Datasets:

keirp
/

tiny_math

Dataset card Data Studio Files Files and versions Community

Split (1)

train · 17.1k rows

url stringlengths 17 172	text stringlengths 44 1.14M	metadata stringlengths 820 832
http://mathhelpforum.com/algebra/170888-simple-problem-just-need-help-formatting-problem-print.html	# Simple problem, just need help formatting the problem. Printable View • February 11th 2011, 02:00 AM snyper1982 Simple problem, just need help formatting the problem. The weekend gross of \$US10.6 million was a record for the series, surpassing what it's predecessor made in five days by 16%. Well I can't for the life of me figure out how to format the problem to figure out what the predecessor made. First off, I can't decide wheter it is simply 16% less than \$US10.6 Mil, or whether \$US10.6Mil is 16% more than what it made. Does that make sense? Am I getting across what i am trying to convey? It is so frustrating, as I used to be fairly good at math, but I have forgotten so much. Anyways, any help will be GREATLY appreciated, as this is bugging the crap out of me. • February 11th 2011, 03:28 AM tonio Quote: Originally Posted by snyper1982 The weekend gross of \$US10.6 million was a record for the series, surpassing what it's predecessor made in five days by 16%. Well I can't for the life of me figure out how to format the problem to figure out what the predecessor made. First off, I can't decide wheter it is simply 16% less than \$US10.6 Mil, or whether \$US10.6Mil is 16% more than what it made. Does that make sense? Am I getting across what i am trying to convey? It is so frustrating, as I used to be fairly good at math, but I have forgotten so much. Anyways, any help will be GREATLY appreciated, as this is bugging the crap out of me. Hint:The quantity A is r% more than the quantity B if $\displaystyle{A=\left(1+\frac{r}{100}\right)B\Long leftrightarrow B=\frac{A}{1+\frac{r}{100}}}$ Tonio All times are GMT -8. The time now is 04:47 PM.	{"extraction_info": {"found_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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.966002881526947, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/25249/another-mixed-mean-inequality/25319	Another mixed mean inequality Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $a_1,$ $a_2,$ $\ldots,$ $a_n$ be positive real numbers. Prove that $$\sqrt{\frac{a_1^2+\left( \frac{a_1+a_2}{2}\right)^2+\cdots +\left(\frac{a_1+a_2+\cdots +a_n}{n}\right)^2}{n}} \le \frac{a_1+\sqrt{\frac{a_1^2+a_2^2}{2}}+\cdots+\sqrt{\frac{a_1^2+a_2^2+\cdots +a_n^2}{n}}}{n}.$$ I have proved this inequality for $n=2$ and $n=3.$ But I still cannot prove it for the general case. Can somone help me? - I came across a monthly article by a graduate student, which has a strong resemblance to the one you ask. but I cannot remember which one. – Sunni May 19 2010 at 16:58 2 You'll probably have more luck at the art of problem solving forums artofproblemsolving.com/Forum/index.php – jc May 19 2010 at 18:42 2 I like the question and am tempted to have a go -- but no time right now. – gowers May 19 2010 at 22:54 1 Answer Mixed mean inequalities have been studied quite a bit, inspired mostly by the inequalities of Carleman and Hardy, starting probably from this article of K. Kedlaya. Note that the following holds if $r < s$ (in your case $r=1, s=2$): $$\left(\frac{1}{n}\sum_{k=1}^n \left(M_k^{[r]}(\mathbf a)\right) ^s\right)^{1/s}\leq \left(\frac{1}{n}\sum_{k=1}^n\left(M_k^{[s]}(\mathbf a)\right) ^r \right)^{1/r}$$ Where $\mathbf a$ denotes a sequence of real numbers, and $M_k^{[r]}$ denotes the $r$-th power mean of the first $k$ variables. You can find a proof of the general form in "Survey on classical inequalities" by T.M. Rassias (page 32, it is the original proof of B. Mond and J. Pecaric which was later extended to matrices and linear operators). - Thanks Gjergji Zaimi for your reference. – can_hang2007 Jun 2 2010 at 1:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9435849785804749, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/105172/what-is-the-conjugate-of-frac12-frac32i/105174	# What is the conjugate of $\frac{1}{2}+ \frac{3}{2}i$? What is the conjugate of $\dfrac{1}{2}+ \dfrac{3}{2}i$? Firstly, what is conjugation? And secondly, can you should the steps to doing this? "$i$" is the imaginary unit. - $\frac{1}{2}-\frac{3}{2}i$ – Potato Feb 3 '12 at 2:22 5 Google is your friend! – JavaMan Feb 3 '12 at 3:48 ## 1 Answer The conjugate of $a+bi$ is $a-bi$. - What is conjugate? How did you arrive at this answer? :) – David Feb 3 '12 at 1:29 4 – Jonas Meyer Feb 3 '12 at 1:32 1 What Joe said is of course correct. If you want a more geometric idea, you can think of it as reflection across the x-axis in the complex plane. – Jackson Walters Feb 3 '12 at 1:34 4 @David There is nowhere to arrive from, conjugation is just defined this way. period. – Christian Rau Feb 3 '12 at 2:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9072457551956177, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/books?page=2&sort=newest&pagesize=30	# Tagged Questions Note: Books questions are off topic now, ask your question in chat instead. Questions asking "What is a good book about X?" or more generally, "What should I read to learn about X?", typically where there is not a single authoritative reference. Please consider adding your question to this list ... 0answers 81 views ### Introduction to quantum mechanics [duplicate] Possible Duplicate: Book recommendations What is a good introductory book on quantum mechanics? I intend to learn quantum mechanics . But I don't have any suggestions about good books to ... 2answers 289 views ### Level of calculus required for physics [closed] First time for me here so kindly let me know if I violate the rules - especially if this is a duplicate. After reading the page how to become a good theoretical phycist, I started a serious revision ... 1answer 189 views ### Texts on field theory in classical physics I need a very good text on field theory and it should provide good understanding of why this concept cant be ignored?I only need that text which will tell me how field theory is an integral part of ... 0answers 77 views ### Are there any good reading materials for variational approach in many-body theory? [closed] I need something like a summary of existing results, including the treatment of BCS Hamiltonian and Hubbard model. Auerbach's book is a good one but I still hope to get more comprehensive review. My ... 0answers 55 views ### Quantum Mechanics Text for Electrical Engineers [duplicate] Possible Duplicate: What is a good introductory book on quantum mechanics? What is a good introductory text on quantum mechanics that could be used to train electrical engineers in device ... 2answers 120 views ### Which book describe all experiments related to quantum mechanic especially entanglement and photon? [closed] Which book describe all experiments related to quantum mechanic especially entanglement and photon? I expect this book to describe where to buy detector and thing to create entanglement, or how to ... 4answers 780 views ### Help an aspiring physicists what to self-study [closed] This is probably not the kind of question you'll often encounter on this forum, but I think a bit of background is needed for this question to make sense and not seem like a duplicate: 2012 has been ... 2answers 194 views ### Electromagnetism for Mathematician I am trying to find a book on electromagnetism for mathematician (so it has to be rigorous). Preferably a book that extensively uses Stoke's theorem for Maxwell's equations (unlike other books that on ... 2answers 771 views ### What are good books for graduates/undergraduates in Astrophysics? There are no book recommendations for Astrophysics here. I will write my own answer, but I am also interested in what are others' views on the question (I will NOT mark my own answer as the best one). ... 4answers 344 views ### Quantum mechanics textbooks that use path integrals I'm looking for a textbook in quantum mechanics that relies heavily on Green functions and the path integral formalism to supplement my QM books. I want to do some calculations using alternative ... 2answers 131 views ### Small Oscillations and matrices: suggestion about text I'm undergraduate and I'm looking for a text about Small Oscillations in which matrices are used. Could you suggest me a book or a PDF file? 0answers 59 views ### Book reviewing current state of research on complex networks [closed] Can anybody recommend a book reviewing the current state of knowledge and active research on complex networks? Not primarily a textbook but a true review of the field - ideally with references to ... 1answer 273 views ### Reference for mathematics of string theory [closed] I have a great interest in the area of string theory, but since I am more focused on mathematics, I was wondering if there is any book out there that covers mathematical aspects of string theory. I ... 3answers 368 views ### Physics book recommendations for transition to PhD study Now in the end of my MSc in physics I've been contemplating in enrolling in a Phd in theoretical physics. Before that I think I would do best to re-study some subjects. My question is then, what books ... 3answers 192 views ### Good books about waves and optics [duplicate] Possible Duplicate: What’s a good textbook to learn about waves and oscillations? Where is a good place to learn (classical) optics? I'm looking for a good book about waves and optics, ... 1answer 218 views ### Mathematical definitions in string theory Does anyone know of a book that has mathematical definitions of a string, a $p$-brane, a $D$-brane and other related topics. All the books I have looked at don't have a precise definition and this is ... 5answers 505 views ### Reading list in topological QFT I'm interested in learning about topological QFT including Chern Simons theory, Jones polynomial, Donaldson theory and Floer homology - basically the kind of things Witten worked on in the 80s. I'm ... 1answer 99 views ### Coverage of QED in introductory QFT books [closed] Do most introductory QFT books contain QED treatment? I do know that most books contain introductory treatment on QCD, but not usre on QED. 3answers 447 views ### What are some of the best books on complex systems? I'm rather interested in getting my feet wet at the interface of complex systems and emergence. Can anybody give me references to some good books on these topics? I'm looking for very introductory ... 3answers 242 views ### Looking for a complete review of the BEC-BCS crossover I'm looking for comprehensive review of the BEC-BCS crossover, both from a theoretical point of view, and from a experimental one. Even something at textbook level, but exhaustive, would be OK, but I ... 0answers 21 views ### Helpful tutorials on force [duplicate] Possible Duplicate: List of good classical physics books I'm having a hard time understanding force in my highschool AP physics class, i looked through youtube but without luck. Could ... 0answers 42 views ### Books dealing with Quantum field theory [duplicate] Possible Duplicate: What is a complete book for quantum field theory? I am looking for good books that deal with Quantum Field theory starting from basics. Please suggest something that ... 3answers 250 views ### What is a good non-technical introduction to theories of everything? I'm not a physicist but I'm interested in unified theories, and I do not know how to start learning about it. What would be a good book to read to start learning about this topic? 4answers 239 views ### Is matrix mechanics or wave mechanics more preferred? Which quantum mechanics formulation is popular: wave mechanics or matrix mechanics? I find this extremely confusing: Some post-quantum mechanics textbooks seem to prefer wave mechanics version, while ... 0answers 104 views ### MIT OpenCourseware vs Real MIT course [closed] Is MIT course much different from MIT OpenCourseWare? I am curious, because as a high schooler, I have some intent to study from MIT OpenCourseWare. Will this allow me to be more comfortable if I am ... 4answers 260 views ### Matrix mechanics for those with wave mechanics background Just curious: Is there any book or resource that teaches matrix mechanics (quantum mechanics) only without wave mechanics stuff - meaning that the book assumes wave mechanics background. 2answers 488 views ### Gentle introduction to twistors When reading about the twistor uprising or trying to follow a corresponding Nima talk, it always annoys me that I have no clue about how twistor space, the twistor formalism, or twistor theory works. ... 4answers 651 views ### Introduction to AdS/CFT AdS/CFT seems like a really hot topic and I'd like to start reading about. I am looking for the best introduction at my level, i.e. I have a background in QFT, CFT and general relativity at the level ... 5answers 500 views ### Crash course in classical thermodynamics I've been working on some projects lately where it would be very handy to know more about thermodynamics than I do, but sadly I never had a chance to take a proper thermodynamics course in college. ... 0answers 23 views ### Can you recommend me a book on the discoveries and properties of these particles? [duplicate] Possible Duplicate: Good book about elementary particles for high school students? Some days ago, I've found this: When I was in high-school, They taught me about electrons, protons and ... 1answer 199 views ### Problem book in Quantum mechanics with emphasis on physical(ly relevant) problems I am a second year undergraduate and studying quantum mechanics from sakurai's 'Modern Quantum Mechanics'. Is it a good idea to solve problems from sakurai, which are mostly mathematical in nature? I ... 10answers 4k views ### What is a good introductory book on quantum mechanics? I'm really interested in the quantum theory and would like to learn all that I can about it. I've followed a few tutorials and read books but none satisfied my need completely. What is the good ... 0answers 43 views ### How to make strong base for fundamental physics? [duplicate] Possible Duplicate: Book recommendations How to make strong base for fundamental physics? I want to learn the fundamentals of physics. But in a simple manners with very general examples. ... 0answers 49 views ### Prerquisites for QFT [duplicate] Possible Duplicate: Book recommendations Is there a book that covers everything you need to know (and possibly more) before starting a course on QFT. Alternatively, I need a list of what ... 1answer 443 views ### Best Materials Science Introduction Book? Why is this here? Sorry, I know that this isn't the best place to put this however there does not seem to be any other stack exchange specific to this so I thought this would be the best place ... 0answers 31 views ### I'm a business major but I'd like to be generally fluent in undergraduate physics. (Fluent; not getting a degree or anything.) [duplicate] Possible Duplicate: Book recommendations Since I've already picked up introductory physics, can someone here help me construct a sequence of textbooks I should read (proof rigorous; ... 1answer 125 views ### What would be the requirement to learn matrix mechanics? What would be the requirement to learn matrix mechanics? More specifically, what math do I need? Can anyone recommend me a book that covers all maths needed for matrix mechanics? 5answers 923 views ### Recommendations for Statistical Mechanics book I saw Book recommendations No reference to Statistical Mechanics there. I learned thermodynamics and the basics of statistical mechanics but I'd like to sit through a good advanced book/books. Mainly ... 2answers 160 views ### Supersymmetry in Quantum Field Theory I have an idea of supersymmetry in quantum mechanics, can you suggest a book on "supersymmetry in quantum field theory", which has sufficient mathematical rigour like "Peskin and Schroeder" 4answers 477 views ### Topology needed for Differential Geometry [duplicate] I am a physics undergrad, and need to study differential geometry ASAP to supplement my studies on solitons and instantons. How much topology do I need to know. I know some basic concepts reading from ... 2answers 342 views ### Book covering Topology required for physics and applications I am a physics undergrad, and interested to learn Topology so far as it has use in Physics. Currently I am trying to study Topological solitons but bogged down by some topological concepts. I am not ... 2answers 389 views ### Where can I find beginner's information about quantum mechanics? [duplicate] Possible Duplicate: Popular books on QM I hope that this question is suitable for this site. I am interested in reading up on quantum theory so that I can reasonably understand the ... 2answers 389 views ### Introduction to string theory I am in the last year of MSc. and would like to read string theory. I have the Zwiebach Book, but along with it what other advanced book can be followed, which can be a complimentary to Zwiebach. I ... 1answer 130 views ### What books, papers, and authors to read to improve your scientific writing skills and style? [closed] Reading well-written phsyics books and papers can improve one's writing skills and style. However, as someone who speaks English as a second language, I'm not sure which books, papers, and authors are ... 0answers 71 views ### Beginning Physics, Advise on Books [duplicate] Possible Duplicate: Beginner Physics Resources? Book recommendations I'm wanting to learn physics. In fact i'm a software engineer and i find physics very fascinating. But everything I ... 0answers 47 views ### Books on representation theory [duplicate] Possible Duplicate: Best books for mathematical background? I'm looking for a textbook on the group/representation theory for a student-physicist. The main questions of interest are ... 0answers 208 views ### Applications of Physics to other fields like Economics, Finance, Political Science, Sociology? [closed] I know that Physics can be applied to other fields like Economics, Finance, Political Science, Sociology,etc. Can someone give me some examples of those applications that have been empirically tested ... 3answers 372 views ### Textbook on group theory to be able to start QFT I am very enthusiastic about learning QFT. How much group theory would I need to master? Please could you recommend me a textbook on group theory, which would help me to start QFT? 2answers 179 views ### Smithsonian Astrophysical Observatory (SAO) star catalog Is a Smithsonian Astrophysical Observatory (SAO) star catalog, 4 volumes, dated 1977, worth keeping in a library or is it too outdated? Can it be used for epoch 200? 1answer 262 views ### some hints/introductions/textbooks for LBM(Lattice Boltzmann methods) fluid simulation? I come from a computer science background and I want to learn details about LBM in fluid simulation. I have been searching around for a concise tutorial that tells me how LBM is used to solve 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": 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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9284644722938538, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/151134/what-is-1-cm-3?answertab=votes	# What is 1 cm$^{ -3}$? I am having trouble with notations like 1 cm$^{-3}$, especially since I am converting them between compound units. Is there a way to express 1 cmcm$^{-3}$ without writing the negative exponent? The particular queston I was asked was: The EC standard for lead in the atmosphere is 2 µg/m$^{−3}$. Express this value in scientific notation in g/cm$^{−3}$, to an appropriate number of significant figures. I have never run into these before, and I have serious problems coming to grips with the notation. $1\cdot10 ^{ -3}$ would be fine, but 1 * cm$^{−3}$ just seems weird. But, I simply have a hard time visualising these quantities. Can someone suggest a way to imagine them, or motivate the need for a negative exponent attached to a number of units? What does the negative exponent actually mean? Also, is there a term for these? ````1 (centimetre^(-3)) = 1 000 000 m^-3 ```` Which is not helpful at all. - Express two micrograms per cubic metre as a number of grams per cubic centimetre. – Mark Bennet May 29 '12 at 12:38 So in particular what google is telling you is if you have $1$ unit (e.g. gram) per cubic centimetre, you have $1,000,000$ units per cubic metre, which is what you expect. – Matt Pressland May 29 '12 at 12:46 6 The notation $\mathrm{g}/\mathrm{cm}^{-3}$ does not mean 'gram per cubic centimeter' but 'gram times cubic centimiter'. Gram per cubic centimeter would be either $\mathrm{g}/\mathrm{cm}^3$ or $\mathrm{g}\,\mathrm{cm}^{-3}$. – Egbert May 29 '12 at 13:04 3 Egbert's right - either you've miscopied the question, or the question is nonsense. – Gerry Myerson May 29 '12 at 13:20 ## 2 Answers I believe the quoted passage contains an error. The EC standard for lead in the atmosphere is 2 µg/m$^{−3}$. Express this value in scientific notation in g/cm$^{−3}$, to an appropriate number of significant figures. It should either be 2 µg m$^{−3}$ or 2 µg/m$^3$, both of which have the same meaning, but not 2 µg/m$^{−3}$. Likewise, g/cm$^{−3}$ should either be g cm$^{−3}$ or g/cm$^3$. The negative exponent is just an alternative notation for division. Think of µg/m$^3$ as "micrograms per cubic meter". This is a density. So a higher number means a greater density of lead. Visualize 2 µg of substance spread out over the volume of a cube with dimensions 1m $\times$ 1m $\times$ 1m. Now as a warmup to doing unit conversions, ask yourself whether 2 µg/m$^3$ is a higher or lower density than 2 µg/cm$^3$. If you think about it, you will realize that the former represents 2 µg of substance spread out over a large cube with dimensions 1m $\times$ 1m $\times$ 1m, while the latter represents the same amount of substance spread out over a tiny cube with dimensions 1cm $\times$ 1cm $\times$ 1cm. So 2 µg/cm$^3$ is the higher density since the same amount of substance is crammed into a much smaller volume. Now lets go back to the original density, 2 µg/m$^3$, and try to write it, first in units of µg/cm$^3$, and then in units of g/cm$^3$. A cube with dimensions 1m $\times$ 1m $\times$ 1m is the same as a cube with dimensions 100cm $\times$ 100cm $\times$ 100cm. Thinking about that cube as composed of many tiny 1cm $\times$ 1cm $\times$ 1cm cubes, how many of those tiny cubes are there? The answer is $100\times100\times100=100^3=10^6=1,000,000$. So if the big cube contains 2 µg of substance, which we imagine to be uniformly distributed, how much substance does each tiny cube contain? Since there are 1,000,000 tiny cubes, each contains $1/1000,000$ of the total amount of substance, or $2\times10^{-6}$ µg. Therefore, a density of 2 µg/m$^3$ is the same as a density of $2\times10^{-6}$ µg/cm$^3$. This insight can be encoded in the following algebraic manipulation: $$2\frac{{\mu}\text{g}}{\text{m}^3}=2\frac{{\mu}\text{g}}{\text{m}^3}\left(\frac{1\,\text{m}}{100\,\text{cm}}\right)^3=2\frac{{\mu}\text{g}}{\text{m}^3}\frac{\text{m}^3}{1,000,000\,\text{cm}^3}=\frac{2}{1,000,000}\frac{{\mu}\text{g}}{\text{cm}^3}.$$ In the first step we are free to multiply by (1 m/100 cm) since that's just a way of writing 1. Now let's deal with the conversion from µg to g. One µg is a tiny amount of material - just 1 millionth of a gram. If you have 1 µg of material in a certain volume, that's the same as having 1/1,000,000 g of material in the same volume. This insight allows us to complete the calculation: $$\frac{2}{1,000,000}\frac{{\mu}\text{g}}{\text{cm}^3}=\frac{2}{1,000,000}\frac{{\mu}\text{g}}{\text{cm}^3}\frac{\frac{1}{1,000,000}\text{g}}{\mu\text{g}}=\frac{2}{10^{12}}\frac{\text{g}}{\text{cm}^3}=2\times10^{-12}\frac{\text{g}}{\text{cm}^3}.$$ Again we are allowed to multiply by (1/1,000,000 g)/(1 µg) in the first step since that's just equal to 1. - 2 This is a great answer! +1 – M Turgeon May 29 '12 at 15:31 You can treat units like ordinary factors. For example $2 \mu g/m^{−3}=2 \cdot \mu g/m^{−3}=2\cdot(0.000001g)/(100cm)^{-3}=2\cdot 0.000001/100^{-3}\cdot g/(cm)^{-3}$ Just do plain algebra and replace $m\to 100\cdot cm$ with no further thinking. EDIT: there is no need to visualize this when you merely want to change units and there are much harder physical constants with more complicating units. - 1 I agree you don't have to visualise - but some people learn better when they can visualise things, until they become confident with the notation. I remember when I was ten having a problem understanding the mathematics of acceleration because I couldn't make immediate or intuitive sense of $s^{-2}$. When I did "get" it, it was easy. – Mark Bennet May 29 '12 at 14:00 When trying to understand these simpler units, try taking them apart. For example $1ms^{-2}$, the unit of acceleration, means a change of velocity per time, i.e. $1ms^{-1}/s$. Velocity in turn is distance per time. – Wormbo May 29 '12 at 16:23 I suggest one should visualize equations and not units (unless they are a/b fraction with no power). If you skip making mental essays about what fractions of unit powers mean, you have much more time to think about stuff that is really matters. Maths is made for providing tool so that you get results without having to visualize complicating stuff like high dimensional space. – Gerenuk May 29 '12 at 17:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9301517605781555, "perplexity_flag": "middle"}
http://physicspages.com/2011/04/14/length-contraction-and-the-pole-in-a-barn-paradox/	Notes on topics in science ## Length contraction and the pole-in-a-barn paradox Required math: algebra, geometry Required physics: basics of relativity Two famous results of relativity are time dilation and length contraction. We looked at time dilation in an earlier post, so we’ll examine length contraction here. Although it is possible to analyze length contraction using space-time diagrams, it is actually easier to derive the effect from the Lorentz transformations, which are \| \| \| \| \|----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\| \| $\displaystyle t_{2}$ \| $\displaystyle =$ \| $\displaystyle \frac{1}{\sqrt{1-v^{2}}}(t_{1}-vx_{1})$ \| \| $\displaystyle x_{2}$ \| $\displaystyle =$ \| $\displaystyle \frac{1}{\sqrt{1-v^{2}}}(x_{1}-vt_{1})$ \| \| $\displaystyle y_{2}$ \| $\displaystyle =$ \| $\displaystyle y_{1}$ \| \| $\displaystyle z_{2}$ \| $\displaystyle =$ \| $\displaystyle z_{1}$ \| These transformations apply to the special case of inertial frames that are aligned so that observer ${O_{2}}$ travels with speed ${v}$ along ${O_{1}}$‘s ${x_{1}}$ axis, with the corresponding ${y}$ and ${z}$ axes parallel in the two systems. Since the quantity ${1/\sqrt{1-v^{2}}}$ occurs so often in relativity, it is usual for it to be given the shorthand notation $\displaystyle \gamma\equiv\frac{1}{\sqrt{1-v^{2}}}$ so the Lorentz transformations become \| \| \| \| \|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\| \| $\displaystyle t_{2}$ \| $\displaystyle =$ \| $\displaystyle \gamma(t_{1}-vx_{1})$ \| \| $\displaystyle x_{2}$ \| $\displaystyle =$ \| $\displaystyle \gamma(x_{1}-vt_{1})$ \| \| $\displaystyle y_{2}$ \| $\displaystyle =$ \| $\displaystyle y_{1}$ \| \| $\displaystyle z_{2}$ \| $\displaystyle =$ \| $\displaystyle z_{1}$ \| Now suppose a rod of length ${l}$ is at rest on the ${x_{1}}$ axis. How long does this rod appear to observer ${O_{2}}$? The key point in answering this seemingly simple question is that ${O_{2}}$ must measure the length of the rod in such a way that the two events defining the location of the two endpoints of the rod occur at the same time in ${O_{2}}$‘s frame. That is, if we require both events to occur at a fixed time ${t_{2}=0}$, from the Lorentz transformation, we must satisfy the condition \| \| \| \| \|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\| \| $\displaystyle t_{2}$ \| $\displaystyle =$ \| $\displaystyle 0=\gamma(t_{1}-vx_{1})$ \| \| $\displaystyle t_{1}$ \| $\displaystyle =$ \| $\displaystyle vx_{1}$ \| Since the two frames coincide at ${t_{1}=t_{2}=x_{1}=x_{2}=0}$, this condition is satisfied for the measurement of one of the end points. At the other end point, ${x_{1}=l}$, so ${t_{1}}$ must be ${t_{1}=vl}$, and using the second Lorentz transformation \| \| \| \| \|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\| \| $\displaystyle x_{2}$ \| $\displaystyle =$ \| $\displaystyle \gamma(l-v^{2}l)$ \| \| $\displaystyle$ \| $\displaystyle =$ \| $\displaystyle \gamma l(1-v^{2})$ \| \| $\displaystyle$ \| $\displaystyle =$ \| $\displaystyle \frac{l}{\gamma}$ \| using the definition of ${\gamma}$. Since ${\gamma\ge1}$, we see that the length of the rod is measured to be shorter by ${O_{2}}$ which gives rise to the phenomenon of length contraction. A popular ‘paradox’ that involves length contraction is the pole-in-a-barn problem. Suppose there is a pole of length 20 m and a barn of length 15 m (both measured in their rest frames). If an athlete picks up the pole and runs at a speed ${0.8}$ (giving ${\gamma=5/3}$) from left to right towards the barn, then (assuming the barn has a front and back door in line with the athlete’s path) to an observer in the barn’s frame of reference, since the pole is contracted to a length of 12 m, it will fit entirely inside the barn. However, in the athlete’s frame of reference, it is the barn which appears contracted to a length of 9 m so the pole will not fit inside the barn. Who is right? The catch is that the problem is phrased in a way which ignores the concept of simultaneity. In each case, the observer must measure the lengths of the pole and the barn at the same time in their respective frames. This can be seen by using a space-time diagram and the Lorentz transformations. We’ll examine the Lorentz transformations first, and work out the problem in general, assuming that the coordinate systems coincide at the event where the right end of the pole enters the left end of the barn. We’ll look specifically at three events: • Event A: the left (back) end of the pole enters the left end of the barn; • Event B: the location of the right end of the pole at the same time as event A, as seen by the athlete; • Event C: as event B, except in the frame of the barn. To work out the coordinates of A, it is easiest to start in the frame of the athlete. The pole’s right end passed the left end of the barn at ${t_{2}=0}$, the pole is length ${l}$ and the athlete is moving at speed ${v}$, so to him it will take a time ${l/v}$ for the entire length of the pole to pass into the barn. Since the pole is at rest in this frame and its right end has coordinate 0, the left end has coordiante ${-l}$. Thus the coordinates of A in the athlete’s frame are $\displaystyle A_{a}=\left(\frac{l}{v},-l\right)=(25,-20)$ We can now use the Lorentz transformations to convert this to the barn’s frame. Remember that the barn is moving at speed ${-v}$ relative to the athlete so we need to use ${-v}$ in the Lorentz formulas. Plugging in the coordinates, we get $\displaystyle A_{b}=\left(\gamma\left(\frac{l}{v}-lv\right),0\right)=(15,0)$ Note as a check that the ${x_{b}}$ coordinate is zero, since the left end of the barn is fixed at that point in the barn’s frame. Note also that in the barn’s frame, event A occurs earlier than in the athlete’s frame. Now we want to find event B. In the athlete’s frame, we want the location of the front of the pole at the same time as event A, that is at ${t_{a}=l/v}$. Since the pole is at rest in that frame and its front end defines the location ${x_{a}=0}$, this is just $\displaystyle B_{a}=\left(\frac{l}{v},0\right)=(25,0)$ Using the Lorentz transformations, we find the coordinates of B in the barn’s frame: $\displaystyle B_{b}=\left(\gamma\frac{l}{v},\gamma l\right)=\left(41.67,33.3\right)$ Note that both observers agree that event B occurs well outside the right end of the barn, since the barn is 15 m long in its rest frame, and event B occurs at ${x_{b}=33.3}$ m. In the athlete’s frame, the barn is only 9 m long and the pole is 20 m long, so the right end of the pole will be 11 m outside the right end of the barn. However, although events A and B occur at the same time to the athlete, they clearly do not occur at the same time to the barn (${t_{b}=15}$ for event A and ${t_{b}=41.67}$ for event B). So this leads us to event C: the location and time of the right end of the pole at the same time as event A, as measured in the barn’s frame. Since the pole appears to be contracted to length ${l/\gamma}$ in this frame, we get $\displaystyle C_{b}=\left(\gamma\left(\frac{l}{v}-lv\right),\frac{l}{\gamma}\right)=(15,12)$ Applying Lorentz, we get the coordinates in the athlete’s frame (this time we’re converting to an observer with speed ${+v}$, so we use ${+v}$ in the Lorentz formulas): $\displaystyle C_{a}=\left(\frac{l}{v}-vl,0\right)=(9,0)$ Again, both observers agree that the right end of the pole is inside the barn at event C, however the athlete believes that the left end of the pole has not yet entered the barn. Remember that the left end of the pole enters the barn at event A, and for that event ${t_{a}=25}$, so event C’s time of 9 is well before event A. The events are summarized in the space-time diagram (remember that the scales along the athlete’s axes are not the same as those along the barn’s axes; you need to use the invariant hyperbolas (which we haven’t drawn here) to calibrate the athlete’s axes): \par The diagram shows the situation in the barn’s frame, so the ends of the barn are represented by the world line of the vertical axis for the left end and the blue line for the right end. The world lines of the ends of the pole are the two solid red lines. The dashed line AB is the position of the pole as seen by the athlete (that is, it corresponds to events A and B above). The solid green line AC shows the pole as seen by the barn’s frame (corresponding to events A and C above). Thus to the athlete, if both ends of the pole are always viewed at the same time in his frame, the pole will never fit inside the barn, but in the barn’s frame, viewing the two ends of the pole at the same time does allow it to fit inside the barn. So in a sense, both conclusions are correct; it just depends on which frame you are in. Actually, the phrasing of the question as to whether the pole fits inside the barn is misleading, since the pole really only can be said to fit inside the barn if they are both at rest, and in this case it clearly won’t fit. The conclusion that the pole somehow fits inside if it is moving very fast is misleading, since the the line AC in the diagram cuts across a set of lines parallel to AB, each of which represents a constant time in the althlete’s frame, so in a sense, the barn observer is sampling a set of infinitesimal slices of the pole, each selected from a different time in the athlete’s frame (the rest frame of the pole). Such a sampling process isn’t really a true reflection of the physical pole itself, so in that sense, the pole doesn’t actually fit inside the barn. About these ads ### Like this: Like Loading... By , on Thursday, 14 April 2011 at 09:09, under Physics, Relativity. Tags: length contraction, Lorentz contraction, Lorentz transformations, pole in a barn. 2 Comments Post a comment or leave a trackback: Trackback URL. ### Trackbacks • By Lorentz transformations in three dimensions « Physics tutorials on Wednesday, 22 June 2011 at 14:47 [...] seen that distances parallel to the direction of motion are contracted by a factor , while distances perpendicular to the motion are unaffected. We can use these facts to [...] • By Compton scattering « Physics tutorials on Tuesday, 5 July 2011 at 10:50 [...] seen that distances parallel to the direction of motion are contracted by a factor , while distances perpendicular to the motion are unaffected. We can use these facts to [...] • ### Meta Cancel Post was not sent - check your email addresses! Email check failed, please try again Sorry, your blog cannot share posts by email. %d bloggers like this:	{"extraction_info": {"found_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": 85, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9268366694450378, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/5342/montgomery-exponentiation-selecting-input-value-r-for-a-given-biginteger/5343	Montgomery Exponentiation - selecting input value R for a given BigInteger I have Montgomery exponentiation working, but it's working quite slow. I suspect there are two reasons for this - I implemented it bit size instead of word size (I didn't realize at the time that software implementation should use word size). The second is how I select R. Given a modulus N with bit length n, I'm calculating R by raising 2 to the power of the bit length of N, n. i.e. This is way all the examples I've seen use, but they all use small numbers. Every number I'm dealing with here has at least 1024 bits. ````BigInteger R = new BigInteger("2").pow(n.bitLength()); ```` I suspect this might be why my Montgomery is running slower than my Right-To-Left binary implementation - can anyone give me a more efficient way of selecting r? (In as simple English as possible please!) Edit - the full code is at - http://codereview.stackexchange.com/questions/18199/optimization-of-exponentiation if anyone want to have a look at it. - (I hope this is just for fun. You'll have huge side channels with such an implementation) – CodesInChaos Nov 13 '12 at 17:02 – Saf Nov 13 '12 at 22:03 1 Answer Montgomery multiplication makes sense only with word sizes. If your word size is $w$ (e.g. $w = 32$ if you have 32-bit words), then $R = 2^{kw}$ for some integer $k$; you choose $k$ as small as possible, given that you must have $R \geq N$. In plain words, if your modulus $N$ has size $n$ bits, then you look for the next multiple of $w$. For a 1024-bit modulus, which is already a multiple of 32, you use $R = 2^{1024}$. For a 1025-bit modulus, you would use $R = 2^{1056}$. You really want to read chapter 14 of the Handbook of Applied Cryptography. - Thank you, this was how I was calculating R. I guess the slowdown is the way I programmed the actual Montgomery methods monPro and modExp (See the code review I example I linked if you are a coder) – Saf Nov 11 '12 at 22:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9188952445983887, "perplexity_flag": "middle"}
http://mathforum.org/mathimages/index.php?title=Transformations_and_Matrices&diff=25869&oldid=25832	# Transformations and Matrices ### From Math Images (Difference between revisions) \| \| \| \| \| \|----------\|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\|----------\|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\| \| \| \| \| \| \| Line 36: \| \| Line 36: \| \| \| \| \| \| \| \| \| ==Transformation Composition Is Matrix Multiplication== \| \| ==Transformation Composition Is Matrix Multiplication== \| \| - \| Transformations are usually not used by themselves, especially in graphics, so you need to have a way to compose transformations, as in <math>g(f(P))</math>. But if G is the matrix for the transformation g, and F is the matrix for the transformation f, then the matrix product GF is the matrix for the composed functions gf.<br /> \| + \| Transformations are usually not used by themselves, especially in graphics, so you need to have a way to compose transformations, as in <math>g(f(P))</math>. But if G is the matrix for the transformation g, and F is the matrix for the transformation f, then the matrix product GF is the matrix for the composed functions gf.<br /><br /> \| \| \| \| + \| For example, we have the translation represented by the matrix <br /> \| \| \| \| + \| ::<math>\begin{bmatrix} 1 & 0 & 2 \\ \| \| \| \| + \| 0 & 1 & 1 \\ \| \| \| \| + \| 0 & 0 & 1 \| \| \| \| + \| \end{bmatrix}</math> which represents a move two units in the x direction and one unit in the y direction. If we want to then rotate the same object with the matrix <br /> \| \| \| \| + \| ::<math>\begin{bmatrix} 0.471 & -0.5 & 0 \\ \| \| \| \| + \| 0.5 & 0.471 & 0 \\ \| \| \| \| + \| 0 & 0 & 1 \| \| \| \| + \| \end{bmatrix}</math> we can represent the combination of the two actions with a single composed matrix. This matrix is found by multiplying the second action by the first action. <br /> \| \| \| \| + \| ::<math>\begin{bmatrix} 0.471 & -0.5 & 0 \\ \| \| \| \| + \| 0.5 & 0.471 & 0 \\ \| \| \| \| + \| 0 & 0 & 1 \| \| \| \| + \| \end{bmatrix} * \begin{bmatrix} 1 & 0 & 2 \\ \| \| \| \| + \| 0 & 1 & 1 \\ \| \| \| \| + \| 0 & 0 & 1 \| \| \| \| + \| \end{bmatrix} = \begin{bmatrix} 0.471 & -0.5 & .442 \\ \| \| \| \| + \| 0.5 & 0.471 & 1.471 \\ \| \| \| \| + \| 0 & 0 & 1 \| \| \| \| + \| \end{bmatrix}</math> So this matrix represents moving, then rotating an object in sequence. \| \| \| \| + \| \| \| \| \| + \| \| \| \| \| \| \| \| \| ==Basic Transformations For Graphics== \| \| ==Basic Transformations For Graphics== \| ## Revision as of 13:18, 18 July 2011 Transformations Field: Geometry Created By: Nordhr Website: [ ] Transformations This picture shows an example of four basic transformations (where the original teapot is a red wire frame). On the top left is a translation, which is essentially the teapot being moved. On the top right is a scaling. The teapot has been squished or stretched in each of the three dimensions. On the bottom left is a rotation. In this case the teapot has been rotated around the x axis and the z axis (veritcal). On the bottom right is a shearing, creating a skewed look. # Basic Description When an object undergoes a transformation, the transformation can be represented as a matrix. Different transformations such as translations, rotations, scaling and shearing are represented mathematically in different ways. One matrix can also represent multiple transformations in sequence when the matrices are multiplied together. ## Linear Transformations Are Matrices A linear transformation on 2D (or 3D) space is a function f from 2D (or 3D) space to itself that has the property that $f(aA + bB) = af(A) + bf(B).$ Since points in 2D or 3D space can be written as $P = xi + yj$ or $P = xi + yj +zk$ with $i$, $j$, and $k$ the coordinate vectors, then we see that $f(P) = xf(i) + yf(j)$ or $f(P) = xf(i) + yf(j) + zf(k)$ This tells us that the linear transformation is completely determined by what it does to the coordinate vectors. Let’s see an example of this: if the transformation has the following action on the coordinates: $\begin{align}f(1,0,0) = f(i) = (2,-2,1) \\ f(0,1,0) = f(j) = (-1,3,2) \\ f(0,0,1) = f(k) = (4,3,-2) \end{align}$ then for any point we have: $\begin{align}f(x,y,z) = (2x,-2x,x)+(-y,3y,2y)+(4z,3z,-2z) \\ = (2x-y+4z, -2x+3y+3z, x+2y-2z) \\ = \begin{bmatrix} 2x - y + 4z \\ -2x + 3y + 3z \\ x + 2y - 2z \end{bmatrix} = \begin{bmatrix} 2 & -1 & 4 \\ -2 & 3 & 3 \\ 1 & 2 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \end{align}$ From this example, we see that the linear transformation is exactly determined by the matrix whose first column is $f(i)$, whose second column is $f(j)$, and whose third column is $f(k)$, and that applying the function f is exactly the same as multiplying by the matrix. So the linear transformation is the matrix multiplication, and we can use the concepts of linear transformation and matrix multiplication interchangeably. ## Transformation Composition Is Matrix Multiplication Transformations are usually not used by themselves, especially in graphics, so you need to have a way to compose transformations, as in $g(f(P))$. But if G is the matrix for the transformation g, and F is the matrix for the transformation f, then the matrix product GF is the matrix for the composed functions gf. For example, we have the translation represented by the matrix $\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ which represents a move two units in the x direction and one unit in the y direction. If we want to then rotate the same object with the matrix $\begin{bmatrix} 0.471 & -0.5 & 0 \\ 0.5 & 0.471 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ we can represent the combination of the two actions with a single composed matrix. This matrix is found by multiplying the second action by the first action. $\begin{bmatrix} 0.471 & -0.5 & 0 \\ 0.5 & 0.471 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0.471 & -0.5 & .442 \\ 0.5 & 0.471 & 1.471 \\ 0 & 0 & 1 \end{bmatrix}$ So this matrix represents moving, then rotating an object in sequence. ## Basic Transformations For Graphics Computer graphics works by representing objects in terms of simple primitives (link to the graphics primitives page) that are manipulated with transformations that preserve some primitives’ essential properties. These properties may include angles, lengths, or basic shapes. Some of these transformations can work on primitives with vertices in standard 2D or 3D space, but some need to have vertices in homogeneous coordinates. The general graphics approach is to do everything in homogeneous coordinates, but we’ll talk about the primitives in terms of both kinds when we can. The most fundamental kinds of transformations for graphics are rotation, scaling, and translation. There are also a few cases when you might want to use shear transformations, so we’ll talk about these as well. # A More Mathematical Explanation Note: understanding of this explanation requires: stacks [Click to view A More Mathematical Explanation] ### Rotation <span class="_togglegroup _toggle_initshow _toggle _toggler toggle-visible" style="dis [...] [Click to hide A More Mathematical Explanation] ### Rotation A 2D rotation transformation rotates everything in 2D space around the origin by a given angle. In order to see what it does, let’s take a look at what a rotation by a positive angle  does to the coordinate axes. Now (x,y) is the result when you apply the transformation to (1,0), which means that $\begin{align} x = cos(\theta ) \\ y = sin(\theta ) \end{align}$ But (x’,y’) is the result when you apply the transformation to (0,1), or $\begin{align} x' = cos(\theta +\frac{p}{2}) = cos(\theta )cos(\frac{p}{2}) - sin(\theta )sin(\frac{p}{2}) = -sin(\theta ) \\ y' = sin(\theta +\frac{p}{2}) = sin(\theta )cos(\frac{p}{2}) + cos(\theta )sin(\frac{p}{2}) = cos(\theta ) \end{align}$ Then as we saw above, the rotation transformation must have the image of (1,0) as the first column and the image of (0,1) as the second column, or $rotate(\theta ) = \begin{bmatrix} cos(\theta ) & -sin(\theta ) \\ sin(\theta ) & cos(\theta ) \end{bmatrix}$ or, as XKCD (http://xkcd.com/184/) sees it (notice that the rotation is by -90° and $sin(-90\,^{\circ}) = -sin(90\,^{\circ})$, <br The situation for 3D rotations is different because a rotation in 3D space must leave a fixed line through the origin. In fact we really only handle the special cases where the fixed line is one of the coordinate axes. Let’s start with the easiest one. A rotation around the Z-axis is a 2D rotation as above with the third dimension fixed. So the matrix for this rotation is pretty clearly $\begin{bmatrix} cos(\theta ) & -sin(\theta ) & 0 \\ sin(\theta ) & cos(\theta ) & 0 \\ 0 & 0 & 1 \end{bmatrix}$ A rotation around the X-axis is pretty similar. If we look down the X-axis, we see the following 2D coordinates: with $Y \times Z = X$, the axis of rotation. This looks like an exact analogue of the XY-plane, and so we can see that the rotation matrix must leave X fixed and operate only on Y and Z as $\begin{bmatrix} 1 & 0 & 0 \\ 1 & cos(\theta ) & -sin(\theta ) \\ 0 & sin(\theta ) & cos(\theta ) \end{bmatrix}$ For rotations around the Y-axis, the view down the Y-axis looks different from the one down the Z-axis; it is Here a positive-angle is from the X-axis towards the Z-axis, but $X \times Z = -Z \times X = -Y$, so the rotation axis dimension is pointing in the opposite direction from the Y-axis. Thus a the angle for the rotation is the negative of the angle we would see in the axes above, and since cos is an even function but sin is odd, we have the rotation matrix $\begin{bmatrix} cos(-\theta ) & 0 & -sin(-\theta ) \\ 0 & 1 & 0 \\ sin(-\theta ) & 0 & cos(-\theta ) \end{bmatrix} = \begin{bmatrix} cos(\theta ) & 0 & sin(\theta ) \\ 0 & 1 & 0 \\ sin(-\theta ) & 0 & cos(\theta ) \end{bmatrix}$ around the Y-axis. When you want to get a rotation around a different line than a coordinate axis, the usual approach is to find two rotations that, when composed, take a coordinate line into the fixed line you want. You can then apply these two rotations, apply the rotation you want around the coordinate line, and apply the inverses of the two rotations (in inverse order) to construct the general rotation. The sequence goes like this: apply a rotation $R_1$ around the Z-axis to move your fixed line into the YZ-plane apply a rotation $R_2$ around the X-axis to move that line to the Y-axis apply the rotation by your desired angle around the Y-axis apply the inverse $R_2^{-1}$ to move your rotation line back into the YZ-plane apply the inverse $R_1^{-1}$ to move your rotation line back to the original line. Whew! This can all be put into a function – or you can simply keep everything in terms of rotations around X, Y, and Z. If we are working in homogeneous coordinates, we see that all of the rotation operations take place in standard 3D space and so the fourth coordinate is not changed. Thus the general pattern for all the rotations in homogeneous coordinates is $\begin{bmatrix} & * & * & 0 \\ [0.3em] * & * & * & 0 \\ [0.3em] * & * & * & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$ where the * terms are the terms from the 3D rotations above. ### Scaling Scaling is the action of multiplying each coordinate of a point by a constant amount. As an example, let $f(x,y,z) = (2x,3y,4z)$. Then $f((x,y,z)+(a,b,c)) = f(x+a,y+b,z+c) = (2(x+a),3(y+b), 4(z+c)) = (2x,3y,4z)+(2a,3b,4c)$ So this is a linear transformation. If we look at what this transformation does to each of the coordinate vectors, we have $\begin{align} f(1,0,0) = (2,0,0) \\ f(0,1,0) = (0,3,0) \\ f(0,0,1) = (0,0,4) \end{align}$ So the matrix for this transformation is $\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$ and, in general, a scaling matrix looks like $\begin{bmatrix} \sigma_x & 0 & 0 \\ 0 & \sigma_y & 0 \\ 0 & 0 & \sigma_z \end{bmatrix}$ where the $\sigma_x , \sigma_y ,\text{ and }\sigma_z$ are the scaling factors for x, y, and z respectively. In case of only 2D transformations, scaling simply scales down to two dimensions and we simply have $\begin{bmatrix} \sigma_x & 0 \\ 0 & \sigma_y \end{bmatrix}$ In case we are working with homogeneous coordinates, a scaling transformation only acts on the three primary components and leaves the homogeneous component alone, so we simply have the matrix $\begin{bmatrix} \sigma_x & 0 & 0 & 0 \\ 0 & \sigma_y & 0 & 0 \\ 0 & 0 & \sigma_z & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$ for the scaling transformation. ### Translation Notice that a translation function cannot be a linear transformation on normal space because it does not take the origin to the origin. These are examples of affine transformations, transformations that are composed of a linear transformation, such as a rotation, scaling, or shear, and a translation. In order to write a translation matrix, we need to use homogeneous coordinates. If we want to add $T_x$ to the X-coordinate and $T_y$ to the Y-coordinate of every point in 2D space, we see that $\begin{bmatrix} 1 & 0 & T_x \\ 0 & 1 & T_y \\ 0 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} x \\ y \\ 1 \end{bmatrix} = \begin{bmatrix} x+T_x \\ y+T_y \\ 1 \end{bmatrix}$ so that the matrix $\begin{bmatrix} 1 & 0 & T_x \\ 0 & 1 & T_y \\ 0 & 0 & 1 \end{bmatrix}$ gives a 2D translation. The 3D case is basically the same, and by the same argument we see that the 3D translation is given by $\begin{bmatrix} 1 & 0 & 0 & T_x \\ 0 & 1 & 0 & T_y \\ 0 & 0 & 1 & T_z \\ 0 & 0 & 0 & 1 \end{bmatrix}$ hese are linear transformations in the space one degree higher than the geometry you are working with. In fact, the main reason for including homogeneous coordinates is the math for graphics is to be able to handle translations (and thus all basic transformations) as linear transformations represented by matrices. ### Shear The shear transformation is not widely used in computer graphics, but can be used for things like the oblique view in engineering drawings. The concept of a shear is to add a multiple of one coordinate to another coordinate of each point, or, for example, $shear(x,y,z) = (x+3y,y,z)$ The matrix for this shear transformation looks like $\begin{bmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. In general, the matrix for a shear transformation will look like the identity matrix with one non-zero element A off the diagonal. If A is in row $i$, column $j$, then the matrix will add A times the $j^{th}$ coordinate of the vector to the $i^{th}$ coordinate. For the oblique view of engineering drawings, we look at the shear matrices that add a certain amount of the z-coordinate to each of the x- and y-coordinates. The matrices are $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ A & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & B & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ A & B & 1 \end{bmatrix}$ that take $(x,y,z) \times (x+Az,y+Bz,z)$. The values of A and B are adjusted to give precisely the view that you want, and the z-term of the result is usually dropped to give the needed 2D view of the 3D object. An example is the classical cabinet view shown below: To experiment with these transformations, we have two interactive applets. The first one lets you apply the 2D transformations to a 2D figure. Transformation Matrix The second applet lets you apply the 3D transformations to a 3D figure. (Currently Unavailable) ### Matrix Inverses In general, getting the inverse of a matrix can be difficult, but for the basic graphics transformations the inverses are easy because we can simply undo the geometric action of the