLVED]Fast Fourier Transform and its inverse
dwsmith
Well-known member
Feb 1, 2012
1,673
Does every FFT have $$i$$ in it?
Given $$u_t = -(u_{xxx} + 6uu_x)$$.
$$f'''(x) = \mathcal{F}^{-1}\left[(ik)^3\mathcal{F}(f(x))\right]$$
$$f'(x) = \mathcal{F}^{-1}\left[(ik)\mathcal{F}(f(x))\right]$$
The only equation I have used the pseudo-spectral method on was the NLS which is
$$u_t = i(u_{xx} + |u|^2u)$$. In this case, I know I will have $$i$$ in the FFT.
Are my transforms for the KdV correct or do I need to remove $$i$$? | {"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": 0.9106096625328064, "perplexity": 1067.2098982865625}, "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-2020-40/segments/1600400249545.55/warc/CC-MAIN-20200926231818-20200927021818-00521.warc.gz"} |
https://amathew.wordpress.com/2009/11/30/a-primer-on-harmonic-functions/ | The topic for the next few weeks will be Riemann surfaces. First, however, I need to briefly review harmonic functions because I will be talking about harmonic forms. I will have more to say about them later, and I actually won’t use most of today’s post even until then. But it’s fun.
Some of this material has also been covered by hilbertthm90 at A Mind for Madness.
Definition
A ${C^2}$ function ${f}$ on an open subset of ${\mathbb{R}^n}$, ${n >1}$, is called harmonic if it satisfies the Laplace equation$\displaystyle \Delta f = \sum \frac{\partial^2f}{\partial x_i^2} = 0.$ For now, we are primarily interested in the case ${n=2}$, and we will identify ${\mathbb{R}^2}$ with ${\mathbb{C}}$. In this case, as is well-known, harmonic functions are locally the real parts of holomorphic functions.
The Poisson Integral
The following fact is well-known: given a continuous function ${f}$ on the circle ${C_1(0)}$, there is a unique continuous function on the closed unit disk ${\overline{U}}$ which is harmonic in the interior and coincides with ${f}$ on the boundary.The idea of the proof is that ${f}$ can be represented as a Fourier series,
$\displaystyle f(e^{it}) = \sum_{n \in \mathbb{Z}} c_n e^{int}$
where the ${c_n}$ are obtained through the orthogonality relations
$\displaystyle c_n = ( f, e^{-int} )$
where the inner product is the ${L^2}$ product taken with respect to the Haar measure on the circle group. This convergence holds in ${L^2}$, because the exponentials form an orthonormal basis for that space. Indeed, orthonormality can be checked by integration, and the Stone-Weierstrass theorem implies their linear combinations are dense in the space of continuous functions on the circle. It is even the case that convergence holds uniformly if ${f}$ is well-behaved (say, ${C^2}$). But this is only for motivational purposes, and I refer anyone interested to, say, Zygmund’s book on trigonometric series for a whole lot fo such results.
Now, it is clear that the functions$\displaystyle z \rightarrow r^n e^{int}, \ z \rightarrow r^n e^{-int}$
are harmonic (where ${t = Arg(z), r = |z|}$) as the real parts of ${z^n, \bar{z}^n}$.
It thus makes sense to define the extended function ${\tilde{f}}$ as$\displaystyle \tilde{f}(re^{it}) = \sum_n c_n r^{|n|} e^{int}.$
Thus, writing ${F(r,t) := \tilde{f(re^{it})}}$, we find
$\displaystyle F(r,t) = \frac{1}{2\pi} \int_0^{2\pi} \sum_n f(e^{ix}) e^{-inx} e^{int} r^{|n|} dx$ which implies
$\displaystyle \boxed{ F(r,t) = \frac{1}{2\pi} \int f(e^{ix}) P_r(t-x) }$
where
$\displaystyle P_r(y) = \sum_n r^{|n|} e^{iny} = \frac{1-r^2}{1-2r\cos y + r^2}$
Theorem 1 The function ${\tilde{f}}$ is continuous in ${\bar{U}}$, harmonic in the interior ${U}$, and equal to ${f}$ on the boundary.
I’m not going to actually fully prove the theorem; the basic idea is that the ${P_r}$ are all of norm ${1}$ (because they are nonnegative and of integral 1), so we have ${||\tilde{f}||_{\infty} \leq ||f||_{\infty}}$. Consequently the result follows from its counterpart on trigonometric polynomials, which is evident since the Fourier series is finite! The approximation result proves useful again.
The Poisson integral shows that it is possible to solve the Dirichlet problem for the disk: that is, one can extend a continuous function on the boundary to a harmonic function. This does not work if ${D}$ is the deleted disk, because there is no harmonic function ${f}$ vanishing on ${\{z: |z|=1\}}$ with ${f(0)=1}$. (Cf. the maximum principle below.)
It is in fact the case that ${\tilde{f}}$ is the only such harmonic function, satisfying the conclusions of the theorem. This follows from the maximum principle—a nonconstant harmonic function has no local maxima, which in turn follows from the Laplace equation and the second derivative test as follows. If ${f}$ is harmonic and has a local maximum at ${0}$, so does ${h:=f + \epsilon(x^2+y^2)}$ for ${\epsilon>0}$ small; however, ${\Delta h > 0}$, a contradiction.
In particular, in view of the expression for the Poisson kernel, a harmonic function is necessarily smooth on its domain; apparently this is more generally true for solutions to elliptic PDEs, but I haven”t learned about them yet. This is probably one of the most important facts for us in the next few posts.
The mean value property
A harmonic function ${f}$ must satisfy on any disk ${D_r(a)}$ in its domain
$\displaystyle \boxed{f(a) = \frac{1}{2\pi} \int_0^{2\pi} f(a + re^{it}) dt.}$
This is in fact a corollary of the Poisson formula and uniqueness above. The mean value property is actually a sufficient condition for harmonicity (together with, say, continuity), but that is not necessary for us, and I refer anyone interested to Rudin’s Real and Complex Analysis.
The Harnack principle
Positive harmonic functions take comparable values on compact subsets of their domains, according to the next result:
Theorem 2 (Harnack)
Let ${D_0}$ be a compact subset of the open region ${D}$. Then there is a positive constant ${c := c(D_0,D)}$ such that for any positive harmonic function on ${D}$ and ${z,z' \in D_0}$$\displaystyle c^{-1} \leq \frac{f(z)}{f(z')} \leq c .$
If ${r<1}$ is sufficiently small, there is a positive constant ${c_0}$ such that$\displaystyle c_0^{-1} \leq P_{r'}(t) \leq c_0, \ \mathrm{when} \ r' \leq r , \forall t.$ This is evident, e.g., from the power series expression. (We actually could have said somewhat more.)
So let ${U}$ be the unit disk and ${U'}$ a small proper subdisk. If ${f}$ is a positive harmonic function on ${U}$, the above observation implies that the values of ${f}$ on ${U'}$ satisfy$\displaystyle d^{-1} \leq \frac{f(z)}{f(z')} \leq d$ for some ${d>0}$. In particular, we get Harnack’s theorem locally. By covering ${D_0}$ with a finite number of overlapping disks, we get the theorem globally. | {"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": 67, "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": 0.9882336854934692, "perplexity": 167.42518880335666}, "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/1627046154085.58/warc/CC-MAIN-20210731074335-20210731104335-00127.warc.gz"} |
https://wiki.kidzsearch.com/wiki/Euler%27s_identity | kidzsearch.com > wiki Explore:images videos games
# Euler's identity
Euler's identity, sometimes called Euler's equation, is a simple equation. It links several important numbers (mathematical constants) in mathematics in an unexpected way. Euler's identity is named after the Swiss mathematician Leonhard Euler, though it is not clear that he did invent it.[1]
Euler's identity is the equation $e^{i\pi} + 1 = 0$.
The special numbers in Euler's Identity, are
• 0: zero, special because zero plus any number is still that same number
• 1: one, special because one times any number is still that same number
• $\pi$: pi, special because it is one of the most common numbers in mathematics, and the distance around the outside of a circle divided by the distance across the circle.
$\pi \approx 3.14159$
• $e$, Euler's Number. Euler's Number appears in calculus and is related to the area between a curve that follows $y = {1 \over x}$ and the line $y = 0$.
$e \approx 2.71828$
• $i$, which is an imaginary number. The number $i = \sqrt{-1}$ and has the property $i \times i = i^2 = -1$.
## Reputation
A reader poll done by Physics World in 2004 called Euler's identity the "greatest equation ever", together with Maxwell's equations. Richard Feynman called Euler's identity "the most beautiful equation". The Identity is well known for its mathematical beauty: It combines the fields of geometry and algebra, and yet does so using only 7 of the most common and important mathematical symbols.
## Mathematical proof using Euler's formula
Euler's Formula is the equation $e^{ix} = \cos(x) + i \sin(x)$. Our variable $x$ can be any real number, but for this proof $x = \pi$. Then $e^{i\pi} = \cos(\pi) + i \sin(\pi)$. Since $\cos(\pi) = -1$ and $\sin(\pi) = 0$, the equation can be changed to read $e^{i\pi} = -1$, which gives the identity $e^{i\pi} + 1 = 0$.
## References
1. Sandifer, C. Edward 2007. Euler's greatest hits. Mathematical Association of America, p. 4. ISBN 978-0-88385-563-8
pl:Wzór Eulera#Tożsamość Eulera | {"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": 0.9872763752937317, "perplexity": 861.7512125472086}, "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/1600400222515.48/warc/CC-MAIN-20200925053037-20200925083037-00792.warc.gz"} |
https://www.physicsforums.com/threads/is-this-equation-conservative-or-non-conservative.878399/ | # Is this equation conservative or non-conservative?
Tags:
1. Jul 10, 2016
### humphreybogart
1. The problem statement, all variables and given/known data
This is the Navier-Stokes equation for compressible flow. nj is the unit normal vector to the surface 'j', and ni is the unit normal vector in the 'i' direction. Is this equation written for a control volume or a material volume?
2. Relevant equations
3. The attempt at a solution
I believe it's for a control volume, since it's in integral form and expressing fluxes out of a cube (taking advantage of conservation of momentum). However, I know that integral forms of non-conservative equations also exist, so I'm not sure.
2. Jul 11, 2016
### Staff: Mentor
Would the first term on the right hand side be present in the material volume form?
3. Jul 14, 2016
### humphreybogart
I'm tempted to say 'no', because no fluid enters or leaves a material volume. So the term would disappear. I'd like to see the integral and differential form for conservative, and the integral and differential form for non-conservative.
4. Jul 14, 2016
### Staff: Mentor
The integral form for material volume is the same as for control volume, except that the first term on the right hand side is absent. The differential forms for both are identical. See this link to see why the integral form of the material volume development reduces to the same differential form as the control volume development: https://en.wikipedia.org/wiki/Reynolds_transport_theorem
5. Jul 15, 2016
### humphreybogart
Thank you.
Great! I seen in another post a reference to Bird's Transport Phenomena book. Thanks.
Draft saved Draft deleted
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https://www.arxiv-vanity.com/papers/1207.0553/ | # [
June Huh Department of Mathematics, University of Michigan
Ann Arbor, MI 48109
USA
###### Abstract
We show that the maximum likelihood degree of a smooth very affine variety is equal to the signed topological Euler characteristic. This generalizes Orlik and Terao’s solution to Varchenko’s conjecture on complements of hyperplane arrangements to smooth very affine varieties. For very affine varieties satisfying a genericity condition at infinity, the result is further strengthened to relate the variety of critical points to the Chern-Schwartz-MacPherson class. The strengthened version recovers the geometric deletion-restriction formula of Denham et al. for arrangement complements, and generalizes Kouchnirenko’s theorem on the Newton polytope for nondegenerate hypersurfaces.
maximum likelihood degree, logarithmic differential form, Chern-Schwartz-MacPherson class.
###### :
14B05, 14C17, 52B40
The maximum likelihood degree of a very affine variety]The maximum likelihood degree of a very affine variety
## 1 Introduction
Maximum likelihood estimation in statistics leads to the problem of finding critical points of a product of powers of polynomials on an algebraic variety [PS05, Section 3.3]. When the polynomials and the variety are linear and defined over the real numbers, the number of critical points is the number of bounded regions in the corresponding arrangement of hyperplanes.
Studying Bethe vectors in statistical mechanics, Varchenko conjectured a combinatorial formula for the number of critical points for complex hyperplane arrangements [Var95]. Let be the complement of hyperplanes in defined by the linear functions . The master function , where the exponents are integral parameters, is a holomorphic function on . We assume that the affine hyperplane arrangement is essential, meaning that the lowest-dimensional intersections of the hyperplanes are isolated points.
###### Varchenko’s conjecture.
If the hyperplane arrangement is essential and the exponents are sufficiently general, then the following hold.
1. has only finitely many critical points in .
2. All critical points of are nondegenerate.
3. The number of critical points is equal to the signed Euler characteristic .
The conjecture was proved by Varchenko in the case where the hyperplanes are defined over the real numbers [Var95], and by Orlik and Terao in general [OT95]. Subsequent works of Silvotti and Damon extended this result to some nonlinear arrangements [Dam99, Dam00, Sil96]. The assumption made on the arrangement is certainly necessary, for there are arrangements violating the inequality .
The principal aim of this paper is to generalize the theorem of Orlik and Terao. The generalization is pursued in two directions. In Theorem 1, we obtain the same conclusion for a wider class of affine varieties than complements of essential arrangements; in Theorem 2, we recover the whole characteristic class from the critical points instead of the topological Euler characteristic. A connection to Kouchnirenko’s theorem on the relation between the Newton polytope and the Euler characteristic is pointed out in Section 4.
The above extensions are motivated by the problem of maximum likelihood estimation in algebraic statistics. Recall that an irreducible algebraic variety is said to be very affine if it is isomorphic to a closed subvariety of an algebraic torus. Very affine varieties have recently received considerable attention due to their central role in tropical geometry [EKL06, Spe05, Tev07]. The complement of an affine hyperplane arrangement is affine, and it is very affine if and only if the hyperplane arrangement is essential. Any complement of an affine hyperplane arrangement is of the form , where is the complement of an essential arrangement.
In view of maximum likelihood estimation, very affine varieties are the natural class of objects generalizing complements of hyperplane arrangements. Consider the projective space with the homogeneous coordinates , where the coordinate represents the probability of the -th event. An implicit statistical model is a closed subvariety . The data comes in the form of nonnegative integers , where is the number of times the -th event was observed.
In order to find the values of on which best explain the given data , one finds critical points of the likelihood function
L(p1…,pn)=pu11⋯punn/(p1+⋯+pn)u1+⋯+un.
Statistical computations are typically done in the affine chart defined by the nonvanishing of , where the sum can be set equal to and the denominator of can be ignored. The maximum likelihood degree of the model is defined to be the number of complex critical points of the restriction of to the projective variety , where we only count critical points that are not poles or zeros of , and are assumed to be sufficiently general [HKS05]. In other words, the maximum likelihood degree is the number of critical points of the likelihood function on the very affine variety
U:={x∈V∣p1⋯pn(p1+⋯+pn)≠0}.
### 1.1 Varchenko’s conjecture for very affine varieties
We extend the theorem of Orlik and Terao to smooth very affine varieties. Let be a smooth very affine variety of dimension . Choose a closed embedding
f:U⟶(C∗)n,f=(f1,…,fn).
The master function , where the exponents are integral parameters, is a holomorphic function on . The maximum likelihood degree of is defined to be the number of critical points of the master function with sufficiently general exponents .
###### Theorem 1.
If the exponents are sufficiently general, then the following hold.
1. has only finitely many critical points in .
2. All critical points of are nondegenerate.
3. The number of critical points is equal to the signed Euler characteristic .
More precisely, there is a nonzero polynomial such that the assertions are valid for with .
Theorem 1 shows that, for instance, the conclusions of [CHKS06, Theorem 20] and [Dam99, Corollary 6] hold for smooth very affine varieties without further assumptions. This has a few immediate corollaries that might be of interest in algebraic geometry and algebraic statistics. First, the maximum likelihood degree does not depend on the embedding of into an algebraic torus. Second, the maximum likelihood degree satisfies the deletion-restriction formula as in the case of a linear model. Third, the sign of the Euler characteristic of a smooth very affine variety depends only on the parity of its dimension.
### 1.2 A geometric formula for the CSM class
The theorem of Orlik and Terao can be further generalized to very affine varieties which admit a good tropical compactification in the sense of Tevelev [Tev07]. See Definition 3.6 for schön very affine varieties. For example, the complement of an essential hyperplane arrangement is schön, and the hypersurface defined by a sufficiently general Laurent polynomial (with respect to its Newton polytope) is schön. The open subset of the Grassmannian given by nonvanishing of all Plücker coordinates is another schön very affine variety, which is of interest in algebraic statistics [SS04].
The generalization is formulated in terms of the variety of critical points of , the totality of critical points of all possible (multivalued) master functions for . More precisely, given a compactification of , the variety of critical points is defined to be the closure
X(U)=¯¯¯¯¯¯¯¯¯¯¯¯¯¯X∘(U)⊆¯¯¯¯U×Pn−1of X∘(U)={n∑i=1ui⋅dlog(fi)(x)=0}⊆U×Pn−1,
where is the projective space with the homogeneous coordinates . The variety of critical points has been studied previously in the context of hyperplane arrangements [CDFV11, DGS12]. See Section 2 for a detailed construction in the general setting.
We relate the variety of critical points to the Chern-Schwartz-MacPherson class of [Mac74]. Let be the intrinsic torus of , an algebraic torus containing whose character lattice is the group of nonvanishing regular functions on modulo nonzero constants. We compactify the intrinsic torus by the projective space , where is the dimension of .
###### Theorem 2.
Suppose that is an -dimensional very affine variety which is not isomorphic to a torus. If is schön, then
[X(U)]=r∑i=0vi[Pr−i×Pn−1−r+i]∈A∗(Pn×Pn−1),
where
cSM(1U)=r∑i=0(−1)ivi[Pr−i]∈A∗(Pn).
Theorem 1 is recovered by considering the number of points in a general fiber of the second projection from , which is the maximum likelihood degree
vr=(−1)r∫cSM(1U)=(−1)rχ(U).
When is the complement of an essential hyperplane arrangement and is the usual compactification of defined by the ratios of homogeneous coordinates , Theorem 2 specializes to the geometric formula for the characteristic polynomial of Denham et al. [DGS12, Theorem 1.1]:
χA(q+1)=r∑i=0(−1)iviqr−i.
The formula is used in [Huh] to verify Dawson’s conjecture on the logarithmic concavity of the -vector of a matroid complex, for matroids representable over a field of characteristic zero [Daw84]. Other implications of the geometric formula are collected in Remark 3.12.
### 1.3 A generalization of Kouchnirenko’s theorem
A schön hypersurface in an algebraic torus is defined by a Laurent polynomial which is nondegenerate in the sense of Kouchnirenko [Kou76]. We generalize Kouchnirenko’s theorem equating the Euler characteristic with the signed volume of the Newton polytope, in the setting of Theorem 2. We hope that the approach of the present paper clarifies an analogy noted in [Var95, Remarks (e)], where Varchenko asks for a connection between Kouchnirenko’s theorem and the conjecture stated in the introduction.
Let be a Laurent polynomial in variables with the Newton polytope , and denote the corresponding hypersurface by
U={g=0}⊆(C∗)n.
Fix the open embedding defined by the ratios of homogeneous coordinates .
We follow the convention of [CLO98, Chapter 7] and write for the -dimensional mixed volume. For example, the -dimensional standard simplex in has the unit volume .
###### Theorem 3.
Let be a nonzero Laurent polynomial in variables with
cSM(1U)=r∑i=0(−1)ivi[Pr−i]∈A∗(Pn).
If is nondegenerate, then
vi=MVn(Δ,…,Δr−i,Δg,…,Δgi+1)for i=0,…,r.
In particular, the maximum likelihood degree of is equal to the normalized volume
vr=(−1)r∫cSM(1U)=Volume(Δg).
Theorem 3 has applications not covered by Kouchnirenko’s theorem. In particular, we get an explicit formula for the degree of the gradient map of a homogeneous polynomial in terms of the Newton polytope; see Corollary 4.6. This shows that many delicate examples discovered in classical projective geometry have a rather simple combinatorial origin.
As an example, we find irreducible homaloidal projective hypersurfaces of given degree and ambient dimension , improving upon previous constructions in [CRS08, FM12]; see Example 4.9.
### 1.4 Organization
We provide a brief overview of the paper.
Section 2 is devoted to the proof of Theorem 1. Along the way we construct the variety of critical points and describe its basic properties.
Section 3 introduces the deletion-restriction for hyperplane arrangements, extending that to very affine varieties. A brief introduction to the Chern-Schwartz-MacPherson class is given, and Theorem 2 is proved.
Section 4 focuses on the maximum likelihood degree of nondegenerate hypersurfaces in algebraic tori. Applications of Theorem 3 to the geometry of projective hypersurfaces are given.
## 2 Proof of Theorem 1
### 2.1 The Gauss map of very affine varieties
An important role will be played by the Gauss map of a very affine variety in its intrinsic torus. Let be a smooth very affine variety of dimension . Choose a closed embedding
f:U⟶(C∗)n,f=(f1,…,fn).
By a theorem of Samuel (see [Sam66]), the group of invertible regular functions is a finitely generated free abelian group. Therefore one may choose to form a basis of . In this case, is a closed embedding of into the intrinsic torus with the character lattice
f:U⟶TU.
Any morphism from to an algebraic torus is a composition of with a homomorphism . The Gauss map of is defined by the pushforward of followed by left-translation to the identity; that is,
In coordinates, the first map is represented by the Jacobian matrix
(∂fi∂xj),1≤i≤n,1≤j≤r,
and the second map is represented by the diagonal matrix with diagonal entries . The composition of the two is the logarithmic Jacobian matrix
(∂logfi∂xj),1≤i≤n,1≤j≤r.
This defines the Gauss map from to the Grassmannian of :
U⟶Grr(T1TU),x⟼TxU⊆T1TU.
Let be the sheaf of differential one-forms on . Consider the complex vector space
W:=MU⊗ZC.
The dependence of on will often be omitted from the notation. We write for (the sheaf sections of) the trivial vector bundle over with the fiber . There is a vector bundle homomorphism , defined by the evaluation of the logarithmic differential forms as follows:
Φ:WU⟶Ω1U,(n∏i=1fuii,x)⟼n∑i=1ui⋅dlog(fi)(x).
At a point , the linear map between the fibers is dual to the injective linear map considered above,
Therefore is surjective and is a vector bundle over .
### 2.2 The variety of critical points
The inclusion of into defines a closed immersion between the projective bundles
X∘(U):=Proj(Sym(kerΦ∨))⟶Proj(Sym(W∨U))≃U×P(W).
Note that the following conditions are equivalent:
1. is injective.
2. is isomorphic to a torus.
3. is empty.
If is not empty, then is a projective bundle over of dimension equal to that of , defined by the equation
n∑i=1ui⋅dlog(fi)(x)=0
where are the homogeneous coordinates of . In short, is the set of critical points of all possible (multivalued) master functions.
###### Definition 2.1.
Given a compactification of , the variety of critical points of is defined to be the closure
XV(U):=¯¯¯¯¯¯¯¯¯¯¯¯¯¯X∘(U)⊆V×P(W).
We denote the variety of critical points by when there is no danger of confusion.
The variety of critical points is irreducible by construction. When is the complement of an essential arrangement of hyperplanes, is the variety of critical points previously considered in the context of hyperplane arrangements [CDFV11, DGS12]. This variety has its origin in [OT95, Proposition 4.1].
We record here the following basic compatibility: If is a morphism between two compactifications of which is the identity on , then the class of maps to the class of under the induced map between the Chow groups
A∗(V1×P(W))⟶A∗(V2×P(W)),[XV1(U)]⟼[XV2(U)].
###### Remark 2.2.
We point out a technical difference between Definition 2.1 and the variety of critical points in the cited literature. This remark is intended for readers familiar with [CDFV11, DGS12] and is independent of the rest of the paper.
Let be the complement of , an essential arrangement of hyperplanes in . The variety of critical points in [DGS12] is defined when is a central arrangement, so we suppose that this is the case. In our notation, it is the quotient under the torus action
˜X(U):=XV(U)/(C∗×1)⊆(Cr×Pn−1)/(C∗×1)=Pr−1×Pn−1,
where is the partial compactification of by the affine space . Since is central, by [OT95, Proposition 3.9] we have
˜X(U)⊆{u1+⋯+un=0}≃Pr−1×Pn−2⊂Pr−1×Pn−1.
The quotient variety is indeed the variety of critical points (of a closely related arrangement) in our sense. Consider a decone of , an affine arrangement obtained by declaring one of the hyperplanes in the projectivization of to be the hyperplane at infinity. The number of hyperplanes and the rank of the decone are one less than the corresponding quantities for . More precisely, we have the following relation between the characteristic polynomials:
χ˜A(q)=χA(q)/(q−1).
Let be the complement of the decone in , and take the obvious compactification of . Then the variety of critical points of is the subvariety considered above,
˜X(U)⊆Pr−1×Pn−2.
The reader is invited to compare the formula of Corollary 3.11 with its cohomology version [DGS12, Theorem 1.1] for essential central arrangements.
### 2.3 Nonvanishing at the boundary
Let be the torus-invariant hyperplanes in defined by the homogeneous coordinates . Fix the open embedding
ι:(C∗)n⟶Pn,
defined by the ratios . Let be the closure of in , and choose a simple normal crossing resolution of singularities
where is an isomorphism over , is smooth, and is a simple normal crossing divisor with the irreducible components . Our goal is to show that a sufficiently general differential form on with logarithmic poles along has a zero scheme which is a finite set of reduced points in .
Each defines a rational function on which is regular on . We have that
ordDj(fi) is {positiveif π(Dj)⊈H0 and π(Dj)⊆Hi,negativeif π(Dj)⊆H0 and π(Dj)⊈Hi.
###### Lemma 2.3.
For each , there is an such that is nonzero.
###### Proof.
Since is irreducible, each is contained in some . The assertion follows from the following set-theoretic reasoning:
1. If , then for some because .
2. If , then for some because .
An integral vector defines a rational function on
φu=n∏i=1fuii,
which is regular on . Note that for each ,
ordDj(φu)=n∑i=1ui⋅ordDj(fi).
Combining this with Lemma 2.3, we have the following result.
###### Lemma 2.4.
For a sufficiently general , is nonzero for all .
More precisely, there are corank- subgroups of such that is nonzero for and .
Consider the sheaf of logarithmic differential one-forms , where . For the definition and the needed properties of the sheaf of logarithmic differential one-forms, we refer to [Del70, Sai80]. We note that is a locally free sheaf of rank , and the rational function defines a global section
dlog(φu)=n∑i=1ui⋅dlog(fi)∈H0(˜V,Ω1˜V(logD)).
###### Lemma 2.5.
For a sufficiently general , does not vanish on .
###### Proof.
Given a point , let be the irreducible components of containing , and let be local defining equations on a small neighborhood of . Clearly, is at least . By replacing with a smaller neighborhood if necessary, we may assume that trivializes over , and that
φu=go11⋯gollhwhere oj=ordDj(φu)
for some nonvanishing holomorphic function on . Over the open set , we have
dlog(φu)=(l∑j=1ordDj(φu)⋅dlog(gj))+ψ,
where is a regular differential one-form. Since the form part of a free basis of a trivialization of over , it follows from Lemma 2.4 that does not vanish on for a sufficiently general . ∎
### 2.4 Proof of Theorem 1
Let be the variety of critical points of in , where is the complex vector space as before. Write for (the sheaf sections of) the trivial vector bundle over with the fiber , and consider the homomorphism defined by evaluation of the logarithmic differential forms
Ψ:W˜V⟶Ω1˜V(logD),(x,u)⟼dlog(φu)(x):=n∑i=1ui⋅dlog(fi)(x).
We do not attempt to give an individual meaning to the multivalued master function when is not integral. Let and be the two projections from , and define the incidence variety of the evaluation by
We drop the subscript from when there is no danger of confusion. By definition, is the unique irreducible component of which dominates under .
Then, for a sufficiently general ,
1. is contained in , by Lemma 2.5, and
2. is a finite set of reduced points, by the Bertini theorem applied to on [Jou83, Theoreme 6.10].
More precisely, there is a nonempty Zariski open subset such that the two assertions are valid for any element in the infinite set .
It follows that the zero scheme of the section ,
pr1(pr−12(u))={x∈˜V∣dlog(φu)(x)=0},
is a finite set of reduced points in . The smoothness of implies that the section is regular, and the above set represents the homology class [Ful98, Example 3.2.16]. Since all the are nonvanishing on , we may identify the critical points of with the zero scheme of .
Therefore all the critical points of are nondegenerate, and the number of critical points is equal to the degree of the top Chern class of . Finally, from the logarithmic Poincaré-Hopf theorem [Kaw78, Nor78, Sil96], we have
∫˜Vcr(Ω1˜V(logD))=(−1)r∫˜Vcr(Ω1˜V(logD)∨)=(−1)rχ(U).
## 3 Deletion-restriction for very affine varieties
In this section we formulate the deletion-restriction for the characteristic polynomial of a hyperplane arrangement in the very affine setting. The role of the characteristic polynomial will be played by a characteristic class for very affine varieties.
This point of view is particularly satisfactory for very affine varieties satisfying a genericity condition at infinity, and gives new insights on the positivity of the coefficients of the characteristic polynomial. For complements of hyperplane arrangements, we recover the geometric formula for the characteristic polynomial of Denham et al. [DGS12, Theorem 1.1].
Let be a very affine variety, and let be the hypersurface of defined by the vanishing of a regular function. The complement is a very affine variety, being a principal affine open subset of a very affine variety.
###### Definition 3.1.
A triple of very affine varieties is a collection of the form .
In the language of hyperplane arrangements, corresponds to the arrangement obtained by deleting the distinguished hyperplane from the arrangement corresponding to . For this reason, we call the deletion of and call the restriction of . As in the case of hyperplane arrangements, we have
dimU1=dimUanddimU0=dimU−1.
By the set-theoretic additivity of the topological Euler characteristics of complex algebraic varieties [Ful93, Section 4.5], we have
χ(U)=χ(U1)+χ(U0).
Therefore, by Theorem 1, the maximum likelihood degrees of a triple of smooth very affine varieties satisfy an additive formula. Write for the maximum likelihood degree of a very affine variety , i.e. the number of critical points of a master function of with sufficiently general exponents.
###### Corollary 3.2.
If is a triple of smooth very affine varieties, then
ML(U)=ML(U1)−ML(U0).
It would be interesting to obtain a direct justification of Corollary 3.2.
###### Remark 3.3.
Let us call a very affine variety primitive if it does not have any deletion. For example, the complement of an essential hyperplane arrangement is primitive if and only if it is the complement of a Boolean arrangement. Note that this is the case exactly when the complement is isomorphic to its intrinsic torus. A distinguished feature of the deletion-restriction for very affine varieties, when compared to that for hyperplane arrangements, is that there are primitive very affine varieties which are not isomorphic to a torus. These very affine varieties are responsible for the noncombinatorial aspect of the extended theory.
When is a triple of hyperplane arrangement complements, Corollary 3.2 is the deletion-restriction formula for the Möbius invariant , where is the characteristic polynomial of an affine hyperplane arrangement . The full deletion-restriction formula between the characteristic polynomials can be formulated in terms of the Chern-Schwartz-MacPherson (CSM) class [Mac74]. Below we give a brief description of the CSM class; Aluffi provides a gentle introduction in [Alu05].
Recall that the group of constructible functions of an algebraic variety is generated by functions of the form , where is a subvariety of . If is a morphism between complex algebraic varieties, then the pushforward of constructible functions is defined by the homomorphism
f∗:C(X)⟶C(Y),1Z⟼(p⟼χ(f−1(p)∩Z)).
If is a compact complex manifold, then the characteristic class of is the Chern class of the tangent bundle , where is the Chow homology group of (see [Ful98]). A generalization is provided by the Chern-Schwartz-MacPherson class, whose existence was once a conjecture of Deligne and Grothendieck. For a construction with emphasis on smooth and possibly noncompact varieties, see [Alu06b].
Let be the functor of constructible functions from the category of complex algebraic varieties (with proper morphisms) to the category of abelian groups.
###### Definition 3.4.
The CSM class is the unique natural transformation
cSM:C⟶A∗
such that when is smooth and complete.
The uniqueness follows from the naturality, the resolution of singularities, and the normalization for smooth and complete varieties. The CSM class satisfies the inclusion-exclusion relation
cSM(1U∪U′)=cSM(1U)+cSM(1U′)−cSM(1U∩U′)
and captures the Euler characteristic as its degree
χ(U)=∫cSM(1U).
When is the complement of an arrangement of hyperplanes in , the CSM class of is the characteristic polynomial . For the definition of the characteristic polynomial of an affine arrangement, see [OT92, Definition 2.52].
###### Theorem 3.5.
Let be the compactification of defined by the hyperplane at infinity . Then
cSM(1U)=r∑i=0(−1)ivi[Pr−i]∈A∗(Pr),
where
χA(q+1)=r∑i=0(−1)iviqr−i.
This is because the recursive formula for a triple of arrangement complements
cSM(1U1)=cSM(1U−1U0)=cSM(1U)−cSM(1U0),
agrees with the usual deletion-restriction formula
χA1(q+1)=χA(q+1)−χA0(q+1)
(see [OT92, Theorem 2.56]). The induction is on the dimension and on the number of hyperplanes. The case of no hyperplanes involves a direct computation of by the inclusion-exclusion formula, and the case of dimension is a special case of the equality
χ(U)=∫cSM(1U)=χA(1).
See [Alu12, Theorem 1.2] and also [Huh12, Remark 26].
Our goal is to relate the variety of critical points to the CSM class. If we restrict our attention to the degree of the CSM class, then the relation recovers the conclusion stated in Varchenko’s conjecture for the Euler characteristic. We prove this for a class of very affine varieties satisfying a genericity condition at infinity.
The genericity condition is commonly expressed using the language of tropical compactifications. If is a subvariety of an algebraic torus , then we consider the closures of in various (not necessarily complete) normal toric varieties of . The closure is complete if and only if the support of the fan of contains the tropicalization of [Tev07, Proposition 2.3]. We say that is a tropical compactification of if it is complete and the multiplication map
m:Tׯ¯¯¯U⟶X,(t,x)⟼tx
is flat and surjective. Tropical compactifications exist, and they are obtained from toric varieties defined by sufficiently fine fan structures on the tropicalization of [Tev07, Section 2].
###### Definition 3.6.
We say that is schön if the multiplication is smooth for some tropical compactification of .
Equivalently, is schön if the multiplication is smooth for every tropical compactification of [Tev07, Theorem 1.4].
###### Remark 3.7.
There are two classes of schön very affine varieties that are of particular interest. The first is the class of complements of essential hyperplane arrangements, and the second is the class of nondegenerate hypersurfaces [Tev07]. What we need from the schön hypothesis is the existence of a simple normal crossings compactification which admits sufficiently many logarithmic differential one-forms. For arrangement complements, such a compactification is provided by the wonderful compactification of De Concini and Procesi [DP95]. For nondegenerate hypersurfaces, and more generally for nondegenerate complete intersections, the needed compactification has been constructed by Khovanskii [Hov77].
Let be a very affine variety of dimension , where is the intrinsic torus of Section 2.1. Let be the closure of in , where is a fixed toric compactification of . We follow Section 2.2 and define the variety of critical points
X(U)⊆V×P(W)⊆Pn×Pn−1where W=MU⊗ZC.
###### Theorem 3.8.
Suppose that is schön and not isomorphic to a torus. Then
[X(U)]=r∑i=0vi[Pr−i×Pn−1−r+i]∈A∗(Pn×Pn−1),
where
cSM(1U)=r∑i=0(−1)ivi[Pr−i]∈A∗(Pn).
###### Proof.
We prove a slightly more general statement. Let be a compactification of obtained by taking the closure in a toric variety of . We will prove the equality
cSM(1U)=r∑i=0(−1)ipr1∗[pr−12(Pr−i)∩X(U)]∈A∗(V),
where and are the two projections from and is a sufficiently general linear subspace of of the indicated dimension. The projection formula shows that this implies the stated version when .
If is a schön very affine variety, then there is a tropical compactification of which has a simple normal crossings boundary divisor. More precisely, there is a smooth toric variety of , obtained by taking a sufficiently fine fan structure on the tropicalization of , such that the closure of in is a smooth and complete variety with the simple normal crossings divisor [Hac08, Proof of Theorem 2.5].
By taking a further subdivision of the fan of if necessary, we may assume that there is a toric morphism preserving . By the functoriality of the CSM class, we have
A∗(˜V)⟶A∗(V),cSM(1U)⟼cSM(1U).
Note also that
By the projection formula, the problem is reduced to the case where .
In this case, we have the following exact sequence induced by the restriction :
Note that the middle term is isomorphic to the trivial vector bundle over with the fiber [Ful93, Section 4.3]. Under this identification, the restriction of the differential one-forms is the evaluation map of Section 2.4,
It follows that the latter sequence is exact, and the projectivization of the kernel
I(U)={(x,u)∈V×P(W)∣dlog(φu)(x)=0}
coincides with the variety of critical points . Since the pullback of to is the canonical line bundle , we have
r∑i=0pr1∗[pr−12(Pr−i)∩X(U)] = pr1∗(r∑i=0c1(pr∗2 OP(W)(1))n−1−r+i∩[X(U)]) = s(N∨V/X)∩[V] = c(Ω1V(logV∖U))∩[V].
Here is the dimension of , and the last equality is the Whitney sum formula. Now the assertion follows from the fact that the CSM class of a smooth variety is the Chern class of the logarithmic tangent bundle; that is,
cSM(1U)=c(Ω1V(logV∖U)∨)∩[V].
This follows from a construction of the CSM class which is most natural from the point of view of this paper [Alu06b, Section 4]. For precursors, see [Alu99, Theorem 1] and also [GP02, Proposition 15.3]. ∎
###### Remark 3.9.
In a refined form, the above proof shows that the equality
cSM(1U)=r∑i=0(−1)ipr1∗[pr−12(Pr−i)∩X(U)]
holds in the -equivariant proChow group of [Alu06a, Alu06b]. This removes the dependence on the compactification from Theorem 2. It should not be expected that the equality holds in the ordinary proChow group of .
###### Remark 3.10.
Theorem 3.8 may fail to hold for a smooth very affine variety. As an example, consider a smooth hypersurface in whose closure in has a node at a torus orbit of codimension . A direct computation on a simple normal crossings compactification of shows that the incidence variety has a two-dimensional component other than , and hence the classes of and are different in .
For complements of hyperplane arrangements, Theorem 3.8 gives a geometric formula for the characteristic polynomial [DGS12, Theorem 1.1]. Let be the complement of an arrangement of distinct hyperplanes
A={f1=0}∪⋯∪{fn=0}⊂Cr.
Then is a very affine variety if and only if is an essential arrangement, meaning that the lowest-dimensional intersections of the hyperplanes of are isolated points. Indeed, taking one of the isolated points as the origin of and choosing linearly independent hyperplanes intersecting at that point reveals to be a principal affine open subset of . For the converse, write as a product , where is the complement of an essential arrangement, and note that the affine line does not admit a closed embedding into an algebraic torus.
Suppose from now on that is an essential arrangement. Then is a Boolean arrangement if and only if is isomorphic to an algebraic torus. The equations of the hyperplanes define a closed embedding
f:U⟶(C∗)n≃TU,f=(f1,…,fn).
The indicated isomorphism follows from the linear independence of the in . We fix the open embedding defined by the ratios of homogeneous coordinates . Then the closure of in is a linear subspace . In this setting, combining Theorems 3.5 and 3.8 gives the following statement.
###### Corollary 3.11.
Suppose that is not a Boolean arrangement. Then
[X(U)]=r∑i=0vi[Pr−i×Pn−1−r+i]∈A∗(Pr×Pn−1),
where
χA(q+1)=r∑i=0(−1)iviqr−i.
###### Remark 3.12.
A sequence of integers is said to be log-concave if for all , and it is said to have no internal zeros if the indices of the nonzero elements are consecutive integers. Write a homology class as the linear combination
ξ=∑iei[Pk−i×Pi].
It can be shown that some multiple of is the fundamental class of an irreducible subvariety if and only if the form a log-concave sequence of nonnegative integers with no internal zeros [Huh12, Theorem 21].
Therefore, by Theorem 3.8, the of a schön very affine variety form a log-concave sequence of nonnegative integers with no internal zeros. In particular, the coefficients of form a sequence with the three properties. This strengthens the previous result that the coefficients of form a log-concave sequence [Huh12, Theorem 3], and answers several questions on sequences associated to a matroid, for matroids representable over a field of characteristic zero:
1. Read’s conjecture predicts that the coefficients of the chromatic polynomial of a graph form a unimodal sequence [Rea68]. This follows from the log-concavity of when is the graphic arrangement of a given graph [Huh12].
2. Hoggar’s conjecture predicts that the coefficients of the chromatic polynomial of a graph form a strictly log-concave sequence [Hog74]. This follows from the log-concavity of when is the graphic arrangement of a given graph [Huh].
3. Welsh’s conjecture predicts that the -vector of a matroid complex forms a unimodal sequence [Wel69]. This follows from the log-concavity of when is an arrangement corresponding to the cofree extension of a given matroid [Len].
4. Dawson’s conjecture predicts that the -vector of a matroid complex forms a log-concave sequence [Daw84]. This follows from the log-concavity of when is an arrangement corresponding to the cofree extension of a given matroid [Huh].
For details on the derivation of the above variations, see [Huh].
###### Remark 3.13.
The characteristic class approach to Varchenko’s conjecture and the generalized deletion-restriction have been pioneered by Damon for nonlinear arrangements on smooth complete intersections [Dam99, Dam00]. In fact, it can be shown that Damon’s higher multiplicities are the degrees of the CSM class of the arrangement complement. See [Huh12] for the connection between the two, for nonlinear arrangements on a projective space.
The CSM point of view successfully deals with several problems considered by Damon in [Dam99, Dam00]. In particular, an affirmative answer to the conjecture in [Dam00, Remark 2.6] follows from the product formula
cSM(1U1⊗1U2)=cSM(1U1)⊗cSM(1U2)∈A∗(¯¯¯¯¯¯U1ׯ¯¯¯¯¯U2).
The above is a refinement of the product formula of Kwieciński in [Kwi92], and can be viewed as a generalization of the product formula for the Euler characteristic,
χ(U1×U2)=χ(U1)χ(U2);
see [Alu06a, Théorème 4.1].
## 4 Nondegenerate hypersurfaces
A nondegenerate Laurent polynomial defines a hypersurface in an algebraic torus which admits a tropical compactification with a simple normal crossings boundary divisor [Hov77, Section 2]. A sufficiently general Laurent polynomial with the given Newton polytope is nondegenerate, and the corresponding hypersurface is schön.
We show that the variety of critical points of a hypersurface defined by a nondegenerate Laurent polynomial is controlled by the Newton polytope. This gives a formula for the CSM class in terms of the Newton polytope, which specializes to Kouchnirenko’s theorem equating the Euler characteristic with the signed volume of the Newton polytope [Kou76].
Let be a nonzero Laurent polynomial in variables
g=∑ucuxu∈C[x±11,…,x±1n].
We are interested in the CSM class of the very affine variety
U={g=0}⊆(C∗)n.
The Newton polytope of , denoted by , is the convex hull of exponents with nonzero coefficient . Write for the Laurent polynomial made up of those terms of which lie in a face of the Newton polytope. We say that is nondegenerate if is nonvanishing on for every face of its Newton polytope.
We follow the convention of [CLO98, Chapter 7] and write for the -dimensional mixed volume. For example, the -dimensional standard simplex in has normalized volume .
In view of later applications to projective hypersurfaces, we state our result for a fixed compactification , where the open embedding is defined by the ratios of homogeneous coordinates . The formulation for other toric compactifications and the extension to complete intersections are left to the interested reader.
###### Theorem 4.1.
Let be a nonzero Laurent polynomial in variables with
cSM(1U)=r∑i=0(−1)ivi[Pr−i]∈A∗(Pn).
If is nondegenerate, then
vi=MVn(Δ,…,Δr−i,Δg,…,Δgi+1)for i=0,…,r.
In particular, the maximum likelihood degree of is equal to the normalized volume
vr=(−1)r∫cSM(1U)=Volume(Δg).
Since the degree of is the Euler characteristic , this recovers Kouchnirenko’s theorem [Kou76, Théorème IV].
###### Proof.
In any case, is the fundamental class of the ambient smooth and complete toric variety. Therefore we may assume that is nonempty and compute the CSM class of instead.
Fix a sufficiently fine subdivison of the fan of on which the support function of is piecewise linear. We may assume that the corresponding toric variety is smooth and the closure of in has simple normal crossings with [Hov77, Section 2]. Note in this case that
cSM(1(C∗)n)−cSM(1U)=c(Ω1X(logD∪V)∨)∩[X]∈A∗(X).
In order to compute the right-hand side, we use the Poincaré-Leray residue map
r\'{e}s:Ω1X(logD∪V)⟶OV,η⋅dlog(z)+ψ⟼η|V,
where is a local defining equation for and is a rational differential one-form which does not have poles along . The restriction of on is uniquely and globally determined, and in particular it does not depend on the choice of . Note that the residue map fits into the exact sequence
Since is a trivial vector bundle, the Whitney sum formula shows that
c(Ω1X(logD∪V)∨)∩[X]=n∑i=0(−1)ic1(OX(V))i∩[X].
Therefore, by the projection formula applied to the birational map , we have
cSM(1(C∗)n)−cSM(1U)=n∑i=0(−1)i(Hn−i⋅V | {"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": 0.9747875332832336, "perplexity": 381.35380383447983}, "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/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00413.warc.gz"} |
https://www.physicsforums.com/threads/rotational-kinematics-of-a-turntable.727715/ | # Rotational Kinematics of a turntable
1. Dec 10, 2013
### ziggo
Good afternoon! I've been mulling over this question for a bit and I can't seem to understand what it is asking. This is a question for an introductory calculus-based physics university course.
1. The Problem:
A uniform disk, such as a record turntable, turns 8.0 rev/s around a frictionless spindle. A non-rotating rod of the same mass as the disk is dropped onto the freely spinning disk so that both turn around the spindle. Determine the angular velocity of the combination in rev/s.
2. Equations used:
I interpreted this as a conservation of angular momentum problem where the radius remains constant:
m r^2 ω = m(disc and rod) r ^2 ω(final)
3. The solution:
Since the radius remains constant and the mass doubles, both the mass and radius^2 can be removed from both sides leaving:
ω(initial) = 2ω(final)
and since the initial angular velocity was 16π Rad/s the final angular velocity would be 8π Rad/s.
Am I in the ballpark here assuming that this question is concerning the conservation of angular momentum? I don't see any other way to incorporate mass other than using Newton's laws, but I'm not sure on that.
Last edited: Dec 10, 2013
2. Dec 10, 2013
### MostlyHarmless
Is the radius of the rod and the disc necessarily equal?
3. Dec 10, 2013
### ziggo
The problem doesn't state it unfortunately.
4. Dec 10, 2013
### ryandaly
The first thing that comes to my mind is to try solving it with energy, since the moments of inertia of a rod and a disc are different. Have you covered rotational kinetic energy yet?
5. Dec 10, 2013
### Mentz114
Conservation of angular momentum gives
$m_dr_d^2\omega_0= \omega_1(m_dr_d^2+m_r r_r^2)$ so $\frac{\omega_0}{\omega_1}=\frac{I_d+I_r}{I_d}$. The subscripts are 'r' for the rod and 'd' for the disc and $I$ is a moment of inertia. I'm assuming the rod and the disc have $I=mr^2/2$.
6. Dec 10, 2013
### ziggo
We have, but I'm not sure how to imply it in this case without any information concerning the rod other than that it has the same mass as the disc and it is now a part of the system.
7. Dec 10, 2013
### ziggo
This is a very good analysis of it, and this is what I would break it down as. I simply solved it for final angular velocity or "omega 1"
Draft saved Draft deleted
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https://stats.stackexchange.com/questions/62896/jaccard-similarity-from-data-mining-book-homework-problem/88296#88296 | # Jaccard Similarity - From Data Mining book - Homework problem
Exercise 3.1.3 : Suppose we have a universal set U of n elements, and we choose two subsets S and T at random, each with m of the n elements.
What is the expected value of the Jaccard similarity of S and T ?
I am reading the book http://infolab.stanford.edu/~ullman/mmds/ch3.pdf
Each item in T has an $\frac{m}{n}$ chance of also being in S. The expected number of items common to S & T is therefore $\frac{m^2}{n}$.
Exp. $\text{Jaccard Similarity} = \dfrac{\text{No. of common items}}{\text{Size of T} + \text{Size of S} - \text{Number of common items}} = \dfrac{m}{2n - m}$ (after simplification.)
• The expected number of items common to S & T is therefore m^2 / n How?+ Dec 16 '13 at 11:41
• Expected value is calculated as Sigma x.p(x), summing over each of the m elements in set T. Each common item you find adds 1 to the common item total, so in our case, x = 1 and p(x) is m / n. Dec 16 '13 at 12:16
• According to the Example 3.2 in the book, the size of the union should always the sum of the sizes of the two sets. So I think the denominator should be 2m. Numerator is m^2/n. So SIM= m/2n. Dec 21 '15 at 0:19
• In my opinion, this is wrong or at least, not sufficiently explained. Reasoning that E[X/Y] = E[X] / E[Y] is not valid in general. Mar 25 '16 at 15:50
• Incorrect. Counter example: m=2, n=3. You can compute by hand that the correct answer is 5/9, this formula yields 1/2. As pointed out by Antoine, the E[X/Y] = E[X] / E[Y] derivation is incorrect. Apr 9 '19 at 19:15
The above answer assumes that an element in $T$ may be repeated several times in $S$ (i.e. $S$ and $T$ are not sets but multisets); else the probability will not be $m/n$ uniformly.
