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http://www.physicsforums.com/showthread.php?p=4185420 | # Prove that a function is continuous on an interval
by DeadOriginal
Tags: continuous, function, interval, prove
P: 274 1. The problem statement, all variables and given/known data I have to prove that $\sqrt{x}$ is continuous on the interval [1,$\infty$). 2. The attempt at a solution Throughout the school semester I believed that to show that a function is continuous everywhere all I need to do was show that $\lim\limits_{h\rightarrow 0}f(x+h)-f(x)=0$ and I never thought much about it. It never really came up as a problem on any of the homeworks or exams so I never had much problem with it. I am now doing review problems for the final and I realized that f(x)=1/x is a clear counterexample to what I stated above. The limit as h approaches 0 of f(x+h)-f(x) is 0 but 1/x is not continuous everywhere. Since this is the case, how do I show that a function is continuous everywhere within an interval?
P: 1,310 To show it is continuous, you could go back to your first principles and use an epsilon delta argument. Also given the interval, you can use the definition of the derivative, it's up to you.
P: 274 Say I didn't know that $\sqrt{x}$ was differentiable. How would I use epsilon delta to prove that it was differentiable everywhere in that interval? The only epsilon delta proof for continuity that I know only proves that it is continuous at one point. I read somewhere that I could choose my delta to not only depend on epsilon but also to depend on x. Is this correct? If this is, would it be right if I said Choose $\delta=\epsilon(|\sqrt{x}+\sqrt{a}|)$. Then $|x-a|<\delta\Rightarrow|x-a|<\epsilon(|\sqrt{x}+\sqrt{a}|)\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}=|\sqrt{x}-\sqrt{a}|<\epsilon$.
HW Helper
Thanks
PF Gold
P: 7,184
## Prove that a function is continuous on an interval
Quote by DeadOriginal Say I didn't know that $\sqrt{x}$ was differentiable. How would I use epsilon delta to prove that it was differentiable everywhere in that interval? The only epsilon delta proof for continuity that I know only proves that it is continuous at one point. I read somewhere that I could choose my delta to not only depend on epsilon but also to depend on x. Is this correct? If this is, would it be right if I said Choose $\delta=\epsilon(|\sqrt{x}+\sqrt{a}|)$. Then $|x-a|<\delta\Rightarrow|x-a|<\epsilon(|\sqrt{x}+\sqrt{a}|)\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}=|\sqrt{x}-\sqrt{a}|<\epsilon$.
No, that isn't right. When you are discussing continuity at the point ##x=a##, the ##\delta## will normally depend on ##\epsilon## and ##a##. If it doesn't depend on ##a## you have uniform continuity. So in your example, ##\delta## can depend on the ##a## but not ##x##. But you have$$|\sqrt x -\sqrt a| = \frac {|x-a|}{|\sqrt x + \sqrt a|}$$Given that you have ##x\ge 1## can you find an overestimate of the right side that doesn't involve ##x## in the denominator, then choose ##\delta## to make it work?
P: 274
Quote by LCKurtz No, that isn't right. When you are discussing continuity at the point ##x=a##, the ##\delta## will normally depend on ##\epsilon## and ##a##. If it doesn't depend on ##a## you have uniform continuity. So in your example, ##\delta## can depend on the ##a## but not ##x##. But you have$$|\sqrt x -\sqrt a| = \frac {|x-a|}{|\sqrt x + \sqrt a|}$$Given that you have ##x\ge 1## can you find an overestimate of the right side that doesn't involve ##x## in the denominator, then choose ##\delta## to make it work?
I'm looking at ##|x-a|<\delta## and I'm thinking that the way I would find a bound for for ##|\sqrt{x}|## is to first make ##\delta_{1}=1## so that I have ##|x-a|<1\Rightarrow -1<x-a<1\Rightarrow a-1<x\Rightarrow \sqrt{a-1}+\sqrt{a}<\sqrt{x}+\sqrt{a}\Rightarrow \frac{1}{\sqrt{x}+\sqrt{a}}<\frac{1}{\sqrt{a-1}+\sqrt{a}}## and since these numbers are positive ##\frac{1}{|\sqrt{x}+\sqrt{a}|}<\frac{1}{|\sqrt{a-1}+\sqrt{a}|}##
What if I then chose ##\delta=\min(\delta_{1},\epsilon(|\sqrt{a-1}+\sqrt{a}|))##?
Then my epsilon delta argument would be:
Choose ##\delta=\min(\delta_{1},\epsilon(|\sqrt{a-1}+\sqrt{a}|))##. Then
##|x-a|<\delta\Rightarrow |x-a|<\epsilon(|\sqrt{a-1}+\sqrt{a}|)\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon##.
Also while we are talking about uniform continuity, if delta doesnt depend on ##a## for uniform continuity then what does delta depend on? Does it depend on x and y in that case? Like say I wanted to prove that ##\sqrt{x}## was uniformly continuous on [1,##\infty##) and I have ##|x-y|<\delta##, can I then choose a ##\delta## that depends on ##x## and ##y## to show that ##|\sqrt{x}-\sqrt{y}|<\epsilon##?
HW Helper
Thanks
PF Gold
P: 7,184
Quote by DeadOriginal I'm looking at ##|x-a|<\delta## and I'm thinking that the way I would find a bound for for ##|\sqrt{x}|## is to first make ##\delta_{1}=1## so that I have ##|x-a|<1\Rightarrow -1
That just makes my head hurt. Why don't you just use my suggestion, since ##x\ge 1##:$$\frac {|x-a|} {\sqrt x + \sqrt a }\le \frac {|x-a|} {1+\sqrt a}$$and use that to figure out ##\delta##?
Then my epsilon delta argument would be: Choose ##\delta=\min(\delta_{1},\epsilon(|\sqrt{a-1}+\sqrt{a}|))##. Then ##|x-a|<\delta\Rightarrow |x-a|<\epsilon(|\sqrt{a-1}+\sqrt{a}|)\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon##. Also while we are talking about uniform continuity, if delta doesnt depend on ##a## for uniform continuity then what does delta depend on? Does it depend on x and y in that case?
No. In that case ##\delta## depends only on ##\epsilon##, not ##x## or ##y##. Look at my suggestion above and see if you can also overestimate getting rid of the ##a## in the denominator to get a ##\delta## that depends only on ##\epsilon##.
Like say I wanted to prove that ##\sqrt{x}## was uniformly continuous on [1,##\infty##) and I have ##|x-y|<\delta##, can I then choose a ##\delta## that depends on ##x## and ##y## to show that ##|\sqrt{x}-\sqrt{y}|<\epsilon##?
No.
P: 274
Quote by LCKurtz That just makes my head hurt. Why don't you just use my suggestion, since ##x\ge 1##:$$\frac {|x-a|} {\sqrt x + \sqrt a }\le \frac {|x-a|} {1+\sqrt a}$$and use that to figure out ##\delta##?
OOOOOOOOOOOhhhhhhhh. So since ##\frac {|x-a|} {\sqrt x + \sqrt a }\le \frac {|x-a|} {1+\sqrt a}## we can choose ##\delta=\epsilon|1+\sqrt{a}|## and then get
##|x-a|<\epsilon|1+\sqrt{a}|\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}<\epsilon|1+\sqrt{a}|\frac{1}{|1+\sqrt{a }|}\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon##. Third times a charm?
No. In that case ##\delta## depends only on ##\epsilon##, not ##x## or ##y##. Look at my suggestion above and see if you can also overestimate getting rid of the ##a## to get a ##\delta## that depends only on ##\epsilon##.
Hmm. Thanks. This helps to clarify a lot! I will do this part on my own time. As long as I get the top part, it seems like this part shouldn't be too much harder to do.
HW Helper
Thanks
PF Gold
P: 7,184
Quote by DeadOriginal OOOOOOOOOOOhhhhhhhh. So since ##\frac {|x-a|} {\sqrt x + \sqrt a }\le \frac {|x-a|} {1+\sqrt a}## we can choose ##\delta=\epsilon|1+\sqrt{a}|## and then get ##|x-a|<\epsilon|1+\sqrt{a}|\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}<\epsilon|1+\sqrt{a}|\frac{1}{|1+\sqrt{a }|}\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon##. Third times a charm? Hmm. Thanks. This helps to clarify a lot! I will do this part on my own time. As long as I get the top part, it seems like this part shouldn't be too much harder to do.
Good, now you are getting somewhere. Your ##\delta## depends on ##\epsilon## and ##a##, so you have continuity for each ##a##. Now, can you find a ##\delta## that doesn't depend on ##a## either? If it depended only on ##\epsilon## you would have shown uniform continuity.
P: 274
Quote by LCKurtz Good, now you are getting somewhere. Your ##\delta## depends on ##\epsilon## and ##a##, so you have continuity for each ##a##. Now, can you find a ##\delta## that doesn't depend on ##a## either? If it depended only on ##\epsilon## you would have shown uniform continuity.
As per your advice I noted that the smallest value of ##a## would also be 1 on the interval so choosing ##\delta=2\epsilon## I would have:
##|x-a|<\delta\Rightarrow |x-a|<2\epsilon\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}<2\epsilon\frac{1}{2}\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon##.
Thank you! This has clarified a lot for me!
HW Helper
Thanks
PF Gold
P: 7,184
Quote by DeadOriginal As per your advice I noted that the smallest value of ##a## would also be 1 on the interval so choosing ##\delta=2\epsilon## I would have: ##|x-a|<\delta\Rightarrow |x-a|<2\epsilon\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}<2\epsilon\frac{1}{2}\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon##. Thank you! This has clarified a lot for me!
OK, you have all the pieces figured out, but you could present it a little more smoothly. Start with
$$|\sqrt x - \sqrt a| = \frac{|x-a|}{\sqrt x + \sqrt a}\le \frac{|x-a|}{2}$$since ##x\ge 1## and ##a\ge 1##. (This is really the "exploratory argument" where you are figuring out what ##\delta## might work.)
Suppose ##\epsilon>0##. Let ##\delta=2\epsilon##. Then if ##|x-a|<\delta = 2\epsilon## we have$$|\sqrt x - \sqrt a| <\frac{2\epsilon} 2=\epsilon$$
Related Discussions Calculus & Beyond Homework 0 Calculus & Beyond Homework 5 Calculus 10 Calculus & Beyond Homework 2 Differential Geometry 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": 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": 0.908025324344635, "perplexity": 1077.2025359343631}, "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-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00080-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/10109/whats-a-closed-countably-infinite-intersection/10111 | # What's a closed countably infinite intersection?
When operating with sigma algebras, what does it mean when we talk about a countably infinite set? And, what are then closed countably infinite interesections?
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A set which is countably infinite. I don't follow. As for the second question, you should quote whatever it is you're having trouble following in full. – Qiaochu Yuan Nov 13 '10 at 12:17
These definitions would be near the front of any textbook on measure theory. – Carl Mummert Nov 13 '10 at 13:13
I expect that you have read that a sigma algebra $\mathcal{A}$ has to be "closed under countably infinite intersections". What this means is that if you have a family $(A_n)$ of elements of $\mathcal{A}$ indexed by the natural numbers $\mathbb{N}$, then the intersection $\bigcap_{n\in\mathbb{N}}A_n$ of all the $A_n$ is also an element of $\mathcal{A}$. | {"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": 0.9529677629470825, "perplexity": 172.2460422971759}, "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/1408500830323.35/warc/CC-MAIN-20140820021350-00030-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://www.jetpletters.ac.ru/ps/1827/article_27931.shtml | Home For authors Submission status Current Archive (English) Archive Volumes 61-80 Volumes 41-60 Volumes 21-40 Volumes 1-20 Volumes 81-92 Volume 92 Volume 91 Volume 90 Volume 89 Volume 88 Volume 87 Volume 86 Volume 85 Volume 84 Volume 83 Volume 82 Volume 81 Search
VOLUME 86 | ISSUE 10 | PAGE 762
Infrared catastrophe in two-quasiparticle collision integral
O. V. Dimitrova+*, V. E. Kravtsov+*
+L.D. Landau Institute for Theoretical Physics RAS, 117940 Moscow, Russia
*The Abdus Salam International Centre for Theoretical Physics, P.O.B. 586, 34100 Trieste, Italy
PACS: 71.23.An, 72.15.Rn, 72.25.Ba, 73.63.Nm
Abstract
Relaxation of a non-equilibrium state in a disordered metal with a spin-dependent electron energy distribution is considered. The collision integral due to the electron-electron interaction is computed within the approximation of a two-quasiparticle scattering. We show that the spin-flip scattering processes with a small energy transfer may lead to the divergence of the collision integral for a quasi one-dimensional wire. This divergence is present only for a spin-dependent electron energy distribution which corresponds to the total electron spin magnetization and only for non-zero interaction in the triplet channel. In this case a non-perturbative treatment of the electron-electron interaction is needed to provide an effective infrared cut-off. | {"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": 0.8266119956970215, "perplexity": 2546.3287079607335}, "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-2019-51/segments/1575541301014.74/warc/CC-MAIN-20191215015215-20191215043215-00242.warc.gz"} |
http://www.nist.gov/manuscript-publication-search.cfm?pub_id=905036 | # Publication Citation: Effect of Diamond Nanolubricant on R134A Pool Boiling Heat Transfer
NIST Authors in Bold
Author(s): Mark A. Kedzierski; Effect of Diamond Nanolubricant on R134A Pool Boiling Heat Transfer December 15, 2009 This paper quantifies the influence of diamond nanoparticles on the pool boiling performance of R134a/polyolester mixtures on a roughened, horizontal, flat surface. Nanofluids are liquids that contain dispersed nano-size particles. A lubricant based nanofluid (nanolubricant) was made by suspending 10 nm diameter diamond particles in a synthetic ester to roughly a 2.6 % volume fraction. For the 0.5 % nanolubricant mass fraction, the nanoparticles caused a heat transfer enhancement relative to the heat transfer of pure R134a/polyolester (99.5/0.5) up to 129 %. A similar enhancement was observed for the R134a/nanolubricant (99/1) mixture, which had a heat flux that was on average 91 % larger than that of the R134a/polyolester (99/1) mixture. Further increase in the nanolubricant mass fraction to 2 % resulted in boiling heat transfer degradation of approximately 19 % for the best performing tests. It was speculated that the poor quality of the nanolubricant suspension caused the performance of the (99.5/0.5), and the (98/2) nanolubricant mixtures to decay over time to, on average, 36 % and 76 % of the of pure R134a/polyolester performance, respectively. Thermal conductivity and viscosity measurements and a refrigerant\lubricant mixture pool-boiling model were used to suggest that increases in thermal conductivity and lubricant viscosity are mainly responsible for the heat transfer enhancement due to nanoparticles. Particle size measurements were used to suggest that particle agglomeration induced a lack of performance repeatability for the (99.5/0.5) and the (98/2) mixtures. From the results of the present study, it is speculated that if a good dispersion of nanoparticles in the lubricant is not obtained, then the agglomerated nanoparticles will not provide interaction with bubbles, which is favorable for heat transfer. 2nd ASME Micro/Nanoscale Heat & Mass Transfer International Conference 10 pp. Shanghai, -1 December 18-21, 2009 additives; boiling; diamond; enhanced heat transfer; nanotechnology; refrigerants; refrigerant/lubricant mixtures High Performance Buildings, Building and Fire Research Click here to retrieve PDF version of paper (372KB) | {"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": 0.8172183036804199, "perplexity": 4173.319547098761}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414637900030.8/warc/CC-MAIN-20141030025820-00141-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://homework.cpm.org/cpm-homework/homework/category/CCI_CT/textbook/Calc/chapter/Ch5/lesson/5.2.5/problem/5-104 | ### Home > CALC > Chapter Ch5 > Lesson 5.2.5 > Problem5-104
5-104.
This is Ana's definition of the derivative. The limit is equivalent to f '(0).So you can use the power rule to find the limit, if it exists.
Refer to the hint in part (b).
Recall that derivatives do not exist at locations where f(x) has a: cusp, endpoint, jump, hole or vertical tangent.
Sketching the graph may help visualize the issue. | {"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": 0.9590875506401062, "perplexity": 2205.4995072472425}, "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-2020-24/segments/1590347385193.5/warc/CC-MAIN-20200524210325-20200525000325-00268.warc.gz"} |
http://mathhelpforum.com/calculus/74003-evaluate-integration-using-integration-parts-print.html | # Evaluate the integration by using integration by parts
$\int\ln^3xdx=\int x'\ln^3xdx=x\ln^3x-\int x\cdot 3\ln^2x\cdot\frac{1}{x}dx=x\ln^3x-3\int\ln^2xdx$ | {"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": 0.9809057712554932, "perplexity": 333.3531029485531}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049288709.66/warc/CC-MAIN-20160524002128-00199-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://www.earthdoc.org/content/papers/10.3997/2214-4609.201410460 | 1887
### Abstract
Directional borehole radars have good potential (or 3-D determination of water-filled fractures and geological boundaries. The choice of polarization of electromagnetic wave used for measurements are quite significant in designing radar systems. Circular polarization has advantages for Ground penetrating radars (GPR), because the reflection efficiency is independent of target orientation to antenna arrangemenis. Radar measurements with circular polarization is possible with antennas such as spiral antennas( lizuka & Freundorfer, 1983) . Another important role of polarization in GPR is isolation of transmitting and receiving antennas. GPR measurements normally use two antennas set closely to each other . Direct coupling often shades small reflection trom targets. A set of cross-polarized antennas increases the antenna isolation and enhances radar reflection. Further possibility of radar signal processing with polarization characteristics are known as radar polarimetry ( Zebker & Zyl 1991) . This paper introduces a possibility of polarimetric radar measurements in a borehole .
/content/papers/10.3997/2214-4609.201410460
1992-06-01
2021-09-18 | {"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": 0.9483867883682251, "perplexity": 4643.4675022028105}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056392.79/warc/CC-MAIN-20210918093220-20210918123220-00523.warc.gz"} |
http://mathhelpforum.com/pre-calculus/159669-sequential-limit-question.html | # Math Help - Sequential Limit Question
1. ## Sequential Limit Question
I'm not entirely sure what cateogry this question falls under I got it in my Probability class and for some reason none of us can answer it lol. Any help would be greatly appreciated
If b is a positive integer n, such that 1/(2^(n-1)) < b?
2. Originally Posted by AlmasKhan01
I'm not entirely sure what cateogry this question falls under I got it in my Probability class and for some reason none of us can answer it lol. Any help would be greatly appreciated
If b is a positive integer n, such that 1/(2^(n-1)) < b?
Try again: there is no question here.
Tonio
3. Well my Professor was asking questions like can you give a number where this inequality is true. That's the best I can explain it?
Or maybe he was asking If b is a positive integer, give an n such that 1/(2^(n-1)) < b?
He was talking about proving the solution but the question he made us write down the one I wrote I don't understand at all.
Actually he may have asked...... Is it that for any real b there is an integer n such that 1/2^{n-1}
4. Originally Posted by AlmasKhan01
I'm not entirely sure what cateogry this question falls under I got it in my Probability class and for some reason none of us can answer it lol. Any help would be greatly appreciated
If b is a positive integer n, such that 1/(2^(n-1)) < b?
There is no verb in that sentence!
"Given a positive integer b, find an integer n such that $\frac{1}{2^{n-1}}< b$"?
There are no negative numbers here and log is an increasing function so this is pretty much the same as solving the equation.
This inequality is the same as asking for n such that $b< 2^{n-1}$. Since the logarithm is an increasing function, log(b)< log(2^{n-1}= (n-1) log(2). Since log(2) is positive (for positive base) $\frac{log(b)}{log(2)}< n-1$, $n> \frac{log(b)}{log(2)}$. | {"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": 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": 0.8712452054023743, "perplexity": 443.07926966159846}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00254-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://mathhelpforum.com/trigonometry/160518-find-sin.html | 1. ## Find sinά
Hi, Can anyone give me an idea how can I solve this?
Find sinά, if sin(45- ά)= - 2/3
2. Have you have any ideas so far?
4. How about applying the arcsin function to both sides? What would that give you?
5. I'm sorry I dont know what arcsin function is:$:$
6. The arcsin function is also written $\sin^{-1}(x).$ It is the functional inverse of sin. Generally speaking, if you have $\sin(x)=y,$ then $x=\arcsin(y)=\sin^{-1}(y),$ modulo the symmetries of the sin function. You follow?
7. Im sorry I havent learned inverse functions..there isnt any other way right..?
Can u please tell me how arcsin function is applied to both sides?
I really appreciate ur help. Thanks
8. It's just like "undoing" any other function. To undo the exponential function, you might have this:
$e^{x}=y$
$\ln(e^{x})=\ln(y)$ (applied logarithm to both sides)
$x=\ln(y).$ (simplified)
Same thing here:
$\sin(x)=y$
$\sin^{-1}(\sin(x))=\sin^{-1}(y)$
$x=\sin^{-1}(y).$
So how could you use this in your situation?
9. 45- ά= arcsin-2/3 am I close..?:\$
10. Right. Now, with your notation, you should make sure you have ά in degrees. What do you get when you solve for it?
11. Sorry for disturbing again, but can you solve that for me please?
Thanks for your patience.I wish my maths techer was like you.
12. If you're doing trig, you should know how to solve an equation like a-x=b, for x. That's what you have here. What do you get?
13. [IMG][/IMG]
I'd like to know if this procedure is allowed in Maths.
Thanks a lot
14. The idea is sound, but you've made a mistake or two in carrying it out. Going from
$\displaystyle\frac{\sqrt{2}}{2}[\cos(\alpha)-\sin(\alpha)]=-\frac{2}{3}$ to
$\displaystyle\cos(\alpha)-\sin(\alpha)=\frac{2\sqrt{2}}{3},$
you lost the minus sign. You should have gotten
$\displaystyle\cos(\alpha)-\sin(\alpha)=-\frac{2\sqrt{2}}{3}.$
This is a mistake that, due to another sign error, you later correct! This is a bit sloppy.
So, I'd say I agree with your last step, but only as the result of two self-correcting sign errors.
I should also point out that you have done one (what we call) irreversible step, when you squared the equation. Squaring an equation can sometimes introduce spurious results. I think in your case it does. So, when you get your answers, double-check them against your original equation, and make sure they satisfy the original equation. You should always double-check the answers to problems anyway, to make sure they make sense. Make sense? | {"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": 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": 0.8088780641555786, "perplexity": 752.3819195011645}, "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/1476988718866.34/warc/CC-MAIN-20161020183838-00321-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://confluence.esss.lu.se/plugins/viewsource/viewpagesrc.action?pageId=244613126 | The size distribution of particles leads to smearing effects. The scattered intensity reads I_{poly}(q,x) = \int_0^{\infty} G(x,r) I(q,r) dr where G(x,r) characterizes the distribution function such that \int G( x, r ) dr = 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": 0.984593391418457, "perplexity": 3802.952050174752}, "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-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00249.warc.gz"} |
https://brilliant.org/discussions/thread/about-0/ | I know that $$0! = 1$$. If you don't, here, I show you.
We use $$!$$ (or factorial) like this:
$$x! = x \mathtt{1} \times x \mathtt{2} \times x \mathtt{2} \times \dots 3 \times 2 \times 1$$
Here's an example:
$$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$
Another way to do this is this. For example, $$4!$$.
$$4! =$$$$\frac{5!}{5}$$$$= \frac{5 \times 4 \times 3 \times 2 \times 1}{5}$$$$= 24$$
Okay now, why $$0! = 1$$? Let's start from $$4!$$ to $$0!$$.
(already did the $$4!$$)
$$3! = \frac{4!}{4} = \frac{4 \times 3 \times 2 \times 1}{4} = 6$$
$$2! = \frac{3!}{3} = \frac{3 \times 2 \times 1}{3} = 2$$
$$1! = \frac{2!}{2} = \frac{2 \times 1}{2} = 1$$
Okay, now... $$0!$$...
$$0! = \frac {1!}{1} = \frac {1}{1} = 1$$
Surprising right? Okay, now you know.
What I want to talk about is that why are we allowed to do $$0!$$? Because if we are, that mean $$0 = 1$$ and that's wrong.
$$0! = 1 = 1!$$
$$0! = 1!$$
$$0 = 1$$
5 years, 2 months ago
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$$1!$$ is the number of ways of arranging 1 item, which is 1. $$0!$$ is the number of ways of arranging 0 items. Since there is only 1 way to do NOTHING, (that is to not do it) we have $$0!=1$$. So by comparing $$0!$$ and $$1!$$, we are comparing the number of ways of arranging 1 and 0 items respectively. Just because the result is same doesn't mean that the number of items has to be same. So $$0\neq1$$.
- 5 years, 2 months ago
From this,we can also infer that if $$x!=y!\Rightarrow x=y if x,y\neq 0,1$$
- 5 years, 2 months ago
But not to do it also the number of ways of arranging 1 items, that means 1! = 2.
- 5 years, 2 months ago
You must be careful, for some function $$f$$, there doesn't always exist a unique inverse $$f^{-1}$$, other than for some domain. In this case, we may assume $$x>0$$, or, alternatively that $$x\ne1$$
- 5 years, 2 months ago
Just because $$x! = y!$$ does not always mean that $$x = y.$$ By defining $$0! = 1,$$ we have created an exception to that rule.
- 5 years, 2 months ago
Actually not an exception but a rule. It is defined as to make whole body of factorials to make sense. Like everything else in maths this is just made up by us and we may tweak it as we like.
- 5 years, 2 months ago
I totally agree with Guillermo, as you see,
If $$f(x_1)=y$$ and $$f(x_2)=y$$, do we say $$x_1=x_2$$? No.
For an example, $$3^2=9$$ and $$(-3)^2=9$$, but $$-3\neq3$$.
As explaining why $$0!=1$$, Bruce has the exact explaination. The concept for $$n!$$ is the number of ways to arrange $$n$$ distinct object. Try imagine an empty space with no object, how many arrangements are there? The image in your mind, which is AN EMPTY SPACE, is the ONLY arrangement for 0 object. Thus, $$0!=1$$
- 5 years, 2 months ago
By the way, it is defined to be so. The basic idea of combinatorics is reducing the large list into a recognisable pattern which doesn't have to repeated time and again. So, when we observe somthing like this, we have to define it in such a way that it agrees with our recognized pattern. There is one way to take a single object.(1!) There is one way Not to take that object.(a possibility).( 0!) So, even though the number of ways are same, it doesn't mean that the ways are same!
- 5 years, 2 months ago
The factorial function is a special case of what is called the Gamma function $\Gamma(x) \; = \; \int_0^\infty t^{x-1}\,e^{-t}\,dt \qquad x > 0$ (Actually, the Gamma function can be defined for all complex numbers except $$0,-1,-2,-3,\ldots$$, but that is more complicated). Integration by parts shows that $\Gamma(x+1) = x\Gamma(x) \qquad x > 0$ and $$\Gamma(1) = 1$$. It is easy to see from these properties that $$\Gamma(n+1) = n!$$ for any nonnegative integer $$n$$ (including $$n=0$$).
- 5 years, 2 months ago
While this is true, the sticking point is that the gamma function was motivated by the definition of the factorial function on the positive integers, so to then define the factorial in terms of the gamma is pedagogically and historically inverted, even though it is perfectly valid from a mathematical standpoint. So, in order to motivate the value of $$0!$$ using the gamma function, we would need to discuss concepts like continuity, limits, and integrability--i.e., most of the fundamentals of calculus on the real line, much in the same way that to motivate the value of $$\Gamma(z)$$ for complex $$z$$, we'd need to talk about analytic continuation.
What the usual combinatorial and elementary definitions of factorial have going for them is that they can more intuitively justify $$0! = 1$$, even though such reasoning may not be rigorously derived from established axioms.
- 5 years, 2 months ago
Firstly, we do not need to use analytic continuation to study the Gamma function for $$\mathfrak{R}z > 0$$, and that is all it takes to handle $$n!$$ for $$n \ge 0$$.
As to your main point, yes and no. There are times where it is both necessary and useful to invert the historical development of a subject - just because something was done first does not mean it was done better that way.
For example, the "definition" of $$\pi$$ as the ratio between the circumference and diameter of a circle is a bad definition of $$\pi$$, since it can only be used to calculate $$\pi$$ by a process of measurement, which is inherently inaccurate. The "definitions" of the trigonometric functions in terms of the sides of a right-angled triangle are similarly flawed. At some stage, you have to change your point of view and define the trigonometric functions in terms of their Maclaurin series; $$\pi$$ then appears in terms of the periodicity of the trig functions, which can be established from the series definitions. You can then go on to prove that the circumference/diameter stuff, or the triangle ratio results, are all consequences of these definitions.
Similarly, at some stage you have to realise that the factorial function is only a portion of the truth and that, seen in the light of the Gamma function, the identity $$0!=1$$ is not in the least mysterious, and does not have to be explained by linguistic sleight of hand of the "there is just one way to do nothing" variety, or by "believe me, it works, and it makes the formulae for the binomial coefficients neat" arguments. If you choose to make the argument that "the value of $$0!$$ is chosen so that the identity $$n! \,=\, n \times (n-1)!$$ is valid for $$n=1$$", then you are doing a little bit of analytic continuation yourself!
It is important pedagogically to offer both, and complement these utilitarian arguments with the observation (if only as a passing comment) that there is a better reason out there.
- 5 years, 2 months ago
0! is a conjecture..it is just like u cannot define a point or define infinity in classical mathematics!it is an assumption that 0!=1. and again the factorial function is one-many as seen in case of 0! and 1!.so while proving both equals 1,u have to do the inverse which doesnt always exist and x!=y! doesnt imply x=y.
- 5 years, 2 months ago | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9766061305999756, "perplexity": 431.618369125035}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583835626.56/warc/CC-MAIN-20190122095409-20190122121409-00046.warc.gz"} |
http://mlwiki.org/index.php?title=Determinants&printable=yes | # ML Wiki
## Determinants
A determinant is a value associated with a square matrix $A$
• it provides important information about invertability of the matrix
• it's denoted as $\text{det } A$ or sometimes $| A |$
## Defining Properties
These properties define what a determinant is (they don't say how to compute it)
### Property 1: Determinant of $I$
• $\text{det } I = 1$
### Property 2: Sign Reversal
• let $A'$ be a matrix $A$ with two rows exchanged, then $\text{det } A' = \text{det } A$
Consequence of property 2:
• for a Permutation Matrix $P$
• $\text{det } P = 1$ if if has even number of row exchanges
• and $\text{det } P = -1$ is it has odd number of exchanges
### Property 3: Linearity
• determinant is a linear function of a row - if all other rows stays the same
• 3a) $\begin{vmatrix} t a_{11} & t a_{12} & t a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{vmatrix} = t \cdot \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{vmatrix}$
• 3b) $\begin{vmatrix} a_{11} + a'_{11} & a_{12} + a'_{12} & a_{13} + a'_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{vmatrix} = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{vmatrix} + \begin{vmatrix} a'_{11} & a'_{12} & a'_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{vmatrix}$
• applies to all the rows: we always can put any row to the first position (property 2), then apply the property 3, and then put the row back
## Other properties
These properties are consequences of the defining properties
### Property 4: Equal Rows
• if 2 rows are equal, then $\text{det } A = 0$
• proof: if we exchange two rows, nothing happens to the matrix, but property 2 says the sign should be reversed
### Property 5: Linear Combinations
• if we subtract a multiple of row $i$ from the row $k$, the determinant remains the same
• for Gaussian Elimination it means that $\text{det } A = \text{det } U$
• $\begin{vmatrix} a_{11} & a_{12} \\ a_{21} - c a_{11} & a_{22} - c a_{12} \\ \end{vmatrix} \ \mathop{=}\limits^{3^{\circ}} \ \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{vmatrix} - c \begin{vmatrix} a_{11} & a_{12} \\ a_{11} & a_{12} \\ \end{vmatrix} \ \mathop{=}\limits^{4^{\circ}} \ \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{vmatrix} - c 0 = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{vmatrix}$
### Property 6: Zero Row
• if we have a rows full of zeros, then $\text{det } A = 0$
• $c \cdot \begin{vmatrix} a_{11} & a_{12} \\ 0 & 0 \\ \end{vmatrix} = \begin{vmatrix} a_{11} & a_{12} \\ c \cdot 0 & c \cdot 0 \\ \end{vmatrix} = \begin{vmatrix} a_{11} & a_{12} \\ 0 & 0 \\ \end{vmatrix}$
• the only possible way for this to be valid is when the det is 0
### Property 7: Determinant of $U$
• for upper-triangular matrix $U$, determinant of $U$ is the product of elements on the main diagonal
• $\begin{vmatrix} d_1 & 0 & 0 \\ a_{21} & d_2 & 0 \\ a_{31} & a_{32} & d_3 \\ \end{vmatrix} = \prod d_i$
Why?
• if we can do $LU$ factorization, then we can do $LDU$ factorization as well
• and by property 5, $\text{det } A = \text{det } U = \text{det } D$
• $\begin{vmatrix} d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \\ \end{vmatrix} \ \mathop{=}\limits^{3^{\circ}} \ d_1 \cdot \begin{vmatrix} 1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \\ \end{vmatrix} = d_1 d_2 \cdot \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & d_3 \\ \end{vmatrix} = d_1 d_2 d_3 \cdot \text{det } I = \prod d_i$
Consequence:
• the easiest way to compute the determinant is to apply $A = LU$ Factorization and then compute $\text{det } U = \prod_i d_i$
• note that you should be careful with row exchanges!
• what if some $d_i = 0$? then $\text{det } U = 0$
### Property 8: Singularity Test
• when $A$ is singular, then $\text{det } A = 0$
• when $A$ is non-singular, then $\text{det } A \ne 0$
• it makes it a good test for invertability
Why?
• directly follows from property 7
• compute $A = LU$ factorization
• if the matrix is singular, then at least one $d_i = 0$, then $\text{det } U = 0$
• if $A$ is not singular, then no pivot is 0, thus $\text{det } U \ne 0$
### Property 9: Product Rule
• $\text{det } AB = \text{det } A \cdot \text{det } B$
Proof:
• consider ratio $D(A) = \cfrac{\text{det } AB}{\text{det } B}$ (for $\text{det } B \ne 0$)
• note that $D(A)$ obeys the properties 1, 2, 3 of $\text{det } A$
• property 1: if $A = I$, then $D(A) = \cfrac{\text{det } IB}{\text{det } B} = 1$
• property 2: if we exchange two rows of $A$, the same rows are exchanged for $AB$, thus $\text{det } AB$ changes the sign, and so does $D(A)$
• property 3:
• 3a) multiply row 1 of $A$ by $c$, then row 1 of $AB$ also gets multiplied by $c$
• 3b) add $[a'_{11}, \ ... \ , a'_{1n}]$ to row 1 of $A$ - then row 1 of $AB$ gets row 1 of $A' B$ (where $A'$ is $A$ with row 1 replaced)
• illustration: $\text{det } \begin{bmatrix} a_{11} + a'_{11} & a_{12} + a'_{12} \\ ... & ... \\ \end{bmatrix} \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{bmatrix} =$ $\begin{vmatrix} (a_{11} + a'_{11}) b_{11} + (a_{12} + a'_{12}) b_{21} & (a_{11} + a'_{11}) b_{12} + (a_{12} + a'_{12}) b_{22}\\ ... & ... \\ \end{vmatrix} =$ $\begin{vmatrix} a_{11} b_{11} + a_{12} b_{21} & a_{11} b_{12} + a_{12} b_{22}\\ ... & ... \\ \end{vmatrix} +$ $\begin{vmatrix} a'_{11} b_{11} + a'_{12} b_{21} & a'_{11} b_{12} + a'_{12} b_{22}\\ ... & ... \\ \end{vmatrix}$
• thus $D(A)$ obeys the same properties as $\text{det } A$, so $D(A) = \text{det } A$ and we have $\text{det } AB = \text{det } A \cdot \text{det } B$
Consequence:
• $I = A^{-1} A$
• $\text{det }I = \text{det }A^{-1} A$
• $\text{det }A^{-1} \cdot \text{det } A = 1$
• $\text{det }A^{-1} = \cfrac{1}{\text{det } A}$
Consequence 2:
• now can take into account the permutation matrix $P$ in the $PA = LU$ decomposition
• $\text{det } PA = \text{det } LU$
• $\text{det } P \cdot \text{det } A = \text{det } L \cdot \text{det } U$
• $\text{det } P = \pm 1$, $\text{det } L = 1$ (elements on the diagonal of $L$ are 1's)
• so $\text{det } A = \text{det } P \cdot \text{det } U$
### Property 10: Transposition
$\text{det } A^T = \text{det } A$
• The transpose of $A$ has the same determinant as $A$
Proof:
• consider $PA = LU$ factorization
• transpose: $A^T P^T = U^T L^T$
• take determinant, apply property 10 and compare $\text{det }P \cdot \text{det }A = \text{det } L \cdot \text{det } U$ with $\text{det }A^T \cdot \text{det }P^T = \text{det }U^T \cdot \text{det } L^T$
• $\text{det } L = \text{det } L^T = 1$ (both have 1's on the diagonal)
• $\text{det } U = \text{det } U^T = \prod d_i$ - they have the same elements on the diagonal
• finally $\text{det } P = \text{det } P^T$ because $P^T P = I$ ($P$ is orthogonal) and by property 9 have $\text{det } P^T \cdot \text{det } P = 1$. That happens only when they agree on the sign.
• thus, $\text{det } A^T = \text{det } A$
Consequence
• all the properties above are applied to rows, but the property #10 says that we can apply them to columns as well
## Calculating Determinants
There are several possible ways to calculate determinants:
• the Determinant Formula
• the Pivot Formula
• Cofactors
## Determinant Formula
Let's try to find out how we can compute the determinant using the properties
### $2 \times 2$ case
• $\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{vmatrix} = \ ...$
• can use the property 3 to divide the problem into smaller parts, and solve them separately
• $... \ = \begin{vmatrix} a_{11} + 0 & 0 + a_{12} \\ a_{21} & a_{22} \\ \end{vmatrix} \ \mathop{=}\limits^{3^{\circ}} \ \begin{vmatrix} a_{11} & 0 \\ a_{21} & a_{22} \\ \end{vmatrix} + \begin{vmatrix} 0 & a_{12} \\ a_{21} & a_{22} \\ \end{vmatrix} = \underbrace{\begin{vmatrix} a_{11} & 0 \\ a_{21} & 0 \\ \end{vmatrix}}_{0} + \begin{vmatrix} a_{11} & 0 \\ 0 & a_{22} \\ \end{vmatrix} + \begin{vmatrix} 0 & a_{12} \\ a_{21} & 0 \\ \end{vmatrix} + \underbrace{\begin{vmatrix} 0 & a_{12} \\ 0 & a_{22} \\ \end{vmatrix}}_{0} = \ ...$
• by property 6 (zero row) and 10 (determinant of transpose), we know that some parts are 0. so we're left with
• $... \ = \begin{vmatrix} a_{11} & 0 \\ 0 & a_{22} \\ \end{vmatrix} + \begin{vmatrix} 0 & a_{12} \\ a_{21} & 0 \\ \end{vmatrix} = \ ...$
• now can change the rows of the second summand by property 2 (sign reversal) and get
• $... \ = \begin{vmatrix} a_{11} & 0 \\ 0 & a_{22} \\ \end{vmatrix} - \begin{vmatrix} a_{21} & 0 \\ 0 & a_{12} \\ \end{vmatrix} = a_{11}a_{22} - a_{21}a_{12}$
### $3 \times 3$ case
Can do the same for $3 \times 3$ matrices
• $\begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ \end{vmatrix} = \ ...$
• we follow the same divide and conquer approach
• most of the terms will go away because they will be equal to 0
• the "survivers" will have one non-zero entry from each row
• so for $3 \times 3$ we have:
• $... \ = \begin{vmatrix} a_{11} & 0 & 0\\ 0 & a_{22} & 0\\ 0 & 0 & a_{33}\\ \end{vmatrix} + \begin{vmatrix} a_{11} & 0 & 0\\ 0 & 0 & a_{23}\\ 0 & a_{32} & 0\\ \end{vmatrix} + \begin{vmatrix} 0 & a_{12} & 0\\ a_{21} & 0 & 0\\ 0 & 0 & a_{33}\\ \end{vmatrix} + \begin{vmatrix} 0 & a_{12} & 0\\ 0 & 0 & a_{23}\\ a_{31} & 0 & 0\\ \end{vmatrix} + \begin{vmatrix} 0 & 0 & a_{13}\\ a_{21} & 0 & 0\\ 0 & a_{32} & 0\\ \end{vmatrix} + \begin{vmatrix} 0 & 0 & a_{13}\\ 0 & a_{22} & 0\\ a_{31} & 0 & 0\\ \end{vmatrix}$
• let's have a closer look at each of them
• $\begin{vmatrix} a_{11} & 0 & 0\\ 0 & a_{22} & 0\\ 0 & 0 & a_{33}\\ \end{vmatrix} = a_{11}a_{22}a_{33}$, diagonal and nice
• $\begin{vmatrix} a_{11} & 0 & 0\\ 0 & 0 & a_{23}\\ 0 & a_{32} & 0\\ \end{vmatrix} = - a_{11}a_{23}a_{32}$ - need 1 row exchange to transform it to $I$-like form
• $\begin{vmatrix} 0 & a_{12} & 0\\ a_{21} & 0 & 0\\ 0 & 0 & a_{33}\\ \end{vmatrix} = - a_{12}a_{21}a_{33}$ - also 1 flip away
• $\begin{vmatrix} 0 & a_{12} & 0\\ 0 & 0 & a_{23}\\ a_{31} & 0 & 0\\ \end{vmatrix} = a_{12}a_{23}a_{31}$ - 2 exchanges,
• $\begin{vmatrix} 0 & 0 & a_{13}\\ a_{21} & 0 & 0\\ 0 & a_{32} & 0\\ \end{vmatrix} = a_{13}a_{21}a_{32}$ - 2 exchanges
• $\begin{vmatrix} 0 & 0 & a_{13}\\ 0 & a_{22} & 0\\ a_{31} & 0 & 0\\ \end{vmatrix} = -a_{13}a_{22}a_{31}$ - 3 exchanges
So, the formula for $3 \times 3$:
• $\begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ \end{vmatrix} = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} - a_{13}a_{22}a_{31}$
• or, schematically:
### $n \times n$ case: "Big Formula"
The big formula:
• we consider all $n!$ possible permutation matrices $P$
• why $n!$? we can choose an element from the row 1 in $n$ ways, an element from the row 2 in $n - 1$ ways, ..., the last - in one way
• $\text{det } A = \sum\limits_{\text{$n!$permutations$P$}} \text{det } P \cdot a_{1\alpha_1} a_{2\alpha_2} ... a_{n\alpha_n}$
• where $\boldsymbol \alpha = (\alpha_1, \ ... \ , \alpha_n)$ is a Permutation of $(1, \ ... \ , n)$
## The Pivot Formula
The easiest way is to use the properties 2, 5, 7 and 9:
• do the factorization $PA = LU$
• know that $\text{det } A = \text{det } P \cdot \text{det } U$
• if one of the pivots is 0, then $\text{det } A = 0$
• if $P$ is $\pm 1$, depending on the number of permutations, and $\text{det } U = \prod d_i$
## Cofactors
Cofactors give a way to break $n \times n$ determinant to $(n - 1) \times (n - 1)$ determinants
### $3 \times 3$ Case: Intuition
Suppose $A$ is a $3 \times 3$ matrix
• $\begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ \end{vmatrix} = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} - a_{13}a_{22}a_{31}$
• let's group them
• $\text{det } A = a_{11} (a_{22}a_{33} - a_{23}a_{32}) + a_{12} (-1) (a_{21}a_{33} - a_{23} a_{33}) + a_{13} (a_{21}a_{32} - a_{22}a_{31})$
• note that now in parentheses we have determinants of smaller matrices!
• for $a_{11}$ we have $a_{22}a_{33} - a_{23}a_{32} = \begin{vmatrix} a_{22} & a_{23}\\ a_{32} & a_{33}\\ \end{vmatrix}$
• for $a_{12}$ we have $- (a_{21}a_{33} - a_{23} a_{33}) = - \begin{vmatrix} a_{21} & a_{23}\\ a_{31} & a_{33}\\ \end{vmatrix}$ (note the $-$ sign!)
• for $a_{13}$ we have $a_{21}a_{32} - a_{22}a_{31} = \begin{vmatrix} a_{21} & a_{22}\\ a_{31} & a_{32}\\ \end{vmatrix}$
• these are co-factors of $a_{11}, a_{12}, a_{13}$ respectively
• so we can write this as
• $\begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ \end{vmatrix} = \begin{vmatrix} a_{11} & 0 & 0\\ 0 & a_{22} & a_{23}\\ 0 & a_{32} & a_{33}\\ \end{vmatrix} - \begin{vmatrix} 0 & a_{12} & 0\\ a_{21} & 0 & a_{23}\\ a_{31} & 0 & a_{33}\\ \end{vmatrix} + \begin{vmatrix} 0 & 0 & a_{13}\\ a_{21} & a_{22} & 0\\ a_{31} & a_{32} & 0\\ \end{vmatrix}$
### Cofactors
a cofactor of $a_{ij}$ is $C_{ij}$
• $C_{ij}$ is a determinant of a $n - 1$ matrix - it's a matrix $A$ with row $i$ and column $j$ removed
• note that we can have a minus sign before some of the cofactors
• we have $C_{ij}$ with $-$ if $i+j$ is odd, and $+$ if $i+j$ is even
• so for $3 \times 3$ matrix we take signs this way: $\begin{vmatrix} + & - & + \\ - & + & - \\ + & - & + \\ \end{vmatrix}$
a minor of $a_{ij}$ is $M_{ij}$
• it's the same as cofactor, but always with the same sign
### The Cofactor Formula
We can take co-factors along any row or column
• suppose we take it along row 1
• then the formula is $\text{det } A = a_{11} C_{11} + a_{12} C_{12} + \ ... \ + a_{1n} C_{1n}$
## Applications
What can we do with determinants?
### Cramer's Rule
through the Cramer's Rule:
### Volume
$\text{det } A$ = volume of a parallelepiped formed by vector-rows of $A$
• $A = \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ \end{bmatrix}$
• $\mathbf r_1 = \Big[a_{11} \ \ a_{12} \ \ a_{13} \Big]$
• $\mathbf r_2 = \Big[a_{21} \ \ a_{22} \ \ a_{23} \Big]$
• $\mathbf r_3 = \Big[a_{31} \ \ a_{32} \ \ a_{33} \Big]$
$| \text{det } A|$ is the volume of the box formed by vectors $\mathbf r_1, \mathbf r_2, \mathbf r_3$
To check if this is indeed true, we need to verify that the volume obeys the 3 defining properties
• $A = I$ works, the volume is 1
• property 2: reversing two rows changes the sign (don't care), but the volume remains the same - true
• linearity:
• 3a. suppose we double one edge: the volume double
• 3b. see pictorially
### Area of Triangle
We know how to compute the area of a square
• so we can compute the area of a triangle!
• let $A$ be $2 \times 2$ matrix, $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix}$
• the area of a triangle that starts in origin is $\cfrac{1}{2} \text{det } A = \cfrac{1}{2} (a_{11} \, a_{22} - a_{12} \, a_{21})$
What is it doesn't start in origin?
• then we calculate it by taking the following determinant:
• $\begin{vmatrix} x_1 & y_1 & 1 \\ x_1 & y_1 & 1 \\ x_1 & y_1 & 1 \\ \end{vmatrix}$ | {"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": 0.9278755784034729, "perplexity": 988.2536025579192}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304217.55/warc/CC-MAIN-20220123081226-20220123111226-00602.warc.gz"} |
http://mathhelpforum.com/advanced-math-topics/145706-rewriting-summations.html | # Math Help - Rewriting summations
1. ## Rewriting summations
Can all double summations be rewritten as a single sum?
For example, can this, $\sum_{i=-1}^{1}\sum_{j=0}^{2}(2i+3j)$, be rewritten as a single sum, and if so, how?
The answer isn't important this is just an example.
2. Originally Posted by dwsmith
Can all double summations be rewritten as a single sum?
For example, can this, $\sum_{i=-1}^{1}\sum_{j=0}^{2}(2i+3j)$, be rewritten as a single sum, and if so, how?
The answer isn't important this is just an example.
Well, I don't know of any general method, but I would attack the above double sum as follows:
$\sum_{i=-1}^{1}\sum_{j=0}^{2}(2i+3j)$
$=\sum_{i=-1}^{1}\left(2i\sum_{j=0}^{2}1+3\sum_{j=0}^{2}j\rig ht)$
$=\sum_{i=-1}^{1}\left((2i)(3)+3\frac{(2)(2+1)}{2}\right)$
$=\sum_{i=-1}^{1}6i+9$
Depending on the sums you're dealing with, similar strategies could be possible.
3. Originally Posted by undefined
Well, I don't know of any general method, but I would attack the above double sum as follows:
$\sum_{i=-1}^{1}\sum_{j=0}^{2}(2i+3j)$
$=\sum_{i=-1}^{1}\left(2i\sum_{j=0}^{2}1+3\sum_{j=0}^{2}j\rig ht)$
$=\sum_{i=-1}^{1}\left((2i)(3)+3\frac{(2)(2+1)}{2}\right)$
$=\sum_{i=-1}^{1}6i+9$
Depending on the sums you're dealing with, similar strategies could be possible.
I was wondering if there was a way of combining the summations without prior to summing the inside summation.
4. Originally Posted by dwsmith
I was wondering if there was a way of combining the summations without prior to summing the inside summation.
Well it's not possible in general to rewrite a double integral as a single integral, without trying to compute the inner one (before or after a change in order of integration), right? So I expect you may be looking for a method that does not exist.
Edit: I stand corrected. But I probably won't be of much further help in this discussion.. I am familiar with Kronecker Delta but don't see how it relates to the question, also it's been years since I used Green's Theorem and I don't really remember it.
5. Originally Posted by undefined
Well it's not possible in general to rewrite a double integral as a single integral, without trying to compute the inner one (before or after a change in order of integration), right? So I expect you may be looking for a method that does not exist.
So the Kronecker Delta is a special case then?
6. There are some ways to write double integrals as single integrals; I should point out that Green's Theorem does this for us.
7. Originally Posted by roninpro
There are some ways to write double integrals as single integrals; I should point out that Green's Theorem does this for us.
How can Green's Theorem be applied to the summations?
8. This mostly just has application to rewriting an infinite double sum as an infinite single sum. I'm not sure if it can be used in the finite case. | {"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": 0.8945924043655396, "perplexity": 305.1163092752233}, "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/1410657104131.95/warc/CC-MAIN-20140914011144-00146-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://getrevising.co.uk/revision-tests/topic_5_security_polices_part_2_1 | # Topic 5: Security Polices Part 2
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S X V O M N T Q O C S Y P G X E T A M S M D Y Y R I X D A R R A Y T M I S Q U H G I N P I A G B T P P B R D S R Y Q I X W N A C S O O R F E Q C E S C D I B W C W O E S R T G C T J R X M T R W H U H A M F A L L A C K D S F I R E W A L L S Y H T G S I Q C S B T F W F I M M V A E W C D V A F B W K S N E P T T O Y I S C O K B V R Y J P F X U J R B K S Y V L S V I L T N Y N T H U R Y J Y T L K J O O I P E K P J E J W M A M J I H R S H Y Q O X T U S E
### Clues
• 2.The machines, wiring, and other physical components of a computer or other electronic system. (8)
• a name or a number that is used to identify a certain user of network or system (4, 2)
• a piece of software or hardware which protects a network from hackers (9)
• a series of characters chosen by the user to access files (8)
• copies of software and data kept so that the data can be recovered should there be a total loss of the ICT system (4, 2)
• The programs and other operating information used by a computer. (8) | {"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": 0.9392431974411011, "perplexity": 1366.5911775984105}, "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/1476988720000.45/warc/CC-MAIN-20161020183840-00120-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://forum.math.toronto.edu/index.php?PHPSESSID=ku2b6j89099b81aqppsb8e8an2&action=profile;u=2369;area=showposts;start=30 | ### Show Posts
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### Messages - RunboZhang
Pages: 1 2 [3] 4
31
##### Quiz 2 / LEC0201 Quiz2 Question2B
« on: October 02, 2020, 01:46:06 PM »
Question and my answer is in the pic attached below.
32
##### Chapter 1 / Re: Question on past quiz problem
« on: September 30, 2020, 09:52:48 PM »
Basically you can substitute z by re^i*theta and use the formula Log(z)=ln|z|+iArg(z)=lnr+i*theta where theta belongs to [-pi, pi). After rearranging you will have a sampled equation and it would be much easier to solve. The pic attached below is my solution, correct me if I am wrong.
33
##### Chapter 1 / Re: Q10 in Section 1.3
« on: September 30, 2020, 09:30:19 PM »
I think we have restricted arg(z) to (pi/2,pi). After expanding z^2 = (e^i*theta)^2, you will observe that the argument changes to (pi,2pi), which indicates that the second quadrant (excluding axis) is getting mapped to the lower half of the complex plane (still excluding the axises). Thus you can easily prove that this set is open and connected. That is what I think about this question, correct me if I am wrong.
34
##### Chapter 1 / Re: how to solve question 19 in section 1.2 in the textbook
« on: September 27, 2020, 09:42:31 AM »
Hello Darren, I have two questions regarding your solution. Firstly, why did you exclude p^2 in equation (1-c^2)x^2-2xp+y^2=0 where p is a real number? And secondly, when you got the solution for eclipse and hyperbola, how did you get 1 on RHS? I thought RHS would be an expression with respect to x and c and it is not necessary to be 1 right?
35
##### Chapter 2 / Question on Week3 Lec3 Integrating Factor
« on: September 25, 2020, 04:27:01 PM »
Hi guys, we went through integrating factors in lecture 3 of this week. When using integrating factor to make a non-exact equation exact, we have different types of it. Namely, we have mu(x,y)=mu(x), mu(x,y)=mu(y), and mu(x,y)=mu(xy). I am particularly confused by them. When we are solving a non-exact equation, shall we try three different types and see which one would be helpful, or is there a way to give an instant intuition?
36
##### Chapter 1 / Re: past quiz problem
« on: September 24, 2020, 09:38:27 AM »
Hi, the picture attached below is my solution to this problem.
37
##### Chapter 1 / Re: how to solve this problem?
« on: September 23, 2020, 06:07:15 PM »
I think there are two ways to solve this problem.
Firstly, you can substitute w by x+iy into the equation. Then organize the equation and put real parts together and imaginary parts together. Now you will have the equation |(x+2)+iy|=|(x-2)+iy|. Then take the square of both sides and organize the equation. Finally you will get x=0, which is the y-axis.
The second way to do this problem is to illustrate it geometrically. |w+2| = |w-2| is same as |w-(-2)|=|w-2|, and it asks you to find all w in complex plane that has equal distances to (2,0) and (-2,0). Thus the answer will be the perpendicular bisector of (2,0) and (-2,0), which is still the y-axis.
38
##### Chapter 1 / Question on Textbook Section1.5 The Log Function Example 3
« on: September 22, 2020, 09:16:34 PM »
Hi everyone, I have went through textbook section 1.5 and found an example quite confusing (pic is attached below). In this example, in order to get the limit (n->∞), we discuss the real part and imaginary part separately. But I don't know how does it solve for the real part. In particular, it says we need to use L'Hopital's rule and we will get x as the result. However, I have been stuck at this computation (the draft in orange color is my computation) and it seems like a dead end. I dont know if I did wrong at the very first step. Besides, when we are using L'Hopital's rule here, are we taking the derivative with respect to n or x? How can we know which is the variable?
39
##### Chapter 1 / Re: Solving roots of complex numbers
« on: September 20, 2020, 07:18:56 PM »
I think if the angle is unfamiliar, we can leave it as z=r[cos theta + i sin theta], otherwise we need to compute the value of sin and cos. Also, I think it has no difference with z=a+bi, it is just in the polar form.
40
##### Chapter 2 / Re: Lec 0101 - 9/15 Question
« on: September 20, 2020, 07:12:43 PM »
Hi, I have plugged it in y and calculated the LHS. Computation is attached below.
