url
stringlengths 15
1.13k
| text
stringlengths 100
1.04M
| metadata
stringlengths 1.06k
1.1k
|
|---|---|---|
https://www.livmathssoc.org.uk/cgi-bin/sews.py?OrbitalVelocity | The speed you must go to be in orbit.
Slightly misleading, as your orbital velocity depends on your orbital height: the higher your orbit, the slower you go.
It's relatively easy to compute orbital velocity for an Earth grazing orbit, and that's intimately related to the calculation of the distance to the horizon.
Escape velocity is larger by a factor of $\sqrt{2},$ which is interesting. | {"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": 0.9517733454704285, "perplexity": 508.43804375470455}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585561.4/warc/CC-MAIN-20211023033857-20211023063857-00589.warc.gz"} |
https://mathoverflow.net/questions/211007/least-prime-for-which-a-square-free-integer-is-a-non-residue | # Least prime for which a square-free integer is a non-residue
Suppose $a$ is a square-free integer and $\left(\frac{a}{p}\right)=1$ for the primes $p\leq k$. I'll call $a$ a quasi-square of order $k$. What I am interested in is the maximum value of $k$ in terms of $a$. For instance if $a$ is a prime which is $1$ mod $4$ then by quadratic reciprocity, we are really asking about the least non-residue mod $a$. So on the GRH we have $k=O((\log a)^2)$. But if we think about these symbols as coin flips, then I would suspect that after about $\log_2 a$ primes we should see a $-1$. So, since the $n$'th prime is about $n\log n$ one might guess that $k$ should be not much larger than $(\log_2 a)(\log\log a)$. Is this reasonable? Is there a well-known conjecture which suggests this? Does it hold on average?
Edit: As further justification, if we look at all $N\leq a\leq 2N$ then checking that $\left(\frac{a}{p}\right)=1$ rules out half of the integers in this range. So we should be run out of integers after about $\log_2 N$ primes. Of course the squares will remain, but that should be it.
• What you call a quasi-square is known in the literature as a pseudosquare. See, e.g., MR2282926 Wooding, Kjell; Williams, Hugh C. Doubly-focused enumeration of pseudosquares and pseudocubes. Algorithmic number theory, 208–221, Lecture Notes in Comput. Sci., 4076, Springer, Berlin, 2006. – Gerry Myerson Jul 8 '15 at 0:41
For fixed $a$, the function $\big( \frac ap \big)$ defines a Dirichlet character (mod $4a$) (and often modulo a smaller modulus). More precisely, the Jacobi symbol $\big( \frac an \big)$ defines such an extension of the Legendre symbol to all odd $n$ (simply by multiplicativity), and that extension is a Dirichlet character.
So everything you know about least nonresidues of Dirichlet characters holds here: unconditionally there is a prime $p \ll a^{1/4\sqrt e+\epsilon}$ for which $\big( \frac ap \big)=-1$, and on GRH there is such a prime $p \ll \log^2 a$, as you've noted. I don't remember the heuristic for what the best possible bound should be, but the order of magnitude $\log a \cdot \log\log a$ or $\log a\cdot(\log\log a)^2$ seems plausible. | {"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": 0.9361667037010193, "perplexity": 224.36223173120806}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657149819.59/warc/CC-MAIN-20200714083206-20200714113206-00568.warc.gz"} |
http://openstudy.com/updates/509ace0fe4b058d4974346ca | Here's the question you clicked on:
55 members online
• 0 viewing
## zerlinav 2 years ago What are the possible number of positive, negative, and complex zeros of f(x) = 3x4 - 5x3 - x2 - 8x + 4 Delete Cancel Submit
• This Question is Open
1. richyw
• 2 years ago
Best Response
You've already chosen the best response.
1
Ok well first of all it's a quadratic function, so the total will be 4. Use Descartes’ Rule of Signs. Your function is $p(x)=3x^4-5x^3-x^2-8x+4$ So count the number of sign changes. A "sign change" is when the sign the coefficient of the $$a^n$$ is different from the sign of the coefficient of $$a^{n-1}$$. So here you have one sign changes. You do the same thing for $$p(-x)$$ and also find one sign change (don't include the sign change for the 4! only the ones that multiply by x matter. Since there must be 4 total you know that it is one positive, one negative and two complex (the complex ones have to be an even number!)
2. richyw
• 2 years ago
Best Response
You've already chosen the best response.
1
sorry not quadratic. haha. 4th degree polynomial
3. richyw
• 2 years ago
Best Response
You've already chosen the best response.
1
4. cruffo
• 2 years ago
Best Response
You've already chosen the best response.
1
I thought there were two sign changes. $p(x)=3x^4-5x^3-x^2-8x+4 \rightarrow \color{red}{+}\color{red}{-}- \color{red}- \color{red}+$
5. cruffo
• 2 years ago
Best Response
You've already chosen the best response.
1
# of possible positive zeros is the number of sigh changes of the polynomial, or less than that number by an even integer. So the number of possible positive zeros is 2 or 0
6. richyw
• 2 years ago
Best Response
You've already chosen the best response.
1
there is only one sign change though. That last one doesn't matter.
7. cruffo
• 2 years ago
Best Response
You've already chosen the best response.
1
8. richyw
• 2 years ago
Best Response
You've already chosen the best response.
1
weird, I might be wrong. But I don't think so.
9. cruffo
• 2 years ago
Best Response
You've already chosen the best response.
1
Take a look at the graph: http://www.wolframalpha.com/input/?i=p%28x%29%3D3x^4-5x^3-x^2-8x%2B4+from+-3+to+3
10. richyw
• 2 years ago
Best Response
You've already chosen the best response.
1
oh yeah you are totally correct!
11. richyw
• 2 years ago
Best Response
You've already chosen the best response.
1
so in this case using the rule of signs all that you can know is one of four possible combinations then right? Also I have been doing this wrong for like 5 years haha. good thing we have calculators...
12. cruffo
• 2 years ago
Best Response
You've already chosen the best response.
1
: ) Sounds right: degree 4 means you will find 4 solutions to p(x) = 0 For this polynomial, just looking at the signs, we can tell that there will be 2 or 0 positive roots, 2 or 0 negative roots. For the complex roots - they always come in pairs. so... case 1: 2 + real zeros, 0 - real zeros, and 2 complex zeros case 2: 0 + real zeros, 2 - real zeros, and 2 complex zeros case 3: 2 + real zeros, 2 - real zeros, and 0 complex zeros case 4: 0 + real zeros, 0 - real zeros, and 4 complex zeros
13. Not the answer you are looking for?
Search for more explanations.
• Attachments:
Find more explanations on OpenStudy
##### spraguer (Moderator) 5→ View Detailed Profile
23
• Teamwork 19 Teammate
• Problem Solving 19 Hero
• You have blocked this person.
• ✔ You're a fan Checking fan status...
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": 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": 0.9913469552993774, "perplexity": 1602.0345454073752}, "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/1435375097861.54/warc/CC-MAIN-20150627031817-00141-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://book.caltech.edu/bookforum/printthread.php?s=5e1b992ee44709409311d4126ac49e55&t=4284 | LFD Book Forum (http://book.caltech.edu/bookforum/index.php)
- Homework 6 (http://book.caltech.edu/bookforum/forumdisplay.php?f=135)
- - What about residual analysis in linear regression? (http://book.caltech.edu/bookforum/showthread.php?t=4284)
jlaurentum 05-13-2013 08:07 PM
What about residual analysis in linear regression?
I've been kind of saving this question, but decided to ask at this point.
Why is there no mention of residual analysis in any of the linear regression topics the course has covered? How does residual analysis fit into the data learning picture (if it fits in at all)?
Specifically: starting with this week's topic of regularization, we've seen how weight decay softens the weights, but in doing so, chages them from the normal weights you'd obtain in linear regression. I would imagine that with weight decay, it would no longer hold that the mean of the errors (as in linear regression errors: ) is equal to zero, so the residuals would not be normally distributed with same variance and zero mean. In other words, with weight decay at least one of the Gauss-Markov assumptions do not hold?
Does that matter?
In general, are the standard tools of linear regression analysis we were taught in school (looking at the determination coefficient, hypothesis testing on the significance of the coefficients, and residual analysis to see if the assumptions that back up the previous elements hold) entirely pointless when you're doing machine learning?
yaser 05-13-2013 08:38 PM
Re: What about residual analysis in linear regression?
Quote:
Originally Posted by jlaurentum (Post 10815) I've been kind of saving this question, but decided to ask at this point. Why is there no mention of residual analysis in any of the linear regression topics the course has covered? How does residual analysis fit into the data learning picture (if it fits in at all)? Specifically: starting with this week's topic of regularization, we've seen how weight decay softens the weights, but in doing so, chages them from the normal weights you'd obtain in linear regression. I would imagine that with weight decay, it would no longer hold that the mean of the errors (as in linear regression errors: ) is equal to zero, so the residuals would not be normally distributed with same variance and zero mean. In other words, with weight decay at least one of the Gauss-Markov assumptions do not hold? Does that matter? In general, are the standard tools of linear regression analysis we were taught in school (looking at the determination coefficient, hypothesis testing on the significance of the coefficients, and residual analysis to see if the assumptions that back up the previous elements hold) entirely pointless when you're doing machine learning?
Residual analysis and other details of linear regression are worthy topics. They are regularly covered in statistics, but often not covered in machine learning. If you recall in Lecture 1, we alluded quickly to the contrast between statistics and machine learning (which do have a substantive overlap) in terms of mathematical assumptions and level of detailed analysis. Linear regression is a case in point for that contrast.
jlaurentum 05-14-2013 07:31 AM
Re: What about residual analysis in linear regression?
Thank you for the quick reply, Professor. I'll review lecture one more closely.
All times are GMT -7. The time now is 05:11 AM. | {"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": 0.8521609306335449, "perplexity": 942.0341169763801}, "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-2020-50/segments/1606141195417.37/warc/CC-MAIN-20201128095617-20201128125617-00469.warc.gz"} |
https://socratic.org/questions/what-is-the-slope-x-intercept-and-y-intercept-of-the-graph-of-3x-y-7 | Algebra
Topics
# What is the slope, x-intercept, and y-intercept of the graph of 3x+y=7?
Mar 5, 2018
The slope m = -3
The y intercept = 7
The x intercept = $\frac{7}{3}$
#### Explanation:
The y intercept form is
$y = m x + b$
Changing the equation
$3 x + y = 7$ to get the y intercept form add $- 3 x$ to both sides
$3 x + \left(- 3 x\right) + y = - 3 x + 7$ which gives the y intercept form
$y = - 3 x + 7$
$m =$ the slope ( think mountain ski slope)$m = - 3$
$b =$ the y intercept ( thing beginning) $b = 7 \mathmr{and} \left(0 , 7\right)$
The x intercept form is
$x = m y + b$
Solve the equation for x
$3 x + y = 7 \text{ }$ subtract y from both sides
$3 x + y - y = - y + 7 \text{ }$ This gives
$3 x = - y + 7 \text{ }$ divide both sides by 3
$\frac{3 x}{3} = \frac{- y}{3} + \frac{7}{3}$ The result is
$x = - \frac{1}{3} \times y + \frac{7}{3}$
The $x$-intercept is $\frac{7}{3} \mathmr{and} \left(\frac{7}{3} , 0\right)$
##### Impact of this question
4705 views around the world | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 18, "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": 0.8379439115524292, "perplexity": 2974.349937607392}, "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-2020-10/segments/1581875145657.46/warc/CC-MAIN-20200222085018-20200222115018-00160.warc.gz"} |
https://pos.sissa.it/350/001/ | Volume 350 - 7th Annual Conference on Large Hadron Collider Physics (LHCP2019) - Posters
Simple explanation of strong suppression offermionic EFT operators & custodial symmetry breaking
A.E. Guevara Escalante,* F. Alvarado, J.J. Sanz Cillero
*corresponding author
Full text: pdf
Pre-published on: 2019 September 03
Published on:
Abstract
The present approach relies on the SM chiral symmetry breaking pattern $SU(2)_L\otimes SU(2)_R\to SU(2)_{L+R}$, with the EW Goldstone bosons given in a non-linear realization and the Higgs boson described by an EW singlet field. In addition, we assume the presence of new physics heavy states around the TeV scale that do not couple to the SM fermions, only to the SM bosonic sector. However, the mixing between gauge bosons and BSM resonances induces a small indirect interaction between the BSM sector and the SM fermions. This leads to an important suppression of the fermionic operators in the low-energy EW effective theory (bilinear and four-fermion operators) in comparison with the purely bosonic ones. This naturally explains the strong experimental bounds on fermionic operators and why these resonances could not be yet detected: even if energies of the order of the TeV can be reached in present and future accelerators, their production from initial SM fermions yields a very small cross section because of this suppression mechanism. On the other hand, they can leave an imprint in SM boson measurements accessible to future experimental runs (e.g., the oblique $S$ and $T$ parameters). Finally, we compare our results with constraints from collider data.
Open Access | {"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": 0.9337505102157593, "perplexity": 1765.4520238229175}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986657586.16/warc/CC-MAIN-20191015055525-20191015083025-00502.warc.gz"} |
https://chemistry.stackexchange.com/questions/72268/experiment-using-hydroxides-of-alkaline-earth-metals | # Experiment using hydroxides of alkaline earth metals
In three different test tubes we add a very small amount of:
1)In the first test tube, we add oxide of magnesium
2)In the second test tube, we add hydroxide of calcium
3)In the third test tube, we add hydroxide of barium
Then we add deionized water in the three test tubes. We shake them and after we filtrate them, we measure the pH of each filtrate. I'm curious why we use hydroxides of calcium and barium but we use oxide of magnesium instead of hydroxide of magnesium?
I would attribute this two 2 reasons:
Reason 1: Insolubility of Magnesium hydroxide.
Magnesium hydroxide is almost insoluble in water and will not easily dissolve in water.
Reason 2: Safety is one of the reason of the choice of compounds.
The formation of hydroxides of alkaline earth metals are generally exothermic and this intensity increases down this group.
1. Reaction of magnesium oxide with water will produce the corresponding hydroxide:
$$\ce{MgO + H2O → Mg(OH)2 (very slight reaction)}$$
(The reaction quickly stops because the magnesium hydroxide formed is almost insoluble in water and forms a barrier on the magnesium preventing further reaction).
2. The reaction of calcium oxide to produce its hydroxide is exothermic:
$$\ce{CaO (s) + H2O (l) ⇌ Ca(OH)2 (aq) (ΔH_r = −63.7 kJ/mol of CaO)}$$
1. Barium oxide is harmful to human skin and may cause irritation on exposure for prolonged periods or in large quantity.
• Its reaction with water is also know to be violent.
So in my opinion it is much better to user hydroxides of Ba and Ca. | {"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": 0.8055558800697327, "perplexity": 4267.703854796195}, "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-2020-10/segments/1581875143805.13/warc/CC-MAIN-20200218180919-20200218210919-00087.warc.gz"} |
https://arxiv.org/abs/1808.03742 | physics.plasm-ph
# Title:Efficient Fourier Basis Particle Simulation
Abstract: The standard particle-in-cell algorithm suffers from grid heating. There exists a gridless alternative which bypasses the deposition step and calculates each Fourier mode of the charge density directly from the particle positions. We show that a gridless method can be computed efficiently through the use of an Unequally Spaced Fast Fourier Transform (USFFT) algorithm. After a spectral field solve, the forces on the particles are calculated via the inverse USFFT (a rapid solution of an approximate linear system). We provide one and two dimensional implementations of this algorithm with an asymptotic runtime of $O(N_p + N_m^D \log N_m^D)$ for each iteration, identical to the standard PIC algorithm (where $N_p$ is the number of particles and $N_m$ is the number of Fourier modes, and $D$ is the spatial dimensionality of the problem) We demonstrate superior energy conservation and reduced noise, as well as convergence of the energy conservation at small time steps.
Comments: 17 pages, 12 figures Subjects: Plasma Physics (physics.plasm-ph); Computational Physics (physics.comp-ph) Cite as: arXiv:1808.03742 [physics.plasm-ph] (or arXiv:1808.03742v3 [physics.plasm-ph] for this version)
## Submission history
From: Matthew Mitchell [view email]
[v1] Sat, 11 Aug 2018 02:35:02 UTC (276 KB)
[v2] Thu, 18 Apr 2019 22:25:30 UTC (307 KB)
[v3] Wed, 26 Jun 2019 01:22:22 UTC (298 KB) | {"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": 0.8437378406524658, "perplexity": 1717.1345641968342}, "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/1566027317817.76/warc/CC-MAIN-20190823020039-20190823042039-00101.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-5-test-page-709/13 | ## Precalculus (6th Edition) Blitzer
$\{\frac{\pi}{2}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} \}$
Step 1. Rewrite the equation as $2sin(x)cos(x)+cos(x)=0$ or $cos(x)(2sin(x)+1)=0$ Step 2. For $cos(x)=0$, we can find $x=\frac{\pi}{2}, \frac{3\pi}{2}$ Step 3. For $sin(x)=-\frac{1}{2}$, we can find $x=\frac{7\pi}{6}, \frac{11\pi}{6}$ Step 4. The solution set is $\{\frac{\pi}{2}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} \}$ | {"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": 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": 0.9734869003295898, "perplexity": 187.35461186352146}, "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/1679296948609.41/warc/CC-MAIN-20230327060940-20230327090940-00347.warc.gz"} |
http://mathhelpforum.com/calculus/153064-maximum-value.html | 1. ## Maximum Value
Find the absolute maximum value of y = 8sin(pix/8).
I understand maximum value but this particular function is giving me some trouble.
2. Originally Posted by bobsanchez
Find the absolute maximum value of y = 8sin(pix/8).
I understand maximum value but this particular function is giving me some trouble.
There is obviously something missing from the question as asked since the maximum attained by sin is 1 the maximum value of y is 8. But I don't think that is what you want to know.
CB
3. That's the problem as written.
4. Then you should have immediately thought "I know that sin(x) has a maximum value of 1 so 8 times sine must have a maximum value of 8". End of problem!
5. Agreed. I need to get better at looking at the simplicity of problems rather than the complexity of them. | {"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": 0.8030037879943848, "perplexity": 380.5327049859213}, "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-2016-40/segments/1474738660350.13/warc/CC-MAIN-20160924173740-00108-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/physics-help.121366/ | # Physics Help
1. May 19, 2006
### Taryn
The main muscle responsible for raising an arm is the deltoid. The deltoid muscle connects at the upper end of the shoulder, extends over the upper arm bone (humerus), and attaches near the elbow. Effectively there are three forces involved in raising the arm: (i) the force of the deltoid muscle, Fm, acting at an angle of approximately 13.1o with respect to the negative x axis, (ii) the force of gravity, Fg, acting effectively at the centre of mass of the arm located close to the elbow, and (iii) the force of the shoulder socket, Fs, acting on the humerus effectively along the x axis. Assuming the mass of the arm is 1.92 kg, calculate the magnitude of Fm in Newton.
#### Attached Files:
• ###### arm.gif
File size:
1.5 KB
Views:
43
2. May 19, 2006
### Taryn
hey this is just a problem I need desperate help with, for some reason I continually get it wrong.... plz help!
3. May 19, 2006
### Taryn
Its okay now.. just figured it out, so simple lol! | {"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": 0.8440931439399719, "perplexity": 2896.904432012651}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257645604.18/warc/CC-MAIN-20180318091225-20180318111225-00469.warc.gz"} |
https://www.physicsforums.com/threads/f-2-f-but-f-1-and-f-0.681583/ | # F^2=f but f=/=1 and f=/=0?
1. Mar 28, 2013
### robertjordan
1. The problem statement, all variables and given/known data
Show there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$ s.t. $f^2=f$ but $f\neq{0,1}$.
2. Relevant equations
Here $f^2=f$ means for arbitrary $a\in{\mathbb{R}}, f(a)^2=f(a)$
3. The attempt at a solution
I came up with the function $f(a)= \begin{cases} 0, & \text{if }a\text{> 0 } \\ 1, & \text{if }a \leq 0 \end{cases}$
What do you guys think? Is this right? I figured the only real numbers r for which r^2=r are r=0 and r=1 so the function f will have to only spit out those values or else there would be some input a for which f(a)^2=/=f(a)
Last edited: Mar 28, 2013
2. Mar 28, 2013
### Staff: Mentor
Make that:
$f(x)= \begin{cases} 0, & \text{if }x > 0 \\ 1, & \text{if }x \leq 0 \end{cases}$
What you have works, but I don't think it's what the write of the problem had in mind. Instead, I think they had a kind of identity function in mind - one whose output is the same as its input.
3. Mar 28, 2013
### robertjordan
But we need $f(a)^2=f(a)$ for all real numbers a. If we make $f(a)=a$, then in order for $f(a)^2=f(a)$, we would need $a^2=a$ which is clearly not true in general...
Can you elaborate some more on what you mean? I think I missed it...
4. Mar 28, 2013
### Stimpon
Assuming that robertjordan hasn't completely misunderstood the question, they definitely aren't looking for an identity function.
Normally $f^{2}=f \circ f$, so I think Mark44 thought that that's what the question writer meant by $f^{2}$. Certainly if the question writer did mean $f \circ f$ then a (the) real identity function would be correct, but you clarified that the writer meant $f \cdot f$ so Mark44 is wrong.
5. Mar 28, 2013
### Staff: Mentor
The notation used here is confusing to me, and appears to be in contradiction to itself.
f2 as used above normally indicates function composition, as in f(f(x)).
To my mind, the notation used immediately above contradicts the meaning at the top of this page. Even if f2 denotes multiplication, it should be written as [f(a)]2 to be clear.
6. Mar 28, 2013
### Stimpon
As an aside, if $f^{2}$ was being used to mean $f \circ f$, then the function given wouldn't work because we would have $f^{2}(x)=f(0)=1$ for $x>0$.
ETA: As well as $f^{2}(x)=f(1)=0$ for $x \leq 0$.
Last edited: Mar 28, 2013
Draft saved Draft deleted
Similar Discussions: F^2=f but f=/=1 and f=/=0? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9400127530097961, "perplexity": 1020.5024023685218}, "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-2017-34/segments/1502886102663.36/warc/CC-MAIN-20170816212248-20170816232248-00107.warc.gz"} |
https://chem.libretexts.org/Courses/Los_Angeles_Trade_Technical_College/Foundations_of_Introductory_Chemistry-1/1.03%3A_Measurements | # 2: Measurements
In 1983, an Air Canada airplane had to make an emergency landing because it unexpectedly ran out of fuel; ground personnel had filled the fuel tanks with a certain number of pounds of fuel, not kilograms of fuel. In 1999, the Mars Climate Orbiter spacecraft was lost attempting to orbit Mars because the thrusters were programmed in terms of English units, even though the engineers built the spacecraft using metric units. In 1993, a nurse mistakenly administered 23 units of morphine to a patient rather than the “2–3” units prescribed (the patient ultimately survived). These incidents occurred because people weren’t paying attention to quantities.
Chemistry, like all sciences, is quantitative. It deals with quantities, things that have amounts and units. Dealing with quantities is very important in chemistry, as is relating quantities to each other. In this chapter, we will discuss how we deal with numbers and units, including how they are combined and manipulated.
• 2.1 Expressing Numbers - Scientific Notation
Scientific notation is a system for expressing very large or very small numbers in a compact manner. It uses the idea that such numbers can be rewritten as a simple number multiplied by 10 raised to a certain exponent, or power. Scientific notation expressed numbers using powers of 10.
• 2.2 Expressing Numbers - Significant Figures
Significant figures properly report the number of measured and estimated digits in a measurement. There are rules for applying significant figures in calculations.
• 2.3 The International System of Units
Recognize the SI base units. Combining prefixes with base units creates new units of larger or smaller sizes.
• 2.4 Converting Units
The ability to convert from one unit to another is an important skill. A unit can be converted to another unit of the same type with a conversion factor.
• 2.5 Other Units - Temperature and Density
Chemistry uses the Celsius and Kelvin scales to express temperatures. A temperature on the Kelvin scale is the Celsius temperature plus 273.15. The minimum possible temperature is absolute zero and is assigned 0 K on the Kelvin scale. Density relates a substance’s mass and volume. Density can be used to calculate volume from a given mass or mass from a given volume. | {"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": 0.8211228847503662, "perplexity": 1336.262532700977}, "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-2021-25/segments/1623487633444.37/warc/CC-MAIN-20210617192319-20210617222319-00061.warc.gz"} |
https://kerodon.net/tag/020U | # Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
Proposition 5.5.6.2. The simplicial set $\operatorname{\mathcal{QC}}_{\ast }$ is an $\infty$-category, and the projection map $\operatorname{\mathcal{QC}}_{\ast } \rightarrow \operatorname{\mathcal{QC}}$ is a left fibration of $\infty$-categories.
Proof. By virtue of Proposition 5.5.4.3, the simplicial set $\operatorname{\mathcal{QC}}$ is an $\infty$-category. It follows that for every object $\operatorname{\mathcal{C}}\in \operatorname{\mathcal{QC}}$, the projection map $\operatorname{\mathcal{QC}}_{\operatorname{\mathcal{C}}/} \rightarrow \operatorname{\mathcal{QC}}$ is a left fibration (Corollary 4.3.6.11). Taking $\operatorname{\mathcal{C}}= \Delta ^{0}$, we conclude that the projection map $\operatorname{\mathcal{QC}}_{\ast } \rightarrow \operatorname{\mathcal{QC}}$ is a left fibration, so that $\operatorname{\mathcal{QC}}_{\ast }$ is an $\infty$-category (Remark 4.2.1.4). $\square$ | {"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": 0.9783269762992859, "perplexity": 180.2069700853913}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662577757.82/warc/CC-MAIN-20220524233716-20220525023716-00128.warc.gz"} |
http://tex.stackexchange.com/questions/38423/want-to-fill-line-with-repeating-string/38424 | # Want to fill line with repeating string
Similar to the \dotfill command, but with your own text, rather than dots.
If the string was: kitty!
The entire line would look like the following: kitty!kitty!kitty!kitty!kitty!kitty!kitty!kitty!kitty!
Is there any way to do this?
-
Have a look at the file latex.ltx. THere you will find the definition of dotfill and hruefil – Marco Daniel Dec 15 '11 at 20:40
## 1 Answer
One way of doing this is using leaders:
\documentclass{article}
\newcommand\kitty{\leavevmode\xleaders\hbox{kitty!}\hfill\kern0pt}
\begin{document}
This is a test. \kitty \par
This is a test.
\noindent \kitty
\end{document}
The general format for constructing leaders is \leaders<box or rule><glue> (which repeats <box or rule>). There are three kinds of leaders that you can use: \leaders, \cleaders and \xleaders. Here is an informal description of each, taken from The Advanced TeXBook:
When \leaders is used, TeX first locates the innermost box A containing the \leader command. It then fills up A, from the left, with copies of the leader. There may be some space left on the right. ...
The \cleaders command centers the leaders in the leader window, regardless of the size of the enclosing box A. There is normally some space left on both sides of the window. The \xleaders is still different. It distributes the window space evenly between the individual copies of the leader.
Here, and just for fun, the difference between using \kitty with \leaders, \cleaders and \xleaders:
\documentclass{article}
\usepackage{showframe}% http://showframe
\begin{document}
\noindent \leaders\hbox{kitty!}\hfill\kern0pt \par
\noindent \cleaders\hbox{kitty!}\hfill\kern0pt \par
\noindent \xleaders\hbox{kitty!}\hfill\kern0pt
\end{document}
Also see the TeXBook for reference (chapter 21 Making Boxes, p 223):
The dots you see before your eyes here . . . . . . . . . . . . are called "leaders" because they lead your eyes across the page; such things are often used in indexes or tables of contents. The general idea is to repeat a box as many times as necessary to fill up some given space. TeX treats leaders as a special case of glue; no, wait, it's the other way around: TeX treats glue as a special case of leaders. Ordinary glue fills space with nothing, while leaders fill space with any desired thing. In horizontal mode you can say
\leaders<box or rule>\hskip<glue>
and the effect will be the same as if you had said just \hskip<glue>, except that the space will be occupied by copies of the specified <box or rule>. The glue stretches or shrinks in the usual way.
-
Exactly what I was looking for, thanks! – Walmart_Hobo Dec 15 '11 at 21:35 | {"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": 0.9320231080055237, "perplexity": 1733.8722207443109}, "config": {"markdown_headings": true, "markdown_code": false, "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-2014-23/segments/1406510272680.42/warc/CC-MAIN-20140728011752-00123-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/189462/consistent-estimate-using-likelihood-function | # Consistent estimate using likelihood function
Let be $X$ a population with normal distribution. Using likelihood function I get the below expression $\hat{\sigma_X}^2 = \sum_{i=1}^{n}{\dfrac{(X_i-\mu)^2}{n}}$ for variance.
I want prove that expression is consistent, i.e. $E[\hat{\sigma_X}^2]=\sigma_X^2$.
I begin ...
$E[\hat{\sigma_X}^2]=\dfrac{1}{n}E[\sum_{i=1}^{n}{(X_i-\mu)^2}]$
$\dfrac{1}{n}(E[(X_1-\mu)^2]+E[(X_2-\mu)^2]\cdots E[(X_n-\mu)^2])$
I don't know what else to do.
pdta: $\mu=\dfrac{1}{n}\sum_{i=0}^{n}{X_i}$
-
Do you intend the parameter $\mu$ to be a fixed constant, of a random variable, function of $X_i$? – Sasha Aug 31 '12 at 23:31
yes, $\mu=\dfrac{1}{n}\sum_{i=0}^{n}{X_i}$ – juaninf Aug 31 '12 at 23:58
In that case, it is a well known result that $\mathsf{E}\left(\hat{\sigma}_X^2\right) = \frac{n-1}{n} \sigma_X^2$. – Sasha Sep 1 '12 at 0:02
but, in mi lecture would have to be $E[\hat{\sigma_X}^2] = \sigma_X^2$ and not $\mathsf{E}\left(\hat{\sigma}_X^2\right) = \frac{n-1}{n} \sigma_X^2$...(for the consistent)... then this estimative is not consistent? – juaninf Sep 1 '12 at 0:08
The proper term is $\hat{\sigma}_X^2$ is a biased estimator. Consistency has to do with large $n$ limit. $\hat{\sigma}_X^2$ is consistent, since the expectation approaches the population value for large $n$. – Sasha Sep 1 '12 at 0:10 | {"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": 0.9411673545837402, "perplexity": 1587.4052415462513}, "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-2015-06/segments/1422122152935.2/warc/CC-MAIN-20150124175552-00120-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://planetmath.org/TeichmullerCharacter | # Teichmüller character
Before we define the Teichmüller character, we begin with a corollary of Hensel’s lemma.
###### Corollary.
Let $p$ be a prime number. The ring of $p$-adic integers (http://planetmath.org/PAdicIntegers) $\mathbb{Z}_{p}$ contains exactly $p-1$ distinct $(p-1)$th roots of unity. Furthermore, every $(p-1)$th root of unity is distinct modulo $p$.
###### Proof.
Notice that $\mathbb{Q}_{p}$, the $p$-adic rationals, is a field. Therefore $f(x)=x^{p-1}-1$ has at most $p-1$ roots in $\mathbb{Q}_{p}$ (see this entry (http://planetmath.org/APolynomialOfDegreeNOverAFieldHasAtMostNRoots)). Moreover, if we let $a\in\mathbb{Z}$ with $1\leq a\leq p-1$ then $f(a)=a^{p-1}-1\equiv 0\mod p$ by Fermat’s little theorem. Since $f^{\prime}(a)=(p-1)\cdot a^{p-2}$ is non-zero modulo $p$, the trivial case of Hensel’s lemma implies that there exist a root of $x^{p-1}-1$ in $\mathbb{Z}_{p}$ which is congruent to $a$ modulo $p$. Hence, there are at least $p-1$ roots in $\mathbb{Z}_{p}$, and we can conclude that there are exactly $p-1$ roots. ∎
###### Definition.
The Teichmüller character is a homomorphism of multiplicative groups:
$\omega\colon\mathbb{F}_{p}^{\times}\to\mathbb{Z}_{p}^{\times}$
such that $\omega(a)$ is the unique $(p-1)$th root of unity in $\mathbb{Z}_{p}$ which is congruent to $a$ modulo $p$ (which exists by the corollary above). The map $\omega$ is sometimes called the Teichmüller lift of $\mathbb{F}_{p}$ to $\mathbb{Z}_{p}$ ($0\mod p$ would lift to $0\in\mathbb{Z}_{p}$).
###### Remark.
Some authors define the Teichmüller character to be the homomorphism:
$\hat{\omega}\colon\mathbb{Z}_{p}^{\times}\to\mathbb{Z}_{p}^{\times}$
defined by
$\hat{\omega}(z)=\lim_{n\to\infty}z^{p^{n}}.$
Notice that for any $z\in\mathbb{Z}_{p}^{\times}$, $\hat{\omega}(z)$ is a $(p-1)$th root of unity:
$(\hat{\omega}(z))^{p}=\left(\lim_{n\to\infty}z^{p^{n}}\right)^{p}=\lim_{n\to% \infty}z^{p^{n+1}}=\hat{\omega}(z).$
Thus, the value $\hat{\omega}(z)$ is the same than $\omega(z\mod p)$.
Title Teichmüller character Canonical name TeichmullerCharacter Date of creation 2013-03-22 15:09:04 Last modified on 2013-03-22 15:09:04 Owner alozano (2414) Last modified by alozano (2414) Numerical id 7 Author alozano (2414) Entry type Definition Classification msc 13H99 Classification msc 11S99 Classification msc 12J99 Synonym Teichmuler character Synonym Teichmuller lift Synonym Teichmüller lift Related topic PAdicIntegers | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 43, "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": 0.9700245261192322, "perplexity": 427.09420782272616}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202526.24/warc/CC-MAIN-20190321152638-20190321174638-00256.warc.gz"} |
http://math.stackexchange.com/questions/38544/finding-the-population-size-from-the-common-elements-of-multiple-samples | # Finding the population size from the common elements of multiple samples
A coworker of mine has taken a certification exam twice and has seen the same ten questions on both exams. I am curious if it is possible to determine the total number of questions in the question pool from the number of questions per exam and the number questions common to both exams.
The exam has fifty-one questions, and in taking it twice for a total of one hundred and two questions, it was determined that ten of the questions appeared on both exams. Also, I am curious if this is even enough information to determine the answer.
Mathematics has never been my strong point.
Thank you!
-
I accepted, then rolled back, a suggested edit, but can't comment to the suggester. I think the suggestion should go on a separate question. Sorry for any confusion. – Ross Millikan May 13 '11 at 23:10
You could argue that 10/51 of the second set of questions had been in the first set, and so assume that with a random selection of questions from a fixed population, a reasonable central estimate is that the same fraction of the whole population was in the first set, i.e. $\hat{n}_\text{pop}\times \frac{10}{51} = 51$, giving an estimate of the whole population size of 260.1. Obviously it will not be exactly that (it is not an integer), but it is broadly indicative of the actual figure if the assumptions are correct. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.821115255355835, "perplexity": 389.6528085694079}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645339704.86/warc/CC-MAIN-20150827031539-00333-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://images.planetmath.org/vectorspacesareisomorphicifftheirbasesareequipollent | # vector spaces are isomorphic iff their bases are equipollent
###### Theorem 1.
Vector spaces $V$ and $W$ are isomorphic iff their bases are equipollent (have the same cardinality).
###### Proof.
($\Longrightarrow$) Let $\phi:V\to W$ be a linear isomorphism. Let $A$ and $B$ be bases for $V$ and $W$ respectively. The set
$\phi(A):=\{\phi(a)\mid a\in A\}$
is a basis for $W$. If
$r_{1}\phi(a_{1})+\cdots+r_{n}\phi(a_{n})=0,$
with $a_{i}\in A$. Then
$\phi(r_{1}a_{1}+\cdots+r_{n}a_{n})=0$
since $\phi$ is linear. Furthermore, since $\phi$ is one-to-one, we have
$r_{1}a_{1}+\cdots+r_{n}a_{n}=0,$
hence $r_{i}=0$ for $i=1,\ldots,n$, since $A$ is linearly independent. This shows that $\phi(A)$ is linearly independent. Next, pick any $w\in W$, then there is $v\in V$ such that $\phi(v)=w$ since $\phi$ is onto. Since $A$ spans $V$, we can write
$v=r_{1}a_{1}+\cdots+r_{n}a_{n},$
so that
$w=\phi(v)=r_{1}\phi(a_{1})+\cdots+r_{n}\phi(a_{n}).$
This shows that $\phi(A)$ spans $W$. As a result, $\phi(A)$ is a basis for $W$. $A$ and $\phi(A)$ are equipollent because $\phi$ is one-to-one. But since $B$ is also a basis for $W$, $\phi(A)$ and $B$ are equipollent. Therefore
$|A|=|\phi(A)|=|B|.$
($\Longleftarrow$) Conversely, suppose $A$ is a basis for $V$, $B$ is a basis for $W$, and $|A|=|B|$. Let $f$ be a bijection from $A$ to $B$. We extend the domain of $f$ to all of $A$, and call this extension $\phi$, as follows: $\phi(a)=f(a)$ for any $a\in A$. For $v\in V$, write
$v=r_{1}a_{1}+\cdots+r_{n}a_{n}$
with $a_{i}\in A$, set
$\phi(v)=r_{1}\phi(a_{1})+\cdots+r_{n}\phi(a_{n}).$
$\phi$ is a well-defined function since the expression of $v$ as a linear combination of elements of $A$ is unique. It is a routine verification to check that $\phi$ is indeed a linear transformation. To see that $\phi$ is one-to-one, let $\phi(v)=0$. But this means that $v=0$, again by the uniqueness of expression of $0$ as a linear combination of elements of $A$. If $w\in W$, write it as a linear combination of elements of $B$:
$w=s_{1}b_{1}+\cdots+s_{m}b_{m}.$
Each $b_{i}\in B$ is the image of some $a\in A$ via $f$. For simplicity, let $f(a_{i})=b_{i}$. Then
$w=s_{1}f(a_{1})+\cdots+s_{m}f(a_{m})=s_{1}\phi(a_{1})+\cdots+s_{m}\phi(a_{m})=% \phi(s_{1}a_{1}+\cdots+s_{m}a_{m}),$
which shows that $\phi$ is onto. Hence $\phi$ is a linear isomorphism between $V$ and $W$. ∎
Title vector spaces are isomorphic iff their bases are equipollent VectorSpacesAreIsomorphicIffTheirBasesAreEquipollent 2013-03-22 18:06:55 2013-03-22 18:06:55 CWoo (3771) CWoo (3771) 7 CWoo (3771) Result msc 13C05 msc 15A03 msc 16D40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 79, "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": 0.9967835545539856, "perplexity": 69.109805492014}, "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/1529267864546.30/warc/CC-MAIN-20180622143142-20180622163142-00284.warc.gz"} |
https://brilliant.org/practice/complex-numbers-argand-plane/ | Algebra
# Complex Numbers - Argand Plane
Suppose that we have two complex numbers $$z_1 = 3 - 2i$$ and $$z_2 = 4 - 3i$$.
In what quadrant is this complex number $$z_1z_2$$ located?
What is the shape of graph of the equation $$\left|z\right| = 4$$ in the complex plane?
In $$\text{units}^2$$, what is the area of the rectangle formed by the vertices $$2 + 3i, 4 + 2i, -i$$, and $$2 - 2i$$?
Suppose the function $$f(z)$$ takes the complex number $$z$$ and rotates it 90 degrees counterclockwise around the origin.
Which function represents this transformation?
For a complex number $$z$$, on which part of the Argand plane does $$\text{Re}(z) + \text{Im}(iz)$$ sit?
×
Problem Loading...
Note Loading...
Set Loading... | {"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": 0.8232409358024597, "perplexity": 477.0736590316037}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676594886.67/warc/CC-MAIN-20180723032237-20180723052237-00221.warc.gz"} |
https://www.physicsforums.com/threads/one-dimensional-motion-with-constant-acceleration.336871/ | # One-dimensional motion with constant acceleration
1. Sep 13, 2009
### tja2468
1. The problem statement, all variables and given/known data
A falling object travels one-fourth of its total distance in the last second of its fall. From what height was it dropped?
2. Relevant equations
equations of motion for constant acceleration
2. Sep 13, 2009
### Hootenanny
Staff Emeritus
You seem to have omitted a vital section,
Similar Discussions: One-dimensional motion with constant acceleration | {"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": 0.8464906811714172, "perplexity": 3152.621043389278}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934807044.45/warc/CC-MAIN-20171123233821-20171124013821-00779.warc.gz"} |
http://mathhelpforum.com/calculus/85490-trig-integral-print.html | trig integral
• Apr 24th 2009, 04:05 PM
alex83
trig integral
integral
dx/[(sin x)(cos x)]
how do i approach such an example
• Apr 24th 2009, 04:08 PM
Jester
Quote:
Originally Posted by alex83
integral
dx/[(sin x)(cos x)]
how do i approach such an example
$\frac{1}{\sin x \cos x} = \frac{2}{\sin 2x} = 2 \csc 2x$ and
$2 \int \csc 2x\, dx$ is a standard integral (if $u = 2x$).