I expect the answer should be more along the following lines:-
Let the number of common elements between $S$ and $T$ be $k$. Then, as mentioned by ack_inc in the comment to his answer, Jaccard similarity $Sim(S,T)=k/(2m-k)$.
Now, $Pr(Sim(S,T)=k/(2m-k))$ will be $\dfrac{{m\choose {k}} {n-m\choose m-k}}{n\choose m}$ since there are $n$ total elements, of which $m$ are in $S$ and $k$ are common. So the number of ways we can choose $m$ elements for $T$ is given by ${m \choose k}$ (choosing the $k$ common elements from $S$) times ${n-m\choose m-k}$ (choosing remaining $m-k$ elements).
Thus, $E(Sim(S,T))=\sum_{k=0}^{m} \dfrac{k}{2m-k} \dfrac{{m\choose {k}} {n-m\choose m-k}}{n\choose m}$.
However, simplifying the above expression is beyond my limited knowledge of combinatorial identities. If anyone can do so, kindly update the answer.
• When size of $T$ is same as the size of $U$ (each of size n), then wny not $S\subset T$?
– CKM
Apr 6 '20 at 6:18
I'm posting an alternative solution.
Jaccard similarity of two sets $S$ and $T$ is defined as the fraction of elements these two sets have in common, i.e. $\text{sim}(S,T)=|S\cap T|/|S\cup T|$. Suppose we chose $m$-element subsets $S$ and $T$ uniformly at random from an $n$-element set. What is the expected Jaccard similarity of these two sets? Suppose the $|S\cap T|=k$ for some $0\le k\le m$. Notice that for the first set, $S$, we have $\binom{n}{m}$ choices, while for $T$ we have $\binom{m}{k}\binom{n-m}{m-k}$ choices, because $k$ elements must be from $S$ and $m-k$ elements must not be from $S$. This gives us $$\Pr[|S\cap T|=k]=\frac{\binom{m}{k}\binom{n-m}{m-k}}{\binom{n}{m}},$$ meaning that $$\text{E}[\text{sim}(S,T)]=\sum_{k=0}^m\frac{\binom{m}{k}\binom{n-m}{m-k}}{\binom{n}{m}}\frac{k}{2m-k}.$$ Even though $\text{E}[|S\cap T|/|S\cup T|]\neq\text{E}[|S\cap T|]/\text{E}[|S\cup T|]=m/(2n-m)$, this expression seems to give good approximation.
Thanks to Mitja Trampus for pointing out an alternate solution, with $$\Pr[|S\cap T|=k]=\binom{m}{k}\frac{\binom{m}{k}}{\binom{n}{k}}\frac{\binom{n-m}{m-k}}{\binom{n}{m}},$$ giving the following expression: $$\text{E}[\text{sim}(S,T)]=\sum_{k=0}^m\binom{m}{k}\frac{\binom{m}{k}}{\binom{n}{k}}\frac{\binom{n-m}{m-k}}{\binom{n}{m}}\frac{k}{2m-k}.$$
(The above expressions are, of course, equivalent.)
EDIT: Regarding the simplification, perhaps applying the following identity (from Aigner's book, page 13) could work: $$\binom{n}{m}\binom{m}{k}=\binom{n}{k}\binom{n-k}{m-k}.$$
• There is a problem. What exactly do you do with Pr [S union T]? How do you go from E [ S int. T / S union T] to the second equation? Apr 21 '14 at 8:23
• Um, which equation exactly? Once I know the size of the intersection is $k$, I know the size of the union must be $2m-k$. So I view the size of the intersection $|S\cap T|$ as a random variable. Once I compute $\Pr[|S\cap T|=k]$, I just use the definition of the expectation: $E[sim(S,T)]=\Pr[|S\cap T|=1]/(2m-1)+\ldots+\Pr[|S\cap T|=m]$. Does this answer your question? Apr 21 '14 at 9:32
I agree with blazs answer - just want to add a small correction (credit to another guy in the course who pointed it out).
The summation does not start at 0. (you'll see it if you make n=100 and m=99)
$$\text{E}[\text{sim}(S,T)]=\sum_{k=max(0, 2m-n)}^m\binom{m}{k}\frac{\binom{m}{k}}{\binom{n}{k}}\frac{\binom{n-m}{m-k}}{\binom{n}{m}}\frac{k}{2m-k}.$$
I just want to add the following:
As pointed out, ack_inc's answer is not correct and can serve only as an approximation. Also, the lower bound should be $$\max\{0, 2m - n\}$$ instead of $$0$$, as GM1313 mentions.
I needed to compute the similarities for $$1\leq m\leq n$$ where $$n = 5000$$ or even bigger, so computing all the probabilities $$P_{n, m}[|S\cap T| = k]$$ (blindly following the definition)
• takes a lot of time,
• gives wrong results due to floating-point arithmetics, e.g., $$p = 0.0$$.
Therefore, I used the fact that $${a \choose b + 1} = \frac{a - b}{b + 1}{a \choose b}$$ to speed up the process, and compute the probabilities recursively, starting with the biggest one.
I also used the approximation $$m / (2n - m)$$ and actually, it is really good, especially for larger values of $$n$$ (curves for $$n\in\{20, 100, 5000\}$$): | {"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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9098352789878845, "perplexity": 350.6562092072229}, "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-2021-43/segments/1634323586043.75/warc/CC-MAIN-20211024142824-20211024172824-00038.warc.gz"} |
https://www.physicsforums.com/threads/calculus-problems-where-to-begin.6859/ | # Calculus problems, where to begin?
1. Oct 7, 2003
### Jeebus
Hello! I have a math problem (mostly proofs) that im stuck
on, partially because I do not know where to begin and partially
because I believe I dont even fully understand the problem. I
was wondering if any of you would be kind enough to show me what to
do? Thank you.
1. Suppose f(x,y) is differentiable for all (x,y), f(x,y)=17 on the
unit circle x^2+y^2=1 and grad f is never zero on the unit circle. For
any real number K, find a unit vector parallel to grad
f(cos(k),sin(k))....grad f stands for the gradient of f. But isnt it contradicting what its saying? It says f(x,y)=17 on the unit circle x^2+y^2. How the...?
I'm just supposing f(x,y) is differentiable for all (x,y), f(x,y)=17 on the unit circle x^2+y^2=1 and grad(f) is never zero on the unit circle(?) So you just find a unit vector parallel to grad f(cos(k),sin(k)), for k real, right?
PS- Do level curves apply to this problem?
2. Oct 8, 2003
### dhris
Hello Jeebus,
You're being asked to find the direction of ∇f on the unit circle (k is just an angle). I think it's easier if we use polar coordinates (r,θ), and their corresponding unit vectors r and θ (don't know how to make the little hats yet). If we look at the gradient on the circle, and dot it with θ: θ dot ∇f, we get the rate of change of f in the θ direction. But what is the rate of change of f in the θ-direction on the unit circle? And what can you conclude about the direction of ∇f from this?
Hope this helps,
dhris
3. Oct 8, 2003
### HallsofIvy
dhris was giving hints that should help you but I got the impression that you really had no idea what was going on (and so need more than hints).
You ask "Do level curves apply to this problem?" Well, yes, of course. You are GIVEN that f(x,y)= 17 on the unit on the unit circle. The point is that f(x,y) is CONSTANT on that circle. The unit circle IS a level curve. Now, what is the relationship between level curves of a function and the gradient of the function at points on a level curve?
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http://www.ganitcharcha.com/view-article-A-Population-Growth-Model-Including-Hermaphrodites.html | # A Population Growth Model Including Hermaphrodites
1. Introduction: Biomathematics or mathematical biosciences are concerned with the applications of mathematical techniques to get an insight into the problems of biosciences. The can well be interpreted in a similar way as in engineering, physical or social science mathematics. However, situations in life sciences are quite complex and complicated and, therefore, we have to look for a situation before a mathematical model is constructed. If a model be formed, then its consequences can be deduced by using mathematical tools and the results so obtained can be compared with observations. The discrepancies between theoretical results and observations suggest further improvement of the model. The process is repeated till the model becomes a realistic one.
Biomathematics include both mathematical modeling in biology and medicine and give useful information’s to enlighten the complex biological situations. Some of the disciplines included in the subject are: botany, zoology, ecology, population dynamics, genetics, epidemiology, pharmacokinetics, physiology, environmental science and so on. The extent to which mathematics has penetrated into different disciplines varies in each instance, but with its own right in exponentially growing literature. The techniques used in biomathematics are: classical, probabilistic and statistical, computer and simulation, operations research etc.
In the present article, it is proposed to enlighten a mathematical model on population dynamics (also called demography). The field entails the study of population growth, population dispersal, effects or immigration, emigration and mixing of populations, effect of age structure on population sizes etc.
The subject is based on the populations of microorganisms so tiny in size to be seen by the naked eye. But they play an important role in (i) fermentation technology, e.g. production of alcohol, beverages, vinegar, biogas etc, (ii) mining technology like leaching out undesirable elements from ores, (iii) sanitary engineering, e.g. removal of pollutants from waste water, (iv) bioconversion of solar energy and soon. The growth difference of individual has a vital role in demography and it occurs due to space location, genetic property, age and size variations etc. of which size-difference plays a significant role due to its wide applications in fishery and forestry.
Another important consideration of population growth is the marriage in society. Kendall [1] proposed a mathematical model on this by taking the marriage rate to be constant. But the rate increases throughout the world and there exists a curvilinear relationship between the marriage rate and time. On the basis of this conception, Mishra [2] and ojha and pandey [3] modified Kendall’s model by taking the marriage rate to be a linear and quadratic function of time. However, in all these problems, the number of hermaphrodites (although a few in number) has been neglected. We, therefore, propose a model by introducing the number of hermaphrodites and taking the marriage rate to be a linear function of time.
2. The Model and Basis Equations:
Suppose M, F, H and Z denote respectively the number of unmarried males, females, hermaphrodites and married couples at any time t. $u_{1}$, $u_{2}$, $u_{3}$ are the male, female and hermaphrodite birth rates per married couples per unit time. $a_{1}$, $a_{2}$, $a_{3}$ are the death rates of unmarried male, female and hermaphrodite per unit time; $a_{4}$, $a_{5}$, $a_{6}$ are the male, female and hermaphrodite death rates per married couple per unit time. We also suppose that the marriage rate is a linear function of time t and is given by $A_{0} + A_{1}.t$, where $A_{0}$, $A_{1}$ are non-negative constant. Then the population model runs as follows:
3. Solutions:
Integrating equation (4) w.r.t. $t$, we get
References:
[1] Kendall, D. G. : Stochastic Model and Population Growth, Demography, Springer Verlag(1977)
[2] Mishra, P. : Progress of Mathematics, Vol. 22(1988). P 20.
[3] Ojha, V.P. and Pandey, H: Jour. Nat. Acad. Math., Vol. 7(1989) p. 99
Dr D. C. Sanyal {Email: dcs_klyuniv[at]yahoo.com} is a retired professor of mathematics of University of Kalyani, India. Prof. Sanyal has published many research papers in several national and international journals and conferences of repute. His areas of expertise include Solid Mechanics, Hydrodynamics, Geophysics and Geidynamics and Bio Mathematics. | {"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": 0.8613725900650024, "perplexity": 1705.3101831086847}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320707.69/warc/CC-MAIN-20170626101322-20170626121322-00570.warc.gz"} |
http://www.scholarpedia.org/article/Expansive_system | # Expansive Systems
(Redirected from Expansive system)
Post-publication activity
Curator: Jorge Lewowicz
A discrete invertible (the case we shall mainly refer to) expansive system is a dynamical system such that every point of the underlying space has a distinctive behaviour. A homeomorphism $$f$$ from the compact metric space $$M$$ onto $$M$$ is expansive if there exists $$\alpha >0\ ,$$ (called expansivity constant of $$f$$) such that if $$x,y\in M$$ and $$dist(f^{n}(x),f^{n}(y))\leq \alpha$$ for every $$n\in Z$$ then, $$x=y\ .$$ Thus, if $$x\neq y\ ,$$ then for some $$n,$$ $$dist(f^{n}(x),f^{n}(y))>\alpha .$$
Expansive systems are then wholly sensitive to initial conditions and therefore, in this sense, chaotic.
Assume the dynamics of $$f$$ is observed with a precision that permits to distinguish points at a distance larger than $$\alpha ,$$ meanwhile, points at a distance less than $$\delta >0,$$ $$\delta <<\alpha,$$ are not distinguished. Then, a $$\delta$$-small neighbourhood of, say, $$x\in M$$ with infinite points, will be seen -at present- as only one point. However, for some $$n\in Z\ ,$$ the $$n$$-iterate through $$f$$ of this point, will show many of them, since points at a distance larger than $$\alpha$$ are distinguished by the observer (see [B]).
Since $$M$$ is compact, on account of the expansiveness of $$f,$$ it is not difficult to show that given a $$\delta$$ like in the preceding paragraph, $$\delta <\alpha /2\ ,$$ there is a $$C^{0}-$$neighbourhood $$N$$ of $$f\ ,$$ such that if $$g\in N\ ,$$ and $$dist(g^{n}(x),g^{n}(y))\leq \alpha$$ for every $$n\in Z,$$ then $$dist(g^{n}(x),g^{n}(y))\leq \delta$$ for all $$n\in Z.$$ Therefore the relation $$\mathcal{R}_{\delta }=\{(x,y)\in M\times M:dist(g^{n}(x),g^{n}(y))\leq \delta ,\;\;n\in Z\}$$ is an equivalence relation on $$M,\ .$$ The canonical projection $$\pi:M\to M/\mathcal{R}_{\delta }$$ is closed and consequently $$M/\mathcal{R}_{\delta }$$ is a Hausdorff compact topological space and therefore the $$M/\mathcal{R}_{\delta }$$ is a compact metrizable space, and $$g^{\ast }: M /\mathcal{R}_{\delta }\rightarrow M /\mathcal{R}_{\delta }$$ defined by $$g^{\ast }(\pi (x))=\pi (g(x))\ ,$$ is an expansive homeomorphism of $$M/ \mathcal{R}_{\delta }.$$
Again, an observer that can not distinguish points at a distance less than $$\delta$$ will see the motion as taking place in $$M/ \mathcal{R}_{\delta }$$ (instead of $$M$$) under the action of $$g^{\ast }\ .$$
Clearly, homeomorphisms conjugate to an expansive one, are also expansive.
## Examples
### The shift
Consider $$\ 2^{Z}=\left\{ \left( a_{n}\right) :a_{n}=0\text{ or }1,n\in Z\right\}$$ and the distance$dist(( a_{n}) ,( b_{n}) )=\sum_{-\infty }^{\infty }| a_{n}-b_{n}| 2^{-| n| }$
With this metric, which induces the product toplogy, $$2^{Z}$$ is compact. Let $$\sigma :2^{Z}\rightarrow 2^{Z}$$ be defined by $$\sigma (a_{n})=(b_{n}),$$ where $$b_{n}=a_{n-1}\ .$$ $$\sigma$$ is the usual shift homeomorphism. If $$(a_{n}) \neq ( b_{n})$$ then, for some $$K\in Z,a_{K}\neq b_{K}\ ,$$ and , therefore
$$dist(\sigma ^{K}(( a_{n})) ,\sigma ^{K} ((b_{n})) )\geq 1,$$
showing that the shift is an expansive homeomorphism.
### The Denjoy map
Take a rotation of $$S^{1}$$ by an angle $$2\pi \alpha\ ,$$
where $$\alpha$$ is irrational, and replace the points of a dense orbit, say $$\left\{ x_{n},n\in Z\right\}\ ,$$with arcs of diameter decreasing with $$| n|$$ in order to get a new space also homeomorphic to $$S^{1}\ .$$ The Denjoy map, $$f\ ,$$ may be defined by assigning to each point not
on the added arcs, the former image under the rotation, and mapping (length) linearly the arc replacing $$x_{n}$$ onto the one replacing $$x_{n+1},n\in Z\ .$$ It is easy to see that this map is a homeomorphism of $$S^{1}\ ,$$ and that the set $$D$$ of points not lying in the interior of the added arcs is compact and invariant under the Denjoy map (in fact this set is homeomorphic to the Cantor set). A non-trivial arc whose end points lie on this set contains some of the added arcs, and, consequently, some iterate of this arc will include the one replacing $$x_{0}$$ of diameter, say $$d\ .$$ Thus, $$d$$ will be an expansivity constant for the restriction of $$f$$ to $$D\ .$$
### Anosov and quasi-Anosov diffeomophisms.
Let $$f$$ be a diffeomorphism of a compact, Riemannian, smooth manifold $$M$$ onto itself; $$f$$ is Anosov if there exitsts $$L>0,0<\lambda <1\ ,$$ and continuous non trivial $$Tf$$ invariant sub-bundles $$S,U$$ of $$TM,\ ,$$ such that $$S\oplus U=TM\ ,$$ $$\left\| Tf^{n}(s)\right\| \leq L\lambda ^{n}$$ for $$s\in S,n\geq 0\ ,$$ and $$\| Tf^{-n}(u)\| \leq L\lambda ^{n}$$ and $$u\in U,n\geq 0\ .$$
If $$A$$ is a compact $$f$$-invariant subset of $$M$$ and the above decomposition holds on $$A\ ,$$ $$A$$ is called a hyperbolic set. The restriction $$f|A\ ,$$ of $$f$$ to a hyperbolic set $$A$$ is also expansive. Anosov difeeomorphisms may also be characterized in a different way (see [L1]). Let $$B:TM\rightarrow R$$ be a continuous quadratic form, i.e, $$B_{x}=B|T_{x}M$$ is a quadratic form on the vector space $$T_{x}M$$ that depends continuously on $$x\in M\ .$$ A diffeomorphism $$f:M\rightarrow M$$ is quasi-Anosov if there exists such a $$B$$ with the property $$B_{f(x)}((Tf)_{x}(v))-B_{x}(v)>0,$$ for every $$x\in M\ ,$$ and each $$v\in T_{x}M,\| v\| \neq 0\ .$$ A diffeomorphism $$f$$ is Anosov if and only if it is quasi-Anosov and $$B_{x}$$ is non-degenerate for all $$x\in M\ .$$ There are quasi-Anosov diffeomorphisms that fail to be Anosov (see [FR], the examples in this paper have a strange attractor and a strange repeller [M] and the motion of most points evolves to the attractor and comes from the reppeller). This characterization of quasi-Anosov (Anosov) diffeomorphisms permits to conclude the existence of a $$C^{1}$$ neighbourhood of $$f$$ such that any finite composition of diffeomorphisms in that neighbourhhod is also quasi-Anosov (Anosov). We shall see below that Anosov and quasi-Anosov diffeomorphisms (and hyperbolic sets) are expansive.
### Pseudo-Anosov homeomorphisms.
Figure 1: A singular point x of a pseudo-Anosov map f and nearby orbits.
Let $$f$$ be a homeomorphism of an oriented compact surface $$M$$ of genus larger than 1 onto itself. The map $$f$$ is pseudo-Anosov if there exist two $$f$$-invariant, transversal foliations with singularities (see figure 1) $$W^{S},W^{U},$$ and also two transversal measures $$\mu _{S},\mu _{U}$$ (defined on the space of (stable, unstable) leaves of $$W^{S},$$ respectively $$W^{U}$$) and $$\lambda >1$$ such that $$f^{\ast }(\mu _{U})=\lambda \mu _{U}$$ and $$f^{\ast }(\mu _{S})=\lambda ^{-1}\mu _{S}\ .$$ The existence and expansivity of these homeomorphisms is proved in [T], [FLP].
### Another example.
Let $$f:T^{2}\rightarrow T^{2}$$ be defined by
$$\tag{1} f(x,y)=(2x+y-\frac 12\pi c\sin (2\pi x),\;x+y-\frac 12\pi c\sin (2\pi x)).$$
For $$0\leq c<1,$$ $$f$$ is Anosov (for $$c=0$$ $$f$$ is linear), but for $$c=1\ ,$$ $$f$$ is expansive but is neither Anosov nor quasi-Anosov since $$Tf_{0}$$ has no non-trivial invariant subspaces.
## General Properties.
### Question: Why not to define expansivity only for the future?
Theorem [U]. Let $$M$$ be a compact metric space and $$f:M\rightarrow M$$ be an homeomorphism such that there is $$\alpha >0$$ with the property that for $$x,y\in M,x\neq y,$$ $$dist(f^{n}(x),f^{n}(y))>\alpha$$ for some $$n\geq 0\ .$$ Then, $$M$$ is finite.
### Stable (unstable) sets
Let $$f:M\rightarrow M$$ be a homeomorphism; for $$x\in M\ ,$$ the stable set of $$x$$ is
$$W^{S}(x)=\{ y\in M:dist(f^{n}(x),f^{n}(y))\rightarrow 0 \mbox{ if } n\rightarrow +\infty \} ,$$ and the unstable set is
$$W^{U}(x)=\{ y\in M:dist(f^{n}(x),f^{n}(y))\rightarrow 0 \mbox{ if } n\rightarrow -\infty \} .$$
The local stable (unstable) sets of $$x,$$ are defined as follows: given $$\varepsilon >0,$$ $$W_{\varepsilon }^{S}(x)=\left\{ y\in M:dist(f^{n}(x),f^{n}(y))\leq\varepsilon ,n\geq 0\right\}$$ $$W_{\varepsilon }^{U}(x)=\left\{ y\in M:dist(f^{n}(x),f^{n}(y))\leq \varepsilon ,n\leq 0\right\}$$
Let now $$f$$ be expansive. May the stable set contain a neighbourhood of $$x$$ for every $$\varepsilon >0\ ?$$ In other words : may $$x$$ be Lyapunov stable in the future? The answer is yes; it is easy to find a shift invariant subset of $$2^{Z}$$ for which $$0$$ is Lyapunov stable in the future. Nevertheless,
Theorem [L2 ]. If $$M$$ is locally connected there are no stable points (either in the future or in the past).
Corollary. If $$M$$ is locally connected, for every $$\varepsilon >0,$$ there is $$r>0\ ,$$ such that for every $$x\in M\ ,$$ $$W_{\varepsilon }^{S}(x)$$ and $$W_{\varepsilon }^{U}(x)$$ contain a compact connected set of diameter larger than $$r\ .$$
(Compare with the Denjoy map $$f|D\ ;$$ for points not lying on the added arcs the local stable (unstable) sets are trivial.)
Application. There are no expansive homeomorphisms of $$S^{1}.$$
Proof: Assume by contradiction that there exist an expansive homoemorphism on $$S^1\ .$$ Then by the previous Corollary there are non-trivial stable open sets (a connected set of $$S^1$$ contains an open arc) and every point of it is a stable point, in contradiction with the above Theorem.
### Expansiveness and Lyapunov Functions.
Theorem [L1]. Let $$f$$ be a homeomophism of $$M\ ,$$ then $$f$$ is expansive if and only if there exist a neighbourhood $$N$$ of the diagonal in $$M\times M$$ and a real continuous function $$V$$ (Lyapunov) defined on $$N\ ,$$ vanishing on the diagonal and such that for $$(x,y)\in N,x\neq y,$$ $$V(f(x),f(y))-V(x,y)>0.$$
In order to proof expansivity for Anosov and quasi-Anosov diffeomorphisms, the quadratic form $$B$$ mentioned in the section Anosov and quasi-Anosov diffeomorphisms, can be used to construct a Lyapunov function. In fact, for $$y$$ close to $$x\ ,$$ the Lyapunov function is $$V(x,y)=B_{x}(u)\ ,$$ where $$\exp _{x}(u)=y\ .$$ The expansivity of pseudo-Anosov maps may be shown also using Lyapunov functions [L2]. For the examples in (Figure 3), choose
$$V(x,y)=V((x_{1},x_{2}),(y_{1},y_{2}))=(y_{1}-y_{2})((x_{1}-y_{1})-(x_{2}-y_{2})).$$
## On Surfaces.
Classification Theorem ([Hi], [L3]). Let $$f$$ be an expansive homeomorphism of a compact connected oriented boundaryless surface $$M\ .$$ Then,
• $$S^{2}$$ does not support such a homeomorphism,
• if $$M=T^{2},$$ $$f$$ is conjugate to an Anosov diffeomorphism
• if the genus of $$M$$ is larger than 1, then $$f$$ is conjugate to a pseudo-Anosov homeomorphism.
($$T^{2}$$ is the unique surface that supports Anosov diffeomorphisms.)
Those properties are consequences of the description of the local stable (unstable) sets of $$f\ .$$
Usually, the study of local stable (unstable) sets are made on the basis of strong assumptions on the dynamics of $$Tf\ ,$$ as for Anosov diffeomorphisms, hyperbolic sets, etc. In our case, even for expansive diffeomorphims, we only have the dialogue between the topology of $$M$$ and the dynamics of $$f\ .$$ Nevertheless, after showing the local connectedness of the connected component containing $$x$$ of $$W_{\varepsilon}^{S}(x)(W_{\varepsilon }^{U}(x))$$ the following theorem is proved.
Theorem. For $$x\in M,\; W_{\varepsilon }^{S}(x)(W_{\varepsilon }^{U}(x))$$ is the union of a finite number $$r$$ of arcs, $$(r\geq 2)$$ that meet only at $$x\ .$$ Stable (unstable) sectors (the sets limited by two consecutive stable (unstable) arcs) are separated by unstable (resp. stable) arcs. If at $$x \in M\ ,$$ $$r\geq 3\ ,$$ $$x$$ is called a singular point; the set of singular points is finite.
When $$r=2\ ,$$ as it is always the case for Anosov diffeomorphisms, $$x$$ has a neighbourhood $$N$$ such that if $$y$$ and $$z$$ belong to $$N\ ,$$ $$W_{\varepsilon }^{S}(y)\cap W_{\varepsilon }^{U}(z)$$ is not void. This is not the case for singular points (see figure 2).
Now a very brief mention of some steps of the proof of the Classification Theorem is given. For $$r\geq2\ ,$$ if $$y$$ and $$z$$ lie in a sector then $$W_{\varepsilon }^{S}(y)$$ and $$W_{\varepsilon }^{U}(z)$$ meet only once. The set of these intersections includes, by the Theorem of invariance of domain, a neighbourhood of $$x$$ in the sector (local product structure). This implies that singular points can not accumulate and then, their number is finite. Let now $$M^{\ast }$$ be the universal cover of $$M\ .$$ It is not difficult to show that the lifting to $$M^{\ast }$$ of a stable or an unstable set is closed and that the union of the lifting of a stable arc and an unstable one can not be homeomorphic to $$S^{1}\ .$$
If $$S^{2}$$ supported an expansive homemorphism, and $$W^{S}(x)$$ does not contain singular points, it is homeomorphic to $$S^{1}\ ,$$ and this in turn, implies the existence of stable points; a contradiction.
Figure 2: At the singular point x the stable manifold through y does not intersect the unstable one through z.
That expansive homeomorphisms $$f$$ of surfaces of genus $$\geq 1\ ,$$ are conjugate to Anosov or to pseudo-Anosov maps follows from the following two Lemmas.
Lemma An expansive homeomorphism $$f$$ on a surface $$M$$ of genus $$\geq 1$$ is isotopic to an Anosov (if $$M=T^2$$) or to a pseudo-Anosov map (genus $$\geq 1$$).
Proof. It follws from [L3] on account of Thurston's results [T].
Definition. Let $$f,g$$ be homeomorphisms of the compact metric space $$M\ ;$$ $$f$$ is semi-conjugate to g if there exists $$h:M\rightarrow M$$ continuous and surjective, such that $$h\circ f=g\circ h\ .$$
Lemma If the expansive homeomorphism $$f$$ of the surface $$M$$ is isotopic to an Anosov diffeomorphism, or to a pseudo-Anosov homeomorphism $$g\ ,$$ then $$f$$ is semi-conjugate to $$g$$
Proof. See [F], [L3].
In both cases, $$h^*:M^*\to M^*\ ,$$ a lifting of the semi-conjugacy $$h$$ is a proper map, and this fact is an essential tool to prove that the semi-conjugacy is, actually, a conjugacy.
## Higher Dimension.
Consider now expansive homeomorphisms $$f$$ defined on compact boundaryless manifolds $$M$$ of dimension larger than 2. In the case of surfaces, it follows from the Classification theorem that periodic points are dense on the surface, and , moreover, that on an open and dense set, $$r=2\ .$$ Thus for points $$x$$ in that set, $$W_{\varepsilon }^{S}(x)$$ includes a topological 1-dimensional manifold and $$W_{\varepsilon }^{U}(x)$$ another such manifold, topologically transversal to the first one at $$x\ .$$ The results concerning $$dim M\geq 3$$ assume the existence of a dense set of periodic points $$p$$ such that $$W_{\varepsilon }^{S}(p)$$ contains a topological manifold of dimension $$d,\;1\leq d<\dim M\ ,$$ and $$W_{\varepsilon }^{U}(p)$$ a manifold of complementary dimension, topologically transversal to $$W_{\varepsilon }^{S}(p)$$ at $$p\ .$$ Points $$x$$ with such a behaviour of $$W_{\varepsilon }^{S}(x)$$ and $$W_{\varepsilon }^{U}(x)$$ are called topologically hyperbolic.(This is the case for Anosov diffeomorphisms at every $$x\in M$$).
Theorem([ABP], [V1], [V2]). Let $$f$$ be an expansive homeomorphism of $$M$$ with a dense set of topologically hyperbolic periodic points. Then there is an open and dense set with local product structure. Furthermore if $$\dim M\geq 3,$$ and for some topologically hyperbolic periodic point $$p\ ,$$ either $$W_{\varepsilon }^{S}(p)$$ or $$W_{\varepsilon }^{U}(p)$$ is one-dimensional, $$M$$ is a torus and $$f$$ is conjugate to a linear Anosov diffeomorphism.
Therefore, in this case, in contrast with what happens for surfaces, there are no singularities. This is, essentially, a consequence of the fact that if $$\dim M\geq 3\ ,$$ say, $$W_{\varepsilon }^{S}(p)$$ separates small balls centered at $$p\ ,$$ meanwhile $$W_{\varepsilon }^{U}(p)$$ does not. Of course, if we do not assume that one of this dimensions is one, the result is false: take the product of two pseudo-Anosov maps.
## $$C^{0}$$-perturbations of expansive systems.
Let $$f$$ be a homeomorphism of a compact metric space $$M$$ onto itself.
### a) Persistence.
$$f$$ is persistent if for any $$\varepsilon >0$$ there exists a $$C^{0}$$-neighbourhood $$N$$ of $$f$$ such that for $$g\in N$$ and $$x\in M\ ,$$ there exists $$y\in M$$ with the following property. $$dist(f^{n}(x),g^{n}(y))\leq \varepsilon ,\;\;n\in Z$$
### b) Topological Stability
$$f$$ is topologically stable if for $$\varepsilon >0\ ,$$ there exists,$$N\ ,$$ a $$C^{0}-$$neighbourhood of $$f\ ,$$ such that any $$g\in N$$ is semi-conjugate to $$f$$ (see 4)) and $$dist(x,h(x))<\varepsilon\ .$$
A $$\delta$$ pseudo-orbit for $$f$$ is a sequence $$\{ x_{n}:n\in Z\}$$ such that $$dist(f(x_{n}),x_{n+1})<\delta\ ,$$ $$n\in Z\ .$$ Such a pseudo-orbit is $$\varepsilon$$ shadowed if there is $$y\in M$$ such that $$dist(x_{n},f^{n}(y))\leq \varepsilon ,n\in Z.$$
Clearly b) implies a) since the semi-conjugacy $$h$$ is surjective, but a) does not imply b). All three properties are invariant under conjugacy. Anosov diffeomorphisms satisfy b) ([W1]) and, since because of the classification theorem, every expansive homeomorphism of $$T^{2}$$ is conjugate to an Anosov, then all expansive homeomorphisms of $$T^{2}$$ sastisfy b). A pseudo-Anosov homeomorphism $$f$$ satisfies a) (see [H]) but not b); because, according to [W2], for expansive systems b) is equivalent to c) and figure 3 shows an $$f$$ pseudo-orbit shadowed by no $$f$$-trajectory; thus $$f$$ does not satisfy c).
The quasi-Anosov diffeomorphisms are not even persistent. However each semi-trajectory is persistent : given $$x\in M\ ,$$ and $$\varepsilon >0$$ there is, $$N_{x}\ ,$$ a $$C^{0}$$-neighbourhood of $$f$$ such that for any $$g\in N_x,$$ there is $$y\in M\ ,$$ with the property $$dist(f^{n}(x),g^{n}(y))\leq \varepsilon,\;\;n\geq 0.$$
Figure 3: The pseudo orbit consisting of the $$f$$ past of $$x_k$$ and the $$f$$-future of $$x_{k+1}$$ $$(x_{k+1}$$ very close to $$x_{k})$$ is not shadowed by an $$f$$-orbit (see fig. 1)
This is the $$f$$ persistence of $$x$$ in the future. We define similarly persistence in the past. A point $$x$$ could be $$f$$ persistent in the future and in the past without being persistent on both sides. This is the case of many points in a quasi-Anosov diffeomorphism. An open question is: are all the semi-trajectories of an expansive system persistent?
## Links with the tangent map.
Let $$M$$ be a compact boundaryless smooth manifold, and let $$E$$ be the set of all expansive diffeomorphisms of $$M\ .$$
Theorem [Ma]. The $$C^{1}$$-interior of $$E$$ is the set of quasi-Anosov diffeomorphisms of $$M\ .$$
On surfaces , quasi- Anosov diffeomorphisms are Anosov, and since in case $$M$$ has genus larger than 1, $$M$$ does not support Anosov diffeomorphisms, the interior mentioned in the theorem, is, in this case, void. Thus, there are expansive diffeomorphisms which are not approximated by Anosov . Consider now the case $$M=T^{2}\ ,$$ where we do have Anosov diffeomorphisms. Since every expansive homeomorphism $$f$$is conjugate to a linear Anosov diffeomorphism $$l\ ;$$ $$f=hlh^{-1}$$ and according to [Mu] $$h$$ may be $$C^{0}$$-approximated by a diffeomorphism $$g$$ it follows easily, as $$glg^{-1}$$ is Anosov, that $$f$$ has arbitrarily $$C^{0}$$-close Anosov diffeomorphisms. However, it is not known, whether the $$C^{1}$$-closure of the $$C^{1}$$-interior of the expansive diffeomorphisms of $$T^{2}$$ includes all the expansive diffeomorphisms of the 2-torus. In other words: Is every expansive diffeomorphism the $$C^{1}$$-limit of Anosov diffeomorphisms? On the other hand , according to the results in [K], it is possible to conclude that such an expansive diffeomorphism has a dense set of periodic hyperbolic points.
## Expansive flows
We consider flows with no equilibrium points. Such a flow $$\varphi _{t}:M\rightarrow M,t\in R,$$ is expansive if there exist $$\alpha ,\sigma >0,$$ such that if $$x,y\in M,$$ and $$dist(\varphi _{t}(x),\varphi _{\tau (t)}(y))\leq \alpha$$ for every $$t\in R,$$ then $$y=\varphi _{t_{0}}(x)$$ for some $$t_{0},0\leq \left| t_{0}\right| \leq \sigma\ .$$ Here $$\tau :R\rightarrow R$$ is a re-parametrization of the flow through $$y\ ,$$ i.e, a surjective homeomorphism with $$\tau (0)=0\ .$$ This definition is somewhat more complicated than the one for discrete expansive systems as a consequence of the fact that we ask for geometric (instead of kinematic) separation. Important examples of expansive flows are geodesic flows on compact smooth Riemannian manifolds of negative curvature.
We mention below a short list of papers concerning expansive flows:
• R. Bowen, P. Walters. On expansive one-parameter flows. J. Diff Eq. 12(1972) 180-193
• M. Brunella. Expansive flows on Seifert manifolds and on Torus bundles.Bol. Soc. Brasil. Mat. (N.S.) 24(1993),89-104
• M. Brunella. Surfaces of section for expansive flows on three-manifolds.J.Math.Soc.Japan 47(1995), 491-501
• K. Moriyasu, K. Sakai, W. Sun. $$C^{1}-$$stably expansive flows. J. Differential Equations 213(2005) 352-367.
• J. Lewowicz. Lyapunov functions and Stability of Geodesic Flows. Springer Lecture Notes in Math. 1007(1981),463-480.
• M. Paternain. Expansive flows and the fundamental group. Bol.Soc.Brasil. Mat.(N.S.)24(1993), 179-199
• M. Paternain. Expansive geodesic flows on surfaces. Ergodic Theory Dynam. Systems 13(1993),153-165
• R. Ruggiero, V. Rosas. On the Pesin set of expansive geodesic flows in manifolds with no conjugate points Bol.Soc. Brasil. Mat. (N.S.)34(2003), 263-274
• R. Ruggiero. The accessibility property of expansive geodesic flows without conjugate points. Ergodic Theory Dynam. Systems 28(2008), 229-244.
## Non-invertible expansive maps
This section refers to continuous maps $$f$$ of a compact metric space $$M$$ to itself that are not necessarily one-to-one. For those maps, a natural analogue to the notion of expansiveness is positive expansiveness.
A map $$f$$ is positively expansive if $$dist(f^{n}(x),f^{n}(y))\leq \alpha \; ;n\geq 0\ ,$$ implies $$x=y\ .$$ A simple example of such a map is $$f:S^{1}\rightarrow S^{1};\ ,$$ $$f(z)=z^{n},$$ $$n>1,$$ where $$S^{1}$$ is the set of complex numbers $$z$$ of modulus 1.
As in the preceding section we mention a short list of papers concerning, mainly, positively expansive maps.
• E.Coven and W. Reddy. Positively expansive maps on compact manifolds. Lecture notes in Math 819, Springer Verlag, 1980, 96-110
• K. Hiraide. Positively expansive open maps of Peano spaces, Topology and its Appl. 37 (1990), 213-220
• K. Hiraide. Nonexistence of positively expansive maps on compact connected manifolds with boundary, Proc. Amer. Math.Soc. 110 (1990), 565-568
• M.Nasu. Endomorphisms of Expansive systems on compact metric spaces and the pseudo-orbit tracing property. Trans. of the Am. Math Soc. 352(2000),10, 4731-4757
• W. Reddy. Expanding maps on compact metric spaces. Toplogy and its Appl. 13 (1982) 327-334
• D. Richeson and J. Wiseman. Positively expansive dynamical systems. Topology and its Appl. 154(3), (2007), 604-613
• M.Shub. Endomorphisms of compact differentiable manifolds. Amer. J. Math 91 (1969), 175-199.
## References
[ABP] A. Artigue, J. Brum, R. Potrie. Local product structure for expansive homeomorphisms. Toplogy and its Applications, (2008) (To appear).
[B] J.L. Borges. Tigres azules. Obras Completas (3). Emece Editores (1989), 381-388
[FLP}] A. Fathi, F. Laudenbach, V. Poenaru. Travaux de Thurston sur les surfaces. Asterisque (1979)66-67
[F] J. Franks. Anosov Diffeomorphisms.Proceedings of the Symposium in pure mathematics. 14(1970), 61-94
[F,R}] J. Franks, C. Robinson. A quasi-Anosov diffeomorphism that is not Anosov. Trans. Am. Math.Soc. 283(1976), 267-278.
[H] M. Handel. Global Shadowing of pseudo-Anosov diffeomorphisms. Ergodic Theory Dynam. Systems 5(1985)373.377
[Hi] K. Hiraide. Expansive diffeomorphisms of surfaces are pseudo-Anosov. OsakaJ. Math.27(1990), 117-162.
[K] A. Katok. Lyapunov exponents, entropy and periodic orbits of diffeomorphisms. Publ. Marh. IHES 51 (1980).
[L1] J. Lewowicz. Lyapunov functions and Topological Stability. Journal of Diff. Equations. 38(1980) 192-209.
[L2] J. Lewowicz. Persistence in expansive systems. Ergodic Theory Dynam. Systems 3(1983), 567-578.
[L3] J. Lewowicz. Expansive Homeomorphisms of surfaces.Bol. Soc. Bras. Math. 20(1989), 113-133.
[Ma] R. Mañe. Expansive Diffeomorphisms. Lecture Notes in Math.468 (1975), 162-174
[Mu] J. Munkres. Obstructions to the smoothing of piece-wise differentiable homeomorphisms. Ann. of Math. 72(3)(1960), 521-554
[T] W. Thurston. On the geometry and dynamics of diffeomorphisms of surfaces. Bull. Am. Math. Soc. 19 (1988), 417-431
[U] W. Utz. Unstable homeomorphisms. Proc. Am. Math. Soc. 1(1950), 769-774
[V1] J. Vieitez.Three Dimensional expansive homeomorphisms. Pitman Research Notes in Math.285(1993),299-323.
[V2] J. Vieitez. Expansive homeomorphisms and hyperbolic diffeomorphisms on three manifolds. Ergodic Theory Dynam. Systems 16(1996), 591-622.
[W1] P. Walters. Anosov diffeomorphisms are topologically stable.Topology 9(1970), 71-78
[W2] P. Walters. On the pseudo-orbit tracing property and its relation to stability. Lecture Notes in Math. 668 (1978), 231-244.
Internal references
• Eugene M. Izhikevich (2007) Equilibrium. Scholarpedia, 2(10):2014.
• Philip Holmes and Eric T. Shea-Brown (2006) Stability. Scholarpedia, 1(10):1838.
### Internal Reference
[M] J. Milnor. Attractor. Scholarpedia 1(11):1815 (2006),1-9.
• A. Katok, B. Hasselblatt. Introduction to the Modern theory of Dynamical Systems. Encyclopedia of Mathematics and its Applications, 54. Cambridge University Press, Cambridge, 1995. ISBN: 0-521-34187-6 | {"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": 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": 0.932223379611969, "perplexity": 431.66220099198074}, "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-2017-22/segments/1495463612018.97/warc/CC-MAIN-20170529053338-20170529073338-00104.warc.gz"} |
http://images.planetmath.org/node/87512 | # Noetherian ring of infinite Krull dimension
Hi,
In Eisenbud’s Commutative Algebra with a view towards algebraic topology there is an example of a such ring. I have a question concerning one of the steps to find one (section 9.2, exercise 9.6, page 229). I’ll do my best to explain myself if you don’t have the book:
Let $k$ be an algebraic closed field and $R=k[x_{1},x_{2},...,x_{r},...]$ be a polynomial ring in infinitely many variables over $k$, and let $P_{1}=(x_{1},...,x_{{d(1)}})$, $P_{2}=(x_{{d(1)+1}},...,x_{{d(2)}})$, …, $P_{m}=(x_{{d(m-1)+1}},...,x_{{d(m)}}),...$ be an infinite collection of prime ideals made from disjoint subsets of the variables.
Let $U=R-\bigcup_{{m=1}}^{{\infty}}P_{m}$. As we know, this is multiplicative set in $R$ and thus we can form the localization of $U$ at $R$, denoted by $S=R[U^{{-1}}]$.
The idea is to prove that $S$ has infinite Krull dimension. To achieve this, the book uses an exercise: If $I\subset R$ is an ideal and $I\subset\bigcup_{{m=1}}^{{\infty}}P_{m}$, then $I\subset P_{n}$ for some $n$. The problem is that this seems false. Here an example:
$I=(x_{{d(1)-1}},x_{{d(1)+1}})\subset P_{1}\cup P_{2}$ but $I$ is not contained in any of the $P_{i}$. Is there something I am missing? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 19, "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": 0.9671661257743835, "perplexity": 40.62904559194956}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2017-47/segments/1510934806310.85/warc/CC-MAIN-20171121021058-20171121041058-00702.warc.gz"} |
https://www.sawaal.com/simplification-questions-and-answers/to-fill-a-tank-25-buckets-of-water-is-required-nbsp-how-many-buckets-of-water-will-be-required-to-fi_2047 | 49
Q:
# To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of bucket is reduced to two-fifth of its present ?
A) 52.5 B) 62.5 C) 72.5 D) 82.5
Explanation:
Let the capacity of 1 bucket = x.
Then, capacity of tank = 25x.
New capacity of bucket = (2/5)x
$\inline \therefore$Required no of buckets = 25x/(2x/5) = 62.5
Q:
If
A) 30° B) 90° C) 60° D) 45°
Explanation:
0 301
Q:
The value of
A) 5 B) 0 C) 2-22 D) 2
Explanation:
1 582
Q:
Solve the following
113 × 87 =?
A) 9831 B) 10026 C) 10169 D) 10000
Explanation:
0 925
Q:
In ΔABC measure of angle B is 90 deg. If tanA = 12/5, and AB = 1cm, then what is the length (in cm) of side BC?
A) 2.6 B) 2.4 C) 1.5 D) 2
Explanation:
0 142
Q:
ΔXYZ is right angled at Y. If m∠X = 60 deg, then find the value of (cotZ + 2)
A) (2√2+1)/2 B) √3+2 C) (√6+1)/√3 D) (2√2+√3)/2
Explanation:
0 608
Q:
If ,then which of the following is correct?
A) a=2197 B) a>2197 C) a<2197 D) a<1728
Explanation:
0 734
Q:
Find the number of solutions a pair of linear equation (4x-9y+13=0 and 2x+3y-13=0) have.
A) 1 B) 2 C) 10 D) infinite
Explanation:
0 631
Q:
Find the area (in sq units) of the triangle formed by lines x - 3y = 0, x – y = 4 and x + y = 4.
A) 1 B) 2 C) 3 D) 4 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.879658043384552, "perplexity": 3035.5345424235707}, "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/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00671.warc.gz"} |
https://mathoverflow.net/questions/288259/how-to-show-that-the-following-function-isnt-a-polynomial-over-q/288288 | # How to show that the following function isn't a polynomial over Q?
Enumerate the rationals as $b_1,b_2,\dots$ and define the (set) function: $$f(x) = (x-b_1)^2 + (x-b_1)^2(x-b_2)^2 + \dots.$$ At any particular $x$, only finitely many terms are non zero so this is perfectly well defined as a (set) function but surely, it is not equal to any polynomial! (or is it?) How do I show that there is no polynomial $p(t) \in \mathbb Q[t]$ such that $p(x) = f(x)$ for all $x \in \mathbb Q$?
If $f(x)$ were defined as $(x-b_1) + (x-b_1)(x-b_2) + \dots$, then this question is not so hard. If $p(x)$ has degree $n$, then testing on $b_1,\dots,b_n$ would show that $p(x)$ is necessarily $(x-b_1) + (x-b_1)(x-b_2) + \dots + (x-b_1)\dots(x-b_n)$ but then $x=b_{n+1}$ derives a contradiction.
I don't know how to adapt this approach. Trying to guess the polynomial seems hard even if we think $p(x)$ is degree $1$.
I posted a follow up to this question here: (Variation of an old question) Are these functions polynomials?.
• It is even conceivable that the question of whether $f(x),\mathcal{O}$ (where $\mathcal{O}$ represents some ordering in the enumeration of the rationals,) is expressible as a polynomial, depends on the choice taken for $\mathcal{O}$. So a good question is whether there is any $\mathcal{O}$ and $p(t)\in\Bbb Q[t]$ for which $p(x) = f(x)$ for all rational $x$. Dec 11 '17 at 19:27
• It's easy to see that then $f_1=f/(x-b_1)^2$ would have to be a polynomial too, but of lower degree. Then again $f_2=(f_1-1)/(x-b_2)^2$ would also be a polynomial of lower degree yet... and so on. Dec 11 '17 at 21:08
• There may be a flow in my scheme: I'm no longer sure that it's obvious that $b_1$ is a zero of $f/(x-b_1)$ and thus that $f/(x-b_1)^2$ is indeed a polynomial... Dec 11 '17 at 21:37
• @ChristianRemling, but (modulo my comment) this doesn't seem to show that the polynomial $p/(x - b_1)$ has a root. That is, we can divide $f$ by $x - b_1$ 'canonically' to obtain a function $g$ (EDIT: oops, sorry, not your $g$), and we can divide $p$ by $x - b_1$ canonically to obtain a polynomial $q$, but it isn't a priori clear to me that $g = q$ (in other words, that these two kinds of division preserve equality). Dec 11 '17 at 23:52
• I agree that $f/(x - b_1)$ is not always well defined for a function vanishing at $b_1$. In this case, though, each term in the defining sum for $f$ is a polynomial divisible by $x - b_1$, and so there is a natural sum of polynomials that deserves to be called $f/(x - b_1)$. Dec 12 '17 at 1:15
For each positive integer $n$ and any rational $x$, we have
$$f(x)\geq (x-b_1)^2(x-b_2)^2\dots(x-b_n)^2.$$ For large $x$, we then have $f(x)\gg x^{2n}$, which implies that if $f$ is a polynomial, it must have degree $≥2n$.
• Just curious: Do you see a way to prove something similar if we replace $\mathbb Q$ by $\overline{ \mathbb F}_p$ ? Dec 11 '17 at 19:50
You can adapt the same approach as follows: Say $p_n(x)$ is an $n$-th degree polynomial matching $f(x)$ at all rational points, then in particular, $$p_n(b_1) = 0\\ p_n(b_2) = (b_2-b_1)^2 \\ \vdots\\ p_n(b_k) = \sum_{i\in\Bbb Z+, i<k}\prod_{j\in\Bbb Z+, j\leq i} (b_k-b_i)^2$$ For a given fixed sequencing of the rationals as $b_1, b_2, \cdots$, and for any given $k$, the latter expression is just some fixed rational number.