41
##### Chapter 2 / Re: section 2.1 practice problem 15
« on: September 20, 2020, 10:21:52 AM »
I am sorry, I think there is something I did not make clear. Since at the very first place we have determined x!=(15^(1/2))/2 due to the zero-division error. Beyond that we have initial condition y(2)=0. That indicates that we deduce our ODE from x=(15^(1/2))/2 to x=2. Then you can observe the graph and find its slope is greater than 0 at every defined point. Moreover, since we deduce our ODE starting from (15^(1/2))/2 to 2, we can only go in the same direction due to singularity. I think more will be expanded in lecture.
42
##### Chapter 1 / Re: Past quiz 1
« on: September 20, 2020, 09:59:52 AM »
I think it is explained by the first reply in that link. |x+yi-i|=Rez is equivalent to x^2 + (y-1)^2=x^2, subtract both sides by x^2 and this equation would be irrelevant to x. So the only restriction is on y and we have to let y=1. Also, if put it in geometry, it does not matter whether Rez has a positive sign or a negative sign since |x+yi-i| means that the distance between (x,y) and (0,1) is fixed and equal to Rez=x, and it has two corresponding points, (x,1) and (-x,1).
43
##### Chapter 2 / Re: section 2.1 practice problem 15
« on: September 19, 2020, 06:51:24 PM »
First of all, it does not equal to 15^(1/2)/2 since the derivative does not exist at x=15^(1/2)/2 (zero devision error). And after you have plotted the curve by part (b) of this question, you would find that y' is greater than 0 at every defined point. Thus only when x>15^(1/2)/2, y>-1/2, its derivative is always positive. Thats why x is strictly greater than 15^(1/2)/2.
44
##### Chapter 2 / Re: LEC 0201 Question
« on: September 19, 2020, 02:59:05 PM » | {"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": 0.9288438558578491, "perplexity": 605.4544703402469}, "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/1664030335286.15/warc/CC-MAIN-20220928212030-20220929002030-00670.warc.gz"} |
https://proceedings.neurips.cc/paper/2020/file/440924c5948e05070663f88e69e8242b-MetaReview.html | NeurIPS 2020
Meta Review
The problem of exploration in RL with function approximation is very important and any advancement on the topic is of interest for the community. The reviewers all agreed about the algorithmic and technical contribution of the paper, in particular the introduction of sensitive sampling and its analysis in the regret proof. This convinced us that the paper deserves acceptance. Nonetheless, I also encourage the authors to improve the current submission. As pointed out by R3, the assumptions used in the paper are quite strong and they may somehow limit the generality of the results. The authors should stress the potential limitations coming from the assumptions and better contrasted it with the related literature. In particular, using the Eluder dimension as a measure of complexity is definitely interesting but, as of today, it lacks of interpretability. In fact, we have very few families of MDPs for which a meaningful bound for the Eluder dimension is available (i.e., linear and GLM). The authors should point this out and possibly clarify whether it is possibly to obtained good bounds for other classes of problems. | {"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": 0.9543271660804749, "perplexity": 459.08868159052554}, "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-2021-39/segments/1631780060677.55/warc/CC-MAIN-20210928092646-20210928122646-00631.warc.gz"} |
http://www.koreascience.or.kr/article/ArticleFullRecord.jsp?cn=DBSHCJ_2013_v28n3_463 | ON NIL GENERALIZED POWER SERIESWISE ARMENDARIZ RINGS
Title & Authors
ON NIL GENERALIZED POWER SERIESWISE ARMENDARIZ RINGS
Ouyang, Lunqun; Liu, Jinwang;
Abstract
We in this note introduce a concept, so called nil generalized power serieswise Armendariz ring, that is a generalization of both S-Armendariz rings and nil power serieswise Armendariz rings. We first observe the basic properties of nil generalized power serieswise Armendariz rings, constructing typical examples. We next study the relationship between the nilpotent property of R and that of the generalized power series ring [[$\small{R^{S,{\leq}}}$]] whenever R is nil generalized power serieswise Armendariz.
Keywords
nil generalized power serieswise Armendariz;generalized power series ring;nilpotent property;
Language
English
Cited by
References
1.
D. D. Anderson and V. Camillo, Armendariz rings and Gaussian rings, Comm. Algebra 26 (1998), no. 7, 2265-2272.
2.
R. Antoine, Nilpotent elements and Armendariz rings, J. Algebra 319 (2008), no. 8, 3128-3140.
3.
E. P. Armendariz, A note on extensions of Baer and p.p.-rings, J. Austral. Math. Soc. 18 (1974), 470-473.
4.
J. A. Beachy and W. D. Blair, Rings whose faithful left ideals are cofaithful, Pacific J. Math. 58 (1975), no. 1, 1-13.
5.
H. Cartan and E. Eilenberg, Homological Algebra, Princeton Landmarks inMathematics, originally published in 1956, Princeton: Princeton University Press, 1956.
6.
C. Faith, Rings with zero intersection property on annihilators: zip rings, Publ. Mat. 33 (1989), no. 2, 329-332.
7.
S. Hizem, A note on nil power serieswise Armendariz rings, Rendiconti del Circolo Mathematico di Palermo 59 (2010), no. 1, 87-99.
8.
N. K. Kim, K. H. Lee, and Y. Lee, Power series rings satisfying a zero divisor property, Comm. Algebra 34 (2006), no. 6, 2205-2218.
9.
T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics, Springer-Verlag, Berlin, 1991.
10.
Z. K. Liu, Special properties of rings of generalized power series, Comm. Algebra 32 (2004), no. 8, 3215-3226.
11.
Z. K. Liu, On weak Armendariz rings, Comm. Algebra 34 (2006), no. 7, 2607-2616.
12.
L. Ouyang, Ore extensions of weak zip rings, Glasg. Math. J. 51 (2009), no. 3, 525-537.
13.
M. B. Rege and S. Chhawchharia, Armendariz rings, Proc. Japan Acad. Ser. A Math. Sci. 73 (1997), no. 1, 14-17.
14.
P. Ribenboim, Rings of generalized power series: Nilpotent elements, Abh. Math. Sem. Univ. Hamburg 61 (1991), 15-33.
15.
P. Ribenboim, Noetherian rings of generalized power series, J. Pure Appl. Algebra 79 (1992), no. 3, 293-312.
16.
P. Ribenboim, Semisimple rings and von Neumann regular rings of generalized power series, J. Algebra 198 (1997), no. 2, 327-338. | {"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": 1, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.917738139629364, "perplexity": 1849.062101651346}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542695.22/warc/CC-MAIN-20161202170902-00505-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/help-with-graphical-technique-of-adding-2-sinusoids.561602/ | # Help with graphical technique of adding 2 sinusoids
1. Dec 20, 2011
### p75213
1. The problem statement, all variables and given/known data
I have a problem understanding how equation 9.11 (see attached) works with Fig. 9.4 (a). Any help would be appreciated.
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
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2. Dec 20, 2011
### Simon Bridge
The two equations (9.11) are being represented as a 2D vector (fig 9.4 a).
The authors have used the phasor representation of the sinusoids and the diagrams represent the vector sums of the phasors. I'd have explained it a bit differently.
Basically, cosine and sine functions have a relative phase of π/2 - so you can use them as x and y axes. The amount each contributes to the result is it's amplitude, which is what is being plotted on the axis. This does not work if the sinusoids have other than nπ/2: n=0,1,2,...
In general - use the phasor representation for arbitrary sinusoids and add the phasor arrows head-to-tail just like vectors.
3. Dec 20, 2011
### JHamm
Try expanding the right hand side, you'll see that it works out :)
Edit: Oops beaten, cutthroat business this physics stuff :P
4. Dec 20, 2011
### p75213
Thanks for that. Makes sense now that I realize they are vectors. Makes you wonder why they didn't make it clearer. Phasors is the subject of the next chapter.
5. Dec 20, 2011
### Simon Bridge
Ah well, they are trying to be kind to you by building up to them.
I always just launch right into it. By now you've had vectors drilled into you so you see them in your sleep so you can handle rotating vectors no problem.
Never mind - skip ahead to phasors and you'll see it makes better sense.
6. Dec 20, 2011
### Staff: Mentor
Remember, wherever you see A.cos(wt) you can replace it by A.sin(wt+Pi/2) because this is an exact equivalence.
By similar reasoning, -cos(wt) = cos(wt+Pi) = cos(wt-Pi)
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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acid/Graphical_Treatment_of_Acid-Base_Systems | # Graphical Treatment of Acid-Base Systems
Most students in General Chemistry are never taken beyond the traditional, algebraic treatment of acid-base equilibria. This is unfortunate for the following reasons:
• It can be an awful lot of work. Have you ever noticed that many of the acid-base systems most commonly encountered (phosphate, citrate, salts such as ammonium acetate, amino acids, EDTA) are rarely treated in standard textbooks? Treating these analytically requires setting up a series of mass- and charge-balance expressions which must be solved simultaneously.
• Most algebraic treatments are approximations anyway. Did you know, for example, that the exact calculation of the pH of a solution of a monoprotic weak acid requires the solution of a cubic equation? Carrying out the same operation for H3PO4 requires solution of a fifth-order polynomial! To simplify the calculations, approximations are made, sometimes unwittingly. The algebraic calculations that we actually carry out are almost never exact in the first place!
• Equilibrium constants are not really. Even if you carry out an exact calculation, the results will never be any more reliable than the equilibrium constants you use. But the values of these vary with the temperature and especially with the ionic content of the solution. The values listed in tables are rarely applicable to practical applications.
• All you get is a number: Algebraic approaches contribute almost nothing to the larger view of how an acid-base system behaves as the pH is changed.
Yes, I still teach my students how to set up a quadratic equation to calculate the pH of an acetic acid solution, but I insist that they also be able to do the same thing graphically. The major advantages of the graphical approach:
• Easy to do: the graphs are easily constructed and generally give pH values as good as those from algebra. Even if you sketch out the graph on the back of an envelope and without a straightedge, you can get results to within half a pH unit. A good way to amaze your friends!
• Provides an overall picture of the acid-base system. A glance at the graph shows you the approximate concentrations of all species present over a range of pH values, thus providing a bird's-eye view of the acid-base system as s whole.
• Reinforces important principles. In learning to sketch out and interpret log-concentration vs. pH plots, the student must consider such things as the nature of the equilibria occurring in solutions of the pure acid and of its conjugate base, conservation of protons, and the significance of the pK.
• Allows treatment of a wider range of systems. Graphical estimation of the pH of solutions of acids, bases and ampholytes of such as phosphate, citrate, amino acids, EDTA etc. are not all that much more difficult than treating a monoprotic or diprotic system.
Log-C vs. pH graphs are plotted on coordinates of the form shown above.Be sure you understand the y-axis; "4", for example, corresponds to a concentration of 10–4 M for whatever species we plot. Note that concentrations increases with height on this graph, and that those less than about 10–8 M are usually negligible for most practical purposes.
The two plots shown on this graph are no more than formal definitions of pH and pOH; for example, when the pH is 4, –log pH = 4 and [H+] = 10–4 M. (Notice that the ordinate is the negative of the log concentration, so the smaller numbers near the top of the scale refer to larger concentrations.)
The above graph of is of no use by itself, but it forms the basis for the construction of other graphs specific to a given acid-base system. You should be able to draw this graph from memory.
The H+ and OH log concentration lines are the same ones that we saw in Figure 1. The other two lines show how the concentrations of CH3COOH and of the the acetate ion vary with the pH of the solution.
How do we construct the plots for [HAc] and [Ac]? If you look carefully at Fig 2, you will observe that each line is horizontal at the top, and then bends to become diagonal. There are thus three parameters that define these two lines: the location of their top, horizontal parts, their crossing points with the other lines, and the slopes of their diagonal parts.
The horizontal sections of these lines are placed at 3 on the ordinate scale, corresponding to the nominal acid concentration of
10–3 M. This value corresponds to
Ca = [HAc] + [Ac]
which you will recognize as the mass balance condition saying that "acetate" is conserved; Ca is the nominal "acid concentration" of the solution, and is to be distinguished from the concentration of the actual acidic species HAc.
At low pH values (strongly acidic solution) the acetate species is completely protonated, so [HAc] = 10–3 M and [Ac]=0. Similarly, at high pH, –log [Ac]=3\$ and [HAc]=0. If the solution had some other nominal concentration, such as 0.1 M or 10–5, we would simply move the pair of lines up or down.
The diagonal parts of the lines have slopes of equal magnitude but opposite sign. It can easily be shown that these slopes d(–log [HAc]}/d{pH} etc.are +-1, corresponding to the slopes of the [OH] and [H+] lines. Using the latter as a guide, the diagonal portions of lines 3 and 4 Can easily be drawn.
The crossing point of the plots for the acid and base forms corresponds to the condition [HAc]=[Ac]. You already know that this condition holds when the pH is the same as the pKa of the acid, so the the pH coordinate of the crossing point must be 4.75 for acetic acid. The vertical location of the crossing point is found as follows: When [HAc] = [Ac], the concentration of each species must be Ca/2 or in this case 0.0005 M. The logarithm of 1/2 is 0.3, so a 50% reduction in the concentration of a species will shift its location down on the log concentration scale by 0.3 unit. The crossing point therefore falls at a log-C value of (–3) – 0.3 = –3.3. Knowing the value of log100.5 is one of the few new "facts" that must be learned in order to construct these graphs.
Of special interest in acid-base chemistry are the pH values of a solution of an acid and of its conjugate base in pure water; as you know, these correspond to the beginning and equivalence points in the titration of an acid with a strong base.
## pH of an acid in pure water
Except for the special cases of extremely dilute solutions or very weak acids in which the autoprotolysis of water is a major contributor to the hydrogen ion concentration, the pH of a solution of an acid in pure water will be determined largely by the extent of the reaction
$\ce{HAc + H2O -> H3O^{+} + Ac^{–}}$
so that at equilibrium, the approximate relation $$\ce{[H3O^{+}] \approx [Ac^{–}]}$$ will hold. The equivalence of these two concentrations corresponds to the point labeled 1 in Fig 3; this occurs at a pH of about 3.7, and this is the pH of a 0.001M solution of acetic acid in pure water.
### pH of a solution of the conjugate base
Now consider a 0.001M solution of sodium acetate in pure water. This, you will recall, corresponds to the composition of a solution of acetic acid that has been titrated to its equivalence point with sodium hydroxide. The acetate ion, being the conjugate base of a weak acid, will undergo hydrolysis according to
$\ce{Ac– + H2O -> HAc + OH–}$
As long as we can neglect the contribution of OH from the autoprotolysis of the solvent, we can expect the relation [HAc]=[OH] to be valid in such a solution. The equivalence of these two concentrations corresponds to the intersection point 3 in Fig 3; a 0.001M solution of sodium or potassium acetate should have a pH of about 8.
## Putting it all together
The top part of this figure is the log-C plot for acetic acid that we saw previously, the blue dashed vertical lines now showing the pH of a 0.001M solution of HAc before and and after its titration with strong base.
The extension of these lines (and the one representing the pK) down to the bottom part of the figure locates the three points that define the titration curve for the acid.
Take a moment to verify that the lower figure is indeed a titration curve; that is, a plot of the pH as a function of the fraction of acid neutralized.
These two plots nicely sum up the complete acid-base chemistry of a monoprotic system.
Diprotic acids are no more complicated to treat graphically than monoprotic acids-- something that definitely cannot be said for exact algebraic treatment! The above example for oxalic acid (H2A = HOOCCOOH) is just the superposition for separate plots of the two acid-base systems H2A-HA and HA-A2– whose pKs are 1.2 and 4.2. The system points corresponding to 0.01M solutions of pure H2A; and A2– are determined just as in the monoprotic case. For a solution of the ampholyte, several equilibria can be written, but the one that dominates is
$2 HA^– \rightleftharpoons H_2A + A^{2–}$
$[HA^–] \approx [A^{2–}]$
Thus fixing the pH at about 2.7. This same result can be obtained from thewell known approximation
$pH = \dfrac{pK_1 + pK_2}{2}$
The quantitative treatment of a solution of a salt of a weak acid and a weak base is algebraically complicated, even when done to the crudest approximation. Graphically, it is a piece of cake! As you can see above, we just construct plots of the two acid-base system on the same graph. The stoichiometry of the salt defines the system point at about pH 6.3, showing that hydrolyses of the two systems does not quite "cancel out". Of course, this is still an approximation that neglects ionization of either the anion or cation, but it is probably as valid as any calculation that is likely to be carried out ordinarily-available data.
## The Phosphate system
... but why stop at diprotic acids? With the phosphate system we have three pKs, and thus two ampholytes H2PO4and HPO42–.
A glance at the graph allows one to estimate the relative amounts of the two major species at any pH.
The pH of a solution of the acid H3PO4 or of the base PO43– is found from the system points 1 and 4 which are constructed in the same way as those for a monoprotic acid.
System points 2 and 3 for solutions of the two ampholytes provide only approximate values pH values because of the effects of competing equilibria, but the accuracy is good enough for most applications.
Here it is -- the most important of all acid-base systems in natural waters and physiology! Nothing new as far as theory goes -- see the explanation given for oxalic acid. What's different here is the two sts of plots. The lower (more dilute) one corresponds to the concentration of dissolved CO2 present in water in equlibrium with the atmosphere having the normal CO2 partial pressure of about .0005 atm. You can see from this that when pure water is exposed to the atmosphere, its pH will fall from 7 to about 5 (but see the last paragraph below); thus all rain is in a sense "acid rain".
Note that around pH 8 (the pH of the ocean and of blood), over 99% of all carbonate is in the form of bicarbonate HCO3. As a result, the ocean acts as a sink for atmospheric CO2 and contains about 50 times as much as does the atmosphere. Similarly, CO2 resulting from cellular respiration is converted to HCO3 for transport through the blood stream, and is converted back to CO2 in the lungs.
The upper (.001M) set of graphs correspond to the approximate carbonate content of natural waters in contact with rocks and sediments (which normally contain carbonate components). It describes the situation in lakes, streams, ground waters and the ocean. Notice how the pH of this more concentrated CO2 solution is lower (about 2, compared to 4.3), which is just what you would expect. This is the reason the pH of an algae-containing pond rises during the day (when CO2 is being consumed by photosynthesis) and falls at night (CO2 restored by respiration.)
The term "closed system" in the caption at the top of the graph means that CT, the sum of the concentrations of all three carbonate species, is assumed constant here. For an open system the plots for the bicarbonate and carbonate concentrations would be dramatically different because an alkaline solution open to the atmosphere has access to an infinite supply of CO2 and will absorb it up to the solubility limit of sodium carbonate. (You may recall that a common test for CO2 is to watch for a precipitate to appear in a saturated solution of calcium hydroxide.) Thus the estimate of pH=5 for pure water exposed to the atmosphere is not quite correct; using an "open system" plot, a somewhat lower pH, around 4.5, will be found. It turns out, however, that for many practical situations, the relatively slow transport of CO2 between air and water makes the "closed system" predictions reasonably accurate within the time scale of interest.
Glycine is the simplest of the amino acids, the building blocks of proteins. It is also a zwitterion (the word comes from the German term for "hermaphrodite" ) which chemists use to denote a species that possesses both an acidic and a basic functional group, both of which can simultaneously exist in their ionized (conjugate) forms. Examination of this graph shows that the double-ionic form (the true zwitterion) reigns supreme between pH 3 and 9; outside this range, the glycinium anion or the glycinate cation prevails. Notice that the completely un-ionized form does not exist in solution.
The log-C vs. pH diagram is constructed as s superposition of plots for each conjugate pair at its respective pKa. Note especially that the pH of a solution of glycine does lie exactly at the crossing point [Gly] = [H+], but is slightly displaced from it according to the proton balance equation shown in the inset on the graph. The other important quantity shown on this graph is the isoelectric point, which is the pH at which the concentrations of the cationic and anionic forms are identical.
It goes without saying that treating this system algebraically would be far more complicated than the results would warrant for most applications.
Note especially that
• the only really important pH on this graph is 8, the approximate pH of the natural ocean.
• The pH of the ocean is controlled mainly by the carbonate system because it is more concentrated than the two minor buffering systems, borate and silicate.
• The surface waters of the ocean are everywhere supersaturated in calcium bicarbonate and in CO2 , as shown by the dotted curves (the "ss[CO2 ]" label is the illegible one-- sorry!)
• Ion-pair complexes such as MgOH+ are significant species in seawater; there are many others. | {"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": 0.8589887022972107, "perplexity": 1102.1345951701378}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487643703.56/warc/CC-MAIN-20210619051239-20210619081239-00134.warc.gz"} |
https://www.physicsforums.com/threads/slope-and-deflection-of-beam.724212/ | # Slope and Deflection of Beam
1. Nov 21, 2013
### Woopydalan
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
For this problem, I think that something is wrong with my moment equation, I am not sue how to find the loading as a function of x for this triangle shape.
Do I need to account for the sections 0 < x < L/2 and L/2 < x < L??
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• ###### 15.10 attempt 1.pdf
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2. Nov 21, 2013
### SteamKing
Staff Emeritus
It's best to split this single load into 2 triangular loads imposed back-to-back.
I think you should also construct the shear diagram. It can help with getting the moment diagram correctly.
Draft saved Draft deleted
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https://socratic.org/questions/how-do-you-use-the-product-rule-to-find-the-derivative-of-x-2-x-3-4 | Calculus
Topics
How do you use the Product Rule to find the derivative of (x^2)(x^3+4)?
Dec 25, 2015
The explanation is given below.
Explanation:
Product rule for differentiation where product of functions is given.
If $f \left(x\right)$ and $g \left(x\right)$ are two functions then their product would be $f \left(x\right) \cdot g \left(x\right)$
The derivative $\left(f \left(x\right) \cdot g \left(x\right)\right) '$ is given by the product rule
$\left(f \left(x\right) g \left(x\right)\right) ' = f \left(x\right) g ' \left(x\right) + g \left(x\right) f ' \left(x\right)$
Derivative rules
$\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$
$\frac{d}{\mathrm{dx}} \left(f \left(x\right) + g \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) + \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)$
$\frac{d}{\mathrm{dx}} \left(c\right) = 0$ Derivative of constant is zero.
Now coming to our problem
$y = \left({x}^{2}\right) \left({x}^{3} + 4\right)$
$f \left(x\right) = {x}^{2}$ and $g \left(x\right) = \left({x}^{3} + 4\right)$
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{2}\right) \implies f ' \left(x\right) = 2 x$
$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{3} + 4\right) = \frac{d}{\mathrm{dx}} \left({x}^{3}\right) + \frac{d}{\mathrm{dx}} \left(4\right)$
$g ' \left(x\right) = 3 {x}^{2} + 0$
$g ' \left(x\right) = 3 {x}^{2}$
Using product rule
$\frac{d}{\mathrm{dx}} \left(\left({x}^{2}\right) \left({x}^{3} + 4\right)\right) = {x}^{2} \frac{d}{\mathrm{dx}} \left({x}^{3} + 4\right) + \left({x}^{3} + 4\right) \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$
$= {x}^{2} \left(3 {x}^{2}\right) + \left({x}^{3} + 4\right) \left(2 x\right)$
$= 3 {x}^{4} + 2 {x}^{4} + 8 x$ simplifying
$= 5 {x}^{4} + 8 x$ Answer
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## Causal interactive links between presence and fear in virtual reality height exposure
Please always quote using this URN: urn:nbn:de:bvb:20-opus-201855
• Virtual reality plays an increasingly important role in research and therapy of pathological fear. However, the mechanisms how virtual environments elicit and modify fear responses are not yet fully understood. Presence, a psychological construct referring to the ‘sense of being there’ in a virtual environment, is widely assumed to crucially influence the strength of the elicited fear responses, however, causality is still under debate. The present study is the first that experimentally manipulated both variables to unravel the causal linkVirtual reality plays an increasingly important role in research and therapy of pathological fear. However, the mechanisms how virtual environments elicit and modify fear responses are not yet fully understood. Presence, a psychological construct referring to the ‘sense of being there’ in a virtual environment, is widely assumed to crucially influence the strength of the elicited fear responses, however, causality is still under debate. The present study is the first that experimentally manipulated both variables to unravel the causal link between presence and fear responses. Height-fearful participants (N = 49) were immersed into a virtual height situation and a neutral control situation (fear manipulation) with either high versus low sensory realism (presence manipulation). Ratings of presence and verbal and physiological (skin conductance, heart rate) fear responses were recorded. Results revealed an effect of the fear manipulation on presence, i.e., higher presence ratings in the height situation compared to the neutral control situation, but no effect of the presence manipulation on fear responses. However, the presence ratings during the first exposure to the high quality neutral environment were predictive of later fear responses in the height situation. Our findings support the hypothesis that experiencing emotional responses in a virtual environment leads to a stronger feeling of being there, i.e., increase presence. In contrast, the effects of presence on fear seem to be more complex: on the one hand, increased presence due to the quality of the virtual environment did not influence fear; on the other hand, presence variability that likely stemmed from differences in user characteristics did predict later fear responses. These findings underscore the importance of user characteristics in the emergence of presence. | {"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": 0.8320056200027466, "perplexity": 3378.4769589976845}, "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-2020-16/segments/1585371660550.75/warc/CC-MAIN-20200406200320-20200406230820-00215.warc.gz"} |
https://www.physicsforums.com/threads/deriving-the-lorentz-transforms.117969/ | # Homework Help: Deriving the lorentz transforms
1. Apr 18, 2006
### Thrice
It's a really easy question, I know, but I must be doing something stupid. Can someone please spell out how to get the right hand side matrix form out of the individual equations?
http://img234.imageshack.us/img234/8497/lorentz25wv.jpg [Broken]
Last edited by a moderator: May 2, 2017
2. Apr 18, 2006
### Andrew Mason
If you multiply the matrix by the 4-vector (t', x', y', z') it should result in the 4-vector (t, x, y, z) as set out on the left side of the arrow. I think the matrix is wrong, though. The numerator of the second term in the top row should be v/c^2 and the numerator of the first term in the second row should be v.
AM
Last edited by a moderator: May 2, 2017
3. Apr 18, 2006
### Thrice
See I thought that as well, but they have the inverse of that matrix in the book too & it matches up with the one on http://en.wikipedia.org/wiki/Lorent...rmation_for_frames_in_standard_configuration". Apparently it's gotten by switching V for (-V). Does it work out if you use the 4 vector (ct', x', y', z') & (ct, x, y, z)?
Edit: Right.. it does.. I knew I was doing something stupid sorry.
Last edited by a moderator: Apr 22, 2017
4. Apr 18, 2006
### Thrice
Wait don't go. I have more foolish questions once I figure out how to post in that .. latex is it?
5. Apr 18, 2006
### Thrice
$$\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } } g_{\it ij} \pd{}{V^k}{} (V^i V^j )= 2 g_{\it kj} V^j$$
Last edited: Apr 18, 2006
6. Apr 18, 2006
### Thrice
K got it. Why does that work? Something to do with the symmetry of the metric....
7. Apr 18, 2006
### nrqed
No need for any special symmetry. It follows from ${\partial V^i \over \partial V^k} = \delta^i_k$ and similarly if i is replaced by j. Then you just need to rename a dummy index in one of the terms and you get the answer provided.
8. Apr 18, 2006
### Thrice
Thanks both of you. I don't know where i'd go when my brain isn't working. :) | {"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": 0.8618906140327454, "perplexity": 1098.3702012083097}, "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-2018-30/segments/1531676591481.75/warc/CC-MAIN-20180720022026-20180720042026-00359.warc.gz"} |
https://www.gradesaver.com/textbooks/math/geometry/geometry-common-core-15th-edition/chapter-8-right-triangles-and-trigonometry-8-2-special-right-triangles-lesson-check-page-503/3 | ## Geometry: Common Core (15th Edition)
$x = 4 \sqrt 2$
In a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle, the hypotenuse is $\sqrt 2$ times each leg. Let's write an equation to solve for $x$, the length of a leg: $8 = \sqrt 2(x)$ Divide each side by $\sqrt 2$ to solve for $x$: $x = \frac{8}{\sqrt 2}$ To simplify this fraction, we need to get rid of radicals in the denominator. We do this by multiplying both the numerator and denominator by the denominator: $x = \frac{8 \sqrt 2}{\sqrt 4}$ Take the square root of the denominator: $x = \frac{8 \sqrt 2}{2}$ Divide both the numerator and denominator by their greatest common factor to simplify: $x = 4 \sqrt 2$ | {"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": 0.9848658442497253, "perplexity": 101.40228377898642}, "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-2022-21/segments/1652662658761.95/warc/CC-MAIN-20220527142854-20220527172854-00223.warc.gz"} |
https://www.physicsforums.com/threads/space-probe-planet-and-sun.10347/ | # Homework Help: Space probe, planet, and sun .
1. Dec 5, 2003
### ConfusedStudent
Space probe, planet, and sun.....
Correcting test again, and I don't know where to start on this problem:
A space prob lies along a line between a planet and the sun so that the sun's gravitational pull on the probe balances the planets pull. The distance of the probe from the planet is 1.0 x 10^11m. The distance between the planet and the sun is 1.5 x10^12m. Find the mass of the planet. Mass of the sun = 2x10^30kg.
I got 0/15 on this problem so nothing I put down was right....help???
2. Dec 5, 2003
### gnome
what is the equation for the gravitational force between two objects?
Fg = ???
3. Dec 5, 2003
### ConfusedStudent
F1,2=[(G *m1*m2)/(r1,2^2)]*r1,2
That's what it says in my book, it looks weird when you type it.
But I tried this problem another way, and tell me what you think:
I used t^2=Cr^3 compared to the earth to get the period of the planet which I got was 6.53x10^15.
Then I used that in the eq:
T^2=(4*pie^2)/(G*Mp)*r^3
6.53x10^15m^2=(39.5)/(6.67x10^-11*Mp)*1.5x10^12m^3
and I got the mass to be 4.69x10^16 kg
Is this completely off??[?]
4. Dec 5, 2003
### gnome
Yes, it's completely off. The M in that equation (Kepler's Third Law) is MS, not MP. It's the mass of the sun. You can't use that equation to find the mass of the planet.
But, back to the other equation, the last r1,2 that you see there has a little ^ on top of it, right? That is a unit vector to indicate the direction of the force. It has no effect on the magnitude of the force, which is all we're concerned with here (because here we know that the two forces act along the same line, in opposite directions).
So, leave off that last r.
Now
1. call the planet X, so it's mass is MX, and the distance from the sun to the planet is rSX and the distance from the planet to the probe is rXp. And let's call the mass of the probe Mp.
2. You know that FSp = FXp
3. Find rSp
4. Can you set up an equation for the two forces using
F = GM1M2/(r122)?
5. Dec 5, 2003
### gnome
By the way, do you realize that r1,2 in the gravitational force equation is just the distance between the two objects?
(In other words, it is not the radius of the objects' orbits around the sun, which is what Kepler's equation is all about.)
6. Dec 5, 2003
### ConfusedStudent
I must really be dense, I've been looking at this screen for 15 min, I still don't get it.
7. Dec 5, 2003
### Staff: Mentor
The force of the sun on the probe equals the force of the planet on the probe. In other words (calling the Sun 1, the planet 2, and the probe 3):
F13 = F23
Now write what each force is using the gravity law:
FAB = G MAMB/(RAB)2
When you plug this into the first equation, G and M3 will cancel. You have the distances and the mass of the sun. The only variable left will be M2; solve for it.
Does this help at all?
8. Dec 6, 2003
### ConfusedStudent
Okay, this is what I got, and I hope it's right:
F(13)=F(23)
GM1M3/(R13^2)=GM2M3/(R23^2)
M1/(R13^2)=M2/(R23^2)
2x10^30/(1.4x10^12^2)=M2/(1.0x10^11^2)
M2= 1.02x10^28
Was it okay to assume the distance from the probe to the sun is the distance from the planet to the sun- the distance from planet to probe???
9. Dec 6, 2003
### Staff: Mentor
Yep.
Of course. I wouldn't call that an assumption; it's a given since they're all in a straight line.
10. Dec 6, 2003
### ConfusedStudent
Thank you so much...you're a life saver... | {"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": 0.8950356245040894, "perplexity": 898.1975914489419}, "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/1524125946120.31/warc/CC-MAIN-20180423164921-20180423184921-00017.warc.gz"} |
http://brb.epr.ch/blog/blog:latex_under_windows | # Latex under windows
This text aims to give people interested in getting started with latex a starting point and recommendations for programs and extensions.
# Programs for windows
• miktex: very good latex implementation. It is recommended to use the 'basic installation', which loads necessary packages from the internet.
• Important: use Programs-Miktex-MikTex Options and install the following packages and add-ons:
• Documentation-Latex2e Reference. Gives you context sensitive help in Winedt using CTRL-F1
• winedt, works very well with miktex. It is shareware, but well worth the price. Some features that set it apart from other editors:
• Understands the structure of latex documents and shows a tree of all connectec documents. This also works if you use macros to include files, after you have configure the proper macros for file inclusion.
• Very good spell checking support
• Can be set up to work as a portable app.
# Document templates
The initial learning curve for LaTeX can be steep. Particularly if you want to force LaTeX to format text exactly to you liking, you can loose a lot of time. Once you have useful template, you can concentrate on the content. As a starting point, here is my masterfile. The idea is to copy and customize this, and include your content as separate files. Another approach is a custom document class. However i don't use this any more, because you have to install the class file every time you change machines. I like self-contained document directories better.
# Books
• Goossens, Mittelbach: Latex companion. Pearson Studium. | {"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": 0.9376712441444397, "perplexity": 3128.8117952329076}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500550967030.98/warc/CC-MAIN-20170728123647-20170728143647-00462.warc.gz"} |
http://wiki.stat.ucla.edu/socr/index.php?title=AP_Statistics_Curriculum_2007_ANOVA_2Way&diff=6659&oldid=6257 | # AP Statistics Curriculum 2007 ANOVA 2Way
(Difference between revisions)
Revision as of 01:57, 7 February 2008 (view source)IvoDinov (Talk | contribs) (New page: == General Advance-Placement (AP) Statistics Curriculum - Two-Way Analysis of Variance (ANOVA) == === Two-Way ANOVA === Example on how to attach images ...)← Older edit Revision as of 01:14, 20 February 2008 (view source)IvoDinov (Talk | contribs) Newer edit → Line 1: Line 1: ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Two-Way Analysis of Variance (ANOVA) == ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Two-Way Analysis of Variance (ANOVA) == - === Two-Way ANOVA === + In the [[AP_Statistics_Curriculum_2007_ANOVA_2Way |previous section]], we discussed statistical inference in comparing ''k'' independent samples separated by a single (grouping) factor. Now we will discuss variance decomposition of data into (independent/orthogonal) components when we have two (grouping) factors. Hence, this procedure called '''Two-Way Analysis of Variance'''. - Example on how to attach images to Wiki documents in included below (this needs to be replaced by an appropriate figure for this section)! + -
[[Image:AP_Statistics_Curriculum_2007_IntroVar_Dinov_061407_Fig1.png|500px]]
+ - ===Approach=== + ===Motivational Example=== - Models & strategies for solving the problem, data understanding & inference. + Suppose 5 varieties of peas are currently being tested by a large agribusiness cooperative to determine which is best suited for production. A field was divided into 20 plots, with each variety of peas planted in four plots. The yields (in bushels of peas) produced from each plot are shown in two identical forms in the tables below. - * TBD +
+ {| class="wikitable" style="text-align:center; width:30%" border="1" + |- + | colspan=5| Variety of Pea + |- + | A || B || C || D || E + |- + | 26.2 || 29.2 || 29.1 || 21.3 || 20.1 + |- + | 24.3 || 28.1 || 30.8 || 22.4 || 19.3 + |- + | 21.8 || 27.3 || 33.9 || 24.3 || 19.9 + |- + | 28.1 || 31.2 || 32.8 || 21.8 || 22.1 + |} +
+ {| class="wikitable" style="text-align:center; width:30%" border="1" + |- + | A || 26.2,24.3,21.8,28.1 + |- + | B || 29.2,28.1,27.3,31.2 + |- + | C || 29.1,30.8,33.9,32.8 + |- + | D || 21.3,22.4,24.3,21.8 + |- + | E || 20.1,19.3,19.9,22.1 + |} +
- ===Model Validation=== + Using the [http://socr.ucla.edu/htmls/SOCR_Charts.html SOCR Charts] (see [[SOCR_EduMaterials_Activities_BoxAndWhiskerChart | SOCR Box-and-Whisker Plot Activity]] and [[SOCR_EduMaterials_Activities_DotChart | Dot Plot Activity]]), we can generate plots that enable us to compare visually the yields of the 5 different types peas. - Checking/affirming underlying assumptions. + - * TBD +
[[Image:SOCR_EBook_Dinov_ANOVA1_021708_Fig1.jpg|500px]]
+ + ===Two-Way ANOVA Calculations=== + + Let's make the following notation: + : $y_{i,j}$ = the measurement from ''group i'', ''observation-index j''. + : k = number of groups + : $n_i$ = number of observations in group ''i'' + : n = total number of observations, $n= n_1 + n_2 + \cdots + n_k$ + : The group mean for group i is: $\bar{y}_{i,.} = {\sum_{j=1}^{n_i}{y_{i,j}} \over n_i}$ + : The grand mean is: $\bar{y}=\bar{y}_{.,.} = {\sum_{i=1}^k {\sum_{j=1}^{n_i}{y_{i,j}}} \over n}$ + + : Two-way Model: $y_{i,j,k} = \mu +\tau_i +\beta_j +\gamma_{i,j} + \epsilon_{i,j,k}$, for all $1\leq i\leq a$, $1\leq j\leq b$ and $1\leq k\leq r$. Here $\mu$ is the overall mean response, $\tau_i$ is the effect due to the $i^{th}$ level of factor A, $\beta_j$ is the effect due to the $j^{th}$ level of factor B and $\gamma_{i,j}$ is the effect due to any interaction between the $i^{th}$ level of factor A and the $j^{th}$ level of factor B. + + When an $a \times b$ factorial experiment is conducted with an equal number of observation per treatment combination, and where AB represents the interaction between A and B, the total (corrected) sum of squares is partitioned as: + +
$SS(Total) = SS(A) + SS(B) + SS(AB) + SSE$
+ + ===Hypotheses=== + There are three sets of hypotheses with the two-way ANOVA. + + ====The null hypotheses for each of the sets==== + * The population means of the first factor are equal. This is like the one-way ANOVA for the row factor. + * The population means of the second factor are equal. This is like the one-way ANOVA for the column factor. + * There is no interaction between the two factors. This is similar to performing a test for independence with contingency tables. + + ====Factors==== + The two independent variables in a two-way ANOVA are called factors (denoted by A and B). The idea is that there are two variables, factors, which affect the dependent variable (Y). Each factor will have two or more levels within it, and the degrees of freedom for each factor is one less than the number of levels. + + ====Treatment Groups==== + Treatement Groups are formed by making all possible combinations of the two factors. For example, if the first factor has 5 levels and the second factor has 6 levels, then there will be $5\times6=30$ different treatment groups. + + ====Main Effect==== + The main effect involves the independent variables one at a time. The interaction is ignored for this part. Just the rows or just the columns are used, not mixed. This is the part which is similar to the one-way analysis of variance. Each of the variances calculated to analyze the main effects are like the between variances + + ====Interaction Effect==== + The interaction effect is the effect that one factor has on the other factor. The degrees of freedom here is the product of the two degrees of freedom for each factor. + + ====Within Variation==== + The Within variation is the sum of squares within each treatment group. You have one less than the sample size (remember all treatment groups must have the same sample size for a two-way ANOVA) for each treatment group. The total number of treatment groups is the product of the number of levels for each factor. The within variance is the within variation divided by its degrees of freedom. The within group is also called the error. + + ====F-Tests==== + There is an F-test for each of the hypotheses, and the F-test is the mean square for each main effect and the interaction effect divided by the within variance. The numerator degrees of freedom come from each effect, and the denominator degrees of freedom is the degrees of freedom for the within variance in each case. + + ====Two-Way ANOVA Table==== + It is assumed that main effect A has a levels (and A = a-1 df), main effect B has b levels (and B = b-1 df), n is the sample size of each treatment, and $N = a\times b\times n$ is the total sample size. Notice the overall degrees of freedom is once again one less than the total sample size. + + Source SS df MS F + Main Effect A given A, + a-1 SS / df MS(A) / MS(W) + Main Effect B given B, + b-1 SS / df MS(B) / MS(W) + Interaction Effect given A*B, + (a-1)(b-1) SS / df MS(A*B) / MS(W) + Within given N - ab, + ab(n-1) SS / df + Total sum of others N - 1, abn - 1 +
+ {| class="wikitable" style="text-align:center; width:50%" border="1" + |- + | Variance Source || Degrees of Freedom (df) || Sum of Squares (SS) || Mean Sum of Squares (MS) || F-Statistics || [http://socr.ucla.edu/htmls/SOCR_Distributions.html P-value] + |- + | Main Effect A || df(A)=a-1 || $SS(A)=r\times b\times\sum_{i=1}^{a}{(\bar{y}_{i,.,.}-\bar{y})^2}$ || ${SS(A)\over df(A)}$ || $F_o = {MS(A)\over MSE}$ || $P(F_{(df(A), df(E))} > F_o)$ + |- + | Main Effect B || df(B)=b-1 || $SS(B)=r\times a\times\sum_{j=1}^{b}{(\bar{y}_{., j,.}-\bar{y})^2}$ || ${SS(B)\over df(B)}$ || $F_o = {MS(B)\over MSE}$ || $P(F_{(df(B), df(E))} > F_o)$ + | A vs.B Interaction || df(AB)=(a-1)(b-1) || $SS(AB)=r\times \sum_{i=1}^{a}{\sum_{j=1}^{b}{((\bar{y}_{i, j,.}-\bar{y}_{i, .,.})+(\bar{y}_{., j,.}-\bar{y}))^2}}$ || ${SS(AB)\over df(AB)}}$ || $F_o = {MS(AB)\over MSE}$ || $P(F_{(df(AB), df(E))} > F_o)$ + |- + | Error || $N-a\times b$ || $\sum_{k=1}^r{\sum_{i=1}^{a}{\sum_{j=1}^{b}{(\bar{y}_{i, j,k}-\bar{y}_{i, j,.})}}}$ || ${SSE\over df(Error)}$ || || + |- + | Total || N-1 || $\sum_{k=1}^r{\sum_{i=1}^{a}{\sum_{j=1}^{b}{(\bar{y}_{i, j,k}-\bar{y}_{., .,.})}}}$ || || || [[SOCR_EduMaterials_AnalysisActivities_ANOVA_2 | ANOVA Activity]] + |} +
+ + To compute the difference between the means, we will compare each group mean to the grand mean. + + + ===SOCR ANOVA Calculations=== + [http://socr.ucla.edu/htmls/SOCR_Analyses.html SOCR Analyses] provide the tools to compute the [[SOCR_EduMaterials_AnalysisActivities_ANOVA_1 |1-way ANOVA]]. For example, the ANOVA for the peas data above may be easily computed - see the image below. Note that SOCR ANOVA requires the data to be entered in this format: +
+ {| class="wikitable" style="text-align:center; width:30%" border="1" + |}
+ +
[[Image:SOCR_EBook_Dinov_ANOVA1_021708_Fig2.jpg|500px]]
- ===Computational Resources: Internet-based SOCR Tools=== - * TBD ===Examples=== ===Examples=== - Computer simulations and real observed data. - * TBD + ====TBD==== - + - ===Hands-on activities=== + - Step-by-step practice problems. + - * TBD + + ===Two-Way ANOVA Conditions=== + The Two-way ANOVA is valid if: + * The populations from which the samples were obtained must be normally or approximately normally distributed. + * The samples must be independent. + * The variances of the populations must be equal. + * The groups must have the same sample size.
+ ===References=== ===References=== - * TBD
## General Advance-Placement (AP) Statistics Curriculum - Two-Way Analysis of Variance (ANOVA)
In the previous section, we discussed statistical inference in comparing k independent samples separated by a single (grouping) factor. Now we will discuss variance decomposition of data into (independent/orthogonal) components when we have two (grouping) factors. Hence, this procedure called Two-Way Analysis of Variance.
### Motivational Example
Suppose 5 varieties of peas are currently being tested by a large agribusiness cooperative to determine which is best suited for production. A field was divided into 20 plots, with each variety of peas planted in four plots. The yields (in bushels of peas) produced from each plot are shown in two identical forms in the tables below.
Variety of Pea A B C D E 26.2 29.2 29.1 21.3 20.1 24.3 28.1 30.8 22.4 19.3 21.8 27.3 33.9 24.3 19.9 28.1 31.2 32.8 21.8 22.1
A 26.2,24.3,21.8,28.1 B 29.2,28.1,27.3,31.2 C 29.1,30.8,33.9,32.8 D 21.3,22.4,24.3,21.8 E 20.1,19.3,19.9,22.1
Using the SOCR Charts (see SOCR Box-and-Whisker Plot Activity and Dot Plot Activity), we can generate plots that enable us to compare visually the yields of the 5 different types peas.
### Two-Way ANOVA Calculations
Let's make the following notation:
yi,j = the measurement from group i, observation-index j.
k = number of groups
ni = number of observations in group i
n = total number of observations, $n= n_1 + n_2 + \cdots + n_k$
The group mean for group i is: $\bar{y}_{i,.} = {\sum_{j=1}^{n_i}{y_{i,j}} \over n_i}$
The grand mean is: $\bar{y}=\bar{y}_{.,.} = {\sum_{i=1}^k {\sum_{j=1}^{n_i}{y_{i,j}}} \over n}$
Two-way Model: yi,j,k = μ + τi + βj + γi,j + εi,j,k, for all $1\leq i\leq a$, $1\leq j\leq b$ and $1\leq k\leq r$. Here μ is the overall mean response, τi is the effect due to the ith level of factor A, βj is the effect due to the jth level of factor B and γi,j is the effect due to any interaction between the ith level of factor A and the jth level of factor B.
When an $a \times b$ factorial experiment is conducted with an equal number of observation per treatment combination, and where AB represents the interaction between A and B, the total (corrected) sum of squares is partitioned as:
SS(Total) = SS(A) + SS(B) + SS(AB) + SSE
### Hypotheses
There are three sets of hypotheses with the two-way ANOVA.
#### The null hypotheses for each of the sets
• The population means of the first factor are equal. This is like the one-way ANOVA for the row factor.
• The population means of the second factor are equal. This is like the one-way ANOVA for the column factor.
• There is no interaction between the two factors. This is similar to performing a test for independence with contingency tables.
#### Factors
The two independent variables in a two-way ANOVA are called factors (denoted by A and B). The idea is that there are two variables, factors, which affect the dependent variable (Y). Each factor will have two or more levels within it, and the degrees of freedom for each factor is one less than the number of levels.
#### Treatment Groups
Treatement Groups are formed by making all possible combinations of the two factors. For example, if the first factor has 5 levels and the second factor has 6 levels, then there will be $5\times6=30$ different treatment groups.
#### Main Effect
The main effect involves the independent variables one at a time. The interaction is ignored for this part. Just the rows or just the columns are used, not mixed. This is the part which is similar to the one-way analysis of variance. Each of the variances calculated to analyze the main effects are like the between variances
#### Interaction Effect
The interaction effect is the effect that one factor has on the other factor. The degrees of freedom here is the product of the two degrees of freedom for each factor.
#### Within Variation
The Within variation is the sum of squares within each treatment group. You have one less than the sample size (remember all treatment groups must have the same sample size for a two-way ANOVA) for each treatment group. The total number of treatment groups is the product of the number of levels for each factor. The within variance is the within variation divided by its degrees of freedom. The within group is also called the error.
#### F-Tests
There is an F-test for each of the hypotheses, and the F-test is the mean square for each main effect and the interaction effect divided by the within variance. The numerator degrees of freedom come from each effect, and the denominator degrees of freedom is the degrees of freedom for the within variance in each case.
#### Two-Way ANOVA Table
It is assumed that main effect A has a levels (and A = a-1 df), main effect B has b levels (and B = b-1 df), n is the sample size of each treatment, and $N = a\times b\times n$ is the total sample size. Notice the overall degrees of freedom is once again one less than the total sample size.
Source SS df MS F Main Effect A given A, a-1 SS / df MS(A) / MS(W) Main Effect B given B, b-1 SS / df MS(B) / MS(W) Interaction Effect given A*B, (a-1)(b-1) SS / df MS(A*B) / MS(W) Within given N - ab, ab(n-1) SS / df Total sum of others N - 1, abn - 1
Variance Source Degrees of Freedom (df) Sum of Squares (SS) Mean Sum of Squares (MS) F-Statistics P-value Main Effect A df(A)=a-1 $SS(A)=r\times b\times\sum_{i=1}^{a}{(\bar{y}_{i,.,.}-\bar{y})^2}$ ${SS(A)\over df(A)}$ $F_o = {MS(A)\over MSE}$ P(F(df(A),df(E)) > Fo) Main Effect B df(B)=b-1 $SS(B)=r\times a\times\sum_{j=1}^{b}{(\bar{y}_{., j,.}-\bar{y})^2}$ ${SS(B)\over df(B)}$ $F_o = {MS(B)\over MSE}$ P(F(df(B),df(E)) > Fo) A vs.B Interaction df(AB)=(a-1)(b-1) $SS(AB)=r\times \sum_{i=1}^{a}{\sum_{j=1}^{b}{((\bar{y}_{i, j,.}-\bar{y}_{i, .,.})+(\bar{y}_{., j,.}-\bar{y}))^2}}$ Failed to parse (syntax error): {SS(AB)\over df(AB)}} $F_o = {MS(AB)\over MSE}$ P(F(df(AB),df(E)) > Fo) Error $N-a\times b$ $\sum_{k=1}^r{\sum_{i=1}^{a}{\sum_{j=1}^{b}{(\bar{y}_{i, j,k}-\bar{y}_{i, j,.})}}}$ ${SSE\over df(Error)}$ Total N-1 $\sum_{k=1}^r{\sum_{i=1}^{a}{\sum_{j=1}^{b}{(\bar{y}_{i, j,k}-\bar{y}_{., .,.})}}}$ ANOVA Activity
To compute the difference between the means, we will compare each group mean to the grand mean.
### SOCR ANOVA Calculations
SOCR Analyses provide the tools to compute the 1-way ANOVA. For example, the ANOVA for the peas data above may be easily computed - see the image below. Note that SOCR ANOVA requires the data to be entered in this format:
### Two-Way ANOVA Conditions
The Two-way ANOVA is valid if:
• The populations from which the samples were obtained must be normally or approximately normally distributed.
• The samples must be independent.
• The variances of the populations must be equal.
• The groups must have the same sample size. | {"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": 21, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8279842734336853, "perplexity": 1654.4631840735206}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221217354.65/warc/CC-MAIN-20180820215248-20180820235248-00659.warc.gz"} |
https://scholars.ttu.edu/en/publications/why-water-makes-2-aminopurine-fluorescent | # Why water makes 2-aminopurine fluorescent?
Mario Barbatti, Hans Lischka
Research output: Contribution to journalArticlepeer-review
22 Scopus citations
## Abstract
2-Aminopurine (2AP) is often chosen as a fluorescent replacement for purine bases and used as a probe in nucleic acid research. The luminescence of this molecule is strongly dependent on the environment. Through computational simulations of isolated 2AP and a series of 2AP-water clusters, we show that the experimentally-observed dependence of the excited-state lifetime of 2AP on the number and location of water molecules is controlled by a barrier for internal conversion between the S1 minimum and a conical intersection. Other possible competing pathways (proton transfer, intersystem crossing, and internal conversion at other intersections) were also investigated but discarded. The tuning of the luminescence of 2AP by water is related to the order of the nπ∗ and ππ∗ states. When a water molecule interacts with the amino group, the pathway from the S1 minimum to the conical intersection requires a nonadiabatic change, thus increasing the energy barrier for internal conversion. As a consequence, a single water molecule hydrogen-bonded to the amino group is sufficient to make 2AP fluorescent.
Original language English 15452-15459 8 Physical Chemistry Chemical Physics 17 23 https://doi.org/10.1039/c5cp01151e Published - Jun 21 2015
## Fingerprint
Dive into the research topics of 'Why water makes 2-aminopurine fluorescent?'. 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": 0.8166835308074951, "perplexity": 3602.527309392693}, "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/1627046152156.49/warc/CC-MAIN-20210726215020-20210727005020-00457.warc.gz"} |
https://www.physicsforums.com/threads/limit-x-0-cos-x-1-sin-x.264640/ | # Homework Help: Limit x -> 0 cos(x) - 1 / sin(x)
1. Oct 15, 2008
### smith007
1. The problem statement, all variables and given/known data
x in the following is actually Theta
Limit as x -> 0 (cos(x) - 1) / sin(x)
2. Relevant equations
none
3. The attempt at a solution
Multiply both sides by cos x +1
[ (cos x - 1) / sin x ] * [ (cos x + 1 )/ (cos x + 1)]
= [ cos2 x -1 ] / sin(x) (cos(x) +1)
= -sin2 x / sin x (cos x + 1)
- sin x / sin x cancels to -1 leaving sin x / (cos x + 1)
Sub in x -> 0
= (-1) (sin(0) / cos(0) + 1)
= (-1) (0/2) = 0
The only step I am not sure about the is cancelation of sin x /sin x. Is this allowed? How does the solution look?
2. Oct 15, 2008
### Dick
The solution looks fine. Sure you can cancel the sin(x)/sin(x).
3. Oct 15, 2008
### smith007
Thank you for the fast response. Your help is much appreciated. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8954674601554871, "perplexity": 2938.660392937927}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039743913.6/warc/CC-MAIN-20181117230600-20181118012600-00183.warc.gz"} |
http://mathhelpforum.com/calculus/167697-finding-extreme-value-fourth-degree-function-given-interval.html | Thread: Finding extreme value of a fourth degree function in a given interval.
1. Finding extreme value of a fourth degree function in a given interval.
Function f(x)=x^4+8x^2-48x+19 has one extreme value for a x-value in 1<x<2 interval. Decide this value with 2 decimals and if that is max/min.
I would be grategul if any body could help me.
2. The point where the derivative vanishes are the roots of the equation $x^{3} + 4 x -12 =0$ and the only real root can be easily found with the Newton Raphson iterations ... for $1< x< 2$ the second derivative is $>0$ so that the point is a minimum...
Kind regards
$\chi$ $\sigma$
3. is x=2 an possible answer because $f'(2) \neq 0$
4. But $x=2$ is not a root of $x^{3} +4 x -12 =0$ ...
Kind regards
$\chi$ $\sigma$
5. So what will be the answer?
6. The Newton Raphson iterations are useful for the search of a root of an equation like $f(x)=0$. The procedure obtain an more and more precise approximation...
$\displaystyle x_{n+1} = x_{n} - \frac{f(x_{n})}{f^{'} (x_{n})}$ (1)
... starting from a given value $x_{0}$. In Your case is...
$f(x)= x^{3} + 4\ x -12 \implies f^{'}(x)= 3\ x^{2} + 4$
... so that the iterations produce...
$x_{0}= 1$
$x_{1}= 2$
$x_{2}= 1.75$
$x_{3}= 1.72274881517...$
$x_{4}= 1.72244823567...$
$x_{5}= 1.72244819948...$
... and for $n>5$ the result pratically doesn't change...
Kind regards
$\chi$ $\sigma$
7. No, x= 2 is not a possible answer, but not because of the fact that it does not make the derivative 0. And endpoint of a closed interval can be a max or min without the derivative being 0 there. However, "2" is NOT in the interval you gave, (1, 2).
8. From what I remember being taught the extreme values are always found where the derivative of the function equals zero, since it means the slope there is zero. I do not understand that long and complicate post. Why is $x_{0}= 1$ a given value?
9. Originally Posted by BugzLooney
... why is $x_{0}= 1$ a given value?...
In general the convergence of the Newton Raphson iteration is not guaranted for all the possible value of $x_{0}$. For example the [trivial] second order equation $x^{2}-1=0$ has roots $\pm 1$ but is $f^{'} (0)=0$ so that starting with $x_{0}$ is not the best possible way ... In our case however is $f^{'} (x)>0$ for all x and any value of $x_{0}$ is valid...
Originally Posted by BugzLooney
... I do not understand that long and complicate post...