• Apr 24th 2009, 04:14 PM
Jameson
alex83 - Once again I'm going to ask that you show some effort before making another thread on trig integral problems. It seems you are going down the list of your homework. We want you to learn how to do these on your own and it doesn't seem you know how to.
• Apr 24th 2009, 04:14 PM
mr fantastic
Quote:
Originally Posted by alex83
integral
dx/[(sin x)(cos x)]
how do i approach such an example
sin x cos x can be re-written as 0.5 sin (2x). So your job is to integrate a cosec function. See here: http://www.mathhelpforum.com/math-he...barrassed.html
You will not get any further help on this question until you show your working and state where you get stuck. You may not think so, but this is actually for your own academic benefit. Maths is not a spectator sport.
• Apr 24th 2009, 04:31 PM
alex83
Quote:
Originally Posted by Jameson
alex83 - Once again I'm going to ask that you show some effort before making another thread on trig integral problems. It seems you are going down the list of your homework. We want you to learn how to do these on your own and it doesn't seem you know how to.
actually im studying for my finals and i have my solution manual with me but when the book shows an answer differnet than mine and i dont understand the way they got to a certain step i seek help from this helpful forum which i learned from more than my class..
thank you anyway but pls. dont get me wrong
• Apr 24th 2009, 04:35 PM
mr fantastic
Quote:
Originally Posted by alex83
actually im studying for my finals and i have my solution manual with me but when the book shows an answer differnet than mine and i dont understand the way they got to a certain step i seek help from this helpful forum which i learned from more than my class..
thank you anyway but pls. dont get me wrong
If that happens you should post the solution you've been given and say what part of it you don't understand. | {"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": 0.8781496286392212, "perplexity": 857.6007938952281}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948542031.37/warc/CC-MAIN-20171214074533-20171214094533-00011.warc.gz"} |
http://nrich.maths.org/6762/solution | ### Strange Numbers
All strange numbers are prime. Every one digit prime number is strange and a number of two or more digits is strange if and only if so are the two numbers obtained from it by omitting either its first or its last digit. Find all strange numbers.
### Stars
Can you find a relationship between the number of dots on the circle and the number of steps that will ensure that all points are hit?
### Factor Trio
Weekly Problem 45 - 2013
Which of the numbers shown is the product of exactly 3 distinct prime factors?
# Three Primes
##### Stage: 3 Short Challenge Level:
Let $p$, $q$ and $r$ be three prime numbers such that $pqr=5(p+q+r)$. Then one of the prime numbers must be $5$, say $r$.
This implies that $5pq=5(p+q+5)\Rightarrow pq=p+q+5\Rightarrow pq-p-q+1=6\Rightarrow (p-1)(q-1)=6$.
Therefore either $p-1=1$ and so $q-1=6$ i.e. $(p,q)=(2,7)$ (or vice versa) or $p-1=2$ and so $q-1=3$ i.e. $(p,q)=(3,4)$ (or vice versa). But $4$ is not prime, so the only triple of primes which satisfies the condition is $(2,5,7)$.
This problem is taken from the UKMT Mathematical Challenges.
View the archive of all weekly problems grouped by curriculum topic
View the previous week's solution
View the current weekly problem | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8500052690505981, "perplexity": 416.1788180419037}, "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-2015-40/segments/1443738008122.86/warc/CC-MAIN-20151001222008-00142-ip-10-137-6-227.ec2.internal.warc.gz"} |
http://fermatslasttheorem.blogspot.com/2006/02/eulers-formula.html | ## Friday, February 24, 2006
### Euler's Formula
Today's proof for Euler's Formula is based on the Taylor's Series. Euler's Formula is the equation:
eix = cosx + isinx
In a previous blog, I spoke about Euler's Identity which is derived from Euler's Formula. Richard Cotes was the first person to provide a proof but the great popularizer of this result was Leonhard Euler. Euler's Identity is used in the construction of cyclotomic integers which are used in Kummer's proof of Fermat's Last Theorem for regular primes.
Lemma 1: Maclaurin's Series
f(x) = f(0) + (x/1!)f'(0) + (x
2/2!)f''(0) + (x3/3!)f'''(0) + ... + (xn/n!)fn(0)
(1) Taylor's series gives us:
f(x) = f(a) + f'(a)(x-a) + [f''(a)/2!](x-a)2 + .... + [f(n)(a)/n!](x-a)n + ....
[See here for its proof]
(2) Now, if a=0, then we have:
f(x) = f(0) + (x/1!)f'(0) + (x2/2!)f''(0) + (x3/3!)f'''(0) + ... + (xn/n!)fn(0).
QED
Lemma 2: ex = 1 + (x/1) + (x2/2!) + (x3/3!) + ....
(1) Let f(x) = ex
(2) From the properties of ex [See here for details]:
f(x) = ex → f(0) = e0 = 1
f'(x) = ex → f'(0) =e0 = 1
fn(x) = ex → fn(0) = e0 = 1
(3) We know that ex is continuous since it has a derivative at each point. [See here for details of why this is true]
(4) By Lemma 1 above, we have:
ex = 1 + (x/1!)(1) + (x2/2!)(1) + ... (xn/n!)(1)
QED
Lemma 3: sinx = x - (x3/3!) + (x5/5!) - (x7/7!) + ...
(1) Let f(x) = sin x
(2) From the properties of sin, we know:
f(0) = sin(0) = 0 [See here for details if needed]
f'(x) = cos x → f'(0) = 1 [See here for proof if needed]
f''(x) = -sin(x) → f'(0) = 0 [See here for proof if needed]
f'''(x) = -cos x → f'(0) = -1
(3) From this, we see that:
fn(0) = 0 if n is even.
fn(0) = 1 if (n-1)/2 is even
fn(0) = -1 if (n-1)/2 is odd
(4) Putting this all together gives us:
sin x = (x/1)(1) + (x2/2!)(0) + (x3/3!)(-1) + ...
QED
Lemma 4: cos x = 1 - (x2)/2! + (x4/4!) - (x6/6!) + ...
(1) Let f(x) = cos x
(2) From the properties of cos, we know:
f(0) = cos(0) = 1 [Details if needed are found here]
f'(x) = -sin x → f'(0) = 0 [Details if needed are found here]
f''(x) = -cos(x) → f'(0) = -1
f'''(x) = sin x → f'(0) = 0
(3) From this, we see that:
fn(0) = 0 if n is odd.
fn(0) = 1 if (n/2) is even
fn(0) = -1 if (n/2) is odd
(4) Putting this all together gives us:
cos x = 1 + (x2/2!)(-1) + (x3/3!)(0) + (x4/4!)(1) + ...
QED
Theorem: Euler's Formula
e
ix = cos x + isin x
(1) From Lemma 2, we have:
eix = 1 + ix + (ix)2/2! + (ix)3/3! + ...
(2) Since i2 = -1 and i4 = 1, this gives us: (for details on i, see here)
eix = (1 - x2/2! + x4/4! + ...) + i(x - x3/3! + x5/5! + ...)
(3) From Lemma 4 above, we see that:
cos(x) = (1 - x2/2! + x4/4! + ...)
(4) From Lemma 3 above, we see that:
isin(x) = i(x - x3/3! + x5/5! + ...)
(5) Combining step #2 with step #3 and step #4 gives us:
eix = cos x + i sin x.
QED
Corollary: De Moivre's Formula
(cos x + isin x)n = cos(nx) + isin(nx)
Proof:
(1) (eix)n = einx
(2) (cos x + i sin x)n = cos(nx) + isin(nx) [Applying Euler's formula above]
QED
professiona said...
This formula can be interpreted as saying that the function eix traces out the unit circle in the complex number plane as x ranges through the real numbers. Here, x is the angle that a line connecting the origin with a point on the unit circle makes with the positive real axis, measured counter clockwise and in radians The formula is valid only if sin and cos take their arguments in radians rather than in degrees.
source: Wikipedia
For x=pi, the equation is
e^ipi = -1 or e^ipi + 1 = 0 or
e^ipi = 0 - 1 ; ln [e^ipi] = ln [0/1]; hence, ipi = ln [0] = 0 ;
e^0 = 1 ; 1 + 1 = 0 (mod 2)= 2
See 'Proof' by David Auburn(p.73-4)
"Let X equal the quantity of all quantities of X. Let X equal the cold ... months [11, 12, 1, 2] ... and four of heat [5, 6, 7, 8] leaving four months of in-/determine temperature [2,3,9,10].
[The months are a unit circle, and the Euler equation is its design. The indetermined area is undefined as is the definition of infinite.]
"Let X equal the month of full bookstores [infinite or undefined as month 10]. The number of books approaches infinity as the number of months of cold approaches four."
The 'proof' is also a statement of:
cos x = e^ix + e^[-ix] (over 2)and
sin x = e^ix - e^[-ix] (over 2i)
solving for both cos x and sin x
cos pi = e^ipi + 1/1 = 2 =
0 (mod 2) = -1
sin pi = 1 - 1 = 0
e^ipi = cos pi + isin pi = -1 + 0
sin pi - cos pi = 2 = 0 (mod 2)
The 'proof' is a restatement of Fermat's Last Theorem for modular formulation of infinite number of primes, in the transformation of finite to infinite, using a modular clock function:
a unit circle in the complex number plane. "Let X equal the number ...."
Mahndisa S. Rigmaiden said...
12 04 06
What a great blog you have here!!! I did a similar derivation of the Inverse tangent function using these same principles. But the explicit representation for Tan^-1 came out algebraically. See here. Meanwhile I will blogroll you:)
Batuhan said...
When compared to other silly blogs, yours seems to me like "God among insects". Keep the good work bro.
Michael Ejercito said...
This theorem resulted in a breakthrough in mathematics. Before, imaginary exponents were undefined. This theorem introduced imaginary exponents, as well as logarithms of numbers that are not positive and real.
Sanju said...
Though this proves the desired result, it doesn't come out of intuition like the previous simpler proof.. | {"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": 0.9305636882781982, "perplexity": 2565.885361653195}, "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/1440644059993.5/warc/CC-MAIN-20150827025419-00005-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://www.sarthaks.com/399891/figure-shows-situations-which-small-block-released-respect-smooth-fixed-wedge-shown-figure | # Figure shows four situations in which a small block of mass 'm' is released from rest (with respect to smooth fixed wedge) as shown in figure.
0 votes
342 views
in Physics
Figure shows four situations in which a small block of mass 'm' is released from rest (with respect to smooth fixed wedge) as shown in figure. Column-II shows work done by normal reaction with respect to an observer who is stationary with respect to ground till block reaches at the bottom of inclined wedge, match the appropreate column (Assume that there is infinite friction between block and floor of cabin) :
## 1 Answer
0 votes
by (60.5k points)
selected
Best answer
(A) – (p, t) ; (B) – (p, t) ; (C) – (s) ; (D) – (q, t)
0 votes
1 answer
0 votes
1 answer
0 votes
1 answer
0 votes
1 answer
0 votes
1 answer | {"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": 0.8891661763191223, "perplexity": 2898.3410227195363}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655886178.40/warc/CC-MAIN-20200704135515-20200704165515-00486.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-9-quadratic-functions-and-equations-chapter-test-page-607/4 | ## Algebra 1: Common Core (15th Edition)
We are given $y=x^2-3x+2$ Let's make a table of values: $x=-1 \rightarrow y=6$ $x=0 \rightarrow y=2$ $x=1 \rightarrow y=6$ Then graph the equation. | {"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": 0.8825974464416504, "perplexity": 1818.5996270941278}, "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/1600402128649.98/warc/CC-MAIN-20200930204041-20200930234041-00628.warc.gz"} |
https://chat.stackexchange.com/transcript/36/2021/9/25 | 12:32 AM
How can I show subgroups of order 6 are conjugate in a group of order 42?
one way is outlined in math.stackexchange.com/questions/3010207/… some of the answer is scattered across the comments
yeah I actually saw that post but it only shows the number of conjugates. But I don't know the number of subgroups of order 6
2 hours later…
2:27 AM
OOP sucks
2:39 AM
yeah
1 hour later…
3:46 AM
@Euler2 I disagree, but this is highly opinion-based
OOP can be useful, so many people like it
nothing wrong with that
4:06 AM
what what
I think I proved it
Does exponential decay have to be related to $ln()$? Can't $r^{-x}$ be considered exponential decay? And in the inverse square case, x is a constant ($x=2$) ie. $r^{-2}$?
What do you people think? Actually the post @leslietownes linked was very helpful.
4:43 AM
oop is no panacea but it is useful. i liked dylan's (and lisp, or course) multi methods.
1 hour later…
5:43 AM
@antimony for the first part, yes, and in applications it’s inevitably so. One speaks of exponential decay having some time constant T, such that $y=y_0 e^{-t/T}$. But that can equally well be written as $(e^{1/T})^{-t}=r^{-t}$
For the second, tho, not really. $r^{-x}$ is an exponential function of $x$, but it’s a power-law function of $r$
It being exponential in $x$ is irrelevant if $x$ isn’t changing, eg when $x=2$ to get inverse-square behavior in $r$
@Semiclassical thanks i've revised my view in terms of the definition of 'exponential decay'. however re. power law i think this seems to be a preferred terminology
checking books.google.com seems to show using that cambridge definition of 'exponential' (rather than 'power law') is going out of fashion
and was more popular back in the 1950s etc
*seems to be a matter of preferred terminology rather than incorrect terminology
i would not use a general purpose dictionary for 'exponential,' the word in common english usage has no meaning
or perhaps, too many distinct meanings
exponential decay is as semiclassical describes it
@leslietownes you will note it says "Mathematics Specialised' its not a general definition
but i agree for 'exponential decay' i was wrong i think
it is a general purpose dictionary
5:59 AM
i'm not sure i agree that it is incorrect about one being able to apply 'exponential' to refer to power law type forms
you can stuff logs in if you want. b = e^(ln(b)) is often used to write base-b expressions in terms of base-e expressions
if you have a preferred source of definitions please direct me to it
i don't think dictionaries are very useful in math. the best source of definitions for anything is the materials in front of you, which might not agree with dictionaries or wikipedia or even what someone with a phd in math says is in common use
i think it is the case that exponential to refer to non-natural-exponential power-law type things is going out of fashion, rather than being "incorrect"
there is too much variation in how people use terminology and it isn't practical to list all of them. wikipedia is OK but not always best at signaling when there are multiple definitions
6:01 AM
right, agreed
wikipedia is funny: "The exponent is usually shown as a superscript to the right of the base. In that case, b^n is called "b raised to the nth power", "b raised to the power of n",[1] "the nth power of b", "b to the nth power",[3] or most briefly as "b to the nth"."
weirdly none of those are how i would say it. "b to the n"
yes i think there is a language relic here since n is (still) referred to as the exponent
the page reads like it was written by a chatterbot. fairly common problem with topics applicable to a broad range of contexts. every editor adds or edits one the sentences they are interested in and you get this frankenstein monster of facts
> It being exponential in $x$ is irrelevant if $x$ isn’t changing
@Semiclassical do you pls know where i can read more about that ^ so i can learn what the distinction is?
haha yes @leslietownes
i was wondering where the complex plots were for complex exponents. whenever anything can be graphed in the complex plane, it should be graphed in the complex plane, according to wikipedia. even if those pictures might not tell you anything, or at least not be one of the first 10 things you learn about a concept. but they're stored separately, in he page for "exponential function"
no offense if anyone around here is like the world curator-in-chief of these wikipedia pages
haha
6:12 AM
hehe
i think there's a business opportunity for us to publish a Mathematical Dictionary
though i don't think i should be the one to write it lol
i will take a pay cheque, however :)
lololol
vandalism in wikpedia is intense
i'm surprised that wasn't auto rolled back
6:44 AM
0
The idea is this: If $T\in L(V,W)$. Let $B=\{u_1,...,u_m\}$ be a basis of $V$ and let $C=\{w_1,...,w_n\}$ the given basis of $W$. Matrix of $T$ with respect to $B$ and $C$ has $Tu_i$'s as columns. Applying elementary column operations on the matrix won't change the column space of $T$ and the co...
3 hours later…
9:32 AM
@antimony How was Euler 2 was able to edit the wikipedia like that?
10:08 AM
Anyone can check my proof who's in interest chat.stackexchange.com/transcript/message/59226872#59226872
10:35 AM
@leslietownes I am against all generalizations.
10:55 AM
can anyone give hint regarding this question math.stackexchange.com/questions/4259698/…
1 hour later…
12:04 PM
Let $N_2 \rtimes G \le N_1 \rtimes G$ be internal semidirect products, where $N_2 \le N_1$. Is there a nice formula for the index $|N_1 \rtimes G : N_2 \rtimes G|$? I think it's equal to $|N_1 : N_2|$; is that right?
I want to check if the following is true : If $n$ is a number of distinct subgroup of order $6$ of some large group $G$, then there are at least $6n-3n+3$ distinct elements in $G$.
suppose the constants a, b, c, d, e, and f are positive. does mean that a,b,c,d,e,f >=0 or a,b,c,d,e,f >0
@Mohcine The second one.
thank you so much
@jasmine I think you should also mention what $(13)$ is.
1 hour later…
1:18 PM
@Prithubiswas anyone can edit wikipedia pages (unless the page is locked), that page is probably locked right now since it was recently edited for mischievous means
@antimony Can you edit the html?
2:04 PM
If all eigenvalues of a matrix $A$ over a field $\Bbb C$ are identical, then $A$ is a constant multiple of identity matrix?
I'll never make this mistake: For any $x\in \mathbb R, x-\sqrt {x^2}$ is equal to $0$.
2:26 PM
@Prithubiswas
the vandalism happened in april 2012
3:28 PM
If we have
$$\mathcal{T}(2^{\log_2 z})=\frac z2+\frac 12\mathcal{T}\left(2^{\log_2{\frac z2}}\right)$$
now calling $\mathbb{T}(\cdot)=\mathcal{T}\left(2^{\log_2{(\cdot)}}\right)$ and $u=\log_2 z$ we follow with
$$\mathbb{T}(u)=2^{u-1}+\frac 12 \mathbb{T}(u-1)$$
Thank you very clear. Last question please, should not we have $\mathbb{T}(2^u)=2^{u-1}+\frac 12 \mathbb{T}(2^{u-1})$ instead of $\mathbb{T}(u)=2^{u-1}+\frac 12 \mathbb{T}(u-1)$ as I did substitutions based on values you wrote for $u=\log{z}$
4:15 PM
Hello
Has anyone seen this definition before: math.stackexchange.com/questions/4260001/…
Literally everyone I asked has not seen this definition of an affine space before, so I'm not able to get help with that problem lol
Hopefully it's just linear algebra
Could someone help?
4:42 PM
it's a definition chase more than it is linear algebra. someone should fill it in pretty quickly. it might help to prove at the outset that d(a,b) = -d(b,a) for any a, b; can you prove that?
I'm confused at affine subset of a vector space V; and affine parallel subset of V?
yeah, the T in the title should be X
I think that both are the same. That is if U is a subspace of V then for any v in V, left coset v+U is affine subset of V
And "affine subset of V" and "affine parallel subset of V" are the same thing.
Background: definition 3.81 in chapter 3.E LADR
actually, something's goofy here.
?
4:55 PM
what is x + V, if V is a subspace of T
what is T?
I am referring to this definition: 1) An affine subset of V is a subset of V of the form v+U for some v in V and subspace U of V.
x is a "set of points," T is a vector space as in the definition of the problem
how do you add an element of X to an element of T
i'm not saying that there isn't some obvious way to do this, but my internal type checker is thrown by this
2) For v in V, and U a subspace of V , the affine subset v+U is said to be parallel to this. My confusion is: are (1) and (2) different? If not, why were they stated in two different points?
oh, i see, it's defined in terms of this bijection thingy
@leslietownes I think you're referring to some other question. I have not used T here. Are you referring to the question that I asked the other way about finding a basis of U?
4:58 PM
the conditions (1) and (2) to be proved are different, one talks only about T, the other requires the restriction to X' to be bijective (when the definition only gives you the restriction to X being bijective)
i'm thinking out loud about epsilon emperor's question, koro
what a misfortune to have X and (1) and (2) having independent meaning in two problems at the same time
Ah, I see. My question has nothing to do with @epsilon's question. I had confusion in definition 3.81 in LADR. I was trying to understand that.
that's a really weird coincidence. his question also has the word affine in it
@shintuku what was the solution to your $y-y^3$ with [-1,1]?
koro, in your setting, (1) defines the term "affine subset", and (2) defines what "parallel to" means, so yes, they're different? they involve the same set of notions, but define different things?
but then what is the difference between "affine subset" and "parallel affine subset"?
5:06 PM
he doesn't define 'parallel affine subset', he defines a relation ("parallel to") between an affine subset and a vector subspace
by his definitions, every affine subset is parallel to a subspace, if that's what you're asking, so "being parallel to a subspace" is not a new property of affine subspaces but one that every affine subspace has by definition
but that's not what he's defining
I ask because just after the above stated definition. There's a section "parallel affine subsets" in which there are two examples.
@leslietownes hmm, I understand now.
Thanks a lot, Leslie. :)
i agree that these definitions aren't "doing" very much, but you do sometimes see this. keeping definitions as simple as possible so they are easier to verify
@leslietownes. Are you familiar with recurrence relations please?
I have one question about some algebric manipulations done in one question
I am very close to get my hands on them, but still they are not easy to work with
Is there a mathematical proof that is not verifiable by any human?
I am just wondering here where did $p$ go inside $2f(m-1)$?
5:20 PM
How can I choose the parametric coordinates of a parabola?
@Wolgwang For $y^2=4ax$, you could choose $x=at^2$ and $y=2at$.
@Koro I have to take $\theta$ as a parameter.
and what is $\theta$?
> The parametric coordinates of any point on the parabola $y^2 = x$ can be
> (A) $\left(\sin ^{2} \theta, \sin \theta\right)$
(B) $\left(\cos ^{2} \theta, \cos \theta\right)$
(C) $\left(\sec ^{2} \theta, \sec \theta\right)$
(D) none of these
Ok. So here, since x can be any non-negative number. The first two are ruled out. (Note that (100, 10) lies on the parabola but sine and cosine have maximum value 1)
and similarly, what can you say about $C$ ?@Wolgwang
5:36 PM
@Koro Its range is $(– \infty, -1] ∪ [1 , \infty)$
@Wolgwang right.
I meant to say any point with abscissa in $[0,1)$
@Koro Thanks :-)
@Wolgwang yeah, so take any such point on the parabola. Is it possible to write that point using $(C)$?
Nope
:)
5:55 PM
@Koro Yeah, I didn't notice that
6:46 PM
@Wolgwang The language on that is confusing. From the options, I assume they are looking for a parameterization of the whole parabola, not just points that lie on the parabola (it would have been more confusing if "all of the above" were a choice). In which case, $(0,0)$ is only in (A) and (B), whereas $(4,2)$ is only in (C).
$\left(\tan^2(\theta),\tan(\theta)\right)$ would do.
7:06 PM
Hello, how are you? Are you well now ? @robjohn
Getting better, but still not all that great.
great. I hope you get well soon, professor Rob.
the language in that parabola question is indeed confusing. I think the question asker's intention was: "the parametric coordinates of all points" on the parabola...
Hyello everyone!
Let $I$ be a nonempty directed set, and $H_i \le G_i$ a directed sequence of groups living in some ambient group. Does the following index formula hold: $$|\bigcup_{i \in I} G_i : \bigcup_{i \in I} H_i| = \lim_{i} |G_i : H_i|?$$
For one-one and onto functions can i bypass the usual methods, by saying for any y=f(x) if there exist dy/dx=0 then the function is not onto and one-one? And also I ensure that the point where dy/dx=0 is a local minima/maxima?
7:18 PM
@Koro that was my guess, but it's not good to be guessing.
@AdilMohammed is that a mellow hyello?
@AdilMohammed think of $y=x^3$...
that is a bijection from $\mathbb{R}$ to $\mathbb{R}$.
yet $y'(0)=0$
@robjohn Nah its better, yellow h-yello (we are talking about the pronunciation right lol)
quite rightly
@leslietownes electrical banana is bound to be the very next phase
7:35 PM
@robjohn ooh the 60s songs... I always picture myself in a hammock when I am listening to them
@robjohn Oh yes true but for one-one, i expanded to see if its local minima/maxima (searching for turning points basically)
here is something i find mildly amusing
for reasons that i will not disclose at the moment, consider function $y=\cos(x)(\sin(x)+\sqrt{\sin(x)^2+a})$ with $a\geq 0$.
If $a=0$, this simplifies to $y=\cos x\sin x$ with max value $y(x=\pi/4)=1/2$
But at a glance, that looks incredibly painful to maximize for arbitrary $a$
And yet, the solution is way nicer than you'd expect: the critical point occurs when $x=\tan^{-1}(1/\sqrt{1+a})$ and yields a max value of $y=\sqrt{1+a}$
i can show that in various ways, including without use of calculus at all. and yet it still just baffles me how the answer can be as simple as it is, starting from such an ugly equation
8:21 PM
how to define operatorname in mathjax?
as in, how to define it so you don't have to keep doing \newoperator{} over and over?
yes
i know i've seen it done
though only on the main site. i think on chatjax it'd be impossible
i was in the middle of posting one question. I wanted to define an operatorname for null as in null T
I got it.
neat
the fix i'm seeing is that \newoperator needs to be done in math mode
so dollar signs around it
8:27 PM
\newcommand{\name} {\operatorname {\name}}
then calling \name works
ahh
another option is \DeclareMathOperator
e.g. \DeclareMathOperator{\End}{End}
so that \End would give $\operatorname{End}$
that's specific to making new operators rather than new commands
But somehow, that's showing an error: Mathjax internal buffer size exceeded, is there a recursive macro call?
no idea
oh
operatorname should just use name, not \name
yeah, I noted that. Now there's a strange problem that I just encountered.
\newcommand{\null}{\operatorname{null}}
when I do this and try calling \null, it prints twice (concatenation) happens
that is, nullnull
8:33 PM
why is it so?
i'm not seeing that error
@Koro you shouldn't have the backslash inside the \operatorname
just as Semiclassical said
yeah, I fixed that but still that concatenation is happening
2 mins ago, by Koro
\newcommand{\null}{\operatorname{null}}
8:34 PM
it looks like \newcommands in mathjax are persistent until the page is reloaded
yeah working now :)
thanks a lot :)
$$u\frac{\partial}{\partial u}\left(u\frac{\partial f}{\partial u}\right) + v\frac{\partial}{\partial v}\left(v\frac{\partial f}{\partial v}\right)=\frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2}\tag{1}$$
is that supposed to be a differential equation?
it's not an identity
yeah it's asking when the two differential equations are equal
if ever
8:47 PM
fair
is it possible to define many operatornames in one go?
mathematica's DSolve command shrugs its hands at it
not with one function, i suspect @koro
what're you trying to define, tho
can you construct a smooth manifold which is a surface built by taking a rotating homotopy that projects onto planar homotopic curves?
plane homotopy has two fixed endpoints
I can only do this for a simple case using rotation matrices to rotate the curve by a certain angle and then projecting down onto the planar curve. then repeating the process
the hard part is if the plane homotopy is say analytic chiral, you have to "deform/twist" the curve s.t. it projects precisely onto the planar curve, probably sacrificing analyticity of the surface
9:04 PM
@Semiclassical in chat, they may be. On the main site, they should only be active for the post in which they were executed.
@Koro you need multiple \newcommands
what i meant specifically was: suppose you define and use a new command in an answer. the wysiwyg editor on the main site will properly format that command
if you delete the definition from the post, the use of the command will persist in the display
yo
yo joe
does anybody know what the probability measure $\mathbb{P}^X$ mean?
9:07 PM
@Koro it works as it should for me.
so if you make an error in creating a mathjax definition, it can persist until you reload the page
ah I see you got it working
I see so multiple \newcommands will be required.
again, what specifically are you trying to do?
me?
I wanted to define operatornames span, null, range etc.
@Semiclassical: I was trying to solve one problem on linear maps. I discussed that problem here also the other day. I have tried to present my understanding here in this answer math.stackexchange.com/questions/1924952/…
I request you to please take a look at it. Thanks.
9:12 PM
ah. yeah, if they're all different things, then gotta define each seperately
I had imagined a matrix and on that used the fact that column operations on a matrix won't affect its column space.
to be clear, i'm only looking at the mathjax :)
ah, okay :)
but the question being: if two matrices have the same null space, then they're equivalent up to an invertible linear transformation?
"they are equivalent up to " ?
9:18 PM
as in, $T_1=ST_2$
ah, okay. That's the question that I posted today.
koro i don't think axler has anything about row or column equivalence of matrices. a more matrix theoretic book would.
oh, i saw that you added a question and thought that was what you meant
i think he does talk a little bit about change of basis and matrices in a later chapter, but maybe only in the context of operators on a single vector space? memory is hazy about this.
9:20 PM
yes, nothing explicit. every linear algebra book has this under the hood but only some of them will have theorems expressly talking about it.
@Semiclassical math.stackexchange.com/questions/1924952/… I request a review of this answer of mine please.
@leslietownes I think knowing a thing or two about matrices helps immensely to think in terms of linear maps.
some books go the route of, the following are equivalent for a linear map T: V to V. [list of about 20 things that are equivalent to "T is invertible"]. then another one for T: V to W possibly of different dimensions with stuff about rank, nullity, left invertibility, right invertibility.
i'm a physicist, so working in bases is pretty common to me
$\require{begingroup}\begingroup\newcommand{\test}{\operatorname{blah}}\test\endgroup\test$ I used the same macro, but limited its scope.
a favorite trick in QM is basically just notation for changing bases
9:23 PM
oh semi what do you work in I never asked
i did my phd in statistical physics/qm
since then i've been doing quantum foundations stuff
nice
koro yeah. note that the other answer solves the problem very quickly and directly. the approach about column operations is a little more 'conceptual,' in that it suggests a whole family of exercises on the same subject and hints at a larger point on the kinds of matrices that can be chosen to represent a given operator.
trying to move over to quantum computing, at least in terms of teaching it
then do you know what the notation $\mathbb{P}^X$ refers to?
not the distribution, by chance?
or just the probability measure associated with the rv $X$? is that it?
9:24 PM
i saw you ask about it earlier, yeah. i would need context
it doesn't resemble any notation i have seen, but notation is far from standardized. i would expect a book or resource to define it somewhere.
@leslietownes yeah, I agree. But I'm glad that you reviewed my answer. Thanks a lot.
not much context unfortunately, I was just hoping it was standard
well, you had to have gotten it from somewhere :P
well,
Calculate $\mathbb{E}(X)$ for the following probability measures $\mathbb{P}^X$...
if youd like
9:26 PM
one might be able to guess at a meaning from a number of examples. a guess still being a guess.
i'm not asking what problem you got it from
oh just a problem sheet
not a book
ah
probably need to ask the prof then
I suspect Varadhan might define it somewhere
but I don't have my book on me
if only there were some way of infringing copyrights in books using the internet.
:)
9:28 PM
perhaps just a way to insist on the fact that $\mathbb{P}$ is the probability measure of $X$? 🤔
as far as I remember
google books has a preview but
preview
@JoeShmo looks like it might be functions from $X$ to $\mathbb{P}$ which I assume is a probability space.
joe: the mathcal P is often used to denote a fixed probability measure on some space. we don't seem to have a notation for that space. here X is a random variable on that space. X does induce a measure on the real numbers: assign the subset E of the real numbers the measure P(X in E).
so maybe P^X is that measure. this is just a guess.
if my guess is right, it strikes me as weird to use a superscript for that purpose. robjohn is interpreting the superscript in a more standard way.
and i'm interpreting it as "not my problem" :P
i think we should come up with a slate of two or three more guesses and decide the issue by majority vote.
semiclassical has provided a third option. i think it would win the vote.
9:36 PM
Hope patient is doing well.
Off to San Jose if I know the way.
sounds like fun!
@robjohn no unfortunately $\mathbb{P}^X$ is a probability measure
the suggestion is that $B^A$ is common notation for functions from $A$ to $B$
in which case being an element of $\mathbb{P}^X$ might be plausible
but that doesn't seem to quite fit i think
I understood that
oh, mystery solved (?)
he writes here explicitly (..blah blah compute expectation for...)
$\mathbb{P}^X = p \delta_a + (1-p) \delta_b$ and
$\mathbb{P}^X(\{n\}) = e^{-\lambda} \dfrac{\lambda^n}{n!}$
looks like they're just measures on the reals? but then i don't understand what X is or why you would decorate P with X
9:51 PM
exactly..
if X is the identity function on the reals, you could compute E(X) for each of those measures, but the measure doesn't depend on X
goofy
but then I guess he didn't define the probability space explicitly, and he wants you to compute $\mathbb{E}(X)$
so he wants you to know that $\mathbb{P}$ is is the probability measure associated with $X$
10:03 PM
Any ideas on how to integrate this beast (a,b are real>0)?:
www.shorturl.at/jlKY7
I have a quick question about logic - how can two things be tautologically equivalent but not equal? The example given in the slides for my CS class were [ not (P and Q) ] and [ (not P) or (not Q) ] - firstly with De Morgan's laws I think these are equal, but even if we're not allowed to use that in this specific field, how can two formulas be equivalent but not equal?
This is the beast in question, (so you dont have to follow the URL):
int exp(-b*x)*sqrt(x(x+a)) dx, from x = 0 to infinity
oh @leslietownes I didn't see your last comment. No, $X$ is a random variable
10:48 PM
joe: i was suggesting that the underlying space might be the reals and X: R -> R the identity function X(t) = t
in this guise, X is a random variable
again, all of this is guessing, i'm not wedded to any of it | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8541591167449951, "perplexity": 954.2863401590976}, "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/1634323585828.15/warc/CC-MAIN-20211023224247-20211024014247-00257.warc.gz"} |
http://dlmf.nist.gov/10.68 | # §10.68(i) Definitions
10.68.1 $\mathop{M_{\nu}\/}\nolimits\!\left(x\right)e^{i\!\mathop{\theta_{\nu}\/}% \nolimits\!\left(x\right)}=\mathop{\mathrm{ber}_{\nu}\/}\nolimits x+i\mathop{% \mathrm{bei}_{\nu}\/}\nolimits x,$
10.68.2 $\mathop{N_{\nu}\/}\nolimits\!\left(x\right)e^{i\!\mathop{\phi_{\nu}\/}% \nolimits\!\left(x\right)}=\mathop{\mathrm{ker}_{\nu}\/}\nolimits x+i\mathop{% \mathrm{kei}_{\nu}\/}\nolimits x,$
where $\mathop{M_{\nu}\/}\nolimits\!\left(x\right)\,(>0)$, $\mathop{N_{\nu}\/}\nolimits\!\left(x\right)\,(>0)$, $\mathop{\theta_{\nu}\/}\nolimits\!\left(x\right)$, and $\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right)$ are continuous real functions of $x$ and $\nu$, with the branches of $\mathop{\theta_{\nu}\/}\nolimits\!\left(x\right)$ and $\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right)$ chosen to satisfy (10.68.18) and (10.68.21) as $x\to\infty$. (See also §10.68(iv).)