So $p(n)$ is fixed by its values at $b_1 \ldots b_n$, and now consider $-f(b_{n+1}-p_n(b_{n+1}))$. Since all the terms past the $n+2$ term in $f(b_{n+1})$ are zero, $$f(b_{n+1}) = p_n(b_{n+1}) + \prod_{i\leq n}(b_{n+1}-b_i)^2 > p_n(b_{n+1})$$ which contradicts the statement that $f$ matches $p_n$ at all rationals.
• why do $n$ values determine a degree $n$ polynomial? Don't you need $n+1$ terms? Dec 11 '17 at 20:20
• Do you really mean to consider $-f(b_{n+1} - p_n(b_{n+1}))$ and not $-f(b_{n+1}) - p_n(b_{n+1})$? Is the idea to use that $f$ is always positive? How do you show that the difference between f(b_{n+1}) and $p_n(b_{n+1})$ is that expression? Dec 11 '17 at 20:29
"If $f(x)$ were defined as $(x−b_1) + (x−b_1)(x−b_2) + …,$ then this question is not so hard."
Does taking the derivative of the given function get you to this simpler case?
"At any particular $x$, only finitely many terms are non zero"
But at irrational values of $x$, none of the terms are zero. Now, you may respond "But I'm talking about it over $\mathbb Q.$" But unless you're going to claim that $f$ isn't continuous, if you take a sequence of irrationals approaching a rational, the function evaluated at that rational must be the limit of the function evaluated at those irrational numbers. You could also construct a sequence of rationals approaching $b_1$ such that there exists some $\epsilon > 0$ such that for all $x$ in the sequence, $f(x) > \epsilon$. If $f(x)$ is continuous, $f(b_1)$ must be $> 0$, but clearly by the definition of $f$, $f(b_1) = 0$. I suppose you'll still have to argue that $f$ must be continuous, but that should follow from it being a polynomial, even with the restriction to $\mathbb Q$.
• It is not clear that the function is continuous, much less continuously extendible to $\mathbb R$, or differentiable. (Also, the termwise derivative doesn't match this easier function.) Even if the function extended continuously, your argument only shows that the same formula doesn't naïvely define the extension. Dec 11 '17 at 23:48
• It's a proof by contradiction. I'm certainly not claiming that it is continuous, merely that if it were defined by a polynomial over Q, then it would be continuous. Dec 12 '17 at 0:04 | {"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": 0.9536929130554199, "perplexity": 101.50568356471258}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320305341.76/warc/CC-MAIN-20220128013529-20220128043529-00322.warc.gz"} |
http://physics.stackexchange.com/questions/54299/semi-conductor-band-gap-and-deformation-potential | # Semi-conductor band-gap and deformation potential
Submitting a semi-conductor to stress leads to a deformation in the energy-bands, roughly described by:$$H_{ij} = {\cal{D}}_{ij}^{\alpha\beta}\;\epsilon_{\alpha\beta}$$ $\epsilon$ being the strain (linked to the stress by Hooke's law), $H$ the perturbation Hamiltonian to the Hamiltonian describing a stress-free semiconductor, $i,j$ being indexing the energy level of the previous "free" Hamiltonian. Still, I lack intuition regarding the apparition of this term, it seems that compression enlarges the band-gap whereas dilation tightens it. Why do we have this behaviour? I have tried thinking on electrostatic arguments, the potential decreasing as $r^{-1}$ we do have an increase of energies in $r \rightarrow \alpha r$ for $\alpha < 1$ or also seeing the dilatation as a renormalization group transformation, basically going to a coarser grain (although there probably other pertinent length scales in an atomic lattice (spreading of the electronic orbitals, ...) which would make this argument wobbly). Cutting to the point, what is your hand-waved way of seeing it? Books on the subject of deformation potential don't really seem to offer intuition on it, more numerical values for specific materials.
-
Isn't that the only term you can write down which is linear in the strain? The assumption would be that higher order terms $\sim\epsilon^2$ are negligible unless for some reason $\mathcal{D}^{\alpha\beta}_{ij}$ vanished. Just a guess. – Michael Brown Feb 18 '13 at 14:48
## 1 Answer
Well, possibly the simplest argument (I have no idea if this is correct) is that compressing the material reduces the configuration space available for the electrons, and increasing confinement means increased energy (differences) in quantum mechanics. Consider just the old particle-in-a-box problem, the energy levels scale as $E_n \sim \frac{n^2}{L^2}$, where $L$ is the size of the box, so a smaller box means bigger spacings between energy levels.
-
Valid quantitative argument, although you're not taking into account that our electrons are not free in this box, and sweeping under the rug all the details of the interactions (another way of explaining my doubts in "renormalization group" arguments. – Learning is a mess Feb 18 '13 at 15:52
The bound electrons aren't free, but the conduction electrons are. – KDN Feb 18 '13 at 16:30
@Learningisamess, I guess you mean "qualitative"? It's certainly not a quantitative argument. The particle in a box is not meant to even be an approximation to the problem you are interested in. I am simply trying to illustrate a completely general feature of quantum mechanics that may be relevant to the problem at hand: greater confinement implies greater energy (differences). Weak interactions won't change this behaviour much; you can imagine making a mean-field approx. where increasing the electron density just increases the single-particle effective potential due to Coulombic repulsion. – Mark Mitchison Feb 18 '13 at 16:37 | {"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": 0.8117648363113403, "perplexity": 695.582970216048}, "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/1432207928078.25/warc/CC-MAIN-20150521113208-00291-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/454801/copy-tabularx-x-column-as-new-vertically-centered-column | # Copy tabularx X column as new vertically centered column
My question is very similar to Vertical alignmnent in tabularx X column type and there seems to be many similar questions so hopefully I haven't missed the answer somewhere.
However, I would like to create a new column that is a copy of tabularx X column, with the difference being the new column Y is an m column compared to a p column (which X is by default as described in the docs).
\tabularxcolumn - The default denition of X is p{#1}.
\def\tabularxcolumn#1{p{#1}}
So I would like to define a new column type like
\newcolumntype{Y}{>{\centering\arraybackslash}X}
and then convert the Y column type to m instead of p as it would currently be. That way I can leave X as it's original definition. If I use
\renewcommand{\tabularxcolumn}[1]{>{\small}m{#1}}
Then X is changed to an m column which is not desired. I would need something like
\renewcommand{Y}[1]{>{\small}m{#1}}
But that doesn't work.
Thanks for any help,
• Do you want/need both X and Y columns in a single table?
– Werner
Oct 11, 2018 at 18:00
• Potentially, I would like to be able to have a p type X column and a m type X column if possible. Oct 11, 2018 at 18:02
you need to patch in a second X-like columntype, this just duplicates the definition of X so you can separately specify \tabularxcolumn for X and \tabularxycolumn for Y
\documentclass[a4paper]{article}
\usepackage{etoolbox,tabularx}
\tracingtabularx
\makeatletter
\newcolumntype{Y}{}
\def\tabularxycolumn#1{m{#1}}
\def\TX@newycol{\newcol@{Y}[0]}
\patchcmd\TX@endtabularx
{\expandafter\TX@newcol}%
{\expandafter\TX@newycol\expandafter{\tabularxycolumn{\TX@col@width}}%
\expandafter\TX@newcol}
{}
{}
\patchcmd\TX@endtabularx
{\def\NC@rewrite@X}%
{\def\NC@rewrite@Y{\NC@rewrite@X}%
\def\NC@rewrite@X}
{}
{}
\makeatother
\begin{document}
\begin{tabularx}{6cm}{XXc}
aa aaa aaa aaa aaa aaa&
bb bb bb bb bb bb bb bb bb bb bb bb bb bb b &
aa aaa
\end{tabularx}
\begin{tabularx}{6cm}{XYc}
aa aaa aaa aaa aaa aaa&
bb bb bb bb bb bb bb bb bb bb bb bb bb bb b &
aa aaa
\end{tabularx}
\begin{tabularx}{6cm}{YYc}
aa aaa aaa aaa aaa aaa&
bb bb bb bb bb bb bb bb bb bb bb bb bb bb b &
aa aaa
\end{tabularx}
\end{document}
• I thought it worked but I was looking at the PDF (which compiled fine) but there were errors. I am using > pdfLaTeX Version 3.14159265-2.6-1.40.18 (MiKTeX 2.9.6500 64-bit). My original use was in knitr, but it failed there, and then I went to TexStudio, but it failed there as well. I started with a blank tex document and copied and pasted your code. Any thoughts? Thanks! Oct 19, 2018 at 23:02
• @Prevost sorry I had omitted another place to patch in the Y version of X columns, the original wasn't counting Y columns in its count of the number of active X columns. code and result image updated Oct 20, 2018 at 0:02
• That's great compiles no errors! Could that code be theoretically created from just your docs? Oct 20, 2018 at 0:19
• @Prevost not from the user level documentation, it's definitely an extension of the design not just "using tabularx". Perhaps it could be derived from the documented code, although actually I only looked at the docstripped code with comments removed, perhaps that's why I missed one place the first time:-) Oct 20, 2018 at 8:12 | {"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": 0.8858765959739685, "perplexity": 1925.6036098474005}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103989282.58/warc/CC-MAIN-20220702071223-20220702101223-00388.warc.gz"} |
http://math.stackexchange.com/questions/164526/closed-form-expression-of-the-following-double-integral | # Closed-form expression of the following double integral
How can I find closed-form expression of the following double integral $$\int_{0}^{\pi/4}\int_{0}^{\infty}{{\rm d}r\,{\rm d}\phi \over u_{1}^{2} + u_{2}^{2}\,r + 2\,u_{1}u_{2}\,\sqrt{r\,}\,\cos\left(\phi\right)}\ {\large ?}$$ Please help me as soon as you can.
-
Are you sure it is convergent? The integral with respect to $r$ diverges at infinity. – Sasha Jun 29 '12 at 12:44
Are you missing a factor of like $1/r$ or $1/r^2$? – KennyTM Jun 29 '12 at 17:49
Hence the answer is: there is a closed-form expression, which is $+\infty$. – Did Jul 1 '12 at 20:36 | {"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": 0.9791617393493652, "perplexity": 542.2392666299686}, "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-2015-06/segments/1422115862207.44/warc/CC-MAIN-20150124161102-00186-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://pballew.blogspot.com/2010/12/more-almost-binomial-distributions.html | ## Thursday, 23 December 2010
### More "Almost Binomial" Distributions
In my recent post I illustrated the extension of the binomial to a Multinomial Distribution. In a similar way, the geometric distribution and the Pascal (aka the Negative Binomial) Distribution are very much like special cases of the binomial.
I will illustrate each with a simple probability example. In a limited version of the game of "greedy pig" you roll a die as many times as you wish each turn and you add the points on the top of the die to your score for that turn...but... if you roll a one, your turn ends and you lose all the points you have earned for that round. One might inquire, what is the probability that you could roll the die n times without rolling a one. Since the probability on each roll is the same, we could handle this using the binomial (or multinomial) distribution with n trials, p=5/6, and the number of successes also equal to n. For n=5 for example, we get (using the notation established in that blog) which simplifies to just (5/6)5 .
But a slightly different question might be, what is the probability that our first failure (rolling a one) would occur on the sixth roll. This is asking for the probability that the first five rolls succeed, and then the final roll is a failure. This is the general model for a geometric distribution. The reason it is called a geometric distribution is clear if you calculate the probability of the first failure happening on the first, second, etc rolls.
Roll...1....2.......3........4...
Prob...1/6..5/36...25/216 ..5^3/6^4..
notice that each probability is the previous probability multiplied by a constant ratio of 5/6. The terms for a geometric sequence (which must sum to one to be a probability distribution......check)
In general, if the probability of failure is q = 1-p.. then the probability of the first failure occurring on the nth trial is given by (p)n-1(q)
It often surprises students that the mean for such a distribution is 1/q where q is the probability of a failure. OK before I confuse someone.. the geometric distribution is sometimes described as the number of trials to the first success, so you may see the expected or mean value as 1/p. In any event, if the probability of an event happening (whether you call it success or failure) is p, the expected number of trials before it happens is 1/p.
Now if you are really clever you can figure out how to do the next problem without me, but let's walk through it anyway, (hey...it's MY blog).
Suppose instead, you could keep rolling until you had three rolls of one..... sort of "three strikes and you're out." Now what is the probability that the third strike comes on the tenth roll.
The idea of course, a collection of 9 rolls with 2 failures anywhere in the string, and then a third failure on the tenth roll. To get the probability of all the possible ways to get 7 successes and 2 failures in the first nine rolls is a straight binomial (multinomial) probability problem.
We just multiply this by a failure on the tenth roll and we have the probability we seek. Since we have a couple of "failures" in that (1/6)^2, we might as well just up it to a three and be done. The final probability is
If you would like to experiment with these distributions, I came across a nice experimental applet here
This experiment uses the trials to k successes instead of failures, and so p and q are switched here (and it seems I could only adjust these in .05 increments). This is a nice routine and you can simulate trials by clicking on the "step" button to see how many trials it took to get three successes.
This applet is part of a nice virtual laboratory created by Kyle Siegrist of the Department of Mathematical Sciences at the University of Alabama in Huntsville. There is lots of nice stuff. See the home page here.
When we deal with integer numbers of failures this is called a Pascal Distribution after Blaise Pascal. It can be extended to any real and is then called a Polya Distribution, after George Polya. This has application for events which are very rare, but related to each other, such as hurricanes. Both are special cases of the general Negative Binomial Distribution. | {"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": 0.9353320598602295, "perplexity": 358.7145910548953}, "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-2020-29/segments/1593655891654.18/warc/CC-MAIN-20200707044954-20200707074954-00114.warc.gz"} |
http://mathhelpforum.com/pre-calculus/129558-vectors.html | 1. ## vectors
I have some questions on a paper that I am typing up, and I've looked through my text book and I cannot find this, or I just don't know where to look.
-Explain how to write a vector in terms of it's magnitude and direction.
2. ## Magnitude and direction of a vector
Hello Chinnie15
Originally Posted by Chinnie15
I have some questions on a paper that I am typing up, and I've looked through my text book and I cannot find this, or I just don't know where to look.
-Explain how to write a vector in terms of it's magnitude and direction.
I'm not sure that there is any particular way to write a vector in terms of its magnitude and direction.
You can describe a vector in terms of its magnitude and direction by giving several examples; for instance:
1. The displacement vector of a point $B$ from a point $A$ can be written as $\vec{AB}$ and would be described in terms of the distance $AB$ (in suitable units) and the angle that the line segment $AB$ measured from a fixed direction. E.g. $AB = 5$ units in a direction making an angle of $30^o$ with the direction of the $x$-axis.
2. The velocity vector of a moving body can be described in terms of the speed of the body and the direction of its motion at a particular instant. E.g. a shell is fired in a north-easterly direction at $300$ m/sec at an angle of $45^o$ above the horizontal.
... and so on. | {"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": 0.8471003174781799, "perplexity": 129.61544527386482}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171900.13/warc/CC-MAIN-20170219104611-00412-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://proceedings.mlr.press/v37/yi15.html | Binary Embedding: Fundamental Limits and Fast Algorithm
Xinyang Yi, Constantine Caramanis, Eric Price ;
Proceedings of the 32nd International Conference on Machine Learning, PMLR 37:2162-2170, 2015.
Abstract
Binary embedding is a nonlinear dimension reduction methodology where high dimensional data are embedded into the Hamming cube while preserving the structure of the original space. Specifically, for an arbitrary N distinct points in \mathbbS^p-1, our goal is to encode each point using m-dimensional binary strings such that we can reconstruct their geodesic distance up to δuniform distortion. Existing binary embedding algorithms either lack theoretical guarantees or suffer from running time O(mp). We make three contributions: (1) we establish a lower bound that shows any binary embedding oblivious to the set of points requires m =Ω(\frac1δ^2\logN) bits and a similar lower bound for non-oblivious embeddings into Hamming distance; (2) we propose a novel fast binary embedding algorithm with provably optimal bit complexity m = O(\frac1 δ^2\logN) and near linear running time O(p \log p) whenever \log N ≪δ\sqrtp, with a slightly worse running time for larger \log N; (3) we also provide an analytic result about embedding a general set of points K ⊆\mathbbS^p-1 with even infinite size. Our theoretical findings are supported through experiments on both synthetic and real data sets. | {"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": 0.937036395072937, "perplexity": 1088.1632089096424}, "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-2019-22/segments/1558232256163.40/warc/CC-MAIN-20190520222102-20190521004102-00473.warc.gz"} |
http://blog.jpolak.org/?tag=invariant-theory | # An Example Using Chevalley Restriction
Here is a classic problem of geometric invariant theory: let $G$ be a reductive linear algebraic group such as $\mathrm{GL}_n$ and let $\mathfrak{g}$ be its Lie algebra. Determine the invariant functions $k[\mathfrak{g}]^G$, where $G$ acts on $\mathfrak{g}$ via the adjoint action. This problem is motivated by the search for quotients: What is the quotient $\mathfrak{g}/G$? Here, the action of $G$ on $\mathfrak{g}$ is given by the adjoint action. More explicitly, an element $g\in G$ acts via the differentiation of $\mathrm{Int}_g$, where $\mathrm{Int}_g$ is conjugation by $g$ on $G$.
For simplicity, we will stay in the realm of varieties over an algebraically closed field $k$ of characteristic zero.
First, we should ask:
What should $\mathfrak{g}/G$ even 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": 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": 0.9679450392723083, "perplexity": 71.51256843617902}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2018-39/segments/1537267155413.17/warc/CC-MAIN-20180918130631-20180918150631-00483.warc.gz"} |
http://blog.decatech.com/how-to-xgek/plus-minus-latex-cfa35c | This is when you state whether a mathematical quantity is either positive or negative. Latex indicator function; Latex plus or minus symbol; Latex symbol for all x; Latex symbol exists; Latex symbol not exists; Latex horizontal space: qquad,hspace, thinspace,enspace; Latex square root symbol; Latex degree symbol; LateX Derivatives, Limits, Sums, Products and Integrals; Latex copyright, trademark, registered symbols; Latex euro symbol Markdown. This is in math. Diesen Post per E-Mail versenden BlogThis! List of LaTeX mathematical symbols. At what height range would you consider a Manlet? how to do the coding with plus minus sign infront of square root? mp When I enter \text{3+2} into the textbox, the preview of the textbox shows plus, but when I click OK the textbox shows 3-2. ~ Thicker line for minus sign and plus sign Typical to Microsoft users, SharePoint seems to be tailored rather to a casual user (mouse interaction) than a power user. begin{tabular}...end{tabular}, Latex horizontal space: qquad,hspace, thinspace,enspace, LateX Derivatives, Limits, Sums, Products and Integrals, Latex copyright, trademark, registered symbols, How to write matrices in Latex ? No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. 2. How to place +/- plus minus operator in text annotation of plot (ggplot2)? The command \section{}marks the beginning of a new section, inside the braces is set the title. The the minus-plus sign, on the other hand, is represented using \\mp; it looks like this: ∓ . Mit der Ausgabe: Plusminus: \pm Minusplus: \mp. This is because a commission is billed by the home on each bet. Share. Latex Kurs Sonntag, 16. It can be used on some websites like Stack Overflow or to write documentations (essentially on GitHub). What is Christmas in July and how did it start? Minus Or Plus Latex. The usual plus-minus symbol is \pm. It indicates a choice between using the plus sign or the minus sign, with two unique solutions. Vote. pm Means a quantity of same magnitude can have either a positive or a negative value. How to type Plus/minus sign ± Hold down the ALT key and type 0177 on the keypad. minus You can choose from a variety of units. LaTeX macro for siunitx x plus/minus y units (requires package siunitx) - sipm_macro.tex matrix, pmatrix, bmatrix, vmatrix, Vmatrix, Horizontal and vertical curly Latex braces: \left\{,\right\},\underbrace{} and \overbrace{}, How to get dots in Latex \ldots,\cdots,\vdots and \ddots, Latex symbol if and only if / equivalence. I don't like how \pm looks. Follow edited Jul 20 '15 at 8:20. Just have to say that -- and --- are not math symbols in LaTeX. If you want it to be with a minus at the top, i.e., minus plus is \ mp. Commands to organize a document vary depending on the document type, the simplest form of organization is the sectioning, available in all formats. etc. These are dashes and should be used in text. LaTeX deals with the + and − signs in two possible ways. Follow 134 views (last 30 days) amanina abdul rahman on 15 Oct 2015. (adsbygoogle = window.adsbygoogle || []).push({}); ... (0.5,0.5, '$\pm$', 'interpreter', 'latex') If you need the minus on top and the plus underneath then \mp. Your email address will not be published. Many LaTeX commands take a length as an argument. Sign in to answer this question. It does not have the excellent online possibilities and tools to edit Latex documents overleaf is providing (to my knowledge), but it provides a robust environment for versioning, and collaborative versioning with branching, merging, etc. How to change the font of math operators, e.g. It will give me the energy and motivation to continue this development. Sorry if this is an obvious question, I've only been using latex for a few weeks. Lengths come in two types. Knowledge base dedicated to Linux and applied mathematics. Textbox changes plus to minus when using Latex. Do you mean something like. When two maths elements appear on either side of the sign, it is assumed to be a binary operator, and as such, allocates some space to either side of the sign. R - Combine non numerical and numerical data together in the same cell in a dataframe. How To Write Plus Minus Symbol In Latex. An online LaTeX editor that's easy to use. 14 Lengths. documentclass{article} usepackage{amsmath} begin{document} $1.23substack{+0.4 \ -0.5}$,pb end{document} (I've been asked about this for siunitx, but have never really worked out what it actually means or a good interface for an 'automated' approach.) Hold down the Shift and Option keys and press =, ± or ± More symbols in the category: How to. The LaTeX commands assume that you are in normal text mode. croober shared this problem 9 months ago . Currently, I'm using these lines of code but I would like to obtain plus-minus sign (±) values preceding a . (Plain TeX calls this a dimen. How to write Latex minus or plus symbol: \mp. A rigid length such as 10pt does not contain a plus or minus component. All other units are converted to the point by a fixed ratio.Here are some less common units. Latex to render mathematical and scientific writing. From OeisWiki. Sign in to comment. The minus–plus sign (also minus-or-plus sign), ∓, is generally used in conjunction with the ± sign, in such expressions as x ± y ∓ z, which can be interpreted as meaning x + y … Blog template built with Bootstrap and Spip by Nadir Soualem @mathlinux. The plus-or-minus sign in LATEX is represented using \pm; it looks like this: ±. Export (png, jpg, gif, svg, pdf) and save & share with note system When rendered, the \setminus command looks identical to \backslash , except that it has a little more space in front and behind the slash, akin to the LaTeX sequence \mathbin{\backslash} . What I don‘t like is that it is in the cloud, which makes me hesitating when it comes to hosting documents there with sensitive/confidential content. Then it's a good reason to buy me a coffee. plus and ... Accounting Blog: What Is Accounting Symbol For Plus Or Minus. (adsbygoogle = window.adsbygoogle || []).push({}); All the versions of this article: This is good is you are still a beginner participant. Is there a way to alter the thickness of the line for the minus sign (and, for matching purposes, the plus sign also)? For Matlab plots, how do I place plus or minus sign as X tick labels in a Matlab plot? There are no approved revisions of this page, so it may not have been reviewed. Plus-Minus. And you need to buy a monthly license to be able to use it to upload and collaborate. LaTeX: What is the command for the plus-minus sign? Latex Barplot one column based on another column from the same dataframe. As you said, the plus or minus sign is quite literally called the plus or minus sign. Show Hide all comments. In math mode there is just one type of minus. Ask Question Asked 5 years ago. Categories Here are the most common ones.The point is the default unit and 1pt is the default length. Open an example in Overleaf Save my name, email, and website in this browser for the next time I comment. Below is the Alt code keyboard shortcut for inserting the plus minus sign.If you are new to ALT codes and need detailed instructions on how to use them, please read How to Use ALT Codes to Enter Special Characters.. For the the complete list of the ASCII based Windows ALT Codes, refer to Windows ALT Codes for Special Characters & Symbols. This website was useful to you? It’s a very simple language that allows you to write HTML in a shortened way. Hold down the Shift and Option keys and press = LaTeX decides whether it's unary or binary and adds proper spacing to distinguish these variants. Overleaf: seems to be the best collaborative tool on the market. Required fields are marked *. Markdown file extension is .md. Jump to: navigation, search. The plus-or-minus sign in L A T E X is represented using \\pm ; it looks like this: ± . Your email address will not be published. As commonly found in square roots such as $$\sqrt[2]{625}= +-25$$. September 2012. latex plusminus Das Plusminus Zeichen $\pm$ gibt es auch als Minusplus Zeichen $\mp$. Last modified 07/24/2020. And there is no plus OR minus symbol in MATLAB, that is, a symbol that indicates EITHER plus or minus. The alternative way is a sign designation. The $'s around a command mean that it has to be used in maths mode, and they will temporarily put you into maths mode if you use them. Section numbering is automatic and can be disabled. All Rights Reserved. Improve this answer. In Twitter freigeben In … 0. All LaTeX units are two-letter abbreviations. plus All the predefined mathematical symbols from the T e X package are listed below. A length is a measure of distance. An interesting discussion on self–hosted online tools is here on Stackexchange: Git(lab) is my favorite tool so far. online LaTeX editor with autocompletion, highlighting and 400 math symbols. The usual plus or minus character is \ pm. LaTeX notation In the LaTeX typesetting language, the command \setminus [8] is usually used for rendering a set difference symbol, which is similar to a backslash symbol. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … , How to write Latex plus or minus symbol: \pm TeX - LaTeX: Most of the times the minus sign in my documents is so thin and so short that I have a hard time seeing it on a printed copy. Everything you need to … )A rubber length (what Plain TeX calls a skip) such as as with 1cm plus0.05cm minus0.01cm can contain either or both of those components. Answered: Walter Roberson on 17 Oct 2015 Accepted Answer: Walter Roberson 1 Comment. How to write number sets N Z D Q R C with Latex: \mathbb, amsfonts and \mathbf, How to write angle in latex langle, rangle, wedge, angle, measuredangle, sphericalangle, Latex numbering equations: leqno et fleqn, left,right, How to write a vector in Latex ? Even if I like Sharepoint very much as a content management platform, it lacks the branching and merging features and the possibilities to interact with the repository from the command line Git offers. If you want it with the minus on top, meaning minus-plus, it’s \mp. 0. See Also. plus or minus ∓ \mp: minus or plus ... detexify: applet for looking up LaTeX symbols by drawing them. https://www.youtube.com/channel/UCmV5uZQcAXUW7s4j7rM0POg?sub_confirmation=1 How to type Plus-Minus & Minus-Plus symbol in Word Excel ... How to annotate() ggplot with latex. The … \boxed, How to write table in Latex ? 0 ⋮ Vote. With hundreds of games literally at your fingertips you can attempt almost any of them in one session. Thank you ! ... or LaTex, docs here and here: \pm is used for plus/minus . Home > Latex > FAQ > Latex - FAQ > Latex plus or minus symbol. IT Jobs Dubai UAE © 2021. The most common is as a binary operator. First, we introduce the LaTeX measurement units. minus-plus sign (plural minus-plus signs) (mathematics) the symbol ∓, meaning "minus or plus", used alongside the plus-minus sign to show that a negative value is to be taken where the positive value is indicated by the plus-minus sign, and vice versa (as in (x ± 1) / (x ∓ 2), which means (x + 1) / (x - 2) and (x - 1) / (x + 2)). You can add spaces manually in GeoGebra via \, \! Eingestellt von Safra um 09:17. If it were self–hosted, I think I would definitively like my company to provide it as a standard tool. SharePoint is not very good in this context to my taste. Is there an easy way to type a plusminus symbol where the plus and minus are not touching? Not a Problem. \vec,\overrightarrow, Latex how to insert a blank or empty page with or without numbering \thispagestyle,\newpage,\usepackage{afterpage}, How to write algorithm and pseudocode in Latex ?\usepackage{algorithm},\usepackage{algorithmic}, How to display formulas inside a box or frame in Latex ? I have been using LaTeX for ten years as a student and graduate student in mathematics. 'M using these lines of code but I would definitively like my company to provide it as standard... Or negative sign and plus sign or the minus sign infront of square?. Indicates a choice between using the plus or minus sign is quite literally called the plus or minus and., pdf ) and save & share with note system do you mean something....: plusminus: \pm is used for Plus/minus ( ± ) values preceding a were self–hosted I... Negative value save & share with note system do you mean something like -- - are not touching Ausgabe. -- - are not math symbols the plus-or-minus sign in L a T E X is represented \\mp! Mathematical quantity is either positive or a negative value the the minus-plus sign, with two unique solutions upload collaborate... Way to type a plusminus symbol where the plus sign an online LaTeX editor that 's easy use. On Stackexchange: Git ( lab ) is my favorite tool so.! In July and how did it start 2 ] { 625 } =$! Overleaf: seems to be the best collaborative tool on the market ) and save & with. A casual user ( mouse interaction ) than a power user key and type 0177 on the other,. Matlab, that is, a symbol that indicates either plus or minus character is \.. Currently, I think I would like to obtain plus-minus sign ( ± ) values preceding.... 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It will give me the energy and motivation to continue this development a plusminus symbol where the plus or! An obvious question, I 've only been using LaTeX for a few weeks as an argument such \$... Continue this development and should be used on some websites like Stack Overflow or to write HTML in shortened! Consider a Manlet were self–hosted, I 'm using these lines of code I... Math symbols in LaTeX is represented using \\mp ; it looks like this:.. Template built with Bootstrap and Spip by Nadir Soualem @ mathlinux non numerical and numerical data together in same. 17 Oct 2015 Accepted Answer: Walter Roberson 1 Comment I 've only been using for... Minus component and collaborate +/- plus minus sign infront of square root marks beginning. Roberson on 17 Oct 2015 of a new section, inside the braces set. Plots, how do I place plus or minus component type a plusminus symbol where the plus minus... Type of minus HTML in a Matlab plot fixed ratio.Here are some less common.., gif, svg, pdf ) and save & share with note system do you mean something like symbol! To change the font of math operators, e.g plus-or-minus sign in LaTeX is using... Days ) amanina abdul rahman on 15 Oct 2015 Accepted Answer: Walter Roberson on 17 Oct..
Japanese Leadership Ww2, Liveaboard Cocos Island, Detailed Lesson Plan In Math Grade 1 Counting Numbers, Detailed Lesson Plan In Math Grade 1 Counting Numbers, Speechify Crossword Clue, Vw Atlas Used Canada, Golden Retrievers Needing New Homes, | {"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": 0.860044538974762, "perplexity": 2855.996731734082}, "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/1627046154457.66/warc/CC-MAIN-20210803092648-20210803122648-00269.warc.gz"} |
https://repository.lboro.ac.uk/articles/Effects_of_bandwidth_limitations_on_the_localized_state_distribution_calculated_from_transient_photoconductivity_data/9560429 | ## Effects of bandwidth limitations on the localized state distribution calculated from transient photoconductivity data
2009-01-16T11:11:16Z (GMT) by
The possible effects of experimental bandwidth limitation on the accuracy of the energy distribution of the density of localized states (DOS) calculated from transient photoconductivity data by the Fourier transform method is examined. An argument concerning the size of missing contributions to the numerical Fourier integrals is developed. It is shown that the degree of distortion is not necessarily large even for relatively small experimental bandwidths. The density of states calculated from transient photodecay measurements in amorphous arsenic triselenide is validated by comparing with modulated photocurrent data. It is pointed out that DOS distributions calculated from transient photoconductivity data at a high photoexcitation density are valid under certain conditions. This argument is used to probe the conduction band tail in undoped a-Si:H to energies shallower than 0.1 eV below the mobility edge. It is concluded that there is a deviation in the DOS from exponential at about 0.15 eV below the mobility edge. | {"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": 0.9420344829559326, "perplexity": 1086.402806064851}, "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-2020-34/segments/1596439739046.14/warc/CC-MAIN-20200813132415-20200813162415-00562.warc.gz"} |
http://slideplayer.com/slide/1424008/ | # THE ATMOSPHERE.
## Presentation on theme: "THE ATMOSPHERE."— Presentation transcript:
tHE ATMOSPHERE
Earth’s Atmosphere The Earth’s atmosphere is a thin layer of air that forms a protective covering around the planet. This layer of gas maintains a balance between the heat absorbed from the sun and the amount released back into space.
Earth’s Atmosphere Earth’s atmosphere is made up of a mixture of gases. 78% Nitrogen 21% Oxygen 4% Water Vapor Other gases include Argon and Carbon Dioxide
Earth’s Atmosphere Earth’s Atmosphere has 5 layers Lower Layers
Troposphere Stratosphere Upper Layers Mesosphere Thermosphere Exosphere
Troposphere This is the lowest layer of the atmosphere.
This is where we are. It contains 99% of the water vapor and 75 of all atmospheric gases.
Troposphere This begins at the surface of Earth and extends up to 10 km. All weather occurs in the troposphere.
Stratosphere This is found directly above the troposphere.
It extends from 10 km to about 50 km above Earth’s surface. Most importantly, the Ozone Layer is found in the Stratosphere.
Ozone Layer Ozone is made up of 3 atoms of oxygen (O3)
It is found at 19 km - 43 km. The ozone layer shields humans from the sun’s harmful ultraviolet radiation. UV radiation can cause skin damage and lead to skin cancer.
Ozone Layer There has been damage done to the Ozone Layer.
Chlorofluorocarbons (CFC) are compounds found in refrigerators, air conditioners, and aerosol sprays that can destroy ozone layers. This can allow more of the sun’s UV rays to reach Earth.
Mesosphere The Mesosphere extends from 50 km to 85 km.
“Meso” means middle. It is the third of five layers. Meteors can be seen when they reach the mesosphere.
Thermosphere This is the largest layer of the atmosphere.
It reaches from 85 km to 500 km. It gets its name from the high temperatures that can be found there.
Ionosphere Through the Mesosphere and Thermosphere is the Ionosphere.
This is a layer of electrically charged particles. This allows radio waves to travel across the country.
Exosphere Beyond 500 km is the Exosphere.
This is where shuttles will be. However, there are so few molecules, the wings do not provide any guidance. Beyond the exosphere is outer space.
aTMOSPHERIC pRESSURE Air Pressure- the measure of the force with which the air molecules push on a surface. Air pressure changes throughout the atmosphere The atmosphere is held by a planet’s gravity
Temperature Altitude- the height of an object above the Earth’s surface. Air temperature also changes as you increase altitude.
tEMPERATURE AND hEAT Temperature- a measure of the average energy of particles in motion. A high temperature means that the particles are moving fast. Heat- transfer of energy between objects at different temperatures.
Energy in the ATMOSPHERE
Radiation- the transfer of energy as electromagnetic waves. (Sunlight) The radiation absorbed by land, water, and atmosphere is changed into thermal (heat) energy Conduction- is the transfer of thermal energy from one material to another by direct contact.
Water Cycle
Energy in the Atmosphere
Convection- the transfer of thermal energy by the circulation or movement of a liquid or gas. The continual process of warm air rising and cool air sinking creates a circular movement of air called convection current.
Greenhouse effect 50% of the radiation that enters the Earth’s atmosphere is absorbed by the Earth’s surface. The Earth’s heating process, in which the gases in the atmosphere trap thermal energy, is known as the greenhouse effect. A rise in average global temperature is called global warming.
ATmospheric Pressure and Winds
Wind is moving air Wind is created by differences in air pressure. | {"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": 0.8043323159217834, "perplexity": 1082.3403858198292}, "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/1558232256778.29/warc/CC-MAIN-20190522083227-20190522105227-00037.warc.gz"} |
https://physics.stackexchange.com/questions/514603/work-done-while-compressing-an-ideal-gas-the-physical-significance-of-int-ma | # Work done while compressing an ideal gas (the physical significance of $\int \mathrm dp\,\mathrm dV$)
Today in our chemistry class we derived the pressure-volume work done on an ideal gas. Our assumption was that $$p_\mathrm{ext}=p_\mathrm{int}+\mathrm dp$$ so that all the time the system remains (approximately) in equilibrium with the surrounding and the process occurs very slowly (it's a reversible process). Now \begin {align} W_\mathrm{ext}&=\int p_\mathrm{ext}\,\mathrm dV\\ \Rightarrow W_\mathrm{ext}&=\int (p_\mathrm{int}+\mathrm dp)\,\mathrm dV\\ W_\mathrm{ext}&=\int p_\mathrm{int}\,\mathrm dV \end{align} (Since $$\mathrm dp\,\mathrm dV$$ is very small $$\Rightarrow \int \mathrm dp\,\mathrm dV =0$$, though it is an approximation I guess.)
Now, the question is:
• In the case of (say) pushing a book the force on the book and that on the pusher form action reaction pair hence their work shows the same energy transfer but such isn't the case here and hence their work done does not represent the same energy transfer. So what does it represent? As in non-approximate case $$W_\mathrm{ext}-W_\mathrm{int}=\int \mathrm dp\,\mathrm dV$$. What does $$\int \mathrm dp\,\mathrm dV$$ mean physically?
[Note that I ain't equalizing the case of book with that of gas but giving (a kind of analogy or something) with respect to which I want the answerer to compare/contrast the compressing situation]
EDIT
I posted a similar on Maths SE to realize the mathematical significance of the term $$\int \mathrm dp\,\mathrm dV$$. I got this answer over there. Though it mostly satisfies what I wanted to know but states that
The last term (I believe is referring to $$\int \Delta p\,\mathrm dV$$) is then the energy “lost” e.g. by friction, that is, it is not reversible.
Now I'm wondering how does this external pressure term incorporate the frictional force in it?
• The product of two differentials is considered insignificant. – Chet Miller Nov 18 '19 at 12:14
• @Aaron What I was trying to( say )is that when we apply some force say $2N$ on a book to move it some distance (say) $3m$ then work done by the push force on book is $6J$ whereas the push force on us by the book (via $3^{rd}$ law) is $-6J$. Here both represent the same energy transfer (i.e., $6J$ from the pusher to the book). But in case of ideal gas this isn't so, so what does that represent? – user238497 Nov 18 '19 at 16:44
• A book is a solid. – Gaurav Nov 19 '19 at 3:17
• @Shreyansh I wasn't equalizing both the case but giving (a kind of analogy or something) with respect to which I want the answerer to compare/contrast the compressing situation. – user238497 Nov 19 '19 at 6:14
...$$W_{ext}-W_{int}=\int dPdV$$. What does $$\int dP dV$$ mean physically?
Note that in the "non-approximate" case, we have assumed that $$P_{ext}\neq P_{int}$$. More precisely $$P_{ext}-P_{int}=dP$$. Now let's assume that the ideal gas is stored in a container with a movable piston(of a finite mass $$m$$, but ignore gravity) of area $$A$$ on top. For now, let's assume that there is no friction. So to do external work, you(or rather, surroundings) are applying a pressure $$P_{ext}$$(which corresponds to a force $$F_1=P_{ext}A$$) and the gas is doing internal work by applying a pressure $$P_{int}$$(which corresponds to a force $$F_2=P_{int}A$$).
Now let's analyze the forces on the piston. So piston has an upward force of $$F_2$$(applied by the gas) and a downward force $$F_1$$ applied by the surroundings. So in this case the net force in the downward direction is,
$$dF_{net}=m(da_{net})=F_1-F_2=P_{ext}A-P_{int}A=(P_{ext}-P_{int})A=dP×A$$
$$\therefore dK = Fds=dP(Ads)=dPdV$$
where $$dK$$ is the infinitesimal change in the kinetic energy of the piston, and $$dV=Ads$$ is the infinitesimal change in the volume.
There you have it. You see, there is an infinitesimally small(yet non-zero) net force on the piston which gives an infinitesimally small(yet non-zero) acceleration to the piston. And this infinitesimal acceleration increases the speed of the piston from $$0$$ to some infinitesimally small velocity. And thus the piston gains an infinitesimal amount of kinetic energy. And the $$\int dPdV$$ term accounts for this change in kinetic energy.
I know the last paragraph is heavily populated with "infinitesimals", but it is just to show you the insignificance of the motion of the piston. Now what if friction would have been present? In that case, the piston won't move in the first place. But if we also assume that the force due to friction is infinitesimally small, then yeah, the piston would move. But this time it would have a lower value of that infinitesimal acceleration. And, also, it will lose some of its kinetic energy in the form of heat(due to frictional losses).
Summary :- The $$\int dPdV$$ term accounts for the infinitesimal change in the kinetic energy of the piston.
I hope this is what you meant by "physical interpretation".
• So have you ever handled such quantities mathematically? – user238497 Dec 9 '19 at 11:30
• I have never come across any problem which requires you to compite the kinetic energy of the piston. However, there are many problems on irreversible processes where you discover the fact that $W_{int}\neq W_{ext}$. I suspect there might be problems where you would have to use this difference to compute the kinetic energy of the piston. – user243267 Dec 9 '19 at 11:34
• And neither have I pondered about this(theoretically) deeply before you asked this question. The only places where I encoutered this was while solving questions. – user243267 Dec 9 '19 at 11:35
• Do you think this might involve multi variable calculus? – user238497 Dec 9 '19 at 11:37
• @TheLastAirbender No, you will only be asked to calculate $\int \Delta P dV$ where $\Delta P$ is a finite pressure difference between the system and the surroundings and since $\int \Delta PdV$ is just a single integral, you really won't need any knowledge about multivariable calculus. The hard part is to express $\Delta P$ as a function $V$. And then obviously, you also should have luck so that the integral formed can be easily integrable. – user243267 Dec 9 '19 at 11:45
Let me try to convince you that $$\int dPdv$$ is almost negligible. As you have said, $$P_{ext} = P_{int} + dP$$ but what $$dP$$ really is? Well I think it is better to assume $$dP$$ as very small number and hence just adding it to $$P_{int}$$ will give a value bigger than $$P_{int}$$ at any moment no matter whatever $$P_{int}$$ is. So, in this sense $$dP$$ is just acting as constant. Let's see what this angle of thinking about $$dP$$ can lead to $$W_{ext} = \int_{V_i}^{V_f} (P_{int}+dP)dV$$ $$W_{ext} = \int_{V_i}^{V_f} P_{int}dV + \int_{V_i}^{V_f}dPdV$$ Now, let's just focus on the $$dP$$ part $$X= dP\int_{V_i}^{V_f}dV$$ as $$dP$$ is constant. $$X= dP (V_i - V_f)$$ We agreed that $$dP$$ is a very small number and hence if we multiply it with any other thing no matter what the result will be very very small and therefore $$X$$ will be a very small number. $$W_{ext} = \int_{V_i}^{V_f} P_{int}dV + X$$ Now, we can neglect $$X$$ and hence write $$W_{ext} = \int_{V_i}^{V_f} P_{int}dV = W_{int}$$. Your argument that $$\int dPdV$$ is negligible is quite sloppy as the integral adds many many pieces of small things ($$f(x)dx$$ is a very small number as $$dx$$ is very very small but adding many many of them would produce a different result).
Even in mechanics, when calculate gravitational potential energy we take the working force to be just a little more than $$mg$$ and hence calculate the work done just plugging the work with $$mg$$, however, the actual force is more than that.
I said that your argument was sloppy because it’s a matter of hyperreal numbers that when and when we cannot consider something negligible, your argument is vey all right if we just accept the rules of differentials.
Your exact question is what is the physical interpretation of $$\int_{V_i}^{V_f}dPdV$$
I will try to explain this without using mathematics. Suppose you have a cylinder with piston at its one end which is free to move and the cylinder is filled with compressed gas at pressure $$P_{int}$$.
Your task here is to keep the piston stationary. You will have to apply exactly same pressure at its other end to keep it stationary and hence maintaining the thermodynamic state of gas at its initial condition.
$$dP$$, work done by you on the gas and work done by the gas on you are all zero in this case. This is the equilibrium state.
However, if your task was to slowly push the piston inwards further compressing the gas, you will have to increase the pressure applied by you on piston. The piston's acceleration will be dictated by how much you increased external pressure. Let's assume external pressure is increased by an amount $$\delta$$.
One assumption in your derivation is that the process occurs very slowly meaning the piston's acceleration is almost zero. Even if we assume that piston is not accelerating at all, we still need to increase the external pressure. Why? Because there is friction between piston and cylinder wall in real life scenario and additional pressure $$\delta$$ is used to overcome this friction.
And, the work done by $$\delta$$ is $$\int_{V_i}^{V_f}\delta.dV$$ And $$\delta$$ is your $$dP$$ .
Therefore, $$\int_{V_i}^{V_f}dPdV$$ represents nothing energy lost to over any dissipative force present in the system due to irreversibility of the process.
Since one of the assumption in your derivation is process is irreversible hence there is no friction and $$\int_{V_i}^{V_f}\delta.dV$$ is Zero. | {"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": 63, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9596875905990601, "perplexity": 286.34350309772157}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735916.91/warc/CC-MAIN-20200805065524-20200805095524-00536.warc.gz"} |
http://mathoverflow.net/questions/101474/system-of-local-coefficients-on-x-locally-constant-sheaves-and-orientation-shea | # system of local coefficients on X, locally constant sheaves and orientation sheaves
Hi,
I try to understand the orientation sheaves. When searching it in the google, i meet new areas such as local coefficient system and locally constant sheaves. I realize that any system of local coefficients on X is a locally constant sheaves. But what is the relation with orientation sheaves. Which refferences are there to read it?
-
Could someone retag, please? Say, some "sheaf-theory" and "at.algebraic-topology". – Anton Fonarev Jul 6 '12 at 11:00
## 1 Answer
These are purely topological notions and have nothing to do with algebraic geometry in particular.
Let $M$ for simplicity be a topological manifold of dimension $n$. Then the orientation sheaf $\mathcal{L}_{or}(M)$ is the sheafification of the presheaf $U\mapsto H_n(M,M-U;\mathbb{Z})$. It's always a locally constant sheaf with stalks equal to $\mathbb{Z}$. One immediately checks that $\mathcal{L}_{or}$ is trivial if and only if $M$ is orientable. This definition can be generalized.
As for the references, I'd suggest checking A.Dimca, Sheaves in Topology or B.Iversen, Cohomology of Sheaves.
-
to be more explicit one could say that the stalks of $L_{or}(M)$ depend only on the presheaf you defined. To see what the stalks are we can reduce to $\mathbb{R}^n$ as $M$ is a manifold. There we can take a direct system of concentric balls. The relative homology then becomes the n-th homology of the sphere $S^n$ which is $\mathbb{Z}$. Hence $L_{or}(M)$ is indeed locally constant. – Yosemite Sam Jul 6 '12 at 15:09
what does the trivial orientation sheaf mean? – zatilokum Aug 1 '12 at 22:59
@zatilokum It means that this is a constant sheaf. – Anton Fonarev Aug 2 '12 at 12:04 | {"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": 0.9600180983543396, "perplexity": 333.3290836671341}, "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-2014-52/segments/1418802770130.120/warc/CC-MAIN-20141217075250-00171-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/whats-1-1-equal-to.527169/ | # What's (±1)(±1) equal to?
1. Sep 4, 2011
### GreenPrint
What does (±1)(±1) equal to, is it just positive 1?
2. Sep 4, 2011
### ArcanaNoir
wouldn't it be (1)(1) = 1
(1)(-1)= -1
(-1)(1) = -1
(-1)(-1) = 1?
3. Sep 4, 2011
### ArcanaNoir
so.. plus or minus 1?
4. Sep 4, 2011
### GreenPrint
I thought it would have to be just positive one because I thought that (±1)(±1) = (1)(1) or (-1)(-1) which both equal 1, I thought they both had to be both either positive or negative at the same time, in which case (±1)(±1) = 1?
5. Sep 4, 2011
### ArcanaNoir
I don't see why they should have to be positive or negative at the same time, unless the reality of the problem dictates it.
6. Sep 4, 2011
### GreenPrint
I thought that if one wanted to distinguish them being either positive or negative at different times you would put (±1)(-+1)
-+ is suppose to be ± rotated 180 degrees?
7. Sep 4, 2011
### gb7nash
That's the way I interpret it. Sometimes you'll see things like:
$$(5 \pm 5) \mp 10$$
where you get 0 or 10. The rule is to take the top operation to get one answer, then take the bottom operation to obtain the second answer. In the OP's problem, both answers would be 1.
8. Sep 4, 2011
### eumyang
You mean this?
$(\pm 1)(\mp 1)$
I've seen the "minus-plus" symbol before. The cosine of a sum and difference can be written in one formula like so:
$\cos (a \pm b) = \cos a \cos b \mp \sin a \sin b$
... indicating that the symbol on the RHS is different from the one on the LHS.
EDIT: gb7nash beat me to it. ;)
9. Sep 4, 2011
### ArcanaNoir
well you've lost me. forget I said anything :P
10. Sep 4, 2011
### uart
Id say both interpretations were possible, depending on the context.
Say X was a two state random variable that could be either +1 or -1. Similarly Y is an independent two state random variable. The product XY is $(\pm 1)(\pm 1)$, but it's certainly not always +1 in this case.
11. Sep 4, 2011
### Staff: Mentor
This would be $\pm 1$. The first factor could be either positive or negative, and so could the second factor. You can't assume (and shouldn't) that if the first factor is positive, so is the second.
12. Sep 4, 2011
### flyingpig
what about $$\mp 1$$?
13. Sep 5, 2011
### uart
I could accept the idea of "coupled" plus or minuses as a shorthand notation in some specific circumstances, eg:
$$\cos(a \pm b) = \cos a \cos b \mp \sin a \sin b$$
$$\sin(a \pm b) = \sin a \cos b \pm \cos a \sin b$$
In general however, without any specific context as in the OP, I would never consider all the ($\pm$)'s in an equation (or set of equations) to be coupled in this way.