May be that someone strongly aspires to became 'Best Mathematician' and thinks that the best way to meet the goal is to exaltate the 'theoretical aspects' of the problems... that isn't the only post where that happens ...
Kind regards
$\chi$ $\sigma$
10. Originally Posted by chisigma
The point where the derivative vanishes are the roots of the equation $x^{3} + 4 x -12 =0$ and the only real root can be easily found with the Newton Raphson iterations ... for $1< x< 2$ the second derivative is $>0$ so that the point is a minimum...
Kind regards
$\chi$ $\sigma$
Since f'(x) is -ve at 1 and +ve at 2 the bisection algorithm can be used on the derivative to locate a root in [1,2].
I would suggest that being given an interval like this indicates that may-be the student is expected to use the bisection algorithm for this problem.
CB
11. Originally Posted by CaptainBlack
Since f'(x) is -ve at 1 and +ve at 2 the bisection algorithm can be used on the derivative to locate a root in [1,2].
I would suggest that being given an interval like this indicates that may-be the student is expected to use the bisection algorithm for this problem.
CB
It is very interesting to compare the 'efficiency' of the bisection and Newton's algorithms for the search of the root between 1 and 2 of the equation $\displaystyle x^{3} + 4\ x -12=0$. The iteration for the Newton Raphson algorithm are reported in my previous post. The first iterations of the bisection algorithm are...
$\overrightarrow{x}_{0}= [1,2]$
$\overrightarrow{x}_{1}= [1.5,2]$
$\overrightarrow{x}_{2}= [1.5,1.75]$
$\overrightarrow{x}_{3}= [1.625,1.75]$
$\overrightarrow{x}_{4}= [1.6875,1.75]$
$\overrightarrow{x}_{5}= [1.71875,1.75]$
$\overrightarrow{x}_{6}= [1.71875,1.734375]$
$\overrightarrow{x}_{7}= [1.71875,1.7265625]$
$\overrightarrow{x}_{8}= [1.71875,1.72265625]$
$\overrightarrow{x}_{9}= [1.720703125,1.72265625]$
Using Newton's algorithm 3 iterations allow three digits precision and 5 iteration allow a twelve digits precision. With the bisection algorithm for the same precision You need 9 and 38 iterations respectively... it seems to be a very time and work spending approach...
Kind regards
$\chi$ $\sigma$
12. Originally Posted by BugzLooney
From what I remember being taught the extreme values are always found where the derivative of the function equals zero, since it means the slope there is zero. I do not understand that long and complicate post. Why is $x_{0}= 1$ a given value?
Chi Sigma is not suggesting that x= 1 may be a solution. But to use Newton's method, it helps to start from a point that you know is close to a solution. "1" is reasonably close to any number in (1, 2). Chi Sigma could have used 2 or any number in (1, 2) as a starting value. I imagine he used "1" because it seemed to give simpler intermediate values.
By the way, "extreme values are always found where the derivative of the function equals zero" is NOT true. On an interval that is not closed and bounded (does not have or does not contain its endpoints) there may be NO extreme values. On an interval that is closed and bounded, extreme values will occur at one of three different kinds of points:
1) Where the derivative is 0.
2) Where the derivative does not exist.
3) At an endpoint of the interval.
Here, the given function is a polynomial so the derivative always exists. And the interval is not closed so does not contain its endpoints so, in this particular example, you do end up with just (1).
13. Originally Posted by chisigma
It is very interesting to compare the 'efficiency' of the bisection and Newton's algorithms for the search of the root between 1 and 2 of the equation $\displaystyle x^{3} + 4\ x -12=0$. The iteration for the Newton Raphson algorithm are reported in my previous post. The first iterations of the bisection algorithm are...
$\overrightarrow{x}_{0}= [1,2]$
$\overrightarrow{x}_{1}= [1.5,2]$
$\overrightarrow{x}_{2}= [1.5,1.75]$
$\overrightarrow{x}_{3}= [1.625,1.75]$
$\overrightarrow{x}_{4}= [1.6875,1.75]$
$\overrightarrow{x}_{5}= [1.71875,1.75]$
$\overrightarrow{x}_{6}= [1.71875,1.734375]$
$\overrightarrow{x}_{7}= [1.71875,1.7265625]$
$\overrightarrow{x}_{8}= [1.71875,1.72265625]$
$\overrightarrow{x}_{9}= [1.720703125,1.72265625]$
Using Newton's algorithm 3 iterations allow three digits precision and 5 iteration allow a twelve digits precision. With the bisection algorithm for the same precision You need 9 and 38 iterations respectively... it seems to be a very time and work spending approach...
Kind regards
$\chi$ $\sigma$
The difference is that for a continuous function the bisection method is guaranteed to find a root (to what ever precision we require) given the product of the function values at each end of the initial interval is negative, also at each step we have an interval containing the root.
There is a bit more fiddling involved with NR to get the same extras (and more work as we need derivatives).
CB | {"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": 66, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8943207859992981, "perplexity": 381.3164048341971}, "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-2016-40/segments/1474738661326.74/warc/CC-MAIN-20160924173741-00133-ip-10-143-35-109.ec2.internal.warc.gz"} |
http://papers.nips.cc/paper/3753-on-learning-rotations | # NIPS Proceedingsβ
## On Learning Rotations
[PDF] [BibTeX]
### Abstract
An algorithm is presented for online learning of rotations. The proposed algorithm involves matrix exponentiated gradient updates and is motivated by the Von Neumann divergence. The additive updates are skew-symmetric matrices with trace zero which comprise the Lie algebra of the rotation group. The orthogonality and unit determinant of the matrix parameter are preserved using matrix logarithms and exponentials and the algorithm lends itself to interesting interpretations in terms of the computational topology of the compact Lie groups. The stability and the computational complexity of the algorithm are discussed. | {"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": 0.9418504238128662, "perplexity": 363.050767250527}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125946120.31/warc/CC-MAIN-20180423164921-20180423184921-00321.warc.gz"} |
https://eccc.weizmann.ac.il/author/019911/ | Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > AUTHORS > RIDWAN SYED:
All reports by Author Ridwan Syed:
TR19-006 | 17th January 2019
Anna Gal, Ridwan Syed
#### Upper Bounds on Communication in terms of Approximate Rank
Revisions: 1
We show that any Boolean function with approximate rank $r$ can be computed by bounded error quantum protocols without prior entanglement of complexity $O( \sqrt{r} \log r)$. In addition, we show that any Boolean function with approximate rank $r$ and discrepancy $\delta$ can be computed by deterministic protocols of complexity ... more >>>
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": 0.9737700819969177, "perplexity": 2476.583228078557}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945282.33/warc/CC-MAIN-20230324113500-20230324143500-00197.warc.gz"} |
https://www.physicsforums.com/threads/the-clausius-inequality.776029/ | # The Clausius Inequality
1. Oct 13, 2014
### Cassiano
Not sure if this was the right place for this but here goes.
Hello all, so I'm trying to get an intuitive grasp of the Clausius Clapeyron relation dP/dT= L/TdelV. Where L is the latent heat of the phase transition. What I've got so far is this; the relation tells you how much extra pressure must be exerted on a system in order to change its phase for every degree it is away from the temperature it changes phase at 1 atm. Is this correct?
Also the Tempertaure used in the equation would be the temperature of the phase change at standard conditions right? Thank you for the help
2. Oct 14, 2014
### Staff: Mentor
Hi Cassiano, welcome to PF!
Not exactly. It is the slope of the boundary between two phases in a PT-diagram. It is not related to any "standard" condition, be it 1 atm or something else. The temperature is simply the temperature at which you are considering the slope. Note also that it is $dP/dT$, not $\Delta P / \Delta T$, which is kind of implied by the way you formulated it. In other words, the relationship is not linear between the pressure and the temperature at which the transition takes place. | {"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": 0.9574922323226929, "perplexity": 290.77670985090606}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824471.6/warc/CC-MAIN-20171020230225-20171021010225-00107.warc.gz"} |
https://www.nickzom.org/blog/tag/horizontal-stress/ | How to Calculate and Solve for Horizontal Stress | Rock Mechanics
The image above represents horizontal stress.
To compute for horizontal stress, two essential parameters are needed and these parameters are Poisson’s ratio (v) and vertical stress (σv).
The formula for calculating the horizontal stress:
σH = v / 1 – v
Where;
σH = Horizontal Stress
v = Poisson’s Ratio
σv = Vertical Stress
Let’s solve an example;
Find the horizontal stress when the Poisson’s ratio is 24 and the vertical stress is 28.
This implies that;
v = Poisson’s Ratio = 24
σv = Vertical Stress = 28
σH = v / 1 – v
σH = 24(28) / (1 – 24)
σH = 672 / -23
σH = -29.217
Therefore, the horizontal stress is 29.217.
Calculating the Poisson’s Ratio when the Horizontal Stress and Vertical stress is Given.
v = σH / σv + σH
Where;
v = Poisson’s Ratio
σH = Horizontal Stress
σv = Vertical Stress
Let’s solve an example;
Find the Poisson’s ratio when the horizontal stress is 12 and the vertical stress is 16.
This implies that;
σH = Horizontal Stress = 12
σv = Vertical Stress = 16
v = σH / σv + σH
v = 12 / 16 + 12
v = 12 / 28
v = 12 / 28
v = 0.428
Therefore, the Poisson’s ratio is 0.428. | {"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": 0.9958930015563965, "perplexity": 2717.24569126908}, "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-2019-51/segments/1575540585566.60/warc/CC-MAIN-20191214070158-20191214094158-00281.warc.gz"} |
https://dl.asminternational.org/istfa/proceedings-abstract/ISTFA2008/30910/375/11382 | ## Abstract
Decapsulation of complex semiconductor packages for failure analysis is enhanced by laser ablation. If lasers are potentially dangerous for Integrated Circuits (IC) surface they also generate a thermal elevation of the package during the ablation process. During measurement of this temperature it was observed another and unexpected electrical phenomenon in the IC induced by laser. It is demonstrated that this new phenomenon is not thermally induced and occurs under certain ablation conditions.
This content is only available as a PDF. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8185338377952576, "perplexity": 1881.6779167429552}, "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/1610704824728.92/warc/CC-MAIN-20210127121330-20210127151330-00370.warc.gz"} |
https://www.askiitians.com/forums/Electrostatics/three-identical-charges-each-of-1-micro-coulomb-ar_202643.htm | # Three identical charges each of 1 micro coulomb are placed at the corners of an equilateral triangle of side 10 cm. Calculate the electric field intensity at the geometric centre of the triangle.[I know that the answer is zero, but as per our school exams we have to find the value of electric field intensity due to every charge. SO PLEASE SOLVE BY FINDING THE VALUE OF ELECTRIC FIELD INTENSITY DUE TO EACH CHARGE !!!!!!]
venkat
105 Points
4 years ago
Distance of each charge from the geometric centre of the equilateral triangle d=10/$\sqrt{3}$
Electric field due to each charge is given by E=1/$q/(4\pi\epsilon _{0})d^{2}$
Using the above formula find the electric field intensity due to each charge and add them vectorially to get the answer. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9745751023292542, "perplexity": 217.99692531465115}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662533972.17/warc/CC-MAIN-20220520160139-20220520190139-00668.warc.gz"} |
http://math.stackexchange.com/questions/141409/comparing-a-riemann-sum-to-its-limit | # Comparing a Riemann sum to its limit
For $n\geq 1$ and $x \gt 0$, define
$$R_n(x)=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{(x+\frac{2k-1}{2n})^2}$$
then $R_n(x)$ is a Riemann sum, which converges to the integral
$$R=\int_{0}^1 \frac{dt}{(x+t)^2}=\frac{1}{x}-\frac{1}{x+1}$$
We have
$$R-R_1(x)=\frac{1}{x(x+1)(2x+1)^2}$$
$$R-R_2(x)=\frac{16x^2+16x+9}{x(x+1)(4x+1)^2(4x+3)^2}$$
$$R-R_3(x)=\frac{3888x^4 + 7776x^3 + 6984x^2 + 3096x + 675}{x(x+1)(6x+1)^2(6x+3)^2(6x+5)^2}$$
Thus for $n\leq 3$, the numerator of $R-R_n(x)$ has positive coefficients. Can anyone show that $R>R_n(x)$ for any $n$ and $x$, by showing those coefficients are always positive, or by any other method?
By the way, this might look like homework, but it isn't.
-
Maybe investigate this: When a function is convex, is there an inequality between the integral and the midpoint Riemann sum for the integral? – GEdgar May 5 '12 at 15:49
As mentioned by GEdgar, the geometry takes care of things. – André Nicolas May 5 '12 at 16:07
GEdgar, André : thanks for answering my stupid question. – Ewan Delanoy May 5 '12 at 17:33
The comments on the OP say it all. Just to make things a little more explicit : if $f$ is a convex, integrable function $[a,b] \to \mathbb R$, then we have $f\left(\frac{a+b}{2}\right) \leq \frac{f(t)+f(a+b-t)}{2}$ for any $t \in [a,b]$. Integrating,we deduce $\int_a^b f(t)dt \geq (b-a)f\left(\frac{a+b}{2}\right)$. Thus, in the mid-point Riemann sum
$$R_n(f,x)=\frac{1}{n}\sum_{k=1}^{n}f\left(x+\frac{2k-1}{2n}\right)$$
each individual term $\frac{1}{n}\cdot f\left(x+\frac{2k-1}{2n}\right)$ is lower than $\int_{\frac{2k-2}{2n}}^{\frac{2k}{2n}} f(t)dt$, and summing we obtain
$$R_n(f,x) \leq R=\int_{0}^{1} f(t)dt$$
The original question is about $f(x)=\frac{1}{x^2}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9274635910987854, "perplexity": 247.95179688765162}, "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/1435375635143.91/warc/CC-MAIN-20150627032715-00078-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/19308/holomorphic-factorization-in-cft-2?answertab=votes | # Holomorphic Factorization in CFT$_2$
Is a CFT$_2$ always holomorphically factorizable? I had this idea because that's what we usually see is taken in string theory e.g (taking $z$ and $\bar{z}$ as independent variables). E.g. Ginsparg explains in "Applied Conformal field theory" paper page 7, that due to the emergence of Virasoro symmetry group, it is "useful" to consider them as independent, though he said that we should take $z^*=\bar{z}$ later according to our convenience. But it would be nice to understand what's going on. E.g, does holomorphic factorizability becomes an essential in some CFT$_2$'s? E.g. if we are considering some field theory in AdS$_3$ (no Chern Simons term say), should we expect the boundary CFT to be factorizable (its then not a pure gravity any more)?
-
What is your definition of the term "holomorphically factorizable"? I think the common definition has something to do with the partition function of the theory factoring into a sum over terms that can be written as products of holomorphic and anti-holomorphic pieces. – joshphysics Nov 27 '13 at 4:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.949843168258667, "perplexity": 267.67290496894736}, "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/1429246659449.65/warc/CC-MAIN-20150417045739-00020-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.intechopen.com/chapters/21845 | Open access peer-reviewed chapter
# Atmospheric Thermodynamics
By Francesco Cairo
Submitted: November 3rd 2010Reviewed: June 16th 2011Published: November 2nd 2011
DOI: 10.5772/19429
## 1. Introduction
Thermodynamics deals with the transformations of the energy in a system and between the system and its environment. Hence, it is involved in every atmospheric process, from the large scale general circulation to the local transfer of radiative, sensible and latent heat between the surface and the atmosphere and the microphysical processes producing clouds and aerosol. Thus the topic is much too broad to find an exhaustive treatment within the limits of a book chapter, whose main goal will be limited to give a broad overview of the implications of thermodynamics in the atmospheric science and introduce some if its jargon. The basic thermodynamic principles will not be reviewed here, while emphasis will be placed on some topics that will find application to the interpretation of fundamental atmospheric processes. An overview of the composition of air will be given, together with an outline of its stratification in terms of temperature and water vapour profile. The ideal gas law will be introduced, together with the concept of hydrostatic stability, temperature lapse rate, scale height, and hydrostatic equation. The concept of an air parcel and its enthalphy and free energy will be defined, together with the potential temperature concept that will be related to the static stability of the atmosphere and connected to the Brunt-Vaisala frequency.
Water phase changes play a pivotal role in the atmosphere and special attention will be placed on these transformations. The concept of vapour pressure will be introduced together with the Clausius-Clapeyron equation and moisture parameters will be defined. Adiabatic transformation for the unsaturated and saturated case will be discussed with the help of some aerological diagrams of common practice in Meteorology and the notion of neutral buoyancy and free convection will be introduced and considered referring to an exemplificative atmospheric sounding. There, the Convective Inhibition and Convective Available Potential Energy will be introduced and examined. The last subchapter is devoted to a brief overview of warm and cold clouds formation processes, with the aim to stimulate the interest of reader toward more specialized texts, as some of those listed in the conclusion and in the bibliography.
## 2. Dry air thermodynamics and stability
We know from experience that pressure, volume and temperature of any homogeneous substance are connected by an equation of state. These physical variables, for all gases over a wide range of conditions in the so called perfect gas approximation, are connected by an equation of the form:
pV=mRTE1
where pis pressure (Pa), Vis volume (m3), mis mass (kg), Tis temperature (K) and Ris the specific gas constant, whose value depends on the gas. If we express the amount of substance in terms of number of moles n=m/Mwhere Mis the gas molecular weight, we can rewrite (1) as:
pV=nR*TE2
where R*is the universal gas costant, whose value is 8.3143 J mol-1 K-1. In the kinetic theory of gases, the perfect gas is modelled as a collection of rigid spheres randomly moving and bouncing between each other, with no common interaction apart from these mutual shocks. This lack of reciprocal interaction leads to derive the internal energy of the gas, that is the sum of all the kinetic energies of the rigid spheres, as proportional to its temperature. A second consequence is that for a mixture of different gases we can define, for each component i, a partial pressure pias the pressure that it would have if it was alone, at the same temperature and occupying the same volume. Similarly we can define the partial volume Vias that occupied by the same mass at the same pressure and temperature, holding Dalton’s lawfor a mixture of gases i:
p= piE3
Where for each gas it holds:
piV=niR*TE4
We can still make use of (1) for a mixture of gases, provided we compute a specific gas constant Ras:
R?=amiRimE5
The atmosphere is composed by a mixture of gases, water substance in any of its three physical states and solid or liquid suspended particles (aerosol). The main components of dry atmospheric air are listed in Table 1.
Gas Molar fraction Mass fraction Specific gas constant(J Kg-1 K-1) Nitrogen (N2) 0.7809 0.7552 296.80 Oxygen (O2) 0.2095 0.2315 259.83 Argon (Ar) 0.0093 0.0128 208.13 Carbon dioxide (CO2) 0.0003 0.0005 188.92
### Table 1.
Main component of dry atmospheric air.
The composition of air is constant up to about 100 km, while higher up molecular diffusion dominates over turbulent mixing, and the percentage of lighter gases increases with height. For the pivotal role water substance plays in weather and climate, and for the extreme variability of its presence in the atmosphere, with abundances ranging from few percents to millionths, it is preferable to treat it separately from other air components, and consider the atmosphere as a mixture of dry gases and water. In order to use a state equation of the form (1) for moist air, we express a specific gas constant Rdby considering in (5) all gases but water, and use in the state equation a virtual temperature Tvdefined as the temperature that dry air must have in order to have the same density of moist air at the same pressure. It can be shown that
Tv=T1-ep1-MwMdE6
Where Mwand Mdare respectively the water and dry air molecular weights. Tvtakes into account the smaller density of moist air, and so is always greater than the actual temperature, although often only by few degrees.
### 2.1. Stratification
The atmosphere is under the action of a gravitational field, so at any given level the downward force per unit area is due to the weight of all the air above. Although the air is permanently in motion, we can often assume that the upward force acting on a slab of air at any level, equals the downward gravitational force. This hydrostatic balanceapproximation is valid under all but the most extreme meteorological conditions, since the vertical acceleration of air parcels is generally much smaller than the gravitational one. Consider an horizontal slab of air between zand z +δz, of unit horizontal surface. If ρis the air density at z, the downward force acting on this slab due to gravity is gρδz. Let pbe the pressure at z, and p+δpthe pressure at z+δz. We consider as negative, since we know that pressure decreases with height. The hydrostatic balance of forces along the vertical leads to:
-ep=g?ezE7
Hence, in the limit of infinitesimal thickness, the hypsometric equationholds:
-?p?z=-g?E8
pz=?zig?dzE9
As we know that p(∞)=0, (9) can be integrated if the air density profile is known.
Two useful concepts in atmospheric thermodynamic are the geopotential Φ, an exact differential defined as the work done against the gravitational field to raise 1 kg from 0 to z, where the 0 level is often taken at sea level and, to set the constant of integration, Φ(0)=0, and the geopotential height Z= Φ/g0, where g0 is a mean gravitational acceleration taken as 9,81 m/s.
We can rewrite (9) as:
Zz=1g0?zigdzE10
Values of zand Zoften differ by not more than some tens of metres.
We can make use of (1) and of the definition of virtual temperature to rewrite (10) and formally integrate it between two levels to formally obtain the geopotential thickness of a layer, as:
?Z=Rdg0?p1p2TvdppE11
The above equations can be integrated if we know the virtual temperature Tvas a function of pressure, and many limiting cases can be envisaged, as those of constant vertical temperature gradient. A very simplified case is for an isothermal atmosphere at a temperature Tv=T0, when the integration of (11) gives:
?Z=RdT0g0lnp1p2=Hylnp1p2E12
In an isothermal atmosphere the pressure decreases exponentially with an e-folding scale given by the scale height Hwhich, at an average atmospheric temperature of 255 K, corresponds roughly to 7.5 km. Of course, atmospheric temperature is by no means constant: within the lowest 10-20 km it decreases with a lapse rateof about 7 K km-1, highly variable depending on latitude, altitude and season. This region of decreasing temperature with height is termed troposphere, (from the Greek “turning/changing sphere”) and is capped by a region extending from its boundary, termed tropopause, up to 50 km, where the temperature is increasing with height due to solar UV absorption by ozone, that heats up the air. This region is particularly stable and is termed stratosphere( “layered sphere”). Higher above in the mesosphere(“middle sphere”) from 50 km to 80-90 km, the temperature falls off again. The last region of the atmosphere, named thermosphere, sees the temperature rise again with altitude to 500-2000K up to an isothermal layer several hundreds of km distant from the ground, that finally merges with the interplanetary space where molecular collisions are rare and temperature is difficult to define. Fig. 1 reports the atmospheric temperature, pressure and density profiles. Although the atmosphere is far from isothermal, still the decrease of pressure and density are close to be exponential. The atmospheric temperature profile depends on vertical mixing, heat transport and radiative processes.
### 2.2. Thermodynamic of dry air
A system is openif it can exchange matter with its surroundings, closedotherwise. In atmospheric thermodynamics, the concept of “air parcel” is often used. It is a good approximation to consider the air parcel as a closed system, since significant mass exchanges between airmasses happen predominantly in the few hundreds of metres close to the surface, the so-called planetary boundary layerwhere mixing is enhanced, and can be neglected elsewhere. An air parcel can exchange energy with its surrounding by work of expansion or contraction, or by exchanging heat. An isolatedsystem is unable to exchange energy in the form of heat or work with its surroundings, or with any other system. The first principle of thermodynamics states that the internal energy Uof a closed system, the kinetic and potential energy of its components, is a state variable, depending only on the present state of the system, and not by its past. If a system evolves without exchanging any heat with its surroundings, it is said to perform an adiabatictransformation. An air parcel can exchange heat with its surroundings through diffusion or thermal conduction or radiative heating or cooling; moreover, evaporation or condensation of water and subsequent removal of the condensate promote an exchange of latent heat. It is clear that processes which are not adiabatic ultimately lead the atmospheric behaviours. However, for timescales of motion shorter than one day, and disregarding cloud processes, it is often a good approximation to treat air motion as adiabatic.
#### 2.2.1. Potential temperature
For adiabatic processes, the first law of thermodynamics, written in two alternative forms:
cvdT + pdv=δqE13
cpdT vdp= δqE14
holds for δq=0, where cpand cvare respectively the specific heats at constant pressure and constant volume, pand vare the specific pressure and volume, and δqis the heat exchanged with the surroundings. Integrating (13) and (14) and making use of the ideal gas state equation, we get the Poisson’s equations:
Tvγ1= constantE15
Tpκ= constantE16
pvγ= constantE17
where γ=cp/cv=1.4 and κ=(γ-1)/γ=R/cp≈ 0.286, using a result of the kinetic theory for diatomic gases. We can use (16) to define a new state variable that is conserved during an adiabatic process, the potential temperature θ, which is the temperature the air parcel would attain if compressed, or expanded, adiabatically to a reference pressure p0, taken for convention as 1000 hPa.
?=Tp0p?E18
Since the time scale of heat transfers, away from the planetary boundary layer and from clouds is several days, and the timescale needed for an air parcel to adjust to environmental pressure changes is much shorter, θcan be considered conserved along the air motion for one week or more. The distribution of θin the atmosphere is determined by the pressure and temperature fields. In fig. 2 annual averages of constant potential temperature surfaces are depicted, versus pressure and latitude. These surfaces tend to be quasi-horizontal. An air parcel initially on one surface tend to stay on that surface, even if the surface itself can vary its position with time. At the ground level θattains its maximum values at the equator, decreasing toward the poles. This poleward decrease is common throughout the troposphere, while above the tropopause, situated near 100 hPa in the tropics and 3-400 hPa at medium and high latitudes, the behaviour is inverted.
An adiabatic vertical displacement of an air parcel would change its temperature and pressure in a way to preserve its potential temperature. It is interesting to derive an expression for the rate of change of temperature with altitude under adiabatic conditions: using (8) and (1) we can write (14) as:
cp dT + g dz=0E19
and obtain the dry adiabatic lapserate Γd:
If the air parcel thermally interacts with its environment, the adiabatic condition no longer holds and in (13) and (14) δq ≠ 0. In such case, dividing (14) by T and using (1) we obtain:
dln?p-y?ydln?p=-eqcpTE21
Combining the logarithm of (18) with (21) yields:
dln??=eqcpTE22
That clearly shows how the changes in potential temperature are directly related to the heat exchanged by the system.
#### 2.2.2. Entropy and potential temperature
The second law of the thermodynamics allows for the introduction of another state variable, the entropys, defined in terms of a quantity δq/Twhich is not in general an exact differential, but is so for a reversible process, that is a process proceeding through states of the system which are always in equilibrium with the environment. Under such cases we may pose ds = (δq/T)rev. For the generic process, the heat absorbed by the system is always lower that what can be absorbed in the reversible case, since a part of heat is lost to the environment. Hence, a statement of the second law of thermodynamics is:
dsoeqTE23
If we introduce (22) in (23), we note how such expression, connecting potential temperature to entropy, would contain only state variables. Hence equality must hold and we get:
dln??=dscpE24
That directly relates changes in potential temperature with changes in entropy. We stress the fact that in general an adiabatic process does not imply a conservation of entropy. A classical textbook example is the adiabatic free expansion of a gas. However, in atmospheric processes, adiabaticity not only implies the absence of heat exchange through the boundaries of the system, but also absence of heat exchanges between parts of the system itself (Landau et al., 1980), that is, no turbulent mixing, which is the principal source of irreversibility. Hence, in the atmosphere, an adiabatic process always conserves entropy.
### 2.3. Stability
The vertical gradient of potential temperature determines the stratification of the air. Let us differentiate (18) with respect to z:
?ln???z=?ln?T?z+Rcp?p0?z-?p?zE25
By computing the differential of the logarithm, and applying (1) and (8), we get:
T????z=?T?z+gcpE26
If Г= - (∂T/∂z)is the environment lapse rate, we get:
Now, consider a vertical displacement δzof an air parcel of mass mand let ρand Tbe the density and temperature of the parcel, and ρ’and T’the density and temperature of the surrounding. The restoring force acting on the parcel per unit mass will be:
fz=-?'-??'gE28
That, by using (1), can be rewritten as:
fz=-T-T'TgE29
We can replace (T-T’)with(Г dδz- Г) if we acknowledge the fact that the air parcel moves adiabatically in an environment of lapse rate Г. The second order equation of motion (29) can be solved in δzand describes buoyancy oscillations with period 2π/Nwhere Nis the Brunt-Vaisala frequency:
It is clear from (30) that if the environment lapse rate is smaller than the adiabatic one, or equivalently if the potential temperature vertical gradient is positive, Nwill be real and an air parcel will oscillate around an equilibrium: if displaced upward, the air parcel will find itself colder, hence heavier than the environment and will tend to fall back to its original place; a similar reasoning applies to downward displacements. If the environment lapse rate is greater than the adiabatic one, or equivalently if the potential temperature vertical gradient is negative, N will be imaginary so the upward moving air parcel will be lighter than the surrounding and will experience a net buoyancy force upward. The condition for atmospheric stability can be inspected by looking at the vertical gradient of the potential temperature: if θincreases with height, the atmosphere is stable and vertical motion is discouraged, if θdecreases with height, vertical motion occurs. For average tropospheric conditions, N≈ 10-2 s-1 and the period of oscillation is some tens of minutes. For the more stable stratosphere, N≈ 10-1 s-1 and the period of oscillation is some minutes. This greater stability of the stratosphere acts as a sort of damper for the weather disturbances, which are confined in the troposphere.
## 3. Moist air thermodynamics
The conditions of the terrestrial atmosphere are such that water can be present under its three forms, so in general an air parcel may contain two gas phases, dry air (d) and water vapour (v), one liquid phase (l) and one ice phase (i). This is an heterogeneous system where, in principle, each phase can be treated as an homogeneous subsystem open to exchanges with the other systems. However, the whole system should be in thermodynamical equilibrium with the environment, and thermodynamical and chemical equilibrium should hold between each subsystem, the latter condition implying that no conversion of mass should occur between phases. In the case of water in its vapour and liquid phase, the chemical equilibrium imply that the vapour phases attains a saturation vapour pressure esat which the rate of evaporation equals the rate of condensation and no net exchange of mass between phases occurs.
The concept of chemical equilibrium leads us to recall one of the thermodynamical potentials, the Gibbs function, defined in terms of the enthalpyof the system. We remind the definition of enthalpy of a system of unit mass:
h=u+pvE31
Where uis its specific internal energy, vits specific volume and pits pressure in equilibrium with the environment. We can think of has a measure of the total energy of the system. It includes both the internal energy required to create the system, and the amount of energy required to make room for it in the environment, establishing its volume and balancing its pressure against the environmental one. Note that this additional energy is not stored in the system, but rather in its environment.
The First law of thermodynamics can be set in a form where his explicited as:
eq=dh-vdpE32
And, making use of (14) we can set:
dh=cpdTE33
By combining (32), (33) and (8), and incorporating the definition of geopotential Φ we get:
eq=d(h+e)E34
Which states that an air parcel moving adiabatically in an hydrostatic atmosphere conserves the sum of its enthalpy and geopotential.
The specific Gibbs free energy is defined as:
g=h-Ts=yu+pv-TsE35
It represents the energy available for conversion into work under an isothermal-isobaric process. Hence the criterion for thermodinamical equilibrium for a system at constant pressure and temperature is that gattains a minimum.
For an heterogeneous system where multiple phases coexist, for the k-th species we define its chemical potential μkas the partial molar Gibbs function, and the equilibrium condition states that the chemical potentials of all the species should be equal. The proof is straightforward: consider a system where nvmoles of vapour (v) and nlmoles of liquid water (l) coexist at pressure eand temperature T, and let G =nvμv+nlμlbe the Gibbs function of the system. We know that for a virtual displacement from an equilibrium condition, dG > 0must hold for any arbitrary dnv(which must be equal to – dnl, whether its positive or negative) hence, its coefficient must vanish and μv= μl.
Note that if evaporation occurs, the vapour pressure echanges by deat constant temperature, and v= vvde, l= vldewhere vvand vlare the volume occupied by a single molecule in the vapour and the liquid phase. Since vv>> vlwe may pose dv- μl) = vvdeand, using the state gas equation for a single molecule, d(μv- μl) = (kT/e) de. In the equilibrium,
μv= μland e = eswhile in general:
av-al=kTlneesE36
holds. We will make use of this relationship we we will discuss the formation of clouds.
### 3.1. Saturation vapour pressure
The value of esstrongly depends on temperature and increases rapidly with it. The celebrated Clausius –Clapeyron equation describes the changes of saturated water pressure above a plane surface of liquid water. It can be derived by considering a liquid in equilibrium with its saturated vapour undergoing a Carnot cycle (Fermi, 1956). We here simply state the result as:
desdT=LvTaE37
Retrieved under the assumption that the specific volume of the vapour phase is much greater than that of the liquid phase. Lvis the latent heat, that is the heat required to convert a unit mass of substance from the liquid to the vapour phase without changing its temperature. The latent heat itself depends on temperature – at 1013 hPa and 0 °C is 2.5*106 J kg-, - hence a number of numerical approximations to (37) have been derived. The World Meteoreological Organization bases its recommendation on a paper by Goff (1957):
Log10 es = 10.79574 (1273.16/T) 5.02800 Log10(T/273.16) ++ 1.50475 104(1 10(8.2969*(T/273.161))) + 0.42873 103 (10(+4.76955*(1273.16/T)) 1) + 0.78614E38
Where Tis expressed in K and esin hPa. Other formulations are used, based on direct measurements of vapour pressures and theoretical calculation to extrapolate the formulae down to low T values (Murray, 1967; Bolton, 1980; Hyland and Wexler, 1983; Sonntag, 1994; Murphy and Koop, 2005) uncertainties at low temperatures become increasingly large and the relative deviations within these formulations are of 6% at -60 °C and of 9% at -70 .
An equation similar to (37) can be derived for the vapour pressure of water over ice esi. In such a case, Lvis the latent heat required to convert a unit mass of water substance from ice to vapour phase without changing its temperature. A number of numerical approximations holds, as the Goff-Gratch equation, considered the reference equation for the vapor pressure over ice over a region of -100 °C to 0 °C:
Log10 esi = 9.09718 (273.16/T 1) 3.56654 Log10(273.16/ T) ++ 0.876793 (1 T/ 273.16) + Log10(6.1071)E39
with Tin K and esiin hPa. Other equations have also been widely used (Murray, 1967; Hyland and Wexler, 1983; Marti and Mauersberger, 1993; Murphy and Koop, 2005).
Water evaporates more readily than ice, that is es> esieverywhere (the difference is maxima around -20 °C), so if liquid water and ice coexists below 0 °C, the ice phase will grow at the expense of the liquid water.
### 3.2. Water vapour in the atmosphere
A number of moisture parameters can be formulated to express the amount of water vapour in the atmosphere. The mixing ratio ris the ratio of the mass of the water vapour mv, to the mass of dry air md, r=mv/mdand is expressed in g/kg-1 or, for very small concentrations as those encountered in the stratosphere, in parts per million in volume (ppmv). At the surface, it typically ranges from 30-40 g/kg-1 at the tropics to less that 5 g/kg-1 at the poles; it decreases approximately exponentially with height with a scale height of 3-4 km, to attain its minimum value at the tropopause, driest at the tropics where it can get as low as a few ppmv. If we consider the ratio of mvto the total mass of air, we get the specific humidity qas q = mv/(mv+md) =r/(1+r). The relative humidity RHcompares the water vapour pressure in an air parcel with the maximum water vapour it may sustain in equilibrium at that temperature, that is RH = 100 e/es(expressed in percentages). The dew point temperature Tdis the temperature at which an air parcel with a water vapour pressure eshould be brought isobarically in order to become saturated with respect to a plane surface of water. A similar definition holds for the frost point temperature Tf, when the saturation is considered with respect to a plane surface of ice.
The wet-bulb temperatureTw is defined operationally as the temperature a thermometer would attain if its glass bulb is covered with a moist cloth. In such a case the thermometer is cooled upon evaporation until the surrounding air is saturated: the heat required to evaporate water is supplied by the surrounding air that is cooled. An evaporating droplet will be at the wet-bulb temperature. It should be noted that if the surrounding air is initially unsaturated, the process adds water to the air close to the thermometer, to become saturated, hence it increases its mixing ratio rand in general T ≥ Tw≥ Td, the equality holds when the ambient air is already initially saturated.
### 3.3. Thermodynamics of the vertical motion
The saturation mixing ratio depends exponentially on temperature. Hence, due to the decrease of ambient temperature with height, the saturation mixing ratio sharply decreases with height as well.
Therefore the water pressure of an ascending moist parcel, despite the decrease of its temperature at the dry adiabatic lapse rate, sooner or later will reach its saturation value at a level named lifting condensation level(LCL), above which further lifting may produce condensation and release of latent heat. This internal heating slows the rate of cooling of the air parcel upon further lifting.
If the condensed water stays in the parcel, and heat transfer with the environment is negligible, the process can be considered reversible – that is, the heat internally added by condensation could be subtracted by evaporation if the parcel starts descending - hence the behaviour can still be considered adiabatic and we will term it a saturated adiabatic process. If otherwise the condensate is removed, as instance by sedimentation or precipitation, the process cannot be considered strictly adiabatic. However, the amount of heat at play in the condensation process is often negligible compared to the internal energy of the air parcel and the process can still be considered well approximated by a saturated adiabat, although it should be more properly termed a pseudoadiabaticprocess.
If within an air parcel of unit mass, water vapour condenses at a saturation mixing ratio rs, the amount of latent heat released during the process will be -Lwdrs. This can be put into (34) to get:
-Lwdrs=cpdT+gdzE40
Dividing by cpdzand rearranging terms, we get the expression of the saturated adiabatic lapse rateГ s:
Whose value depends on pressure and temperature and which is always smaller than Г d, as should be expected since a saturated air parcel, since condensation releases latent heat, cools more slowly upon lifting.
#### 3.3.2. Equivalent potential temperature
If we pose δq = - Lwdrsin (22) we get:
d??=-LwdrscpT?-dLwrscpTE42
The approximate equality holds since dT/T << drs/rsand Lw/cpis approximately independent of T. So (41) can be integrated to yield:
?e=?yexpLwrscpTE43
That defines the equivalent potential temperature θe(Bolton, 1990) which is constant along a pseudoadiabatic process, since during the condensation the reduction of rsand the increase of θact to compensate each other.
### 3.4. Stability for saturated air
We have seen for the case of dry air that if the environment lapse rate is smaller than the adiabatic one, the atmosphere is stable: a restoring force exist for infinitesimal displacement of an air parcel. The presence of moisture and the possibility of latent heat release upon condensation complicates the description of stability.
If the air is saturated, it will cool upon lifting at the smaller saturated lapse rate Г sso in an environment of lapse rate Г, for the saturated air parcel the cases Г < Г s, Г = Г s, Г > Г sdiscriminates the absolutely stable, neutral and unstable conditions respectively. An interesting case occurs when the environmental lapse rate lies between the dry adiabatic and the saturated adiabatic, that is Г s< Г < Г d. In such a case, a moist unsaturated air parcel can be lifted high enough to become saturated, since the decrease in its temperature due to adiabatic cooling is offset by the faster decrease in water vapour saturation pressure, and starts condensation at the LCL. Upon further lifting, the air parcel eventually get warmer than its environment at a level termed Level of Free Convection (LFC)above which it will develop a positive buoyancy fuelled by the continuous release of latent heat due to condensation, as long as there is vapour to condense. This situation of conditional instabilityis most common in the atmosphere, especially in the Tropics, where a forced finite uplifting of moist air may eventually lead to spontaneous convection. Let us refer to figure 4 and follow such process more closely. In the figure, which is one of the meteograms discussed later in the chapter, pressure decreases vertically, while lines of constant temperature are tilted 45 rightward, temperature decreasing going up and to the left.
The thick solid line represent the environment temperature profile. A moist air parcel initially at rest at point A is lifted and cools at the adiabatic lapse rate Г dalong the thin solid line until it eventually get saturated at the Lifting Condensation Level at point D. During this lifting, it gets colder than the environment. Upon further lifting, it cools at a slower rate at the pseudoadiabatic lapse rate Г salong the thin dashed line until it reaches the Level of Free Convection at point C, where it attains the temperature of the environment. If it gets beyond that point, it will be warmer, hence lighter than the environment and will experience a positive buoyancy force. This buoyancy will sustain the ascent of the air parcel until all vapour condenses or until its temperature crosses again the profile of environmental temperature at the Level of Neutral Buoyancy (LNB).Actually, since the air parcel gets there with a positive vertical velocity, this level may be surpassed and the air parcel may overshoot into a region where it experiences negative buoyancy, to eventually get mixed there or splash back to the LNB. In practice, entrainment of environmental air into the ascending air parcel often occurs, mitigates the buoyant forces, and the parcel generally reaches below the LNB.
If we neglect such entrainment effects and consider the motion as adiabatic, the buoyancy force is conservative and we can define a potential. Let ρ and ρ’ be respectively the environment and air parcel density. From Archimede’s principle, the buoyancy force on a unit mass parcel can be expressed as in (29), and the increment of potential energy for a displacement δzwill then be, by using (1) and (8):
dP=fb=T'-TTgez=RT'-TdlogpE44
Which can be integrated from a reference level p0to give:
dPp=-R?p0pT'-Tdlogp=-RA(p)E45
Referring to fig. 4, A(p)represent the shaded area between the environment and the air parcel temperature profiles. An air parcel initially in A is bound inside a “potential energy well” whose depth is proportional to the dotted area, and that is termed Convective Inhibition (CIN). If forcedly raised to the level of free convection, it can ascent freely, with an available potential energy given by the shaded area, termed CAPE (Convective Available Potential Energy).
In absence of entrainment and frictional effects, all this potential energy will be converted into kinetic energy, which will be maximum at the level of neutral buoyancy. CIN and CAPE are measured in J/Kg and are indices of the atmospheric instability. The CAPE is the maximum energy which can be released during the ascent of a parcel from its free buoyant level to the top of the cloud. It measures the intensity of deep convection, the greater the CAPE, the more vigorous the convection. Thunderstorms require large CAPE of more than 1000 Jkg-1.
CIN measures the amount of energy required to overcome the negatively buoyant energy the environment exerts on the air parcel, the smaller, the more unstable the atmosphere, and the easier to develop convection. So, in general, convection develops when CIN is small and CAPE is large. We want to stress that some CIN is needed to build-up enough CAPE to eventually fuel the convection, and some mechanical forcing is needed to overcome CIN. This can be provided by cold front approaching, flow over obstacles, sea breeze.
CAPE is weaker for maritime than for continental tropical convection, but the onset of convection is easier in the maritime case due to smaller CIN.
We have neglected entrainment of environment air, and detrainment from the air parcel, which generally tend to slow down convection. However, the parcels reaching the highest altitude are generally coming from the region below the cloud without being too much diluted.
Convectively generated clouds are not the only type of clouds. Low level stratiform clouds and high altitude cirrus are a large part of cloud cover and play an important role in the Earth radiative budget. However convection is responsible of the strongest precipitations, especially in the Tropics, and hence of most of atmospheric heating by latent heat transfer.
So far we have discussed the stability behaviour for a single air parcel. There may be the case that although the air parcel is stable within its layer, the layer as a whole may be destabilized if lifted. Such case happen when a strong vertical stratification of water vapour is present, so that the lower levels of the layer are much moister than the upper ones. If the layer is lifted, its lower levels will reach saturation before the uppermost ones, and start cooling at the slower pseudoadiabat rate, while the upper layers will still cool at the faster adiabatic rate. Hence, the top part of the layer cools much more rapidly of the bottom part and the lapse rate of the layer becomes unstable. This potential(or convective) instabilityis frequently encountered in the lower leves in the Tropics, where there is a strong water vapour vertical gradient.
It can be shown that condition for a layer to be potentially unstable is that its equivalent potential temperature θedecreases within the layer.
### 3.5. Tephigrams
To represent the vertical structure of the atmosphere and interpret its state, a number of diagrams is commonly used. The most common are emagrams, Stüve diagrams, skew T- log pdiagrams, and tephigrams.
An emagramis basically a T-zplot where the vertical axis is log pinstead of height z. But since log pis linearly related to height in a dry, isothermal atmosphere, the vertical coordinate is basically the geometric height.
In the Stüve diagramthe vertical coordinate is p(Rd/cp)and the horizontal coordinate is T: with this axes choice, the dry adiabats are straight lines.
A skew T- log p diagram, like the emagram, has log pas vertical coordinate, but the isotherms are slanted. Tephigramslook very similar to skew T diagrams if rotated by 45 , have Tas horizontal and log θas vertical coordinates so that isotherms are vertical and the isentropes horizontal (hence tephi, a contraction of Tand Φ, where Φ = cplog θstands for the entropy). Often, tephigrams are rotated by 45 so that the vertical axis corresponds to the vertical in the atmosphere.
A tephigram is shown in figure 5: straight lines are isotherms (slope up and to the right) and isentropes (up and to the left), isobars (lines of constant p) are quasi-horizontal lines, the dashed lines sloping up and to the right are constant mixing ratio in g/kg, while
the curved solid bold lines sloping up and to the left are saturated adiabats.
Two lines are commonly plotted on a tephigram – the temperature and dew point, so the state of an air parcel at a given pressure is defined by its temperature Tand Td, that is its water vapour content. We note that the knowledge of these parameters allows to retrieve all the other humidity parameters: from the dew point and pressure we get the humidity mixing ratio w; from the temperature and pressure we get the saturated mixing ratio ws, and relative humidity may be derived from 100*w/ws, when w and ws are measured at the same pressure.
When the air parcel is lifted, its temperature Tfollows the dry adiabatic lapse rate and its dew point Tdits constant vapour mixing ratio line - since the mixing ratio is conserved in unsaturated air - until the two meet a t the LCL where condensation may start to happen. Further lifting follows the Saturated Adiabatic Lapse Rate. In Figure 5 we see an air parcel initially at ground level, with a temperature of 30 and a Dew Point temperature of 0 (which as we can see by inspecting the diagram, corresponds to a mixing ratio of approx. 4 g/kg at ground level) is lifted adiabatically to 700 mB which is its LCL where the air parcel temperature following the dry adiabats meets the air parcel dew point temperature following the line of constant mixing ratio. Above 700 mB, the air parcel temperature follows the pseudoadiabat. Figure 5 clearly depicts the Normand’s rule: The dry adiabatic through the temperature, the mixing ratio line through the dew point, and the saturated adiabatic through the wet bulb temperature, meet at the LCL. In fact, the saturated adiabat that crosses the LCL is the same that intersect the surface isobar exactly at the wet bulb temperature, that is the temperature a wetted thermometer placed at the surface would attain by evaporating - at constant pressure - its water inside its environment until it gets saturated.
Figure 6 reports two different temperature sounding: the black dotted line is the dew point profile and is common to the two soundings, while the black solid line is an early morning sounding, where we can see the effect of the nocturnal radiative cooling as a temperature inversion in the lowermost layer of the atmosphere, between 1000 and 960 hPa. The state of the atmosphere is such that an air parcel at the surface has to be forcedly lifted to 940 hP to attain saturation at the LCL, and forcedly lifted to 600 hPa before gaining enough latent heat of condensation to became warmer than the environment and positively buoyant at the LFB. The temperature of such air parcel is shown as a grey solid line in the graph.
The blue solid line is an afternoon sounding, when the surface has been radiatively heated by the sun. An air parcel lifted from the ground will follow the red solid line, and find itself immediately warmer than its environment and gaining positive buoyancy, further increased by the release of latent heat starting at the LCL at 850 hPa. Notice however that a second inversion layer is present in the temperature sounding between 800 hPa and 750 hPa, such that the air parcel becomes colder than the environment, hence negatively buoyant between 800 hPa and 700 hPa. If forcedly uplifted beyond this stable layer, it again attains a positive buoyancy up to above 300 hPa.
As the tephigram is a graph of temperature against entropy, an area computed from these variables has dimensions of energy. The area between the air parcel path is then linked to the CIN and the CAPE. Referring to the early morning sounding, the area between the black and the grey line between the surface and 600 hPa is the CIN, the area between 600 hPa and 400 hPa is the CAPE.
## 4. The generation of clouds
Clouds play a pivotal role in the Earth system, since they are the main actors of the atmospheric branch of the water cycle, promote vertical redistribution of energy by latent heat capture and release and strongly influence the atmospheric radiative budget.
Clouds may form when the air becomes supersaturated, as it can happen upon lifting as explained above, but also by other processes, as isobaric radiative cooling like in the formation of radiative fogs, or by mixing of warm moist air with cold dry air, like in the generation of airplane contrails and steam fogsabove lakes.
Cumulusor cumulonimbusare classical examples of convective clouds, often precipitating, formed by reaching the saturation condition with the mechanism outlined hereabove.
Other types of clouds are alto-cumuluswhich contain liquid droplets between 2000 and 6000m in mid-latitudes and cluster into compact herds. They are often, during summer, precursors of late afternoon and evening developments of deep convection.
Cirrusare high altitude clouds composed of ice, rarely opaque. They form above 6000m in mid-latitudes and often promise a warm front approaching. Such clouds are common in the Tropics, formed as remains of anvils or by in situ condensation of rising air, up to the tropopause. Nimbo-stratusare very opaque low clouds of undefined base, associated with persistent precipitations and snow. Strato-cumulusare composed by water droplets, opaque or very opaque, with a cloud base below 2000m, often associated with weak precipitations.
Stratusare low clouds with small opacity, undefined base under 2000m that can even reach the ground, forming fog. Images of different types of clouds can be found on the Internet (see, as instance, http://cimss.ssec.wisc.edu/satmet/gallery/gallery.html).
In the following subchapters, a brief outline will be given on how clouds form in a saturated environment. The level of understanding of water cloud formation is quite advanced, while it is not so for ice clouds, and for glaciation processes in water clouds.
### 4.1. Nucleation of droplets
We could think that the more straightforward way to form a cloud droplet would be by condensation in a saturated environment, when some water molecules collide by chance to form a cluster that will further grow to a droplet by picking up more and more molecules from the vapour phase. This process is termed homogeneous nucleation. The survival and further growth of the droplet in its environment will depend on whether the Gibbs free energy of the droplet and its surrounding will decrease upon further growth. We note that, by creating a droplet, work is done not only as expansion work, but also to form the interface between the droplet and its environment, associated with the surface tensionat the surface of the droplet of area A. This originates from the cohesive forces among the liquid molecules. In the interior of the droplet, each molecule is equally pulled in every direction by neighbouring molecules, resulting in a null net force. The molecules at the surface do not have other molecules on all sides of them and therefore are only pulled inwards, as if a force acted on interface toward the interior of the droplet. This creates a sort of pressure directed inward, against which work must be exerted to allow further expansion. This effect forces liquid surfaces to contract to the minimal area.
Let σbe the energy required to form a droplet of unit surface; then, for the heterogeneous system droplet-surroundings we may write, for an infinitesimal change of the droplet:
We note that dmv= - dml= - nldVwhere nlis the number density of molecules inside the droplet. Considering an isothermal-isobaric process, we came to the conclusion that the formation of a droplet of radius rresults in a change of Gibbs free given by:
?G=4ar2a-43ar3nlKTlneesyE47
Where we have used (36). Clearly, droplet formation is thermodynamically unfavoured for
e < es, as should be expected. If e > es, we are in supersaturated conditions, and the second term can counterbalance the first to give a negative ΔG.
Figure 7 shows two curves of ΔGas a function of the droplet radius r, for a subsaturated and supersaturated environment. It is clear that below saturation every increase of the droplet radius will lead to an increase of the free energy of the system, hence is thermodynamically unfavourable and droplets will tend to evaporate. In the supersaturated case, on the contrary, a critical value of the radius exists, such that droplets that grows by casual collision among molecules beyond that value, will continue to grow: they are said to get activated. The expression for such critical radiusis given by the Kelvin’s formula:
r0=2anlKTlneesE48
The greater ewith respect to es, that is the degree of supersaturation, the smaller the radius beyond which droplets become activated.
It can be shown from (48) that a droplet with a radius as small as 0.01 μm would require a supersaturation of 12% for getting activated. However, air is seldom more than a few percent supersaturated, and the homogeneous nucleation process is thus unable to explain the generation of clouds. Another process should be invoked: the heterogeneous nucleation. This process exploit the ubiquitous presence in the atmosphere of particles of various nature (Kaufman et al., 2002), some of which are soluble (hygroscopic) or wettable (hydrophilic) and are called Cloud Condensation Nuclei (CCN). Water may form a thin film on wettable particles, and if their dimension is beyond the critical radius, they form the nucleus of a droplet that may grow in size. Soluble particles, like sodium chloride originating from sea spray, in presence of moisture absorbs water and dissolve into it, forming a droplet of solution. The saturation vapour pressure over a solution is smaller than over pure water, and the fractional reduction is given by Raoult’s law:
f=e'eE49
Where ein the vapour pressure over pure water, and e’is the vapour pressure over a solution containing a mole fraction f(number of water moles divided by the total number of moles) of pure water.
Let us consider a droplet of radius rthat contains a mass mof a substance of molecular weight Msdissolved into iions per molecule, such that the effective number of moles in the solution is im/Ms.The number of water moles will be ((4/3)πr3ρ - m)/Mwwhere ρand Mware the water density and molecular weight respectively. The water mole fraction fis:
f=43ar3?-mMw43ar3?-mMw-imMs=1+imMwMs43ar3?-m-1E50
Eq. (49) and (50) allows us to express the reduced value e’of the saturation vapour pressure for a droplet of solution. Using this result into (48) we can compute the saturation vapour pressure in equilibrium with a droplet of solution of radius r:
e'es=exp2anKTr1+imMwMs43ar3?-m-1E51
The plot of supersaturation e’/es-1 for two different values of mis shown in fig. 8, and is named Köhler curve.
Figure 8 clearly shows how the amount of supersaturation needed to sustain a droplet of solution of radius ris much lower than what needed for a droplet of pure water, and it decreases with the increase of solute concentration. Consider an environment supersaturation of 0.2%. A droplet originated from condensation on a sphere of sodium chloride of diameter 0.1 μm can grow indefinitely along the blue curve, since the peak of the curve is below the environment supersaturation; such droplet is activated. A droplet originated from a smaller grain of sodium chloride of 0.05 μm diameter will grow until when the supersaturation adjacent to it is equal to the environmental: attained that maximum radius, the droplet stops its grow and is in stable equilibrium with the environment. Such haze dropledis said to be unactivated.
### 4.2. Condensation
The droplet that is able to pass over the peak of the Köhler curve will continue to grow by condensation. Let us consider a droplet of radius rat time t, in a supersaturated environment whose water vapour density far from the droplet is ρv(∞), while the vapour density in proximity of the droplet is ρv(r).The droplet mass Mwill grow at the rate of mass flux across a sphere of arbitrary radius centred on the droplet. Let Dbe the diffusion coefficient, that is the amount of water vapour diffusing across a unit area through a unit concentration gradient in unit time, and ρv(x)the water vapour density at a distance x > rfrom the droplet. We will have:
dMdt=4ax2Dd?vxdxE52
Since in steady conditions of mass flow this equation is independent of x, we can integrate it for xbetween rand to get:
dMdt?ridxx2=??vr?vid?vxE53
Or, expliciting Mas (4/3)πr3ρl:
drdt=Dr?l?vi-?vr=D?vir?leiei-erE54
Where we have used the ideal gas equation for water vapour. We should think of e(r)as given by e’in (49), but in fact we can approximate it with the saturation vapour pressure over a plane surface es, and pose (e(∞)-e(r))/e(∞)roughly equal to the supersaturation S=(e(∞)-es)/esto came to:
drdtr=D?vi?lSE55
This equation shows that the radius growth is inversely proportional to the radius itself, so that the rate of growth will tend to slow down with time. In fact, condensation alone is too slow to eventually produce rain droplets, and a different process should be invoked to create droplet with radius greater than few tens of micrometers.