original transformation. The inverse of the scaling matrix $\begin{bmatrix} \sigma_x & 0 & 0 \\ 0 & \sigma_y & 0 \\ 0 & 0 & \sigma_z \end{bmatrix}$ is clearly $\begin{bmatrix} \frac{1}{\sigma_x } & 0 & 0 \\ 0 & \frac{1}{\sigma_y } & 0 \\ 0 & 0 & \frac{1}{\sigma_z } \end{bmatrix}$ The inverse of a rotation transformation by angle $\theta$ is clearly the rotation around the same line by the angle $-\theta$. The inverse of the translation matrix $\begin{bmatrix} 1 & 0 & 0 & T_x \\ 0 & 1 & 0 & T_y \\ 0 & 0 & 1 & T_z \\ 0 & 0 & 0 & 1 \end{bmatrix}$ is clearly $\begin{bmatrix} 1 & 0 & 0 & -T_x \\ 0 & 1 & 0 & -T_y \\ 0 & 0 & 1 & -T_z \\ 0 & 0 & 0 & 1 \end{bmatrix}$ The inverse of the simple shear transformation is also straightforward. Since a simple shear adds a multiple of one vector component to another component, the inverse only needs to subtract that multiple. So we have $\begin{bmatrix} 1 & A \\ 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -A \\ 0 & 1 \end{bmatrix}$ and the 3D case is a simple extension of this. So we have a major observation: If any transformation is the product of basic graphics transformations, it is easy to find the inverse of its matrix (and hence its inverse transformation) as the product of the inverses of the components in reverse order. Or: $(A \times B \times C)^{-1} = C^{-1} \times B^{-1} \times A^{-1}$ ## Transformations and Graphics Environments Attention – this concept needs a bit of programming background; it involves stacks. When you are defining the geometry for a graphics image, you will sometimes want to model your scene as a hierarchy of simpler objects. You might have a desk, for example, that is made up of several parts (legs, drawers, shelves); the drawers may have handles or other parts; you may want to put several things on top of the desk; and so on. It’s common to define general models for each simple part, and then to put the pieces together in a common space, called the “world space.” Each simple part will be defined in its own “model space,” and then you can apply transformations that move all the parts into the right place in the more complex part. In turn, that whole more complex part may be moved into another position, and so on – you can build up quite complex models this way. One common technique for this kind of hierarchical modeling is to build a “scene graph” that shows how everything is assembled and the transformations that are used in the assembly. As an example, consider the simple picture of a bunny head, basically made up of several spheres. Each sphere is scaled (making it an ellipsoid of the right size), rotated into the right orientation, and then translated into the proper place. The tree next to the picture shows how this is organized. In order to make this work, you have to apply each set of transformations to its own sphere and then “forget” those transformations so you can apply the transformations for the next piece. You could, of course, use inverses to undo the transformations, but that’s slow and invites roundoff errors from many multiplications. instead, it is common to maintain a “transformation stack” that holds the history of every place you want to get back to – all the transformations you have saved. This is a stack of 4x4 matrices that implement the transformations. You also have an active transformation to which you apply any new transformations by matrix multiplication. To save a transformation to get back to later, you push a copy of the current active transformation (as a 4x4 matrix) onto the stack. Later, when you have applied whatever new transformations you need and want to get back to the last saved transformation, you pop the top matrix off the stack and make it the current active transformation. Presto – all the transformations you had used since the corresponding push operation are gone. So let’s get back to the rabbit. We want to create the rabbit head, and we have whatever active transformation was in place when we wanted to draw the head. Then we have push scale sphere for main part pop translate scale sphere for left eye pop Translate Scale sphere for right eye pop Translate Rotate Scale sphere for left ear pop Translate Rotate Scale sphere for right ear pop Notice something important: the transformations are written in the order they are applied, with the one closest to the geometry to be applied first. The right ear operations are really Translate(Rotate(Scale(sphere-for-right-ear))) If you are not familiar with stacks, this won’t make much sense, but you don't need to understand this to understand basic transformation concepts. A simple way to look at these stacks is to notice that a transformation is a 4x4 matrix or, equivalently, a 16-element array, so maintaining a stack is simply a matter of building an array float transStack[N][16]; or float transStack[N][4][4]; where N is the number of transformations one wants to save. # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page. Categories: \| \| \| \|	{"extraction_info": {"found_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": 60, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8942251801490784, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/52308/a-hot-object-exposed-to-low-temperature-in-a-vacuum-doesnt-lose-heat/52313	A hot object exposed to low temperature in a vacuum doesn't lose heat? I heard somewhere that if the human body were exposed to outer space where the temperature is extremely low, the human won't actually feel cold because in a vacuum, the heat energy doesn't have another physical object to travel too. So my question is, i thought heat is just a form of radiation, and that radiation can travel through a vacuum. So then, why doesn't a body lose heat in a vacuum the same way our sun emits a whole bunch of UV radiation? In which case, the body should be losing energy and feeling cold. - 1 – Qmechanic♦ Jan 27 at 14:20 Correct - This is what 'vacuum flasks/ thermoses' try to imitate. – user12345 Jan 27 at 14:24 Well, "heat" is not radiation. But a body tranfers "heat" as radiation. – user774025 Jan 27 at 14:50 2 Answers There are many ways to carry heat. The first is conduction, which is about the "vibration" of atoms on one material passing to another by simple physical contact. (Example: you touch something hot and get hurt). The second is convection: hot molecules simply move from one place to another (Example, water starts to boil in the bottom of a pan, but moves on the top because is lighter). The third is radiation and is precisely what you say: a warm body emits electromagnetic radiation. At "normal" temperatures (an oven, a human body), it's Infrared radiation, but it can be of higher frequency at higher temperatures, according to Planck's blackbody radiation law. Notice, though, that the power emitted by radiation only is proportional to the fourth power of temperature. So the effect is very relevant in sun, but negligible for a human body. It should be around 500 W/m$^2$, which OK, is not small, but probably the most heat is transferred by conduction when the human is in air. This is how garments work: they create a small layer of warm air around your skin, avoiding contact with constantly renewed cold air. - Given that people are normally assumed to have a skin area of 2m^2 and at-rest produce around 100W I think a 1000W cooling rate would be considerable! – Martin Beckett Jan 28 at 2:38 Yes, actually I am wondering if it is not a bit too much. I just used Stefan-Boltzmann's law. However, a human body is not a black body... – Bzazz Jan 28 at 10:25 500W is correct for 37C and e=0.5. My guess is that unless you were in very cold water the radiation losses in space are higher than convection losses on Earth – Martin Beckett Jan 28 at 17:29 But I thought radiation wouldnt change between earth or space... or you mean that in the meanwhile you absorb radiation from the earth and air surrounding you? (It seems right, you can feel very well the temperature of a fireplace withoug getting close to it) In this case, we should know the difference between the absorption and emission coefficient of our body. – Bzazz Jan 28 at 19:07 1 Radiation less is normally approximated as temperature_diff^4. In fact you lose exactly the same rate, but also absorb the energy from the surroundings which is also emitting as t^4 - so on earth you don't notice. – Martin Beckett Jan 28 at 23:25 The body will lose heat when it's in vacuum in the same way the sum emits thermal radiation. In fact all matter with temperature above absolute zero emits radiation and human body is no exception.The power emitted by radiation per unit surface is given by the Stefan-Boltzmann law $$j= \epsilon \sigma T^4$$ where, $j$ that the total energy radiated per unit surface area per unit time, $\sigma$ is the Stefan-Boltzman constant, $\epsilon$ is the emissivity of the body and $T$ is the absolute temperatue. Also, "heat" is not radiation. Heat or thermal energy is the energy which matter posseses because of the random motion of its molecules. And temperature of a macroscopic body is the measure of the average kinetic energy of its molecules. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9461233019828796, "perplexity_flag": "middle"}
http://mathhelpforum.com/number-theory/159639-multiplicative-function-questions.html	# Thread: 1. ## Multiplicative Function Questions 1) Show that if p is prime, $2^ap+1$ is composite for a = 1,2,....,r and p is not a Fermat prime, where r is a positive integer, then $\phi(n) = 2^rp$ has no solution. 2) The arithmetic funtion g is said to be the inverse of the arithmetic function f if f * g = g * f = i. Show that the arithmetic function f has an inverse if and only if f(1) does not equal 0. Show that if f has an inverse it is unique. (Hint: When f(1) is not equal to 0, find the inverse $f^{-1}$ of f by calculating $f^{-1}(n)$ recursively, using the fact that i(n) = summation $(f(d)f^{-1}(n/d)$.) 2. 1) seems like it shouldn't be too difficult. I don't get what I'm trying to prove though. How would the value of $2^rp$ be able to show that it is not solvable? I know $\phi(n)$ is always even when $n \geq 2$. But $2^rp$ is always going to be even, so I'm confused as to how I prove this. 2) I don't even understand how I recursively go back and do this. f*f{-1} = i, and set i = summation....	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9534348845481873, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2007/09/13/free-enriched-categories/?like=1&source=post_flair&_wpnonce=c1c9c39881	# The Unapologetic Mathematician ## Free Enriched Categories Now we’re going to assume that our monoidal category $\mathcal{V}$ is also cocomplete. In particular, we’ll assume that it has all small coproducts. This is enough to ensure that the “underlying set” functor $\hom_{\mathcal{V}_0}(\mathbf{1},\underline{\hphantom{X}})$ has a left adjoint $(\underline{\hphantom{X}})\cdot\mathbf{1}$ that sends a set $E$ to the coproduct of a bunch of copies of $\mathbf{1}$ indexed by $E$. The adjunction says that $\hom_{\mathcal{V}_0}(\coprod\limits_E\mathbf{1},X)\cong\prod_E\hom_{\mathcal{V}_0}(\mathbf{1},X)$ is naturally isomorphic to $\hom_\mathbf{Set}(E,\hom_{\mathcal{V}_0}(\mathbf{1},X))$. That is, a function from $E$ to the underlying set of $X$ is the same as an $E$-indexed collection of elements of the underlying set of $X$. It’s straightforward from here to verify that this adjunction interchanges the cartesian product on $\mathbf{Set}$ and the monoidal structure on $\mathcal{V}$. That is, $(E\times F)\cdot\mathbf{1}\cong(E\cdot\mathbf{1})\otimes(F\cdot\mathbf{1})$ and ${*}\cdot\mathbf{1}\cong\mathbf{1}$. And now the 2-functor $(\underline{\hphantom{X}})_0:\mathcal{V}\mathbf{-Cat}\rightarrow\mathbf{Cat}$ has a left 2-adjoint $(\underline{\hphantom{X}})_\mathcal{V}$. Starting with an ordinary category $\mathcal{C}$ (with hom-sets) we get the “free $\mathcal{V}$-category” $\mathcal{C}_\mathcal{V}$ with the same objects as $\mathcal{C}$, and with the hom-objects given by $\hom_{\mathcal{C}_\mathcal{V}}(C,C')=\hom_\mathcal{C}(C,C')\cdot\mathbf{1}$. Compositions and identities for this $\mathcal{V}$-category are induced by the above exchange of cartesian and monoidal structures. The actions of this 2-functor on functors and natural transformations are straightforwardly defined. For example, when $\mathcal{V}=\mathbf{Ab}$, we just replace each hom-set $\hom_\mathcal{C}(C,C')$ by the free abelian group on $\hom_\mathcal{C}(C,C')$. We extend the composition and identity maps from the ordinary category $\mathcal{C}$ by linearity. In particular, if $\mathcal{C}$ has only one object — if it’s a monoid $M$ — then $\mathcal{C}_\mathbf{Ab}$ is the free ring $\mathbb{Z}[M]$ on $M$. To finish off, let $\mathcal{C}$ be a small category, so $\mathcal{C}_\mathcal{V}$ is a small $\mathcal{V}$-category. Then we can define functor categories and functor $\mathcal{V}$-categories. Verify that $\left(\mathcal{B}^{\mathcal{C}_\mathcal{V}}\right)_0\cong\left(\mathcal{B}_0\right)^\mathcal{C}$ by the above 2-adjunction. ### Like this: Posted by John Armstrong \| Category theory No comments yet. « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for 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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 38, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8967811465263367, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/62476/list	## Return to Answer 2 minor edit I think your argument is essentially correct. Here is a proof for the algebraic closure of a finite field. It is enough to deal with the units of the ring of truncated Witt vectors $W_n(k)$ for all $n$. But then this is an algebraic group over a finite field and a theorem of Lang (Amer J Math 1956) states that $x \mapsto \sigma(x)x^{-1}$ is surjective for any algebraic group over such a field. I think from the algebraic closure of a finite field, the result follows for any algebraically closed field of positive characteristic. It's not going to hold for any finite field, as you'll get the elements of norm one only. For $\mathbb{F}_p$ you don't get the identity but the function identically equal to $1$. 1 I think your argument is essentially correct. Here is a proof for the algebraic closure of a finite field. It is enough to deal with the units of the ring of truncated Witt vectors $W_n(k)$ for all $n$. But then this is an algebraic group over a finite field and a theorem of Lang (Amer J Math 1956) states that $x \mapsto \sigma(x)x^{-1}$ is surjective for any algebraic group over such a field. I think from the algebraic closure of a finite field, the result follows for any algebraically closed field of positive characteristic. It's not going for any finite field, as you'll get the elements of norm one. For $\mathbb{F}_p$ you don't get the identity but the function identically equal to $1$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9499821662902832, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/140577/double-sum-involving-binomial-coefficients-relating-to-number-of-record-lows/140638	# Double sum involving binomial coefficients, relating to number of record lows Is there any way to simplify the following expression? $$\sum_{d = 1}^k \left(\sum_{i=1}^d \frac{1}{i}\right) \frac{{n-t \choose d}{t \choose k-d}}{{n \choose k}}$$ This formula comes from the expected number of record lows over the first $k$ elements in a permutation of $[1,n]$, given some minimum threshold $t$ below which the elements don't count as a record low. Let $L$ be the number of record lows, and let $d$ be the number of elements above the threshold. $$E[L] = E[E[L\|d]] = \sum_{d=1}^k E[L\|d]\cdot P(d)$$ where $P(d)$ is a hypergeometric distribution, with n-t success states, population size $n$, and $k$ draws. This comes up for example in this question: Expected number of cards in the stack? - 1 Shouldn't it be $t$ at the top right instead of $n$? – joriki May 3 '12 at 20:46 – Phira May 3 '12 at 20:48 @joriki thanks for catching my typo – Joe May 3 '12 at 21:42 ## 2 Answers Forget about my other answer; this is actually much simpler than that. Think of the elements below the threshold as above a threshold above all other elements instead. The crucial thing to notice is that the events "the $j$-th element is the lowest of the first $j$ elements" and "at least one of the first $j$ elements is below the threshold" are independent. The event "the $j$-th element is a record low" is simply the intersection of these events, so its probability is their product, $1/j$ times $1-{s\choose j}/{n\choose j}$, and the expected number of record lows is just the sum of these probabilities: $$\sum_{j=1}^k\frac1j\left(1-\frac{s\choose j}{n\choose j}\right)\;.$$ Simplifying Wolfram\|Alpha's take on this leads to the "closed form" $$H_k+\psi(s-n)-\psi(-n)+\frac{\binom s{k+1}\,_3F_2(1,k+1,k-s+1;k+2,k-n+1;1)}{(k+1)\binom n{k+1}}\;,$$ where $H_k$ is the $k$-th harmonic number, $\psi$ is the digamma function and $F$ is the generalized hypergeometric function. - – joriki May 4 '12 at 16:23 The $jth$ element is a record low if $it$ is below the threshold, not if at least one of the first $j$ elements is below the threshold. – Joe May 4 '12 at 22:45 @Joe: What I said was that it's a record low if it's the lowest of the first $j$ elements and at least one of the first $j$ elements is below the threshold. I don't see how what you wrote contradicts that. If an element is the lowest of the first $j$ elements and at least one of them is below the threshold, then necessarily that element is below the threshold. – joriki May 5 '12 at 11:09 I'm not sure whether this is the sort of simplification you're looking for, but you can determine the same expected number directly. The $j$-th element has probability $$\frac1n\binom{n-1}{j-1}^{-1}\sum_{m=1}^d\binom{n-m}{j-1}$$ of being a record low, so the expected number is $$\frac1n\sum_{j=1}^k\binom{n-1}{j-1}^{-1}\sum_{m=1}^d\binom{n-m}{j-1}\;.$$ Wolfram\|Alpha gives a closed form for the inner sum: $$\sum_{m=1}^d\binom{n-m}{j-1}=\frac1j\left(n\binom{n-1}{j-1}-(n-d)\binom{n-d-1}{j-1}\right)\;.$$ The first term gives the usual expected number of record lows for the case without a threshold, and the second term gives a correction: $$\sum_{j=1}^k\frac1j-\left(1-\frac dn\right)\sum_{j=1}^k\frac{\binom{n-d-1}{j-1}}{j\binom{n-1}{j-1}}\;.$$ Note that the cases $d=0$ and $d=n$ come out right. Wolfram\|Alpha also gives a "closed form" for the correction term, but it seems unlikely to help. P.S.: I just noticed that the result can actually be considerably simplified. With the number $s=n-d$ of values below the threshold, we have $$\sum_{j=1}^k\frac1j-\frac sn\sum_{j=1}^k\frac1j\frac{\binom{s-1}{j-1}}{\binom{n-1}{j-1}}=\sum_{j=1}^k\frac1j-\sum_{j=1}^k\frac1j\frac{\binom sj}{\binom nj}\;.$$ That should make us wonder whether there isn't a simpler way to derive that result. - This definitely helps. I don't fully follow the record low probability. ${n -1 \choose j -1}$ is the total number of choices for the first j -1 elements. ${n - m \choose j - 1}$ is the number of choices for the first $j - 1$ elements out of the $n - m$ elements that are below the threshold? – Joe May 3 '12 at 21:53 1 @Joe: $m$ runs over the possible above-threshold values of the $j$-th element, and ${n - m \choose j - 1}$ is the number of choices for the first $j-1$ elements that would make that value a record low. – joriki May 3 '12 at 21:56 That makes sense. So we have number of choices that would make element $j$ a record low divided by total number of choices. Then where does the $\frac{1}{n}$ come from? – Joe May 3 '12 at 22:08 @Joe: The probability of the $j$-th element to have that particular value. – joriki May 3 '12 at 22:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9312487244606018, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/226706/topological-dual-of-x-sigmax-phi	# Topological dual of $(X,\sigma(X,\Phi))$ Let $X$ be a normed vector space and $\Phi$ a vector subspace of $X^$. I would like to prove that if a linear functional $\phi$ on $X$ is continuous for the topology $\sigma(X,\Phi)$ (that is the initial topology for $\Phi$) then $\phi \in \Phi$. If $\phi$ is such a linear functional then for all $\epsilon>0$ there exists $\zeta_i$ and $\epsilon_i$ such that $\phi^{-1}(]-\epsilon,\epsilon[) \supset \bigcap\limits_{i=1}^n \zeta_i^{-1}(]-\epsilon_i,\epsilon_i[)$. So I hope that the inclusion, for some $\epsilon>0$, implies that $\phi$ is a linear combination of the $\zeta_i$'s. Is it true? If so how to prove it? - ## 1 Answer Let $\varepsilon=1$: we can find an integer $N$ and $f_i\in X^$, $\varepsilon_i>0$ such that $$\bigcap_{i=1}^Nf_i^{-1}((-\varepsilon_i,\varepsilon_i))\subset \phi^{-1}((-1,1)).$$ Let $x\in\bigcap_{j=1}^N\ker f_i$. Then for all $\alpha\in\Bbb R$, $\alpha x\in\bigcap_{j=1}^N\ker f_i$. This gives $\|\phi(\alpha x)\|<1$ and $\|\phi(x)\|\leq \|\alpha\|^{-1}$ for all $\alpha$. So $\phi(x)=0$. We have proved that $\bigcap_{j=1}^N\ker f_i\subset\ker \phi$. We conclude by this result. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9441083669662476, "perplexity_flag": "head"}
http://mathoverflow.net/questions/25977/how-to-understand-the-concept-of-compact-space/25986	## How to understand the concept of compact space [closed] ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) the definition of compact space is: A subset K of a metric space X is said to be compact if every open cover of K contains finite subcovers. What is the meaning of defining a space is "compact". I found the explanation on wikipedia : "In mathematics, more specifically general topology and metric topology, a compact space is an abstract mathematical space in which, intuitively, whenever one takes an infinite number of "steps" in the space, eventually one must get arbitrarily close to some other point of the space. " I can not understand it, or at least I can not get this by its definition. Can anybody help me? - 1 Well suppose you have compact metric space $X$, consider for any $\varepsilon>0$ the covering $\{B_\varepsilon(x)\|x\in X\}$. It has a finite subcovering. Now take any infinite sequence $(x_i)_{i \in \mathbb{N}}$. Each $x_i$ is contained in one open set of this finite subcovering. Hence (for fixed $\varepsilon>0$) there must be a point $x$ such that infinitely many $x_i$ are contained in $B_\varepsilon(x)$. – HenrikRüping May 26 2010 at 8:51 See a standard textbook. There are several different definitions for compactness. The definition wikipedia is refering to is the following. A metric space is compact if any sequence admits a converging subsequence. The equivalence with the covering definition is not obvious. Have a look at the Heine-Borel theorem in wikipedia. – coudy May 26 2010 at 8:54 The intuition of the idea comes from metric space theory: suppose you took an 'infinite number of steps' to get from point A to point B, then each step could be covered by a ball around the centre of the step. The worst you could have done to take an infinite number of steps would have been to 'traverse the whole space' (so your balls cover the space), but then compactness would guarantee that a finite number of them would have done the job anyway. Very roughly. Really what you're after is the equivalence of sequential and covering compactness. See: any metric space textbook ever. – Tom Boardman May 26 2010 at 9:00 Thanks very much! Your comments do make sense to me:) – jkjium May 26 2010 at 9:13 How the standard definition of compactness came about is a bit mysterious to me as well, although perhaps less so than the definition of a topology itself: mathoverflow.net/questions/19152/… – Minhyong Kim May 26 2010 at 23:00 show 1 more comment ## 6 Answers Some heuristic remarks are helpful only to a subset of readers. (Maybe that's true of all heuristics, as a meta-heuristic - if everyone accepts a rough explanation, it's something rather more than that.) Non-compactness is about being able to "move off to infinity" in some way in a space. On the real line you can do that to the left, or right: but bend the line round to fill all but one point on a circle (which is compact) and you see the difference having the "other point" near which you end up. This example of real line versus circle is too simple, really. Another way you can "go off to infinity" in a space is by having paths branching out infinitely (as in König's lemma, which supplies another kind of intuition). Compactness is a major topological concept because the various ways you might try to "trap" movement within a space to prevent "escape" to infinity can be summed up in a single idea (for metric spaces, let's say). The definition by open sets is cleaner, but the definition by sequences having to accumulate on themselves (not necessarily to converge, but to have at least one convergent subsequence) is somewhat quicker to say. If you restrict attention to spaces that are manifolds, you can think of continuous paths and whether they have to wind back close to themselves or not. - Your answer gives me a very intuitive view to compactness! Thanks very much! – jkjium May 26 2010 at 10:07 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Terry Tao has a nice explanation in the Princeton companion to maths. The article's also on his website: http://www.math.ucla.edu/~tao/preprints/compactness.pdf - +1. I have been searching, for a while, for a good motivation of compactness. I read the article by Terry Tao and it is really illuminating. Thanks for the link! – user11000 Aug 16 2011 at 17:03 A very nice exposition was given by Hewitt in this article in the Monthly; the central thesis is that compactness facilitates proving results that are nigh on trivial for finite sets in wider settings. - Yep. I've heard people informally describe a compact set as one that is "almost finite". – Nate Eldredge May 26 2010 at 15:15 Etymology may help. Compact, from the Latin Compactus, past participle of the verb Compingere: "to pack together closely and firmly". You may have an idea of how strong is the hypothesis of compactness if you look at what happens when even total boundedness is lacking. Just consider e.g. the possibly most familiar infinite dimensional object, the separable Hilbert space H=l2. Its unit closed ball is not compact. A continuous real valued function on it may be unbounded, or bounded without minimum and maximum value. A linear form on H may be everywhere discontinuous and locally unbounded. There is a continuous transformation of the unit closed ball with no fixed points. The ball itself retracts on its boundary. Infinitely many disjoint unit open balls may be packed within a ball of radius $1+ \sqrt{2}$. The linear group GL(H) is connected. There are injective linear continuous transformations of H that are not surjective.... - In elementary analysis one learns that every continuous function from a closed bounded interval to $\mathbb{R}$ is bounded, but this is not the case for open or unbounded intervals. A little later one learns that each continuous function from a closed bounded subset of $\mathbb{R}^n$ to $\mathbb{R}$ is bounded, but this fails for other subsets of $\mathbb{R}^n$. The key properties are that a subset of $\mathbb{R}^n$ is compact iff it is closed and bounded (Heine-Borel theorem) and that the image of a compact space under a continuous map is compact. Compactness is a sufficient condition on a space to ensure that all continuous functions to $\mathbb{R}$; moreover compactness is a purely topological property, definable in terms of open sets. The wikipedia quotation is a bit vague, but it refers to a property called sequential compactness, which all compact metric spaces have. This means that for any sequence of points in the space $X$ there is a subsequence converging to a point of $X$. A metric space is compact if and only if it is sequentially compact, but this does not hold for non-metrizable topological spaces. I don't like the wikipedia quote, as it sort of suggests that sequences in compact spaces must be convergent; this is not so, though they must have convergent subsequences. - In some way, you can ALWAYS think about compactness as "sequential compactness" provided you allow for things a bit more general than sequences- nets. Compactness is equivalent to the statement that every net as a convergent subnet. If a space is metrizable, this is equivalent to the analogous statement, where you only need to consider sequences. I prefer, in fact, to think in terms of ultrafilters and say that a space is compact if and only if every ultrafilter has a limit point. I suggest reading up about nets and ultrafilters. I found http://www.math.ksu.edu/~nagy/real-an/1-02-convergence.pdf a nice quick introduction to give you the idea (but it does not classify compactness). A more comprehensive reference might be http://math.uga.edu/~pete/convergence.pdf (but note here compact means compact Hausdorff, while quasi-compact means compact and possibly not Hausdorff). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9474722146987915, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Specific_entropy	# Entropy (Redirected from Specific entropy) It has been suggested that be merged into this article. (Discuss) Proposed since February 2013. For a generally accessible and less technical introduction to the topic, see Introduction to entropy. Thermodynamics The classical Carnot heat engine Branches State Processes Cycles Functions of state (Conjugate variables in italics) Temperature / Entropy Introduction to entropy Specific heat capacity $c=$ $T$ $\partial S$ $N$ $\partial T$ Compressibility $\beta=-$ $1$ $\partial V$ $V$ $\partial p$ Thermal expansion $\alpha=$ $1$ $\partial V$ $V$ $\partial T$ • Internal energy $U(S,V)$ • Enthalpy $H(S,p)=U+pV$ $A(T,V)=U-TS$ $G(T,p)=H-TS$ History / Culture Philosophy History Theories Key publications Timelines Art Education Scientists Entropy articles Introduction History Classical Statistical Entropy in statistical mechanics is a measure of the number of specific ways in which a system may be arranged, often taken to be a measure of "disorder"; the higher the entropy, the higher the disorder. The entropy of an isolated system never decreases, because isolated systems spontaneously evolve towards thermodynamic equilibrium – the state of maximum entropy. (see Second law of thermodynamics) Entropy is a mathematically-defined thermodynamic quantity that helps to account for the flow of energy through a thermodynamic process. Entropy was originally defined for a thermodynamically reversible process as $\Delta S = \int \frac {dQ_{rev}}T$ where the entropy (S) is found from the uniform temperature (T) of a closed system divided into an incremental reversible transfer of heat energy into that system (dQ). The above definition is sometimes called the macroscopic definition of entropy because it can be used without regard to any microscopic picture of the contents of a system. In thermodynamics, entropy has been found to be more generally useful and it has several other reformulations. Entropy was discovered when it was noticed via mathematics to be a quantity that behaves as a function of state. Entropy is an extensive property, but it is often given as an intensive property of specific entropy as entropy per unit mass or entropy per mole. In statistical mechanics, entropy is often related to the notions of order and disorder. In the modern microscopic interpretation of entropy in statistical mechanics, entropy is the amount of additional information needed to specify the exact physical state of a system, given its thermodynamic specification. Various thermodynamic processes thus can be reduced to a description of how that information changes as the system evolves from its initial condition. It is often said that entropy is an expression of disorder or randomness, although those concepts lack clear definitions except in terms of entropy itself. The second law is now often seen as an expression of the fundamental postulate of statistical mechanics via the modern definition of entropy. ## History (1822 – 1888), originator of the concept of entropy Main article: History of entropy The analysis which led to the concept of entropy began with the work of French mathematician Lazare Carnot who in his 1803 paper Fundamental Principles of Equilibrium and Movement proposed that in any machine the accelerations and shocks of the moving parts represent losses of moment of activity. In other words, in any natural process there exists an inherent tendency towards the dissipation of useful energy. Building on this work, in 1824 Lazare's son Sadi Carnot published Reflections on the Motive Power of Fire which posited that in all heat-engines whenever "caloric", or what is now known as heat, falls through a temperature difference, work or motive power can be produced from the actions of the "fall of caloric" between a hot and cold body. He made the analogy with that of how water falls in a water wheel. This was an early insight into the second law of thermodynamics.[1] Carnot based his views of heat partially on the early 18th century "Newtonian hypothesis" that both heat and light were types of indestructible forms of matter, which are attracted and repelled by other matter, and partially on the contemporary views of Count Rumford who showed (1789) that heat could be created by friction as when cannon bores are machined.[2] Carnot reasoned that if the body of the working substance, such as a body of steam, is returned to its original state at the end of a complete engine cycle, that "no change occurs in the condition of the working body". The first law of thermodynamics, formalized based on the heat-friction experiments of James Joule in 1843, deals with the concept of energy, which is conserved in all processes; the first law, however, is unable to quantify the effects of friction and dissipation. In the 1850s and 1860s, German physicist Rudolf Clausius objected to the supposition that no change occurs in the working body, and gave this "change" a mathematical interpretation by questioning the nature of the inherent loss of usable heat when work is done, e.g. heat produced by friction.[3] Clausius described entropy as the transformation-content, i.e. dissipative energy use, of a thermodynamic system or working body of chemical species during a change of state.[3] This was in contrast to earlier views, based on the theories of Isaac Newton, that heat was an indestructible particle that had mass. Later, scientists such as Ludwig Boltzmann, Josiah Willard Gibbs, and James Clerk Maxwell gave entropy a statistical basis. In 1877 Boltzmann visualized a probabilistic way to measure the entropy of an ensemble of ideal gas particles, in which he defined entropy to be proportional to the logarithm of the number of microstates such a gas could occupy. Henceforth, the essential problem in statistical thermodynamics, i.e. according to Erwin Schrödinger, has been to determine the distribution of a given amount of energy E over N identical systems. Carathéodory linked entropy with a mathematical definition of irreversibility, in terms of trajectories and integrability. ## Definitions and descriptions Any method involving the notion of entropy, the very existence of which depends on the second law of thermodynamics, will doubtless seem to many far-fetched, and may repel beginners as obscure and difficult of comprehension. Willard Gibbs, Graphical Methods in the Thermodynamics of Fluids[4] There are two related definitions of entropy: the thermodynamic definition and the statistical mechanics definition. Historically, the classical thermodynamics definition developed first, and it has more recently been extended in the area of non-equilibrium thermodynamics. Entropy was defined from a classical thermodynamics viewpoint, in which the detailed quantum constituents are not directly considered, with their behavior only showing up in macroscopically averaged properties, e.g. heat capacity.. Later, thermodynamic entropy was more generally defined from a statistical thermodynamics viewpoint, in which the detailed quantum constituents (photons, phonons, spins, etc.) are explicitly considered. ### Function of state There are many thermodynamic properties that are functions of state. This means that at a particular state, these properties have a certain value. Often, if two properties have a particular value, then the state is determined and the other properties values are set. For instance, an ideal gas, at a particular temperature and pressure, has a particular volume according to the ideal gas equation. That entropy is a function of state is what makes it very useful. As another instance, a pure substance of single phase at a particular uniform temperature and pressure (and thus a particular state) is at not only a particular volume but also at a particular entropy.[5] In the Carnot cycle, the working fluid returns to the same state at particular part of the cycle, hence the loop integral equaling zero. ### Reversible process Entropy is defined for a reversible process and for a system that, at all times, can be treated as being at a uniform state and thus at a uniform temperature. Reversibility is an ideal that some real processes approximate and that is often presented in study exercises. For a reversible process, entropy behaves as a conserved quantity and no change occurs in total entropy. More specifically, total entropy is conserved in a reversible process and not conserved in an irreversible process.[6] One has to be careful about system boundaries. For example, in the Carnot cycle, while the heat flow from the hot reservoir to the cold reservoir represents an increase in entropy, the work output, if reversibly and perfectly stored in some energy storage mechanism, represents a decrease in entropy that could be used to operate the heat engine in reverse and return to the previous state, thus the total entropy change is still zero at all times if the entire process is reversible. Any process that does not meet the requirements of a reversible process must be treated as an irreversible process, which is usually a complex task. An irreversible process increases entropy.[7] Heat transfer situations require two or more non-isolated systems in thermal contact. In irreversible heat transfer, heat energy is irreversibly transferred from the higher temperature system to the lower temperature system, and the combined entropy of the systems increases. Each system, by definition, must have its own absolute temperature applicable within all areas in each respective system in order to calculate the entropy transfer. Thus, when a system at higher temperature TH transfers heat dQ to a system of lower temperature TC, the former loses entropy dQ/TH and the latter gains entropy dQ/TC. The combined entropy change is dQ/TC-dQ/TH which is positive, reflecting an increase in the combined entropy. When calculating entropy, the same requirement of having an absolute temperature for each system in thermal contact exchanging heat also applies to the entropy change of an isolated system having no thermal contact. ### Carnot cycle The concept of entropy arose from Rudolf Clausius's study of the Carnot cycle.[8] In a Carnot cycle, heat ($Q_1$) is absorbed from a 'hot' reservoir, isothermally at the higher temperature $T_1$, and given up isothermally to a 'cold' reservoir, $Q_2$, at a lower temperature, $T_2$. According to Carnot's principle, work can only be done when there is a temperature difference, and the work should be some function of the difference in temperature and the heat absorbed. Carnot did not distinguish between $Q_1$ and $Q_2$, since he was working under the incorrect hypothesis that caloric theory was valid, and hence heat was conserved (the incorrect assumption that Q1 and Q2 were equal) when, in fact, $Q_1 - Q_2 > 0$.[9] Through the efforts of Clausius and Kelvin, it is now known that the maximum work that can be done is the product of the Carnot efficiency and the heat absorbed at the hot reservoir: $W=\left(1-\frac{T_2}{T_1}\right)Q_1.$ In order to derive the Carnot efficiency, $1-\frac{T_2}{T_1}$, Kelvin had to evaluate the ratio of the work done to the heat absorbed in the isothermal expansion with the help of the Carnot-Clapeyron equation which contained an unknown function, known as the Carnot function. The fact that the Carnot function could be the temperature, measured from zero, was suggested by Joule in a letter to Kelvin, and this allowed Kelvin to establish his absolute temperature scale.[10] It is also known that the work is the difference in the heat absorbed at the hot reservoir and rejected at the cold one: $W=Q_1-Q_2.$ Since the latter is valid over the entire cycle, this gave Clausius the hint that at each stage of the cycle, work and heat would not be equal, but rather their difference would be a state function that would vanish upon completion of the cycle. The state function was called the internal energy and it became the first law of thermodynamics.[11] Now equating the two expressions gives $\frac{Q_1}{T_1}-\frac{Q_2}{T_2}=0 .$ If we allow $Q_2$ to incorporate the algebraic sign, this becomes a sum and implies that there is a function of state which is conserved over a complete cycle. Clausius called this state function entropy. One can see that entropy was discovered through mathematics rather than through laboratory results. It is a mathematical construct and has no easy physical analogy. This makes the concept somewhat obscure or abstract, akin to how the concept of energy arose. Then Clausius asked what would happen if there would be less work done than that predicted by Carnot's principle. The right-hand side of the first equation would be the upper bound of the work, which would now be converted into an inequality $W<\left(1-\frac{T_2}{T_1}\right)Q_1.$ When the second equation is used to express the work as a difference in heats, we get $Q_1-Q_2<\left(1-\frac{T_2}{T_1}\right)Q_1$ or $Q_2>\frac{T_2}{T_1}Q_1$ So more heat is given off to the cold reservoir than in the Carnot cycle. If we denote the entropies by $S_i=Q_i/T_i$ for the two states, then the above inequality can be written as a decrease in the entropy $S_1-S_2<0$ The wasted heat implies that irreversible processes must have prevented the cycle from carrying out maximum work. ### Classical thermodynamics Main article: Entropy (classical thermodynamics) The thermodynamic definition was developed in the early 1850s by Rudolf Clausius and essentially describes how to measure the entropy of an isolated system in thermodynamic equilibrium. Clausius created the term entropy in 1865 as an extensive thermodynamic variable was shown to be useful in characterizing the Carnot cycle. Heat transfer along the isotherm steps of the Carnot cycle was found to be proportional to the temperature of a system (known as its absolute temperature). This relationship was expressed in increments of entropy equal to the ratio of incremental heat transfer divided by temperature, which was found to vary in the thermodynamic cycle but eventually return to the same value at the end of every cycle. Thus it was found to be a function of state, specifically a thermodynamic state of the system. Clausius based the term entropy on the Greek εντροπία [entropía], a turning toward, from εν- [en-] (in) and τροπή [tropē] (turn, conversion).[12][note 1] While Clausius based his definition on a reversible process, there are also irreversible processes that change entropy. Following the second law of thermodynamics, entropy of an isolated system always increases. The difference between an isolated system and closed system is that heat may not flow to and from an isolated system, but heat flow to and from a closed system is possible. Nevertheless, for both closed and isolated systems, and indeed, also in open systems, irreversible thermodynamics processes may occur. According to the Clausius equality, for a reversible cyclic process: $\oint \frac{\delta Q}{T}=0$. This means the line integral $\int_L \frac{\delta Q}{T}$ is path independent. So we can define a state function S called entropy, which satisfies: $dS = \frac{\delta Q}{T} \!$ With this we can only obtain the difference of entropy by integrating the above formula. To obtain the absolute value, we need the Third Law of Thermodynamics, which states that S=0 at absolute zero for perfect crystals. From a macroscopic perspective, in classical thermodynamics the entropy is interpreted as a state function of a thermodynamic system: that is, a property depending only on the current state of the system, independent of how that state came to be achieved. In any process where the system gives up energy ΔE, and its entropy falls by ΔS, a quantity at least TR ΔS of that energy must be given up to the system's surroundings as unusable heat (TR is the temperature of the system's external surroundings). Otherwise the process will not go forward. In classical thermodynamics, the entropy of a system is defined only if it is in thermodynamic equilibrium. ### Statistical mechanics The statistical definition was developed by Ludwig Boltzmann in the 1870s by analyzing the statistical behavior of the microscopic components of the system. Boltzmann showed that this definition of entropy was equivalent to the thermodynamic entropy to within a constant number which has since been known as Boltzmann's constant. In summary, the thermodynamic definition of entropy provides the experimental definition of entropy, while the statistical definition of entropy extends the concept, providing an explanation and a deeper understanding of its nature. The interpretation of entropy in statistical mechanics is the measure of uncertainty, or mixedupness in the phrase of Gibbs, which remains about a system after its observable macroscopic properties, such as temperature, pressure and volume, have been taken into account. For a given set of macroscopic variables, the entropy measures the degree to which the probability of the system is spread out over different possible microstates. In contrast to the macrostate, which characterizes plainly observable average quantities, a microstate specifies all molecular details about the system including the position and velocity of every molecule. The more such states available to the system with appreciable probability, the greater the entropy. In statistical mechanics, entropy is a measure of the number of ways in which a system may be arranged, often taken to be a measure of "disorder" (the higher the entropy, the higher the disorder).[13][14][15] This definition describes the entropy as being proportional to the natural logarithm of the number of possible microscopic configurations of the individual atoms and molecules of the system (microstates) which could give rise to the observed macroscopic state (macrostate) of the system. The constant of proportionality is the Boltzmann constant. Specifically, entropy is a logarithmic measure of the number of states with significant probability of being occupied: $S = - k_{\mathrm{B}}\sum_i P_i \ln P_i \, ,$ where kB is the Boltzmann constant, equal to 1.38065×10−23 J K−1. The summation is over all the possible microstates of the system, and Pi is the probability that the system is in the ith microstate.[16] This definition assumes that the basis set of states has been picked so that there is no information on their relative phases. In a different basis set, the more general expression is $S = - k_{\mathrm{B}} Tr(\widehat{\rho} \ln(\widehat{\rho}))$ where $\widehat{\rho}$ is the density matrix. This density matrix formulation is not needed in cases of thermal equilibrium so long as the basis states are chosen to be energy eigenstates. For most practical purposes, this can be taken as the fundamental definition of entropy since all other formulas for S can be mathematically derived from it, but not vice versa. In what has been called the fundamental assumption of statistical thermodynamics or the fundamental postulate in statistical mechanics, the occupation of any microstate is assumed to be equally probable (i.e. Pi=1/Ω where Ω is the number of microstates); this assumption is usually justified for an isolated system in equilibrium.[17] Then the previous equation reduces to: $S = k_{\mathrm{B}} \ln \Omega \, ,$ In thermodynamics, such a system is one in which the volume, number of molecules, and internal energy are fixed (the microcanonical ensemble). The most general interpretation of entropy is as a measure of our uncertainty about a system. The equilibrium state of a system maximizes the entropy because we have lost all information about the initial conditions except for the conserved variables; maximizing the entropy maximizes our ignorance about the details of the system.[18] This uncertainty is not of the everyday subjective kind, but rather the uncertainty inherent to the experimental method and interpretative model. The interpretative model has a central role in determining entropy. The qualifier "for a given set of macroscopic variables" above has deep implications: if two observers use different sets of macroscopic variables, they will observe different entropies. For example, if observer A uses the variables U, V and W, and observer B uses U, V, W, X, then, by changing X, observer B can cause an effect that looks like a violation of the second law of thermodynamics to observer A. In other words: the set of macroscopic variables one chooses must include everything that may change in the experiment, otherwise one might see decreasing entropy![19] Entropy can be defined for any Markov processes with reversible dynamics and the detailed balance property. In Boltzmann's 1896 Lectures on Gas Theory, he showed that this expression gives a measure of entropy for systems of atoms and molecules in the gas phase, thus providing a measure for the entropy of classical thermodynamics. ### Entropy of a system A temperature–entropy diagram for steam. The vertical axis represents uniform temperature, and the horizontal axis represents specific entropy. Each dark line on the graph represents constant pressure, and these form a mesh with light gray lines of constant volume. (Dark-blue is liquid water, light-blue is boiling water, and faint-blue is steam. Grey-blue represents supercritical liquid water.) Entropy is the above-mentioned unexpected and, to some, obscure integral that arises directly from the Carnot cycle. It is reversible heat divided by temperature. It is, remarkably, a function of state and it is fundamental and very useful. In a thermodynamic system, pressure, density, and temperature tend to become uniform over time because this equilibrium state has higher probability (more possible combinations of microstates) than any other; see statistical mechanics. As an example, for a glass of ice water in air at room temperature, the difference in temperature between a warm room (the surroundings) and cold glass of ice and water (the system and not part of the room), begins to be equalized as portions of the thermal energy from the warm surroundings spread to the cooler system of ice and water. Over time the temperature of the glass and its contents and the temperature of the room become equal. The entropy of the room has decreased as some of its energy has been dispersed to the ice and water. However, as calculated in the example, the entropy of the system of ice and water has increased more than the entropy of the surrounding room has decreased. In an isolated system such as the room and ice water taken together, the dispersal of energy from warmer to cooler always results in a net increase in entropy. Thus, when the "universe" of the room and ice water system has reached a temperature equilibrium, the entropy change from the initial state is at a maximum. The entropy of the thermodynamic system is a measure of how far the equalization has progressed. Thermodynamic entropy is a non-conserved state function that is of great importance in the sciences of physics and chemistry.[13][20] Historically, the concept of entropy evolved in order to explain why some processes (permitted by conservation laws) occur spontaneously while their time reversals (also permitted by conservation laws) do not; systems tend to progress in the direction of increasing entropy.[21][22] For isolated systems, entropy never decreases.[20] This fact has several important consequences in science: first, it prohibits "perpetual motion" machines; and second, it implies the arrow of entropy has the same direction as the arrow of time. Increases in entropy correspond to irreversible changes in a system, because some energy is expended as waste heat, limiting the amount of work a system can do.[13][14][23][24] Unlike many other functions of state, entropy cannot be directly observed but must be calculated. Entropy can be calculated for a substance as the standard molar entropy from absolute zero (also known as absolute entropy) or as a difference in entropy from some other reference state which is defined as zero entropy. Entropy has the dimension of energy divided by temperature, which has a unit of joules per kelvin (J/K) in the International System of Units. While these are the same units as heat capacity, the two concepts are distinct.[25] Entropy is not a conserved quantity: for example, in an isolated system with non-uniform temperature, heat might irreversibly flow and the temperature become more uniform such that entropy increases. The second law of thermodynamics, states that a closed system has entropy which may increase or otherwise remain constant. Chemical reactions cause changes in entropy and entropy plays an important role in determining in which direction a chemical reaction spontaneously proceeds. One dictionary definition of entropy is that it is "a measure of thermal energy per unit temperature that is not available for useful work". For instance, a substance at uniform temperature is at maximum entropy and cannot drive a heat engine. A substance at non-uniform temperature is at a lower entropy (than if the heat distribution is allowed to even out) and some of the thermal energy can drive a heat engine. A special case of entropy increase, the entropy of mixing, occurs when two or more different substances are mixed. If the substances are at the same temperature and pressure, there will be no net exchange of heat or work – the entropy change will be entirely due to the mixing of the different substances. At a statistical mechanical level, this results due to the change in available volume per particle with mixing.