# §10.68(ii) Basic Properties
10.68.3 $\displaystyle\mathop{\mathrm{ber}_{\nu}\/}\nolimits x$ $\displaystyle=\mathop{M_{\nu}\/}\nolimits\!\left(x\right)\mathop{\cos\/}% \nolimits\mathop{\theta_{\nu}\/}\nolimits\!\left(x\right),$ $\displaystyle\mathop{\mathrm{bei}_{\nu}\/}\nolimits x$ $\displaystyle=\mathop{M_{\nu}\/}\nolimits\!\left(x\right)\mathop{\sin\/}% \nolimits\mathop{\theta_{\nu}\/}\nolimits\!\left(x\right),$
10.68.4 $\displaystyle\mathop{\mathrm{ker}_{\nu}\/}\nolimits x$ $\displaystyle=\mathop{N_{\nu}\/}\nolimits\!\left(x\right)\mathop{\cos\/}% \nolimits\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right),$ $\displaystyle\mathop{\mathrm{kei}_{\nu}\/}\nolimits x$ $\displaystyle=\mathop{N_{\nu}\/}\nolimits\!\left(x\right)\mathop{\sin\/}% \nolimits\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right).$
10.68.5 $\displaystyle\mathop{M_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=({\mathop{\mathrm{ber}_{\nu}\/}\nolimits^{2}}x+{\mathop{\mathrm{% bei}_{\nu}\/}\nolimits^{2}}x)^{\ifrac{1}{2}},$ $\displaystyle\mathop{N_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=({\mathop{\mathrm{ker}_{\nu}\/}\nolimits^{2}}x+{\mathop{\mathrm{% kei}_{\nu}\/}\nolimits^{2}}x)^{\ifrac{1}{2}},$
10.68.6 $\displaystyle\mathop{\theta_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\mathop{\mathrm{Arctan}\/}\nolimits\!\left(\mathop{\mathrm{bei}_% {\nu}\/}\nolimits x/\mathop{\mathrm{ber}_{\nu}\/}\nolimits x\right),$ $\displaystyle\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\mathop{\mathrm{Arctan}\/}\nolimits\!\left(\mathop{\mathrm{kei}_% {\nu}\/}\nolimits x/\mathop{\mathrm{ker}_{\nu}\/}\nolimits x\right).$
10.68.7 $\displaystyle\mathop{M_{-n}\/}\nolimits\!\left(x\right)$ $\displaystyle=\mathop{M_{n}\/}\nolimits\!\left(x\right),$ $\displaystyle\mathop{\theta_{-n}\/}\nolimits\!\left(x\right)$ $\displaystyle=\mathop{\theta_{n}\/}\nolimits\!\left(x\right)-n\pi.$
With arguments $(x)$ suppressed,
10.68.8 ${\mathop{\mathrm{ber}_{\nu}\/}\nolimits^{\prime}}x=\tfrac{1}{2}\mathop{M_{\nu+% 1}\/}\nolimits\mathop{\cos\/}\nolimits\!\left(\mathop{\theta_{\nu+1}\/}% \nolimits-\tfrac{1}{4}\pi\right)-\tfrac{1}{2}\mathop{M_{\nu-1}\/}\nolimits% \mathop{\cos\/}\nolimits\!\left(\mathop{\theta_{\nu-1}\/}\nolimits-\tfrac{1}{4% }\pi\right)=(\nu/x)\mathop{M_{\nu}\/}\nolimits\mathop{\cos\/}\nolimits\mathop{% \theta_{\nu}\/}\nolimits+\mathop{M_{\nu+1}\/}\nolimits\mathop{\cos\/}\nolimits% \!\left(\mathop{\theta_{\nu+1}\/}\nolimits-\tfrac{1}{4}\pi\right)=-(\nu/x)% \mathop{M_{\nu}\/}\nolimits\mathop{\cos\/}\nolimits\mathop{\theta_{\nu}\/}% \nolimits-\mathop{M_{\nu-1}\/}\nolimits\mathop{\cos\/}\nolimits\!\left(\mathop% {\theta_{\nu-1}\/}\nolimits-\tfrac{1}{4}\pi\right),$
10.68.9 ${\mathop{\mathrm{bei}_{\nu}\/}\nolimits^{\prime}}x=\tfrac{1}{2}\mathop{M_{\nu+% 1}\/}\nolimits\mathop{\sin\/}\nolimits\!\left(\mathop{\theta_{\nu+1}\/}% \nolimits-\tfrac{1}{4}\pi\right)-\tfrac{1}{2}\mathop{M_{\nu-1}\/}\nolimits% \mathop{\sin\/}\nolimits\!\left(\mathop{\theta_{\nu-1}\/}\nolimits-\tfrac{1}{4% }\pi\right)=(\nu/x)\mathop{M_{\nu}\/}\nolimits\mathop{\sin\/}\nolimits\mathop{% \theta_{\nu}\/}\nolimits+\mathop{M_{\nu+1}\/}\nolimits\mathop{\sin\/}\nolimits% \!\left(\mathop{\theta_{\nu+1}\/}\nolimits-\tfrac{1}{4}\pi\right)=-(\nu/x)% \mathop{M_{\nu}\/}\nolimits\mathop{\sin\/}\nolimits\mathop{\theta_{\nu}\/}% \nolimits-\mathop{M_{\nu-1}\/}\nolimits\mathop{\sin\/}\nolimits\!\left(\mathop% {\theta_{\nu-1}\/}\nolimits-\tfrac{1}{4}\pi\right).$
10.68.10 $\displaystyle{\mathop{\mathrm{ber}\/}\nolimits^{\prime}}x$ $\displaystyle=\mathop{M_{1}\/}\nolimits\mathop{\cos\/}\nolimits\!\left(\mathop% {\theta_{1}\/}\nolimits-\tfrac{1}{4}\pi\right),$ $\displaystyle{\mathop{\mathrm{bei}\/}\nolimits^{\prime}}x$ $\displaystyle=\mathop{M_{1}\/}\nolimits\mathop{\sin\/}\nolimits\!\left(\mathop% {\theta_{1}\/}\nolimits-\tfrac{1}{4}\pi\right).$
10.68.11 ${\mathop{M_{\nu}\/}\nolimits^{\prime}}=(\nu/x)\mathop{M_{\nu}\/}\nolimits+% \mathop{M_{\nu+1}\/}\nolimits\mathop{\cos\/}\nolimits\!\left(\mathop{\theta_{% \nu+1}\/}\nolimits-\mathop{\theta_{\nu}\/}\nolimits-\tfrac{1}{4}\pi\right)=-(% \nu/x)\mathop{M_{\nu}\/}\nolimits-\mathop{M_{\nu-1}\/}\nolimits\mathop{\cos\/}% \nolimits\!\left(\mathop{\theta_{\nu-1}\/}\nolimits-\mathop{\theta_{\nu}\/}% \nolimits-\tfrac{1}{4}\pi\right),$
10.68.12 ${\mathop{\theta_{\nu}\/}\nolimits^{\prime}}=(\mathop{M_{\nu+1}\/}\nolimits/% \mathop{M_{\nu}\/}\nolimits)\mathop{\sin\/}\nolimits\!\left(\mathop{\theta_{% \nu+1}\/}\nolimits-\mathop{\theta_{\nu}\/}\nolimits-\tfrac{1}{4}\pi\right)=-(% \mathop{M_{\nu-1}\/}\nolimits/\mathop{M_{\nu}\/}\nolimits)\mathop{\sin\/}% \nolimits\!\left(\mathop{\theta_{\nu-1}\/}\nolimits-\mathop{\theta_{\nu}\/}% \nolimits-\tfrac{1}{4}\pi\right).$
10.68.13 $\displaystyle{\mathop{M_{0}\/}\nolimits^{\prime}}$ $\displaystyle=\mathop{M_{1}\/}\nolimits\mathop{\cos\/}\nolimits\!\left(\mathop% {\theta_{1}\/}\nolimits-\mathop{\theta_{0}\/}\nolimits-\tfrac{1}{4}\pi\right),$ $\displaystyle{\mathop{\theta_{0}\/}\nolimits^{\prime}}$ $\displaystyle=(\mathop{M_{1}\/}\nolimits/\mathop{M_{0}\/}\nolimits)\mathop{% \sin\/}\nolimits\!\left(\mathop{\theta_{1}\/}\nolimits-\mathop{\theta_{0}\/}% \nolimits-\tfrac{1}{4}\pi\right).$
10.68.14 $\displaystyle\ifrac{d(x{\mathop{M_{\nu}\/}\nolimits^{2}}{\mathop{\theta_{\nu}% \/}\nolimits^{\prime}})}{dx}$ $\displaystyle=x{\mathop{M_{\nu}\/}\nolimits^{2}},$ $\displaystyle x^{2}{\mathop{M_{\nu}\/}\nolimits^{\prime\prime}}+x{\mathop{M_{% \nu}\/}\nolimits^{\prime}}-\nu^{2}\mathop{M_{\nu}\/}\nolimits$ $\displaystyle=x^{2}\mathop{M_{\nu}\/}\nolimits{{\mathop{\theta_{\nu}\/}% \nolimits^{\prime}}^{2}}.$
Equations (10.68.8)–(10.68.14) also hold with the symbols $\mathop{\mathrm{ber}\/}\nolimits$, $\mathop{\mathrm{bei}\/}\nolimits$, $\mathop{M\/}\nolimits$, and $\mathop{\theta\/}\nolimits$ replaced throughout by $\mathop{\mathrm{ker}\/}\nolimits$, $\mathop{\mathrm{kei}\/}\nolimits$, $\mathop{N\/}\nolimits$, and $\mathop{\phi\/}\nolimits$, respectively. In place of (10.68.7),
10.68.15 $\displaystyle\mathop{N_{-\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\mathop{N_{\nu}\/}\nolimits\!\left(x\right),$ $\displaystyle\mathop{\phi_{-\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right)+\nu\pi.$
# §10.68(iii) Asymptotic Expansions for Large Argument
When $\nu$ is fixed, $\mu=4\nu^{2}$, and $x\to\infty$
10.68.16 $\displaystyle\mathop{M_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\frac{e^{x/\sqrt{2}}}{(2\pi x)^{\frac{1}{2}}}\left(1-\frac{\mu-1% }{8\sqrt{2}}\frac{1}{x}+\frac{(\mu-1)^{2}}{256}\frac{1}{x^{2}}-\frac{(\mu-1)(% \mu^{2}+14\mu-399)}{6144\sqrt{2}}\frac{1}{x^{3}}+\mathop{O\/}\nolimits\!\left(% \frac{1}{x^{4}}\right)\right),$ 10.68.17 $\displaystyle\mathop{\ln\/}\nolimits\mathop{M_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\frac{x}{\sqrt{2}}-\frac{1}{2}\mathop{\ln\/}\nolimits\!\left(2% \pi x\right)-\frac{\mu-1}{8\sqrt{2}}\frac{1}{x}-\frac{(\mu-1)(\mu-25)}{384% \sqrt{2}}\frac{1}{x^{3}}-\frac{(\mu-1)(\mu-13)}{128}\frac{1}{x^{4}}+\mathop{O% \/}\nolimits\!\left(\frac{1}{x^{5}}\right),$ 10.68.18 $\displaystyle\mathop{\theta_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=\frac{x}{\sqrt{2}}+\left(\frac{1}{2}\nu-\frac{1}{8}\right)\pi+% \frac{\mu-1}{8\sqrt{2}}\frac{1}{x}+\frac{\mu-1}{16}\frac{1}{x^{2}}-\frac{(\mu-% 1)(\mu-25)}{384\sqrt{2}}\frac{1}{x^{3}}+\mathop{O\/}\nolimits\!\left(\frac{1}{% x^{5}}\right).$ 10.68.19 $\displaystyle\mathop{N_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=e^{-x/\sqrt{2}}\left(\frac{\pi}{2x}\right)^{\frac{1}{2}}\left(1+% \frac{\mu-1}{8\sqrt{2}}\frac{1}{x}+\frac{(\mu-1)^{2}}{256}\frac{1}{x^{2}}+% \frac{(\mu-1)(\mu^{2}+14\mu-399)}{6144\sqrt{2}}\frac{1}{x^{3}}+\mathop{O\/}% \nolimits\!\left(\frac{1}{x^{4}}\right)\right),$ 10.68.20 $\displaystyle\mathop{\ln\/}\nolimits\mathop{N_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=-\frac{x}{\sqrt{2}}+\frac{1}{2}\mathop{\ln\/}\nolimits\!\left(% \frac{\pi}{2x}\right)+\frac{\mu-1}{8\sqrt{2}}\frac{1}{x}+\frac{(\mu-1)(\mu-25)% }{384\sqrt{2}}\frac{1}{x^{3}}-\frac{(\mu-1)(\mu-13)}{128}\frac{1}{x^{4}}+% \mathop{O\/}\nolimits\!\left(\frac{1}{x^{5}}\right),$ 10.68.21 $\displaystyle\mathop{\phi_{\nu}\/}\nolimits\!\left(x\right)$ $\displaystyle=-\frac{x}{\sqrt{2}}-\left(\frac{1}{2}\nu+\frac{1}{8}\right)\pi-% \frac{\mu-1}{8\sqrt{2}}\frac{1}{x}+\frac{\mu-1}{16}\frac{1}{x^{2}}+\frac{(\mu-% 1)(\mu-25)}{384\sqrt{2}}\frac{1}{x^{3}}+\mathop{O\/}\nolimits\!\left(\frac{1}{% x^{5}}\right).$
# §10.68(iv) Further Properties
Additional properties of the modulus and phase functions are given in Young and Kirk (1964, pp. xi–xv). However, care needs to be exercised with the branches of the phases. Thus this reference gives $\mathop{\phi_{1}\/}\nolimits\!\left(0\right)=\tfrac{5}{4}\pi$ (Eq. (6.10)), and $\lim_{x\to\infty}(\mathop{\phi_{1}\/}\nolimits\!\left(x\right)+(x/\sqrt{2}))=-% \tfrac{5}{8}\pi$ (Eqs. (10.20) and (Eqs. (10.26b)). However, numerical tabulations show that if the second of these equations applies and $\mathop{\phi_{1}\/}\nolimits\!\left(x\right)$ is continuous, then $\mathop{\phi_{1}\/}\nolimits\!\left(0\right)=-\tfrac{3}{4}\pi$; compare Abramowitz and Stegun (1964, p. 433). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 201, "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": 0.9908902645111084, "perplexity": 1195.5695764677453}, "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/1438042990611.52/warc/CC-MAIN-20150728002310-00110-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://bugzilla.mozilla.org/show_bug.cgi?id=500293 | Open Opened 11 years ago Updated 11 years ago
# Built-up (vulgar) fractions using the Unicode fraction slash (or other methods) not supported
Not set
UNCONFIRMED
## Details
User-Agent: Mozilla/5.0 (Macintosh; U; PPC Mac OS X 10_5_7; fr-fr) AppleWebKit/530.18 (KHTML, like Gecko) Version/4.0.1 Safari/530.18
Build Identifier:
Unicode contains only a handful of precomposed vulgar fractions (viz, ½; ¼, ¾; ⅓, ⅔; ⅕, ⅖, ⅗, ⅘; ⅙, ⅚; ⅛, ⅜, ⅝, ⅞). In particular, 16hts and 32nds, which are frequently used with imperial units, are not included.
The preferred solution, as suggested in an e-mail [1] sent to the WhatWG mailing list and supported by Jonas Sickling [2], would be to let a sequence of one or more digits 0--9 followed by a fraction slash [U+2044] followed by one or more digits 0--9 be rendered as a built-up fraction. This behaviour is defined in Unicode, and Unicode Technical Report No. 20 explicitly says that this solution is appropriate for use in mark-up languages and not only in plain text.
Ideally, digits specifically designed for use in built-up fractions should be used when available in the font used. Otherwise, scaling normal digits to 60% height and 65% width, as suggested in the PostScript
Language Cookbook, would seem to be appropriate as fall-back.
It should also be noted that MathML does not seem to support vulgar fractions at all, neither as specified nor as implemented in Firefox.
[1] <http://lists.whatwg.org/pipermail/whatwg-whatwg.org/2009-April/019340.html>
[2] <http://lists.whatwg.org/pipermail/whatwg-whatwg.org/2009-June/020120.html>
Reproducible: Always
Component: General → Layout: Text
Product: Firefox → Core
QA Contact: general → layout.fonts-and-text
Version: unspecified → Trunk
Would be great if you could link to the actual specifications for how this is supposed to work as well.
The correct behaviour is specified in Unicode 5.0, Chapter 6 Punctuation' [1], Section 6.1 General Punctuation', heading General Punctuation: U+2000--U+206F', subheading Other Punctuation', subsubheading Fraction Slash' (p. 154):
“Fraction Slash.
“U+2044 FRACTION SLASH is used between digits to form numeric fractions, such as 2/3, 3/9, and so on. The standard form of a fraction built using the fraction slash is defined as follows: Any sequence of one or more decimal digits, followed by the fraction slash, followed by any sequence of one or more decimal digits. Such a fraction should be displayed as a unit, such as ¾ or $\frac34$ [vertical built-up fraction]. The precise choice of display can depend upon additional formatting information.
“If the displaying software is incapable of mapping the fraction to a unit, then it can also be displayed as a simple linear sequence as a fallback (for example, 3/4). If the fraction is to be separated from a previous number, then a space can be used, choosing the appropriate width (normal, thin, zero width, and so on). For example, 1 + ZERO WIDTH SPACE + 3 + FRACTION SLASH + 4 is displayed as 1¾.”
[1] <http://unicode.org/book/ch06.pdf>
This would be a nice feature, and not too hard to implement at the gfxTextRunWordCache level where all textrun processing goes through ... as long as you don't expect this to work when the fraction contains multiple font styles or direction changes. Alternatively we could implement it in Harfbuzz...
I don't think it will be high enough priority to demand attention anytime soon, though. Patches welcome.`
You need to log in before you can comment on or make changes to this bug. | {"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": 0.8342733383178711, "perplexity": 2908.2058218537436}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370511408.40/warc/CC-MAIN-20200410173109-20200410203609-00108.warc.gz"} |
https://www.physicsforums.com/threads/continuity-problems-for-my-analysis-class.472523/ | # Continuity problems for my Analysis class
• #1
12
0
I am having a lot of difficulty on my continuity problems for my Analysis class.
1. Prove that (f O g)(x) = f(g(x)) is continuous at any point p in R in three ways a.) Using the episolon delta definition of continuity, b.) using the sequence definition of continuity, and c.) using the open set definition of continuity.
2. Prove that if U is an open set in R, then its inverse is open.
• #2
tiny-tim
Homework Helper
25,832
251
Hi LauraLovies!
(I assume f and g are both continuous? and have a delta: δ and an epsilon: ε )
Show us what you've tried, and where you're stuck, and then we'll know how to help!
• #3
12
0
f and g are both continuous so i know that there exists some $$\epsilon > 0 and greater than zero that fulfills the continuity definition. It just seems to obvious to me that i dont even know where to start. \delta$$
• #4
tiny-tim
Homework Helper
25,832
251
f and g are both continuous so i know that there exists some \epsilon > 0 and greater than zero that fulfills the continuity definition. It just seems to obvious to me that i dont even know where to start. \delta
Start with an εf and δf, and an εg and δg, and then construct a proof using a new ε and δ based on them.
• #5
lurflurf
Homework Helper
2,440
138
Which sequence definition? that lim f(x_n)=f(lim x_m)?
The general idea adaptible to the cases is that we desire
f(g(x+h))-f(x)=f(g(x)+[g(x+h)-g(x)])-f(x)
be small when h is
• Last Post
Replies
3
Views
1K
• Last Post
Replies
4
Views
1K
• Last Post
Replies
2
Views
1K
• Last Post
Replies
5
Views
5K
• Last Post
Replies
11
Views
2K
• Last Post
Replies
3
Views
866
• Last Post
Replies
3
Views
3K
• Last Post
Replies
2
Views
3K
• Last Post
Replies
8
Views
2K
• Last Post
Replies
6
Views
431 | {"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": 0.9418038129806519, "perplexity": 2022.9939473178536}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988837.67/warc/CC-MAIN-20210508031423-20210508061423-00306.warc.gz"} |
https://eprints.utas.edu.au/view/authors/Stephens=3AThomas=3A=3A.html | # Items where Author is "Stephens, Thomas"
Up a level
Export as ASCII CitationBibTeXDublin CoreDublin CoreEP3 XMLEndNoteHTML CitationJSONMETSMultiline CSVObject IDsOpenURL ContextObjectRDF+N-TriplesRDF+N3RDF+XMLReferReference Manager
Group by: Item Type | No Grouping
Number of items: 18.
## Article
Stephens, Thomas 1889 , Papers & Proceedings of the Royal Society of Tasmania , p. 54 .
Stephens, Thomas 1909 , Papers and Proceedings of the Royal Society of Tasmania , pp. 170-174 .
Stephens, Thomas 1897 , Papers & Proceedings of the Royal Society of Tasmania , pp. 189-196 .
Stephens, Thomas 1873 , Monthly Notices of Papers & Proceedings of the Royal Society of Tasmania , pp. 36-38 .
Stephens, Thomas 1878 , Papers & Proceedings and Report of the Royal Society of Tasmania , pp. 63-64 .
Stephens, Thomas 1881 , Papers & Proceedings and Report of the Royal Society of Tasmania , pp. 24-25 .
Stephens, Thomas 1897 , Papers & Proceedings of the Royal Society of Tasmania , pp. 54-58 .
Stephens, Thomas 1884 , Papers & Proceedings of the Royal Society of Tasmania , pp. 217-219 .
Stephens, Thomas 1885 , Papers & Proceedings of the Royal Society of Tasmania , pp. 403-406 .
Stephens, Thomas 1897 , Papers & Proceedings of the Royal Society of Tasmania , pp. 92-96 .
Stephens, Thomas 1912 , Papers and Proceedings of the Royal Society of Tasmania , pp. 85-86 .
Stephens, Thomas 1887 , Papers & Proceedings of the Royal Society of Tasmania , pp. 107-109 .
Stephens, Thomas 1909 , Papers and Proceedings of the Royal Society of Tasmania , pp. 82-84 .
Stephens, Thomas 1869 , Monthly Notices of Papers & Proceedings of the Royal Society of Tasmania , pp. 55-57 .
Stephens, Thomas 1881 , Papers & Proceedings and Report of the Royal Society of Tasmania , pp. 36-38 .
Stephens, Thomas 1869 , Monthly Notices of Papers & Proceedings of the Royal Society of Tasmania , pp. 17-21 .
Stephens, Thomas 1889 , Papers & Proceedings of the Royal Society of Tasmania , pp. 92-94 .
Stephens, Thomas 1900 , Papers & Proceedings of the Royal Society of Tasmania , pp. 42-44 .
This list was generated on Tue Jul 27 01:50:39 2021 AEST.
TOP | {"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": 0.9527719616889954, "perplexity": 3051.3531268465335}, "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/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00432.warc.gz"} |
https://brilliant.org/problems/power-of-derivatives/ | # Power of Derivatives
Calculus Level 3
$f(x) = e^x \cdot \sin x$
For non-zero values of $$f(x)$$, simplify the expression below. $\log_{2} \dfrac{f^{(2016)} (x)}{f(x)}$
Notation: $$f^{(n)}(x)$$ denotes the $$n^\text{th}$$ derivative of $$f(x)$$.
× | {"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": 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": 0.9989118576049805, "perplexity": 2728.444767442412}, "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/1476988719416.57/warc/CC-MAIN-20161020183839-00541-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.zbmath.org/?q=ai%3Awang.yujuan+se%3A00007424 | ×
# zbMATH — the first resource for mathematics
First kind weak total chromatic numbers of graphs $$C_m\vee F_n$$, $$C_m\vee W_n$$ and $$C_m\vee C_n$$. (Chinese. English summary) Zbl 1313.05137
Summary: This paper provides programs about the colorings of cycles and fans, cycles and wheels, and cycles and cycles: (1) for $$C_m\vee F_n$$, $$\chi_{fwt}(C_m\vee F_n)=m+n+1$$, (2) for $$C_m\vee W_n$$, $$\chi_{fwt}(C_m\vee W_n)=m+n+1$$, (3) for $$C_m\vee C_n$$, $$\chi_{fwt}(C_m\vee C_n)=m+n$$. We also give the proof of these results.
##### MSC:
05C15 Coloring of graphs and hypergraphs | {"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": 0.8153532147407532, "perplexity": 1418.5477830793247}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703517559.41/warc/CC-MAIN-20210119011203-20210119041203-00242.warc.gz"} |
https://www.scholars.northwestern.edu/en/publications/black-hole-genealogy-identifying-hierarchical-mergers-with-gravit | # Black Hole Genealogy: Identifying Hierarchical Mergers with Gravitational Waves
Chase Kimball*, Colm Talbot, Christopher P. Christopher, Matthew Carney, Michael Zevin, Eric Thrane, Vicky Kalogera
*Corresponding author for this work
Research output: Contribution to journalArticlepeer-review
68 Scopus citations
## Abstract
In dense stellar environments, the merger products of binary black hole mergers may undergo additional mergers. These hierarchical mergers are naturally expected to have higher masses than the first generation of black holes made from stars. The components of hierarchical mergers are expected to have significant characteristic spins, imprinted by the orbital angular momentum of the previous mergers. However, since the population properties of first-generation black holes are uncertain, it is difficult to know if any given merger is first-generation or hierarchical. We use observations of gravitational waves to reconstruct the binary black hole mass and spin spectrum of a population including the possibility of hierarchical mergers. We employ a phenomenological model that captures the properties of merging binary black holes from simulations of globular clusters. Inspired by recent work on the formation of low-spin black holes, we include a zero-spin subpopulation. We analyze binary black holes from LIGO and Virgo's first two observing runs, and find that this catalog is consistent with having no hierarchical mergers. We find that the most massive system in this catalog, GW170729, is mostly likely a first-generation merger, having a 4% probability of being a hierarchical merger assuming a 5 × 105 M o˙ globular cluster mass. Using our model, we find that 99% of first-generation black holes in coalescing binaries have masses below 44 M o˙, and the fraction of binaries with near-zero component spins is less than 0.16 (90% probability). Upcoming observations will determine if hierarchical mergers are a common source of gravitational waves.
Original language English (US) 177 Astrophysical Journal 900 2 https://doi.org/10.3847/1538-4357/aba518 Published - Sep 10 2020
## ASJC Scopus subject areas
• Astronomy and Astrophysics
• Space and Planetary Science
## Fingerprint
Dive into the research topics of 'Black Hole Genealogy: Identifying Hierarchical Mergers with Gravitational Waves'. 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": 0.822828471660614, "perplexity": 3615.9344846980453}, "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-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00698.warc.gz"} |
https://www.physicsforums.com/threads/standing-wave-energy.285419/ | Standing wave energy
1. Jan 17, 2009
atavistic
1. The problem statement, all variables and given/known data
A standing wave is maintained in a homogenous string of cross-sectional area a and density $$\rho$$ . It is formed by the superposition of two waves travelling in opposite directions given by the equations
y1 = Asin(wt-kx)
y2 = 2Asin(wt +kx)
Find the total mechanical energy confined between the section corresponding to the adjacent antinodes.
2. The attempt at a solution
The wave is given by y = Asin(wt-kx) + 2Asin(wt +kx)
KE of a small element is 1/2 u dx v^2 , where u = mass per unit length.
I find v by differentiating y wrt t.
Then I integrate with proper limits. But the integration looks outrageous looking at the simple answer. Is there any shortcut ? Also using above method we only get the KE, what about the PE?
2. Jan 17, 2009
tiny-tim
Hi atavistic!
Show us!
3. Jan 17, 2009
atavistic
Well can you tell me what I did is right or not and the thing about PE. | {"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": 0.9732652902603149, "perplexity": 798.0875131036508}, "config": {"markdown_headings": false, "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-2017-04/segments/1484560279248.16/warc/CC-MAIN-20170116095119-00113-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/1594253/difference-between-tensoring-and-wedging/1594316 | # Difference Between Tensoring and Wedging.
Let $V$ be a vector space and $\omega\in \otimes^k V$. There are $2$ ways (at least) of thinking about $\omega\otimes \omega$.
1) We may think of $\otimes^k V$ as a vector space $W$, and $\omega\otimes \omega$ as a member of $W\otimes W$.
2) We may think of $\omega\otimes \omega$ as a member of $\otimes^{2k}V$.
The two interpretations are "same" because $W\otimes W$ is naturally isomorphic to $\otimes^{2k} V$.
However, the situation is a bit different when talking about "wedging".
Let $\eta\in \Lambda^k V$. We want to wonder about $\eta\wedge \eta$.
1) Let $W=\Lambda^k V$ and think of $\eta\wedge \eta$ as a member of $\Lambda^2 W$. Then $\eta\wedge \eta=0$ by super-commutativity of the wedge-product.
2) Think of $\eta\wedge\eta$ as a member of $\Lambda^{2k}V$. Then $\eta\wedge \eta$ may not be $0$.
Perhaps this confusion would not arise if we write $\wedge_V$ rather that $\wedge$, for when wedging we must remember the base space. Moreover, there is no such thing as taking the wedge product of two vector spaces, thought we can talk about tensor product of two vector spaces.
Admittedly, my mind is not completely clear here. Can somebody throw some more light on the different behaviours of tensoring and wedging.
• I'm not quite sure what the question is, but one nice thing about $\otimes$ is that is makes the category of vector spaces to a (symmetric) monoidal category, i.e. $\otimes$ is "associative" (and "commutative"). As you noticed, $\wedge$ is not even defined for two different vector spaces, and so is best used as a (graded-commutative) associative product on $\bigwedge V$. – user8268 Dec 30 '15 at 19:00
This is not really a direct answer but is just too long for a comment.
One curious fact about the wedge construction is that $\bigwedge^n V$ can be (functorially) realized either as a subspace of $\bigotimes^n V$ or as a quotient. (These realizations are canonically isomorphic when the characteristic of the underlying field is $0$ or greater than $n$, but otherwise are non-canonically isomorphic.)
Although the quotient construction tends to be more natural, it's often useful to think about the subspace construction. The little wedge symbol means different things, depending on your construction.
In the quotient construction, $\bigwedge^n V$ is the quotient of $\bigotimes^n V$ by the subspace generated by symbols with repeated vectors, and the symbol $v_1 \wedge \ldots \wedge v_n$ means "the image of $v_1 \otimes \ldots \otimes v_n$ under the quotient map."
In the subspace construction, on the other hand, $\bigwedge^n V$ is the subspace of $\bigotimes^n V$ on which $S_n$ acts via the sign character, and the symbol $v_1 \wedge \ldots \wedge v_n$ means either (depending on your convention and on whether $n! = 0$ in your field) $$\sum_{\sigma \in S_n} (-1)^{\text{sign}(\sigma)} v_{\sigma(1)}\otimes \ldots \otimes v_{\sigma(n)}$$ or $$\frac{1}{n!} \sum_{\sigma \in S_n} (-1)^{\text{sign}(\sigma)} v_{\sigma(1)}\otimes \ldots \otimes v_{\sigma(n)}.$$ (The second convention has the advantage that under the natural map "subspace intepretation to quotient interpretation" is compatible with the notation, and the disadvantage that it's only available in characteristic prime to $n!$).
Now let's think about $\bigwedge^2 \bigwedge^n V$ versus $\bigwedge^{2n} V$ in terms of the subspace interpretation. The former is the subspace of $\bigotimes^2 \bigwedge^n V$ on which $S_2$ acts by the sign character, which is the subspace of $\bigotimes^2 \bigotimes^n V$ on which $S_2$ and $S_n$ act independently via the sign character. Under the natural "unravelling map" $$\bigotimes^2(\bigotimes^n V) \to \bigotimes^{2n} V$$ we get the subspace of $\bigotimes^{2n} V$ on which $(S_n \times S_n) \rtimes S_2 < S_{2n}$ acts via the product of the sign characters. But this is a weaker, and different, demand than demanding that all of $S_{2n}$ act via the sign character, and so this subspace is bigger.
(EDIT: see comments for more details.) In terms of representation theory, your observation could be restated as follows: Write $G = (S_n \times S_n) \rtimes S_2.$ Then there is a natural embedding $G \hookrightarrow S_{2n}$. Now for a field $k$ there is a unique character $G \to k^\times$ which restricts to the sign character each $S_n$ and to the sign character on $S_2$. There is also a character $G \to k^\times$ using the embedding $G \hookrightarrow S_{2n}$ followed by the sign character of $S_{2n}$. These characters aren't the same.
• While I like this approach, I am not sure about the action of $S_n$ of $\bigotimes^2 \bigotimes^n V$: If we just let act $S_n$ on both copies of $\bigotimes^n V$ simultaneously via the sign representation (i.e. if we take the usual tensor product of representations), then it acts trivially on $\bigotimes^2 \bigotimes^n V$ because the sings cancel out. I feel like we should instead let $S_n \times S_n$ act on $\bigotimes^2 \bigotimes^n V$, each copy of $S_n$ on the corresponding copy of $\bigotimes^n V$. – Jendrik Stelzner Dec 30 '15 at 21:07
• Thank you. This does make things clearer. One think I am yet not able to see is how do we get the semidirect product $S_n\rtimes S_2$. – caffeinemachine Dec 31 '15 at 3:19
• @JendrikStelzner Oops, the above two comments are correct. It is $(S_n \times S_n) \rtimes S_2$ that is acting, not $S_n \rtimes S_2$ as I erroneously stated (and here it is clear what the semidirect product structure is). Then the requirement is that both $S_n$ factors individually act via the sign character, and that $S_2$ does as well. Sorry! Will edit it into the answer when I have more energy. – hunter Dec 31 '15 at 3:32
• (edit to above comment: I am confused about what the requirement is now!) – hunter Dec 31 '15 at 3:39
• Right. Each $S_n$ has to act via the sign character, and so does $S_2$. This is not the same as demanding that they act via the restriction of the sign character from the ambient $S_{2n}$ to $(S_n \times S_n) \rtimes S_2$, which is maybe a group-theoretic restatement of the phenomenon OP was observing in the first place. – hunter Dec 31 '15 at 3:50 | {"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": 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": 0.8970144391059875, "perplexity": 127.95693544871862}, "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/1573496669967.80/warc/CC-MAIN-20191119015704-20191119043704-00387.warc.gz"} |
http://mathhelpforum.com/calculus/134038-check-some-derivatives-please.html | # Math Help - Check some derivatives please.
1. ## Check some derivatives please.
Hi:
Could somebody check this for me please.
thanks you
Attached Thumbnails
2. Hello stealthmaths
Originally Posted by stealthmaths
Hi:
Could somebody check this for me please.
thanks you
They look fine to me, except that in (i) you have found $\frac{d^2y}{dx^2}$.
3. ahh yes. I put the first derivative instead of the second.
ahh wait. I still need to find the second derivative of (ii)
4. Hi:
part iib)
Attached Thumbnails
5. Hello stealthmaths
Originally Posted by stealthmaths
Hi:
part iib)
You have a sign wrong. See the attached - should be a minus sign at the point I've indicated.
I should set it out like this. First notice that it's easier to divide by $3$ at the start, and then factorise where you can:
$\tfrac{1}{3}y= \frac{x+1}{x^2-9}$
$\Rightarrow \frac13\frac{dy}{dx}= \frac{(x^2-9)(1)-(x+1)(2x)}{(x^2-9)^2}$
$=\frac{-x^2-2x-9}{(x^2-9)^2}$
$\Rightarrow \frac13\frac{d^2y}{dx^2}= \frac{(x^2-9)^2(-2x-2)-(-x^2-2x-9)2(x^2-9)(2x)}{(x^2-9)^4}$
$=2\frac{(x^2-9)^2(-x-1)+(x^2+2x+9)(x^2-9)(2x)}{(x^2-9)^4}$
$=2\frac{(x^2-9)(-x-1)+(x^2+2x+9)(2x)}{(x^2-9)^3}$
$=2\frac{-x^3-x^2+9x+9+2x^3+4x^2+18x}{(x^2-9)^3}$
$\Rightarrow \frac{d^2y}{dx^2}=\frac{6(x^3+3x^2+27x+9)}{(x^2-9)^3}$
But I get things wrong sometimes, too, so check my working!
Attached Thumbnails
I'm confused. You divide by 3 and just take the 1/3 out of the problem (on the top line you gave me)?
7. Somehow I doubt you rarely make a mistake Grandad(not unintentionally anyway)
I have gone through the work and I think it looks fine - except I'm not sure about the last line where you have a 6 - should this not be a 2?
I have marked it in red:
Thank you
8. Hello stealthmaths
I'm sure you'll kick yourself. I began by dividing both sides of the equation by $3$ to keep the numbers easier - hence the $\tfrac13y$, the $\frac13\frac{dy}{dx}$ and the $\frac13\frac{d^2y}{dx^2}$.
So at the end, I've simply multiplied by $3$ to get back to what we want, namely $\frac{d^2y}{dx^2}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "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": 0.8434153199195862, "perplexity": 905.2732680190073}, "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-2015-18/segments/1429246659319.74/warc/CC-MAIN-20150417045739-00244-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://www.maa.org/publications/periodicals/american-mathematical-monthly/american-mathematical-monthly-october-1997 | # American Mathematical Monthly: October, 1997
Click on the months above to see summaries of articles in the MONTHLY.
An archive for all the 1997 issues is now available
## October, 1997
Areas and Intersections in Convex Domains
by Norbert Peyerimhoff
[email protected]
The article grew out of the author's playing around with randomly chosen line segments in a convex domain and the probability that they intersect. It turned out that this probability is connected to an old problem posed by Sylvester: What is the probability that four independently chosen points in a convex bounded domain span a quadrilateral?
The author derives a surprising relationship between areas of particular subsets of an arbitrary bounded convex domain by interpreting Sylvester's probability in two different ways. He also considers a three-dimensional analogue of his original question: Assume a triangle and a line segment are chosen at random in the 3-dimensional unit ball. What is the probability that they intersect?
Newman's Short Proof of the Prime Number Theorem
by Don Zagier
[email protected]
The prime number theorem, now celebrating its 100th birthday, is one of the most famous results in number theory, but has always retained an aura of mystery because there were no really easy proofs: the original proofs given by Hadamard and by de la Vallee Poussin in 1896, and many subsequent variants, involved complicated and required proving non-trivial lower bounds on the size of the Riemann zeta-function on the line Re(s)=1; other analytic proofs avoided such estimates but needed instead subtle Tauberian theorems like the Wiener-Ikehara theorem; and the "elementary" proof given by Selberg and Erdos in 1949 was far more complicated than the analytic proofs and, despite various subsequent simplifications, has remained so to this day. In 1980 a very simple replacement of the needed Tauberian argument was discovered by D.J. Newman. The resulting proof of the prime number theorem is short, beautiful, and understandable without any knowledge of number theory or complex function theory beyond Cauchy's theorem; it should be known to every mathematician. The present article is intended as a step towards this goal.
An Exploratory Approach to Kaplansky's Lemma Leads to a Generalized Resultant
by David Callan
[email protected]
A generalization of the classical notion of resultant from two to several polynomials is given. An immediate consequence is Kaplansky's Lemma on linear operators for arbitrary fields, at least in the finite-dimensional case.
Metric Spaces in Which All Triangles Are Degenerate
by Bettina Richmond and Thomas Richmond
[email protected]
[email protected]
Clearly any subspace of the real line with the Euclidean metric is a metric space in which all triangles are degenerate. Can you find the other metric spaces in which all triangles are degenerate? Is the usual topology on the plane generated by any metric in which all triangles are degenerate? Is the usual topology on the real line generated by any metric which has no degenerate triangles with more than two distinct vertices? These questions are answered in our article.
Partitions of Unity for Countable Covers
by Albert Fathi
[email protected]
Existence of partitions of unity for metric spaces is usually proved using the (equivalent) concept of paracompactness. Although the standard proof of paracompactness of metric spaces is the one given by M.E. Rudin, in his 1965 PhD thesis, Michael Mather showed that it is easier, for metric spaces, to show directly the existence of locally finite partitions of unity for arbitrary covers. We adapt Mather's argument to prove existence of a partition of unity subordinated to a countable open cover of a metric space. An advantage of the method is that the same proof can be used in the smooth category.
Calculus: A Modern Perspective
by Jeff Knisley
[email protected]
Curve-fitting, cubic splines, regression lines, symbolic derivatives -- it's as if the computer was made for doing calculus. Anywhere calculus is found, a computer is not far away -- except, that is, in a calculus course. Could it be that our calculus curriculum is out of step with the science and mathematics of this century? If so, should not that curriculum be changed? And if it is to be changed, which topics should go and which should stay? This essay addresses such questions as it investigates what should constitute a modern calculus curriculum.
Pro Choice
by Paul Zorn and Arnold Ostebee
[email protected]
[email protected]
Although self-styled reformers tend to agree on a few common broad goals and tenets, such as the primacy of conceptual understanding, there is no single calculus reform "party line." On the contrary, different reformers make significantly different choices --- indeed, reformed approaches offer more diversity of choice than do traditional texts. Calculus reform is not a single alternative to a standard diet; reform offers a diverse menu of different but carefully-considered choices. The greatest lasting value of calculus reform may well be in highlighting and forcing important choices among competing goods --- choices that have always been present but too little acknowledged. In the end, choosing everything means choosing nothing.
Rethinking Calculus: Learning and Thinking
by James J. Kaput
[email protected]
This paper addresses a new level of K-16 Calculus reform by first distinguishing between that peculiar American web of habits, expectations, and structures termed "Calculus the Institution" and that splendid historical achievement, "Calculus the System of Knowledge and Technique." Calculus Reform has been about remodeling the former while leaving the larger K-12 educational structures in place, which in turn continues to deny most students (90%) access to the key aspects of the latter. We argue that a deeper reform is in order, one that deeply rethinks the subject matter of calculus in terms of learning and thinking, that makes intelligent and innovative use of potent technologies, and that integrates the mathematics of change and variation with other important mathematics across grades K-12. The author illustrates efforts in these directions through the NSF funded SimCalc Project (http://www.simcalc.umassd.edu), which is directed towards democratizing access to the mathematics of change and variation beginning in the middle grades and even earlier.
What Do We Do About Calculus? First, Do No Harm
Calculus is an old subject whose teaching has been refined through the ages. Some of the lessons of the past seem to have been forgotten. Examples are given.
The Fifty-Seventh William Lowell Putnam Mathematical Competition
by Leonard F. Klosinski, Gerald L. Alexanderson, and Loren C. Larson
[email protected]
NOTES
by Joseph Kupka
A Discrete Form of the Beckman-Quarles Theorem
by Apoloniusz Tyszka
UNSOLVED PROBLEMS
David Gale's Subset Take-Away Game
by J. Daniel Christensen and Mark Tilford
PROBLEMS AND SOLUTIONS
REVIEWS
An Introduction to Difference Equations. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8115720748901367, "perplexity": 1027.9671718450481}, "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/1408500823598.56/warc/CC-MAIN-20140820021343-00077-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://teachingcalculus.com/tag/alternating-series/ | # Infinite Sequences and Series – Unit 10
Unit 10 covers sequences and series. These are BC only topics (CED – 2019 p. 177 – 197). These topics account for about 17 – 18% of questions on the BC exam.
### Topics 10.1 – 10.2
Topic 10.1: Defining Convergent and Divergent Series.
Topic 10. 2: Working with Geometric Series. Including the formula for the sum of a convergent geometric series.
### Topics 10.3 – 10.9 Convergence Tests
The tests listed below are tested on the BC Calculus exam. Other methods are not tested. However, teachers may include additional methods.
Topic 10.3: The nth Term Test for Divergence.
Topic 10.4: Integral Test for Convergence. See Good Question 14
Topic 10.5: Harmonic Series and p-Series. Harmonic series and alternating harmonic series, p-series.
Topic 10.6: Comparison Tests for Convergence. Comparison test and the Limit Comparison Test
Topic 10.7: Alternating Series Test for Convergence.
Topic 10.8: Ratio Test for Convergence.
Topic 10.9: Determining Absolute and Conditional Convergence. Absolute convergence implies conditional convergence.
### Topics 10.10 – 10.12 Taylor Series and Error Bounds
Topic 10.10: Alternating Series Error Bound.
Topic 10.11: Finding Taylor Polynomial Approximations of a Function.
Topic 10.12: Lagrange Error Bound.
### Topics 10.13 – 10.15 Power Series
Topic 10.13: Radius and Interval of Convergence of a Power Series. The Ratio Test is used almost exclusively to find the radius of convergence. Term-by-term differentiation and integration of a power series gives a series with the same center and radius of convergence. The interval may be different at the endpoints.
Topic 10.14: Finding the Taylor and Maclaurin Series of a Function. Students should memorize the Maclaurin series for $\displaystyle \frac{1}{{1-x}}$, sin(x), cos(x), and ex.
Topic 10.15: Representing Functions as Power Series. Finding the power series of a function by, differentiation, integration, algebraic processes, substitution, or properties of geometric series.
### Timing
The suggested time for Unit 9 is about 17 – 18 BC classes of 40 – 50-minutes, this includes time for testing etc.
### Previous posts on these topics:
Before sequences
Amortization Using finite series to find your mortgage payment. (Suitable for pre-calculus as well as calculus)
A Lesson on Sequences An investigation, which could be used as early as Algebra 1, showing how irrational numbers are the limit of a sequence of approximations. Also, an introduction to the Completeness Axiom.
Everyday Series
Convergence Tests
Reference Chart
Which Convergence Test Should I Use? Part 1 Pretty much anyone you want!
Which Convergence Test Should I Use? Part 2 Specific hints and a discussion of the usefulness of absolute convergence
Good Question 14 on the Integral Test
Sequences and Series
Graphing Taylor Polynomials Graphing calculator hints
Introducing Power Series 1
Introducing Power Series 2
Introducing Power Series 3
New Series from Old 1 substitution (Be sure to look at example 3)
New Series from Old 2 Differentiation
New Series from Old 3 Series for rational functions using long division and geometric series
Geometric Series – Far Out An instructive “mistake.”
A Curiosity An unusual Maclaurin Series
Synthetic Summer Fun Synthetic division and calculus including finding the (finite)Taylor series of a polynomial.