14. Sep 5, 2011
### Staff: Mentor
In what context did you encounter this?
15. Sep 5, 2011
### Mentallic
When I used to deal with equations that involved $\pm$ that were both dependent and independent of others, I would label them with numbers such as $\pm_1, \pm_2$ for example. I think later on when I saw them being used in formal writing, they were denoted by dashes, such as what you see when dealing with derivatives, $\pm', \pm''$ etc.
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https://brilliant.org/problems/2016-amc-sample-question/ | # 2016 AMC (Sample Question)
Let $$k$$ be a positive integer. Bernado and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $$k+1$$ digits. Every time Bernardo writes a number, Silvia erases the last $$k$$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $$k$$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let $$f(k)$$ be the smallest positive integer not written on the board. For example, if $$k=1$$, then the numbers that Bernardo writes are $$16,25,36,49,64$$, and the numbers showing on the board after Silvia erases are $$1,2,3,4,5,6$$, and thus $$f(1) = 5$$. What is the sum of digits of $$f(2) + f(4) + f(6) + \cdots + f(2016)$$?
× | {"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": 0.8532286286354065, "perplexity": 374.95574195666455}, "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-2017-51/segments/1512948512584.10/warc/CC-MAIN-20171211071340-20171211091340-00189.warc.gz"} |
http://link.springer.com/article/10.1023%2FA%3A1005266731458 | Article
Studia Logica
, Volume 63, Issue 2, pp 245-268
First online:
# Proof-Theoretic Modal PA-Completeness II: The Syntactic Countermodel
• Paolo GentiliniAffiliated withIstituto per la Matematica Applicata del Consiglio Nazionale delle Ricerche (IMA-CNR)
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## Abstract
This paper is the second part of the syntactic demonstration of the Arithmetical Completeness of the modal system G, the first part of which is presented in [9]. Given a sequent S so that ⊢GL-LIN S, ⊬G S, and given its characteristic formula H = char(S), which expresses the non G-provability of S, we construct a canonical proof-tree T of ~ H in GL-LIN, the height of which is the distance d(S, G) of S from G. T is the syntactic countermodel of S with respect to Gand is a tool of general interest in Provability Logic, that allows some classification in the set of the arithmetical interpretations.
Proof-Theory Provability Logic countermodel of a sequent classification of arithmetical interpretations of modal logic | {"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": 0.9336269497871399, "perplexity": 2458.45555260052}, "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/1454701147841.50/warc/CC-MAIN-20160205193907-00128-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://blog.evanchen.cc/2014/12/25/representation-theory-part-2-schurs-lemma/ | # Representation Theory, Part 2: Schur’s Lemma
Merry Christmas!
In the previous post I introduced the idea of an irreducible representation and showed that except in fields of low characteristic, these representations decompose completely. In this post I’ll present Schur’s Lemma at talk about what Schur and Maschke tell us about homomorphisms of representations.
1. Motivation
Fix a group ${G}$ now, and consider all isomorphism classes of finite-dimensional representations of ${G}$. We’ll denote this set by ${\mathrm{Irrep}(G)}$. Maschke’s Theorem tells us that any finite-dimensional representation ${\rho}$ can be decomposed as
$\displaystyle \bigoplus_{\rho_\alpha \in \mathrm{Irrep}(G)} \rho_{\alpha}^{\oplus n_\alpha}$
where ${n_\alpha}$ is some nonnegative integer. This begs the question: what is ${n_\alpha}$? Is it even uniquely determined by ${\rho}$?
To answer this I first need to compute ${\mathrm{Hom}_G(\rho, \pi)}$ for any two distinct irreducible representations ${\rho}$ and ${\pi}$. One case is easy.
Lemma 1 Let ${\rho}$ and ${\pi}$ be non-isomorphic irreducible representations (not necessarily finite dimensional). Then there are no nontrivial homomorphisms ${\phi : \rho \rightarrow \pi}$. In other words, ${\mathrm{Hom}_G(\rho, \pi) = \{0\}}$.
I haven’t actually told you what it means for representations to be isomorphic, but you can guess — it just means that there’s a homomorphism of ${G}$-representations between them which is also a bijection of the underlying vector spaces.
Proof: Let ${\phi : \rho_1 \rightarrow \rho_2}$ be a nonzero homomorphism. We can actually prove the following stronger results.
• If ${\rho_2}$ is irreducible then ${\phi}$ is surjective.
• If ${\rho_1}$ is irreducible then ${\phi}$ is injective.
Exercise Prove the above two results. (Hint: show that ${\text{Im } \phi}$ and ${\ker \phi}$ give rise to subrepresentations.)
Combining these two results gives the lemma because ${\phi}$ is now a bijection, and hence an isomorphism. $\Box$
2. Schur’s Lemma
Thus we only have to consider the case ${\rho \simeq \pi}$. The result which relates these is called Schur’s Lemma, but is important enough that we refer to it as a theorem.
Theorem 2 (Schur’s Lemma) Assume ${k}$ is algebraically closed. Let ${\rho}$ be a finite dimensional irreducible representation. Then ${\mathrm{Hom}_{G} (\rho, \rho)}$ consists precisely of maps of the form ${v \mapsto \lambda v}$, where ${\lambda \in k}$; the only possible maps are multiplication by a scalar. In other words,
$\displaystyle \mathrm{Hom}_{G} (\rho, \rho) \simeq k$
and ${\dim \mathrm{Hom}_G(\rho, \rho) = 1}$.
This is NOT in general true without the algebraically closed condition, as the following example shows.
Example Let ${k = {\mathbb R}}$, let ${V = {\mathbb R}^2}$, and let ${G = {\mathbb Z}_3}$ act on ${V}$ by rotating every ${\vec x \in {\mathbb R}^2}$ by ${120^{\circ}}$ around the origin, giving a representation ${\rho}$. Then ${\rho}$ is a counterexample to Schur’s Lemma.
Proof: This representation is clearly irreducible because the only point that it fixes is the origin, so there are no nontrivial subrepresentations.
We can regard now ${\rho}$ as a map in ${{\mathbb C}}$ which is multiplication by ${e^{\frac{2\pi i}{3}}}$. Then for any other complex number ${\xi}$, the map “multiplication by ${\xi}$” commutes with the map “multiplication by ${e^{\frac{2\pi i}{3}}}$”. So in fact
$\displaystyle \mathrm{Hom}_G(\rho, \rho) \simeq {\mathbb C}$
which has dimension ${2}$. $\Box$
Now we can give the proof of Schur’s Lemma.
Proof: Clearly any map ${v \mapsto \lambda v}$ respects the ${G}$-action.
Now consider any ${T \in \mathrm{Hom}_G(\rho, \rho)}$. Set ${\rho = (V, \cdot_\rho)}$. Here’s the key: because ${k}$ is algebraically closed, and we’re over a finite dimensional vector space ${V}$, the map ${T}$ has an eigenvalue ${\lambda}$. Hence by definition ${V}$ has a subspace ${V_\lambda}$ over which ${T}$ is just multiplication by ${\lambda}$.
But then ${V_\lambda}$ is a ${G}$-invariant subspace of ${V}$! Since ${\rho}$ is irreducible, this can only happen if ${V = V_\lambda}$. That means ${T}$ is multiplication by ${\lambda}$ for the entire space ${V}$, as desired. $\Box$
3. Computing dimensions of homomorphisms
Since we can now compute the dimension of the ${\mathrm{Hom}_G}$ of any two irreducible representations, we can compute the dimension of the ${\mathrm{Hom}_G}$ for any composition of irreducibles, as follows.
Corollary 3 We have
$\displaystyle \dim \mathrm{Hom}_G \left( \bigoplus_\alpha \rho_\alpha^{\oplus n_\alpha}, \bigoplus_\beta \rho_\beta^{\oplus m_\beta} \right) = \sum_{\alpha} n_\alpha m_\alpha$
where the direct sums run over the isomorphism classes of irreducibles.
Proof: The ${\mathrm{Hom}}$ just decomposes over each of the components as
\displaystyle \begin{aligned} \mathrm{Hom}_G \left( \bigoplus_\alpha \rho_\alpha^{\oplus n_\alpha}, \bigoplus_\beta \rho_\beta^{\oplus m_\beta} \right) &\simeq \bigoplus_{\alpha, \beta} \mathrm{Hom}_G(\rho_\alpha^{\oplus n_\alpha}, \rho_\beta^{\oplus m_\beta}) \\ &\simeq \bigoplus_{\alpha, \beta} \mathrm{Hom}_G(\rho_\alpha, \rho_\beta)^{\oplus n_\alpha m_\alpha}. \end{aligned}
Here we’re using the fact that ${\mathrm{Hom}_G(\rho_1 \oplus \rho_2, \rho) = \mathrm{Hom}_G(\rho_1, \rho) \oplus \mathrm{Hom}_G(\rho_2, \rho)}$ (obvious) and its analog. The claim follows from our lemmas now. $\Box$
As a special case of this, we can quickly derive the following.
Corollary 4 Suppose ${\rho = \bigoplus_\alpha \rho_\alpha^{n_\alpha}}$ as above. Then for any particular ${\beta}$,
$\displaystyle n_\beta = \dim \mathrm{Hom}_G(\rho, \rho_\beta).$
Proof: We have
$\displaystyle \dim \mathrm{Hom}_G(\rho, \rho_\beta) = n_\beta \mathrm{Hom}_G(\rho_\beta, \rho_\beta) = n_\beta$
as desired. $\Box$
This settles the “unique decomposition” in the affirmative. Hurrah!
It might be worth noting that we didn’t actually need Schur’s Lemma if we were solely interested in uniqueness, since without it we would have obtained
$\displaystyle n_\beta = \frac{\dim \mathrm{Hom}_G(\rho, \rho_\beta)}{\dim \mathrm{Hom}_G(\rho_\beta, \rho_\beta)}.$
However, the denominator in that expression is rather unsatisfying, don’t you think?
4. Conclusion
In summary, we have shown the following main results for finite dimensional representations of a group ${G}$.
• Maschke’s Theorem: If ${G}$ is finite and ${\text{char } k}$ does not divide ${\left\lvert G \right\rvert}$, then any finite dimensional representation is a direct sum of irreducibles. This decomposition is unique up to isomorphism.
• Schur’s Lemma: If ${k}$ is algebraically closed, then ${\mathrm{Hom}_G(\rho, \rho) \simeq k}$ for any irreducible ${\rho}$, while there are no nontrivial homomorphisms between non-isomorphic irreducibles.
In the next post I’ll talk about products of irreducibles, and use them in the fourth post to prove two very elegant results about the irreducibles, as follows.
1. The number of (isomorphsim classes) of irreducibles ${\rho_\alpha}$ is equal to the number of conjugacy classes of ${G}$.
2. We have ${ \left\lvert G \right\rvert = \sum_\alpha \left( \dim \rho_\alpha \right)^2 }$.
Thanks to Dennis Gaitsgory, who taught me this in his course Math 55a. My notes for Math 55a can be found at my website. | {"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": 94, "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": 0.9929470419883728, "perplexity": 216.19169876337295}, "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-2021-43/segments/1634323583408.93/warc/CC-MAIN-20211016013436-20211016043436-00188.warc.gz"} |
https://wiki.synfig.org/index.php?title=Dev:Bline_Speed&diff=next&oldid=5760 | Difference between revisions of "Dev:Bline Speed"
Bline's parameter's "speed"
If you have played enough with Blines, "BLine Vertex" and "BLine Tangent" converts you perhaps have noticed that a change in the "Amount" parameter doesn't always step forward/backwards the same amount. For example, adding 0.1 to "Amount" doesn't move a "Bline Vertex" by the same distance all the time.
Near the bline's vertices (or near the curved parts) you'll notice that evenly spaced "Amount" values are either compressed together or expanded away from each other. Trying to make an object follow a bline will lead to the object changing speeds as it goes along it.
The problem lies in how Blines are defined and how a position in the Bline changes as the "Amount" parameter changes. I'll refer to the rate of change as the Bline's "speed".
Why does "speed" changes?
Firstly, a Synfig Bline is composed of several bezier sections. Each segment is a cubic bezier line. This sections are joined back to back, allowing for arbitrarily complex shapes. All the properties that for a single section, also hold true for any number of sections. So I'm gonna focus on Blines with a single section, in other words, Blines with only two vertexes.
A Bline with a single section reduces to a Cubic Bezier defined like this:
$\displaystyle \mathbf{B}(t)=(1-t)^3\mathbf{P}_0+3t(1-t)^2\mathbf{P}_1+3t^2(1-t)\mathbf{P}_2+t^3\mathbf{P}_3 \mbox{ , } t \in [0,1].$
This equation describes the shape of the curve. As the $\displaystyle t\,\!$ parameter increases from zero up to one, the point defined by the equation moves from the Bezier's start towards it's end. The rate of the motion to as $\displaystyle t\,\!$ increases describes the curve's "speed". Taking the derivative of this equation yields the "speed":
$\displaystyle \frac{d\mathbf{B}(t)}{dt}= (1-t)^2 [ 3 ( \mathbf{P}_1 - \mathbf{P}_0 ) ] + 2t(1-t) [ 3 ( \mathbf{P}_2 - \mathbf{P}_1 ) ] + t^2 [ 3 ( \mathbf{P}_3 - \mathbf{P}_2 ) ] \mbox{ , }t \in [0,1].$
You may have noticed that this equation is equivalent to a Quadratic Bezier. This tells us that the "speed" can and does change as the $\displaystyle t\,\!$ parameter changes.
Our objective is now to compensate the derivative to achieve a desired "speed". We cannot change the control points to the curve without changing it's shape. The only other thing we can change is the parameter $\displaystyle t\,\!$ . Therefore, we define a function $\displaystyle g(t)\,\!$ so that:
$\displaystyle \frac{d\mathbf{B}(g(t))}{dt}=\boldsymbol{s}(t)\,\,$
Where $\displaystyle \boldsymbol{s}(t)$ is a vector $\displaystyle (s_x(t),s_y(t))\,\!$ that defines the desired speed as a function of $\displaystyle t\,\!$ . The curve needs to move in a whole range of directions as the curve describes its shape. Our objective is only to control its magnitude. This magnitude condition can be expressed as:
$\displaystyle s_x^2(t)+s_y^2(t)=s_{mag}^2(t)$
Where $\displaystyle s_{mag}(t)\,\!$ is a function defining the desired "speed" magnitude.
We can expand our first equation a bit:
$\displaystyle \frac{d\mathbf{B}(g(t))}{dt}=\frac{d\mathbf{B}(g(t))}{d(g(t))}\,\frac{dg(t)}{dt}=\boldsymbol{s}(t)\,\,$
Expanding the equation like this lets us use the original Bline's derivative definition, by replacing $\displaystyle t\,\!$ with $\displaystyle g(t)\,\!$ . Next we replace the x and y components into the magnitude condition equation:
$\displaystyle \Bigg[\frac{dB_x(g(t))}{d(g(t))}\,\frac{dg(t)}{dt}\Bigg]^2+\Bigg[\frac{dB_y(g(t))}{d(g(t))}\,\frac{dg(t)}{dt}\Bigg]^2=s_{mag}^2(t)$
Rearranging we obtain an ordinary non-linear differential equation:
$\displaystyle \frac{dg(t)}{dt}=\frac{s_{mag}(t)}{\sqrt{\Big[\frac{dB_x(g(t))}{d(g(t))}\Big]^2+\Big[\frac{dB_y(g(t))}{d(g(t))}\Big]^2}}$
Solving this equation yields a function $\displaystyle g(t)\,\!$ such that the curve's "speed" is dictated by the function $\displaystyle s_{mag}(t)\,\!$ .
Solving the equation
All that is left is to solve the equation. It is quite complex and as I said before, the differential equation that we got is non-linear. This makes it hard to find $\displaystyle g(t)\,\!$ in a clear formula.
But even such a complex equation is easy to solve numerically. Keeping in mind that what we want is simply the value of $\displaystyle g(t)\,\!$ so we can plug it into "Amount". A numerical solution of the equation gives us just that, the value of $\displaystyle g(t)\,\!$ at certain intervals.
The Runge-Kutta method serves this purpose quite well, and it's quite simple also. All we need is to evaluate the derivative of the function that we need to find, and feed the values into the Runge-Kutta method.
Let's try a simple case, constant speed. If $\displaystyle s_{mag}(t)\,\!$ is a constant value, then it would need to be equal to the Bline's length, so that as the $\displaystyle t\,\!$ goes from 0.0 up to 1.0, the curve moves from the start to the end. Too little speed and the curve won't reach the end when $\displaystyle t\,\!$ reaches 1.0. Too much and the curve will go past the end when $\displaystyle t\,\!$ reaches 1.0.
Conveniently, this method also allows to find a Bline's length. If we assume $\displaystyle s_{mag}(t)=1\,\!$ then the curve will reach it's end when $\displaystyle t=LENGTH\,\!$ , where LENGTH is the Bline's length. | {"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": 0, "texerror": 29, "math_score": 0.8910622596740723, "perplexity": 404.6497903316781}, "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-2019-30/segments/1563195525500.21/warc/CC-MAIN-20190718042531-20190718064531-00550.warc.gz"} |
http://zbmath.org/?q=an:1059.62581 | # zbMATH — the first resource for mathematics
##### Examples
Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.
##### Operators
a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
Multihypothesis sequential probability ratio tests. II: Accurate asymptotic expansions for the expected sample size. (English) Zbl 1059.62581
Summary: For Part I, see ibid. 45, 2448–2461 (1999; Zbl 1131.62313).
We proved in Part I that two specific constructions of multihypothesis sequential tests, which we refer to as multihypothesis sequential probability ratio tests (MSPRTs), are asymptotically optimal as the decision risks (or error probabilities) go to zero. The MSPRTs asymptotically minimize not only the expected sample size but also any positive moment of the stopping time distribution, under very general statistical models for the observations. In this paper, based on nonlinear renewal theory we find accurate asymptotic approximations (up to a vanishing term) for the expected sample size that take into account the “overshoot” over the boundaries of decision statistics. The approximations are derived for the scenario where the hypotheses are simple, the observations are independent and identically distributed (i.i.d.) according to one of the underlying distributions, and the decision risks go to zero. Simulation results for practical examples show that these approximations are fairly accurate not only for large but also for moderate sample sizes. The asymptotic results given here complete the analysis initiated by C. W. Baum and V. V. Veeravalli [see IEEE Trans. Inf. Theory 40, No. 6, 1994–2007 (1994; Zbl 0828.62070), where first-order asymptotics were obtained for the expected sample size under a specific restriction on the Kullback-Leibler distances between the hypotheses
##### MSC:
62L10 Sequential statistical analysis | {"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": 0.8097127079963684, "perplexity": 4037.873372724856}, "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/1398223206647.11/warc/CC-MAIN-20140423032006-00225-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://vincenttam.github.io/blog/2014/09/08/another-way-of-writing-piecewise-functions/ | # Another Way of Writing Piecewise Functions
## Background
I changed my way of writing block equations for RSS.1 However, in my old post about the Contraction Mapping Principles, there’s an inequality on the rate of convergence of a point in a complete metric space to the unique fixed point of the Lipschitz mapping with a Lipschitz constant strictly less than one.2
\begin{aligned} d(x^*,x_n) =& d\left( \lim_{k \to \infty} x_k,x_n\right) \\\ =& \lim_{k \to \infty} d(x_k,x_n) \\\ \le& \lim_{k \to \infty} \frac{q^n}{1 - q} d(x_1,x_0) \\\ =& \frac{q^n}{1 - q} d(x_1,x_0) \end{aligned}
Starting from the second line in the above block equation, at the left of the binary relation symbols there’s a whitespace character.
### Visual effects in pages under “/posts/” or the index page
Note that due to the development of Octopress, I now see three backslashes in the “MathJax Equation Source”.
It doesn’t matter much in block equations.
## Problem
However, it does matter if I have to define a piecewise function. Take one defined in one of my old posts as an example.3
(Added on DEC 12TH, 2014) (Revised on SEP 3RD, 2015)
Note: As you can see from the above piecewise function, the problem is now gone.
$f(x,y) = \begin{cases} 0 &\text{if } (x,y) \in \vect{I} \land y \ge x\\\ 1 &\text{if } (x,y) \in \vect{I} \land y < x \end{cases}$
### Visual effects in pages under “/posts/” or the index page
Note that due to the development of Octopress, I now see three backslashes in the “MathJax Equation Source”.
## Solution
Now, I know how to tell kramdown to ignore MathJax code. This is much more convenient than the method described below.
After I observed that the two equations, which are aligned by the aligned environment, at the bottom of the post cited in footnote #3, I used the $\rm \LaTeX$ commands \left\\{ and \right. in the Markdown source file for posts to construct the left curly brace only.
f(x,y) = \left\{ \begin{aligned} 0 &\text{ if } (x,y) \in \vect{I} \land y \ge x\\\ 1 &\text{ if } (x,y) \in \vect{I} \land y < x, \end{aligned} \right. | {"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": 0.9884640574455261, "perplexity": 1254.0550777695003}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320679.64/warc/CC-MAIN-20170626050425-20170626070425-00573.warc.gz"} |
http://tex.stackexchange.com/questions/35356/strange-looking-table | # Strange looking table
I am trying to print this rather odd looking table, below
I know there is an package called exam, though I sort of want to do this "my own style"
I did try to make the table below, but I was not even close in doing so.
Most importantly is that the table looks good, not that it is an exact replica of the table. Could anyone be so kind as to help me? I have started learning latex recently, and stumbling down the path of learning =)
EDIT: One could remove the vertical lines furthest to the left and right aswell, then the table would be more inline witht the standard of tables.
Edit2:
\documentclass[10pt,a4paper]{article}
\usepackage{mathtools}
\usepackage{booktabs}
\usepackage{multirow}
\usepackage{multicol}
\begin{document}
\noindent \begin{tabular*}{\textwidth}{@{}clccccccccccccccc@{}}
\toprule
& & & & & & Sum \\
\cmidrule{1-1}\cmidrule(l){2-16}\cmidrule(l){17-17}
\multirow{2}{*}{Part 1} & Problem & 1a1) & 1a2) & 1b1) & 1b2) & 1c & 1d1) & 1d2) & 1e & 1f & 2a & 2b & 2c \\
& Score & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 24 \\
\midrule
\multirow{2}{*}{Part 1} & Problem & 1a1) & 1a2) & 1b1) & 1b2) & 1c & 1d1) & 1d2) & 1e & 1f & 2a & 2b & 2c \\
& Score & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 24 \\
\cmidrule{1-1}\cmidrule(l){2-17}\cmidrule(l){17-17}
& \multicolumn{5}{r}{Total number of points} & 14 \\
\bottomrule
\end{tabular*}
\end{document}
-
Can you post what you tried so we have something to start with. This should be compilable in that it should include the \documentclass and all the appropriate packages needed fir your version. – Peter Grill Nov 19 '11 at 0:19
This table is not at all strange looking. :) – Werner Nov 19 '11 at 0:19
I tried looking into the multirow and multicolumn packages as it seems this would work best. Will update with an minimal example. Reason why I did not do this, is as all my attempts to create the table would not even compile. Also using booktabs for extra nice tables. – N3buchadnezzar Nov 19 '11 at 0:31
Updated again. Atleast it works for me now, but itss still broken. – N3buchadnezzar Nov 19 '11 at 3:59
@N3buchadnezzar: what do you mean with "broken"? – Gonzalo Medina Nov 19 '11 at 4:25
I would try changing the table layout, first of all, suppressing the vertical rules. Using the features provided by the booktabs package you can improve your tables layout. Here are two possibilities:
\documentclass{article}
\usepackage{multirow}
\usepackage{booktabs}
\begin{document}
\noindent\begin{tabular}{@{}clccccc@{}}
\toprule
& & & & & & Sum \\
\cmidrule{1-1}\cmidrule(l){2-6}\cmidrule(l){7-7}
\multirow{2}{*}{Part 1} & Problem & 1a & 2b & 3a & 5b & \\
& Score & 2 & 2 & 1 & 2 & 7 \\
\midrule
\multirow{2}{*}{Part 1} & Problem & 6a & 6b & 7a & 8b & \\
& Score & 2 & 2 & 1 & 2 & 7 \\
\cmidrule{1-1}\cmidrule(l){2-6}\cmidrule(l){7-7}
& \multicolumn{5}{r}{Total number of points} & 14 \\
\bottomrule
\end{tabular}
\vspace{2cm}
\noindent\begin{tabular}{@{}lccccccccc@{}}
\toprule
& \multicolumn{8}{c}{Part} & Total \\
\cmidrule{2-9}
& \multicolumn{4}{c}{1} & \multicolumn{4}{c}{2} \\
\cmidrule(r){1-1}\cmidrule(lr){2-5}\cmidrule(lr){6-9}\cmidrule(l){10-10}
Problem & 1a & 2b & 3a & 5b & 6a & 6b & 7a & 8b \\
Score & 2 & 2 & 1 & 2 & 2 & 2 & 1 & 2 \\
\cmidrule(r){1-1}\cmidrule(lr){2-5}\cmidrule(lr){6-9}\cmidrule(l){10-10}
Sum & \multicolumn{4}{c}{7} & \multicolumn{4}{c}{7} & 14 \\
\bottomrule
\end{tabular}
\end{document}
Here's a third variant:
\documentclass{article}
\usepackage{multirow}
\usepackage{booktabs}
\begin{document}
\noindent\begin{tabular}{@{}cccc@{}}
\toprule
Part & Problem & Score & Sum \\
\cmidrule(r){1-1}\cmidrule(lr){2-2}\cmidrule(lr){3-3}\cmidrule(l){4-4}
\multirow{4}{*}{1} & 1a & 2 & \multirow{4}{*}{7} \\
& 2b & 2 & \\
& 3a & 1 & \\
& 5b & 2 & \\
\midrule
\multirow{4}{*}{2} & 6a & 2 & \multirow{4}{*}{7} \\
& 6b & 2 & \\
& 7a & 1 & \\
& 8b & 2 & \\
\midrule
& \multicolumn{2}{c}{Total} & 14 \\
\bottomrule
\end{tabular}
\end{document}
-
That looks really good =) I think I will use the bottom one. I have tried for almost 30 minutes now to produce something good. Could you perhaps take a look at my code and tell me why it will not compile? If i remove the multicol arguments in the second row, it compiles. I guess I need to learn this stuff too and not just copy =) Great job once again! – N3buchadnezzar Nov 19 '11 at 1:24
@N3buchadnezzar: the second argument of \multicolumn can only take one value: l, r, c, p{<length>} (or any other column type, but only one value). – Gonzalo Medina Nov 19 '11 at 1:32
@N3buchadnezzar: also, the third argument of \multicolumn cannot contain alignment characters &. The third argument is used for the material (text, in most cases) that will span the selected number of columns. – Gonzalo Medina Nov 19 '11 at 2:33
Gah! Your code breaks down, with the number of elements I need in that table... I updated my first post. Perhaps your last solution is the best? – N3buchadnezzar Nov 19 '11 at 3:37
@N3buchadnezzar: there are some strategies to reduce a table width: 1) Change \tabcolsep (default value = 6pt); you can say, for example, \setlength\tabcolsep{3pt} just before \begin{tabular}. 2) Reduce the font size; for this, you can use one of th font switches \small, \footnotesize; again this goes right before \begin{tabular}. To keep those changes local, you can use a pair of braces: {\small\begin{tabular}{...}...\end{tabular}}`. Of course you can try one or both methods. – Gonzalo Medina Nov 19 '11 at 3:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8874163627624512, "perplexity": 787.2728226798628}, "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/1435375095677.90/warc/CC-MAIN-20150627031815-00309-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://www.snowland.se/tag/function/ | Change a folder LastWriteTime based on files within it
A few days ago I wrote a script that copies some files. Did notice that everything except the date on the folders were ok. So I added a few more lines of powershell code.
Did find a few suggestions on the web, but I like this one…. Since I wrote it. 😛
```Function Set-FolderDate {
Param (
[string] \$Path
)
Trap [Exception] {
Write-Debug \$("TRAPPED: " + \$_.Exception.Message);
Write-Verbose "Could not change date on folder (Folder open in explorer?)"
Continue
}
# Get latest filedate in folder
\$LatestFile = Get-ChildItem \$Path | Sort-Object LastWriteTime -Descending | Select-Object -First 1
# Change the date, if needed
\$Folder = Get-Item \$path
if (\$LatestFile.LastWriteTime -ne \$Folder.LastWriteTime) {
Write-Verbose "Changing date on folder '\$(\$Path)' to '\$(\$LatestFile.LastWriteTime)' taken from '\$(\$LatestFile)'"
\$Folder.LastWriteTime = \$LatestFile.LastWriteTime
}
Return \$Folder
}
Set-FolderDate -Path "D:\temp"
``` | {"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": 0.84536212682724, "perplexity": 1100.8538981455115}, "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-2018-13/segments/1521257647681.81/warc/CC-MAIN-20180321180325-20180321200325-00541.warc.gz"} |
https://motls.blogspot.com/2015/03/dwarf-galaxy-37-sigma-evidence-for-dark.html | ## Tuesday, March 10, 2015 ... //
### Dwarf galaxy 3.7-sigma evidence for dark matter claimed
A new astro-ph paper cross-linked to hep-ph,
Evidence for Gamma-ray Emission from the Newly Discovered Dwarf Galaxy Reticulum 2
by Geringer-Sameth and 6 co-authors (Carnegie-Mellon/Brown/Cambridge), boasts that they have looked into the Fermi-LAT telescope data more carefully than the Fermi folks and found something that looks like a 3.7-sigma (or, with model-independent background estimates, 2.3-sigma) deviation from the expected gamma-ray backgrounds at frequencies $2$-$10\GeV$.
The approximate shape of Reticulum 2, a dwarf galaxy, and the light it emits.
If that deviation is a sign of new physics, and it is a huge "If" indeed, the interpretation is that the signal is due to a pairwise self-annihilating particle whose mass is a few times $100\GeV$.
The signal-resembling observations have been made in Reticulum 2, a dwarf galaxy. One must realize that Reticulum 2 was "cherry-picked" as the "most signal-like dwarf galaxy" among a few dozens of dwarf galaxies which should reduce your faith (or certainty) that the signal is real.
An excess: in the middle of the picture, the red crosses seem to be above the solid black line.
Reticulum means a "small reticle" in Latin, and a "reticle" is a web, typically a system of thin lines (with a cross) in the circular visual field of a telescope. More importantly, Reticulum 2 (Ret2) is a dwarf galaxy which is really close – it is about 30 kpc away from Pilsen. ;-) It's further than Prague but it is very close – almost exactly equal to the diameter of the Milky Way (according to Wolfram Alpha). Ret2 is the third closest dwarf galaxy after Segue 1 (Seg1, 23 kpc) and Sagittarius (24 kpc).
Ret2 is one of the 9 new low-luminosity objects. It (or they) may be viewed as "satellites of the Milky Way". Among the 9 guys, it's probably the most obvious one. Along with Eridanus 2 and no one else, it shows a statistically significant elongation (others are round) which may perhaps increase the density at some points and improve the odds that self-annihilation of dark matter takes place there.
The signal-like optimistic interpretation is that a new particle, such as a superpartner, pairwise annihilates first into a pair of Standard Model $X\bar X$ particles before they create some gamma-rays. We don't know what the $X\bar X$ pair is. Depending on whether $X\bar X$ is $b\bar b$ or $\tau^+\tau^-$ or $hh$ or $\mu^+\mu^-$, the mass $m$ and $J\langle\sigma v \rangle$ of the dark matter particle may have different values – the mass is distributed in rather wide intervals roughly between $50$ and $1,000\GeV$.
It's obviously incorrect to say that the dark matter has been discovered. After all, the significance level wouldn't be sufficient for a hard discovery even if one forgot about all the other possible reasons to question the bold claim suggested in this paper (Ret2 was cherry-picked; Fermi folks haven't noticed that; even according to these 7 authors, it may be due to astrophysical sources, etc.). Nevertheless, the evidence has some beef and it's plausible that the signal is real and in the future, with the hindsight, these folks will be quoted as those who already saw the actual proof of the dark matter.
Thanks to Matt Buckley
#### snail feedback (12) :
I'm not a physicist, so please forgive my ignorant comment: I suspect dark matter is just as likely to exist as Gravity is likely to be in inisotropic. Maybe the fundamental solution needs to be revisited.
"between 50 and 1000GeV" Between a chromium atom and 5.77 top quarks. The particle is stable for the current life of the universe and has no interaction except gravitation. It felicitously self-annihilates without providing universal background radiation or a beacon near galactic central black holes.
Does it also fart rainbows?
I've never quite understood what thermal equilibrium of the universe means physically if dark matter essentially only interacts gravitationally, or the consequences on cosmic expansion. Or how dark matter gets cold. So in trying to picture dark matter decay, it seems like dark matter should be fairly hot, implying many heavy unstable particles still around to decay. From that it seems like the difficulty in spotting the radiation is already fairly strong evidence against the predominance of primordial dark matter. Is there some reason why decay of unstable dark matter would be faster?
Refresher course for the simple minded. If dark matter does not interact except gravitationally how can they produce photons? Also, is it possible that the dark matter world is as complex -- with atoms, molecules, etc., or something comparable -- as the world we know?
no, skittles.
Why do you suspect that these possibilities are equal? I'd say that's because you don't understand them enough to feel the difference.
You can't just state that the fundamental principles should be changed. You should actually construct a viable model. And the fundamental principles are fundamental because they are very hard to dismiss. That's the theoretical problem with MOND (putting aside the observations) it's not so trivial to make consistent fundamental theory for it as you may think. And be sure, you will have to introduce new fields and particles.
On the other hand some dark matter candidates like SUSY particles would be the truly revolutionary discovery
Interesting article. But I do not understand the reason for cherry picking low luminosity objects. I would think that emission of lot of visible light would not obscure detection of gamma rays!
I like your depiction of the dwarf galaxy, Reticulum 2 ;). I think this is what we call a Leprechaun in Ireland. There was another preprint yesterday from the Fermi-LAT and DES collaborations where they reported no excess of gamma ray radiation from 8 dwarf galaxies found by the Dark Energy Survey (http://arxiv.org/pdf/1503.02632v1.pdf). In this preprint Reticulum 2 is DES J0335.65403.
Low luminosity objects will have fewer astrophysical backgrounds like neutron stars, et cetera. The concentration of baryons to dark matter seems to drop in these objects, so the signal to noise of any dark matter signal could go up, all other things being equal.
Thanks for the reply. I understand their hypothesis now. Then, if a convincing signal of dark matter is not found in these objects, their hypothesis of a smaller baryon to dark matter in these objects would be proven wrong. | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8344977498054504, "perplexity": 1040.8641954814111}, "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-2019-39/segments/1568514574532.44/warc/CC-MAIN-20190921145904-20190921171904-00314.warc.gz"} |
http://tex.stackexchange.com/questions/67552/in-tikz-how-to-make-the-text-on-an-edge-have-the-same-direction-as-the-edges | # In Tikz, how to make the text on an edge have the same direction as the edge's?
I have created a simple node-edge graph in which I would like to have the texts on each edge aligned to the direction of the edge.
\documentclass[class=minimal,border=0pt]{standalone}
\usepackage{pgf}
\usepackage{tikz}
\usetikzlibrary{arrows,automata}
\usepackage[latin1]{inputenc}
\begin{document}
\begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=2.8cm,
semithick]
\tikzstyle{every state}=[fill=red,draw=none,text=white]
\node[state] (D) {$D$};
\node[state] (C) [below right of=D] {$C$};
\node[state] (B) [above right of=D] {$B$};
\node[state] (A) [below right of=B] {$A$};
\path (D) edge node {tdc} (C)
(B) edge node {tbc} (C)
(C) edge node {tca} (A);
\end{tikzpicture}
\end{document}
For example, I would like the text tbc rotates -90 degrees automatically to be aligned to the direction of the edge connects the node B to the node C. Is that possible?
-
Try adding the key sloped to the relevant nodes (PGF Manual Section 16.8). Does that do what you want? – Loop Space Aug 17 '12 at 12:24
Thanks! That's the solution! I was trying to figure out how to use text along path... however, the text is not aligned to the center of edge. any solution? – Rasoul Aug 17 '12 at 12:30
The answer to the question of how to get the text to go along the path is to use the sloped key. This rotates the node to match the tangent of the path at the point at which the node would be placed (see Section 16.8 of the PGF Manual).
However, as noted in the comments, this doesn't quite work as desired. This is because of the auto key. This shifts the node off the path. It does so by placing one of the node anchors at the placement point, the anchor chosen appropriately for the tangent of the path (I haven't looked at the code so I'm guessing as to how it chooses it). The problem comes with the fact that the anchor is chosen first and then the node is rotated. What ought to happen is that the node is rotated and then the anchor is chosen. But actually, this doesn't need any complicated code since the anchor will always be one of north or south (if allow upside down is set then it will always be south). So simply setting anchor=south instead of auto will do.
Here's a simple example demonstrating the above analysis. The first three nodes are not sloped, the second three are. The first of the triples has no (other) options, the second is auto and the third is anchor=south. It's clear from the 2nd and 5th that the node has effectively been rotated around its anchor point.
Here's your code with this change (plus a couple of minor stylistic changes: tikz loads pgf automatically, and \tikzstyle is depreciated).
\documentclass{standalone}
%\url{http://tex.stackexchange.com/q/67552/86}
\usepackage{tikz}
\usetikzlibrary{arrows,automata}
\usepackage[latin1]{inputenc}
\begin{document}
\begin{tikzpicture}[
->,
>=stealth',
shorten >=1pt,
auto,
node distance=2.8cm,
semithick,
every state/.style={fill=red,draw=none,text=white},
]
\node[state] (D) {$D$};
\node[state] (C) [below right of=D] {$C$};
\node[state] (B) [above right of=D] {$B$};
\node[state] (A) [below right of=B] {$A$};
\path[every node/.style={sloped,anchor=south,auto=false}]
(D) edge node {tdc} (C)
(B) edge node {tbc} (C)
(C) edge node {tca} (A);
\end{tikzpicture}
\end{document}
(NB the auto=false isn't necessary as the anchor=south overrides the auto key)
And here's the result:
-
Thanks for your detailed answer. It helped me to get it to work properly. – Rasoul Aug 18 '12 at 16:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8515960574150085, "perplexity": 973.5528059383994}, "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/1438042988924.75/warc/CC-MAIN-20150728002308-00214-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/yo-yo-and-torque.146284/ | # Yo-yo and torque
1. Dec 1, 2006
### igor123d
1. A yo-yo lies on a frictionless table. If you apply a horizontal force F to the string to the right, how would the yo-yo move (linearly and rotationally)
2. EF=ma ET=I(alpha)
3. Initially I thought that the yo-yo would move to the right and rotate counter-clockwise (because the string lies below the com). On the test in class that was the right answer apparently. But I looked online and one study guide said that there would be no rotation and the yo-yo would be moving to the right. (which is how I had ultimately answered the question). The idea is that the sum of all forces is ma and that the only external force is the pulling force, F. So if it would rotate, some kinetic energy would go to rotational thus linear would be smaller and a would not be F/m. But how could there be no torque if the force is applied at a point that is not the center of mass and has a component perpendicular to the distance?
Last edited: Dec 1, 2006
2. Dec 1, 2006
### OlderDan
I assume this is a qualitative question. Is the string pulling right or left?
3. Dec 1, 2006
### turdferguson
The answer is all of the above, it depends on the angle from horizontal. It has to do with where the torque is applied. If you pick the correct angle theta, the force is on the instantaneous axis of rotation and produces no torque, only translation. If its below (in your case it was horizontal), then your answer is correct. If the angle is greater than the axis, the yoyo will move in the opposite direction from which you pulled it.
4. Dec 1, 2006
### igor123d
Oh sorry forgot to mention, to the right.
5. Dec 1, 2006
### igor123d
Well but in this specific example, the force actos horizontally on a point below the center of mass, so there would have to be torque, right?
6. Dec 1, 2006
### igor123d
By the way here is an image of a similar situation (look at b):
7. Dec 1, 2006
### OlderDan
Then you are correct. The only horizontal force acting is F, so there will be an accleration F/m to the right. If the string is below the CM there will be a torqe CCW of F*r where r is the radius of the inner spool, so there will be a CCW rotation with angular acceleration.
If there were sufficient friction to prevent slipping instead of no friction, the yo-yo would wind itself up on the string as it rolled to the right. Friction would oppose F and produce a greater CW torque than the CCW torque of the string.
8. Dec 1, 2006
### igor123d
But would this work with conservation of energy? I mean if the only force is F then the total work would have to be Fx. But in this case, you would also have rotational energy, so you would have Fx+K rotational. How could this work?
9. Dec 1, 2006
### OlderDan
It is a bit tricky to calculate the work. Think of you doing work on the string, and the string being an energy transmitter to the yo-yo. You apply a force over a distance. How does that distance compare to the distance the yo-yo moves?
10. Dec 2, 2006
### igor123d
Well I guess since the string moves around the yo-yo, its distance would be R*theta. Then would it also have a translational part x, so that d=R*theta+x
But since the rotation is not what's casuing the linear motion, how would x be related to r*theta?
11. Dec 2, 2006
### Hootenanny
Staff Emeritus
Some really useful comments you've been posting recently....
12. Dec 2, 2006
### OlderDan
The relationship between x and r*θ will depend on the moment of inertia of the yo-yo. What is important is the relationship between W = F*(x + r*θ) and the total kinetic energy. Does F*x equal the translational KE and F*r*θ equal the rotational KE?
13. Dec 2, 2006
### igor123d
Yes, I guess they will. Ok thanks for your help Dan, this makes sense now. | {"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": 0.8900046348571777, "perplexity": 719.775559825865}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719033.33/warc/CC-MAIN-20161020183839-00288-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/finding-the-electric-field-magnitude-in-each-region-of-a-sphere.771855/ | # Finding the electric field magnitude in each region of a sphere
1. Sep 19, 2014
### Joa Boaz
1. The problem statement, all variables and given/known data
A charged insulating spherical shell has winner radius of a/3 an an outer radius of a. The cross section is as shown on the picture. The outer shell has a non-constant volume charge density of ρ = 6*α*r^3. Find the electric field magnitude in each region (outside the object, in the shell itself, and in the cavity contained in the shell). Find the electric potential in each region, and sketch a graph of the electric potential as a function of radius for r = 0 to r = 2a and label the potential at r = a/3, a, and 2a. The reference point for the potential is that V = 0 to r = ∞
2. Relevant equations
Not sure but,
E = (1/4∏*ε) * Q/r^2
E = (1/4∏*ε) * (Q/r^3) * r
3. The attempt at a solution
q(inside) = -Q(sphere) = - 4/3 * ∏ * (a^3) ρ = (-4/3)*∏*((a^6)/729)*6α = (-8/729)*(a^6)*∏*α
q(outise) = - q(inside) = (8/729)*∏*α
σ(outer) = q(outside)/4∏*a^2 = (2/729)*(a^4)*∏*α
Not sure if the above is correct, but I am not sure what I have to do after this. Please can anyone help, thank you.
#### Attached Files:
• ###### IMG_0113.jpg
File size:
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2. Sep 20, 2014
### vanhees71
I'm not so sure about your notation. In such cases it's much easier to integrate the electrostatic Poisson equation,
$$\Delta \Phi=-\frac{1}{\epsilon_0} \rho.$$
In this case you have spherical symmetry and thus $\Phi=\Phi(r)$, where $r=|\vec{r}|$ is the distance from the origin (the center of your spheres). Just look up the Laplace operator in spherical coordinates and solve the differential equation in the different regions. The constants are determined by the appropriate boundary conditions.
3. Sep 20, 2014
### Joa Boaz
Thank you for your help. But I am not sure if I am allowed to use the Laplace operator since the professor has not gone over the Laplace transform. Is there any other way to tackle the above question without using Laplace transform?
Thank you.
4. Sep 20, 2014
### vanhees71
It's not about the Laplace transform but the Laplace operator. In Cartesian coordinates it reads
$$\Delta \Phi=\partial_x^2 \Phi + \partial_y^2 \Phi + \partial_z^2 \Phi.$$
5. Sep 20, 2014
### vela
Staff Emeritus
It's not very clear at all what you're doing. As far as I can tell, you found a similar example in the book or your notes and just plugged in the values from this problem into the equations from the example. That approach doesn't require any real understanding, and it won't help you in the long run. In this case, you seemed to have picked an example that doesn't apply to the described situation.
Try again, this time justifying why you're taking each step. To find the electric field, use Gauss's law. You have spherical symmetry here, so use it. Because the charge density is non-uniform, you'll need to integrate to find the enclosed charge.
6. Sep 20, 2014
### Joa Boaz
I do really appreciate your help. It seems that my level of physics is very low because I just don't see how I can use partial derivatives on this problem. But I really do appreciate your help.
7. Sep 20, 2014
### Dr Transport
This problem is solvable by Gauss's Law since it is spherically symmetric.
Last edited: Sep 20, 2014
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https://kar.kent.ac.uk/53812/ | # A preconditioner for the Ohta-Kawasaki equation
Farrell, Patrick E, Pearson, John W (2017) A preconditioner for the Ohta-Kawasaki equation. SIAM Journal on Matrix Analysis and Applications, 38 (1). pp. 217-225. ISSN 0895-4798. E-ISSN 1095-7162. (doi:10.1137/16M1065483)
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Official URL
http://dx.doi.org/10.1137/16M1065483
## Abstract
We propose a new preconditioner for the Ohta-Kawasaki equation, a nonlocal Cahn-Hilliard equation that describes the evolution of diblock copolymer melts. We devise a computable approximation to the inverse of the Schur complement of the coupled second-order formulation via a matching strategy. The preconditioner achieves mesh independence: as the mesh is refined, the number of Krylov iterations required for its solution remains approximately constant. In addition, the preconditioner is robust with respect to the interfacial thickness parameter if a timestep criterion is satisfied. This enables the highly resolved finite element simulation of three-dimensional diblock copolymer melts with over one billion degrees of freedom.
Item Type: Article 10.1137/16M1065483 Q Science > QA Mathematics (inc Computing science) > QA297 Numerical analysisQ Science > QA Mathematics (inc Computing science) > QA377 Partial differential equationsQ Science > QD Chemistry > QD473 Physical properties in relation to structure Faculties > Sciences > School of Mathematics Statistics and Actuarial Science > Applied Mathematics John Pearson 20 Jan 2016 17:21 UTC 29 May 2019 16:54 UTC https://kar.kent.ac.uk/id/eprint/53812 (The current URI for this page, for reference purposes) | {"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": 0.9321808218955994, "perplexity": 3388.6288311531716}, "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/1568514574039.24/warc/CC-MAIN-20190920134548-20190920160548-00445.warc.gz"} |
http://engineersphere.com/sine-functions | # Sine Functions
### Why do we need Sine functions?
Much of what you will work with in electronics requires alternating current, meaning you will encounter sine waves, square waves, and triangle waves. It might have been a few years since trigonometry, but you will quickly realize that mastering the basics of trigonometry is essential for your success in electrical engineering. Sine functions are one thing you will be seeing a lot of in electrical engineering. There is no avoiding them. Later in the curriculum, you will even learn how to break down square waves and triangle waves into series of sine waves.
The general equation for sinusoidal voltage signal is given by
$v(t) = A \sin(\omega t + \Phi) + V_{DC} = A \sin( 2 \pi f t + \Phi) + V_{DC}$
Where A is the amplitude, $\omega$ is the angular frequency, t is time, $\Phi$ is phase shift, and $V_{DC}$ is a constant voltage.
Likewise, current follows the same general equation.
$i(t) = A \sin(\omega t + \Phi) + I_{DC} = A \sin( 2 \pi f t + \Phi) + I_{DC}$
The equation for the sine wave in Figure 1 is $v(t) = 4\cdot\sin(2 \pi \cdot 1000t) + 2V$. The following sections will explain how to derive the equation for this sine wave as well as any other sine function that you may encounter.
### Amplitude
The amplitude (A) is found by taking the difference between the highest point and the lowest point of a waveform and dividing by two. In Figure 2, the highest point (VH) is 6 V while the lowest point (VL) is -2 V. The amplitude, therefore, is 4 V.
$A = \frac{V_{H} - V_{L}}{2} = \frac{6-(-2)}{2} = 4V$
### Period, Frequency, and Angular Frequency
The period (T) of a waveform is defined as the time it takes for that waveform to complete one cycle. Figure 2 shows two periods, or cycles, of a sine wave. For this sine wave, the period is equal to 1 ms. Identifying the period length allows you to derive the frequency (f) and angular frequency (ω) of the signal.
$\omega = 2 \pi \cdot f$ $\omega = \frac{2\pi}{T} = \frac{2\pi}{.001} = 6283 rad/s$ $f = \frac{1}{T} = \frac{1}{.001} = 1000Hz = 1kHz$
Frequency is defined as the number of cycles a waveform completes in one second. One cycle per second is known as a Hertz (Hz). The waveform above has a frequency of 1000 cycles per second, or 1 kHz.
The units for angular frequency are radians per second (rad/s). Angular frequency is similar to the frequency measured in Hertz, except now we are measuring radians rather than cycles. Because one cycle is equal toradians (360°), the angular frequency is derived by multiplying the frequency by. A common error in homework problems occurs when students neglect to convert one measure of frequency to another. Pay attention to which form you should be using and always remember the equation $\omega = 2\pi \cdot f$.