[26] ## Second law of thermodynamics Main article: Second law of thermodynamics The second law of thermodynamics states that in general the total entropy of any system will not decrease other than by increasing the entropy of some other system. Hence, in a system isolated from its environment, the entropy of that system will tend not to decrease. It follows that heat will not flow from a colder body to a hotter body without the application of work (the imposition of order) to the colder body. Secondly, it is impossible for any device operating on a cycle to produce net work from a single temperature reservoir; the production of net work requires flow of heat from a hotter reservoir to a colder reservoir, or a single expanding reservoir undergoing adiabatic cooling, which performs adiabatic work. As a result, there is no possibility of a perpetual motion system. It follows that a reduction in the increase of entropy in a specified process, such as a chemical reaction, means that it is energetically more efficient. It follows from the second law of thermodynamics that the entropy of a system that is not isolated may decrease. An air conditioner, for example, may cool the air in a room, thus reducing the entropy of the air of that system. The heat expelled from the room (the system), which the air conditioner transports and discharges to the outside air, will always make a bigger contribution to the entropy of the environment than will the decrease of the entropy of the air of that system. Thus, the total of entropy of the room plus the entropy of the environment increases, in agreement with the second law of thermodynamics. In mechanics, the second law in conjunction with the fundamental thermodynamic relation places limits on a system's ability to do useful work.[27] The entropy change of a system at temperature T absorbing an infinitesimal amount of heat $\delta q$ in a reversible way, is given by $\frac{\delta q}{T}$. More explicitly, an energy TRS is not available to do useful work, where TR is the temperature of the coldest accessible reservoir or heat sink external to the system. For further discussion, see Exergy. Statistical mechanics demonstrates that entropy is governed by probability, thus allowing for a decrease in disorder even in an isolated system. Although this is possible, such an event has a small probability of occurring, making it unlikely.[28] ## Applications ### The fundamental thermodynamic relation Main article: Fundamental thermodynamic relation The entropy of a system depends on its internal energy and the external parameters, such as the volume. In the thermodynamic limit this fact leads to an equation relating the change in the internal energy to changes in the entropy and the external parameters. This relation is known as the fundamental thermodynamic relation. If the volume is the only external parameter, this relation is: $dU = T dS - P dV$ Since the internal energy is fixed when one specifies the entropy and the volume, this relation is valid even if the change from one state of thermal equilibrium to another with infinitesimally larger entropy and volume happens in a non-quasistatic way (so during this change the system may be very far out of thermal equilibrium and then the entropy, pressure and temperature may not exist). The fundamental thermodynamic relation implies many thermodynamic identities that are valid in general, independent of the microscopic details of the system. Important examples are the Maxwell relations and the relations between heat capacities. ### Entropy in chemical thermodynamics Thermodynamic entropy is central in chemical thermodynamics, enabling changes to be quantified and the outcome of reactions predicted. The second law of thermodynamics states that entropy in an isolated system – the combination of a subsystem under study and its surroundings – increases during all spontaneous chemical and physical processes. The Clausius equation of δqrev/T = ΔS introduces the measurement of entropy change, ΔS. Entropy change describes the direction and quantifies the magnitude of simple changes such as heat transfer between systems – always from hotter to cooler spontaneously. The thermodynamic entropy therefore has the dimension of energy divided by temperature, and the unit joule per kelvin (J/K) in the International System of Units (SI). Thermodynamic entropy is an extensive property, meaning that it scales with the size or extent of a system. In many processes it is useful to specify the entropy as an intensive property independent of the size, as a specific entropy characteristic of the type of system studied. Specific entropy may be expressed relative to a unit of mass, typically the kilogram (unit: Jkg−1K−1). Alternatively, in chemistry, it is also referred to one mole of substance, in which case it is called the molar entropy with a unit of Jmol−1K−1. Thus, when one mole of substance at about 0K is warmed by its surroundings to 298K, the sum of the incremental values of qrev/T constitute each element's or compound's standard molar entropy, an indicator of the amount of energy stored by a substance at 298K.[29][30] Entropy change also measures the mixing of substances as a summation of their relative quantities in the final mixture.[31] Entropy is equally essential in predicting the extent and direction of complex chemical reactions. For such applications, ΔS must be incorporated in an expression that includes both the system and its surroundings, ΔSuniverse = ΔSsurroundings + ΔS system. This expression becomes, via some steps, the Gibbs free energy equation for reactants and products in the system: ΔG [the Gibbs free energy change of the system] = ΔH [the enthalpy change] −T ΔS [the entropy change].[29] ### Entropy change formulas for an ideal gas When an ideal gas undergoes a change, its entropy may also change. For cases where the specific heat does not change and either volume, pressure or temperature is also constant, the change in entropy can be easily calculated.[32] In this section $C_v$ is the specific heat at constant volume, $C_p$ is the specific heat at constant pressure, $R$ is the ideal gas constant, and $n$ is the number of moles of gas. $T_0, V_0, P_0$ are the initial temperature, volume and pressure respectively for the change, and $T, V, P$ are the final temperature, volume and pressure. When specific heat and volume are constant, the change in entropy is given by: $\Delta S = n C_v \ln \frac{T}{T_0}$. When specific heat and pressure are constant, the change in entropy is given by: $\Delta S = n C_p \ln \frac{T}{T_0}$. The above equations also apply for incompressible materials such as some liquids and solids, where Cv and Cp are equal. Note that near absolute zero, heat capacities of solids quickly drop off to near zero, so the assumption of constant heat capacity does not apply.[33] When specific heat and temperature are constant, the change in entropy is given by: $\Delta S = n R \ln \frac{V}{V_0} = - n R \ln \frac{P}{P_0} .$ The above equations also apply for expansion into a finite vacuum or a throttling process, where the temperature, internal energy and enthalpy for an ideal gas remain constant. For some other transformations, not all of these properties (specific heat, volume, pressure or temperature) are constant. In these cases, for only 1 mole of an ideal gas, the change in entropy can be given by[34] either: $\Delta S = C_v \ln \frac{T}{T_0} + R \ln \frac{V}{V_0}$ or $\Delta S = C_p \ln \frac{T}{T_0} - R \ln \frac{P}{P_0}$. ### Entropy balance equation for open systems During steady-state continuous operation, an entropy balance applied to an open system accounts for system entropy changes related to heat flow and mass flow across the system boundary. In chemical engineering, the principles of thermodynamics are commonly applied to "open systems", i.e. those in which heat, work, and mass flow across the system boundary. In a system in which there are flows of both heat ($\dot{Q}$) and work, i.e. $\dot{W}_S$ (shaft work) and P(dV/dt) (pressure-volume work), across the system boundaries, the heat flow, but not the work flow, causes a change in the entropy of the system. This rate of entropy change is $\dot{Q}/T,$ where T is the absolute thermodynamic temperature of the system at the point of the heat flow. If, in addition, there are mass flows across the system boundaries, the total entropy of the system will also change due to this convected flow. To derive a generalized entropy balanced equation, we start with the general balance equation for the change in any extensive quantity Θ in a thermodynamic system, a quantity that may be either conserved, such as energy, or non-conserved, such as entropy. The basic generic balance expression states that dΘ/dt, i.e. the rate of change of Θ in the system, equals the rate at which Θ enters the system at the boundaries, minus the rate at which Θ leaves the system across the system boundaries, plus the rate at which Θ is generated within the system. Using this generic balance equation, with respect to the rate of change with time of the extensive quantity entropy S, the entropy balance equation for an open thermodynamic system is:[35] $\frac{dS}{dt} = \sum_{k=1}^K \dot{M}_k \hat{S}_k + \frac{\dot{Q}}{T} + \dot{S}_{gen}$ where $\sum_{k=1}^K \dot{M}_k \hat{S}_k$ = the net rate of entropy flow due to the flows of mass into and out of the system (where $\hat{S}$ = entropy per unit mass). $\frac{\dot{Q}}{T}$ = the rate of entropy flow due to the flow of heat across the system boundary. $\dot{S}_{gen}$ = the rate of entropy production within the system. Note, also, that if there are multiple heat flows, the term $\dot{Q}/T$ is to be replaced by $\sum \dot{Q}_j/T_j,$ where $\dot{Q}_j$ is the heat flow and $T_j$ is the temperature at the jth heat flow port into the system. ### Entropy and other forms of energy beyond work The fundamental equation of thermodynamics for a system containing n constituent species, with the i-th species having Ni particles is, with additional terms: $dU = T dS - P dV + \sum_i^n \mu_i dN_i + \Phi dQ + v dp$ U is internal energy, T is temperature, P is pressure, V is volume, $\mu_i$ and Ni are the chemical potential and number of molecules of the chemical, $\Phi$ and Q are electric potential and charge, v and p are velocity and momentum. Solving for the change in entropy we get: $dS = \frac{1}{T} dU + \frac{P}{T} dV - \sum_i^n \frac{\mu_i}{T} dN_i - \frac{\Phi}{T} dQ - \frac{v}{T} dp$ There is a minuscule change in internal energy for any change in entropy (ds will change by 1/T*dU). But in theory, the entropy of a system can be changed without changing its energy. That is done by keeping all variables constant, including temperature (isothermally) and entropy (adiabatically). That is easy to see, but typically the energy of the system will change. e.g. You can attempt to keep volume constant but you will always do work on the system, and work changes the energy.[36] Other potentials such as the gravitational potential can also be taken into account. ## Approaches to understanding entropy As a fundamental aspect of thermodynamics and physics, several different approaches to entropy beyond that of Clausius and Boltzmann are valid. ### Standard textbook definitions The following is a list of additional definitions of entropy from a collection of textbooks: • a measure of energy dispersal at a specific temperature. • a measure of disorder in the universe or of the availability of the energy in a system to do work.[37] • a measure of a system’s thermal energy per unit temperature that is unavailable for doing useful work.[38] In Boltzmann's definition, entropy is a measure of the number of possible microscopic states (or microstates) of a system in thermodynamic equilibrium. Consistent with the Boltzmann definition, the second law of thermodynamics needs to be re-worded as such that entropy decreases over time, though the underlying principle remains the same. ### Order and disorder Main article: Entropy (order and disorder) Entropy has often been loosely associated with the amount of order, disorder, and/or chaos in a thermodynamic system. The traditional qualitative description of entropy is that it refers to changes in the status quo of the system and is a measure of "molecular disorder" and the amount of wasted energy in a dynamical energy transformation from one state or form to another. In this direction, several recent authors have derived exact entropy formulas to account for and measure disorder and order in atomic and molecular assemblies.[39][40][41] One of the simpler entropy order/disorder formulas is that derived in 1984 by thermodynamic physicist Peter Landsberg, based on a combination of thermodynamics and information theory arguments. He argues that when constraints operate on a system, such that it is prevented from entering one or more of its possible or permitted states, as contrasted with its forbidden states, the measure of the total amount of “disorder” in the system is given by:[40][41] $\mbox{Disorder}={C_D\over C_I}.\,$ Similarly, the total amount of "order" in the system is given by: $\mbox{Order}=1-{C_O\over C_I}.\,$ In which CD is the "disorder" capacity of the system, which is the entropy of the parts contained in the permitted ensemble, CI is the "information" capacity of the system, an expression similar to Shannon's channel capacity, and CO is the "order" capacity of the system.[39] ### Energy dispersal Main article: Entropy (energy dispersal) The concept of entropy can be described qualitatively as a measure of energy dispersal at a specific temperature.[42] Similar terms have been in use from early in the history of classical thermodynamics, and with the development of statistical thermodynamics and quantum theory, entropy changes have been described in terms of the mixing or "spreading" of the total energy of each constituent of a system over its particular quantized energy levels. Ambiguities in the terms disorder and chaos, which usually have meanings directly opposed to equilibrium, contribute to widespread confusion and hamper comprehension of entropy for most students.[43] As the second law of thermodynamics shows, in an isolated system internal portions at different temperatures will tend to adjust to a single uniform temperature and thus produce equilibrium. A recently developed educational approach avoids ambiguous terms and describes such spreading out of energy as dispersal, which leads to loss of the differentials required for work even though the total energy remains constant in accordance with the first law of thermodynamics[44] (compare discussion in next section). Physical chemist Peter Atkins, for example, who previously wrote of dispersal leading to a disordered state, now writes that "spontaneous changes are always accompanied by a dispersal of energy".[45] ### Relating entropy to energy usefulness Following on from the above, it is possible (in a thermal context) to regard entropy as an indicator or measure of the effectiveness or usefulness of a particular quantity of energy.[46] This is because energy supplied at a high temperature (i.e. with low entropy) tends to be more useful than the same amount of energy available at room temperature. Mixing a hot parcel of a fluid with a cold one produces a parcel of intermediate temperature, in which the overall increase in entropy represents a “loss” which can never be replaced. Thus, the fact that the entropy of the universe is steadily increasing, means that its total energy is becoming less useful: eventually, this will lead to the "heat death of the Universe". ### Entropy and adiabatic accessibility A definition of entropy based entirely on the relation of adiabatic accessibility between equilibrium states was given by E.H.Lieb and J. Yngvason in 1999.[47] This approach has several predecessors, including the pioneering work of Constantin Carathéodory from 1909 [48] and the monograph by R. Giles from 1964.[49] In the setting of Lieb and Yngvason one starts by picking, for a unit amount of the substance under consideration, two reference states $X_0$ and $X_1$ such that the latter is adiabatically accessible from the former but not vice versa. Defining the entropies of the reference states to be 0 and 1 respectively the entropy of a state $X$ is defined as the largest number $\lambda$ such that $X$ is adiabatically accessible from a composite state consisting of an amount $\lambda$ in the state $X_1$ and a complementary amount, $(1-\lambda)$, in the state $X_0$. A simple but important result within this setting is that entropy is uniquely determined, apart from a choice of unit and an additive constant for each chemical element, by the following properties: It is monotonic with respect to the relation of adiabatic accessibility, additive on composite systems, and extensive under scaling. ### Entropy in quantum mechanics Main article: von Neumann entropy In quantum statistical mechanics, the concept of entropy was developed by John von Neumann and is generally referred to as "von Neumann entropy", $S = - k_\mathrm{B}\mathrm{Tr} ( \rho \log \rho ) \!$ where $\rho$ is the density matrix and Tr is the trace operator. This upholds the correspondence principle, because in the classical limit, when the phases between the basis states used for the classical probabilities are purely random, this expression is equivalent to the familiar classical definition of entropy, $S = - k_\mathrm{B}\sum_i p_i \, \log \, p_i$, i.e. in such a basis the density matrix is diagonal. Von Neumann established a rigorous mathematical framework for quantum mechanics with his work Mathematische Grundlagen der Quantenmechanik. He provided in this work a theory of measurement, where the usual notion of wave function collapse is described as an irreversible process (the so-called von Neumann or projective measurement). Using this concept, in conjunction with the density matrix he extended the classical concept of entropy into the quantum domain. ### Information theory I thought of calling it 'information', but the word was overly used, so I decided to call it 'uncertainty'. [...] Von Neumann told me, 'You should call it entropy, for two reasons. In the first place your uncertainty function has been used in statistical mechanics under that name, so it already has a name. In the second place, and more important, nobody knows what entropy really is, so in a debate you will always have the advantage.' Conversation between Claude Shannon and John von Neumann regarding what name to give to the attenuation in phone-line signals[50] When viewed in terms of information theory, the entropy state function is simply the amount of information (in the Shannon sense) that would be needed to specify the full microstate of the system. This is left unspecified by the macroscopic description. In information theory, entropy is the measure of the amount of information that is missing before reception and is sometimes referred to as Shannon entropy.[51] Shannon entropy is a broad and general concept which finds applications in information theory as well as thermodynamics. It was originally devised by Claude Shannon in 1948 to study the amount of information in a transmitted message. The definition of the information entropy is, however, quite general, and is expressed in terms of a discrete set of probabilities $p_i$: $H(X) = -\sum_{i=1}^n {p(x_i) \log p(x_i)}.$ In the case of transmitted messages, these probabilities were the probabilities that a particular message was actually transmitted, and the entropy of the message system was a measure of the average amount of information in a message. For the case of equal probabilities (i.e. each message is equally probable), the Shannon entropy (in bits) is just the number of yes/no questions needed to determine the content of the message.[16] The question of the link between information entropy and thermodynamic entropy is a debated topic. While most authors argue that there is a link between the two,[52][53][54] a few argue that they have nothing to do with each other.[16][55] The expressions for the two entropies are similar. The information entropy H for equal probabilities $p_i = p = 1/n$ is $H = k\, \log(1/p),$ where k is a constant which determines the units of entropy.[56] There are many ways of demonstrating the equivalence of "information entropy" and "physics entropy", that is, the equivalence of "Shannon entropy" and "Boltzmann entropy". Nevertheless, some authors argue for dropping the word entropy for the H function of information theory and using Shannon's other term "uncertainty" instead.[57] ## Interdisciplinary applications of entropy Although the concept of entropy was originally a thermodynamic construct, it has been adapted in other fields of study, including information theory, psychodynamics, thermoeconomics/ecological economics, and evolution.[39][58][59] ### Thermodynamic and statistical mechanics concepts • Entropy unit – a non-S.I. unit of thermodynamic entropy, usually denoted "e.u." and equal to one calorie per Kelvin per mole, or 4.184 Joules per Kelvin per mole.[60] • Gibbs entropy – the usual statistical mechanical entropy of a thermodynamic system. • Boltzmann entropy – a type of Gibbs entropy, which neglects internal statistical correlations in the overall particle distribution. • Tsallis entropy – a generalization of the standard Boltzmann-Gibbs entropy. • Standard molar entropy – is the entropy content of one mole of substance, under conditions of standard temperature and pressure. • Residual entropy – the entropy present after a substance is cooled arbitrarily close to absolute zero. • Entropy of mixing – the change in the entropy when two different chemical substances or components are mixed. • Loop entropy – is the entropy lost upon bringing together two residues of a polymer within a prescribed distance. • Conformational entropy – is the entropy associated with the physical arrangement of a polymer chain that assumes a compact or globular state in solution. • Entropic force – a microscopic force or reaction tendency related to system organization changes, molecular frictional considerations, and statistical variations. • Free entropy – an entropic thermodynamic potential analogous to the free energy. • Entropic explosion – an explosion in which the reactants undergo a large change in volume without releasing a large amount of heat. • Entropy change – a change in entropy dS between two equilibrium states is given by the heat transferred dQrev divided by the absolute temperature T of the system in this interval. • Sackur-Tetrode entropy – the entropy of a monatomic classical ideal gas determined via quantum considerations. ### The arrow of time Main article: Entropy (arrow of time) Entropy is the only quantity in the physical sciences that seems to imply a particular direction of progress, sometimes called an arrow of time. As time progresses, the second law of thermodynamics states that the entropy of an isolated system never decreases. Hence, from this perspective, entropy measurement is thought of as a kind of clock. ### Cosmology Main article: Heat death of the universe Since a finite universe is an isolated system, the Second Law of Thermodynamics states that its total entropy is constantly increasing. It has been speculated, since the 19th century, that the universe is fated to a heat death in which all the energy ends up as a homogeneous distribution of thermal energy, so that no more work can be extracted from any source. If the universe can be considered to have generally increasing entropy, then – as Sir Roger Penrose has pointed out – gravity plays an important role in the increase because gravity causes dispersed matter to accumulate into stars, which collapse eventually into black holes. The entropy of a black hole is proportional to the surface area of the black hole's event horizon.[61] Jacob Bekenstein and Stephen Hawking have shown that black holes have the maximum possible entropy of any object of equal size. This makes them likely end points of all entropy-increasing processes, if they are totally effective matter and energy traps. Hawking has, however, recently changed his stance on this aspect.[citation needed] The role of entropy in cosmology remains a controversial subject. Recent work has cast some doubt on the heat death hypothesis and the applicability of any simple thermodynamic model to the universe in general. Although entropy does increase in the model of an expanding universe, the maximum possible entropy rises much more rapidly, moving the universe further from the heat death with time, not closer. This results in an "entropy gap" pushing the system further away from the posited heat death equilibrium.[62] Other complicating factors, such as the energy density of the vacuum and macroscopic quantum effects, are difficult to reconcile with thermodynamical models, making any predictions of large-scale thermodynamics extremely difficult.[63] The entropy gap is widely believed to have been originally opened up by the early rapid exponential expansion of the universe. ## Notes 1. A machine in this context includes engineered devices as well as biological organisms. ## References 1. "Carnot, Sadi (1796–1832)". Wolfram Research. 2007. Retrieved 2010-02-24. 2. McCulloch, Richard, S. (1876). Treatise on the Mechanical Theory of Heat and its Applications to the Steam-Engine, etc. D. Van Nostrand. 3. ^ a b Clausius, Rudolf (1850). On the Motive Power of Heat, and on the Laws which can be deduced from it for the Theory of Heat. Poggendorff's Annalen der Physick, LXXIX (Dover Reprint). ISBN 0-486-59065-8. 4. The scientific papers of J. Willard Gibbs in Two Volumes 1. Longmans, Green, and Co. 1906. p. 11. Retrieved 2011-02-26. 5. B. H. Lavenda, "A New Perspective on Thermodynamics" Springer, 2009, Sec. 2.3.4, 6. S. Carnot, "Reflexions on the Motive Power of Fire", translated and annotated by R. Fox, Manchester University Press, 1986, p. 26; C. Truesdell, "The Tragicomical History of Thermodynamics, Springer, 1980, pp. 78–85 7. J. Clerk-Maxwell, "Theory of Heat", 10th ed. Longmans, Green and Co., 1891, pp. 155–158. 8. R. Clausius, "The Mechanical Theory of Heat", translated by T. Archer Hirst, van Voorst, 1867, p. 28 9. "Entropy". Online Etymology Dictionary. Retrieved 2008-08-05. 10. ^ a b c McGraw-Hill Concise Encyclopedia of Chemistry, 2004 11. ^ a b Sethna, J. Statistical Mechanics Oxford University Press 2006 p. 78 12. Barnes & Noble's Essential Dictionary of Science, 2004 13. ^ a b c 14. Schroeder, Daniel V. An Introduction toThermal Physics. Addison Wesley Longman, 1999, p. 57 15. "EntropyOrderParametersComplexity.pdf www.physics.cornell.edu" (PDF). Retrieved 2012-08-17. 16. ^ a b Sandler S. I., Chemical and Engineering Thermodynamics, 3rd Ed. Wiley, New York, 1999 p. 91 17. McQuarrie D. A., Simon J. D., Physical Chemistry: A Molecular Approach, University Science Books, Sausalito 1997 p. 817 18. Haynie, Donald, T. (2001). Biological Thermodynamics. Cambridge University Press. ISBN 0-521-79165-0. 19. Oxford Dictionary of Science, 2005 20. de Rosnay, Joel (1979). . Harper & Row, Publishers. ISBN 0-06-011029-5. 21. Ben-Naim, Arieh, On the So-Called Gibbs Paradox, and on the Real Paradox, Entropy, 9, pp. 132–136, 2007 Link 22. Daintith, John (2005). Oxford Dictionary of Physics. Oxford University Press. ISBN 0-19-280628-9. 23. ^ a b Moore, J. W.; C. L. Stanistski, P. C. Jurs (2005). Chemistry, The Molecular Science. Brooks Cole. ISBN 0-534-42201-2. 24. Jungermann, A.H. (2006). "Entropy and the Shelf Model: A Quantum Physical Approach to a Physical Property". Journal of Chemical Education 83 (11): 1686–1694. Bibcode:2006JChEd..83.1686J. doi:10.1021/ed083p1686. 25. Levine, I. N. (2002). Physical Chemistry, 5th ed. McGraw-Hill. ISBN 0-07-231808-2. 26. "GRC.nasa.gov". GRC.nasa.gov. 2000-03-27. Retrieved 2012-08-17. 27. "GRC.nasa.gov". GRC.nasa.gov. 2008-07-11. Retrieved 2012-08-17. 28. Sandler, Stanley, I. (1989). Chemical and Engineering Thermodynamics. John Wiley & Sons. ISBN 0-471-83050-X. 29. Dincer, Cengel (2001). Energy, entropy and exergy concepts and their roles in thermal engineering. pp. 116–149. 30. Gribbin's Q Is for Quantum: An Encyclopedia of Particle Physics, Free Press ISBN 0-684-85578-X , 2000 31. ^ a b c Brooks, Daniel, R.; Wiley, E.O. (1988). Evolution as Entropy– Towards a Unified Theory of Biology. University of Chicago Press. ISBN 0-226-07574-5. 32. ^ a b Landsberg, P.T. (1984). “Is Equilibrium always an Entropy Maximum?” J. Stat. Physics 35, pp. 159–169 33. ^ a b Landsberg, P.T. (1984). “Can Entropy and “Order” Increase Together?” Physics Letters 102A, pp. 171–173 34. Carson, E. M. and J. R. Watson (Department of Educational and Professional Studies, Kings College, London), , University Chemistry Education – 2002 Papers, Royal Society of Chemistry 35. Atkins, Peter (1984). The Second Law. Scientific American Library. ISBN 0-7167-5004-X. 36. Sandra Saary (Head of Science, Latifa Girls’ School, Dubai) (23 February 1993). "Book Review of "A Science Miscellany"". Khaleej Times (Galadari Press, UAE): XI. 37. Constantin Carathéodory: Untersuchungen über die Grundlagen der Thermodynamik, Math. Ann., 67, pp. 355–386, 1909 38. Robin Giles: Mathematical Foundations of Thermodynamics", Pergamon, Oxford 1964 39. Brillouin, Leon (1956). Science and Information Theory. ISBN 0-486-43918-6. 40. Georgescu-Roegen, Nicholas (1971). The Entropy Law and the Economic Process. Harvard University Press. ISBN 0-674-25781-2. 41. Chen, Jing (2005). The Physical Foundation of Economics – an Analytical Thermodynamic Theory. World Scientific. ISBN 981-256-323-7. 42. "Edwin T. Jaynes – Bibliography". Bayes.wustl.edu. 1998-03-02. Retrieved 2009-12-06. 43. Schneider, Tom, DELILA system (Deoxyribonucleic acid Library Language), (Information Theory Analysis of binding sites), Laboratory of Mathematical Biology, National Cancer Institute, FCRDC Bldg. 469. Rm 144, P.O. Box. B Frederick, MD 21702-1201, USA 44. Avery, John (2003). Information Theory and Evolution. World Scientific. ISBN 981-238-399-9. 45. Yockey, Hubert, P. (2005). Information Theory, Evolution, and the Origin of Life. Cambridge University Press. ISBN 0-521-80293-8. 46. IUPAC, , 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "Entropy unit". 47. von Baeyer, Christian, H. (2003). Information–the New Language of Science. Harvard University Press. ISBN 0-674-01387-5. Srednicki M (August 1993). "Entropy and area". Phys. Rev. Lett. 71 (5): 666–669. arXiv:hep-th/9303048. Bibcode:1993PhRvL..71..666S. doi:10.1103/PhysRevLett.71.666. PMID 10055336. Callaway DJE (April 1996). "Surface tension, hydrophobicity, and black holes: The entropic connection". Phys Rev E Stat Phys Plasmas Fluids Relat Interdiscip Topics 53 (4): 3738–3744. arXiv:cond-mat/9601111. Bibcode:1996PhRvE..53.3738C. doi:10.1103/PhysRevE.53.3738. PMID 9964684. 48. Stenger, Victor J. (2007). God: The Failed Hypothesis. Prometheus Books. ISBN 1-59102-481-1. 49. Benjamin Gal-Or (1981, 1983, 1987). Cosmology, Physics and Philosophy. Springer Verlag. ISBN 0-387-96526-2. ## Further reading • Atkins, Peter; Julio De Paula (2006). Physical Chemistry, 8th ed. Oxford University Press. ISBN 0-19-870072-5. • Baierlein, Ralph (2003). Thermal Physics. Cambridge University Press. ISBN 0-521-65838-1. • Ben-Naim, Arieh (2007). Entropy Demystified. World Scientific. ISBN 981-270-055-2. • Callen, Herbert, B (2001). Thermodynamics and an Introduction to Thermostatistics, 2nd Ed. John Wiley and Sons. ISBN 0-471-86256-8. • Chang, Raymond (1998). Chemistry, 6th Ed. New York: McGraw Hill. ISBN 0-07-115221-0. • Cutnell, John, D.; Johnson, Kenneth, J. (1998). Physics, 4th ed. John Wiley and Sons, Inc. ISBN 0-471-19113-2. • Dugdale, J. S. (1996). Entropy and its Physical Meaning (2nd ed.). Taylor and Francis (UK); CRC (US). ISBN 0-7484-0569-0. • Fermi, Enrico (1937). Thermodynamics. Prentice Hall. ISBN 0-486-60361-X. • Goldstein, Martin; Inge, F (1993). The Refrigerator and the Universe. Harvard University Press. ISBN 0-674-75325-9. • Gyftopoulos, E.P.; G.P. Beretta (1991, 2005, 2010). Thermodynamics. Foundations and Applications. Dover. ISBN 0-486-43932-1. • Haddad, Wassim M.; Chellaboina, VijaySekhar; Nersesov, Sergey G. (2005). Thermodynamics – A Dynamical Systems Approach. Princeton University Press. ISBN 0-691-12327-6. • Kroemer, Herbert; Charles Kittel (1980). Thermal Physics (2nd ed.). W. H. Freeman Company. ISBN 0-7167-1088-9. • Lambert, Frank L.; entropysite.oxy.edu • Penrose, Roger (2005). The Road to Reality: A Complete Guide to the Laws of the Universe. New York: A. A. Knopf. ISBN 0-679-45443-8. • Reif, F. (1965). Fundamentals of statistical and thermal physics. McGraw-Hill. ISBN 0-07-051800-9. • Schroeder, Daniel V. (2000). Introduction to Thermal Physics. New York: Addison Wesley Longman. ISBN 0-201-38027-7. • Serway, Raymond, A. (1992). Physics for Scientists and Engineers. Saunders Golden Subburst Series. ISBN 0-03-096026-6. • Spirax-Sarco Limited, Entropy – A Basic Understanding A primer on entropy tables for steam engineering • vonBaeyer; Hans Christian (1998). . Random House. ISBN 0-679-43342-2. • Entropy for beginners – a wikibook • An Intuitive Guide to the Concept of Entropy Arising in Various Sectors of Science – a wikibook	{"extraction_info": {"found_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": 92, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8977822065353394, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Dual_space	# Dual space In mathematics, any vector space, V, has a corresponding dual vector space (or just dual space for short) consisting of all linear functionals on V. Dual vector spaces defined on finite-dimensional vector spaces can be used for defining tensors. When applied to vector spaces of functions (which typically are infinite-dimensional), dual spaces are employed for defining and studying concepts like measures, distributions, and Hilbert spaces. Consequently, the dual space is an important concept in the study of functional analysis. There are two types of dual spaces: the algebraic dual space, and the continuous dual space. The algebraic dual space is defined for all vector spaces. When defined for a topological vector space there is a subspace of this dual space, corresponding to continuous linear functionals, which constitutes a continuous dual space. ## Algebraic dual space Given any vector space V over a field F, the dual space V* is defined as the set of all linear maps φ: V → F (linear functionals). The dual space V* itself becomes a vector space over F when equipped with the following addition and scalar multiplication: $\begin{align} & (\varphi + \psi)(x) = \varphi(x) + \psi(x) \\ & (a \varphi)(x) = a \left(\varphi(x)\right) \end{align}$ for all φ, ψ ∈ V, x ∈ V, and a ∈ F. Elements of the algebraic dual space V are sometimes called covectors or one-forms. The pairing of a functional φ in the dual space V* and an element x of V is sometimes denoted by a bracket: φ(x) = [φ,x] [1] or φ(x) = ⟨φ,x⟩.[2] The pairing defines a nondegenerate bilinear mapping[3] [·,·] : V* × V → F. ### Finite-dimensional case If V is finite-dimensional, then V* has the same dimension as V. Given a basis {e1, ..., en} in V, it is possible to construct a specific basis in V, called the dual basis. This dual basis is a set {e1, ..., en} of linear functionals on V, defined by the relation $\mathbf{e}^i(c_1 \mathbf{e}_1+\cdots+c_n\mathbf{e}_n) = c_i, \quad i=1,\ldots,n$ for any choice of coefficients ci ∈ F. In particular, letting in turn each one of those coefficients be equal to one and the other coefficients zero, gives the system of equations $\mathbf{e}^i(\mathbf{e}_j) = \delta_{ij}$ where δij is the Kronecker delta symbol. For example if V is R2, and its basis chosen to be {e1 = (1, 0), e2 = (0, 1)}, then e1 and e2 are one-forms (functions which map a vector to a scalar) such that e1(e1) = 1, e1(e2) = 0, e2(e1) = 0, and e2(e2) = 1. (Note: The superscript here is the index, not an exponent). In particular, if we interpret Rn as the space of columns of n real numbers, its dual space is typically written as the space of rows of n real numbers. Such a row acts on Rn as a linear functional by ordinary matrix multiplication. If V consists of the space of geometrical vectors (arrows) in the plane, then the level curves of an element of V form a family of parallel lines in V. So an element of V* can be intuitively thought of as a particular family of parallel lines covering the plane. To compute the value of a functional on a given vector, one needs only to determine which of the lines the vector lies on. Or, informally, one "counts" how many lines the vector crosses. More generally, if V is a vector space of any dimension, then the level sets of a linear functional in V* are parallel hyperplanes in V, and the action of a linear functional on a vector can be visualized in terms of these hyperplanes.[4] ### Infinite-dimensional case If V is not finite-dimensional but has a basis[5] eα indexed by an infinite set A, then the same construction as in the finite-dimensional case yields linearly independent elements eα (α ∈ A) of the dual space, but they will not form a basis. Consider, for instance, the space R∞, whose elements are those sequences of real numbers which have only finitely many non-zero entries, which has a basis indexed by the natural numbers N: for i ∈ N, ei is the sequence which is zero apart from the ith term, which is one. The dual space of R∞ is RN, the space of all sequences of real numbers: such a sequence (an) is applied to an element (xn) of R∞ to give the number ∑anxn, which is a finite sum because there are only finitely many nonzero xn. The dimension of R∞ is countably infinite, whereas RN does not have a countable basis. This observation generalizes to any[5] infinite-dimensional vector space V over any field F: a choice of basis {eα : α ∈ A} identifies V with the space (FA)0 of functions f : A → F such that fα = f(α) is nonzero for only finitely many α ∈ A, where such a function f is identified with the vector $\sum_{\alpha\in A} f_\alpha\mathbf{e}_\alpha$ in V (the sum is finite by the assumption on f, and any v ∈ V may be written in this way by the definition of the basis). The dual space of V may then be identified with the space FA of all functions from A to F: a linear functional T on V is uniquely determined by the values θα = T(eα) it takes on the basis of V, and any function θ : A → F (with θ(α) = θα) defines a linear functional T on V by $T\biggl(\sum_{\alpha\in A} f_\alpha \mathbf{e}_\alpha\biggr) = \sum_{\alpha \in A} f_\alpha T(e_\alpha) = \sum_{\alpha\in A} f_\alpha \theta_\alpha.$ Again the sum is finite because fα is nonzero for only finitely many α. Note that (FA)0 may be identified (essentially by definition) with the direct sum of infinitely many copies of F (viewed as a 1-dimensional vector space over itself) indexed by A, i.e., there are linear isomorphisms $V\cong (F^A)_0\cong\bigoplus_{\alpha\in A} {F}.$ On the other hand FA is (again by definition), the direct product of infinitely many copies of F indexed by A, and so the identification $V^* \cong \biggl(\bigoplus_{\alpha\in A}F\biggr)^* \cong \prod_{\alpha\in A}F^* \cong \prod_{\alpha\in A}F \cong F^A$ is a special case of a general result relating direct sums (of modules) to direct products. Thus if the basis is infinite, then the algebraic dual space is always of larger dimension than the original vector space. This is in marked contrast to the case of the continuous dual space, discussed below, which may be isomorphic to the original vector space even if the latter is infinite-dimensional. ### Bilinear products and dual spaces If V is finite-dimensional, then V is isomorphic to V. But there is in general no natural isomorphism between these two spaces.[6] Any bilinear form ⟨•,•⟩ on V gives a mapping of V into its dual space via $v\mapsto \langle v, \cdot\rangle$ where the right hand side is defined as the functional on V taking each w ∈ V to 〈v,w〉. In other words, the bilinear form determines a linear mapping $\Phi_{\langle\cdot,\cdot\rangle} : V\to V^$ defined by $[\Phi_{\langle\cdot,\cdot\rangle}(v),w] = \langle v, w\rangle.$ If the bilinear form is assumed to be nondegenerate, then this is an isomorphism onto a subspace of V. If V is finite-dimensional, then this is an isomorphism onto all of V. Conversely, any isomorphism Φ from V to a subspace of V* (resp., all of V) defines a unique nondegenerate bilinear form ⟨•,•⟩Φ on V by $\langle v,w \rangle_\Phi = (\Phi (v))(w) = [\Phi (v),w].\,$ Thus there is a one-to-one correspondence between isomorphisms of V to subspaces of (resp., all of) V and nondegenerate bilinear forms on V. If the vector space V is over the complex field, then sometimes it is more natural to consider sesquilinear forms instead of bilinear forms. In that case, a given sesquilinear form ⟨•,•⟩ determines an isomorphism of V with the complex conjugate of the dual space $\Phi_{\langle\cdot,\cdot\rangle} : V\to \overline{V}^.$ The conjugate space V can be identified with the set of all additive complex-valued functionals f : V → C such that $f(\alpha v) = \overline{\alpha}f(v).$ ### Injection into the double-dual There is a natural homomorphism Ψ from V into the double dual V*, defined by (Ψ(v))(φ) = φ(v) for all v ∈ V, φ ∈ V. This map Ψ is always injective;[5] it is an isomorphism if and only if V is finite-dimensional. Indeed, the isomorphism of a finite-dimensional vector space with its double dual is an archetypal example of a natural isomorphism. Note that infinite-dimensional Hilbert spaces are not a counterexample to this, as they are isomorphic to their continuous duals, not to their algebraic duals. ### Transpose of a linear map If f : V → W is a linear map, then the transpose (or dual) f* : W* → V* is defined by $f^(\varphi) = \varphi \circ f \,$ for every φ ∈ W. The resulting functional f(φ) in V is called the pullback of φ along f. The following identity holds for all φ ∈ W* and v ∈ V: $[f^(\varphi),\, v] = [\varphi,\, f(v)],$ where the bracket [•,•] on the left is the duality pairing of V with its dual space, and that on the right is the duality pairing of W with its dual. This identity characterizes the transpose,[7] and is formally similar to the definition of the adjoint. The assignment f ↦ f produces an injective linear map between the space of linear operators from V to W and the space of linear operators from W* to V; this homomorphism is an isomorphism if and only if W is finite-dimensional. If V = W then the space of linear maps is actually an algebra under composition of maps, and the assignment is then an antihomomorphism of algebras, meaning that (fg) = gf. In the language of category theory, taking the dual of vector spaces and the transpose of linear maps is therefore a contravariant functor from the category of vector spaces over F to itself. Note that one can identify (f) with f using the natural injection into the double dual. If the linear map f is represented by the matrix A with respect to two bases of V and W, then f* is represented by the transpose matrix AT with respect to the dual bases of W* and V, hence the name. Alternatively, as f is represented by A acting on the left on column vectors, f is represented by the same matrix acting on the right on row vectors. These points of view are related by the canonical inner product on Rn, which identifies the space of column vectors with the dual space of row vectors. ### Quotient spaces and annihilators Let S be a subset of V. The annihilator of S in V, denoted here So, is the collection of linear functionals f ∈ V such that [f, s] = 0 for all s ∈ S. That is, So consists of all linear functionals f : V → F such that the restriction to S vanishes: f\|S = 0. The annihilator of a subset is itself a vector space. In particular, ∅o = V* is all of V* (vacuously), whereas Vo = 0 is the zero subspace. Furthermore, the assignment of an annihilator to a subset of V reverses inclusions, so that if S ⊂ T ⊂ V, then $0 \subset T^o \subset S^o \subset V^.$ Moreover, if A and B are two subsets of V, then $(A \cap B)^o \supseteq A^o + B^o,$ and equality holds provided V is finite-dimensional. If Ai is any family of subsets of V indexed by i belonging to some index set I, then $\left(\bigcup_{i\in I} A_i\right)^o = \bigcap_{i\in I} A_i^o.$ In particular if A and B are subspaces of V, it follows that $(A + B)^o = A^o \cap B^o.\,$ If V is finite-dimensional, and W is a vector subspace, then $W^{oo} = W \,$ after identifying W with its image in the second dual space under the double duality isomorphism V ≈ V. Thus, in particular, forming the annihilator is a Galois connection on the lattice of subsets of a finite-dimensional vector space. If W is a subspace of V then the quotient space V/W is a vector space in its own right, and so has a dual. By the first isomorphism theorem, a functional f : V → F factors through V/W if and only if W is in the kernel of f. There is thus an isomorphism $(V/W)^ \cong W^o.$ As a particular consequence, if V is a direct sum of two subspaces A and B, then V* is a direct sum of Ao and Bo. ## Continuous dual space When dealing with topological vector spaces, one is typically only interested in the continuous linear functionals from the space into the base field. This gives rise to the notion of the "continuous dual space" which is a linear subspace of the algebraic dual space V, denoted V′. For any finite-dimensional normed vector space or topological vector space, such as Euclidean n-space, the continuous dual and the algebraic dual coincide. This is however false for any infinite-dimensional normed space, as shown by the example of discontinuous linear maps. The continuous dual V′ of a normed vector space V (e.g., a Banach space or a Hilbert space) forms a normed vector space. A norm \|\|φ\|\| of a continuous linear functional on V is defined by $\\|\varphi\\| = \sup \{ \|\varphi(x)\| : \\|x\\| \le 1 \}.$ This turns the continuous dual into a normed vector space, indeed into a Banach space so long as the underlying field is complete, which is often included in the definition of the normed vector space. In other words, this dual of a normed space over a complete field is necessarily complete. The continuous dual can be used to define a new topology on V, called the weak topology. ### Examples Let 1 < p < ∞ be a real number and consider the Banach space ℓ p of all sequences a = (an) for which $\\|\mathbf{a}\\|_p = \left ( \sum_{n=0}^\infty \|a_n\|^p \right) ^{1/p}$ is finite. Define the number q by 1/p + 1/q = 1. Then the continuous dual of ℓ p is naturally identified with ℓ q: given an element φ ∈ (ℓ p)′, the corresponding element of ℓ q is the sequence (φ(en)) where en denotes the sequence whose n-th term is 1 and all others are zero. Conversely, given an element a = (an) ∈ ℓ q, the corresponding continuous linear functional φ on ℓ p is defined by φ(b) = ∑n anbn for all b = (bn) ∈ ℓ p (see Hölder's inequality). In a similar manner, the continuous dual of ℓ 1 is naturally identified with ℓ ∞ (the space of bounded sequences). Furthermore, the continuous duals of the Banach spaces c (consisting of all convergent sequences, with the supremum norm) and c0 (the sequences converging to zero) are both naturally identified with ℓ 1. By the Riesz representation theorem, the continuous dual of a Hilbert space is again a Hilbert space which is anti-isomorphic to the original space. This gives rise to the bra-ket notation used by physicists in the mathematical formulation of quantum mechanics. ### Transpose of a continuous linear map If T : V → W is a continuous linear map between two topological vector spaces, then the (continuous) transpose T′ : W′ → V′ is defined by the same formula as before: $T'(\varphi) = \varphi \circ T, \quad \varphi \in W'. \,$ The resulting functional T′(φ) is in V′. The assignment T → T′ produces a linear map between the space of continuous linear maps from V to W and the space of linear maps from W′ to V′. When T and U are composable continuous linear maps, then $(U \circ T)' = T' \circ U'. \,$ When V and W are normed spaces, the norm of the transpose in L(W′, V′) is equal to that of T in L(V, W). Several properties of transposition depend upon the Hahn–Banach theorem. For example, the bounded linear map T has dense range if and only if the transpose T′ is injective. When T is a compact linear map between two Banach spaces V and W, then the transpose T′ is compact. This can be proved using the Arzelà–Ascoli theorem. When V is a Hilbert space, there is an antilinear isomorphism iV from V onto its continuous dual V′. For every bounded linear map T on V, the transpose and the adjoint operators are linked by $i_V \circ T^ = T' \circ i_V. \,$ When T is a continuous linear map between two topological vector spaces V and W, then the transpose T′ is continuous when W′ and V′ are equipped with"compatible" topologies: for example when, for X = V and X = W, both duals X′ have the strong topology β(X′, X) of uniform convergence on bounded sets of X, or both have the weak-∗ topology σ(X′, X) of pointwise convergence on X. The transpose T′ is continuous from β(W′, W) to β(V′, V), or from σ(W′, W) to σ(V′, V). ### Annihilators Assume that W is a closed linear subspace of a normed space V, and consider the annihilator of W in V′, $W^\perp = \{ \varphi \in V' : W \subset \ker \varphi\}. \,$ Then, the dual of the quotient V / W can be identified with W⊥, and the dual of W can be identified with the quotient V′ / W⊥.[8] Indeed, let P denote the canonical surjection from V onto the quotient V / W ; then, the transpose P′ is an isometric isomorphism from (V / W )′ into V′, with range equal to W⊥. If j denotes the injection map from W into V, then the kernel of the transpose j′ is the annihilator of W: $\ker (j') = W^\perp$ and it follows from the Hahn–Banach theorem that j′ induces an isometric isomorphism V′ / W⊥ → W′. ### Further properties If the dual of a normed space V is separable, then so is the space V itself. The converse is not true: for example the space ℓ 1 is separable, but its dual is ℓ ∞ is not. ### Double dual In analogy with the case of the algebraic double dual, there is always a naturally defined continuous linear operator Ψ : V → V′′ from a normed space V into its continuous double dual V′′, defined by $\Psi(x)(\varphi) = \varphi(x), \quad x \in V, \ \varphi \in V'. \,$ As a consequence of the Hahn–Banach theorem, this map is in fact an isometry, meaning \|\|Ψ(x)\|\| = \|\|x\|\| for all x in V. Normed spaces for which the map Ψ is a bijection are called reflexive. When V is a topological vector space, one can still define Ψ(x) by the same formula, for every x ∈ V, however several difficulties arise. First, when V is not locally convex, the continuous dual may be equal to {0} and the map Ψ trivial. However, if V is Hausdorff and locally convex, the map Ψ is injective from V to the algebraic dual V′* of the continuous dual, again as a consequence of the Hahn–Banach theorem.[9] Second, even in the locally convex setting, several natural vector space topologies can be defined on the continuous dual V′, so that the continuous double dual V′′ is not uniquely defined as a set. Saying that Ψ maps from V to V′′, or in other words, that Ψ(x) is continuous on V′ for every x ∈ V, is a reasonable minimal requirement on the topology of V′, namely that the evaluation mappings $\varphi \in V' \mapsto \varphi(x), \quad x \in V, \,$ be continuous for the chosen topology on V′. Further, there is still a choice of a topology on V′′, and continuity of Ψ depends upon this choice. As a consequence, defining reflexivity in this framework is more involved than in the normed case. ## See also • Duality (mathematics) • Duality (projective geometry) • Pontryagin duality • Reciprocal lattice – dual space basis, in crystallography • Covariance and contravariance of vectors • Dual norm ## Notes 1. In many areas, such as quantum mechanics, ⟨·,·⟩ is reserved for a sesquilinear form defined on V × V. 2. ^ a b c Several assertions in this article require the axiom of choice for their justification. The axiom of choice is needed to show that an arbitrary vector space has a basis: in particular it is needed to show that RN has a basis. It is also needed to show that the dual of an infinite-dimensional vector space V is nonzero, and hence that the natural map from V to its double dual is injective. 3. If V is locally convex but not Hausdorff, the kernel of Ψ is the smallest closed subspace containing {0}. ## References • Bourbaki, Nicolas (1989), Elements of mathematics, Algebra I, Springer-Verlag, ISBN 3-540-64243-9 • Halmos, Paul (1974), Finite-dimensional Vector Spaces, Springer, ISBN 0-387-90093-4 • Lang, Serge (2002), Algebra, Graduate Texts in Mathematics 211 (Revised third ed.), New York: Springer-Verlag, ISBN 978-0-387-95385-4, Zbl 0984.00001, MR1878556 • MacLane, Saunders; Birkhoff, Garrett (1999), Algebra (3rd ed.), AMS Chelsea Publishing, ISBN 0-8218-1646-2 . • Misner, Charles W.; Thorne, Kip S.; Wheeler, John A. (1973), Gravitation, W. H. Freeman, ISBN 0-7167-0344-0 • Rudin, Walter (1991), Functional analysis, McGraw-Hill Science, ISBN 978-0-07-054236-5	{"extraction_info": {"found_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": 30, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9028092622756958, "perplexity_flag": "head"}