Error Bounds
Error Bounds Error bounds in general and the alternating Series error bound, and the Lagrange error bound
The Lagrange Highway The Lagrange error bound.
What’s the “Best” Error Bound?
Review Notes
Type 10: Sequences and Series Questions
# 2019 CED Unit 10: Infinite Sequences and Series
Unit 10 covers sequences and series. These are BC only topics (CED – 2019 p. 177 – 197). These topics account for about 17 – 18% of questions on the BC exam.
### Timing
The suggested time for Unit 9 is about 17 – 18 BC classes of 40 – 50-minutes, this includes time for testing etc.
### Previous posts on these topics :
Introducing Power Series 1
# Power Series 2
This is a BC topic
Good Question 16 (11-30-2018) What you get when you substitute.
Geometric Series – Far Out (2-14-2017) A very interesting and instructive mistake
Synthetic Summer Fun (7-10-2017) Finding the Taylor series coefficients without differentiating
Error Bounds (2-22-2013) The alternating series error bound, and the Lagrange error bound
The Lagrange Highway (5-20-15) a metaphor for the error bound
REVIEW NOTES Type 10: Sequence and Series Questions (4-6-2018) A summary for reviewing sequences and series.
# What Convergence Test Should I Use? Part 2
In last Friday’s post I really didn’t answer this question. Rather, I tried to show that there is not only one convergence test that must be used on a given series. Nevertheless, the form of a series suggests a test that is likely to work. In this post, I’ll try to give some suggestions as to what test to try first based on the form of the series.
For reference, click here for a table summarizing the common convergence tests.
The goal is for students to be able to decide which test to start with at a glance.
Start with the nth-term test for divergence. If the limit of the general term as n goes to infinity is not zero, the sequence will diverge. The $\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{a}_{n}}=0$ is a necessary condition for convergence. It is not sufficient; if the limit is zero then the series may converge. Look for a convergence test.
If the series alternates plus and minus signs, it is an alternating series and if it satisfies the other hypotheses use the Alternating Series Test. If the series contains positive and negative signs that do not alternate, or one of the other hypotheses is not met, then a different test must be used.
If the series is geometric then the Geometric Series Test may be used. If the common ratio (the number multiplied by each term to get the next term) is between –1 and 1 the series converges. If the common ratio is greater than or equal to 1, or less than or equal to –1, the series diverges.
The remaining tests are for series with all positive terms. They are tests for absolute convergence. If you series has negative terms then you may ignore the signs and try one of the following tests. If your series is absolutely convergent, then it is convergent. (If not, it may still be convergent.)
If the general term (written with x’s) looks like something that you can integrate, use the Integral Test.
The Direct Comparison Test and the Limit Comparison Test are used if you can find a test to compare them with.
A p-series, $\sum\limits_{{i=1}}^{\infty }{{\frac{1}{{{{n}^{p}}}}}}$ converges if $p>1$ and diverges if $p\le 1$. A p-series is often a good test to use for comparison in the next two tests. However, any series whose convergence you are sure of may be used.
The Direct Comparison Test is used with fraction expressions. “Extra” factors in the denominator can often be ignored. Some examples
• $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{1}{{{{5}^{n}}\sqrt{n}}}}}$ would be a geometric series except for the radical. Compare it with the geometric series $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{1}{{{{5}^{n}}}}}}$
• $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{{{{n}^{2}}}}{{{{n}^{4}}+2n+1}}}}$ can be compared with the p-series $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{1}{{{{n}^{2}}}}}}$. The hint here is that ignoring the lower power terms in the denominator and reducing we see that the original series looks like $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{1}{{{{n}^{2}}}}}}$. Both series converge. But be careful $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{{{{n}^{2}}}}{{{{n}^{4}}-2n-1}}}}$ while similar, has terms greater than the terms of $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{1}{{{{n}^{2}}}}}}$.)
• The terms of the series $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{1}{{{{{\left( {{{n}^{2}}+2} \right)}}^{{1/3}}}}}}}$ are larger than the harmonic series $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{1}{n}}}$ a divergent p-series, so this series diverges.
The Limit Comparison Test may be used with the same kinds of series that are messy to use with direct comparison.
• Returning to $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{{{{n}^{2}}}}{{{{n}^{4}}+2n+1}}}}$, try the limit comparison test with $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{1}{{{{n}^{2}}}}}}$. The limit is 1, so both series converge.
• $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{1}{{\sqrt{{{{n}^{2}}+3}}}}}}$ Series with radicals also are candidates for the limit comparison test. Since the general terms is approximately $\displaystyle {\frac{1}{n}}$ Compare this with $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{1}{n}}}$. Both series diverge.
More complicated series, perhaps with exponential factors and/or factorials can be examined with the Ratio Test.
• $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{{{{3}^{n}}}}{{n!}}}}$ or $\displaystyle \sum\limits_{{i=1}}^{\infty }{{\frac{{{{n}^{3}}}}{{{{5}^{n}}}}}}$ are candidates for the Ratio Test. Both Converge.
• $\displaystyle \sum\limits_{{i=1}}^{\infty }{{{{{\left( {-1} \right)}}^{n}}\frac{{n!}}{{{{{500}}^{n}}}}}}$ appears to be a candidate for the alternating series test. However, for large values of n > 530 the terms increase in absolute vale, so the alternating series test cannot be applied. The ratio test works here, but since the terms do not approach 0 as n increases, the nth-term test for divergence also works. This series diverges.
Practice, Practice, Practice
The AP Calculus BC exams rarely, if ever, specify which test to use. Often these are multiple-choice questions. If students can see whether the series converges or diverges, that is enough. But here again the key is practice, practice, practice.
As you teach the various tests, pause to look at the form of the series in the exercises for each test that your book provides. Most books also have mixed sets of exercises where tests other than the one in that section are needed. One of the things you can do is assign these entire sets with the directions that students should determine what test they would try, and, for their comparison tests, to which series they would compare it. Discuss their opinions especially if there is more than one suggested or suggest others. Work only those those students are confused about or those for which they have divergent opinions; try to converge on a good test for each.
Revised July 18, 2021
# Which Convergence Test Should I Use? Part 1
One common question from students first learning about series is how to know which convergence test to use with a given series. The first answer is: practice, practice, practice. The second answer is that there is often more than one convergence test that can be used with a given series.
I will illustrate this point with a look at one series and the several tests that may be used to show it converges. This will serve as a review of some of the tests and how to use them. For a list of convergence tests that are required for the AP Calculus BC exam click here.
To be able to use these tests the students must know the hypotheses of each test and check that they are met for the series in question. On multiple-choice questions students do not need to how their work, but on free-response questions (such as checking the endpoints of the interval of convergence of a Taylor series) they should state them and say that the series meets them.
For our example we will look at the series $\displaystyle 1-\frac{1}{3}+\frac{1}{9}-\frac{1}{{27}}+\frac{1}{{81}}-+\ldots =\sum\limits_{{n=1}}^{\infty }{{{{{\left( {-\frac{1}{3}} \right)}}^{{n-1}}}}}$
Spoiler: Except for the first two tests to be considered, the other tests are far more work than is necessary for this series. The point is to show that several tests may be used for a given series, and to practice the other tests.
The Geometric Series Test is the obvious test to use here, since this is a geometric series. The common ratio is (–1/3) and since this is between –1 and 1 the series will converge.
The Alternating Series Test (the Leibniz Test) may be used as well. The series alternates signs, is decreasing in absolute value, and the limit of the nth term as n approaches infinity is 0, therefore the series converges.
The Ratio Test is used extensively with power series to find the radius of convergence, but it may be used to determine convergence as well. To use the test, we find
$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{\left| {{{{\left( {-\frac{1}{3}} \right)}}^{{n+1}}}} \right|}}{{\left| {{{{\left( {-\frac{1}{3}} \right)}}^{n}}} \right|}}=\frac{1}{3}$ Since the limit is less than 1, we conclude the series converges.
Absolute Convergence
A series, $\sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}$, is absolutely convergent if, and only if, the series $\sum\limits_{{n=1}}^{\infty }{{\left| {{{a}_{n}}} \right|}}$ converges. In other words, if you make all the terms positive, and that series converges, then the original series also converges. If a series is absolutely convergent, then it is convergent. (A series that converges but is not absolutely convergent is said to be conditionally convergent.)
The advantage of going for absolute convergence is that we do not have to deal with the negative terms; this allows us to use other tests.
Applied to our example, if the series $\sum\limits_{{n=1}}^{\infty }{{{{{\left( {\frac{1}{3}} \right)}}^{{n-1}}}}}$ converges, then our series $\sum\limits_{{n=1}}^{\infty }{{{{{\left( {-\frac{1}{3}} \right)}}^{{n-1}}}}}$ will converge absolutely and converge.
The Geometric Series Test can be used again as above.
The Integral Test says if the improper integral $\displaystyle {{\int_{1}^{\infty }{{\left( {\frac{1}{3}} \right)}}}^{x}}dx$ converges, then our original series will converge absolutely.
$\displaystyle \int\limits_{1}^{\infty }{{{{{\left( {\frac{1}{3}} \right)}}^{x}}}}dx=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\int\limits_{1}^{n}{{{{{\left( {\frac{1}{3}} \right)}}^{x}}}}dx=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {\frac{{{{{\left( {\frac{1}{3}} \right)}}^{n}}}}{{\ln \left( {1/3} \right)}}-\frac{{{{{\left( {\frac{1}{3}} \right)}}^{1}}}}{{\ln \left( {1/3} \right)}}} \right)=0-\frac{{1/3}}{{\ln \left( {1/3} \right)}}$
$\displaystyle =-\frac{{1/3}}{{\ln \left( {1/3} \right)}}>0$ since ln(1/3) < 0.
The limit is finite, so our series converges absolutely, and therefore converges.
The Direct Comparison Test may also be used. We need to find a positive convergent series whose terms are term-by-term greater than the terms of our series. The geometric series $\sum\limits_{{n=1}}^{\infty }{{{{{\left( {\frac{1}{2}} \right)}}^{{n-1}}}}}$ meets these two requirements. Therefore, the original series converges absolutely and converges.
The Limit Comparison Test is another possibility. Here we need a positive series that converges; we can use $\sum\limits_{{n=1}}^{\infty }{{{{{\left( {\frac{1}{2}} \right)}}^{{n-1}}}}}$ again. We look at
$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{{{{\left( {1/3} \right)}}^{n}}}}{{{{{\left( {1/2} \right)}}^{n}}}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {\frac{2}{3}} \right)}^{n}}=0$ and since the series in the denominator converges, our series converges absolutely.
So, for this example all the convergences that may be tested on the AP Calculus BC exam may be used with the single exception of the p-series Test which cannot be used with this series.
Teaching suggestions
1. While the convergence of the series used here can be done all these ways, other series lend themselves to only one. Stress the form of the series that works with each test. For example, the Limit Comparison Test is most often used for rational expressions with the numerator of lower degree than the denominator and for expressions involving radicals of polynomials. The comparison is made with a p-series of whatever degree will make the numerator and denominator the same degree allowing the limit to be found.
2. Most textbooks, after explaining each test and giving exercises on them, include a series of mixed exercises that require all the test covered up to that point. A good way to use this set is to assign students to state which test they would try first on each series. Discuss the opinions of the class and work any questions that students are unsure of or on which several ways are suggested.
3. Give your students the series above, or a similar one, and have them prove its convergence using each of the convergence tests as was done above.
4. Divide your class into groups and assign each group the series and one of the convergence tests. Ask them to use the test to prove convergence and then discuss the results as a group.
Of course, I didn’t really answer the question, did I? Check What Convergence Test Should I use Part 2
Updated February 23, 2013
# Error Bounds
Whenever you approximate something, you should be concerned about how good your approximation is. The error, E, of any approximation is defined to be the absolute value of the difference between the actual value and the approximation. If Tn(x) is the Taylor/Maclaurin approximation of degree n for a function f(x) then the error is $E=\left| f\left( x \right)-{{T}_{n}}\left( x \right) \right|$. This post will discuss the two most common ways of getting a handle on the size of the error: the Alternating Series error bound, and the Lagrange error bound.
Both methods give you a number B that will assure you that the approximation of the function at $x={{x}_{0}}$ in the interval of convergence is within B units of the exact value. That is,
$\left( f\left( {{x}_{0}} \right)-B \right)<{{T}_{n}}\left( {{x}_{0}} \right)<\left( f\left( {{x}_{0}} \right)+B \right)$
or
${{T}_{n}}\left( {{x}_{0}} \right)\in \left( f\left( {{x}_{0}} \right)-B,\ f\left( {{x}_{0}} \right)+B \right)$.
Stop for a moment and consider what that means: $f\left( {{x}_{0}} \right)-B$ and $f\left( {{x}_{0}} \right)+B$ are the endpoints of an interval around the actual value and the approximation will lie in this interval. Ideally, B is a small (positive) number.
Alternating Series
If a series $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ alternates signs, decreases in absolute value and $\underset{n\to \infty }{\mathop{\lim }}\,\left| {{a}_{n}} \right|=0$ then the series will converge. The terms of the partial sums of the series will jump back and forth around the value to which the series converges. That is, if one partial sum is larger than the value, the next will be smaller, and the next larger, etc. The error is the difference between any partial sum and the limiting value, but by adding an additional term the next partial sum will go past the actual value. Thus, for a series that meets the conditions of the alternating series test the error is less than the absolute value of the first omitted term:
$\displaystyle E=\left| \sum\limits_{k=1}^{\infty }{{{a}_{k}}}-\sum\limits_{k=1}^{n}{{{a}_{k}}} \right|<\left| {{a}_{n+1}} \right|$.
Example: $\sin (0.2)\approx (0.2)-\frac{{{(0.2)}^{3}}}{3!}=0.1986666667$ The absolute value of the first omitted term is $\left| \frac{{{(0.2)}^{5}}}{5!} \right|=0.26666\bar{6}\times {{10}^{-6}}$. So our estimate should be between $\sin (0.2)\pm 0.266666\times {{10}^{-6}}$ (that is, between 0.1986666641 and 0.1986719975), which it is. Of course, working with more complicated series, we usually do not know what the actual value is (or we wouldn’t be approximating). So an error bound like $0.26666\bar{6}\times {{10}^{-6}}$ assures us that our estimate is correct to at least 5 decimal places.
The Lagrange Error Bound
Taylor’s Theorem: If f is a function with derivatives through order n + 1 on an interval I containing a, then, for each x in I , there exists a number c between x and a such that
$\displaystyle f\left( x \right)=\sum\limits_{k=1}^{n}{\frac{{{f}^{\left( k \right)}}\left( a \right)}{k!}{{\left( x-a \right)}^{k}}}+\frac{{{f}^{\left( n+1 \right)}}\left( c \right)}{\left( n+1 \right)!}{{\left( x-a \right)}^{n+1}}$
The number $\displaystyle R=\frac{{{f}^{\left( n+1 \right)}}\left( c \right)}{\left( n+1 \right)!}{{\left( x-a \right)}^{n+1}}$ is called the remainder.
The equation above says that if you can find the correct c the function is exactly equal to Tn(x) + R. Notice the form of the remainder is the same as the other terms, except it is evaluated at the mysterious c. The trouble is we almost never can find the c without knowing the exact value of f(x), but; if we knew that, there would be no need to approximate. However, often without knowing the exact values of c, we can still approximate the value of the remainder and thereby, know how close the polynomial Tn(x) approximates the value of f(x) for values in x in the interval, i.
Corollary – Lagrange Error Bound.
$\displaystyle \left| \frac{{{f}^{\left( n+1 \right)}}\left( c \right)}{\left( n+1 \right)!}{{\left( x-a \right)}^{n+1}} \right|\le \left( \text{max}\left| {{f}^{\left( n+1 \right)}}\left( x \right) \right| \right)\frac{{{\left| x-a \right|}^{n+1}}}{\left( n+1 \right)!}$
The number $\displaystyle \left( \text{max}\left| {{f}^{\left( n+1 \right)}}\left( x \right) \right| \right)\frac{{{\left| x-c \right|}^{n+1}}}{\left( n+1 \right)!}\ge \left| R \right|$ is called the Lagrange Error Bound. The expression $\left( \text{max}\left| {{f}^{\left( n+1 \right)}}\left( x \right) \right| \right)$ means the maximum absolute value of the (n + 1) derivative on the interval between the value of x and c. The corollary says that this number is larger than the amount we need to add (or subtract) from our estimate to make it exact. This is the bound on the error. It requires us to, in effect, substitute the maximum value of the n + 1 derivative on the interval from a to x for ${{f}^{(n+1)}}\left( x \right)$. This will give us a number equal to or larger than the remainder and hence a bound on the error.
Example: Using the same example sin(0.2) with 2 terms. The fifth derivative of $\sin (x)$ is $-\cos (x)$ so the Lagrange error bound is $\displaystyle \left| -\cos (0.2) \right|\frac{\left| {{\left( 0.2-0 \right)}^{5}} \right|}{5!}$, but if we know the cos(0.2) there are a lot easier ways to find the sine. This is a common problem, so we will pretend we don’t know cos(0.2), but whatever it is its absolute value is no more than 1. So the number $\left( 1 \right)\frac{\left| {{\left( 0.2-0 \right)}^{5}} \right|}{5!}=2.6666\bar{6}\times {{10}^{-6}}$ will be larger than the Lagrange error bound, and our estimate will be correct to at least 5 decimal places.
This “trick” is fairly common. If we cannot find the number we need, we can use a value that gives us a larger number and still get a good handle on the error in our approximation.
FYI: $\displaystyle \left| -\cos (0.2) \right|\frac{\left| {{\left( 0.2-0 \right)}^{5}} \right|}{5!}\approx 2.61351\times {{10}^{-6}}$
Corrected: February 3, 2015, June 17, 2022 | {"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": 58, "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": 0.9372532367706299, "perplexity": 381.57765156937955}, "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/1669446711552.8/warc/CC-MAIN-20221209213503-20221210003503-00088.warc.gz"} |
https://rbspgway.jhuapl.edu/biblio?keyword=wind | # Bibliography
## Found 1 entries in the Bibliography.
### Showing entries from 1 through 1
2017 Model-observation comparison for the geographic variability of the plasma electric drift in the Earth\textquoterights innermost magnetosphere Plasmaspheric rotation is known to lag behind Earth rotation. The causes for this corotation lag are not yet fully understood. We have used more than two years of Van Allen Probe observations to compare the electric drift measured below L~2 with the predictions of a general model. In the first step, a rigid corotation of the ionosphere with the solid Earth was assumed in the model. The results of the model-observation comparison are twofold: (1) radially, the model explains the average observed geographic variability of the ... Lejosne, ène; Maus, Stefan; Mozer, F.; YEAR: 2017 DOI: 10.1002/2017GL074862
1 | {"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": 0.8718240857124329, "perplexity": 3278.8248782948817}, "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-2021-21/segments/1620243991378.52/warc/CC-MAIN-20210515192444-20210515222444-00108.warc.gz"} |
https://arxiv-export-lb.library.cornell.edu/abs/2108.09585 | math.OC
(what is this?)
# Title: Sequential Stochastic Optimization in Separable Learning Environments
Abstract: We consider a class of sequential decision-making problems under uncertainty that can encompass various types of supervised learning concepts. These problems have a completely observed state process and a partially observed modulation process, where the state process is affected by the modulation process only through an observation process, the observation process only observes the modulation process, and the modulation process is exogenous to control. We model this broad class of problems as a partially observed Markov decision process (POMDP). The belief function for the modulation process is control invariant, thus separating the estimation of the modulation process from the control of the state process. We call this specially structured POMDP the separable POMDP, or SEP-POMDP, and show it (i) can serve as a model for a broad class of application areas, e.g., inventory control, finance, healthcare systems, (ii) inherits value function and optimal policy structure from a set of completely observed MDPs, (iii) can serve as a bridge between classical models of sequential decision making under uncertainty having fully specified model artifacts and such models that are not fully specified and require the use of predictive methods from statistics and machine learning, and (iv) allows for specialized approximate solution procedures.
Comments: 30 pages (Main), 12 pages (Figures, References, Appendices), 5 figures Subjects: Optimization and Control (math.OC); Machine Learning (cs.LG); Machine Learning (stat.ML) Cite as: arXiv:2108.09585 [math.OC] (or arXiv:2108.09585v1 [math.OC] for this version)
## Submission history
From: Reid Bishop [view email]
[v1] Sat, 21 Aug 2021 21:29:04 GMT (746kb,D)
Link back to: arXiv, form interface, contact. | {"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": 0.837847888469696, "perplexity": 1840.236788358203}, "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/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00052.warc.gz"} |
http://www.math.tifr.res.in/putabstract.php?date=2011-01-06 | Colloquium abstracts
H'el`ene Esnault
University of Essen, Germany
January 6, 2011
Rational Points and Motivic Integration: We give a positive answer to a question by Serre over char. 0 strictly henselian fields asserting the existence of fixed rational points of a finite $\ell$-group action on the affine space, using motivic integration. Here $\ell$ is different from the residue characteristic. (Joint with Johannes Nicaise) | {"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": 0.9244410395622253, "perplexity": 1441.0858521264572}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583515555.58/warc/CC-MAIN-20181022222133-20181023003633-00385.warc.gz"} |
http://proceedings.mlr.press/v51/sadhanala16.html | # Graph Sparsification Approaches for Laplacian Smoothing
Veeru Sadhanala, Yu-Xiang Wang, Ryan Tibshirani ;
Proceedings of the 19th International Conference on Artificial Intelligence and Statistics, PMLR 51:1250-1259, 2016.
#### Abstract
Given a statistical estimation problem where regularization is performed according to the structure of a large, dense graph G, we consider fitting the statistical estimate using a \it sparsified surrogate graph \mathbfG, which shares the vertices of G but has far fewer edges, and is thus more tractable to work with computationally. We examine three types of sparsification: spectral sparsification, which can be seen as the result of sampling edges from the graph with probabilities proportional to their effective resistances, and two simpler sparsifiers, which sample edges uniformly from the graph, either globally or locally. We provide strong theoretical and experimental results, demonstrating that sparsification before estimation can give statistically sensible solutions, with significant computational savings. | {"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": 0.8936296105384827, "perplexity": 1350.6740179685949}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039744368.74/warc/CC-MAIN-20181118114534-20181118140534-00289.warc.gz"} |
https://www.universetoday.com/1204/the-milky-way-and-the-seven-dwarfs/ | # The Milky Way and the Seven Dwarfs
There’s no easy way to put this, our home galaxy is a killer. It’s torn up galaxies in the past, and it’s going to do it again in the future. Each galaxy we consume makes us larger. If you need evidence that this is still going on, you only need to look at the conveyor belt of dwarf galaxies orbiting the Milky Way; each of which will eventually get torn apart, its stars assimilated.
Researchers have turned up seven (or maybe eight) new dwarf galaxies orbiting the Milky Way. They’re relatively intact now, but they’ll eventually spiral into our galaxy and get torn apart by gravity. Each dwarf galaxy contains a million stars at most, and a few already appear on the edge of being torn apart.
The discovery was made by analyzing groups of stars in the Sloan Digital Sky Survey. This is a detailed database of images covering a fifth of the sky that astronomers can use for their research.
Astronomers theorize that most, or maybe all of the stars in the galactic halo and disk of the Milky Way started out in smaller dwarf galaxies, which then merged together to form the Milky Way. All these dwarf galaxies are just leftovers from that galactic feast.
Original Source: PSU News Release | {"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": 0.8104259967803955, "perplexity": 868.1221371552248}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00736.warc.gz"} |
http://en.wikibooks.org/wiki/Electrodynamics/Magnetization | # Electrodynamics/Magnetization
## Magnetization Field
We can define two new vector fields to help aide us in further studies of magnetics. The first new field is the magnetization field M. This field represents the magnetic moment per unit volume of a magnetized substance. The second, H is known occasionally as the "magnetic field strength" and is used here primarily as a simplifying agent.
## Magnetization
We can define the magnetic moment in terms of the magnetization as:
$\mathbf{m} = \mathbf{M}(\mathbf{r}')dV'$
$\mathbf{A}(\mathbf{r}') = -\int \mathbf{M}(\mathbf{r})' \times \nabla \frac{1}{|\mathbf{r} - \mathbf{r}'|}dV'$
$\mathbf{A}(\mathbf{r}) = \int \frac{\nabla \times \mathbf{M}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}dV'$
$\mathbf{j}_M = \nabla \times \mathbf{M}$
## H Field
$\mathbf{H} = \mathbf{B} - 4\pi\mathbf{M}$
$\nabla \times \mathbf{H} = 4 \pi \mathbf{j}_M$
$\oint_C \mathbf{H} \cdot d\mathbf{l} = 4 \pi \int_C\mathbf{j}_F \cdot \mathbf{n}dA$ | {"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": 7, "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": 0.9574143290519714, "perplexity": 1538.8469729287565}, "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-2014-10/segments/1393999654758/warc/CC-MAIN-20140305060734-00059-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/66418/tilde-over-a-symbol-phi/66423 | # tilde over a symbol Phi
Please suggest me a way to put tilde over a symbol phi. I can easily put it over a letter but coming to the case of symbols, it is giving problems. Thanks in advance.
• What about $\tilde{\phi}$? – Claudio Fiandrino Aug 9 '12 at 6:39
• More generally: accents in math mode use different commands to accents in text mode. – Niel de Beaudrap Aug 9 '12 at 9:07
Does \widetilde{\phi} work for you?
If you enter this in http://www.codecogs.com/latex/eqneditor.php, it produces a nice Phi with Tilde.
• In my case, I wanted a tilde over an M_{A}. I had to use \widetilde{M}_{A} so that the tilde is the full width of the M without extending over the subscript as well. – Paul Mar 18 '14 at 23:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9313512444496155, "perplexity": 1509.7751014062555}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347391309.4/warc/CC-MAIN-20200526191453-20200526221453-00460.warc.gz"} |
https://socratic.org/questions/what-is-the-ph-of-a-073-m-hcl-solution | Chemistry
Topics
# What is the pH of a .073 M HCl solution?
Aug 13, 2016
The solution has a pH of 1.14
#### Explanation:
Since HCl is a strong acid, meaning that it completely ionizes when dissolved in water, the pH can be obtained directly from the concentration using the formula below:
Take the -logarithm of the concentration of hydronium ions that are in the solution:
$p H = - \log \left(0.073 M\right) = 1.14$
##### Impact of this question
778 views around the world
You can reuse this answer | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "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": 0.925934374332428, "perplexity": 2963.066227047251}, "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-2020-40/segments/1600400221382.33/warc/CC-MAIN-20200924230319-20200925020319-00332.warc.gz"} |
https://arxiv.org/abs/1705.07200 | cs.GT
(what is this?)
# Title: Smoothed and Average-case Approximation Ratios of Mechanisms: Beyond the Worst-case Analysis
Abstract: The approximation ratio has become one of the dominant measures in mechanism design problems. In light of analysis of algorithms, we define the \emph{smoothed approximation ratio} to compare the performance of the optimal mechanism and a truthful mechanism when the inputs are subject to random perturbations of the worst-case inputs, and define the \emph{average-case approximation ratio} to compare the performance of these two mechanisms when the inputs follow a distribution. For the one-sided matching problem, \citet{FFZ:14} show that, amongst all truthful mechanisms, \emph{random priority} achieves the tight approximation ratio bound of $\Theta(\sqrt{n})$. We prove that, despite of this worst-case bound, random priority has a \emph{constant smoothed approximation ratio}. This is, to our limited knowledge, the first work that asymptotically differentiates the smoothed approximation ratio from the worst-case approximation ratio for mechanism design problems. For the average-case, we show that our approximation ratio can be improved to $1+e$. These results partially explain why random priority has been successfully used in practice, although in the worst case the optimal social welfare is $\Theta(\sqrt{n})$ times of what random priority achieves. These results also pave the way for further studies of smoothed and average-case analysis for approximate mechanism design problems, beyond the worst-case analysis.
Comments: This paper is to appear in the 42nd International Symposium on Mathematical Foundations of Computer Science (MFCS 2017) Subjects: Computer Science and Game Theory (cs.GT) Cite as: arXiv:1705.07200 [cs.GT] (or arXiv:1705.07200v2 [cs.GT] for this version)
## Submission history
From: Jie Zhang [view email]
[v1] Fri, 19 May 2017 21:41:53 GMT (21kb)
[v2] Wed, 21 Jun 2017 21:10:31 GMT (21kb) | {"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": 0.8767284154891968, "perplexity": 1461.9985662221327}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812913.37/warc/CC-MAIN-20180220070423-20180220090423-00282.warc.gz"} |
https://www.physicsforums.com/threads/mass-of-soil-scooped.748203/ | Mass of soil scooped
1. Apr 11, 2014
ExoP
1. The problem statement, all variables and given/known data
Hi,
I really need help with a problem. I have a plate and need to calculate how much soil it can dig.
2. Relevant equations
3. The attempt at a solution[/b
I first calculated the volume of the scoop (a square scoop) and after that I'm stuck. Anyone have an idea?
Volume of scoop: 0.009 m3 (I calculated this)
density of soil: 1400 kg/m3
How much soil can it hold?
2. Apr 11, 2014
BvU
If the volume were 1 m3, how many kg of soil would go in ?
If it were 0.1 m3 ? 0.2 ? 0.01 ? 0.009 ?
How can a square scoop have a volume ?
3. Apr 11, 2014
BvU
Hey, you were my 1000 th p: ! Congrats!
4. Apr 11, 2014
ExoP
That's true. Well, imagine a box with no top. It has a thickness, width, length and height.
From that, how am I going to calculate how much soil it can contain? Maybe should start with that
hehe, np
5. Apr 11, 2014
BvU
I still don't understand where you encounter a problem here: You have a volume and you have a weight per unit volume, so calculating the weight of this particular volume is straightforward. Right ?
If 1 m3 weighs 1400 kg, how much does 2 m3 weigh? And 0.5 m3 ? 0.1 m3 ? 0.2 ? 0.01 ? 0.009 ?
Draft saved Draft deleted
Similar Discussions: Mass of soil scooped | {"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": 0.9442816376686096, "perplexity": 3181.888933296731}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084886739.5/warc/CC-MAIN-20180116204303-20180116224303-00101.warc.gz"} |
https://encyclopediaofmath.org/wiki/Prenex_formula | # Prenex formula
A formula from the restricted predicate calculus having the form
$$Q _ {1} x _ {1} \dots Q _ {n} x _ {n} \Psi ,$$
where $Q _ {i}$ denotes the universal quantifier $\forall$ or the existential quantifier $\exists$, the variables $x _ {i} , x _ {j}$ are distinct for $i \neq j$, and $\psi$ is a formula without quantifiers. Prenex formulas are also called prenex normal forms or prenex forms.
For each formula $\phi$ of the language of the restricted predicate calculus there is a prenex formula that is logically equivalent to $\phi$ in the classical predicate calculus. The procedure of finding a prenex formula is based on the following equivalences, which can be deduced in the classical predicate calculus:
$$( \forall x \phi ( x) \supset \Psi ) \equiv \ \exists x ^ \prime ( \phi ( x ^ \prime ) \supset \Psi ) ,$$
$$\exists x \phi ( x) \supset \Psi \equiv \ \forall x ^ \prime ( \phi ( x ^ \prime ) \supset \Psi ) ,$$
$$\Psi \supset \forall x \phi ( x) \equiv \ \forall x ^ \prime ( \Psi \supset \phi ( x ^ \prime ) ) ,$$
$$\Psi \supset \exists x \phi ( x) \equiv \ \exists x ^ \prime ( \Psi \supset \phi ( x ^ \prime ) ) ,$$
$$\neg \forall x \phi \equiv \exists x \neg \phi ,\ \ \neg \exists x \phi \equiv \forall x \neg \phi ,$$
$$Q y \forall x \phi \equiv \forall x \phi ,\ Q y \exists x \phi \equiv \exists x \phi ,$$
where $x ^ \prime$ is any variable not appearing as a free variable in $\phi ( x)$ or $\psi$, and $\phi ( x ^ \prime )$ can be obtained from $\phi ( x)$ by changing all free appearances of $x$ to $x ^ \prime$; the variable $y$ does not appear as a free variable in $\forall x \phi$ or $\exists x \phi$. To use the above equivalences one has to first express all logical operators by $\supset$ and $\neg$ and then move all quantifiers to the left by applying the equivalences. The prenex formula thus obtained is called the prenex form of the given formula.
#### References
[1] E. Mendelson, "Introduction to mathematical logic" , v. Nostrand (1964) MR0164867 Zbl 1173.03001 Zbl 0915.03002 Zbl 0681.03001 Zbl 0534.03001 Zbl 0498.03001 Zbl 0192.01901 | {"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": 0.9238925576210022, "perplexity": 214.90028335259896}, "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-20220926020051-20220926050051-00634.warc.gz"} |
http://mathhelpforum.com/calculus/83743-solving-complex-variables-equation.html | # Math Help - Solving a complex variables equation
1. complex numbers aren't my strongest side of math.
It says solve this, I have no idea what to do
$\left(\frac{x-i}{x+i}\right)^3+\left(\frac{x-i}{x+i}\right)^2+\frac{x-i}{x+i}+1=0$
2. Originally Posted by hlolli
ok thank you, I was nervous that it was possible to simplify this to Euler formula or other hard complex formula. I appreciate all your help.
Solve this $z^3 +z^2 + z +1 =0$
It appears that $x=-1,0,1$ works.
3. hahahahah
Been doing the algebra for about 7 minutes on the computer (It's part of a final assignment)
But I likes this advise better also! It's no harm done, maybe I can ask you one more question or two later instead
thanks again | {"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": 0.8173823356628418, "perplexity": 1223.703224382126}, "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/1440644065324.41/warc/CC-MAIN-20150827025425-00205-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://spinningnumbers.org/a/preparing-to-study-ee.html | A learner asked, “What math and science prerequisites and skills might be considered the minimum for this electrical engineering course?”
Good question! Here is a summary of the math and science preparation that will help you have the best experience learning the electrical engineering topics taught here. The links take you to the relevant topics on Khan Academy.
## Trigonometry
• Definitions of sine, cosine, and tangent from the sides of a triangle.
There’s lots more at trigonometry.
SOH CAH TOA
Here's a way to remember the definitions of $\sin$, $\cos$, and $\tan$.
$\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}\qquad \textbf S\text{ine is } \textbf O\text{pposite over } \textbf H\text{ypotenuse}$
$\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}\qquad \textbf C\text{osine is } \textbf A\text{djacent over } \textbf H\text{ypotenuse}$
$\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}\quad\qquad \textbf T\text{angent is } \textbf O\text{pposite over } \textbf A\text{djacent}$
## A few beginning concepts from calculus
It really helps to get to know these two ideas from the start of calculus. You don’t have to become an expert, but check out these links to get a basic idea of a what a derivative is.
Notation for derivatives
#### d notation
A popular derivative notation developed by Gottfried Leibniz is $\bold{d}$ notation. If $y$ is some function of the variable $x$, meaning $y = f(x)$, then the derivative of $y$ with respect to variable $x$ is
$\dfrac{dy}{dx}$
When you say it out loud, say it like this, "$dy\:dx$", not "$dy$ over $dx$".
The style of Leibniz's notation gives us a hint that derivatives can be treated like fractions. This comes up when you study the **chain rule**. You will also hear this called **differential notation**, where the individual terms $dy$ and $dx$ are called **differentials**.
You can write Leibniz's notation to make $\dfrac{d}{dx}$ look like an operator, like this $\dfrac{d}{dx} \,y$.
Second-order and higher derivatives using Leibniz notation will remind you of exponent notation:
The second derivative $\left( \dfrac{d}{dx}\right )^2 y\quad$ is the same as $\quad \dfrac{d^2 y}{dx^2}$
Fun fact: Leibnitz also invented the elongaged $\int$ we use for the integral symbol.
#### prime notation
The prime notation was introduced by Joseph-Louis Lagrange. The function $f^\prime(x)$ stands for the first derivative of $f(x)$ with respect to $x$. Say this as "f prime of $x$." If $y = f(x)$, then $y^\prime = f^\prime(x)$.
To indicate second-order and higher derivatives you just add prime symbols. For example, the second derivative of $y$ with respect to $x$ is written as
$y^{\prime\prime}(x)$
#### dot notation
Isaac Newton gave us dot notation where the derivative of $x$ is written as $\dot{x}$.
Say this as "$x$ dot."
These math fundamentals, plus this little bit of terminology from calculus will get you all the way through the sections on DC circuits and circuit analysis methods.
## Calculus
When you move beyond resistor circuits and include capacitors or inductors, these circuits change with time. We need to use the beautiful methods of calculus to get meaningful solutions. You don’t need to have a complete calculus background to get started, but it becomes more and more helpful as you go along. You can think of calculus as a corequisite in parallel with electrical engineering. Many students learn calculus at the same time as introductory electrical engineering classes.
These are the calculus concepts we use in electrical engineering:
Differential equations: When we need to solve first-order differential equations, we will walk through the solution step by step (example: the RC natural response), so you don’t need to have already studied differential equations. The most advanced problems involve second-order differential equations, and again, we go through the solution step by step.
Electrostatics: The electrostatics section is most advanced topics covered here in electrical engineering. This sequence develops precise definitions of electric field and voltage. My goal is to have you appreciate (but not recreate) how calculus helps us define the meaning of voltage and the electric fields near a point, line, and plane of charge.
## Physics
High school physics:
## Chemistry
High school chemistry:
## Classics
Engineering equations make more sense if you recognize the Greek alphabet.
alpha, beta, gamma, ...
## Summary
Welcome to the study of electrical engineering. Good luck! | {"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": 0.9477241039276123, "perplexity": 595.6919592859211}, "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/1524125936833.6/warc/CC-MAIN-20180419091546-20180419111546-00368.warc.gz"} |
https://fullnitrous.com/posts/FxUlGM5JxG8Gpn16PEAo/ | # Inverted Pendulum — Math and Simulation
The mathematics and computer code for simulating an inverted pendulum.
May 16, 2019
# Introduction
The inverted pendulum is a physical system which can be defined by its equations of motion. The equation of motion (singular) in this case, is a differential equation. More specifically, a second order non-linear ordinary differential equation. The solution to this equation is a function which takes time as its argument and returns an angle as its return value. Theoretically, one would just solve the equation and calculate the angle at specific time points, thus simulate the system. Practically, this is a bad choice for several reasons, mainly difficulty of solving the equation and simulation performance. This becomes even more prevalent for more complex systems like simulating a rocket for which the inverted pendulum is the precursor.
# Solution
The alternative to solving the differential equations is to use a numerical approach instead. This has some disadvantages to the purely mathematical approach but which are insignificant when simulating which is why it is a good choice. The properties of using a purely mathematical approach are listed below.
• The calculations are exact without an error margin
• Avoids a state system – Calculations are independent of each other
• Non-continuous calculations are significantly faster
• The solutions can be extremely difficult to find or impossible
• Continuous calculations are usually slower
While the approach for the numerical approach has the following properties.
• Continuous calculations are usually faster
• The solutions are extremely simple to find
• Requires a state system – Calculations depend on each other
• Non-continuous calculations are extremely slow
## Discrete Derivative
### First Order
In order to use the numerical approach one must first understand what a discrete derivative is. A discrete derivative is simply a derivative which is finite and calculated as it is defined. Below is the discrete derivative for any first order derivative function.
$y' = \frac{y_{i + 1} - y_i}{h}$
The purpose of the expression above is to be substituted into a first order differential equation such as the following.
$y' = f(y)$
Which can be numerically solved by simple substitution.
$\frac{y_{i + 1} - y_i}{h} = f(y)$
$y_{i + 1} - y_i = f(y)h$ $y_{i + 1} = f(y)h + y_i$
The above is equivalent to the solution of the differential $$y(x)$$. The variable $$y_{i + 1}$$ and $$y_i$$ are the next and current value from the function $$y(x)$$ respectively.
### Second Order
As shown before the discrete derivative is equivalent to the solution to a differential. The same can be done for a second order derivative.
$y'' = \frac{\frac{y_{i + 1} - y_i}{h} - \frac{y_i - y_{i - 1}}{h}}{h}$
$y'' = \frac{\frac{y_{i + 1} - y_i - (y_i - y_{i - 1})}{h}}{h}$
$y'' = \frac{y_{i + 1} - y_i - y_i + y_{i - 1}}{h^2}$
$y'' = \frac{y_{i + 1} - 2y_i + y_{i - 1}}{h^2}$
A simple substitution can be done for any second order differential as below.
$y'' = f(y)$
$\frac{y_{i + 1} - 2y_i + y_{i - 1}}{h^2} = f(y)$
$y_{i + 1} - 2y_i + y_{i - 1} = f(y)h^2$
$y_{i + 1} = f(y)h^2 - y_{i - 1} + 2y_i$
In the case of a second order differential a third term $$y_{i - 1}$$ is added. This simply the previous value of $$y(x)$$.