### 4.3. Collision and coalescence
The droplet of density ρland volume Vis suspended in air of density ρso that under the effect of the gravitational field, three forces are acting on it: the gravity exerting a downward force ρlVg, the upward Archimede’s buoyancy ρVand the drag forcethat for a sphere, assumes the form of the Stokes’ drag 6πηrvwhere ηis the viscosity of the air and vis the steady state terminal fall speed of the droplet. In steady state, by equating those forces and assuming the droplet density much greater than the air, we get an expression for the terminal fall speed:
v=29?lgr2?E56
Such speed increases with the droplet dimension, so that bigger droplets will eventually collide with the smaller ones, and may entrench them with a collection efficiency Edepending on their radius and other environmental parameters, as for instance the presence of electric fields. The rate of increase of the radius r1of a spherical collector drop due to collision with water droplets in a cloud of liquid water content wl, that is is the mass density of liquid water in the cloud, is given by:
drdt=v1-v2wlE4?l?v1wlE4?l?E57
Since v1increases with r1, the process tends to speed up until the collector drops became a rain drop and eventually pass through the cloud base, or split up to reinitiate the process.
### 4.4. Nucleation of ice particles
A cloud above 0 is said a warm cloudand is entirely composed of water droplets. Water droplet can still exists in cold cloudsbelow 0 , although in an unstable state, and are termed supecooled. If a cold cloud contains both water droplets and ice, is said mixed cloud; if it contains only ice, it is said glaciated.
For a droplet to freeze, a number of water molecules inside it should come together and form an ice embryothat, if exceeds a critical size, would produce a decrease of the Gibbs free energy of the system upon further growing, much alike the homogeneous condensation from the vapour phase to form a droplet. This glaciations process is termed homogeneous freezing, and below roughly -37 °C is virtually certain to occur. Above that temperature, the critical dimensions of the ice embryo are several micrometers, and such process is not favoured. However, the droplet can contain impurities, and some of them may promote collection of water droplets into an ice-like structure to form a ice-like embryo with dimension already beyond the critical size for glaciations. Such particles are termed ice nucleiand the process they start is termed heterogeneous freezing. Such process can start not only within the droplet, but also upon contact of the ice nucleus with the surface of the droplet (contact nucleation) or directly by deposition of ice on it from the water vapour phase (deposition nucleation). Good candidates to act as ice nuclei are those particle with molecular structure close to the hexagonal ice crystallography. Some soil particles, some organics and even some bacteria are effective nucleators, but only one out of 103-105 atmospheric particles can act as an ice nucleus. Nevertheless ice particles are present in clouds in concentrations which are orders of magnitude greater than the presence of ice nuclei. Hence, ice multiplication processes must be at play, like breaking of ice particles upon collision, to create ice splinterings that enhance the number of ice particles.
### 4.5. Growth of ice particles
Ice particles can grow from the vapour phase as in the case of water droplets. In a mixed phase cloud below 0 °C, a much greater supersaturation is reached with respect to ice that can reach several percents, than with respect to water, which hardly exceed 1%. Hence ice particles grows faster than droplets and, since this deplete the vapour phase around them, it may happen that around a growing ice particle, water droplets evaporate. Ice can form in a variety of shapes, whose basic habits are determined by the temperature at which they grow. Another process of growth in a mixed cloud is by riming, that is by collision with supercooled droplets that freeze onto the ice particle. Such process is responsible of the formation of hailstones.
A process effective in cold clouds is the aggregationof ice particles between themselves, when they have different shapes and/or dimension, hence different fall speeds.
## 5. Conclusion
A brief overview of some topic of relevance in atmospheric thermodynamic has been provided, but much had to remain out of the limits of this introduction, so the interested reader is encouraged to further readings. For what concerns moist thermodynamics and convection, the reader can refer to chapters in introductory atmospheric science textbooks like the classical Wallace and Hobbs (2006), or Salby (1996). At a higher level of deepening the classical reference is Iribarne and Godson (1973). For the reader who seeks a more theoretical approach, Zdunkowski and Bott (2004) is a good challenge. Convection is thoughtfully treated in Emmanuel (1994) while a sound review is given in the article of Stevens (2005). For what concerns the microphysics of clouds, the reference book is Pruppacher and Klett (1996). A number of seminal journal articles dealing with the thermodynamics of the general circulation of the atmosphere can be cited: Goody (2003), Pauluis and Held (2002), Renno and Ingersoll (2008), Pauluis et al. (2008) and references therein. Finally, we would like to suggest the Bohren (2001) delightful book, for which a scientific or mathematical background is not required, that explores topics in meteorology and basic physics relevant to the atmosphere.
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Francesco Cairo (November 2nd 2011). Atmospheric Thermodynamics, Thermodynamics - Interaction Studies - Solids, Liquids and Gases, Juan Carlos Moreno-Pirajan, IntechOpen, DOI: 10.5772/19429. Available from:
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https://oneclass.com/class-notes/us/nku/sta/sta-205/2292201-sta-205-lecture-1.en.html | # STA 205 Lecture 1: Packet 1-Normal Distributions
41 views8 pages
19 Dec 2018
School
Department
Course
Packet 1: Normal Distributions
After completing this material, you should be able to:
describe what a normal distribution is.
use the Empirical rule to describe normal distributions.
calculate a z-score and interpret its meaning.
find probabilities ( or percentages) of normal curves above a given observation, below a given observation, and
between two observations.
determine the value of a normal distribution corresponding to any given percentage (or probability).
determine when observations are unusual based upon probabilities and explain the reasoning.
What exactly is a normal distribution? In the space below, briefly define a normal distribution and draw several normal
curves.
A normal distribution is controlled by two features:
1.
2.
The ___________________ of the distribution controls the
center of the distribution (see Figure 1 below).
The ______________________________________ of the
distribution controls how spread out the curve is, which in
turn controls its scale (see the Figure 2 below).
(You must remember this notation!!)
Figure 1.
Blue Curve:
Red Curve:
Figure 2.
Blue Curve:
Red Curve:
Figure 3.
Blue Curve:
Red Curve:
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Packet 1: Normal Distributions page 2
STA 205 Notes Buckley Spring 2017
Example: Last year, the opening price for one share of Google stock has been approximately normally distributed with a
mean price of \$665.45 and a standard deviation of \$36.31.
Draw and label the normal distribution corresponding to opening prices. Use the distribution to answer the following:
Example: Delta reports that the average age of their pilots is 42.6 years with a standard deviation of 6.3 years. Assume
that the distribution of ages is normally distributed.
Draw and label the normal model which describes the ages of Delta pilots. What percentage of Delta pilots is older than
50? Why can we not use the Empirical Rule to arrive at an exact answer to this question? What is the best we can say?
To help understand how normal distributions work, we can use the
Empirical Rule
(which is also referred to as the
68-95-99.7 Rule – I’ll let you figure out where this name comes from!). This rule states the following:
________ % of the values in a normal distribution fall within __________ standard deviation of the mean.
________ % of the values in a normal distribution fall within __________ standard deviations of the mean.
________ % of the values in a normal distribution fall within __________ standard deviations of the mean.
Between what two values do we expect the opening price of one share of
Google stock to fall 95% of the time?
What percentage of the time was the opening price more than \$701.76?
What percentage of the time is an opening price between \$629.14 and
\$738.07?
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Packet 1: Normal Distributions page 3
STA 205 Notes Buckley Spring 2017
Finding Normal Probabilities (also known as “working forwards”)
In the Delta pilot example, we ran into a problem the age we were interested in wasn’t exactly 1, 2, or 3 standard
deviations from the mean. This meant the Empirical Rule could no longer be used to find the percentage of the curve
below the observation. We need a way to find probabilities (or percentages) associated with any observation.
In order to find probabilities, we first need a way to determine the exact number of standard deviations above or below
the mean an observation falls. This process is known as standardizing the observation.
Let’s return to the pilot example we looked at last class meeting. The age of Delta pilots is normally distributed with a
mean of 42.6 years and a standard deviation of 6.3 years. We were interested in an age of 50 calculate and interpret
the z-score associated with this quantity.
How will standardizing the observation by calculating its z-score help us find the percentage of pilots older than 50? To
answer that question, let’s fill in the normal model below:
This formula will be given on the
formula sheet.
What information would we like to know?
Using the normal table, what percentage of the curve is below
an age of 50?
M42.6 F6.3
t
DzScore for an
observation OR
i
Standardize 2N age of 50
50 42.6
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1it abovethe Mean
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https://www.physicsforums.com/threads/slightly-interesting-triangle-formula.108028/ | # Slightly interesting triangle formula
1. Jan 25, 2006
### StatusX
Here's a strange identity I found. Nothing spectacular, and it should be clear from the form of it how I derived it, but I though it was kind of cool, and I've never seen this type of geometric identity.
Given a triangle with side lengths A, B, and C, (with B<A, and maybe some other restrictions I've missed) and angle $\gamma$ opposite C, then the following identity holds:
$$\ln \left( \frac{A}{C} \right) = \frac{B}{A} \cos (\gamma) + \frac{1}{2}\left( \frac{B}{A} \right)^2 \cos (2\gamma) + ...$$
Anyone ever seen this before? Do you know if it can be put in a more pleasing form, or applied to anything useful? I know it quickly gives a few taylor series with certain values for the sides and angle. It was derived algebraically, but maybe there's a really clever geometric proof?
EDIT: Actually, I was thinking, wouldn't this give a method for constructing (ie, with a compas and straightedge) the logs of numbers, albeit in an infinite number of steps? I don't know much about constructing numbers, but it does seem to be a conducive form.
Last edited: Jan 25, 2006 | {"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": 0.9250660538673401, "perplexity": 420.2614423381397}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125946011.29/warc/CC-MAIN-20180423125457-20180423145457-00231.warc.gz"} |
http://mathhelpforum.com/algebra/65025-height-ball.html | # Math Help - Height of Ball
1. ## Height of Ball
The formula for the height of a ball as a function of time is given by the equation h = -16t^2 + vt + h, where h is the height of a ball in feet, v is the initial velocity of the ball in feet per second, h is the initial height of the ball in feet, and t is the time in seconds after the ball was thrown.
If a ball is thrown from an initial height of 5 feet at an initial velocity of 20 feet per second, what is its height after 1 second?
Do I plug and chug here?
2. Originally Posted by magentarita
The formula for the height of a ball as a function of time is given by the equation h = -16t^2 + vt + h, where h is the height of a ball in feet, v is the initial velocity of the ball in feet per second, h is the initial height of the ball in feet, and t is the time in seconds after the ball was thrown.
If a ball is thrown from an initial height of 5 feet at an initial velocity of 20 feet per second, what is its height after 1 second?
Do I plug and chug here?
Yes that's right but remember to plug in h which is the initial height into the correct h in the equation as the equation has 2 h's in it.
3. ## ok...
Originally Posted by tester85
Yes that's right but remember to plug in h which is the initial height into the correct h in the equation as the equation has 2 h's in it.
So, h = 5 in the function.
Plug the rest for v and t and simplify, right?
4. Originally Posted by magentarita
So, h = 5 in the function.
Plug the rest for v and t and simplify, right?
Correct. The equation you should get should be $h\left(t\right)=-16t^2+20t+5$
Now, what do you think $h\left(1\right)$ should be? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9629947543144226, "perplexity": 224.60207387213373}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701160822.87/warc/CC-MAIN-20160205193920-00296-ip-10-236-182-209.ec2.internal.warc.gz"} |
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# Equilibrium statistical mechanics of bipartite spin systems
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### Abstract
Aim of this paper is to give an extensive treatment of bipartite mean field spin systems, ordered and disordered: at first, bipartite ferromagnets are investigated, achieving an explicit expression for the free energy trough a new minimax variational principle. Furthermore via the Hamilton-Jacobi technique the same free energy structure is obtained together with the existence of its thermodynamic limit and the minimax principle is connected to a standard max one. The same is investigated for bipartite spin-glasses: By the Borel-Cantelli lemma a control of the high temperature regime is obtained, while via the double stochastic stability technique we get also the explicit expression of the free energy at the replica symmetric level, uniquely defined by a minimax variational principle again. A general results that states that the free energies of these systems are convex linear combinations of their independent one party model counterparts is achieved too. For the sake of completeness we show further that at zero temperature the replica symmetric entropy becomes negative and, consequently, such a symmetry must be broken. The treatment of the fully broken replica symmetry case is deferred to a forthcoming paper. As a first step in this direction, we start deriving the linear and quadratic constraints to overlap fluctuations.
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### Broken Replica Symmetry Bounds in the Mean Field Spin Glass Model
(2002)
By using a simple interpolation argument, in previous work we have proven the existence of the thermodynamic limit, for mean field disordered models, including the Sherrington-Kirkpatrick model, and the Derrida p-spin model. Here we extend this argument in order to compare the limiting free energy with the expression given by the Parisi Ansatz, and including full spontaneous replica symmetry breaking. Our main result is that the quenched average of the free energy is bounded from below by the value given in the Parisi Ansatz uniformly in the size of the system. Moreover, the difference between the two expressions is given in the form of a sum rule, extending our previous work on the comparison between the true free energy and its replica symmetric Sherrington-Kirkpatrick approximation. We give also a variational bound for the infinite volume limit of the ground state energy per site.
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### General properties of overlap probability distributions in disordered spin systems. Toward Parisi ultrametricity
(1998)
For a very general class of probability distributions in disordered Ising spin systems, in the thermodynamical limit, we prove the following property for overlaps among real replicas. Consider the overlaps among s replicas. Add one replica s+1. Then, the overlap q(a,s+1) between one of the first s replicas, let us say a, and the added s+1 is either independent of the former ones, or it is identical to one of the overlaps q(a,b), with b running among the first s replicas, excluding a. Each of these cases has equal probability 1/s.
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• Record: found
• Abstract: found
• Article: found
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### An Extended Variational Principle for the SK Spin-Glass Model
(2003)
The recent proof by F. Guerra that the Parisi ansatz provides a lower bound on the free energy of the SK spin-glass model could have been taken as offering some support to the validity of the purported solution. In this work we present a broader variational principle, in which the lower bound, as well as the actual value, are obtained through an optimization procedure for which ultrametic/hierarchal structures form only a subset of the variational class. The validity of Parisi's ansatz for the SK model is still in question. The new variational principle may be of help in critical review of the issue.
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### Author and article information
###### Journal
1012.1261
10.1088/1751-8113/44/24/245002
Theoretical physics | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9084928631782532, "perplexity": 776.3694580314614}, "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/1593655886706.29/warc/CC-MAIN-20200704201650-20200704231650-00558.warc.gz"} |
https://unapologetic.wordpress.com/2009/11/17/ | # The Unapologetic Mathematician
## Cramer’s Rule
We’re trying to invert a function $f:X\rightarrow\mathbb{R}^n$ which is continuously differentiable on some region $X\subseteq\mathbb{R}^n$. That is we know that if $a$ is a point where $J_f(a)\neq0$, then there is a ball $N$ around $a$ where $f$ is one-to-one onto some neighborhood $f(N)$ around $f(a)$. Then if $y$ is a point in $f(N)$, we’ve got a system of equations
$\displaystyle f^j(x^1,\dots,x^n)=y^j$
that we want to solve for all the $x^i$.
We know how to handle this if $f$ is defined by a linear transformation, represented by a matrix $A=\left(a_i^j\right)$:
\displaystyle\begin{aligned}f^j(x^1,\dots,x^n)=a_i^jx^i&=y^j\\Ax&=y\end{aligned}
In this case, the Jacobian transformation is just the function $f$ itself, and so the Jacobian determinant $\det\left(a_i^j\right)$ is nonzero if and only if the matrix $A$ is invertible. And so our solution depends on finding the inverse $A^{-1}$ and solving
\displaystyle\begin{aligned}Ax&=y\\A^{-1}Ax&=A^{-1}y\\x&=A^{-1}y\end{aligned}
This is the approach we’d like to generalize. But to do so, we need a more specific method of finding the inverse.
This is where Cramer’s rule comes in, and it starts by analyzing the way we calculate the determinant of a matrix $A$. This formula
$\displaystyle\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)a_1^{\pi(1)}\dots a_n^{\pi(n)}$
involves a sum over all the permutations $\pi\in S_n$, and we want to consider the order in which we add up these terms. If we fix an index $i$, we can factor out each matrix entry in the $i$th column:
$\displaystyle\sum\limits_{j=1}^na_i^j\sum\limits_{\substack{\pi\in S_n\\\pi(i)=j}}\mathrm{sgn}(\pi)a_1^{\pi(1)}\dots\widehat{a_i^j}\dots a_n^{\pi(n)}$
where the hat indicates that we omit the $i$th term in the product. For a given value of $j$, we can consider the restricted sum
$\displaystyle A_j^i=\sum\limits_{\substack{\pi\in S_n\\\pi(i)=j}}\mathrm{sgn}(\pi)a_1^{\pi(1)}\dots\widehat{a_i^j}\dots a_n^{\pi(n)}$
which is $(-1)^{i+j}$ times the determinant of the $i$$j$ “minor” of the matrix $A$. That is, if we strike out the row and column of $A$ which contain $a_i^j$ and take the determinant of the remaining $(n-1)\times(n-1)$ matrix, we multiply this by $(-1)^{i+j}$ to get $A_j^i$. These are the entries in the “adjugate” matrix $\mathrm{adj}(A)$.
What we’ve shown is that
$\displaystyle A_j^ia_i^j=\det(A)$
(no summation on $i$). It’s not hard to show, however, that if we use a different row from the adjugate matrix we find
$\displaystyle\sum\limits_{j=1}^nA_j^ka_i^j=\det(A)\delta_i^k$
That is, the adjugate times the original matrix is the determinant of $A$ times the identity matrix. And so if $\det(A)\neq0$ we find
$\displaystyle A^{-1}=\frac{1}{\det(A)}\mathrm{adj}(A)$
So what does this mean for our system of equations? We can write
\displaystyle\begin{aligned}x&=\frac{1}{\det(A)}\mathrm{adj}(A)y\\x^i&=\frac{1}{\det(A)}A_j^iy^j\end{aligned}
But how does this sum $A_j^iy^j$ differ from the one $A_j^ia_i^j$ we used before (without summing on $i$) to calculate the determinant of $A$? We’ve replaced the $i$th column of $A$ by the column vector $y$, and so this is just another determinant, taken after performing this replacement!
Here’s an example. Let’s say we’ve got a system written in matrix form
$\displaystyle\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}u\\v\end{pmatrix}$
The entry in the $i$th row and $j$th column of the adjugate matrix is calculated by striking out the $i$th column and $j$th row of our original matrix, taking the determinant of the remaining matrix, and multiplying by $(-1)^{i+j}$. We get
$\displaystyle\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$
and thus we find
$\displaystyle\begin{pmatrix}x\\y\end{pmatrix}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\begin{pmatrix}u\\v\end{pmatrix}=\frac{1}{ad-bc}\begin{pmatrix}ud-bv\\av-uc\end{pmatrix}$
where we note that
\displaystyle\begin{aligned}ud-bv&=\det\begin{pmatrix}u&b\\v&d\end{pmatrix}\\av-uc&=\det\begin{pmatrix}a&u\\c&v\end{pmatrix}\end{aligned}
In other words, our solution is given by ratios of determinants:
\displaystyle\begin{aligned}x&=\frac{\det\begin{pmatrix}u&b\\v&d\end{pmatrix}}{\det\begin{pmatrix}a&b\\c&d\end{pmatrix}}\\y&=\frac{\det\begin{pmatrix}a&u\\c&v\end{pmatrix}}{\det\begin{pmatrix}a&b\\c&d\end{pmatrix}}\end{aligned}
and similar formulae hold for larger systems of equations.
November 17, 2009 Posted by | Algebra, Linear Algebra | 8 Comments
## Another Lemma on Nonzero Jacobians
Sorry for the late post. I didn’t get a chance to get it up this morning before my flight.
Brace yourself. Just like last time we’ve got a messy technical lemma about what happens when the Jacobian determinant of a function is nonzero.
This time we’ll assume that $f:X\rightarrow\mathbb{R}^n$ is not only continuous, but continuously differentiable on a region $X\subseteq\mathbb{R}^n$. We also assume that the Jacobian $J_f(a)\neq0$ at some point $a\in X$. Then I say that there is some neighborhood $N$ of $a$ so that $f$ is injective on $N$.
First, we take $n$ points $\{z_i\}_{i=1}^n$ in $X$ and make a function of them
$\displaystyle h(z_1,\dots,z_n)=\det\left(\frac{\partial f^i}{\partial x^j}\bigg\vert_{x=z_i}\right)$
That is, we take the $j$th partial derivative of the $i$th component function and evaluate it at the $i$th sample point to make a matrix $\left(a_{ij}\right)$, and then we take the determinant of this matrix. As a particular value, we have
$\displaystyle h(a,\dots,a)=J_f(a)\neq0$
Since each partial derivative is continuous, and the determinant is a polynomial in its entries, this function is continuous where it’s defined. And so there’s some ball $N$ of $a$ so that if all the $z_i$ are in $N$ we have $h(z_1,\dots,z_n)\neq0$. We want to show that $f$ is injective on $N$.
So, let’s take two points $x$ and $y$ in $N$ so that $f(x)=f(y)$. Since the ball is convex, the line segment $[x,y]$ is completely contained within $N\subseteq X$, and so we can bring the mean value theorem to bear. For each component function we can write
$\displaystyle0=f^i(y)-f^i(x)=df^i(\xi_i)(y-x)=\frac{\partial f^i}{\partial x^j}\bigg\vert_{\xi_i}(y^j-x^j)$
for some $\xi_i$ in $[x,y]\subseteq N$ (no summation here on $i$). But like last time we now have a linear system of equations described by an invertible matrix. Here the matrix has determinant
$\displaystyle\det\left(\frac{\partial f^i}{\partial x^j}\bigg\vert_{\xi_i}\right)=h(\xi_1,\dots,\xi_n)\neq0$
which is nonzero because all the $\xi_i$ are inside the ball $N$. Thus the only possible solution to the system of equations is $x^i=y^i$. And so if $f(x)=f(y)$ for points within the ball $N$, we must have $x=y$, and thus $f$ is injective.
November 17, 2009 Posted by | Analysis, Calculus | 2 Comments | {"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": 106, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9892368316650391, "perplexity": 188.95823606254285}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375098059.60/warc/CC-MAIN-20150627031818-00065-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/145547/the-collection-of-continuous-functions-between-any-2-compact-hausdorff-spaces-fo | # The collection of continuous functions between any 2 compact Hausdorff spaces forms a set
I would like to show precisely what I have stated in the title (assuming that it is correct; I have reason to suspect it is, thanks to a tricky past exam paper I'm trying to surmount); namely, that given any 2 compact Hdf topological spaces $T_1$ and $T_2$, the class of continuous functions between them forms a set (rather than, say, a class which is too large to be a set).
I suspect this result might be obvious in a much more general sense, and just wanted to check my thinking: is it actually valid to say that for any function between 2 fixed sets $S_1$ and $S_2$, the function is simply a set of ordered pairs $(a,f(a))$ and therefore contained in the power set of the union $S_1 \cup S_2$, so is again a set (by standard set-theoretic axioms for power sets and unions)?
-
Please forgive my ignorance of set theory, but working solely from the fact that $\mathsf{Set}$ is a category, my understanding is that the collection of all functions from a set $A$ to a set $B$ always forms a set, and the continuous functions between two topological spaces are of course just a subset of the set of all functions between them. Perhaps there is a different foundation you are working from where this is not the case? – Zev Chonoles May 15 '12 at 18:03
Yes, that is essentially what I was confirming. – Spyam May 15 '12 at 18:10
@Zev: The only possible way your argument could fail is if the set theory in question doesn't have a sufficiently strong separation axiom or if it doesn't have function spaces. The former is highly unlikely in any reasonable set theory, and it's very hard to do any point set topology in the latter (since there would be no powersets!). – Zhen Lin May 15 '12 at 19:20
One usually takes the ordered pair $(a, b)$ to be an abbreviation for the set $\{\{a\}, \{a, b\}\}$, so if $a\in A$ and $b\in B$, then $\def\pow#1{{\mathcal P}{(#1)}} (a,b)\in\pow{\pow{A\cup B}}$, and the cartesian product $A\times B$ is a subset of $\pow{\pow{A\cup B}}$, which is a set by the union and power set axioms.
A function $f:A\to B$ is a subset of $A\times B\subset \pow{\pow{A\cup B}}$, and so is a set and also is an element of $\pow{\pow{\pow{A\cup B}}}$. The set $\mathcal F$ of all functions $f:A\to B$ is therefore a subset of $\pow{\pow{\pow{A\cup B}}}$ and is a set. The set of all continuous functions is a subset of $\mathcal F$, and is therefore a set, an element of $\pow{\pow{\pow{\pow{A\cup B}}}}$. | {"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": 0.9011722207069397, "perplexity": 95.93432860400365}, "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/1368710366143/warc/CC-MAIN-20130516131926-00037-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://python-advanced.quantecon.org/dyn_stack.html | # 40. Stackelberg Plans¶
In addition to what’s in Anaconda, this lecture will need the following libraries:
!pip install --upgrade quantecon
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## 40.1. Overview¶
This notebook formulates and computes a plan that a Stackelberg leader uses to manipulate forward-looking decisions of a Stackelberg follower that depend on continuation sequences of decisions made once and for all by the Stackelberg leader at time $$0$$.
To facilitate computation and interpretation, we formulate things in a context that allows us to apply dynamic programming for linear-quadratic models.
Technically, our calculations are closely related to ones described this lecture.
From the beginning, we carry along a linear-quadratic model of duopoly in which firms face adjustment costs that make them want to forecast actions of other firms that influence future prices.
import numpy as np
import numpy.linalg as la
import quantecon as qe
from quantecon import LQ
import matplotlib.pyplot as plt
%matplotlib inline
## 40.2. Duopoly¶
Time is discrete and is indexed by $$t = 0, 1, \ldots$$.
Two firms produce a single good whose demand is governed by the linear inverse demand curve
$p_t = a_0 - a_1 (q_{1t}+ q_{2t} )$
where $$q_{it}$$ is output of firm $$i$$ at time $$t$$ and $$a_0$$ and $$a_1$$ are both positive.
$$q_{10}, q_{20}$$ are given numbers that serve as initial conditions at time $$0$$.
By incurring a cost of change
$\gamma v_{it}^2$
where $$\gamma > 0$$, firm $$i$$ can change its output according to
$q_{it+1} = q_{it} + v_{it}$
Firm $$i$$’s profits at time $$t$$ equal
$\pi_{it} = p_t q_{it} - \gamma v_{it}^2$
Firm $$i$$ wants to maximize the present value of its profits
$\sum_{t=0}^\infty \beta^t \pi_{it}$
where $$\beta \in (0,1)$$ is a time discount factor.
### 40.2.1. Stackelberg Leader and Follower¶
Each firm $$i=1,2$$ chooses a sequence $$\vec q_i \equiv \{q_{it+1}\}_{t=0}^\infty$$ once and for all at time $$0$$.
We let firm 2 be a Stackelberg leader and firm 1 be a Stackelberg follower.
The leader firm 2 goes first and chooses $$\{q_{2t+1}\}_{t=0}^\infty$$ once and for all at time $$0$$.
Knowing that firm 2 has chosen $$\{q_{2t+1}\}_{t=0}^\infty$$, the follower firm 1 goes second and chooses $$\{q_{1t+1}\}_{t=0}^\infty$$ once and for all at time $$0$$.
In choosing $$\vec q_2$$, firm 2 takes into account that firm 1 will base its choice of $$\vec q_1$$ on firm 2’s choice of $$\vec q_2$$.
### 40.2.2. Statement of Leader’s and Follower’s Problems¶
We can express firm 1’s problem as
$\max_{\vec q_1} \Pi_1(\vec q_1; \vec q_2)$
where the appearance behind the semi-colon indicates that $$\vec q_2$$ is given.
Firm 1’s problem induces the best response mapping
$\vec q_1 = B(\vec q_2)$
(Here $$B$$ maps a sequence into a sequence)
The Stackelberg leader’s problem is
$\max_{\vec q_2} \Pi_2 (B(\vec q_2), \vec q_2)$
whose maximizer is a sequence $$\vec q_2$$ that depends on the initial conditions $$q_{10}, q_{20}$$ and the parameters of the model $$a_0, a_1, \gamma$$.
This formulation captures key features of the model
• Both firms make once-and-for-all choices at time $$0$$.
• This is true even though both firms are choosing sequences of quantities that are indexed by time.
• The Stackelberg leader chooses first within time $$0$$, knowing that the Stackelberg follower will choose second within time $$0$$.
While our abstract formulation reveals the timing protocol and equilibrium concept well, it obscures details that must be addressed when we want to compute and interpret a Stackelberg plan and the follower’s best response to it.
To gain insights about these things, we study them in more detail.
### 40.2.3. Firms’ Problems¶
Firm 1 acts as if firm 2’s sequence $$\{q_{2t+1}\}_{t=0}^\infty$$ is given and beyond its control.
Firm 2 knows that firm 1 chooses second and takes this into account in choosing $$\{q_{2t+1}\}_{t=0}^\infty$$.
In the spirit of working backward, we study firm 1’s problem first, taking $$\{q_{2t+1}\}_{t=0}^\infty$$ as given.
We can formulate firm 1’s optimum problem in terms of the Lagrangian
$L=\sum_{t=0}^{\infty}\beta^{t}\{a_{0}q_{1t}-a_{1}q_{1t}^{2}-a_{1}q_{1t}q_{2t}-\gamma v_{1t}^{2}+\lambda_{t}[q_{1t}+v_{1t}-q_{1t+1}]\}$
Firm 1 seeks a maximum with respect to $$\{q_{1t+1}, v_{1t} \}_{t=0}^\infty$$ and a minimum with respect to $$\{ \lambda_t\}_{t=0}^\infty$$.
We approach this problem using methods described in [LS18], chapter 2, appendix A and [Sar87], chapter IX.
First-order conditions for this problem are
\begin{aligned} \frac{\partial L}{\partial q_{1t}} & = a_0 - 2 a_1 q_{1t} - a_1 q_{2t} + \lambda_t - \beta^{-1} \lambda_{t-1} = 0 , \quad t \geq 1 \cr \frac{\partial L}{\partial v_{1t}} & = -2 \gamma v_{1t} + \lambda_t = 0 , \quad t \geq 0 \cr \end{aligned}
These first-order conditions and the constraint $$q_{1t+1} = q_{1t} + v_{1t}$$ can be rearranged to take the form
\begin{aligned} v_{1t} & = \beta v_{1t+1} + \frac{\beta a_0}{2 \gamma} - \frac{\beta a_1}{\gamma} q_{1t+1} - \frac{\beta a_1}{2 \gamma} q_{2t+1} \cr q_{t+1} & = q_{1t} + v_{1t} \end{aligned}
We can substitute the second equation into the first equation to obtain
$(q_{1t+1} - q_{1t} ) = \beta (q_{1t+2} - q_{1t+1}) + c_0 - c_1 q_{1t+1} - c_2 q_{2t+1}$
where $$c_0 = \frac{\beta a_0}{2 \gamma}, c_1 = \frac{\beta a_1}{\gamma}, c_2 = \frac{\beta a_1}{2 \gamma}$$.
This equation can in turn be rearranged to become the second-order difference equation
(40.1)$q_{1t} + (1+\beta + c_1) q_{1t+1} - \beta q_{1t+2} = c_0 - c_2 q_{2t+1}$
Equation (40.1) is a second-order difference equation in the sequence $$\vec q_1$$ whose solution we want.
It satisfies two boundary conditions:
• an initial condition that $$q_{1,0}$$, which is given
• a terminal condition requiring that $$\lim_{T \rightarrow + \infty} \beta^T q_{1t}^2 < + \infty$$
Using the lag operators described in [Sar87], chapter IX, difference equation (40.1) can be written as
$\beta(1 - \frac{1+\beta + c_1}{\beta} L + \beta^{-1} L^2 ) q_{1t+2} = - c_0 + c_2 q_{2t+1}$
The polynomial in the lag operator on the left side can be factored as
(40.2)$(1 - \frac{1+\beta + c_1}{\beta} L + \beta^{-1} L^2 ) = ( 1 - \delta_1 L ) (1 - \delta_2 L)$
where $$0 < \delta_1 < 1 < \frac{1}{\sqrt{\beta}} < \delta_2$$.
Because $$\delta_2 > \frac{1}{\sqrt{\beta}}$$ the operator $$(1 - \delta_2 L)$$ contributes an unstable component if solved backwards but a stable component if solved forwards.
Mechanically, write
$(1- \delta_2 L) = -\delta_{2} L (1 - \delta_2^{-1} L^{-1} )$
and compute the following inverse operator
$\left[-\delta_{2} L (1 - \delta_2^{-1} L^{-1} )\right]^{-1} = - \delta_2 (1 - {\delta_2}^{-1} )^{-1} L^{-1}$
Operating on both sides of equation (40.2) with $$\beta^{-1}$$ times this inverse operator gives the follower’s decision rule for setting $$q_{1t+1}$$ in the feedback-feedforward form.
(40.3)$q_{1t+1} = \delta_1 q_{1t} - c_0 \delta_2^{-1} \beta^{-1} \frac{1}{1 -\delta_2^{-1}} + c_2 \delta_2^{-1} \beta^{-1} \sum_{j=0}^\infty \delta_2^j q_{2t+j+1} , \quad t \geq 0$
The problem of the Stackelberg leader firm 2 is to choose the sequence $$\{q_{2t+1}\}_{t=0}^\infty$$ to maximize its discounted profits
$\sum_{t=0}^\infty \beta^t \{ (a_0 - a_1 (q_{1t} + q_{2t}) ) q_{2t} - \gamma (q_{2t+1} - q_{2t})^2 \}$
subject to the sequence of constraints (40.3) for $$t \geq 0$$.
We can put a sequence $$\{\theta_t\}_{t=0}^\infty$$ of Lagrange multipliers on the sequence of equations (40.3) and formulate the following Lagrangian for the Stackelberg leader firm 2’s problem
(40.4)\begin{aligned} \tilde L & = \sum_{t=0}^\infty \beta^t\{ (a_0 - a_1 (q_{1t} + q_{2t}) ) q_{2t} - \gamma (q_{2t+1} - q_{2t})^2 \} \cr & + \sum_{t=0}^\infty \beta^t \theta_t \{ \delta_1 q_{1t} - c_0 \delta_2^{-1} \beta^{-1} \frac{1}{1 -\delta_2^{-1}} + c_2 \delta_2^{-1} \beta^{-1} \sum_{j=0}^\infty \delta_2^{-j} q_{2t+j+1} - q_{1t+1} \} \end{aligned}
subject to initial conditions for $$q_{1t}, q_{2t}$$ at $$t=0$$.
Comments: We have formulated the Stackelberg problem in a space of sequences.
The max-min problem associated with Lagrangian (40.4) is unpleasant because the time $$t$$ component of firm $$1$$’s payoff function depends on the entire future of its choices of $$\{q_{1t+j}\}_{j=0}^\infty$$.
This renders a direct attack on the problem cumbersome.
Therefore, below, we will formulate the Stackelberg leader’s problem recursively.
We’ll put our little duopoly model into a broader class of models with the same conceptual structure.
## 40.3. Stackelberg Problem¶
We formulate a class of linear-quadratic Stackelberg leader-follower problems of which our duopoly model is an instance.
We use the optimal linear regulator (a.k.a. the linear-quadratic dynamic programming problem described in LQ Dynamic Programming problems) to represent a Stackelberg leader’s problem recursively.
Let $$z_t$$ be an $$n_z \times 1$$ vector of natural state variables.
Let $$x_t$$ be an $$n_x \times 1$$ vector of endogenous forward-looking variables that are physically free to jump at $$t$$.
In our duopoly example $$x_t = v_{1t}$$, the time $$t$$ decision of the Stackelberg follower.
Let $$u_t$$ be a vector of decisions chosen by the Stackelberg leader at $$t$$.
The $$z_t$$ vector is inherited physically from the past.
But $$x_t$$ is a decision made by the Stackelberg follower at time $$t$$ that is the follower’s best response to the choice of an entire sequence of decisions made by the Stackelberg leader at time $$t=0$$.
Let
$\begin{split} y_t = \begin{bmatrix} z_t \\ x_t \end{bmatrix} \end{split}$
Represent the Stackelberg leader’s one-period loss function as
$r(y, u) = y' R y + u' Q u$
Subject to an initial condition for $$z_0$$, but not for $$x_0$$, the Stackelberg leader wants to maximize
(40.5)$-\sum_{t=0}^\infty \beta^t r(y_t, u_t)$
The Stackelberg leader faces the model
(40.6)$\begin{split}\begin{bmatrix} I & 0 \\ G_{21} & G_{22} \end{bmatrix} \begin{bmatrix} z_{t+1} \\ x_{t+1} \end{bmatrix} = \begin{bmatrix} \hat A_{11} & \hat A_{12} \\ \hat A_{21} & \hat A_{22} \end{bmatrix} \begin{bmatrix} z_t \\ x_t \end{bmatrix} + \hat B u_t\end{split}$
We assume that the matrix $$\begin{bmatrix} I & 0 \\ G_{21} & G_{22} \end{bmatrix}$$ on the left side of equation (40.6) is invertible, so that we can multiply both sides by its inverse to obtain
(40.7)$\begin{split}\begin{bmatrix} z_{t+1} \\ x_{t+1} \end{bmatrix} = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} \begin{bmatrix} z_t \\ x_t \end{bmatrix} + B u_t\end{split}$
or
(40.8)$y_{t+1} = A y_t + B u_t$
### 40.3.1. Interpretation of Second Block of Equations¶
The Stackelberg follower’s best response mapping is summarized by the second block of equations of (40.7).
In particular, these equations are the first-order conditions of the Stackelberg follower’s optimization problem (i.e., its Euler equations).
These Euler equations summarize the forward-looking aspect of the follower’s behavior and express how its time $$t$$ decision depends on the leader’s actions at times $$s \geq t$$.
When combined with a stability condition to be imposed below, the Euler equations summarize the follower’s best response to the sequence of actions by the leader.
The Stackelberg leader maximizes (40.5) by choosing sequences $$\{u_t, x_t, z_{t+1}\}_{t=0}^{\infty}$$ subject to (40.8) and an initial condition for $$z_0$$.
Note that we have an initial condition for $$z_0$$ but not for $$x_0$$.
$$x_0$$ is among the variables to be chosen at time $$0$$ by the Stackelberg leader.
The Stackelberg leader uses its understanding of the responses restricted by (40.8) to manipulate the follower’s decisions.
### 40.3.2. More Mechanical Details¶
For any vector $$a_t$$, define $$\vec a_t = [a_t, a_{t+1} \ldots ]$$.
Define a feasible set of $$(\vec y_1, \vec u_0)$$ sequences
$\Omega(y_0) = \left\{ (\vec y_1, \vec u_0) : y_{t+1} = A y_t + B u_t, \forall t \geq 0 \right\}$
Please remember that the follower’s Euler equation is embedded in the system of dynamic equations $$y_{t+1} = A y_t + B u_t$$.
Note that in the definition of $$\Omega(y_0)$$, $$y_0$$ is taken as given.
Although it is taken as given in $$\Omega(y_0)$$, eventually, the $$x_0$$ component of $$y_0$$ will be chosen by the Stackelberg leader.
### 40.3.3. Two Subproblems¶
Once again we use backward induction.
We express the Stackelberg problem in terms of two subproblems.
Subproblem 1 is solved by a continuation Stackelberg leader at each date $$t \geq 0$$.
Subproblem 2 is solved by the Stackelberg leader at $$t=0$$.
The two subproblems are designed
• to respect the protocol in which the follower chooses $$\vec q_1$$ after seeing $$\vec q_2$$ chosen by the leader
• to make the leader choose $$\vec q_2$$ while respecting that $$\vec q_1$$ will be the follower’s best response to $$\vec q_2$$
• to represent the leader’s problem recursively by artfully choosing the state variables confronting and the control variables available to the leader
Subproblem 1
$v(y_0) = \max_{(\vec y_1, \vec u_0) \in \Omega(y_0)} - \sum_{t=0}^\infty \beta^t r(y_t, u_t)$
Subproblem 2
$w(z_0) = \max_{x_0} v(y_0)$
Subproblem 1 takes the vector of forward-looking variables $$x_0$$ as given.
Subproblem 2 optimizes over $$x_0$$.
The value function $$w(z_0)$$ tells the value of the Stackelberg plan as a function of the vector of natural state variables at time $$0$$, $$z_0$$.
## 40.4. Two Bellman Equations¶
We now describe Bellman equations for $$v(y)$$ and $$w(z_0)$$.
Subproblem 1
The value function $$v(y)$$ in subproblem 1 satisfies the Bellman equation
(40.9)$v(y) = \max_{u, y^*} \left\{ - r(y,u) + \beta v(y^*) \right\}$
where the maximization is subject to
$y^* = A y + B u$
and $$y^*$$ denotes next period’s value.
Substituting $$v(y) = - y'P y$$ into Bellman equation (40.9) gives
$-y' P y = {\rm max}_{ u, y^*} \left\{ - y' R y - u'Q u - \beta y^{* \prime} P y^* \right\}$
which as in lecture linear regulator gives rise to the algebraic matrix Riccati equation
$P = R + \beta A' P A - \beta^2 A' P B ( Q + \beta B' P B)^{-1} B' P A$
and the optimal decision rule coefficient vector
$F = \beta( Q + \beta B' P B)^{-1} B' P A$
where the optimal decision rule is
$u_t = - F y_t$
Subproblem 2
We find an optimal $$x_0$$ by equating to zero the gradient of $$v(y_0)$$ with respect to $$x_0$$:
$-2 P_{21} z_0 - 2 P_{22} x_0 =0,$
which implies that
$x_0 = - P_{22}^{-1} P_{21} z_0$
## 40.5. Stackelberg Plan¶
Now let’s map our duopoly model into the above setup.
We will formulate a state space system
$y_t = \begin{bmatrix} z_t \cr x_t \end{bmatrix}$
where in this instance $$x_t = v_{1t}$$, the time $$t$$ decision of the follower firm 1.
### 40.5.1. Calculations to Prepare Duopoly Model¶
Now we’ll proceed to cast our duopoly model within the framework of the more general linear-quadratic structure described above.
That will allow us to compute a Stackelberg plan simply by enlisting a Riccati equation to solve a linear-quadratic dynamic program.
As emphasized above, firm 1 acts as if firm 2’s decisions $$\{q_{2t+1}, v_{2t}\}_{t=0}^\infty$$ are given and beyond its control.
### 40.5.2. Firm 1’s Problem¶
We again formulate firm 1’s optimum problem in terms of the Lagrangian
$L=\sum_{t=0}^{\infty}\beta^{t}\{a_{0}q_{1t}-a_{1}q_{1t}^{2}-a_{1}q_{1t}q_{2t}-\gamma v_{1t}^{2}+\lambda_{t}[q_{1t}+v_{1t}-q_{1t+1}]\}$
Firm 1 seeks a maximum with respect to $$\{q_{1t+1}, v_{1t} \}_{t=0}^\infty$$ and a minimum with respect to $$\{ \lambda_t\}_{t=0}^\infty$$.
First-order conditions for this problem are
\begin{aligned} \frac{\partial L}{\partial q_{1t}} & = a_0 - 2 a_1 q_{1t} - a_1 q_{2t} + \lambda_t - \beta^{-1} \lambda_{t-1} = 0 , \quad t \geq 1 \cr \frac{\partial L}{\partial v_{1t}} & = -2 \gamma v_{1t} + \lambda_t = 0 , \quad t \geq 0 \cr \end{aligned}
These first-order order conditions and the constraint $$q_{1t+1} = q_{1t} + v_{1t}$$ can be rearranged to take the form
\begin{aligned} v_{1t} & = \beta v_{1t+1} + \frac{\beta a_0}{2 \gamma} - \frac{\beta a_1}{\gamma} q_{1t+1} - \frac{\beta a_1}{2 \gamma} q_{2t+1} \cr q_{t+1} & = q_{1t} + v_{1t} \end{aligned}
We use these two equations as components of the following linear system that confronts a Stackelberg continuation leader at time $$t$$
$\begin{bmatrix} 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr \frac{\beta a_0}{2 \gamma} & - \frac{\beta a_1}{2 \gamma} & -\frac{\beta a_1}{\gamma} & \beta \end{bmatrix} \begin{bmatrix} 1 \cr q_{2t+1} \cr q_{1t+1} \cr v_{1t+1} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 1 \cr 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \cr q_{2t} \cr q_{1t} \cr v_{1t} \end{bmatrix} + \begin{bmatrix} 0 \cr 1 \cr 0 \cr 0 \end{bmatrix} v_{2t}$
Time $$t$$ revenues of firm 2 are $$\pi_{2t} = a_0 q_{2t} - a_1 q_{2t}^2 - a_1 q_{1t} q_{2t}$$ which evidently equal
$z_t' R_1 z_t \equiv \begin{bmatrix} 1 \cr q_{2t} \cr q_{1t} \end{bmatrix}' \begin{bmatrix} 0 & \frac{a_0}{2}& 0 \cr \frac{a_0}{2} & -a_1 & -\frac{a_1}{2}\cr 0 & -\frac{a_1}{2} & 0 \end{bmatrix} \begin{bmatrix} 1 \cr q_{2t} \cr q_{1t} \end{bmatrix}$
If we set $$Q = \gamma$$, then firm 2’s period $$t$$ profits can then be written
$y_t' R y_t - Q v_{2t}^2$
where
$y_t = \begin{bmatrix} z_t \cr x_t \end{bmatrix}$
with $$x_t = v_{1t}$$ and
$R = \begin{bmatrix} R_1 & 0 \cr 0 & 0 \end{bmatrix}$
We’ll report results of implementing this code soon.
But first, we want to represent the Stackelberg leader’s optimal choices recursively.
It is important to do this for several reasons:
• properly to interpret a representation of the Stackelberg leader’s choice as a sequence of history-dependent functions
• to formulate a recursive version of the follower’s choice problem
First, let’s get a recursive representation of the Stackelberg leader’s choice of $$\vec q_2$$ for our duopoly model.
## 40.6. Recursive Representation of Stackelberg Plan¶
In order to attain an appropriate representation of the Stackelberg leader’s history-dependent plan, we will employ what amounts to a version of the Big K, little k device often used in macroeconomics by distinguishing $$z_t$$, which depends partly on decisions $$x_t$$ of the followers, from another vector $$\check z_t$$, which does not.
We will use $$\check z_t$$ and its history $$\check z^t = [\check z_t, \check z_{t-1}, \ldots, \check z_0]$$ to describe the sequence of the Stackelberg leader’s decisions that the Stackelberg follower takes as given.
Thus, we let $$\check y_t' = \begin{bmatrix}\check z_t' & \check x_t'\end{bmatrix}$$ with initial condition $$\check z_0 = z_0$$ given.
That we distinguish $$\check z_t$$ from $$z_t$$ is part and parcel of the Big K, little k device in this instance.
We have demonstrated that a Stackelberg plan for $$\{u_t\}_{t=0}^\infty$$ has a recursive representation
\begin{aligned} \check x_0 & = - P_{22}^{-1} P_{21} z_0 \cr u_t & = - F \check y_t, \quad t \geq 0 \cr \check y_{t+1} & = (A - BF) \check y_t, \quad t \geq 0 \end{aligned}
From this representation, we can deduce the sequence of functions $$\sigma = \{\sigma_t(\check z^t)\}_{t=0}^\infty$$ that comprise a Stackelberg plan.
For convenience, let $$\check A \equiv A - BF$$ and partition $$\check A$$ conformably to the partition $$y_t = \begin{bmatrix}\check z_t \cr \check x_t \end{bmatrix}$$ as
$\begin{bmatrix}\check A_{11} & \check A_{12} \cr \check A_{21} & \check A_{22} \end{bmatrix}$
Let $$H^0_0 \equiv - P_{22}^{-1} P_{21}$$ so that $$\check x_0 = H^0_0 \check z_0$$.
Then iterations on $$\check y_{t+1} = \check A \check y_t$$ starting from initial condition $$\check y_0 = \begin{bmatrix}\check z_0 \cr H^0_0 \check z_0\end{bmatrix}$$ imply that for $$t \geq 1$$
$x_t = \sum_{j=1}^t H_j^t \check z_{t-j}$
where
\begin{aligned} H^t_1 & = \check A_{21} \cr H^t_2 & = \check A_{22} \check A_{21} \cr \ \ \vdots \ \ & \ \ \quad \vdots \cr H^t_{t-1} & = \check A_{22}^{t-2} \check A_{21} \cr H^t_t & = \check A_{22}^{t-1}(\check A_{21} + \check A_{22} H^0_0 ) \end{aligned}
An optimal decision rule for the Stackelberg’s choice of $$u_t$$ is
$u_t = - F \check y_t \equiv - \begin{bmatrix} F_z & F_x \cr \end{bmatrix} \begin{bmatrix}\check z_t \cr x_t \cr \end{bmatrix}$
or
(40.10)$u_t = - F_z \check z_t - F_x \sum_{j=1}^t H^t_j z_{t-j} = \sigma_t(\check z^t)$
Representation (40.10) confirms that whenever $$F_x \neq 0$$, the typical situation, the time $$t$$ component $$\sigma_t$$ of a Stackelberg plan is history-dependent, meaning that the Stackelberg leader’s choice $$u_t$$ depends not just on $$\check z_t$$ but on components of $$\check z^{t-1}$$.
### 40.6.1. Comments and Interpretations¶
After all, at the end of the day, it will turn out that because we set $$\check z_0 = z_0$$, it will be true that $$z_t = \check z_t$$ for all $$t \geq 0$$.
Then why did we distinguish $$\check z_t$$ from $$z_t$$?
The answer is that if we want to present to the Stackelberg follower a history-dependent representation of the Stackelberg leader’s sequence $$\vec q_2$$, we must use representation (40.10) cast in terms of the history $$\check z^t$$ and not a corresponding representation cast in terms of $$z^t$$.
## 40.7. Dynamic Programming and Time Consistency of Follower’s Problem¶
Given the sequence $$\vec q_2$$ chosen by the Stackelberg leader in our duopoly model, it turns out that the Stackelberg follower’s problem is recursive in the natural state variables that confront a follower at any time $$t \geq 0$$.
This means that the follower’s plan is time consistent.
To verify these claims, we’ll formulate a recursive version of a follower’s problem that builds on our recursive representation of the Stackelberg leader’s plan and our use of the Big K, little k idea.
### 40.7.1. Recursive Formulation of a Follower’s Problem¶
We now use what amounts to another “Big $$K$$, little $$k$$” trick (see rational expectations equilibrium) to formulate a recursive version of a follower’s problem cast in terms of an ordinary Bellman equation.
Firm 1, the follower, faces $$\{q_{2t}\}_{t=0}^\infty$$ as a given quantity sequence chosen by the leader and believes that its output price at $$t$$ satisfies
$p_t = a_0 - a_1 ( q_{1t} + q_{2t}) , \quad t \geq 0$
Our challenge is to represent $$\{q_{2t}\}_{t=0}^\infty$$ as a given sequence.
To do so, recall that under the Stackelberg plan, firm 2 sets output according to the $$q_{2t}$$ component of
$y_{t+1} = \begin{bmatrix} 1 \cr q_{2t} \cr q_{1t} \cr x_t \end{bmatrix}$
which is governed by
$y_{t+1} = (A - BF) y_t$
To obtain a recursive representation of a $$\{q_{2t}\}$$ sequence that is exogenous to firm 1, we define a state $$\tilde y_t$$
$\tilde y_t = \begin{bmatrix} 1 \cr q_{2t} \cr \tilde q_{1t} \cr \tilde x_t \end{bmatrix}$
that evolves according to
$\tilde y_{t+1} = (A - BF) \tilde y_t$
subject to the initial condition $$\tilde q_{10} = q_{10}$$ and $$\tilde x_0 = x_0$$ where $$x_0 = - P_{22}^{-1} P_{21}$$ as stated above.
Firm 1’s state vector is
$X_t = \begin{bmatrix} \tilde y_t \cr q_{1t} \end{bmatrix}$
It follows that the follower firm 1 faces law of motion
(40.11)$\begin{split}\begin{bmatrix} \tilde y_{t+1} \\ q_{1t+1} \end{bmatrix} = \begin{bmatrix} A - BF & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \tilde y_{t} \\ q_{1t} \end{bmatrix} + \begin{bmatrix} 0 \cr 1 \end{bmatrix} x_t\end{split}$
This specification assures that from the point of the view of a firm 1, $$q_{2t}$$ is an exogenous process.
Here
• $$\tilde q_{1t}, \tilde x_t$$ play the role of Big K
• $$q_{1t}, x_t$$ play the role of little k
The time $$t$$ component of firm 1’s objective is
$\tilde X_t' \tilde R x_t - x_t^2 \tilde Q = \begin{bmatrix} 1 \cr q_{2t} \cr \tilde q_{1t} \cr \tilde x_t \cr q_{1t} \end{bmatrix}' \begin{bmatrix} 0 & 0 & 0 & 0 & \frac{a_0}{2} \cr 0 & 0 & 0 & 0 & - \frac{a_1}{2} \cr 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 \cr \frac{a_0}{2} & -\frac{a_1}{2} & 0 & 0 & - a_1 \end{bmatrix} \begin{bmatrix} 1 \cr q_{2t} \cr \tilde q_{1t} \cr \tilde x_t \cr q_{1t} \end{bmatrix} - \gamma x_t^2$
Firm 1’s optimal decision rule is
$x_t = - \tilde F X_t$
and it’s state evolves according to
$\tilde X_{t+1} = (\tilde A - \tilde B \tilde F) X_t$
under its optimal decision rule.
Later we shall compute $$\tilde F$$ and verify that when we set
$X_0 = \begin{bmatrix} 1 \cr q_{20} \cr q_{10} \cr x_0 \cr q_{10} \end{bmatrix}$
we recover
$x_0 = - \tilde F \tilde X_0$
which will verify that we have properly set up a recursive representation of the follower’s problem facing the Stackelberg leader’s $$\vec q_2$$.
### 40.7.2. Time Consistency of Follower’s Plan¶
Since the follower can solve its problem using dynamic programming its problem is recursive in what for it are the natural state variables, namely
$\begin{bmatrix} 1 \cr q_{2t} \cr \tilde q_{10} \cr \tilde x_0 \end{bmatrix}$
It follows that the follower’s plan is time consistent.
## 40.8. Computing Stackelberg Plan¶
Here is our code to compute a Stackelberg plan via a linear-quadratic dynamic program as outlined above
# Parameters
a0 = 10
a1 = 2
β = 0.96
γ = 120
n = 300
tol0 = 1e-8
tol1 = 1e-16
tol2 = 1e-2
βs = np.ones(n)
βs[1:] = β
βs = βs.cumprod()
# In LQ form
Alhs = np.eye(4)
# Euler equation coefficients
Alhs[3, :] = β * a0 / (2 * γ), -β * a1 / (2 * γ), -β * a1 / γ, β
Arhs = np.eye(4)
Arhs[2, 3] = 1
Alhsinv = la.inv(Alhs)
A = Alhsinv @ Arhs
B = Alhsinv @ np.array([[0, 1, 0, 0]]).T
R = np.array([[0, -a0 / 2, 0, 0],
[-a0 / 2, a1, a1 / 2, 0],
[0, a1 / 2, 0, 0],
[0, 0, 0, 0]])
Q = np.array([[γ]])
# Solve using QE's LQ class
# LQ solves minimization problems which is why the sign of R and Q was changed
lq = LQ(Q, R, A, B, beta=β)
P, F, d = lq.stationary_values(method='doubling')
P22 = P[3:, 3:]
P21 = P[3:, :3]
P22inv = la.inv(P22)
H_0_0 = -P22inv @ P21
# Simulate forward
z0 = np.array([[1, 1, 1]]).T
x0 = H_0_0 @ z0
y0 = np.vstack((z0, x0))
yt, ut = lq.compute_sequence(y0, ts_length=n)[:2]
π_matrix = (R + F. T @ Q @ F)
for t in range(n):
π_leader[t] = -(yt[:, t].T @ π_matrix @ yt[:, t])
# Display policies
print("Computed policy for Stackelberg leader\n")
print(f"F = {F}")
Computed policy for Stackelberg leader
F = [[-1.58004454 0.29461313 0.67480938 6.53970594]]
## 40.9. Time Series for Price and Quantities¶
The following code plots the price and quantities
q_leader = yt[1, :-1]
q_follower = yt[2, :-1]
q = q_leader + q_follower # Total output, Stackelberg
p = a0 - a1 * q # Price, Stackelberg
fig, ax = plt.subplots(figsize=(9, 5.8))
ax.plot(range(n), q_follower, 'r-', lw=2, label='follower output')
ax.plot(range(n), p, 'g-', lw=2, label='price')
ax.set_title('Output and prices, Stackelberg duopoly')
ax.legend(frameon=False)
ax.set_xlabel('t')
plt.show()
### 40.9.1. Value of Stackelberg Leader¶
We’ll compute the present value earned by the Stackelberg leader.