### Offset Voltage
When a DC voltage is applied to an AC voltage signal, the AC signal is said to have a DC offset voltage. A signal’s offset voltage can also be seen as the signal’s average voltage. You can use integration to find the average value of a time-varying function, but for sine functions, it is easier to use observation.
Consider the sine wave in Figure 3. The equation for this sine wave is $v(t) = 4 \sin(6283t)$. The waveform is oscillating around zero and has no offset. In contrast, the sine wave in Figure 4 has an offset of 2 V. This time, the waveform is oscillating around 2 and its equation is $v(t) = 4 \sin(6283t) + 2V$.
When determining the offset of a sine function, remember that an offset is a constant value. Adding a constant to a sine function only changes the level about which the sine wave oscillates.
### Phase Shift
A sine function has a phase shift when it does not begin its cycle at t = 0. A phase shift is illustrated in the figure to the right. To calculate phase shift, use the equations,
$\Phi_{degrees} = \frac{t_{1} - t_{2}}{T} \cdot 360 ^ {\circ}$ $\Phi_{radians} = \frac{t_{1} - t_{2}}{T} 2\pi rad$
where t1 is the waveform’s original starting point, t2 is where the wave’s starting point has been shifted to, and T is the period.
Example: Find the phase shift of the sine function in Figure 5.
Solution: The sine wave has been shifted from its original starting position (t1) of 0 ms to its new starting point (t2) at -0.4 ms. Therefore,
t1 = 0 ms, t2 = -0.4 ms, and T = 1 ms. Solving for degrees, the phase shift is equal to 144°.
$\phi = \frac{0-(-0.4 ms)}{1 ms} \cdot 360^ {\circ} = 144^ {\circ}$
Defining phase shift with respect to t = 0 works fine in mathematics, but real life won’t tell you when time is equal to zero. In labs this semester you will be observing signals on an oscilloscope. For a single signal, you can assume that there is no phase shift. Phase shift becomes important when more than one signal is involved. There are times when knowing the phase shift between two signals is crucial (the design of an audio system is one instance).
In instances with multiple signals, the phase shift of one signal is measured relative to another signal. One wave is your reference point from which you measure all other waves. Figure 6 shows V1 as the reference and V2 as the wave that has been shifted. Figure 7 shows the opposite. Once you decide on a reference, the same formula given above applies.
$\phi = \frac{t_{1} - t_{2}}{T} \cdot 360^{\circ}$
The sign of the phase shift will vary depending on which signal you use as a reference. The phase shift from V1 to V2 in Figure 6 is positive because V2 is shifted to an earlier time. This is a confusing concept for many people. Though it appears that V1 is ahead of V2, this is not the case. V1 starts its cycle a full 3 ms before V2 does. V1, therefore, is leading V2.
Another way of explaining this is by using the equation. In Figure 6, t1 – t2 must be positive because t1 is larger. In Figure 7, t1 – t2 must be negative.
$\phi_{figure6} = \frac{3 ms}{15 ms} \cdot 360^ {\circ} = 72^ {\circ}$
$\phi_{figure7} = \frac{-3 ms}{15 ms} \cdot 360^ {\circ} = -72^ {\circ}$
Written by Ryan Eatinger ([email protected]). Thank you! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 19, "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": 0.8754701018333435, "perplexity": 673.8797890430773}, "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-2017-43/segments/1508187824293.62/warc/CC-MAIN-20171020173404-20171020193404-00118.warc.gz"} |
https://www.arxiv-vanity.com/papers/hep-ph/9406360/ | Saclay, T94/078 Orsay, LPTHE 94/58 (June 1994)
INTERMITTENCY AND EXOTIC CHANNELS
A. Bialas On leave from: Institute of Physics, Jagellonian University, Reymonta 4, PL-30 059 Cracow, POLAND
LPTHE, Bâtiment 211, Université Paris-Sud
F-91405 Orsay, FRANCE
R. Peschanski
CEA, Service de Physique Théorique, CE de Saclay
F-91191 Gif-sur-Yvette Cedex, FRANCE
ABSTRACT
It is pointed out that accurate measurements of short-range two-particle correlations in like-charge and in channels should be very helpful in determining the origin of the “intermittency” phenomenon observed recently for the like-charge pion pairs. Submitted to Phys. Rev. D (Brief Reports)
It is now well established that the correlations between two pions at small relative momenta observed in different hadron-induced processes of multiparticle production are drastically different for the like and unlike pairs. The like pairs show the so-called “intermittency” i.e. a strong (probably power-law) rise of the correlation function at small where
Q2=(p1−p2)2=(p1+p2)2−4m2π
is the square of the difference of particle momenta. This phenomenon finds a natural explanation in terms of quantum interference between identical particles (HBT correlations which is now commonly accepted
This interpretation implies that the observed correlations have “geometrical” origin, i.e. they are related to the space-time structure of the volume from which the pions are emitted Before this picture is accepted, however, it is important to discuss if there perhaps exists other dynamical explanations for this phenomenon. In fact, it was already a long time ago pointed out and recalled recently that these effects can be related to another remarkable difference between and channels, namely their resonance structure. While the system contains many resonances, they are absent in the “exotic” channels.
It is of course an entirely open question if this difference implies or not a different behaviour of correlations at very low invariant mass. However, there exist an argument suggesting that this may actually be the case It can be summarized as follows.
In the intermediate energy region, the amplitudes describing the cross-section are known to be dominated by exchange of Regge singularities in the -channel. The principle of duality implies that in the exotic channel the contribution of leading Regge poles exactly cancel each other We do not consider the Pomeron exchange contribution, as it is irrelevant for short-range correlations we are interested here.. Consequently, an exotic channel is dominated by exchange of singularities with low intercept It follows that the energy-dependent part of the cross-section is expected to fall with increasing invariant mass of the system following the power law
σ∼(M2)α−1=(p1+p2)2(α−1)
with
Since there are no resonances in the exotic channel, this smooth behaviour is expected to hold even at relatively low energies. It is therefore perhaps not unreasonable to speculate that it continues even down to the threshold Of course the phase-space factors at the threshold must be corrected for. This, however, is to a large extent taken care of, when one considers normalized correlation functions .. This possibility is particularly suggestive, when compared to the data of Refs. which show precisely the same power law A closer look at the data shows that they follow a power law in rather than This is not a serious difficulty: the asymptotic formula (2), justified at large cannot distinguish these two possibilities. at small and medium . In contrast, for non-exotic channel the leading Regge poles do contribute, the intercept is much higher and, consequently, the expected dependence much weaker than in (2). Even more important, the presence of prominent resonances in the channel disturbs significantly the simple power-law behaviour and thus precludes its trivial continuation to the threshold.
This completes the argument. Admittedly, it is not very convincing as it requires a rather bold extrapolation. We feel, however, that it should not be entirely dismissed, particularly in view of the experimental evidence shown in and Indeed, if one can (at least partly) explain the difference between like-change and unlike-change correlations without invoking quantum (HBT) interference, the interpretation of the existing data may be profoundly modified. Therefore it seems important to check carrefully this possibility.
The main point of the present note is to indicate that there exists a feasible way to test these ideas and perhaps even to give access to a relative strength of the effects of quantum (HBT) interference and those of the exotic nature of the like-charge interactions. The idea is there exist particle combinations which are either
(a) exotic but not identical
(b) identical but non-exotic
In the category (a) one can list and pairs. Since the particles are not identical, quantum interference cannot be responsible for short-range correlations. Therefore if the effect is present in channels it is probably due to their “exoticity”.
In the category (b) one has channel. It is formed by identical particles, so if quantum interference is at work we expect a pattern similar to channel. On the other hand system forms with probability and with probability Only state is exotic and thus if the resonance structure is responsible we expect a weaker short-range correlation effect than in system.
The early data of NA22 coll. and the more recent data of Aleph coll. indicate that the very short range correlations in the exotic channels, if present at all, are substantially weaker than in the channel. However, we feel that a precise comparison of the behavior at very small and at medium is necessary to assess correctly the role of exoticity in the channel.
To summarize, we suggest that accurate measurements of short-range correlations in and channels, should help to determine the origin of the phenomenon of “intermittency”. It seems to us important to clarify this point before one can safely conclude that the present data can be attributed solely to the effect of quantum (HBT) interference.
ACKNOWLEDGEMENTS
We would like to thank Wolfram Kittel for the correspondence about data and Andrzej Krzywicki for interesting comments. This research was supported in part by the KBN grant
REFERENCES
1 F. Mandl and B. Buschbeck, Vienna-preprint HEPHY-PUB-590-93 (May 1993); and Proc. XXII Int. Symp. on Multiparticle Dynamics, Santiago de Compostela Spain, 1992, p.561. A. Pajares (ed.) World Scientific, Singapore.
2 N.A. Agababyan et al., NA 22 coll., Z. Phys. C59 (1993) 405.
3 A. Bialas and R. Peschanski, Nucl. Phys. B273 (1986) 703. For a recent review, see E.A. DeWolf, I.M. Dremin and W. Kittel, to be published in Phys. Rep. C.
4 R. Hanbury-Brown and R. Q. Twiss : Phil Mag. 45 (1954) 663; For a review see D.H. Boal et al., Rev. Mod. Phys. 62 (1990) 553.
5 H. Bächler et al., NA 35 Coll., Zeit. Phys. C61 (1994) 551.
6 P. Grassberger, Nucl. Phys. B120 (1977) 231.
7 A. Bialas, Acta Phys. Pol. B23 (1992) 561.
8 E.L. Berger, R. Singer, G.M. Thomas and T. Kafka, Phys. Rev. D15 (1977) 206.
9 I.V. Ajinenko et al., NA22 Coll., Zeit. Phys. C61 (1994) 567.
10 M. Adamus et al., NA22 Coll., Zeit. Phys. C37 (1988) 347, fig.3 e,f.
11 D. Decamp et al., NA22 Coll., Zeit. Phys. C54 (1992) 75, fig.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": 0.8952919244766235, "perplexity": 1388.6212039334548}, "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-2021-17/segments/1618038056325.1/warc/CC-MAIN-20210416100222-20210416130222-00105.warc.gz"} |
https://gitlab.math.unistra.fr/patapon/patapon/-/commit/48bf31eea8ac7aea526c285566689667442f9a73 | Commit 48bf31ee by ph
### up
parent 5c6b6a87
... ... @@ -851,7 +851,7 @@ \subsection{Smooth vortex (performance test)} The smooth vortex test is a classical test for MHD codes. It is described for instance in \cite{dumbserJCP2016}. Because this is a exact solution, for instance in \cite{dumbserJCP2016}. Because this is an exact solution, it allows to assess the accuracy of the solver. Here we also used this test to evaluate the efficiency of the parallel implementation. The test case is built upon a single vortex, which is a stationary ... ... @@ -938,7 +938,7 @@ \end{tabular}\caption{Convergence and performance study\label{tab:Convergence-and-performance}. Some tests are done in single precision (float32) and others in double precision (float64). The efficiency'' is a comparison for N=1024 with the slowest device. "CU" means "Compute Units": it is the number of activated cores in a CPU computation or of OpenCL processing elements for a GPU computation. } with the slowest device. "CU" means "Compute Units": it is the number of activated cores in a CPU computation or of OpenCL compute units for a GPU computation. } \end{table} ... ...
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http://mathhelpforum.com/advanced-algebra/177458-degree-basis.html | 1. ## Degree/Basis
1.Find the degree and basis for Q(3^1/2,7^1/2) over Q.
2.For any positive integers a, b, show that Q(a^1/2+b^1/2)=Q(a^1/2,b^1/2)
Ideas:
1. Well I know if I looked at (3)^1/2 over Q
Then (3)^1/2 has minimal polynomial x^2-3, so degree 2 over Q
(7)^1/2 has minimal polynomial x^2-7 so degree 2 over Q
so entire thing has degree 2.
2.I started by computing the minimal polynomial of a^1/2+b^1/2 over Q
x=a^1/2+b^1/2
x-a^1/2=b^1/2
x^2+2(a)^(1/2)x+a=3
x^2+a-3=2(a)^1/2
x^4+x^2a-3x^2+x^2a+a^2-3a-3x^2-3a+9=4a
x^4+2x^2a-6x^2+a^2-6a+9-4a=0
I don't know how that would hekp me, though
2. For 1.
Ideas:
1. Well I know if I looked at (3)^1/2 over Q
Then (3)^1/2 has minimal polynomial x^2-3, so degree 2 over Q
a basis is: $1, \sqrt{3}$
(7)^1/2 has minimal polynomial x^2-7 so degree 2 over Q
a basis is $1, \sqrt{7}$
so entire thing has degree 2.
this is wrong.
a basis for the extension $\mathbb Q (\sqrt{3}, \sqrt{7})/ \mathbb Q$ is: $1, \sqrt{3}, \sqrt{7}, \sqrt{21}$ so the degree is 4.
3. 1. are you sure about this? remember Q(√3) is a subfield of R. the factorization of x^2 - 7 in R is (x + √7)(x - √7). if Q(√3,√7) has degree 2 over Q, wouldn't that force Q(√3,√7) to have degree 1 over Q(√3)? isn't this the same as saying √7 = a+b√3 for some rational a,b?
i think you should look for a polynomial of degree 4 as the minimal polynomial. the basis elements should be obvious to write down.
it is my contention that Q(√3,√7) = [Q(√3)](√7).
2. think about the basis elements you (hopefully) found for part (1). can you write √a + √b in terms of those basis elements, adapted to a,b instead of 3,7?
if so, then Q(√a+√b) is contained in Q(√a,√b). look at your minimal polynomial. what can you say about the dimension over Q of Q(√a+√b)? can you use a theorem from linear algebra about subspaces here?
4. for 2, I guess I should show (a)^1/2 + (b)^1/2 is in Q((a)^1/2, (b)^1/2) and a^1/2, b^1/2 in Q((a)^1/2 + (b)^1/2)
would this involve degrees and basises?
a basis for Q((a)^1/2 , (b)^1/2) is 1, a^1/2, b^1/2, (ab)^1/2
a^1/2+ b^1/2 in Q((a)^1/2 , (b)^1/2)
Q((a)^1/2 + (b)^1/2) has basis 1, (a)^1/2 + (b)^1/2?
Not quite sure about finding this basis
5. 1.basis elements are 1, root 3, root 7, and root(21)
2.root a + root b= root a+ root b since root a and root b would be basis elements
looking at the minimal polynomial, has dimension 4
Would I use this theorem: Let V ba an n dimensional vector space. If W is any subspace of V, then W=V iff W has dimension n?
6. that would be the one. you could show that you could express √a and √b in terms of powers of (√a+√b), but that seems like more work to me.
7. √a and √b in terms of powers of (√a+√b), but that seems like more work to me.
I want to look at both ways, so if I were to do it this way:
x=(√a+√b)
x^2=a+2√ab+b
???
8. well, the elements 1, √a+√b, (√a+√b)^2 = (a+b) + 2√a√b, (√a+√b)^3 = (2a+b)√a + (a+2b)√b all have to be linearly independent, since [Q(√a+√b):Q] = 4.
so if we set √a+√b = u, √a = (1/(a-b))(u^3 - (a+2b)u), and √b = (1/(b-a))(u^3 - (2a+b)u) (you might want to double-check my algebra).
this shows that Q(√a,√b) is contained in Q(√a+√b) (note as well, that this gives a different basis for Q(√a,√b):
(it is an interesting exercise to show that if we set √a = x, √b = y, that the set {1,x+y,a+b+2xy,(2a+b)x+(a+2b)y} is linearly independent if {1,x,y,xy} is).
9. ok that makes sense
10. Ok this question is related to 1.
What about degree and basis for Q(2^1/2,2^1/3) over Q
2^1/2 has minimal polynomial x^2-2
a basis would be 1,2^1/2
2^1/3 has minimal polynomial x^3-2
basis would be 1, 2^1/3,2^2/3
basis for Q(2^1/2,2^1/3) has degree 5 and basis 1, 2^1/3,2^2/3, 2^1/2,2^1/6?
11. no, 2*3 is not 5.
x^3 - 2 still doesn't have any roots in Q(√2). any root still lies in a cubic extension of Q(√2).
look at your set of "powers" of two: {0, 1/6, 1/3, 1/2, 2/3}. can you see what is missing from that list?
hint: 1/3 = 2/6, 1/2 = 3/6, 2/3 = 4/6....
12. degree 6
we need 5/6
13. Originally Posted by Deveno
well, the elements 1, √a+√b, (√a+√b)^2 = (a+b) + 2√a√b, (√a+√b)^3 = (2a+b)√a + (a+2b)√b all have to be linearly independent, since [Q(√a+√b):Q] = 4.
so if we set √a+√b = u, √a = (1/(a-b))(u^3 - (a+2b)u), and √b = (1/(b-a))(u^3 - (2a+b)u) (you might want to double-check my algebra).
this shows that Q(√a,√b) is contained in Q(√a+√b) (note as well, that this gives a different basis for Q(√a,√b):
(it is an interesting exercise to show that if we set √a = x, √b = y, that the set {1,x+y,a+b+xy,(2a+b)x+(a+2b)y} is linearly independent if {1,x,y,xy} is).
I thing I understand the first part
Now for showing Q(√a+√b) is contained in Q(√a,√b)
Let √a = x, √b = y
I have x+y=√a+√b
(x+y)^2=a+b+2xy
14. my bad, that was a typo, it should be a+b+2xy | {"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": 4, "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": 0.8998640179634094, "perplexity": 2593.462330852159}, "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/1386164954485/warc/CC-MAIN-20131204134914-00094-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-much-is-70-degrees-fahrenheit-in-celsius | Chemistry
Topics
# How much is 70 degrees Fahrenheit in Celsius?
Mar 12, 2017
${70}^{\circ} F = 21. {\overline{1}}^{\circ} C$
#### Explanation:
To convert Fahrenheit to Celsius, the normal method is to subtract $32$ then divide by $1.8$ (i.e. $\frac{9}{5}$).
So in our example:
$\frac{5}{9} \left(70 - 32\right) = \frac{5}{9} \left(38\right) = \frac{190}{9} = 21. \overline{1}$
Alternatively, since $- {40}^{\circ} F = - {40}^{\circ} C$, we can convert by adding $40$, multiplying by $\frac{5}{9}$, then subtracting $40$:
$\frac{5}{9} \left(70 + 40\right) - 40 = \frac{5}{9} \left(110\right) - 40 = \frac{550}{9} - 40 = 61. \overline{1} - 40 = 21. \overline{1}$
The reverse conversion from Celsius to Fahrenheit can be done by either of the following:
• Multiply by $\frac{9}{5}$ then add $32$.
• Add $40$, multiply by $\frac{9}{5}$, then subtract $40$.
##### Impact of this question
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http://physics.stackexchange.com/questions/9296/how-does-carbon-dioxide-or-water-vapour-absorb-thermal-infra-red-radiation-from | # How does carbon dioxide or water vapour absorb thermal infra red radiation from the sun?
We are all told at school water vapour and carbon dioxide are the top two greenhouse gases, and that they absorb thermal infra red radiation, trap heat and warm up the Earth. My question is how do they do that? Why can't Oxygen or Nitrogen or any other gas not absorb infra red radiation as well as water vapour or CO2?
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As you can see on these absorption spectra for $\textrm{H}_2\textrm{O}$ and $\textrm{C}\textrm{O}_2$, both molecules have moderate to strong absorbtion in the mid-IR wavelengths, with the absorption of $\textrm{C}\textrm{O}_2$ extending out into the longer wavelengths.
Other molecules common in the atmosphere don't have such strong absorption at the wavelengths given off by thermal radiation. If you are asking why that is, I'm afraid I can't give you a very detailed answer, except to say that the absorption spectra of molecules (and atoms) is governed by quantum mechanics. Maybe somebody else can explain how they would be calculated from principles, but that is beyond my education.
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The absorption of IR radiation is due to vibrations of molecules. When a vibration causes change in charge distribution (or dipole moment to be more specific) the IR radiation is absorbed.
Generally, hetero-polar molecules, like $H_2O \; and \; CO_2$, have permanent dipole moment. The external oscillating electric field in this case perturbs the Hamiltonian and causes IR absorption. Hence, they contribute to "Green House effect" by absorbtion of heat.
$H_2O$, which is non-linear molecule, has three fundamental modes of vibrations. Symmetrical Stretching, asymmetrical Stretching and scissoring (bending). $CO_2$, which is linear molecule, has four fundamental modes of vibrations. Symmetrical stretching, asymmetrical stretching and two degenerate scissoring modes, in planes perpendicular. The symmetric stretching mode in $CO_2$ does not produce or absorb any IR, as it does not cause change in dipole moment, but other modes do change charge distribution causing absorption of IR.
Well, the homo-polar molecules, like $N_2 \; and \; O_2$, does not have any permanent dipole moment. The external oscillating electric field does not perturb the Hamiltonian for nuclear motion, and does not absorb IR, though it perturbs Hamiltonian for electronic motion. Hence, do not contribute for "Green House effect".
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Chutsu was seen here last time 1st of may. And homo-polar is a nonsenseword. This words are used in chemistry for different types of bonds, but for polarity of molecules this words are nonsense. – Georg Oct 27 '11 at 16:50
In some quantity, everything in the air including nitrogen and oxygen absorbs and emits black body radiation at frequencies which overlap the frequencies absorbed by CO2. In fact, the only reason why there is IR in the air is because the surface of the earth emits black body radiation in proportion to its temperature. The air then does the same thing at some level. The question then is, in what quantity is the atmosphere absorbing and emitting black body radiation. The emissivity of nitrogen and oxygen gasses should be closed to 100%, since they do not reflect IR significantly. But the larger question is how does the quantity of black body absorption compare to the fingerprint absorption of CO2. Actual measurements and numbers do not seem to exist. So promoters use computer models to divide up the heat of the atmosphere between pollutants such as CO2 and everything else. They then pull such numbers out of the hat which say increases in CO2 levels will create a global temperature increase of about 6 deg. This is about 20% of the 33 deg. which the atmosphere is said to contribute to the temperature of the globe.There are about 30 times as many water vapor molecules in the air as CO2 molecules, and water vapor has a more effective fingerprint spectrum. It is also much more variable. This means water vapor will swamp whatever CO2 does. It is obviously not being honest to say CO2 does twenty percent of the heating, when there is a hundred times as much water vapor doing the same thing.
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The real fact is that all of the atmosphere is principally heated by thermal conduction and convection - heating the atmosphere by absorbing infrared radiation is negligible.
Water vapour is accepted to be about 1% of a normal atmosphere. At the Earth's surface the mass of a cubic metre of air is about 1.205 kg. As we go up in the atmosphere pressure decreases and thus so does the mass of air per cubic metre.
At 1% water vapour concentration by volume or about 1.5% by weight there are about 19 grams per cubic metre at sea level - significantly less at higher altitudes. Kiehl & Trenberth claim greenhouse gases produce "backradiation" of 324 Watts / square metre.
A watt is equivalent to a Newton metre per second. A newton is a mass of 1 kilogram accelerated by gravity - 9.8 metres per second.
Hence 324 Watts per square metre is equivalent to 324 Newton metres per second.
324 Newtons is 33 kg subjected to gravity's acceleration.
If anyone can explain to me how a mere 12 grams of water vapour in every cubic metre of normal air at sea level can produce that sort of energy I would feel much enlightened - I simply say it is impossible without furnace like temperatures.
As for Carbon Dioxide at 0.04% by volume which is 0.06% by weight it constitutes approximately 0.732 grams per cubic metre - again significantly less at high altitudes as it is heavier than air.
By similar logic if any one can explain how such an insignificant mass of Carbon Dioxide can produce any detectable quantity of radiation I would feel much enlightened !!
As data from the Moon orbiting satellites a heated surface radiating to space cools at a very slow rate.
True the Moon can become very cold on the night side but that lunar night lasts 354 Earth hours and the surface cools about 290 degrees during that 354 hours - less than a degree per hour.
At that rate the dark side of the Earth would "cool" less than 12 degrees before the Sun rose again.
It is the period of rotation of the Earth that is significant into determining how much the Earth cools.
It is the power of the Solar radiation that determines the maximum daytime temperature.
The atmosphere and oceans reduce the temperature the Solar radiation is capable of causing and there is simple proof in that the Moon's surfaces are heated to over the boiling point of water whilst the Earth never approaches those extremes except when conduction/convection is suppressed or prevented entirely.
Finally do you really believe that although Nitrogen and Oxygen do not significantly absorb Infrared radiation they do not radiate Infrared radiation ?
A well known fact of physics is that everything that has a temperature above absolute zero radiates electromagnetic radiation !
At ambient air temperatures well established physics, Wien's law, predicts that the intensity of the emitted electromagnetic radiation is at a peak well within the Infrared radiation wavelength band.
Almost all of the electromagnetic radiation in the atmosphere is coming from Nitrogen and Oxygen which constitute about 99% of the atmosphere.
The only way Nitrogen and Oxygen would not constitute the vast bulk of this atmospheric radiation is if they were at absolute zero and all the warmth you feel came from 19 grams per cubic metre water vapour and 0.732 grams per cubic metre CO2.
I think that proposition is absurd !
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https://brilliant.org/discussions/thread/new-profiles/ | # New profiles!
Hey everyone, One of the feature requests that we've seen frequently is that when you come across an amazing user-submitted solution or discussion post, you want to see other solutions or posts by that same user. We've made this and much more possible in our recent update to everyone's profile pages. Check it out and let us know what you think!
Cheers, The team at Brilliant
Note by Suyeon Khim
6 years, 11 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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Furthermore,
Note that you can now attempt to solve any problem solved by a person who's profile you click on, regardless of what level it is in. So if you have ever been curious what people in other levels are solving you can now find out by clicking on their profiles.
Staff - 6 years, 11 months ago
Thank you so much for this update!!! I love being able to solve so many more problems :)
- 6 years, 11 months ago
Do you think you can develop some sort of search bar. So you can search for people you know? That would be awesome!
- 6 years, 11 months ago
Great work! I love the new profiles.
Now that you are tweaking, can you do a little addition to "Past Solutions" section? I have been a member of this site from March 2013. Each week I left some of the problems unsolved. Recently, I have thought of returning back to those problems and try them again but it takes a lot of time to reach the previous problem sets. I mean I have to scroll down a lot and click "Load Previous Week" a numerous times to reach the first problem set I attempted. If possible, can you make this job a little easier for us, for example by adding a drop-down menu which lists all the problem sets?
EDIT: Moreover, I wanted to report that the scroll bar which appears in the Past Solutions section, does not function at all in Google Chrome. I often need to switch to Mozilla Firefox for that.
Thanks!
- 6 years, 11 months ago
Yes, the scroll bar doesn't work for me too. I opened a discussion thread about that but got no response.
- 6 years, 11 months ago
Just what I've been waiting for, thank you so much! I also like the details about every little thing, such as number of problems submitted. The only thing I'm still waiting for is the section which shows how many solutions have been featured. This came on Brilliant and was removed for some reason.
Also, instead of featured solutions, voting goes on forever, which I think is less of a good idea than the featured solution idea.
- 6 years, 11 months ago
This is awesome. It's also great how we can even submit solutions for problems not in our problem set. Here are some minor suggestions:
• Just like the solution vote count is shown in green, it would be cool if that would be done too for the topic vote count and comment vote count.
• Maybe you can divide all activitiy of a user in tabs, e.g.: "Solves", "Solutions", "Comments", "Topics", and "Votes" for a better overview.
• When visiting a problem from somebody else's problem set, it would be rather handy to have a badge or something similar indicate from which level it is.
- 6 years, 11 months ago
I think your first point is a good idea, and we will probably do it soonish.
With your second point, are you asking for a way to easily filter your feed, or another person's feed? For example If I went to your profile and only wanted to see your discussions comments, I could click a button and it would only display your discussions comments chronologically.
The last point is a little more complicated because there is so much overlap between levels, but we can definitely find someway to communicate what the difficulty of the problem is in the scheme of Brilliant.
Staff - 6 years, 11 months ago
I was talking about just any feed, actually, so I could see a list of just the problems someone has solved recently, or the comment's he's made, etc.
Oh yes, I forgot about the overlap. Maybe the number of point that could be earned could be displayed, although that might be confusing, because you can't actually earn points on a problem not for your problem set. Maybe you could show the problem sets it appears in.
- 6 years, 11 months ago
excellent update!!!
- 6 years, 11 months ago
One thing, though, I don't know if I like the lifetime points disappearing under quick hover-profiles. I used that a lot. Unless Brilliant is trying to get us away from profiling and stereotyping by lifetime points, which makes sense since "lifetime" is relative.
- 6 years, 11 months ago
I see the lifetime points section.
- 6 years, 11 months ago
"quick hover-over profiles"... You must have special vision enabling you to see what doesn't exist.
- 6 years, 11 months ago
These unlabeled colorless bars are different...
- 6 years, 11 months ago
So many changes everywhere all in 24 hours... My brain... aaahhhh... This is not correct discussion but I don't want to start another one- My friends are empty because I don't have Facebook. I just have anonymous blurred imaginary friends. That is kind of sad and unnerving, knowing that blur will always be there.
- 6 years, 11 months ago
Hahaha, Brilliant is always growing and changing... :)
We'll make the "blurry friends" smarter soon—you are probably going to have to tolerate it through the weekend though...
Staff - 6 years, 11 months ago
Now there are lots of colorful bars and problems are now called "a problem". I liked seeing the title of the problem.
- 6 years, 11 months ago
It says "a problem" in an event stream only when there isn't a name for the problem. This usually happens when the problem is in the Practice section...
Staff - 6 years, 11 months ago
Oooohhhhhhh... I'm not very smart. Thanks for revealing the obvious.
- 6 years, 11 months ago
Great work done by brilliant. I really appreciate that. Can you also add a section for Logical Reasoning. That would be great.
- 6 years, 11 months ago
I like the possibilities for many prompts instead of "intellectually exciting discoveries" and such. It leaves it open for more information.
- 6 years, 11 months ago
Keep in mind, you can also write your own prompt. You don't have to be limited by the random prompts we supply.
Staff - 6 years, 11 months ago
We can update our status too :) Thanks, Brilliant!!
- 6 years, 11 months ago
Your status displays in the little box when people hover over your profile with a cursor, so everybody can get a quick sense of what's going on with you.
Staff - 6 years, 11 months ago
Nice, all updates on Brilliant, Amazing!!
- 6 years, 11 months ago
from its too?but we can't to mention people but this is amazing
- 6 years, 11 months ago
It is possible that the ability to mention people in conversations you think are relevant to them will come out in the future. Like all things on Brilliant, your profiles will evolve.
Staff - 6 years, 11 months ago
This is the Amaziing of Brilliant Future sir.Two Thumbs For Brilliant (y)
- 6 years, 11 months ago
In my homepage, i am shown that i am in level 4 in number theory but on my profile it shows me that i am at level 3.. Please help..
- 6 years, 11 months ago
Its fixed now...Thanks.
- 6 years, 11 months ago
This is beyond amazing. Good job guys :)
- 6 years, 11 months ago
In my profile, i am shown that i am in level 5 in number theory but on problem page, shows me that i am at level 4.. Please help..Thank's
- 6 years, 11 months ago
You probably was well in ur last answers, so now you've been promoted. In the next update u will answer the questions of level 5. :)
- 6 years, 11 months ago
I love it!
- 6 years, 11 months ago
The new profiles are brilliant
- 6 years, 11 months ago
ohh its wonderful....thanx...
- 6 years, 11 months ago
Feature Request: We could add a messageboard to everyone's profile that would be much like a talk page
- 5 years, 9 months ago
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http://superuser.com/users/196505/edi9999?tab=activity | # edi9999
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bio website javascript-ninja.fr location New York, United States age 22 member for 1 year, 6 months seen Aug 5 at 15:46 profile views 7
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Jun12 comment How do I modify a style so that it will appear as a section heading in a Table of Contents? I don't think they is a way to do this. The right way to do that is to `add 1` to all of your Headings Jun2 awarded Commentator Jun2 comment Is there a way to embed hidden information in a paragraph in Microsoft Word Can't you use the commenting functionnality ? Feb6 awarded Teacher Feb6 answered Copying all files from one folder to an other with command line in windows Feb6 comment Copying all files from one folder to an other with command line in windows I needed it to be done in command line because it's a task I wanted to do often (it's part off the creation of an installer) Feb6 comment Copying all files from one folder to an other with command line in windows Without the /s it works, except that the folders aren't copied Feb6 comment Copying all files from one folder to an other with command line in windows This command : `copy /s C:\wamp\www\poker-tell\server\*.* C:\wamp\www\poker-tell\build\window` raises the same error Feb6 comment Copying all files from one folder to an other with command line in windows I would like to copy subdirectories too... Feb6 comment Copying all files from one folder to an other with command line in windows I get "The syntax of the command is incorrect. " My command : `copy C:\\wamp\\www\\poker-tell\\server\\*.* C:\\wamp\\www\\poker-tell\\build\\window /s` Feb6 comment Copying all files from one folder to an other with command line in windows I get "insufficient memory" (seems that the content of the paths of some of the files exceed 254) Feb6 comment Copying all files from one folder to an other with command line in windows Actually this doesn't work either because I would like to copy the folders that are in src too Feb6 awarded Student Feb6 asked Copying all files from one folder to an other with command line in windows Sep26 awarded Autobiographer Sep10 comment converting html into word doc or pdf preserving styles (or how to use git for word documents) So you mean automatically using one command line ? Sep10 answered converting html into word doc or pdf preserving styles (or how to use git for word documents) Jul31 awarded Editor Jul31 revised how to use codefolding feature of notepad++ for normal text documents Image Inline, corrected typos Jul31 suggested suggested edit on how to use codefolding feature of notepad++ for normal text documents | {"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": 0.8833510279655457, "perplexity": 4593.745164563457}, "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-2014-35/segments/1409535917663.12/warc/CC-MAIN-20140901014517-00001-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://www.bartleby.com/solution-answer/chapter-111-problem-4e-precalculus-mathematics-for-calculus-6th-edition-6th-edition/9780840068071/11e50de8-c2b1-11e8-9bb5-0ece094302b6 | # The equation relating x , y to z for the given conditions.
BuyFind
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
#### Solutions
Chapter 1.11, Problem 4E
To determine
Expert Solution
## Answer to Problem 4E
The quantities x,y and z are related by equation z=12xy .
### Explanation of Solution
Given:
z is proportional to the product of x and y and z is 10 when x is 4 and y is 5.
Definition:
If the quantities x,y and z are related by an equation,
z=kxy (1)
For some constant k0 , we say that z is proportional to the product of x and y. The constant k is called the constant of proportionality.
Calculation:
It is given that z is proportional to the product of x and y.
Compare the above equation with the definition of proportionality in equation(1).
z=kxy (2)
Substitute 10 for z, 4 for x and 5 for y in equation (2).
10=k×4×510=20k
Simplify the above equation to find the value of k.
10=20kk=1020k=12
Substitute the value of k in equation (2)
z=12xy
Therefore, the quantities x,y and z are related by equation z=12xy .
### Have a homework question?
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!
© 2021 bartleby. All Rights Reserved. | {"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": 0.8204250335693359, "perplexity": 2051.9526924159527}, "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/1631780057775.50/warc/CC-MAIN-20210925202717-20210925232717-00395.warc.gz"} |
http://mathhelpforum.com/geometry/217122-calculating-radius-circle-simple-trigonometry-geometry.html | # Math Help - Calculating the radius of a circle - simple trigonometry/geometry
1. ## Calculating the radius of a circle - simple trigonometry/geometry
From the circle detailed in this video:
To find the radius of the inner circle, why cannot we take the line down from OC to form a right angled triangle, and then do
tan(pi/6)=r/6?
Since tan is opposite over adjacent, why would this not work?
Apparently the answer is r=2, that's now what you would get doing the above. Why is it wrong?
2. ## Re: Calculating the radius of a circle - simple trigonometry/geometry
Hey Mukilab.
I don't know what triangle you are referring to, but it isn't in any way reflecting the radius of the circle. What the video did is the best way to figure out the problem IMO.
3. ## Re: Calculating the radius of a circle - simple trigonometry/geometry
Ah nevermind, I started drawing the circle/triangle on paint and I realised my mistake.
Thank you for taking the time to post anyway | {"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": 0.8830304145812988, "perplexity": 562.6610181098512}, "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-2015-40/segments/1443736682773.27/warc/CC-MAIN-20151001215802-00114-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/kinematics-driving-question.260961/ | # Kinematics, driving question
1. Oct 1, 2008
### Jennifer001
1. The problem statement, all variables and given/known data
you're driving down the highway late one night at 20ms when a deep steps onto the road 35m infront of you. your reaction time before steeping on the brakes is 0.5s and the max deceleration of ur car is 10m/s^2
what is the max speed you could have and still not hit the deer?
deltaX=35m V=?
a=-10m/s^2 reaction time = 0.5s
2. Relevant equations
Xf= Xi+ViT+1/2aT^2
Vf-Vi=aT
Vf^2=Vi^2+1/2a(deltaX)
3. The attempt at a solution
so this is what i tried
i pluged the numbers into the first equation to get
35=0+ViT+1/2(-10)T^2
and i have 2 missing variables so i tried to get it from the other equation
Vf-Vi=aT
0-Vi=(-10)T then that didnt work so i tried the other equation
Vf^2=Vi^2+1/2a(deltaX)
0=Vi^2+1/2(-10)35
Vi= 13.23m/s
and i know that Vi can't be right because i did the question before it and it had Vi of 20m/s and it was 5m from hitting the deer...i dont know what i did wrong can someone help please?
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Oct 1, 2008
### HallsofIvy
Staff Emeritus
In order to JUST miss hitting the deer, it must take all 35 meters to slow to 0 m/s. So you have TWO of the equations you just gave: 35=0+ViT+1/2(-10)T^2 for the 35 m and 0-Vi=(-10)T for the time to slow to 0. Now you have two equations in the two "unknown" numbers Vi and T. 0-Vi= -10T tells you that Vi= 10T. Replace Vi by that in the first equation to get a simple equation for T alone. Once you have found T, put that into Vi= 10T to find Vi.
Similar Discussions: Kinematics, driving question | {"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": 0.8263382315635681, "perplexity": 1483.949907741207}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187828189.71/warc/CC-MAIN-20171024071819-20171024091819-00188.warc.gz"} |
http://sachinashanbhag.blogspot.com/2012/03/nonsingular-and-nondefective-random.html | ## Thursday, March 1, 2012
### Nonsingular and Nondefective Random Matrices
In a previous post, I posed the folllowing question:
Consider again, a random square n by n matrix A, whose entries are restricted to the set of integers {-p, -p+1, ..., 0, ... p-1, p}. Each of the 2p+1 values are equally probable.
• What is the probability that this matrix is nonsingular?
• What is the probability that this matrix is nondefective?
For n = 3 and 4, using a simple Monte Carlo method, this is what I get. As p increases, we approach the continuous distribution of entries asymptotically.
Blue lines are for n=3, and green for n=4. Triangles and squares denote the probability that a random matrix is nonsingular, and nondefective, respectively
As we can see, both singular and defective matrices are extremely rare for large p. Between the two, defectiveness is a rarer feature. | {"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": 0.9356228113174438, "perplexity": 858.5306296866261}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886102891.30/warc/CC-MAIN-20170817032523-20170817052523-00430.warc.gz"} |
https://cracku.in/blog/number-series-questions-for-ssc-gd-pdf/ | 0
429
# Number Series Questions For SSC GD PDF
Question 1: In the following question, select the missing number from the given series:
6, 19, 54, 167, 494, ?
a) 1491
b) 1553
c) 1361
d) 1642
Question 2: In the following question, select the missing number from the given series:
67, 70, 74, 77, 81, 84 ?
a) 87
b) 88
c) 86
d) 89
Question 3: In the following question, a series is given with one or more number (s) missing. Choose the correct alternative from the given options.
22, 22, 23, 20, 16, 17, 17, ?, ?, 8
a) 18, 9
b) 12, 13
c) 10, 9
d) 13, 10
Question 4: In the following question, a series is given with one or more number (s) missing. Choose the correct alternative from the given options.
?, 5, 30, 186, 1309, 10480
a) 0.25
b) 0.75
c) 1.00
d) 0
Question 5: In the following question, a series is given with one or more number (s) missing. Choose the correct alternative from the given options.
7, 51, 8, 65, 9, ?
a) 79
b) 80
c) 81
d) 82
Question 6: In the following question, a series is given with one or more number (s) missing. Choose the correct alternative from the given options.
0.2, 0.16, 0.072, 0.0256, ?
a) 0.0016
b) 0.004
c) 0.00512
d) 0.008
Question 7: Which of the following replaces ‘?’ in the following series?
10,21,45,84,500,?
a) 1050
b) 1045
c) 985
d) 1025
Instructions
In the following question, select the missing number from the given series.
Question 8: 84, 42, 44, 22, 24, 12, ?
a) 20
b) 14
c) 24
d) 28
Question 9: 1, 4, 13, 40, 121, ?
a) 284
b) 286
c) 364
d) 396
Instructions
In the following question, select the missing number from the given series.
Question 10: 2, 7, 22, 67, ?
a) 197
b) 198
c) 200
d) 202
Question 11: 1, 8, 29, 92, 281, ?
a) 567
b) 628
c) 776
d) 848
Instructions
In the following question, select the missing number from the given series.
Question 12: 1, 7, 3, 9, 6, 12, 10, 16, 15, ?
a) 18
b) 15
c) 20
d) 21
Question 13: 7, 10, 14, 19, 25, ?
a) 32
b) 36
c) 38
d) 40
Question 14: In the following question, select the missing number from the given series.
?, 5, 15, 45, 113
a) 1
b) 2
c) 3
d) 4
Question 15: In the following question, select the missing number from the given series.
1, 1, 3, 4, 5, 9, 7, 16, 9, 25, 11, ?
a) 17
b) 36
c) 49
d) 37
Instructions
In the following question, select the missing number from the given series.
Question 16: 2.2, 14.8, 40, 90.4, ?
a) 191.2
b) 194.4
c) 196.2
d) 208.4
Question 17: 84, 42, 28, 21, ?
a) 10.5
b) 16.8
c) 18.4
d) 19.6
Question 18: In the following series find 20th number
9, 5, 1, -3 -7, -11,……..
a) -64
b) -75
c) -70
d) -67
Instructions
A series is given, with one from missing, Choose the correct alternative from the given ones that will complete the series.
Question 19: 16, 61, 25, 52, 36, 63, 49, ?
a) 36
b) 94
c) 72
d) 46
Question 20: Find the wrong number in the given series ?
15, 28, 30, 39, 48
a) 28
b) 15
c) 30
d) 39
Difference between the numbers is in the form of $[x][x^{2}]$
$10+[1][1^{2}]=10+11=21$
$21+[2][2^{2}]=21+24=45$
$45+[3][3^{2}]=45+39=84$
$84+[4][4^{2}]=84+416=500$
Next number:
$500+[5][5^{2}]=500+525=1025$
The given series is an arithmetic progression with first term $a=9$ and common difference $d=-4$
$n^{th}$ term in an A.P. = $A_n=a+(n-1)d$
=> $A_{20}=9+(20-1)\times(-4)$
= $9+(19)(-4)$
= $9-76=-67$
=> Ans – (D) | {"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": 0.8662180304527283, "perplexity": 388.16329215612336}, "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-2021-04/segments/1610703644033.96/warc/CC-MAIN-20210125185643-20210125215643-00167.warc.gz"} |
https://www.physicsforums.com/threads/classical-equations-and-light.935851/ | # B Classical equations and light
1. Jan 1, 2018
### JulianM
But momentum = mass x velocity so that part of the equation is E2 = c.(mv)
If the mass is zero then that formulation also yields zero energy (which we know is not true)
Last edited by a moderator: Jan 1, 2018
2. Jan 1, 2018
### DrStupid
This definition is based on another concept of mass. Today "mass" means invariant mass and that results in another equation for momentum.
3. Jan 1, 2018
### PeroK
The momentum of a photon is given by $p = hf/c$. Not by $p = mv$.
In fact, $p = mv$ is a non-relativistic approximation of the momentum of a massive particle. The correct equation is:
$p = \gamma mv$, where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$
4. Jan 1, 2018
### JulianM
So lightarrow was wrong?
5. Jan 1, 2018
### DrStupid
No, he was and is still right.
6. Jan 1, 2018
### JulianM
Regardless of whether you include gamma this equation still contains mass.
Anything multiplied by zero is still zero.
7. Jan 1, 2018
### PeroK
For a photon $E = cp$. So, he is right and you are not.
$p = mv$, as I have already explained, is a classical equation for the momentum of a particle that is not valid in relativity - for any particle, massless or otherwise.
Last edited by a moderator: Jan 1, 2018
8. Jan 1, 2018
### DrStupid
Again, this "m" is an outdated concept of mass. It dates back to classical mechanics and is not used anymore because it is frame dependent in relativity. With the modern concept of mass (invariant mass) the equations for momentum and energy are
$p = \frac{{m \cdot v}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$
$E = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$
Of course they cannot be used for photons (because v=c), but they result in equations like
$p = \frac{{E \cdot v}}{{c^2 }}$
and
$E^2 = m^2 \cdot c^4 + p^2 c^2$
which are valid for v=c and m=0.
Last edited by a moderator: Jan 1, 2018
9. Jan 1, 2018
### PeroK
That equation is for a massive particle. It doesn't apply for a photon. In fact, it is not $0$ for a photon, it is undefined. Since the photon has zero mass and speed $c$ you would get:
$p = \frac{mc}{0} = \frac00$
Which is undefined. So, you have to look elsewhere for the equation that gives the momentum of a photon.
10. Jan 1, 2018
### JulianM
So what is the equation in relativity? It can't be just multiplying momentum by gamma as PeroK did (above). That still gives zero energy.
11. Jan 1, 2018
### PeroK
The general equation for momentum is:
$p^2c^2 = E^2 - m^2c^4$
That works for both massive and massless particles.
For a massive particle the energy is $E = \gamma mc^2$
And, for a photon the energy is $E = hf$.
12. Jan 1, 2018
### DrStupid
But the limit for $v \to c$ (and therefore $m \to 0$) is defined.
13. Jan 1, 2018
### JulianM
How does
$p^2c^2 = E^2 - m^2c^4$
work for a massless particle if p= m.v
That would give (for v = c)
( mc^2).c^2 = E^2 - m^2.c^4
is equivalent to
E^2 = m^2.c^4 - m^2.c^4
There's something wrong, surely it cannot apply to a massless particle which posses energy
14. Jan 1, 2018
### PeroK
What will it take for you to accept that $p = mv$ is not a valid equation?
15. Jan 1, 2018
### Orodruin
Staff Emeritus
Again, as you have been told repeatedly, p = mv is not a valid equation relativistically. Repeating the same mistake does not make it true.
16. Jan 1, 2018
### JulianM
Well now you have me very confused. How do we know when to apply relativistic equations or not?
17. Jan 1, 2018
### phinds
For a massless particle, you have no choice. For massive particles, the relativistic equation is always true to but the slower the particle is, the less difference there is between the results of the classical equation and the relativistic equation. At human-scale speeds, the classical equation always gives results that are useful.
18. Jan 1, 2018
### PeroK
In principle, the relativistic equations are always valid and the classical equations are an approximation. If that approximation is good enough, then you can apply the classical equations.
If the velocities involved are a significant proportion of the speed of light, then you definitely need the relativistic equations.
In practice, it depends how accurate you need your answer. Time calculations for GPS satellites must be so accurate that relativistic equations are needed despite the modest speeds. But, normal engineering calculations are usually more than accurate enough with classical equations.
19. Jan 1, 2018
### ZapperZ
Staff Emeritus
No, in THIS case, momentum is defined via the wavenumber, i.e.
p=ħk
So there is no requirement for a "mass" here.
Zz.
20. Jan 1, 2018
### Staff: Mentor
You can always apply the relativistic equations, and you will always get the right answer. They're more exact than the non-relativistic ones.
However, the relativistic equations are also more complicated and using them is more work, so we don't use them when the relativistic effects are small enough to ignore. For example: what is the kinetic energy and the momentum of a one-kilogram mass moving at ten meters per second? Try calculating that using the classical formulas and using the relativistic ones. Which was more work? And what difference did it make?
The classical formulas can be used any time the speeds involved are small compared with the speed of light (equivalently, $\gamma$ is so close to 1 that you're OK with the approximation $\gamma=1$). Clearly this is not the case for particles travelling at the speed of light, so you know that the classical formulas can't be used here. | {"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": 0.9619095921516418, "perplexity": 964.8592539168455}, "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-2018-17/segments/1524125945037.60/warc/CC-MAIN-20180421051736-20180421071736-00354.warc.gz"} |
http://math.stackexchange.com/questions/241439/how-do-i-determine-coefficients-of-a-square-function-in-this-example | # How do I determine coefficients of a square function in this example?
This point goes into plot of the function: P (-8, 6.25)
I also know that: C= 11/12
The unknowns are A and B.
Do I have enough data to solve this problem?
-
$$\;\;\;\;\;a(-8)^2+b(-8)+\frac{11}{12}=\frac{25}{4}=6.25\Longleftrightarrow 64a-8b+\frac{11}{12}=\frac{25}{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": 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": 0.8639755845069885, "perplexity": 1869.3689323055914}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049272823.52/warc/CC-MAIN-20160524002112-00150-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://farside.ph.utexas.edu/teaching/329/lectures/node83.html | Next: 2-d problem with Neumann Up: The diffusion equation Previous: An improved 1-d solution
## 2-d problem with Dirichlet boundary conditions
Let us consider the solution of the diffusion equation in two dimensions. Suppose that
(214)
for , and . Suppose that satisfies mixed boundary conditions in the -direction:
(215)
at , and
(216)
at . Here, , , etc., are known functions of , whereas , are known functions of and . Furthermore, suppose that satisfies the following simple Dirichlet boundary conditions in the -direction:
(217)
As before, we discretize in time on the uniform grid , for . Furthermore, in the -direction, we discretize on the uniform grid , for , where . Finally, in the -direction, we discretize on the uniform grid , for , where . Adopting the Crank-Nicholson temporal differencing scheme discussed in Sect. 6.6, and the second-order spatial differencing scheme outlined in Sect. 5.2, Eq. (214) yields
(218)
where . The discretized boundary conditions take the form
(219) (220)
plus
(221)
Here, , etc., and , etc.