http://motls.blogspot.com/2012/05/how-20-scft-little-string-theory-and.html?m=1	# The Reference Frame Our stringy Universe from a conservative viewpoint ## Monday, May 21, 2012 ### How the (2,0) SCFT, little string theory, and others arise from string theory We often say that the primary reason why string/M-theory is so essential for modern physics is that it is the only known – and most likely, the only mathematically possible – consistent theory of gravity. Everyone who believes that he or she can do state-of-the-art research of quantum gravity without string theory is an unhinged crank, a barbarian, and a conspiracy theorist of the same kind as those who believe that Elvis Presley lives on the Moon. But another reason why string/M-theory is indispensable for the 21st century theoretical and particle physics is that many of the "ordinary", important, non-gravitational quantum field theories and some of their non-field-theoretical but still non-gravitational generalizations are tightly embedded as limits in string theory. In this way, a theory whose main strength is to provide us with robust quantum rules governing gravity is important for our knowledge of contexts that avoid gravity, too. Because of the dense network of relationships within string theory that link ideas, concepts, and equations that used to be considered independent – and I mostly mean dualities but not only dualities – each of the "ordinary" non-gravitational theories may be analyzed from new perspectives. In particular, extreme limits of the old theories in which a quantity is sent to infinity (or zero) could have been very mysterious but many of the mysteries go away as string/M-theory allows us to use new descriptions. Among the new insights that we're learning from the stringy network of ideas, rules, equations, and maps, we also encounter new quantum field theories – and some other non-gravitational generalizations of these theories which are not quantum field theories – i.e. theories that are not full-fledged string vacua and that we shouldn't have overlooked in the past but we have. What are they? In March, I discussed the maximally supersymmetric gauge theory in four dimensions. It's arguably the most far-reaching or at least the most widely studied example of the point I made in the second paragraph. The $$\NNN=4$$ gauge theory in $$d=4$$ is a gauge theory with 16 real supercharges. If you write it in terms of components, it's a gauge theory with a gauge group – it can be $$SU(N)$$, $$O(N)$$, $$OSp(2k)$$, $$E_6$$, or any other compact Lie group – which is coupled to four Weyl neutrinos in the adjoint representation of the same group and six Hermitian scalars in the same representation. When the interactions are appropriately chosen, we discover that the theory has those 16 supersymmetries even at the interacting level. Nima Arkani-Hamed would call this theory a harmonic oscillator of the 21st century. Andy Strominger reserves this term for black holes but it's true that these two theoretical constructs are perhaps even more important if they work as a team and they often do. String theory tells us lots of things about the seemingly ordinary gauge theory which wasn't known to have any direct connection to strings. In fact, we have known for almost 15 years that this gauge theory is string theory. The $$SU(N)$$ maximally supersymmetric gauge theory is totally equivalent to the superselection sector of type IIB string theory respecting the asymptotic conditions of $$AdS_5\times S^5$$. This relationship is, of course, the most famous example of Juan Maldacena's AdS/CFT correspondence. However, the remarkable relationship was found – and may be "almost proven" – by less shocking relationships between this gauge theory and string theory. In particular, the simplest representation of the gauge theory is the dynamics of D3-branes in type IIB string theory at very long distances. Some properties of the gauge theory may be deduced out of this realization immediately. In particular, the theory inherits the $$SL(2,\ZZ)$$ S-duality group – which includes the $$g\to 1/g$$ exchange of the weak coupling with the strong coupling – from the full type IIB string theory. In the type IIB string theory, the S-duality group may also be motivated by representing type IIB string theory as a 12-dimensional theory, F-theory, compactified on a two-torus. This toroidal proof of the S-duality group may also be realized by another embedding: the gauge theory may also be viewed as a long-distance limit of the $$d=6$$ $$(2,0)$$ superconformal field theory compactified on a two-torus; the logic is the same. You should appreciate that the S-duality is an extremely complicated relationship if you want to construct it or prove it by hand. In fact, it replaces point-like elementary oscillations that are weakly coupled with extended objects such as magnetic monopoles that are strongly coupled. They look like very different physical objects and the proof of the equivalence can't be made in perturbative expansion – because it is not a duality that holds order-by-order in this expansion – but it's still true. But of course, all tests you can fully calculate work: the gauge theory seems to possess the non-trivial S-duality group. In its stringy incarnation, the S-duality may be seen within a second. Also, Maldacena's holographic duality boils down to the construction of the gauge theory involving D3-branes, too. The low-energy limit of the D3-branes' internal interactions has to be an interacting theory with 16 supercharges – because they aren't being broken by anything – and that has a field content that may be obtained from the counting of open string excitations attached to the D3-branes. You will find out that the theory has to be a gauge theory with the degrees of freedom I enumerated above; the supersymmetries and consistency dictate the interactions uniquely. In the long-distance limit, only the massless open strings i.e. gauge fields and their superpartners matter; closed strings (especially gravity) is decoupled because the energy density per Planck volume is very low in this limit. So we really do have a non-gravitational theory. On the other hand, the D3-branes in string theory are real objects, lively animals that manifest themselves in many other physical ways. In particular, they have a gravitational field that extends to the transverse dimensions. Much like D0-branes would be particles that would behave as black holes, D3-branes are extended versions of the same objects, extended black holes. We call them black branes or black $$p$$-branes. They are black 3-branes, in this case. Just to be sure, in the previous paragraph, I stated that the gravitational force between the open string interactions may be neglected; but the gravitational field from their substrate – the static D3-branes in which the open strings live – still curves the 10-dimensional spacetime of type IIB string theory. A funny thing is that if you adopt the full 10-dimensional perspective, the low-energy excitations have another interpretation: they are physical states that are located near the event horizon of the black branes. The relationship between the adjectives "low-energy" and "near-horizon" holds because near the horizon, it's where the excitations that look "very red" from the global viewpoint (of an observer at infinity) may be created in generic processes. That's because of the gravitational red shift, of course. If you ask which degrees of freedom are kept if you simply consider all low-energy excitations of those 3-branes, you have two methods to answer: you either realize that the 3-branes may be described as D3-branes whose dynamics is governed by interactions of open strings and the low-energy limit of the open strings' interactions is nothing else than the gauge theory; or you may imagine that the D3-branes are actual solutions of a gravitational theory – an extension of general relativity – and low-energy states are the states of all objects that move near the event horizon. Each of these operations is a valid method to isolate the low-energy states; so the two theories obtained by these methods must be exactly equivalent. That's an elegant proof of the AdS/CFT correspondence, a non-technical, non-constructive proof that avoids almost all mathematics (although one should still add some mathematics in order to show that it really deserves to be called the "proof"). The near-horizon geometry of the black 3-branes is nothing else than $$AdS_5\times S^5$$ and gravitational – well, type IIB stringy – phenomena within this spacetime must therefore be exactly described by a four-dimensional gauge theory. Of course, this successful union of string theory and gauge theory may be extended to other gauge groups, less supersymmetric gauge theories corresponding to less symmetric compactifications of the gravitational side, and even to other dimensions. Lots of objects on both sides of the equivalence may be given new interpretations using the other description, and so on. But the main goal of this text is to describe new field theories and new non-gravitational non-field theories that arise from similar constructions. The most supersymmetric example of the first category is the so-called $$(2,0)$$ superconformal field theory in 6 dimensions. M5-branes and their dynamics In the case of the D3-branes above, we considered objects in string theory in ten dimensions. In the usual weakly coupled approach, these theories are parameterized by the string coupling constant $$g_s$$ which is the exponential of the (stringy) dilaton; greetings. The coupling constant is adjustable in the simplest vacua; all values are equally good but the choice isn't a parameter representing inequivalent possibilities. Instead, because the coupling is an exponential of the dilaton and the dilaton is a dynamical field, different values of the coupling constant correspond to different environments that may be achieved in a single theory. In realistic compactifications, a potential for the dilaton is generated (much like the potential for all other moduli) and string theory picks a preferred value of the string coupling which is at least in principle but – to a large extent – also in practice calculable (much like the detailed shape of the extra dimensions etc.). However, there exists a vacuum of string/M-theory that has no dilaton-like scalar field that would label inequivalent environments. Of course, it's the 11-dimensional M-theory. The field content of the eleven-dimensional supergravity only includes the graviton, some spin-3/2 gravitino, and spin-1 three-form generalizing electromagnetism. No spin-0 scalar fields here. That's kind of nice because the theories we may obtain from M-theory in similar ways as the theories obtained from type II or type I or heterotic string theory have an unusual property: they have no adjustable dimensionless coupling constants. This is something we're not used to from the quantum field theory courses taught at schools. In those courses, we first start with a free theory and interactions are added as a voluntary deformation. All these interactions may be chosen to be weak because the coupling constants are adjustable and the free, non-interacting limit is assumed to be OK. However, for theories obtained from M-theory, we can't turn off the interactions at all! These theories inevitably force their degrees of freedom to interact with a particular vigor that cannot be reduced at all. Because the coupling constants may be measured as the strength of the "quantum processes" – how much the one-loop diagrams where virtual pairs exist for a while are important relatively to the tree-level "classical" processes – we may also say that the theories extracted from M-theory are intrinsically quantum and they have no classical limit. Are there any? You bet. As I mentioned in my discussion of 11D SUGRA, the theory has to contain a three-form potential $$C_3$$. One may add terms in the Lagrangian where $$C_3$$ is integrated over a 3-dimensional world volume in the spacetime. This term generalizes the $$\int \dd x^\mu A_\mu$$ coupling of the electromagnetic fields with world lines of charged particles (in the limit in which they're treated as particles with clear world lines, not as fields). And indeed, M-theory does allow such terms; the 3-dimensional world volumes are those of M2-branes, or membranes, objects with 2 spatial and 1 temporal dimensions. Also, the exterior derivative of the $$C_3$$ potential is a four-form $$F_4$$ field strength. By using the epsilon symbol in eleven dimensions, this may get mapped to a Hodge-dual seven-form $$F_7$$ potential which is locally, in the vacuum, the exterior derivative of a six-form "dual potential" $$C_6$$. So M-theory also admits couplings of this $$C_6$$ and indeed, the 6-dimensional world volume we integrate over is the world volume of M5-branes, the electromagnetic dual partners of M2-branes. Just like string theories contain fundamental strings, F1-branes, and lots of heavy D-branes of various dimensions, M-theory contains no strings or 1-branes but it has M2-branes and M5-branes which have different dimensions but are "comparably heavy" as long as their typical mass scale goes. A nice thing is that just like you may study the long-distance dynamics of D3-branes which led to the very important maximally supersymmetric gauge theory, you may also study the long-distance limit of the dynamics inside M2-branes and M5-branes. Both of them give you some new interesting theories. The theories related to the M2-branes were the subject of the recent "membrane minirevolution"; this was my name for the intense research of some supersymmetric 3-dimensional gauge theories extending the Chern-Simons theory. Some new ways to see the hidden symmetries of these theories were found; the most obvious "clearly new" development of the minirevolution were the ABJM theories extending the long-distance of the membranes to more complicated compactifications. The membrane minirevolution has surprised many people who had thought that such M(ysterious) field theories would never be written in terms of ordinary Lagrangians. They could have been written. People could only discover these very interesting and special Lagrangians once they were forced by string/M-theory to look for them. When you consider the low-energy limit of the M5-branes, you get a six-dimensional theory: 5 dimensions of space and 1 dimension of time. It is useful to mention how spinors work in 6 dimensions. In 4 dimensions, the minimal spinor is a Weyl spinor (or, equivalently – when it comes to the counting of fields – the Majorana spinor). But there's only one kind: if you include a left-handed Weyl spinor, the theory immediately possesses the Hermitian conjugate right-handed one, too. So you only need to know how many spinors your theory has. For example, the $$\NNN=4$$ theory has supercharges that may be organized into 4 Weyl or Majorana spinors. However, things are a bit different in $$d=6$$. Because it is an even number, one still distinguishes left-handed and right-handed Weyl spinors. But in spacetime dimensions of the form $$4k+2$$, the left-handed and right-handed spinors are actually not complex conjugates to each other. You may incorporate them independently of each other. The same comment holds for supersymmetries; if you want to accurately describe how the spinors of supersymmetric transform, you must specify how many left-moving and how many right-moving Weyl spinors there are in the list of supercharges. In ten dimensions, we use the "shortened" terms type I, type IIA, type IIB for $$(1,0)=(0,1)$$ supersymmetric theories, $$(1,1)$$ supersymmetric theories, and $$(2,0)=(0,2)$$ supersymmetric theories, respectively. The permutation of the two labels is immaterial. The type I and type IIB theories are inevitably left-hand-asymmetric i.e. chiral; type IIA is left-right-symmetric i.e. non-chiral, as expected from the fact that it may be produced as a compactification of an 11-dimensional theory. In six dimensions, there's a similar classification. The $$(1,1)$$ theories are non-chiral and typically include some gauge fields. On the other hand, the $$(2,0)$$ theories are chiral. The $$(2,0)$$ theory we find in the long-distance limit of the M5-branes is non-chiral not only when it comes to the fermions in the field content. Because the labels $$(2,0)$$ are "very asymmetric" between the first and second digit, the left-right asymmetry actually inevitably gets imprinted to the bosonic spectrum, too. If we're explicit, it's because the theory contains "self-dual field strength fields" i.e. 3-form(s) $$H_3$$ generalizing $$F_2$$ in Maxwell's theory that however obey $$H_3=H_3$$. Note that this is possible in 6 dimensions but not in 4 dimensions because $$()^2=+1$$ in 6 dimensions but $$(*)^2=-1$$ in 4 dimensions. Because the $$(2,0)$$ theory must allow a generalization of the gauge field whose field strength is however constrained by the self-duality condition, it's hard to write an explicit Lagrangian definition of the theory, at least if we want it to be manifestly Lorentz-symmetric one. It's a part of the unproven lore that this can't be done. However, you must be careful about such widely held beliefs. In particular, the membrane minirevolution has shown that various Lagrangians that would be thought of as impossible are actually totally possible and you never know whether someone will find a clever trick by which this explicit construction may be extended to 6 dimensions. So the six-dimensional theory can't be constructed as a "quantization" of a classical theory. It's a point that I discussed in less specific contexts in several recent articles about the foundations of quantum mechanics. We see many independent reasons why it's natural that no such "master classical theory" may exist in this case. First, the quantum theory requires the coupling constant to be "one" in some normalization: it can't be adjusted to be close to zero so studying the theory as the deformation of a free theory would be similar to studying $$\pi$$ using the $$\pi\to 0$$ limit. Second, we have mentioned that the theory contains self-dual fields and it's hard to write a Lagrangian for a potential if you also want its field strength to be self-dual. Third, and it is related, you would have a problem to write renormalizable interactions in a theory in 6 or more dimensions, anyway. A $$\phi^3$$ cubic coupling for a scalar would be the "maximum" that would still be renormalizable but it would create instabilities. By denying that there exists a way to represent the full quantum field theory as a quantization of a classical theory (with a polynomial Lagrangian), string/M-theory finds the loophole in all these arguments that a sloppy person could offer as an excuse that such a non-trivial 6-dimensional theory shouldn't exist. However, this theory still exists as an interacting, non-gravitational theory with all the things you expect from a local quantum field theory. One may define local fields $$\Phi_k(x^\mu)$$ and these fields have various correlation functions and may be evolved according to some well-defined Heisenberg equations, and so on. It may be hard or impossible to use the perturbative (and other) techniques we know from the gauge theory but the resulting product – Green's functions etc. – is conceptually identical to the product in the gauge theory. You may be ignorant about methods how to compute these physical answers in the $$(2,0)$$ theory; but one may actually prove – using the consistency of string theory as a main tool or assumption – that these answers exist and have the same useful properties as similar answers in gauge theory. However, in gauge theory, we may calculate a whole 1-parameter or 2-parameter family of the "collection of Green's functions"; the families are parameterized by the coupling constant (and the axion). In the $$(2,0)$$ case, there are no such parameters. It's just an isolated theory – one isolated set of Green's functions encoding all the evolution and interactions – without continuously adjustable dimensionless parameters. Much like the $$\NNN=4$$ gauge theory is equivalent to type IIB string theory in $$AdS_5\times S^5$$ which we could have derived as the near-horizon geometry of a stack of the D3-branes, the $$(2,0)$$ theory in six dimensions may be shown to be equivalent to M-theory on $$AdS_7\times S^4$$, the near-horizon geometry of a stack of the M5-branes in M-theory. Just to be sure, there is a similar case involving a 3-dimensional Chern-Simons-like theory andd M-theory on $$AdS_4\times S^7$$ – note that the labels four and seven got exchanged – which is the near-horizon geometry of a stack of M2-branes in M-theory. So while the perturbative, weakly coupled methods don't exist for this six-dimensional theory, the holographic AdS/CFT methods work as well as they do for the gauge theory. Also, this six-dimensional theory is as important for Matrix theory, a non-gravitational way to describe some simple enough compactifications of string/M-theory on flat backgrounds, as the gauge theory is. In particular, if you compactify the $$(2,0)$$ theory on a five-torus (times the real line for time), you get a matrix description for M-theory on a four-torus. Perturbatively, the $$\NNN=4$$ gauge theory with the $$SU(N)$$ gauge group seems to have the number of degrees of freedom – independent elementary fields – that scales like $$N^2$$. That's because the adjoint representation may be viewed as a square matrix, of course. There are actually different, independent methods to derive this power law, too, in particular a holographic one that is based on the entropy of a dual bulk black hole. The holographic methods may also be used for the M2-based 3-dimensional theory and the M5-based 6-dimensional theory. They tell you that the number of degrees of freedom in these two theories should scale like $$N^{3/2}$$ and $$N^3$$ in $$d=3$$ and $$d=6$$, respectively. The first case, a fractional power, doesn't even produce an integer but it has still been motivated in various ways. The 6-dimensional case is even more intriguing because the integral exponent does suggest that there could exist a "constructive explanation" – some formulation that uses fields with three "fundamental gauge indices", if you wish. Many authors have tried to shed light on this strange power law. A month ago, Sav Sethi and Travis Maxfield offered a brand new calculation of the "conformal anomaly" (what was interpreted as the number of degrees of freedom) which also produces the right $$N^3$$ scaling. There's still a significant activity addressing this 6-dimensional theory and its less supersymmetric cousins. A few days ago, Elvang, Freedman, Myers, and 3 more colleagues wrote an interesting paper about the a-theorem in six dimensions. You should realize that despite the absence of an old-fashioned, "textbook" Lagrangian classical-based construction of the theory, the amount of knowledge has been growing for more than 15 years. Let me pick my 1998 paper with Ori Ganor as some "relatively early" research of physical effects that occur in this theory. So the $$(2,0)$$ theory is conformal and therefore scale-invariant (it is a "fixed point" of the renormalization group) which is why it may occur as the low-energy limit of other physical theories in 6 dimensions; I will mention one momentarily. It has a qualitatively well-understood holographic dual and it appears in a matrix description of M-theory on a four-torus. Some fields, especially the "supersymmetry preserving ones", may be isolated and some of their correlation functions may be calculated purely from SUSY, and so on. The theory has various topological solutions that may be interpreted by various "perspectives" to look at this theory that string/M-theory offers. This six-dimensional theory is also an "ancestor" of the maximally supersymmetric gauge theory; the $$\NNN=4$$ gauge theory may be obtained from a compactification of the six-dimensional theory on a two-torus. There are interesting modifications and projections of this theory, too. For example, there are $$(1,0)$$ theories in six dimensions which respect an $$E_8$$ global symmetry. This global symmetry is inherited from the $$E_8$$ gauge symmetry that lives on the domain walls (ends-of-the-world) in M-theory whenever the M5-branes are places on such a boundary. I can't say everything that is interesting about this theory but be sure that there would be lots of other things just to enumerate – and lots of interesting details if I were to fully "teach you" about those things. One of the broader points is that physics is making progress and finding "conceptually new ways" how to think about old theories, how to calculate their predictions, and how to related previous unrelated physical mechanisms and insights. Quantum field theory is essential in all this research; however, we know that quantum field theory isn't just some mechanical exercise starting from a classical theory and adding interactions to a free limit by perturbative interactions. There are lots of nonperturbative processes and insights that may be obtained without explicit perturbative calculations, too. Little string theory I have mentioned that the $$(2,0)$$ superconformal field theory discussed above was a quantum field theory whose Green's functions are as real as those coming from a gauge theory; they satisfy the same consistency, unitarity, and locality conditions, too. But it's a "fixed point", a scale-invariant theory that may be identified as the "ultimate long-distance limit" of some other theories. Are there any other theories of this kind? Yes, you bet. But the most interesting ones aren't gauge theories. They're "little string theories". A little string theory is a type of a theory in spacetime that is something in between a quantum field theory in the spacetime; and the full gravitating string theory in the same spacetime. They're not local because we may say that their elementary degrees of freedom or elementary building blocks arise from strings much like in the full string theory; however, an appropriate limit is taken so that the gravitational force between the strings decouples. This seemingly contradicts the lore that every theory constructed from interacting strings inevitably includes gravity; however, there's actually no congtradiction because while the little string theories contain strings and they are interacting theories, they actually cannot be constructed out of these "elementary strings" by following the usual constructive methods of the full string theory. Fine, so what is the little string theory? The simplest little string theories carry the same $$(2,0)$$ supersymmetry in $$d=6$$ as the superconformal quantum field theory I was discussing at the beginning. In fact, the long-distance limit of these little string theories (they are parameterized by discrete labels such as the number of 5-branes) produce the superconformal field theory we have already discussed. But these little string theories are not superconformal or scale-invariant. In fact, they are not local quantum field theories at all. In this sense, they are just a generalization of a quantum field theory in a similar sense as the full string theory is a generalization of a quantum field theory. How can we obtain them? The most straightforward way to obtain the $$(2,0)$$ superconformal field theories above were a stack of M5-branes in M-theory. Are there some other objects in string theory that are not M5-branes but that look as M5-branes in the low-energy limit? The answer is Yes. M-theory may be obtained as the strong coupling limit of type IIA string theory. Type IIA string theory also contains 5-branes. But they are not D5-branes which may be found in type IIB string theory; type IIB D5-branes produce $$(1,1)$$ supersymmetric theories in six dimensions, not $$(2,0)$$: their world volume is exactly as left-right-symmetric as the type IIB spacetime fails to be. There are also NS5-branes in type IIB string theory which have the same SUSY as the D5-branes, because of S-duality that relates them. Type IIA string theory only contains D-even-branes, not D5-branes, but it still allows NS5-branes, the electromagnetic duals of fundamental strings. And while type IIA is left-right-symmetric in the spacetime, its NS5-branes are left-right asymmetric; not that there is an anticorrelation between the chirality of the spacetime and the chirality of the NS5-brane world volume. The dilaton of type IIA string theory has a value that depends on the distance from the NS5-branes; this contrasts with the behavior of D3-branes in type IIB string theory that preserve the constant dilaton (and string coupling) in the whole spacetime. This depends of the dilaton – it goes to infinity near the NS5-branes' core – means that the ultimate low-energy limit of the dynamics of NS5-branes is the same one as it is for M5-branes in M-theory: the new 11th dimension really emerges if you're close enough to the NS5-branes. On the other hand, one may define a different scaling limit of dynamics inside the type IIA NS5-branes in which the gravity in between the excitations of the NS5-branes is sent to zero; but which is not the ultimate long-distance, scale-invariant limit yet. Such a theory inherits a privileged length scale, the string scale, from the "parent" type IIA string theory. But it doesn't preserve the dilaton or the coupling constant because it's scaled to infinity. The resulting theory of this limit, the little string theory, has no gravitational force but it has string-like excitations. It is not a local quantum field theory but its low energy limit is a quantum field theory. The theory – which has a "qualitatively higher level of conceptual complexity than the $$(2,0)$$ superconformal field theory" – also enters Matrix theory; its compactification on a five-torus is the matrix description of M-theory compactified on a five-torus. All the usual limits and dualities between the toroidally compactified string/M-theoretical backgrounds may be deduced from the matrix description, too: these dualities may be reduced to relationships between their non-gravitational matrix descriptions. The little string theories have various other relationships to quantum field theories and vacua of the full string theory, too. Again, I can't say everything that is known about them and everything that makes them important. Let me emphasize that none of these theories – neither the new superconformal field theories nor the little string theories – has any adjustable continuous dimensionless parameters. They still have discrete parameters – counting the number of 5-branes in the stack and/or whether or not these 5-branes were positioned at some end-of-the-world boundaries or other singular loci in the parent spacetime. But the absence of the continuously adjustable parameters allows us to say that all these quantum theories are "islands" of a sort. They're obviously important islands. If you want to study consistent non-gravitational interacting theories in 6 dimensions, these islands may be as important as Hawaii or the Greenland or Polynesia or Africa – it's hard to quantify their importance accurately in this analogy. However, the importance is clearly "finite" and can't go to zero. Hawaii, the Greenland, Polynesia, or Africa inevitably enters many people's lives. Finally, I want to end up with a more general comment. New exceptional theories that were previously overlooked but that obey all the "quality criteria" that were satisfied by the more well-known theories; and all the new perspectives and "pictures" that allow us to say something or calculate something about these as well as the more ordinary theories are important parts of the genuine progress in theoretical physics and everyone who actually likes theoretical physics must be thrilled by this kind of progress and by the new "concise ways" how some previously impenetrable technical insights may be explained or proved. There exists a class of people with a very low intelligence, no creativity, no imagination, and no ability to see the "big picture" who are only capable of learning some very limited rules and who are devastated by every new powerful technique or technology that physics learns. These human feces often concentrate around Shmoit-and-Shmolin kind of aggressive sourball crackpot forums. I hope that all readers with IQ above 100 have managed to understand why the text above is enough as a proof of the simple assertion that all these Shmoits-and-Shwolins are just intellecutally worthless dishonest scum. And that's the memo. ## Who is Lumo? Luboš Motl Pilsen, Czech Republic View my complete profile ← by date	{"extraction_info": {"found_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": 72, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9388725757598877, "perplexity_flag": "head"}
http://mathoverflow.net/questions/82926/spherical-building-of-an-exceptional-group-of-lie-type/82935	Spherical building of an exceptional group of Lie type Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I've read that one of Tits' original motivations for studying buildings was that he wanted to give a unified description of algebraic groups that would allow the definition of exceptional groups such as $E_6$, $E_7$, $E_8$, $F_4$ and $G_2$ over any field. Now, Chevalley managed to construct the corresponding groups of Lie type over finite fields; but is there a way one can instead first construct the buildings and then define the group as a particular subgroup of the automorphism group of the building? What do these buildings look like? The building for $\mathrm{SL}_3(\mathbb{F}_2)$, for instance, is the Heawood graph (I can't embed images yet); one can count the 28 apartments making up this complex, which are just hexagons with alternating colors. This graph is the incidence graph for the Fano plane $\mathbb{P}^2/\mathbb{F}_2$. One recovers the full group $\mathrm{SL}_2(\mathbb{F}_2)$ as the group of type-preserving automorphisms of this graph (preserving the distinction points vs lines of the geometry, i.e. the colours in the above graph). One also recovers the corresponding BN-pair: pick any chamber (any edge) and any apartment containing it (a hexagon with alternating colours); then B consists of the subgroup of automorphisms that fix the chosen chamber, and N consists of the subgroup of automorphism preserving the chosen apartment. Obviously, one could use Chevalley's construction to get a BN-pair and then get back the corresponding building, but I'm really looking for something that starts with the building and gets back the corresponding group. Are there similar descriptions for the exceptional groups and their corresponding buildings? (I'm aware that such neat pictorial representations aren't really going to be possible, given that already $G_2(3)$ has order of several million.) (Crossposted from Math Stackexchange.) - the link to the MS post is broken – José Figueroa-O'Farrill Dec 8 2011 at 2:00 Sorry; fixed now. – Will Dec 8 2011 at 2:04 The lines leading up to the question here are not quite correct as written. Chevalley's Tohoku paper already gave a uniform way to study simple (adjoint type) algebraic groups of all types over an arbitrary field (not just a finite field). I think the study of the groups in terms of their actions on buildings evolved from his earlier interest in geometries arising from group structure and how the groups act on them. This was exploited for instance in classifying simple groups. Soon the buildings took on a life of their own, especially over local fields. – Jim Humphreys Dec 19 2011 at 0:24 2 Answers In the case of groups of rank 2, such as your examples $\mathrm{SL}_3(\mathbb{F}_2)$ or $\mathsf{G}_2(3)$, the building is rather easy to describe (either as an incidence geometry or as a bipartite graph); it is a so-called generalized polygon. Spherical buildings of higher rank are completely determined by their rank $2$ residues, but perhaps that point of view is not explicit enough for your purposes. There exist several more "geometric" characterizations of the exceptional buildings in the literature, notably by people like Arjeh Cohen and Bruce Cooperstein; see for instance here. The split buildings of type $\mathsf{G}_2$, $\mathsf{F}_4$ and $\mathsf{E}_6$ can also be described in terms of algebraic structures (but then the corresponding groups can also more easily be described in terms of this structure). For instance, the split groups of type $\mathsf{F}_4$ are precisely the automorphism groups of split Albert algebras (these are certain $27$-dimensional non-associative algebras); the corresponding building can be described in terms of isotropic subspaces of this Albert algebra. A similar description is possible for $\mathsf{E}_6$, but is much more difficult for $\mathsf{E}_7$ and to the best of my knowledge unknown for $\mathsf{E}_8$. - 1 I knew you were going to answer, Tom! (:-) – Alain Valette Dec 8 2011 at 10:57 @Alain: Indeed, I felt obligated to answer this question ;-) – Tom De Medts Dec 8 2011 at 14:57 You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. For spherical buildings of type $F_4$, a complete description appears in J. Tits, Buildings of spherical type and finite BN-pairs'', Springer LNM 386. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9481868147850037, "perplexity_flag": "head"}
http://mathoverflow.net/questions/tagged/young-tableaux	## Tagged Questions 0answers 70 views ### Expression of basis vectors of permutation modules in different bases. This is a cross-post from math.se, because I did not get any answer there: Write $[n]:=\{1,\ldots,n\}$. For a partition $\lambda\vdash n$, I will write $[\lambda]$ for the Specht … 2answers 172 views ### How exactly does Schützenberger promotion relate to Striker-Williams promotion? Schützenberger promotion, studied (for example) in Richard Stanley, Promotion and Evacuation, 2009, is a permutation of the set of all linear extensions of a finite poset. Since on … 1answer 279 views ### Generating Random Young Tableaux: A peculiar probability identity In the paper by Greene, Nijenhuis and Wilf, an algorithm is proposed for generating uniformly random Young tableaux of shape $\lambda$. The algorithm is to uniformly randomly pick … 3answers 396 views ### bijection between number of partitions of 2n satisfying certain conditions with number of partitions of n Suppose $\lambda = (\lambda_1,\lambda_2,.....,\lambda_k)$ is a partition of $2n$ where $n \in \mathbb N$ satisfying the following conditions: (1) $\lambda_{k} = 1$. (2) \$\lambda_ … 1answer 107 views ### Representations of Sym(n) and SL_d Irreducible representations of the symmetric group Sym$(n)$, and degree-$n$ algebraic representations of SL$_d(\mathbb C)$ for $d\ge n$, can both be classified by Young diagrams wi … 0answers 154 views ### Explicit description of isomorphism when decomposing into irreps I had a question which is slightly more general than this one on mathoverflow: I am looking for an explicit description of the isomorphism \$\mathbb S_\nu(V\otimes W) \cong \bigoplu … 0answers 110 views ### An inequality for the ratio of standard Young tableau with {1,2,…,k} in the first row For a partition $\lambda \vdash n$, define $\dim \lambda$ to be the number of standard Young tableaux of shape $\lambda$, and $\dim \lambda/(k)$ as the number of standard Young tab … 2answers 224 views ### Viennot-type geometric description for dual RSK correspondence? Is a geometric construction of the dual RSK correspondence along the lines of Viennot's "light and shadows construction" written up somewhere? This is a bijective correspondence be … 1answer 387 views ### Analogues of the Knuth and Forgotten equivalences on permutations: have they been studied? Consider a totally ordered alphabet $A$ of $n$ letters. Let $W$ be the set of all words over $A$ which have no two letters equal. Then, for example, we can define the Knuth equival … 3answers 387 views ### RS to RSK correspondence The RS correspondence is a correspondence which associates to each permutation a pair of standard Young tableaux of the same shape. The RSK correspondence associates to each integ … 1answer 399 views ### Bruhat order and the Robinson-Schensted correspondence The Robinson-Schensted correspondence is a bijection between elements of the symmetric group $S_n$ and pairs of standard tableaux of the same shape. The symmetric group is partial … 1answer 290 views ### Categorifying the equality of product and coproduct of symmetric functions Littlewood-Richardson coefficients are both multiplicities of $GL_n$ tensor products, and of restrictions of $GL_{m+n}$ representations to $GL_m \times GL_n$. I want to turn this e … 1answer 146 views ### Box-dual of a partition - what is it called? Fix natural numbers $n,m\in\mathbb{N}$. Given a partition $\lambda\vdash d$ with at most $n$ rows (and at most $m$ columns), we can define a partition \$\lambda^\ast=(\lambda^\ast_1 … 0answers 202 views ### The number of rows in a tableau generated by the RSK algorithm. It is well known that the number of rows in the semistandard Young tableaux correspondent to a two-line array via RSK is equal to the length of the longest (strictly) decreasing su … 1answer 156 views ### Dimension of spaces of invariants/tableaux functions The Hook lenght formula gives the number of standard Young tableaux on a given diagram. A variant gives the number of semistandard tableuax. Does there exist a formula for coun … 15 30 50 per page	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 28, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8378322720527649, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/135916/what-am-i-describing-discrete-continuity	# What am I describing - discrete continuity? What area of math best describes continuous, but discrete changes in an object? For example. Example. If the start vector $(1,1,1)$ changed from itself to $(1,4,2)$ in one step, then since the 2nd component jumped up by 3, that change is not continuous, where as if it went to $(1,0,2)$ it is continuous. Now imagine if there was some final vector $(x,y,z)$. Then is there a chain of continuous changes from the start to final? In this case, yes, always. However for more exotic objects there isn't always, especially when we add the constraint that a function of the changing object must satisfy some criterion. So for such objects, knowing whether there is a path of small discrete changes from the object to a final object is useful, and since we've limited the possible changes from one object, we have less potential changes to look at than if we were allowed to jump anywhere. The objects I was thinking of were algorithms. - Given that your objects are algorithms, what are the (discrete) changes? – Quinn Culver Apr 23 '12 at 18:35 On second thought, applying this to algorithms directly would not work so well. – Enjoys Math Apr 27 '12 at 16:41 ## 1 Answer You are using what is sometimes called "The Discrete Intermediate Value Theorem." If you type $$\rm discrete\ intermediate\ value$$ into Google I think you'll find some useful links appear. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.965757429599762, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/97275/list	## Return to Question 2 Edited title: Szőkefalvi-Nagy is a joint family name, it is not appropriate to abbreviate. 1 # Sz.-Nagy's unitarizability theorem in the Calkin algebra? Here's a research problem, which I think interesting. Suppose that $t$ is an invertible element in the Calkin algebra $\mathcal{Q} = \mathcal{B}(\ell_2)/\mathcal{K}(\ell_2)$ which satisfies $\sup_{n \in \mathbb{Z}} \\|t^n\\|<+\infty$. Does it follow $t$ is similar to a unitary element, i.e., is there an invertible element $s \in \mathcal{Q}$ such that $s t s^{-1}$ is unitary? Sz.-Nagy has proved that it is the case when one replaces $\mathcal{Q}$ with $\mathcal{B}(\ell_2)$ (or any other von Neumann algebra, e.g., $\mathcal{Q}^{\ast\ast}$). I'm asking this question, because a counterexample (which I believe exists) would provide an amenable operator algebra which is not isomorphic to a $\mathrm{C}^\ast$-algebra---The existence of such an example is an open problem. I will explain why a counterexample provides such an example. Let $\pi\colon \mathcal{B}(\ell_2)\to \mathcal{Q}$ be the quotient map, $G$ an abelian group (which is $\mathbb{Z}$ in the above question) and $u\colon G\to \mathcal{Q}$ a uniformly bounded homomorphism. Then, the operator algebra $\mathcal{A} := \pi^{-1}( \overline{\mathrm{span}}\ u(G) )$ is amenable. If $\mathcal{A}$ is isomorphic to a $\mathrm{C}^\ast$-algebra, then by the solution of the similarity problem for amenable $\mathrm{C}^\ast$-algebras, there is $S \in \mathcal{B}(\ell_2)$ such that $S \mathcal{A} S^{-1}$ is a $\mathrm{C}^\ast$-subalgebra. Thus $s \pi(\mathcal{A}) s^{-1}$ (where $s=\pi(S)$) is an abelian $\mathrm{C}^\ast$-subalgebra of $\mathcal{Q}$ and hence it consists of normal elements. Since $u$ is uniformly bounded, $s u(\cdot)s^{-1}$ is a unitary homomorphism. (The converse is also true: if $u$ is similar to a unitary homomorphism, then $\mathcal{A}$ is isomorphic to a $\mathrm{C}^\ast$-algebra.) The obvious thing one should try is to see whether $H^1_b(G,\mathcal{Q}(\ell_2G))\neq0$. For a starter, I looked at $H^1_b(G,\ell_\infty(G)/c_0(G))$, but it was zero for every countable exact group $G$. (Whether it is zero for every group $G$ is unclear.)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9436177015304565, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/P-value	# p-value In statistical significance testing the p-value is the probability of obtaining a test statistic at least as extreme as the one that was actually observed, assuming that the null hypothesis is true.[1] One often "rejects the null hypothesis" when the p-value is less than the predetermined significance level which is often 0.05[2][3] or 0.01, indicating that the observed result would be highly unlikely under the null hypothesis. Many common statistical tests, such as chi-squared tests or Student's t-test, produce test statistics which can be interpreted using p-values. The p-value is a key concept in the approach of Ronald Fisher, where he uses it to measure the weight of the data against a specified hypothesis, and as a guideline to ignore data that does not reach a specified significance level. Fisher's approach does not involve any alternative hypothesis, which is instead the Neyman–Pearson approach. The p-value should not be confused with the Type I error rate (false positive rate) α in the Neyman–Pearson approach – though α is also called a "significance level" and is often 0.05, these terms have different meanings, these are incompatible approaches, and the numbers p and α cannot meaningfully be compared. There is a great deal of confusion and misunderstanding on this point, and many misinterpretations, discussed below.[4] Fundamentally, the p-value does not in itself allow reasoning about the probabilities of hypotheses (this requires a prior, as in Bayesian statistics), nor choosing between different hypotheses (this is instead done in Neyman–Pearson statistical hypothesis testing) – it is simply a measure of how likely the data is to have occurred by chance, assuming the null hypothesis is true. Despite the above caveats, statistical hypothesis tests making use of p-values are commonly used in many fields of science and social sciences, such as economics, psychology,[5] biology, criminal justice and criminology, and sociology,[6] though this is criticized (see below). ## Examples Computing a p-value requires a null hypothesis, a test statistic (together with deciding if one is doing one-tailed test or a two-tailed test), and data. A few simple examples follow, each illustrating a potential pitfall. One roll of a pair of dice Rolling a pair of dice once, assuming a null hypothesis of fair dice, the test statistic of "total value of numbers rolled" (one-tailed), and with data of both dice showing 6 (so a test statistic of 12, the total) yields a p-value of 1/36, or about 0.028 (most extreme value out of 6×6 = 36 possible outcomes). At the 0.05 significance level, one rejects the hypothesis that the dice are fair (not loaded towards 6). This illustrates the danger with blindly applying p-value without considering experiment design – a single roll of a pair of dice is a very weak basis (insufficient data) to draw any meaningful conclusion. Five heads in a row Flipping a coin five times, assuming a null hypothesis of a fair coin, a test statistic of "total number of heads" (one-tailed or two-tailed), and with data of all heads (HHHHH) yields a test statistic of 5. In a one-tailed test, this is the unique most extreme value (out of 32 possible outcomes), and yields a p-value of 1/25 = 1/32 ≈ 0.03, which is significant at the 0.05 level. In a two-tailed test, all tails (TTTTT) is as extreme, and thus the data of HHHHH yields a p-value of 2/25 = 1/16 ≈ 0.06, which is not significant at the 0.05 level. These correspond respectively to testing if the coin is biased towards heads, or if the coin is biased either way. This demonstrates that specifying a direction (on a symmetric test statistic) halves the p-value (increases the significance) and can mean the difference between data being considered significant or not. Sample size dependence Flipping a coin n times, assuming a null hypothesis of a fair coin, a test statistic of "total number of heads" (two-tailed), and with data of all heads yields a test statistic of n and a p-value of 2/2n = 2−(n−1). If one has a two-headed coin and flips the coin 5 times (obtaining heads each time, as it is two-headed), the p-value is 0.0625 > 0.05, but if one flips the coin 10 times (obtaining heads each time), the p-value is ≈ 0.002 < 0.05. In both cases the data suggest that the null hypothesis is false, but changing the sample size changes the p-value and significance level. In the first case the sample size is not large enough to allow the null hypothesis to be rejected at the 0.05 level (in fact, in this example the p-value cannot be below 0.05 given a sample size of 5). In cases when a large sample size produces a significant result, a smaller sample size may produce a result that is not significant, simply because the sample size is too small to detect the effect. This demonstrates that in interpreting p-values, one must also know the sample size, which complicates the analysis. Alternating coin flips Flipping a coin ten times, assuming a null hypothesis of a fair coin, a test statistic of "total number of heads" (two-tailed), and with data of alternating heads/tails (HTHTHTHTHT) yields a test statistic of 5 and a p-value of 1 (completely unexceptional), as this is exactly the expected number of heads. However, using the subtler test statistic of "number of alternations" (times when H is followed by T or T is followed by H), again two-tailed, yields a test statistic of 9, which is extreme, and has a p-value of $1/2^8 = 1/256 \approx 0.0039,$ which is extremely significant. The expected number of alternations is 4.5 (there are 9 gaps, and each has a 0.5 chance of being an alternation), the values as extreme as this are 0 and 9, and there are only 4 sequences (out of 1024 possible outcomes) this extreme: all heads, all tails, alternating starting from heads (this case), or alternating starting from tails. This data indicates that, in terms of this test statistic, the data set is extremely unlikely to have occurred by chance, though it does not suggest that the coin is biased towards heads or tails. There is no "alternative hypothesis", only rejection of the null hypothesis, and such data could have many causes – the data may instead be forged, or the coin flipped by a magician who intentionally alternated outcomes. This example demonstrates that the p-value depends completely on the test statistic used, and illustrates that p-values are about rejecting a null hypothesis, not about considering other hypotheses. Impossible outcome and very unlikely outcome Flipping a coin two times, assuming a null hypothesis of a two-headed coin, a test statistic of "total number of heads" (one-tailed), and with data of one head and one tail (HT) yields a test statistic of 1, and a p-value of 0. In this case the data is inconsistent with the hypothesis – for a two-headed coin, a tail can never come up. In this case the outcome is not simply unlikely in the null hypothesis, but in fact impossible, and the null hypothesis can be definitely rejected as false. In practice such experiments almost never occur, as all data that could be observed would be possible in the null hypothesis, albeit unlikely. If the null hypothesis were instead that the coin came up heads 99% of the time (otherwise the same setup), the p-value would instead be[a] $0.0199 \approx 0.02.