We’ll compute it two ways (they give identical answers – just a check on coding and thinking)
v_leader_forward = np.sum(βs * π_leader)
v_leader_direct = -yt[:, 0].T @ P @ yt[:, 0]
# Display values
print("Computed values for the Stackelberg leader at t=0:\n")
Computed values for the Stackelberg leader at t=0:
v_leader_forward(forward sim) = 150.0316
v_leader_direct (direct) = 150.0324
# Manually checks whether P is approximately a fixed point
P_next = (R + F.T @ Q @ F + β * (A - B @ F).T @ P @ (A - B @ F))
(P - P_next < tol0).all()
True
# Manually checks whether two different ways of computing the
# value function give approximately the same answer
v_expanded = -((y0.T @ R @ y0 + ut[:, 0].T @ Q @ ut[:, 0] +
β * (y0.T @ (A - B @ F).T @ P @ (A - B @ F) @ y0)))
(v_leader_direct - v_expanded < tol0)[0, 0]
True
## 40.10. Time Inconsistency of Stackelberg Plan¶
In the code below we compare two values
• the continuation value $$- y_t P y_t$$ earned by a continuation Stackelberg leader who inherits state $$y_t$$ at $$t$$
• the value of a reborn Stackelberg leader who inherits state $$z_t$$ at $$t$$ and sets $$x_t = - P_{22}^{-1} P_{21}$$
The difference between these two values is a tell-tale sign of the time inconsistency of the Stackelberg plan
# Compute value function over time with a reset at time t
yt_reset = yt.copy()
yt_reset[-1, :] = (H_0_0 @ yt[:3, :])
for t in range(n):
vt_leader[t] = -yt[:, t].T @ P @ yt[:, t]
vt_reset_leader[t] = -yt_reset[:, t].T @ P @ yt_reset[:, t]
fig, axes = plt.subplots(3, 1, figsize=(10, 7))
axes[0].plot(range(n+1), (- F @ yt).flatten(), 'bo',
axes[0].plot(range(n+1), (- F @ yt_reset).flatten(), 'ro',
label='continuation leader at t', ms=2)
axes[0].set(title=r'Leader control variable $u_{t}$', xlabel='t')
axes[0].legend()
axes[1].plot(range(n+1), yt[3, :], 'bo', ms=2)
axes[1].plot(range(n+1), yt_reset[3, :], 'ro', ms=2)
axes[1].set(title=r'Follower control variable $x_{t}$', xlabel='t')
axes[2].plot(range(n), vt_leader, 'bo', ms=2)
axes[2].plot(range(n), vt_reset_leader, 'ro', ms=2)
axes[2].set(title=r'Leader value function $v(y_{t})$', xlabel='t')
plt.tight_layout()
plt.show()
## 40.11. Recursive Formulation of Follower’s Problem¶
We now formulate and compute the recursive version of the follower’s problem.
We check that the recursive Big $$K$$ , little $$k$$ formulation of the follower’s problem produces the same output path $$\vec q_1$$ that we computed when we solved the Stackelberg problem
A_tilde = np.eye(5)
A_tilde[:4, :4] = A - B @ F
R_tilde = np.array([[0, 0, 0, 0, -a0 / 2],
[0, 0, 0, 0, a1 / 2],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[-a0 / 2, a1 / 2, 0, 0, a1]])
Q_tilde = Q
B_tilde = np.array([[0, 0, 0, 0, 1]]).T
lq_tilde = LQ(Q_tilde, R_tilde, A_tilde, B_tilde, beta=β)
P_tilde, F_tilde, d_tilde = lq_tilde.stationary_values(method='doubling')
y0_tilde = np.vstack((y0, y0[2]))
yt_tilde = lq_tilde.compute_sequence(y0_tilde, ts_length=n)[0]
# Checks that the recursive formulation of the follower's problem gives
# the same solution as the original Stackelberg problem
fig, ax = plt.subplots()
ax.plot(yt_tilde[4], 'r', label="q_tilde")
ax.plot(yt_tilde[2], 'b', label="q")
ax.legend()
plt.show()
Note: Variables with _tilde are obtained from solving the follower’s problem – those without are from the Stackelberg problem
# Maximum absolute difference in quantities over time between
# the first and second solution methods
np.max(np.abs(yt_tilde[4] - yt_tilde[2]))
6.661338147750939e-16
# x0 == x0_tilde
yt[:, 0][-1] - (yt_tilde[:, 1] - yt_tilde[:, 0])[-1] < tol0
True
### 40.11.1. Explanation of Alignment¶
If we inspect the coefficients in the decision rule $$- \tilde F$$, we can spot the reason that the follower chooses to set $$x_t = \tilde x_t$$ when it sets $$x_t = - \tilde F X_t$$ in the recursive formulation of the follower problem.
Can you spot what features of $$\tilde F$$ imply this?
Hint: remember the components of $$X_t$$
# Policy function in the follower's problem
F_tilde.round(4)
array([[-0. , 0. , -0.1032, -1. , 0.1032]])
# Value function in the Stackelberg problem
P
array([[ 963.54083615, -194.60534465, -511.62197962, -5258.22585724],
[ -194.60534465, 37.3535753 , 81.97712513, 784.76471234],
[ -511.62197962, 81.97712513, 247.34333344, 2517.05126111],
[-5258.22585724, 784.76471234, 2517.05126111, 25556.16504097]])
# Value function in the follower's problem
P_tilde
array([[-1.81991134e+01, 2.58003020e+00, 1.56048755e+01,
1.51229815e+02, -5.00000000e+00],
[ 2.58003020e+00, -9.69465925e-01, -5.26007958e+00,
-5.09764310e+01, 1.00000000e+00],
[ 1.56048755e+01, -5.26007958e+00, -3.22759027e+01,
-3.12791908e+02, -1.23823802e+01],
[ 1.51229815e+02, -5.09764310e+01, -3.12791908e+02,
-3.03132584e+03, -1.20000000e+02],
[-5.00000000e+00, 1.00000000e+00, -1.23823802e+01,
-1.20000000e+02, 1.43823802e+01]])
# Manually check that P is an approximate fixed point
(P - ((R + F.T @ Q @ F) + β * (A - B @ F).T @ P @ (A - B @ F)) < tol0).all()
True
# Compute P_guess using F_tilde_star
F_tilde_star = -np.array([[0, 0, 0, 1, 0]])
P_guess = np.zeros((5, 5))
for i in range(1000):
P_guess = ((R_tilde + F_tilde_star.T @ Q @ F_tilde_star) +
β * (A_tilde - B_tilde @ F_tilde_star).T @ P_guess
@ (A_tilde - B_tilde @ F_tilde_star))
# Value function in the follower's problem
-(y0_tilde.T @ P_tilde @ y0_tilde)[0, 0]
112.65590740578058
# Value function with P_guess
-(y0_tilde.T @ P_guess @ y0_tilde)[0, 0]
112.6559074057807
# Compute policy using policy iteration algorithm
F_iter = (β * la.inv(Q + β * B_tilde.T @ P_guess @ B_tilde)
@ B_tilde.T @ P_guess @ A_tilde)
for i in range(100):
# Compute P_iter
P_iter = np.zeros((5, 5))
for j in range(1000):
P_iter = ((R_tilde + F_iter.T @ Q @ F_iter) + β
* (A_tilde - B_tilde @ F_iter).T @ P_iter
@ (A_tilde - B_tilde @ F_iter))
# Update F_iter
F_iter = (β * la.inv(Q + β * B_tilde.T @ P_iter @ B_tilde)
@ B_tilde.T @ P_iter @ A_tilde)
dist_vec = (P_iter - ((R_tilde + F_iter.T @ Q @ F_iter)
+ β * (A_tilde - B_tilde @ F_iter).T @ P_iter
@ (A_tilde - B_tilde @ F_iter)))
if np.max(np.abs(dist_vec)) < 1e-8:
dist_vec2 = (F_iter - (β * la.inv(Q + β * B_tilde.T @ P_iter @ B_tilde)
@ B_tilde.T @ P_iter @ A_tilde))
if np.max(np.abs(dist_vec2)) < 1e-8:
F_iter
else:
print("The policy didn't converge: try increasing the number of \
outer loop iterations")
else:
print("P_iter didn't converge: try increasing the number of inner \
loop iterations")
# Simulate the system using F_tilde_star and check that it gives the
# same result as the original solution
yt_tilde_star = np.zeros((n, 5))
yt_tilde_star[0, :] = y0_tilde.flatten()
for t in range(n-1):
yt_tilde_star[t+1, :] = (A_tilde - B_tilde @ F_tilde_star) \
@ yt_tilde_star[t, :]
fig, ax = plt.subplots()
ax.plot(yt_tilde_star[:, 4], 'r', label="q_tilde")
ax.plot(yt_tilde[2], 'b', label="q")
ax.legend()
plt.show()
# Maximum absolute difference
np.max(np.abs(yt_tilde_star[:, 4] - yt_tilde[2, :-1]))
0.0
## 40.12. Markov Perfect Equilibrium¶
The state vector is
$z_t = \begin{bmatrix} 1 \cr q_{2t} \cr q_{1t} \end{bmatrix}$
and the state transition dynamics are
$z_{t+1} = A z_t + B_1 v_{1t} + B_2 v_{2t}$
where $$A$$ is a $$3 \times 3$$ identity matrix and
$B_1 = \begin{bmatrix} 0 \cr 0 \cr 1 \end{bmatrix} , \quad B_2 = \begin{bmatrix} 0 \cr 1 \cr 0 \end{bmatrix}$
The Markov perfect decision rules are
$v_{1t} = - F_1 z_t , \quad v_{2t} = - F_2 z_t$
and in the Markov perfect equilibrium, the state evolves according to
$z_{t+1} = (A - B_1 F_1 - B_2 F_2) z_t$
# In LQ form
A = np.eye(3)
B1 = np.array([[0], [0], [1]])
B2 = np.array([[0], [1], [0]])
R1 = np.array([[0, 0, -a0 / 2],
[0, 0, a1 / 2],
[-a0 / 2, a1 / 2, a1]])
R2 = np.array([[0, -a0 / 2, 0],
[-a0 / 2, a1, a1 / 2],
[0, a1 / 2, 0]])
Q1 = Q2 = γ
S1 = S2 = W1 = W2 = M1 = M2 = 0.0
# Solve using QE's nnash function
F1, F2, P1, P2 = qe.nnash(A, B1, B2, R1, R2, Q1,
Q2, S1, S2, W1, W2, M1,
M2, beta=β, tol=tol1)
# Simulate forward
AF = A - B1 @ F1 - B2 @ F2
z = np.empty((3, n))
z[:, 0] = 1, 1, 1
for t in range(n-1):
z[:, t+1] = AF @ z[:, t]
# Display policies
print("Computed policies for firm 1 and firm 2:\n")
print(f"F1 = {F1}")
print(f"F2 = {F2}")
Computed policies for firm 1 and firm 2:
F1 = [[-0.22701363 0.03129874 0.09447113]]
F2 = [[-0.22701363 0.09447113 0.03129874]]
q1 = z[1, :]
q2 = z[2, :]
q = q1 + q2 # Total output, MPE
p = a0 - a1 * q # Price, MPE
fig, ax = plt.subplots(figsize=(9, 5.8))
ax.plot(range(n), q, 'b-', lw=2, label='total output')
ax.plot(range(n), p, 'g-', lw=2, label='price')
ax.set_title('Output and prices, duopoly MPE')
ax.legend(frameon=False)
ax.set_xlabel('t')
plt.show()
# Computes the maximum difference between the two quantities of the two firms
np.max(np.abs(q1 - q2))
7.771561172376096e-15
# Compute values
u1 = (- F1 @ z).flatten()
u2 = (- F2 @ z).flatten()
π_1 = p * q1 - γ * (u1) ** 2
π_2 = p * q2 - γ * (u2) ** 2
v1_forward = np.sum(βs * π_1)
v2_forward = np.sum(βs * π_2)
v1_direct = (- z[:, 0].T @ P1 @ z[:, 0])
v2_direct = (- z[:, 0].T @ P2 @ z[:, 0])
# Display values
print("Computed values for firm 1 and firm 2:\n")
print(f"v1(forward sim) = {v1_forward:.4f}; v1 (direct) = {v1_direct:.4f}")
print(f"v2 (forward sim) = {v2_forward:.4f}; v2 (direct) = {v2_direct:.4f}")
Computed values for firm 1 and firm 2:
v1(forward sim) = 133.3303; v1 (direct) = 133.3296
v2 (forward sim) = 133.3303; v2 (direct) = 133.3296
# Sanity check
Λ1 = A - B2 @ F2
lq1 = qe.LQ(Q1, R1, Λ1, B1, beta=β)
P1_ih, F1_ih, d = lq1.stationary_values()
v2_direct_alt = - z[:, 0].T @ lq1.P @ z[:, 0] + lq1.d
(np.abs(v2_direct - v2_direct_alt) < tol2).all()
True
## 40.13. Comparing Markov Perfect Equilibrium and Stackelberg Outcome¶
It is enlightening to compare equilbrium quantities for firms 1 and 2 under two alternative settings:
• A Markov perfect equilibrium like that described in this lecture
• A Stackelberg equilbrium
The following code performs the required computations.
vt_MPE = np.zeros(n)
vt_follower = np.zeros(n)
for t in range(n):
vt_MPE[t] = -z[:, t].T @ P1 @ z[:, t]
vt_follower[t] = -yt_tilde[:, t].T @ P_tilde @ yt_tilde[:, t]
fig, ax = plt.subplots()
ax.plot(vt_MPE, 'b', label='MPE')
ax.plot(vt_follower, 'g', label='Stackelberg follower')
ax.set_title(r'MPE vs. Stackelberg Value Function')
ax.set_xlabel('t')
ax.legend(loc=(1.05, 0))
plt.show()
# Display values
print("Computed values:\n")
print(f"vt_follower(y0) = {vt_follower[0]:.4f}")
print(f"vt_MPE(y0) = {vt_MPE[0]:.4f}")
Computed values:
# Compute the difference in total value between the Stackelberg and the MPE
-3.970942562087714 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.8287011981010437, "perplexity": 4470.711298827244}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00699.warc.gz"} |
http://mathhelpforum.com/calculus/189928-limit-help.html | 1. ## limit help
hi
how can I evaluate this limit, using the epsilon delta way if possible. I read my textbook about the method but didn't really understand.
lim
(x,y) -> (0,0) of [x-xy+3]/[x^2y+5xy-y^3]
plugging in the point you can see its discontinuous.
so if I move along the x axis in the end I will get x-3/0 and along the y axis I will get -3/y^3, in the end both of those will be -3/0. If I move along the line y = x I also get the same answer. So I can say the limit does nit exist? Now how can I show this using the epsilon delta method for limits?
2. ## Re: limit help
Firstly: taking into account that $\lim_{(x,y)\to (0,0)}(x-xy-3)=-3$ and $\lim_{(x,y)\to (0,0)}(x^2+5xy-y^3)=0$, by a well-known theorem $\lim_{(x,y)\to (0,0)}\frac{x-xy-3}{x^2+5xy-y^3}=\infty$ . Then, you can use an "indecent trick": transport the arguments $\epsilon-\delta$ of that theorem to your problem.
3. ## Re: limit help
hi. Thanks for the reply but I'm not sure if i still understand fully. This theorem, is it the limit of a rational function is the limit of the numerator over the denominator? I am also unsure about this trick I can use to apply the epsilon delta method to my problem.
thanks
4. ## Re: limit help
Originally Posted by Kuma
This theorem, is it the limit of a rational function is the limit of the numerator over the denominator?
In general,
$\lim_{(x,y)\to (x_0,y_0)}f(x,y)=k\neq 0,\;\wedge\; \lim_{(x,y)\to (x_0,y_0)}g(x,y)= 0\Rightarrow$
$\lim_{(x,y)\to (x_0,y_0)}\frac{f(x,y)}{g(x,y)}=\infty$
I mean $\infty$ not $+\infty$.
I am also unsure about this trick I can use to apply the epsilon delta method to my problem.
Could you transcribe from your book the exact formulation of the problem? That is, does the problem say to find out the limit using the definition?
5. ## Re: limit help
Originally Posted by FernandoRevilla
Could you transcribe from your book the exact formulation of the problem? That is, does the problem say to find out the limit using the definition?
It does not, but I thought it would be the way to start since my book only explains evaluating limits through that method.
However your explanation helped! So I can conclude that the limit goes to infinity.
6. ## Re: limit help
Originally Posted by Kuma
It does not, but I thought it would be the way to start since my book only explains evaluating limits through that method.
That makes no sense. The "epsilon-delta" is a way of proving the limit after you have found it, not a way of evaluating limits.
However your explanation helped! So I can conclude that the limit goes to infinity.
More correctly, the limit is infinity (or simply does not exist). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9747326374053955, "perplexity": 384.3148519000738}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982292697.31/warc/CC-MAIN-20160823195812-00200-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://www.javatpoint.com/latex-formats | LATEX FORMATS
Formatting Arguments
The formatting arguments are used to provide a unique look to your file or document. There are various methods to format the document, which are listed below:
1. Optional arguments
The popular optional arguments are:
• twocolumn - it is 2 column pages
• titlepage- \maketitle generates a title page
• openright- the chapter begins on the right-hand page, if two side is used
• landscape- used to display in landscape
• legno- it puts the equation number on the left side
• flegn- left align equations versus center
• twoside- it prints on both sides of the paper.
a) Class
The documents types used here are articles in presentation and scientific journals, report, book, slides, a0poster proc (a class based on article class), letters, beamer (for writing presentations), etc.
The class is used at the beginning of the program, {article} is the commonly used class.
The document class defines the overall layout of the document.
The other classes are also enclosed in the curly brackets, which are given below:
• report- it is used for chapter containing longer documents.
• thesis- for preparing a thesis with Latex.
• letter- for letters writing
• books- for the books
• slides- for the transparencies.
b) Groups
To keep the scope limited and the state to local, the particular part can be enclosed locally in the curly braces. But in some cases, if braces are not possible, then you can use \bgroup and \egroup to begin and end a group respectively.
For example, normal text {xyz} more normal text
This can be written as normal text \bgroup xyz \egroup {} more normal text.
c) Packages
It is a method which provides or adds extra formatting features to the document such as pictures, bibliography and links. It also adds new functions to the Latex. All the packages must be included in the preamble only. To load a package, the command used is:
It can also be used as \usepackage[options]{package}
Some of the commonly used packages are:
• color: provides a way to use colors.
• rotating: it is used for the rotation especially figures and tables.
• fancyhdr: it customizes headers and footers.
• graphicx: it is the command used to include graphics.
• setspace: it is an easy way to change the linespacing.
2. Structure formats
• Title
The title is used in most of the paper formats like a letter, articles, report, book class, etc. To create a title, you need to specify the text for the author, title, and date, then with the use of \maketitle command, the title page is generated by the Latex. The \begin{document} must be used at the starting. The command used to make the title is \maketitle. The code for creating the title is given below:
The output is shown in the below image:
If you want no date at all, you can use the \date{} command.
To display the current date, you can use the \date{\today} command.
• Sections
You can divide your document into sections, sub-sections. The commands used to create the sections are:
• \section{..}
• \subsection{..}
• \subsubsection{..}
• \paragraph{..}
• \subparagraph{..}
These are the popular commands for the articles class. You can also use \chapter{..} command for the books and the report class.
The code for the following section is given below:
The output is shown in the below image:
• Labeling
To refer to other parts of the document, you can label any of the section commands. You can label the section with \label{labelname}.
If you want to refer to the section or page number of the label; you can use the \ref{labelname} and \pageref{labelname}.
The code is shown below:
Output
For the multiple pages or the file, you can easily refer to the page number and the section according to the requirements.
• Page Numbering
This format is used to number the pages before writing the document. It also ensures that the main document starts on page number 1. The page numbering can be switched between roman and Arabic. The command used to number the pages is \pagenumbering{..}, which is declared after the \maketitle command.
The code for the above method is given below:
The output is given in the below image:
appears in the head or the foot sections. The output screen consists of a head, body, and foot. The page number is printed by default by all the classes except the letter.
To modify the default actions, you can use the \pagestyle command, which is placed just after the \chapter or the command similar to this.
The types of standard page styles are given below:
• \pagestyle{plain}
It is the default style for the report classes and the articles. The page number is listed on the foot while the head is empty.
In this style, the foot is empty. The section heading and the page number is put in the head.
To specify the information used in the head; the following commands are used.
\markboth is used for a two-sided file or document, while \markright is used for a one-sided document.
• \pagestyle{empty}
In this style, both the header and foot are empty.
• \thispagestyle{style}
It is used to change the style for the current page only. For example, \thispagestyle{empty} is used to have nothing in the foot and head. This command will be executed only for the current page without affecting other pages.
If you want to format or customize your header and footers in another way. You can use the fancyhdr command. This command in Latex is used as \usepackage{fancyhdr}. The package used here can align your header and footers to the left, right, and the center. You can also define multi-line headers and filters, separate headers used for the even and odd pages, etc.
\pagestyle{fancy} is also used here.
The following commands can be included in the {} ( curly brackets): \lhead{}, \chead{}, \rhead{}, \lfoot{}, \cfoot{}, \rfoot{}.
If you want to suppress the horizontal line under the header, which is drawn by default, you can use the \renewcommand{\headrulewidth}{0pt} command.
g. Margins
The commands are used to set the margins manually. The margins internally are considered 1 inch by default. Hence, if you want to set a margin of 0.5 inches, you have to mention that margin -0.5 inches. You can also set the height and width of the text area by using the textheight and textwidth commands.
The available commands to set margins are given below in the table:
Margins command
left margin
(for odd pages if you are using single-sided)
\oddsidemargin
left margin
(for the even pages if you are using double-sided)
\evensidemargin
right margin \textwidth
top margin \topmargin
bottom margin \textheight
The commands used to set the inches for margins are:
• To set the top margin 0.7 inch, \setlength{\topmargin}{-0.7in}
• To set margin 1 inch on right pages, \setlength{\oddsidemargin}{0in}
• To set margin 1 inch on left pages in two-sided document, \setlength{\evensidemargin}{0in}
• To leave the space for right margin, \setlength{\textwidth}{0in}
• To reserve 9 inches for the text , \setlength{\textheight}{9in}
h. Paragraph
To start a paragraph, you can use the control sequence \par or can either leave a blank space. The indent of the paragraph is 1.5em by default (1.5 times the point size of the current font). There is no extra space inserted between the paragraphs. To control the paragraph separation and indention, the commands \parindent and \parskip are used. Let-s consider an example below:
This example contains a simple paragraph. The code is given below:
The output is given in the image below:
This example contains a paragraph with the paragraph separation and indention commands. The code is given below:
This example contains a simple paragraph. The code is given below:
The output is given in the image below:
You can notice the difference between the above two examples.
i. References
It is also called as cross-references used for sections, figures, equations, or tables. The commands used are given below:
• \pageref{marker} it includes the page number related to the \label command.
• \label{marker} is used to set the marker for future references.
• \ref{marker} consists of the number of sections, figures, etc. related to the \label command.
3. Drawing rules
If you want to draw the horizontal or vertical line on the page of your document, you can use the \rule command. In Latex, it will be written as:
The lift is the optional parameter, which is the amount raised above the baseline. The width is the horizontal dimension, while height is the vertical dimension. For example, if you want to draw the line given below:
You can use the \rule{\textwidth}{2pt} command.
The code or program for the above example is mentioned below:
The output is shown in the image below:
You can draw many lines according to the requirements.
For example, to draw the two lines, the code is given below:
The output for the above example, is shown in the image below:
• Footnotes
The Latex numbers the footnotes automatically. The command \footnote is placed exactly where you want the footnote to be appeared. There should be no extra space between the \footnote command and the text before it. The command is written as \footnote{the note text is written here}.
The text inside the curly brackets is displayed at the bottom. For example, consider the code below:
The output is shown in the image below:
The footnotes will appear at the bottom. For the above example, the footnotes are shown in the image below:
• Centering
If you have only one line to place in the center; then you can use the \centerline command. For example,
\centerline{ the line mentioned here will be displayed in the center}.
Let-s consider an example below:
The output is shown in the image below:
If you have several lines to display, then you can write in the following way.
• Quotations
The quotes can be used in the environment with:
In the paragraphs, there is no blank line separation and intent between the quotes in a section.
Next TopicLatex Table | {"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": 0.9394309520721436, "perplexity": 1882.4810933844565}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144498.68/warc/CC-MAIN-20200220005045-20200220035045-00474.warc.gz"} |
http://mathhelpforum.com/calculus/206303-limit-infinity-print.html | # Limit to infinity
• Oct 29th 2012, 03:58 AM
PhizKid
Limit to infinity
$\lim_{x \to \infty} \frac{\sqrt{9x^6 - x}}{x^3 + 1}\\
\frac{\sqrt{9x^6 - x}}{x^3 + 1} \cdot \frac{\frac{1}{x^3}}{\frac{1}{x^3}} = \\
\frac{\frac{\sqrt{9x^6 - x}}{x^3}}{1 + \frac{1}{x^3}} = \\
\frac{(1 + \frac{1}{x^3})(\sqrt{9x^6 - x})}{x^3} = \\
\frac{(1 + \frac{1}{x^3})(\sqrt{9x^6 - x})}{x^3} \cdot \frac{\frac{1}{x^3}}{\frac{1}{x^3}} = \\
\frac{(1 + \frac{1}{x^3})(\sqrt{9x^6 - x})}{x^3}$
And it just repeats over and over again and I can't find anything to divide by without destroying the work I've already done. What am I supposed to do in a loop and there's nothing to divide by?
• Oct 29th 2012, 04:26 AM
a tutor
Re: Limit to infinity
Take your denominator into the square root $\sqrt{\frac{9x^6-x}{(x^3+1)^2}}$.
You may see that the numerator and denominator both have 6 as their highest power of x.
Divide top and bottom by x^6. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9365233182907104, "perplexity": 592.1616709768556}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917119356.19/warc/CC-MAIN-20170423031159-00247-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://clay6.com/qa/21939/two-metal-balls-of-same-radii-are-charged-to-5-and-25-units-of-electricity- | Two metal balls of same radii are charged to 5 and -25 units of electricity . They are brought in contact with each other and then again seperated to original distance. Ratio of magnitude of force b/w two balls after and before contact is
$(A)\;5:4\\ (B)\;4:5 \\ (C)\; 1:5 \\ (D)\;5:1$
$F_{\large before }=\large\frac{1}{4 \pi \in _0} \frac{(5) \times (-25)}{r^2}$
Charge after $=q_1=q_2=\large\frac{5+ (-25)}{2}$
$\qquad=-10$
$F_{\large after }=\large\frac{1}{4 \pi \in _0} \frac{(-10) \times (-10)}{2}$
$\large\frac{|F_{\Large after}|}{|F_{\Large before}|}=\bigg| \large\frac{(-10) \times (-10)}{(5) \times (-25)}\bigg|$
$\qquad= \large\frac{4}{5}$
$\qquad=4:5$
Hence B is the correct answer. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8642146587371826, "perplexity": 562.8002002120328}, "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/1516084887600.12/warc/CC-MAIN-20180118190921-20180118210921-00661.warc.gz"} |
http://opencyto.org/reference/prior_flowclust.html | We elicit data-driven prior parameters from a flowSet object for specified channels. For each sample in the flowSet object, we apply the given prior_method to elicit the priors parameters.
prior_flowclust(
flow_set,
channels,
prior_method = c("kmeans"),
K = 2,
nu0 = 4,
w0 = c(10, 10),
shrink = 1e-06,
...
)
## Arguments
flow_set a flowSet object a character vector containing the channels in the flowSet from which we elicit the prior parameters for the Student's t mixture the method to elicit the prior parameters the number of mixture components to identify prior degrees of freedom of the Student's t mixture components. the number of prior pseudocounts of the Student's t mixture components. (only the first element is used and the rest is ignored at the moment) the amount of eigenvalue shrinkage to add in the case the prior covariance matrices are singular. See details. Additional arguments passed to the prior elicitation method selected
## Value
list of the necessary prior parameters
## Details
Currently, we have implemented only two methods. In the case that one channel is given, we use the kernel-density estimator (KDE) approach for each sample to obtain K peaks from which we elicit prior parameters. Otherwise, if more than one channel is specified, we apply K-Means to each of the samples in the flowSet and aggregate the clusters to elicit the prior parameters.
In the rare case that a prior covariance matrix is singular, we shrink the eigenvalues of the matrix slightly to ensure that it is positive definite. For instance, if the flow_set has two samples, this case can occur. The amount of shrinkage is controlled in shrink. | {"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": 0.8061817288398743, "perplexity": 1176.1256419788656}, "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/1656103034930.3/warc/CC-MAIN-20220625095705-20220625125705-00608.warc.gz"} |
https://byjus.com/question-answer/a-thin-hollow-sphere-of-mass-m-is-completely-filled-with-non-viscous-liquid-of/ | Question
# A thin hollow sphere of mass m is completely filled with non-viscous liquid of mass m. When the sphere rolls on horizontal ground such that centre moves with velocity v, kinetic energy of the system is equal to :
A
mv2
B
43mv2
C
45mv2
D
None of these
Solution
## The correct options are B 43mv2Non-viscous liquid has only translational kinetic energy. ∴ KEsystem= (Translational + Rotational) K.E of hollow sphere + Translational K.E of liquid =12mv2+12×I(ω)2+12mv2 For pure rolling v=ωR ∴ KEsystem=12mv2+12×23mR2(vR)2+12mv2=43mv2 Physics
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View More | {"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": 0.8492072224617004, "perplexity": 2168.432838195607}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300849.28/warc/CC-MAIN-20220118122602-20220118152602-00482.warc.gz"} |
http://mathoverflow.net/questions/153887/can-aperiodic-tilings-be-non-hierarchical-and-confusion-over-domino-problem | # Can aperiodic tilings be non-hierarchical? and confusion over domino problem
Anyone experienced with the undecidability of aperiodic tiling?
It's related to the halting problem which Turing proved was undecidable in the 30's and basically superimposes tiles onto other tiles which can be used to construct a Turing machine which would tile the plane aperiodically.
The proof given in Robinson (see below) shows that because you can construct such a machine the tiling problem is equal to the halting problem and therefore undecidable. I follow this through the paper and feel I understand the conclusion.
The problem is my girlfriend says this is all fine but only works for hierarchical tilings. I say well all aperiodic tiles are hierarchical and she says prove it.
Problem is I can't so I think I'm reading the proof wrong.
The paper is here
http://lipn.univ-paris13.fr/~fernique/qc/robinson.pdf
-
I don't understand what "only works for hierarchical tilings" means. Let's say we've agreed on what it means for a set of tiles to be "hierarchical". The proof reduces the halting problem to the tiling problem. Do you want a reduction of the halting problem to "the tiling problem for non-hierarchical tile sets"? – Anton Malyshev Jan 7 at 23:55
The proof, as I understand it, reduces the halting problem to the tiling problem by specifying that the Turing machine runs within one of the finite sized "red borders" which can be infinitely large. If no such hierarchical structure exists, that is to say no set of borders (of some shape or form) of increasing size can be constructed, then over what space would the Turing machine be running? Cont... – Scott C Jan 8 at 0:26
Assuming that the tiling problem is in fact undecidable, then there must exist aperiodic tile sets for which there is no way to prove their aperiodicity therefore clearly not all aperiodic tile sets have this hierarchical structure, else the problem would be decidable. We are aware she is wrong but we need to know why. – Scott C Jan 8 at 0:30
Say A is the collection of tile sets which don't tile the plane periodically, and T is the collection of tile sets which tile the plane. The set of aperiodic tile sets is their intersection, AT. You're right: the argument in the paper can be used to show that there are tile sets which belong to AT, but we can't prove they belong to AT. However, it's still possible that all tile sets in AT are hierarchical, and that for each of them we can prove it belongs to A. It's just that there are some which we can't prove belong to T. In other words, AT is not recursively enumerable, but perhaps A is. – Anton Malyshev Jan 8 at 2:13
I am confused by the question. Is the word "hierarchical" being used here with a precise mathematical meaning? – Joel David Hamkins Jan 8 at 4:18
It sounds like you're misunderstanding the proof: the main theorem is
There is a computable function $f$ (such things are in general called many-one reductions) from $\mathbb{N}$ to $\{$finite sets of tiles$\}$ (construed appropriately) such that $\Phi_e(e)\uparrow$ if and only if the set of tiles $f(e)$ can tile the plane.
From this we conclude that the set of finite sets of shapes which tile the plane is not computable, since otherwise - pulling back along $f$ - we could compute the halting problem.
At this point, maybe (I haven't read the proof in detail in a long time) it turns out that every set of tiles of the form $f(e)$ which does tile the plane, does so in a "hierarchical" fashion (whatever that means). This might be the case; however, it would be wrong to say that the proof "only works for hierarchical tilings," since the tilings aren't the 'inputs' to the theorem. Additionally, I'd be very surprised if this were true: I have a vague picture in my head of what "hierarchical" ought to mean to me, and my recollection is that the tilings constructed in the proof of undecidability of the tiling problem appear to be not hierarchical.
It might be the case that all the images of all such reductions $f$ all have to mostly consist of hierarchically tilable sets of tiles; for example, it is conceivable that
There is no computable function $f$ from $\mathbb{N}$ to $\{$finite sets of tiles$\}$, such that (i) $f(e)$ can tile the plane if and only if $\Phi_e(e)\uparrow$, and (ii) each set of tiles $f(e)$ (for $\Phi_e(e)\uparrow$) can not tile the plane hierarchically.
Depending on the definition of "hierarchical," I'd be surprised by that result, but in principle it could be true. However, if true it would require considerable work to prove, probably more than proving the original result.
None of the above, by the way, relies on aperiodicity. Where aperiodicity arises is in the fact that, if every set of tiles which could tile the plane could do so periodically, then the tiling problem would be computable (see the bottom of the first page of the article), so we must be using mostly sets of tiles which do not tile the plane periodically. That is, a lemma standing 'morally' (at least I think so) at odds to the undecidability of the tiling problem is:
There is no computable function $f$ from $\mathbb{N}$ to $\{$finite sets of tiles$\}$ such that $(i)$ $f(e)$ can tile the plane if and only if $\Phi_e(e)\uparrow$, and $(ii)$ if $\Phi_e(e)\uparrow$ then $f(e)$ can tile the plane periodically.
So aperiodicity is really an after-the-fact thing: because the tiling problem is unsolvable, it therefore follows that there are finite sets of tiles which can tile the plane, but can not do so periodically. (This is the first full sentence on page 178 of Robinson's article.)
I'm a little unclear, though, exactly what your question is, so I might not have addressed it well; is this what you were looking for?
-
By the way, I think this question is borderline for MO; certainly it could be clearer. On the other hand, issues about reductions in logic are notoriously subtle when encountered for the first time, which is why I've tried to answer it. – Noah S Jan 8 at 1:48
I just read the paper, so it's fresh for me: the tile sets used in the proof are "hierarchical" in the sense that they're built on top of a particular tile set which tiles the plane (only) hierarchically. The fact that none of those tile sets can tile the plane periodically follows easily from that structure. – Anton Malyshev Jan 8 at 2:26
I guess I'm still confused: what does "hierarchical" mean? – Noah S Jan 8 at 2:26
As far as I can tell, when Scott C says a tile set is hierarchical, he means that it some specific proof technique can be sued to show that it doesn't tile the plane periodically. I think the important thing is that the hierarchical tile sets are recursively enumerable, can't tile the plane periodically, and include the tile sets used in the proof of undecidability. There's a suitable definition for "hierarchical" that makes that true. – Anton Malyshev Jan 8 at 2:41
Well, that can't be it - if they were recursively enumerable, that would solve the tiling problem. Maybe what is meant is "the intersection of the set of tiling tile sets with an r.e. set?" – Noah S Jan 8 at 2:45 | {"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": 0.8012068867683411, "perplexity": 477.42527328104063}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768050.31/warc/CC-MAIN-20141217075248-00105-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://www.yesterdayscoffee.de/2017/09/08/include-pages-from-a-pdf-into-a-latex-beamer-presentation/ | # Include pages from a pdf into a LaTeX beamer presentation
As you know, I do basically everything with LaTeX. But, I have colleagues who work with other tools and sometimes we exchange slides. Fortunately by now people have realized that I don’t like to get weird formats, so they send me pdfs. Yay!
It is actually really easy to include pages from a presentation in pdf format into a LaTeX beamer presentation. You will need the package pdfpages and then just write:
{
\setbeamercolor{background canvas}{bg=}
\includepdf[pages=3-8]{slides.pdf}
}
The first line is necessary, because it seems like otherwise the pdf slides end up being inserted behind the background of the slides, which doesn’t make so much sense to me, but anyway.
You can also include one pdf page into a beamer-slide (“frame”). This is useful if you want to edit the slide a bit, for example to hack your own footer back into the slide to get consistent page numbering:
{
\setbeamercolor{background canvas}{bg=}
\begin{frame}[t]
\includepdf[pages=3]{slides.pdf}
\vspace{0.81\paperheight} % go down to where we want the footer
\hspace*{0.31\paperwidth} % space to the left
\begin{minipage}{0.6\paperwidth} % insert my footer
\tiny\colorbox{white}{~\insertshortauthor: \insertshorttitle} | {"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": 0.9660826921463013, "perplexity": 1552.4908117826728}, "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/1596439735867.93/warc/CC-MAIN-20200804102630-20200804132630-00348.warc.gz"} |
https://convert.ehehdada.com/newtontorankine | # Newton To Rankine
Calculates the Rankine temperature from the given Newton scale value
Type what you want to convert in the box below or
(autosubmits, max. 1MB)
<- ups! invalid URL! please, delete it or correct it!
## Newton scale to Rankine
The Newton temperature scale was defined by Isaac Newton in 1701 setting as 0 on this scale "the heat of air in winter at which water begins to freeze", or in other words, 0 as in Celsius scale, and the value 33 for "heat at which water begins to boil", so around 100 ℃, being exactly 100 the value commonly used for conversions between both scales. The Newton is represented as °N after the value.
The Rankine temperature scale was proposed by William John Macquorn Rankine in 1859 similarly to Kelvin temperature scale. The zero absolute at Rankine scale is also 0 K, and degree of Rankine is the same than Fahrenheit degree. The Rankine is represented as °R after the value, and sometimes like °RA.
The Rankine scale degree values are calculated from Newton values using the formula $$Newton × {60 \over 11} + 491.67$$ | {"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": 0.8967580795288086, "perplexity": 2753.5475346010558}, "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/1656103917192.48/warc/CC-MAIN-20220701004112-20220701034112-00469.warc.gz"} |
https://ir.lib.uwo.ca/etd/255/ | #### Degree
Doctor of Philosophy
Computer Science
Marc Moreno Maza
#### Abstract
Finding the solutions of a polynomial system is a fundamental problem with numerous applications in both the academic and industrial world. In this thesis, we target on computing symbolically both the real and the complex solutions of nonlinear polynomial systems with or without parameters. To this end, we improve existing algorithms for computing triangular decompositions. Based on that, we develop various new tools for solving polynomial systems and illustrate their effectiveness by applications.
We propose new algorithms for computing triangular decompositions of polynomial systems incrementally. With respect to previous works, our improvements are based on a weakened notion of a polynomial GCD modulo a regular chain, which permits to greatly simplify and optimize the sub-algorithms. Extracting common work from similar expensive computations is also a key feature of our algorithms.
We adapt the concepts of regular chain and triangular decomposition, originally designed for studying the complex solutions of polynomial systems, to describing the solutions of semi-algebraic systems. We show that any such system can be decomposed into finitely many regular semi-algebraic systems. We propose two specifications (full and lazy) of such a decomposition and present corresponding algorithms. Under some assumptions, the lazy decomposition can be computed in singly exponential time w.r.t. the number of variables.
We introduce the concept of {\em comprehensive triangular decomposition} for solving parametric polynomial systems. It partitions the parametric space into disjoint cells such that the complex or real solutions of a polynomial system depend continuously on the parameters in each cell. In the real case, we rely on cylindrical algebraic decomposition (CAD) to decompose a cell into connected components. CAD itself is one of the most important tools for computing with semi-algebraic sets. We present a brand new algorithm for computing it based on triangular decomposition.
COinS | {"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": 0.8879773616790771, "perplexity": 380.1226764423712}, "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-20180622143142-20180622163142-00553.warc.gz"} |
http://blog.computationalcomplexity.org/2010/04/possible-np-intermediary-problem.html | Wednesday, April 28, 2010
A possible NP-Intermediary Problem
(REMINDERS: STOC Early Registration closes on April 30. CCC Early Registration closes May 3. EC Early Registration closes on May 6. )
(ADDED LATER- AS STATED the problem below has problems with it. After reading see Eric Allenders comment in the comments.)
Here is a problem whose complexity has probably not been studied. I think it is NP-Intermediary. I will also give a version that is likely NPC. I am proposing that you either prove it is in P or NPC or show that if it is NPC then PH collapses (or something unlikely happens).
DEFINITION: We call a coloring of the x by y grid proper if there are no rectangles with all four corners the same color.
Let L be The set of all (x,y,c) in UNARY such that there is a proper c-coloring of the x by y grid.
1. Clearly in NP: verifying that a c-coloring of x by y is proper can be done in time poly in x,y,c.
2. Let Lc be the set of all (x,y) such that (x,y,c) ∈ L. Lc has a finite obstruction set and hence is in O(|x|+|y|) thought the constant depends on c. This can be proven by Well-Quasi-Order theory which yields nonconstructive bounds on the size of the obs set, or there is a proof with reasonable O(c2) bounds. (For ALL info on this problem and links to more info see the links below.) Hence the problem is Fixed Parameter Tractable.
3. I suspect that the problem L is NP-intermediary. Why? I think its NOT NPC since there is not much to play with- just 3 numbers. I think its not in P because my co-authors and I have not been able to do much more than ad-hoc colorings (this is not that good a reason- however the 17x17 challenge (linked to below) has lead other people to think about the problem and not come up with clean solutions.)
4. It is likely that the following is NPC: The set of all (x,y,c,f) where f is a partial c-coloring of the x by y grid such that f can be extended to a proper c-coloring.
5. I suspect that whatever is true for rectangles is true if you replace rectangles by other shapes such as a squares. There are also other Ramsey-Theoretic functions that could be studied (though Ramsey's theorem itself not-so-much--- verifying is hard).
6. This question is related to the (still unresolved) question I posed here and that Brian Hayes explained better here. However, I don't think proving the general problem NPC will shed light on why determining if (17,17,4) ∈ L is hard. That's just one instance.
1. ...or show that if it is NPC then PH collapses
Since this problem (as stated) can be efficiently encoded as a *tally* language (since all three inputs (x,y,c) are in unary), this is a sparse set, and can't be NP-complete unless P=NP by Mahaney's Theorem.
I guess that the interesting question arises if we consider the *binary* encoding of this language. In this setting, the problem is in NE, and the question is: Is it complete for NE? I won't claim to have any intuition about whether it's complete. Probably it's easier to consider the candidate NP-complete problem (where the partial coloring f is also given).
2. Mahaney's Theorem, right?
3. Sune Kristian Jakobsen12:27 PM, April 28, 2010
"I think its not in P because my co-authors and I have not been able to do much more than ad-hoc colorings (this is not that good a reason- however the 17x17 challenge (linked to below) has lead other people to think about the problem and not come up with clean solutions.)"
Is your point that a 4-coloring of the 17x17 grid is impossible, or that the colorings of maximal grids seems to be difficult to find?
It might be that all the lowest possible upper bounds (on grid size for fixed c) can be proven using a simple counting argument, but that there don't exist any proof that these upper bounds are the lowest possible. In this case the problem would be in P, but we wouldn't be able to prove it. In particular, if this problem is not in P it not only implies that maximal colorings are hard to find, but also that there is no simple reason that larger grid can't be c-colored.
4. Eric- THANKS for stating the version of my problem that makes sense.
Sune K. J.- DETERMINIng if 17x17 is
4-colorable seems to be hard. I had many
techniques for showing that a grid was NOT c-colorable but they PROVABLY did not work on 4-coloring 17x17. AND trying to show that 17x17 IS 4-colorable ALSO seems hard as nobody has been able to find it
(of course, it might not exist). | {"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": 0.8272445201873779, "perplexity": 772.6089241448078}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931011477.80/warc/CC-MAIN-20141125155651-00064-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://repository.uantwerpen.be/link/irua/135515 | Publication
Title
Effective ionisation coefficients and critical breakdown electric field of $CO_{2}$ at elevated temperature : effect of excited states and ion kinetics
Author
Abstract
Electrical breakdown by the application of an electric field occurs more easily in hot gases than in cold gases because of the extra electron-species interactions that occur as a result of dissociation, ionization and excitation at higher temperature. This paper discusses some overlooked physics and clarifies inaccuracies in the evaluation of the effective ionization coefficients and the critical reduced breakdown electric field of CO2 at elevated temperature, considering the influence of excited states and ion kinetics. The critical reduced breakdown electric field is obtained by balancing electron generation and loss mechanisms using the electron energy distribution function (EEDF) derived from the Boltzmann transport equation under the two-term approximation. The equilibrium compositions of the hot gas mixtures are determined based on Gibbs free energy minimization considering the ground states as well as vibrationally and electronically excited states as independent species, which follow a Boltzmann distribution with a fixed excitation temperature. The interaction cross sections between electrons and the excited species, not reported previously, are properly taken into account. Furthermore, the ion kinetics, including electronion recombination, associative electron detachment, charge transfer and ion conversion into stable negative ion clusters, are also considered. Our results indicate that the excited species lead to a greater population of high-energy electrons at higher gas temperature and this affects the Townsend rate coefficients (i.e. of electron impact ionization and attachment), but the critical reduced breakdown electric field strength of CO2 is only affected when also properly accounting for the ion kinetics. Indeed, the latter greatly influences the effective ionization coefficients and hence the critical reduced breakdown electric field at temperatures above 1500 K. The rapid increase of the dissociative electron attachment cross-section of molecular oxygen with rising vibrational quantum number leads to a larger electron loss rate and this enhances the critical reduced breakdown electric field strength in the temperature range where the concentration of molecular oxygen is relatively high. The results obtained in this work show reasonable agreement with experimental results from literature, and are important for the evaluation of the dielectric strength of CO2 in a highly reactive environment at elevated temperature.
Language
English
Source (journal)
Plasma sources science and technology / Institute of Physics. - Bristol, 1992, currens
Publication
Bristol : Institute of Physics, 2016
ISSN
0963-0252
Volume/pages
25:5(2016), p. 1-22
ISI
000385494000006
Full text (Publisher's DOI)
Full text (open access)
Full text (publisher's version - intranet only)
UAntwerpen
Faculty/Department Research group Publication type Subject Affiliation Publications with a UAntwerp address | {"extraction_info": {"found_math": true, "script_math_tex": 1, "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": 0.8043790459632874, "perplexity": 1311.8031341951005}, "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/1502886108264.79/warc/CC-MAIN-20170821095257-20170821115257-00196.warc.gz"} |
http://aas.org/archives/BAAS/v25n2/aas182/abshtml/S6801.html | Externally-Driven Gravitational Torques on Disk/Star Systems
Session 68 -- Star Formation
Oral presentation, Thursday, 10:30-12:00, Zellerbach Playhouse Room
## [68.01] Externally-Driven Gravitational Torques on Disk/Star Systems
E.C. Ostriker (UCB)
Using linear perturbation theory, I investigate the torques between a centrifugally-supported gas/dust disk surrounding a young star and a second star. I consider both the cases of a binary companion orbiting beyond the disk, and a passing member of a stellar cluster. Extending the Goldreich-Tremaine theory of resonant wave excitation, I develop a theory for excitation of one-armed disturbances in the broad near-inner resonance of a Keplerian potential. These waves can be important when excitation of $m\ge 2$-armed waves is impossible (or much reduced). This occurs in a bound system when an annulus is cleared in the disk around the binary companion, like the gaps seen near moons in Saturn's rings. The external torque on the disk is negative, encouraging accretion. I compute numerically and analytically accretion timescales $t_{\rm acc}$ in the disk due to near-resonantly driven waves. For the example of a $0.5 {\rm M}_{\sun}$ disk in a solar-mass scale binary with separation 100 AU, $t_{\rm acc} \sim 10^7$ yrs, with $t_{\rm acc}$ proportional to the binary period.
For unbound systems, I investigate the angular momentum $L$ and energy transferred during parabolic perturber passages, which are most common for typical disk sizes and cluster velocity dispersions. I compute the torques for arbitrary inclination angle and longitude of the pertuber's orbit by integrating over all horizontal and vertical resonances. I present both numerical computations and analytic formulae for cases with closest perturber approach $x_{\rm min}$ beyond two disk radii $R_{\rm D}$. I find that $\Delta L$ transferred to the disk is negative, inducing accretion. The fractional angular momentum removed from a disk is $\sim 10\%$ at $x_{\rm min}/R_{\rm D}=2$, averaged over orbit orientations. But the applied torque drops off exponentially with increasing $x_{\rm min}/R_{\rm D}$, due to the mismatch of high disk rotation frequency and low perturber orbital frequency. When the incoming perturber's orbit is polar or retrograde, energy is removed from the orbit and given to the disk, permitting capture. The probability of capture is quite small, however, because the positive energy transferred to the disk in an encounter is a tiny fraction of the typical kinetic energy of a star in a cluster. | {"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": 0.9726621508598328, "perplexity": 1974.2296925963801}, "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-2015-14/segments/1427131298020.57/warc/CC-MAIN-20150323172138-00211-ip-10-168-14-71.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/207467-inequalities-print.html | # Inequalities
• November 13th 2012, 08:08 AM
Petrus
Inequalities
What is the smallest value of a for which the inequality https://webwork.math.su.se/webwork2_...5445c9ff01.png is observed for x> 0?
what i thought is that i put x = 1 and solve it but did not work well :(
• November 13th 2012, 08:26 AM
ebaines
Re: Inequalities
You can rearrange to $2 \ln x - 3x^2 - 6x \le a$, then find the max value for the left hand side in the interval x>0. That max value is the value for a.
• November 13th 2012, 08:29 AM
Petrus
Re: Inequalities
Quote:
Originally Posted by ebaines
You can rearrange to $2 \ln x - 3x^2 - 6x \le a$, then find the max value for the left hand side in the interval x>0. That max value is the value for a.
hmmm i still dont get any progress :S
• November 13th 2012, 08:38 AM
ebaines
Re: Inequalities
Quote:
Originally Posted by Petrus
hmmm i still dont get any progress :S
Do you know how to find the max value of a function? Hint - start by finding where the derivative = 0....
• November 13th 2012, 08:44 AM
Petrus
Re: Inequalities
Quote:
Originally Posted by ebaines
Do you know how to find the max value of a function? Hint - start by finding where the derivative = 0....
ok i got these
x=1/6(sqrt(21)-3)
x=1/6(-3-sqrt(21))
is this correct? ajnd what shall i do next
• November 13th 2012, 09:24 AM
ebaines
Re: Inequalities
Quote:
Originally Posted by Petrus
is this correct?
Yes, those are the values of x where the derivative is 0. But remember you only need worry about x>0.
Quote:
Originally Posted by Petrus
what shall i do next
Next step - what is the value of $2 lnx - 3x^2 -6x$ at this value of x? That's what 'a' is equal to. | {"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": 0.8824885487556458, "perplexity": 862.6539359079535}, "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-2015-22/segments/1432207928350.51/warc/CC-MAIN-20150521113208-00146-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/integration-of-rational-function-by-partial-functions-the-last-step-confuses-me.639887/ | Homework Help: Integration of rational function by partial functions. The last step confuses me!
1. Sep 29, 2012
randoreds
Ok all save y'all a little reading.
Worked out the problem. Got
X^2+2x-1=A(2x-1)(x+2)+bx(x+2)+cx(2x-1)
Ok then you write it standard form for a polynomial. Then use can use there coefficients to write new equations at you get
2a+b+c=1
3a+2b-c=2 and finally
-2a=1
Now you solve for a,b,c and this is where I'm confused.
To solve for a is easy, it is just 1/2. But the book doesn't show how to solve for b and c.
B =1/5 and C =-1/10
I'm totally confused how they got those numbers so if you could explain it. That would be so helpful!!
2. Sep 29, 2012
voko
This is a 3x3 system of linear equations. Are you saying you don't know how to solve it?
3. Sep 29, 2012
randoreds
I guess yes. I totally don't remember how to solve these
4. Sep 29, 2012
randoreds
But thanks, Ill look it up. I couldn't remember what they were called. Thanks
5. Sep 29, 2012
voko
6. Sep 29, 2012
HallsofIvy
This last is incorrect. It should be -2a= -1 or 2a= 1. That's why a= 1/2 rather than -1/2.
Putting a= 1/2 into the other two equations, 2a+ b+ c= 1+ b+ c= 1 so b+ c= 0 and 3a+ 2b- c= 3/2+ 2b- c= 2 so that 2b- c= 1/2. If you don't like fractions multiply both sides by 2 to get 4b- c= 1. From b+ c= 0, b= -c so 4b= -4c. 4b- c= -4c- c= -5c= 7.
Of course, you can set each corresponding coefficients equal because x^2+2x-1=a(2x-1)(x+2)+bx(x+2)+cx(2x-1) is true for all x. So you can get the three equations to solve for a, b, and c by taking three values for x. And choosing those value cleverly can simplify the resulting equations! For example, if you take x= 0, x(x+2) and x(2x-1) are both equal to 0. The equation, for x= 0, becomes 0^2+ 2(0)-1= a(2(0)-1)(0+2)+ b(0)+ c(0)= -2a. That is, -2a= -1 or 2a= 1. Taking x= -2, both (2x-1)(x+2) and x(x+2) are 0 so the equation becomes 2^2- 2(2)- 1= a(0)+ b(0)+c(2)(2(2)-1) or 6c= -1. Finally, if you take x= 1/2, (2x-1)(x+2) and x(2x-1) are 0 and the equation is (1/2)^2+2(1/2)- 1= b(1/2)(1/2+ 2) so 1/4= (5/4)b.
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": 0.9341263175010681, "perplexity": 2750.723092661925}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267866888.20/warc/CC-MAIN-20180624063553-20180624083553-00004.warc.gz"} |
http://reciprocal.systems/phpBB3/viewtopic.php?t=590&p=3528 | ## Division Algebras and Motion
Discussion concerning the first major re-evaluation of Dewey B. Larson's Reciprocal System of theory, updated to include counterspace (Etheric spaces), projective geometry, and the non-local aspects of time/space.
blaine
Posts: 40
Joined: Mon Jan 16, 2017 9:44 am
### Division Algebras and Motion
When Dewey Larson discussed ratios of space and time, he was probably thinking in terms of the field of reals, $\mathbb{R}$. This is a mathematical object which contains the multiplication operation, and its inverse, division.
Division is required for motion (s/t, t/s) so its reasonable to say that any mathematical system which is able to express division will be useful for describing aspects of reality in a universe of motion. Thus, we should focus efforts towards the mathematical expression of such a universe in terms of the division algebras. Specifically, we are interested in finite dimensional division algebras in order to describe finite aspects of the universe.
There is a well known mathematical theorem called the Frobenius theorem which states that there are 3 "representations" (isomorphisms) of finite-dimensional associative division algebras over the real numbers. These are the real numbers $\mathbb{R}$, the complex numbers $\mathbb{C}$, and the quaternions $\mathbb{H}$.