### Summary
Differential Discrete Solution
$$y' = f(y)$$ $$y_{i + 1} = f(y)h + y_i$$
$$y'' = f(y)$$ $$y_{i + 1} = f(y)h^2 - y_{i - 1} + 2y_i$$
The solution will look a bit different depending on how the differential looks but the steps are the same as presented before.
# Simulation
In this section the equations of motion will be derived, code will be written, and a simulation will be executed. Below is a diagram of the inverted pendulum of interest.
The arm of the inverted pendulum is defined to be massless and the mass at the end is defined to be a single point mass. There is no joint friction as well as no air resistance.
## Equations of Motion (E.O.M)
The following calculations can be made given the diagram of the inverted pendulum.
$F = mg \leftarrow [F_g = mg]$
$sin(\theta) = \frac{x}{F} = \frac{x}{mg} \leftarrow \left[sin(v) = \frac{opposite}{hypotenuse}\right]$
$x = sin(\theta)mg$
$\tau = Lx \leftarrow [\tau = r \times F]$
$I = mL^2 \leftarrow [I = mr^2]$
$Lx = mL^2\alpha \leftarrow [\tau = I\alpha]$
\begin{aligned} \alpha &= \frac{Lx}{mL^2} \\ &= \frac{Lsin(\theta)mg}{mL^2} \\ &= \frac{sin(\theta)g}{L} \\ &= \frac{g}{L}sin(\theta) \\ \end{aligned}
If we denote the angular acceleration $$\alpha$$ as the double derivative of the angle – $$\ddot{\theta}$$ the following equation of motion can be obtained for the inverted pendulum.
$\ddot{\theta} = \frac{g}{L}sin(\theta)$
The steps to solve the equation above are extremely complicated and so is the solution itself. However, there is one step to get the discrete solution, by simple substitution.
$y_{i + 1} = \left(\frac{g}{L}sin(\theta)\right)h^2 - y_{i - 1} + 2y_i$
## Programming
Since this article exists on a webpage it would be nice to have a live example of the simulation. Therefore the webstack will be used to create the simulation. The programming language which will be used is JavaScript which is easy to read and easy to translate to any other language. Regarding a simulation which is needed for actual research purposes a language like JavaScript is not a valid choice. The only accepted programming languages are either C or C++. Choosing anything else only means you do not know C or C++. By the way, fuck off with Rust.
Proceeding, some basic functions required for the simulation to be run need to be defined. These are the functions that calculate the new $$y$$ value and the $$f(y)$$ function.
const g = 9.82; //unit: m / s^2
const L = 1; //unit: m
const h = 0.001; //unit: seconds
function f(y)
{
return (g / L) * Math.sin(y)
}
function y_next(y_last, y_now)
{
return f(y_now) * h * h - y_last + 2 * y_now;
}
While the simulation runs virtual time passes. The time $$h$$ passes for each calculation which means that for each calculation the program must be paused for $$h$$ seconds before calculating again. This is so that the simulation runs in real time.
One problem arises though, since $$h$$ is very small the actual pause will probably be smaller than the time it actually takes to call the pause function. Therefore $$h$$ should be summed up until it reaches a certain limit $$h_{limit}$$, where it is reset back to $$h = 0$$ but the program now pauses for $$h_{limit}$$ seconds.
function simulate()
{
//y0 = y(i - 1)
//y1 = y(i)
//y2 = y(i + 1)
const h_limit = 0.1; //seconds
var h_sum = 0; //seconds
//initializers, define start angular velocity
y0 = 0; //initial angle
y1 = 0.0001; //angle after h seconds
function loop()
{
y2 = y_next(y0, y1);
y0 = y1;
y1 = y2;
h_sum += h;
//quit the simulation if the angle is 90 degrees
if(y2 >= Math.PI / 2) return;
if(h_sum >= h_limit)
{
h_sum = 0;
//multiply with 1000 because setTimeout takes milliseconds
setTimeout(loop, h_limit * 1000);
}
else
{
loop();
}
return;
}
loop();
return;
}
In order to make the graphical part of the simulation some HTML and CSS are added in separate files and the javascript is modified to work together with the graphical part. The code is available on https://github.com/fullnitrous/inverted-pendulum-example. Below is a embedded version of the simulation, you view the simulation at https://fullnitrous.com/inverted-pendulum-example/. Embedded version is only visible if you are reading this in a web-browser.
• My balls | {"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": 0.9364782571792603, "perplexity": 657.2387597002636}, "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-2020-10/segments/1581875143784.14/warc/CC-MAIN-20200218150621-20200218180621-00052.warc.gz"} |
https://www.physicsforums.com/threads/n-of-possible-answers-on-a-filtered-test-with-different-numbers-of-passing-options.620115/ | N of possible answers on a filtered test with different numbers of passing options
• #1
3
0
Main Question or Discussion Point
Hopefully I'm posting this to the appropriate section. I am trying to figure out a formula for describing the number of possible answer "paths" that may be taken by an individual who is administered a filtered test format. In the filtered test format, the individual must select a passing response to item j in order to proceed to item j + 1. If the individual selects a failing option, the test is over. My question is, given that each of the items on the test can have different numbers of passing and failing options (e.g., item j might have two passing options and three failing options, item j + 1 might have one passing option and two failing options), how can I even begin to figure out how to put this down mathematically on paper? My math background is rather weak, so I'm not really even sure where to start. Any help would be much appreciated!
Related Set Theory, Logic, Probability, Statistics News on Phys.org
• #2
mfb
Mentor
34,387
10,475
You can calculate it step by step: For the first question, there are n1 answers, c1 of them are correct. Therefore, you have n1-c1 ways to fail the test, and c1 ways to proceed.
The second question has n2 answers, c2 of them are correct. For each way to reach this question, you have nn-cn ways to fail the test, and c2 ways to proceed.
Does that help?
You can get a general formula using this approach, too.
• #3
3
0
Thanks for the reply. After I posted this I actually worked out a formula, but it's really clunky and I couldn't really say how I derived it. Here's what I have (by the way I'm uploading a JPG for this because I know if I tried to type it in I would screw it up real nice).
Explanation of notation:
M = total number of answer possibilities (number of "paths")
k = 1,...,K is the number of items on the test
V is the total number of answer choices on item k
VkP is the number of passing answers on item k
VkF is the number of failing answers on item k
Hence, VK is the total number of answer choices on the last (Kth) item on the test, and V1F is the number of failing answers on the first item (k = 1) on the test. (And in the last term in the formula, n is used as an arbitrary index.)
Any advice for making this prettier/more efficient?
Thanks!
You can calculate it step by step: For the first question, there are n1 answers, c1 of them are correct. Therefore, you have n1-c1 ways to fail the test, and c1 ways to proceed.
The second question has n2 answers, c2 of them are correct. For each way to reach this question, you have nn-cn ways to fail the test, and c2 ways to proceed.
Does that help?
You can get a general formula using this approach, too.
Attachments
• 11.6 KB Views: 345
• #4
mfb
Mentor
34,387
10,475
I did not check all indices, but the formula looks good. I think you can include V_1^F in the last expression, using the convention that an empty product (from n=1 to n=0) is 1.
In the same way, you can include the case "n-1 correct answers and failed at the last one" there, which simplifies the first part a bit.
• #5
3
0
Great, thanks for the tips!
I did not check all indices, but the formula looks good. I think you can include V_1^F in the last expression, using the convention that an empty product (from n=1 to n=0) is 1.
In the same way, you can include the case "n-1 correct answers and failed at the last one" there, which simplifies the first part a bit.
• Last Post
Replies
12
Views
2K
• Last Post
Replies
0
Views
2K
• Last Post
Replies
10
Views
945
• Last Post
Replies
3
Views
2K
• Last Post
Replies
3
Views
920
• Last Post
Replies
16
Views
8K
• Last Post
Replies
3
Views
9K
• Last Post
Replies
0
Views
1K
• Last Post
Replies
3
Views
752
• Last Post
Replies
11
Views
727 | {"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": 0.8206610083580017, "perplexity": 831.3705547126941}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655937797.57/warc/CC-MAIN-20200711192914-20200711222914-00272.warc.gz"} |
https://motls.blogspot.com/2005/12/electric-dipole-moments.html?showComment=1296517930004&m=1 | ## Wednesday, December 07, 2005
### Electric dipole moments
Adam Ritz (CERN & Victoria) just gave a talk about the electric dipole moments of elementary particles and their constraints on new physics.
Do elementary particles have a dipole moment? Something analogous to two opposite charges separated by a distance? A dipole moment is a vector, and by rotational symmetry, the only direction that the dipole could take is the direction of the spin.
A special comment for Quantoken. By elementary particles, I mean leptons, quarks and gauge bosons (and possible the Higgs and graviton) - and their simple enough bound states that have no particle physics reason to have a significant (=beyond the CKM contribution) dipole moment, such as the neutron. More complicated structures such as molecules of course usually have a dipole moment whose origin we understand very well. That's also the case of some atomic states where the dipole arises from various relativistic corrections.
You know that the magnetic moment of a particle is always proportional to the spin with a fixed coefficient that depends on the particle type only. So if the particles also have an electric dipole, then there is some correlation between the electric moment and the magnetic moment. But which sign should this relative factor have? You know that E is a vector but B is a pseudovector, and therefore the electric dipole moments of the particles violate P (parity) much like CP (parity combined with charge conjugation).
The Standard Model predicts certain small electric dipole moments of particles such as neutron - because of the CP-violating phases in the CKM matrix used for quark masses. New theories of physics including supersymmetry typically predict more significant violations of the CP symmetry which also means significant dipole moments.
Experiments try to measure the dipole moments in various types of materials - liquid Xenon is a crazy example. No dipole moments of the elementary particles have been found so far which puts strong constraints on the parameters of new theories such as MSSM (minimal supersymmetric Standard Model) as well as baryogenesis (but no constraints on leptogenesis).
Various other experiments are underway and some of them are being prepared. One of them may cost as much as 10 million dollars, if you want to have an idea about the budget. Many of them want to access the limit
• d = 10^{-29} electron.centimeter
where "electron.centimeter" is a natural unit of the dipole moment. It is the same dipole as an electron-positron pair separated by 1 centimeter. You see that the corresponding distance "10^{-29}" centimeters is tiny; it is much shorter than the typical "radius" of the particles. Once the experiments get to this level, the subject of EDMs will be over because this is the scale of the dipole moment predicted by the Standard Model. Well, we should eventually see the basic dipole moments predicted by the Standard Model but it is virtually impossible to measure the EDMs accurately. In other words, it would be impossible to distinguish new contributions from the Standard Model background.
The negative results of the experiments looking for the dipole moments is a bad news for more or less any theory of new physics. No really convincing explanation why the CP violating terms should vanish has been given in the case of supersymmetry. It is also a bad news for the anthropic principle because the unnaturally small values of the dipole moments are apparently not required for any mechanism underlying life.
Of course, it is good news for everyone who is quite happy with the Standard Model. It is even better than we thought.
1. Dear Lumos,
Fundamental particles are electric monopoles with a magnetic dipole moment.
Forget electric dipole and magnetic monopole speculation. Maxwell's equations have never been violated in this sense. Gauss' law of the electric monopole applies to the electron, and at very high energy collisions you simply break through the polarised vacuum veil around the core and see more of the strong core charge.
Nobody has ever come up with evidence for it being anything other than an electric monopole with a magnetic dipole associated with spin. The Heaviside trapped TEM wave mechanism for the electron is illustrated here and in more detail for magnetic dipole mechanism here.
If you want to test something, figure out some way of testing string theory, but leave Maxwell's equations alone.
The only error Maxwell made was the subtle assumption that energy instantly spreads along a whole capacitor plate before displacement current travels between the plates. This fouls up his theory of radio waves and light and the real energy flow of displacement current is more complex, a transverse radio wave, with no need to add in Faradays law of induction to create a loop or cyclical oscillatory theory..
Best wishes,
Nigel
2. (I'm talking fermions when I discuss fundamental particles)
3. Lubos, you asked a dumb question. It all depends on how exactly you define "elementary" particles. If you define "elementary" as one that is without any intrisic internal structure, then of course there will be no electric dipole since electric dipole requires a structure. Structure-less particles so then is without electric dipole.
On another side, many particles, like netron, does not fit such a definition of "elementary" particle. Neutron has internal structure, and so it of course has electric dipole. You know what it is, quarks that carry different charges.
The very fact that neutrons do absorb and emit gamma ray photons tells you that it has dipoles, because you need dipole momentum to interact with electromagnetic waves. That basic electromagnetism, any radiation and absorption of EM waves is associated with dipole momentums. I hope you have not forgotten your physics 101.
4. Quantoken, for an electron, 99.884% of the magnetic dipole moment comes from the core and only 0.116% comes from the polarised vacuum which couples a virtual fermion to the core, increasing the total magnetism as Schwinger and Feynman found.
For a neutron, the contribution from the fundamental quarks is smaller, and much, much more of the magnetic dipole moment comes from the polarised vacuum. You also have complex interactions of field quanta ("gluons").
I think the key problem in electromagnetism is understanding the photon. The photon is an electric dipole, of course, but Maxwell's theory gets the angle wrong. He has the variation from positive to negative electric field taking place along the line of propagation, when in fact it is transverse to the line of propagation.
This is because of his error in assuming that displacement current has no transverse component, due to neglecting the obvious fact that electric energy takes time to spread across the capacitor plate (or radio transmitter aerial) while energy propagates from one capacitor plate to the other (or from a radio transmitter aerial to a parallel radio receiver aerial across empty space).
Best wishes,
nigel
5. Science News has a progress report on the EDM search:
http://www.sciencenews.org/view/feature/id/69229/title/Sizing_up_the_Electron
It lists 10^-38 as the Standard Model prediction. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8795314431190491, "perplexity": 569.1388906359884}, "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/1634323585171.16/warc/CC-MAIN-20211017082600-20211017112600-00151.warc.gz"} |
http://ablconnect.harvard.edu/filter_by/expos20 | # Paper Structure Exercise
When students are in the revision stage of their paper, Jerusha Achterberg uses fruits to teach students how to structure their papers so that the organization coordinates with the thesis. The idea behind this activity is to break the 5 paragraph mold students bring from high school.
# Writing the Methods Section
In her Expos section, Jerusha Achterberg teaches how to clearly describe the methods that will be used in a subsequent paper. This activity was motivated by the fact that students were having trouble writing the methods section in their final paper proposals.
# Workshop Conferences
For the third paper of the semester, Jerusha Achterberg has her students do small-group workshops where they read and provide feedback for each other in groups of 2-3.
# Student Paper Workshop
In Jerusha Achterberg's first and second Expos workshops of the semester, she chooses two paper drafts from the section and all the other students read and comment on those two papers. The authors also serve as the moderators for each other's discussion. | {"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": 0.888650119304657, "perplexity": 2987.278240457775}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170404.1/warc/CC-MAIN-20170219104610-00427-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://www.math.princeton.edu/events/counterexamples-min-oos-conjecture-2010-10-15t190010 | # Counterexamples to Min-Oo's Conjecture
-
Simon Brendle, Stanford University
Fine Hall 314
Consider a compact Riemannian manifold $M$ of dimension $n$ whose boundary $\partial M$ is totally geodesic and is isometric to the standard sphere $S^{n-1}$. A natural conjecture of Min-Oo asserts that if the scalar curvature of $M$ is at least $n(n-1)$, then $M$ is isometric to the hemisphere $S_+^n$ equipped with its standard metric. This conjecture is inspired by the positive mass theorem in general relativity, and has been verified in many special cases. I will present joint work with F.C. Marques and A. Neves which shows that Min-Oo's conjecture fails in dimension $n \geq 3$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.84989994764328, "perplexity": 223.81365333928852}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812855.47/warc/CC-MAIN-20180219231024-20180220011024-00157.warc.gz"} |
http://www.physicsimplified.com/2016/09/26-gravitational-potential.html | ## Pages
### 2.6 Gravitational potential.
In the previous section we discussed about the gravitational field. In this section we'll discuss another important concept in physics called the concept of potential, in particular the gravitational potential.
Firstly, the definition of the gravitational potential. I can't emphaisze enough to remind how important it is to understand the definition before we go ahead. "Gravitational potential at any point is the work done to bring a unit mass from infinity to the point". Its SI unit is joules per kilogram (Jkg$^{-1}$)
By this definition the gravitational potential is zero at infinity and negative near the point source. The reson for it being negative can be understood on the basis of the fact that the gravitaional force is attractive force which means that the mass does the work in moving from infinity to the point. The potential $\phi$ in the field of a point mass $M$ is given by the equation.
$$\phi=-\frac{GM}{r}$$
where, $\phi$=gravitational potential, in Jkg$^{-1}$
$M$=mass, in kg
$r$= distance from the point mass
The above definition can be understood as coming from the following derivation,
$$\phi=\frac{Work done}{mass}=\frac{\int_{-\infty}^{r}F_{G}dx}{m}=-\frac{GM}{r}$$
The expression for gravitational force has been used in the above derivation. The plot of the gravitational potential can be seen below.
The change in the gravitational potential energy, $E_{p}$ of a mass $m$ that is moved through a change in gravitational potential of $\Delta \phi$, is the work done on the mass to produce the move, and is given by:
$$E_{p}=m\Delta \phi$$
Ex. Calculate the gain in the potential energy of a iron ball of 30kg when on the surface vs when at a height of 10$^{5}$m above the ground. Given earth radius=$6.38\times 10^{6}$ and mass of the earth =$5.98\times 10^{24}$
Soln: gravitational potential= $9.18\times 10^{5} Jkg^{-1}$
gravitational potential energy=m$\Delta \phi$=$2.75\times10^7J$ | {"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": 0.9893320798873901, "perplexity": 259.6459104628201}, "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/1664030337421.33/warc/CC-MAIN-20221003133425-20221003163425-00594.warc.gz"} |
https://www.spp2026.de/members-guests/01-member-pages/dr-gye-seon-lee/ | # Members & Guests
## Dr. Gye Seon Lee
Ruprecht-Karls-Universität Heidelberg
E-mail: lee(at)mathi.uni-heidelberg.de
Telephone: +49-6221-54-14219
Homepage: https://www.mathi.uni-heidelberg.de/~lee…
## Project
1Hitchin components for orbifolds
## Publications within SPP2026
In order to obtain a closed orientable convex projective four-manifold with small positive Euler characteristic, we build an explicit example of convex projective Dehn filling of a cusped hyperbolic four-manifold through a continuous path of projective cone-manifolds.
Related project(s):
1Hitchin components for orbifolds
For d = 4, 5, 6, 7, 8, we exhibit examples of $$\mathrm{AdS}^{d,1}$$ strictly GHC-regular groups which are not quasi-isometric to the hyperbolic space $$\mathbb{H}^d$$, nor to any symmetric space. This provides a negative answer to Question 5.2 in a work of Barbot et al. and disproves Conjecture 8.11 of Barbot-Mérigot [Groups Geom. Dyn. 6 (2012), pp. 441-483]. We construct those examples using the Tits representation of well-chosen Coxeter groups. On the way, we give an alternative proof of Moussong's hyperbolicity criterion (Ph.D. Thesis) for Coxeter groups built on Danciger-Guéritaud-Kassel's 2017 work and find examples of Coxeter groups W such that the space of strictly GHC-regular representations of W into $$\mathrm{PO}_{d,2}(\mathbb{R})$$ up to conjugation is disconnected.
Journal Transactions of the American Mathematical Society Volume 372 Pages 153-186 Link to preprint version Link to published version
Related project(s):
1Hitchin components for orbifolds
We extend the notion of Hitchin component from surface groups to orbifold groups and prove that this gives new examples of higher Teichmüller spaces. We show that the Hitchin component of an orbifold group is homeomorphic to an open ball and we compute its dimension explicitly. We then give applications to the study of the pressure metric, cyclic Higgs bundles, and the deformation theory of real projective structures on 3-manifolds. | {"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": 0.8750839829444885, "perplexity": 1097.1018544659223}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655884012.26/warc/CC-MAIN-20200704042252-20200704072252-00202.warc.gz"} |
http://mathhelpforum.com/calculus/112130-even-odd-functions.html | Math Help - even and odd functions
1. even and odd functions
Is there a function such that it is both even and odd?
Is it f(x)=0? any others?
2. A function is odd if $\forall x, \ f(-x) = -f(x)$
A function is even if $\forall x, \ f(-x) = f(x)$
Assume we have a function g that is both even and odd, then:
$g(-x) = g(x) = -g(-x) \Rightarrow g(-x) = -g(-x) \Rightarrow 2g(-x) = 0$
$\Rightarrow g(x) \equiv 0$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 0.9474064707756042, "perplexity": 545.4493604637647}, "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-2015-27/segments/1435375096780.24/warc/CC-MAIN-20150627031816-00064-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://socratic.org/questions/what-is-the-derivative-of-e-x-1-e-x | Calculus
Topics
# What is the derivative of [e^x / (1 - e^x)]?
Jul 31, 2015
${y}^{'} = {e}^{x} / {\left(1 - {e}^{x}\right)}^{2}$
#### Explanation:
The quotient rule will work just fine for this function because you can write it as
$y = {e}^{x} / \left(1 - {e}^{x}\right) = f \frac{x}{g} \left(x\right)$
In such cases, the derivative of the function can be found by
color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2, with $g \left(x\right) \ne 0$.
Apart from this, all you really need to know is that
$\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$
So, the derivative of $y$ will be
$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left({e}^{x}\right)\right] \cdot \left(1 - {e}^{x}\right) - {e}^{x} \cdot \frac{d}{\mathrm{dx}} \left(1 - {e}^{x}\right)}{1 - {e}^{x}} ^ 2$
${y}^{'} = \frac{{e}^{x} \cdot \left(1 - {e}^{x}\right) - {e}^{x} \cdot \left(- {e}^{x}\right)}{1 - {e}^{x}} ^ 2$
${y}^{'} = \frac{{e}^{x} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{2 x}}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{2 x}}}}}{1 - {e}^{x}} ^ 2$
${y}^{'} = \textcolor{g r e e n}{{e}^{x} / {\left(1 - {e}^{x}\right)}^{2}}$
##### Impact of this question
463 views around the world | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "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": 0.8906901478767395, "perplexity": 1015.0375262659547}, "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/1610703549416.62/warc/CC-MAIN-20210124141945-20210124171945-00094.warc.gz"} |
http://mathhelpforum.com/number-theory/117511-euler-phi-function-proof-print.html | # Euler phi function proof
• November 29th 2009, 09:24 PM
MichaelG
Euler phi function proof
Assuming that d divides n, prove that Φ(d) divides Φ(n)
hint; work with the prime factorization of d and n
• November 29th 2009, 09:56 PM
Bruno J.
Use the fact that $\phi(n)=n\prod_{p\mid n}(1-p^{-1})$. What can you say about $\phi(n)/\phi(d)$ using this?
• November 29th 2009, 09:58 PM
chiph588@
Suppose $d = p_1^{\alpha_1}p_2^{\alpha_2}\cdot\cdot\cdot p_k^{\alpha_k}$ and $n = p_1^{\beta_1}p_2^{\beta_2}\cdot\cdot\cdot p_k^{\beta_k}p_{k+1}^{\beta_{k+1}} \cdot\cdot\cdot p_r^{\beta_r}$ where $p_1$ through $p_k$ are the same for both $d$ and $n$.
Note $\beta_i \geq \alpha_i$.
$\phi(n) = \phi(p_1^{\beta_1}p_2^{\beta_2}\cdot\cdot\cdot p_k^{\beta_k})\phi(p_{k+1}^{\beta_{k+1}} \cdot\cdot\cdot p_r^{\beta_r}) = C \cdot p_1^{\beta_1-1}p_2^{\beta_2-1}\cdot\cdot\cdot p_k^{\beta_k-1}(p_1-1)(p_2-1)\cdot\cdot\cdot(p_k-1)$
$\phi(d) = \phi(p_1^{\alpha_1}p_2^{\alpha_2}\cdot\cdot\cdot p_k^{\alpha_k}) = p_1^{\alpha_1-1}p_2^{\alpha_2-1}\cdot\cdot\cdot p_k^{\alpha_k-1}(p_1-1)(p_2-1)\cdot\cdot\cdot(p_k-1)$
Now just match up terms to see indeed $\phi(d) \mid \phi(n)$. | {"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": 12, "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": 0.8866453170776367, "perplexity": 1397.143386486231}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860128071.22/warc/CC-MAIN-20160428161528-00082-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://proofwiki.org/wiki/Symbols:Arithmetic_and_Algebra | # Symbols:Arithmetic and Algebra
## Symbols used in Arithmetic and Algebra
$+$
A binary operation on two numbers or variables.
Its $\LaTeX$ code is + .
### Positive Quantity
$+$
A unary operator prepended to a number to indicate that it is positive.
For example:
$+5$
If a number does not have either $+$ or $-$ prepended, it is assumed to be positive by default.
The $\LaTeX$ code for $+5$ is +5 .
### Subtraction
$-$
Minus, or subtract.
A binary operation on two numbers or variables.
Its $\LaTeX$ code is - .
### Negative Quantity
$-$
A unary operator prepended to a number to indicate that it is negative.
For example:
$-6$
The $\LaTeX$ code for $-6$ is -6 .
### Multiplication (Arithmetic)
$\times$
Times, or multiplied by.
A binary operation on two numbers or variables.
Usually used when numbers are involved (as opposed to variables) to avoid confusion with the use of $\cdot$ which could be confused with the decimal point.
The symbol $\times$ is cumbersome in the context of algebra, and may be confused with the letter $x$.
Its $\LaTeX$ code is \times .
### Multiplication (Algebra)
$\cdot$
$x \cdot y$ means $x$ times $y$, or $x$ multiplied by $y$.
A binary operation on two variables.
Usually used when variables are involved (as opposed to numbers) to avoid confusion with the use of $\times$ which could be confused with the symbol $x$ when used as a variable.
It is preferred that the symbol $\cdot$ is not used in arithmetic between numbers, as it can be confused with the decimal point.
Its $\LaTeX$ code is \cdot .
### Division
$\div$, $/$
A binary operation on two numbers or variables.
$x \div y$ and $x / y$ both mean $x$ divided by $y$, or $x \times y^{-1}$.
$x / y$ can also be rendered $\dfrac x y$ (and often is -- it tends to improve comprehension for complicated expressions).
$x \div y$ is rarely seen outside grade school.
Their $\LaTeX$ codes are as follows:
The $\LaTeX$ code for $x \div y$ is x \div y .
The $\LaTeX$ code for $x / y$ is x / y .
The $\LaTeX$ code for $\dfrac {x} {y}$ is \dfrac {x} {y} .
### Plus or Minus
$\pm$
Plus or minus.
$a \pm b$ means $a + b$ or $a - b$, often seen when expressing the two solutions of a quadratic equation.
Its $\LaTeX$ code is \pm .
### Absolute Value
$\size x$
The absolute value of the variable $x$, when $x \in \R$.
$\size x = \begin {cases} x & : x > 0 \\ 0 & : x = 0 \\ -x & : x < 0 \end {cases}$
The $\LaTeX$ code for $\size x$ is \size x .
### Binomial Coefficent
$\dbinom n m$
The binomial coefficient, which specifies the number of ways you can choose $m$ objects from $n$ (all objects being distinct).
Formally defined as:
$\dbinom n m = \begin {cases} \dfrac {n!} {m! \, \paren {n - m}!} & : m \le n \\ 0 & : m > n \end {cases}$
The $\LaTeX$ code for $\dbinom {n} {m}$ is \dbinom {n} {m} or \displaystyle {n} \choose {m}. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9518663287162781, "perplexity": 1205.3032602037229}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250608062.57/warc/CC-MAIN-20200123011418-20200123040418-00500.warc.gz"} |
http://glaucia.servers.ec-lyon.fr/spip.php?article1087 | # Laboratoire de Mécanique des Fluides et d’Acoustique - UMR 5509
LMFA - UMR 5509
Laboratoire de Mécanique des Fluides et d’Acoustique
Lyon
France
## Nos partenaires
Article dans J. Fluid Mech. (2018)
## Instability of pressure-driven gas–liquid two-layer channel flows in two and three dimensions
Lennon Ó Náraigh & Peter D. M. Spelt
We study unstable waves in gas–liquid two-layer channel flows driven by a pressure gradient, under stable stratification, not assumed to be set in motion impulsively. The basis of the study is direct numerical simulation (DNS) of the two-phase Navier–Stokes equations in two and three dimensions for moderately large Reynolds numbers, accompanied by a theoretical description of the dynamics in the linear regime (Orr–Sommerfeld–Squire equations). The results are compared and contrasted across a range of density ratios $r=\rho_\mathit{liquid}\,/\,\rho_\mathit{gas}$. Linear theory indicates that the growth rate of small-amplitude interfacial disturbances generally decreases with increasing ; at the same time, the cutoff wavenumbers in both streamwise and spanwise directions increase, leading to an ever-increasing range of unstable wavenumbers, albeit with diminished growth rates. The analysis also demonstrates that the most dangerous mode is two-dimensional in all cases considered. The results of a comparison between the DNS and linear theory demonstrate a consistency between the two approaches : as such, the route to a three-dimensional flow pattern is direct in these cases, i.e. through the strong influence of the linear instability. We also characterize the nonlinear behaviour of the system, and we establish that the disturbance vorticity field in two-dimensional systems is consistent with a mechanism proposed previously by Hinch (J. Fluid Mech., vol. 144, 1984, p. 463) for weakly inertial flows. A flow-pattern map constructed from two-dimensional numerical simulations is used to describe the various flow regimes observed as a function of density ratio, Reynolds number and Weber number. Corresponding simulations in three dimensions confirm that the flow-pattern map can be used to infer the fate of the interface there also, and show strong three-dimensionality in cases that exhibit violent behaviour in two dimensions, or otherwise the development of behaviour that is nearly two-dimensional behaviour possibly with the formation of a capillary ridge. The three-dimensional vorticity field is also analysed, thereby demonstrating how streamwise vorticity arises from the growth of otherwise two-dimensional modes. | {"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": 0.8941141963005066, "perplexity": 1121.583927312224}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267157216.50/warc/CC-MAIN-20180921151328-20180921171728-00136.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/200221-root-unity-sum.html | # Thread: Root of unity sum
1. ## Root of unity sum
Let X be one of the nth roots of unity.
Find the sum: $1 + 2x + 3x^2 + ... + nx^{n-1}$
I've been trying to figure this out but nothing...
I know that sum of roots of unity is 0. But I have no clue about this one.
2. ## Re: Root of unity sum
Let $f(x)=x+x^2+...+x^n$
This is geometric progression with first member x and q=x.
Therefore, its sum is $f(x)=\frac {x(x^n-1)}{x-1}$
We obtained equality: $x+x^2+...+x^n=\frac {x(x^n-1)}{x-1}$
Differentiating with respect to x yields:
$1+2x+...+nx^{n-1}=\frac {nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$
Since x is n-th root of unity then $x^n=1$ and we can write above equality as
$1+2x+...+nx^{n-1}=\frac {nx^{n+1}-(n+1)*1+1}{(x-1)^2}$
Or
$1+2x+...+nx^{n-1}=n \frac {x^{n+1}-1}{(x-1)^2}$
Again, since $x^n=1$ then $n \frac {x^{n+1}-1}{(x-1)^2}=n \frac {x x^n-1}{(x-1)^2}=n \frac {x-1}{(x-1)^2}=\frac {n}{x-1}$
Finally,
$1+2x+...+nx^{n-1}=\frac {n}{x-1}$
As can be seen from formula x=1 can't be applied to the formula.
When x=1 $1+2x+...+nx^{n-1}=1+2+3+...+n=\frac{n(n+1)}{2}$
Therefore, if x=1 $1+2x+...+nx^{n-1}=\frac{n(n+1)}{2}$, otherwise $1+2x+...+nx^{n-1}=\frac {n}{x-1}$
3. ## Re: Root of unity sum
Hello, gregx22!
$\text{Let }x\text{ be one of the }n^{th}\text{ roots of unity.}$
$\text{Find the sum: }\:S \;=\;1 + 2x + 3x^2 + \hdots + nx^{n-1}$
$\begin{array}{ccccccc}\text{We have:} & S \;=\; 1 + 2x + 3x^2 + 4x^3 + \hdots + nx^{n-1} \qquad\qquad\quad \\ \text{Multiply by }x\!: & xS \;=\; \quad\; x + 2x^2 + 3x^3 + \hdots + (n-1)x^{n-1} + nx^n \end{array}$
$\text{Subtract: }\:S - xS \;=\;\underbrace{1 + x + x^2 + x^3 + \hdots + x^{n-1}}_{\text{geometric series}} - nx^n$
$\text{The geometric series has: first term }a = 1\text{, common ratio }r = x\text{, and }n\text{ terms.}$
. . $\text{Its sum is: }\:\frac{1 - x^n}{1-x}$
$\text{But }x\text{ is an }n^{th}\text{ root of unity. }\;\text{Hence, }1 - x^n \:=\:0$
$\text{So we have: }\:S(1-x) \;=\;-nx^n \quad\Rightarrow\quad S \;=\;\frac{nx^n}{x-1} \;=\;\frac{n}{x-1}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 22, "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": 0.9281958937644958, "perplexity": 1175.4531287989785}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170609.0/warc/CC-MAIN-20170219104610-00512-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/more-thermodynamics-please-help.70783/ | 1. Apr 9, 2005
### flythisforme
(Re-posted)
A 50g block of copper having a temperature of 400K is placed in an insulating box with a 100g block of lead having a temperature of 200K.
a) What is the equilibrium temperature of this two-block system?
b) What is the change in the internal energy of the two-block system as it goes from the initial condition to the equilibrium condition?
c) What is the change in the entropy of the two-block system?
I set the heat loss of the copper block equal to the heat gain of the lead block to find the equilibrium temperature using the equation deltaQ=mcdeltaT and got 318K as the answer for part (a). Is that correct?
Also, for part (b), would I find the change in internal energy of each block and add the two numbers together? Is there an equation to directly get changes in internal energy, or would I have to figure out the work first? (I know that deltaU = deltaQ - W)
2. Apr 10, 2005
### so-crates
a) What is the heat capacity of the copper block and the lead block ? You cannot solve the problem without these two quantities. Heat gained by lead = heat lost by copper (in magnitude, ignoring s sign).
b) First of all delta U = q + w for a closed system, not delta q - w. You can find the sum of the two quantities (remember to keep your signs right), but there is a much easier way to find the answer. Hint (how much heat is flowing in from the outside
c) Whats your definition of entropy?
3. Apr 10, 2005
### flythisforme
a) Well, since you said that, is what I showed in my original post correct?
b) There is no heat flowing in from the outside. So the change in internal energy would be zero?
c) entropy = degree of disorder
I have a general equation for change in entropy and then equations for isothermal, adiabatic, isochoric, and isobaric processes. Is this isochoric, since the volume inside the box does not change?
4. Apr 10, 2005
### so-crates
a) I can't verify your numbers without the heat capacity of copper and lead.
b) Yes . Since there is no heat flow and no work done from outside the change in internal energy is zero.
c) Have you had calculus yet? dS = dQ/T. When you substitute dQ = Cp dT, and integrate this equation from some final to initial temperature. You wind up with delta S = Cp ln(Tf/Ti) But remember, that is for a single block so you will have to sum the two quantities.
Since we are talking about solids and not gases, there really isn't a distinction made between an isobaric and isochoric processes, since solids don't expand with temperature very much, and pressure doesn't really have an effect on there thermodynamic behavor.
5. Apr 10, 2005
### flythisforme
a) oh sorry... for copper it is 386 J/kg*K and for lead it is 128 J/kg*K
b) yay
c) Okay, that makes sense. So I just use that equation for each block and add the two results together.
And on the topic of the gases, for another problem there is a cycle (PV-diagram) where I have to calculate various things for each leg of the cycle. One part is an adiabatic process. Based on values I figured out for other parts of the cycle, I found that for an adiabatic process, the change in temperature is zero. So then that made the work and change in internal energy zero. That doesn't seem to make sense to me. So is that definitely wrong? If so, I need to go back and find where I went wrong with other things.
6. Apr 10, 2005
### al_201314
Sorry to go a little off topic, this question brought up a question in me..
I know dU = dQ + dW
i.e, it's a process. How do you measure internal energy of of the 50g of copper block at 400K in this case? And does this equation only apply to gases?
thanks.
7. Apr 10, 2005
### al_201314
flythisforme,
I can't really know what went wrong without the question but anyway hopefully this can be of some help to you.
In gases, the first law states that,
dU = dQ + dW.
In an adiabatic process, is one which takes place such that no heat flows in or out of the system, so dQ = 0. First Law becomes
dU = -dW. (-ve because work is done by gas)
This implies that work is done at the expense of internal energy of the system.
===> dW > 0 (adiabatic expansion
dU < 0 (internal energy decreases)
8. Apr 10, 2005
### flythisforme
Okay, so then most definitely the work done and change in internal energy CANNOT be zero?
Well I'll have to go through the other parts of the problem again. Hopefully I'll figure out where I went wrong.
9. Apr 10, 2005
### flythisforme
P.S. THANKS for all the help. I'm going to bed now but if you have anymore input I'll definitely check back in the morning.
Thanks again!!
10. Apr 10, 2005
### so-crates
No, this expression applies to all systems. It is a fundamental law of thermodynamics (the First Law). You cannot intrinsically measure the internal energy of a system. Only changes are meaningful, and those changes are the heat added or lost plus and the work done on or by the system. Heat and work are actually just two forms of kinetic energy, so what you are really measuring is the net change in kinetic energy of the entire system. Heat is a more "random" type of energy, and so it reasonably follows that it is a less usable form of energy, and from there you can see the justification for the Second Law of Thermodynamics.
Last edited: Apr 10, 2005
11. Apr 10, 2005
### so-crates
No, internal energy is unchanged in the SEALED, ISOLATED container. Internal energy changes for the copper block, the lead block, and the air in between them(assumed to be unimportant), but the system as a whole, has no net change in internal energy. Because work is not done, and heat does not flow out of, the container. The container cannot expand, so it cannot do expansion work. The pressure may increase or decrease, but no work can be done since dw = - p dV and dV = 0. The lead block will expand slightly, and the copper block will shrink slightly, but the question doesn't ask you to calculate the work done for those components.
12. Apr 10, 2005
### al_201314
Thanks for the clarification so-crates. That piece of information was based on flythisforme's question on gases. But thanks once again for the help on the first law.
13. Apr 10, 2005
### al_201314
flythisforme,
Based on adiabatic process itself, the change in Internal Energy of the system (dU) is dependent on work done BY or ON the gas (dW). So, work done by or on the gas cannot be 0, since no heat was flows in or out of the system, unless dU is 0.
Last edited: Apr 10, 2005
14. Apr 10, 2005
### flythisforme
So it is possible for dU to be zero in an adiabatic process? And then the work would be 0? Because I am getting zero temperature change, and this causes the deltaQ, deltaW, and deltaU to be zero.
[EDIT] Maybe I should clarify a bit. I guess what I am asking is that can a process be adiabatic AND isothermal?
Last edited: Apr 10, 2005
15. Apr 11, 2005
### al_201314
If dU is 0, it means that there is no change in internal energy of the system, i.e, dW is also 0, which implies not doing anything to the system at all.
Adiabatic processes will definately result in a change in temperature. Since dU = dW, when work is done by the gas, dU is -ve -> final temperature is lower than the initial, since total internal energy decreases in the system.
When work is done on the gas, dU is +ve -> final temperature is higher than initial, since total internal engery increases in the system.
To justify this, look to the topic on ideal gases,
1/2 N m <c^2> = 3/2 N K T
where N is no. of molecules, m is mass of one molecule, <c^2> the mean square speed of the molecules, K is boltszmann constant and T is temperature is Kelvins.
So, since 1/2 N m and 3/2 N K are constants, K.E of gas is directly proportional to Temp.
Hope this helps. | {"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": 0.8618090748786926, "perplexity": 612.3961566345585}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218193716.70/warc/CC-MAIN-20170322212953-00423-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/geometry/elementary-geometry-for-college-students-5th-edition/chapter-1-review-exercises-page-65/19a | ## Elementary Geometry for College Students (5th Edition)
Based on appearance, the angle is an acute angle since its measure is less than 90$^{\circ}$ and greater than 0$^{\circ}$. | {"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": 0.8112968802452087, "perplexity": 1308.774440326823}, "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/1524125947939.52/warc/CC-MAIN-20180425174229-20180425194229-00435.warc.gz"} |
http://physics.stackexchange.com/questions/44238/fock-picture-of-bosonification-in-condensates?answertab=votes | # Fock picture of bosonification in condensates
I want to understand how bosonification in a condensate must be interpreted in the Fock states picture
Say i have uncoupled fermions in a set of states $E_1$, $E_2$ ... over the vacuum $E_0$. They occupy all the levels up to the Fermi energy.