$ In this case the null hypothesis could not definitely be ruled out – this outcome is unlikely in the null hypothesis, but not impossible – but the null hypothesis would be rejected at the 0.05 level, and in fact at the 0.02 level, since the outcome is less than 2% likely in the null hypothesis. ### Coin flipping Main article: Checking whether a coin is fair As an example of a statistical test, an experiment is performed to determine whether a coin flip is fair (equal chance of landing heads or tails) or unfairly biased (one outcome being more likely than the other). Suppose that the experimental results show the coin turning up heads 14 times out of 20 total flips. The null hypothesis is that the coin is fair, so the p-value of this result is the chance of a fair coin landing on heads at least 14 times out of 20 flips. This probability can be computed from binomial coefficients as $\begin{align} & \operatorname{Prob}(14\text{ heads}) + \operatorname{Prob}(15\text{ heads}) + \cdots + \operatorname{Prob}(20\text{ heads}) \\ & = \frac{1}{2^{20}} \left[ \binom{20}{14} + \binom{20}{15} + \cdots + \binom{20}{20} \right] = \frac{60,\!460}{1,\!048,\!576} \approx 0.058 \end{align}$ This probability is the p-value, considering only extreme results which favor heads. This is called a one-tailed test. However, the deviation can be in either direction, favoring either heads or tails. We may instead calculate the two-tailed p-value, which considers deviations favoring either heads or tails. As the binomial distribution is symmetrical for a fair coin, the two-sided p-value is simply twice the above calculated single-sided p-value; i.e., the two-sided p-value is 0.115. Example of a p-value computation. The vertical coordinate is the probability density of each outcome, computed under the null hypothesis. The p-value is the area under the curve past the observed data point. In the above example we thus have: • Null hypothesis (H0): The coin is fair; Prob(heads) = 0.5 • Observation O: 14 heads out of 20 flips; and • p-value of observation O given H0 = Prob(≥ 14 heads or ≥ 14 tails) = 2*(1-Prob(< 14)) = 0.115. The calculated p-value exceeds 0.05, so the observation is consistent with the null hypothesis, as it falls within the range of what would happen 95% of the time were the coin in fact fair. Hence, we fail to reject the null hypothesis at the 5% level. Although the coin did not fall evenly, the deviation from expected outcome is small enough to be consistent with chance. However, had one more head been obtained, the resulting p-value (two-tailed) would have been 0.0414 (4.14%). This time the null hypothesis – that the observed result of 15 heads out of 20 flips can be ascribed to chance alone – is rejected when using a 5% cut-off. ## Definition In brief, the (left-tailed) p-value is the quantile of the value of the test statistic, with respect to the sampling distribution under the null hypothesis. The right-tailed p-value is one minus the quantile, while the two-tailed p-value is twice whichever of these is smaller. This is elaborated below. Computing a p-value requires a null hypothesis, a test statistic (together with deciding if one is doing one-tailed test or a two-tailed test), and data. The key preparatory computation is computing the cumulative distribution function (CDF) of the sampling distribution of the test statistic under the null hypothesis; this may depend on parameters in the null distribution and the number of samples in the data. The test statistic is then computed for the actual data, and then its quantile computed by inputting it into the CDF. This is then normalized as follows: • one-tailed (left tail): quantile, value of cumulative distribution function (since values close to 0 are extreme); • one-tailed (right tail): one minus quantile, value of complementary cumulative distribution function (since values close to 1 are extreme: 0.95 becomes 0.05); • two-tailed: twice p-value of one-tailed, for whichever side value is on (since values close to 0 or 1 are both extreme: 0.05 and 0.95 both have a p-value of 0.10, as one adds the tails on both sides). Even though computing the test statistic on given data may be easy, computing the sampling distribution under the null hypothesis, and then computing its CDF is often a difficult computation. Today this computation is done using statistical software, often via numeric methods (rather than exact formulas), while in the early and mid 20th century, this was instead done via tables of values, and one interpolated or extrapolated p-values from these discrete values. Rather than using a table of p-values, Fisher instead inverted the CDF, publishing a list of values of the test statistic for given fixed p-values; this corresponds to computing the quantile function (inverse CDF). ## Interpretation Hypothesis tests, such as Student's t-test, typically produce test statistics whose sampling distributions under the null hypothesis are known. For instance, in the above coin-flipping example, the test statistic is the number of heads produced; this number follows a known binomial distribution if the coin is fair, and so the probability of any particular combination of heads and tails can be computed. To compute a p-value from the test statistic, one must simply sum (or integrate over) the probabilities of more extreme events occurring. For commonly used statistical tests, test statistics and their corresponding p-values are often tabulated in textbooks and reference works. Traditionally, following Fisher, one rejects the null hypothesis if the p-value is less than or equal to a specified significance level,[1] often 0.05,[3] or more stringent values, such as 0.02 or 0.01. These numbers should not be confused with the Type I error rate α in Neyman–Pearson-style statistical hypothesis testing; see misunderstandings, below. A significance level of 0.05 would deem extraordinary any result that is within the most extreme 5% of all possible results under the null hypothesis. In this case a p-value less than 0.05 would result in the rejection of the null hypothesis at the 5% (significance) level. ## History While the modern use of p-values was popularized by Fisher in the 1920s, computations of p-values date back to the 1770s, where they were calculated by Pierre-Simon Laplace:[7] In the 1770s Laplace considered the statistics of almost half a million births. The statistics showed an excess of boys compared to girls. He concluded by calculation of a p-value that the excess was a real, but unexplained, effect. The p-value was first formally introduced by Karl Pearson in his Pearson's chi-squared test,[8] using the chi-squared distribution and notated as capital P.[8] The p-values for the chi-squared distribution (for various values of χ2 and degrees of freedom), now notated as P, was calculated in (Elderton 1902), collected in (Pearson 1914, pp. xxxi–xxxiii, 26–28, Table XII) The use of the p-value in statistics was popularized by Ronald Fisher,[9] and it plays a central role in Fisher's approach to statistics.[10] In the influential book Statistical Methods for Research Workers (1925), Fisher proposes the level p = 0.05, or a 1 in 20 chance of being exceeded by chance, as a limit for statistical significance, and applies this to a normal distribution (as a two-tailed test), thus yielding the rule of two standard deviations (on a normal distribution) for statistical significance – see 68–95–99.7 rule.[11][b][3] He then computes a table of values, similar to Elderton, but, importantly, reverses the roles of χ2 and p. That is, rather than computing p for different values of χ2 (and degrees of freedom n), he computes values of χ2 that yield specified p-values, specifically 0.99, 0.98, 0.95, 0,90, 0.80, 0.70, 0.50, 0.30, 0.20, 0.10, 0.05, 0.02, and 0.01.[12] This allowed computed values of χ2 to be compared against cutoffs, and encouraged the use of p-values (especially 0.05, 0.02, and 0.01) as cutoffs, instead of computing and reporting p-values themselves. The same type of tables were then compiled in (Fisher & Yates 1938), which cemented the approach.[3] As an illustration of the application of p-values to the design and interpretation of experiments, in his following book The Design of Experiments (1935), Fisher presented the lady tasting tea experiment,[13] which is the archetypal example of the p-value. To evaluate a lady's claim that she (Muriel Bristol) could distinguish by taste how tea is prepared (first adding the milk to the cup, then the tea, or first tea, then milk), she was sequentially presented with 8 cups: 4 prepared one way, 4 prepared the other, and asked to determine the preparation of each cup (knowing that there were 4 of each). In this case the null hypothesis was that she had no special ability, the test was Fisher's exact test, and the p-value was $1/\binom{8}{4} = 1/70 \approx 0.014,$ so Fisher was willing to reject the null hypothesis (consider the outcome highly unlikely to be due to chance) if all were classified correctly. (In the actual experiment, Bristol correctly classified all 8 cups.) Fisher reiterated the p = 0.05 threshold and explained its rationale, stating:[14] It is usual and convenient for experimenters to take 5 per cent. as a standard level of significance, in the sense that they are prepared to ignore all results which fail to reach this standard, and, by this means, to eliminate from further discussion the greater part of the fluctuations which chance causes have introduced into their experimental results. He also applies this threshold to the design of experiments, noting that had only 6 cups been presented (3 of each), a perfect classification would have only yielded a p-value of $1/\binom{6}{3} = 1/20 = 0.05,$ which would not have met this level of significance.[14] Fisher also underlined the frequentist interpretation of p, as the long-run proportion of values at least as extreme as the data, assuming the null hypothesis is true. In later editions, Fisher explicitly contrasted the use of the p-value for statistical inference in science with the Neyman–Pearson method, which he terms "Acceptance Procedures".[15] Fisher emphasizes that while fixed levels such as 5%, 2%, and 1% are convenient, the exact p-value can be used, and the strength of evidence can and will be revised with further experimentation. In contrast, decision procedures require a clear-cut decision, yielding an irreversible action, and the procedure is based on costs of error, which he argues are inapplicable to scientific research. ## Misunderstandings Despite the ubiquity of p-value tests, this particular test for statistical significance has been criticized for its inherent shortcomings and the potential for misinterpretation. The data obtained by comparing the p-value to a significance level will yield one of two results: either the null hypothesis is rejected, or the null hypothesis cannot be rejected at that significance level (which however does not imply that the null hypothesis is true). In Fisher's formulation, there is a disjunction: a low p-value means either that the null hypothesis is true and a highly improbable event has occurred, or that the null hypothesis is false. However, people interpret the p-value in many incorrect ways, and try to draw other conclusions from p-values, which do not follow. The p-value does not in itself allow reasoning about the probabilities of hypotheses; this requires multiple hypotheses or a range of hypotheses, with a prior distribution of likelihoods between them, as in Bayesian statistics, in which case one uses a likelihood function for all possible values of the prior, instead of the p-value for a single null hypothesis. The p-value refers only to a single hypothesis, called the null hypothesis, and does not make reference to or allow conclusions about any other hypotheses, such as the alternative hypothesis in Neyman–Pearson statistical hypothesis testing. In that approach one instead has a decision function between two alternatives, often based on a test statistic, and one computes the rate of Type I and type II errors as α and β. However, the p-value of a test statistic cannot be directly compared to these error rates α and β – instead it is fed into a decision function. There are several common misunderstandings about p-values.[16][17] 1. The p-value is not the probability that the null hypothesis is true, nor is it the probability that the alternative hypothesis is false – it is not connected to either of these. In fact, frequentist statistics does not, and cannot, attach probabilities to hypotheses. Comparison of Bayesian and classical approaches shows that a p-value can be very close to zero while the posterior probability of the null is very close to unity (if there is no alternative hypothesis with a large enough a priori probability and which would explain the results more easily). This is Lindley's paradox. But there are also a priori probability distributions where the posterior probability and the p-value have similar or equal values.[18] 2. The p-value is not the probability that a finding is "merely a fluke." As the calculation of a p-value is based on the assumption that a finding is the product of chance alone, it patently cannot also be used to gauge the probability of that assumption being true. This is different from the real meaning which is that the p-value is the chance of obtaining such results if the null hypothesis is true. 3. The p-value is not the probability of falsely rejecting the null hypothesis. This error is a version of the so-called prosecutor's fallacy. 4. The p-value is not the probability that a replicating experiment would not yield the same conclusion. Quantifying the replicability of an experiment was attempted through the concept of p-rep (which is heavily criticized) 5. The significance level, such as 0.05, is not determined by the p-value. Rather, the significance level is decided before the data are viewed, and is compared against the p-value, which is calculated after the test has been performed. (However, reporting a p-value is more useful than simply saying that the results were or were not significant at a given level, and allows readers to decide for themselves whether to consider the results significant.) 6. The p-value does not indicate the size or importance of the observed effect (compare with effect size). The two do vary together however – the larger the effect, the smaller sample size will be required to get a significant p-value. ## Criticisms Main article: Statistical hypothesis testing#Controversy Critics of p-values point out that the criterion used to decide "statistical significance" is based on an arbitrary choice of level (often set at 0.05).[19] If significance testing is applied to hypotheses that are known to be false in advance, a non-significant result will simply reflect an insufficient sample size; a p-value depends only on the information obtained from a given experiment. The p-value is incompatible with the likelihood principle, and p-value depends on the experiment design, or equivalently on the test statistic in question. That is, the definition of "more extreme" data depends on the sampling methodology adopted by the investigator;[20] for example, the situation in which the investigator flips the coin 100 times yielding 50 heads has a set of extreme data that is different from the situation in which the investigator continues to flip the coin until 50 heads are achieved yielding 100 flips.[21] This is to be expected, as the experiments are different experiments, and the sample spaces and the probability distributions for the outcomes are different even though the observed data (50 heads out of 100 flips) are the same for the two experiments. Some regard the p-value as the main result of statistical significance testing, rather than the acceptance or rejection of the null hypothesis at a pre-prescribed significance level. Fisher proposed p as an informal measure of evidence against the null hypothesis. He called on researchers to combine p in the mind with other types of evidence for and against that hypothesis, such as the a priori plausibility of the hypothesis and the relative strengths of results from previous studies.[4] Many misunderstandings concerning p arise because statistics classes and instructional materials ignore or at least do not emphasize the role of prior evidence in interpreting p. A renewed emphasis on prior evidence could encourage researchers to place p in the proper context, evaluating a hypothesis by weighing p together with all the other evidence about the hypothesis.[1] ## Related quantities A closely related concept is the E-value,[22] which is the average number of times in multiple testing that one expects to obtain a test statistic at least as extreme as the one that was actually observed, assuming that the null hypothesis is true. The E-value is the product of the number of tests and the p-value. The 'inflated' (or adjusted) p-value,[23] is when a group of p-values are changed according to some multiple comparisons procedure so that each of the adjusted p-values can now be compared to the same threshold level of significance (α), while keeping the type I error controlled. The control is in the sense that the specific procedures controls it, it might be controlling the familywise error rate, the false discovery rate, or some other error rate. ## Notes 1. Odds of TT is $(0.01)^2,$ odds of HT and TH are $0.99 \times 0.01$ and $0.01 \times 0.99,$ which are equal, and adding these yield $0.01^2 + 2\times 0.01 \times 0.99 = 0.0199$ 2. To be precise the p = 0.05 corresponds to about 1.96 standard deviations for a normal distribution (two-tailed test), and 2 standard deviations corresponds to about a 1 in 22 chance of being exceeded by chance, or p ≈ 0.045; Fisher notes these approximations. ## References 1. ^ a b c Goodman, SN (1999). "Toward Evidence-Based Medical Statistics. 1: The P Value Fallacy.". Annals of Internal Medicine 130: 995–1004. 2. ^ a b c d Dallal 2012, Note 31: Why P=0.05?. 3. ^ a b 4. Wetzels, R.; Matzke, D.; Lee, M. D.; Rouder, J. N.; Iverson, G. J.; Wagenmakers, E. -J. (2011). "Statistical Evidence in Experimental Psychology: An Empirical Comparison Using 855 t Tests". Perspectives on Psychological Science 6 (3): 291. doi:10.1177/1745691611406923. 5. Babbie, E. (2007). The practice of social research 11th ed. Thomson Wadsworth: Belmont, CA. 6. ^ a b 7. Fisher 1925, p. 47, Chapter III. Distributions. 8. ^ a b 9. Sterne, J. A. C.; Smith, G. Davey (2001). "Sifting the evidence–what's wrong with significance tests?". BMJ (Clinical research ed.) 322 (7280): 226–231. doi:10.1136/bmj.322.7280.226. PMC 1119478. PMID 11159626. 10. Schervish, M. J. (1996). "P Values: What They Are and What They Are Not". 50 (3). doi:10.2307/2684655. JSTOR 2684655. 11. Casella, George; Berger, Roger L. (1987). "Reconciling Bayesian and Frequentist Evidence in the One-Sided Testing Problem". Journal of the American Statistical Association 82 (397): 106–111. 12. Sellke, Thomas; Bayarri, M. J.; Berger, James O. (2001). "Calibration of p Values for Testing Precise Null Hypotheses". 55 (1): 62–71. doi:10.1198/000313001300339950. JSTOR 2685531. 13. Casson, R. J. (2011). "The pesty P value". Clinical & Experimental Ophthalmology 39 (9): 849–850. doi:10.1111/j.1442-9071.2011.02707.x. 14. Johnson, D. H. (1999). "The Insignificance of Statistical Significance Testing". Journal of Wildlife Management 63 (3): 763–772. doi:10.2307/3802789. 15. Hochberg, Y.; Benjamini, Y. (1990). "More powerful procedures for multiple significance testing". Statistics in Medicine 9 (7): 811–818. doi:10.1002/sim.4780090710. PMID 2218183. (page 815, second paragraph) • • Elderton, William Palin (1902). "Tables for Testing the Goodness of Fit of Theory to Observation". Biometrika 1 (2): 155–163. doi:10.1093/biomet/1.2.155. • Fisher, Ronald (1925). . Edinburgh: Oliver & Boyd. ISBN 0-05-002170-2. • Fisher, Ronald A. (1971) [1935]. The Design of Experiments (9th ed.). Macmillan. ISBN 0-02-844690-9. • Fisher, R. A.; Yates, F. (1938). Statistical tables for biological, agricultural and medical research. London. • Stigler, Stephen M. (1986). The history of statistics : the measurement of uncertainty before 1900. Cambridge, Mass: Belknap Press of Harvard University Press. ISBN 0-674-40340-1. • Hubbard, Raymond; Bayarri, M. J. (November 2003), P Values are not Error Probabilities, a working paper that explains the difference between Fisher's evidential p-value and the Neyman–Pearson Type I error rate α. • Hubbard, Raymond; Armstrong, J. Scott (2006). "Why We Don't Really Know What Statistical Significance Means: Implications for Educators". Journal of Marketing Education 28 (2): 114. doi:10.1177/0273475306288399. • Hubbard, Raymond; Lindsay, R. Murray (2008). "Why P Values Are Not a Useful Measure of Evidence in Statistical Significance Testing". Theory & Psychology 18 (1): 69–88. doi:10.1177/0959354307086923. • Stigler, Stephen (December 2008). "Fisher and the 5% level". Chance 21 (4): 12. doi:10.1007/s00144-008-0033-3. • ## Further reading • 12 Misconceptions, good overview given in following Article	{"extraction_info": {"found_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": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9073182940483093, "perplexity_flag": "middle"}
http://nrich.maths.org/1944	### LOGO Challenge 5 - Patch Using LOGO, can you construct elegant procedures that will draw this family of 'floor coverings'? ### LOGO Challenge - Triangles-squares-stars Can you recreate these designs? What are the basic units? What movement is required between each unit? Some elegant use of procedures will help - variables not essential. ### LOGO Challenge - Tilings Three examples of particular tilings of the plane, namely those where - NOT all corners of the tile are vertices of the tiling. You might like to produce an elegant program to replicate one or all of these. # Napoleon's Theorem ##### Stage: 4 and 5 Challenge Level: Triangle $ABC$ has equilateral triangles drawn on its edges. Points $P$, $Q$ and $R$ are the centres of the equilateral triangles. You can change triangle $ABC$ below by dragging the vertices and observe what happens to triangle $PQR$. What can you prove about the triangle $PQR$? NOTES AND BACKGROUND There are many ways of proving this result. One way you might like to try involves tessellation. (1) Draw any triangle, with angles $A, B$ and $C$ say. (2) Draw equilateral triangles $T_1, T_2$ and $T_3$ on the three sides of $\Delta ABC$. (3) Fit copies of the original triangle and $T_1, T_2$ and $T_3$ into a tessellation pattern so that, at each vertex of the tessellation, the angles are $A, B$ and $C$ and three angles of $60^o$ making an angle sum of $360^o$. (4) Napoleon's Theorem can be proved by simple geometry using a small part of this pattern without even assuming that this tessellation extends indefinitely in all directions, which is intuitively obvious but requires advanced mathematics to prove it. This text is usually replaced by the Flash movie. Van Aubel's Theorem is related to Napoleon's Theorem. Van Aubel's Theorem states that if four squares are drawn on the edges of any quadrilateral then the lines joining the centres of the squares on opposite edges are equal in length and perpendicular. For an animated proof of Van Aubel's Theorem see http://agutie.homestead.com/files/vanaubel.html The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found 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": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9006264209747314, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=4165440	Physics Forums Thread Closed Page 10 of 10 « First < 7 8 9 10 ## Light Clock Problem Quote by PeterDonis Yes, that's true, but you did not specify the upward speed of the projectile in the unprimed (observer) frame. You specified it in the primed (clock) frame. (Btw, I mis-stated this somewhat in my previous post; I said that you specified $v_p$, but I should have said that you specified $v'_p$. I can go back and continue the analysis I was doing in my last post with that corrected, but it may not be worth bothering.) No problem, that was an honest mistake. Don't bother correcting it. And I didn't give the upward velocity in the unprimed frame because the goal is to find the speed of the beam in the first place. And we can find it since I gave the distances and times you need to do so. So before you do this vector addition, you have to first transform the upward speed of the projectile from the primed to the unprimed frame. You don't need to do this at all, since we are calculating speeds from given distances and times, not from different speed vectors. The upward distance (AB in the primed frame; BC in the unprimed frame, since the clock is moving in that frame) does not change when you change frames, but the time does, because of time dilation Once again, the distance does change between frames. The distance marks are clear and visible for both frames. From the clock's measurements, AB = CB = y'. For the distant observer, event B1 occurs h' lightseconds away from A. This is direct visual data. Also, time dilation effects in this setup occur because of the constant speed of the signal from B to A, not the speed of the beam - and that is the unknown we are seeking. so the upward speed of the projectile (i.e., the upward component of its velocity) is different in the unprimed frame than in the primed frame. So you do need to know the relative velocity of the light clock and the observer; without that you can't transform the upward velocity in the primed frame to the upward velocity component in the unprimed frame. But you don't have to transform upward components to find the observed speed of the beam! All you have to do is plot the given distance over the observed time, and since we know the time of the event B as seen from the light clock's frame, we can add the time the signal takes to reach A from B to find the observed time of event B as seen from A. You don't need to separate the motion into vector components at all, you have the time and distance of the event at B, you can calculate the observed speed straight from that. Yes, let's do that. We have a light beam traveling from A to B, and a second light beam (the one that is emitted at the instant the first one strikes the mirror) traveling from B to A. The round-trip travel time is measured by the observer at A, and he already knows the distance AB because he measured it beforehand (and then controlled the speed of the light clock to ensure that the mirror was just passing B at the instant the first beam hit it). So we have two light beams each covering the same distance; if we assume that both beams travel at the same speed in the unprimed frame (even if we don't assume that that speed is c), then we can simply divide the round-trip time by the round-trip distance (2 * AB) to get the beam speed. Fine. See below for further comment. But you can't assume both beams travel at the same speed! That's the point of the setup. You have to SHOW that. Otherwise you are assuming what you are trying to prove. If you are at rest in A relative to B and send a beam towards B, it will take 1.12s to get there, but if you are at A moving along with a mirror 1 ls away and send a beam towards the mirror, it takes 1s to reach it, even if it is coincidentally at B. The bottom line is that the time of event B in the primed frame is 1 and that same event is observed at A when light from that event reaches A, after crossing BA. If the event at B = 1s was self generated (if it wasn't the outcome of any reflection event, like manually turning on a flash of light), how would you find the observed time for event B as seen from A? Only in the sense that we assume that both light beams (the one from A to B and the one from B to A) travel at the same speed. Do you challenge that assumption? Both beams are "observed" in your sense--one endpoint of each beam is directly observed by the observer at A. It's impossible for both endpoints of either beam to be directly observed by the same observer, so if that's your criterion for a beam being "directly observed", then no beam is ever directly observed. But if you accept that receiving a beam counts as directly observing it, then emitting a beam should also count as directly observing it; either one gives the observer direct knowledge of one endpoint of the beam. You can't calculate the speed of light from the time of emission, you need a distance, a time of emission and a time of reception. The reception is the observation, not the emission. And although emission gives you knowledge of the coordinates of one endpoint of a beam (since you can determine the place and time of emission as you please), it is far from enough to determine any speed, any distance travelled and any timr of travel, so we absolutely need the reception coordinates. You see, the observer moving along with the bottom mirror in the clock only knows T'B because it is half the return time (T'D), so you need both the coordinates of emission and reception to determine anything. Now, the observer at A only knows the emission coordinates for the beam (0,0,0). The next piece of information he receives is the light coming from event B, from which he must calculate the coordinates of reception! But how do we know the time it takes for that event to be seen at A? Are you assuming that the beam from B to A travels at c? Yes! If so, then why not also assume that the beam from A to B travels at c? What makes a received beam any different from an emitted beam? Finally the fundamental question! What makes it different is the operation. When we determine the speed if light, we take the distance from the emitter over the time of detection, which is measured locally relative to the point of detection. We are always at rest relative to the point of detection (this is also true for emission). But the detector at B is in relative motion wrt the observer at A, so he is not ar rest relative to the point of detection. This is the fundamental reason, I think, that the speeds are not the same. If you are teavelling along with the clock, you are at rest relative to the point of detection and of emission, so light is always going at c, since all motion is given to the light. If the detector is in relative motion, you are not at rest relative to the point of detection, so you can't give all motion to light. Does this make any sense? Do you ser how this does not violate the light postulate? Light is constant relative to source or detector, but the observer at A is neither source nor detector of the beam going from A to B - the source is the bottom mirror and the detector is the top mirror, and they are both moving relative to A. By contrast, I am only assuming that the two beams (A to B and B to A) travel at the same* speed, without assuming what that speed is (we calculate that by dividing round-trip distance by round-trip time, as above). That seems like a much more reasonable approach, since it does not require assuming that there is any difference between an emitted beam and a received beam. Assuming the speed is the same as seen from A is the problem. You can't make that assumption. You can assume the speed from A to B as seen fron B is the same as the speed from B to A as seen from A, though, but this is not what the setup demands. But only one endpoint of the light is directly detected. Why should received light count as directly detected but not emitted light? See above. No, they are both "observed" (by any reasonable definition of "observed") at the same time, when the beam from B to A is received and its time of reception is observed. At that point the observer knows the round-trip travel time and the round-trip distance and can calculate the beam speed. That is what I meant. I just point out that the roundtrip can't be split in half to determine any speed here. Theory and experiment clearly shows that the speed from B to A must be c, the rest must be given to AB. So must T_BA. The observer doesn't directly observe the emission of the beam from B to A, any more than he directly observes the reception of the beam from A to B. He has to calculate the times of both those events. The way he does that is to use the fact that both events occur at the same instant, by construction. Correct, but he knows the distance AB and he knows that directly observed light must travel at c. T_BA is simply AB/c. This is where the postulate of SR makes the problem possible to solve. Quote by PeterDonis Yes, that's true, but you did not specify the upward speed of the projectile in the unprimed (observer) frame. You specified it in the primed (clock) frame. (Btw, I mis-stated this somewhat in my previous post; I said that you specified $v_p$, but I should have said that you specified $v'_p$. I can go back and continue the analysis I was doing in my last post with that corrected, but it may not be worth bothering.) No problem, that was an honest mistake. Don't bother correcting it. And I didn't give the upward velocity in the unprimed frame because the goal is to find the speed of the beam in the first place. And we can find it since I gave the distances and times you need to do so. So before you do this vector addition, you have to first transform the upward speed of the projectile from the primed to the unprimed frame. You don't need to do this at all, since we are calculating speeds from given distances and times, not from different speed vectors. The upward distance (AB in the primed frame; BC in the unprimed frame, since the clock is moving in that frame) does not change when you change frames, but the time does, because of time dilation Once again, the distance does change between frames. The distance marks are clear and visible for both frames. From the clock's measurements, AB = CB = y'. For the distant observer, event B1 occurs h' lightseconds away from A. This is direct visual data. Also, time dilation effects in this setup occur because of the constant speed of the signal from B to A, not the speed of the beam - and that is the unknown we are seeking. so the upward speed of the projectile (i.e., the upward component of its velocity) is different in the unprimed frame than in the primed frame. So you do need to know the relative velocity of the light clock and the observer; without that you can't transform the upward velocity in the primed frame to the upward velocity component in the unprimed frame. But you don't have to transform upward components to find the observed speed of the beam! All you have to do is plot the given distance over the observed time, and since we know the time of the event B as seen from the light clock's frame, we can add the time the signal takes to reach A from B to find the observed time of event B as seen from A. You don't need to separate the motion into vector components at all, you have the time and distance of the event at B, you can calculate the observed speed straight from that. Yes, let's do that. We have a light beam traveling from A to B, and a second light beam (the one that is emitted at the instant the first one strikes the mirror) traveling from B to A. The round-trip travel time is measured by the observer at A, and he already knows the distance AB because he measured it beforehand (and then controlled the speed of the light clock to ensure that the mirror was just passing B at the instant the first beam hit it). So we have two light beams each covering the same distance; if we assume that both beams travel at the same speed in the unprimed frame (even if we don't assume that that speed is c), then we can simply divide the round-trip time by the round-trip distance (2 * AB) to get the beam speed. Fine. See below for further comment. But you can't assume both beams travel at the same speed! That's the point of the setup. You have to SHOW that. Otherwise you are assuming what you are trying to prove. If you are at rest in A relative to B and send a beam towards B, it will take 1.12s to get there, but if you are at A moving along with a mirror 1 ls away and send a beam towards the mirror, it takes 1s to reach it, even if it is coincidentally at B. The bottom line is that the time of event B in the primed frame is 1 and that same event is observed at A when light from that event reaches A, after crossing BA. If the event at B = 1s was self generated (if it wasn't the outcome of any reflection event, like manually turning on a flash of light), how would you find the observed time for event B as seen from A? Only in the sense that we assume that both light beams (the one from A to B and the one from B to A) travel at the same speed. Do you challenge that assumption? Both beams are "observed" in your sense--one endpoint of each beam is directly observed by the observer at A. It's impossible for both endpoints of either beam to be directly observed by the same observer, so if that's your criterion for a beam being "directly observed", then no beam is ever directly observed. But if you accept that receiving a beam counts as directly observing it, then emitting a beam should also count as directly observing it; either one gives the observer direct knowledge of one endpoint of the beam. You can't calculate the speed of light from the time of emission, you need a distance, a time of emission and a time of reception. The reception is the observation, not the emission. And although emission gives you knowledge of the coordinates of one endpoint of a beam (since you can determine the place and time of emission as you please), it is far from enough to determine any speed, any distance travelled and any timr of travel, so we absolutely need the reception coordinates. You see, the observer moving along with the bottom mirror in the clock only knows T'B because it is half the return time (T'D), so you need both the coordinates of emission and reception to determine anything. Now, the observer at A only knows the emission coordinates for the beam (0,0,0). The next piece of information he receives is the light coming from event B, from which he must calculate the coordinates of reception! But how do we know the time it takes for that event to be seen at A? Are you assuming that the beam from B to A travels at c? Yes! If so, then why not also assume that the beam from A to B travels at c? What makes a received beam any different from an emitted beam? Finally the fundamental question! What makes it different is the operation. When we determine the speed if light, we take the distance from the emitter over the time of detection, which is measured locally relative to the point of detection. We are always at rest relative to the point of detection (this is also true for emission). But the detector at B is in relative motion wrt the observer at A, so he is not ar rest relative to the point of detection. This is the fundamental reason, I think, that the speeds are not the same. If you are teavelling along with the clock, you are at rest relative to the point of detection and of emission, so light is always going at c, since all motion is given to the light. If the detector is in relative motion, you are not at rest relative to the point of detection, so you can't give all motion to light. Does this make any sense? Do you ser how this does not violate the light postulate? Light is constant relative to source or detector, but the observer at A is neither source nor detector of the beam going from A to B - the source is the bottom mirror and the detector is the top mirror, and they are both moving relative to A. By contrast, I am only assuming that the two beams (A to B and B to A) travel at the same* speed, without assuming what that speed is (we calculate that by dividing round-trip distance by round-trip time, as above). That seems like a much more reasonable approach, since it does not require assuming that there is any difference between an emitted beam and a received beam. Assuming the speed is the same as seen from A is the problem. You can't make that assumption. You can assume the speed from A to B as seen fron B is the same as the speed from B to A as seen from A, though, but this is not what the setup demands. But only one endpoint of the light is directly detected. Why should received light count as directly detected but not emitted light? See above. No, they are both "observed" (by any reasonable definition of "observed") at the same time, when the beam from B to A is received and its time of reception is observed. At that point the observer knows the round-trip travel time and the round-trip distance and can calculate the beam speed. That is what I meant. I just point out that the roundtrip can't be split in half to determine any speed here. Theory and experiment clearly shows that the speed from B to A must be c, the rest must be given to AB. So must T_BA. The observer doesn't directly observe the emission of the beam from B to A, any more than he directly observes the reception of the beam from A to B. He has to calculate the times of both those events. The way he does that is to use the fact that both events occur at the same instant, by construction. But he knows the distance AB and he knows that directly observed light must travel at c. T_BA is simply* Unbelievable; you now admit this, yet you were arguing that we could not assume this before. This applies if the observer and the point B are at rest wrt each other. See above. Which it is; the time of event B, in the unprimed frame, is 1.12s (if we allow v, the velocity of the light clock relative to the observer, to be set appropriately to 0.45 instead of 0.5, per the comments of DaleSpam, harrylin, and myself). The time of event B, in the primed frame, is 1s; but that's not what the observer at A is interested in. He's interested in the time of event B in his frame, the unprimed frame, and that time is different from the time of event B in the primed frame because of time dilation. Which, of course, requires you to know the velocity of the light clock relative to the observer, contrary to your repeated erroneous claim that you don't. I think I have answered this above, but just for good measure, notice I said it must be 1s as seen from B, which would be in the primed frame. But we know that the mirror at B must detect the beam at T=1s measuring from its internal clock, and si the event B must be seen by the observer at A at T=T'AB+TBA. Notice that T'AB = CB and TAB is not even measurable from A. TAB is strictly a measurement made in the clock's frame, or more precisely, in the top mirror's frame, since this detection only happens there. The observer at A detects the signal from B to A, and from the given distance and the light speed postulate, he can subtract the time it took light to reach him from event B and find the time of the event in the primed frame. The times are different indeed, but I don't know if I should call the reason "time dilation". Anyway, I'll wait for your follow-up. Mentor Quote by altergnostic Giving 1.12 to the time it takes the beam to travel from A to B as seen from A is a mistake. That would only be true if the time of event B were also 1.12s. If you send a light beam from A to B, B would receive it at t'=1.12s. You are mixing up quantities in different frames. B receives it at t=1.12 (wrt the frame where the clock is moving). B receives it at t'=1 (wrt the frame where the clock is stationary). That is the whole point of the exercise. Quote by altergnostic This thread is way too long already and I think that if we haven't reached an agreement yet, we won't reach it anytime. Yes, that is true. However, since the universe disagrees with your position, I will stick with mine for now. I would rather disagree with you than disagree with the universe. Blog Entries: 9 Recognitions: Gold Member Science Advisor Quote by altergnostic Once again, the distance does change between frames. Only distances parallel to the relative motion are affected by length contraction. The distance A'B' (in the primed frame) and CB (in the unprimed frame) are perpendicular to the relative motion, so they don't change, and A'B' = CB by the geometry of the problem. Quote by altergnostic The distance marks are clear and visible for both frames. From the clock's measurements, AB = CB = y'. You don't measure distance directly from clock measurements. You have to have a speed. You are basically allowing the speed of light beams to be c when you want it to be, but insisting that the speed of other light beams is not c when you don't want it to be. Also, shouldn't this be A'B' = CB? If both of these distances are in the unprimed frame, they are not equal; they can't be, because AB is the hypotenuse of a right triangle and CB is one of its legs. I don't understand what you're doing here. Quote by altergnostic For the distant observer, event B1 occurs h' lightseconds away from A. This is direct visual data. Huh? Who is the "distant observer"? Is he located at B and at rest relative to the observer at A (i.e,. at rest in the unprimed frame)? If so, why are you assigning a distance in the primed frame to what he observes? Quote by altergnostic Also, time dilation effects in this setup occur because of the constant speed of the signal from B to A Time dilation is not a "travel time delay" effect. It is what is left over after you have subtracted out all effects from signal travel time delay. Quote by altergnostic since we know the time of the event B as seen from the light clock's frame, we can add the time the signal takes to reach A from B to find the observed time of event B as seen from A. You can't add times in different frames and get a meaningful answer. How many times does this need to be repeated? Quote by altergnostic You don't need to separate the motion into vector components at all, you have the time and distance of the event at B, you can calculate the observed speed straight from that. In the primed frame, yes. In the unprimed frame, no. Quote by altergnostic But you can't assume both beams travel at the same speed! That's the point of the setup. You have to SHOW that. Otherwise you are assuming what you are trying to prove. This really gets to the issue of what counts as a "directly observed" light beam, since you agree that any directly observed beam does travel at c. I claim that both beams in this case are directly observed; but you want to say that only one is. See below for further comment. Quote by altergnostic If you are at rest in A relative to B and send a beam towards B, it will take 1.12s to get there, but if you are at A moving along with a mirror 1 ls away and send a beam towards the mirror, it takes 1s to reach it, even if it is coincidentally at B. Unbelievable. Now you agree that T_AB (in the unprimed frame) is 1.12s, and T'_AB (in the primed frame) is 1s. But you were arguing before that this is not true. Do you read your own posts? Quote by altergnostic The bottom line is that the time of event B in the primed frame is 1 and that same event is observed at A when light from that event reaches A, after crossing BA. If the event at B = 1s was self generated (if it wasn't the outcome of any reflection event, like manually turning on a flash of light), how would you find the observed time for event B as seen from A? I would find the observed time for event B as seen from A using the standard SR formulas, which I have already done several times in this thread. You don't appear to like that method, so let's try this one: the beam from B to A is directly detected by A; therefore its speed should be c, as seen by A, according to your own claim that any "directly detected" light beam travels at c. A also knows the distance from B to A; it is 1.12 light seconds (for the numbers we are currently using). Therefore the time of event B, according to A, is 1.12s before the time he receives the light beam. As far as I can tell, you agree with the above; but then you want to go on and claim that, if A starts his clock at zero when the first light beam (from A to B) is emitted, he will receive the second light beam (from B to A) at t = 2.12s, because the first light beam will take 1s to travel, according to A. But that beam also covers 1.12 light seconds of distance, according to A, so it covers 1.12 light seconds in 1 second, according to A; so you are claiming that light can travel faster than light. Is this what you're claiming? Quote by altergnostic You can't calculate the speed of light from the time of emission, you need a distance, a time of emission and a time of reception. The reception is the observation, not the emission. In other words, you are claiming that it is impossible to directly observe the emission of light? Quote by altergnostic And although emission gives you knowledge of the coordinates of one endpoint of a beam (since you can determine the place and time of emission as you please), it is far from enough to determine any speed, any distance travelled and any timr of travel, so we absolutely need the reception coordinates. The same would be true if we only had the reception coordinates, wouldn't it? We would need both sets of coordinates, in the same frame, to calculate the speed, right? Oh, wait: Quote by altergnostic You see, the observer moving along with the bottom mirror in the clock only knows T'B because it is half the return time (T'D), so you need both the coordinates of emission and reception to determine anything. And yet you are claiming that the observer at A somehow "directly knows" that the beam he receives from B took 1.12s to get to him, even though he doesn't know the coordinates of emission. In other words, again you are picking and choosing when you can assume light travels at c and when you can't; but yet you also claim that you can't calculate a speed at all unless you know both the emission and the reception coordinates. You seem to me to be contradicting yourself. Quote by altergnostic Now, the observer at A only knows the emission coordinates for the beam (0,0,0). The next piece of information he receives is the light coming from event B, from which he must calculate the coordinates of reception! But you did not state what the coordinates of event A2 (the event where the observer at A receives the light signal from B) were; you have claimed to calculate them by adding 1s to 1.12s. You need to justify this calculation. Alternately, if you change your ground and claim that T_A2 = 2.12s is part of your statement of the problem, you need to show how that is consistent with the other givens. You haven't done any of that. Again, you're picking and choosing numbers to suit your claims, without backing them up. Quote by altergnostic What makes it different is the operation. When we determine the speed if light, we take the distance from the emitter over the time of detection, which is measured locally relative to the point of detection. What does "measured locally relative to the point of detection" mean? Quote by altergnostic We are always at rest relative to the point of detection (this is also true for emission). Then why is emission treated differently from detection? Quote by altergnostic But the detector at B is in relative motion wrt the observer at A, so he is not ar rest relative to the point of detection. This is the fundamental reason, I think, that the speeds are not the same. If you are teavelling along with the clock, you are at rest relative to the point of detection and of emission, so light is always going at c, since all motion is given to the light. If the detector is in relative motion, you are not at rest relative to the point of detection, so you can't give all motion to light. Does this make any sense? For a "projectile" moving slower than light, yes; it amounts to saying that the observed speed of the projectile depends on your motion relative to its source. For light, no; the speed of light is independent of the motion of the source. That is what the null result for the Michelson-Morley experiment means; and that experiment has been repeated with greater and greater accuracy, and the null result continues to hold. If you ran a "Michelson-Morley experiment" using slower-than-light projectiles in the apparatus instead of light, you would not* get a null result; you would observe different speeds for the two projectiles (moving on perpendicular trajectories) if you were in motion relative to the apparatus. Quote by altergnostic I just point out that the roundtrip can't be split in half to determine any speed here. Theory and experiment clearly shows that the speed from B to A must be c, the rest must be given to AB. But you have to determine what "the rest" is, and you are trying to do it by mixing numbers from different frames. Quote by altergnostic I think I have answered this above, but just for good measure, notice I said it must be 1s as seen from B, which would be in the primed frame. Yes. Quote by altergnostic But we know that the mirror at B must detect the beam at T=1s measuring from its internal clock, and si the event B must be seen by the observer at A at T=T'AB+TBA. No; you're adding numbers from different frames, which is meaningless. What you should be doing is T=T_AB + T_BA = gamma (T'_AB) + T_BA, where gamma is the gamma factor corresponding to the light clock velocity relative to the observer. Quote by altergnostic Notice that T'AB = CB and TAB is not even measurable from A. TAB is strictly a measurement made in the clock's frame, or more precisely, in the top mirror's frame, since this detection only happens there. This applies to T'_AB, but not to T_AB; the mirror can't "directly measure" a time in the unprimed frame, only in the primed frame. Quote by altergnostic The times are different indeed, but I don't know if I should call the reason "time dilation". Time dilation is the reason that T'_AB does not equal T_AB; it is the reason why I wrote T_AB = gamma (T'_AB) above. Working things out from the geometry of the problem that you gave shows that gamma (T'_AB) = T_AB = T_BA, which means that both light beams (A to B and B to A) travel at c as seen from the unprimed frame. We already know that the beam from A to B travels at c as seen from the primed frame; we then have to calculate the coordinates of event A2 (where the beam from B to A is received by A) in the primed frame to show that the beam from B to A travels at c as seen from the primed frame. Having said all that, I want to go back to your original claim. Your original claim is different from what you appear to be claiming now. Your original claim was that the standard SR picture of the light clock is inconsistent. I have now produced several analyses that show that the standard SR picture is consistent. Now you have shifted your ground, and you are saying that the standard SR picture makes unwarranted assumptions (that the speed of all light beams is c, not just directly detected ones). I have also shown that the assumptions aren't what you are claiming they are (they are things like translation and rotation invariance; things like the speed of all light beams being c are derived claims, not fundamental assumptions, if you start with the assumption of translation and rotation invariance). But regardless of the assumptions, the fact is that SR matches experiments, as DaleSpam pointed out. Your claims, such as the light beam from A to B taking only 1s (in the unprimed frame) to travel 1.12 light seconds (in the unprimed frame), do not match experiments; if we actually ran a light clock experiment in which the distance from A to B, in the unprimed