Adopting the Fourier method introduced in Sect. 5.7, we write the in terms of their Fourier-sine harmonics:
(222)
which automatically satisfies the boundary conditions (221). The above expression can be inverted to give (see Sect. 5.9)
(223)
When Eq. (218) is written in terms of the , it reduces to
(224)
for , and . Here, , and . Moreover, the boundary conditions (219) and (220) yield
(225) (226)
where
(227)
etc. Equations (224)--(226) constitute a set of uncoupled tridiagonal matrix equations for the , with one equation for each separate value of .
In order to advance our solution by one time-step, we first Fourier transform the and the boundary conditions, according to Eqs. (223) and (227). Next, we invert the tridiagonal equations (224)--(226) to obtain the . Finally, we reconstruct the via Eq. (222).
Next: 2-d problem with Neumann Up: The diffusion equation Previous: An improved 1-d solution
Richard Fitzpatrick 2006-03-29 | {"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": 0.9880627393722534, "perplexity": 1363.7504573924564}, "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-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00315.warc.gz"} |
http://math.stackexchange.com/questions/58914/a2-i-det-a0-implies-ai-is-non-singular | # $A^2=I,\det A>0$ implies $A+I$ is non-singular
Question:
If a square matrix $A$ satisfies $A^2=I$ and $\det A>0$, show that $A+I$ is non-singular.
I have tried to suppose a non-zero vector $x$ s.t. $Ax=x$ but fail to make a contradiction.
And I tried to find the inverse matrix of $A+I$ directly, suppose $(A+I)^{-1}=\alpha I +\beta A$, but it still doesn't work.
(Update: According to the two answers, this question itself is incorrect.)
-
This seems to be false: consider the $3 \times 3$ diagonal matrix with diagonal entries $1,-1,-1$. Similarly, taking an $n \times n$ diagonal matrix with two entries $-1$ and all the rest equal to $1$ gives a counterexample for any $n \geq 2$.
(A comment on how I came up with this: the matrix $A+I$ is singular iff $-I-A$ is singular iff $-1$ is an eigenvalue of $A$. The condition $A^2 = I$ forces each of the eigenvalues to be $\pm 1$ and the condition $\operatorname{det} A > 0$ forces the number of instances of $-1$ to be even, but this is not enough to give the result.)
Of course the above reasoning would also lead to Ricky Demer's counterexample, and probably should have. For some reason I thought of the above first.
-
$-I$ with even size is a counterexample.
-
$(I+A)(I-A)=0$, so $I+A$ is invertible (for $A$ satisfying $A^2=I$) if and only if $A=I$.
This works for $A$ in any ring with unit [in which $2$ is invertible], not only a ring of square matrices.
-
Why $I+A$ is invertible implies $A=I$? There may be $A$ s.t. $A^2=I$ and $A\not=I,-I$. – NGY Aug 22 '11 at 11:03
If I+A has inverse B, (I+A)(I-A)=0 implies B(I+A)(I-A)=B.0 hence I-A=0. – Did Aug 22 '11 at 11:58
@Didier: thanks. Also, to go in the other direction, from A=I to (I+A) being invertible, I should have specified that there be no 2-torsion. – zyx Aug 23 '11 at 3:11
A counter example is take matrix $A$ with diagonal entries 1 and -1. – srijan May 12 '12 at 6:27 | {"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": 0.97335284948349, "perplexity": 276.19685402671604}, "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/1418802770324.129/warc/CC-MAIN-20141217075250-00029-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://icsecbsemath.com/2019/02/10/class-9-theorems-on-area-lecture-notes/ | Area Axioms
Area Axiom: Every polygonal region has an area and is measured in square units.
Congruent Area Axiom: If $\triangle ABC$ and $\triangle DEF$ are two congruent triangles, then $ar(\triangle ABC) = ar(\triangle DEF)$. ie. two congruent regions have equal area.
Rectangle Area Axiom: If $ABCD$ is a rectangular region such that $AB = a$ units and $BC = b$ units, then $ar(ABCD) = ab$ square units.
Parallelograms on the same base and between the same parallels
Theorem 1: A diagonal of a parallelogram divides it into two triangles of same area
In this case $ar (\triangle ABC) = ar (\triangle ADC)$
Also $ar (\triangle ABD) = ar (\triangle BCD)$
Theorem 2: Parallelograms on the same base and between the same parallels are equal in area.
$ar (ABCD) = AB \times h$
$ar (ABFE) = AB \times h$
$\therefore ar (ABCD) = ar (ABFE)$
Theorem 3: Area of a parallelogram is the product of its base and the corresponding altitude.
Let the two adjacent sides of the parallelogram be $a$ and $b$. Area $= Base \times Height$
Triangle Area Axiom
Theorem 4: The area of a triangle is half the product of any of its sides and the corresponding altitude.
$a, b, c$ denotes the sides of the Triangle. Then:
$\displaystyle \text{Area } = \frac{1}{2} \times Base \times Height = \frac{1}{2} bh$
Area $= \sqrt{s(s-a)(s-b)(s-c)}$. This is known as Heron’s Theorem.
Theorem 5: If a triangle and parallelogram are on the same base and between the same parallels, the area of the triangle is equal to the half of the parallelogram.
$ar (ABCD) = AB \times h$
$\displaystyle ar ( \triangle ABE) = \frac{1}{2} \times AB \times h$
$ar ( \triangle ABE) = ar (ABCD)$
Trapezium Area Axiom
Theorem 6: The area of a trapezium is half the product of its heights and the sum of parallel sides.
A trapezium is a quadrilateral two of whose sides are parallel. A trapezium whose non-parallel sides are equal is known as an isosceles trapezium.
Let $a$ and $b$ be the parallel sides and $h$ be the distance between the parallel sides. $\displaystyle \text{Then Area } = \frac{1}{2} (a+b) \times h$
Triangles on the same base and between the same parallels
Theorem 7: Triangles on the same base and the same parallels have the same area.
$\displaystyle ar ( \triangle ABD) = \frac{1}{2} \times AB \times h$
$\displaystyle ar ( \triangle ABC) = \frac{1}{2} \times AB \times h$
$\displaystyle \therefore ar ( \triangle ABD) = ar ( \triangle ABC)$
Theorem 8: Triangles having equal areas and having one side of one of the triangles equal to one side of the other triangle, have their corresponding altitudes the same.
Theorem 9: Two triangles having the same bases (or equal bases) and equal area lie between the same parallels. | {"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": 56, "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": 0.8732918500900269, "perplexity": 342.26660405096226}, "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-2021-43/segments/1634323585504.90/warc/CC-MAIN-20211022084005-20211022114005-00279.warc.gz"} |
https://mathhelpboards.com/threads/testing-latex-powered-by-mathjax.26/ | ### Welcome to our community
• Moderator
• #1
#### Chris L T521
##### Well-known member
Staff member
Jan 26, 2012
995
$\int_0^{\infty}e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}$
This is inline: $$\frac{\sqrt{b^2-4ac}}{2a}$$
Other centered equation
$$\Gamma(x)=\int_0^{\infty}e^{-t}t^{x-1}\,dt$$
Last edited:
#### The Chaz
##### Member
Jan 26, 2012
24
So LaTex in brackets is centered? Not sure I get the syntax, though I'm trying to scroll through it on the mobile...
#### Ackbach
##### Indicium Physicus
Staff member
Jan 26, 2012
4,203
So \ ( and \ ), with no spaces, gives you inline. \ [ and \ ], with no spaces, gives you displayed. $$x^{2}+4x-4=0$$, but $x^{2}+4x-4=0.$
The first is inline, the second displayed.
#### Prove It
##### Well-known member
MHB Math Helper
Jan 26, 2012
1,469
How odd. I was able to do centred fine using the dollar signs, but the others wouldn't work...
#### Ackbach
##### Indicium Physicus
Staff member
Jan 26, 2012
4,203
How odd. I was able to do centred fine using the dollar signs, but the others wouldn't work...
Yeah, it looks like the double-dollar signs are enabled for displayed equations, but the single are not enabled for inline, although MathJax is capable of doing those as well. Chris mentioned to me in an email that the reason for that is there are quite a few algebra problems concerning money, and newbies would have a hard time escaping those characters.
#### ThePerfectHacker
##### Well-known member
Jan 26, 2012
236
$$\Gamma(x)=\int_0^{\infty}e^{xt}t^{x-1}\,dt$$
Apparently LaTeX does not check for math mistakes.
#### Prove It
##### Well-known member
MHB Math Helper
Jan 26, 2012
1,469
Yeah, it looks like the double-dollar signs are enabled for displayed equations, but the single are not enabled for inline, although MathJax is capable of doing those as well. Chris mentioned to me in an email that the reason for that is there are quite a few algebra problems concerning money, and newbies would have a hard time escaping those characters.
I actually meant starting with the square and round brackets didn't work. Maybe I wasn't doing it right...
Test: $$(a + b)^n = \sum_{r = 0}^n {n\choose{r}}a^{n-r}b^r$$
$\lim_{h \to 0}\frac{\sin{h}}{h} = 1$
Hmm, now it works ><
#### CaptainBlack
##### Well-known member
Jan 26, 2012
890
Test:
Some text then inline $$f(x)=\sin(x^2/23)$$ OK
$f(x)=\frac{1}{\sin(x)}$ OK
$$\displaystyle f(x)=\frac{1}{\sin(x)}$$.....$$f(x)=\frac{1}{\sin(x)}$$ - just messing with the inline form
more text.
The thing that I don't like about the MathJax \ [ ... \ ] delimiters is the amount of white space inserted before and after the LaTeX
Note the [ noparse] [ /noparse] taggs are not working
CB
• Moderator
• #9
#### Chris L T521
##### Well-known member
Staff member
Jan 26, 2012
995
Test:
Some text then inline $$f(x)=\sin(x^2/23)$$ OK
$f(x)=\frac{1}{\sin(x)}$ OK
$$\displaystyle f(x)=\frac{1}{\sin(x)}$$.....$$f(x)=\frac{1}{\sin(x)}$$ - just messing with the inline form
more text.
The thing that I don't like about the MathJax \ [ ... \ ] delimiters is the amount of white space inserted before and after the LaTeX
Note the [ noparse] [ /noparse] taggs are not working
CB
Since this being called from the MathJax js file, there was no need to create a BBcode for it. Hence, noparse will have no effect on it.
#### Ackbach
##### Indicium Physicus
Staff member
Jan 26, 2012
4,203
Since this being called from the MathJax js file, there was no need to create a BBcode for it. Hence, noparse will have no effect on it.
So question: how do we output raw LaTeX code? Verbatim?
• Moderator
• #11
#### Chris L T521
##### Well-known member
Staff member
Jan 26, 2012
995
So question: how do we output raw LaTeX code? Verbatim?
It turns out you can do that using the code tags. For example,
Code:
$\displaystyle\int_{\partial M} \omega$
would compile to $$\large{\displaystyle{\int_{\partial M} \omega}}$$.
Last edited by a moderator:
#### sbhatnagar
##### Active member
Jan 27, 2012
95
Writing Inline LaTeX
LaTeX without "\displaystyle"
Code:\int_{0}^{\infty}\frac{\ln(x)}{(x-a)(x-b)}dx
$$\int_{0}^{\infty}\frac{\ln(x)}{(x-a)(x-b)}dx$$
LaTeX with "\displaystyle"
Code:\displaystyle \int_{0}^{\infty}\frac{\ln(x)}{(x-a)(x-b)}dx
$$\displaystyle \int_{0}^{\infty}\frac{\ln(x)}{(x-a)(x-b)}dx$$
#### Markov
##### Member
Feb 1, 2012
149
Re: Writing Inline LaTeX
I just realized the LaTeX is showing itself nicely! That's good.
Perhaps a little suggestion: I love to render LaTeX by using the dollar sign, but, is there a way to render fractions, sums, integrals, etc by not using \displaystyle? It's annoying to write it everytime!
#### Ackbach
##### Indicium Physicus
Staff member
Jan 26, 2012
4,203
Re: Writing Inline LaTeX
Try double-dollar signs (two to open, two to close).
#### Markov
##### Member
Feb 1, 2012
149
Re: Writing Inline LaTeX
Well it didn't work that well, because by using doble dollar sign, it will center the content, and I don't always need to center the content.
#### afwings
##### New member
Mar 7, 2012
9
Just thought I'd test out some things too. Here's an inline equation $$u = \phi \cdot \exp \left\{ {{\textstyle{1 \over 2}}\sigma \left( {x + y} \right)} \right\}$$ and here's a display (centered) equation ${\rm{Bond Value}} = {\rm{C}}\left[ {\frac{{1 - \frac{1}{{{{\left( {1 + {\rm{r}}} \right)}^{\rm{t}}}}}}}{{\rm{r}}}} \right] + \frac{{{\rm{FV}}}}{{{{\left( {1 + {\rm{r}}} \right)}^{\rm{t}}}}}$
So, how'd that work out?
#### afwings
##### New member
Mar 7, 2012
9
Just thought I'd test out some things too. Here's an inline equation $$u = \phi \cdot \exp \left\{ {{\textstyle{1 \over 2}}\sigma \left( {x + y} \right)} \right\}$$ and...
Ok, that's something I just noticed -- following the MathJax close delimiter -- \) -- even if I add a space, the space doesn't show up in the post. Wonder what happens if I add 2 spaces? Here's a shot at it: $${b^2} - 4ac % MathType!MTEF!2!1!+- % feaagGart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbdfgBPj % MCPbqedmvETj2BSbqefm0B1jxALjhiov2DGCKCLv2AGW0B3bqefqvA % Tv2CG4uz3bIuV1wyUbqee0evGueE0jxyaibaieYhf9irFfeu0dXdh9 % vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqai-hGuQ8kuc9pgc9 % q8qqaq-dir-f0-yqaiVgFr0xfr-xfr-xb9adbaqaaeGaciGaaiaabe % qaamaaeaqbaaGcbaGaamOyamaaCaaaleqabaGaaGOmaaaakiabgkHi % TiaaisdacaWGHbGaam4yaaaa!4201!$$ ok, I added 2 spaces, and still one showed up. Nice to know. (If it's already been pointed out, sorry I missed it.)
Trying something else. Here's a simple matrix: $\left[ {\begin{array}{*{20}{c}}{10}&{20}\\{30}&4\end{array}} \right] % MathType!MTEF!2!1!+- % feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbdfgBPj % MCPbqedmvETj2BSbqefm0B1jxALjhiov2DGCKCLv2AGW0B3bqefqvA % Tv2CG4uz3bIuV1wyUbqee0evGueE0jxyaibaieYhf9irFfeu0dXdh9 % vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqai-hGuQ8kuc9pgc9 % q8qqaq-dir-f0-yqaiVgFr0xfr-xfr-xb9adbaqaaeGaciGaaiaabe % qaamaaeaqbaaGcbaWaamWaaeaafaqabeGacaaabaGaaGymaiaaicda % aeaacaaIYaGaaGimaaqaaiaaiodacaaIWaaabaGaaGinaaaaaiaawU % facaGLDbaaaaa!43CF!$
#### Jameson
Staff member
Jan 26, 2012
4,093
I'm not following what's going on in the matrix you made. Is that the intended output or did something go wrong?
#### afwings
##### New member
Mar 7, 2012
9
I'm not following what's going on in the matrix you made. Is that the intended output or did something go wrong?
Jameson, thanks for following up on that. There are really 2 issues shown in my post, neither of which I believe to be MHB's error.
1. When viewing the forum in FF16 for Windows (and only in that configuration), the character immediately following an inline equation rendered by MathJax overlaps the right edge of the equation. If a period follows the equation, the period ends up inside the equation, like this one (again, this effect only happens on Windows, only in Firefox, and only with Math Renderer > MathML): $$\left( {x,y} \right) % MathType!MTEF!2!1!+- % feaagGart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbdfgBPj % MCPbqedmvETj2BSbqefm0B1jxALjhiov2DGCKCLv2AGW0B3bqefqvA % Tv2CG4uz3bIuV1wyUbqee0evGueE0jxyaibaieYhf9irFfeu0dXdh9 % vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqai-hGuQ8kuc9pgc9 % q8qqaq-dir-f0-yqaiVgFr0xfr-xfr-xb9adbaqaaeGaciGaaiaabe % qaamaaeaqbaaGcbaWaaeWaaeaacaWG4bGaaiilaiaadMhaaiaawIca % caGLPaaaaaa!40E2!$$. I've filed this as a MathJax bug, but it may be a Firefox bug. There are other things going on in Firefox 16. (And it may already be corrected; hard to tell.)
2. The second issue is related to how MathType codes matrices with the Math Help Boards translator, which will be added to Cut & Copy Preferences when MathType 6.9 is released (no projected release date yet). We've already corrected this.
Bob
#### Jameson
Staff member
Jan 26, 2012
4,093
Hi Bob,
Glad it's all working then. I notice in your post how a long string of code renders a very short $$\displaystyle \left( {x,y} \right)$$. I'm guessing this is part of the Math Type plugin?
Anyway, it sounds like you're more than on top of things so I won't ask you to explain it all to me. Just let me know if there's anything I can do and if any of the issues are on our end
Best,
Jameson
#### Jay
##### New member
Aug 19, 2014
14
So question: how do we output raw LaTeX code? Verbatim?
You insert LaTeX code verbatim like this using CODE and /CODE tags enclosed within square brackets. Everything between the tags will be listed verbatim.
Code:
$$X = \sum_{n=0}^5\left(\left(\sum_{j=1}^k A_{X,n,j} \cdot {cos(B_{X,n,j} + t \cdot C_{X,n,j})}\right) \cdot t^n\right)$$
which means:
$$X = \sum_{n=0}^5\left(\left(\sum_{j=1}^k A_{X,n,j} \cdot {cos(B_{X,n,j} + t \cdot C_{X,n,j})}\right) \cdot t^n\right)$$
Last edited: | {"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": 2, "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": 0.8584708571434021, "perplexity": 4662.702133482664}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00493.warc.gz"} |
http://mathhelpforum.com/number-theory/202559-investigate-equality.html | # Math Help - Investigate Equality
1. ## Investigate Equality
Problem: Investigate following equality:
$\frac{2}{\pi }=\frac{\sqrt{2}}{2} \frac{\sqrt{2+\sqrt{2}}}{2} \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2} \text{...}$
Thanks.
2. ## Re: Investigate Equality
What is there to investigate? Do you want to prove the equality? Do you want to formulate a partial product and see if it converges?
4. ## Re: Investigate Equality
My thorough investigation revealed that this is Viète's formula. Now what?
5. ## Re: Investigate Equality
Thanks a lot Vlasev, you are truly a good n efficient investigator...good to know the origin of the problem.
6. ## Re: Investigate Equality
You can define a circle to be a regular polygon with an infinite number of sides. We can define \displaystyle \begin{align*} \pi \end{align*} as the circumference of a circle of diameter equal to 1 unit. Therefore, we can get an approximation for \displaystyle \begin{align*} \pi \end{align*} by evaluating the perimeter of said regular polygon.
Some algebraic manipulation of this will give the result you are trying to prove.
7. ## Re: Investigate Equality
This explanation is truly enlightening. Thanks a lot ProveIt.
8. ## Re: Investigate Equality
That's indeed a magnificent proof! | {"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": 0.994012176990509, "perplexity": 2370.008897634011}, "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-2014-52/segments/1418802777454.142/warc/CC-MAIN-20141217075257-00068-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://thebeardsage.com/short-primer-on-probability/ | # Short Primer on Probability
## Fundamentals
A random variable is denoted in capital, and the values it can take is denoted in small .
Consider a collection of random variables . Random variables can be thought of as features of a particular domain of interest.
For example, the result of a coin toss can be represented using a single random variable, . This variable can take either of the categorical values or . If the same coin is tossed times, this can be represented using variables . Each of these values can be either or .
## Joint Probability
An expression of the form is called a joint probability function over the variables . The joint probability is defined when the values of are respectively. This is denoted by the expression
This is sometimes abbreviated as .
For a fair coin toss, . If a fair coin is tossed five times, .
The joint probability function satisfies
## Marginal Probability
The marginal probability of one of the random variables can be computed if the values of all of the joint probabilities for a set of random variables are known.
For example, the marginal probability is defined to be the sum of all those joint probabilities for which
When dealing with propositional variables (True/False) TrueFalse is denoted as .
## Conditional Probabilities
The conditional probability of given is denoted by .
where is the joint probability of and and is the marginal probability of . Thus
Joint conditional probabilities of several variables conditioned on several other variables is expressed as
A joint probability can be expressed in terms of a chain of conditional probabilities.
The general form of this chain rule is
## Bayes Rule
Different possible orders give different expressions but they all have the same value for the same set of variable values. Since the order of variables is not important
Which gives Bayes’ Rule
## Probabilistic Inference
In set notation , , The variables having the values respectively is denoted by , where and are ordered lists.
For a set , the variables in a subset of are given as evidence.
For example, consider . The evidence is being false. In other words equates to .
Thus need not be computed.
## Conditional Independence
A variable is conditionally independent of a set of variables given a set if
(1)
(2)
Saying that is conditionally independent of given also means that is conditionally independent of given . The same result also applies to sets and .
As a generalization of pairwise independence, the variables are mutually conditionally independent, given a set if each of the variables is conditionally independent of all of the others given .
(3)
When is empty
This implies that the variables are unconditionally independent.
### Thank You
• Mark – for pointing out typos and errors
## 2 Comments →Short Primer on Probability
1. Mark
Another great article! Nice, clear explanations.
I do have some suggestions for improvement though:
It seems there is an error in the Bayes Rule section. It should be P(V_i|V_j) = P(V_j|V_i)P(V_i)/P(V_j)
I would also recommend explaining how you introduced the P and not P random variables in the Probabilistic Inference section as it is not quite clear how that works out.
Finally, it seems your Latex implementation had trouble rendering for the fourth paragraph in the Conditional Independence section (\textbf{mutually conditionally independent}), right above equation (3).
1. TheBeard
I have expanded the Probabilistic Inference section and fixed the errors you have pointed out. Thank you. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.9611799716949463, "perplexity": 577.5965722055848}, "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/1664030334644.42/warc/CC-MAIN-20220926020051-20220926050051-00671.warc.gz"} |
http://openstudy.com/updates/503e0c29e4b0c29115be4cfe | Here's the question you clicked on:
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## anonymous 3 years ago y varies inversely as twice x. When x = 4, y = 1. Find y when x = 5. Delete Cancel Submit
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1. anonymous
• 3 years ago
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y varies inversely as twice x will be something like $y = \frac{k}{2x}$ does that help? or do you need more hints?
2. anonymous
• 3 years ago
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i need more hints
3. anonymous
• 3 years ago
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good day vegas14, this is an example of a proportionality problem. what you need to do is read the problem word for word, each word will be giving you hint as to how to approach the problem. I will list the possible cases in this type of problem, I hope that it will help you. caseI: if the problem states that "y varies directly to x" it means that y is directly proportional to x "but", it will be proportional only if you multiply a proportionality constant "k" $y \alpha x$ this equation (eq1) symbolizes the direct proportionality of y to x, to be able to replace the proportionality symbol alpha to an equal (=) sign you must multiply x to a proportionality constant which is "k". therefore transforming the equation to, $y=k(x)$
4. anonymous
• 3 years ago
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caseII: (this is the case of your problem) when the problem states that "y varies inversely as x" it means that your equation would be like this, $y \alpha \frac{ 1 }{ x }$ from this equation you should again have to replace the proportionality symbol "alpha" to an equal (=) sign. transforming the equation to be,$y=k(\frac{ 1 }{ x })$ which if simplified will become $y=\frac{ k }{ x }$
5. anonymous
• 3 years ago
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of course problems may vary in this cases, for example "it may state twice as x therefore you should use "2x" or thrice as x use "3x" or may be as the square of x use $x^{2}$ it will now depend on your own analysis of the problem.
6. anonymous
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i really wanted to solve the problem for you, but it might be best if you do it on your own, just to test yourself if you really understand the logic behind =) but just to give you more tips: 1) Apply CaseII 2) Find the proportionality constant "k" ---- just this I will not give you which on the given you will use to solve for "k" =) 3) Solve for y(new) I hope this posts will help you. will check your answers soon. . . =)
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy. | {"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": 0.973931610584259, "perplexity": 1390.408053582421}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049270555.40/warc/CC-MAIN-20160524002110-00096-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://gorchilin.com/articles/energy/RLC_3?lang=en | 2017-07-27
Research website of Vyacheslav Gorchilin
All articles
Parametric change of inductance in a RL-circuit. Back EMF
In this note first we prove the theoretical possibility of obtaining additional energy from the back EMF in the inductor core. The proof is based on the classic formulas of electrical engineering. According to the classification here will be considered the generator of the first kind of first order partial cycle PCCIE.
In our time, divorced many myths and legends about the inexhaustible possibilities of back EMF (AEDS) in the inductor. According to some researchers, OADS can give more energy than it expended, and their experiences, in some cases, confirm this. Theorists explain such powers by the ether theory or the unused energy of the atoms of ferromagnetic materials. We will try to draw conclusions based on mathematics and the theory of electrical circuits [1], which worked well and can fully display the necessary processes. Thus, the author reiterates the idea that issledovatelyam it is not necessary to go into the jungle of other theories and hypotheses, enough of a different angle to look at the classics.
Next, we will show that the school or even high school — a nonparametric OATS — can increase the efficiency of the second kind $$(K_{\eta2})$$, so the outset that we consider only parametric coil which changes its inductance depending on the magnetic field $$(H)$$ in it. And since this tension is directly proportional to the current, as parameter we will have to address it. To convert the current back to $$H$$ is not extremely difficult, you need to know the parameters of the coil and core, i.e. the parameters of a particular device.
In the previous section we showed that a parametric change in capacitance in the full cycle: charge-discharge, does not increase $$K_{\eta2}$$. The same can be said of inductance: it is enough to replace the voltage on the currents and to change some of the ratios in the diff. equation. But with the coil inductance, which will contain a ferromagnetic core, we can do a little differently.
A typical graph of the magnetic permeability of the core is $$(\mu)$$ on the magnetic field $$H$$ to the left [2]. But the tension is proportional to the current, and the permeability of the inductance, this means that we can build a graph $$L(I)$$, which is proportional to $$\mu(H)$$. Thus, we obtain the parametric dependence of the inductance $$L$$ from the current $$I$$ flowing in it, and with this dependence, we will continue to work.
As an example, compare the two charts to the actual measured characteristics of the ferrite from the flyback transformer: graph 1, graph 2. On the second chart the y coordinate $$M(I)$$ shows how changing the inductance of the coil compared to the original, and $$I$$ is the current in the coil.
Recall that in this note we investigate AEDS, therefore, assume that the initial current $$I_0$$ in the coil already, and it will pass through the electric circuit consisting of inductance and resistance. How there came a shock we will discuss later, but for now we will use the classical formula for the calculation of the energy in the coil at the time of closure of the SW key: $W_L = \frac{L_0 I_0^2}{2} \qquad (3.1)$ where $$L_0$$ is the inductance of the coil at the time of closure of key SW. Now our task is this: to properly use this energy. It is necessary to find a mode in which the current in the circuit will produce the most work on active loading. The classic formula for this energy is: $W_R = R \int_0^T I(t)^2 dt \qquad (3.2)$ Where $$I(t)$$ is the current in the circuit depending on time, which is unknown to us. In order to find it, it is necessary to compile the differential equation for the transition process in our scheme [1]: ${L \over R} I(t)' + I(t) = 0 \qquad (3.3)$ where the inductance $$L$$ is a parameter from the current, and the current in turn — on time. For the solution of the equation it remains to determine the dependence of the inductance from the current for which we can take the polynomial $$M(I)$$ with coefficients $$..k_1 k_4$$. Its convenience — flexibility for various dependencies: $L = L_S\, M(I), \quad M(I) = {1 + k_1 I(t) + k_2 I(t)^2 \over 1 + k_3 I(t) + k_4 I(t)^3} \qquad (3.4)$ where $$L_S$$ is the initial inductance of the coil (no current). $$M(I)$$ must be proportional to $$\mu$$ from $$H$$ to the real core.
We introduce the time constant of the circuit $$\tau = L_S/R$$, which will record the final shape of the diff. equation: $\tau\, M(I)\, I' + I = 0 , \quad I=I(t) \qquad (3.5)$ the Analytical solution of this equation in this form is complicated, so we use mathematical editor MathCAD and get it in a numerical form. In the editor we appreciate the real time operation of the circuit after closing of the key, we denote it $$T$$ and make this time a finite number.
Why do we need time?
Indeed, if we want to obtain the dependence of the increase in efficiency of $$\mu$$ (see top graph), why should we use one additional variable — the time? Will try to fix it. We rewrite (3.6) in another form: $-\tau\, M(I)\, I' = I, \quad I=I(t) \qquad (3.6)$ and substitute into the formula (3.2): $W_R = R \int_0^T \left[\tau\, M(I)\, I'\right]^2 dt \qquad (3.7)$ After some mathematical operations, the author obtained a rather unusual integral in which energy dissipation on resistance does not depend on $$t$$, and even from the resistance: $W_R = \frac{L_S}{2} \int_0^{I_0^2} M(\sqrt{J})\ dJ , \quad J = I^2 \qquad (3.8)$ Here you need to pay attention to the fact that in paginegialle functions are all $$I$$ change for $$J$$ by the rule: $$I=\sqrt{J}$$. To find the final formula calculates the energy gain, divide the energy dispersed on the active resistance on the starting energy in the coil is: $K_{\eta2} = {W_R \over W_L} \qquad (3.9)$ Given that $W_L = {L_0\, I_0^2 \over 2} = {L_S\, M(I_0)\, I_0^2 \over 2}, \quad M(I_0)= {1 + k_1 I_0 + I_0 k_2^2 \over 1 + I_0 k_3 + k_4 I_0^3} \qquad (3.10)$ substitute the previously obtained results and we get an incredibly interesting pattern that does not depend on the time coordinate: $K_{\eta2} = \frac{1}{M(I_0)\, I_0^2} \int_0^{I_0^2} M(\sqrt{J})\, dJ , \quad J = I^2 \qquad (3.11)$ Or this same formula in another form: $K_{\eta2} = \frac{2}{M(I_0)\, I_0^2} \int_0^{I_0} M(I) \, I \, dI \qquad (3.12)$ Well, if we all want the same thing to Express through the magnetic permeability of the core and the magnetic field $$\mu(H)$$, then the formula (3.12), simply replace the appropriate letters: $K_{\eta2} = \frac{2}{\mu (H_0)\, H_0^2} \int_0^{H_0} \mu (H) \, H \, dH \qquad (3.13)$
Equations (3.11-3.13) and are the mathematical expression of free energy for back EMF in the coil core! A more General approach for the derivation of this formula see here
From it immediately clear that if a parametric dependency is missing, i.e. $$M(I)=M(J)=1$$, then a raise is not: $$K_{\eta2}=1$$. We have to learn to use this formula and find out whether there can be $$K_{\eta2}$$ is greater than one.
What is the result?
For example, take $$M(I)$$ from formulas (3.4) and substitute it into (3.11) replacing $$I$$ to $$J$$ by the rule: $$I=\sqrt{J}$$. The time dependence of ex and obtain: $K_{\eta2} = \frac{1}{M(I_0)\, I_0^2} \int_0^{I_0^2} {1 + k_1 J^{0.5} + k_2 J \over 1 + k_3 ^{0.5} + k_4 ^{1.5}}\, dJ \qquad (3.14)$ Again for example, enter the following current, and the coefficients are $$I_0=1.4, k_1=10, k_2=0, k_3=0, k_4=5$$. The left side shows a graph of the inductance of the coil from the current in accordance with these coefficients. As you can see, we work in growing, and in the drop-down field magnetization curve of the core and primary inductance is less than the maximum about 4 times, which is close to the real values of permeability of ferromagnets. Solving the integral (3.14) we get $$K_{\eta2}=2.07$$ is greater than one!
To substitute other factors and to get your results you can in the editor MathCAD downloading there this program or to test it in the calculator. It a visible pattern, which confirms the assumption that the drop down part of the curve $$\mu$$ from $$H$$ gives the largest contribution to the energy increase. In addition to choosing the work area on the chart, to achieve a positive result, it is necessary that the device provide access to the coil for the desired mode of current and magnetic field strength.
The rationale for the occurrence of additional energy in such devices, see here
What in practice?
You need to understand that we get according to equations (3.11-3.13) is the marginal value. Depending on the material of the core, when working on the right drop-down plot of $$M(I)$$, you may experience a variety of losses, mostly heat. But they will not always be equal to this increase, and therefore is described here the way to super devices still remains open. Izvestia of materials, good performance $$K_{\eta2}$$ should give Permalloy, on which chart, you can easily find the working site, even better — Metglas [8]. A little worse, in this sense, the situation with ferrites. Promising look some modern steels and certain core construction [5], however, we must not forget that part of the energy can be spent on Foucault.
Permalloy is working in a relatively small frequency and therefore the device running on it cannot develop large capacities. Ferrite can withstand orders of magnitude for large frequencies, but it is much more fragile. However, as a result, it can operate at high capacity.
The question remains open: how do you receive the initial current $$I_0$$ in the coil and how long it takes for the energy? For answer there are different approaches. One of them suggests that the core need to podmanivaya perpendicular to the primary field [3], and it can be done with short pulse whose energy is several times lower than required due to the inertia of a ferromagnet. The second approach [4-6] involves the ramp-up without additional perpendicular field. The third offers mechanical excitation of the core is a permanent magnet, and thus the initial current appears in the coil [7].
Given all the above it seems to us a very real seredinny receiving devices based on coils with ferromagnetic core. As you know, if math gave the green light, then the practical implementation will not take long.
The results of this work was done by a specialized calculatorthat allows you to find the dependence of magnetic permeability of any core on the magnetic field and to calculate it is potentially achievable increment of efficiency of the second kind.
The materials used | {"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": 0.8721423149108887, "perplexity": 298.5374576890251}, "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-2018-34/segments/1534221211935.42/warc/CC-MAIN-20180817084620-20180817104620-00475.warc.gz"} |
http://mathhelpforum.com/calculus/11275-calculus-vector-valued-func.html | # Thread: Calculus of Vector-Valued Func.
1. ## Calculus of Vector-Valued Func.
1.) Given: r(t)=tcos(t)i+e(t^2)j+ln(t)k,
what is r'(t)?
2.) Suppose r(t) is a vector-valued func.,
What geometric/graphical info does r'(a) tell us?
3.) Let r(t)=cos(t)i+sin(t)j and s(t)=sin(5t)i+cos(5t)j.
(a) What does the graphs of r(t) and s(t) look like?
(b) Suppose the graphs of two vector-valued func. r(t) and s(t) are the same, then must r'(0)=s'(0)? (Explain whether this is a new result, or was it also true for functions f(x) and g(x)?)
2. Hello, fifthrapiers!
Here's some help . . .
1) Given: . $\vec{r}(t)\;=\;t\cos(t)\vec{i} + e^{t^2}\vec{j} + \ln(t)\vec{k}$. . Find $\vec{r'}(t)$
$\vec{r'}(t)\;=\;(\cos t - t\sin t)\vec{i} + \left(2te^{t^2}\right)\vec{j} + \left(\frac{1}{t}\right)\vec{k}$
3) Let $r(t)\:=\:\cos(t)i + \sin(t)j$ and $s(t)\:=\:\sin(5t)i + \cos(5t)j$
(a) What does the graphs of $r(t)$ and $s(t)$ look like?
Both are unit circles, centered at the origin.
(b) Suppose the graphs of two vector-valued func. $r(t)$ and $s(t)$ are the same,
. . .then must $r'(0) = s'(0)$? . no
Consider: . $\begin{array}{cc}\vec{r}(t) \;= & \cos(t)\vec{i} + \sin(t)\vec{j} \\ \vec{s}(t) \:= & \cos(t + \pi)\vec{i} + \sin(t + \pi)\vec{j} \end{array}$
When $t = 0:$
. . $\begin{array}{ccc}r(0) = \langle 1,\,0\rangle & \text{ and } & r'(0) = \langle 0,\,1\rangle\!:\;\uparrow \\ \\
s(0) = \langle \text{-}1,0\rangle & \text{ and } &s'(0) = \langle 0,\,\text{-}1\rangle\!: \;\downarrow\end{array}$
. . . different derivatives
Explain whether this is a new result,
or was it also true for functions $f(x)$ and $g(x)$ ?
This is a new result.
With rectangular functions, $y = f(x)$ and $y = g(x)$,
. . if their graphs are identical, then the functions are identical.
. . Hence, their derivatives are equal.
With parametric functions, a graph can be generated in a number of ways.
. . As seen in part (b), their derivatives need not be equal. | {"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": 17, "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": 0.9071699976921082, "perplexity": 1937.1617174127648}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988725451.13/warc/CC-MAIN-20161020183845-00090-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/361856/why-is-quantum-mechanics-called-01-dimensional-qft/361858 | # Why is quantum mechanics called 0+1 dimensional QFT?
I know that quantum mechanics is sometimes called 0+1 dimensional quantum field theory. What is the meaning? How should we understand it?
In field theory, a field can be thought of as a map from the spacetime $M$, usually a Lorentzian manifold---a particularly popular choice is $\Bbb R^{1,n-1}$ (Minkowski space)---to some other space. For instance, a scalar field $\phi$ can be viewed as a map $\phi:M\to \Bbb R$, or equivalently as a global section of the trivial (real) line bundle over $M$. The spacetime $M^n$ has one timelike direction and $n-1$ spacelike directions, and one can say that one studies $(n-1)+1$-dimensional field theory.
When doing mechanics, what do we use for a "field"? The position of the particle(s)! The position depends only on time, and hence we have maps $x_i:\Bbb R\to \Bbb R^n$ (in case the space in which the particles move is not simply $\Bbb R^n$, the target may be some other Riemannian manifold of dimension $n$) and fitting this into the general QFT picture we can identify $\Bbb R$ as our "spacetime", where we now have no spatial directions and only a timeline direction. Thus, we may call the theory a $0+1$-dimensional field theory. After quantization, one obtains a $0+1$-dimensional quantum field theory. | {"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": 0.8996157646179199, "perplexity": 252.17069335200685}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669755.17/warc/CC-MAIN-20191118104047-20191118132047-00089.warc.gz"} |
https://enacademic.com/dic.nsf/enwiki/26564/Thermodynamic_temperature | # Thermodynamic temperature
Thermodynamic temperature is the absolute measure of temperature and is one of the principal parameters of thermodynamics. Thermodynamic temperature is an “absolute” scale because it is the measure of the fundamental property underlying temperature: its "null" or zero point, absolute zero, is the temperature at which the particle constituents of matter have minimal motion and can be no colder.
## Overview
[
Fig. 1 The "translational motion" of fundamental particles of nature such as atoms and molecules gives a substance its temperature. Here, the size of helium atoms relative to their spacing is shown to scale under 1950 atmospheres of pressure. These room-temperature atoms have a certain, average speed (slowed down here two trillion fold). At any given instant however, a particular helium atom may be moving much faster than average while another may be nearly motionless. Five atoms are colored red to facilitate following their motions.] Temperature arises from the random submicroscopic vibrations of the particle constituents of matter. These motions comprise the kinetic energy in a substance. More specifically, the thermodynamic temperature of any bulk quantity of matter is the measure of the average kinetic energy of a certain kind of vibrational motion of its constituent particles called "translational motions." Translational motions are ordinary, whole-body movements in three-dimensional space whereby particles move about and exchange energy in collisions. "Fig. 1 "at right shows translational motion in gases; " "below shows translational motion in solids. Thermodynamic temperature’s null point, absolute zero, is the temperature at which the particle constituents of matter are as close as possible to complete rest; that is, they have motion, retaining only quantum mechanical motion. While scientists are achieving temperatures ever closer to absolute zero, they can not fully achieve a state of "“zero”" temperature. However, even if scientists could remove "all" kinetic heat energy from matter, quantum mechanical "zero-point energy" (ZPE) causes particle motion that can never be eliminated. Encyclopedia Britannica Online [http://britannica.com/eb/article-9078341 defines zero-point] energy as the “"vibrational energy that molecules retain even at the absolute zero of temperature."” ZPE is the result of all-pervasive energy fields in the vacuum between the fundamental particles of nature; it is responsible for the Casimir effect and other phenomena. See " [http://calphysics.org/zpe.html Zero Point Energy and Zero Point Field] ", which is an excellent explanation of ZPE by Calphysics Institute. See also " [http://www.phys.ualberta.ca/~therman/lowtemp/projects1.htm Solid Helium] " by the University of Alberta’s Department of Physics to learn more about ZPE’s effect on Bose–Einstein condensates of helium.
Although absolute zero ("T"=0) is not a state of zero molecular motion, it "is "the point of zero temperature and, in accordance with the Boltzmann constant, is also the point of zero particle kinetic energy and zero kinetic velocity. To understand how atoms can have zero kinetic velocity and simultaneously be vibrating due to ZPE, consider the following thought experiment: two "T"=0 helium atoms in zero gravity are carefully positioned and observed to have an average separation of 620 pm between them (a gap of ten atomic diameters). It’s an “average” separation because ZPE causes them to jostle about their fixed positions. Then one atom is given a kinetic kick of precisely 83 yoctokelvin (1 yK = 1 × 10–24 K). This is done in a way that directs this atom’s velocity vector at the other atom. With 83 yK of kinetic energy between them, the 620-pm gap through their common barycenter would close at a rate of 719 pm/s and they would collide after 0.862 second. This is the same speed as shown in the "Fig. 1 "animation above. Before being given the kinetic kick, both "T"=0 atoms had zero kinetic energy and zero kinetic velocity because they could persist indefinitely in that state and relative orientation even though both were being jostled by ZPE. At "T"=0, no kinetic energy is available for transfer to other systems. The Boltzmann constant and its related formulas describe the realm of particle kinetics and velocity vectors whereas ZPE is an energy field that jostles particles in ways described by the mathematics of quantum mechanics. In atomic and molecular collisions in gases, ZPE introduces a degree of "chaos", i.e., unpredictability, to rebound kinetics; it is as likely that there will be "less" ZPE-induced particle motion after a given collision as "more." This random nature of ZPE is why it has no net effect upon either the pressure or volume of any "bulk quantity" (a statistically significant quantity of particles) of "T">0 K gases. However, in "T"=0 condensed matter; e.g., solids and liquids, ZPE causes inter-atomic jostling where atoms would otherwise be perfectly stationary. Inasmuch as the real-world effects that ZPE has on substances can vary as one alters a thermodynamic system (for example, due to ZPE, helium won’t freeze unless under a pressure of at least 25 bar), ZPE is very much a form of heat energy and may properly be included when tallying a substance’s internal energy.
Note too that absolute zero serves as the baseline atop which thermodynamics and its equations are founded because they deal with the exchange of heat energy between "“systems”" (a plurality of particles and fields modeled as an average). Accordingly, one may examine ZPE-induced particle motion "within" a system that is at absolute zero but there can never be a net outflow of heat energy from such a system. Also, the peak emittance wavelength of black-body radiation shifts to infinity at absolute zero; indeed, a peak no longer exists and black-body photons can no longer escape. Due to the influence of ZPE however, "virtual" photons are still emitted at "T"=0. Such photons are called “virtual” because they can’t be intercepted and observed. Furthermore, this "zero-point radiation" has a unique "zero-point spectrum." However, even though a "T"=0 system emits zero-point radiation, no net heat flow "Q" out of such a system can occur because if the surrounding environment is at a temperature greater than "T"=0, heat will flow inward, and if the surrounding environment is at "T"=0, there will be an equal flux of ZP radiation both inward and outward. A similar "Q "equilibrium exists at "T"=0 with the ZPE-induced “spontaneous” emission of photons (which is more properly called a "stimulated" emission in this context). The graph at upper right illustrates the relationship of absolute zero to zero-point energy. The graph also helps in the understanding of how zero-point energy got its name: it is the vibrational energy matter retains at the "“zero kelvin point.”" Citation: "Derivation of the classical electromagnetic zero-point radiation spectrum via a classical thermodynamic operation involving van der Waals forces", Daniel C. Cole, Physical Review A, Third Series 42, Number 4, 15 August 1990, Pg. 1847–1862.] Zero kinetic energy remains in a substance at absolute zero (see "Heat energy at absolute zero", below).
Throughout the scientific world where measurements are made in SI units, thermodynamic temperature is measured in kelvins (symbol: K). Many engineering fields in the U.S. however, measure thermodynamic temperature using the Rankine scale.
By [http://www1.bipm.org/en/si/si_brochure/chapter2/2-1/2-1-1/kelvin.html international agreement,] the unit “kelvin” and its scale are defined by two points: absolute zero, and the triple point of Vienna Standard Mean Ocean Water (water with a specified blend of hydrogen and oxygen isotopes). Absolute zero—the coldest possible temperature—is defined as being precisely 0 K "and" −273.15 °C. The triple point of water is defined as being precisely 273.16 K "and" 0.01 °C. This definition does three things:
#It fixes the magnitude of the kelvin unit as being precisely 1 part in 273.16 parts the difference between absolute zero and the triple point of water;
#It establishes that one kelvin has precisely the same magnitude as a one-degree increment on the Celsius scale; and
#It establishes the difference between the two scales’ null points as being precisely 273.15 kelvins (0 K = −273.15 °C and 273.16 K = 0.01 °C).
Temperatures expressed in kelvins are converted to degrees Rankine simply by multiplying by 1.8 as follows: "T"K × 1.8 = "T"°R, where "T"K and "T"°R are temperatures in kelvins and degrees Rankine respectively. Temperatures expressed in Rankine are converted to kelvins by "dividing" by 1.8 as follows: "T"°R ÷ 1.8 = "T"K.
## Table of thermodynamic temperatures
The full range of the thermodynamic temperature scale and some notable points along it are shown in the table below.
A For Vienna Standard Mean Ocean Water at one standard atmosphere (101.325 kPa) when calibrated strictly per the two-point definition of thermodynamic temperature.
B The 2500 K value is approximate. The 273.15 K difference between K and °C is rounded to 300 K to avoid false precision in the Celsius value.
C For a true blackbody (which tungsten filaments are not). Tungsten filaments’ emissivity is greater at shorter wavelengths, which makes them appear whiter.
D Effective photosphere temperature. The 273.15 K difference between K and °C is rounded to 273 K to avoid false precision in the Celsius value.
E The 273.15 K difference between K and °C is ignored to avoid false precision in the Celsius value.
F For a true blackbody (which the plasma was not). The Z machine’s dominant emission originated from 40 MK electrons (soft x–ray emissions) within the plasma.
The relationship of temperature, motions, conduction, and heat energy
The nature of kinetic energy, translational motion, and temperature
At its simplest, “temperature” arises from the kinetic energy of the vibrational motions of matter’s particle constituents (molecules, atoms, and subatomic particles). The full variety of these kinetic motions contribute to the total heat energy in a substance. The relationship of kinetic energy, mass, and velocity is given by the formula "Ek" = frac|2"m" • "v" 2. [At non-relativistic temperatures of less than about 30 GK, classical mechanics are sufficient to calculate the velocity of particles. At 30 GK, individual neutrons (the constituent of neutron stars and one of the few materials in the universe with temperatures in this range) have a 1.0042 γ (gamma or Lorentz factor). Thus, the classic Newtonian formula for kinetic energy is in error less than half a percent for temperatures less than 30 GK.] Accordingly, particles with one unit of mass moving at one unit of velocity have precisely the same kinetic energy—and precisely the same temperature—as those with four times the mass but half the velocity.
.
The extent to which the kinetic energy of translational motion of an individual atom or molecule (particle) in a gas contributes to the pressure and volume of that gas is a proportional function of thermodynamic temperature as established by the Boltzmann constant (symbol: "kB"). The Boltzmann constant also relates the thermodynamic temperature of a gas to the mean kinetic energy of an individual particle’s translational motion as follows:
:"Emean" = frac|3|2"kBT" ::where…::"Emean" is the mean kinetic energy in joules (symbol: J)::"kB" = val|1.3806504|(24)|e=-23|u=J/K::"T" is the thermodynamic temperature in kelvins
While the Boltzmann constant is useful for finding the mean kinetic energy of a particle, it’s important to note that even when a substance is isolated and in thermodynamic equilibrium (all parts are at a uniform temperature and no heat is going into or out of it), the translational motions of individual atoms and molecules occurs across a wide range of speeds (see animation in "Fig. 1 "above). At any one instant, the proportion of particles moving at a given speed within this range is determined by probability as described by the Maxwell–Boltzmann distribution. The graph shown here in "Fig. 2 " shows the speed distribution of 5500 K helium atoms. They have a "most probable" speed of 4.780 km/s (0.2092 s/km). However, a certain proportion of atoms at any given instant are moving faster while others are moving relatively slowly; some are momentarily at a virtual standstill (off the "x"–axis to the right). This graph uses "inverse speed" for its "x"–axis so the shape of the curve can easily be compared to the curves in "" below. In both graphs, zero on the "x"–axis represents infinite temperature. Additionally, the "x" and "y"–axis on both graphs are scaled proportionally.