Now, i introduce some coupling between them so, they become effectively boson pairs. Are the $E_i$ energy states out of the picture for these bosons? My random guess is that fermions would couple in states like
$$| E_1 \times E_2 \rangle - | E_2 \times E_1 \rangle$$ $$| E_1 \times E_2 \rangle + | E_2 \times E_1 \rangle$$ $$| E_3 \times E_4 \rangle - | E_4 \times E_3 \rangle$$ $$| E_3 \times E_4 \rangle + | E_4 \times E_3 \rangle$$
And so on, but this is the aspect i'm not sure and i want clarification
What are the states where the boson pairs are condensing into??
- | {"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": 0.9264240860939026, "perplexity": 1015.6815922219096}, "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-2014-10/segments/1393999650916/warc/CC-MAIN-20140305060730-00091-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://horsebitbro.com/watermark3.php?image=auwebcp/pictures/288299600_270_pic_3.jpg | JFIF>CREATOR: gd-jpeg v1.0 (using IJG JPEG v62), default quality C \$.' ",# (7),01444 '9=82<.342C 2! !22222222222222222222222222222222222222222222222222" }!1AQa"q2#BR\$3br %&'()*456789:CDEFGHIJSTUVWXYZcdefghijstuvwxyz w!1AQaq"2B #3Rbr \$4%&'()*56789:CDEFGHIJSTUVWXYZcdefghijstuvwxyz ?( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( (\$r^21@ ̋ހ5h+}"[=EZ\$(ɨI̓M[I6sSPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEPEJD?S@Ot^MPyYI((((UFrZu5NH>Xu\$UU;ɷ6*ET@7,XzVU>8PBFԳdrsQPEPEPEPVk.)ZH@(((((((((((((((((((((((((((((((((((((((((((((((((((((((6ɬ;нlGΠ`0 V!\$rhX;{p6VeN&"((((v;OJJhh *ᴚ@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@]1 ߅eջC6U(Ŭ[Ajb #ƲIɭÈk6 ( avBTuv`5ROhG̕Tè4)[c84(PJos k>XȠ *i7*j((((((((((((((((((((((((((((((((((((((((((((((((((((*IjjxС.y@ŘB p S0x}V}hތŚΠ(YTexj@aKE1V\b5= \2t@F5@#\nJE) N2zZD#@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@g]Yvw5J+nX:JмEXJϠ hڴèaXp:֥f8hnt\$VIV\0 4Ҥuֵ֯JCxr@((F`\$mlrE9bD)ܘQzXaF(K}+1cI@Q@OUڣaժQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQE# )5+ZRk42q@8 ҷQA5JZ(( w6AsoK@h I@Q@T[h:rɠ %dժvK@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@R8Veag5/ Q4q.duQkQT%i ((((((((((((((()XD094{'h;ic *֛S ƽ9 Zb ֩ːOrEzl_7cO:cX 7|S" 4r1EӗTn2?ӿ )LQ ǩ%2}Q.XO Ϥz£1I*P=\LZ\AW-ʑ[?+1#-Ec1 пZ"Juq4>(((((((((((((((()(iv 8S_rw7zo}V \MD(fAߢúlܕ^GJ =QƼ}@tuiz|r)91Iz ]-,h-DQܘ_\aW1 +upf=`C?ʺh< @[{8 Sˋ[e.dU T\_E@jK yFqqm)Zy42/zu0x3s Gl!Z\,xA*y^|4?QP-!oǚA6 qZ1ڶM|дy3:'*\ (D݈ ~s Ǝ<%V c?Hy}>e]R5F_.o', nLV5ԣc.P ̗1~L+ y?ۑ%OƝ gތFI*yͷ\}S,va=%^{45#T1\$nB*JEQ@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@Q@E eR@͝wZEb"nZF h܇XVkGZѶ.,FF P||*m6NUXȭyî iɸ lYjSZB:`:;gnX韧)@'*HmYaW5A@ El'ާWZZ6IcgZ.IQqFY"BW@vy&5X]#Y4ۊ6^yҹ-z; @XOo,I7(i ?MIPT%bKyL%Y[?J&/. OʣK'YZ?W-|1]7,=ZoʋHΚ ^_v-I> SH_|>rAƸjx3DN]nd?LqG0NNMi1~(5\X]lb |o57N:i.`Wт`,?~U rγ, ƯZƌe~d#-1- HZ ) 2ҚW%F` YZκռ0:'!~ɕy~=Ghך;a)i5{2Eh%;Z|Ŭ--f]f?G]}J'ZLI様 68i"7Swk8@=bg&Ԕ5-1MEQHI4ɥ&OZ̖Vi A@ڦG +LU[\$ڥ7RamiNOE_5 F 61 X'4u&x j\$2W/}z"J=Qu8m,#=*l[A& 69s??nCܟOjinSt%o2-1[F1W HI V0lnlywn#A]e, WB!EFAQ46UCb;h#@CLOrj,JCi4_\ (FW_"(m4m"y 8歘) ǿtG&< < wf^K EgN5,ym m-Օ ew'Zo ӭuwp? ^i&d|*lc55ϳ۱"7 Zv훖 a (9sӊ-`b}\~˭\첈^Zfأ =>? -ڥl#ˋڮ K,׳֯A8,5UZVeE@Q\$PJÙM4Bjp Ǟ2AVR jc R,B ND# ()v@ Fui6ӨSLb*U WM((- r rhU,tG. (ܶ.! 6#*rAϥKnN? Ǯ+ҥ% A CGyertC&n=Vd ZO۬^_nxwP3[!Ouh9TQEQEQEQEQEQEQEQE!!FOJϹ.v/FhHV& ՎEx4泝J1OJ˺P\$\$-YR\_Ldud@_+5Ba:S- ltyZ (w'5\$f , &Oj#Yaja J+H".)h=)PvѴS>`1JJҹ ևڡhuVr:φ C8uޣG2 jtmRp?Нj \)sI12jl]ɵ]ʽZoݘN߾O+DKGqk-k2o2 k 9Q'GalF>V?*ZgIE ((((((((((+/\I^s!c +GWZFs 'jОQ eu1}nۑq! a` Ms]~Qv63X🆆!?e;>T*: (b8Zi95 95"GX\*@689T REQEQEQEQEQEQEQE! =Z(# yߺF?vO u[er>4g9y } % XZEL3pI w k?O>F H[XCc TM0R:2Aj`s\GvEn[\Gqx0n-nOEU((((((:Z(\$bE*Ee D=+Z\$bD zk({r>J/\ajP9#5C@Ru(Oj\$Ҳ<-[t#ڲj!H=VPґ#XS( ՋƫR1`Vހ P>((t&# 8*XIp]Zrف)f?(Y^'ͧkrl TInivڗXI\$'@qUk87F VYZ6n=j8tdxڅ6{»y UtQ[ 8n@MX,E {U! U8QJ: QEQEQEQEQEQEQESJu =JLe`jL[\.A[Z+>}:ff+dڹcS/ gPG1]p®O wp M5YJ cE=X# 4F=J9ӆ# = _ jY̜C7Oc]ڶj-2J((((((((( jM ; 5b|V[<--{J˰lJ=iǹ2ab:ZӅl̟wP?mm9,8#[{xN5 90M>45& JEZU".H(((((((((((p¸>ݯtpmA]/Lnpxϝ'5Ŋ6a~b*כJ㊥`TѶrhЏiJ8KaptA1_AS5@I;MZ|fͦ3i[ 2zO2,5dOFm uTq')xKGC i'I1i]\$V{W.glc" 6ִ%uTWq|M9 @{32:F_qQG! ʻ(+ėZ^vKKRwR0qٗSߨa5?iJ#tMVVHR\$_SPU#/yŭRĒRid[B6S Zմ,@k5֦ b=jbykU)y"v G T\$2Y|R)hiy; RElĵDC]ێ2w]T 憦Ve:TEPEPEPEPEPEPEPEPEPEPJGVi1f4T\F` 7Νc0=inA!'WLp Q} ~| :6r=NvpwOqQ Ke쥶=a~?rӪ"(((((((((4(s~rb^y ż ?3ٵQ؆TУLPWk,6c) RKbhRĸ-\QHc-PEPEPEPEPEPEPEPEPEPEP7%rzb lC?GpMsvS*Y7 KM{yAEg!EP=MKo}~I"/-ᓌVfXi*(p7\$w<@|kFǗL}I'"+gS3[Y\$ϲuY l{EV;fe W?&ahAE C2[KFjϊMOeL2./hN<mq *[lqxjw kvt+SF| 2*փPUW(\ۛJI2fYZ[TYlLS oȯ4 1P;cWe](,\$J^Z1#=pzrv~Aq?y\$Լ'\ o XjTQ@BZx~P`Q-Am=zn._|)hPn ٮf =NkQ6qi[fص#HǼG S\ lKYؒ%>'寭4|Agg?'m>t; A״3:`|#w2\c6 z6Hʇ+-Rb+\$>Okz{3hPϽhS& /O腯-PgRDqu?CR!itȬ}^-&x䳻\x'>\l:8L[0 YeF5 wU?ONb19*njfڲq:?p^Z8z rޏuu ?d%<+5RQػ#CASԌ(((((((((((E9!{D#'x|>LI&n=ݼWj:Ɉ\$jY@_6s_A1ZlGoA\NrӞѲ0%>4lʼnQ[?I.EG& ]O@:Iu6>bxgH!rG3Sg}G 3Vr1 COOo 6hK4n9 ڼu͚ * +,t%pf,K# bRr \$vf^~\t+/H?3x,ZϹ> 瞇A]UI9>"Xa\X)wlq]8>¦[;>8'*} 5pQud_62?Z"/b fraz֟TOOf3 }zזIGICl{͖Db̀zzVJ?5-wgx"XǪRmxpېOymmy3\R+4ܞAl.Ov9iXnbMSD؍NY~ yby t٣xncc p_ r`Ia \$bп9RH̅ 槱䲩^0t a@ 5`jk|=fQ,NxS^qD3:to t.285z۲ nC+On.SϖNg:*J@]V;7sԎ-})- #Zu=b9-s Yry@\$s! *% mkcx]}7DvΙ+gs^yja>/p<ֻsO]R8`ڹ3pD@0@kR<[ f8ʹ\$^kZj:s¶nrxn.s.ޕM<3I'ݻTz u)U08 PKya~V-o!H|uz]&jk9Fc-ԴYVNFcq}r#N]>~X]kܶ敳s>=5Jw`go SV/Տڷ"p]YmXV4k-g>;vDU_@?l0kGI ̺N+=Vd"{6Hsӵc, :2gl(7c\$jhzTyqq*yy gv6>ݨ[ :)# ]_ʪ]6W! {2pGҖ F &YIIޱ{6\_ii V+Itۏ!rQEx5/眬NUҲE|\:S@X:Ov/th[ \<3q5p vk? n屖XdO9UVGk IeNYDHxPoF}E| SE\$Lw v-ˍ?u@Ru[)sEhU( \^f:(((((((((((dK ] zy8Wcn'xvX٬)ϢZu tA[[ؔwEhFxzք]* ,Q@EQEQEQEQEQEQEPAC^G8L+.w+|UCC _Sߎ1ՂF :p} hhu_cWY- k=+ӧY01_a>!R*(OU58PXQEQEQEQEQEQEQEQEQEQERW3@\$J fm~Ҹ*ʈ 9 TO6xD-ʪ@8EUԉ>{ ɦy"gwG X{Y-Ahz6GsJl8hdkY9X}MtTSf8Z&{PHS5d6 <_, iS@s'jo,ҮY5r;c NKfvWGX["(.;q`Ԋ-jā\$]3Ҳ nc [%C7NGJ`PHY7PE]fSwiȭd VOsWly5 ԩ @R=R aOibb ua gytYG Gc_i_[ג][D =kZ\$XYʚohR{؞n z}ܫҺOT "/" w>۟mљ a)aG*덮q]dEakdO-Eii&ac3~`~ Ly*# %(((((((((( ei-ÕڟS\.XڮP: f(O͎ 㭥ێVԙ3] kwV -o닾CVNU>gma705R`zք=*#Zt(tҊ(((((((i: kEʇ&kg={ Yӊ/1eoJ< T3&8u u-c :t:d7Q!٣=Z^d^{VS6z u[SJ5k/dw Q j4}f N,_ Pi((((((((Tu RKPlip\$F;(ʾwu\$6jry MG%Ē [U \NK\b̖' ZegO(7 ;o xR;X&ѺnE xy4hdkm d};Բ6M5W&l3V*tQU nFN#ր_EhU{ڀ31W-kUB0pi(Et\ANTf(6j; r@@%!YS (\Y9-L+xScc hVK*6 `^}T#@7j&-hgzPY4ǭ51 Teq[bOS@8]Nb]Z#jɑA*y f@cN)ai[ǭiƸ#SbREb(0(-hKE0C\$yfFh59Ohk{ y}kZ0jŪME" vIh Pzc.! Qִi4{`yjݤW+'q]: C;Xí/9OGN (0x{tm@NOhi:)ȒlXdޟRPQEQEQEQEQEQEQEQEQE,a&E =jιAc\tQ<>6en@4%q7b; v敹X:q)~UO_5D7@ *Z]gm Mڴ% O49%]N8_^rXPXGإ|叫U5 ܤEhF8/J>(((((((("HEq(f ?Ƹ ,3J >>r>"ٺlpC8M; a 0{VR!9 럒~,76)AMP֠u-䙢SYPCj&ǫi~!Y[_ ҷmu8# /+wN.,-盟GqQYZyRI85F{T-Q@Q@Q@Q@Phj)!MJ\KLD Pu,p+\y1ЛEsr_F+oP`f6TFi:O X% Nn/fk}*Z#PvDʩMo)2H4"0\+;Ub[% -[xEc]iM\$"^ >c}JmWls[K)&Be֢M!_zm*x{Û;JQ u?K\$PZ@&E 68Wb<81N GurƳgi[ڀ%UJT'xJEdTzPmVBH aoZ@4R; >XHUyT@5N)+橻l@ \$Bep;V E (QE\$XԳY Fjʚfzvi¡Ѳb84f_f GuLϘ)" UR2ha\$#6) n ^zn,"n9h@) 6I[Ut)W 27QF&C=i(zA/,,ç֭;ѭiށ uvi 9Xz%ޞrZ.p=I0!p +o-0Ee_(Pюj&QE!Q@Q@Q@Q@Q@ŖeЮwY,vtxOsYw>lx <6zGuv&PǩmƥNNjXW&C "d*\ ( ( ( ( ( ( ( ( (5Zhc.h=oLn;e_\ 0 mp2rXڳ!%6*"<8OvRQ#E:r hFHˤʹwC{VTZޓ6ɖ[G(jW%Sm|v<ִfağ0i7(孿iyӵP="LGUPqj/w+B i< bnz'4`r: j#0FF^V;PqRk .Q z~uQ zG7? 9H FPF2"Zݮ敾P=|E 5 z ;#\$v 7~m4Ԯp3xNᮠ_dK \?h.vP46Y2ʒ7p;V~VnKZ١Zy`nn;RHW-K 3Iˁrjet眿2 XZD" \$=_5-꯹"dSɖ˧jM[qq7(-no1C,Dq/Vm] N; A[ /a*yOMlr7G3*M04*jJ5ڋwVGt% ݜa: cGK VC=f wM`t7чkɡqnbІ0fVrJocrG|Օ+ 1\o\$zٟo f ԁMf5T 棕5[SCɸ\ _θ翴`j}]o:g<<|tƣvt˨CYc+(5Ѭofԥ取/ʩ\$!s1Iw'ur5Vk-! 7<Z?lm+X`^Mki>g)˷چ gjb, JF1 c'@{#y}=Z[[-!Hv Α5 ܤ;43@RjqRGWLScڬ@((((((((((c.jUi\[U.- OH.keVne!/e,vU+"úհ^ cMz DGj^²<5alGv{ܮ\$f6ҫV ë1ESZL{F):0O'dSҤ-}^C5+G.`b? /.`Q<\$gA5~\$ ,HҽB-٫Cք~TFDv8 _Wj ;&CҶܛ.?F}(ͯ|9[m# |}6H>}ֽi^ ?_4cca0j#iGR _Z=wZ'y[oOV< e` Vw3krM# W8o'[K̂B [c5?0 H ƟZ91l"o!r;4 yr*~ֱ֟{g WV#@EAff -44 }l[ H[GHA>vo+NOӬ-!tb?.aZ&10O/r[o6{ˀ~54@ڣ2I؎MMv2s53IQP-JMK ^f8}(q=cPQEQEQEQEQEQEQEQEQEQEQEQEQE!J- _ SJ hMFb5bhEf4ݦ R旚{RyҀ*Y=D>W]ա\=+J7(aXbS㱠 >1Ȣ̅ꌹ/a@Co#Vi ( p*O0i&@j 6[븶685L1\$eq@8mjNEՊ((((((((((((()-LM YPh}i D f0Њ@0irjكژa .RiI?&Ry4@0jS T)[T O "LR( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( LR@ HREG劒˥IEF#)Pv▊1EPEPE `ihpS"]AWQEQEQEQQ]5nHċUfeI,qW-9A9#Z(*^/OC@QEQEi܌2MU,r(((((((((((((((((()6Ө O,TP!I䊱F()|S˧yu%ݴE-b(((((((((((((((((((((((((7Y'OZrY3Dѹ 7Zh"ż6@ ZF G"⡵i1.QEQEQHd)o eүUK8 EPEPU%7FF Ao2`ȩ2|NÊt7l14h"M: ) *)w0E,C j1=ě Z*TPEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQEQE5"M:͖^Wހ&eL %\$ 梭.@6 1⥸v%Pv;N@sdI~aCLlHH&fq | {"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": 0.9801874160766602, "perplexity": 410.5553238415003}, "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-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00695.warc.gz"} |
http://math.stackexchange.com/questions/261298/why-za-is-an-equivalence-relation-on-a?answertab=oldest | # Why $Z(A)$ is an equivalence relation on $A$?
For every algebra $A$, the center $Z(A)$ is a congruence on $A$. Why is $Z(A)$ an equivalence relation on $A$?
-
I don't understand. If "congruence" means what I think it means then I don't know what you mean when you say that a subset of $A$ (as opposed to a subset of $A \times A$) is a congruence. – Qiaochu Yuan Dec 18 '12 at 8:32
So perhaps your confusion arises from talking about the center of a group rather than a general or "universal" algebra. In that case, for a group $G$, the center $Z(G)$ is not really a congruence relation of $G$. However, as a normal subgroup, $Z(G)$ is a single class of a congruence relation of $G$ -- namely, the class containing the identity element of $G$.
In general, each normal subgroup $N$ of a group $G$ can be identified with a congruence relation, $\theta_N$, of $G$. The congruence classes of $\theta_N$ are the cosets of $N$. Thus, the lattice of normal subgroups of $G$ is isomorphic to the lattice of congruence relations of $G$. It is because of this identification that a normal subgroup is sometimes loosely referred to as a "congruence" of the group. | {"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": 0.9539819359779358, "perplexity": 74.36269373964404}, "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-2015-48/segments/1448398467979.1/warc/CC-MAIN-20151124205427-00352-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/152815-symmtric-difference-question.html | # Thread: A Symmtric Difference Question
1. ## A Symmtric Difference Question
I have to admit that I am completely lost on this one.
For any three sets $A, B, C,$ show that $A \bigtriangleup B = C$ if and only if $A=B \bigtriangleup C$.
2. Originally Posted by akolman
I have to admit that I am completely lost on this one.
For any three sets $A, B, C,$ show that $A \bigtriangleup B = C$ if and only if $A=B \bigtriangleup C$. | {"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": 6, "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": 0.8928259015083313, "perplexity": 184.05520775417554}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698540839.46/warc/CC-MAIN-20161202170900-00048-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://mathhelpforum.com/math-topics/32626-need-some-help-alevel-maths.html | # Math Help - need some help in Alevel maths
1. ## need some help in Alevel maths
hey everyone i have attached the problems i will try to see what how u guys can solve them and later i will ask questions thanks a lot.
Attached Thumbnails
hey everyone i have attached the problems i will try to see what how u guys can solve them and later i will ask questions thanks a lot.
4a). The two reaction forces are equal (you are told so) and as the plank is
statric their sum must equal the total load which is (60+90)g Newtons, so
the reaction of the plank at B is 75g N.
RonL
hey everyone i have attached the problems i will try to see what how u guys can solve them and later i will ask questions thanks a lot.
4b) You do this by taking moments of the forces about A.
Let $x$ be the distance of the centre of mass of the plank from A. Then
the sum of the moments of the three forces acting (the weight of the woman,
the weight of the plank and the reaction at B - the reaction at A has zero moment about A so we ignore it) is:
$2 \times 60 \times g + x \times 90 \times g - 6 \times R_B=0$
where $R_B$ is the reaction at B found in the first part of this question.
RonL | {"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": 0.9376109838485718, "perplexity": 521.2002601002027}, "config": {"markdown_headings": true, "markdown_code": false, "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-2016-07/segments/1454701158609.98/warc/CC-MAIN-20160205193918-00045-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://eccc.weizmann.ac.il/eccc-reports/2006/TR06-146/index.html | Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > DETAIL:
### Paper:
TR06-146 | 30th September 2006 00:00
#### On the Subgroup Distance Problem
TR06-146
Authors: Christoph Buchheim, Peter J Cameron, Taoyang Wu
Publication: 1st December 2006 22:13
Keywords:
Abstract:
We investigate the computational complexity of finding an element of
a permutation group~$H\subseteq S_n$ with a minimal distance to a
given~$\pi\in S_n$, for different metrics on~$S_n$. We assume
that~$H$ is given by a set of generators, such that the problem
cannot be solved in polynomial time by exhaustive enumeration. For
the case of the Cayley Distance, this problem has been shown to be
NP-hard, even if~$H$ is abelian of exponent two~\cite{pinch06}. We
present a much simpler proof for this result, which also works for
the Hamming Distance, the $l_p$ distance, Lee's Distance, Kendall's
tau, and Ulam's Distance. Moreover, we give an NP-hardness proof
for the $l_\infty$ distance using a different reduction idea.
Finally, we settle the complexity of the corresponding
fixed-parameter and maximization problems.
ISSN 1433-8092 | Imprint | {"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": 0.9270020723342896, "perplexity": 2351.5128785756356}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578759182.92/warc/CC-MAIN-20190426033614-20190426055614-00261.warc.gz"} |
http://mathhelpforum.com/pre-calculus/182944-help-me-limit-problem.html | # Math Help - Help me with a limit problem
1. ## Help me with a limit problem
Could anyone help me with this problem:
I have got a wrong result
But I find that if I divided the fraction's upper part and lower part by -x instead of x at the same time, I can get the correct result which is 1/4.
But could anyone explain that, why I can't divided the fraction's upper part and lower part by x at the same time?
2. 1) Never wrote "1/0" again. It doesn't mean anything.
2) Your problem is the sign. When you square under the radical, it is not the same.
At least two ways to proceed.
a) Do you know l'Hospital? That will help.
b) (-x)^2 = x^2. See if you can figure out why changing the sign on the radical, when moving the x^2 inside, is the right thing to do.
3. Originally Posted by piscoau
Could anyone help me with this problem:
I have got a wrong result
But I find that if I divided the fraction's upper part and lower part by -x instead of x at the same time, I can get the correct result which is 1/4.
But could anyone explain that, why I can't divided the fraction's upper part and lower part by x at the same time?
$\lim_{x\to - \infty} \sqrt{4x^2-x}+2x= \lim_{x\to \infty} \sqrt{4x^2+x}-2x$ | {"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": 0.8841500878334045, "perplexity": 597.4193026766876}, "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-2015-06/segments/1422115857200.13/warc/CC-MAIN-20150124161057-00182-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://physics.aps.org/articles/v5/87 | # Viewpoint: Transition Path Times for DNA and RNA Folding from Force Spectroscopy
Physics 5, 87
Experiments show that the times required to cross the barrier for the folding and unfolding of different nucleic acids are consistently about a few microseconds, despite many orders of magnitude differences in rate coefficients.
The folding of proteins and nucleic acids from disordered to highly ordered structures serves as a prime example of complex conformational dynamics in the condensed phase. Beyond establishing the connection between the one-dimensional information contained in the sequences of these biopolymers and their three-dimensional structures in the folded state, a central problem in biological physics is the development of a rigorous and quantitative understanding of folding mechanisms. All the mechanistic information on how a biopolymer folds is contained in its transition paths, the tiny segments of an equilibrium trajectory during which the molecule actually crosses the barrier between folded and unfolded states, after a very large number of failed attempts (see Fig. 1).
Folding transition paths are a single-molecule property and have not been observed by experiment for either proteins or nucleic acids, but this situation is rapidly changing [1,2]. Using single-molecule fluorescence experiments, Chung et al. [2] measured average transition path times that are within a factor of $5$ for proteins whose folding rates differ by $4$ orders of magnitude. Given the complexity of the structural rearrangements involved in folding, this finding is surprising but, interestingly, can be readily explained by the theory for diffusion of a single particle on a one-dimensional free-energy surface [3,4]. Now, in a paper in Physical Review Letters, Krishna Neupane at the University of Alberta, Canada, and colleagues [1] exploit this theory to greatly extend the work of Chung et al. [2]. In technologically demanding optical-tweezer experiments analyzed using energy landscape theory, they find remarkably similar low-microsecond transition times for a wide range of nucleic acids, whose theoretically extrapolated unfolding rates in the absence of force range over $18$ orders of magnitude [1]. These single-molecule fluorescence and force spectroscopic experiments represent a major step forward toward the goal of observing transition paths in experiments, i.e., of watching proteins and nucleic acids fold one molecule at a time.
Chung et al. [2] determined the transition path time from a photon-by-photon analysis [5] of hundreds of transition events observed in fluorescence resonant energy transfer experiments. In the new work, Neupane et al. [1] studied folding transition paths by mechanical pulling on the ends of nucleic acids tethered by DNA handles to optically trapped beads. With this setup, a constant piconewton (pN) force was exerted on the nucleic acid, and the folding and unfolding events were detected by monitoring the bead distance, which reported on the nanometer (nm) changes in the molecular extension. Impressive statistics only partly explain their success, since the time resolution of the experiments is inherently limited. For a laser trap with a spring constant $k∼0.1$ pN/nm and a bead of radius $R∼1$ micrometers, the diffusional relaxation time, assuming Stokes-Einstein friction, is on the order of $100$ microseconds, setting a limit on what can be resolved directly without using smaller beads and/or stiffer traps. To overcome this limit, Neupane et al. [1] employed formalisms of energy landscape theory.
Theory [3], simulation [6], and ensemble experiments [7] for proteins with two-state behavior support the postulate of the energy landscape formalism that (un)folding transitions can be described by diffusive dynamics on a low-dimensional, free-energy surface with a single barrier separating the folded and unfolded states. For the two-dimensional free-energy surface in Fig. 1, $x$ is the molecular extension probed by the pulling experiments [1], and $y$ is an unresolved coordinate that helps separate the two states, folded and unfolded. Along the transition path, which may include transient visits to low-population intermediates, the mechanistically relevant changes occur; so the transition path is by far the most important trajectory segment.
For dynamics described by the Smoluchowski diffusion equation, an analytic expression relates the duration of the transition path between two states on a one-dimensional free-energy surface to the barrier height $ΔG≠$ and diffusivity $D$ [4]. An accurate approximation derived by Szabo [8] shows that the average transition path time $τTP¯$ grows only logarithmically with $ΔG≠$ for a high and parabolic free-energy barrier, in contrast to the exponentially growing average time that a molecule resides or “waits” in a state before a “lucky” series of thermal kicks results in passage over the barrier. The rate coefficient, the reciprocal of this average time, provides the conventional measure of speed of a process in kinetic experiments.
To employ this formalism, Neupane et al. [1] monitored fluctuations in the extension $x$ of the molecules in presence of a constant external force. The molecular extensions are a reasonable choice of reaction coordinate for the folding by zipping and unfolding by unzipping of nucleic acid hairpins and similar structures (see Fig. 1). The corresponding one-dimensional free-energy surface, $G(x)$ vs $x$ (schematically shown in Fig. 1, bottom) was estimated from the observed distributions, corrected for the effects of the trap and DNA handle and instrument response by deconvolution. Knowing $G(x)$, the diffusivity $D$ could be estimated from the measured rate coefficients for (un)folding (the reciprocal of the mean residence time), and then from $G(x)$ and $D$, the mean transition path time $τTP¯$ could be calculated (see also Fig. 1).
Neupane et al. [1] validated their estimates of $ΔG≠$ and $D$ by measuring unfolding times under a continuously increasing force, interpreted with a different, but related theory [9]. By fitting theoretical expressions to the measured data, Neupane et al. obtained independent estimates of the unfolding barrier height $ΔG≠$ and the diffusivity $D$ that led to consistent transition path times for the DNA hairpins. For the more complex RNA structures, only this latter technique was used to estimate $τTP¯$. Neupane et al. further backed up their calculated low-microsecond transition path times by estimating an upper bound of $∼50$ microseconds directly from the observed bead trajectories, consistent with the rough estimate of the bead relaxation time above. Moreover, the observed diffusion coefficients $D$ are consistent with earlier estimates for DNA hairpin zipping and unzipping.
Measuring average transition path times is only a first step. Future studies will aim to characterize the evolution of structure during transition paths. At present, researchers can only address this by molecular simulations [10,11]. However, we are optimistic that observation of individual transition paths will soon become possible in experiment, through continued improvements in time resolution and the application of methods to probe several intramolecular distances simultaneously.
## References
1. K. Neupane, D. B. Ritchie, H. Yu, D. A. N. Foster, F. Wang, and M. T. Woodside, ”Transition Path Times for Nucleic Acid Folding Determined from Energy-Landscape Analysis of Single-Molecule Trajectories,” Phys. Rev. Lett. 109, 068102 (2012)
2. H. S. Chung, K. McHale, J. M. Louis, and W. A. Eaton, “Single-Molecule Fluorescence Experiments Determine Protein Folding Transition Path Times,” Science 335, 981 (2012)
3. J. D. Bryngelson and P. G. Wolynes, “Intermediates and Barrier Crossing in a Random Energy Model (with Applications to Protein Folding),” J. Phys. Chem. 93, 6902 (1989)
4. G. Hummer, “From Transition Paths to Transition States and Rate Coefficients,” J. Chem. Phys. 120, 516 (2004)
5. I. V. Gopich and A. Szabo, “Decoding the Pattern of Photon Colors in Single-Molecule FRET,” J. Phys. Chem. B 113, 10965 (2009)
6. R. B. Best and G. Hummer, “Coordinate-Dependent Diffusion in Protein Folding,” Proc. Natl. Acad. Sci. USA 107, 1088 (2010)
7. J. Kubelka, E. R. Henry, T. Cellmer, J. Hofrichter, and W. A. Eaton, “Chemical, Physical, and Theoretical Kinetics of an Ultrafast Folding Protein,” Proc. Natl. Acad. Sci. USA 105, 18655 (2008)
8. H. S. Chung, J. M. Louis, and W. A. Eaton, “Experimental Determination of Upper Bound for Transition Path Times in Protein Folding from Single-Molecule Photon-by-Photon Trajectories,” Proc. Natl. Acad. Sci. USA 106, 11837 (2009)
9. O. K. Dudko, G. Hummer, and A. Szabo, “Theory, Analysis, and Interpretation of Single-Molecule Force Spectroscopy Experiments,” Proc. Natl. Acad. Sci. USA 105, 15755 (2008)
10. P. G. Bolhuis, D. Chandler, C. Dellago, and P. L. Geissler, “Transition Path Sampling: Throwing Ropes over Rough Mountain Passes, in the Dark,” Annu. Rev. Phys. Chem. 53, 291 (2002)
11. K. Lindorff-Larsen, S. Piana, R. O. Dror, and D. E. Shaw, “How Fast-Folding Proteins Fold,” Science 334, 517 (2011)
Gerhard Hummer received his Ph.D. in physics for work done jointly at the University of Vienna and at the Max Planck Institute for Biophysical Chemistry. He is a Senior Investigator in the Laboratory of Chemical Physics at the National Institutes of Health, and a recipient of the 2010 Raymond and Beverly Sackler International Prize in Biophysics. His research in biological physics focuses on the theory and simulation of biomolecular systems, including the theory of force spectroscopy.
William Eaton received his M.D. and Ph.D. in molecular biology from the University of Pennsylvania. He is an NIH Distinguished Investigator in the Laboratory of Chemical Physics at the National Institutes of Health, and recipient of the 2012 Max Delbrück Prize in Biological Physics of the American Physical Society. His research in biological physics focuses on single molecule fluorescence and statistical mechanical models for protein folding.
## Related Articles
Biological Physics
### Synopsis: Better Signals from Electronic Body Implants
The transmission distance of a wireless implant could be tripled by carefully tuning the frequency of the electromagnetic signal it emits. Read More »
Fluid Dynamics
### Synopsis: How Hairy Tongues Help Bats Drink Nectar
Experiments and theory show that hairs on a bat’s tongue allow the animal to drink 10 times more nectar than it could if its tongue were smooth. Read More »
Nuclear Physics
### Viewpoint: Heaviest Element Has Unusual Shell Structure
Calculations of the structure in oganesson—the element with the highest atomic number—reveal a uniform, gas-like distribution of its electrons and nucleons. Read More » | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 48, "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": 0.851123034954071, "perplexity": 2299.7799267822893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257646952.38/warc/CC-MAIN-20180319140246-20180319160246-00777.warc.gz"} |
http://math.stackexchange.com/questions/129744/factors-of-a-polynomial-in-several-variables | # Factors of a polynomial in several variables
Fix an embedding of $\overline{\mathbb{Q}}$ into $\mathbb{C}$. Suppose you have a polynomial in several variables, with algebraic coefficients: $P\in \overline{\mathbb{Q}}[z_1, \ldots, z_n]$. Also suppose we find some factorization $P=QR$ for some $Q, R\in \mathbb{C}[z_1, \ldots, z_n]$ (with, a priori, just complex coefficients).
It seems intuitively clear to me that $Q$ and $R$ would need to have algebraic coefficients as well. (Well, almost -- we might need to do some scaling first. Let's say we pick some $\alpha\in\mathbb{C}^\times$ such that $\alpha Q$ has at least one nonzero, algebraic coefficient. Then replace $Q$ and $R$ with $\alpha Q$ and $\frac{1}{\alpha} R$.) Can anyone remind me how you would go about proving this?
(I'm sure this has nothing to do with the particulars of $\overline{\mathbb{Q}}$ and $\mathbb{C}$, really -- we could probably replace them with any pair of nested, algebraically closed fields.)
- | {"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": 0.928853452205658, "perplexity": 179.31995909922992}, "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-2015-32/segments/1438042988250.59/warc/CC-MAIN-20150728002308-00087-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.electricalexams.co/compilers-mcq-quiz/ | # Compilers MCQ Quiz – Objective Question with Answer for Compilers
1. Which of the following helps in reducing the energy consumption of the embedded system?
A. compilers
B. simulator
C. debugger
D. emulator
The compilers can reduce the energy consumption of the embedded system and the compilers performing the energy optimizations are available.
2. Which of the following help to meet and prove real-time constraints?
A. simulator
B. debugger
C. emulator
D. compiler
There are several reasons for designing the optimization and compilers and one such is that it could help to meet and prove the real-time constraints.
3. Which of the following is an important ingredient of all power optimization?
A. energy model
B. power model
C. watt model
D. power compiler
Saving energy can be done at any stage of the embedded system development. The high-level optimization techniques can reduce power consumption and similarly compiler optimization also can reduce the power consumption the most important thing in power optimization is the power model.
4. Who proposed the first power model?
A. Jacome
B. Russell
C. Tiwari
D. Russell and Jacome
Tiwari proposed the first power model in the year 1974. The model includes the so-called bases and the inter-instruction instructions. Base costs of the instruction correspond to the energy consumed per instruction execution when an infinite sequence of that instruction is executed. Inter instruction costs model the additional energy consumed by the processor if instructions change.
5. Who proposed the third power model?
A. Tiwari
B. Russell
C. Jacome
D. Russell and Jacome
The third model was proposed by Russell and Jacome in the year 1998.
6. Which compiler is based on the precise measurements of two fixed configurations?
A. first power model
B. second power model
C. third power model
D. fourth power model
The third model was proposed by Russell and Jacome in the year 1998 and is based on the precise measurements of the two fixed configurations.
7. What does SPM stand for?
B. sensor parity machine
D. sensor parity memories
The smaller memories provide faster access and consume less energy per access and SPM or scratch pad memories are a kind of small memory that access fastly and consumes less energy per access and it can be exploited by the compiler.
8. Which model is based on precise measurements using real hardware?
A. encc energy-aware compiler
B. first power model
C. third power model
D. second power model
The encc-energy-aware compiler uses the energy model by Steinke et al. it is based on the precise measurements of the real hardware. The power consumption of the memory, as well as the processor, is included in this model.
9. What is the solution to the knapsack problem?
A. many-to-many mapping
B. one-to-many mapping
C. many-to-one mapping
D. one-to-one mapping
The knapsack problem is associated with the size constraints, that is the size of the scratch pad memories. This problem can be solved by one-to-one mapping which was presented in an integer programming model by Steinke et al.
10. How can one compute the power consumption of the cache?
A. Lee power model
B. First power model
C. Third power model
D. CACTI | {"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": 0.8651686906814575, "perplexity": 2407.173460307168}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00725.warc.gz"} |
https://tex.stackexchange.com/questions/209909/latex-doesnt-break-spaces-in-texttt-correctly-overfull-hbox | # Latex doesn't break spaces in \texttt correctly (Overfull \hbox)
I'm having some text in an \texttt environment and it turns out that latex breaks the lines in these environments sometimes correctly and sometimes not.
A working minimum example
\documentclass{scrartcl}
\begin{document}
\section{Test}
This is some text text and even more text \texttt{and some variable names which are written in typewriter font
\end{document}
And a non-working one (leading to an overfull \hbox)
\documentclass{scrartcl}
\begin{document}
\section{Test}
This is some text text and even more text \texttt{and some variable names which are writtenin typewriter font}
\end{document}
It seems like if there is an too long word at the end of the line LaTeX isn't capable of setting a proper line break? I've found some similar problems with line breaks in \texttt, but the problem was always that there was no space in the \texttt, this is definitely not the problem here.
Is there a workaround for this problem?
• LaTeX sets things so hyphenation is not tried for words in typewriter font. – egreg Oct 31 '14 at 16:41
LaTeX doesn't hyphenate tt fonts, which makes it hard to set the paragraph.
You could use sloppypar to allow white space to stretch more to compensate, or you could reset the hyphen for the tt font (which is a global change, affecting tt for the rest of the document).
\documentclass{scrartcl}
\begin{document}
\section{Test1}
This is some text text and even more text \texttt{and some variable names which are writtenin typewriter font}
\section{Test2}
\begin{sloppypar}
This is some text text and even more text \texttt{and some variable names which are writtenin typewriter font}
\end{sloppypar}
\section{Test3}
{{\ttfamily \hyphenchar\the\font=\-}%
This is some text text and even more text \texttt{and some variable names which are writtenin typewriter font}\par}
\end{document}
Because LaTeX doesn't hyphenate words that are typeset in "teletype" font (aka typewriter font, monospaced font), you may need to give up on full justification of the paragraph in question. Instead, you may wish to typeset it in \raggedright mode. To keep the scope of \raggedright local to just one paragraph, be sure to encase it in curly braces, making sure to insert \par before the closing curly brace.
\documentclass{scrartcl}
\begin{document}
\section{Test}
{\raggedright
This is some text text and even more text \texttt{and some variable names
which are writtenin typewriter font}.
\par}
\end{document}
` | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9291388392448425, "perplexity": 2700.3760269145873}, "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/1570987829507.97/warc/CC-MAIN-20191023071040-20191023094540-00146.warc.gz"} |
https://brilliant.org/discussions/thread/a-probe-to-the-sun/ | # A probe to the sun
For the purpose of collescting data related to the space near the sun ,a probe is to be prepared that would be sending solar data back to the earth as it rushes towards it's demise.The design plan for our probe is that we will release it from our spacecraft in such a way that it has zero velocity relative to the sun.the probe has a mass $200$ kg and no rocket engine ,so it simply falls into the sun from rest due to the gravitational force from the sun.All other gravitational forces on the probe from other planets are neglected (for simplyfication) We will model the probe as a system consisting of a single particle.The gravitational force from the sun acts as the interaction with the environment ,doing work on thr probe .The result of this work is the increasing kinetic energy of the probe as it falls towards the sun .Thus situation is suitable for application of work energy theorem.