frame, was 1.12 light seconds, we would not get the timing results you have claimed; we would get the results I derived using the standard SR formulas. So I'm not sure where you are going with this thread. Your original claim has been shown to be wrong; the standard SR model of a light clock is consistent. Your claims about the event coordinates are wrong, because they don't match the standard SR model, which agrees with experiment. What now? Mentor altergnostic, several people have repeated the claim that the experimental evidence supports SR. In fact, if you are uncomfortable with the light-speed postulate of SR then you could easily NOT assume it, make a general theory of all possible transformations between inertial frames, and use experimental data to set the parameters in such a general theory. In Robertsons famous paper he did exactly that and demonstrated that SR could be deduced to within about 0.1% from the Michelson Morely, Ives Stillwell, and Kennedy Thorndike experiments even without making Einstein's assumptions. See: http://rmp.aps.org/abstract/RMP/v21/i3/p378_1 I strongly recommend that you read that paper as well as the wealth of experimental evidence listed here: http://www.physicsforums.com/showthread.php?t=229034 I hold my claim that the standard light clock diagram is inconsistent. If I understand every claim correctly, the basic disagreement is that, if the distance AB is 1.12ls, then light should take 1.12s to travel that distance regardless of direction, which is basically saying that the speed of light is c regardless of the speed of the source. What seems hard to grasp is that I am not advocating against this assumption. It has been said that if I simply do the MM interferometer analysis I will have solved this setup without any contradiction. I disagree. It has also been said that the speed from A to B has been calculated repeatedly in this thread, but I disagree, and in many places it has been admitted that that speed was assumed based on the distance AB (not calculated), but the purpose of my setup is precisely to have the necessary givens so we can calculate that speed and check that assumption. Another argument is that I am mixing frames by adding primed and unprimed times and that that is not allowed. Although that is generally true, sometimes it is allowed. Of course we can't mix variables that are being measured, but we can add a given with an observation. We are simply taking the proper time of an event and taking into account how long it takes for that event to be seen at another point AB away. If the time (as shown on a clock at B) of an event at B is 0s and light takes 1.12s to get to A, than the observed time for that event as seen from A is 1.12s. Likewise, if the time of the event is 1s at B, the observed time will be 2.12s. That is all I am saying. It has been said that B receives the beam at 1.12s in the observer at A's frame, and 1s in the clock's frame (or the frame of the mirror at B). But that last is a given, and the former is an assumption. If we have an observer at B, at t'=1s the beam appears to come from the stationary mirror at C. At t=1s, B sees the light of the beam on the mirror at C, 1 light-second away. The path of detected light in the primed frame is CB. But from the point of view of the observer at A, it is the signal, not the beam, that comes from B, 1.12 light-seconds away. The path of the detected light in the unprimed frame is BA. This is what is observed. The speed of whatever is describing AB, so far, is unknown, as a matter of observations. The question is if the observer at A can assume that the time the beam takes to go from A to B is 1.12s based on his observations. And what are his observations? He knows the distance AB and the time light takes to reach him from any event at the point B. He receives a direct signal from B, but what is the observed time of the event at B? In a real experiment, we would find that number directly, but here we have to use logic to determine it. I say that, from A, the observed time time of event B is simply the time light takes to cross from B to A (tBA) + the proper time of emission: tB = tBA + t'B or t'B = tB – tBA which is t' = t – ct which is simply a way to find the time on a distant synchronized clock: t'B = 0.5(tA1 + tA2) t'B = 0.5(0 + 2.24) = 1.12s Hence every event at B will be observed 1.12s later at A: t' = t – AB and if the reflection event at the point B occurs at t'=1s, then it will be observed in A at: 1 = t – 1.12 t = 2.12s So I hope you see there is no frame mixing. We are not adding quantities that are being measured in different frames. t'B is not (only) being measured in the primed frame, but it is the time of the clock at B given by the standard synchronization method. The observer at A can only conclude that the time of the detection at B is 1s, not 1.12s. Please note that event B is observed at A when t=2.12s and that the beam is reflected from the mirror at B when t=1s. This method applies to anything traveling at any speed between the mirrors (or to nothing traveling at all) because it is related only to the observation of events, regardless of the cause of these events. For an observer at B: V'beam = tCB = c For an observer at A: Vsignal = tBA = c and apparent Vbeam = AB/(tB – tA) = 1.12/2.12 = 0.528c Now, he knows that the time of detection at B is not 2.12s, but 1s, so he could try to calculate the speed of the beam like this: calculated Vbeam = AB/(t'B - t'A) = 1.12/1 = 1.12c But that is a mixing of frames after all. We are no longer using given instants in time of events or observations, but elapsed periods between events, and a period is a measurement, and that measurement does not belong to the observer A - he has transformed an observed period into a directly detected period, or the period as measured by A into the period as measured by B - but hasn't transformed the distance. Therefore, AB no longer applies, since he has to use the distance as observed by B also. That distance can be easily found either by directly using the given primed distance (CB or y' or 1ls) or by turning back to the light postulate and finding the distance from the primed time: x' = ct' x' = 1 = y' So everything's double checked. If at point B there was a clock and a camera, the signal from B to A could carry a photograph of the clock at B and its surroundings and the observer at A would receive direct visual information showing the clock at B marking t'=1s and the mirror reflecting off the beam, and everything would be triple checked. Now, concerning the interferometer, the path AB is not observed nor calculated like in our setup at all. AB was assumed to be the path described by the beam in the ether frame, where light travels at c. It is the path of the beam as seen from an absolute frame. But of course, from the interferometer's point of view, light always reaches the detector directly, like I have been saying. The beam's speed is not being determined by distant events, but by local detections. If our setup was equivalent to the interferometer's, the detected beam would describe CBC and we would have no evidence of path AB at all (just like Michelson and Morley didn't), since we wouldn't be able to determine relative motion. And if we wanted to, we would have to place an observer in a moving frame, just like we are doing here, and we fall back to my analysis. We can't simply assume that path AB is being described by the beam at c as seen by the observer at A because, unlike the ether, he is not present everywhere and must directly receive incoming light to determine any coordinates. Received light is always measured to travel at c, which was thought to imply an absolute ether, but it sounds much more Einsteinian to specify c as the speed of light as measured by any observer (i.e.: by direct detection). Einstein's original claim, that the speed of light is the same regardless of the speed of the source, sounds perfect to me, and I think it is absurd to conclude that the postulate applies to the speed of undetected or indirectly detected light, since Einstein never even tried to describe it that way. He actually required that we place observers everywhere, so that light going in any arbitrary direction was actually directly detected, and it is to that light that we assign c. Objects and events are seen with light and it is with this light that we determine their velocities. If a beam traveling in an arbitrary direction is to be observed with light bouncing off of it (just like an ordinary object) and not directly detected as usual, it is not your standard light anymore, it is the v in v/c, since only the incoming light is that c - it is being detected, not being measured. That c in the denominator is the tool with which we measure other things, and the tool with which we are measuring the speed of the beam from A to B. And we can't have c/c. The denominator is the light with which we see and the numerator is the speed of what we are seeing with that light. The last thing that I think may need clarification is why I ignore the speed of the light clock in my setup. If we are given that number, we will tend to solve like it is always done and not really think the whole problem through. My givens are analogous to a given velocity, since we have fixed times and distances, but I require that the observer's frame is aligned with the frame of the source at B and that both frame's x-axis are in line with AB, since that is the easiest way to find unknowns for AB (we can simply ignore y and z and use the coordinates of each event to find AB). I am simply requiring B's frame of reference to be obtained from A's frame by a standard Lorentz transformation. I hope I made it clear how this setup differs from the assumptions and calculations of the standard light-clock and MM diagrams. Blog Entries: 9 Recognitions: Gold Member Science Advisor Quote by altergnostic which is basically saying that the speed of light is c regardless of the speed of the source. What seems hard to grasp is that I am not advocating against this assumption. Oh, really? You mean I can freely use that assumption when analyzing any of these scenarios? Great! That makes things a lot simpler. Quote by altergnostic We are simply taking the proper time of an event and taking into account how long it takes for that event to be seen at another point AB away. The proper time of the event only matches coordinate time in one particular frame, so you still can't add it to a coordinate time in some other frame and expect to get a meaningful answer. Quote by altergnostic If the time (as shown on a clock at B) of an event at B is 0s and light takes 1.12s to get to A, than the observed time for that event as seen from A is 1.12s. Only if the clock at B and the clock at A are mutually at rest and synchronized. But the "clock at B" that you have been referring to is moving relative to the observer at A. Quote by altergnostic Likewise, if the time of the event is 1s at B, the observed time will be 2.12s. That is all I am saying. And it's still wrong no matter how many times you repeat it. See above and further comments below. Quote by altergnostic It has been said that B receives the beam at 1.12s in the observer at A's frame, and 1s in the clock's frame (or the frame of the mirror at B). But that last is a given, and the former is an assumption. Yes, the last is a given. No, the former is not an assumption. It's a calculation. Since you have said you agree that the speed of light in an inertial frame is c regardless of the motion of the source, the calculation is simpler than what I have posted before. The givens are: (1) Light pulse 1 is emitted by the light clock source/detector at the instant that it passes the observer at A. That event occurs at time 0 in both the observer frame and the light clock frame. (2) The distance from A to B is 1.12 light seconds in the observer frame. Therefore, we can immediately calculate: (3) Light pulse 1 reaches B at time t = 1.12s in the observer frame, since it is covering a distance of 1.12 light seconds and travels at c. Since you already agree that light pulse 2 takes 1.12s to travel from B back to A in the observer frame, that makes it clear that light pulse 2 arrives at A at time t = 2.24s in the observer frame, not t = 2.12s. I won't even bother commenting on the rest of your post; the error you are making should be clear from the above. Blog Entries: 9 Recognitions: Gold Member Science Advisor On re-reading, there is one other error that seems to me to be worth commenting on: Quote by altergnostic If we have an observer at B, at t'=1s the beam appears to come from the stationary mirror at C. At t=1s, B sees the light of the beam on the mirror at C, 1 light-second away. The path of detected light in the primed frame is CB. No, it isn't. The triangle diagram you drew is a diagram of the spatial geometry in the unprimed frame, not the primed frame. The point you have labeled C is the location that the light clock source/detector is at, in the unprimed frame, at the time when light pulse 1 hits the mirror. It is not the location of the source/detector in the primed frame, because in the primed frame the source/detector and the mirror (i.e., the entire light clock assembly) are at rest. The spatial geometry in the primed frame would be a different diagram, in which the positions of the observer at A would lie on a line going out to the left, the spatial path of light pulse 1 would be a line straight up, and the spatial path of light pulse 2 would be the hypotenuse of a right triangle whose legs are the path of light pulse 1 and the path of the observer at A. Any reasoning about spatial paths and lengths in the primed frame would have to be done using this other diagram, not the diagram you drew. Quote by altergnostic [..] It has been said that if I simply do the MM interferometer analysis I will have solved this setup without any contradiction. I disagree. [..] Please do so - even if I was wrong and one or two issues remain, it will enormously clear up this discussion. It still appears that you disagree about almost everything, which makes it difficult to untangle the mess. Mentor Quote by altergnostic If I understand every claim correctly, the basic disagreement is that, if the distance AB is 1.12ls, then light should take 1.12s to travel that distance regardless of direction, which is basically saying that the speed of light is c regardless of the speed of the source. What seems hard to grasp is that I am not advocating against this assumption. If the distance is 1.12 ls and the speed is c regardless of direction then the round trip time is clearly 2.24 s. Furthermore, even without assuming the speed is c, experiments confirm the round trip time (aka the two way speed of light) so any value you choose other than 2.24 s is contrary to experiment. See the Robertson paper I posted earlier. Quote by altergnostic I am simply requiring B's frame of reference to be obtained from A's frame by a standard Lorentz transformation. Please post the Lorentz transform from A to B then. You will see that your claims are inconsistent with the Lorentz transform. Quote by DaleSpam If the distance is 1.12 ls and the speed is c regardless of direction then the round trip time is clearly 2.24 s. Furthermore, even without assuming the speed is c, experiments confirm the round trip time (aka the two way speed of light) so any value you choose other than 2.24 s is contrary to experiment. See the Robertson paper I posted earlier. A beam describing ABA would indeed take 2.24s for the round trip, as is confirmed by experiment, but this is not what is happening here. The observer at A is not the emitter, and from the point of view of the emitter, the path is CBC, not ABA. The observer at A only has the distance AB and a signal from B to A, but he has no direct information on the path from A to B, he must calculate it from the time he receives the signal from B, and we are disagreeing on that reception time, so we have to clear this up. Would you agree that, if a stationary clock placed at B is synchronized with the clock at A, we can find the reception time by taking the time of the event as shown on the clock at B and adding the time it takes the signal from that event to get to A? Or, likewise, we can find the time of the event on the stationary clock at B by taking the reception time and subtracting the time it takes for light to cross the distance AB? Isn't this how we find the times of stationary synchronized clocks? For instance, to synchronize the clocks at A and B we would take a roundtrip measurement (2.24s, like you said) and half that value to get the time separation of the clock at B, so we know that a tick of the clock at B is observed 1.12s later at A. After that synchronization, we can calculate the time of any event at B from A simply by subtracting 1.12s from the reception time, don't you agree? Since the observer at A receives the signal from B 1.12s later than the time of emission, and the time of emission (according to the clock at B - which is stationary and synchronized with the clock at A) is 1s, the observer at A must receive that signal at 2.12s. I can't see it any other way. If the reception time was 2.24s than the emission time would have to be 1.12s and that is not in agreement with what is shown on the clock at B. If you are right, either the clocks at A and B are out of sync, or the clock at B is time dilated (and I don't see how that can be true) or the event at B occurs at 1.12s from the point of view of B, which would then be in disagreement with the light-clock time for that event, which is insane since the event at B happens at the same spacetime coordinates for both the light-clock and the clock-signaler at B. Please post the Lorentz transform from A to B then. You will see that your claims are inconsistent with the Lorentz transform. You misunderstand me (or I wasn't clear). What I meant is that this frame alignment is what would allow us to apply standard lorentz transforms. I am simply requiring that both frames x-axis are aligned with AB, no calculations needed for this requirement, it is standard procedure in most SR problems and it helps to simplify things a bit. Blog Entries: 9 Recognitions: Gold Member Science Advisor Quote by altergnostic The observer at A is not the emitter Doesn't matter. The observer at A is co-located with the emitter at the instant of emission. So the path of the light, in the unprimed frame, is ABA. Quote by altergnostic and from the point of view of the emitter, the path is CBC No, it's C'B'A', because "from the point of view of the emitter" means in the primed frame, not the unprimed frame. The geometry in the primed frame is different; it does not look like the triangle diagram you drew. Quote by altergnostic Would you agree that, if a stationary clock placed at B is synchronized with the clock at A, we can find the reception time by taking the time of the event as shown on the clock at B and adding the time it takes the signal from that event to get to A? Yes, if the clock at B is at rest relative to the clock at A and synchronized with it. Quote by altergnostic Or, likewise, we can find the time of the event on the stationary clock at B by taking the reception time and subtracting the time it takes for light to cross the distance AB? For a stationary clock at B, yes. Quote by altergnostic the time of emission (according to the clock at B - which is stationary and synchronized with the clock at A) No, it isn't. Your "clock at B" that reads 1s when light pulse 1 arrives is moving with the light clock; it's the "clock" on the light clock's mirror that records when light pulse 1 arrives. It's not stationary and synchronized with the clock at A. A stationary clock at B, that was synchronized with observer A's clock, would record a time t = 1.12s when light pulse 1 arrives at the mirror. Quote by altergnostic the event at B happens at the same spacetime coordinates for both the light-clock and the clock-signaler at B. I don't know what you mean by "the clock-signaler at B". Do you mean an observer stationary at B, who emits light pulse 2 back to A? If so, then your statement is incorrect; the clock-signaler's clock will read t = 1.12s when light pulse 1 arrives at the mirror, as I said above. The clock-signaler's clock is not synchronized with the moving mirror's clock, nor are his spatial coordinates the same as those of the light clock. The event is the same, but it is described by different coordinates (i.e., a different set of 4 numbers t', x', y', z') in the primed frame than in the unprimed frame. You apparently do not understand how frames in relative motion work. The only event that has the same spacetime coordinates (i.e., the same set of 4 numbers t, x, y, z) in both frames is the origin, the event at which light pulse 1 is emitted by the light clock source/detector at the instant it is spatially co-located with the observer at A. Every other event in the spacetime has different coordinates in the two frames. I suggest a review of a basic relativity textbook, such as Taylor & Wheeler's Spacetime Physics. Quote by harrylin Please do so - even if I was wrong and one or two issues remain, it will enormously clear up this discussion. It still appears that you disagree about almost everything, which makes it difficult to untangle the mess. Harry, that analysis does not apply to this setup at all, that is my point. The only possible way to do that is like it is always done: giving c to AB, so by doing it my arguments and givens are useless. You see, I am putting that very assumption into question with my setup, so I can only do what you ask me to do by giving up what I am concerned about in the first place. Also, it is done all over the place and we can simply google "light clock time dilation interferometer, etc" and we will have tons of results. You can't use that analysis to answer "what is the speed from A to B" since that is a given in that analysis in the first place. Also, MM analysis don't have an observer at A like I have here, and also no possible communication from reflection events to that observer (or any other moving observer for that matter). So I can't possibly relate the interferometer analysis to my setup. If we assume that light travels at c as seen from an absolute ether or in any direction for any frame, my analysis is meaningless since it is intended to check those assumptions. If we rerun my analysis with a projectile going 0.5c from A to B instead of the beam, would you agree that it stands? Do you agree on A's reception time of 2.12s for the signal from B? If not, are you saying that the time of emission of the signal from B shown on the clock at B is 1.12s? If so, how can that be since at t'=1s the light-clock is reflecting the beam at B? Is that event happening at t'=1s or t'=1.12s, as seen from B? Are you saying we can have two different times for the same event, as seen from B? If so, please explain how, because I can't see how that is possible in this problem. Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Thread locked pending moderation. Thread Closed Page 10 of 10 « First < 7 8 9 10 Tags light clock, special relativity, time dilation Thread Tools \| \| \| \| \|------------------------------------------\|------------------------------\|---------\| \| Similar Threads for: Light Clock Problem \| \| \| \| Thread \| Forum \| Replies \| \| \| Special & General Relativity \| 8 \| \| \| Special & General Relativity \| 55 \| \| \| Special & General Relativity \| 14 \| \| \| Special & General Relativity \| 31 \| \| \| Special & General Relativity \| 7 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9639703631401062, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/7391/kolmogorov-smirnov-and-lattice-paths	# Kolmogorov-Smirnov and lattice paths I have seen asserted that the problem of computing the null distribution of Kolmogorov's $D_n^+$ statistic for a finite sample size maps onto the problem of computing the number of lattice paths that stay below the diagonal, and thus can be solved by the ballot theorem. I am familiar with lattice paths and the ballot problem. I am also familiar the expression of the distribution of $D_n^+$ as a series of integrals. But I don't see how one problem maps onto the other. Can someone explain or point me to an ariticle or book that does? I also see the claim the the null distribution of the Kolmogorov-Smirnov $D_n = \max(D_n^+,D_n^-)$ maps onto another lattice path problem that could be solved by a "two-sided ballot theorem". I don't know what a "two-sided" version of the ballot problem would be. Again, can someone explain or point me to an explanation? Finally, is there a general framework around all of this? Can the Kuiper statistic be mapped to yet another lattice path problem? The two-sample KS test? The AD statistic? - 2 I'm not sure about the two-sample problem, but as for references for the former, see: A. P. Dempster, Generalized $D_n^+$ Statistics, Ann. Math. Statist. Volume 30, Number 2 (1959), 593-597. Also, here, here and here might have some material of interest. The first and last links should be to publicly available articles. Sadly, the middle two aren't. – cardinal Feb 19 '11 at 5:42 ## 1 Answer To add to @Cardinal 's answers in the comments, I think there is work that addresses the "claim the null distribution of the Kolmogorov-Smirnov maps onto another lattice path problem that could be solved by a "two-sided ballot theorem" and "is there a general framework around all of this? The two-sample KS test?": This paper (preprint) is concerned with r-sample Kolmogorov-Smirnov tests and they derive the exact null distribution by counting lattice paths by using a generalization of the classical reflection principle. In Section 2, I find that they lay out nicely how lattice path counting comes into play when deriving the null hypothesis. In the introduction the paper also features a discussion on how lattice path counting and the reflection principle tie in here by reviewing ideas started with Kiefer 1959 and David 1958. They also briefly discuss how it can be seen as an r-ballot counting problem, referring to Filaseta 1985. They provide a lattice path counting framework for KS type tests for any number of samples. From the paper: We consider the problem of testing whether $r ≥ 2$ samples are drawn from the same continuous distribution $F(x)$. As a test statistic we will use the circular differences $\delta_r (n) = \max [\delta_{1,2} > (n), \delta_{2,3} (n), . . . , \delta_{r−1,r} (n), \delta_{r,1} (n)],$ where $\delta_{ij} (n) = \sup_x [F_{n,i} (x) − F_{n,j} (x)]$, and $F_{n,i} (x), i = 1, 2, . . . , r$ denote the empirical distribution functions of these samples. We derive the null distribution of $\delta_r(n)$ by considering lattice paths in $r$-dimensional space with standard steps in the positive direction, i.e., steps are given by the unit vectors $e_i , i = 1, 2, . . . , r$. By a simple transformation we show that for some positive integer $k$ the number of ways the event $\{n\delta_r (n) < k\}$ can occur is just the number of paths $X$ with the property that for each point $X_m$ on the path there holds the chain of inequalities $x_{1,m} > x_{2,m} > . . . > > x_{r,m} > x_{1,m − rk}$. Indeed, the enumeration of such paths is a well studied problem in combinatorics. Again the reflection principle comes into play as we have to count paths in alcoves of affine (and therefore infinite) Weyl groups; for references on the technical background of this topic see Gessel and Zeilberger 1992, Grabiner 2002 and Krattenthaler 2007. Hopefully this is a good starting point for further investigations into the Kuiper and AD statistics for example. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9211875200271606, "perplexity_flag": "head"}
http://mathoverflow.net/questions/58320/intersections-of-conjugates-of-lie-subgroups	## Intersections of conjugates of Lie subgroups ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $G$ be a closed, connected Lie group, and let $H$ be a closed (and therefore Lie) subgroup. There is a natural action of $G$ on the space of left cosets $G/H$, for which the stabiliser of $aH$ is the conjugate subgroup ${}^aH:=aHa^{-1}$. Now let $G$ act diagonally on $G/H\times G/H$. The stabiliser of $(aH,bH)$ is the intersection ${}^aH\cap {}^bH$ of conjugate subgroups. My question is, what can be said about ${}^aH\cap {}^bH$? I know that ${}^aH={}^bH$ if and only if $ab^{-1}\in N(H)$, the normaliser of $H$ in $G$. Beyond this I couldn't find much via Google. For instance can ${}^aH\cap {}^bH$ be trivial? Or must it be a conjugate of $H$? Apologies if this is too elementary. Edit: Thanks for the answers, which show that not much can be said at this level of generality. Now I am looking at a specific example - the icosahedral group $I\cong A_5$ inside $SO(3)$ (where the space of left cosets is the Poincaré sphere). This subgroup has the properties that $I$ is finite, $N(I)=I$ and $I/[I,I]$ is trivial. Does this allow me to conclude that conjugates of $I$ are either equal or intersect in the identity? More generally, what can be said if I add the assumption $N(H)=H$ to the original question? Edit 2: I've now asked about the icosahedral group in another question. - 1 Excuse me, but in the last line, shouldn't there be ${}^aH\cap{}^bH$ instead of ${}^aH={}^bH$ ? or not? – Giuseppe Mar 13 2011 at 9:56 Yes, thanks, duly edited. – Mark Grant Mar 13 2011 at 10:00 1 If you consider groups whose cardinality is less than or equal to $\aleph_0$ and endow them with the discrete topology, then you get 0-dimensional Lie groups. Now take $G=S_n$ the symmetric group of $\{1,\ldot,n\},$ and $H_i$ the subgroup of the permutations fixing $i,$ for $i=1,\ldots,n$ these are obviously conjugate each other. Trivially $H_1\cap H_2$ is not conjugate to $H_1$. Excuse me if this answer is not what you wanted. – Giuseppe Mar 13 2011 at 10:19 Thanks Giuseppe. In my applications $G$ is connected, but your comment shows I was being too optimistic (as does Jack's connected answer). – Mark Grant Mar 13 2011 at 13:25 ## 3 Answers Suppose G=SL(2,C) and let H be the stabilizer of a line (so a Borel subgroup). The matrix $$\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ in G acts on projective space by $$z \mapsto \frac{az+b}{cz+d}$$ The stabilizer of ∞ is the matrices with c=0, a Borel subgroup. The stabilizer of both ∞ and 0 is the matrices with b=c=0, a maximal torus. In particular, a two point stabilizer is abelian (the intersection of two Borel subgroups), and a Borel subgroup is non-abelian. Hence they are not isomorphic. This is just a connected version of Giuseppe's answer. If you consider the Borel subgroup to be the Lie group itself, then you get an example where the intersection of the conjugates is trivial. If G is the group of 2×2 matrices with c=0 and d=1 acting on projective space, then the stabilizer of 0 has b=0, and the stabilizer of both 0 and 1 has a=1 and b=0. In particular, G=AGL(1) and H is a maximal torus, and the intersection of two conjugates of H is the identity. When the intersection is the identity, this is called being sharply two-transitive or having a regular stabilizer. - @Mark Grant: More exotic intersections are certainly possible: Take G=Alt(6) wr Sym(2), H=Alt({1..6})×Alt({7..11}), then H is self-normalizing. The conjugates of H are just point stabilizers, and the intersection of conjugates are just 2-point stabilizers. The stabilizer of 11, 12 is Alt({1..6})×Alt({7..10)}≅A6×A4. The stabilizer of 6, 12 is Alt({1..5})×Alt({7..11})≅A5×A5. Similar ideas happen in any wreath product, but this one was chosen so that H is self-normalizing (so that a point stabilizer moves all other points) and perfect. I don't know a connected version of the wreath product. – Jack Schmidt Mar 18 2011 at 16:43 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The motivation for the question is unclear to me, but it seems much too broad to have an interesting answer even if you limit consideration to connected algebraic subgroups of a general linear group (complex or real). As Jack points out, there are easy examples showing various possible outcomes when you intersect a group with a conjugate. The structure theory of semisimple or other Lie/algebraic groups has been extensively studied and makes it easy to find lots of further examples of this type. For instance, a Borel subgroup intersects a conjugate in at least a maximal torus when the ambient group is reductive; the intersection may be precisely a maximal torus if the Borel subgroups are "opposite". Concretely, this is seen when intersecting the upper triangular and lower triangular matrices in the `$n \times$` matrix group: the result is the group of nonsingular diagonal matrices. At another extreme, intersecting this diagonal group with a typical conjugate by an upper triangular unipotent matrix will typically produce a finite group. - 1 I disagree with your premise. Clearly, if you have a transitive action of a Lie group, it is of interest to study the fixator of two points. This motivates the study of $G/H$ and the intersection of two conjugates of $H$. In the case of a two-transitive action, there is even a very interesting answer available: these actions have been classified by Tits (and Borel). See L. Kramer, Two-transitive Lie groups, for a beautiful exposition and simplified proof. – Guntram Mar 13 2011 at 14:36 1 @Guntram: There are certainly such well-focused special cases which are interesting, but the question as asked is too broad and the third paragraph there illustrates the need for cautionary examples. – Jim Humphreys Mar 13 2011 at 17:54 Excuse my misunderstanding. Please accept in substitution a connected example. Let $G=SE(2)$ be the special euclidean group of the plane, and $H_x$ the subgroup stabilizer of $x$, for any $x\in\mathbb{R}^2$. These latter ones are conjugate each other, but $H_x\cap H_y$ is trivial for any pair of distinct points $x$ and $y$ in the plane. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 44, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9196545481681824, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/222312/least-norm-solution-to-ax-b	Least norm solution to $Ax = b$ How to prove that if you have $x^$ such that $x^=\text{psuedoinverse}(A) b$, and $Ay=b$, then $$\Vert x^* \Vert_2 \leq \Vert y \Vert_2$$ - Some people define the pseudoinverse as the mapping that takes $b$ as input and returns the vector $x$ of least norm such that $Ax=\hat{b}$, where $\hat{b}$ is the projection of $b$ onto the range of $A$. Formulas for the pseudoinverse are then derived from this more conceptual definition. – littleO Oct 27 '12 at 23:57 @Marcus. You don't specify, but should it be assumed that $A$ has more columns than rows and has full row rank? – Daryl Oct 28 '12 at 8:29 3 Answers Let me try a one liner solution. $Ay=b$ $\Rightarrow$ $A(y-x^)=0$ $\Rightarrow$ $\langle x^,y-x^\rangle=0$ $\Rightarrow$ $\\|y\\|^2=\\|y-x^\\|^2+\\|x\\|^2\ge \\|x\\|^2$. Remark. $R=A^t(AA^t)^{-1}$, $x^=Rb$. Then we have $\langle x^,y-x^\rangle=\langle Rb,y-x^\rangle=$ $\langle (AA^t)^{-1}b,A(y-x^)\rangle=0$. - Let $A: \mathbb{U} \mapsto \mathbb{V}$, then in terms of SVD, we can write $A$ as $$A=\sum_{n=1}^R\sigma_nv_nu_n^{\dagger},$$ where $\sigma_n$ is a nonzero singular value; $u_n$ and $v_n$ are the right and left singular vector, respectively; $R$ is the rank of $A$. Since {$u_n$} form an orthonormal basis of $\mathbb{U}$, we can expand $y$ as $$y=\sum_{m=1}^M\alpha_m u_m,$$ where $M\ge R$ is the dimension of $\mathbb{U}$. So $$b=Af=\sum_{n=1}^R\sigma_nv_nu_n^{\dagger}\sum_{m=1}^M\alpha_mu_m=\sum_{n=1}^R\alpha_n\sigma_nv_n$$ Also, the pseudoinverse is $$A^+=\sum_{m=1}^R\frac{1}{\sigma_m}u_mv_m^{\dagger}$$ Then $$x^=A^+b=\sum_{m=1}^R\frac{1}{\sigma_m}u_mv_m^{\dagger}\sum_{n=1}^R\alpha_n\sigma_nv_n=\sum_{m=1}^R\alpha_mu_m$$. Finally, we can see that $$\\|y\|\|_2=(\sum_{m=1}^M\alpha_m^2)^{1/2},$$ $$\\|x^\|\|_2=(\sum_{m=1}^R\alpha_m^2)^{1/2}.$$ Therefore, $$\\|x^\|\|_2 \le \\|y\|\|_2,$$ where equality holds when $M=R$ or $\alpha_m=0$ for $m=R+1, \cdots, M$. In other words, if we define $$\\|y_{null}\\|_2=(\sum_{m=R+1}^M\alpha_m^2)^{1/2},$$ we have $$\\|y\\|_2^2=\\|x^\|\|_2^2+\\|y_{null}\\|_2^2.$$ - You essentially want to find the solution to he following optimization problem. $$\min_{x}\Vert x \Vert_2 \text{ such that } Ax = b$$ Using Lagrange multipliers, we get that $$\min_{x, \lambda} \dfrac{x^Tx}2 + \lambda^T (Ax - b)$$ Differentiate with respect to $x$ and $\lambda$ to get that $$x^ = \underbrace{A^T(AA^T)^{-1}}_{\text{pseudoinverse}}b$$ Proof: $$\dfrac{d \left(\dfrac{x^Tx}2 + \lambda^T (Ax - b) \right)}{dx} = 0 \implies x^* + A^T \lambda = 0 \implies x^* = -A^T \lambda$$ We also have $$Ax^* = b \implies AA^T \lambda = -b \implies \lambda = - \left( AA^T\right)^{-1}b$$ Hence, $$x^* = \underbrace{A^T(AA^T)^{-1}}_{\text{pseudoinverse}}b$$ - But how does that prove that it is less than the 2-norm of y? – Marcus Oct 27 '12 at 21:29 @Marcus We have found the $x$ satisfying $Ax=b$ such that $\Vert x \Vert_2$ is minimum. – user17762 Oct 27 '12 at 21:30 Of course, sorry for being a bit dense there. Thank you – Marcus Oct 27 '12 at 21:36 Your pseudoinverse is only true if $A$ has full row rank, so that $AA^T$ has full rank. (I.e. it is a right inverse of $A$ only.) – Daryl Oct 28 '12 at 3:46 @Daryl True. This is assuming that $A$ has full rank. – user17762 Oct 28 '12 at 3:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 46, "mathjax_display_tex": 17, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9344974160194397, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/13596/list	3 edited tags 2 edited tags 1 # Continuous automorphism groups of normed vector spaces? Consider the metric space on, say, ℝ2 induced by the various $L^p$ norms, and the group of isometries from that space into itself that preserve the origin. When $p=2$ I get the continuous group of rotations, but when $p\in\{1,3,4,5,...\infty\}$ it looks like I just get $D_8$, the symmetry group of the square. Question: what's going on here? Why is 2 so special? Are there other natural norms on ℝ2 (or on ℝn) besides the euclidean one that give interesting isometry groups?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9320389628410339, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/226477/laplacian-of-a-function-implies-the-function-cannot-have-max-or-min	# Laplacian of a function implies the function cannot have max or min. If $\bigtriangledown^2f = 0$ in some region in the space, then $f$ cannot have maximum or minimum on that region. My approach was to assume $f$ has a maximum and then use the second derivative test to obtain a contradiction. Is this a right approach? Is there an easy way to tackle this problem? - It's hard to know if your approach is right, since you didn't give any of the details. – Nate Eldredge Nov 1 '12 at 3:43 2 haha i don't wanna be 'that guy' but do you want $f$ nonconstant, and the region connected :p? – uncookedfalcon Nov 1 '12 at 4:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9306234121322632, "perplexity_flag": "head"}
http://cms.math.ca/Reunions/ete12/abs/tp	Réunion d'été SMC 2012 Hôtels Regina Inn et Ramada (Regina~Saskatchewan), 2 -4 juin 2012 www.smc.math.ca//Reunions/ete12 Positivité totale Org: Shaun Fallat (Regina) et Michael Gekhtman (Notre Dame) [PDF] ALVARO BARRERAS, Universidad de Zaragoza Signed bidiagonal decompositions [PDF] Matrices with a bidiagonal decomposition satisfying some sign restrictions are analyzed. They include all nonsingular totally positive matrices, their matrices opposite in sign and their inverses, as well as tridiagonal nonsingular $H$-matrices. Properties of these matrices are presented and the bidiagonal factorization can be used to perform coputations with high relative accuracy. SHAUN FALLAT, University of Regina Rank Deficiency and Shadows in Totally Nonnegative Matrices [PDF] An $m \times n$ matrix is called totally nonnegative (TN) if all of its minors are nonnegative. It is a simple consequence of this definition to deduce that if $A = [a_{ij} ]$ is TN with no zero rows or columns, and if $a_{kl} = 0$, then $A$ will contain a block of zeros determined by the $(k, l)$ position. In this talk, I will present a generalization of this phenomenon to larger sized rank deficient blocks, discuss some related results on row and column inclusion for TN matrices, and connections to the distribution of zero minors in a TN matrix. STEPHANE LAUNOIS, University of Kent (United Kingdom) Deleting derivations algorithm and TNN matrices [PDF] In this talk, I will present the deleting derivations algorithm, which was first developed in the context of quantum algebras, and explain the significance of this algorithm for the study of TNN matrices. MAHMOUD MANJEGANI, Isfahan University of Technology and University of Regina Hadamard Powers of Totally Positive Matrices [PDF] Let $A=(a_{ij})$ be a totally positive $n\times n$ matrix. Is the $A^{(\alpha)}$ totally positive? In this talk we try to show that under some conditions on $\alpha$ the Hadamard power $A^{(\alpha)}$ is totally positive. (Joint work with Shaun M. Fallat) SHAHLA NASSERASR, University of Regina TP$_k$ completion of partial matrices with one unspecified entry [PDF] An $m\times n$ matrix is called TP$_k$ if every minor of size at most $k$ is positive. The TP$_k$ completion problem for patterns of specified entries is considered. For a given pattern with one unspecified entry, the minimal set of conditions characterizing TP$_k$ completability is given. These conditions are finitely many polynomial inequalities in the specified entries of the pattern. This is joint work with C. Johnson and V. Akin. ## Commandites Nous remercions chaleureusement ces commanditaires de leur soutien. © Société mathématique du Canada © Société mathématique du Canada : http://www.smc.math.ca/	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8455810546875, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/121634/whats-wrong-with-this-proof-that-ei-theta-e-i-theta/121637	# What's wrong with this proof that $e^{i\theta} = e^{-i\theta}$? I recently learned that $\cos{\theta} = \frac{e^{i\theta} + e^{-i\theta}}{2}$ and $\sin{\theta} = \frac{e^{i\theta} - e^{-i\theta}}{2}$ Based on this, I managed to "prove" that: $$e^{i\theta} = e^{-i\theta}$$ Since $e^{i\theta} = \cos{\theta} + i\sin{\theta}$, we can substitute the above two identities to get: $$e^{i\theta} = \frac{e^{i\theta} + e^{-i\theta}}{2} + i\frac{e^{i\theta} - e^{-i\theta}}{2}$$ Simplifying, I get $$(1-i)e^{i\theta} = (1-i)e^{-i\theta}$$ which implies $$e^{i\theta} = e^{-i\theta}$$ for all real $\theta$. Obviously, this isn't true in general, but I'm having a hard time seeing what's wrong. Can someone please point out the flaw in the above "proof"? - As a side note, I wasn't too sure what the relevant tags are for this question. I assume (fake-proofs) and (complex-numbers) are definitely relevant, but perhaps there are better tags? – Herman Chau Mar 18 '12 at 8:18 3 You actually want $\sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i}$. – Qiaochu Yuan Mar 18 '12 at 8:19 2 Oh, my textbook misprinted it as $\sin{\theta} = \frac{e^{i\theta} - e^{-i\theta}}{2}$ and I just accepted it without bothering to check. -facepalm- So should I just delete this question? – Herman Chau Mar 18 '12 at 8:23 ## 1 Answer You don't have $\sin\theta = e^{i \theta}/2- e^{-i\theta} /2$. Actually, $$\sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2 i}.$$ - 1 One might add that the presence of $i$ in the formula for the sine is obviously necessary: $e^{i\theta}$ and $e^{-i\theta}$ are complex conjugates, so while their sum is real, their difference is purely imaginary. Since one knows that $\sin x$ is real for real $x$, one needs to divide out the $i$ to get from the imaginary to the real axis. This simple argument may help memorize the formulas. – Marc van Leeuwen Mar 18 '12 at 10:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9534080624580383, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/tagged/representation-theory+number-theory	# Tagged Questions 5answers 185 views ### Applications of Character Theory Some of the applications of character theory are the proofs of Burnside $p^aq^b$ theorem, , Frobenius theorem and factorization of the group determinant (the problem which led Frobenius to character ... 0answers 73 views ### Representations of $\text{GL}_2(\mathbb{Q})$ Let's say that as a representation theorist I am naively interested in representations of $G(\mathbb{Q})$, where $G$ is an algebraic group defined over $\mathbb{Q}$. For the purposes of this question, ... 1answer 46 views ### Some questions about representation theory in the modular case I'm working on a paper which uses representation theory in order to compute some characters and deduce arithmetical statements about certain field extensions. Let $\Delta$ be a group of order prime ... 1answer 277 views ### Undergraduate roadmap for Langlands program and its geometric counterpart What are the topics which an undergraduate with knowledge of algebra, galois theory and analysis learn to understand Langlands program and its goemetric counterpart? I would also like to know what are ... 0answers 57 views ### A natural way of thinking of the definition of an Artin $L$-function? Emil Artin knew that given a finite extension of $L/\mathbb{Q}$, the local factor of the zeta function $\zeta_{L/\mathbb{Q}}$ at the prime $p$ should be \$\displaystyle\prod_{\mathfrak{p}\|p}\frac{1}{1 ... 2answers 115 views ### Property (T) and Number Theory? Pardon the wild question, but Are there any known connections between Kazhdan's property (T) and number theory, or number theoretic consequences of property (T)? Edit: To clarify, I'd like to ... 1answer 254 views ### What is the idea of a monodromy? Is there a connexion between : 1) The monodromy group of a topological space. 2) The $\ell$-adic monodromy theorem of Grothendieck. 3) The $p$-adic monodromy conjecture of Fontaine (which is now ... 1answer 66 views ### Why does rigidity hold only if rank >1? In simple words, why does Margulis' superrigidity and arithemiticit only hold for lattices in Lie groups of rank $>1$? E.g. what is the reason for it to fail for $SL(2,R)$? 1answer 166 views ### Does the $p$-torsion of an elliptic curve with good reduction over a local field always determine whether the reduction is ordinary or supersingular? Let $K$ be a finite extension of $\mathbb{Q}_p$ and $E/K$ an elliptic curve with good reduction. Does the $\mathbb{F}_p[\mathrm{Gal}(\overline{K})]$-module $E[p](\overline{K})$ determine whether the ... 1answer 166 views ### 2-dimensional $\ell$-adic representations [closed] In an assignment, I have to give an example of a 2-dimensional $\ell$-adic representation of the absolute Galois group of $\mathbb{Q}$, bu I am faced with the problem that I do not a lot of these. Or ... 2answers 230 views ### What is the standard definition of an ordinary (local) $p$-adic Galois representation? Let $V$ be a $n$-dimensional $\mathbf{Q}_p$-vector space with a continuous action of $\operatorname{Gal}(\bar{L}/L)$, where $L$ is a complete discretely valued field of characteristic zero with ... 1answer 75 views ### Where can I find rigorous statements about the spectral decomposition of reductive groups? Given a global field $F$ and a reductive group $G$, where can I find the spectral decomposition of $$L^2( Z(\mathbb{A}) G(F) \backslash G( \mathbb{A})).$$ I will need the result in this generality, ... 2answers 66 views ### Eisenstein spectrumfor $GL(n)$ Fix a global field $F$. Does every automorphic representation of $GL(n)$ appear as an arbitrary twists in the continuous spectrum of $GL(m)$, $m>n$? What happens for the automorphic ... 3answers 248 views ### Complex Galois Representations are Finite In A First Course in Modular Forms, Diamond and Shurman leave as an exercise ($9.3.3$) that every complex Galois representation is finite. While I think I have worked through this exercise here, this ... 1answer 214 views ### What is a description for the following number theoretic object? The title couldn't quite contain the question, so I didn't attempt to make it precise. I should note that this is the third or fourth question I've asked these past two days about problems I've been ... 1answer 132 views ### Questions about p-adic representations In a paper I'm currently reading, they have the following situation: $k$ is some number field that doesn't have a primitive $p^{th}$ root of unity, and $k(\zeta_p)$ a field above it with Galois group ... 1answer 230 views ### torsion representation Let $\mathbb{Z}_p$ be te ring of p-adic integers and let $T$,$T'$ be two free $\mathbb{Z}_p$-module with a continuous action of $G_{\mathbb{Q}_p}$ (the absolute Galois group of $\mathbb{Q}_p$). It is ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 46, "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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9137802720069885, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/51453/an-example-which-contradict-to-newtons-3rd-law	# An example which contradict to Newton's 3rd law? Let a,b be two charged particles. $$\vec{r}_a(0)=\vec{0}$$ $$\vec{r}_b(0)=r\hat{j}$$ $$\vec{v}_a(t)=v_a \hat{i}$$ $$\vec{v}_b(t)=v_b\hat{j}$$ In which both $v_a$ and $v_b$ $<<c$. Then $$\vec{E}_{ab}(0)=\frac{q_a}{4\pi \epsilon r^2}\hat{j}$$ $$\vec{B}_{ab}(0)=\frac{\mu q_av_a}{4\pi r^2} \hat{k}$$ $$\vec{E}_{ba}(0)=-\frac{q_b}{4\pi \epsilon r^2}\hat{j}$$ $$\vec{B}_{ba}(0)=\vec{0}$$ Note that $v_a$ and $v_b$ $<<c$ thus a and b almost obey Coulomb's law. Moreover, $\vec{j_i}(\vec{r})=q_i\delta(\vec{r}-\vec{r}_i)\vec{v}_i$ hence BS law can be applied. Hence $$\vec{F}_{ab}(0)=q_b(\vec{E}_{ab}+\vec{v}_b \times \vec{B}_{ab})$$ $$=\frac{q_a q_b}{4\pi \epsilon r^2}\hat{j}-\frac{\mu q_av_a v_b}{4\pi r_b^2} \hat{i}$$ But $$\vec{F}_{ba}(0)=-\frac{q_aq_b}{4\pi \epsilon r^2}\hat{j}$$ Consequently $$\vec{F}_{ab} \ne -\vec{F}_{ba}$$ This result contradict to Newton's 3rd law!! But I cannot find any error... It troubled me. - 2 your first equation about r_a should say r_a=r, because you state it for all t ( time) as it is a cannot have any velocity. The difference vector between r_a ( which you wrongly call r) and r_b should be in the denominator for the force between the two. – anna v Jan 17 at 6:27 1 to be clear, r_a can be 0 only at t=0 or another specific time if it is moving. And correct the above, r_a= r_i ( not r) to be correct. then r should be the vector difference between r_i and r_j. I think you are ignoring that in moving systems only an initial value at a specific time, t=0 can be given for space location. – anna v Jan 17 at 7:01 As @MarkEichenlaub points out, there are some issues with your analysis, but +1---a nice question and astute attempt none-the-less! – zhermes Jan 17 at 7:13 @annav You are right. I'm going to correct it. – Popopo Jan 17 at 13:49 ## 2 Answers The details of your analysis are not quite right - that's not what the electric field of a moving charge looks like, for example. This is probably because you haven't learned all the rules of electromagnetism yet. Still, the spirit of your question is hitting at an important point. Charges do not conserve momentum and don't obey Newton's third law. You have to include the momentum of the electromagnetic field to see conservation laws hold. There's an accessible discussion in section 8.2 of Griffiths "Introduction to Electrodynamics" if you would like a little more math. - Oh right, moving charges do not obey Coulomb&Biot-Savart's law... So I have added a condition that they are both moving in low velocity. Besides, many thanks to your book. – Popopo Jan 17 at 15:00 Okay, I have read section 8.2. It seems the interaction in classical field theory is localized. Hence the forces charges feel are given by EM-field rather than other charges. So the result is not $\vec{F}_{ab} \ne -\vec{F}_{ba}$ but $\vec{F}_{Fa}\ne -\vec{F}_{Fb}$ thus does not violate the 3rd law. – Popopo Jan 18 at 3:15 Well, we need to include the EM-field to avoid this problem, but it seems EM-field obeys more Relativistic mechanics rather than Newton's. So does the 3rd law also holds in Relativistic mechanics? Besides, it seems quantities such as 'force' and 'acceleration' are not so clear and important in Relativistic mechanics... – Popopo Mar 19 at 2:34 The main thing you have to start out with is to define the current of a single electron using a Delta function: $j(\vec r')=-e\,\delta(\vec r'- \vec{r}(t)) \dot{\vec{r}}(t)$, where $r'(t)$ is the position of the particle. Then everything else (Maxwell equations, Biot-Savart law), should work. - Thank you very much. – Popopo Jan 19 at 13:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 12, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9370323419570923, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/34989/horn-clauses-and-satisfiability/34991	## Horn clauses and satisfiability ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is well known that satisfiability of Horn formulae can be checked in polynomial time using unit propagation. But suppose we relax the condition for horn clauses from at most one un-negated literals to two un-negated literals. Then is it possible to prove that satisfiability of such a formula can be checked in time polynomial in the size of the formula? - ## 2 Answers I think 3SAT can be reduced to your problem, since ($a_1$ OR $a_2$ OR $a_3$) AND ($b_1$ OR $b_2$ OR $b_3$) AND ($c_1$ OR $c_2$ OR $c_3$) AND ... is satisfiable iff (NOT $A_1$ OR $a_2$ OR $a_3$) AND ($A_1$ OR $a_1$) AND (NOT $B_1$ OR $b_2$ OR $b_3$) AND ($B_1$ OR $b_1$) AND (NOT $C_1$ OR $c_2$ OR $c_3$) AND ($C_1$ OR $c_1$) AND ... is. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. In the paper The complexity of satisfiability problems MR0521057, Tom Schaefer characterizes exactly which general classes of satisfiability problems are in P and which are NP-complete. Those problems which are in P fall into six cases: • Every relation in S is satisfied when all variables are 0. • Every relation in S is satisfied when all variables are 1. • Every relation in S is definable by a CNF formula in which each conjunct has at most one negated variable. • Every relation in S is definable by a CNF formula in which each conjunct has at most one unnegated variable. • Every relation in S is definable by a CNF formula having at most 2 literals in each conjunct. • Every relation in S is the set of solutions of a system of linear equation over the two-element field {0,1}. Here, S is a set of boolean relations that one takes as primitives for the language; the associated satisfiability problem is then deciding the satisfiability of a finite conjunction of such primitives. Schaefer moreover shows that any set of relations which does not fall into one of the above has a NP-complete satisfiability problem. In your example, S would be a set of boolean relations definable by a CNF formulas in which each conjunct has at most two unnegated variables. This is not in the above list, so the corresponding satisfiability problem is NP-complete. - 1 It does not seem to be widely known that Schaefer’s dichotomy theorem is wrong as stated in his paper, it only works if $S$ does not contain the empty relation $\varnothing$. In general, the first two cases have to be modified to “Every relation in $S-\{\varnothing\}$ is satisfied when all variables are 0” and “Every relation in $S-\{\varnothing\}$ is satisfied when all variables are 1”, respectively. For example, consider $S=\{\varnothing,x\lor y\lor\neg u\lor\neg v\}$. Then $S$ does not fall into any of the six original cases, nevertheless $\mathrm{SAT}(S)$ is trivially in P: ... – Emil Jeřábek Jun 14 at 14:40 1 given an instance of $\mathrm{SAT}(S)$, either the instance includes the empty relation, in which case it is unsatisfiable, or it does not, in which case it is satisfied by the assignment of all variables to 0 (or to 1, if you prefer). – Emil Jeřábek Jun 14 at 14:42 Thanks for the correction, Emil! I did not know about this... – François G. Dorais♦ Jun 14 at 14:54 The underlying error in the proof is due to a widespread confusion in the literature on Boolean clone and co-clone theory about whether nullary functions are admitted in clones. Many results on clones (e.g., descriptions of Post’s lattice) assume functions have nonzero arity, whereas the empty relation is usually not excluded from co-clones. This breaks down the duality of clones and co-clones: the minimal co-clone and the co-clone generated by the empty relation are distinct, but the only functions that are polymorphisms of the former and not of the latter are the nullary functions. – Emil Jeřábek Jun 14 at 14:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9440141320228577, "perplexity_flag": "head"}