There are many examples where quaternions are important in the various physics fields. Bruce has discussed these in several posts. But while quaternions generalize the imaginaries and the reals (allowing for the representation of the noncommutative 3-d rotations), they still are not sufficient in describing more complex phenomena. In RS2 it is posited that Octonions are needed to describe life physics. While the quaternions are associative but not commutative, the octonions are non-associative and non-commutative. This implies that the order of operations matters, even if they are displayed in the same order. In the Reciprocal System, and as recently discussed in a particle physics paper, the variables in this equation are both the states and the operators - in other words, it is all motion transferring between different types of motion. Thus, if the states are non-commutative, then they must also be non associative. We know that 3-d rotations are non commutative so this combined with the previous statement would imply that we are not generically interested in associative algebras. This is why octonions and potentially other algebras could be important.
There is another mathematical theorem that could provide guidance as to what other division algebras may be of import. This is Hurwitz's theorem, which states that
In mathematics, Hurwitz's theorem is a theorem of Adolf Hurwitz (1859–1919), published posthumously in 1923, solving the Hurwitz problem for finite-dimensional unital real non-associative algebras endowed with a positive-definite quadratic form. The theorem states that if the quadratic form defines a homomorphism into the positive real numbers on the non-zero part of the algebra, then the algebra must be isomorphic to the real numbers, the complex numbers, the quaternions and the octonions.
The key here is unital algebras, because physics often deals with conservation laws. Unitary transformations are useful for representing these. Also, a consequence of this theorem is that
The only Euclidean Hurwitz algebras are the real numbers, the complex numbers, the quaternions and the octonions.
. Hence, the octonions encompass the non-commutative and non-associative possibilities of motion. Since it can be shown via the Cayley–Dickson construction that octonions can be constructed by composing quaternions in a special way (just as imaginaries can be constructed from reals, and quaternions from imaginaries and so on). This would seem to imply that determining the mathematical foundations of describing a universe of motion would necessitate the utilization of octonions.
bperet
Posts: 1446
Joined: Thu Jul 22, 2004 1:43 am
Location: 7.5.3.84.70.24.606
Contact:
### Re: Division Algebras and Motion
Good summary.
My original take on this subject is posted here: The Reality of the Imaginary.
Every dogma has its day...
dbundy
Posts: 134
Joined: Mon Dec 17, 2012 9:14 pm
### Re: Division Algebras and Motion
There is something important to consider here: Using imaginary numbers in an attempt to unify mathematics and geometry can be very useful, but misleading at the same time. To avoid confusion and unnecessary complication, it is useful to combine our understanding of the dimensions of Euclidean geometry with the numbers of the Greek tetraktys.
There are several aspects of this endeavor that we should consider carefully. First, the four numbers of the tetraktys (1, 2, 3 & 4) correspond to the four dimensional objects of the geometry (point (requires 1 point), line (requires 2 points), area (requires 3 points), volume (requires 4 points)). Second, each geometric dimension has two reciprocal "directions," with respect to the point, i.e. forward/backward, left/right, up/down (the point itself also has two reciprocal "directions," i.e. in/out.)
When we assign these dual geometric "directions," to the four numbers of the tetraktys, we get 20 = 1, 21 = 2, 22 = 4 and 23 = 8. This is known as the binomial expansion of the tetraktys, and it is highly significant and relevant, not only to the division algebras, but also to the fundamentals of the reciprocal system.
In the case of the four normed division algebras, the reals correspond to dimension 0 (20 = 1), the complexes to dimension 1 (21 = 2), the quaternions to dimension 2 (22 = 4) and the octonians to dimension 3 (23 = 8), following the geometric dimensions of the tetraktys. However, mathematicians, not recognizing that each of these dimensions has two "directions," mistook the "directions" of the algebras for their dimensions, introducing the ensuing confusion and hiding the wonderful connection between the tetraktys and geometry and, ultimately, what we might call multi-dimensional scalar algebra.
Most physicists would balk at such terminology, because they are taught that scalar magnitudes have no direction, as opposed to vectors. They are quantities of magnitude only, applied to vectors. However, it's easy for the students of the reciprocal system to understand that each dimension of Euclidean geometry, and the Greek tetraktys, has two scalar "directions," which are analogous to the two scalar "directions" of numbers, in each of the four dimensions of algebra.
The 0 dimensional Reals, of course, are easy to characterize, especially, using rational numbers. The positive numbers are simply the reciprocals of the negative numbers and the identity number of the group (1/1) can be thought of as unity or zero (i.e. zero displacement, to use Larson's terminology).
The 1 dimensional Complexes (with 21 = 2 "directions") introduced the incredible world of rotation to mathematics, but the 2d quaternions, with 22 = 4 "directions") have only become useful in modern times, with the use of computers, and the 3d octonions (with 23 = 8 "directions") have become objects of research relatively recently, in conjunction with string theory, mostly.
The progressive loss of algebraic properties with increasing dimension in the division algebras (order (the 1d complexes have no intrinsic order), commutativity (the 2d quaternions have no intrinsic order and are not commutative) and associativity (the 3d octonions have no order, are not commutative nor associative), has plagued the physicists for many decades now.
The essence of the problem, believe it or not, is due to the limitations of vector motion, which is the only motion the LST community recognizes. Vector motion can only exist in one "direction" of one dimension at a time, and when we algebraically combine magnitudes in each of these dimensions, the algebraic pathology just described emerges.
However, there is an algebra corresponding to the physics of expanding/contracting lines, areas and volumes and it is a scalar algebra, with no known pathology, due to multiple dimensions. We only need to recognize it. When we do, we can easily explain the concept of quantum spin, and, along with it, the 2d magnetic moment of charged particles, and soon, hopefully, the 3d mass of all particles.
Just saying.
dbundy
Posts: 134
Joined: Mon Dec 17, 2012 9:14 pm
### Re: Division Algebras and Motion
I'd like to make further comparisons of multi-dimensional division algebra and multi-dimensional scalar algebra, if I may.
The Euclidean "norm," as used in the term "normed division algebras," simply refers to a magnitude, or length, of a vector. It's most usefully thought of in terms of the Pythagorean theorem, where the hypotenuse of a right triangle is equal to the square root of the sum of the squares of the sides.
However, as I noted in the comment above, algebraic dimensions, for mathematicians, do not refer to geometric dimensions, creating a disconnect between Euclidean geometry and normed division algebras of the Hurwitz theorem and mathematics in general.
Moreover, Euclidean geometry is limited to three non-zero dimensions, while algebra is not. Indeed, mathematicians consider finite dimensional algebras as only a subset of infinite dimensional algebras. This has caused some consternation in Wikipedia authors trying to sort out the nuances of the consequences (e.g. See here).
Again, it's important to recognize the role of the LST's concept of motion in the development of multi-dimensional division algebras. It is strictly a concept of vector motion, the norm being the length of the vector, representing a magnitude, many times a "unit" magnitude, as in the unit circle, where rotations of the unit magnitude, in either direction around the circle, are used to algebraically manipulate complex numbers, such as z=(a+ib).
Because the number z consists of two terms, it is considered two-dimensional, when, geometrically speaking, it corresponds to a one-dimensional line. This disconnect famously caused Hamilton years of grief, when he sought to multiply what he thought were three-dimensional equations; that is equations with three terms.
It wasn't until he realized that he needed four terms for three dimensional equations that the spark, leading to the modern idea of normed division algebras, fired in his brain. However, what is still not recognized by modern mathematicians, as far as I can determine, is the fact that the multiple terms in the algebras do not correspond to geometric dimensions, but to the "directions" of the non-zero dimensions.
When a point expands in one dimension, it necessarily expands in that dimension's two "directions" <------0------>. This is outward scalar motion in one dimension. To algebraically manipulate it, we need only express it as an increasing (decreasing) magnitude. Specifying direction, or orientation, in a fixed reference system is not necessary. Similarly, we can express a 2d expansion (contraction) of magnitude only, over time, with a simple equation, and likewise for the 3d case.
These scalar equations use expanding, contracting, Euclidean norms in one, two and three dimensions, which correspond to the changing radius (diameter) of 1d, 2d and 3d magnitudes, which can be defined by the linear, quadratic and cubic magnitudes of numbers, by recognizing that these also define corresponding geometric equivalents.
Had Hamilton known of the connection between the Greek tetraktys, which he adored so much, and the geometry generated by the duality of geometric dimensions, together with the reality of scalar motion, it's conceivable our world of theoretical physics and mathematics would have grown into something much more magical and wonderful than what we have today.
dbundy
Posts: 134
Joined: Mon Dec 17, 2012 9:14 pm
### Re: Division Algebras and Motion
One of the major advantages of the new multi-dimensional division algebra (i.e. scalar) over the traditional multi-dimensional division algebra (i.e. vector) is that the dimensions can be seen as an integrated whole; that is to say, if we begin with three non-zero dimensions and reduce them to 1, rather than begin with 1 and build them up to 3, we can gain an important insight: the lower degrees exist within the highest degree.
Let's use the usual polarity symbols to illustrate building them up:
1) 0 degree (20) = + or -
2) 1 degree (21) = + & -
3) 2 degree (22) = ++ & +- & -- & -+
4) 3 degree (23) = +++ & +-+ & +-- & ++- & -++ & --+ & --- & -+-
It's easy to see that the second degree consists of combinations of the first degree and that the third degree consists of combinations of the first and second degrees. Another mathematical way to look at the same thing is:
20 = 1
21 = 2x1 = 2;
22 = 2x2 = 4;
23 = 2x4 = 8.
However, if we start with the geometry of 23 = 8, by assigning orthogonal dimensions to each degree, we get a 2x2x2 stack of 8 unit cubes, and within that stack, we can identify three radii, or norms, using the Pythagorean theorem:
1) √(12 + 12 + 12) = √3
2) √(12 + 12 + 02) = √2
3) √(12 + 02 + 02) = √1
That each of these three norms forms the basis of an n-dimensional division algebra, isomorphic, as they say, to that of the Reals, seems to me to be straightforward, when one recognizes the existence of scalar motion.
Remarkably, when expanding/contracting (oscillating) as a 3d whole, over time, each norm expands to unity, or contracts to zero, at the same point in time (space), even though their magnitudes are not the same. It's mathematically impossible to be otherwise.
Hence, we conclude that this new set of scalar division algebras, unlike the traditional set of vector division algebras, is a well integrated set, with no algebraic pathology, and thus should be well suited for researching the scalar motion combinations, and relations between them, in the new reciprocal system of physical theory.
bperet
Posts: 1446
Joined: Thu Jul 22, 2004 1:43 am
Location: 7.5.3.84.70.24.606
Contact:
### Re: Division Algebras and Motion
blaine wrote:
Sun Sep 23, 2018 10:44 am
The key here is unital algebras, because physics often deals with conservation laws. Unitary transformations are useful for representing these. Also, a consequence of this theorem is that
The only Euclidean Hurwitz algebras are the real numbers, the complex numbers, the quaternions and the octonions.
. Hence, the octonions encompass the non-commutative and non-associative possibilities of motion. Since it can be shown via the Cayley–Dickson construction that octonions can be constructed by composing quaternions in a special way (just as imaginaries can be constructed from reals, and quaternions from imaginaries and so on). This would seem to imply that determining the mathematical foundations of describing a universe of motion would necessitate the utilization of octonions.
Here's what I have found on this topic:
There exist two sets of dimensions, yang (linear) and yin (angular). We are familiar with the yang dimensions of line (1D), plane (2D) and volume (3D). The yin dimensions correspond to the same concepts but using angles instead of lines, resulting in the complex (1D), quaternion (2D) and octonian (3D) structures.
Might be confused here as to why the yin dimensions are "off"... first, I am only looking at the "vector" component, not the scalar (real scalar + <imaginary vector>). So a complex number only needs a single magnitude to express one, angular velocity. A quaternion only needs TWO magnitudes to express a sphere (latitude, longitude, for example)--you only need the 3-vector when you treat an angle as a line. A octonian only needs THREE magnitudes to express a hypersphere, which decomposes into a 7-vector in linear algebra.
Unital algebras only apply to a compound/composite motion when it is in balance (net motion has zero displacement, unit speed). When a motion is not in balance, such as an atom or molecule that is ionized, you end up with extraneous dimensions (3, 5, 6, 7). This causes the system to seek balance through the pushes and pulls of scalar relationships, until it reaches dimensional stability (1, 2, 4 or 8). In RS2, the combining of angular velocities cause dimensional reduction or expansion, such as Nehru's concept of birotation. This is how the dimensionality of a motion becomes unital.
The dimensional datum is 4 for a sector (w, x, y, z). The material is s3/t and the cosmic is t3/s. In a yang, linear form, it is called a "homogeneous coordinate" and in the yin, angular form, a "quaternion."
The projection of the inverse sector has a dimensional datum of 2, a complex quantity that expresses the net motion across the unit speed boundary (real = linear push/pull, imaginary = spin). Larson only considered the real aspect in his work, the push-pull, ignoring the spin component.
This means that the sectors are composed of complex and quaternion structures. So what of the "1" and "8" dimensions? They are "boundary conditions."
1-dimensional motion is literally "scalar" in that it can only contain a single magnitude, nothing else. The linear form of 1D in space is "distance" and in time, "duration." This is what Larson calls "scalar speed," the ratio of distance/duration, since it has no structure, only magnitude. And when this value is unity, it is the progression of the natural reference system--the "local" datum of the Reciprocal System.
The other extreme, the octonian, was never considered by Larson or Nehru. But we know the RS is about symmetry... so if the 1D motion is the natural datum, the "center" of the system, the 8D octonian (a 3D angular velocity) is the inverse of the center--the "sphere at infinity," which we see as a volume.
So in summary, the dimensional relations for the inanimate level of existence are (D=linear, d=angular):
1D/0d: progression, the datum or "center" speed
2D/1d: the projection of motion of one sector on the other
4D/2d: the structure of motion that we perceive as "stuff"
8D/3d: gravitation, the inverse datum or "sphere/plane at infinity"
Every dogma has its day... | {"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": 4, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.836889922618866, "perplexity": 781.91580563302}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823618.14/warc/CC-MAIN-20181211104429-20181211125929-00553.warc.gz"} |
https://andrewcharlesjones.github.io/journal/generalized-pca-2.html | # Generalized PCA: an alternative approach
Principal component analysis is a widely-used dimensionality reduction technique. However, PCA has an implicit connection to the Gaussian distribution, which may be undesirable for non-Gaussian data. Here, we’ll see a second approach for generalizing PCA to other distributions introduced by Andrew Landgraf in 2015.
## Introduction
In its traditional form, PCA makes very few assumptions. Let $\mathbf{X} \in \mathbb{R}^{n \times p}$ be our data matrix. Then, to find the first component, we seek a vector $\mathbf{u} \in \mathbb{R}^p$ such that the variance of the linear combination $\mathbf{X}\mathbf{u}$ is maximized. In particular, we solve
$\mathbf{u} = \text{arg}\max_{||u|| = 1} \text{var}(\mathbf{X}\mathbf{u}) = \text{arg}\max_{||u|| = 1} \mathbf{u}^\top \mathbf{S}_n \mathbf{u}$
where $\textbf{S}_n$ is the sample covariance matrix of $\mathbf{X}$. To find subsequent PCs, we would solve a similar optimization problem for $\mathbf{u}_2, \dots, \mathbf{u}_p$, with the additional constraint that the subsequent PCs are orthogonal to the preceding ones.
There are several other interpretations and solutions for PCA, such as casting it as an eigendecomposition of $\mathbf{X}^\top \mathbf{X}$, or as a minimization of the Frobenius norm of the data projected onto an orthogonal basis:
$\hat{\mathbf{U}} = \text{arg}\max_\mathbf{U} ||XB||$
with the constraint $\mathbf{U}^\top \mathbf{U} = \mathbf{I}$.
In the above formulations, the primary assumption is linearity.
Consider one last interpretation of PCA: as the maximum likelihood solution to a probabilistic model. This was Tipping and Bishop’s 1999 approach when they formulated probabilistic PCA.
In particular, probabilistic PCA assumes that there exist some latent, lower-dimensional variables $\mathbf{z}_1, \dots, \mathbf{z}_n \in \mathbb{R}^k$ where $k < p$, such that the data $\mathbf{x}_1, \dots, \mathbf{x}_n$ can be faithfully represented in this latent variables. Under a Gaussian model we would assume that a data vector $\mathbf{x} \in \mathbb{R}^p$ has the distribution
$\mathbf{x}_i | \mathbf{z}_i \sim \mathcal{N}(\mathbf{U} \mathbf{z}_i, \sigma^2 \mathbf{I})$
where $\mathbf{U} \in \mathbb{R}^{p \times k}$ in this case, and $\mathbf{z}_i \sim \mathcal{N}(0, \mathbf{I})$.
It can be shown that the maximum likelihood solution to this probabilistic model and the solution for traditional PCA are nearly identical. Thus, traditional PCA is strongly related to an assumption of the data being Gaussian distributed.
However, if the data follow a non-Gaussian distribution, this may be undesirable. There have been a few proposals for generalizations of PCA to non-Gaussian distributions, one of which we saw in an earlier post. Here, we’ll see a second approach to generalizing PCA to the exponential family.
## Generalized PCA
Recall the general form of the exponential family of distributions:
$f(x) = \exp\left(\frac{x \theta}{a(\phi)} + c(x, \phi)\right)$
where $\theta$ is the canoical natural parameter, and $\phi$ is a dispersion parameter.
In the earlier approach, we saw how Collins et al. used the theory of generalized linear models to easily substitute in any exponential family likelihood in order to generalize PCA. In particular, they factorized the parameter matrix $\mathbf{\Theta}$ as $\mathbf{\Theta} = \mathbf{A} \mathbf{B}^\top$, and then maximized the negative log-likelihood, where any exponential family likelihood can be substituted in.
## Generalized PCA as a projection
As an alternative generalization of PCA, Landgraf framed PCA as a projection of the natural parameters to a lower-dimensional space.
As we saw above, PCA can be seen as finding the matrix $\mathbf{U}$ (where $\mathbf{U}^\top \mathbf{U} = \mathbf{I}$) that minimizes the following:
$||\mathbf{X} - \mathbf{X} \mathbf{U} \mathbf{U}^\top||_F^2.$
In terms of a GLM, the above formulation is equivalent to minimizing the deviance of a Gaussian model with known variance. Recall that the deviance essentially measures the log-likelihood difference between the “saturated” (full) model and the fitted model.
In the Gaussian case, the natural parameter is equal to the data: $\mathbf{\Theta} = \mathbf{X}$, and the link function is the identity. Thus, the deviance has the form of a sum of squares:
$D = \frac{1}{\sigma^2} \sum\limits_{i=1}^n (\mathbf{x}_i - \mathbf{x}_i \hat{\theta}_i)^2$
where $\hat{\theta}_i$ are the estimated natural parameters.
Now, we’ll consider estimating the natural parameter matrix as
$\Theta = \widetilde{\Theta} \mathbf{U} \mathbf{U}^\top$
where $\widetilde{\Theta}$ are the natural parameters of the saturated model.
Importantly, notice that in this approach, we aren’t completely decomposing $\Theta$ into two submatrices, but rather projecting it onto an orthogonal basis.
We can then formulate the objective function as minimizing the deviance between the PCA model and the saturated model:
\begin{align} D(\mathbf{X}; \widetilde{\Theta} \mathbf{U} \mathbf{U}^\top) &= -2 \underbrace{\log f(\mathbf{X}; \widetilde{\Theta} \mathbf{U} \mathbf{U}^\top)}_{\text{$LL$ for PCA model}} \;\; + \underbrace{2\log f(X; \widetilde{\Theta})}_{\text{$LL$ for saturated model}} \\ &\propto -\langle \mathbf{X}, \widetilde{\Theta} \mathbf{U} \mathbf{U}^\top \rangle + \sum\limits_{i=1}^n \sum\limits_{j=1}^p b\left([\mathbf{U} \mathbf{U}^\top \widetilde{\theta}_i]_j \right) \\ \end{align}
where $b(\cdot)$ depends on the chosen exponential family model.
The primary advantage of Landgraf’s approach is that the formulation only needs to solve for the PC loadings $\mathbf{U}$ without worrying at all about the PC scores, such as is done in the Collins approach. In other words, instead of decomposing the natural parameters as
$\Theta = \mathbf{A}\mathbf{B}^\top$
we decompose them as
$\Theta = \widetilde{\Theta}\mathbf{U}\mathbf{U}^\top.$
This implies another advantage: if we have some held-out data $\mathbf{x}^*$, we can calculate the PC scores with simple matrix computation:
$\hat{U}^\top \mathbf{\widetilde{\theta}}^*$
where $\mathbf{\widetilde{\theta}}^$ are the natural parameters for $\mathbf{x}^$ under the saturated model. In contrast, under the Collins approach, solving for the PC scores on held-out data would require re-running an entire optimization problem.
## References
• Hotelling, Harold. “Analysis of a complex of statistical variables into principal components.” Journal of educational psychology 24.6 (1933): 417.
• Tipping, Michael E., and Christopher M. Bishop. “Probabilistic principal component analysis.” Journal of the Royal Statistical Society: Series B (Statistical Methodology) 61.3 (1999): 611-622.
• Prof. Jonathan Pillow’s notes on PCA
• Landgraf, Andrew J. Generalized principal component analysis: dimensionality reduction through the projection of natural parameters. Diss. The Ohio State University, 2015. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9757059216499329, "perplexity": 883.6383500095353}, "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-2021-31/segments/1627046152085.13/warc/CC-MAIN-20210805224801-20210806014801-00239.warc.gz"} |
https://01law.wordpress.com/2013/02/22/567/ | # Monte Carlo computation for European call price
Given a risk-neutral stock price in the form of geometric brownian motion (GBM),
$\displaystyle d S_t = S_t (r dt + \sigma d W_t),$
we will demonstrate the computation of the call premium. Let the option maturity be ${T}$ and strike ${K}$ Then, the discretely monitored lookback put premium is given by the formula
$\displaystyle P_{m} = e^{-rT}\mathbb{E}[(S_{T} - K)^{+}].$
Matlab code is here. Let
$\displaystyle S_{0} = 100, r = 0.02, \sigma = 0.2, T = 1.$
The exact answer should be ${C_{0} = 14.8065}$ according to the BS formula.
The method is simple:
step1. Use the standard Euler method with ${n}$ intervals in ${[0, T]}$ to simulate the path of ${S_{t}}$, compute discounted payoff of the sample path;
step 2. Repeat the path simulation of step 1 ${k}$ times, and take the average.
The result shows that if ${n = 10000}$ and ${k = 10000}$, then the option price is ${14.6734}$ with ${95\%}$ confidence interval ${[14.3444, 15.0023]}$. The total running time on my desktop PC is ${6.57}$ seconds. | {"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": 16, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9068912863731384, "perplexity": 506.2587256607002}, "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/1529267866926.96/warc/CC-MAIN-20180624083011-20180624103011-00554.warc.gz"} |
http://swmath.org/software/6322 | # BECOOL
BECOOL: Ballooning eigensolver with COOL finite elements: An incompressible variational ideal ballooning mode equation is discretized with the COOL finite element discretization scheme using basis functions composed of variable order Legendre polynomials. This reduces the second order ordinary differential equation to a special block pentadiagonal matrix equation that is solved using an inverse vector iteration method. A benchmark test of BECOOL (Ballooning Eigensolver using COOL finite elements) with second order Legendre polynomials recovers precisely the eigenvalues computed by the VVBAL shooting code. Timing runs reveal the need to determine an optimal lower order case. Eigenvalue convergence runs show that cubic Legendre polynomials construct the optimal ballooning mode equation for intensive computations
This software is also peer reviewed by journal TOMS.
## References in zbMATH (referenced in 1 article , 1 standard article )
Showing result 1 of 1.
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http://analog.intgckts.com/noise/thermal-noise-of-a-resistor/ | # Thermal Noise of a Resistor
Figure 1. Thermal noise model of a resistor
The thermally agitated charge carriers introduce random fluctuations in the voltage across the conductor even the average current through it is zero. This noise was first reported by Johnson and explained the cause as the Brownian movement of charge carriers by Nyquist. Hence this noise is also called Johnson noise or Nyquist noise. The thermal noise of a resistor is modeled as a rms voltage source () in series with its resistance() as illustrated in Figure 1.
Noise power in a resistor at temperature ‘T’ is . Maximum power is transferred to load, when load has same resistance as the noise source. The noise power delivered to load under this condition is called available noise power .
(1)
or voltage spectral density is
(2)
where,
J/K is Boltzman constant
T is absolute temperature of resistor in Kelvin.
is the measurement bandwidth
The spectral density of thermal noise is independent of frequency. Hence it is called white noise.
Rule of thumb:
At 300K, a 50 resistor
• has a spectral density of
• generates noise voltage.
• noise power = -173.83 dBm over 1Hz BW.
Norton equivalent representation of thermal noise is illustrated in Figure 2. and the noise current is given as
Figure 2. Norton equivalent model of thermal noise
Current spectral density is given by
(3)
Intutively, it is the voltage one would see if measured across a resistor held at a temperature T using using a measurement system with a bandwidth of . Therefore the noise increases with system bandwidth. A wideband system has higher noise floor than a narrow band system. Therefore sensitivity of Rx degrades as the bandwidth of the system increases. Eg. GSM Vs WLAN.
Depending on the circuit topology, one model may lead to simpler calculation than the other. Once the polarity is chosen it must be retained throughout the analysis of the circuit.
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https://brilliant.org/problems/inequality-right/ | # Inequality. Right?
Algebra Level 4
$\large\ P = \frac { { a }^{ 2 } }{ 2{ a }^{ 2 } + bc } + \frac { { b }^{ 2 } }{ 2b^{ 2 } + ca } + \frac { { c }^{ 2 } }{ 2c^{ 2 } + ab }$
Let $$a$$, $$b$$ and $$c$$ be real numbers satisfying $$a + b + c = 0$$, find the value of $$P$$ such that $$P$$ is defined.
× | {"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": 0.9209915995597839, "perplexity": 344.4833095698463}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280761.39/warc/CC-MAIN-20170116095120-00034-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://www.jiskha.com/questions/69200/write-down-the-wave-equation-in-symbol-form | # science
write down the wave equation in symbol form?
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3. 👁 130
1. Thank you for using the Jiskha Homework Help Forum. Have a look here:
http://en.wikipedia.org/wiki/Electromagnetic_wave_equation
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https://en.wikipedia.org/wiki/Models_of_DNA_evolution | # Models of DNA evolution
A number of different Markov models of DNA sequence evolution have been proposed. These substitution models differ in terms of the parameters used to describe the rates at which one nucleotide replaces another during evolution. These models are frequently used in molecular phylogenetic analyses. In particular, they are used during the calculation of likelihood of a tree (in Bayesian and maximum likelihood approaches to tree estimation) and they are used to estimate the evolutionary distance between sequences from the observed differences between the sequences.
## Introduction
These models are phenomenological descriptions of the evolution of DNA as a string of four discrete states. These Markov models do not explicitly depict the mechanism of mutation nor the action of natural selection. Rather they describe the relative rates of different changes. For example, mutational biases and purifying selection favoring conservative changes are probably both responsible for the relatively high rate of transitions compared to transversions in evolving sequences. However, the Kimura (K80) model described below merely attempts to capture the effect of both forces in a parameter that reflects the relative rate of transitions to transversions.
Evolutionary analyses of sequences are conducted on a wide variety of time scales. Thus, it is convenient to express these models in terms of the instantaneous rates of change between different states (the Q matrices below). If we are given a starting (ancestral) state at one position, the model's Q matrix and a branch length expressing the expected number of changes to have occurred since the ancestor, then we can derive the probability of the descendant sequence having each of the four states. The mathematical details of this transformation from rate-matrix to probability matrix are described in the mathematics of substitution models section of the substitution model page. By expressing models in terms of the instantaneous rates of change we can avoid estimating a large numbers of parameters for each branch on a phylogenetic tree (or each comparison if the analysis involves many pairwise sequence comparisons).
The models described on this page describe the evolution of a single site within a set of sequences. They are often used for analyzing the evolution of an entire locus by making the simplifying assumption that different sites evolve independently and are identically distributed. This assumption may be justifiable if the sites can be assumed to be evolving neutrally. If the primary effect of natural selection on the evolution of the sequences is to constrain some sites, then models of among-site rate-heterogeneity can be used. This approach allows one to estimate only one matrix of relative rates of substitution, and another set of parameters describing the variance in the total rate of substitution across sites.
## DNA evolution as a continuous-time Markov chain
### Continuous-time Markov chains
Continuous-time Markov chains have the usual transition matrices which are, in addition, parameterized by time, ${\displaystyle t\ }$. Specifically, if ${\displaystyle E_{1},E_{2},E_{3},E_{4}\ }$ are the states, then the transition matrix
${\displaystyle P(t)={\big (}P_{ij}(t){\big )}}$ where each individual entry, ${\displaystyle P_{ij}(t)\ }$ refers to the probability that state ${\displaystyle E_{i}\ }$ will change to state ${\displaystyle E_{j}\ }$ in time ${\displaystyle t\ }$.
Example: We would like to model the substitution process in DNA sequences (i.e. Jukes–Cantor, Kimura, etc.) in a continuous-time fashion. The corresponding transition matrices will look like:
${\displaystyle P(t)={\begin{pmatrix}p_{AA}(t)&p_{GA}(t)&p_{CA}(t)&p_{TA}(t)\\p_{AG}(t)&p_{GG}(t)&p_{CG}(t)&p_{TG}(t)\\p_{AC}(t)&p_{GC}(t)&p_{CC}(t)&p_{TC}(t)\\p_{AT}(t)&p_{GT}(t)&p_{CT}(t)&p_{TT}(t)\end{pmatrix}}}$
where the top-left and bottom-right 2 × 2 blocks correspond to transition probabilities and the top-right and bottom-left 2 × 2 blocks corresponds to transversion probabilities.
Assumption: If at some time ${\displaystyle t_{0}\ }$, the Markov chain is in state ${\displaystyle E_{i}\ }$, then the probability that at time ${\displaystyle t_{0}+t\ }$, it will be in state ${\displaystyle E_{j}\ }$ depends only upon ${\displaystyle i\ }$, ${\displaystyle j\ }$ and ${\displaystyle t\ }$. This then allows us to write that probability as ${\displaystyle p_{ij}(t)\ }$.
Theorem: Continuous-time transition matrices satisfy:
${\displaystyle P(t+\tau )=P(t)P(\tau )\ }$
Note: There is here a possible confusion between two meanings of the word transition. (i) In the context of Markov chains, transition is the general term that refers to the change between two states. (ii) In the context of nucleotide changes in DNA sequences, transition is a specific term that refers to the exchange between either the two purines (A ↔ G) or the two pyrimidines (C ↔ T) (for additional details, see the article about transitions in genetics). By contrast, an exchange between one purine and one pyrimidine is called a transversion.
### Deriving the dynamics of substitution
Consider a DNA sequence of fixed length m evolving in time by base replacement. Assume that the processes followed by the m sites are Markovian independent, identically distributed and constant in time. For a fixed site, let
${\displaystyle \mathbf {P} (t)=(p_{A}(t),\ p_{G}(t),\ p_{C}(t),\ p_{T}(t))^{T}}$
probabilities of states ${\displaystyle A,\ }$ ${\displaystyle \ G,\ }$ ${\displaystyle \ C,\ }$ and ${\displaystyle \ T\ }$ at time ${\displaystyle t\ }$. Let
${\displaystyle {\mathcal {E}}=\{A,\ G,\ C,\ T\}}$
be the state-space. For two distinct
${\displaystyle x,y\in {\mathcal {E}}}$, let ${\displaystyle \mu _{xy}\ }$
be the transition rate from state ${\displaystyle x\ }$ to state ${\displaystyle y\ }$. Similarly, for any ${\displaystyle x\ }$, let:
${\displaystyle \mu _{x}=\sum _{y\neq x}\mu _{xy}}$
The changes in the probability distribution ${\displaystyle p_{A}(t)\ }$ for small increments of time ${\displaystyle \Delta t\ }$ are given by:
${\displaystyle p_{A}(t+\Delta t)=p_{A}(t)-p_{A}(t)\mu _{A}\Delta t+\sum _{x\neq A}p_{x}(t)\mu _{xA}\Delta t}$
In other words, (in frequentist language), the frequency of ${\displaystyle A\ }$'s at time ${\displaystyle t+\Delta t\ }$ is equal to the frequency at time ${\displaystyle t\ }$ minus the frequency of the lost ${\displaystyle A\ }$'s plus the frequency of the newly created ${\displaystyle A\ }$'s.
Similarly for the probabilities ${\displaystyle p_{G}(t),\ p_{C}(t),\ \mathrm {and} \ p_{T}(t)}$. We can write these compactly as:
${\displaystyle \mathbf {P} (t+\Delta t)=\mathbf {P} (t)+Q\mathbf {P} (t)\Delta t}$
where,
${\displaystyle Q={\begin{pmatrix}-\mu _{A}&\mu _{GA}&\mu _{CA}&\mu _{TA}\\\mu _{AG}&-\mu _{G}&\mu _{CG}&\mu _{TG}\\\mu _{AC}&\mu _{GC}&-\mu _{C}&\mu _{TC}\\\mu _{AT}&\mu _{GT}&\mu _{CT}&-\mu _{T}\end{pmatrix}}}$
or, alternately:
${\displaystyle \mathbf {P} '(t)=Q\mathbf {P} (t)}$
where, ${\displaystyle Q\ }$ is the rate matrix. Note that by definition, the columns of ${\displaystyle Q\ }$ sum to zero. For a stationary process, where ${\displaystyle Q\ }$ does not depend upon time t, this differential equation is solvable using matrix exponentiation:
${\displaystyle P(t)=\exp(Qt)}$ and
${\displaystyle \mathbf {P} (t)=P(t)\mathbf {P} (0)=\exp(Qt)\mathbf {P} (0)\,.}$
### Ergodicity
If all the transition probabilities, ${\displaystyle \mu _{xy}\ }$ are positive, i.e. if all states ${\displaystyle x,y\in {\mathcal {E}}\ }$ communicate, then the Markov chain has a unique stationary distribution ${\displaystyle \mathbf {\Pi } =\{\pi _{x},\ x\in {\mathcal {E}}\}}$ where each ${\displaystyle \pi _{x}\ }$ is the proportion of time spent in state ${\displaystyle x\ }$ after the Markov chain has run for infinite time. Such a Markov chain is called, ergodic. In DNA evolution, under the assumption of a common process for each site, the stationary frequencies, ${\displaystyle \pi _{A},\pi _{G},\pi _{C},\pi _{T}\ }$ correspond to equilibrium base compositions.
When the current distribution ${\displaystyle \mathbf {P} (t)}$ is the stationary distribution ${\displaystyle \mathbf {\Pi } }$, then it follows that ${\displaystyle Q\mathbf {\Pi } =0}$ using the differential equation above,
${\displaystyle Q\mathbf {\Pi } =Q\mathbf {P} (t)={\frac {d\mathbf {P} (t)}{dt}}=0\,.}$
### Time reversibility
Definition: A stationary Markov process is time reversible if (in the steady state) the amount of change from state ${\displaystyle x\ }$ to ${\displaystyle y\ }$ is equal to the amount of change from ${\displaystyle y\ }$ to ${\displaystyle x\ }$, (although the two states may occur with different frequencies). This means that:
${\displaystyle \pi _{x}\mu _{xy}=\pi _{y}\mu _{yx}\ }$
Not all stationary processes are reversible, however, almost all DNA evolution models assume time reversibility, which is considered to be a reasonable assumption.
Under the time reversibility assumption, let ${\displaystyle s_{xy}=\mu _{xy}/\pi _{y}\ }$, then it is easy to see that:
${\displaystyle s_{xy}=s_{yx}\ }$
Definition The symmetric term ${\displaystyle s_{xy}\ }$ is called the exchangeability between states ${\displaystyle x\ }$ and ${\displaystyle y\ }$. In other words, ${\displaystyle s_{xy}\ }$ is the fraction of the frequency of state ${\displaystyle x\ }$ that is the result of transitions from state ${\displaystyle y\ }$ to state ${\displaystyle x\ }$.
Corollary The 12 off-diagonal entries of the rate matrix, ${\displaystyle Q\ }$ (note the off-diagonal entries determine the diagonal entries, since the rows of ${\displaystyle Q\ }$ sum to zero) can be completely determined by 9 numbers; these are: 6 exchangeability terms and 3 stationary frequencies ${\displaystyle \pi _{x}\ }$, (since the stationary frequencies sum to 1).
### Scaling of branch lengths
By comparing extant sequences, one can determine the amount of sequence divergence. This raw measurement of divergence provides information about the number of changes that have occurred along the path separating the sequences. The simple count of differences (the Hamming distance) between sequences will often underestimate the number of substitution because of multiple hits (see homoplasy). Trying to estimate the exact number of changes that have occurred is difficult, and usually not necessary. Instead, branch lengths (and path lengths) in phylogenetic analyses are usually expressed in the expected number of changes per site. The path length is the product of the duration of the path in time and the mean rate of substitutions. While their product can be estimated, the rate and time are not identifiable from sequence divergence.
The descriptions of rate matrices on this page accurately reflect the relative magnitude of different substitutions, but these rate matrices are not scaled such that a branch length of 1 yields one expected change. This scaling can be accomplished by multiplying every element of the matrix by the same factor, or simply by scaling the branch lengths. If we use the β to denote the scaling factor, and ν to denote the branch length measured in the expected number of substitutions per site then βν is used the transition probability formulae below in place of μt. Note that ν is a parameter to be estimated from data, and is referred to as the branch length, while β is simply a number that can be calculated from the rate matrix (it is not a separate free parameter).
The value of β can be found by forcing the expected rate of flux of states to 1. The diagonal entries of the rate-matrix (the Q matrix) represent -1 times the rate of leaving each state. For time-reversible models, we know the equilibrium state frequencies (these are simply the πi parameter value for state i). Thus we can find the expected rate of change by calculating the sum of flux out of each state weighted by the proportion of sites that are expected to be in that class. Setting β to be the reciprocal of this sum will guarantee that scaled process has an expected flux of 1:
${\displaystyle \beta =1/\left(-\sum _{i}\pi _{i}\mu _{ii}\right)}$
For example, in the Jukes-Cantor, the scaling factor would be 4/(3μ) because the rate of leaving each state is 3μ/4.
## Most common models of DNA evolution
### JC69 model (Jukes and Cantor, 1969)[1]
JC69 is the simplest substitution model. There are several assumptions. It assumes equal base frequencies ${\displaystyle \left(\pi _{A}=\pi _{G}=\pi _{C}=\pi _{T}={1 \over 4}\right)}$ and equal mutation rates. The only parameter of this model is therefore ${\displaystyle \mu }$, the overall substitution rate. As previously mentioned, this variable becomes a constant when we normalize the mean-rate to 1.
${\displaystyle Q={\begin{pmatrix}{*}&{\mu \over 4}&{\mu \over 4}&{\mu \over 4}\\{\mu \over 4}&{*}&{\mu \over 4}&{\mu \over 4}\\{\mu \over 4}&{\mu \over 4}&{*}&{\mu \over 4}\\{\mu \over 4}&{\mu \over 4}&{\mu \over 4}&{*}\end{pmatrix}}}$
Probability ${\displaystyle P_{ij}}$ of changing from initial state ${\displaystyle i}$ to final state ${\displaystyle j}$ as a function of the branch length (${\displaystyle \nu }$) for JC69. Red curve: nucleotide states ${\displaystyle i}$ and ${\displaystyle j}$ are different. Blue curve: initial and final states are the same. After a long time, probabilities tend to the nucleotide equilibrium frequencies (0.25: dashed line).
${\displaystyle P={\begin{pmatrix}{{1 \over 4}+{3 \over 4}e^{-t\mu }}&{{1 \over 4}-{1 \over 4}e^{-t\mu }}&{{1 \over 4}-{1 \over 4}e^{-t\mu }}&{{1 \over 4}-{1 \over 4}e^{-t\mu }}\\\\{{1 \over 4}-{1 \over 4}e^{-t\mu }}&{{1 \over 4}+{3 \over 4}e^{-t\mu }}&{{1 \over 4}-{1 \over 4}e^{-t\mu }}&{{1 \over 4}-{1 \over 4}e^{-t\mu }}\\\\{{1 \over 4}-{1 \over 4}e^{-t\mu }}&{{1 \over 4}-{1 \over 4}e^{-t\mu }}&{{1 \over 4}+{3 \over 4}e^{-t\mu }}&{{1 \over 4}-{1 \over 4}e^{-t\mu }}\\\\{{1 \over 4}-{1 \over 4}e^{-t\mu }}&{{1 \over 4}-{1 \over 4}e^{-t\mu }}&{{1 \over 4}-{1 \over 4}e^{-t\mu }}&{{1 \over 4}+{3 \over 4}e^{-t\mu }}\end{pmatrix}}}$
When branch length, ${\displaystyle \nu }$, is measured in the expected number of changes per site then:
${\displaystyle P_{ij}(\nu )=\left\{{\begin{array}{cc}{1 \over 4}+{3 \over 4}e^{-4\nu /3}&{\mbox{ if }}i=j\\{1 \over 4}-{1 \over 4}e^{-4\nu /3}&{\mbox{ if }}i\neq j\end{array}}\right.}$
It is worth noticing that ${\displaystyle \nu ={3 \over 4}t\mu =({\mu \over 4}+{\mu \over 4}+{\mu \over 4})t}$ what stands for sum of any column (or row) of matrix ${\displaystyle Q}$ multiplied by time and thus means expected number of substitutions in time ${\displaystyle t}$ (branch duration) for each particular site (per site) when the rate of substitution equals ${\displaystyle \mu }$.
Given the proportion ${\displaystyle p}$ of sites that differ between the two sequences the Jukes-Cantor estimate of the evolutionary distance (in terms of the expected number of changes) between two sequences is given by
${\displaystyle {\hat {d}}=-{3 \over 4}\ln({1-{4 \over 3}p})={\hat {\nu }}}$
The ${\displaystyle p}$ in this formula is frequently referred to as the ${\displaystyle p}$-distance. It is a sufficient statistic for calculating the Jukes-Cantor distance correction, but is not sufficient for the calculation of the evolutionary distance under the more complex models that follow (also note that ${\displaystyle p}$ used in subsequent formulae is not identical to the "${\displaystyle p}$-distance").
### K80 model (Kimura, 1980)[2]
The K80 model distinguishes between transitions (A <-> G, i.e. from purine to purine, or C <-> T, i.e. from pyrimidine to pyrimidine) and transversions (from purine to pyrimidine or vice versa). In Kimura's original description of the model the α and β were used to denote the rates of these types of substitutions, but it is now more common to set the rate of transversions to 1 and use κ to denote the transition/transversion rate ratio (as is done below). The K80 model assumes that all of the bases are equally frequent (πTCAG=0.25).
Rate matrix ${\displaystyle Q={\begin{pmatrix}{*}&{\kappa }&{1}&{1}\\{\kappa }&{*}&{1}&{1}\\{1}&{1}&{*}&{\kappa }\\{1}&{1}&{\kappa }&{*}\end{pmatrix}}}$
The Kimura two-parameter distance is given by:
${\displaystyle K=-{1 \over 2}\ln((1-2p-q){\sqrt {1-2q}})}$
where p is the proportion of sites that show transitional differences and q is the proportion of sites that show transversional differences.
### F81 model (Felsenstein 1981)[3]
Felsenstein's 1981 model is an extension of the JC69 model in which base frequencies are allowed to vary from 0.25 (${\displaystyle \pi _{T}\neq \pi _{C}\neq \pi _{A}\neq \pi _{G}}$)
Rate matrix:
${\displaystyle Q={\begin{pmatrix}{*}&{\pi _{C}}&{\pi _{A}}&{\pi _{G}}\\{\pi _{T}}&{*}&{\pi _{A}}&{\pi _{G}}\\{\pi _{T}}&{\pi _{C}}&{*}&{\pi _{G}}\\{\pi _{T}}&{\pi _{C}}&{\pi _{A}}&{*}\end{pmatrix}}}$
When branch length, ν, is measured in the expected number of changes per site then:
${\displaystyle \beta =1/(1-\pi _{A}^{2}-\pi _{C}^{2}-\pi _{G}^{2}-\pi _{T}^{2})}$
${\displaystyle P_{ij}(\nu )=\left\{{\begin{array}{cc}e^{-\beta \nu }+\pi _{j}\left(1-e^{-\beta \nu }\right)&{\mbox{ if }}i=j\\\pi _{j}\left(1-e^{-\beta \nu }\right)&{\mbox{ if }}i\neq j\end{array}}\right.}$
### HKY85 model (Hasegawa, Kishino and Yano 1985)[4]
The HKY85 model can be thought of as combining the extensions made in the Kimura80 and Felsenstein81 models. Namely, it distinguishes between the rate of transitions and transversions (using the κ parameter), and it allows unequal base frequencies (${\displaystyle \pi _{T}\neq \pi _{C}\neq \pi _{A}\neq \pi _{G}}$). [ Felsenstein described a similar (but not equivalent) model in 1984 using a different parameterization;[5] that latter model is referred to as the F84 model.[6] ]
Rate matrix ${\displaystyle Q={\begin{pmatrix}{*}&{\kappa \pi _{C}}&{\pi _{A}}&{\pi _{G}}\\{\kappa \pi _{T}}&{*}&{\pi _{A}}&{\pi _{G}}\\{\pi _{T}}&{\pi _{C}}&{*}&{\kappa \pi _{G}}\\{\pi _{T}}&{\pi _{C}}&{\kappa \pi _{A}}&{*}\end{pmatrix}}}$
If we express the branch length, ν in terms of the expected number of changes per site then:
${\displaystyle \beta ={\frac {1}{2(\pi _{A}+\pi _{G})(\pi _{C}+\pi _{T})+2\kappa [(\pi _{A}\pi _{G})+(\pi _{C}\pi _{T})]}}}$
${\displaystyle P_{AA}(\nu ,\kappa ,\pi )=\left[\pi _{A}\left(\pi _{A}+\pi _{G}+(\pi _{C}+\pi _{T})e^{-\beta \nu }\right)+\pi _{G}e^{-(1+(\pi _{A}+\pi _{G})(\kappa -1.0))\beta \nu }\right]/(\pi _{A}+\pi _{G})}$
${\displaystyle P_{AC}(\nu ,\kappa ,\pi )=\pi _{C}\left(1.0-e^{-\beta \nu }\right)}$
${\displaystyle P_{AG}(\nu ,\kappa ,\pi )=\left[\pi _{G}\left(\pi _{A}+\pi _{G}+(\pi _{C}+\pi _{T})e^{-\beta \nu }\right)-\pi _{G}e^{-(1+(\pi _{A}+\pi _{G})(\kappa -1.0))\beta \nu }\right]/\left(\pi _{A}+\pi _{G}\right)}$
${\displaystyle P_{AT}(\nu ,\kappa ,\pi )=\pi _{T}\left(1.0-e^{-\beta \nu }\right)}$
and formula for the other combinations of states can be obtained by substituting in the appropriate base frequencies.
### T92 model (Tamura 1992)[7]
T92 is a simple mathematical method developed to estimate the number of nucleotide substitutions per site between two DNA sequences, by extending Kimura’s (1980) two-parameter method to the case where a G+C-content bias exists. This method will be useful when there are strong transition-transversion and G+C-content biases, as in the case of Drosophila mitochondrial DNA. (Tamura 1992)
One frequency only ${\displaystyle \pi _{GC}}$
${\displaystyle \pi _{G}=\pi _{C}={\pi _{GC} \over 2}}$
${\displaystyle \pi _{A}=\pi _{T}={(1-\pi _{GC}) \over 2}}$
Rate matrix ${\displaystyle Q={\begin{pmatrix}{*}&{\kappa (1-\pi _{GC})/2}&{(1-\pi _{GC})/2}&{(1-\pi _{GC})/2}\\{\kappa \pi _{GC}/2}&{*}&{\pi _{GC}/2}&{\pi _{GC}/2}\\{(1-\pi _{GC})/2}&{(1-\pi _{GC})/2}&{*}&{\kappa (1-\pi _{GC})/2}\\{\pi _{GC}/2}&{\pi _{GC}/2}&{\kappa \pi _{GC}/2}&{*}\end{pmatrix}}}$
The evolutionary distance between two noncoding sequences according to this model is given by
${\displaystyle d=-h\ln(1-{p \over h}-q)-{1 \over 2}(1-h)\ln(1-2q)}$
where ${\displaystyle h=2\theta (1-\theta )}$ where ${\displaystyle \theta \in (0,1)}$ is the GC content.
### TN93 model (Tamura and Nei 1993)[8]
The TN93 model distinguishes between the two different types of transition - i.e. (A <-> G) is allowed to have a different rate to (C<->T). Transversions are all assumed to occur at the same rate, but that rate is allowed to be different from both of the rates for transitions.
TN93 also allows unequal base frequencies (${\displaystyle \pi _{T}\neq \pi _{C}\neq \pi _{A}\neq \pi _{G}}$).
Rate matrix ${\displaystyle Q={\begin{pmatrix}{*}&{\kappa _{1}\pi _{C}}&{\pi _{A}}&{\pi _{G}}\\{\kappa _{1}\pi _{T}}&{*}&{\pi _{A}}&{\pi _{G}}\\{\pi _{T}}&{\pi _{C}}&{*}&{\kappa _{2}\pi _{G}}\\{\pi _{T}}&{\pi _{C}}&{\kappa _{2}\pi _{A}}&{*}\end{pmatrix}}}$
### GTR: Generalised time-reversible (Tavaré 1986)[9]
GTR is the most general neutral, independent, finite-sites, time-reversible model possible. It was first described in a general form by Simon Tavaré in 1986.[9]
The GTR parameters consist of an equilibrium base frequency vector, ${\displaystyle \Pi =(\pi _{T},\pi _{C},\pi _{A},\pi _{G})}$, giving the frequency at which each base occurs at each site, and the rate matrix
${\displaystyle Q={\begin{pmatrix}{-(\alpha \pi _{C}+\beta \pi _{A}+\gamma \pi _{G})}&{\alpha \pi _{C}}&{\beta \pi _{A}}&{\gamma \pi _{G}}\\{\alpha \pi _{T}}&{-(\alpha \pi _{T}+\delta \pi _{A}+\epsilon \pi _{G})}&{\delta \pi _{A}}&{\epsilon \pi _{G}}\\{\beta \pi _{T}}&{\delta \pi _{C}}&{-(\beta \pi _{T}+\delta \pi _{C}+\eta \pi _{G})}&{\eta \pi _{G}}\\{\gamma \pi _{T}}&{\epsilon \pi _{C}}&{\eta \pi _{A}}&{-(\gamma \pi _{T}+\epsilon \pi _{C}+\eta \pi _{A})}\end{pmatrix}}}$
Where
{\displaystyle {\begin{aligned}\alpha =r(T\rightarrow C)=r(C\rightarrow T)\\\beta =r(T\rightarrow A)=r(A\rightarrow T)\\\gamma =r(T\rightarrow G)=r(G\rightarrow T)\\\delta =r(C\rightarrow A)=r(A\rightarrow C)\\\epsilon =r(C\rightarrow G)=r(G\rightarrow C)\\\eta =r(A\rightarrow G)=r(G\rightarrow A)\end{aligned}}}
are the transition rate parameters.
Therefore, GTR (for four characters, as is often the case in phylogenetics) requires 6 substitution rate parameters, as well as 4 equilibrium base frequency parameters. However, this is usually eliminated down to 9 parameters plus ${\displaystyle \mu }$, the overall number of substitutions per unit time. When measuring time in substitutions (${\displaystyle \mu }$=1) only 8 free parameters remain.
In general, to compute the number of parameters, one must count the number of entries above the diagonal in the matrix, i.e. for n trait values per site ${\displaystyle {{n^{2}-n} \over 2}}$, and then add n for the equilibrium base frequencies, and subtract 1 because ${\displaystyle \mu }$ is fixed. One gets
${\displaystyle {{n^{2}-n} \over 2}+n-1={1 \over 2}n^{2}+{1 \over 2}n-1.}$
For example, for an amino acid sequence (there are 20 "standard" amino acids that make up proteins), one would find there are 209 parameters. However, when studying coding regions of the genome, it is more common to work with a codon substitution model (a codon is three bases and codes for one amino acid in a protein). There are ${\displaystyle 4^{3}=64}$ codons, but the rates for transitions between codons which differ by more than one base is assumed to be zero. Hence, there are ${\displaystyle {{20\times 19\times 3} \over 2}+64-1=633}$ parameters.
## References
1. ^ Jukes TH & Cantor CR (1969). Evolution of Protein Molecules. New York: Academic Press. pp. 21–132.
2. ^ Kimura M (1980). "A simple method for estimating evolutionary rates of base substitutions through comparative studies of nucleotide sequences". Journal of Molecular Evolution. 16 (2): 111–120. doi:10.1007/BF01731581. PMID 7463489.
3. ^ Felsenstein J (1981). "Evolutionary trees from DNA sequences: a maximum likelihood approach". Journal of Molecular Evolution. 17 (6): 368–376. doi:10.1007/BF01734359. PMID 7288891.
4. ^ Hasegawa M, Kishino H, Yano T (1985). "Dating of human-ape splitting by a molecular clock of mitochondrial DNA". Journal of Molecular Evolution. 22 (2): 160–174. doi:10.1007/BF02101694. PMID 3934395.
5. ^ Kishino H, Hasegawa M (1989). "Evaluation of the maximum likelihood estimate of the evolutionary tree topologies from DNA sequence data, and the branching order in hominoidea". Journal of Molecular Evolution. 29 (2): 170–179. doi:10.1007/BF02100115. PMID 2509717.
6. ^ Felsenstein J, Churchill GA (1996). "A Hidden Markov Model approach to variation among sites in rate of evolution, and the branching order in hominoidea". Molecular Biology and Evolution. 13 (1): 93–104. doi:10.1093/oxfordjournals.molbev.a025575. PMID 8583911.
7. ^ Tamura K (1992). "Estimation of the number of nucleotide substitutions when there are strong transition-transversion and G+C content biases". Molecular Biology and Evolution. 9 (4): 678–687. PMID 1630306.
8. ^ Tamura K, Nei M (1993). "Estimation of the number of nucleotide substitutions in the control region of mitochondrial DNA in humans and chimpanzees". Molecular Biology and Evolution. 10 (3): 512–526. PMID 8336541.
9. ^ a b Tavaré S (1986). "Some Probabilistic and Statistical Problems in the Analysis of DNA Sequences" (PDF). Lectures on Mathematics in the Life Sciences. American Mathematical Society. 17: 57–86. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 129, "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": 0.9703863859176636, "perplexity": 758.3744827282745}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982950764.62/warc/CC-MAIN-20160823200910-00096-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://rd.springer.com/chapter/10.1007/978-3-319-29130-7_9 | Concurrent Scale Interactions in the Far-Field of a Turbulent Mixing Layer
Conference paper
Part of the Springer Proceedings in Physics book series (SPPHY, volume 165)
Abstract
The interaction between large- and small-scale fluctuations in turbulent flow is not only of great fundamental interest but an understanding of these interactions is fundamental to the modelling of the sub-grid scale (SGS) stresses in a large eddy simulation (LES). Particle image velocimetry (PIV) data is acquired with two different spatial resolutions, simultaneously, in the self-similar region of a turbulent planar mixing layer. The SGS activity is observed to be amplified by concurrent large-scale low momentum fluctuations and attenuated by high momentum fluctuations with both the sign and magnitude of the large-scale fluctuations being of significance. Further, regions in which the orientation of the large-scale Reynolds stress tensor is aligned perpendicularly to the direction of the mean shear of the flow are shown to lead to an increased level of small-scale activity.
Keywords
Particle Image Velocimetry Taylor Microscale Open Area Ratio Free Shear Flow High Speed Side
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.
References
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Bandyopadhyay, P., Hussain, A.: The coupling between scales in shear flows. Phys. Fluids 27(9), 2221–2228 (1984)
2. 2.
Buxton, O.: Fine scale features of turbulent shear flows. PhD thesis (2011)Google Scholar
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Kolmogorov, A.: The local structure of turbulence in incompressible viscous fluid for very large Reynolds numbers. C.R. Acad. Sci. URSS 30(4), 301–305 (1941)
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Mathis, R., Hutchins, N., Marusic, I.: Large-scale amplitude modulation of the small-scale structures in turbulent boundary layers. J. Fluid Mech. 628, 311–337 (2009)
5. 5.