The initial separation of the probe and the sun is the radius of earth's orbit , $1.50\times 10^{11}$ m .The final separation is the radius of the sun $7\times 10^8$.The amount of work done by the gravitational force of the sun on the probe can be found using
$\displaystyle\int_{r_i}^{r_f}$$F_r$ $dr$
As the force is in radial direction , further
$W=\displaystyle\int_{r_i}^{r_f}$$F_r$$dr$$= -\displaystyle\int_{r_i}^{r_f}$$\dfrac{G\times{M_{sun}}\times{m_{probe}}}{r^2}=$$G\times{M_{sun}}$$\times{m_{probe}}\dfrac{r_i-r_f}{r_f\times{r_i}}$
Where G is univerasl gravitational constant = $6.7\times{10^{-11}}$N$m^2$/$kg^2$
mass of sun = $2.8\times {10^30}$,using $W=\triangle {K}$ , final speed can be calculated.
This is part of the set A probe to Sun
Note by Parth Lohomi
5 years, 3 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
try the problems here
- 5 years, 3 months ago
There is some problem in the values given. Please rectify the problem asap
- 4 years, 8 months ago
The whole set is problematic. Suggestion is, escape this Set now. I've Reported this set to the Brilliant Physics. As soon as its reviewed and corrected, I will let you know. Please drop your E-mail address.
- 4 years, 6 months ago
[email protected]
- 4 years, 4 months ago
OK. You have an E-mail to check.
- 4 years, 4 months ago
The answer to your question " A probe to sun-3 " comes out to be 16564.26877 m/s. But the answer given is 17000. Why? Though that's close but isn't the exact answer 16564 m/s. Please correct me if I am wrong somewhere.
- 4 years, 7 months ago
The whole set is problematic. Suggestion is, escape this Set now. I've Reported this set to the Brilliant Physics. As soon as its reviewed and corrected, I will let you know. Please drop your E-mail address.
- 4 years, 6 months ago | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 26, "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": 0.9650456309318542, "perplexity": 1034.6050592638226}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347399830.24/warc/CC-MAIN-20200528170840-20200528200840-00313.warc.gz"} |
https://docs.edhecinfra.com/display/IM/Sharpe+Ratio | Go to start of banner
# Sharpe Ratio
The Sharpe ratio is calculated by dividing the mean excess return of the index by its volatility, annualized over the horizon under consideration. In some years, the risk-free rate used to compute excess returns can be negative. The higher the Sharpe ratio, the higher the excess returns for a unit of risk.
where:
$//$ denotes the annualised mean Excess Returns of the index.
$//$ denotes the annualised Index Return Volatility measure.
We compute sharpe ratios depending on the choice of currency to report returns, assuming that for the 'risk-free' asset for any given investor is the domestic 3-month risk-free asset. A Sharpe Ratio based on local currency returns and risk-free rates is also computed using local currency excess returns, as described here.
We also compute an Adjusted Sharpe ratio to account for the skewness and excess kurtosis in the returns distribution
where:
$//$ is the skewness of the return distribution
$//$ is the excess kurtosis of the return distribution
• No labels | {"extraction_info": {"found_math": true, "script_math_tex": 4, "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": 0.8548091650009155, "perplexity": 1865.6685257122217}, "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-2023-06/segments/1674764499967.46/warc/CC-MAIN-20230202070522-20230202100522-00665.warc.gz"} |
https://www.physicsforums.com/threads/find-the-final-angular-speed-of-the-target.288231/ | # Homework Help: Find the final angular speed of the target
1. Jan 28, 2009
### pentazoid
1. The problem statement, all variables and given/known data
A fairgorund target consists of a uniform circular disk of mass M and radius a that can turn freely about its diamerter which is fixed in a vertical position. Iniitally the target is at rest . A bullet of mass m is moving with speed u along a horizontal line at right angles to the target . The bullet embeds itself in the target at a point distance from the rotation axis . Find the final angular speed of the target.(the moment of inertica of the disk about its roation axis is Ma^2/4
2. Relevant equations
Equations for rotational energy
3. The attempt at a solution
T_trans., initial + T_rot., initial, + V_initial=T_rot, final . There is no translational energy now that the bullet is embedded into the block.
1/2*m*u^2++0+0=1/2(Ma^2/4
m*b^2/4)*omega^2? Is my equation correct? There is no potential energy.
2. Jan 29, 2009
### tiny-tim
inelastic collision
Hi pentazoid!
Nooo … energy is not usually conserved in collisions …
and certainly not when one body embeds itself in the other!
Hint: momentum is always conserved in collisions …
and that includes angular momentum.
3. Jan 29, 2009
### pentazoid
Re: inelastic collision
yes I know energy isn't conserved because the collision is not elastic since the bullet becomes embedded into the block.
L_initial=L_final
m*v_bullet+0=(M+m)v_new?
and then I can plug my v_new into my energy equation?
4. Jan 29, 2009
### tiny-tim
No, that's linear momentum …
you need angular momentum.
(and what energy equation? )
5. Jan 29, 2009
### pentazoid
right , how silly of me.
m*v_bullet*a=(m+M)*v_final*(a+b)? Would I then plug v_final into my energy equation.
a being the horizontal distance and a+b being the horizontal distance plus the radius of the disk?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | {"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": 0.9814805388450623, "perplexity": 2094.6562520519938}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948285.62/warc/CC-MAIN-20180426144615-20180426164615-00158.warc.gz"} |
https://global-sci.org/intro/article_detail/cicp/7845.html | Volume 3, Issue 1
Theoretical and Experimental Studies of Seismoelectric Conversions in Boreholes
DOI:
Commun. Comput. Phys., 3 (2008), pp. 109-120.
Published online: 2008-03
Preview Full PDF 387 1089
Export citation
Cited by
• Abstract
We present theoretical and experimental studies on the effects of formation properties on seismoelectric conversions in fluid-filled boreholes. First, we derive the theoretical formulations for seismoelectric responses for an acoustic source in a borehole. Then, we compute the electric fields in boreholes penetrating formations with different permeability and porosity, and then we analyze the sensitivity of the converted electric fields to formation permeability and porosity. We also describe the laboratory results of the seismoelectric and seismomagnetic fields induced by an acoustic source in borehole models to confirm our theoretical and numerical developments qualitatively. We use a piezoelectric transducer to generate acoustic waves and a point electrode to receive the localized seismoelectric fields in layered boreholes and the electric component of electromagnetic waves in a fractured borehole model. Numerical results show that the magnitude ratio of the converted electric wave to the acoustic pressure increases with the porosity and permeability increases in both fast and slow formations. Furthermore, the converted electric signal is sensitive to the formation permeability for the same source frequency and formation porosity. Our experiments validate our theoretical results qualitatively. An acoustic wave at a fracture intersecting a borehole induces a radiating electromagnetic wave.
• Keywords | {"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": 0.8203442096710205, "perplexity": 1927.6960563360913}, "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/1596439738816.7/warc/CC-MAIN-20200811150134-20200811180134-00051.warc.gz"} |
http://math.stackexchange.com/questions/209985/is-there-are-specific-way-to-solve-coupled-first-order-differential-equations-wi/210003 | # is there are specific way to solve coupled first-order differential equations with coefficients varying?
suppose I have "n" coupled differential equation represented by the matrix,
Y = A Y
, where Y is the column matrix containing first derivatives, namely, y1(t), y2(t), ... yn(t) . A is a square matrix whose each element contains some function dependent on "t" (not constants) and Y is the column matrix containing the solution set, namely, y1(t), y2(t), ... yn(t) .
If A, contained constants, then its easy to solve by Matrix Exponential method or Eigen-Value method. But, if it contains some varying functions, then is there any approach to solve this. Please, direct me to a good reference, if possible, with an example.
-
Sounds kind of like covariant differentiation. – ben Oct 9 '12 at 17:59
@ben am sorry, I didn't get u, but actually this is not a derivative, but a set of coupled differential equations, represented in a matrix form. But, till now I have seen the case for which 'A' is a square matrix containing constants, but what about the case if 'A' is a square matrix, which are functions of 't' – tsndiffopera Oct 9 '12 at 18:05
If your matrices commute at different times, that is $A(t)A(s) = A(s)A(t)$, then the solution is $Y(t) = \exp \Big( \int_0^t A(s) \mathrm{d}s \Big) Y(0)$. Here $\exp$ is just an ordinary matrix exponential.
If they do not commute, then you might be able to use Dyson formula (Dyson series). In the realm of quantum mechanics you have the following proposition:
Let $\mathscr{H}$ be a Hilbert space, $A: \mathbb{R} \to \mathscr{B}(\mathscr{H})$ a strongly continuous Hermitian operator valued function (i.e. each $A(t)$ is bounded and Hermitian). Then there is a unique solution of $$i \partial_t \psi(t) = A(t) \psi(t), \quad \psi(0) \in \mathscr{H}.$$ If $A$ is continuous w.r.t. the operator norm, then the solution is of the form $$\psi(t) = T \exp \Big( - i \int_0^t A(\tau) \mathrm{d}\tau \Big) \psi(0),$$ where $T \exp$ is the so called "time-ordered exponential".
I suspect that the assumption of Hermiticity is not important (you need it for unitarity of the solution). Boundedness, on the other hand, is essential. But this is no problem in yout case of a matrix.
-
Wow, its great that you've tried to help me. Million Thanks for that. Can you please provide me a reference for above discussion, so that I can look into it in more detail. And also can you please tell me what Dyson series is. I'm very new to it, but very much acquainted with Taylor series though. Thanks again for your attempt(a whole-hearted appreciation). – tsndiffopera Oct 9 '12 at 23:20
excuse me sir, is it "0" in place of "s" in the above expression like Psi(t) = Texp(-i*int(A(tau).d(tau))_0-to-t)*Psi(0) ? – tsndiffopera Oct 9 '12 at 23:48
I have improved the answer above. Hope it helps. I am not able to give you additional exact reference as I am away from my office. But I guess this should be a fairly standard stuff from any textbook on mathematical methods of quantum mechanics. I would also recommend you to look for some general theory of differential equations in Banach spaces. I will try to give you some references soon. – kalvotom Oct 9 '12 at 23:50
also, please provide me a reference, where can I get the background knowledge for it. Because, I myself, am new to Matrix Algebra, but quite good @ Calculus. – tsndiffopera Oct 9 '12 at 23:51
Yes, it was a typo. You can have a different value of the inital time, though. – kalvotom Oct 9 '12 at 23:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.900574803352356, "perplexity": 221.3649063002553}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207928923.85/warc/CC-MAIN-20150521113208-00075-ip-10-180-206-219.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/54317-binomial-theorem.html | # Math Help - Binomial Theorem
1. ## Binomial Theorem
A computer programmer claims that he generated six real numbers a_1, a_2,...,a_6 so that the sum of any four consecutive a_i is positive,
but the sum of any 3 consecutive a; is negative. Prove that his claim
is false.
thanx 4 the help
2. $a_1+a_2+a_3+a_4>0$ and $a_1+a_2+a_3<0$ therefore $a_4>0$. Following this idea you get $a_5>0$ and $a_6>0$. But then how can $a_4+a_5+a_6$ be negative?! | {"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": 6, "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": 0.9231361746788025, "perplexity": 1082.2353831572768}, "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-2014-41/segments/1410657133078.21/warc/CC-MAIN-20140914011213-00020-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
http://math.stackexchange.com/questions/289022/exist-automorphism-from-an-element-to-its-conjugate | # Exist automorphism from an element to its conjugate
I was asked to prove the galois group of a given normal extension is non-abelian. My original solution was to use isomorphism extension theorem but that was not taught in class. So, in my new attempt to solve it, I hit a road bum.
To show non-abelian, I need 2 automorphisms. I already know the 2 automorphisms exist but I don't know how to prove one of them exist.
Let $K$ be a normal extension of $\mathbb{Q(\sqrt[3]7)}$. How would I prove that given any element $a\in K$ and $b$, a conjugate of $a$, there exist an automorphism $\alpha\in Gal(K/\mathbb{Q})$ such that $\alpha(a)=b$? Let's assume $K$ is a finite dimension over $\mathbb{Q(\sqrt[3]7)}$ because infinite dimension was not taught in class.
Thanks!
EDIt: Added a few more details
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9413520097732544, "perplexity": 150.6393597975417}, "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/1438042986444.39/warc/CC-MAIN-20150728002306-00245-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://stephens999.github.io/fiveMinuteStats/summarize_interpret_posterior.html | Last updated: 2019-03-31
Checks: 6 0
Knit directory: fiveMinuteStats/analysis/
This reproducible R Markdown analysis was created with workflowr (version 1.2.0). The Report tab describes the reproducibility checks that were applied when the results were created. The Past versions tab lists the development history.
Great! Since the R Markdown file has been committed to the Git repository, you know the exact version of the code that produced these results.
Great job! The global environment was empty. Objects defined in the global environment can affect the analysis in your R Markdown file in unknown ways. For reproduciblity it’s best to always run the code in an empty environment.
The command set.seed(12345) was run prior to running the code in the R Markdown file. Setting a seed ensures that any results that rely on randomness, e.g. subsampling or permutations, are reproducible.
Great job! Recording the operating system, R version, and package versions is critical for reproducibility.
Nice! There were no cached chunks for this analysis, so you can be confident that you successfully produced the results during this run.
Great! You are using Git for version control. Tracking code development and connecting the code version to the results is critical for reproducibility. The version displayed above was the version of the Git repository at the time these results were generated.
Note that you need to be careful to ensure that all relevant files for the analysis have been committed to Git prior to generating the results (you can use wflow_publish or wflow_git_commit). workflowr only checks the R Markdown file, but you know if there are other scripts or data files that it depends on. Below is the status of the Git repository when the results were generated:
Ignored files:
Ignored: .Rhistory
Ignored: .Rproj.user/
Ignored: analysis/.Rhistory
Ignored: analysis/bernoulli_poisson_process_cache/
Untracked files:
Untracked: _workflowr.yml
Untracked: analysis/CI.Rmd
Untracked: analysis/gibbs_structure.Rmd
Untracked: analysis/libs/
Untracked: analysis/results.Rmd
Untracked: analysis/shiny/tester/
Untracked: docs/MH_intro_files/
Untracked: docs/citations.bib
Untracked: docs/figure/MH_intro.Rmd/
Untracked: docs/figure/hmm.Rmd/
Untracked: docs/hmm_files/
Untracked: docs/libs/
Untracked: docs/shiny/tester/
Note that any generated files, e.g. HTML, png, CSS, etc., are not included in this status report because it is ok for generated content to have uncommitted changes.
These are the previous versions of the R Markdown and HTML files. If you’ve configured a remote Git repository (see ?wflow_git_remote), click on the hyperlinks in the table below to view them.
File Version Author Date Message
html 34bcc51 John Blischak 2017-03-06 Build site.
Rmd 5fbc8b5 John Blischak 2017-03-06 Update workflowr project with wflow_update (version 0.4.0).
html fb0f6e3 stephens999 2017-03-03 Merge pull request #33 from mdavy86/f/review
html 0713277 stephens999 2017-03-03 Merge pull request #31 from mdavy86/f/review
Rmd d674141 Marcus Davy 2017-02-27 typos, refs
html dc8a1bb stephens999 2017-01-28 add html, figs
Rmd c3e49d5 stephens999 2017-01-28 Files commited by wflow_commit.
# Overview
This vignette illustrates how to summarize and interpret a posterior distribution that has been computed analytically.
You should be familiar with simple analytic calculations of the posterior distribution of a parameter, such as for a binomial proportion.
# Summarising and interpreting a posterior
Suppose we have a parameter $$q$$, whose posterior distribution we have computed to be Beta(31, 71) (as here for example). What does this mean? What statements can we make about $$q$$? How do we obtain interval estimates and point estimates for $$q$$?
Remember that the posterior distribution represents our uncertainty (or certainty) in $$q$$, after combining the information in the data (the likelihood) with what we knew before collecting data (the prior).
To get some intuition, we could plot the posterior distribution so we can see what it looks like.
q = seq(0,1,length=100)
plot(q, dbeta(q, 31,71), main="Posterior for $q$", ylab="density", type="l")
Version Author Date
34bcc51 John Blischak 2017-03-06
dc8a1bb stephens999 2017-01-28
Based on this plot we can visually see that this posterior distribution has the property that $$q$$ is highly likely to be less than 0.4 (say) because most of the mass of the distribution lies below 0.4. In Bayesian inference we quantify statements like this – that a particular event is “highly likely” – by computing the “posterior probability” of the event, which is the probability of the event under the posterior distribution.
For example, in this case we can compute the (posterior) probability that $$q<0.4$$, or $$\Pr(q <0.4 | D)$$. Since we know the posterior distribution is a Be(31,71) distribution, this probability is easy to compute using the pbeta function:
pbeta(0.4,31,71)
[1] 0.9792202
So we would say “The posterior probability that $$q<0.4$$ is 0.98”.
## Interval estimates
We can extend this idea to assess the certainty (or confidence) that $$q$$ lies in any interval. For example, from the plot it looks like $$q$$ will very likely lie in the interval [0.2,0.4] because most of the posterior distribution mass lies between these two numbers. To quantify how likely we compute the (posterior) probability that $$q$$ lies in the interval $$[0.2,0.4]$$, $$\Pr(q \in [0.2,0.4] | D)$$. Again, this can be computed using the pbeta function:
pbeta(0.4,31,71) - pbeta(0.2,31,71)
[1] 0.9721229
Thus, based on our prior and the data, we would be highly confident (probability approximately 97%) that $$q$$ lies between 0.2 and 0.4. That is, $$[0.2,0.4]$$ is a 97% Bayesian Confidence Interval for $$q$$. (Bayesian Confidence Intervals are often referred to as “Credible Intervals”, and also often abbreviated to CI.)
In practice, it is more common to compute Bayesian Confidence Intervals the other way around: specify the level of confidence we want to achieve and find an interval that achieves that level of confidence. This can be done by computing the quantiles of the posterior distribution. For example, the 0.05 and 0.95 quantiles of the posterior would define a 90% Bayesian Confidence Interval.
In our example, these quantiles of the Beta distribution can be computed using the qbeta function, like this:
qbeta(0.05,31,71)
[1] 0.2315858
qbeta(0.95,31,71)
[1] 0.38065
So [0.23, 0.38] is a 90% Bayesian Confidence Interval for $$q$$. (It is 90% because there is a 5% chance of it being above 0.23 and 5% of it being above 0.38).
## Point Estimates
In some cases we might be happy to give our “best guess” for $$q$$, rather than worrying about our uncertainty. That is, we might be interested in giving a “point estimate” for $$q$$. Essentially this boils down to summarizing the posterior distribution by a single number.
When $$q$$ is a continuous-valued variable, as here, the most common Bayesian point estimate is the mean (or expectation) of the posterior distribution, which is called the “posterior mean”. The mean of the Beta(31,71) distribution is 31/(31+71) = 0.3. So we would say “The posterior mean for $$q$$ is 0.3.”
An alternative to the mean is the median. The median of the Beta(31,71) distribution can be found using qbeta:
qbeta(0.5, 31,71)
[1] 0.3026356
So we would say “The posterior median for $$q$$ is 0.3”.
The mode of the posterior (“posterior mode”) is another possible summary, although this perhaps makes more sense in settings where $$q$$ is a discrete variable rather than a continuous variable as here.
## Summary
• The most common summaries of a posterior distribution are interval estimates and point estimates.
• Interval estimates can be obtained by computing quantiles of the posterior distribution. Bayesian Confidence intervals are often called “Credible Intervals”.
• Point estimates are typically obtained by computing the mean or median (or mode) of the posterior distribution. These are called the “posterior mean” or the “posterior median” (or “posterior mode”).
# Exercise
Suppose you are interested in a parameter $$\theta$$ and obtain a posterior distribution for $$\theta$$ to be normal with mean 0.2 and standard deviation 0.4. Find
1. a 90% Credible Interval for $$\theta$$.
2. a 95% Credible Interval for $$\theta$$.
3. a point estimate for $$\theta$$.
sessionInfo()
R version 3.5.2 (2018-12-20)
Platform: x86_64-apple-darwin15.6.0 (64-bit)
Running under: macOS Mojave 10.14.1
Matrix products: default
BLAS: /Library/Frameworks/R.framework/Versions/3.5/Resources/lib/libRblas.0.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/3.5/Resources/lib/libRlapack.dylib
locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] workflowr_1.2.0 Rcpp_1.0.0 digest_0.6.18 rprojroot_1.3-2
[5] backports_1.1.3 git2r_0.24.0 magrittr_1.5 evaluate_0.12
[9] stringi_1.2.4 fs_1.2.6 whisker_0.3-2 rmarkdown_1.11
[13] tools_3.5.2 stringr_1.3.1 glue_1.3.0 xfun_0.4
[17] yaml_2.2.0 compiler_3.5.2 htmltools_0.3.6 knitr_1.21
This site was created with R Markdown | {"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": 0.8455650806427002, "perplexity": 1851.9634043104215}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570730.59/warc/CC-MAIN-20220807211157-20220808001157-00498.warc.gz"} |
https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-2-10/v/derivatives-of-tanx-and-cotx | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
# Derivatives of tan(x) and cot(x)
AP.CALC:
FUN‑3 (EU)
,
FUN‑3.B (LO)
,
FUN‑3.B.3 (EK)
## Video transcript
- [Voiceover] We already know the derivatives of sine and cosine. We know that the derivative with respect to x of sine of x is equal to cosine of x. We know that the derivative with respect to x of cosine of x is equal to negative sine of x. And so what we want to do in this video is find the derivatives of the other basic trig functions. So, in particular, we know, let's figure out what the derivative with respect to x, let's first do tangent of x. Tangent of x, well this is the same thing as trying to find the derivative with respect to x of, well, tangent of x is just sine of x, sine of x over cosine of x. And since it can be expressed as the quotient of two functions, we can apply the quotient rule here to evaluate this, or to figure out what this is going to be. The quotient rule tells us that this is going to be the derivative of the top function, which we know is cosine of x times the bottom function which is cosine of x, so times cosine of x minus, minus the top function, which is sine of x, sine of x, times the derivative of the bottom function. So the derivative of cosine of x is negative sine of x, so I can put the sine of x there, but where the negative can just cancel that out. And it's going to be over, over the bottom function squared. So cosine squared of x. Now, what is this? Well, what we have here, this is just a cosine squared of x, this is just sine squared of x. And we know from the Pythagorean identity, and this is really just out of, comes out of the unit circle definition, the cosine squared of x plus sine squared of x, well that's gonna be equal to one for any x. So all of this is equal to one. And so we end up with one over cosine squared x, which is the same thing as, which is the same thing as, the secant of x squared. One over cosine of x is secant, so this is just secant of x squared. So that was pretty straightforward. Now, let's just do the inverse of the, or you could say the reciprocal, I should say, of the tangent function, which is the cotangent. Oh, that was fun, so let's do that, d dx of cotangent, not cosine, of cotangent of x. Well, same idea, that's the derivative with respect to x, and this time, let me make some sufficiently large brackets. So now this is cosine of x over sine of x, over sine of x. But once again, we can use the quotient rule here, so this is going to be the derivative of the top function which is negative, use that magenta color. That is negative sine of x times the bottom function, so times sine of x, sine of x, minus, minus the top function, cosine of x, cosine of x, times the derivative of the bottom function which is just going to be another cosine of x, and then all of that over the bottom function squared. So sine of x squared. Now what does this simplify to? Up here, let's see, this is sine squared of x, we have a negative there, minus cosine squared of x. But we could factor out the negative and this would be negative sine squared of x plus cosine squared of x. Well, this is just one by the Pythagorean identity, and so this is negative one over sine squared x, negative one over sine squared x. And that is the same thing as negative cosecant squared, I'm running out of space, of x. There you go.
AP® is a registered trademark of the College Board, which has not reviewed this resource. | {"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": 0.9806839227676392, "perplexity": 309.0465779570505}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104668059.88/warc/CC-MAIN-20220706060502-20220706090502-00584.warc.gz"} |
http://mathhelpforum.com/differential-equations/168686-limit-series-1-a.html | # Math Help - limit of series 1 A
1. ## limit of series 1 A
i have dy/dx=x+y
y0=1
y1=1+x+(x^2)/2
y2=1+x+(x^2)/2+(x^3)/6
what is the limit of the series yn
?
1. It looks an awful lot like the exponential series.
2. Theoretically, assuming certain smoothness conditions are met, which it looks like they are, the series should converge to the solution of the DE. You're essentially doing Picard iterations there, if I'm not mistaken.
3. yes its itterations and i need to find yn where
n goes to infinity
?
4. Why can't you just solve the DE exactly? It's first order linear...
5. So, just to be clear, you're setting up the following iteration scheme:
$\displaystyle y_{n+1}(x)=\int(x+y_{n}(x))\,dx,$ right? If so, I don't get what you get. I'm seeing
$y_{0}=1,$
$y_{1}=\displaystyle\int(x+1)\,dx=x+\frac{x^{2}}{2} ,$
$\displaystyle y_{2}=\int\left(x+x+\frac{x^{2}}{2}\right)\,dx=x^{ 2}+\frac{x^{3}}{6},\dots$
Is there an initial condition or something I should know about?
6. yes y(0)=1
the whole question is about approximating
so i cant solve it like first order linear
7. Ok. So y(0) = 1 implies that
$y_{0}=1,$
$y_{1}=\displaystyle\int(x+1)\,dx=1+x+\frac{x^{2}}{ 2},$
$y_{2}=\displaystyle\int\left(x+1+x+\frac{x^{2}}{2} \right)\,dx=1+x+x^{2}+\frac{x^{3}}{6},$
$y_{3}=\displaystyle\int\left(x+1+x+x^{2}+\frac{x^{ 3}}{6}\right)dx=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6 }+\frac{x^{4}}{24},\dots$
That's what you got, I believe. The sequence $\{y_{n}\}$ converges to $e^{x}.$ To convince yourself of that fact, just expand $e^{x}$ in a Taylor series about 0, and you'll see that it's the exact same series as you're getting here.
How rigorous a proof does this require?
8. as much rigorous as you can give me
my orof is very strict
and i only started with converging serieses
9. Try a proof by induction.
10. ok
what expressio for n=k to take
and what expression for n=k+1 to take
11. Well, it looks like you want to show that, aside from the first term,
$\displaystyle y_{n}=\sum_{j=0}^{n+1}\frac{x^{j}}{j!}.$
What ideas does that give you?
12. $y_{n+1}\displaystyle =1+\int (t+ \sum_{j=0}^{n}\frac{t^{j}}{j!})dt$
how to prove that it equals $\displaystyle y_{n}=\sum_{j=0}^{n+1}\frac{x^{j}}{j!}$
13. An induction proof has two parts: the base case, and the inductive step. What's your base case?
14. aahh i understand now
thanks
15. Originally Posted by transgalactic
i have dy/dx=x+y
y0=1
y1=1+x+(x^2)/2
y2=1+x+(x^2)/2+(x^3)/6
what is the limit of the series yn
?
We have the first order linear DE...
$\displaystyle y^{'} = y + x\ \ ,\ y(0)=1$ (1)
... and we have 'forgotten' the procedure to solve it... no problem because we can use an alternative approach!...
Let's suppose that y(x) is analytic in x=0 so that is...
$\displaystyle y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n}$ (2)
But the $y^{(n)} (0)$ can be derived from (1) as follows...
$\displaystyle y^{(0)} (0) = 1$
$\displaystyle y^{(1)}(x)= y+x \implies y^{(1)}(0)= 1$
$\displaystyle y^{(2)}(x) = 1 + y^{(1)}(x) = 1+x+y \implies y^{(2)}(0)= 2$
$\displaystyle y^{(3)}(x) = 1+x+y \implies y^{(3)}(0)= 2$
$\dots$
$\displaystyle y^{(n)}(x) = 1+x+y \implies y^{(n)}(0)= 2$
$\dots$
... so that is...
$\displaystyle y(x)= 1 + x + 2\ \sum_{n=2}^{\infty} \frac{x^{n}}{n!} = 2\ e^{x} -1 -x$ (3)
Kind regards
$\chi$ $\sigma$
Page 1 of 2 12 Last | {"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": 27, "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": 0.922461748123169, "perplexity": 1074.255496937986}, "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/1435375097204.8/warc/CC-MAIN-20150627031817-00131-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://amathew.wordpress.com/2012/05/25/computing-the-complex-cobordism-ring/ | We finally have all the computational tools in place to understand Milnor’s computation of ${\pi_* MU}$, and the goal of this post is to complete it. Let’s recall what we have done so far.
1. We described the Adams spectral sequence, which ran
$\displaystyle \mathrm{Ext}^{s,t}_{\mathcal{A}_p^{\vee}}(\mathbb{Z}/p, H_*(MU; \mathbb{Z}/p)) \implies \widehat{\pi_{t-s}(MU)},$
where the hat denotes ${p}$-adic completion.
2. We worked out the homology of ${MU}$. ${H_*(MU; \mathbb{Z}/p)}$, as a comodule over ${\mathcal{A} _p^{\vee}}$, is ${P \otimes \mathbb{Z}/p[y_i]_{i + 1 \neq p^k}}$ where ${P}$ is a suitable subHopf-algebra of ${\mathcal{A}_p^{\vee}}$.When ${p = 2}$, we had that ${P = \mathbb{Z}/p[\zeta_1^2, \zeta_2^2, \dots ] \subset \mathcal{A}_2^{\vee}}$. When ${p}$ is odd, ${P = \mathbb{Z}/p[\zeta_1, \zeta_2, \dots]}$.
3. We worked out a general “change-of-rings isomorphism” for ${\mathrm{Ext}}$ groups.
Now it’s time to put these all together. The ${E_2}$ page of the Adams spectral sequence for ${MU}$ is, as a bigraded algebra,
$\displaystyle \mathrm{Ext}^{s,t}_{\mathcal{A}_p^{\vee}}(\mathbb{Z}/p, P ) \otimes \mathbb{Z}/p[y_i]_{i + 1 \neq p^k}.$
The polynomial ring can just be pulled out, since it’s not relevant to the comodule structure. Consequently, the ${y_i}$ are in bidegree ${(0, 2i)}$: the first bidegree comes from the ${\mathrm{Ext}}$, and the second is because we are in a graded category. The ${y_i}$ only contribute in the second way.
Now, by the change-of-rings result from last time, we have the ${E_2}$ page:
$\displaystyle E_2^{s,t} = \mathrm{Ext}^{s,t}_{\mathcal{A}_p^{\vee} // P}(\mathbb{Z}/p, \mathbb{Z}/p ) \otimes \mathbb{Z}/p[y_i]_{i + 1 \neq p^k}.\ \ \ \ \ (1)$
This is actually an isomorphism of bigraded algebras. It’s not totally obvious, but the Adams spectral sequence is a spectral sequence of algebras for a ring spectrum like ${MU}$.
Now observe that ${\mathcal{A}_p^{\vee} // P}$ is an exterior algebra ${E}$ on the generators ${\tau_0, \tau_1, \dots}$ for ${p}$ odd, and on the ${\zeta_i}$ for ${p = 2}$. Using the formulas for the codiagonal in ${\mathcal{A}_p^{\vee}}$ in this post, we find that the generators for ${E}$ are primitive.
1. The cohomology of an exterior algebra
So what we need now is compute ${\mathrm{Ext}^{s,t}_E(\mathbb{Z}/p, \mathbb{Z}/p)}$: that is, compute the cohomology of an exterior algebra. That turns out to be fairly straightforward. We will work it out in the case of an exterior algebra on one generator.
Proposition 1 Let ${E}$ be an exterior algebra over ${\mathbb{Z}/p}$ on one primitive generator ${x}$ in an odd degree ${n}$. Then
$\displaystyle \mathrm{Ext}^{s,t}(\mathbb{Z}/p, \mathbb{Z}/p)$
is a polynomial algebra on a generator in bidegree ${(1, n)}$.
In fact, this result suffices to prove an analogous result for a tensor product of exterior algebras on primitive generators (by the Künneth theorem).
To prove it, we can write down an explicit resolution of ${\mathbb{Z}/p}$ by ${E}$-free comodules. It looks like
$\displaystyle 0 \rightarrow \mathbb{Z}/p \rightarrow E \rightarrow E[n] \rightarrow E[2n] \rightarrow E[3n] \rightarrow \dots.$
Here ${E[n]}$ denotes ${E}$ shifted by ${n}$, and the maps are the obvious ones. Taking maps of ${\mathbb{Z}/p}$ into each piece, we get the additive structure of the ${\mathrm{Ext}}$ groups: in fact, if ${C^\bullet}$ is the complex ${E \rightarrow E[n] \rightarrow \dots}$, then ${\hom(\mathbb{Z}/p, C^\bullet)}$ is the complex
$\displaystyle \mathbb{Z}/p \rightarrow \mathbb{Z}/p[n] \rightarrow \mathbb{Z}/p[2n] \rightarrow \dots.$
We can also put a multiplicative structure on this resolution to get the multiplicative structure on ${\mathrm{Ext}}$. In fact, we can think of the resolution as the algebra
$\displaystyle E(x) \otimes \mathbb{Z}/p[y], \quad \deg x = (0, n), \deg y = (1, n).$
We make this algebra into a differential graded algebra via ${dx = y, dy = 0}$. Then this is a dga resolution of ${\mathbb{Z}/p}$, and applying ${\hom_E(\mathbb{Z}/p, \cdot)}$ and taking cohomology gives ${\mathbb{Z}/p[y]}$, as desired.
2. The ${E_2}$ page, completely
We saw that the ${E_2}$ page of the Adams spectral sequence was, by (1),
$\displaystyle E_2^{s,t} = \mathrm{Ext}^{s,t}_{\mathcal{A}_p^{\vee} // P}(\mathbb{Z}/p, \mathbb{Z}/p ) \otimes \mathbb{Z}[y_i]_{i + 1 \neq p^k},$
where ${y_i}$ is in bidegree ${(0, 2i)}$. As we’ve seen, the quotient Hopf algebra
$\displaystyle \mathcal{A}_p^{\vee}//P$
is an exterior algebra on primitive generators of degrees:
1. When ${p = 2}$, degrees ${2^i -1}$ for ${i = 1, 2, \dots}$.
2. When ${p }$ is odd, degrees ${2p^i - 1}$ for ${i = 0, 1, \dots}$.
For the rest of the computation, we will assume that ${p}$ is odd; the case ${p = 2}$ is similar. We find that, by the computation of ${\mathrm{Ext}}$ of an exterior algebra,
$\displaystyle E_2 = \mathbb{Z}/p[y_i, z_j]_{i + 1 \neq p^k} \quad \deg y_i = (0, 2i), \deg z_j = (1, 2p^j - 1). \ \ \ \ \ (2)$
In particular, we find that the entire spectral sequence is concentrated in even degrees: thus, the ASS degenerates.
3. Determination of the structure at a prime
We are now ready to use the spectral sequence to compute the ${p}$-adic completion ${\widehat{\pi_* MU}}$. In fact, first let’s note that there is an element ${a_0 \in E_2^{1,1}}$ given by the element ${z_0}$ in (2). This element corresponds to an element of Adams filtration one in the associated graded of ${\widehat{\pi_0 MU} = \mathbb{Z}_p}$: in particular, it must represent ${p}$ or a unit times ${p}$. This we know just from ${\pi_0 MU}$.
With this in mind, choose elements of ${\pi_* MU}$ representing each of the generators ${y_i}$ and ${z_j, j > 0}$; let’s call them ${\widetilde{y_i}}$ and ${\widetilde{z_j}}$. Then ${\widetilde{y_i}}$ must live in degree ${2i}$ (where ${i + 1 \neq p^k}$) and ${\widetilde{z_j}}$ lives in degree ${2(p^j - 1)}$.
Proposition 2 ${\widehat{\pi_* MU}}$ is a polynomial ring over ${\mathbb{Z}_p}$ on the ${\widetilde{y_i}}$ and ${\widetilde{z_j}, j > 0}$.
Proof: In fact, fix ${k}$, and choose an element ${x \in \widehat{\pi_k MU}}$. Then the image of ${x}$ in the associated graded of ${\widehat{\pi_k MU}}$ that comes from the Adams filtration is a polynomial in the ${y_i, z_j}$ (where the ${i}$, ${j}$ admissible are controlled by ${k}$). So there exists a polynomial ${Q}$ such that
$\displaystyle x - a_0^r Q(y_i, z_j)$
is in one step lower in the Adams filtration. We can keep doing this (subtract ${p^r Q(\widetilde{y}_i, \widetilde{z}_j)}$ from ${x}$) to approximate ${x}$ arbitrarily closely by a polynomial, and eventually make ${x }$ equal to a polynomial with ${\mathbb{Z}_p}$-coefficients in the ${y_i, z_j}$.
It follows that the map
$\displaystyle \mathbb{Z}_{p}[\widetilde{y_i}, \widetilde{z}_j] \rightarrow \widehat{\pi_* MU}$
is a surjection, and it is an injection because, again, we can use the associated graded to see algebraic independence of the ${\widetilde{y}_i, \widetilde{z}_j}$.
$\Box$
It follows now that for each prime ${p}$, we have a sequence ${\left\{x_i\right\} \subset \pi_* MU}$ such that after ${p}$-adic completion, the ${\left\{x_i\right\}}$ become polynomial generators for ${\pi_* MU}$.
Corollary 3 For any prime ${p}$, ${\pi_* MU}$ has no ${p}$-torsion.
In particular, if we consider the Hurewicz map
$\displaystyle \pi_* MU \rightarrow H_* MU,$
it must be injective: in fact, it is an isomorphism after tensoring with ${\mathbb{Q}}$, and we have just seen that the map ${\pi_* MU \rightarrow \pi_* MU \otimes \mathbb{Q}}$ is an injection.
4. Determining the global structure
We’re still not completely there. We know that ${\pi_* MU}$ right now is very close to looking like a polynomial ring on even-degree generators. We also have seen that ${\pi_* MU \rightarrow H_* MU}$ is injective.
Let ${I}$ be the augmentation ideal in ${\pi_* MU}$, and consider the indecomposable quotient ${I/I^2}$. The degree ${k}$ part ${(I/I^2)_k}$ maps to the corresponding part ${(J/J^2)_k}$ of the indecomposable quotient ${J/J^2}$ of ${H_* MU}$.
After ${p}$-adic completion, ${I/I^2}$ becomes a free module of rank one (we’ve seen that ${\widehat{\pi_* MU}}$ is a polynomial ring), and as it is finitely generated we conclude that ${I/I^2}$ is itself isomorphic to ${\mathbb{Z}}$. Moreover, the map
$\displaystyle \mathbb{Z} \simeq (I/I^2)_k \rightarrow (J/J^2)_k \simeq \mathbb{Z}$
is an injection, since it is an isomorphism mod torsion.
For each ${k}$, choose an element ${x_k \in \pi_{2k} MU}$ which generates the indecomposable quotient ${(I/I^2)_k}$. Then, we have:
Theorem 4 (Milnor) ${\pi_* MU}$ is a polynomial ring on the ${x_k}$, that is
$\displaystyle \pi_* MU = \mathbb{Z}[x_1, x_2 , \dots ] \quad \deg x_i = 2i.$
Proof: This is now going to be straightforward from the analysis already done. We have a map
$\displaystyle \mathbb{Z}[x_1, x_2 , \dots ] \rightarrow \pi_* MU,$
which is a surjection since all the indecomposables are hit. Moreover, it is an injection because it is an injection after tensoring with ${\mathbb{Q}}$. After tensoring with ${\mathbb{Q}}$, it is a surjection of graded algebras of the same dimension in each degree. $\Box$ | {"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": 151, "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": 0.9732992649078369, "perplexity": 166.11022487791433}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218187113.46/warc/CC-MAIN-20170322212947-00638-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://curriculum.illustrativemathematics.org/MS/teachers/1/6/17/preparation.html | Lesson 17
Two Related Quantities, Part 2
Lesson Narrative
In this second lesson on representing relationships between two quantities, walking at a constant rate provides the context for writing an equation that represents the relationship. Students use and make connections between tables, graphs, and equations that represent the relationship between time and distance. They use their representations to compare rates and consider how each of the representations would change if the independent and dependent variables were switched.
Learning Goals
Teacher Facing
• Create a table, graph, and equation to represent the relationship between distance and time for an object moving at a constant speed.
• Identify (in writing) the independent and dependent variable in a equation.
• Interpret (orally and in writing) an equation that represents the relationship between distance and time for an object moving at a constant speed.
Student Facing
Let’s use equations and graphs to describe stories with constant speed.
Student Facing
• I can create tables and graphs to represent the relationship between distance and time for something moving at a constant speed.
• I can write an equation with variables to represent the relationship between distance and time for something moving at a constant speed.
Glossary Entries
• coordinate plane
The coordinate plane is a system for telling where points are. For example. point $$R$$ is located at $$(3, 2)$$ on the coordinate plane, because it is three units to the right and two units up.
• dependent variable
The dependent variable is the result of a calculation.