http://nrich.maths.org/1047/note?nomenu=1	## Twenty Divided Into Six Katie had a pack of twenty cards numbered from $1$ to $20$. She arranged the cards into six unequal piles. The numbers on the cards in each pile added to the same total. What was the total and how could this be done? ### Why do this problem? This problem is one that can be accessed easily - everyone can make a start. It offers opportunities for learners to practise addition and subtraction, along with some multiplication and division, and requires a systematic approach. ### Possible approach You could start by asking the group to work on the problem in pairs with digit cards numbered from $1$ to $20$ without saying very much else at this stage. (You could print and cut out cards from this sheet if you do not have enough to go round.) Learners might find it useful to make jottings on mini-whiteboards or paper as they explore the problem. After some time, draw the class together to find out how they are getting on. Invite some pairs to share their approach so far with the whole group. Some children might be using trial and improvement, some may have worked out what the total of each pile needs to be and then used trial and improvement. You may need to talk about how they work out the total of each pile if this does not come up naturally. Can they think of a quick way of doing it without a calculator? They could then continue to work in pairs on the problem. After the initial calculations the problem is a fairly simple one of adding and building the piles but there are many ways of doing it. It would be interesting and instructive to listen to the way that the various pairs are working on the problem, and you may like to gather solutions as a whole class on the board. In a plenary at the end of the lesson, you could talk about how they have found the different solutions and you may want to ask whether they think they have got them all. This might be a good opportunity to share ways of working systematically so that they could convince you they would be able to find every solution. ### Key questions What number must all the piles add to? What is the total of each pile? How do you know you have all the solutions? ### Possible extension Learners could find as many completely different solutions to this problem as possible and some children will be able to suggest a way to find them all. ### Possible support If you want to focus on finding all possibilities, some learners might benefit from using a calculator so they are not held up by the mental arithmetic.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.967845618724823, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/2476/cipher-feedback-mode?answertab=oldest	# Cipher Feedback Mode I can't understand what CFB really is. It said in Wikipedia that CFB is same as CBC, but I find that CFB is more difficult than CBC. Can someone explain to me how CFB works. Such as how Initialization Vector(IV) work inside the Encryption Algorithm process together with the Key, and also how does the ciphertext message working to the next block cipher encryption. And also my other question, is it okay to to use CFB with AES? I am confused because the plaintext message should enter AES algorithm to encrypt the plaintext message, but here in CFB, only Initialization Vector and Secret key will permuted by the block cipher. Here is my another issue with CFB: "To use CFB to make a self-synchronizing stream cipher that will synchronize for any multiple of $x$ bits lost, start by initializing a shift register the size of the block size with the initialization vector. This is encrypted with the block cipher, and the highest $x$ bits of the result are XOR'ed with $x$ bits of the plaintext to produce $x$ bits of ciphertext. These $x$ bits of output are shifted into the shift register, and the process repeats with the next $x$ bits of plaintext. Decryption is similar, start with the initialization vector, encrypt, and XOR the high bits of the result with $x$ bits of the ciphertext to produce $x$ bits of plaintext. Then shift the $x$ bits of the ciphertext into the shift register. This way of proceeding is known as CFB-8 or CFB-1 (according to the size of the shifting). In notation, where $S_i$ is the $i$-th state of the shift register, $a \ll x$ is $a$ shifted up $x$ bits, $\operatorname{head}(a, x)$ is the $x$ highest bits of $a$ and $n$ is number of bits of IV: $$\begin{aligned} C_i &= \operatorname{head}(E_K (S_{i-1}), x) \oplus P_i \\ P_i &= \operatorname{head}(E_K (S_{i-1}), x) \oplus C_i \\ S_i &= ((S_{i-1} \ll x) + C_i) \bmod 2^n \\ S_{0} &= \operatorname{IV} \\ \end{aligned}$$ If $x$ bits are lost from the ciphertext, the cipher will output incorrect plaintext until the shift register once again equals a state it held while encrypting, at which point the cipher has resynchronized. This will result in at most one blocksize of output being garbled." I can't understand the formula and also what is the use of shift register. Can someone help me? - ## 1 Answer Well, with CFB mode, the encryption process is "take the most recent ciphertext block, pass it through the block cipher, and then exclusive-or that with the plaintext block to generate the next ciphertext block". As for the IV, that's used as "the most recent ciphertext block" when encrypting the first plaintext block (where you don't have a most recent ciphertext block yet). Now, the advantage that CFB brings to the table is error recovery, including errors that add or delete ciphertext blocks. That is, is a ciphertext block is garbled (or a block is inserted or removed), the corresponding decrypted plaintext block and the one after that is also garbled; however, after that, the decryption process resyncs, and then it decrypts the rest of the text correctly. Now, CBC also has this same property in regards to entire blocks; however, if you take the last 128 bits of ciphertext, encrypt that, and then use (say) only 8 bits of each AES block output, and exclusive-or that is 8 bits of plaintext to form the next 8 bits of ciphertext, well, now you have something that is resilient against changes that inserts or removes individual bytes from the ciphertext (at the cost of having to run the block cipher N times to encrypt N bytes). Nowadays, this is not considered relevent (we generally want to encrypt and MAC entire messages, and if anything was changed, we reject the entire message); historically, this was considered a nice feature if (say) you were encrypting data going over an RS-232 interface (which really could add/drop individual bytes). The other advantage that CFB mode has (over, say, CBC) is that the decryption process also uses the block cipher in encryption mode; depending on how different the block cipher encryption and decryption is, this can be convienent.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9443114995956421, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/2093/interesting-families-of-sparse-graphs/2104	## Interesting families of sparse graphs? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm interested in graph families which are sparse, and by sparse I mean the number of edges is linear in the number of vertices. \|E\| = O(\|V\|). Besides non-trivial minor-closed families of graphs (these turn out to be sparse), I don't know any other families. Can anyone suggest any interesting graph families (which are not minor-closed) which are sparse? Please don't suggest the trivial family ("the family of sparse graphs"). EDIT: Thanks to the first few people who replied, I realized that bounded degree graphs (max degree < k) also form an interesting and large class of sparse graphs. So perhaps I'll refine my question to exclude those too. Any interesting sparse graph families where the max degree isn't bounded? For example the family of star graphs is sparse and not bounded degree. (But they're minor-closed.) - 2 That depends a lot on what you mean by "interesting." Are you a combinatorialist? A computer scientist? A quiver representation theorist? – Qiaochu Yuan Oct 23 2009 at 14:32 Computer scientist. I'm thinking about graph algorithms for sparse graph families. Hope that helps. – Rune Oct 23 2009 at 15:11 ## 10 Answers 1) Graphs of degree at most d. There are a myriad results known about them. Among them there are regular graphs (graphs in which every vertex is of degree exactly d). 2) Graphs of bounded degeneracy. These are the graphs in which every subgraph has a vertex of degree at most d. For example, stars have unbounded degree, but degeneracy 1. These share some common properties with graphs of bounded degree. For example, there are several results in Ramsey theory that have been extended from graphs of bounded degree to graphs of bounded degeneracy. 3) Random graphs of edge density c/n for constant c. Strictly speaking these are not a 'family of graphs', but since for properties of interest a random graph usually either satisfies it with high probability or fails it with high probability, it is fair to add them. There are also various other models of random graph that yield sparse graphs. 4) On a lower level, there are particular subfamilies of the above, such as stars, grids, etc. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Expander graphs are regular, hence sparse, and have a lot of interesting properties and applications. Elementary Number Theory, Group Theory, and Ramanujan Graphs by Davidoff, Sarnak, and Valette is an excellent introduction. A Google search for expander graphs also turns up a lot of material. - Cayley graphs of groups generated by 2 (or any fixed number) of elements? - Well, for any d, the family of d-regular graphs is sparse in your sense and not minor closed. I don't suppose this is what you're after, so perhaps you can refine the question? EDIT: naturally, there are variations on this theme - graphs of bounded degree, graphs of bounded degree with additional degree restrictions, graphs which cover (in the topological sense) a fixed base graph, ... - That was helpful, thanks! I guess my question is a bit vague, but all you said was helpful. – Rune Oct 23 2009 at 15:38 Although they're not as commonly studied, you can define expander families where the max degree isn't bounded. You can also approach expanders from a spectral perspective (expansion properties have to do with how small the second-largest eigenvalue is), and if you do that, then you can also consider graphs with small third-largest eigenvalue, fourth-largest eigenvalue, etc. These aren't nearly as interesting or important as expanders, but they occasionally have their uses. - Minimally rigid graphs in d dimensions. For d=2 these are the Laman graphs but I don't think any kind of clean combinatorial description of these graphs is known for higher dimensions. - A topological graph which is k-quasi-planar is a topological graph which does not contain k pairwise crossing edges. It is conjectured that for a fixed k, such a graph is sparse. This conjecture is known to hold if k=2 (in that case the graph is planar) and if k=3,4. The conjecture is open as far as I know for larger k. - Let me recommend the book J. Nešetřil and P. Ossona de Mendez, Sparsity: Graphs, Structures, and Algorithms, Springer 2012. It is the first systematic study of sparse graphs and related topics. - You need examples, right? Internet, local networks, social connections :) - One family of graphs that bears unlimited degrees of computational fruit has been dubbed (by me) painted and rooted cacti (PARCs). Painted just means that you can associate any subset of a finite palette $\mathcal{P} = \lbrace p_1, \ldots, p_k \rbrace$ of colors $p_j$ with each node of the underlying rooted cactus. PARCs can be interpreted as propositional calculus formulas in a couple of ways that are logically dual to each other, extending the graphical syntax for propositional logic that C.S. Peirce called his Alpha graphs, with their entitative and existential readings. I'll dig up some links … I'd completely forgotten — here's a sparse exposition of cactus calculus (in the so-called existential interpretation) that I'd already posted on another question. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9546445608139038, "perplexity_flag": "middle"}
http://citizendia.org/Cross-entropy_method	The cross-entropy (CE) method attributed to Reuven Rubinstein is a general Monte Carlo approach to combinatorial and continuous multi-extremal optimization and importance sampling. Monte Carlo methods are a class of Computational Algorithms that rely on repeated Random sampling to compute their results Combinatorial optimization is a branch of optimization. Its domain is optimization problems where the set of Feasible solutions is discrete or can be reduced Continuous optimization is a branch of optimization in Applied mathematics. In Mathematics, the term optimization, or mathematical programming, refers to the study of problems in which one seeks to minimize or maximize a real function In Statistics, importance sampling is a general technique for estimating the properties of a particular distribution, while only having samples generated from a different The method originated from the field of rare event simulation, where very small probabilities need to be accurately estimated, for example in network reliability analysis, queueing models, or performance analysis of telecommunication systems. The CE method can be applied to static and noisy combinatorial optimization problems such as the traveling salesman problem, the quadratic assignment problem, DNA sequence alignment, the max-cut problem and the buffer allocation problem, as well as continuous global optimization problems with many local extrema. The Travelling salesman problem ( TSP) in Operations research is a problem in discrete or Combinatorial optimization. The quadratic assignment problem ( QAP) is one of fundamental Combinatorial optimization problems in the branch of optimization or Operations research In Bioinformatics, a sequence alignment is a way of arranging the Primary sequences of DNA, RNA, or Protein to identify regions of In Graph theory, a cut is a partition of the vertices of a graph into two sets Global optimization is a branch of Applied mathematics and Numerical analysis that deals with the optimization of a function or a set In Mathematics, maxima and minima, known collectively as extrema, are the largest value (maximum or smallest value (minimum that In a nutshell the CE method consists of two phases: 1. Generate a random data sample (trajectories, vectors, etc. ) according to a specified mechanism. 2. Update the parameters of the random mechanism based on the data to produce a "better" sample in the next iteration. This step involves minimizing the cross-entropy or Kullback-Leibler divergence. In Information theory, the cross entropy between two Probability distributions measures the average number of Bits needed to identify an event from a set In Probability theory and Information theory, the Kullback–Leibler divergence (also information divergence, information gain, or relative ## Estimation via importance sampling Consider the general problem of estimating the quantity $\ell = \mathbb{E}_{\mathbf{u}}[H(\mathbf{X})] = \int H(\mathbf{x})\, f(\mathbf{x}; \mathbf{u})\, \textrm{d}\mathbf{x}$, where H is some performance function and $f(\mathbf{x};\mathbf{u})$ is a member of some parametric family of distributions. Using importance sampling this quantity can be estimated as $\hat{\ell} = \frac{1}{N} \sum_{i=1}^N H(\mathbf{X}_i) \frac{f(\mathbf{X}_i; \mathbf{u})}{g(\mathbf{X}_i)}$, where $\mathbf{X}_1,\dots,\mathbf{X}_N$ is a random sample from $g\,$. In Statistics, importance sampling is a general technique for estimating the properties of a particular distribution, while only having samples generated from a different For positive H, the theoretically optimal importance sampling density (pdf)is given by $g^(\mathbf{x}) = H(\mathbf{x}) f(\mathbf{x};\mathbf{u})/\ell$. In Mathematics, a probability density function (pdf is a function that represents a Probability distribution in terms of Integrals Formally a probability This, however, depends on the unknown $\ell$. The CE method aims to approximate the optimal pdf by adaptively selecting members of the parametric family that are closest (in the Kullback-Leibler sense) to the optimal pdf g . In Probability theory and Information theory, the Kullback–Leibler divergence (also information divergence, information gain, or relative ## Generic CE algorithm 1. Choose initial parameter vector $\mathbf{v}^{(0)}$; set t = 1. 2. Generate a random sample $\mathbf{X}_1,\dots,\mathbf{X}_N$ from $f(\cdot;\mathbf{v}^{(t-1)})$ 3. Solve for $\mathbf{v}^{(t)}$, where $\mathbf{v}^{(t)} = \mathop{\textrm{argmax}}_{\mathbf{v}} \frac{1}{N} \sum_{i=1}^N H(\mathbf{X}_i)\frac{f(\mathbf{X}_i;\mathbf{u})}{f(\mathbf{X}_i;\mathbf{v}^{(t-1)})} \log f(\mathbf{X}_i;\mathbf{v})$ 4. If convergence is reached then stop; otherwise, increase t by 1 and reiterate from step 2. In several cases, the solution to step 3 can be found analytically. Situations in which this occurs are • When $f\,$ belongs to the natural exponential family • When $f\,$ is discrete with finite support • When $H(\mathbf{X}) = \mathrm{I}_{\{\mathbf{x}\in A\}}$ and $f(\mathbf{X}_i;\mathbf{u}) = f(\mathbf{X}_i;\mathbf{v}^{(t-1)})$, then $\mathbf{v}^{(t)}$ corresponds to the maximum likelihood estimator based on those $\mathbf{X}_k \in A$. In probability and Statistics, an exponential family is a class of Probability distributions sharing a certain form which is specified below In Topology, a discrete space is a particularly simple example of a Topological space or similar structure one in which the points are " isolated " In Mathematics, the support of a function is the set of points where the function is not zero or the closure of that set Maximum likelihood estimation ( MLE) is a popular statistical method used for fitting a mathematical model to some data ## Continuous optimization—example The same CE algorithm can be used for optimization, rather than estimation. Suppose the problem is to maximize some function S(x), for example, $S(x) = \textrm{e}^{-(x-2)^2} + 0.8\,\textrm{e}^{-(x+2)^2}$. To apply CE, one considers first the associated stochastic problem of estimating $\mathbb{P}_{\boldsymbol{\theta}}(S(X)\geq\gamma)$ for a given level $\gamma\,$, and parametric family $\left\{f(\cdot;\boldsymbol{\theta})\right\}$, for example the 1-dimensional Gaussian distribution, parameterized by its mean $\mu_t\,$ and variance $\sigma_t^2$ (so $\boldsymbol{\theta} = (\mu,\sigma^2)$ here). The normal distribution, also called the Gaussian distribution, is an important family of Continuous probability distributions applicable in many fields Hence, for a given $\gamma\,$, the goal is to find $\boldsymbol{\theta}$ so that $D_{\mathrm{KL}}(\textrm{I}_{\{S(x)\geq\gamma\}}\\|f_{\boldsymbol{\theta}})$ is minimized. This is done by solving the sample version (stochastic counterpart) of the KL divergence minimization problem, as in step 3 above. It turns out that parameters that minimize the stochastic counterpart for this choice of target distribution and parametric family are the sample mean and sample variance corresponding to the elite samples, which are those samples that have objective function value $\geq\gamma$. The worst of the elite samples is then used as the level parameter for the next iteration. This yields the following randomized algorithm for this problem. ### Pseudo-code `1. mu:=-6; sigma2:=100; t:=0; maxits=100; // Initialize parameters2. N:=100; Ne:=10; //3. while t < maxits and sigma2 > epsilon // While not converged and maxits not exceeded4. X = SampleGaussian(mu,sigma2,N); // Obtain N samples from current sampling distribution5. S = exp(-(X-2)^2) + 0. 8 exp(-(X+2)^2); // Evaluate objective function at sampled points6. X = sort(X,S); // Sort X by objective function values (in descending order)7. mu = mean(X(1:Ne)); sigma2=var(X(1:Ne)); // Update parameters of sampling distribution8. t = t+1; // Increment iteration counter9. return mu // Return mean of final sampling distribution as solution` ## References • De Boer, P-T. Simulated annealing (SA is a generic probabilistic Meta-algorithm for the Global optimization problem namely locating a good approximation to the A genetic algorithm (GA is a Search technique used in Computing to find exact or Approximate solutions to optimization and Search Estimation of Distribution Algorithms (EDA sometimes called Probabilistic Model-Building Genetic Algorithms (PMBGA are an outgrowth of Genetic algorithms Tabu search is a mathematical optimization method belonging to the class of local search techniques In Information theory, the cross entropy between two Probability distributions measures the average number of Bits needed to identify an event from a set In Probability theory and Information theory, the Kullback–Leibler divergence (also information divergence, information gain, or relative A randomized algorithm or probabilistic algorithm is an Algorithm which employs a degree of randomness as part of its logic In Statistics, importance sampling is a general technique for estimating the properties of a particular distribution, while only having samples generated from a different , Kroese, D. P, Mannor, S. and Rubinstein, R. Y. (2005). A Tutorial on the Cross-Entropy Method. Annals of Operations Research, 134 (1), 19--67. • Rubinstein, R. Y. (1997). Optimization of Computer simulation Models with Rare Events, European Journal of Operations Research, 99, 89-112. • Rubinstein, R. Y. , Kroese, D. P. (2004). The Cross-Entropy Method: A Unified Approach to Combinatorial Optimization, Monte-Carlo Simulation, and Machine Learning, Springer-Verlag, New York.	{"extraction_info": {"found_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": 29, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8763611912727356, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/commutator+hamiltonian-formalism	Tagged Questions 1answer 56 views Quantum mechanical analogue of conjugate momentum In classical mechanics, we define the concept of canonical momentum conjugate to a given generalised position coordinate. This quantity is the partial derivative of the Lagrangian of the system, with ... 1answer 115 views Canonical transformation and Hamilton's equations I was trying to prove, that for a transformation to be Canonical, one must have a relationship: $$\left\{ Q_a,P_i \right\} = \delta_{ai}$$ Where $Q_a = Q_a(p_i,q_i)$ and $P_a = P_a(p_i,q_i)$. Now ... 3answers 348 views Generalizing Heisenberg Uncertainty Priniciple Writing the relationship between canonical momenta $\pi _i$ and canonical coordinates $x_i$ $$\pi _i =\text{ }\frac{\partial \mathcal{L}}{\partial \left(\frac{\partial x_i}{\partial t}\right)}$$ ... 1answer 174 views Expectation of a commutation relation Is there any significance to: $\langle[H,\hat{O}]\rangle =0$ (which can easily be shown) where $H$ is the Hamiltonian, $\hat{O}$ is an arbitrary operator? Thanks.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8774936199188232, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/154960/wieners-lemma-on-locally-compact-abelian-groups	# Wiener's Lemma on Locally Compact Abelian Groups The version of Wiener's lemma that I know, from Katznelson's Introduction to Harmonic Analysis, is $\lim_{N\to\infty}\frac{1}{2N}\sum_{-N}^{N} \|\hat{\mu}(n)\|^2 =\sum_{t\in\mathbb{T}}\|\mu(\{t\})\|^2$. This is about the group $\mathbb{Z}$ and its dual $\mathbb{T}$. Can this be extended to other groups? I expect/hope this is true for locally compact abelian groups but I have no luck when I search. Wiener has many different lemmas and theorems and formulas, so searching is difficult for me. Any references would be greatly appreciated! By the way, there is this posting on MO where they claim it is true for $\mathbb{R}$: http://mathoverflow.net/questions/64173/a-complex-borel-measure-whose-fourier-transform-goes-to-zero - I'm sure Katznelson does it for $\mathbb R$. It depends on $\mu \bar {\mu}(0) = \sum \mu(t)^2$ and $\hat {\mu \bar{\mu}} = \vert \hat {\mu} \vert^2$ where $\bar {\mu}(x) = \mu(-x)$ ... sec 6.2.9 of Katznelson – mike Jun 7 '12 at 11:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8954735994338989, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2007/07/	# The Unapologetic Mathematician ## Adjoints with parameters Now that we know how to transform adjoints, we can talk about whole families of adjoints parametrized by some other category. That is, for each object $P$ of the parametrizing category $\mathcal{P}$ we’ll have an adjunction $F_P\dashv G_P:\mathcal{C}\rightarrow\mathcal{D}$, and for each morphism of $\mathcal{P}$ we’ll have a transformation of the adjunctions. Let’s actually approach this from a slightly different angle. Say we have a functor $F:\mathcal{C}\times\mathcal{P}\rightarrow\mathcal{D}$, and that for each $P\in\mathcal{P}$ the functor $F(\underline{\hphantom{X}},P):\mathcal{C}\rightarrow\mathcal{D}$ has a right adjoint $G(P,\underline{\hphantom{X}})$. Then I claim that there is a unique way to make $G$ into a functor from $\mathcal{P}^\mathrm{op}\times\mathcal{D}$ to $\mathcal{C}$ so that the bijection $\phi_{C,P,D}:\hom_\mathcal{D}(F(C,P),D)\rightarrow\hom_\mathcal{C}(C,G(P,D))$ is natural in all three variables. Note that $G$ must be contravariant in $P$ here to make the composite functors have the same variance in $P$. If we hold $P$ fixed, the bijection is already natural in $C$ and $D$. Let’s hold $C$ and $D$ fixed and see how to make it natural in $P$. The components $\phi_P:\hom_\mathcal{D}(F(C,P),D)\rightarrow\hom_\mathcal{C}(C,G(P,D))$ are already given in the setup, so we can’t change them. What we need are functions $\hom_\mathcal{D}(F(1_C,p),1_D):\hom_\mathcal{D}(F(C,P'),D)\rightarrow\hom_\mathcal{D}(F(C,P),D)$ and $\hom_\mathcal{C}(1_C,G(p,1_D)):\hom_\mathcal{C}(C,G(P',D))\rightarrow\hom_\mathcal{C}(C,G(P,D))$ for each arrow $p:P\rightarrow P'$. For naturality to hold, we need $\hom_\mathcal{C}(1_C,G(p,1_D))\circ\phi_{P'}=\phi_P\circ\hom_\mathcal{D}(F(1_C,p),1_D)$. But from what we saw last time this just means that the pair of natural transformations $(F(1_C,p),G(p,1_D))$ forms a conjugate pair from $F(\underline{\hphantom{X}},P)\dashv G(P,\underline{\hphantom{X}})$ to $F(\underline{\hphantom{X}},P')\dashv G(P',\underline{\hphantom{X}})$. And this lets us define $G(p,1_D)$ uniquely in terms of $F(1_C,p)$, the counit $\epsilon'$ of $F(\underline{\hphantom{X}},P')\dashv G(P',\underline{\hphantom{X}})$, and the unit $\eta$ of $F(\underline{\hphantom{X}},P)\dashv G(P,\underline{\hphantom{X}})$ by using the first of the four listed equalities. From here, it’s straightforward to show that this definition of how $G$ acts on morphisms of $\mathcal{P}$ makes it functorial in both variables, proving the claim. We can also flip back to the original viewpoint to define an adjunction between categories $\mathcal{C}$ and $\mathcal{D}$ parametrized by the category $\mathcal{P}$ as a functor from $\mathcal{P}$ to the category $\mathbf{Adj}(\mathcal{C},\mathcal{D})$ of adjunctions between those two categories. Posted by John Armstrong \| Category theory \| 4 Comments ## Transformations of Adjoints And now we go back to adjoints. Like every other structure out there, we want to come up with some analogue of a homomorphism between two adjunctions. Let’s consider the adjunctions $F\dashv G:\mathcal{C}\rightarrow\mathcal{D}$ and $F'\dashv G':\mathcal{C}'\rightarrow\mathcal{D}'$, and try to find a good notion of a transformation from the first to the second. We’ll proceed by considering an adjunction to consist of the pair of categories $(\mathcal{C},\mathcal{D})$ with the functors giving extra structure. Down in the land of groups and rings and such, we’d consider sets with extra structure and functions that preserved that structure. So here, naturally, we want to consider functors which preserve this extra structure. That is, a map of adjunctions consists of a pair of functors $K:\mathcal{C}\rightarrow\mathcal{C}'$ and $L:\mathcal{D}\rightarrow\mathcal{D}'$. These must preserve the structure in that $K\circ G=G'\circ L$ and $L\circ F=F'\circ K$. But hold up a second, we’ve forgotten something else that goes into an adjunction: the isomorphism $\phi:\hom_\mathcal{D}(F(C),D)\rightarrow\hom_\mathcal{C}(C,G(D))$. Here’s a diagram showing how the map of adjunctions should play nicely with them: Equivalently we can specify an adjunction by its unit and counit. In this case the compatibility in question is a pair of equations of natural transformations: $1_K\circ\eta=\eta'\circ1_K$ and $1_L\circ\epsilon=\epsilon'\circ1_L$. What if we’re looking at two different adjunctions between the same pair of categories? Well then we may as well try to use the appropriate identity functors for $K$ and $L$. But then it’s sort of silly to insist that $G=1_\mathcal{C}\circ G=G'\circ\mathcal{D}=G'$ on the nose, and similarly for $F'$. Instead, as we do so often, let’s weaken this equality to just a natural transformation. We’ll say that a pair of natural transformations $\sigma:F\rightarrow F'$ and $\tau:G'\rightarrow G$ are “conjugate” if $\hom_\mathcal{C}(1_C,\tau_D)\circ\phi'_{C,D}=\phi_{C,D}\circ\hom_\mathcal{D}(\sigma_C,1_D)$. This is equivalent, in terms of the unit and counit, to any one of the following four equalities: • $\tau=(1_G\circ\epsilon')\cdot(1_G\circ\sigma\circ1_{G'})\cdot(\eta\circ1_{G'})$ • $\sigma=(\epsilon\circ1_{F'})\cdot(1_F\circ\tau\circ1_{F'})\cdot(1_F\circ\eta')$ • $\epsilon\cdot(1_F\circ\tau)=\epsilon'\cdot(\sigma\circ1_{G'})$ • $(1_G\circ\sigma)\cdot\eta=(\tau\circ1_{F'})\cdot\eta'$ Now it’s easily verified that given a pair of categories $(\mathcal{C},\mathcal{D})$ we can build a category whose objects are adjunctions $F\dashv G:\mathcal{C}\rightarrow\mathcal{D}$ and whose morphisms are conjugate pairs of natural transformations, which we write out in full as $(\sigma,\tau):(F,G,\phi)\rightarrow(F',G',\phi'):\mathcal{C}\rightarrow\mathcal{D}$. We compose conjugate pairs in this category in the obvious way, which we write $(\sigma',\tau')\cdot(\sigma,\tau)$. On the other hand, if we have a pair $(\sigma,\tau):(F,G,\phi)\rightarrow(F',G',\phi'):\mathcal{C}\rightarrow\mathcal{D}$ and another $(\bar{\sigma},\bar{\tau}):(\bar{F},\bar{G},\bar{\phi})\rightarrow(\bar{F}',\bar{G}',\bar{\phi}'):\mathcal{D}\rightarrow\mathcal{E}$, then we can form the composite $(\bar{\sigma}\circ\sigma,\tau\circ\bar{\tau}):(\bar{F}\circ F,G\circ\bar{G},\bar{\phi}\cdot\phi)\rightarrow(\bar{F}'\circ F',G'\circ\bar{G}',\bar{\phi}'\cdot\phi'):\mathcal{C}\rightarrow\mathcal{E}$, which we’ll write as $(\bar{\sigma},\bar{\tau})\circ(\sigma,\tau)$. Notice the similarity of this situation with the two different compositions of natural transformations between functors. Posted by John Armstrong \| Category theory \| 3 Comments ## The Simplicial Category There’s another approach to the theory of monoids which finds more direct application in topology and homology theory (which, yes, I’ll get to eventually) — the “simplicial category” $\mathbf{\Delta}$. Really it’s an isomorphic category to $\mathrm{Th}(\mathbf{Mon})$, but some people think better in these other terms. I personally like the direct focus on the algebra, coupled with the diagrammatics so reminiscent of knot theory, but for thoroughness’ sake I’ll describe the other approach. Note that the objects of $\mathrm{Th}(\mathbf{Mon})$ correspond exactly with the natural numbers. Each object is the monoidal product of some number of copies of the generating object $M$. We’re going to focus here on the model of $\mathbb{N}$ given by the ordinal numbers. That is, the object $M^{\otimes n}$ corresponds to the ordinal number $\mathbf{n}$, which is a set of $n$ elements with its unique (up to isomorphism) total order. In fact, we’ve been implicitly thinking about an order all along. When we draw our diagrams, the objects consist of a set of marked points along the upper or lower edge of the diagram, which we can read in order from left to right. Let’s pick a specific representation of each ordinal to be concrete about this. The ordinal $\mathbf{n}$ will be represented by the set of natural numbers from ${0}$ to $n-1$ with the usual order relation. The monoidal structure will just be addition — $\mathbf{m}\otimes\mathbf{n}=\mathbf{m+n}$. The morphisms between ordinals are functions which preserve the order. A function $f:X\rightarrow Y$ between ordinals satisfies this property if whenever $i\leq j$ in $X$ then $f(i)\leq f(j)$ in $Y$. Note that we can send two different elements of $X$ to the same element of $Y$, just as long as we don’t pull them past each other. So what sorts of functions do we have to play with? Well, we have a bunch of functions from $\mathbf{n}$ to $\mathbf{n+1}$ that skip some element of the image. For instance, we could send $\mathbf{3}$ to $\mathbf{4}$ by sending ${0}$ to ${0}$, skipping $1$, sending $1$ to $2$, and sending $2$ to $3$. We’ll say $\delta^n_i:\mathbf{n}\rightarrow\mathbf{n+1}$ for the function that skips $i$ in its image. The above function is then $\delta^3_1$. For a fixed $n$, the index $i$ can run from ${0}$ to $n$. We also have a bunch of functions from $\mathbf{n+1}$ to $\mathbf{n}$ that repeat one element of the image. For example, we could send $\mathbf{4}$ to $\mathbf{3}$ by sending ${0}$ to ${0}$, $1$ and $2$ both to $1$, and $3$ to $2$. We’ll say $\sigma^n_i:\mathbf{n+1}\rightarrow\mathbf{n}$ for the function that repeats $i$ in its image. The above function is then $\sigma^3_1$. Again, for a fixed $n$, the index $i$ can run from ${0}$ to $n-1$. Notice in particular that “skipping” and “repeating” are purely local properties of the function. For instance, $\delta^0_0$ is the unique function from $\mathbf{0}$ (the empty set) to $\mathbf{1}$, which clearly skips $0\in\mathbf{1}$. Then $\delta^n_i$ can be written as $1_i\otimes\delta^0_0\otimes1_{n-i}$, since it leaves the numbers from ${0}$ to $i-1$ alone, sticks in a new $i$, and then just nudges over everything from (the old) $i$ to $n$. Similarly, $\sigma^1_0$ is the unique function from $\mathbf{2}$ to $\mathbf{1}$ that sends both elements in its domain to $0\in\mathbf{1}$. Then all the other $\sigma^n_i$ can be written as $1_i\otimes\sigma^0_0\otimes1_{n-i-1}$. Now every order-preserving function is determined by the set of elements of the range that are actually in the image of the function along with the set of elements of its domain where it does not increase. That is, if we know where it skips and where it repeats, we know the whole function. This tells us that we can write any function as a composition of $\delta$ and $\sigma$ functions. These basic functions satisfy a few identities: • If $i\leq j$ then $\delta^{n+1}_i\circ\delta^n_j=\delta^{n+1}_{j+1}\circ\delta^n_i$. • If $i\leq j$ then $\sigma^{n-1}_j\circ\sigma^n_i=\sigma^{n-1}_i\circ\sigma^n_{j+1}$. • If $i<j$ then $\sigma^n_j\circ\delta^n_i=\delta^{n-1}_i\circ\sigma^{n-1}_{j-1}$. • If $i=j$ or $i=j+1$ then $\sigma^n_j\circ\delta^n_i=1$. • If $i>j+1$ then $\sigma^n_j\circ\delta^n_i=\delta^{n-1}_{i-1}\circ\sigma^{n-1}_j$. We could check all these by hand, and if you like that sort of thing you’re welcome to it. Instead, I’ll just assume we’ve checked the second one for $n=2$ and the fourth one for $n=1$. What’s so special about those conditions? Well, notice that $\sigma^1_0:\mathbf{1}\otimes\mathbf{1}\rightarrow\mathbf{1}$ takes two copies of $\mathbf{1}$ to one copy, and that the second relation becomes the associativity condition for this morphism. Then also $\delta^0_0:\mathbf{0}\rightarrow\mathbf{1}$ takes zero copies to one copy, and the fourth relation becomes the left and right identity conditions. That is, $\mathbf{1}$ with these two morphisms is a monoid object in this category! Now we can verify all the other relations by using our diagrams rather than a lot of messy calculations! We can also go back the other way, breaking any of our diagrams into basic pieces and translating each piece into one of the $\delta$ or $\sigma$ functions. The category of ordinal numbers not only contains a monoid object, it is actually isomorphic to the “theory of monoids” functor — it contains the “universal” monoid object. So why bother with this new formulation at all? Well, for one thing it’s always nice to see the same structure instantiated in many different ways. Now we have it built from the ground up as $\mathrm{Th}(\mathbf{Mon})$, we have it implemented as a subcategory of $\mathcal{OTL}$, we have it as the category of ordinal numbers, and thus we also have it as a full subcategory of $\mathbf{Cat}$ — the category of all small categories (why?). There’s another reason, though, which won’t really concern us for a while yet. The morphisms $\delta^n_i$ and $\sigma^n_i$ turn out to be very well-known to topologists as “face” and “degeneracy” maps when working with shapes they call “simplicial complexes”. Not only is this a wonderful oxymoron, it’s the source of the term “simplicial category”. If you know something about topology or homology, you can probably see how these different views start to tie together. If not, don’t worry — I’ll get back to this stuff. Posted by John Armstrong \| Category theory \| 5 Comments ## The General Associative Law For any monoid object we have an associative law for the multiplication: $\mu\circ(\mu\otimes1_X)=\mu\circ(1_X\otimes\mu)$. This basically says that the two different ways of multiplying together three inputs to give one output are the same. Let’s call the result $\mu_3:X^{\otimes3}\rightarrow X$. In fact, we might go so far as to say $\mu_2=\mu:X^{\otimes2}\rightarrow X$, $\mu_1=1_X:X^{\otimes1}\rightarrow X$, and even $\mu_0=1:X^{\otimes0}\rightarrow X$. This generalizes a lot. We want to say that there’s a unique way (called $\mu_n$) to multiply together $n$ inputs. The usual way is to pick some canonical form and show that everything can be reduced to that form. This ends up being a lot like the Coherence Theorem. In fact, if we take a monoid object in the category $\mathbf{Cat}$ of small categories, this is the Coherence Theorem for a strict monoidal category. But there’s an easier way than walking through that big proof again, and it uses our diagrammatic approach! The first thing we need to realize is that if we can show this rule holds in $\mathrm{Th}(\mathbf{Mon})$, then it will hold for all monoid objects. That’s why the “theory of monoids” category is so nice — it exactly encodes the structure of a monoid. Anything that is true for all monoids can be proved by just looking at this category and proving it there! So how do we show that the general associative law holds in $\mathrm{Th}(\mathbf{Mon})$? Now we need to notice that the functor that makes $\downarrow\otimes\uparrow$ into a monoid object is faithful. That is, if two Temperley-Lieb diagrams in the image are the same, then they must come from the same morphism in $\mathrm{Th}(\mathbf{Mon})$. But if two diagrams are equivalent they differ by either sliding loops arcs around in the plane — which uses the monoidal structure to pull cups or caps past each other — or by using the zig-zag identities — which encode the left and right identity laws. Thus any equalities that hold in the image of the functor must come from equalities already present in $\mathrm{Th}(\mathbf{Mon})$! Now any way of multiplying together $n$ inputs to give one output is a morphism $f:M^{\otimes n}\rightarrow M$ in $\mathrm{Th}(\mathbf{Mon})$, which will be sent to a diagram $F(f):(\downarrow\otimes\uparrow)^{\otimes n}\rightarrow\downarrow\otimes\uparrow$ in $\mathcal{OTL}$. It’s not too hard to see that there’s really only one of these diagrams that could be in the image of the functor (up to equivalence of diagrams). So all such multiplications are sent to the same diagram. By the faithfulness above, this means that they were all the same morphism in $\mathrm{Th}(\mathbf{Mon})$ to begin with, and we’re done. By the way, you should try playing around with the oriented Temperley-Lieb diagrams to verify the claim I made of uniqueness. Try to work out exactly what diagrams are in $\hom_\mathcal{OTL}((\downarrow\otimes\uparrow)^{\otimes n},\downarrow\otimes\uparrow)$, and then which ones can possibly be in the image of the functor. Playing with the diagrams like this should give you a much better intuition for how they work. If nothing else, drawing a bunch of pictures is a lot more fun than algebra homework from back in school. ## More monoid diagrams Let’s pick up with the diagrams for monoid objects from yesterday. In fact, let’s draw the multiplication and unit diagrams again, but this time let’s make the lines really thick. Now we’re looking at something more like a region of the plane than a curve. We really don’t need all that inside part, so let’s rub it out and just leave the outline. Of course, whenever we go from a blob to its outline we like to remember where the blob was. We do this by marking a direction on the outline so if we walk in that direction the blob would be on our left. Those of you who have taken multivariable calculus probably have a vague recollection of this sort of thing. Don’t worry, though — we’re not doing calculus here. Okay, now the outline diagrams look like this: That’s odd. These diagrams look an awful lot like Temperley-Lieb diagrams. And in fact they are! In fact, we get a functor from $\mathrm{Th}(\mathbf{Mon})$ to $\mathcal{OTL}$ that sends $M$ to $\downarrow\otimes\uparrow$. That is, a downward-oriented strand next to an upward-oriented strand makes a monoid object on $\mathcal{OTL}$! But to be sure of this, we need to check that the associativity and identity relations hold. Here’s associativity: Well that’s pretty straightforward. It’s just sliding the arcs around in the plane. How about the identity relations? The right identity relation holds because of one of the “zig-zag” relations for duals, and the left identity relation holds because of the other! Now you should be able to find a comonoid object in $\mathcal{OTL}$ in a very similar way. Posted by John Armstrong \| Category theory \| 3 Comments ## Diagrammatics for Monoid Objects I don’t know about you, but all this algebraic notation starts to blur together. Wouldn’t it be nice if we could just draw pictures? Well luckily for use we can! Just like we had diagrams for braided categories, categories with duals, and braided categories with duals, we have certain diagrammatics to help us talk about monoid objects. First off, we think of our generating object as a point on a line. As we tensor copies of this object together, we just add more points. Then our morphisms will be diagrams in the plane. At the bottom of the diagram is the incoming object — a bunch of marked points — and at the top is the outgoing object — another bunch of marked points. In between, we have morphisms we can build from the two basic pieces we added: multiplication and unit. See? For multiplication, two points come in. They move together and multiply, leaving one point to go out. For the unit, a point comes “out of nowhere” to leave the diagram. As before, we set two diagrams side-by-side for the monoidal product and stack them top-to-bottom for composition. Now, what do those associativity and identity relations look like? Neat! Associativity just means we can pull the branch in the middle to either side of the threefold multiplication, while identity means we can absorb a dangling free end. I haven’t bothered to render a diagram for symmetry, but we can draw it by just having lines cross through each other. The naturality of the symmetry means that we can pull any morphism from one side of a crossing line to the other. And now what about comonoid objects? We’ve got diagrams to talk about them too! Here’s a comultiplication and a counit. We just flip the multiplication and unit upside-down to dualize them. And we do the same thing for the coassociativity and coidentity relations. The one thing we have to take careful note of here is that everything in sight is strict. These diagrams don’t make any distinction between $(M\otimes M)\otimes M$ and $M\otimes(M\otimes M)$; or between $M$, $\mathbf{1}\otimes M$, and $M\otimes\mathbf{1}$. Posted by John Armstrong \| Category theory \| 2 Comments ## Examples of Monoid Objects It’s all well and good to define monoid objects, but it’s better to see that they subsume a lot of useful concepts. The basic case is, of course, that a monoid object in $\mathbf{Set}$ is a monoid. Another example we’ve seen already is that a ring with unit is a monoid object in $\mathbf{Ab}$ — the category of abelian groups with the tensor product of abelian groups as the monoidal structure. Similarly, given a commutative ring $K$, a monoid object in the category $K\mathbf{-mod}$ with tensor product of $K$-modules as its monoidal structure is a $K$-algebra with unit. For extra credit, how would we get rings and $K$-algebras without units? Here’s one we haven’t seen (and which I’ll talk more about later): given any category $\mathcal{C}$, the category of “endofunctors” $\mathcal{C}^\mathcal{C}$ has a monoidal structure given by composition of functors from $\mathcal{C}$ to itself. This is the one I was thinking of that doesn’t have a symmetry, by the way. A monoid object in this category consists of a functor $T:\mathcal{C}\rightarrow\mathcal{C}$ along with natural transformations $\mu:T\circ T\rightarrow T$ and $\eta:1_\mathcal{C}\rightarrow T$. These turn out to be all sorts of useful in homology theory, and also in theoretical computer science. In fact, the programming language Haskell makes extensive and explicit use of them. And now for a really interesting class of examples. Let’s say we start with a monoidal category $\mathcal{C}$ with monoidal structure $\otimes$. We immediately get a monoidal structure $\otimes^\mathrm{op}$ on the opposite category $\mathcal{C}^\mathrm{op}$. Just define $A\otimes^\mathrm{op}B=A\otimes B$ for objects. For morphisms we take $f:A\rightarrow C$ and $g:B\rightarrow D$ (which are in $\hom_{\mathcal{C}^\mathrm{op}}(C,A)$ and $\hom_{\mathcal{C}^\mathrm{op}}(D,B)$, respectively), and define $f\otimes^\mathrm{op}g=f\otimes g$, which is in $\hom_{\mathcal{C}^\mathrm{op}}(C\otimes^\mathrm{op}D,A\otimes^\mathrm{op}B)$. So what’s a monoid object in $\mathcal{C}^\mathrm{op}$? It’s a contravariant functor from $\mathrm{Th}(\mathbf{Mon})$ to $\mathcal{C}$. Equivalently, we can write it as a covariant functor from $\mathrm{Th}(\mathbf{Mon})^\mathrm{op}$ to $\mathcal{C}$. It will be easier to just write down explicitly what this opposite category is. So we need to take $\mathrm{Th}(\mathbf{Mon})$ and reverse all the arrows. It’s enough to just reverse the arrows we threw in to generate the category, and their composites will be reversed as well. We’ll also have to dualize the relations we imposed to make everything work out right. So we’ll have an arrow $\Delta:M\rightarrow M\otimes M$ called comultiplication and another arrow $\epsilon:M\otimes\mathbf{1}$ called the counit. These we require to satisfy the coassociative condition $(\Delta\otimes1_M)\circ\Delta=(1_M\otimes\Delta)\circ\Delta$ and the left and right coidentity conditions $(\epsilon\otimes1_M)\circ\Delta=1_M=(1_M\otimes\epsilon)\circ\Delta$. Now a functor from this category to another monoidal category picks out an object $C\in\mathcal{C}$ and arrows (reusing the names) $\epsilon:C\rightarrow\mathbf{1}$ and $\Delta:C\rightarrow C\otimes C$ satisfying coassociativity and coidentity conditions. We call such an object with extra structure a “comonoid object” in $\mathcal{C}$. In $\mathbf{Set}$ we call them “comonoids”. In $\mathbf{Ab}$ we call them “corings” (with counit), in $K\mathbf{-mod}$ we call them “coalgebras” (with counit), and in $\mathcal{C}^\mathcal{C}$ we call them “comonads”. In general, we call this new model category $\mathrm{Th}(\mathbf{CoMon})$ — the “theory of comonoids”. Posted by John Armstrong \| Category theory \| 4 Comments ## Tell it to Einstein! I’m adding another new link, this time to God Plays Dice. This one is run by a mysterious and shadowy figure known only as “The Probabilist”. I don’t know why, though. There’s a lot of great stuff here, very accessible to a general audience. In fact, it’s rather like another direction I could have gone with this site six months ago, but I think “The Probabilist” does a better job of it than I would have. So let this also be a call for “The Probabilist” to unmask and accept credit for this work! I’ve already figured out the secret, and I imagine others have as well, so we’re all just waiting for the other shoe to drop. However, I will respect “The Probabilist”‘s pseudonymity, however little I understand it. Posted by John Armstrong \| Uncategorized \| 1 Comment ## Monoid Objects Now it’s time to start getting into the fun things we can do with monoidal categories. For my first trick, I’m going to build a neat monoidal category $\mathcal{M}$ and show you what we can do with it. Any monoidal category has an “identity” object $\mathbf{1}$, so to make it a bit more interesting let’s throw in a single non-identity object $M$. Then we get for free all the monoidal products built with $M$ and $\mathbf{1}$. Let’s make our lives easier by saying our category is strict. Then all our objects look like $M^{\otimes n}$ — the monoidal product of $n$ copies of $M$. We can see that $\mathbf{1}=M^{\otimes0}$, and that $M^{\otimes n_1}\otimes M^{\otimes n_2}=M^{\otimes(n_1+n_2)}$. This is all well and good, but we still don’t really have much going on here. All the morphisms in sight are identities. We don’t even have associators or unit isomorphisms because our category is strict. So let’s throw in a couple morphisms, and of course all the other ones we can build from them. First let’s make our category symmetric. That is, we’ll add a “twist” $\tau:M\otimes M\rightarrow M\otimes M$ that swaps the copies of $M$. We’ll insist that it satisfy $\tau^2=1_{M\otimes M}$. We can then build a braiding $\beta_{M^{\otimes n_1},M^{\otimes n_2}}:M^{\otimes n_1}\otimes M^{\otimes n_2}$ by swapping the copies of $M$ one at a time. This seems a little silly at first glance. If $M$ had any additional structure — if it was a set, for instance — this would be clearly useful. As it stands, though, the use isn’t apparent. Don’t worry, we’ll get to it. Next, let’s add a morphism $e:\mathbf{1}\rightarrow M$. From this we can get a bunch of other morphisms. For example, $1_M\otimes e:M\otimes\mathbf{1}\rightarrow M\otimes M$ or $e\otimes1_M:\mathbf{1}\otimes M\rightarrow M\otimes M$. We can use this one to increase the number of copies of $M$ in a product in many different ways, depending on where we stick the new copy of $M$. But we could also add a new copy of $M$ in one place and use the symmetric structure to move it to a different place. For example, instead of adding a copy on the right with $1_M\otimes e$, we could instead use $\tau\circ(e\otimes1_M)$ to add a copy on the left and then swap the two. Notice also that $1_M=\beta_{\mathbf{1},M}$ and $\tau=\beta_{M,M}$, which means that these two morphisms are $(1_M\otimes e)\circ\beta_{\mathbf{1},M}$ and $\beta_{M,M}\circ(e\otimes1_M)$. The naturality of $\beta$ says that these two are really the same. So, adding a new copy