Priyadarshana, P., Klewicki, J., Treat, S., Foss, J.: Statistical structure of turbulent-boundary-layer velocity—vorticity products at high and low Reynolds numbers. J. Fluid Mech. 570, 307–346 (2007)
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Richardson, L.: Atmospheric diffusion shown on a distance-neighbour graph. Proc. R. Soc. Lond. A 110(756), 709–737 (1926) | {"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": 0.8573459982872009, "perplexity": 2631.190678103223}, "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-2018-22/segments/1526794867977.85/warc/CC-MAIN-20180527004958-20180527024958-00186.warc.gz"} |
https://physics.aps.org/articles/v8/45 | Viewpoint
# A More Precise Higgs Boson Mass
Physics 8, 45
A new value for the Higgs boson mass will allow stronger tests of the standard model and of theories about the Universe’s stability.
A great insight of twentieth-century science is that symmetries expressed in the laws of nature need not be manifest in the outcomes of those laws. Consider the snowflake. Its structure is a consequence of electromagnetic interactions, which are identical from any direction, but a snowflake only looks the same when rotated by multiples of $60∘$ about a single axis. The full symmetry is hidden by the particular conditions under which the water molecules crystallize. Similarly, a symmetry relates the electromagnetic and weak interactions in the standard model of particle physics, but we know it must be concealed because the weak interactions appear much weaker than electromagnetism. To learn what distinguishes electromagnetism from the weak interactions was an early goal of experiments at CERN’s Large Hadron Collider (LHC). A big part of the answer was given in mid-2012, when the ATLAS and CMS Collaborations at the LHC announced the discovery of the Higgs boson in the study of proton–proton collisions [1]. Now the discovery teams have pooled their data analyses to produce a measurement of the Higgs boson mass with $0.2%$ precision [2]. The new value enables physicists to make more stringent tests of the electroweak theory and of the Higgs boson’s properties.
The electroweak theory [3] is a key element of the standard model of particle physics that weaves together ideas and observations from diverse areas of physics [4]. In the theory, interactions are prescribed by gauge symmetries. If nature displayed these symmetries explicitly, the force particles would all be massless, whereas we know experimentally that the weak interactions must—because they are short-ranged—be mediated by massive particles. The so-called Higgs field was introduced to the electroweak theory to hide the gauge symmetry, leading to weak force particles ( $W±$ and $Z0$) that have mass but a photon that is massless.
The Higgs boson is a spin-zero excitation of the Higgs field and the “footprint” of the mechanism that hides the electroweak gauge symmetry in the standard model. The Higgs boson’s interactions are fully specified in terms of known couplings and masses of its decay products, but the theory does not predict its mass. Instead, experimentalists must measure the energies and momenta of the Higgs boson’s decay products and determine its mass using kinematical equations. Once that mass is known, the rates at which the Higgs boson decays into different particles can be predicted with high precision, and compared with experiment. For a mass in the neighborhood of $125$ giga-electron-volts ( $GeV$), the electroweak theory foresees a happy circumstance in which several decay paths occur at large enough rates to be detected.
ATLAS and CMS are large, broad-acceptance detectors located in multistory caverns about $100$ meters below ground [5]. In the discovery run of the LHC, the ATLAS and CMS Collaborations searched for decays of a Higgs boson into bottom-quark–antiquark pairs, tau-lepton pairs, and pairs of electroweak gauge bosons: two photons, $W+W-$, and $Z0Z0$. The actual discovery was based primarily on mass peaks associated with either the two-photon final states or $Z0Z0$ pairs decaying to four-lepton (electrons or muons) final states. These channels, for which the ATLAS and CMS detectors have the best mass resolution, form the basis of their new report.
Both of the “high-resolution” final states are relatively rare: the standard model predicts that only about $1/4%$ of Higgs boson decays produce two-photon states; the four-lepton rate is predicted to be nearly 20 times smaller. The two-photon channel exhibits a narrow resonance peak that contains several hundred events per experiment; the $Z0Z0$ to four-lepton channel yields only a few tens of signal events per experiment. To see these events in the first run of the LHC, the ATLAS and CMS collaborations chose different detector technologies, and therefore different measurement and calibration methods [2]. These differences make pooling the data complicated, but also allow the experimentalists to cross-check systematic uncertainties in their separate measurements. Their combined analyses yield a Higgs boson mass of $125.09±0.24GeV$, the precision of which is limited by statistics and by uncertainties in the energy or momentum scale of the ATLAS and CMS detectors.
The first consequence of the new, precise mass value is sharper predictions, within the standard model, for the relative probabilities of different Higgs boson decay modes and production rates [6]. So far, the measured decay modes and production rates agree with standard-model predictions. The current uncertainties in the measured rates are large, but they will be narrowed in the coming runs at the LHC and at possible future colliders. Evidence of any deviation would suggest that the Higgs boson does not follow the standard model textbook, or that new particles or new forces are implicated in its decays.
With a precisely known Higgs boson mass $MH$, theorists can also make more refined predictions of the quantum corrections to many observables, such as the $Z0$ decay rates. These predictions test the consistency of the electroweak theory as a quantum field theory. Figure 1 illustrates a telling example [7]. The diagonal blue ellipses show the values of the $W$ boson and top quark masses required to reproduce a selection of electroweak observables once $MH$ is fixed. (The narrow and wide ellipses represent $68%$ and $95%$ confidence levels, respectively.) The range of masses depends on $MH$, and the precision with which it is known controls the width of the blue ellipses. The preferred range overlaps the green ellipses, which show the directly measured values of the $W$ boson and top quark masses. In the future, more precise values for the masses of the Higgs boson, $W$ boson, and top quark could unveil a discrepancy that might lead to the discovery of new physics.
The specific value of $MH$ constrains speculations about physics beyond the standard model, including supersymmetric or composite models. Perhaps most provocative of all is the possibility that the measured value of the mass is special. Quantum corrections influence not just the predictions for observable quantities, but also the shape of the Higgs potential that lies behind electroweak symmetry breaking in the standard model. According to recent analyses, the newly reported value of the Higgs boson mass corresponds to a near-critical situation in which the Higgs vacuum does not lie at the state of lowest energy, but in a metastable state close to a phase transition [8]. This might imply that our Universe is living on borrowed time, or that the electroweak theory must be augmented in some way.
With LHC Run 2 about to commence, now at higher energies, particle physicists can look forward to a new round of exploration, searches for new phenomena, and refined measurements. Combined analyses and critical evaluations, such as the measurement of the Higgs boson mass discussed here, will help make the most of the data. We still have much to learn about the Higgs boson, the electroweak theory, and beyond.
Acknowledgments
Fermilab is operated by Fermi Research Alliance, LLC, under Contract No. DE-AC02-07CH11359 with the United States Department of Energy. I thank the Fondation Meyer pour le développement culturel et artistique for generous support.
This research is published in Physical Review Letters.
Correction (18 May 2015): In an earlier version of this article, the axes for Figure 1 were incorrectly labeled. The horizontal axis should say “Top quark mass” and the vertical axis should say “W boson mass”.
## References
1. G. Aad et al. (ATLAS Collaboration), “Observation of a New Particle in the Search for the Standard Model Higgs Boson with the ATLAS Detector at the LHC,” Phys. Lett. B 716, 1 (2012); S. Chatrchyan et al. (CMS Collaboration), “Observation of a New Boson at a Mass of 125 GeV with the CMS Experiment at the LHC,” Phys. Lett. B 716, 30 (2012)
2. G. Aad et al. (ATLAS Collaboration, CMS Collaboration), “Combined Measurement of the Higgs Boson Mass in $p\phantom{\rule{0}{0ex}}p$ Collisions at $\sqrt{s}=7$ and 8 TeV with the ATLAS and CMS Experiments,” Phys. Rev. Lett. 114, 191803 (2015)
3. The electroweak theory was developed from a proposal by S. Weinberg, “A Model of Leptons,” Phys. Rev. Lett. 19, 1264 (1967); A. Salam “Weak Electromagnetic Interactions,” in Elementary Particle Theory: Relativistic Groups and Analyticity (Nobel Symposium No. 8), edited by N. Svartholm (Almqvist and Wiksell, Stockholm, 1968), p. 367; http://j.mp/r9dJOo; The theory is built on the SU(2)${}_{L}\otimes$U(1)${}_{Y}$ gauge symmetry investigated by S. L. Glashow, “Partial Symmetries of Weak Interactions,” Nucl. Phys. 22, 579 (1961)
4. C. Quigg, “Electroweak Symmetry Breaking in Historical Perspective,” Ann. Rev. Nucl. Part. Sci. (to be published); arXiv:1503.01756
5. ATLAS Collaboration, “The ATLAS Experiment at the CERN Large Hadron Collider,” JINST 3, S08003 (2008); CMS Collaboration, “The CMS Experiment at the CERN Large Hadron Collider,” 3, S08004 (2008)
6. S. Heinemeyer et al. (LHC Higgs Cross Section Working Group), Handbook of LHC Higgs Cross Sections: 3. Higgs Properties, Report No. CERN-2013-004; Tables of Higgs boson branching fractions are given at http://j.mp/1OrjQL0
7. M. Baak et al. (Gfitter Group), “The global electroweak fit at NNLO and prospects for the LHC and ILC,” Eur. Phys. J. C 74, 3046 (2014); a more detailed version of Figure 1 may be found at http://j.mp/1cvuXGQ
8. D. Buttazzo, G. Degrassi, P. P. Giardino, G. F. Giudice, F. Sala, A. Salvio, and A. Strumia, ”Investigating the Near-Criticality of the Higgs Boson,” J. High Energy Phys. 1312, 089 (2013)
Chris Quigg is Distinguished Scientist Emeritus at the Fermi National Accelerator Laboratory. His research spans many topics in particle physics, from heavy quarks through cosmic neutrinos. His work on electroweak symmetry breaking and supercollider physics, which was recognized by the 2011 J. J. Sakurai Prize of the American Physical Society for outstanding achievement in particle theory, charted the course for exploration at Fermilab’s Tevatron and CERN’s Large Hadron Collider. His current research centers on experiments at the LHC. He is completing a book on particle physics for inquisitive readers. Quigg served as co-chair of Snowmass 2001.
## Subject Areas
Particles and Fields
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A method for detecting dark matter using tiny levitated spheres could reach an unprecedented sensitivity to light dark matter particles. Read More » | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 40, "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": 0.9388019442558289, "perplexity": 1051.1624374805394}, "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/1652662521041.0/warc/CC-MAIN-20220518021247-20220518051247-00424.warc.gz"} |
https://en.wikibooks.org/wiki/LMIs_in_Control/Stability_Analysis/Continuous_Time/Transient_State_Bound_for_Non-Autonomous_LTI_Systems | # LMIs in Control/Stability Analysis/Continuous Time/Transient State Bound for Non-Autonomous LTI Systems
## The System
For a continuous-time LTI system with a state-space representation of:
${\displaystyle {\dot {x}}=Ax+Bu}$
where ${\displaystyle A\in \mathbb {R} ^{nxn}}$, ${\displaystyle B\in \mathbb {R} ^{nxm}}$ and x(0) = x0,
the transient bound can be evaluated with the following LMI.
## The Data
${\displaystyle A\in \mathbb {R} ^{nxn}}$, ${\displaystyle B\in \mathbb {R} ^{nxm}}$ and x(0) = x0.
## The LMI:
The Euclidean norm of the state satisfies:
${\displaystyle \lVert x(T)\rVert _{2}^{2}\leq \gamma ^{2}(\lVert x_{0}\rVert _{2}^{2}+\lVert u\rVert _{2T}^{2}),\forall T\in \mathbb {R} _{\geq 0}}$
if there exists ${\displaystyle P\in \mathbb {S} ^{n}}$ and ${\displaystyle \gamma \in \mathbb {R} _{>0}}$, where P > 0, such that:
• ${\displaystyle P-\gamma 1\leq 0,}$
• ${\displaystyle {\begin{bmatrix}P&1\\*&\gamma 1\end{bmatrix}}\geq 0,}$
• ${\displaystyle {\begin{bmatrix}PA+A^{T}P&PB\\*&-\gamma 1\end{bmatrix}}\leq 0.}$
if x0 = 0 and u is a unit-energy input (${\displaystyle \lVert u\rVert _{2T}\leq 1,\forall T\in \mathbb {R} _{\geq 0}}$), then the above LMIs ensure that ${\displaystyle \lVert x(T)\rVert _{2}\leq \gamma ,\forall T\in \mathbb {R} _{\geq 0}}$
## Conclusion
By using this LMI the transient state bound can be analyzed for a given non-autonomous LTI system.
## Implementation
The LMI given above can be implemented and solved using a tool such as YALMIP, along with an LMI solver such as MOSEK. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 13, "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": 0.9554923176765442, "perplexity": 2427.820071873425}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320305288.57/warc/CC-MAIN-20220127193303-20220127223303-00669.warc.gz"} |
https://www.britannica.com/science/electromagnetic-radiation/Relation-between-electricity-and-magnetism | ## Relation between electricity and magnetism
As early as 1760 the Swiss-born mathematician Leonhard Euler suggested that the same ether that propagates light is responsible for electrical phenomena. In comparison with both mechanics and optics, however, the science of electricity was slow to develop. Magnetism was the one science that made progress in the Middle Ages, following the introduction from China into the West of the magnetic compass, but electromagnetism played little part in the scientific revolution of the 17th century. It was, however, the only part of physics in which very significant progress was made during the 18th century. By the end of that century the laws of electrostatics—the behaviour of charged particles at rest—were well known, and the stage was set for the development of the elaborate mathematical description first made by the French mathematician Siméon-Denis Poisson. There was no apparent connection of electricity with magnetism, except that magnetic poles, like electric charges, attract and repel with an inverse-square law force.
Following the discoveries in electrochemistry (the chemical effects of electrical current) by the Italian investigators Luigi Galvani, a physiologist, and Alessandro Volta, a physicist, interest turned to current electricity. A search was made by the Danish physicist Hans Christian Ørsted for some connection between electric currents and magnetism, and during the winter of 1819–20 he observed the effect of a current on a magnetic needle. Members of the French Academy learned about Ørsted’s discovery in September 1820, and several of them began to investigate it further. Of these, the most thorough in both experiment and theory was the physicist André-Marie Ampère, who may be called the father of electrodynamics.
The list of four fundamental empirical laws of electricity and magnetism was made complete with the discovery of electromagnetic induction by Faraday in 1831. In brief, a change in magnetic flux through a conducting circuit produces a current in the circuit. The observation that the induced current is in a direction to oppose the change that produces it, now known as Lenz’s law, was formulated by a Russian-born physicist, Heinrich Friedrich Emil Lenz, in 1834. When the laws were put into mathematical form by Maxwell, the law of induction was generalized to include the production of electric force in space, independent of actual conducting circuits, but was otherwise unchanged. On the other hand, Ampère’s law describing the magnetic effect of a current required amendment in order to be consistent with the conservation of charge (the total charge must remain constant) in the presence of changing electric fields, and Maxwell introduced the idea of “displacement current” to make the set of equations logically consistent. As a result, he found on combining the equations that he arrived at a wave equation, according to which transverse electric and magnetic disturbances were propagated with a velocity that could be calculated from electrical measurements. These measurements were available to Maxwell, having been made in 1856 by the German physicists Rudolph Hermann Arndt Kohlrausch and Wilhelm Eduard Weber, and his calculation gave him a result that was the same, within the limits of error, as the speed of light in vacuum. It was the coincidence of this value with the velocity of the waves predicted by his theory that convinced Maxwell of the electromagnetic nature of light.
## The electromagnetic wave and field concept
Faraday introduced the concept of field and of field lines of force that exist outside material bodies. As he explained it, the region around and outside a magnet or an electric charge contains a field that describes at any location the force experienced by another small magnet or charge placed there. The lines of force around a magnet can be made visible by iron filings sprayed on a paper that is held over the magnet. The concept of field, specifying as it does a certain possible action or force at any location in space, was the key to understanding electromagnetic phenomena. It should be mentioned parenthetically that the field concept also plays (in varied forms) a pivotal role in modern theories of particles and forces.
Besides introducing this important concept of electric and magnetic field lines of force, Faraday had the extraordinary insight that electrical and magnetic actions are not transmitted instantaneously but after a certain lag in time, which increases with distance from the source. Moreover, he realized the connection between magnetism and light after observing that a substance such as glass can rotate the plane of polarization of light in the presence of a magnetic field. This remarkable phenomenon is known as the Faraday effect.
The Human Body
As noted above, Maxwell formulated a quantitative theory that linked the fundamental phenomena of electricity and magnetism and that predicted electromagnetic waves propagating with a speed, which, as well as one could determine at that time, was identical with the speed of light. He concluded his paper “On the Physical Lines of Force” (1861–62) by saying that electricity may be disseminated through space with properties identical with those of light. In 1864 Maxwell wrote that the numerical factor linking the electrostatic and the magnetic units was very close to the speed of light and that these results “show that light and magnetism are affections of the same substance, and that light is an electromagnetic disturbance propagated through the field according to [his] electromagnetic laws.”
What more was needed to convince the scientific community that the mystery of light was solved and the phenomena of electricity and magnetism were unified in a grand theory? Why did it take 25 more years for Maxwell’s theory to be accepted? For one, there was little direct proof of the new theory. Furthermore, Maxwell not only had adopted a complicated formalism but also explained its various aspects by unusual mechanical concepts. Even though he stated that all such phrases are to be considered as illustrative and not as explanatory, the French mathematician Henri Poincaré remarked in 1899 that the “complicated structure” which Maxwell attributed to the ether “rendered his system strange and unattractive.”
The ideas of Faraday and Maxwell that the field of force has a physical existence in space independent of material media were too new to be accepted without direct proof. On the Continent, particularly in Germany, matters were further complicated by the success of Carl Friedrich Gauss and Wilhelm Eduard Weber in developing a potential field theory for the phenomena of electrostatics and magnetostatics and their continuing effort to extend this formalism to electrodynamics.
It is difficult in hindsight to appreciate the reluctance to accept the Faraday–Maxwell theory. The impasse was finally removed by Hertz’s work. In 1884 Hertz derived Maxwell’s theory by a new method and put its fundamental equations into their present-day form. In so doing, he clarified the equations, making the symmetry of electric and magnetic fields apparent. The German physicist Arnold Sommerfeld spoke for most of his learned colleagues when, after reading Hertz’s paper, he remarked, “the shades fell from my eyes,” and admitted that he understood electromagnetic theory for the first time. Four years later, Hertz made a second major contribution: he succeeded in generating electromagnetic radiation of radio and microwave frequencies, measuring their speed by a standing-wave method and proving that these waves have the properties of reflection, diffraction, refraction, and interference common to light. He showed that such electromagnetic waves can be polarized, that the electric and magnetic fields oscillate in directions that are mutually perpendicular and transverse to the direction of motion, and that their velocity is the same as the speed of light, as predicted by Maxwell’s theory.
Hertz’s ingenious experiments not only settled the theoretical misconceptions in favour of Maxwell’s electromagnetic field theory but also opened the way for building transmitters, antennas, coaxial cables, and detectors for radio-frequency electromagnetic radiation. In 1896 Marconi received the first patent for wireless telegraphy, and in 1901 he achieved transatlantic radio communication.
The Faraday–Maxwell–Hertz theory of electromagnetic radiation, which is commonly referred to as Maxwell’s theory, makes no reference to a medium in which the electromagnetic waves propagate. A wave of this kind is produced, for example, when a line of charges is moved back and forth along the line. Moving charges represent an electric current. In this back-and-forth motion, the current flows in one direction and then in another. As a consequence of this reversal of current direction, the magnetic field around the current (discovered by Ørsted and Ampère) has to reverse its direction. The time-varying magnetic field produces perpendicular to it a time-varying electric field, as discovered by Faraday (Faraday’s law of induction). These time-varying electric and magnetic fields spread out from their source, the oscillating current, at the speed of light in free space. The oscillating current in this discussion is the oscillating current in a transmitting antenna, and the time-varying electric and magnetic fields that are perpendicular to one another propagate at the speed of light and constitute an electromagnetic wave. Its frequency is that of the oscillating charges in the antenna. Once generated, it is self-propagating because a time-varying electric field produces a time-varying magnetic field, and vice versa. Electromagnetic radiation travels through space by itself. The belief in the existence of an ether medium, however, was at the time of Maxwell as strong as at the time of Plato and Aristotle. It was impossible to visualize ether because contradictory properties had to be attributed to it in order to explain the phenomena known at any given time. In his article Ether in the ninth edition of the Encyclopædia Britannica, Maxwell described the vast expanse of the substance, some of it possibly even inside the planets, carried along with them or passing through them as the “water of the sea passes through the meshes of a net when it is towed along by a boat.”
If one believes in the ether, it is, of course, of fundamental importance to measure the speed of its motion or the effect of its motion on the speed of light. One does not know the absolute velocity of the ether, but, as Earth moves through its orbit around the Sun, there should be a difference in ether velocity along and perpendicular to Earth’s motion equal to its speed. If such is the case, the velocity of light and of any other electromagnetic radiation along and perpendicular to Earth’s motion should, predicted Maxwell, differ by a fraction that is equal to the square of the ratio of Earth’s velocity to that of light. This fraction is one part in 100 million.
Michelson set out to measure this effect and, as noted above, designed for this purpose the interferometer sketched in Figure 4. If it is assumed that the interferometer is turned so that half beam A is oriented parallel to Earth’s motion and half beam B is perpendicular to it, then the idea of using this instrument for measuring the effect of the ether motion is best explained by Michelson’s words to his children:
Two beams of light race against each other, like two swimmers, one struggling upstream and back, while the other, covering the same distance, just crosses the river and returns. The second swimmer will always win, if there is any current in the river.
An improved version of the interferometer, in which each half beam traversed its path eight times before both were reunited for interference, was built in 1887 by Michelson in collaboration with Morley. A heavy sandstone slab holding the interferometer was floated on a pool of mercury to allow rotation without vibration. Michelson and Morley could not detect any difference in the two light velocities parallel and perpendicular to Earth’s motion to an accuracy of one part in four billion. This negative result did not, however, shatter the belief in the existence of an ether because the ether could possibly be dragged along with Earth and thus be stationary around the Michelson-Morley apparatus. Hertz’s formulation of Maxwell’s theory made it clear that no medium of any sort was needed for the propagation of electromagnetic radiation. In spite of this, ether-drift experiments continued to be conducted until about the mid-1920s. All such tests confirmed Michelson’s negative results, and scientists finally came to accept the idea that no ether medium was needed for electromagnetic radiation.
## Speed of light
Much effort has been devoted to measuring the speed of light, beginning with the aforementioned work of Rømer in 1676. Rømer noticed that the orbital period of Jupiter’s first moon, Io, is apparently slowed as Earth and Jupiter move away from each other. The eclipses of Io occur later than expected when Jupiter is at its most remote position. This effect is understandable if light requires a finite time to reach Earth from Jupiter. From this effect, Rømer calculated the time required for light to travel from the Sun to Earth as 11 minutes. In 1728 James Bradley, an English astronomer, determined the speed of light from the apparent orbital motion of stars that is produced by Earth’s orbital motion. He computed the time for light to reach Earth from the Sun as 8 minutes 12 seconds. The first terrestrial measurements were made in 1849 by Fizeau and a year later by Foucault. Michelson improved on Foucault’s method and obtained an accuracy of one part in 100,000.
Any measurement of velocity requires, however, a definition of the measure of length and of time. Current techniques allow a determination of the velocity of electromagnetic radiation to a substantially higher degree of precision than permitted by the unit of length that scientists had applied earlier. In 1983 the value of the speed of light was fixed at exactly 299,792,458 metres per second, and this value was adopted as a new standard. As a consequence, the metre was redefined as the length of the path traveled by light in a vacuum over a time interval of 1/299,792,458 of a second. Furthermore, the second—the international unit of time—has been based on the frequency of electromagnetic radiation emitted by a cesium-133 atom.
## Development of the quantum theory of radiation
After a long struggle electromagnetic wave theory had triumphed. The Faraday-Maxwell-Hertz theory of electromagnetic radiation seemed to be able to explain all phenomena of light, electricity, and magnetism. The understanding of these phenomena enabled one to produce electromagnetic radiation of many different frequencies which had never been observed before and which opened a world of new opportunities. No one suspected that the conceptional foundations of physics were about to change again.
## Radiation laws and Planck’s light quanta
The quantum theory of absorption and emission of radiation announced in 1900 by Planck ushered in the era of modern physics. He proposed that all material systems can absorb or give off electromagnetic radiation only in “chunks” of energy, quanta E, and that these are proportional to the frequency of that radiation E = hν. (The constant of proportionality h is, as noted above, called Planck’s constant.)
Planck was led to this radically new insight by trying to explain the puzzling observation of the amount of electromagnetic radiation emitted by a hot body and, in particular, the dependence of the intensity of this incandescent radiation on temperature and on frequency. The quantitative aspects of the incandescent radiation constitute the radiation laws.
The Austrian physicist Josef Stefan found in 1879 that the total radiation energy per unit time emitted by a heated surface per unit area increases as the fourth power of its absolute temperature T (Kelvin scale). This means that the Sun’s surface, which is at T = 6,000 K, radiates per unit area (6,000/300)4 = 204 = 160,000 times more electromagnetic energy than does the same area of Earth’s surface, which is taken to be T = 300 K. In 1889 another Austrian physicist, Ludwig Boltzmann, used the second law of thermodynamics to derive this temperature dependence for an ideal substance that emits and absorbs all frequencies. Such an object that absorbs light of all colours looks black, and so was called a blackbody. The Stefan-Boltzmann law is written in quantitative form W = σT4, where W is the radiant energy emitted per second and per unit area and the constant of proportionality is σ = 0.136 calories per metre2-second-K4.
The wavelength or frequency distribution of blackbody radiation was studied in the 1890s by Wilhelm Wien of Germany. It was his idea to use as a good approximation for the ideal blackbody an oven with a small hole. Any radiation that enters the small hole is scattered and reflected from the inner walls of the oven so often that nearly all incoming radiation is absorbed and the chance of some of it finding its way out of the hole again can be made exceedingly small. The radiation coming out of this hole is then very close to the equilibrium blackbody electromagnetic radiation corresponding to the oven temperature. Wien found that the radiative energy dW per wavelength interval dλ has a maximum at a certain wavelength λm and that the maximum shifts to shorter wavelengths as the temperature T is increased, as illustrated in Figure 8. He found that the product λmT is an absolute constant: λmT = 0.2898 cm-K.
Wien’s law of the shift of the radiative power maximum to higher frequencies as the temperature is raised expresses in a quantitative form commonplace observations. Warm objects emit infrared radiation, which is felt by the skin; near T = 950 K a dull red glow can be observed; and the colour brightens to orange and yellow as the temperature is raised. The tungsten filament of a lightbulb is T = 2,500 K hot and emits bright light, yet the peak of its spectrum is still in the infrared according to Wien’s law. The peak shifts to the visible yellow when the temperature is T = 6,000 K, like that of the Sun’s surface.
where k is Boltzmann’s constant, well known from thermodynamics. With c = λν, Planck’s radiation law then becomes
This is in superb agreement with Wien’s experimental results when the value of h is properly chosen to fit the results. It should be pointed out that Planck’s quantization refers to the oscillators of the blackbody or of heated substances. These oscillators of frequency ν are incapable of absorbing or emitting electromagnetic radiation except in energy chunks of size . To explain quantized absorption and emission of radiation, it seemed sufficient to quantize only the energy levels of mechanical systems. Planck did not mean to say that electromagnetic radiation itself is quantized, or, as Einstein later put it, “The sale of beer in pint bottles does not imply that beer exists only in indivisible pint portions.” The idea that electromagnetic radiation itself is quantized was proposed by Einstein in 1905, as described in the subsequent section.
## Photoelectric effect
Hertz discovered the photoelectric effect (1887) quite by accident while generating electromagnetic waves and observing their propagation. His transmitter and receiver were induction coils with spark gaps. He measured the electromagnetic field strength by the maximum length of the spark of his detector. In order to observe this more accurately, he occasionally enclosed the spark gap of the receiver in a dark case. In doing so, he observed that the spark was always smaller with the case than without it. He concluded, correctly, that the light from the transmitter spark affected the electrical arcing of the receiver. He used a quartz prism to disperse the light of the transmitter spark and found that the ultraviolet part of the light spectrum was responsible for enhancing the receiver spark. Hertz took this discovery seriously because the only other effect of light on electrical phenomena known at that time was the increase in electrical conductance of the element selenium with light exposure.
A year after Hertz’s discovery, it became clear that ultraviolet radiation caused the emission of negatively charged particles from solid surfaces. Thomson’s discovery of electrons (1897) and his ensuing measurement of the ratio m/e (the ratio of mass to charge) finally made it possible to identify the negative particles emitted in the photoelectric effect with electrons. This was accomplished in 1899 by J.J. Thomson and independently by Philipp Lenard, one of Hertz’s students. Lenard discovered that for a given frequency of ultraviolet radiation the maximum kinetic energy of the emitted electrons depends on the metal used rather than on the intensity of the ultraviolet light. The light intensity increases the number but not the energy of emitted electrons. Moreover, he found that for each metal there is a minimum light frequency that is needed to induce the emission of electrons. Light of a frequency lower than this minimum frequency has no effect regardless of its intensity.
In 1905 Einstein published an article entitled “On a Heuristic Point of View about the Creation and Conversion of Light.” Here he deduced that electromagnetic radiation itself consists of “particles” of energy . He arrived at this conclusion by using a simple theoretical argument comparing the change in entropy of an ideal gas caused by an isothermal change in volume with the change in entropy of an equivalent volume change for electromagnetic radiation in accordance with Wien’s or Planck’s radiation law. This derivation and comparison made no references to substances and oscillators. At the end of this paper, Einstein concluded that if electromagnetic radiation is quantized, the absorption processes are thus quantized too, yielding an elegant explanation of the threshold energies and the intensity dependence of the photoelectric effect. He then predicted that the kinetic energy of the electrons emitted in the photoelectric effect increases with light frequency ν proportional to P, where P is “the amount of work that the electron must produce on leaving the body.” This quantity P, now called work function, depends on the kind of solid used, as discovered by Lenard.
Einstein’s path-breaking idea of light quanta was not widely accepted by his peers. Planck himself stated as late as 1913 in his recommendation for admitting Einstein to the Prussian Academy of Sciences “the fact that he [Einstein] may occasionally have missed the mark in his speculations, as, for example, with his hypothesis of light quanta, ought not to be held too much against him, for it is impossible to introduce new ideas, even in the exact sciences, without taking risks.” In order to explain a quantized absorption and emission of radiation by matter, it seemed sufficient to quantize the possible energy states in matter. The resistance against quantizing the energies of electromagnetic radiation itself is understandable in view of the incredible success of Maxwell’s theory of electromagnetic radiation and the overwhelming evidence of the wave nature of this radiation. Moreover, a formal similarity of two theoretical expressions, in Einstein’s 1905 paper, of the entropy of an ideal gas and the entropy of electromagnetic radiation was deemed insufficient evidence for a real correspondence.
Einstein’s prediction of the linear increase of the kinetic energy of photoemitted electrons with the frequency of light, P, was verified by Arthur Llewelyn Hughes, Owen Williams Richardson, and Karl Taylor Compton in 1912. In 1916 Robert Andrews Millikan measured both the frequency of the light and the kinetic energy of the electron emitted by the photoelectric effect and obtained a value for Planck’s constant h in close agreement with the value that had been arrived at by fitting Planck’s radiation law to the blackbody spectrum obtained by Wien.
## Compton effect
Convincing evidence of the particle nature of electromagnetic radiation was found in 1922 by the American physicist Arthur Holly Compton. While investigating the scattering of X-rays, he observed that such rays lose some of their energy in the scattering process and emerge with slightly decreased frequency. This energy loss increases with the scattering angle, θ, measured from the direction of an unscattered X-ray. This so-called Compton effect can be explained, according to classical mechanics, as an elastic collision of two particles comparable to the collision of two billiard balls. In this case, an X-ray photon of energy and momentum /c collides with an electron at rest. The recoiling electron was observed and measured by Compton and Alfred W. Simon in a Wilson cloud chamber. If one calculates the result of such an elastic collision by using the relativistic formulas for the energy and momentum of the scattered electron, one finds that the wavelength of an X-ray after (λ′) and before (λ) the scattering event differs by λ′ − λ = (h/mc)(1 − cos θ). Here m is the rest mass of the electron and h/mc is called Compton wavelength. It has the value 0.0243 angstrom. The energy of a photon of this wavelength is equal to the rest mass energy mc2 of an electron. One might argue that electrons in atoms are not at rest, but their kinetic energy is very small compared with that of energetic X-rays and can be disregarded in deriving Compton’s equation.
## Resonance absorption and recoil
During the mid-1800s the German physicist Gustav Robert Kirchhoff observed that atoms and molecules emit and absorb electromagnetic radiation at characteristic frequencies and that the emission and absorption frequencies are the same for a given substance. Such resonance absorption should, strictly speaking, not occur if one applies the photon picture due to the following argument. Since energy and momentum have to be conserved in the emission process, the atom recoils to the left as the photon is emitted to the right, just as a cannon recoils backward when a shot is fired. Because the recoiling atom carries off some kinetic recoil energy ER, the emitted photon energy is less than the energy difference of the atomic energy states by the amount ER. When a photon is absorbed by an atom, the momentum of the photon is likewise transmitted to the atom, thereby giving it a kinetic recoil energy ER. The absorbing photon must therefore supply not only the energy difference of the atomic energy states but the additional amount ER as well. Accordingly, resonance absorption should not occur because the emitted photon is missing 2ER to accomplish it.
Nevertheless, ever since Kirchhoff’s finding, investigators have observed resonance absorption for electronic transitions in atoms and molecules. This is because for visible light the recoil energy ER is very small compared with the natural energy uncertainty of atomic emission and absorption processes. The situation is, however, quite different for the emission and absorption of gamma-ray photons by nuclei. The recoil energy ER is more than 10,000 times as large for gamma-ray photons as for photons of visible light, and the nuclear energy transitions are much more sharply defined because their lifetime can be one million times longer than for electronic energy transitions. The particle nature of photons therefore prevents resonance absorption of gamma-ray photons by free nuclei.
In 1958 the German physicist Rudolf Ludwig Mössbauer discovered that recoilless gamma-ray resonance absorption is, nevertheless, possible if the emitting as well as the absorbing nuclei are embedded in a solid. In this case, there is a strong probability that the recoil momentum during absorption and emission of the gamma photon is taken up by the whole solid (or more precisely by its entire lattice). This then reduces the recoil energy to nearly zero and thus allows resonance absorption to occur even for gamma rays.
## Wave-particle duality
How can electromagnetic radiation behave like a particle in some cases while exhibiting wavelike properties that produce the interference and diffraction phenomena in others? This paradoxical behaviour came to be known as the wave-particle duality. Bohr rejected the idea of light quanta, and he searched for ways to explain the Compton effect and the photoelectric effect by arguing that the momentum and energy conservation laws need to be satisfied only statistically in the time average. In 1923 he stated that the hypothesis of light quanta excludes, in principle, the possibility of a rational definition of the concepts of frequency and wavelength that are essential for explaining interference.
The following year the conceptual foundations of physics were shaken by the French physicist Louis de Broglie, who suggested in his doctoral dissertation that the wave-particle duality applies not only to light but to a particle as well. De Broglie proposed that any object has wavelike properties. In particular, he showed that the orbits and energies of the hydrogen atom, as described by Bohr’s atomic model, correspond to the condition that the circumference of any orbit precisely matches an integral number of wavelengths λ of the matter waves of electrons. Any particle such as an electron moving with a momentum p has, according to de Broglie, a wavelength λ = h/p. This idea required a conceptual revolution of mechanics, which led to the wave and quantum mechanics of Erwin Schrödinger, Werner Heisenberg, and Max Born.
De Broglie’s idea of the wavelike behaviour of particles was quickly verified experimentally. In 1927 Clinton Joseph Davisson and Lester Germer of the United States observed diffraction and hence interference of electron waves by the regular arrangement of atoms in a crystal of nickel. That same year S. Kikuchi of Japan obtained an electron diffraction pattern by shooting electrons with an energy of 68 keV through a thin mica plate and recording the resultant diffraction pattern on a photographic plate. The observed pattern corresponded to electron waves having the wavelength predicted by de Broglie. The diffraction effects of helium atoms were found in 1930, and neutron diffraction has today become an indispensable tool for determining the magnetic and atomic structure of materials.
The interference pattern that results when a radiation front hits two slits in an opaque screen is often cited to explain the conceptual difficulty of the wave-particle duality. Consider an opaque screen with two openings A and B, called a double slit, and a photographic plate or a projection screen, as shown in Figure 9. A parallel wave with a wavelength λ passing through the double slit will produce the intensity pattern on the plate or screen as shown at the right of the figure. The intensity is greatest at the centre. It falls to zero at all locations x0, where the distances to the openings A and B differ by odd-number multiples of a half wavelength, as, for instance, λ/2, 3λ/2, and 5λ/2. The condition for such destructive interference is the same as for Michelson’s interferometer illustrated in Figure 4. Whereas a half-transparent mirror in Figure 4 divides the amplitude of each wave train in half, the division in Figure 9 through openings A and B is spatial. The latter is called division of wave front. Constructive interference or intensity maxima are observed on the screen at all positions whose distances from A and B differ by zero or an integer multiple of λ. This is the wave interpretation of the observed double-slit interference pattern.
The description of photons is necessarily different because a particle can obviously only pass through opening A or alternatively through opening B. Yet, no interference pattern is observed when either A or B is closed. Both A and B must be open simultaneously. It was thought for a time that one photon passing through A might interfere with another photon passing through B. That possibility was ruled out after the British physicist Geoffrey Taylor demonstrated in 1909 that the same interference pattern can be recorded on a photographic plate even when the light intensity is so feeble that only one photon is present in the apparatus at any one time.
Another attempt to understand the dual nature of electromagnetic radiation was to identify the photon with a wave train whose length is equal to its coherence length cτ, where τ is the coherence time, or the lifetime of an atomic transition from a higher to a lower internal atomic energy state, and c is the light velocity. This is the same as envisioning the photon to be an elongated wave packet, or “needle radiation.” Again, the term “photon” had a different meaning for different scientists, and wave nature and quantum structure remained incompatible. It was time to find a theory of electromagnetic radiation that would fuse the wave theory and the particle theory. Such a fusion was accomplished by quantum electrodynamics (QED).
## Quantum electrodynamics
Among the most convincing phenomena that demonstrate the quantum nature of light are the following. As the intensity of light is dimmed further and further, one can see individual quanta being registered in light detectors. If the eyes were about 10 times more sensitive, one would perceive the individual light pulses of fainter and fainter light sources as fewer and fewer flashes of equal intensity. Moreover, a movie has been made of the buildup of a two-slit interference pattern by individual photons, such as shown in Figure 9. Photons are particles, but they behave differently from ordinary particles like billiard balls. The rules of their behaviour and their interaction with electrons and other charged particles, as well as the interactions of charged particles with one another, constitute QED.
Photons are created by perturbations in the motions of electrons and other charged particles; in reverse, photons can disappear and thereby create a pair of oppositely charged particles, usually a particle and its antiparticle (e.g., an electron and a positron). A description of this intimate interaction between charged particles and electromagnetic radiation requires a theory that includes both quantum mechanics and special relativity. The foundations of such a theory, known as relativistic quantum mechanics, were laid beginning in 1929 by Paul A.M. Dirac, Heisenberg, and Wolfgang Pauli.
The discussion that follows explains in brief the principal conceptual elements of QED. Further information on the subject can be found in subatomic particle: The development of modern theory and quantum mechanics.
Many phenomena in nature do not depend on the reference scale of scientific measurements. For instance, in electromagnetism the difference in electrical potentials is relevant but not its absolute magnitude. During the 1920s, even before the emergence of quantum mechanics, the German physicist Hermann Weyl discussed the problem of constructing physical theories that are independent of certain reference bases or absolute magnitudes of certain parameters not only locally but everywhere in space. He called this property “Eichinvarianz,” which is the conceptual origin of the term “gauge invariance” that plays a crucial role in all modern quantum field theories.
In quantum mechanics all observable quantities are calculated from so-called wave functions, which are complex mathematical functions that include a phase factor. The absolute magnitude of this phase is irrelevant for the observable quantities calculated from these wave functions; hence, the theory describing, for example, the motion of an electron should be the same when the phase of its wave function is changed everywhere in space. This requirement of phase invariance, or gauge invariance, is equivalent to demanding that the total charge is conserved and does not disappear in physical processes or interactions. Experimentally one does indeed observe that charge is conserved in nature.
It turns out that a relativistic quantum theory of charged particles can be made gauge invariant if the interaction is mediated by a massless and chargeless entity which has all the properties of photons. Coulomb’s law of the force between charged particles can be derived from this theory, and the photon can be viewed as a “messenger” particle that conveys the electromagnetic force between charged particles of matter. In this theory, Maxwell’s equations for electric and magnetic fields are quantized.
The range of a force produced by a particle with nonzero mass is its Compton wavelength h/mc, which for electrons is about 2 × 10−10 cm. Since this length is large compared with distances over which stronger nuclear forces act, QED is a very precise theory for electrons.
Despite the conceptual elegance of the QED theory, it proved difficult to calculate the outcome of specific physical situations through its application. Richard P. Feynman and, independently, Julian S. Schwinger and Freeman Dyson of the United States and Tomonaga Shin’ichirō of Japan showed in 1948 that one could calculate the effects of the interactions as a power series in which the coupling constant is called the fine structure constant and has a value close to 1/137. A serious practical difficulty arose when each term in the series, which had to be summed to obtain the value of an observed quantity, turned out to be infinitely large. In short, the results of the calculations were meaningless. It was eventually found, however, that these divergences could be avoided by introducing “renormalized” couplings and particle masses, an idea conceived by the Dutch physicist Hendrik A. Kramers. Just as a ship moving through water has an enhanced mass due to the fluid that it drags along, so will an electron dragging along and interacting with its own field have a different mass and charge than it would without it. By adding appropriate electromagnetic components to the bare mass and charge—that is, by using renormalized quantities—the disturbing infinities could be removed from the theory. Using this method of renormalization and the perturbation theory, Feynman developed an elegant form for calculating the likelihood of observing processes that are related to the interaction of electromagnetic radiation with matter to any desired degree of accuracy. For example, the passage of an electron or a photon through the double slit illustrated in Figure 9 will, in this QED formalism, produce the observed interference pattern on a photographic plate because of the superposition of all the possible paths these particles can take through the slits.
The success of unifying electricity, magnetism, and light into one theory of electromagnetism and then with the interaction of charged particles into the theory of quantum electrodynamics suggests the possibility of understanding all the forces in nature (gravitational, electromagnetic, weak nuclear, and strong nuclear) as manifestations of a grand unified theory (GUT). The first step in this direction was taken during the 1960s by Abdus Salam, Steven Weinberg, and Sheldon Glashow, who formulated the electroweak theory, which combines the electromagnetic force and the weak nuclear force. This theory predicted that the weak nuclear force is transmitted between particles of matter by three messenger particles designated W+, W, and Z, much in the way that the electromagnetic force is conveyed by photons. The three new particles were discovered in 1983 during experiments at the European Organization for Nuclear Research (CERN), a large accelerator laboratory near Geneva. This triumph for the electroweak theory represented another stepping stone toward a deeper understanding of the forces and interactions that yield the multitude of physical phenomena in the universe.
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https://benediktehinger.de/blog/science/scatterplots-regression-lines-and-the-first-principal-component/ | # Scatterplots, regression lines and the first principal component
I made some graphs that show the relation between X1~X2 (X2 predicts X1), X2~X1 (X1 predicts X2) and the first principal component (direction with highest variance, also called total least squares).
The line you fit with a principal component is not the same line as in a regression (either predicting X2 by X1 [X2~X1] or X1 by X2 [X1~X2]. This is quite well known (see references below).
With regression one predicts X2 based on X1 (X2~X1 in R-Formula writing) or vice versa. With principal component (or total least squares) one tries to quantify the relation between the two. To completely understand the difference, image what quantity is reduced in the three cases.
In regression, we reduce the residuals in direction of the dependent variable. With principal components, we find the line, that has the smallest error orthogonal to the regression line. See the following image for a visual illustration.
For me it becomes interesting if you plot a scatter plot of two independent variables, i.e. you would usually report the correlation coefficient. The ‘correct’ line accompaniyng the correlation coefficient would be the principal component (‘correct’ as it is also agnostic to the order of the signals).
#### Further information:
How to draw the line, eLife 2013
Gelman, Hill – Data Analysis using Regression, p.58
Also check out the nice blogpost from Martin Johnsson doing practically the same thing but three years earlier 😉
#### Source
“R
library(ggplot2)
library(magrittr)
library(plyr)
set.seed(2)
corrCoef = 0.5 # sample from a multivariate normal, 10 datapoints
dat = MASS::mvrnorm(10,c(0,0),Sigma = matrix(c(1,corrCoef,2,corrCoef),2,2))
dat[,1] = dat[,1] – mean(dat[,1]) # it makes life easier for the princomp
dat[,2] = dat[,2] – mean(dat[,2])
dat = data.frame(x1 = dat[,1],x2 = dat[,2])
# Calculate the first principle component
# see http://stats.stackexchange.com/questions/13152/how-to-perform-orthogonal-regression-total-least-squares-via-pca
v = dat%>%prcomp%\$%rotation
x1x2cor = bCor = v[2,1]/v[1,1]
x1tox2 = coef(lm(x1~x2,dat))
x2tox1 = coef(lm(x2~x1,dat))
slopeData =data.frame(slope = c(x1x2cor,1/x1tox2[2],x2tox1[2]),type=c(‘Principal Component’,’X1~X2′,’X2~X1′))
# We want this to draw the neat orthogonal lines.
pointOnLine = function(inp){
# y = a*x + c (c=0)
# yOrth = -(1/a)*x + d
# yOrth = b*x + d
x0 = inp[1] y0 = inp[2] a = x1x2cor
b = -(1/a)
c = 0
d = y0 – b*x0
x = (d-c)/(a-b)
y = -(1/a)*x+d
return(c(x,y))
}
points = apply(dat,1,FUN=pointOnLine)
segmeData = rbind(data.frame(x=dat[,1],y=dat[,2],xend=points[1,],yend=points[2,],type = ‘Principal Component’),
data.frame(x=dat[,1],y=dat[,2],yend=dat[,1]*x2tox1[2],xend=dat[,1],type=’X2~X1′),
data.frame(x=dat[,1],y=dat[,2],yend=dat[,2],xend=dat[,2]*x1tox2[2],type=’X1~X2′))
ggplot(aes(x1,x2),data=dat)+geom_point()+
geom_abline( data=slopeData,aes(slope = slope,intercept=0,color=type))+
theme_minimal(20)+coord_equal()
ggplot(aes(x1,x2),data=dat)+geom_point()+
geom_abline( data=slopeData,aes(slope = slope,intercept=0,color=type))+
geom_segment(data=segmeData,aes(x=x,y=y,xend=xend,yend=yend,color=type))+facet_grid(.~type)+coord_equal()+theme_minimal(20)
“
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http://mathhelpforum.com/advanced-statistics/187218-continuous-time-markov-chain-immigration-birth-process.html | # Math Help - Continuous Time Markov Chain - Immigration-Birth Process
1. ## Continuous Time Markov Chain - Immigration-Birth Process
An immigration-birth process with arrival rate λ and birth rate β may be described by the probability statement
P(X(t +δt) = x + 1| X(t) = x)=(λ +βx)δt + 0(δt)
Suppose that at time 0 the size of a population growing according to the above rule is 4. An ‘event’ is said to have occurred when the population size increases by one (and it is immaterial whether this increase is due to a random arrival or to a birth).
Write down an expression for the expected waiting time until the nth event (that is until the population size reaches n + 4).
Since X(0) = 4,
should the birthrate be β or 4β? I am confused here.
I am leaning towards 4β, so am I correct that the expected waiting time should be:
1/(λ + 4β) + 1/(λ + 8β) + 1/(λ + 12β) + ... + 1/(λ + 4nβ)
--------------------------------------------------------------------------------
If the above is correct, for the p.g.f. of a simple birth process with X(0) = 2 and birthrate β, should it be:
Π(s, t) = [(se^(-βt))/(1-s(1-e^(-βt)))]^2
or
Π(s, t) = [(se^(-2βt))/(1-s(1-e^(-2βt)))]^2
--------------------------------------------------------------------------------
Using the former (which I think is correct), is the following workings to find the probabilities at X(t) = 3 correct?
Π(3, t) = [(3e^(-βt))/(1-3(1-e^(-βt)))]^2
= (9e^(-2βt))/(-2+3e^(-βt))^2
= (9e^(-2βt))/(4-12e^(-βt)+9e^(-2βt))
2. ## Re: Continuous Time Markov Chain - Immigration-Birth Process
Any help will be appreciated. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9329202771186829, "perplexity": 1418.8585108800883}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783402479.21/warc/CC-MAIN-20160624155002-00054-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://slideplayer.com/slide/2377550/ | # 9-1 Square Roots and the Pythagorean Theorem The square of 7 is 7 2 = 49 Also,( -7) 2 =49 The reverse of the square of a number is the square root of a.
## Presentation on theme: "9-1 Square Roots and the Pythagorean Theorem The square of 7 is 7 2 = 49 Also,( -7) 2 =49 The reverse of the square of a number is the square root of a."— Presentation transcript:
9-1 Square Roots and the Pythagorean Theorem The square of 7 is 7 2 = 49 Also,( -7) 2 =49 The reverse of the square of a number is the square root of a number. The square root of a number is one of its equal factors. The square of 7 is 7 2 = 49 Also,( -7) 2 =49 The reverse of the square of a number is the square root of a number. The square root of a number is one of its equal factors.
All positive real #’s have 2 sq. roots. The positive sq.root is called the principal root, and the other is the negative sq. root. Sq. roots are written with the radical symbol. The # inside is called the radicand. (read plus or minus) is used to write both sq. roots of a positive #. Perfect squares- #’s whose sq. roots are integers. Irrational #’s- sq. roots of #’s that aren’t perfect. All positive real #’s have 2 sq. roots. The positive sq.root is called the principal root, and the other is the negative sq. root. Sq. roots are written with the radical symbol. The # inside is called the radicand. (read plus or minus) is used to write both sq. roots of a positive #. Perfect squares- #’s whose sq. roots are integers. Irrational #’s- sq. roots of #’s that aren’t perfect.
positive sq. root neg. sq. root zero only has 1 sq. root is undefined Neg. #’s have no sq. root positive sq. root neg. sq. root zero only has 1 sq. root is undefined Neg. #’s have no sq. root
Examples
Completing work for Calculate b 2 Calculate 4 a c Subtract Do square root Calculate b 2 Calculate 4 a c Subtract Do square root
Examples 1)a=3 b= 4 c=1 2)a=3 b=-2 c=-5 1)a=3 b= 4 c=1 2)a=3 b=-2 c=-5
Assignment
Download ppt "9-1 Square Roots and the Pythagorean Theorem The square of 7 is 7 2 = 49 Also,( -7) 2 =49 The reverse of the square of a number is the square root of a."
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http://www.physicsforums.com/showthread.php?t=341907 | ## Homeomorphism between unit square and unit disc
1. The problem statement, all variables and given/known data
I want to find a bijective function from [0,1] x [0,1] -> D, where D is the closed unit disc.
2. Relevant equations
3. The attempt at a solution
I have been able to find two continuous surjective functions, but neither is injective. they are $$f_1(s,t)=\left((1-s)\cos{(2\pi t)}+s,(1-s)\sin{(2\pi t)}\right)$$ and $$f_2(s,t)=\left((1-s)\cos{(2\pi t)},(1-s)\sin{(2\pi t)}\right)$$. I can't think of one that is injective, but there should be one because these two spaces are homeomorphic, right?
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I even have another: $$f_3(s,t)=\left(\sin{(2\pi s)}\cos{(2\pi t)},\sin{(2\pi s)}\sin{(2\pi t)}\right)$$. However, this is also not injective since all points with s=0 are mapped to the origin.
Blog Entries: 1 Recognitions: Homework Help Imagine drawing the unit circle inside of the unit disc. For a given angle t from the x-axis, by what factor do you need to multiply the unit vector in that direction to get it to reach the unit square? Once you map the boundary to the boundary, map the interior to the interior by just stretching a point in the circle by the same factor that boundary point was stretched. You've probably heard of topology as 'rubber sheet geometry' and this is the most basic case of that; you literally want to stretch the circle to cover the square EDIT: Whoops.. I realized my square is too big here. For some reason I was thinking the unit square would have size two. The basic idea is the same though, center the two at the origin and ask what you need to multiply the boundary of the circle by to make it map to the boundary of the square
## Homeomorphism between unit square and unit disc
Thanks for the help!! I got it now. | {"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": 0.818573534488678, "perplexity": 549.1018069389386}, "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/1368700264179/warc/CC-MAIN-20130516103104-00056-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://chemistry.stackexchange.com/questions/130828/solution-of-a-first-and-second-order-consecutive-reaction-scheme | # Solution of a first- and second-order consecutive reaction scheme
I have a compartmental model system X, a simplified version of which, looks like this:
$$\text{System }x: \enspace \text{STO} \xrightleftharpoons[k_2]{k_1} \text{blood} \xrightarrow[]{k_3} \text{urine} \equiv x_1 \xrightleftharpoons[k_2]{k_1} x_2 \xrightarrow[]{k_3} x_3$$
The system of differential equations is then: $$\frac{dx_1}{dt} = -k_1 x_1 + k_2 x_2 \\ \frac{dx_2}{dt} = k_1 x_1 - k_2 x_2 - k_3 x_2\\ \frac{dx_3}{dt} = k_3 x_2 \\$$
I can solve this system of equations using the following matrix approach.
First, I write the rate matrix [R]. From [R] one can obtain a new matrix [A] by first replacing each diagonal element of [R] by the negative of the sum of each of row elements, and then transposing it: $$[R] = \begin{bmatrix} 0 & k_2 & 0 \\ k_1 & 0 & k_3 \\ 0 & 0 & 0 \end{bmatrix} \\ [A'] = \begin{bmatrix} -k_2 & k_2 & 0 \\ k_1 & -(k_1 + k_3) & k_3 \\ 0 & 0 & 0 \end{bmatrix} \\ [A] = \begin{bmatrix} -k_2 & k_1 & 0 \\ k_1 & -(k_1 + k_3) & 0 \\ 0 & k_3 & 0 \end{bmatrix} \\$$
I can calculate the amount in each compartment by doing the following:
$$x(t) = e^{[A]t}x(0) \enspace \text{where} \enspace x(0) = \begin{bmatrix} 0 \\ 1 \\0 \end{bmatrix}$$
In python:
RMatrix = model_matrix.as_matrix()
row, col = np.diag_indices_from(RMatrix)
RMatrix[row, col] = -(RMatrix.sum(axis=1)-RMatrix[row,col])
AMatrix = RMatrix.T
def content(t):
cont = np.dot(linalg.expm(t*AMatrix), x0))
This method is working well for me.
Now, I have a little more complicated model where reactants in compartments 1 and 2 of Systems X and Y combine to get product in System Z.
$$X + Y \rightarrow Z$$, with a reaction constant of $$k_R$$.
\begin{align} &x_1 & \xrightleftharpoons[k_2]{k_1} \enspace & x_2 &\xrightarrow[]{k_3} & x_3 \\ &+ & \ & + \\ &y_1 & \xrightleftharpoons[k_5]{k_4} \enspace & y_2 &\xrightarrow[]{k_6} & y_3 \\ &\downarrow^{k_R} & \ & \downarrow^{k_R} \\ &z_1 & \xrightleftharpoons[k_8]{k_7} \enspace & z_2 &\xrightarrow[]{k_9} & z_3 \\ \end{align}
, and the corresponding system of differential equations would be:
\begin{align} \frac{dx_1}{dt} &= -k_1 x_1 + k_2 x_2 - k_R x_1 y_1 \\ \frac{dx_2}{dt} &= k_1 x_1 - k_2 x_2 - k_3 x_2 - k_R x_2 y_2\\ \frac{dx_3}{dt} &= k_3 x_2 \\ \frac{dy_1}{dt} &= -k_4 y_1 + k_5 y_2 - k_R x_1 y_1 \\ \vdots& \\ \frac{dz_z}{dt} &= -k_7 z_1 + k_8 z_2 + k_R x_1 y_1 \\ \end{align}
I am struggling with a method to solve this system of differential equations (1st and 2nd order) to calculate the amount in each compartment at a certain time t, given the initial conditions, $$k_R$$, and the transfer rates $$k_1$$, $$k_2$$, $$k_3$$, etc...