For example, a boat travels at a constant speed of 25 miles per hour. The equation $$d=25t$$ describes the relationship between the boat's distance and time. The dependent variable is the distance traveled, because $$d$$ is the result of multiplying 25 by $$t$$.
• independent variable
The independent variable is used to calculate the value of another variable.
For example, a boat travels at a constant speed of 25 miles per hour. The equation $$d=25t$$ describes the relationship between the boat's distance and time. The independent variable is time, because $$t$$ is multiplied by 25 to get $$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": 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": 0.8976042866706848, "perplexity": 359.46315865107334}, "config": {"markdown_headings": false, "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-2021-21/segments/1620243991413.30/warc/CC-MAIN-20210512224016-20210513014016-00377.warc.gz"} |
http://math.stackexchange.com/questions/4358/for-s-subseteq-t-subseteq-mathbbq-it-is-possible-for-sup-t-s-to-exist-b | # For $S\subseteq T\subseteq\mathbb{Q}$, it is possible for $\sup_T S$ to exist, but $\sup_\mathbb{Q} S$ does not?
I've been reading a bit about how the set of bounds changes for a set depending on what superset one works with. I considered the sets $S\subseteq T\subseteq\mathbb{Q}$ and worked out a few contrived examples:
If $S=T=$ {$x\in\mathbb{Q}\ | \ x^2\lt 2$}, so here $S$ is not bounded above in $T$, but it is bounded about in $\mathbb{Q}$, with $2$ being a possibility.
Also, if $S=$ {$x\in\mathbb{Q}\ | \ x^2\lt 1$} and $T=$ {$x\in\mathbb{Q}\ | \ x\lt 2 \ \text{and}\ x\neq 1$}, then $S$ is bounded in $T$ and $\sup_\mathbb{Q} S=1$ exists, but $\sup_T S$ does not exist.
My question is, is it possible for $S$ to be bounded in $T$ where $\sup_T S$ exists, but $\sup_\mathbb{Q} S$ does not? And moreover, can both $\sup_T S$ and $\sup_\mathbb{Q} S$ exist, but not be equal? Any example of this would be much appreciated.
-
Don't forget to accept an answer for your questions if you are satisfied with them. You do not seem to have accepted any answers for any of your seven questions so far. – Arturo Magidin Sep 10 '10 at 16:43
Yes, it is possible for $\sup_T(S)$ to exist, but $\sup_\mathbb{Q}(S)$ not to exist. Yes, it's possible for both to exist and be distinct.
Here is an example of the latter. Take $S=\{x\in\mathbb{Q}|0\leq x\lt 1\}$. Let $T=S\cup{{2}}$. Then $S$ is bounded above in $T$, and has a supremum, namely $2$. Indeed, for every $t\in T$, if $t\lt 2$, then there exists $s\in S$ such that $t\lt s\leq 2$, so $2$ is the supremum of $S$. And of course, the supremum of $S$ in $\mathbb{Q}$ is $1$.
For the former, take $S=\{x\in\mathbb{Q}|0\leq x \lt \sqrt{2}\}$ (or any irrational number you please), and take $T=S\cup\{2\}$ (or any number greater than $\sqrt{2}$).
In general, if you have partially ordered sets $P\subseteq Q$, and a subset $A$ of $P$, you can have that $A$ has a supremum in $P$ but not in $Q$; has a supremum in $Q$ but not in $P$; has suprema in neither; has the same supremum in both; or has suprema in both but they are distinct. The one thing you can say is that if both suprema exist, then $\sup_Q(A)\leq\sup_P(A)$. So in your situation, you do know that if they both exist you will have $\sup_{\mathbb{Q}}(S)\leq\sup_{T}(S)$, but you can have strict inequality.
-
Thank you also for the examples, as well as the more general note at the end. – yunone Sep 10 '10 at 6:38
For the first question, let $S$ be the set of all rationals less than $\pi$. Then $S$ has no least upper bound in $\mathbb{Q}$. On the other hand, if $T = S\cup {4}$, then 4 is the least upper bound of $S$ inside $T$.
For the second question, let $S$ be the set of all rationals strictly less than 1, and let $T = \mathbb{Q}\cap \left([-\infty, 1)\cup [2, \infty)\right)$. Then the least upper bound of $S$ inside $T$ is 2. But of course the least upper bound of $S$ in $\mathbb{Q}$ is 1.
Thank you for this example, it didn't occur to me to make a new set $T$ by taking the union of $S$ and some element larger that everything in $S$. It seems fairly simple to do now. – yunone Sep 10 '10 at 6:36 | {"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": 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": 0.9715999960899353, "perplexity": 79.77796366381548}, "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-2015-48/segments/1448398466178.24/warc/CC-MAIN-20151124205426-00226-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.chapters.indigo.ca/en-ca/books/from-classical-to-quantum-fields/9780387400945-item.html | # From Classical to Quantum Fields: An Introduction to the Path Integral Formalism
## byRoland Seneor
not yet rated|write a review
This book is an introduction to the modern ways of teaching classical and quantum field theories. A key tool is symmetries. For the resolution of classical theories, special attention is given to the definition of advanced or retarded potentials to ease the understanding of path integrals. The Path integral is used as the conceptual tool for defining the quantum field theories. The classical formalism is presented as a useful way to concretely compute observables that one defines in the path integral framework. The book contains special chapters which are devoted to new domains which have not been presented in other texts. They include constructive quantum field theories and topological field theory.
### Pricing and Purchase Info
\$97.50
Pre-order online
Ships free on orders over \$25
From the Publisher
This book is an introduction to the modern ways of teaching classical and quantum field theories. A key tool is symmetries. For the resolution of classical theories, special attention is given to the definition of advanced or retarded potentials to ease the understanding of path integrals. The Path integral is used as the conceptual to...
From the Jacket
Quantum field theory has enormously matured during the last two decades. It provides an extremely powerful tool to describe a large variety of fundamental physical phenomenoa over many scales of distances. Once a specialized field in elementary particle physics, quantum field theory has since been dramatically extended to cover many ...
Format:HardcoverDimensions:300 pages, 9.25 × 6.1 × 0.9 inPublished:June 10, 2018Publisher:Springer New YorkLanguage:English
The following ISBNs are associated with this title:
ISBN - 10:038740094X
ISBN - 13:9780387400945
Look for similar items by category:
Extra Content | {"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": 0.868983268737793, "perplexity": 819.2754738314397}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607046.17/warc/CC-MAIN-20170522190443-20170522210443-00346.warc.gz"} |
https://www.quizover.com/physics/test/what-is-the-tension-in-a-tightrope-by-openstax | # 4.5 Normal, tension, and other examples of forces (Page 4/10)
Page 4 / 10
Consider a person holding a mass on a rope as shown in [link] .
Tension in the rope must equal the weight of the supported mass, as we can prove using Newton’s second law. If the 5.00-kg mass in the figure is stationary, then its acceleration is zero, and thus ${\mathbf{\text{F}}}_{\text{net}}=0$ . The only external forces acting on the mass are its weight $\mathbf{\text{w}}$ and the tension $\mathbf{\text{T}}$ supplied by the rope. Thus,
${F}_{\text{net}}=T-w=0,$
where $T$ and $w$ are the magnitudes of the tension and weight and their signs indicate direction, with up being positive here. Thus, just as you would expect, the tension equals the weight of the supported mass:
$T=w=\text{mg}.$
For a 5.00-kg mass, then (neglecting the mass of the rope) we see that
$T=\text{mg}=\left(\text{5.00 kg}\right)\left(9\text{.}{\text{80 m/s}}^{2}\right)=\text{49.0 N}.$
If we cut the rope and insert a spring, the spring would extend a length corresponding to a force of 49.0 N, providing a direct observation and measure of the tension force in the rope.
Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a finger joint, or a bicycle brake cable. If there is no friction, the tension is transmitted undiminished. Only its direction changes, and it is always parallel to the flexible connector. This is illustrated in [link] (a) and (b).
## What is the tension in a tightrope?
Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in [link] .
Strategy
As you can see in the figure, the wire is not perfectly horizontal (it cannot be!), but is bent under the person’s weight. Thus, the tension on either side of the person has an upward component that can support his weight. As usual, forces are vectors represented pictorially by arrows having the same directions as the forces and lengths proportional to their magnitudes. The system is the tightrope walker, and the only external forces acting on him are his weight $\mathbf{\text{w}}$ and the two tensions ${\mathbf{\text{T}}}_{\text{L}}$ (left tension) and ${\mathbf{\text{T}}}_{\text{R}}$ (right tension), as illustrated. It is reasonable to neglect the weight of the wire itself. The net external force is zero since the system is stationary. A little trigonometry can now be used to find the tensions. One conclusion is possible at the outset—we can see from part (b) of the figure that the magnitudes of the tensions ${T}_{\text{L}}$ and ${T}_{\text{R}}$ must be equal. This is because there is no horizontal acceleration in the rope, and the only forces acting to the left and right are ${T}_{\text{L}}$ and ${T}_{R}$ . Thus, the magnitude of those forces must be equal so that they cancel each other out.
A wrecking ball is being used to knock down a building. One tall unsupported concrete wall remains standing. If the wrecking ball hits the wall near the top, is the wall more likely to fall over by rotating at its base or by falling straight down? Explain your answer. How is it most likely to fall if it is struck with the same force at its base? Note that this depends on how firmly the wall is attached at its base.
Will the current used on the load side affect the current on the supply side.
wave is a disturbance that travels through a medium and transfer energy from one place to another without causing any permanent displacement of the medium itself
what is wave
wave is a disturbance that travels through a medium and transfer energy from one place to another without causing any permanent displacement of the medium itself
Bello
measurements of the period of a simple pendulum in an experiment are 20 oscillations for a give length of the pendulum 30.50secs
patient with a pacemaker is mistakenly being scanned for an MRI image. A 10.0-cm-long section of pacemaker wire moves at a speed of 10.0 cm/s perpendicular to the MRI unit’s magnetic field and a 20.0-mV Hall voltage is induced. What is the magnetic field strength
fundamental and derived quantities
is any thing that has weight or mass and can occupy space
explain why the earth is an exact shere
because the pull of gravity is equal on all sides
Paul
how long did sun take to rich the earth
Billy
A tennis ball is projected vertically up with a velocity of 100m/s the top of a tower 50m high. Determine the a). height reach from the ground. b). Time to reach the hmax.
100×50=5000
what are the formula and calculation
what are the formula for resistance, ohm's law, and electric current
Brianca
I=V/R and so R=V/I
Akiiza
what is gamma rays
gamma ray is an energetic ray .it does not have acharge
Merkew
what is wave
What happens to the mass of water in a pot when it cools, assuming no molecules escape or are added? Is this observable in practice? Explain.
please explain to me,coz I have no idea
Brianca
what is a magnetic field
magnetic field is the area around a magnet where the magnetic force is exerted or felt.
Paul
what is the first Faraday's law
Abba
Formula for Newton 1st law | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 19, "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": 0.8894907236099243, "perplexity": 522.4680819981653}, "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/1664030337307.81/warc/CC-MAIN-20221002083954-20221002113954-00705.warc.gz"} |
http://math.stackexchange.com/questions/3275/generic-sets-in-zfc | # Generic sets in ZFC
I'm reading Shelah's book "Proper and Improper Forcing" (the first two chapters were recommended for learning the basics of forcing)
Given a quasi-order $P$ we say that $\mathcal{I}$ is a dense subset of $P$ if $$(\forall p \in P) (\exists q \in \mathcal{I}) (p\le q)$$
We say that it is open if for any $p,q \in P$ we have that $p \in \mathcal{I} \wedge p\le q$ then $q \in \mathcal{I}$
$G$ is called directed if every two elements in $G$ has an upper bound in $G$.
$G$ is called downward closed if for every $p \in G$ and $q\in P$ if $q \le p$ then $q \in G$
A subset $G$ of $P$ is called generic over $V$ if it is directed, downward closed and for any dense and open subset of $P$ that is in $V$, the intersection with $G$ is non-empty.
Now they give proof that if $P$ has no trivial branches then a generic set cannot exist in the universe, saying that if $G \in V$ then $P\backslash G \in V$ and it is dense and open, how come? I can't figure that out.
-
Can you remind me what trivial branch is, please? – Jason DeVito Aug 25 '10 at 14:57
Nevermind - I found it: above any element, there are two elements with no common upper bound. (My edit timer ran out as I was typing this!) – Jason DeVito Aug 25 '10 at 15:14
Also, in Shelah's book, it seems that "directed" has a different meaning: that any two elements have a common upperbound (see page 3, at least on the projecteuclid version) (again the edit timer ran out as I was typing this!). If the mods can (and would like to) combine all these into a single comment, I'd be more than happy with that.) – Jason DeVito Aug 25 '10 at 15:22
I was always taught that a directed set is a set that every two elements has an upper bound. – Asaf Karagila Aug 25 '10 at 15:31
Ah, I see what you're talking about ;) I'll fix that. (man that comments edit timer is really short) – Asaf Karagila Aug 25 '10 at 15:37
I'm going to use the def of "directed" that I found in Shelah's book1: A set $G$ is directed if it is downward closed (if $p$ in G and if $q\leq p$ then $q\in G$) and every two memebers of $G$ have a common upperbound in $G$.
To see that $P-G$ is open is not too hard: Let $p\in P-G$ and suppose $q\in G$ with $p\leq q$. If $q$ were in $G$, then since $G$ is downwardly closed, we'd have $p\in G$, but that contradicts $p\in P-G$. Thus, $q\notin G$ so $q\in P-G$.
Now, why is $P-G$ dense? Well, let $p\in P$. We want to find a $q\in P-G$ with $p\leq q$. Since there are no trivial branches, we know that above $p$ and are two points $r$ and $s$ with no common upperbound. Since $G$ is directed, $r$ and $s$ cannot both be in $G$. Hence, say, $r\notin G$. Then $r\in P-G$, so we're done.
1I'm using chapter 1 and finding the definition on page 3 of the .pdf.
-
Just one minor technicality which I think I got but not really sure - why is $P\backslash G \in V$ if $G \in V$? – Asaf Karagila Aug 25 '10 at 15:47
Since $V$ is a model of $ZFC$, it satisfies the axiom of comprehension. That's the axiom (schema) which allows you to form sets of the form $\{x\in A| \phi(x)\}$, where $\phi$ is some first order formula (with parameters). In this case, $P-G = \{p\in P | p\notin G\}$ which uses the parameter $G$. – Jason DeVito Aug 25 '10 at 15:59
Just what I thought myself. I'm all so new to the idea of taking models of ZFC and it's all very confusing. – Asaf Karagila Aug 25 '10 at 16:12
Thank you for fixing my link Grigory! – Jason DeVito Aug 25 '10 at 16:21 | {"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": 0.971077024936676, "perplexity": 310.99909943121446}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510256757.9/warc/CC-MAIN-20140728011736-00359-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/451453/quadratic-closure-in-characteristic-2 | # Quadratic closure in characteristic 2
Let $F$ be a field of characteristic 2. I need to show the existence of a quadratic polynomial in $F[t]$ which cannot be solved by adjoining all square roots of elements in the field.
Attempt:
For $F=\mathbb{Z}_2$, $f(t)=t^2+t+1$ works since $\mathbb{Z}_2$ is closed taking square roots.
I don't know how to do the general case. I think that the same polynomial could work.
Edit: Robert shows a counterexample in comments.
• In a field $\mathbb F$ of characteristic $2$, every element has a square root, and so adjoining the square roots of all the elements of $\mathbb F$ to $\mathbb F$ does not help in the least. What you need to do is find a quadratic polynomial that is irreducible over $\mathbb F$. – Dilip Sarwate Jul 25 '13 at 0:36
• @DilipSarwate $X$ does not have a square root in ${\bf F}_2(X)$. The point is that forming $L/F$ by adjoining square roots should be expected to hurt our chances of finding a quadratic in $F[t]$ irreducible over $L$, in fact, in characteristic $\ne2$ it's fatal since then the quadratic formula applies, and the exercise asks us to overcome these apparent grim chances. – anon Jul 25 '13 at 0:40
• @anon What is $T$? That is, what properties do we ascribe to $T$ other than it is something that we can add to or multiply by $0$ and $1$? And how can we be sure that $\mathbb F_2(T)$ is a field? For example, $\mathbb F_2[x]$, the set of all polynomials in an indeterminate $x$ with coefficients in $\mathbb F_2$ is a ring and not a field. – Dilip Sarwate Jul 25 '13 at 0:47
• @DilipSarwate I am using standard notation to refer to the field of rational functions in the variable $X$ (I originally had the letter $T$ in my comment) with coefficients from ${\bf F}_2$, the field with just $0$ and $1$. The relevant property we ascribe to $X$ (or $T$, or whatever capital letter we want to use) is that it is transcendental over ${\bf F}_2$. Showing the ring of rational functions with coefficients from a field is itself a field is a basic exercise for students. I can't say I understand why you're asking me about these things. – anon Jul 25 '13 at 0:51
• The polynomial $f(t)=t^2+t+1$ doesn't work if we consider $F$ to be its splitting field. Also, do there need to be more hypotheses? If $F$ is algebraically closed then there exists no such $f$. – rfauffar Jul 25 '13 at 1:00
If $F$ is algebraically closed, then there exists no such $f$, since any $f$ splits. The polynomial $f(t)=t^2+t+1$ works for $\mathbb{Z}_2$, but not in general (take $F$ to be the splitting field of this polynomial). Maybe there are other conditions you can impose on $F$ so that this is true?
• $F$ finite is an enough extra-condition. If $F$ is finite all elements have square roots in $F$ since $t\mapsto t^2$ is onto. But $t\mapsto t+t^2$ is not onto, then there is some $a\in F$ such that $t^2+t+a$ has no roots. – Gaston Burrull Jul 30 '13 at 6:01
• Nice! That's enough then... – rfauffar Jul 30 '13 at 12:28 | {"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": 0.8254764080047607, "perplexity": 110.6576778027586}, "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-2019-51/segments/1575540543850.90/warc/CC-MAIN-20191212130009-20191212154009-00333.warc.gz"} |
http://physics.stackexchange.com/questions?page=3&sort=active | # All Questions
67 views
### In a 2-body problem, when is the moving path closed?
In a 2-body problem is it true that, in all situations, the moving path is closed? In which cases are the paths closed? Fixing the coordinate system or fixing one of the bodies gives us different ...
566 views
### Analytic continuation of imaginary time Greens function in the time domain
Consider the imaginary time Greens function of a fermion field $\Psi(x,τ)$ at zero temperature $$G^τ = -\langle \theta(τ)\Psi(x,τ)\Psi^\dagger(0,0) - \theta(-τ)\Psi^\dagger(0,0)\Psi(x,τ) \rangle$$ ...
36 views
### Accelerometer validation trace
I am attempting to validate an accelerometer. I have a triaxial accelerometer and placed it on a wheel which goes at various frequencies from 1 to 3 Hz. Am I right in thinking that I would expect the ...
67 views
### Does gravity stack?
If I am standing on a planet on the opposite side from its sun will I feel the downward pull of both the planet's gravity and the star's gravity? In other words, does gravity "stack" (Any D&D ...
2k views
### Why isn't the sunset/sunrise rainbow-colored
When the sun is rising/setting, it goes through a phase where the light is bending from the atmosphere. I believe this image will explain much better than I ever could. Now, if light goes through ...
30 views
### Correlation in electron gas
In the textbooks that I read (namely Ashcroft/Mermin , Marder, etc.) it seems that a distinction is made between the correlations in electron gas and a Couloumb interaction between the electrons. What ...
73 views
### Mathematical model that allows the existence of an infinite number of smallest, but different particles
Does there exist a mathematical formalism (model) in particle physics that assumes the existence of an infinite number of different, yet smallest particles (building blocks)?
27 views
### Luminosity distance from angular diameter distance
Consider that I know the cosmological angular diameter distance at a given redshift : $$D_{A}\left(z\right)=\frac{x_{object}}{\theta_{observer}}$$ Is there a general formula to compute the ...
25 views
### Reference on electrodynamics with tempered distributions
Back in my undergrad I had a course on classical electrodynamics where the fields had values in the space of tempered distributions. In this way one could correctly treat self-interaction and ...
25 views
### Schottky Barrier - Why energy band levels at interface are assumed to remain the same that bulk
I have been chewing up some time ago the Schottky-Mott theory of Schottky Barrier height (which ignores the surface states). All the deduction seems to ground on fundamental thermodynamical principles ...
19 views
### P-N junction voltage under polarization
All the books that deal with the p-n junction under applied bias asume that the same equations used for the equilibrium case (no bias) can be used for the biased case provided that the juntcion ...
70 views
### Electric field generated by a point charge moving at the speed of light
As you see, this is the electric field generated by a point charge moving at constant speed v. I know that when $v$ -> 0, $E$ is just the Coloumb Law. But how do you interpret $E$ when $v$ -> $c$ ? ...
37 views
34 views
### Is there a textile like material that can act as a spring? [on hold]
Imagine some sort of textile (as in very light, not metallic) that is tightly rolled up, can be easily unrolled by applying some force (i.e pulling it) but when released will eventually return to it's ...
144 views
### Saturation of the Cauchy-Schwarz Inequality
Going to as little details as possible, here is a statement from Wald's text on QFT in curved spacetimes(I am not quoting the book) He considers two vector spaces ${\cal S}$ and ${\cal H}$. Note ...
769 views
### How to turn water opaque by pouring the smallest quantity of matter into it? [on hold]
Consider a glass of water and a glass of coffee. Their contents differ by no more than a few grams of particles coming from the roasted and ground coffee, yet the former lets almost all visible light ...
139 views
### Entanglement entropy of 1D chiral Fermion
I was told that the entanglement entropy $S_E$ on the ground state of a (1+1)D conformal field theory (CFT) follows the logarithmic behavior $S_E=\frac{c}{12}\ln L$ where $L$ is the length scale ...
645 views
### How does non-Abelian gauge symmetry imply the quantization of the corresponding charges?
I read an unjustified treatment in a book, saying that in QED charge an not quantized by the gauge symmetry principle (which totally clear for me: Q the generator of $U(1)$ can be anything in ...
71 views
### Coupling constant in electroweak theory
Electroweak theory has two coupling constants before and after Spontaneous Symmetry Breaking (SSB) each one for $SU(2)_L$ and $U(1)_Y$, though they are connected by Weinberg angle after SSB. My ...
66 views
### Calculus used by Gauss to find Ceres' position?
With very little data, the great mathematician Gauss was able to predict the exact location of where to find the asteroid Ceres. If possible I would like a reconstruction of his work, from the data ...
111 views
### Why supra-conductivity became super-conductivity?
The original article by the Kamerlingh Onnes team in Leiden does not give a name to the new effect: Kamerlingh Onnes, H. Further experiments with liquid helium. C. On the change of electric ...
187 views
### What will be the relative speed of a photon in a light ray to another photon of opposite direction light ray?
If two light rays start simultaneously in the space from exactly opposite ends in opposite direction that is separated by a distance of 600000 km in a way they meet at the mid point (300000 km from ...
536 views
### Two photons transition
if an atom in its ground state is coupled to an electromagnetic field it can absorb a photon if the EM field contains one with the right frequency. These transitions depends on $⟨f|H_i|i⟩$ (from ...
243 views
### Double slit experiment and single particles. Is the wave function just a mathematical model?
I really do want to apologize in advance, I know this question has been 'answered' before. I have this 'problem' of feeling negatively toward much of today's 'mystical' interpretations of physics, ...
When quantizing bosonic string theory by means of the path integral, one inverts the Faddeev-Popov determinant by going to Grassmann variables, yielding: S_{\mathrm{ghosts}} = ... | {"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": 0.8711697459220886, "perplexity": 781.8348766375193}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997874283.19/warc/CC-MAIN-20140722025754-00219-ip-10-33-131-23.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/61127/use-of-hiderowcolors-showrowcolors-within-lyx | # Use of \hiderowcolors & \showrowcolors within LyX
Loading the xcolor package with the `[table]` option provides the command(s) `\rowcolors`(`*`) for alternate table row coloring. Any attempt, however, to use `\hiderowcolors` and `\showrowcolors` inside table rows within from LyX -- both in- and outisde a LyX table-float -- results in failing to compile a PDF. The error message is: `Undefined control sequence`.
It seems that there is no discussion/question upon this on either LyX-Users' mailing list or tex.stackexchange itself. It certainly is preferable to resolve this in a clean way, i.e. with a correct usage of the above mentioned commands.
-
I suspect that you simply haven't left any space between the ERT containing `\hiderowcolors`/`\showrowcolors`, so that TeX reads the letter(s) after the command as part of the command. E.g. if you put `\hiderowcolors` at the start of a table cell, and then write text right after the ERT, without spaces, TeX will see the command `\hiderowcolorstext`, which is undefined.
Solution: Put a pair of empty braces after `\hiderowcolors`, within the ERT. (You could also add a space after the ERT, but that will lead to wrong alignment.)
I (remember that I) did try that! Will re-try ASAP. – Nikos Alexandris Jun 27 '12 at 12:18 Grrr... it works! (...I really used a pair of empty braces... :-?) – Nikos Alexandris Jun 27 '12 at 12:20 Wouldn't it be nice though to be able to instruct `\rowcolors{no. of starting row}{no. of ending row}{1st color}{2nd color}`? – Nikos Alexandris Jun 27 '12 at 12:24 @NikosAlexandris You could always ask a new question about that (is it possible/how to do it). – Torbjørn T. Jun 27 '12 at 12:29 I think this could be seen as an enhancement request to xcolor... So it'd better be directed to `xcolor` directly. This is why I did not already asked for that. – Nikos Alexandris Jun 27 '12 at 12:37 | {"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": 0.9745567440986633, "perplexity": 1873.5399748473646}, "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/1368706009988/warc/CC-MAIN-20130516120649-00059-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://space.stackexchange.com/questions/23255/physical-meaning-of-perigee-advance | # Physical meaning of perigee advance
I'm starting to study orbital perturbations and I can't find any physical explanation about the perigee advance (apsidal precession) when considering Earth oblateness effects. Can someone give a physical explanation or recommend a book/article to read?.
PS: From the equations it is quite clear that it must happen, but I am looking for some sort of physical meaning similar to the change in angular momentum due to equatorial bulges when studying nodes regression.
• This is an interesting question! If you can mention or add a link to what you are currently reading, it might be helpful for an answer to be written with similar language/terminology. I noticed my first few clicks into Wikipedia failed to find a derivation or even a formula, but I was surprised to learn that the Moon's apsidal precession period is only 8.85 years, which means every orbit around the Earth advances its periapsis by about 3 degrees! It's even faster than the nodal precession period of 18.60 years. Amazing! – uhoh Oct 4 '17 at 16:03
• I have a hunch that the math for an artificial point-mass satellite will turn up something much slower than for the moon with it's strong tidal effects. By the way, you may want to peruse the references in this excellent answer! – uhoh Oct 4 '17 at 16:04
In the subsection The deviations of Earth's gravitational field from that of a homogeneous sphere of the Wikipedia article on Geopotential model you can see that the $J_2$ or quadrupole moment of the Earth's gravitational potential falls off much more rapidly with distance than the monopole term. In the Earth's equatorial plane, the acceleration due to the monopole and quadrupole moments are given as:
$$a_0 = -\frac{GM_E}{r^2},$$
$$a_2 = -\frac{3}{2} J_2 \frac{GM_E R_E^2}{r^4},$$
where the unit-less value of Earth's $J_2$ is about 0.0010825 and $R_E$ is the normalizing radius of the Earth of 6378136.3 meters, and the standard gravitational parameter of the Earth $GM_E$ is about 3.986E+14 m^3/s^2.
You can read a little more about Earth's $J_2$ and it's effect on gravity at the equator and poles in David Hammen's nice table.
On the Earth's surface, at the equator, the values for these two are 9.7983 and 0.0159 m/s^2 respectively, but remember that they fall of with distance as $1/r^2$ and $1/r^4$ respectively as well.
So a satellite orbiting in Earth's equatorial plane in an elliptical orbit will "think" that the Earth's gravity is stronger at periapsis than at apoapsis, even taking $1/r^2$ into account.
Since the Earth (or any oblate spheroid) "pulls harder" as the satellite swings closest to the planet, it sort-of wraps the orbit tighter. The following apoapsis will come a bit later and advance around the planet, as will the periapsis.
Here is a Python simulation run for a satellite in a very elliptical LEO orbit with a periapsis altitude of about 400km and apoapsis altitude of about 32,000 km. I've run it for Earths normal $J_2$, and again for ten times larger $J_2$ to magnify the effect so that each orbit clearly advances. In addition to the advancement you can see that the semimajor axis is slightly smaller for the larger $J_2$ because the average gravitational force is slightly larger.
def deriv(X, t):
x, v = X.reshape(2, -1)
acc0 = -GMe * x * ((x**2).sum())**-1.5
acc2 = -1.5 * GMe * J2 * Re**2 * x * ((x**2).sum())**-2.5
return np.hstack([v, acc0 + acc2])
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint
# David Hammen's nice table https://physics.stackexchange.com/a/141981/83380
# See http://www.iag-aig.org/attach/e354a3264d1e420ea0a9920fe762f2a0/51-groten.pdf
# https://en.wikipedia.org/wiki/Geopotential_model#The_deviations_of_Earth.27s_gravitational_field_from_that_of_a_homogeneous_sphere
GMe = 3.98600418E+14 # m^3 s^-2
J2e = 1.08262545E-03 # unitless
Re = 6378136.3 # meters
X0 = np.hstack([6778000.0, 0.0, 0.0, 10000.]) # x, y, vx, vy
time = np.arange(0, 300001, 100)
J2 = J2e # correct J2
answerJ2, info = ODEint(deriv, X0, time, full_output=True)
J2 = 10*J2e # 10x larger J2
answer10xJ2, info = ODEint(deriv, X0, time, full_output=True)
if 1 == 1:
plt.figure()
x, y = answerJ2.T[:2]
plt.plot(x, y, '-b')
x, y = answer10xJ2.T[:2]
plt.plot(x, y, '-r')
plt.plot([0], [0], 'or')
plt.show()
• Your penultimate paragraph was what I was looking for, nice explanation, thanks!. My class demonstration involved taking Lagrange planetary equations terms as mean values (first order approximation), compute mean value of J2 potential term for 1 revolution and introducing this computed term in the equations which leads to have only non-null time derivatives of RAAN, perigee argument and mean anomaly. – Julio Oct 5 '17 at 7:25
• That's good to hear, thank you for your comment! I didn't know why this happens either. There are other effects that can cause apsidal precession including more complex tidal effects, perturbations from other bodies, and general relativity, but this is the simplest one to understand. – uhoh Oct 5 '17 at 7:56
Sutton, (note - 4th edition!), page 156, has this to say:
[the figure] shows an exaggerated shift of the apsidal line with the center of the earth remaining as a focus point. This perturbation may be visualized as the movement of the prescribed elliptical orbit in a fixed plane. Obviously, both the apogee and perigee points change in position, the rate of change being a function of the satellite altitude and plane inclination angle. At inclinations of 63.4° and 116.6°, the rate of shifting of the apsidal line, also called apsidal drift, is zero. At an apogee altitude of 1000 nautical miles (n.m.) and a perigee of 100 n.m. in an equatorial orbit, the apsidal drift is approximately 10°/day.
More descriptive than explanatory but perhaps it's of interest. | {"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": 0.8460779786109924, "perplexity": 1201.6002754605079}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496664808.68/warc/CC-MAIN-20191112074214-20191112102214-00407.warc.gz"} |
http://www.ams.org/joursearch/servlet/PubSearch?f1=msc&onejrnl=tran&pubname=one&v1=22E15&startRec=1 | # American Mathematical Society
Publications Meetings The Profession Membership Programs Math Samplings Policy and Advocacy In the News About the AMS
You are here: Home > Publications
AMS eContent Search Results
Matches for: msc=(22E15) AND publication=(tran) Sort order: Date Format: Standard display
Results: 1 to 30 of 35 found Go to page: 1 2
[1] Eberhard Kaniuth and Ali Ülger. The Bochner-Schoenberg-Eberlein property for commutative Banach algebras, especially Fourier and Fourier-Stieltjes algebras. Trans. Amer. Math. Soc. 362 (2010) 4331-4356. MR 2608409. Abstract, references, and article information View Article: PDF [2] Adam R. Lucas. Small unitary representations of the double cover of $\operatorname{SL}(m)$. Trans. Amer. Math. Soc. 360 (2008) 3153-3192. MR 2379792. Abstract, references, and article information View Article: PDF This article is available free of charge [3] Jinpeng An and Zhengdong Wang. Nonabelian cohomology with coefficients in Lie groups. Trans. Amer. Math. Soc. 360 (2008) 3019-3040. MR 2379785. Abstract, references, and article information View Article: PDF This article is available free of charge [4] Toshihiko Matsuki. Equivalence of domains arising from duality of orbits on flag manifolds III. Trans. Amer. Math. Soc. 359 (2007) 4773-4786. MR 2320651. Abstract, references, and article information View Article: PDF This article is available free of charge [5] Toshihiko Matsuki. Equivalence of domains arising from duality of orbits on flag manifolds. Trans. Amer. Math. Soc. 358 (2006) 2217-2245. MR 2197441. Abstract, references, and article information View Article: PDF This article is available free of charge [6] Philip Foth and Jiang-Hua Lu. Poisson structures on complex flag manifolds associated with real forms. Trans. Amer. Math. Soc. 358 (2006) 1705-1714. MR 2186993. Abstract, references, and article information View Article: PDF This article is available free of charge [7] Bernhard Krötz and Michael Otto. Lagrangian submanifolds and moment convexity. Trans. Amer. Math. Soc. 358 (2006) 799-818. MR 2177041. Abstract, references, and article information View Article: PDF This article is available free of charge [8] Hee Oh and Dave Witte Morris. Cartan-decomposition subgroups of $\operatorname{SO}(2,n)$. Trans. Amer. Math. Soc. 356 (2004) 1-38. MR 2020022. Abstract, references, and article information View Article: PDF This article is available free of charge [9] Wolfgang Bertram. Complexifications of symmetric spaces and Jordan theory. Trans. Amer. Math. Soc. 353 (2001) 2531-2556. MR 1814081. Abstract, references, and article information View Article: PDF This article is available free of charge [10] Ya'acov Peterzil, Anand Pillay and Sergei Starchenko. Simple algebraic and semialgebraic groups over real closed fields. Trans. Amer. Math. Soc. 352 (2000) 4421-4450. MR 1779482. Abstract, references, and article information View Article: PDF This article is available free of charge [11] Y. Peterzil, A. Pillay and S. Starchenko. Definably simple groups in o-minimal structures. Trans. Amer. Math. Soc. 352 (2000) 4397-4419. MR 1707202. Abstract, references, and article information View Article: PDF This article is available free of charge [12] A. G. Helminck and G. F. Helminck. A class of parabolic $k$-subgroups associated with symmetric $k$-varieties. Trans. Amer. Math. Soc. 350 (1998) 4669-4691. Abstract, references, and article information View Article: PDF This article is available free of charge [13] Jan Kubarski. Bott's vanishing theorem for regular Lie algebroids. Trans. Amer. Math. Soc. 348 (1996) 2151-2167. MR 1357399. Abstract, references, and article information View Article: PDF This article is available free of charge [14] Bradley N. Currey. The structure of the space of coadjoint orbits of an exponential solvable Lie group . Trans. Amer. Math. Soc. 332 (1992) 241-269. MR 1046014. Abstract, references, and article information View Article: PDF This article is available free of charge [15] Ta Sun Wu. The union of compact subgroups of an analytic group . Trans. Amer. Math. Soc. 331 (1992) 869-879. MR 1044968. Abstract, references, and article information View Article: PDF This article is available free of charge [16] Karl-Hermann Neeb. The duality between subsemigroups of Lie groups and monotone functions . Trans. Amer. Math. Soc. 329 (1992) 653-677. MR 1024775. Abstract, references, and article information View Article: PDF This article is available free of charge [17] Ali Nesin and Anand Pillay. Some model theory of compact Lie groups . Trans. Amer. Math. Soc. 326 (1991) 453-463. MR 1002922. Abstract, references, and article information View Article: PDF This article is available free of charge [18] Joram Hirschfeld. The nonstandard treatment of Hilbert's fifth problem . Trans. Amer. Math. Soc. 321 (1990) 379-400. MR 967314. Abstract, references, and article information View Article: PDF This article is available free of charge [19] Harry F. Hoke. Lie groups that are closed at infinity . Trans. Amer. Math. Soc. 313 (1989) 721-735. MR 935533. Abstract, references, and article information View Article: PDF This article is available free of charge [20] Jiri Dadok. Polar coordinates induced by actions of compact Lie groups . Trans. Amer. Math. Soc. 288 (1985) 125-137. MR 773051. Abstract, references, and article information View Article: PDF This article is available free of charge [21] Joachim Hilgert and Karl H. Hofmann. Semigroups in Lie groups, semialgebras in Lie algebras . Trans. Amer. Math. Soc. 288 (1985) 481-504. MR 776389. Abstract, references, and article information View Article: PDF This article is available free of charge [22] T. Christine Stevens. Weakening the topology of a Lie group . Trans. Amer. Math. Soc. 276 (1983) 541-549. MR 688961. Abstract, references, and article information View Article: PDF This article is available free of charge [23] T. Christine Stevens. Decisive subgroups of analytic groups . Trans. Amer. Math. Soc. 274 (1982) 101-108. MR 670922. Abstract, references, and article information View Article: PDF This article is available free of charge [24] B. Bonnard, V. Jurdjevic, I. Kupka and G. Sallet. Transitivity of families of invariant vector fields on the semidirect products of Lie groups . Trans. Amer. Math. Soc. 271 (1982) 525-535. MR 654849. Abstract, references, and article information View Article: PDF This article is available free of charge [25] Dragomir Ž. Djoković. Closures of conjugacy classes in classical real linear Lie groups. II . Trans. Amer. Math. Soc. 270 (1982) 217-252. MR 642339. Abstract, references, and article information View Article: PDF This article is available free of charge [26] Sheldon Rothman. The von Neumann kernel and minimally almost periodic groups . Trans. Amer. Math. Soc. 259 (1980) 401-421. MR 567087. Abstract, references, and article information View Article: PDF This article is available free of charge [27] David Zerling. Dense subgroups of Lie groups. II . Trans. Amer. Math. Soc. 246 (1978) 419-428. MR 515548. Abstract, references, and article information View Article: PDF This article is available free of charge [28] Don Coppersmith. Deformations of Lie subgroups . Trans. Amer. Math. Soc. 233 (1977) 355-366. MR 0457621. Abstract, references, and article information View Article: PDF This article is available free of charge [29] David Zerling. Some theorems on $({\rm CA})$ analytic groups . Trans. Amer. Math. Soc. 205 (1975) 181-192. MR 0364548. Abstract, references, and article information View Article: PDF This article is available free of charge [30] Bon Yao Chu. Symplectic homogeneous spaces . Trans. Amer. Math. Soc. 197 (1974) 145-159. MR 0342642. Abstract, references, and article information View Article: PDF This article is available free of charge
Results: 1 to 30 of 35 found Go to page: 1 2 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.883495569229126, "perplexity": 1570.1709381610056}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131296456.82/warc/CC-MAIN-20150323172136-00014-ip-10-168-14-71.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/204597-integration-trigonometric-substitution-print.html | # integration by trigonometric substitution.
• October 3rd 2012, 05:32 PM
bkbowser
integration by trigonometric substitution.
$\int \frac{\1}{(x^2*\sqrt{x^2-9}}dx$
$3sec\theta = x$
$\int \frac{3sec\theta tan\theta}{sec^2\theta\sqrt{9tan^2\theta}}dx$
$\int sec\theta dx = sin\theta+c$
since secant is hypotenuse over adjacent sine theta must be
$\frac{\sqrt{x^2-9}}{x}$
the book has a nine in the denominator and I can't figure out where it comes from.
• October 3rd 2012, 06:25 PM
Soroban
Re: integration by trigonometric substitution.
Hello, bkbowser!
You dropped a $\tfrac{1}{9}$ along the way.
Quote:
$\int \frac{dx}{x^2\sqrt{x^2-9}}$
Let $x = 3\sec\theta \quad\Rightarrow\quad dx = 3\sec\theta\tan\theta\,d\theta$
Substitute: . $\int \frac{3\sec\theta\tan\theta\,d\theta}{(3\sec\theta )^2\sqrt{9\tan^2\theta}} \;=\;\int\frac{3\sec\theta\tan\theta\,d\theta}{9 \sec^2\theta\cdot3\tan\theta} \;=\;$
. . . . . . . . . $=\;{\color{red}\frac{1}{9}}\int\frac{d\theta}{\sec \theta}$ | {"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": 0.9943663477897644, "perplexity": 2960.020711529181}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860121534.33/warc/CC-MAIN-20160428161521-00154-ip-10-239-7-51.ec2.internal.warc.gz"} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.