The high speeds of translational motion
Although very specialized laboratory equipment is required to directly detect translational motions, the resultant collisions by atoms or molecules with small particles suspended in a fluid produces Brownian motion that can be seen with an ordinary microscope. The translational motions of elementary particles are "very" fast [Even room–temperature air has an average molecular translational "speed" (not vector-isolated velocity) of 1822 km/hour. This is relatively fast for something the size of a molecule considering there are roughly 2.42 × 1016 of them crowded into a single cubic millimeter. Assumptions: Average molecular weight of wet air = 28.838 g/mol and "T" = 296.15 K. Assumption’s primary variables: An altitude of 194 meters above mean sea level (the world–wide median altitude of human habitation), an indoor temperature of 23 °C, a dewpoint of 9 °C (40.85% relative humidity), and 760 mmHg (101.325 kPa) sea level–corrected barometric pressure.] and temperatures close to absolute zero are required to directly observe them. For instance, when scientists at the NIST achieved a record-setting cold temperature of 700 nK (billionths of a kelvin) in 1994, they used optical lattice laser equipment to adiabatically cool caesium atoms. They then turned off the entrapment lasers and directly measured atom velocities of 7 mm per second to in order to calculate their temperature. [Citation: "Adiabatic Cooling of Cesium to 700 nK in an Optical Lattice", A. Kastberg "et al"., Physical Review Letters 74, No. 9, 27 Feb. 1995, Pg. 1542. It’s noteworthy that a record cold temperature of 450 pK in a Bose–Einstein condensate of sodium atoms (achieved by A. E. Leanhardt "et al". of MIT) equates to an average vector-isolated atom velocity of 0.4 mm/s and an average atom speed of 0.7 mm/s.] Formulas for calculating the velocity and speed of translational motion are given in the following footnote.The rate of translational motion of atoms and molecules is calculated based on thermodynamic temperature as follows:
:$ilde\left\{v\right\} = sqrt\left\{frac$k_Bover 2} cdot T}mover 2}::where…::$ilde\left\{v\right\}$ is the vector-isolated mean velocity of translational particle motion in m/s::"kB" (Boltzmann constant) = 1.380 6504(24) × 10−23 J/K::"T" is the thermodynamic temperature in kelvins::"m" is the molecular mass of substance in kg/particle
In the above formula, molecular mass, "m", in kg/particle is the quotient of a substance’s molar mass (also known as "atomic weight", "atomic mass", "relative atomic mass", and "unified atomic mass units") in g/mol or daltons divided by 6.022 141 79(30) × 1026 (which is the Avogadro constant times one thousand). For diatomic molecules such as H2, N2, and O2, multiply atomic weight by two before plugging it into the above formula.
The mean "speed" (not vector-isolated velocity) of an atom or molecule along any arbitrary path is calculated as follows:
:$ilde\left\{s\right\} = ilde\left\{v\right\} cdot sqrt\left\{3\right\}$::where…::$ilde\left\{s\right\}$ is the mean speed of translational particle motion in m/s
Note that the mean energy of the translational motions of a substance’s constituent particles correlates to their mean "speed", not velocity. Thus, substituting $ilde\left\{s\right\}$ for "v" in the classic formula for kinetic energy, "Ek" = frac|2"m" • "v" 2 produces precisely the same value as does "Emean" = 3/2"kBT" (as shown in the section titled "The nature of kinetic energy, translational motion, and temperature)".
Note too that the Boltzmann constant and its related formulas establish that absolute zero is the point of both zero kinetic energy of particle motion and zero kinetic velocity (see also "Note 1" above).]
The internal motions of molecules and specific heat
[
Fig. 3 Molecules have internal structure because they are composed of atoms that have different ways of moving within molecules. Being able to store kinetic energy in these "internal degrees of freedom" contributes to a substance’s "specific heat capacity", allowing it to contain more heat energy at the same temperature.] There are other forms of heat energy besides the kinetic energy of translational motion. As can be seen in the animation at right, molecules are complex objects; they are a population of atoms and thermal agitation can strain their internal chemical bonds in three different ways: via rotation, bond length, and bond angle movements. These are all types of "internal degrees of freedom". This makes molecules distinct from "monatomic" substances (consisting of individual atoms) like the noble gases helium and argon, which have only the three translational degrees of freedom. Kinetic energy is stored in molecules’ internal degrees of freedom, which gives them an "internal temperature." Even though these motions are called “internal,” the external portions of molecules still move—rather like the jiggling of a stationary water balloon. This permits the two-way exchange of kinetic energy between internal motions and translational motions with each molecular collision. Accordingly, as heat is removed from molecules, both their kinetic temperature (the kinetic energy of translational motion) and their internal temperature simultaneously diminish in equal proportions. This phenomenon is described by the equipartition theorem, which states that for any bulk quantity of a substance in equilibrium, the kinetic energy of particle motion is evenly distributed among all the active degrees of freedom available to the particles. Since the internal temperature of molecules are usually equal to their kinetic temperature, the distinction is usually of interest only in the detailed study of non-local thermodynamic equilibrium (LTE) phenomena such as combustion, the sublimation of solids, and the diffusion of hot gases in a partial vacuum.
The kinetic energy stored internally in molecules allows a substance to contain more heat energy at a given temperature (and in the case of gases, at a given pressure and volume), and to absorb more of it for a given temperature increase. This is because any kinetic energy that is, at a given instant, bound in internal motions is not at that same instant contributing to the molecules’ translational motions. [The internal degrees of freedom of molecules cause their external surfaces to vibrate and can also produce overall spinning motions (what can be likened to the jiggling and spinning of an otherwise stationary water balloon). If one examines a "single" molecule as it impacts a containers’ wall, some of the kinetic energy borne in the molecule’s internal degrees of freedom can constructively add to its translational motion during the instant of the collision and extra kinetic energy will be transferred into the container’s wall. This would induce an extra, localized, impulse-like contribution to the average pressure on the container. However, since the internal motions of molecules are random, they have an equal probability of "destructively" interfering with translational motion during a collision with a container’s walls or another molecule. Averaged across any bulk quantity of a gas, the internal thermal motions of molecules have zero net effect upon the temperature, pressure, or volume of a gas. Molecules’ internal degrees of freedom simply provide additional locations where kinetic energy is stored. This is precisely why molecular-based gases have greater specific heat capacity than monatomic gases (where additional heat energy must be added to achieve a given temperature rise).] This extra kinetic energy simply increases the amount of heat energy a substance absorbs for a given temperature rise. This property is known as a substance’s specific heat capacity.
Different molecules absorb different amounts of heat energy for each incremental increase in temperature; that is, they have different specific heat capacities. High specific heat capacity arises, in part, because certain substances’ molecules possess more internal degrees of freedom than others do. For instance, room-temperature nitrogen, which is a diatomic molecule, has "five" active degrees of freedom: the three comprising translational motion plus two rotational degrees of freedom internally. Not surprisingly, in accordance with the equipartition theorem, nitrogen has five-thirds the specific heat capacity per mole (a specific number of molecules) as do the monatomic gases. [When measured at constant-volume since different amounts of work must be performed if measured at constant-pressure. Nitrogen’s "CvH" (100 kPa, 20 °C) equals 20.8 J mol–1 K–1 vs. the monatomic gases, which equal 12.4717 J mol–1 K–1. Citations: [http://www.whfreeman.com/ W.H. Freeman’s] "Physical Chemistry", Part 3: Change ( [http://www.whfreeman.com/college/pdfs/pchem8e/PC8eC21.pdf 422 kB PDF, here] ), Exercise 21.20b, Pg. 787. Also [http://www.gsu.edu/ Georgia State University’s] " [http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/shegas.html Molar Specific Heats of Gases] ".] Another example is gasoline (see table showing its specific heat capacity). Gasoline can absorb a large amount of heat energy per mole with only a modest temperature change because each molecule comprises an average of 21 atoms and therefore has many internal degrees of freedom. Even larger, more complex molecules can have dozens of internal degrees of freedom.
The diffusion of heat energy: Entropy, phonons, and mobile conduction electrons
[
Fig. 4 The temperature-induced translational motion of particles in solids takes the form of "phonons. "Shown here are phonons with identical amplitudes but with wavelengths ranging from 2 to 12 molecules.] "Heat conduction "is the diffusion of heat energy from hot parts of a system to cold. A “system” can be either a single bulk entity or a plurality of discrete bulk entities. The term “bulk” in this context means a statistically significant quantity of particles (which can be a microscopic amount). Whenever heat energy diffuses within an isolated system, temperature differences within the system decrease (and entropy increases).
One particular heat conduction mechanism occurs when translational motion—the particle motion underlying temperature—transfers momentum from particle to particle in collisions. In gases, these translational motions are of the nature shown above in "Fig. 1. "As can be seen in that animation, not only does momentum (heat) diffuse throughout the volume of the gas through serial collisions, but entire molecules or atoms can advance forward into new territory, bringing their kinetic energy with them. Consequently, temperature differences equalize throughout gases very quickly—especially for light atoms or molecules; convection speeds this process even more. [The "speed" at which thermal energy equalizes throughout the volume of a gas is very rapid. However, since gases have extremely low density relative to solids, the "heat flux"—the thermal power conducting through a unit area—through gases is comparatively low. This is why the dead-air spaces in multi-pane windows have insulating qualities.]
Translational motion in "solids "however, takes the form of "phonons "(see "Fig. 4" at right). Phonons are constrained, quantized wave packets traveling at the speed of sound for a given substance. The manner in which phonons interact within a solid determines a variety of its properties, including its thermal conductivity. In electrically insulating solids, phonon-based heat conduction is "usually" inefficient [Diamond is a notable exception. Due to the highly quantized modes of phonon vibration occurring in its rigid crystal lattice, not only does diamond have exceptionally "poor" specific heat capacity, it also has exceptionally "high" thermal conductivity.] and such solids are considered "thermal insulators" (such as glass, plastic, rubber, ceramic, and rock). This is because in solids, atoms and molecules are locked into place relative to their neighbors and are not free to roam.
Metals however, are not restricted to only phonon-based heat conduction. Heat energy conducts through metals extraordinarily quickly because instead of direct molecule-to-molecule collisions, the vast majority of heat energy is mediated via very light, mobile "conduction electrons." This is why there is a near-perfect correlation between metals’ thermal conductivity and their electrical conductivity. [Correlation is 752 (W m−1 K−1) / (MS•cm), σ = 81, through a 7:1 range in conductivity. Value and standard deviation based on data for Ag, Cu, Au, Al, Ca, Be, Mg, Rh, Ir, Zn, Co, Ni, Os, Fe, Pa, Pt, and Sn. Citation: Data from "CRC Handbook of Chemistry and Physics", 1st Student Edition and [http://www.webelements.com/ this link] to Web Elements’ home page.] Conduction electrons imbue metals with their extraordinary conductivity because they are "delocalized," i.e. not tied to a specific atom, and behave rather like a sort of “quantum gas” due to the effects of "zero-point energy" (for more on ZPE, see "Note 1" below). Furthermore, electrons are relatively light with a rest mass only frac|1836th that of a proton. This is about the same ratio as a .22 Short bullet (29 grains or 1.88 g) compared to the rifle that shoots it. As Isaac Newton wrote with his ,
:"“Law #3: All forces occur in pairs, and these two forces":" are equal in magnitude and opposite in direction.”"
However, a bullet accelerates faster than a rifle given an equal force. Since kinetic energy increases as the square of velocity, nearly all the kinetic energy goes into the bullet, not the rifle, even though both experience the same force from the expanding propellant gases. In the same manner—because they are much less massive—heat energy is readily borne by mobile conduction electrons. Additionally, because they’re delocalized and "very" fast, kinetic heat energy conducts extremely quickly through metals with abundant conduction electrons.
The diffusion of heat energy: Black-body radiation
", above).
Black-body radiation diffuses heat energy throughout a substance as the photons are absorbed by neighboring atoms, transferring momentum in the process. Black-body photons also easily escape from a substance and can be absorbed by the ambient environment; kinetic energy is lost in the process.
As established by the Stefan–Boltzmann law, the intensity of black-body radiation increases as the fourth power of absolute temperature. Thus, a black body at 824 K (just short of glowing dull red) emits "60 times" the radiant power as it does at 296 K (room temperature). This is why one can so easily feel the radiant heat from hot objects at a distance. At higher temperatures, such as those found in an incandescent lamp, black-body radiation can be the principal mechanism by which heat energy escapes a system.
The heat of phase changes
The kinetic energy of particle motion is just one contributor to the total heat energy in a substance; another is "phase transitions", which are the potential energy of molecular bonds that can form in a substance as it cools (such as during condensing and freezing). The heat energy required for a phase transition is called "latent heat." This phenomenon may more easily be grasped by considering it in the reverse direction: latent heat is the energy required to "break" chemical bonds (such as during evaporation and melting). Most everyone is familiar with the effects of phase transitions; for instance, steam at 100 °C can cause severe burns much faster than the 100 °C air from a hair dryer. This occurs because a large amount of latent heat is liberated as steam condenses into liquid water on the skin.
Even though heat energy is liberated or absorbed during phase transitions, pure chemical elements, compounds, and eutectic
At one specific thermodynamic point, the melting point (which is 0 °C across a wide pressure range in the case of water), all the atoms or molecules are—on average—at the maximum energy threshold their chemical bonds can withstand without breaking away from the lattice. Chemical bonds are quantized forces: they either hold fast, or break; there is no in-between state. Consequently, when a substance is at its melting point, every joule of added heat energy only breaks the bonds of a specific quantity of its atoms or molecules, [Water’s enthalpy of fusion (0 °C, 101.325 kPa) equates to 0.062284 eV per molecule so adding one joule of heat energy to 0 °C water ice causes 1.0021 × 1020 water molecules to break away from the crystal lattice and become liquid.] converting them into a liquid of precisely the same temperature; no kinetic energy is added to translational motion (which is what gives substances their temperature). The effect is rather like popcorn: at a certain temperature, additional heat energy can’t make the kernels any hotter until the transition (popping) is complete. If the process is reversed (as in the freezing of a liquid), heat energy must be removed from a substance.
As stated above, the heat energy required for a phase transition is called "latent heat." In the specific cases of melting and freezing, it’s called "enthalpy of fusion" or "heat of fusion." If the molecular bonds in a crystal lattice are strong, the heat of fusion can be relatively great, typically in the range of 6 to 30 kJ per mole for water and most of the metallic elements. [Water’s enthalpy of fusion is 6.0095 kJ mol−1 K−1 (0 °C, 101.325 kPa). Citation: "Water Structure and Science, Water Properties, Enthalpy of fusion, (0 °C, 101.325 kPa)" (by London South Bank University). [http://www.lsbu.ac.uk/water/data.html Link to Web site.] The only metals with enthalpies of fusion "not" in the range of 6–30 J mol−1 K−1 are (on the high side): Ta, W, and Re; and (on the low side) most of the group 1 (alkaline) metals plus Ga, In, Hg, Tl, Pb, and Np. Citation: [http://www.webelements.com/ This link] to Web Elements’ home page.] If the substance is one of the monatomic gases, (which have little tendency to form molecular bonds) the heat of fusion is more modest, ranging from 0.021 to 2.3 kJ per mole. [Xenon value citation: [http://www.webelements.com/webelements/elements/text/Xe/heat.html This link] to WebElements’ xenon data (available values range from 2.3 to 3.1 kJ mol−1). It is also noteworthy that helium’s heat of fusion of only 0.021 kJ mol−1 is so weak of a bonding force that zero-point energy prevents helium from freezing unless it is under a pressure of at least 25 atmospheres.] Relatively speaking, phase transitions can be truly energetic events. To completely melt ice at 0 °C into water at 0 °C, one must add roughly 80 times the heat energy as is required to increase the temperature of the same mass of liquid water by one degree Celsius. The metals’ ratios are even greater, typically in the range of 400 to 1200 times. [Citation: Data from "CRC Handbook of Chemistry and Physics", 1st Student Edition and [http://www.webelements.com/ this link] to Web Elements’ home page.] And the phase transition of boiling is much more energetic than freezing. For instance, the energy required to completely boil or vaporize water (what is known as "enthalpy of vaporization") is roughly "540 times" that required for a one-degree increase. [H2O specific heat capacity, "Cp" = 0.075327 kJ mol−1 K−1 (25 °C); Enthalpy of fusion = 6.0095 kJ mol−1 (0 °C, 101.325 kPa); Enthalpy of vaporization (liquid) = 40.657 kJ mol−1 (100 °C). Citation: "Water Structure and Science, Water Properties" (by London South Bank University). [http://www.lsbu.ac.uk/water/data.html Link to Web site.] ]
Water’s sizable enthalpy of vaporization is why one’s skin can be burned so quickly as steam condenses on it (heading from red to green in "Fig. 7 "above). In the opposite direction, this is why one’s skin feels cool as liquid water on it evaporates (a process that occurs at a sub-ambient wet-bulb temperature that is dependent on relative humidity). Water’s highly energetic enthalpy of vaporization is also an important factor underlying why “solar pool covers” (floating, insulated blankets that cover swimming pools when not in use) are so effective at reducing heating costs: they prevent evaporation. For instance, the evaporation of just 20 mm of water from a 1.29-meter-deep pool chills its water 8.4 degrees Celsius.
Internal energy
The total kinetic energy of all particle motion—including that of conduction electrons—plus the potential energy of phase changes, plus zero-point energy comprise the "internal energy" of a substance, which is its total heat energy. The term "internal energy" mustn’t be confused with "internal degrees of freedom." Whereas the "internal degrees of freedom of molecules" refers to one particular place where kinetic energy is bound, the "internal energy of a substance" comprises all forms of heat energy.
Heat energy at absolute zero
As a substance cools, different forms of heat energy and their related effects simultaneously decrease in magnitude: the latent heat of available phase transitions are liberated as a substance changes from a less ordered state to a more ordered state; the translational motions of atoms and molecules diminish (their kinetic temperature decreases); the internal motions of molecules diminish (their internal temperature decreases); conduction electrons (if the substance is an electrical conductor) travel "somewhat" slower; [ Mobile conduction electrons are "delocalized," i.e. not tied to a specific atom, and behave rather like a sort of “quantum gas” due to the effects of zero-point energy. Consequently, even at absolute zero, conduction electrons still move between atoms at the "Fermi velocity" of about 1.6 × 106 m/s. Kinetic heat energy adds to this speed and also causes delocalized electrons to travel farther away from the nuclei.] and black-body radiation’s peak emittance wavelength increases (the photons’ energy decreases). When the particles of a substance are as close as possible to complete rest and retain only ZPE-induced quantum mechanical motion, the substance is at the temperature of absolute zero ("T"=0).
Note that whereas absolute zero is the point of zero thermodynamic temperature and is also the point at which the particle constituents of matter have minimal motion, absolute zero is not necessarily the point at which a substance contains zero heat energy; one must be very precise with what one means by “heat energy.” Often, all the phase changes that "can" occur in a substance, "will" have occurred by the time it reaches absolute zero. However, this is not always the case. Notably, "T"=0 helium remains liquid at room pressure and must be under a pressure of at least 25 bar to crystallize. This is because helium’s heat of fusion—the energy required to melt helium ice—is so low (only 21 J mol−1) that the motion-inducing effect of zero-point energy is sufficient to prevent it from freezing at lower pressures. Only if under at least 25 bar of pressure will this latent heat energy be liberated as helium freezes while approaching absolute zero. A further complication is that many solids change their crystal structure to more compact arrangements at extremely high pressures (up to millions of bars). These are known as "solid-solid phase transitions" wherein latent heat is liberated as a crystal lattice changes to a more thermodynamically favorable, compact one.
The above complexities make for rather cumbersome blanket statements regarding the internal energy in "T"=0 substances. Regardless of pressure though, what "can" be said is that at absolute zero, all solids with a lowest-energy crystal lattice such those with a "closest-packed arrangement" (see "Fig. 8," above left) contain minimal internal energy, retaining only that due to the ever-present background of zero-point energy. [No other crystal structure can exceed the 74.048% packing density of a "closest-packed arrangement." The two regular crystal lattices found in nature that have this density are "hexagonal close packed" (HCP) and "face-centered cubic" (FCC). These regular lattices are at the lowest possible energy state. Diamond is a closest-packed structure with an FCC crystal lattice. Note too that suitable crystalline chemical "compounds", although usually composed of atoms of different sizes, can be considered as “closest-packed structures” when considered at the molecular level. One such compound is the common mineral known as "magnesium aluminum spinel" (MgAl2O4). It has a face-centered cubic crystal lattice and no change in pressure can produce a lattice with a lower energy state.] One can also say that for a given substance at constant pressure, absolute zero is the point of lowest "enthalpy" (a measure of work potential that takes internal energy, pressure, and volume into consideration). [Nearly half of the 92 naturally occurring chemical elements that can freeze under a vacuum also have a closest-packed crystal lattice. This set includes beryllium, osmium, neon, and iridium (but excludes helium), and therefore have zero latent heat of phase transitions to contribute to internal energy (symbol: "U)". In the calculation of enthalpy (formula: "H" = "U" + "pV)", internal energy may exclude different sources of heat energy—particularly ZPE—depending on the nature of the analysis. Accordingly, all "T"=0 closest-packed matter under a perfect vacuum has either minimal or zero enthalpy, depending on the nature of the analysis. Citation: "Use Of Legendre Transforms In Chemical Thermodynamics", Robert A. Alberty, Pure Appl.Chem., 73, No.8, 2001, 1349–1380 ( [http://iupac.org/publications/pac/2001/pdf/7308x1349.pdf 400 kB PDF, here] ).] Lastly, it is always true to say that all "T"=0 substances contain zero kinetic heat energy.
## Practical applications for thermodynamic temperature
Thermodynamic temperature is useful not only for scientists, it can also be useful for lay-people in many disciplines involving gases. By expressing variables in absolute terms and applying Gay–Lussac’s law of temperature/pressure proportionality, the solutions to familiar problems are straightforward. For instance, how is the pressure in an automobile tire affected by temperature? If the tire has a “cold” pressure of 200 kPa-gage , then in absolute terms—relative to a vacuum—its pressure is 300 kPa-absolute. [Pressure also must be in absolute terms. The air still in a tire at 0 kPa-gage expands too as it gets hotter. It’s not uncommon for engineers to overlook that one must work in terms of absolute pressure when compensating for temperature. For instance, a dominant manufacturer of aircraft tires published a document on temperature-compensating tire pressure, which used gage pressure in the formula. However, the high gage pressures involved (180 psi ≈ 12.4 bar) means the error would be quite small. With low-pressure automobile tires, where gage pressures are typically around 2 bar, failing to adjust to absolute pressure results in a significant error. Referenced document: "Aircraft Tire Ratings" ( [http://airmichelin.com/pdfs/05%20-%20Aircraft%20Tire%20Ratings.pdf 155 kB PDF, here] ).] [Regarding the spelling “gage” vs. “gauge” in the context of pressures measured relative to atmospheric pressure, the preferred spelling varies by country and even by industry. Further, both spellings are often used "within" a particular industry or country. Industries in British English-speaking countries typically use the spelling “gauge pressure” to distinguish it from the pressure-measuring instrument, which in the U.K., is spelled “pressure gage.” For the same reason, many of the largest American manufacturers of pressure transducers and instrumentation use the spelling “gage pressure”—the convention used here—in their formal documentation to distinguish it from the instrument, which is spelled “pressure gauge.” (see "Honeywell-Sensotec’s" [http://sensotec.com/pressurefaq.shtml FAQ page] and Fluke Corporation’s [http://us.fluke.com/usen/Home/Search.asp?txtSearchBox=%22gage+pressure%22&x=0&y=0 product search page] ).] [A difference of 100 kPa is used here instead of the 101.325 kPa value of one standard atmosphere. In 1982, the International Union of Pure and Applied Chemistry (IUPAC) recommended that for the purposes of specifying the physical properties of substances, “"the standard pressure"” (atmospheric pressure) should be defined as precisely 100kPa (≈750.062Torr). Besides being a round number, this had a very practical effect: relatively few people live and work at precisely sea level; 100kPa equates to the mean pressure at an altitude of about 112 meters, which is closer to the 194–meter, worldwide median altitude of human habitation. For especially low-pressure or high-accuracy work, true atmospheric pressure must be measured. Citation: IUPAC.org, Gold Book, " [http://goldbook.iupac.org/S05921.html Standard Pressure] "] Room temperature (“cold” in tire terms) is 296 K. What would the tire pressure be if was 20 °C hotter? The answer is frac|316 K|296 K = 6.8% greater thermodynamic temperature "and" absolute pressure; that is, a pressure of 320 kPa-absolute and 220 kPa-gage.
## The origin of heat energy on Earth
Earth’s proximity to the Sun is why most everything near Earth’s surface is warm with a temperature substantially above absolute zero. [The deepest ocean depths (3 to 10 km) are no colder than about 274.7 – 275.7 K (1.5 – 2.5 °C). Even the world-record cold surface temperature established on July 21, 1983 at Vostok Station, Antarctica is 184 K (a reported value of −89.2 °C). The residual heat of gravitational contraction left over from earth’s formation, tidal friction, and the decay of radioisotopes in earth’s core provide insufficient heat to maintain earth’s surface, oceans, and atmosphere “substantially above” absolute zero in this context. Also, the qualification of “most-everything” provides for the exclusion of lava flows, which derive their temperature from these deep-earth sources of heat.] Solar radiation constantly replenishes heat energy that Earth loses into space and a relatively stable state of equilibrium is achieved. Because of the wide variety of heat diffusion mechanisms (one of which is black-body radiation which occurs at the speed of light), objects on Earth rarely vary too far from the global mean surface and air temperature of 287 to 288 K (14 to 15 °C). The more an object’s or system’s temperature varies from this average, the more rapidly it tends to come back into equilibrium with the ambient environment.
## History of thermodynamic temperature
* 1702–1703: Guillaume Amontons (1663 – 1705) published two papers that may be used to credit him as being the first researcher to deduce the existence of a fundamental (thermodynamic) temperature scale featuring an absolute zero. He made the discovery while endeavoring to improve upon the air thermometers in use at the time. His J-tube thermometers comprised a mercury column that was supported by a fixed mass of air entrapped within the sensing portion of the thermometer. In thermodynamic terms, his thermometers relied upon the volume / temperature relationship of gas under constant pressure. His measurements of the boiling point of water and the melting point of ice showed that regardless of the mass of air trapped inside his thermometers or the weight of mercury the air was supporting, the reduction in air volume at the ice point was always the same ratio. This observation led him to posit that a sufficient reduction in temperature would reduce the air volume to zero. In fact, his calculations projected that absolute zero was equivalent to −240 degrees on today’s Celsius scale—only 33.15 degrees short of the true value of −273.15 °C.
* 1742: (1701 – 1744) created a “backwards” version of the modern Celsius temperature scale whereby zero represented the boiling point of water and 100 represented the melting point of ice. In his paper "Observations of two persistent degrees on a thermometer," he recounted his experiments showing that ice’s melting point was effectively unaffected by pressure. He also determined with remarkable precision how water’s boiling point varied as a function of atmospheric pressure. He proposed that zero on his temperature scale (water’s boiling point) would be calibrated at the mean barometric pressure at mean sea level.
* 1744: ) [http://www.bipm.org/en/committees/cipm/cipm-1948.html formally adopted] “degree Celsius” (symbol: °C) in 1948.According to "The Oxford English Dictionary" (OED), the term “Celsius’s thermometer” had been used at least as early as 1797. Further, the term “The Celsius or Centigrade thermometer” was again used in reference to a particular type of thermometer at least as early as 1850. The OED also cites this 1928 reporting of a temperature: “My altitude was about 5,800 metres, the temperature was 28° Celsius.” However, dictionaries seek to find the earliest use of a word or term and are not a useful resource as regards the terminology used throughout the history of science. According to several writings of Dr. Terry Quinn CBE FRS, Director of the BIPM (1988 – 2004), including "Temperature Scales from the early days of thermometry to the 21st century" ( [http://www.imeko.org/publications/tc12-2004/PTC12-2004-PL-001.pdf 148 kB PDF, here] ) as well as "Temperature" (2nd Edition / 1990 / Academic Press / 0125696817), the term "Celsius" in connection with the centigrade scale was not used whatsoever by the scientific or thermometry communities until after the CIPM and CGPM adopted the term in 1948. The BIPM wasn’t even aware that “degree Celsius” was in sporadic, non-scientific use before that time. It’s also noteworthy that the twelve-volume, 1933 edition of OED didn’t even have a listing for the word "Celsius" (but did have listings for both "centigrade" and "centesimal" in the context of temperature measurement). The 1948 adoption of "Celsius" accomplished three objectives::1) All common temperature scales would have their units named after someone closely associated with them; namely, Kelvin, Celsius, Fahrenheit, Réaumur and Rankine.
2) Notwithstanding the important contribution of Linnaeus who gave the Celsius scale its modern form, Celsius’s name was the obvious choice because it began with the letter C. Thus, the symbol °C that for centuries had been used in association with the name "centigrade" could continue to be used and would simultaneously inherit an intuitive association with the new name.
3) The new name eliminated the ambiguity of the term “centigrade,” freeing it to refer exclusively to the French-language name for the unit of angular measurement.]
* 1777: In his book "Pyrometrie" (Berlin: [http://www.spiess-verlage.de/html/haude___spener.html Haude & Spener,] 1779) completed four months before his death, Johann Heinrich Lambert (1728 – 1777)—sometimes incorrectly referred to as Joseph Lambert—proposed an absolute temperature scale based on the pressure / temperature relationship of a fixed volume of gas. This is distinct from the volume / temperature relationship of gas under constant pressure that Guillaume Amontons discovered 75 years earlier. Lambert stated that absolute zero was the point where a simple straight-line extrapolation reached zero gas pressure and was equal to −270 °C.
* Circa 1787: Notwithstanding the work of Guillaume Amontons 85 years earlier, Jacques Alexandre César Charles (1746 – 1823) is often credited with “discovering”, but not publishing, that the volume of a gas under constant pressure is proportional to its absolute temperature. The formula he created was "V"1/"T"1 = "V"2/"T"2.
* 1802: Joseph Louis Gay-Lussac (1778 – 1850) published work (acknowledging the unpublished lab notes of Jacques Charles fifteen years earlier) describing how the volume of gas under constant pressure changes linearly with its absolute (thermodynamic) temperature. This behavior is called Charles’s Law and is one of the gas laws. His are the first known formulas to use the number “273” for the expansion coefficient of gas relative to the melting point of ice (indicating that absolute zero was equivalent to −273 °C).
* 1848: , (1824 – 1907) also known as Lord Kelvin, wrote in his paper, " [http://zapatopi.net/kelvin/papers/on_an_absolute_thermometric_scale.html On an Absolute Thermometric Scale] ," of the need for a scale whereby “infinite cold” (absolute zero) was the scale’s null point, and which used the degree Celsius for its unit increment. Like Gay-Lussac, Thomson calculated that absolute zero was equivalent to −273 °C on the air thermometers of the time. This absolute scale is known today as the Kelvin thermodynamic temperature scale. It’s noteworthy that Thomson’s value of “−273” was actually derived from 0.00366, which was the accepted expansion coefficient of gas per degree Celsius relative to the ice point. The inverse of −0.00366 expressed to five significant digits is −273.22 °C which is remarkably close to the true value of −273.15 °C.
* 1859: William John Macquorn Rankine (1820 – 1872) proposed a thermodynamic temperature scale similar to William Thomson’s but which used the degree Fahrenheit for its unit increment. This absolute scale is known today as the Rankine thermodynamic temperature scale.
* 1877 - 1884: (1844 – 1906) made major contributions to thermodynamics through an understanding of the role that particle kinetics and black-body radiation played. His name is now attached to several of the formulas used today in thermodynamics.
* Circa 1930s: Gas thermometry experiments carefully calibrated to the melting point of ice and boiling point of water showed that absolute zero was equivalent to −273.15 °C.
* 1948: [http://www.bipm.fr/en/CGPM/db/9/3/ Resolution 3] of the 9th CGPM (Conférence Générale des Poids et Mesures, also known as the General Conference on Weights and Measures) fixed the triple point of water at precisely 0.01 °C. At this time, the triple point still had no formal definition for its equivalent kelvin value, which the resolution declared “will be fixed at a later date.” The implication is that "if" the value of absolute zero measured in the 1930s was truly −273.15 °C, then the triple point of water (0.01 °C) was equivalent to 273.16 K. Additionally, both the CIPM (Comité international des poids et mesures, also known as the International Committee for Weights and Measures) and the CGPM [http://www.bipm.org/en/committees/cipm/cipm-1948.html formally adopted] the name “Celsius” for the “degree Celsius” and the “Celsius temperature scale.”
* 1954: [http://www.bipm.fr/en/CGPM/db/10/3/ Resolution 3] of the 10th CGPM gave the Kelvin scale its modern definition by choosing the triple point of water as its second defining point and assigned it a temperature of precisely 273.16 kelvin (what was actually written 273.16 “degrees Kelvin” at the time). This, in combination with Resolution 3 of the 9th CGPM, had the effect of defining absolute zero as being precisely zero kelvin and −273.15 °C.
* 1967/1968: [http://www.bipm.fr/en/CGPM/db/13/3/ Resolution 3] of the 13th CGPM renamed the unit increment of thermodynamic temperature “kelvin”, symbol K, replacing “degree absolute”, symbol °K. Further, feeling it useful to more explicitly define the magnitude of the unit increment, the 13th CGPM also decided in [http://www.bipm.fr/en/CGPM/db/13/4/ Resolution 4] that “The kelvin, unit of thermodynamic temperature, is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water.”
* 2005: The CIPM (Comité International des Poids et Mesures, also known as the International Committee for Weights and Measures) [http://www.bipm.fr/en/si/si_brochure/chapter2/2-1/kelvin.html affirmed] that for the purposes of delineating the temperature of the triple point of water, the definition of the Kelvin thermodynamic temperature scale would refer to water having an isotopic composition defined as being precisely equal to the nominal specification of Vienna Standard Mean Ocean Water.
## Derivations of thermodynamic temperature
Strictly speaking, the temperature of a system is well-defined only if its particles (atoms, molecules, electrons, photons) are at equilibrium, so that their energies obey a Boltzmann distribution (or its quantum mechanical counterpart). There are many possible scales of temperature, derived from a variety of observations of physical phenomena. The thermodynamic temperature can be shown to have special properties, and in particular can be seen to be uniquely defined (up to some constant multiplicative factor) by considering the efficiency of idealized heat engines. Thus the "ratio" "T"2/"T"1 of two temperatures "T"1 and"T"2 is the same in all absolute scales.
Loosely stated, temperature controls the flow of heat between two systems, and the universe as a whole, as with any natural system, tends to progress so as to maximize entropy. This suggests that there should be a relationship between temperature and entropy. To elucidate this, consider first the relationship between heat, work and temperature. One way to study this is to analyse a heat engine, which is a device for converting heat into mechanical work, such as the Carnot heat engine. Such a heat engine functions by using a temperature gradient between a high temperature "T"H and a low temperature "T"C to generate work, and the work done (per cycle, say) by the heat engine is equal to the difference between the heat energy "q"H put into the system at the high temperature the heat "q"C ejected at the low temperature (in that cycle). The efficiency of the engine is the work divided by the heat put into the system or
:$extrm\left\{efficiency\right\} = frac \left\{w_\left\{cy${q_H} = frac{q_H-q_C}{q_H} = 1 - frac{q_C}{q_H} qquad (1)
where wcy is the work done per cycle. Thus the efficiency depends only on qC/qH. Because "q"C and "q"H correspond to heat transfer at the temperatures "T"C and "T"H, respectively, the ratio "q"C/"q"H should be a function "f" of these temperatures:
:$frac\left\{q_C\right\}\left\{q_H\right\} = f\left(T_H,T_C\right)qquad \left(2\right).$
Carnot’s theorem states that all reversible engines operating between the same heat reservoirs are equally efficient. Thus, a heat engine operating between temperatures "T"1 and "T"3 must have the same efficiency as one consisting of two cycles, one between "T"1 and another (intermediate) temperature "T"2, and the second between "T"2 and "T"3. This can only be the case if
:$f\left(T_1,T_3\right) = frac\left\{q_3\right\}\left\{q_1\right\} = frac\left\{q_2 q_3\right\} \left\{q_1 q_2\right\} = f\left(T_1,T_2\right)f\left(T_2,T_3\right).$
Now specialize to the case that $T_1$ is a fixed reference temperature: the temperature of the triple point of water. Then for any "T"2 and "T"3,:$f\left(T_2,T_3\right) = frac\left\{f\left(T_1,T_3\right)\right\}\left\{f\left(T_1,T_2\right)\right\} = frac\left\{273.16 cdot f\left(T_1,T_3\right)\right\}\left\{273.16 cdot f\left(T_1,T_2\right)\right\}.$Therefore if thermodynamic temperature is defined by:$T = 273.16 cdot f\left(T_1,T\right) ,$then the function "f", viewed as a function of thermodynamic temperature, is simply:$f\left(T_2,T_3\right) = frac\left\{T_3\right\}\left\{T_2\right\},$and the reference temperature "T"1 will have the value 273.16. (Of course any reference temperature and any positive numerical value could be used — the choice here corresponds to the Kelvin scale.)
It follows immediately that:$frac\left\{q_C\right\}\left\{q_H\right\} = f\left(T_H,T_C\right) = frac\left\{T_C\right\}\left\{T_H\right\}.qquad \left(3\right).$
Substituting Equation 3 back into Equation 1 gives a relationship for the efficiency in terms of temperature:
:$extrm\left\{efficiency\right\} = 1 - frac\left\{q_C\right\}\left\{q_H\right\} = 1 - frac\left\{T_C\right\}\left\{T_H\right\}qquad \left(4\right).$
Notice that for "T"C=0 the efficiency is 100% and that efficiency becomes greater than 100% for "T"C<0. Since an efficiency greater than 100% violates the first law of thermodynamics, this requires that zero must be the minimum possible temperature. This has an intuitive interpretation: temperature is the motion of particles, so no system can, on average, have less motion than the minimum permitted by quantum physics. In fact, as of June 2006, the coldest man-made temperature was 450 pK.
Subtracting the right hand side of Equation 4 from the middle portion and rearranging gives
:$frac \left\{q_H\right\}\left\{T_H\right\} - frac\left\{q_C\right\}\left\{T_C\right\} = 0,$
where the negative sign indicates heat ejected from the system. This relationship suggests the existence of a state function "S" (i.e., a function which depends only on the state of the system, not on how it reached that state) defined (up to an additive constant) by
:$dS = frac \left\{dq_mathrm\left\{rev\left\{T\right\}qquad \left(5\right),$
where the subscript indicates heat transfer in a reversible process. The function "S" corresponds to the entropy of the system, mentioned previously, and the change of "S" around any cycle is zero (as is necessary for any state function). Equation 5 can be rearranged to get an alternative definition for temperature in terms of entropy and heat:
:$T = frac\left\{dq_mathrm\left\{rev\left\{dS\right\}.$
For a system in which the entropy "S" is a function "S"("E") of its energy "E", the thermodynamic temperature "T" is therefore given by
:$frac\left\{1\right\}\left\{T\right\} = frac\left\{dS\right\}\left\{dE\right\},$ so that the reciprocal of the thermodynamic temperature is the rate of increase of entropy with energy.
* Absolute zero
* Black body
* Boiling
* Boltzmann constant
* Brownian motion
* Carnot heat engine
* Celsius
* Chemical bond
* Condensation
* Convection
* Degrees of freedom
* Delocalized electron
* Diffusion
* Elastic collision
* Electron
* Energy
* Energy conversion efficiency
* Enthalpy
* Entropy
* Evaporation
* Fahrenheit
* First law of thermodynamics
* Freezing
* Gas laws
* Heat
* Heat conduction
* Heat engine
* Internal energy
* ITS-90
* Ideal gas law
* Joule
* Kelvin
* Kinetic energy
* Latent heat
* Laws of thermodynamics
* Maxwell–Boltzmann distribution
* Melting
* Mole
* Molecule
* Orders of magnitude (temperature)
* Phase transition
* Phonon
* Planck’s law of black body radiation
* Potential energy
* Quantum mechanics:
** Introduction to quantum mechanics
** Quantum mechanics (main article)
* Rankine scale
* Specific heat capacity
* Standard enthalpy change of fusion
* Standard enthalpy change of vaporization
* Stefan–Boltzmann law
* Sublimation
* Temperature
* Temperature conversion formulas
* Thermal conductivity
* Thermodynamic equations
* Thermodynamic equilibrium
* Thermodynamics
*
* Timeline of temperature and pressure
measurement technology
* Triple point
* Universal gas constant
* Vienna Standard Mean Ocean Water (VSMOW)
* Wien’s displacement law
* Work (Mechanical)
* Work (thermodynamics)
* Zero-point energy
## Notes
"In the following notes, wherever numeric equalities are shown in ‘concise form’—such as" 1.85487(14) × 1043"—the two digits between the parentheses denotes the uncertainty at "1σ" standard deviation "(68%" confidence level) in the two least significant digits of the significand."
* " [http://www.chm.davidson.edu/ChemistryApplets/KineticMolecularTheory/index.html Kinetic Molecular Theory of Gases.] " An excellent explanation (with interactive animations) of the kinetic motion of molecules and how it affects matter. By David N. Blauch, [http://www.chm.davidson.edu/ Department of Chemistry] , [http://www2.davidson.edu/index.asp Davidson College] .
* " [http://www.calphysics.org/zpe.html Zero Point Energy and Zero Point Field.] " A Web site with in-depth explanations of a variety of quantum effects. By Bernard Haisch, of [http://www.calphysics.org/index.html Calphysics Institute] .
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### Look at other dictionaries:
• thermodynamic temperature — termodinaminė temperatūra statusas T sritis Standartizacija ir metrologija apibrėžtis Vienas iš pagrindinių SI dydžių, kurio matavimo vienetas kelvinas (K) yra vandens trigubojo taško termodinaminės temperatūros 1/273,16 dalis. atitikmenys: angl … Penkiakalbis aiškinamasis metrologijos terminų žodynas
• thermodynamic temperature — termodinaminė temperatūra statusas T sritis fizika atitikmenys: angl. thermodynamic temperature vok. thermodynamische Temperatur, f rus. термодинамическая температура, f pranc. température thermodynamique, f … Fizikos terminų žodynas
• thermodynamic temperature — termodinaminė temperatūra statusas T sritis Energetika apibrėžtis Temperatūra, apibūdinanti medžiagos dalelių šiluminio judėjimo intensyvumą ir nepriklausanti nuo pasirinktos medžiagos ir matavimo būdo. Žymima T. Termodinaminės temperatūros… … Aiškinamasis šiluminės ir branduolinės technikos terminų žodynas
• thermodynamic temperature — noun Temperature defined in terms of the laws of thermodynamics rather than the properties of a real material: expressed in kelvins … Wiktionary
• thermodynamic temperature scale — termodinaminė temperatūros skalė statusas T sritis Energetika apibrėžtis Nepriklauso nuo termometrinės medžiagos ir turi vieną atskaitos tašką – vandens trigubąjį tašką, kuriam suteikta T = 273,16 K vertė. Termodinaminė temperatūros skalė… … Aiškinamasis šiluminės ir branduolinės technikos terminų žodynas
• thermodynamic temperature scale — noun a temperature scale of which the unit is the kelvin (K) which is ¹⁄₂₇₃.₁₆ of the thermodynamic temperature of the triple point of water … Australian English dictionary
• thermodynamic temperature scale — termodinaminė temperatūros skalė statusas T sritis Standartizacija ir metrologija apibrėžtis Temperatūros skalė, pagrįsta absoliučiuoju nuliu, t. y. žemiausia temperatūra, kurią teoriškai galima būtų pasiekti ir kuri yra 273,16 °C žemiau ledo… … Penkiakalbis aiškinamasis metrologijos terminų žodynas
• thermodynamic temperature scale — termodinaminės temperatūros skalė statusas T sritis fizika atitikmenys: angl. thermodynamic temperature scale vok. thermodynamische Temperaturskala, f rus. термодинамическая шкала температур, f pranc. échelle thermodynamique des températures, f … Fizikos terminų žodynas | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 18, "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": 0.8314574360847473, "perplexity": 2009.9445610117443}, "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-2020-34/segments/1596439738950.61/warc/CC-MAIN-20200813014639-20200813044639-00029.warc.gz"} |
https://stats.libretexts.org/Bookshelves/Applied_Statistics/Book%3A_Biological_Statistics_(McDonald)/1.0 | # 1: Basics
• 1.1: Data analysis steps
• 1.2: Types of Biological Variables
One of the first steps in deciding which statistical test to use is determining what kinds of variables you have. When you know what the relevant variables are, what kind of variables they are, and what your null and alternative hypotheses are, it's usually pretty easy to figure out which test you should use. I classify variables into three types: measurement variables, nominal variables, and ranked variables.
• 1.3: Probability
When dealing with probabilities in biology, you are often working with theoretical expectations, not population samples. For example, in a genetic cross of two individual Drosophila melanogaster that are heterozygous at the vestigial locus, Mendel's theory predicts that the probability of an offspring individual being a recessive homozygote (having teeny-tiny wings) is one-fourth, or 0.25. This is equivalent to saying that one-fourth of a population of offspring will have tiny wings.
• 1.4: Basic Concepts of Hypothesis Testing
The technique used by the vast majority of biologists, and the technique that most of this handbook describes, is sometimes called "frequentist" or "classical" statistics. It involves testing a null hypothesis by comparing the data you observe in your experiment with the predictions of a null hypothesis. You estimate what the probability would be of obtaining the observed results, or something more extreme, if the null hypothesis were true.
• 1.5: Confounding Variables
A confounding variable is a variable that may affect the dependent variable. This can lead to erroneous conclusions about the relationship between the independent and dependent variables. You deal with confounding variables by controlling them; by matching; by randomizing; or by statistical control.
## Contributor
• John H. McDonald (University of Delaware) | {"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": 0.8294841647148132, "perplexity": 1270.4354430074698}, "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-2019-43/segments/1570986666959.47/warc/CC-MAIN-20191016090425-20191016113925-00410.warc.gz"} |
https://www.physicsforums.com/threads/a-n-th-lg-n.35904/ | # A[n] = Θ(lg n)?
1. Jul 20, 2004
### e(ho0n3
a[n] = Θ(lg n)!?
Suppose {a[n]} is an increasing sequence and whenever m divides n, then
a[n] = a[n/m] + d
where m is a positive integer and d is a real number. Show that a[n] = Θ(lg n).
a[n/m] + d ≤ a[n/2] + d, and with this I can show that a[n] = O(lg n). All I need to do is show that a[n] = Ω(lg n). This is where I'm stuck. Any hints?
2. Jul 20, 2004
### Hurkyl
Staff Emeritus
To show that something is in omega(log x), it suffices to find some sequence {x_i} such that a[x_i] grows like log x.
By the way, I'm disconcerted with your reasoning that it is in O(log x); what if n is odd? I suppose you're just abbreviating the complete proof you have.
3. Jul 21, 2004
### e(ho0n3
Sorry. What I meant was
$$a_n = a_{\lfloor n/2 \rfloor} + d$$
My head is a total blank. I can't think of anything. | {"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": 0.8493074774742126, "perplexity": 2402.271343606918}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988718278.43/warc/CC-MAIN-20161020183838-00036-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://hiking-calculator.deadunicornz.org/ | # Hiking calculator
## Build the path by clicking on the map to get range and time estimates
Metric unit system
Imperial unit system
$V = V_0 * K_{slope} * K_{elevation} * K_{backpack}$
Where $V_0$ is a maximum safe and stable speed. Hiker achieves it descending -2.86° slope. Default $V_0 = 6 {km \over h}$ translates into 5 $km \over h$ on a flat terrain.
$K_{slope} = e^{-3.5 \left|{h \over l} + 0.05\right|}$ is defined by Tobler's hiking function. $h$ is elevation gain/loss and $l$ is a distance.
$K_{elevation} = e^{-9.80665 * 0.0289644 * (h - h_{acc}) \over {8.3144598 * 288.15}} = e^{-0.00011855(h - h_{acc})}$ and derived from the Barometric formula for atmosphere density. The idea here is that speed is decreased with atmosphere density decreasing, $h$ is elevation and $h_{acc}$ the elevation you are acclimatized for.
$K_{backpack} = (1 - {m_b \over 100})$ coefficient determines how backpack mass slows you down. $m_b$ is backpack mass to body mass ratio, percents.
Disclaimer: The calculator is intended for ballpark estimates. If you're not hiking in a ballpark use your discretion. | {"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": 0.8059558272361755, "perplexity": 3531.812269070359}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948517845.16/warc/CC-MAIN-20171212173259-20171212193259-00043.warc.gz"} |
https://crypto.stackexchange.com/questions/29713/inverting-rsa-function/29729 | # Inverting RSA function
I am in high school and I am writing a paper on RSA. I want to show that low values of the public key exponent can make it easy to 'invert' the function so that the encrypted message can be recovered. How is this done? I have tried to read the Handbook of Applied Cryptography but it's not making sense...
Would this be inversion: Taking cipher text, and continuously adding the modulus, then taking the eth root of the sum. If the result is an integer, then it is the plaintext message.
• PS. My paper is just covering plain old RSA, no padding or anything :) – EVELYN Oct 9 '15 at 2:32
• I suppose it should be the private key and not the public one. – user27950 Oct 9 '15 at 4:54
• RSA without padding is vulnerable, and should not be presented in introductory material without mention that it must not be used for serious purposes (except for random messages almost as wide as the modulus). Combined with low value of the public exponent, RSA without padding is even more vulnerable, but usually not because it is easier to invert the public function (the exception to that being in the end of poncho's answer) – fgrieu Oct 9 '15 at 13:09
I want to show that low values of the public key exponent can make it easy to 'invert' the function so that the encrypted message can be recovered.