Can I solve it using the matrix method like the one above for a system of first order differential equations? What other options in Python do I have? | {"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": 12, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000056028366089, "perplexity": 927.0194010750187}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487623596.16/warc/CC-MAIN-20210616093937-20210616123937-00609.warc.gz"} |
https://math.stackexchange.com/questions/2365512/proving-the-multivariate-central-limit-theorem-using-cram%C3%A9r-wold-device | # Proving the Multivariate Central Limit Theorem using Cramér–Wold Device
I am told that the MV CLT can be proved using the Cramér–Wold device.
The theorem is as follows (from Flury's "A First Course in Multivariate Statistics") Suppose $\bf{X}_1, \bf{X}_2, \ldots$, $\bf{X}_n$ are independent, identically, distributed, p-variate random vector, with mean vectors $\bf{µ}=E[\bf{X}_i]$ and covariance matrices $\sigma = Cov[\bf{X}_i]$. Let $\bf{\overline{X}_N} = \frac{1}{N}\sum_{i=1}^n\bf{X}_i$. Then $\sqrt{N}(\bf{\overline{X}_N}-\bf{\mu}) \overset{d}{\to} N(0, \sigma))$.
Cramér–Wold says I can show the above result by showing
$$a^t\sqrt{N}(\bf{\overline{X}_N}-\bf{\mu}) \overset{d}{\to} a^t N(0,\sigma) = N(0, a\sigma a^t) \forall a \in \mathbb{R}^p$$
Through properties of expectation and variance I have shown
$$E\left(a^t\sqrt{N}(\bf{\overline{X}_N}-\bf{\mu})\right) = 0$$ and $$Var\left(a^t\sqrt{N}(\bf{\overline{X}_N}-\bf{\mu})\right) = a\sigma a^t$$
So whatever $a^t\sqrt{N}(\bf{\overline{X}_N}-\bf{\mu})$ converges to, it must have the properties I want. How can I guarantee it is a univariate normal?
It occurs to me that it might be useful to write
$$a^t\sqrt{N}(\bf{\overline{X}_N}-\bf{\mu}) = \sum_{i=1}^p\left(a_i \sqrt{n} (\overline{x}_{ni} - \mu_i)\right)$$
Then by the univariate central limit theorem, each term converges in distribution to some normal. Can we then say that their sum converges in distribution to some normal (we already know it will have the mean and variance we are looking for)? If not, what else shouldI be looking at? | {"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": 0.9917314648628235, "perplexity": 379.1487060614638}, "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/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00461.warc.gz"} |
http://www.physicsforums.com/showthread.php?s=c8dacbe0d269fe8778da6ca94cd27228&p=4323554 | equal probability principle
by Abigale
Tags: statistical mechanic, statistical physics, thermodynamics
P: 44 Hey Guys, I have got a question. English is not my first language. Is the "equal-probability-principle" the same as "the equal-a-priori-probability-postulate"? Do both names describe the same process? Thx Abby
P: 1,611
Quote by Abigale Hey Guys, I have got a question. English is not my first language. Is the "equal-probability-principle" the same as "the equal-a-priori-probability-postulate"? Do both names describe the same process? Thx Abby
I have not heard the phrase "equal-probability-principle", but I believe that the "equal-a-priori-probability-postulate" refers to the assumption in statistical mechanics that for a system in equilibrium, all reachable microscopic states of the same energy are equally probable.
Related Discussions Calculus 8 Advanced Physics Homework 1 Set Theory, Logic, Probability, Statistics 20 Set Theory, Logic, Probability, Statistics 6 General Physics 4 | {"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": 0.9007357358932495, "perplexity": 1204.7959161983717}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394010547566/warc/CC-MAIN-20140305090907-00086-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/109715/prove-that-there-cant-be-985-divisors-of-123456-19841985 | # Prove that there can't be 985 divisors of $123456…19841985$
Numbers from 1 to 1985 are written one after another so they form a new number, $n=123456\ldots19841985$. Prove that there can't be 985 divisors of $n$.
This should be solved on paper, without using programming methods.
-
Do you mean there can't be exactly $985$ divisors, or there can't be at least that many? – Chris Eagle Feb 15 '12 at 19:28
It's actually unclear, but my guess goes for exactly. – Lazar Ljubenović Feb 15 '12 at 19:30
Assuming you mean there can't be exactly $985$ distinct divisors of $n$... $985=5\times 197$. If $n=p_1^{a_1}\cdots p_r^{a_r}$, with $p_1\lt\cdots\lt p_r$ primes, then the number of divisors of $n$ is $(a_1+1)\cdots(a_n+1)$, so the only way $n$ could have exactly $985$ divisors is if $n=p_1^4p_2^{196}$ for two distinct primes $p_1$ and $p_2$; or $n=p^{984}$ with $p$ a prime.
Since $n$ is divisible by $5$, one of the primes must be $5$; hence, $n$ would be divisible by $25$. However, all multiples of $25$ end in $00$, $25$, $50$, or $75$, which $n$ does not; so $n$ cannot have exactly $985$ distinct divisors.
Robert Israel rightly points out the much simpler argument: $n$ is divisible by $5$ but not $5^2$, so from the formula for number of divisors we immediately see that the number of divisors of $n$ is even, hence not $985$. And even if you don't know the formula for divisors, you can partition the divisors into multiples of $5$ and coprime to $5$, and pair them up to see the number of divisors is even.
-
Maybe slightly more simply: the number $n$ is divisible by $5$ but not by $25$. Any such number has an even number of divisors, because they occur in pairs: if $d$ is not divisible by $5$, $d$ divides $n$ if and only if $5d$ divides $n$. – Robert Israel Feb 15 '12 at 19:42
@Robert: Of course; that's much simpler... <chagrin>. (Or you can just use the formula I mentioned, since you'll have $5$ in the prime factorization, so $2$ as a factor of the number of divisors...) – Arturo Magidin Feb 15 '12 at 19:43
Or else, a variant of Robert Israel's solution, only perfect squares have an odd number of divisors. – André Nicolas Feb 15 '12 at 19:49
By the way, $n$ is also divisible by $3$ but not by $9$, so its number of divisors is divisible by $4$. $n/15$ is a composite number of $6831$ digits, probably too big for current technology to factor. – Robert Israel Feb 15 '12 at 19:50
@sven.hedin: The sum of the digits is the same as the sum of the numbers $1$ through $1985$, which is given by $\frac{(1985)(1986)}{2}$. The numerator is congruent to $5\times 6 = 3$ modulo $9$, hence the numerator is divisible by $3$ but not by $9$. – Arturo Magidin Feb 15 '12 at 21:07 | {"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": 0.9463091492652893, "perplexity": 145.95817033800316}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418815948154.77/warc/CC-MAIN-20141217113228-00133-ip-10-231-17-201.ec2.internal.warc.gz"} |
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##### Electronic balance & a rubber balloon
« on: March 22, 2015, 08:44:10 PM »
My Friend and I got into an argument yesterday. The argument was about - if a deflated rubber balloon is weighed in an electronic balance and then the rubber balloon is inflated and weighed again then will the weight change?
I said that, since more air is inside the balloon now, the weight will increase but he says that weight will remain same and explains something that I don't understand. So, what do you think, will the weight change upon inflation (even if it is miniscule; also let's not get into the topic where an inflated balloon can't be properly weighed in an electronic balance)?
#### Arkcon
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##### Re: Electronic balance & a rubber balloon
« Reply #1 on: March 22, 2015, 09:06:25 PM »
Remove the technology from the equation. You have two balloons full of air, balancing each other in a simple lever balance, maybe even a ruler on a string. Once balanced, you pop one. What happens to the balance?
https://www.teachengineering.org/view_activity.php?url=collection/cub_/activities/cub_air/cub_air_lesson01_activity2.xml
Hey, I'm not judging. I just like to shoot straight. I'm a man of science.
#### Borek
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##### Re: Electronic balance & a rubber balloon
« Reply #2 on: March 23, 2015, 03:56:35 AM »
You have two balloons full of air, balancing each other in a simple lever balance, maybe even a ruler on a string. Once balanced, you pop one. What happens to the balance?
https://www.teachengineering.org/view_activity.php?url=collection/cub_/activities/cub_air/cub_air_lesson01_activity2.xml
I don't like the fact they ignore buoyancy. Even if the system behaves the way they describe it, the explanation is IMHO wrong.
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#### Furanone
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##### Re: Electronic balance & a rubber balloon
« Reply #3 on: March 23, 2015, 04:45:28 AM »
I think about it this way:
If a balloon has hydrogen or helium pumped into it to inflate it relative to the atmospheric gas on the outside (ie ~78% nitrogen 21% Oxygen, 0.5% CO2, 0.5% other gases) then it will weigh lighter as it will float up not even allowing weighing. If you pump it full of argon, a much heavier gas, it will weigh more on the scale. So knowing this what would you think would be the result if you pumped the same exact mix of gases into the balloon as what is the equilibrium atmospheric conditions?
Now what about if you take that balloon with regular atmosphere gases and climb a 5000 metre high mountain, what happens to the balloon? Does it weigh the same, less or more?
"The true worth of an experimenter consists in pursuing not only what he seeks in his experiment, but also what he did not seek."
--Sir William Bragg (1862 - 1942)
#### Corribus
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##### Re: Electronic balance & a rubber balloon
« Reply #4 on: March 23, 2015, 10:09:57 AM »
I don't like the fact they ignore buoyancy. Even if the system behaves the way they describe it, the explanation is IMHO wrong.
I have to agree. I read the webpage, and my first thought was, "That's a totally wrong explanation."
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent? - Richard P. Feynman
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##### Re: Electronic balance & a rubber balloon
« Reply #5 on: March 23, 2015, 06:54:29 PM »
Today I had a heartier talk with my friend. His logic is that if a water balloon is immersed in water, then it will loose weight equal to the weight of the volume of water replaced by the balloon (which is equal to the volume of water inside the balloon). Since electronic balances determine mass based on weight, if we could somehow determine the weight of that balloon underwater, we would only get the weight of the rubber part.
Similarly, when a balloon filled with air is being weighed in normal environment, the balloon is actually immersed in air and thus Due to that Archimedes principle, weight will be lost and only the rubber part can be weighed by the Electronic balance. This is his logic.
I couldn't disagree and conceded defeat. But how can I determine the mass of air inside a balloon? Also it seems that even if a gas much heavier than air was used to inflate the balloon, we still would not get the actual mass of the gas due to buoyancy.
#### Borek
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##### Re: Electronic balance & a rubber balloon
« Reply #6 on: March 24, 2015, 04:16:23 AM »
Today I had a heartier talk with my friend. His logic is that if a water balloon is immersed in water, then it will loose weight equal to the weight of the volume of water replaced by the balloon (which is equal to the volume of water inside the balloon). Since electronic balances determine mass based on weight, if we could somehow determine the weight of that balloon underwater, we would only get the weight of the rubber part.
Yes, that's the buoyancy at work.
Quote
Similarly, when a balloon filled with air is being weighed in normal environment, the balloon is actually immersed in air and thus Due to that Archimedes principle, weight will be lost and only the rubber part can be weighed by the Electronic balance. This is his logic.
Yes, that's the buoyancy at work.
If you could find a way of measuring the balloon volume, you can always account for the buoyancy.
Quote
But how can I determine the mass of air inside a balloon?
You mean - if you wanted to, for whatever reason? This is a very good question. TBH I can't think of any simple approach. Weighing in vacuum is out of the question. You could try freezing it down and weighing the liquid. You could try to pump the air from the balloon and to weigh it compressed , so that buoyancy doesn't matter much.
Quote
Also it seems that even if a gas much heavier than air was used to inflate the balloon, we still would not get the actual mass of the gas due to buoyancy.
Yes.
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##### Re: Electronic balance & a rubber balloon
« Reply #7 on: July 06, 2015, 09:08:32 PM »
Sorry to jump in so late, but one thing you seem to be overlooking in your use of the Archimedes principle is that you are not compressing the water inside the balloon. It was the same density as the water around it. This isn’t the case with a balloon filled with air. When you blow up the balloon, you are compressing more air into the balloon than would be there at STP. This is very simple to show if you have a balance that can go to enough decimal places. It is best If you have one that has the air shields. Weigh the balloon empty, inflate and weigh again. It will have gained mass. This is why a balloon that you inflate with your lungs falls to the ground; it is denser than the surrounding gas.
#### Borek
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##### Re: Electronic balance & a rubber balloon
« Reply #8 on: July 07, 2015, 03:25:53 AM »
When you blow up the balloon, you are compressing more air into the balloon than would be there at STP. This is very simple to show if you have a balance that can go to enough decimal places. It is best If you have one that has the air shields. Weigh the balloon empty, inflate and weigh again. It will have gained mass. This is why a balloon that you inflate with your lungs falls to the ground; it is denser than the surrounding gas.
While it is true that the pressure inside the balloon is slightly higher, you don't need that to explain why it falls down. Weight of the filled balloon is always weight of the trapped air plus weight of the balloon itself. If the pressures inside and outside are identical, weight of the trapped gas and its buoyancy cancel out, but you are still left with the weight of the balloon itself, so it has to fall.
Fact that the air is compressed works in the same direction (makes the balloon slightly heavier), but is not necessary to explain the result.
« Last Edit: July 07, 2015, 05:23:59 AM by Borek »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info | {"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": 0.840778112411499, "perplexity": 1672.4485780966854}, "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/1566027322170.99/warc/CC-MAIN-20190825021120-20190825043120-00085.warc.gz"} |
https://www.physicsforums.com/threads/momentum-car-crash.139983/ | Momentum - Car Crash
1. Oct 25, 2006
bobby3280
A police officer is investigating the scene of an accident where two cars collided at an intersection. One car with a mass of 1135 kg moving west had collided with a 1300 kg car moving north. The two cars, stuck together, skid at an angle of 30° north of west for a distance of 25 m. The coefficient of kinetic friction between the tires and the road is 0.80.
What was the speed of the lighter car?
______ km/h
What was the speed of the heavier car?
______ km/h
I know this is a perfectly inelastic collistion and that Vf will be equal.
so x-dir m1vi1 = (m1 + m2) Vf cos 30
y-dir m2vi2 = (m1 + m2) Vf sin 30
so i can't solve too many unknowns.
How do i use the 25m and kinetic friction??
2. Oct 25, 2006
Staff: Mentor
Use this info to find the acceleration during the skid and then the speed of the cars immediately after the collision. Hint: What's the force of kinetic friction?
3. Oct 25, 2006
bobby3280
F = .8 m g
F = .8 * 2435 * 9.8
F = 19090.4 N
a = F/m
a = 19090.4 / 2435
a = 7.84 m/s
I think this is right but not sure after.
Vf^2 - Vi^2 = 2ax
setting Vf to 0 I get a Vi = 19.8
and i know this Velocity is too slow???
4. Oct 25, 2006
Staff: Mentor
All good. (Acceleration has units of m/s^2, not m/s.)
What makes you think this is too slow? (Convert to miles/hour to compare with typical speeds.)
5. Oct 25, 2006
bobby3280
The last part of the question states The speed limit for each car was 70 km/h. And I know that only the lighter car was speeding.
So V1i > 70 km/s
When i solve for my initial velocities using 19.8 and this formula
V1i = ((m1 + m2) / m1) Vf cos 30 putting in 19.8 i get
V1i = 36.8 km/s
Which is considerably lower that 70 km/s
So either my formula or my Vf is wrong but after many hours I still can't find where i'm wrong. Any other help??? Thanks so far.
6. Oct 25, 2006
Staff: Mentor
The speed limit is 70 km/hr, not 70 km/s.
Also, standard units will give the speed in m/s, not km/s.
7. Oct 25, 2006
bobby3280
Wow don't know how i missed that but thanks alot!!!
Similar Discussions: Momentum - Car Crash | {"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": 0.8745438456535339, "perplexity": 2361.2395614578045}, "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-2017-22/segments/1495463608665.33/warc/CC-MAIN-20170526124958-20170526144958-00365.warc.gz"} |
https://chemistry.stackexchange.com/questions/97340/is-avogadros-number-an-integer?noredirect=1 | # Is Avogadro's number an integer? [duplicate]
I have heard that Avogadro's number, $N_\mathrm A=6.022 \times 10^{23}$, is the number of atoms contained in $12$ grams of $\ce{^{12}C}$. I think it should be an integer, but I couldn't find the exact number. Is it an integer?
• The old (1971 I guess) definition of mole that you stated surely leads to floating point numbers (although only if you measure weights with more than 23 significant digits). The new definition (2018) fixes one mole to be exactly $6.02214076 \times 10^{23}$, which is an integer. – Felipe S. S. Schneider May 22 '18 at 16:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9388737082481384, "perplexity": 413.92417235559606}, "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/1573496670268.0/warc/CC-MAIN-20191119222644-20191120010644-00013.warc.gz"} |
https://math.stackexchange.com/questions/461600/is-this-a-valid-proof-of-why-a-linearly-dependent-matrix-has-a-0-determinant | # Is this a valid proof of why a linearly dependent matrix has a $0$ determinant?
If a $3\times{3}$ matrix has a non-trivial solution to $B\vec{x}=\vec{0}$, then it has linearly dependent rows. Looking at Cramer's rule:
$$x_1=\frac{\begin{vmatrix}0 & b_{12} & b_{13} \\0 & b_{22} & b_{23} \\ 0 & b_{32} & b_{33}\end{vmatrix}}{\det(B)}$$
$$\det(B)x_1=\begin{vmatrix}0 & b_{12} & b_{13} \\0 & b_{22} & b_{23} \\ 0 & b_{32} & b_{33}\end{vmatrix}$$
Since $\begin{vmatrix}0 & b_{12} & b_{13} \\0 & b_{22} & b_{23} \\ 0 & b_{32} & b_{33}\end{vmatrix}$ is clearly equal to $0$, then either $x_1$ has to be $0$, or $\det{B}$ does. Applying the same reasoning to $x_2$ and $x_3$, for $\vec{x}$ to be non-trivial, at least one of the components must be non-zero, meaning that $\det{B}$ must be $0$.
So here's where I have doubts on this proof. Cramer's rule is defined only valid if $\det{B}\ne0$. However, the second equation isn't Cramer's rule, but something that I'm postulating may be a valid statement that follows from Cramer's rule. Since it doesn't result in a division by $0$, I suspect it may be valid, but I'm not entirely sure.
Is this proof valid?
• You proved that if $\det(B) \neq 0$, then $Bx=0$ has only a trivial solution. So just use a contraposition argument. – Seirios Aug 7 '13 at 4:35
• When you write a supposed mathematical expression of the form $\,x_i=\frac ab\;$ , we're implicitly (and even automatically) assuming $\,b\ne 0\,$, otherwise that is not a mathematical expression. Thus, either you can reach $\,\det(B)\,x_i=...\;$ without first having the division by $\,\det(B)\,$ , or else you can prove that $\,\det(B)\ne 0\,$ upon request...which, of course, you can't, otherwise the argument is, imo, logically non-sound. – DonAntonio Aug 7 '13 at 6:39
I think a much easier proof (which avoids the issues you are encountering) runs as follows. If $B$ is a matrix with linearly dependent rows, then we have an elementary row operation matrix such that $AB = B'$, where $B'$ is the same as $B$ except it has a row of zeroes in place of one of the linearly dependent rows. Clearly, $\det(B') = 0$. But by the multiplicativity of determinants, we have $\det(B') = \det(AB) = \det(A)\det(B)$. Since every elementary matrix $A$ has the property $\det(A) = 1$, we have $\det(B) = 0$.
(Of course, if your question is deliberately on this specific method of proof and this answer is not useful to you, I will happily delete it. My apologies if this is the case.)
• My question is specifically about the proof in my question, but this is a good proof. I appreciate your input :) – Ataraxia Aug 7 '13 at 4:36
• Fair enough! I will leave it up then. Thanks for taking the time to let me know :) – Alex Wertheim Aug 7 '13 at 4:38
If you aren't familiar with rings, just think of each $R$ in what follows as the real numbers (or some other field). Throughout what follows, $M_n(R)$ denotes the set of $n\times n$ matrices with entries in $R$. Also, $v^T$ denotes the transpose of $v$ for any $m \times n$ matrix.
Proposition: Let $R$ be a commutative ring with unity. Let $A = (C_1 \mid \cdots \mid C_n) \in \text{M}_n(R)$, $x = (x_1,\ldots,x_n)^T$ and $b = (b_1,\ldots,b_n)^T$ in $R^n$ such that $$Ax = b.$$ Then for all $k \in \{1,\ldots,n\}$ $$\text{det}(A_k) = \text{det}(C_1,\ldots,C_{k-1},Ax,C_{k+1},\ldots,C_n) = x_k \text{det}(A)$$ where $A_k = (C_1 \mid \cdots \mid C_{k-1} \mid Ax \mid C_{k+1} \mid \cdots \mid C_n)$.
Proof: Let $j \in \{1,\ldots,n\}$. Note the first equality is trivial. For the second, observe that $x = x_1 e_1 + \ldots + x_n e_n = \sum_{i = 1}^n x_ie_i$ where $e_1 = (1,0,\ldots,0),\ldots,e_n=(0,\ldots,0,1)$. Hence $$Ax = \sum_{i = 1}^n x_i Ae_i = \sum_{i = 1}^n x_i C_i.$$ Thus by the multilinearity of the determinant \begin{align} \det(A_j) &= \det(C_1,\ldots,C_{j-1},Ax,C_{j+1},\ldots,C_n) \\ &= \det(C_1,\ldots,C_{j-1},\sum_{i = 1}^n x_i C_i,C_{j+1},\ldots,C_n) \\ &= \sum_{i=1}^n x_i \det(C_1,\ldots,C_{j-1},C_i,C_{j+1},\ldots,C_n).\qquad\qquad(\star) \end{align} An easy corollary of the fact that $\det$ is an alternating multilinear function is that $\det(M) = 0$ whenever the columns of $M$ are $R$-linearly dependent in $R^n$ (in particular, when $D_i = D_j$ for any $i,j$ with $1 \leq i < j \leq n$). Thus within $(\star)$, only the $j$-th summand $x_j \det(C_1,\ldots,C_{j-1},C_j,C_{j+1},\ldots,C_n)$ is nonzero. Hence \begin{align*} \det(A_j) = x_j \det(C_1,\ldots,C_{j-1},C_j,C_{j+1},\ldots,C_n) = x_j \det(A) \end{align*} as claimed. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9985774755477905, "perplexity": 164.32230788672504}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627997533.62/warc/CC-MAIN-20190616022644-20190616044644-00544.warc.gz"} |
http://mathhelpforum.com/advanced-applied-math/198907-friction-coefficient-incline-forces-physics-help-print.html | # Friction coefficient/ Incline/ Forces - Physics help?
• May 16th 2012, 07:26 PM
patrickmanning
Friction coefficient/ Incline/ Forces - Physics help?
A 1200kg mass rests on a track inclined at 30 degrees to the horizontal.
A force P inclined at 18 degrees TO THE TRACK (48 degrees to the horizontal) causes the carriage to move up the track at constant velocity (i.e. no acceleration). Coefficient of friction is 0.2; determine P. Use gravity as 9.8 ms^-1.
---
Constant velocity means no net forces act on the mass.
Parallel forces: Gravity = (m)(g)(sin30) = 5880N
Frictional force = (0.2)(Normal force)
That's all I've got so far :/
---
I wasn't taught this in my horrible school so I have only a rough idea of how to work these type of questions. Could you include all formulae used and just solve it for me? I can learn from the working.
~Thanks.
• May 16th 2012, 09:22 PM
Moebius
1 Attachment(s)
Re: Friction coefficient/ Incline/ Forces - Physics help?
Attachment 23889
1.) Draw a picture (aka- free body diagram) to help visualize the problem.
2.) Break up the forces into necessary components (in green). Note: I have chosen the coordinate system, as indicated in the picture, to simplify the problem.
3.) Newton's Second Law (for x):
$\sum F_{x}=ma_{x}$
$\sum F_{x}=0$
$F_{p,x}-F_{g,x}-f_{k}=0$
$F_{p,x}=F_{g,x}+f_{k}$
$F_{p}\cos(18^{\circ})=F_{g}\sin(30^{\circ})+\mu_{k }n$
4.) Newton's Second Law (for y):
$\sum F_{y}=ma_{y}$
$\sum F_{y}=0$
$n+F_{p,y}-F_{g,y}=0$
$n=F_{g,y}-F_{p,y}$
$n=F_{g}\cos(30^{\circ})-F_{p}\sin(30^{\circ})$
5. Substitute back into the equation derived for $\sum F_{x}$:
$F_{p}\cos(18^{\circ})=F_{g}\sin(30^{\circ})+\mu_{k }[F_{g}\cos(30^{\circ})-F_{p}\sin(30^{\circ})]$
6. Solve for $F_{p}$:
$F_{p}=F_{g}[\frac{\sin(30^{\circ})+\mu_{k}\cos(30^{\circ})}{ \cos(18^{\circ})+\mu_{k}\sin(18^{\circ})}]=mg[\frac{\sin(30^{\circ})+\mu_{k}\cos(30^{\circ})}{ \cos(18^{\circ})+\mu_{k}\sin(18^{\circ})}]$
Hopefully that will clear things up! Good luck! :)
• May 16th 2012, 09:37 PM
patrickmanning
Re: Friction coefficient/ Incline/ Forces - Physics help?
There was typo in part 4; last line should have sin18, not sin30.
This was great, though! Thanks SO MUCH! I was terrible at this topic, as my high-school teacher did nothing all year, so I had to try and self-teach. I think your one comment has taught me the entire thing perfectly. Rep! :D
• May 16th 2012, 09:55 PM
Moebius
Re: Friction coefficient/ Incline/ Forces - Physics help?
Lol sorry about the typo! I'm glad it helped! If you have anymore questions feel free to send me a message, and I'll do my best to help.
• May 16th 2012, 10:59 PM
patrickmanning
Re: Friction coefficient/ Incline/ Forces - Physics help?
Last thing, just to make sure I've got this. I'll post here in case someone else ever makes use of this. Same type of question, but the mass = 2kg, angle of incline of plane = 20 degrees, and there's a force Y on the mass that is 25 degrees TO THE PLANE. Coefficient of friction is still 0.2. Just want to make sure my math is right; is the answer approximately 10.5N? This isn't really vital, so no need to bother if you're busy.
Here's what I really want to know: if the mass is just on the verge of slipping DOWN the plane (but still stationary, therefore net force = 0), is everything is still the same, except in the equation for parallel forces you'd put:
Ycos25 - mg(sin20) + (0.2)(n) = 0?
The change is just to add friction, instead of subtracting it, right?
• May 17th 2012, 05:05 AM
Moebius
Re: Friction coefficient/ Incline/ Forces - Physics help?
That's right. Friction always opposes motion, so in this case it will be pointing up the plane because the mass "wants" to move down. Another change, however, is that we no longer use the kinetic friction force, but instead, the static friction force because the block is not moving yet.
• May 17th 2012, 11:52 AM
patrickmanning
Re: Friction coefficient/ Incline/ Forces - Physics help?
Thanks, man. You really saved the day on this one! :) | {"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": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.897835910320282, "perplexity": 1659.472454633372}, "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/1476988719784.62/warc/CC-MAIN-20161020183839-00565-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://joaoff.com/tags/calculational/ | # Calculational
## An Improved Proof of the Handshaking Lemma
In 2009, I posted a calculational proof of the handshaking lemma, a well-known elementary result on undirected graphs. I was very pleased about my proof because the amount of guessing involved was very small (especially when compared with conventional proofs). However, one of the steps was too complicated and I did not know how to improve it. In June, Jeremy Weissmann read my proof and he proposed a different development. His argument was well structured, but it wasn’t as goal-oriented as I’d hoped for.
## Multiples in the Fibonacci Series
I found the following problem on K. Rustan M. Leino’s puzzles page: [Carroll Morgan told me this puzzle.] Prove that for any positive K, every Kth number in the Fibonacci sequence is a multiple of the Kth number in the Fibonacci sequence. More formally, for any natural number n, let F(n) denote Fibonacci number n. That is, F(0) = 0, F(1) = 1, and F(n+2) = F(n+1) + F(n). | {"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": 0.9478578567504883, "perplexity": 689.2886079425524}, "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/1656104660626.98/warc/CC-MAIN-20220706030209-20220706060209-00184.warc.gz"} |
http://mathoverflow.net/questions/45061/linear-ordering-of-color-balls/45063 | # linear ordering of color balls
Suppose that $n+m$ balls of which $n$ are red and $m$ are blue, are arranged in a linear order, we know there are $(n+m)!$ possible orderings. If all red balls are alike and all blue ball are alike, we know there are $\frac{(n+m)!}{n!m!}$ possible orderings.
For example, 2 red and 3 blue balls:
R1 R2 B1 B2 B3
R2 R1 B2 B3 B1
The above two orderings are equivalent and can be denoted as:
R R B B B
Now here is the problem: what if we further concentrate on the color, and record consecutive balls of the same color with the just ONE color code?
For example the color code for the afore-mentioned example would be:
R B
How many possible color code orderings are there?
-
Should be tagged co.combinatorics instead of pr.probability? – Emil Nov 6 '10 at 12:27
Such a color-code ordering starts with either R or B and continues with strictly alternating R and B. The string can be of any length up to the smaller of $n$ or $m$, meaning it can be twice that smaller value, but that can be followed by one more character if there are enough of the other color. Moreover, every such string is a color-code ordering for some linear ordering of balls. There are a couple of special cases, namely that if either $n$ or $m$ is zero then there is exactly one color-code ordering and there aren't any if both are zero. Also, if neither is zero, we must have at least one instance of each letter.
So:
If $n = m = 0$, the answer is 0.
If exactly one of $n$ and $m$ is zero, the answer is 1.
If $n = m > 0$, the answer is $4n - 2$.
Otherwise, let $p$ be the minimum of $n$ and $m$. The answer is $4p-1$.
-
Without loss of generality, assume $n \leq m$. Such a colour code ordering is just a sequence of alternating $R$ and $B$ letters. There are four types of such sequences, depending which letter they start and end with. Say a sequence is of type $(X,Y)$ if it begins with $X$ and ends with $Y$.
So, there are
1. $n$ sequences of type $(R,B)$
2. $n$ sequences of type $(B,R)$
3. $n-1$ sequences of type $(R,R)$
4. $n$ sequences of type $(B,B)$ (and only $n-1$ of them if $n=m$).
Thus, the answer is $4n-1$ if $n < m$, and $4n-2$ if $n=m$.
Edit. As Larry Denenberg mentions, in the degenerate case of $n=0$, the answer is always 1 (I count the empty string if $n=m=0$).
- | {"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": 0.8638597130775452, "perplexity": 312.64813844469904}, "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/1435375096706.9/warc/CC-MAIN-20150627031816-00082-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://nag.com/numeric/CL/nagdoc_cl23/html/F07/f07ftc.html | f07 Chapter Contents
f07 Chapter Introduction
NAG C Library Manual
# NAG Library Function Documentnag_zpoequ (f07ftc)
## 1 Purpose
nag_zpoequ (f07ftc) computes a diagonal scaling matrix $S$ intended to equilibrate a complex $n$ by $n$ Hermitian positive definite matrix $A$ and reduce its condition number.
## 2 Specification
#include #include
void nag_zpoequ (Nag_OrderType order, Integer n, const Complex a[], Integer pda, double s[], double *scond, double *amax, NagError *fail)
## 3 Description
nag_zpoequ (f07ftc) computes a diagonal scaling matrix $S$ chosen so that
$sj=1 / ajj .$
This means that the matrix $B$ given by
$B=SAS ,$
has diagonal elements equal to unity. This in turn means that the condition number of $B$, ${\kappa }_{2}\left(B\right)$, is within a factor $n$ of the matrix of smallest possible condition number over all possible choices of diagonal scalings (see Corollary 7.6 of Higham (2002)).
## 4 References
Higham N J (2002) Accuracy and Stability of Numerical Algorithms (2nd Edition) SIAM, Philadelphia
## 5 Arguments
1: orderNag_OrderTypeInput
On entry: the order argument specifies the two-dimensional storage scheme being used, i.e., row-major ordering or column-major ordering. C language defined storage is specified by ${\mathbf{order}}=\mathrm{Nag_RowMajor}$. See Section 3.2.1.3 in the Essential Introduction for a more detailed explanation of the use of this argument.
Constraint: ${\mathbf{order}}=\mathrm{Nag_RowMajor}$ or Nag_ColMajor.
2: nIntegerInput
On entry: $n$, the order of the matrix $A$.
Constraint: ${\mathbf{n}}\ge 0$.
3: a[$\mathit{dim}$]const ComplexInput
Note: the dimension, dim, of the array a must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{pda}}×{\mathbf{n}}\right)$.
The $\left(i,j\right)$th element of the matrix $A$ is stored in
• ${\mathbf{a}}\left[\left(j-1\right)×{\mathbf{pda}}+i-1\right]$ when ${\mathbf{order}}=\mathrm{Nag_ColMajor}$;
• ${\mathbf{a}}\left[\left(i-1\right)×{\mathbf{pda}}+j-1\right]$ when ${\mathbf{order}}=\mathrm{Nag_RowMajor}$.
On entry: the matrix $A$ whose scaling factors are to be computed. Only the diagonal elements of the array a are referenced.
4: pdaIntegerInput
On entry: the stride separating row or column elements (depending on the value of order) in the array a.
Constraint: ${\mathbf{pda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
5: s[n]doubleOutput
On exit: if NE_NOERROR, s contains the diagonal elements of the scaling matrix $S$.
6: sconddouble *Output
On exit: if NE_NOERROR, scond contains the ratio of the smallest value of s to the largest value of s. If ${\mathbf{scond}}\ge 0.1$ and amax is neither too large nor too small, it is not worth scaling by $S$.
7: amaxdouble *Output
On exit: $\mathrm{max}\left|{a}_{ij}\right|$. If amax is very close to overflow or underflow, the matrix $A$ should be scaled.
8: failNagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_INT
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}\ge 0$.
On entry, ${\mathbf{pda}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{pda}}>0$.
NE_INT_2
On entry, ${\mathbf{pda}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{pda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
NE_MAT_NOT_POS_DEF
The $〈\mathit{\text{value}}〉$th diagonal element of $A$ is not positive (and hence $A$ cannot be positive definite).
## 7 Accuracy
The computed scale factors will be close to the exact scale factors.
The real analogue of this function is nag_dpoequ (f07ffc).
## 9 Example
This example equilibrates the Hermitian positive definite matrix $A$ given by
$A = (3.23 -(1.51-1.92i 1.90+0.84i×1050 -0.42+2.50i (1.51+1.92i -(3.58 -0.23+1.11i×1050 -1.18+1.37i 1.90-0.84i×105 -0.23-1.11i×105 -4.09×1010 (2.33-0.14i×105 (0.42-2.50i (-1.18-1.37i 2.33+0.14i×1050 -4.29 .$
Details of the scaling factors and the scaled matrix are output.
### 9.1 Program Text
Program Text (f07ftce.c)
### 9.2 Program Data
Program Data (f07ftce.d)
### 9.3 Program Results
Program Results (f07ftce.r) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 44, "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": 0.996370255947113, "perplexity": 3020.330538929348}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414637898119.0/warc/CC-MAIN-20141030025818-00052-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://infoscience.epfl.ch/record/83790 | ## The Scrambled Net Filter
The standard Kalman filter is a powerful and widely used tool to perform prediction, filtering and smoothing in the fields of linear Gaussian state-space models. In its standard setting it has a simple recursive form which implies high computational efficiency. As the latter is essentially a least squares procedure optimality properties can be derived easily. These characteristics of the standard Kalman filter depend strongly on distributional and linearity assumptions of the model. If we consider nonlinear non-Gaussian state-space models all these properties and characteristics are no longer valid. Consequently there are different approaches on the robustification of the Kalman filter. One is based on the the ideas of minimax problems and influence curves. Others use numerical integration and Monte Carlo methods. Herein we propose a new filter by implementing a method of numerical integration, called scrambled net quadrature, which consists of a mixture of Monte Carlo methods and quasi-Monte Carlo methods, providing an integration error of order of magnitude $N^{-3/2}\log(N)^{(r-1)/2}$ in probability, where $r$ denotes dimension. We show that the point-wise bias of the posterior density estimate is of order of magnitude $N^{-3}\log(N)^{r-1}$ but grows linearly with time.
Year:
2002
Keywords:
Laboratories: | {"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": 0.9090425968170166, "perplexity": 331.1913435319305}, "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-2018-22/segments/1526794868132.80/warc/CC-MAIN-20180527072151-20180527092151-00618.warc.gz"} |
https://www.physicsforums.com/threads/density-volume-and-slicing-problem.383421/ | # Density, Volume and Slicing Problem
1. Mar 3, 2010
### Eonfluxx
1. The problem statement, all variables and given/known data
A compressible liquid has density which varies with height. AT the level of h meters above the bottom, the density is 40(5 - h) kg/m3
a) The liquid is put in the containers below. The cross sections of the container are isosceles triangles. It has straight sides and looks like a triangular prism. How many kg will it hold when placed as shown on the left, resting on one triangular side?
^There is a picture of the object.
2. Relevant equations
Mass = Volume times Density
3. The attempt at a solution
My theory is that since M = V*D, and you're given the density, should I slice and solve for volume, then evaluate the integral of volume and the given integral for density then multiply the results to get the mass?
$$\int^{4}_{0}2.5 dh$$ and $$40\int^{4}_{0}5-h dh$$
I got to the first integral by taking the volume of the first slice (triangular prism):
$$\sum\frac{1}{2}b*l*\Delta h$$ $$\rightarrow$$ $$\sum\frac{1}{2}2*2.5\Delta h$$ $$\rightarrow$$ lim as $$\Delta h$$ $$\rightharpoonup$$$$\int^{4}_{0}2.5 dh$$
To find the mass, according to the equation M=VD, should I solve both integrals , one being V and one being D, then multiply?
2. Mar 3, 2010
### Eonfluxx
Ah... It's starting to make more sense now, thanks. So since density is varying with the height we make it all one integral rather than separate ones. Part B is the same question except with the 4m on the ground on its point with the rectangle side up. I figured once I'd managed A, B would be simple. Thanks!
3. Mar 3, 2010
### cronxeh
Last edited by a moderator: Apr 24, 2017
Similar Discussions: Density, Volume and Slicing 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": 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": 0.9814971685409546, "perplexity": 1075.5340207247834}, "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/1502886105976.13/warc/CC-MAIN-20170820053541-20170820073541-00089.warc.gz"} |
https://physics.stackexchange.com/questions/340295/why-is-this-a-good-definition-of-temperature | # Why is this a good definition of temperature?
I have started studying thermodynamics and they defined temperature in a statistical way:
$$\frac{1}{k_BT}=\frac{d \ln(\Omega)}{dE}$$ where $k_B$ is the Boltzmann constant, $T$ is the temperature, $\Omega$ is the number of micro-states for a given energy and $E$ is the energy.
I don't really see how this corresponds to our general idea of temperature. So if someone can explain how this definition encompasses how we, colloquially, define temperature, I would be grateful
• Possibly relevant: physics.stackexchange.com/q/65229 Jun 19, 2017 at 20:34
• Your title and question body seem to be asking two slightly different things. Are you asking how to see that this definition of temperature corresponds with a more conventional one, or are you asking why it is "good"? In the former case, which more conventional idea of temperature do you have in mind? In other words, what do you mean by "our general idea of temperature"? Or in the latter case, what makes a definition "good"? Jun 19, 2017 at 20:37
• More on the definition of temperature. Jun 19, 2017 at 21:27
A "good" definition (although the term is vague) must replicate at least some of our major intuitions about the notion that is being defined.
A very fundamental everyday thing that a scientist groping for a notion of temperature might be seeking is that the nett rate of heat flow between two bodies of the same temperature must be nought. When a hot body is in contact with a cold one, the former will heat the latter until they are the same "temperature", whatever that means.
Now consider two bodies, with energy $E_1$ and $E_2$. Let the number of microstates compatible with the macroscopic state of each body be $\Omega_1$ and $\Omega_2$. The total number of microstates of the system as a whole compatible with the system's macrostate is $\Omega_1\,\Omega_2$. What happens when we put the bodies into contact with one another, and assume that they are otherwise isolated?
In my answer here I argue that it is overwhelmingly likely is that any system will wind up in a microstate that is very near to the maximum likelihood microstate, just by undergoing a random walk because, for large ensembles, the set of microstates contains states that look very like the maximum likelihood microstate and almost nothing else (I do some simple calculations with the binomial distribution to show this in my other answer).
So, we conclude that heat is going to flow between the bodies until the system finds its maximum likelihood microstate. Assuming all microstates equally likely, this will be the microstate that maximizes $\Omega_1\,\Omega_2$. At equilibrium, $\Omega_1$ and $\Omega_2$ are functions $\Omega(E)$ of the internal energy of each subsystem. So we write down an equation that maximizes $\Omega(E_1)\,\Omega(E_2)$, subject to the constraint that $E_1+E_2=const$ (assuming the two systems are isolated).
We get, of course:
$$\left.\frac{\mathrm{d}\Omega}{\mathrm{d} E}\right|_{E=E_j}+\lambda=0$$
for $j=1,\,2$ and $\lambda$ our Lagrange multiplier. Well, that's our answer: heat will stop flowing between the bodies when and only when:
$$\left.\frac{\mathrm{d}\Omega}{\mathrm{d} E}\right|_{E=E_1}=\left.\frac{\mathrm{d}\Omega}{\mathrm{d} E}\right|_{E=E_2}=-\lambda$$
so if we define temperature to be some function of $\frac{\partial\Omega}{\partial E}$, we replicate the fundamental property that heat will flow between two bodies of different temperature in contact with one another, whilst no heat flows between two bodies of the same temperature.
Okay, then, what about the initial direction of heat flow? Further careful reasoning like the above then shows that it is the body with greater $\frac{\partial\Omega}{\partial E}$ that gains heat. Therefore, temperature has to be a monotonically decreasing function of $\frac{\partial\Omega}{\partial E}$.
The simplest choice is $T^{-1} \propto \frac{\partial\Omega}{\partial E}$, although, from the reasoning above alone, this choice is far from unique. Ultimately the definition was refined after our knowledge of statistical mechanics grew until we realized that many things could be explained by assuming that the statistical concept of entropy is the same as the classical thermodynamic notion of Carnot and Clausius, as I explain in my answer here. If we do this, then indeed the temperature has to be the reciprocal of $\frac{\partial\Omega}{\partial E}$, if we are to replicate Carnot's definition of temperature in terms of ideal heat engine efficiency. See also some further ideas on temperature and its definition in my recent answer here. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8969128727912903, "perplexity": 268.38210431626425}, "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/1659882571086.77/warc/CC-MAIN-20220809185452-20220809215452-00267.warc.gz"} |
http://mathoverflow.net/feeds/question/38581 | Proof in the literature of an equality involving the prime counting function - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:46:49Z http://mathoverflow.net/feeds/question/38581 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38581/proof-in-the-literature-of-an-equality-involving-the-prime-counting-function Proof in the literature of an equality involving the prime counting function alext87 2010-09-13T14:17:51Z 2010-09-14T04:27:23Z <p>Let $$R(x) = \sum_{k=1}^{\infty}\frac{\mu(k)}{k}li(x^{1/k})$$ where $\mu$ is the Mobius function and $$li(x) = \int_0^x \frac{dt}{\log t}$$ Is there a proof in the literature of $$\pi(x)=R(x)-\sum_{\rho}R(x^{\rho})$$ where $\pi$ is prime counting function and the sum is over all complex zeros of $\zeta(s)$. The literature seems to treat it as fact while stating no proof is available - a strange situation.</p> <p>Thanks in advance. </p> http://mathoverflow.net/questions/38581/proof-in-the-literature-of-an-equality-involving-the-prime-counting-function/38595#38595 Answer by Micah Milinovich for Proof in the literature of an equality involving the prime counting function Micah Milinovich 2010-09-13T16:04:08Z 2010-09-13T16:04:08Z <p>It may be useful to read Section 10 of Chapter V of Ingham's "The Distribution of Prime Numbers."</p> <p>Let $\Pi(x)=\pi(x)+\frac{1}{2}\pi(x^{1/2})+...$, then Moebius proved that</p> <p>$$\pi(x) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \Pi(x^{1/n}).$$</p> <p>However, this isn't overly illuminating because it shows that </p> <p>$$\pi(x) = \Pi(x) + O(\sqrt{x}/\log x ).$$</p> <p>and Littlewood showed that </p> <p>$$\Pi(x) - \ell i(x) = \Omega_\pm(\sqrt{x} \log\log\log x/\log x).$$ </p> http://mathoverflow.net/questions/38581/proof-in-the-literature-of-an-equality-involving-the-prime-counting-function/38646#38646 Answer by Gerry Myerson for Proof in the literature of an equality involving the prime counting function Gerry Myerson 2010-09-14T04:27:23Z 2010-09-14T04:27:23Z <p>Stopple, A Primer of Analytic Number Theory, proves a theorem which looks something like the one under discussion. On page 248, he has $$\pi(x)=R(x)+\sum_{\rho}R(x^{\rho})+\sum_{n=1}^{\infty}{\mu(n)\over n}\int_{x^{1/n}}^{\infty}{dt\over t(t^2-1)\log t}$$</p> <p>You say that the literature treats your formula as a fact, but you give no citation. Where in the literature do you find your formula? </p> | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 3, "x-ck12": 0, "texerror": 0, "math_score": 0.9729008078575134, "perplexity": 996.3230890587065}, "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/1368711515185/warc/CC-MAIN-20130516133835-00099-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/eigenvalues-of-an-operator.279573/ | Eigenvalues of an operator
1. Dec 14, 2008
soul
Hi, everyone!
While I was studying for my midterm, I encountered this question.
------
Consider the hermitian operator H that has the property that
H4 = 1
What are the eigenvalues of the operator H?
What are the eigenvalues if H is not restricted to being Hermitian?
------
What I am going to ask is that does it matter this operator have a different power like H6 or H5? I mean what is the role of the power in this question? Also, I couldn't figure out that how I can find the eigenvalues of such operator when it is not Hermitian?
Thank you.
2. Dec 14, 2008
dextercioby
Assume a finite dimensional vector space on which H acts. Can you write the characteristic equation for H ?
3. Dec 14, 2008
student111
I think you can do it simple like this:
Let E be an eigenvector of H with eigenvalue e:
H E = e E
Then H^4 E = e H^3 E = e^2 H^2 E = ....
And use H^4 E = E since H^4 = 1.
This gives you an general equation for the eigenvalues. Now when H is hermitian the eigenvalues must be real, and this restricts the solutions to the equation obtained.
4. Dec 15, 2008
soul
Isn't it H*$$\Psi$$ = E * $$\Psi$$?
5. Dec 15, 2008
soul
Also, can I say that since H^4 = 1, then one eigenvalue of it is 1 and if H were not a Hermitian operator, one eigenvalue would be i, since H^2 = -1 or 1?
6. Dec 15, 2008
student111
Let me just post the solution, u are very close anyway:
Let E be an eigenvector of H with eigenvalue e:
H E = e E
Then we have:
H^4 E = e H^3 E = e^2 H^2 E = e^3 H E = e^4 E.
You simply let one H act on E in each step.
Since you have H^4 = 1 we have:
H^4 E = E = e^4 E.
So all you need to solve is:
e^4 = 1.
If H is hermitian it has 4 solutions. But if H is not Hermitian the complex eigenvalues are ruled out and you only have the real solutions.
7. Dec 15, 2008
soul
Student111, thanks for your answer, but I guess H has 2 values if H is hermitian, if it isn't, it has 4?
8. Dec 15, 2008
student111
exactly, i mistyped.
Nonhermitian: 1, -1, i, -i.
And only 1,-1 if it is hermitian.
9. Dec 15, 2008
turin
That's a good thing. Not all operators are guarunteed to be diagnonalizable, but Hermitian is an example that is. Of course, since a power of the operator is unity (i.e. the identity operator), they are hinting that H is at least unitary, which is also diagonalizable. | {"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": 0.964641809463501, "perplexity": 1042.0747700010775}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281419.3/warc/CC-MAIN-20170116095121-00299-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://thinktibits.blogspot.com/2011/01/gate-2006-computer-science-question.html | # GATE 2006 - Computer Science Question Paper
GATE - Computer Science and Engineering - 2006 Question Paper | {"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": 0.843278169631958, "perplexity": 2517.9805017458534}, "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-2018-30/segments/1531676590046.11/warc/CC-MAIN-20180718021906-20180718041906-00482.warc.gz"} |
https://elementaryassessments.com/sentence-stems-for-math/ | # 61 Great Sentence Stems for Math
To encourage math talk during math block, test out a few of these simple yet effective sentence stems for math.
It can be difficult for kids to explain how they arrived at a particular answer or to articulate why they chose to utilize a certain strategy.
That’s why it’s beneficial to support their instruction by providing sentence stems for math.
By using these sentence stems, students will “talk math” confidently and with comprehension.
## Sentence Stems for Math
2. While working on the problem, I noticed that…
3. This is how I can check my answer…
4. A math topic that I am still struggling to understand deeply is…
5. I believe the first step would be to…
6. Another way this problem could have been solved is…
7. Let me explain what I am thinking…
8. I think this way because…
10. I believe my answer is correct because…
11. I noticed that…, so I…
12. Let me explain why I did this…
13. The problem is asking me to… I know this because…
14. A question I still have about (insert any math topic) is…
15. Here is how I started solving the problem…
16. I know the answer is reasonable because…
17. The first step I took was…, Then I…, Next…, Finally…
18. I discovered that…
19. The clues I used to arrive at a final solution…
20. To get my answer, I…
21. I would like to clarify…
22. Next time I solve a problem similar to this one, I will…
23. I infer that…
24. This makes sense because….
25. This does not make sense because…
26. Let me explain how…
27. Today in math class, I learned that…
29. This is different from how I solved another problem because…
30. I wonder what would happen if…
31. The strategy that guided me in solving the math problem was…
32. My answer is… I figured it out by…
33. I chose this method because…
34. Here is how I show my work…
35. This clue is important because…
36. I predict that…
37. The strategies I used to solve the problem…
38. I wonder…
39. One thing that is very important to remember…
40. This problem reminded me of …, so I …
41. My strategy is similar/different to…
42. I tried this method, but a better strategy would be…
43. The first thing I did to solve the problem was…
44. I proved my thinking by…
45. After solving, I still wonder…
46. This is how I knew how to solve the problem….
47. I agree with the solution because…
48. I disagree with the outcome because…
49. What I did is similar to…
50. I arrived at this solution by…
51. A math word I learned today is…
52. I chose to divide, subtract, add, or multiply because…
53. Three key things I learned in math today…
54. One math concept that I’m still struggling with after today is…
55. The answer is incorrect because…
56. In order to solve the problem, one has to already know that…
57. Key information needed to solve the problem includes…
58. Two sentence stems for math I think my teacher might like…
59. The main idea of the world problem is…
60. Here’s how I found my answer…
61. First I thought _____ . However, after reading more carefully, I see that…
Final Thoughts
Now you’re all set with this quality collection of sentence stems for math that is sure to prompt math discussion among students.
If you liked these sentence stems, you might also likeexit ticket ideas for math. | {"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": 0.8783965110778809, "perplexity": 1548.3632955880707}, "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/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00285.warc.gz"} |
http://libros.duhnnae.com/2017/aug6/150291798645-Dyadic-Sets-Maximal-Functions-and-Applications-on-ax-b-Groups-Mathematics-Classical-Analysis-and-ODEs.php | # Dyadic Sets, Maximal Functions and Applications on $ax b$ -Groups - Mathematics > Classical Analysis and ODEs
Dyadic Sets, Maximal Functions and Applications on $ax b$ -Groups - Mathematics > Classical Analysis and ODEs - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.
Abstract: Let $S$ be the Lie group $\mathrm{R}^n\ltimes \mathrm{R}^+$ endowed with theleft-invariant Riemannian symmetric space structure and the right Haar measure$ho$, which is a Lie group of exponential growth. Hebisch and Steger inMath. Z. 2452003, 37-61 proved that any integrable function on $S, ho$admits a Calder\-on-Zygmund decomposition which involves a particular familyof sets, called Calder\-on-Zygmund sets. In this paper, we first show theexistence of a dyadic grid in the group $S$, which has {nice} propertiessimilar to the classical Euclidean dyadic cubes. Using the properties of thedyadic grid we shall prove a Fefferman-Stein type inequality, involving thedyadic maximal Hardy-Littlewood function and the dyadic sharp dyadic function.As a consequence, we obtain a complex interpolation theorem involving the Hardyspace $H^1$ and the $BMO$ space introduced in Collect. Math. 602009,277-295.
Autor: Liguang Liu, Maria Vallarino, Dachun Yang
Fuente: https://arxiv.org/ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9565799236297607, "perplexity": 3971.841076026781}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676596204.93/warc/CC-MAIN-20180723090751-20180723110751-00046.warc.gz"} |
http://www.ck12.org/probability/Tree-Diagrams/lecture/Probability-Part-3/r1/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Tree Diagrams
## Multiply probabilities along the branches and add probabilities in columns
0%
Progress
Practice Tree Diagrams
Progress
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Probability Part 3
Learn how to write out all the possible outcomes of a tree diagram by hand and how to calculate the probability of an outcome. | {"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": 0.9836464524269104, "perplexity": 2947.3069436736396}, "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/1448398459875.44/warc/CC-MAIN-20151124205419-00237-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://sciencehq.com/chemistry/daltons-law-of-partial-pressure.html | # Dalton’s Law of Partial Pressure
Dalton’s Law of Partial Pressure (1807):
This law states that, “The total pressure developed by a mixture of gases is equal to the sum of the partial pressure developed by the individual gases”.
I. e.
$P = p_1 + p_2 + p_3 + p_4 + \cdots$
Where P is the total pressure and $p_1, p_2, p_3, p_4 \cdots$ are the partial pressures of the different non-reacting gases. The partial pressure may be defined as the pressure developed non-reacting gases.
The partial pressure may be defined as the pressure developed by a gas when it is present alone in the same container at the same temperature.
I. e.
Conversion of factors and constants
$\dfrac{F- 32}{9} = \dfrac{C}{5} \\ TK = (t^0C + 273) \\[3mm] 1 atm = 76 cm Hg = 760 mm Hg \\[3mm] = 760 torr \\[3mm] = 101324 Pa = 101325 Nm^{-2} \\[3mm] = 1.011325 bar = 14.7 Ib in^{-2} \\[3mm] 1J = 10^7 erg = 0.239 cal \\[3mm] 1 g mole = 22.4 L \text{of a gas at STP} \\[3mm] R = 0.0821 L atm K^{-1} mol^{-1} \\[3mm] =8.314 JK^{-1} mol^{-1} \\[3mm] 8.314 \times 10^7 erg K^{-1} mol^{-1} \\[3mm] = 1.987 = 2 cal K^{-1} mol^{-1}$
$\text{Partial pressure} = \dfrac{\text{Initial volume} \times \text{Initial pressure}}{\text{Total pressure}}$
$\% \text{of gas in mixture} = \dfrac{\text{Partial pressure}}{\text{total pressure}} \times 100$
$\text{in terms of number of moles},$
$P_A = \dfrac{\text{Moles of A}}{\text{Total moles}} \times P_{mix.}$
$P_{mix.} = \dfrac{(n_1 + n_2 + n_3 \cdots )}{V}RT$
$\left( \text{as} n = \dfrac{w}{m} \right)$
Where $n_1, n_2, n_3 \cdots$ are the number of the moles of different gases present in the mixture.
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