That is not known to be true; as long as the modulus is large enough to make factorization infeasible, there is no known way to compute e-th roots in general.
Now, if the plaintext $p$ is small enough that $p^e < N$, then it is easy to recover $p$ (just take the e-th root over the integers, which is an easy problem). That's as close as we know to the result you're trying to get.
Would this be inversion: Taking cipher text, and continuously adding the modulus, then taking the eth root of the sum. If the result is an integer, then it is the plaintext message.
If $N$ is small enough to make work in a practical amount of time, then $N$ is small enough to factor.
• I'm reading the last paragraph of the question as a simple extension of the e-th root attack, working when $p^e<k\cdot N$ for moderate $k$. I'm sure $k\approx2^{20}$ is feasible. I wish I knew how large $k$ needs to be in order to avoid a more sophisticated attack. This answer to a significantly different question does not tell. – fgrieu Oct 9 '15 at 16:26
• Hello, could you please explain k? I have not encountered a constant or variable k in my readings. – EVELYN Oct 10 '15 at 9:57
• @EVELYN : $\:$ k is a positive integer, and in this case that's all it is. $\;\;\;\;$ – user991 Oct 11 '15 at 8:39
While Poncho's answer is fully valid I'll hereby extend the part on the proposed idea being inferior to factoring.
I want to show that low values of the public key exponent can make it easy to 'invert' the function so that the encrypted message can be recovered.
As poncho said this is only feasible if the message's length is smaller than the modulus length divided by e, i.e. when no modular reduction takes place.
Would this be inversion: Taking cipher text, and continuously adding the modulus, then taking the eth root of the sum. If the result is an integer, then it is the plaintext message.
The problem of this approach is the run-time. For this approach to be viable you need to beat the General Number Field Sieve (GNFS) in terms of performance at least for some messages. So let's assume the best case scenario, where your algorithm starts operating: if $||m||\geq ||N||/e$, so for example for a 1025-bit message for 3072-bit RSA with $e=3$. To be beat the GNFS, you need to be able to carry out the attack on less than $2^{256}$ operations for this scenario. Clearly, the ciphertext will have length $||c||=3072$ bits with high probability and the term you're looking for $m^e$ has $1025*3=3075$ bits. So you have to test every single (or at least half) of all the multiples of $N$ in the 3075 bit range, which are $2^3=8$, which is indeed feasible. But which area of messages can you break using this technique? You can break any message for which the above equation holds and for which $(||m||-||N||/e)*e<2^{60}$, meaning for $e=3$ you "can break 20 more bits" (f.ex. 1025-1045). If you consider your technique "good" if it stays below factoring, you can even extend that to 80 bits. But in practice this kind of attack is already countered by randomized padding.
• The approach in the question remains feasible when the message's length is slightly higher than the modulus length divided by $e$, at least up to $20/e$ extra bits, with modest means. I do not know what the upper limit is; this answer to a significantly different question does not tell. – fgrieu Oct 9 '15 at 16:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8214862942695618, "perplexity": 398.86665974370806}, "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-2021-17/segments/1618038075074.29/warc/CC-MAIN-20210413213655-20210414003655-00283.warc.gz"} |
http://clay6.com/qa/11146/the-linear-momentum-p-of-a-moving-body-varies-with-time-according-to-equati | Browse Questions
# The linear momentum p of a moving body varies with time according to equation $p=a+bt^2$ where a,b are constants. Net force acting on the body is
a) Constant
b) proportional to $t^2$
c) Inversely proportional to t
d) proportional to t
$F=\large\frac{dp}{dt}=\frac{d}{dt}$$(a+bt^2)$
$F=2bt$
$f \;\alpha\; t$
Net force acting on the body is proportional to t
Hence d is the correct answer.
edited Jan 26, 2014 by meena.p
+1 vote | {"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": 0.8251183032989502, "perplexity": 1410.9605479675272}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988720962.53/warc/CC-MAIN-20161020183840-00559-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://papers.nips.cc/paper/1989/hash/92c8c96e4c37100777c7190b76d28233-Abstract.html | #### Authors
Gintaras Reklaitis, Athanasios Tsirukis, Manoel Tenorio
#### Abstract
A nonlinear neural framework, called the Generalized Hopfield network, is proposed, which is able to solve in a parallel distributed manner systems of nonlinear equations. The method is applied to the general nonlinear optimization problem. We demonstrate GHNs implementing the three most important optimization algorithms, namely the Augmented Lagrangian, Generalized Reduced Gradient and Successive Quadratic Programming methods. The study results in a dynamic view of the optimization problem and offers a straightforward model for the parallelization of the optimization computations, thus significantly extending the practical limits of problems that can be formulated as an optimization problem and which can gain from the introduction of nonlinearities in their structure (eg. pattern recognition, supervised learning, design of content-addressable memories).
1 To whom correspondence should be addressed.
356
Reklaitis, Tsirukis and Tenorio
1 RELATED WORK The ability of networks of highly interconnected simple nonlinear analog processors (neurons) to solve complicated optimization problems was demonstrated in a series of papers by Hopfield and Tank (Hopfield, 1984), (Tank, 1986). The Hopfield computational model is almost exclusively applied to the solution of combinatorially complex linear decision problems (eg. Traveling Salesman Problem). Unfortunately such problems can not be solved with guaranteed quality, (Bruck, 1987), getting trapped in locally optimal solutions. Jeffrey and Rossner, (Jeffrey, 1986), extended Hopfield's technique to the nonlinear unconstrained optimization problem, using Cauchy dynamics. Kennedy and Chua, (Kennedy, 1988), presented an analog implementation of a network solving a nonlinear optimization problem. The underlying optimization algorithm is a simple transformation method, (Reklaitis, 1983), which is known to be relatively inefficient for large nonlinear optimization problems.
2 LINEAR HOPFIELD NETWORK (LHN) The computation in a Hopfield network is done by a collection of highly interconnected simple neurons. Each processing element, i, is characterized by the activation level, Ui, which is a function of the input received from the external environment, Ii, and the state of the other neurons. The activation level of i is transmitted to the other processors, after passing through a filter that converts Ui to a 0-1 binary value, Vi' The time behavior of the system is described by the following model:
~ T·V· - -' + I· ' ~ 'J J J | {"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": 0.8430560231208801, "perplexity": 1272.461616608657}, "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/1664030336921.76/warc/CC-MAIN-20221001195125-20221001225125-00255.warc.gz"} |
https://cstheory.stackexchange.com/questions/35991/np-hardness-on-cayley-graphs/35993 | # NP-hardness on Cayley graphs
What is known about complexity of NP-hard problems on Cayley graphs?
Suppose that the graph is given explicitly as the multiplication table of the group and the list of generators. So the input length is the size of the graph. Can we solve NP-complete problems on such graphs (maximum clique/max-cut) in polynomial time?
What about some special cases of groups? For example, $\mathbb{Z}_n$ (a.k.a. circulant graphs) or $\mathbb{Z}_2^{\log(n)}$. That is, the input to the problem is the set of generators (and $1^n$ to represent the size of the graph). | {"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": 0.9370612502098083, "perplexity": 282.1104436839532}, "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/1566027316783.70/warc/CC-MAIN-20190822042502-20190822064502-00541.warc.gz"} |
http://www.math.u-szeged.hu/mathweb/index.php/hu/component/jevents/icalrepeat.detail/2021/09/23/1509/-/sadegh-marzban-university-of-szeged-a-hybrid-pde-abm-model-for-infection-dynamics-study-on-stochastic-variability-application-to-sars-cov-2-and-influe?Itemid=110 | Év szerint Hónap szerint Ugrás a hónaphoz
## Sadegh Marzban (University of Szeged): A hybrid PDE-ABM model for infection dynamics: study on stochastic variability, application to SARS-COV-2 and influenza, and exploring some treatment options
Csütörtök, 23. Szeptember 2021, 11:00 - 12:30
Abstract.
We propose a hybrid partial differential equation -- agent-based (PDE--ABM) model to describe the spatio-temporal viral dynamics in a cell population. The virus concentration is considered as a continuous variable and virus movement is modelled by diffusion, while changes in the states of cells (i.e. healthy, infected, dead) are represented by a stochastic agent-based model. The two subsystems are intertwined: the probability of an agent getting infected in the ABM depends on the local viral concentration, and the source term of viral production in the PDE is determined by the cells that are infected.
We develop a computational tool that allows us to study the hybrid system and the generated spatial patterns in detail. We systematically compare the outputs with a classical ODE system of viral dynamics, and find that the ODE model is a good approximation only if the diffusion coefficient is large.
We demonstrate that the model is able to predict SARS--CoV--2 infection dynamics, and replicate the output of in vitro experiments. Applying the model to influenza as well, we can gain insight into why the outcomes of these two infections are different.
Hely : Riesz Lecture Hall, 1st Floor, Bolyai Institute, Aradi Vértanúk tere 1., Szeged | {"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": 0.8074097037315369, "perplexity": 1814.6329929075528}, "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/1674764499634.11/warc/CC-MAIN-20230128121809-20230128151809-00379.warc.gz"} |
http://mathhelpforum.com/calculus/17256-parametric-equations.html | Math Help - Parametric equations
1. Parametric equations
Hi,
I'm completely lost in this problem .... can someone help me pls?
Consider the parametric equations x= 4*cos^2(θ) and y = 2*sin(θ)
and fill this chart.
When t = -pi/2, x = ? and y = ?
When t = -pi/4, x = ? and y = ?
When t = pi/4, x = ? and y = ?
When t = pi/2, x = ? and y = ?
I also have to graph the equation in my pgrahing calculator. I tried to do it but I don't know if I did it right.
- Isabel
2. Originally Posted by cuteisa89
Hi,
I'm completely lost in this problem .... can someone help me pls?
Consider the parametric equations x= 4*cos^2(θ) and y = 2*sin(θ)
and fill this chart.
When t = -pi/2, x = ? and y = ?
When t = -pi/4, x = ? and y = ?
When t = pi/4, x = ? and y = ?
When t = pi/2, x = ? and y = ?
I also have to graph the equation in my pgrahing calculator. I tried to do it but I don't know if I did it right.
- Isabel
When you say "When t = -pi/2" do you mean "When $\theta$ = -pi/2?"
Just plug the angle into the given formulas for x and y.
For example:
$\theta = \pi/2 \implies x = 4cos^2(\pi/2) = 0 \text{ and }y = 2sin(\pi/2) = 2$
-Dan
3. Yeah.... t= θ.
Thanks I think I've got it now. I just wasn't sure how to do it properly.
Thanks Dan!
4. Originally Posted by cuteisa89
Hi,
I'm completely lost in this problem .... can someone help me pls?
Consider the parametric equations x= 4*cos^2(θ) and y = 2*sin(θ)
and fill this chart.
When t = -pi/2, x = ? and y = ?
When t = -pi/4, x = ? and y = ?
When t = pi/4, x = ? and y = ?
When t = pi/2, x = ? and y = ?
I also have to graph the equation in my pgrahing calculator. I tried to do it but I don't know if I did it right.
- Isabel
x=4(cost)^2=4(1-(sint)^2)=4-(2sint)^2=4-y^2
So it is a parabola like this:
Attached Thumbnails | {"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": 0.8724790215492249, "perplexity": 1207.8493070874908}, "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-2014-23/segments/1404776438539.21/warc/CC-MAIN-20140707234038-00003-ip-10-180-212-248.ec2.internal.warc.gz"} |
http://mathhelpforum.com/number-theory/227581-help-i-m-stuck-question.html | # Math Help - Help I'm stuck with this question:
1. ## Help I'm stuck with this question:
Prove that:
If n is any integer that is not divisible by 2 or 3 then n^2 mod 12 =1?
help i don't know what to do...
2. ## Re: Help I'm stuck with this question:
If $n\not| 2$ or $n\not| 3$, then $n^2(\text{mod} 12) \equiv 1$
Is this what you mean?
3. ## Re: Help I'm stuck with this question:
Originally Posted by mendax05
Prove that:
If n is any integer that is not divisible by 2 or 3 then n^2 mod 12 =1?
help i don't know what to do...
$\exists\ p \in \mathbb Z\ such\ that\ n = 3p + k,\ where\ k = 1\ or\ k = 2.$
$CASE\ A:\ k = 1.$
Assume p is odd. Then 3p is odd. So 3p + 1 is even. Thus n is even. But by hypothesis n is not evenly divisible by 2 and so is not even. Contradiction. So p is even.
$\exists\ q \in \mathbb Z\ such\ that\ 2q = p \implies n = 3p + 1 = 6q + 1 \implies$
$n^2 = 36q^2 + 12q + 1 \implies \dfrac{n^2}{12} = \dfrac{36q^2 + 12q + 1}{12} = 3q^2 + q + \dfrac{1}{12}.$ QED
$CASE\ B:\ k = 2.$
Assume p is even. Then 3p is even. So 3p + 2 is even. Thus n is even. But by hypothesis n is not evenly divisible by 2 and so is not even. Contradiction. So p is odd.
$\exists\ q \in \mathbb Z\ such\ that\ 2q + 1 = p \implies n = 3p + 2 = 3(2q + 1) + 2 = 6q + 5 \implies$
$n^2 = 36q^2 + 60q + 25 \implies \dfrac{n^2}{12} = \dfrac{36q^2 + 60q + 25}{12} = 3q^2 + 5q + 2 + \dfrac{1}{12}.$ QED
4. ## Re: Help I'm stuck with this question:
It finally dawned on me that I made this proof MUCH harder than it needed to be.
What we need is this lemma: for any three successive integers, exactly one is evenly divisible by 3.
$n\ not\ evenly\ divisible\ by\ 2 \implies n + 1\ and\ n - 1\ evenly\ divisible\ by\ 2.$
$n\ not\ evenly\ divisible\ by\ 3 \implies either\ n + 1\ or\ else\ n - 1\ evenly\ divisible\ by\ 3.$
$\therefore n \pm 1\ is\ evenly\ divisible\ by\ 2\ and\ 3 \implies n \pm 1\ is\ evenly\ divisible\ by\ 6 \implies$
$\exists\ p \in \mathbb Z\ such\ that\ 6p = n \pm 1 \implies n = 6p \pm 1 \implies n^2 = 36p^2 \pm 12p + 1 \implies$
$\dfrac{n^2}{12} = \dfrac{36p^2 \pm 12p + 1}{12} = 3p^2 \pm p + \dfrac{1}{12} = an\ integer + \dfrac{1}{12}. QED.$
5. ## Re: Help I'm stuck with this question:
please don't post multiple copies of the same question. I answered this question here as well. | {"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": 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": 0.937960684299469, "perplexity": 462.8406222471629}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644064538.31/warc/CC-MAIN-20150827025424-00204-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/linear-algebra-and-its-applications-5th-edition/chapter-1-linear-equations-in-linear-algebra-1-1-exercises-page-10/18 | # Chapter 1 - Linear Equations in Linear Algebra - 1.1 Exercises: 18
No. The planes do not have a point of intersection; i.e., the system of equations is not consistent.
#### Work Step by Step
Simplifying the augmented matrix to triangular form provides the relations: $x_1=\frac{-14}{6}$, $x_2=\frac{-5}{6}$, and $0=\frac{-5}{6}$. This final condition is not true, so there is no $(x_1, x_2)$ such that the equations for the three specified planes are satisfied.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | {"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": 0.8061339855194092, "perplexity": 414.07023493374857}, "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-2018-30/segments/1531676589618.52/warc/CC-MAIN-20180717070721-20180717090721-00254.warc.gz"} |
https://proceedings.neurips.cc/paper/2020/hash/09ccf3183d9e90e5ae1f425d5f9b2c00-Abstract.html | #### Authors
Qian Huang, Horace He, Abhay Singh, Yan Zhang, Ser Nam Lim, Austin R. Benson
#### Abstract
<p>Incorporating relational reasoning into neural networks has greatly expanded their capabilities and scope. One defining trait of relational reasoning is that it operates on a set of entities, as opposed to standard vector representations. Existing end-to-end approaches for relational reasoning typically extract entities from inputs by directly interpreting the latent feature representations as a set. We show that these approaches do not respect set permutational invariance and thus have fundamental representational limitations. To resolve this limitation, we propose a simple and general network module called Set Refiner Network (SRN). We first use synthetic image experiments to demonstrate how our approach effectively decomposes objects without explicit supervision. Then, we insert our module into existing relational reasoning models and show that respecting set invariance leads to substantial gains in prediction performance and robustness on several relational reasoning tasks. Code can be found at github.com/CUAI/BetterSetRepresentations.</p> | {"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": 0.8799557685852051, "perplexity": 3366.089477514165}, "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-2021-10/segments/1614178359497.20/warc/CC-MAIN-20210227204637-20210227234637-00519.warc.gz"} |
https://andjournal.sgu.ru/en/rubrika/determinirovannyy-haos?page=4 | ISSN 0869-6632 (Print)
ISSN 2542-1905 (Online)
# Детерминированный хаос
## Experimental realization of lorenz model of liquid’s convective instability in vertical toroidal loop
Stable and unstable regimes of glycerine convection in vertical toroidal loop are investigated experimentally. The results of Fourier-analysis, DFA, wavelet-, and correlation analysis of liquid’s motion peculiarities are presented. Chaotic attractor with Lorenz-attractor signs is constructed.
## Verification of hyperbolicity conditions for a chaotic attractor in a system of coupled nonautonomous van der pol oscillators
We present a method and results of numerical computations on veri?cation of hyperbolic nature for the chaotic attractor in a system of two coupled nonautonomous van der Pol oscillators (Kuznetsov, Phys. Rev. Lett., 95, 2005, 144101). At selected parameter values, we indicate a toroidal domain in four-dimensional phase space of Poincar? e map (topologically, a direct product of a circle and a three-dimensional ball), which is mapped into itself and contains the attractor we analyze.
## Statistical properties of the intermittent transition to chaos in the quasi-periodically forced system
By the example of the quasi-periodically forced logistic map we investigate statistical properties of the transition from strange nonchaotic attractor to chaos in the system with intermittent dynamics. The probability characteristics of laminar and chaotic phase distributions, as well as scaling laws for distributions of local Lyapunov exponents are studied at parameter values near the transition point.
## Influence of noise on chaotic self-sustained oscillations in the regime of spiral attractor
In the present paper we analyze the in?uence of white and colored noise on chaotic selfsustained oscillations in the regime of spiral attractor. We study characteristics of instantaneous phase and spectra of noisy chaotic oscillations. The phenomenon of chaos synchronization by external narrow-band noise has been estimated. Synchronization phenomena under the in?uence of narrow-band noise signals with equal spectra and di?erent probability densities are compared.
## Critical behavior of asymmetrically coupled noisy driven nonidentical systems with period-doublings
We investigated the in?uence of external noise on the critical behavior typical to nonidentical coupled systems with period-doubling. We obtained the numerical value of the scaling factor for noise amplitude by means of the renormalization group analysis. Also we demonstrated the selfsimilar structure of the parameter plane near the critical point in the model system of two noisy driven coupled logistic maps.
## Peculiarities of complex dynamics and transitions to chaotic regimes in the model of two interacting systems with phase control
The work is devoted to investigation of complex dynamics in the model of two interacting systems with phase and delay control. Stability conditions of synchronous regime are determined. The processes of excitement of nonsynchronous regimes and transitions between them are considered. Scenarios of development of nonsynchronous regimes under variation of the systems parameters are determined. Routes to chaotic behavior of the model are discussed. Results are presented in the form of one-parameter bifurcation diagrams and phase portraits of the model attractors.
## About scaling properties of identical coupled logistic maps with two types of coupling without noise and under influence of external noise
In this paper the in?uence of noise in system of identical coupled logistic maps with two types of coupling – dissipative and inertial – is discussed. The corresponding renormalization group analysis is presented. Scaling property in the presence of noise is considered, and necessary illustrations in «numerical experiment style» are given.
## Analytical solution of spectral problem for the perron – frobenius operator of piece-wise linear chaotic maps
Spectral properties of the linear non-self-adjoint Perron – Frobenius operator of piece-wise linear chaotic maps having regular structure are investigated. Eigenfunctions of the operator are found in the form of Bernoulli and Euler polynomials. Corresponding eigenvalues are presented by negative powers of number of map brunches. The solution is obtained in general form by means of generating functions for eigenfunctions of the operator. Expressions for eigenfunctions and eigenvalues are di?erent for original and inverse maps having even and odd number of branches.
## Synchronization in systems with bimodal dynamics
Considering model with bimodal dynamics we investigate the synchronization of di?erent time scales. Transition between mode-locked and mode-unlocked chaotic attractors is investigated. It is shown that this transition involves a situation in which the synchronized chaotic attractor loses its band structure.
## Computation of Lyapunov exponents for spatially extended systems: advantages and limitations of various numerical methods
Problems emerging in computations of Lyapunov exponents for spatially extended systems are considered. We concentrate on the incorrect orthogonalization of high sized ill conditioned matrices appearing in course of the computation, and on large errors emerging under certain conditions if the finite difference numerical method is applied to solve equations. The practical guidelines helping to avoid the mentioned problems are represented. | {"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": 0.8818835020065308, "perplexity": 1714.8783367307035}, "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/1627046154459.22/warc/CC-MAIN-20210803124251-20210803154251-00322.warc.gz"} |
https://www.maplesoft.com/support/help/maple/view.aspx?path=Units/volume_flow&L=E | volume flow - Maple Help
Units of Volume Flow
Description
• Volume flow has the dimension length cubed over time. The SI composite unit of volume flow is the cubic meter per second.
• Maple knows the units of volume flow listed in the following table.
Name Symbols Context Alternate Spellings Prefixes gallon_per_minute gpm US_liquid * gallons_per_minute UK US_dry cubic_foot_per_minute cfm standard * cubic_feet_per_minute
An asterisk ( * ) indicates the default context, an at sign (@) indicates an abbreviation, and under the prefixes column, SI indicates that the unit takes all SI prefixes, IEC indicates that the unit takes IEC prefixes, and SI+ and SI- indicate that the unit takes only positive and negative SI prefixes, respectively. Refer to a unit in the Units package by indexing the name or symbol with the context, for example, gallon_per_minute[US_liquid] or cfm[standard]; or, if the context is indicated as the default, by using only the unit name or symbol, for example, gallon_per_minute or cfm.
The units of volume flow are defined as follows.
A US liquid, US dry, or UK gallon per minute is defined as a gallon of the same context per minute.
A standard cubic foot per minute is defined as a cubic standard foot per minute.
Examples
> $\mathrm{convert}\left('\mathrm{gpm}','\mathrm{dimensions}','\mathrm{base}'=\mathrm{true}\right)$
$\frac{{{\mathrm{length}}}^{{3}}}{{\mathrm{time}}}$ (1)
> $\mathrm{convert}\left(1,'\mathrm{units}','\mathrm{gpm}',\frac{'\mathrm{gal}'}{'\mathrm{min}'}\right)$
${1}$ (2)
> $\mathrm{convert}\left(1,'\mathrm{units}','\mathrm{gpm}\left[\mathrm{UK}\right]',\frac{'\mathrm{gal}\left[\mathrm{UK}\right]'}{'\mathrm{min}'}\right)$
${1}$ (3)
> $\mathrm{convert}\left(1,'\mathrm{units}','\mathrm{gpm}\left[\mathrm{US_dry}\right]',\frac{'\mathrm{gal}\left[\mathrm{US_dry}\right]'}{'\mathrm{min}'}\right)$
${1}$ (4)
> $\mathrm{convert}\left(1,'\mathrm{units}','\mathrm{cfm}',\frac{{'\mathrm{ft}'}^{3}}{'\mathrm{min}'}\right)$
${1}$ (5) | {"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": 0.930768609046936, "perplexity": 2979.0493041167238}, "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/1637964363125.46/warc/CC-MAIN-20211204215252-20211205005252-00361.warc.gz"} |
http://nag.com/numeric/fl/nagdoc_fl24/html/D03/d03pyf.html | D03 Chapter Contents
D03 Chapter Introduction
NAG Library Manual
# NAG Library Routine DocumentD03PYF
Note: before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.
## 1 Purpose
D03PYF may be used in conjunction with either D03PDF/D03PDA or D03PJF/D03PJA. It computes the solution and its first derivative at user-specified points in the spatial coordinate.
## 2 Specification
SUBROUTINE D03PYF ( NPDE, U, NBKPTS, XBKPTS, NPOLY, NPTS, XP, INTPTS, ITYPE, UP, RSAVE, LRSAVE, IFAIL)
INTEGER NPDE, NBKPTS, NPOLY, NPTS, INTPTS, ITYPE, LRSAVE, IFAIL REAL (KIND=nag_wp) U(NPDE,NPTS), XBKPTS(NBKPTS), XP(INTPTS), UP(NPDE,INTPTS,ITYPE), RSAVE(LRSAVE)
## 3 Description
D03PYF is an interpolation routine for evaluating the solution of a system of partial differential equations (PDEs), or the PDE components of a system of PDEs with coupled ordinary differential equations (ODEs), at a set of user-specified points. The solution of a system of equations can be computed using D03PDF/D03PDA or D03PJF/D03PJA on a set of mesh points; D03PYF can then be employed to compute the solution at a set of points other than those originally used in D03PDF/D03PDA or D03PJF/D03PJA. It can also evaluate the first derivative of the solution. Polynomial interpolation is used between each of the break points ${\mathbf{XBKPTS}}\left(\mathit{i}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NBKPTS}}$. When the derivative is needed (${\mathbf{ITYPE}}=2$), the array ${\mathbf{XP}}\left({\mathbf{INTPTS}}\right)$ must not contain any of the break points, as the method, and consequently the interpolation scheme, assumes that only the solution is continuous at these points.
None.
## 5 Parameters
Note: the parameters U, NPTS, NPDE, XBKPTS, NBKPTS, RSAVE and LRSAVE must be supplied unchanged from either D03PDF/D03PDA or D03PJF/D03PJA.
1: NPDE – INTEGERInput
On entry: the number of PDEs.
Constraint: ${\mathbf{NPDE}}\ge 1$.
2: U(NPDE,NPTS) – REAL (KIND=nag_wp) arrayInput
On entry: the PDE part of the original solution returned in the parameter U by the routine D03PDF/D03PDA or D03PJF/D03PJA.
3: NBKPTS – INTEGERInput
On entry: the number of break points.
Constraint: ${\mathbf{NBKPTS}}\ge 2$.
4: XBKPTS(NBKPTS) – REAL (KIND=nag_wp) arrayInput
On entry: ${\mathbf{XBKPTS}}\left(\mathit{i}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NBKPTS}}$, must contain the break points as used by D03PDF/D03PDA or D03PJF/D03PJA.
Constraint: ${\mathbf{XBKPTS}}\left(1\right)<{\mathbf{XBKPTS}}\left(2\right)<\cdots <{\mathbf{XBKPTS}}\left({\mathbf{NBKPTS}}\right)$.
5: NPOLY – INTEGERInput
On entry: the degree of the Chebyshev polynomial used for approximation as used by D03PDF/D03PDA or D03PJF/D03PJA.
Constraint: $1\le {\mathbf{NPOLY}}\le 49$.
6: NPTS – INTEGERInput
On entry: the number of mesh points as used by D03PDF/D03PDA or D03PJF/D03PJA.
Constraint: ${\mathbf{NPTS}}=\left({\mathbf{NBKPTS}}-1\right)×{\mathbf{NPOLY}}+1$.
7: XP(INTPTS) – REAL (KIND=nag_wp) arrayInput
On entry: ${\mathbf{XP}}\left(\mathit{i}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{INTPTS}}$, must contain the spatial interpolation points.
Constraints:
• ${\mathbf{XBKPTS}}\left(1\right)\le {\mathbf{XP}}\left(1\right)<{\mathbf{XP}}\left(2\right)<\cdots <{\mathbf{XP}}\left({\mathbf{INTPTS}}\right)\le {\mathbf{XBKPTS}}\left({\mathbf{NBKPTS}}\right)$;
• if ${\mathbf{ITYPE}}=2$, ${\mathbf{XP}}\left(\mathit{i}\right)\ne {\mathbf{XBKPTS}}\left(\mathit{j}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{INTPTS}}$ and $\mathit{j}=2,3,\dots ,{\mathbf{NBKPTS}}-1$.
8: INTPTS – INTEGERInput
On entry: the number of interpolation points.
Constraint: ${\mathbf{INTPTS}}\ge 1$.
9: ITYPE – INTEGERInput
On entry: specifies the interpolation to be performed.
${\mathbf{ITYPE}}=1$
The solution at the interpolation points are computed.
${\mathbf{ITYPE}}=2$
Both the solution and the first derivative at the interpolation points are computed.
Constraint: ${\mathbf{ITYPE}}=1$ or $2$.
10: UP(NPDE,INTPTS,ITYPE) – REAL (KIND=nag_wp) arrayOutput
On exit: if ${\mathbf{ITYPE}}=1$, ${\mathbf{UP}}\left(\mathit{i},\mathit{j},1\right)$, contains the value of the solution ${U}_{\mathit{i}}\left({x}_{\mathit{j}},{t}_{\mathrm{out}}\right)$, at the interpolation points ${x}_{\mathit{j}}={\mathbf{XP}}\left(\mathit{j}\right)$, for $\mathit{j}=1,2,\dots ,{\mathbf{INTPTS}}$ and $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$.
If ${\mathbf{ITYPE}}=2$, ${\mathbf{UP}}\left(\mathit{i},\mathit{j},1\right)$ contains ${U}_{\mathit{i}}\left({x}_{\mathit{j}},{t}_{\mathrm{out}}\right)$ and ${\mathbf{UP}}\left(\mathit{i},\mathit{j},2\right)$ contains $\frac{\partial {U}_{\mathit{i}}}{\partial x}$ at these points.
11: RSAVE(LRSAVE) – REAL (KIND=nag_wp) arrayCommunication Array
The array RSAVE contains information required by D03PYF as returned by D03PDF/D03PDA or D03PJF/D03PJA. The contents of RSAVE must not be changed from the call to D03PDF/D03PDA or D03PJF/D03PJA. Some elements of this array are overwritten on exit.
12: LRSAVE – INTEGERInput
On entry: the size of the workspace RSAVE, as in D03PDF/D03PDA or D03PJF/D03PJA.
13: IFAIL – INTEGERInput/Output
On entry: IFAIL must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of IFAIL on exit.
On exit: ${\mathbf{IFAIL}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6 Error Indicators and Warnings
If on entry ${\mathbf{IFAIL}}={\mathbf{0}}$ or $-{\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF).
Errors or warnings detected by the routine:
${\mathbf{IFAIL}}=1$
On entry, ${\mathbf{ITYPE}}\ne 1$ or $2$, or ${\mathbf{NPOLY}}<1$, or ${\mathbf{NPDE}}<1$, or ${\mathbf{NBKPTS}}<2$, or ${\mathbf{INTPTS}}<1$, or ${\mathbf{NPTS}}\ne \left({\mathbf{NBKPTS}}-1\right)×{\mathbf{NPOLY}}+1$, or ${\mathbf{XBKPTS}}\left(\mathit{i}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NBKPTS}}$, are not ordered.
${\mathbf{IFAIL}}=2$
On entry, the interpolation points ${\mathbf{XP}}\left(\mathit{i}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{INTPTS}}$, are not in strictly increasing order, or when ${\mathbf{ITYPE}}=2$, at least one of the interpolation points stored in XP is equal to one of the break points stored in XBKPTS.
${\mathbf{IFAIL}}=3$
You are attempting extrapolation, that is, one of the interpolation points ${\mathbf{XP}}\left(i\right)$, for some $i$, lies outside the interval [${\mathbf{XBKPTS}}\left(1\right),{\mathbf{XBKPTS}}\left({\mathbf{NBKPTS}}\right)$]. Extrapolation is not permitted.
## 7 Accuracy
See the documents for D03PDF/D03PDA or D03PJF/D03PJA. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 61, "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": 0.9940730929374695, "perplexity": 2969.950064281214}, "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-2014-49/segments/1416931009004.88/warc/CC-MAIN-20141125155649-00133-ip-10-235-23-156.ec2.internal.warc.gz"} |
http://quantumskeptic.blogspot.com/ | Saturday, September 27, 2014
My new paper on relative spin phase modulated magnetic force
When I was thinking it was necessary to take a difference between retarded and advanced fields to have a time-symmetric electrodynamics that's Lorentz covariant, I investigated whether a magnetic interaction between point charges moving in little circles at or near the speed of light could play the role of the Coulomb force in binding atoms. Really it can't, but it's remarkable how close it comes.
I needed a force that was radial and of the same strength and range dependence as the Coulomb force. Rivas shows in his book Kinematical Theory of Spinning Particles how the electric acceleration field of a point charge moving in a circle of an electron Compton wavelength radius has an average that is inverse square like the Coulomb force and of the same strength. (That is pretty surprising given that the acceleration field falls off explicitly only directly inversely with distance, hence it's characterization as the radiation field.) But, since the magnetic field in Gaussian units is just the electric field crossed by a unit vector, the magnetic acceleration field for a charge doing the relativistic circular motion field has the same strength as the electric field.
If we have a magnetic field that's just as strong as a Coulomb field, and the charge moving in it is moving at the speed of light, then v/c for the charge is one and the resulting magnetic force via the Lorentz force law is just as strong as the Coulomb force. So, if we suppose the test charge is going in the same type of little circle as the field-generating charge, if the two motions are exactly aligned and in phase, then it turns out there's a purely radial component of the force that's constant in time and just as strong as the Coulomb force. If the circular motions of the two particles are out of phase, then the strength varies sinusoidally with the phase difference, and so it could either double or cancel the electric force. But the phase difference has to include the time delay of propagation from one circulating particle to the other. (It's important to realize that the charges aren't orbiting around each other, but each doing their little circular motions separately, with the circle centers many (Compton wavelength) diameters apart.) So there can be a very substantial influence on the motion of the center of the test charge's circular motion due to the magnetic force, if it is already also moving because of the electric field, due to the relative orientation of the charges plane of motion (which translates to the spin polarization) and the difference in phase of the internal motions of their spins. This kind of phase difference has been proposed by David Hestenes to correspond to the phase of the wave function of quantum mechanics.
So, in the process of putting the story together about this magnetic force as an alternative to the Coulomb force in the time-symmetric picture, I realized (see the previous post) that I missed a sign change in going from retardation to advancement of the magnetic field, so that it does not change sign, and so the magnetic force does not need to replace the Coulomb force but only augment it, in order to plausibly explain quantum behavior.
I'm looking forward to understanding how this picture plays out, but that will take a while, so for now I am putting out what I have. It will appear on arxiv tomorrow, but I have already posted it to Researchgate here.
Wednesday, September 24, 2014
Correction
A few days ago I realized I'd overlooked a different way out of the problem that time-advanced magnetic fields want to reverse sign compared to retarded ones, and so causing magnetic effects to cancel out in the time-symmetric picture. The cancellation doesn't happen because the sign on the unit vector from the source to field point, that is crossed onto the electric field to obtain the magnetic field, also changes sign in going from retardation to advancement. That is, in the retarded case we have B = n x E, but in the advanced case this becomes B = -n x E. The unit vector n here changes sign because it originates as the gradient of the retarded or advanced time, so the sign change that changes retardation to advancement applies directly to it.
This means the problem I've been working to overcome for the last eight or ten months does not even exist. In particular, the cancellation of the strong magnetic force that I was getting in the time symmetric picture does not occur. So, it can account for preon binding (for charged preons, at least) just fine. What I have been saying recently, that Lorentz covariance requires magnetic forces be preserved in the time-symmetric picture, if there is one, is perfectly true, but it is also completely consistent with electrodynamics and time-symmetric electrodynamics under the usual assumption that the retarded and advanced solutions are summed rather than differenced. Writing it up for publication forced me to realize this, because when I went through, as a simple demonstration of the problem, calculating the magnetic moment from a static current loop from the retarded potentials (see, e.g., Landau and Lifshitz Eq. 66.2) I discovered that it absolutely does not change sign when going over to the advanced case.
Now, everything is better except the idea that the electron g factor of (about) 2 can be explained as the difference between retarded only and time symmetric electrodynamics. I still think this is a very attractive idea, but the story isn't as compelling because it isn't true as I said that in the conventional time-symmetric picture the g factor would be zero, i.e., that the electron magnetic moment would vanish. In the usual picture (attributable to Dirac 1938) the time symmetric field is the mean of the retarded and advanced fields, which leads to a g-factor of one.
Maybe I should give up my obsession with the electron g factor, now that I am routinely thinking of the electron as a composite particle. If it has a positive charge part with opposite but smaller spin, doesn't that give a g factor larger than unity? Maybe g factor two is easy to get in the preon model. I am still used to thinking of an electron as a structureless object so this kind of thinking is not natural. Maybe g-factor two is simply a confirmation that it's a composite particle.
Later I think I will look more at the g factor of a composite electron, but for right now I'm trying to complete something entirely new to put on arxiv and hopefully soon after submit to a journal. Hopefully I will upload it to arxiv within a day or two. Before that happens, I am also planning on revising my kinematics arxiv paper to remove the new section I just added a few weeks ago. Probably I will do that later today. I can't let it stay up there long knowing it is dead wrong.
Sunday, August 10, 2014
Why the electron g-factor is 2
Today I realized (I think) that if electrodynamics is time-symmetric, and if the magnetic force does not flip its sign for the advanced magnetic forces compared to retarded magnetic forces, then this will naturally double the strength of the magnetic forces at scales where the retarded and advanced interactions are experienced close to in phase. So the electron magnetic field, if produced by moving charge, will be twice what it would be for retarded only forces. So perhaps the electron g-factor being (about) 2 can be taken as confirmation that electrodynamics is time-symmetric.
After mulling it over for a while today, I decided to do a quick update to my arxiv paper to include this observation. I added a new section V that consists of 3 paragraphs with no equations. It will post tomorrow (as v8) if I don't change it further and reset the clock. (I also corrected Eq. (8), which did not affect any subsequent results. It may become relevant in the next update (v9) however. I have a lot of material towards a new version beyond v7/v8, but it was inconclusive until the new ideas of the last few weeks, which tentatively seem to be panning out nicely. I have only been working on it again for the last few days, though. Prior to that I was unusually tied up for several weeks with my engineering job on a hot project.)
Another thing I want to mention, that I was being coy about in my last post, is how it might be possible to have electrical velocity fields invert sign between retardation and advancement, and still have apparent electrostatic forces between (apparently) stationary charges. The way it might possible is if what we take as electrostatic electric fields and forces are actually time averages of electric acceleration fields. Martin Rivas (citation will be in the new posted version, and is in some of the earlier versions already) has already shown how if the electron is modeled as a circulating point charge moving at the speed of light (and it will still be true for asymptotically close to the speed of light), then the time-averaged acceleration field is Coulomb-like by several Compton radii away (when the radius of the circular motion is the Compton wavelength). Also, for the ultra-relativistic charge, the velocity field collapses to a point and so doesn't contribute to the average Coulomb-like field.
I spent today trying to modify Rivas' calculation to see if I can get a similar result in spite of switching the sign of the advanced forces. It seems intuitively that it couldn't but I am encouraged, as far as I got today. It doesn't make it vanish identically, as my intuition predicts it should have. This is a very preliminary observation, so maybe it will fall apart, but it shouldn't take long to get an answer one way or the other. I have another reason to be optimistic, though, because I also tried yesterday adding sign-reversed (compared to the usual) advanced fields into my attempted derivation of anti-Euler forces from the velocity magnetic field terms, and now it does seem to be emerging. I have spent six long months trying to get this with no prior success, so it seems very encouraging. This is also only a preliminary observation that could evaporate. I still have a lot of work to do before I can have something to submit to a journal, but I feel like I'm making serious progress again finally, after months of getting nowhere fast.
Monday, July 28, 2014
In time-symmetric electrodynamics, it is always assumed that the sign of the electric force is the same for the advanced force as for the retarded force. This must be so (one would think) because in the inertial reference frame where an infinitely heavy charge is at rest, a test charge held initially at rest and then released would experience no net Coulomb force if it were otherwise.
It follows from this seemingly necessary choice that the sign of the magnetic force in a different inertial reference frame, where the heavy charge is in motion, will invert for advanced compared to retarded magnetic forces. So, in time symmetric electrodynamics, magnetic forces tend to cancel out. This appears at least at first glance to keep the magnetic force that seems to correspond to my predicted anti-centrifugal force from being able to overcome Coulomb repulsion.
On account of the considerations above, I have recently been going carefully over how to get the time-advanced and time-retarded fields and forces by Lorentz transforming from the reference frame where the field-source charge is stationary. (The retarded case is already analyzed quite a bit in the appendices of my arxiv paper.) There haven't been any surprises there, but a few days ago I began to realize that the derivation of the magnetic force as the anti-Coriolis force of the Thomas precession doesn't care whether the field or force is retarded or advanced. It can't change sign from retardation to advancement because neither has entered into the derivation in any way. Thus, if the anti-Coriolis force is a real force, then either time-advanced electromagnetic forces cannot exist, or the sign of the Coulomb force must flip between retardation and advancement. Consistency of the force law with relativistic kinematics demands this, if I am correct.
Saturday, April 5, 2014
Superstrong electromagnetic interactions
Since I've given up trying to put out a separate paper quickly on the superstrong magnetic force between highly accelerating ultrarelativistic charges, as I said in the previous post, and have gone back to trying to finish my more general paper on the relationship between electrodynamics and relativistic kinematics, I should report out a little on what I found and didn't find.
The strong attractiveness of the magnetic force in the retarded magnetic acceleration field is already shown in the version posted on arxiv. What I was trying to determine was whether there's an obvious way there can be net attraction in the time symmetric case, as considered by Schild, where the magnetic force due to the advanced field tends to cancel the force due to the retarded field. My idea was that in the ultrarelativistic case the delay and advance angles approach 90 degrees, so maybe it might be possible to change the phase relationship so that the net force is strongly attractive on average.
It turned out that although I was able show tentatively that the retarded and advanced forces don't have to exactly cancel and can easily exceed in magnitude Coulomb repulsion, I wasn't able to generate a net attractive force. I may try further later but for now I have gone back to trying to finish the more general argument.
What I did was to assume two charges were circularly orbiting each other in an approximately circular orbit with an orbit diameter smaller than one one-hundredth the size of a proton, and at a velocity very close to the speed of light. I wrote a matlab program to calculate the full retarded and advanced, and non-radiative and radiative, electric and magnetic fields at the position of one particle due to the other, and accounting for delay and advancement, where the motion was assumed to be circular and periodic, but allowing the accelerations to depart from the strict centripetal acceleration of a pure circular orbit. That is, I let the non-centripetal acceleration affect the fields but not the orbit. Then I looked at the induced acceleration of one particle due to the other, and attempted to construct a configuration where the motions of each particle induced by the other would be consistent. I totally ignored radiation damping, as did Schild, although it's enormous in this configuration.
It turned out to be pretty simple to build a configuration where the motions seem approximately consistent in the time-symmetric electrodynamic sense. A lot more work would be needed to determine if this is a real or meaningful result, and I don't mean to assert that it is. If I had more confidence I could build something convincingly meaningful in a reasonable amount of time, I'd continue to work on it, but for now I think my time is better spent elsewhere.
To illustrate what I'm trying to describe, I captured a plot from my matlab program, see below. Clicking on the figure should expand it. The top two strip plots are what is used to calculate the full em field at the position of the second (test) particle, and then the bottom two plots are the acceleration induced on the test particle. The scales are not very meaningful because the magnitudes depend on how close the velocity is to the speed of light, and the orbital radius, and the invariant particle masses, and in a complicated way.
The top two strip plots show the motion of the charge the generates the field that the test particle moves in. That's the field source particle. The field that the test particle generates isn't allowed to affect the source particle here. The top plot shows that I arbitrarily imposed a strong radial oscillating (at the orbital period) acceleration on top of the constant radial centripetal acceleration in it's circular circular orbit. The second plot is showing that there's no comparatively significant motion in the tangential or axial directions. That's just the noise level when some large positive and negative numbers got added together, at matlab default precision.
Then the time retarded and advanced fields at the test particle position, moving oppositely in a nominally circular orbit, are calculated and the corresponding acceleration of the test particle due to their sum is plotted on the two lower strips. What's interesting to me is that the oscillating radial acceleration of the source particle has induced a similar radial acceleration in the test particle, out of phase such that if it were allowed to act back on the source particle has a hope of leading to a consistent periodic motion, perhaps. There is also a tangential acceleration induced, but it's much smaller in magnitude. The smaller magnitude is in part at least due to the difference in relativistic mass along track versus cross-track.
This would be a rabbit hole to pursue seriously, that one might never emerge from. But it would be fun.
Wednesday, January 8, 2014
Magnetism as the Origin of Preon Binding
A week or so ago I googled "preon binding force" and turned up an article by Jogesh Pati, the originator of the term "preon," according to wikipedia:
Magnetism as the origin of preon binding
Physics Letters B, Volume 98, Issue 1-2, p. 40-44.
It is argued that ordinary ``electric''-type forces - abelian or nonabelian - arising within the grand unification hypothesis are inadequate to bind preons to make quarks and lepton unless we proliferate preons. It is therefore suggested that the preons carry electric and magnetic charges and that their binding force is magnetic. Quarks and leptons are magnetically neutral. Possible consistency of this suggestion with the known phenomena and possible origin of magnetic charges are discussed.
So, apparently, I am not the first to think preons might be bound magnetically. However, in order to achieve magnetic binding, the above article postulates that preons possess magnetic charges, which are not required by the mechanism I propose.
I decided to write a short paper on how electrical charges even of like polarity can be magnetically bound according classical electrodynamics, without going extensively into the relativistic kinematics arguments, to submit to a journal as soon as possible. I thought I could just excerpt that part of my paper as it's currently posted on arxiv, but now I'm wanting to elaborate a little bit further, taking better account of retardation and perhaps looking at how it acts in time symmetric electrodynamics (i.e., allowing for time-advanced as well as time-retarded interaction). Properly accounting for retardation makes things much more complicated and possibly intractable, but it is impossible to argue that it's negligible in this case. It is thus not going as quickly as I'd initially hoped.
Wednesday, November 13, 2013
A new version of my magnetic force paper on Arxiv
It's here. It isn't the final version, but it has significant improvements compared to previous. Section IIb is improved in the sense that there are no leftover terms in the magnetic force derived as a Coriolis effect of the relative rotation of the lab frame relative to the field source particle rest frame as seen by the test particle co-moving observer (TPCMO). This is a result of having the correct sign on the Thomas precession as observed by the TPCMO, which is opposite of that seen by an inertial observer of an accelerated frame, as usually is provided in textbooks. The explanation of how this happens is at the end of new Appendix A.
The new Appendix A also has a complete derivation of the Thomas precession using very elementary analysis that I hope is more transparent than other derivations, and may be unique in its own right. I needed such a derivation because unlike other derivations that focus on the precession of a spinning particle, this one is focused on kinematics more generally, I'd say, and so obtains directly standard kinematical effects of rotation, such as that the velocity of a particle in a rotating frame is the velocity in the non-rotating plus an angular velocity of the rotation crossed with the radius vector to the particle from the center of rotation. This is particularly important because it has been argued previously (by Bergstrom) that even though the magnetic force is clearly a Coriolis effect of the Thomas precession, it cannot give rise to an anticentrifugal forces because it applies only at a point and not more globally. Bergstrom invents an interpretation that there is a "mosaic" of transformations between non-inertial and inertial reference frames such that the rotation applies only at the center of rotation, but I believe this interpretation is without real basis, and furthermore is disproved by the analysis in my Appendix A of version 7. It seems pretty clear that the sole purpose of Bergstrom's interpretation is to avoid the otherwise obvious conclusion that if the Thomas precession causes a Coriolis effect as the magnetic force, then it must also cause a centrifugal-like force. So, I believe this clears the way for a convincing relativistic argument that there need to be anti-centrifugal and anti-Euler forces.
I also used this update as an opportunity to introduce for the first time on arxiv the hypothesis that the anti-centrifugal force is the ultra-strong force that binds preons to from quarks.
The improvements to section IIb make it fully consistent with that part of the talk I gave at the PIERS conference last August. Unfortunately due to confusion related to finding a sign error at the last minute and the deadline for the paper, they didn't get into the paper published in the conference proceedings. I discussed that sign error in at least one previous post. Later on perhaps I will make a corrected version of that and post it on Reasearchgate. The charts I gave as the talk for the PIERS conference are already posted there. The talk also has an overview of the analysis that is now in Appendix A, but Appendix A is more advanced and more rigorous, in particular in how the partial derivative of time in the TPCMO's frame with respect to source particle rest frame time should be obtained. The version in the talk gets the right result but the reasoning behind it is not quite right. Getting it through a defensible derivation is a very significant improvement, I feel.
The path should now be clear to complete the analysis and obtain a relativistically exact (to order v^2/c^2) derivation of the magnetic force as a Coriolis effect of the Thomas precession. This should also bring along an anti-Euler force of the Thomas precession, if one exists as I think necessary. The anti-centrifugal force with be strongly implied, but can't be proven until the analysis is extended to order v^4/c^4. But of course, as mentioned previously, it can already be found in Maxwell-Lorentz electrodynamics, if one knows where to look. | {"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": 0.8688641786575317, "perplexity": 581.7449702531596}, "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/1414637897717.20/warc/CC-MAIN-20141030025817-00001-ip-10-16-133-185.ec2.internal.warc.gz"} |
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