of $M$ and then moving it around immediately to another position is the same as just adding it in the new position right away. Now let’s add a way to reduce the number of copies. We’ll use a morphism $m:M\otimes M\rightarrow M$. Of course, we get for free such compositions as $(m\otimes1_M)\circ(1_M\otimes\tau):M^{\otimes3}\rightarrow M^{\otimes2}$ and $m\circ(e\otimes1_M):M\rightarrow M$. There will be some equalities arising from the naturality of $\beta$, but nothing too important yet. So let’s throw in a few more equalities. Let’s say that $m\circ(m\otimes1_M)=m\circ(1_M\otimes m)$ and that $m\circ(e\otimes1_M)=1_M=m\circ(1_M\otimes e)$. And of course there are other equalities we can build from these. The whole thing should start looking a bit familiar by this point. Okay, so we’ve got ourselves a strict monoidal category $\mathcal{M}$ with a bunch of objects and a few morphisms satisfying some equations. So what? Well, let’s start looking at symmetric monoidal functors from $\mathcal{M}$ into other symmetric monoidal categories. The first monoidal category we’ll look at is $\mathbf{Set}$, which uses the cartesian product as its monoidal structure. What does a monoidal functor $F:\mathcal{M}\rightarrow\mathbf{Set}$ look like? Well, $F(M)$ is some set $X$, and by monoidality we see that $F(M^{\otimes n})=X^{\times n}$ — the cartesian product of $n$ copies of $X$. In particular, $F(\mathbf{1})=\{\}$: a set with a single element. The symmetry for $\mathbf{Set}$ is the natural isomorphism $\beta_{A,B}:A\times B\rightarrow B\times A$ defined by $\beta_{A,B}: (a,b)\mapsto(b,a)$. In particular, we get $F(\tau)=\beta_{X,X}: (x_1,x_2)\mapsto(x_2,x_1)$. The morphism $e:\mathbf{1}\rightarrow M$ now becomes $F(e)=1:\{\}\rightarrow X$, which picks out a particular point of $X$. Let’s call this point $1$, just like the function that picks it out. The morphism $m:M\otimes M\rightarrow M$ is now a function $F(m)=\mu:X\otimes X\rightarrow X$. The equations that we imposed in $\mathcal{M}$ must still apply here: $\mu\circ(\mu\times1_X)=\mu\circ(1_X\times\mu)$ and $\mu\circ(1\otimes1_X)=1_X=\mu\circ(1_X\otimes1)$. Since we’re in the category of sets, let’s just write these all out as functions and see what they do to particular elements. The first equation is between two functions with source $X\times X\times X$, so let’s pick an arbitary element $(x_1,x_2,x_3)$ to follow. The left side of the equation sends this to $\mu(\mu(x_1,x_2),x_3)$, while the right sends it to $\mu(x_1,\mu(x_2,x_3))$. The equation now reads $\mu(\mu(x_1,x_2),x_3)=\mu(x_1,\mu(x_2,x_3))$. But that’s just the associative law for a composition! The second equation is between three functions that all have source $X$. Starting with an arbitrary element $x$ we read off the equation $\mu(1,x)=x=\mu(x,1)$. And that’s the left and right unit law for a composition! So what we see here is that $X=F(M)$ gets the structure of a monoid. And given any monoid $X$ we can construct such a symmetric monoidal functor with $F(M)=X$ and sending $m$ and $e$ to the multiplication and identity functions. Can we do better? Sure we can. Let’s say we’ve got a homomorphism between two monoids $f:X_1\rightarrow X_2$. We can consider this to be a function between their underlying sets. Immediately we get $f^{\times n}:X_1^{\times n}\rightarrow X_2^{\times n}$ as well, applying $f$ to each entry of the product. This is clearly symmetric. Saying that $f$ preserves the multiplication of these monoids is just the same as saying that $f\circ F_1(m)=F_2(m)\circ(f\times f)$, which is the naturality square for $m$. Similarly, preserving the identities is the same as making the naturality square for $e$ commute. So a monoid homomorphism is the same as a natural transformation between these functors! Let’s back up a bit and give our toy category a better name. Let’s call it $\mathrm{Th}(\mathbf{Mon})$ — the “theory of monoids”. What we’ve just seen is that our familiar category $\mathbf{Mon}$ of monoids is “really” the category $\mathbf{Set}^{\mathrm{Th}(\mathbf{Mon})}$ of symmetric monoidal functors from the “theory of monoids” to sets. We now slightly shift our terminology and instead of calling such a set-with-extra-structure a “monoid”, we call it a “monoid object in $\mathbf{Set}$“. And now the road is clear to generalize. Given any symmetric monoidal category $\mathcal{C}$ we can take the category $\mathcal{C}^{\mathrm{Th}(\mathbf{Mon})}$ of “monoid objects in $\mathcal{C}$“. [UPDATE]: On reflection, the symmetric property isn’t really essential. That is, we can just consider the category of monoidal functors from $\mathrm{Th}(\mathbf{Mon})$ to $\mathcal{C}$. In fact, there’s one example I’ll be getting to that doesn’t have a symmetry. In general, though, when the target category has a symmetry we’ll usually ask that our functors preserve that structure as well. [UPDATE]: You know what? Scrap that whole symmetry bit altogether. Sometimes the target category will have symmetry and sometimes that will be helpful, but it’s just not worth it in the general theory. I’m almost sorry I brought it up in the first place. Posted by John Armstrong \| Category theory \| 13 Comments ## Adjoints Preserve Limits We can easily see that limits commute with each other, as do colimits. If we have a functor $F:\mathcal{J}_1\times\mathcal{J}_2\rightarrow\mathcal{C}$, then we can take the limit $\varprojlim_{\mathcal{J}_1\times\mathcal{J}_2}F$ either all at once, or one variable at a time: $\varprojlim_{\mathcal{J}_1}\varprojlim_{\mathcal{J}_2}F=\varprojlim_{\mathcal{J}_2}\varprojlim_{\mathcal{J}_1}F$. That is, if the category $\mathcal{C}$ has $\mathcal{J}$-limits, then the functor $\varprojlim_{J}$ preserves all other limits. But now we know that limit functors are right adjoints. And it turns out that any functor which has a left adjoint (and thus is a right adjoint) preserves all limits. Dually, any functor which has a right adjoint (and thus is a left adjoint) preserves all colimits. First we need to note that we can compose adjunctions. That is, if we have adjunctions $F_1\dashv F_2:\mathcal{C}\rightarrow\mathcal{D}$ and $G_1\dashv G_2:\mathcal{D}\rightarrow\mathcal{E}$ then we can put them together to get an adjunction $G_1\circ F_1\dashv F_2\circ G_2:\mathcal{C}\rightarrow\mathcal{E}$. Indeed, we have $\hom_\mathcal{E}(G_1(F_1(C)),E)\cong\hom_\mathcal{D}(F_1(C),G_2(E))\cong\hom_\mathcal{C}(C,F_2(G_2(E)))$ We also need to note that adjoints are unique up to natural isomorphism. That is, if $F\dashv G_1:\mathcal{C}\rightarrow\mathcal{D}$ and $F\dashv G_2:\mathcal{C}\rightarrow\mathcal{D}$ then there is a natural isomorphism $G_1\cong G_2$. This is essentially because adjunctions are determined by universal arrows, and universal arrows are unique up to isomorphism. Okay, now we can get to work. We start with an adjunction $F\dashv G:\mathcal{C}\rightarrow\mathcal{D}$. Given another (small) category $\mathcal{J}$ we can build the functor categories $\mathcal{C}^\mathcal{J}$ and $\mathcal{D}^\mathcal{J}$. It turns out we get an adjunction here too. Define $F^\mathcal{J}(S)=F\circ S$ for each functor $S:\mathcal{J}\rightarrow\mathcal{C}$. The unit $\eta:1_\mathcal{C}\rightarrow G\circ F$ induces a unit $\eta^\mathcal{J}_S=\eta\circ1_S:S\rightarrow G\circ F\circ S$. We can similarly define $G^\mathcal{J}$ and $\epsilon^\mathcal{J}$, and show that they determine an adjunction $F^\mathcal{J}\dashv G^\mathcal{J}:\mathcal{C}^\mathcal{J}\rightarrow\mathcal{D}^\mathcal{J}$ Now let’s say that $\mathcal{C}$ and $\mathcal{D}$ both have $\mathcal{J}$-limits. Then we have an adjunction $\Delta\dashv\varprojlim_\mathcal{J}:\mathcal{C}\rightarrow\mathcal{C}^\mathcal{J}$ and a similar one for $\mathcal{D}$. We can thus form the composite adjunctions $F^\mathcal{J}\circ\Delta\dashv\varprojlim_\mathcal{J}\circ G^\mathcal{J}:\mathcal{C}\rightarrow\mathcal{D}^\mathcal{J}$ $\Delta\circ F\dashv G\circ\varprojlim_\mathcal{J}:\mathcal{C}\rightarrow\mathcal{D}^\mathcal{J}$ So what is $F^\mathcal{J}(\Delta(C))$? Well, $\Delta(C)$ is the functor that sends every object of $\mathcal{J}$ to $C$ and every morphism to $1_C$. Then composing this with $F$ gives the functor that sends every object of $\mathcal{J}$ to $F(C)$ and every morphism to $1_{F(C)}$. That is, we get $\Delta(F(C))$. So $F^\mathcal{J}\circ\Delta=\Delta\circ F$. But these are the two left adjoints listed above. Thus the two right adjoints listed above are both right adjoint to the same functor, and therefore must be naturally isomorphic! We have $\varprojlim_\mathcal{J}G\circ T\cong G(\varprojlim_\mathcal{J}T)$ for every functor $T:\mathcal{J}\rightarrow\mathcal{D}$. And thus $G$ preserves $\mathcal{J}$-limits. Posted by John Armstrong \| Category theory \| 2 Comments ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for 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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 377, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9265561699867249, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/22442/list	## Return to Answer 3 (spelling) changed Noncomutative to Noncommutative The following pearl by Jacobson can under no circumstances be left out from the list: Let $\mathbf{R}$ be a ring with center $\mathrm{Z}$. Let us suppose that you can find $n \in \mathbb{N}_{>1}$ such that $x^{n}-x \in \mathrm{Z}$ for every $x \in \mathbf{R}$. Then $\mathbf{R}$ is a commutative ring. 2 added 21 characters in body; deleted 8 characters in body; deleted 11 characters in body The following pearl by Jacobson can under no circumstances be left out from the list: Let $\mathbf{R}$ be a ring with center $\mathrm{Z}$ \mathrm{Z}$. Let us suppose that you can find$n \in \mathbb{N}_{>1}$such that$x^{n}-x \in \mathrm{Z}$for every$x \in \mathbf{R}$and a fixed natural number$n$greater than$1$. mathbf{R}$. Then $\mathbf{R}$ is a commutative ring. In Herstein's Noncomutative rings you can find a delightful treatment of Let $\mathbf{R}$ be a ring with center $\mathrm{Z}$ such that $x^{n}-x \in \mathrm{Z}$ for every $x \in \mathbf{R}$ and a fixed natural number $n$ greater than $1$. Then $\mathbf{R}$ is a commutative ring.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 22, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8874104022979736, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/152065-induction.html	# Thread: 1. ## Induction Is the the correct approach to work through this problem 23n – 1 is divisible by 7 for all n>=1 n(1) 23(1) – 1 = (222) – 1 = 7 23n – 1 = 7m n(k+1) 23(k+1) – 1 = 2.23k-1 = 2 * (7m+1) – 1 = 14m + 2 -1 = 14m-1 =7(2m-1) Where (2m-1) is an int. So 23n – 1 is divisible by 7. Therefore, by P.M.I 23n – 1 is divisible by 7 for n>=1. 2. First, this is extremely difficult to read. Please learn some LaTeX. You are trying to prove $2^{3n} - 1$ is divisible by $7$ for all $n \geq 1$ (I assume $n$ is also an integer)... Base Step: $n = 1$. Then $2^{3\cdot 1} - 1 = 8 - 1$ $=7$ which is divisible by $7$. Inductive Step: Assume this statement is true for $n = k$. Then $2^{3k} - 1 = 7m$ where $m$ is an integer. Now let $n = k + 1$ we have $2^{3(k + 1)} - 1 = 2^{3k + 3} - 1$ $= 2^{3k}\cdot 2^3 - 1$ $= 8\cdot 2^{3k} + 8 - 7$ $= 8(2^{3k} + 1) - 7$ $= 8\cdot 7m - 7$ $= 7(8m - 1)$ which is divisible by $7$. Therefore $2^{3n}- 1$ is divisible by $7$ for all integer $n \geq 1$. 3. Thanks ProveIt, I completely forgot about Latex for a minute and just pasted in the equation I was working on in word, so sorry! How does the negative 1 in this line $= 2^{3k}\cdot 2^3 - 1$ change to negative 7? $=8\cdot 2^{3k} + 8 - 7$ 4. Sorry, that should actually read $8\cdot 2^{3k} - 8 + 7$ $= 8(2^{3k} - 1) + 7$. The rest follows. 5. ok, thanks But how/where does the seven come from? 6. $-8 + 7 = -1$. It just takes a bit of practice, I knew I needed to create a $7$ in order to have this value be divisible by $7$. 7. Please excuse my ignorance, as mathematical Induction and I just aren't getting along. But did you just place the "7" in the equation because you needed it. It didn't come from any other part of the equation? 8. That's correct. And I also realised I'd need to take out a common factor of $8$. Like I said, it just comes from experience.	{"extraction_info": {"found_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": 30, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9571608304977417, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Context-sensitive_language	# Context-sensitive language In theoretical computer science, a context-sensitive language is a formal language that can be defined by a context-sensitive grammar. That is one of the four types of grammars in the Chomsky hierarchy. ## Computational properties Computationally, a context-sensitive language is equivalent with a linear bounded nondeterministic Turing machine, also called a linear bounded automaton. That is a non-deterministic Turing machine with a tape of only kn cells, where n is the size of the input and k is a constant associated with the machine. This means that every formal language that can be decided by such a machine is a context-sensitive language, and every context-sensitive language can be decided by such a machine. This set of languages is also known as NLIN-SPACE, because they can be accepted using linear space on a non-deterministic Turing machine. The class LIN-SPACE is defined the same, except using a deterministic Turing machine. Clearly LIN-SPACE is a subset of NLIN-SPACE, but it is not known whether LIN-SPACE=NLIN-SPACE. It is widely suspected they are not equal[citation needed]. ## Examples An example of a context-sensitive language that is not context-free is L = { ap : p is a prime number }. L can be shown to be a context-sensitive language by constructing a linear bounded automaton which accepts L. The language can easily be shown to be neither regular nor context free by applying the respective pumping lemmas for each of the language classes to L. Simpler example would be $L = \{ a^nb^nc^n : n \ge 1 \}$. An example of recursive language that is not context-sensitive is any recursive language whose decision is an EXPSPACE-hard problem, say, the set of pairs of equivalent regular expressions with exponentiation. ## Properties of context-sensitive languages • The union, intersection, concatenation and Kleene star of two context-sensitive languages is context-sensitive. • The complement of a context-sensitive language is itself context-sensitive. • Every context-free language is context-sensitive. • Membership of a string in a language defined by an arbitrary context-sensitive grammar, or by an arbitrary deterministic context-sensitive grammar, is a PSPACE-complete problem. ## See also • Linear bounded automaton • Chomsky hierarchy • Noncontracting grammars – generate exactly the context-sensitive languages • Indexed languages – a strict subset of the context-sensitive languages ## References • Sipser, M. (1996), Introduction to the Theory of Computation, PWS Publishing Co. • Immerman, Neil (1988). "Nondeterministic space is closed under complementation". SIAM J. Comput. 17 (5): 935–938. doi:10.1137/0217058.	{"extraction_info": {"found_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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8299716114997864, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/184545-finding-subspace-continuous-function.html	Thread: 1. Finding subspace/continuous function Prove that the set U={f ∈ C([0,1]): f(1/2)=f(1)} is a subspace of C([0,1]). I am not sure how to do this. I know what the necessary conditions are for a subspace but I can't quite figure out how to show them given a continuous function. Help please? 2. Re: Finding subspace/continuous function Originally Posted by steph3824 Prove that the set U={f ∈ C([0,1]): f(1/2)=f(1)} is a subspace of C([0,1]). I am not sure how to do this. I know what the necessary conditions are for a subspace but I can't quite figure out how to show them given a continuous function. Help please? You have to prove the subspace axioms. U is nonempty since $f\in U$ Let $f,g\in U$ $(f+g)(1/2)=f(1/2)+g(1/2)=f(1)+g(1)=(f+g)(1)$ How about the last axiom now. 3. Re: Finding subspace/continuous function Would it be (af)(1/2) = a·f(1/2) = a·f(1)? 4. Re: Finding subspace/continuous function Originally Posted by steph3824 Prove that the set U={f ∈ C([0,1]): f(1/2)=f(1)} is a subspace of C([0,1]). I am not sure how to do this. I know what the necessary conditions are for a subspace but I can't quite figure out how to show them given a continuous function. Help please? I don't suppose you know the fact that if $T:V\to W$ is a linear equation, then $\ker T$ is a subspace of $V$. Well, I'm sure you can note that $C[0,1]\to\mathbb{R}:f\mapsto f(1)-f(\tfrac{1}{2})$ is a linear transformation and $\ker f=U$. 5. Re: Finding subspace/continuous function Originally Posted by steph3824 Would it be (af)(1/2) = a·f(1/2) = a·f(1)? Good.	{"extraction_info": {"found_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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9284361004829407, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/68574-equivalence-relation-quotient-space.html	# Thread: 1. ## equivalence relation, quotient space (a) Deﬁne an equivalence relation on the plane $X=\mathbb{R}^2$ as follows: $x_0 \times y_0$ ~ $x_1 \times y_1$ if $x_0 + y_0^2=x_1 + y_1^2.$ Let $X^*$ be the corresponding quotient space. It is homeomorphic to a familiar space; what is it? [Hint: Set $g(x \times y)=x+y^2.$] (b) Repeat (a) for the equivalence relation $x_0 \times y_0$ ~ $x_1 \times y_1$ if $x_0^2+y_0^2=x_1^2+y_1^2.$ [this is Munkres §22.4] 2. The quotient space defined by an equivalence relation has, as points, the equivalence classes. Further, a set of such equivalence classes is open if and only if their union is open in the original space. Here, I notice that, since (0,0) satisfies $x+ y^2= 0+ 0^2= 0$ the equivalence class of (0,0) contains all (x,y) such that $x+ y^2= 0$, a parabola. In fact, given any point $(x_0,y_0)$, its equivalence class consists of all points on the parablola $x+ y^2= x_0+ y_0^2$ Every such parabola has a unique vertex, $(x_0+y_0^2, 0)$. I haven't worked out the details (I'll leave that to you) but I suspect that the function that identifies each such equivalence class with its vertex is a homeomorphism and so the quotient space is homeomorphic to a straight line. Similarly, for (b), (x, y) is equivalent to a given point $(x_0, y_0)$ if and only if $x^2+ y^2= x_0^2+ y_0^2$, a circle with center (0,0) and radius $\sqrt{x_0^2+ y_0^2}$. Thus every such equivalence class can be identified with the non-negative number $\sqrt{x_0^2+ y_0^2}$ and so the quotient space is homeomorphic to the half open interval $[0, \infty)$. Your job, now, is to show that those identifications are homeomorphisms. 3. Another way to prove this use: if f $: X \rightarrow Y$ is a quotient map, then the relation $x$~ $y$ iff $f(x)=f(y)$ is an equivalence relation on $X$, and $Y$ is homeomorphic to $X/$~. I leave the details to you. But either way works.	{"extraction_info": {"found_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": 26, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9118579030036926, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=4029876	Physics Forums ## I am having problems with lots of rigor. Hello everyone, Essentially, I am looking for advice as to what to do/how to improve/cope. It is my first time learning calculus and I am working through Spivak's Calculus. For the first few chapters the problems weren't that hard (up through derivatives and the fundamental theorem of calculus). Recently (i.e. the chapter on calculating integrals) the problems have gotten much more difficult and I cannot solve the ones that do not involve something beyond ordinary computation. In general, I'm having a problem with mathematical rigor: I find it difficult and unmotivated to prove things that are obvious to me. For example, the cauchy schwarz inequality: it should be obvious that the dot product of two vectors is less than or equal to the products of their lengths, but proving it (at least for me, who doesn't know linear algebra) is nonintuitive and I can't remember it. Ultimately, I feel like I'm losing my confidence in my ability to do math. By the way, I'm a high school student going into junior year. What advice do you have? Should I just move on to physics? My intuition behind the math I know is very good. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Is your main interest in math or physics. If its in physics don't worry about rigor. Their are math proofs in physics. Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus First of all, if your only goal is to do physics then you don't need to know Spivak's calculus. Spivak's calculus is quite a difficult math text and is more real analysis than calculus. You should do Spivak if your interest is in learning rigorous math and learning the foundations of the math you use. If your only interest is in solving physics questions, then Spivak is not necessary. It is true that rigorous math texts sometimes focus on proving obvious things. If you feel that this is unnecessary, then you're missing the whole point of mathematics. Mathematics provides a system of axioms and tries to derive entire mathematics from that. For example, Spivak uses in his entire book only 13 axioms and derives entire calculus from that. The point in proving elementary things is to show that our axioms are good enough. For example, it should be obvious that $f(x)=x^2$ is continuous. The rigorous epsilon-delta definition of continuity can be used to show that this is continuous. If we were not able to show that f were continuous using epsilon-delta, then this shows us that there is something wrong with our definition of continuity. We did not invent epsilon-delta to prove that silly functions like $f(x)=x^2$ are continuous. We invented it to prove more advanced stuff that isn't intuitive. The whole point of showing that f is continuous is to show that our current definitions do correspond with our intuition: that is, we get the results we intuitively want. Eventually, there will come a time that you want to prove a mathematical result that might not be intuitively true. If you want to be succesful at this, you better have the correct axioms and definitions. One way to make sure you have the correct machinery is to prove the elementary results. The same with Cauchy-Schwarz. It is a result we intuitively want to be true. If we were not able to show this, then we must get different axioms. Also, Cauchy-Schwarz isn't as obvious as it may appear. Do you think that $$\int_a^b f(t)g(t)dt \leq \sqrt{\int_a^b f(t)^2dt}\sqrt{\int_a^b g(t)^2dt}$$ is obvious? I don't think it's very obvious. Nevertheless, its prove is very analogous to the prove of the Cauchy-Schwarz inequality. So if you can prove an elementary result, then maybe the proof of the elementary result can be used to prove more advanced stuff like this integral version of Cauchy-Schwarz. ## I am having problems with lots of rigor. I like math, and physics as well - I just don't really care too much about the precision that is required required in analysis, nor am I good dealing with it. Essentially the problem comes down to this: In trying to find the surface area of the curve created by the function f, from a to b, "rotated" about the x-axis (http://www.math.wpi.edu/Course_Mater...rev/node1.html), you can partition the interval into $$\{a=t_0, t_1, ... , t_n = b \}$$, then using some simple geometry (which Spivak considers "fudging"), you can approximate the surface area by the surface areas of successive frustums, a frustum for each interval $$[t_{i-1},t_i]$$. The sum is $$\pi \sum_{i=1}^n [f(t_{i-1}) + f(t_i)]\sqrt{[f(t_i) - f(t_{i-1})]^2 + (t_i - t_{i-1})^2} = \pi \sum_{i=1}^n [f(t_{i-1}) + f(t_i)]\sqrt{f'(x_i)^2 +1}(t_i - t_{i-1})$$ Justified by the mean value theorem. I'm sure that you could use uniform continuity and then show that this is a riemann sum of some sort, and get to the result. But for me, it's totally satisfactory (and much more efficient) to just say that as $$t_i - t_{i-1}$$ gets small, then $$f(t_i) + f(t_{i+1})$$ is essentially equal to $2f(t_i)$, and a similar thing for the $$f'(x_i)^2$$, and the sum becomes $$2\pi\int_a^b f(x)\sqrt{f'(x) +1} dx$$. Does this mean that math isn't for me? Or does this just mean I'm not as into analysis? Does this same kind of thing carry over into other domains like algebra/combinatorics? Quote by micromass First of all, if your only goal is to do physics then you don't need to know Spivak's calculus. Spivak's calculus is quite a difficult math text and is more real analysis than calculus. You should do Spivak if your interest is in learning rigorous math and learning the foundations of the math you use. If your only interest is in solving physics questions, then Spivak is not necessary. It is true that rigorous math texts sometimes focus on proving obvious things. If you feel that this is unnecessary, then you're missing the whole point of mathematics. Mathematics provides a system of axioms and tries to derive entire mathematics from that. For example, Spivak uses in his entire book only 13 axioms and derives entire calculus from that. The point in proving elementary things is to show that our axioms are good enough. For example, it should be obvious that $f(x)=x^2$ is continuous. The rigorous epsilon-delta definition of continuity can be used to show that this is continuous. If we were not able to show that f were continuous using epsilon-delta, then this shows us that there is something wrong with our definition of continuity. We did not invent epsilon-delta to prove that silly functions like $f(x)=x^2$ are continuous. We invented it to prove more advanced stuff that isn't intuitive. The whole point of showing that f is continuous is to show that our current definitions do correspond with our intuition: that is, we get the results we intuitively want. Eventually, there will come a time that you want to prove a mathematical result that might not be intuitively true. If you want to be succesful at this, you better have the correct axioms and definitions. One way to make sure you have the correct machinery is to prove the elementary results. The same with Cauchy-Schwarz. It is a result we intuitively want to be true. If we were not able to show this, then we must get different axioms. Also, Cauchy-Schwarz isn't as obvious as it may appear. Do you think that $$\int_a^b f(t)g(t)dt \leq \sqrt{\int_a^b f(t)^2dt}\sqrt{\int_a^b g(t)^2dt}$$ is obvious? I don't think it's very obvious. Nevertheless, its prove is very analogous to the prove of the Cauchy-Schwarz inequality. So if you can prove an elementary result, then maybe the proof of the elementary result can be used to prove more advanced stuff like this integral version of Cauchy-Schwarz. Thank you for the response, Micromass. I understand what you are saying - I guess I just need to look at more motivating examples, such as the integral version of the Cauchy-Schwarz inequality. For me, it is not obvious either. Quote by micromass Also, Cauchy-Schwarz isn't as obvious as it may appear. Do you think that $$\int_a^b f(t)g(t)dt \leq \sqrt{\int_a^b f(t)^2dt}\sqrt{\int_a^b g(t)^2dt}$$ is obvious? I don't think it's very obvious. Nevertheless, its prove is very analogous to the prove of the Cauchy-Schwarz inequality. So if you can prove an elementary result, then maybe the proof of the elementary result can be used to prove more advanced stuff like this integral version of Cauchy-Schwarz. If you make the leap between finite and infinite sums, the connection is not that hard to see. Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Quote by chiro If you make the leap between finite and infinite sums, the connection is not that hard to see. Well, if you're happy with that explanation... Quote by micromass Well, if you're happy with that explanation... A riemann sum is an infinite sum, just as a valid sum in l2(R) (little l) is an infinite-sum. When you end up going from there to L^2(R) and you do all the formal stuff, the connection still remains the same when the template of the vector spaces and geometry is used but instead of dealing with finite-sums (and finite-dimensional spaces), you are now dealing with infinite ones, with special properties (convergence of the inner products being a big one, but the nature of infinity makes things really nasty in other ways). Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Quote by chiro A riemann sum is an infinite sum, just as a valid sum in l2(R) (little l) is an infinite-sum. When you end up going from there to L^2(R) and you do all the formal stuff, the connection still remains the same when the template of the vector spaces and geometry is used but instead of dealing with finite-sums (and finite-dimensional spaces), you are now dealing with infinite ones, with special properties (convergence of the inner products being a big one, but the nature of infinity makes things really nasty in other ways). That's a good intuition, but it's not a good rigorous explanation. Not every result about finite sums is true for integrals. Also, the integrals in $L^2(\mathbb{R})$ are not Riemann sums. Quote by micromass That's a good intuition, but it's not a good rigorous explanation. Not every result about finite sums is true for integrals. Also, the integrals in $L^2(\mathbb{R})$ are not Riemann sums. Yes true, but even then you can still use the characteristic function to get the right analog (in terms of a "sum" and not a Riemann sum so to speak). After all, that's all an integral is: you sum stuff with respect to some measure: so if it's not an area like a Riemann, but it's more like a Riemann-Stieltjes Integral or a Lebesgue integral, then you find the analog of where the sums make the correspondence. A Riemann is a lot easier than the general L^2(R), but you can make the same analog for the other spaces. AlwaysCurious, I wouldn't worry about it. You are a high school junior and tackling Spivak on your own. That is quite a high level! If you are finding it difficult take a break and study something else for a while. What you find boring now may become very interesting to you in the future. I hated math in High School and did poorly in it but after I got out of the military 8 years later I am pursuing a dual degree in Math and Electrical Engineering. I have loved my Analysis courses but in my opinion I enjoyed and benifited from it far more by having a Professor guide me through it. Don't discount mathematics from your future just because it's boring. Make sure you take an analysis course in college and see how you feel about it after that. Quote by chiro If you make the leap between finite and infinite sums, the connection is not that hard to see. Not to step in on the argument of sorts, but yeah after seeing the integral form of cauchy schwarz I thought about it - it should be intuitive in terms of riemann sums. Quote by srl17 AlwaysCurious, I wouldn't worry about it. You are a high school junior and tackling Spivak on your own. That is quite a high level! If you are finding it difficult take a break and study something else for a while. What you find boring now may become very interesting to you in the future. I hated math in High School and did poorly in it but after I got out of the military 8 years later I am pursuing a dual degree in Math and Electrical Engineering. I have loved my Analysis courses but in my opinion I enjoyed and benifited from it far more by having a Professor guide me through it. Don't discount mathematics from your future just because it's boring. Make sure you take an analysis course in college and see how you feel about it after that. Thank you for the advice - it is something similar to what I'm thinking at the moment. Maybe a love of precision will come with maturity. Also, to go on the record, I do like math - it's just recently the demands of Spivak have been a bit much. Quote by AlwaysCurious Does this same kind of thing carry over into other domains like algebra/combinatorics? While micromass' description of why rigour is important is true for all of mathematics, each field has its own distinct flavour. Personally, I have always found analysis to be less intuitive than algebra or combinatorics. Even (point-set) topology, which is really abstracted analysis, was easier for me. I blame the evil real number system... Take home message: Don't get discouraged if some parts of math are a bit harder than others. The rigour gets easier as you see the connections between the different fields. This is the mythical "mathematical maturity" that just takes time and work to gain. Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Quote by Sankaku While micromass' description of why rigour is important is true for all of mathematics, each field has its own distinct flavour. Personally, I have always found analysis to be less intuitive than algebra or combinatorics. Even (point-set) topology, which is really abstracted analysis, was easier for me. I blame the evil real number system... Take home message: Don't get discouraged if some parts of math are a bit harder than others. The rigour gets easier as you see the connections between the different fields. This is the mythical "mathematical maturity" that just takes time and work to gain. Oh, yes, this is very true. Each field of math is a bit different and has its own techniques. Which field you like best is really very personal. Some people are excellent in analysis, but don't like algebra. Other people are exactly opposite!! Spivak is analysis. Maybe you just don't like analysis very much?? Thread Tools \| \| \| \| \|---------------------------------------------------------------\|----------------------------------\|---------\| \| Similar Threads for: I am having problems with lots of rigor. \| \| \| \| Thread \| Forum \| Replies \| \| \| General Math \| 1 \| \| \| Precalculus Mathematics Homework \| 39 \| \| \| Electrical Engineering \| 5 \| \| \| Introductory Physics Homework \| 3 \| \| \| Introductory Physics Homework \| 7 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 11, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.961689829826355, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/25071/the-birthday-problem-revisited	# The Birthday Problem, revisited…? I'm currently facing a problem I have been stuck in for almost half a day and I really can't keep this because I have a lot of work to do! It obviously has to do with the birthday problem. The original birthday problem solves the question of "what is the probability of having at least two persons with the same birthday in a room of $N$ persons", and I'm trying to get to this in a different way of that of the canonical solution, which is pretty easy to understand. I started asking "what is the probability of exactly $k$ persons, out of $N$, having the same birthday as me". Having the same "birthday as me" is the same as asking "birthday on January 1st" or "birthday on December 31", which have $p=1/365$. To me, this looked like a binomial problem (i.e. tossing a coin with probability $p$ of being heads, where I ask for exactly $k$ heads out of $N$ tosses), so my answer to this question is, simply, \begin{equation} p(k\|N)=\binom{N}{k}p^{k}(1-p)^{N-k}\text{.} \end{equation} I think here's where my argument starts to fall appart: I say that if I want to ask "what is the probability of exactly $k$ persons, out of $N$, having a birthday on January 1st" OR "having a birthday on January 2nd"...OR "having a birthday on December 31th", is \begin{equation} 365p(k\|N)=365\binom{N}{k}p^{k}(1-p)^{N-k}\text{.} \end{equation} Finally, in order to answer the question "what is the probability of having at least two persons with the same birthday in a room of $N$ persons", I ask "what is the probability of having $k=2$ persons with the same birthday in the room" OR "$k=3$"...OR "\$k=N", so the probability would be: \begin{equation} 365\sum_{k=2}^{k=N}p(k\|N)\text{.} \end{equation} However, this is not the answer, as when I compare the results with the canonical answer it makes no sense. Can anyone help me with this? - Do you think the independence assumption is valid? – Dason Mar 22 '12 at 15:22 – whuber♦ Mar 22 '12 at 15:23 Néstor, rather than updating your question as "solved", please consider accepting @Martin's answer (check the green mark below his response votes) if you feel that solved your problem. I'm giving you +1 so that you can even vote on his reply. – chl♦ Mar 22 '12 at 16:10 Done! Thanks @chi :-). – Néstor Mar 22 '12 at 16:27 ## 1 Answer The events "$k$ people have a birthday on Jan 1" and "$k$ people have a birthday on Jan 2" are not mutually exclusive, and so the probability that either of them occurs is not equal to the sum of their individual probabilities. You would need to subtract the probability that both events occur, but this is likely to become extremely complex once you scale up to all 365 days. - Can't believe I overlooked this fact! It was on my tougths at first, but I discarded it in order to try to come up with a quick answer. I'll be more careful from now on :-), thanks! – Néstor Mar 22 '12 at 15:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9670395255088806, "perplexity_flag": "head"}
http://polymathprojects.org/2011/05/12/possible-new-polymath-project/?like=1&source=post_flair&_wpnonce=d42062ed33	# The polymath blog ## May 12, 2011 ### Possible new polymath project Filed under: polymath proposals — Terence Tao @ 5:56 pm Richard Lipton has just proposed on his blog to discuss the following conjecture of Erdos as a polymath project: that there are no natural number solutions to the equation $1^k + \ldots + (m-1)^k = m^k$ with $k \geq 2$. Previous progress on this problem (including, in particular, a proof that any solution to this equation must have an extremely large value of $m$, and specifically that $m \geq 10^{10^9}$) can be found 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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 4, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9353609085083008, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/256309/square-of-the-sum-of-n-positive-numbers?answertab=active	# Square of the sum of n positive numbers I have a following problem When we want to write $a^2 + b^2$ in terms of $(a \pm b)^2$ we can do it like that $$a^2 +b^2 = \frac{(a+b)^2}{2} + \frac{(a-b)^2}{2}.$$ Can we do anything similar for $a_1^2 + a_2^2 + \ldots + a_n^2$ ? I can add the assumption that all $a_i$ are positive numbers. I mean to express this as combination of their sums and differences. I know that this question is a little bit naive but I'm curious whether it has an easy answer. - Is sunflower's answer enough? – Amr Dec 11 '12 at 15:04 ## 2 Answers Yes. You have to sum over all of the possibilities of $a\pm b\pm c$: $$4(a^2+b^2+c^2)=(a+b+c)^2+(a+b-c)^2+(a-b+c)^2+(a-b-c)^2$$ This can be extended to n factors by: $$\sum_{k=1}^n a_k^2=\sum_{\alpha=(1,-1,...,-1)\; \|a_i\|=1}^{(1,...,1)}\frac{\big(\sum_{i=1}^{n}\alpha_ia_i\big)^2}{2^{n-1}}$$ ($\alpha$ is a multiindex with values that are either -1 or 1, except the first that is always 1) - Another way would be: $$\frac{\sum_{x\in \{-1,1\}^n}(x_1a_1+x_2a_2+....+x_na_n)^2}{2^n}$$ Where $x_i$ is the ith component of the vector $x$ - I think this is correct but I still need to check. Unfortunately I have to leave now. I will check it later – Amr Dec 11 '12 at 15:09 nice way to write it too ^.^ I wrote an answer using multiindices... this is probably nicer. But should it not be $2^{n-1}$? – CBenni Dec 11 '12 at 15:12 No. It should be $2^n$ – Amr Dec 11 '12 at 15:19 Try the case when $n=2$ or 1 – Amr Dec 11 '12 at 15:20 nah, sorry. I was looking at the solutions for n=2 or 3 where the first index is 1. (These work just fine too). For those, we have only half the terms and thus, only $2^{n-1}$ but you are correct. – CBenni Dec 11 '12 at 15:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.954710841178894, "perplexity_flag": "head"}
http://dsp.stackexchange.com/questions/6084/difference-of-gaussians	# Difference of Gaussians I use the following formula for DoG: $$\frac{1}{\sigma} (\frac{x^2}{2\sigma^2}-1.0) e^{\frac{-x^2}{2\sigma^2}}$$ What is the relationship between this formula and the difference of two Gaussian functions? Can someone show me how to apply DoG on a 1D signal in MATLAB? My input can be considered as a 1D real-valued function, and I applied the above formula to detect blobs. The results make sense except that I don't know why I get many local minima in a relatively flat region (i.e., function values at points in the region almost stay constant). Since I don't have enough signal processing/image processing background, I want to see what a constant signal look like after being convolved with a DoG filter. - 2 I suggest you to look for "difference-of-gaussians" in Google IMAGES. The graphical representation of these abstract concepts sometimes can be very enlightening, specially regarding detection and other signal processing applications related to "the real world" (instead of pure mathematical formulas). – heltonbiker Nov 24 '12 at 16:05 Let me try to answer my first question. DoG is known to be an approximation of Laplacian of Gaussian(LoG). The formula is very close to the second-order derivative of a Gaussian function. Still hope someone who can share some matlab code which would allow me to play with convolution with DoG. – user11869 Nov 24 '12 at 20:57 1 @user11869 Welcome to dsp.SE ! You are allowed to edit you own questions, so, as a general rule you'll want to add all additional information about the problem as well as all the major explanations in to your question, instead of leaving them in the comments -- like you did (any new information / things you try should also be added that arise through your own research while waiting for an answer). Comments are usually for short explanations and remarks. In any case, have fun here :) – penelope Nov 26 '12 at 15:01	{"extraction_info": {"found_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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9443878531455994, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/28018/on-distributions-over-mathbb-r-whose-derivatives-vanishes?answertab=votes	# On distributions over $\mathbb R$ whose derivatives vanishes Let $I \subset \mathbb R$ be open, $u \in \mathcal D'(I)$ be a distribution whose distributional derivatives vanishes (i.e. is zero for all test functions, which we may assume to be complex valued ). We show $\forall c \in \mathbb C: \forall \phi \in \mathcal D(I) : u(\phi) = \int c\cdot\phi dx$. (EDIT: Correctly, $c$ should be quantified with $\exists$. My question has been why the following proof doesn't allow for arbitrary complex $c$, which explains the preceding statement.) Proof: Let $\Psi \in D(I)$. $\Psi$ is the derivative of a test function iff $\int \phi dx = 0$. In that case $u(\Psi) = 0$. Let $h \in D(I)$ be arbitrary with $\int h dx = 1$. Now for every test function $\phi \in D(I)$ we see: $\phi - \int \phi dx \cdot h \in D(I)$ and $\int ( \phi - \int \phi dx h ) dx = 0$. therefore $u( \phi - \int \phi dx h ) = 0$, i.e. $u(\phi) = u(h) \int \phi dx$ $\square$ If this is not wrong, how can I interpret the fact that $h$ has been arbitrary? Note: This is part of a larger proof, which shows the same for non-one-dimensional domains. - ## 2 Answers The function $h$ is not arbitrary; it is arbitrary with respect to the condition that $\int h dx = 1$. (And this latter condition is certainly necessary for the proof to go through.) The proof given (which seems correct to me) shows that $u(\phi)$ only depends on the value of $\int \phi$. In particular, one sees (after the proof is done) that $u(h)$ only depends on the value of $\int h dx$, which was fixed to be $1$. Thus $u(h)$ is independ of the choice of $h$ (as long as $\int h dx = 1$), and is equal to the constant $c$ in the statement of the theorem. [Added: As Theo points out in his answer, you have an incorrect universal quantifier on the constant $c$ in the statement of the theorem; it should read for some $c$. I didn't notice this when I read the question!] - The statement you're actually proving is: If the distributional derivatives of $u$ vanish then there exists $c$ such that $u(\phi) = c \int \phi$. (You're stating a completely different property: $\forall c$...) If you suppose for a moment that $u(\phi) = c \int \phi$, how can you recover $c$? Well, by evaluating $u$ at any test function $h$ with $\int h = 1$ (this is a restriction but not a severe one). Apart from the slip of confusing $\forall$ with $\exists$ your argument is correct. In fact, the trick of using $h$ with $\int h = 1$ is used quite often in the theory of distributions and the fact that it doesn't matter which $h$ you take is one of the big strengths of the theory of distributions. For instance, if $h$ is such that $\int h = 1$ then $u_{n}(\phi) = \int nh(nx) \phi(x)$ approximates the Dirac $\delta_{0}$-distribution in the sense that $u_{n}(\phi) \to \phi(0) = \delta_{0}(\phi)$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 52, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9466062188148499, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/3436/if-a-1kg-mass-was-accelerated-close-to-the-speed-of-light-would-it-turn-into-a-b	If a 1kg mass was accelerated close to the speed of light would it turn into a black hole? I'm a big fan of the podcast Astronomy Cast and a while back I was listening to a Q&A episode they did. A listener sent in a question that I found fascinating and have been wondering about ever since. From the show transcript: Arunus Gidgowdusk from Lithuania asks: “If you took a one kilogram mass and accelerated it close to the speed of light would it form into a black hole? Would it stay a black hole if you then decreased the speed?” Dr. Gay, an astrophysicist and one of the hosts, explained that she'd asked a number of her colleagues and that none of them could provide a satisfactory answer. I asked her more recently on Facebook if anyone had come forward with one and she said they had not. So I thought maybe this would be a good place to ask. - 4 Yet another reason why the concept of relativistic mass is not a good one. ;-) physics.stackexchange.com/q/2516/724#2518 – Bruce Connor Jan 20 '11 at 20:25 1 Yeah, though for the purposes of this question, it's mostly just a vocabulary issue: for matter, gravity does couple to relativistic mass density (i.e., energy density), so the question still makes sense in terms of rel. mass. But gravity also couples to momentum (and pressure, etc.), which tends to cancel the effect of higher energy density. – Stan Liou Jan 20 '11 at 20:53 2 – MBN Jan 20 '11 at 22:26 4 Answers The answer is no. The simplest proof is just the principle of relativity: the laws of physics are the same in all reference frames. So you can look at that 1-kg mass in a reference frame that's moving along with it. In that frame, it's just the same 1-kg mass it always was; it's not a black hole. - 3 An addendum: It's worth pausing to ask why one might have thought it would form a black hole, and why those reasons are incorrect. Presumably the thought is that a combination of Lorentz contraction and relativistic "mass increase" squeeze the object below its Schwarzschild radius. So what's wrong with that reasoning? The main thing is just that the derivation of the Schwarzschild radius only applies under certain conditions. At the very least, it only applies in the object's rest frame (since it assumes spherical symmetry -- i.e., no preferred direction). – Ted Bunn Jan 20 '11 at 20:27 but if 1kg mass is rotated? – voix Feb 19 '11 at 13:24 2 The gravitational field of a rotating 1-kg mass is different from that of a non-rotating mass. I don't remember the details, which are complicated, but the gravitational pull probably does get stronger because the rotational kinetic energy gravitates. If you start with a mass that's larger than its Schwarzschild radius, I don't know whether you can make it turn into a black hole by supplying rotational kinetic energy. – Ted Bunn Feb 19 '11 at 14:48 @TedBunn "If you start with a mass that's larger than its Schwarzschild radius, I don't know whether you can make it turn into a black hole by supplying rotational kinetic energy." But even kinetic energy is relative, if I accelerate to the same velocity as your object, your object don't have any kinetic energy relative to me. – Calmarius Apr 22 at 15:59 No, a 1kg mass would not turn into a black hole, even if it were zipping past you at very close to the speed of light. The principle of relativity is a fundamental idea in physics, and one consequence of it is that we can understand the physics of something that's moving by imagining we're moving alongside it. For example, you are watching people play pool on a train as it rushes past you. You want to know whether a certain shot that's just been made will sink the 8-ball. You figure it out by imagining you're inside the train and calculating everything you'd expect to happen from that simpler viewpoint where the pool table is stationary. If the 8-ball goes into a certain pocket from that point of view, you can rest assured it will go into the same pocket if you analyze the situation again from your original vantage point on terra firma. Applying the same principle to the 1kg mass, we see that moving along side it, it just looks like a normal mass, not a black hole. Hence, from another point of view in which it moves close to the speed of light, it still looks like a normal mass, not a black hole. - So does it then follow that the relative speed of a mass has no bearing on the gravitational force felt by a nearby stationary mass? That is if a mass flew by me at .1 c, would I feel the same tug as if it flew by at .999c? Would there be some sort of equivalence given the time it takes the object to pass? That is, would the total force felt over time be the same; sort of like how the area covered by an orbit is the same over a given time? – shopsinc Jan 20 '11 at 20:39 @shops Your question can't be answered using simply the principle of relativity because it's asking about different types of relative motion. You might try asking it as a separate question on the main site. I don't have a good, concise answer to that question. – Mark Eichenlaub Jan 20 '11 at 20:50 While good, I think the other answers are currently missing one ingredient, so I'll post this answer. For particles traveling at constant velocity there is no event horizon, and so they act nothing like a black hole. Light from other regions of space will eventually reach it, unlike a black hole. Further, the forces between atoms in what ever matter constitutes the mass are co-moving and so there is no increased gravitational interaction between them. While the distances between them appear to change to an outside observer (as the mass is accelerated) once it reaches constant velocity they are fixed. What has not been mentioned in other answers is the effect of acceleration. When a particle is continuously accelerated there is an apparent event horizon. See the relevant Wikipedia page here. So this has some features that we associate with a black hole, however there are still major differences. An object undergoing constant acceleration does indeed behave like it is static in a constant gravitational field. However, in the case of such an object the direction of the equivalent field is constant (and in a constant direction) throughout the object. This is not true for the gravitational field of a black hole, which is spherically symmetric. Of course once the particle stops accelerating the apparent horizon disappears. - True, but the apparent horizon in this situation is very different from a black hole horizon. In the case of the accelerating particle, the stuff that's "behind" the apparent horizon is far away from the particle -- just the opposite of the black hole's event horizon. That is, heuristically, a black hole's event horizon says that once you're sufficiently close to the black hole you can't get far away, whereas the accelerated particle's horizon says that once you're sufficiently far from the particle you can't get close. – Ted Bunn Jan 21 '11 at 15:03 I am presuming the idea is the 1kg mass will length contract to below the Planck length. It is either that or the relativistic energy (mass) $E~=~\gamma mc^2$ would be so large it would gravitationally implode. The question though can be thought of according to what would happen to an observer on the mass. The question could be turned around: Would the universe implode? If a mass $M$ passes by a smaller mass $m~<<~M$ then one might think that $M$ could become a black hole and the small mass $m$ if close enough would become trapped in the black hole. However, from the frame of the big mass $M$ the small mass is not a black hole. This is a contradiction. A ultra-relativistic mass will behave similar to a gravity wave as it passes another reference point. This Aichelburg-Sexl ultraboost has a plane wave pulse of spacetime. The relativistic mass will result in a gravity wave pulse as detected by a stationary observer. So there is a gravitational implication to such extreme relativistic boosts. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9543818235397339, "perplexity_flag": "head"}
http://mathoverflow.net/questions/84671/minimum-distance-between-two-data-sets	## Minimum distance between two data sets ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose we have two sets of data, $X$ and $Y$, each of which contains $10$ positive numbers. Now let us order the data sets `$X=\left\{ x_{1},\cdots,x_{10}\right\}$`, $x_{1}\ge\cdots\ge x_{10}>0$ and `$Y=\left\{ y_{1},\cdots,y_{10}\right\}$`, $y_{1}\ge\cdots\ge y_{10}>0$ and define $d:=\sum_{k=1}^{10}\left\|x_{i_{k}}-y_{j_{k}}\right\|$, that is the sum of the distances of the numbers in pairs from the two data sets. Does anyone know how to prove that $d$ achieves its minimum when $i_{k}=j_{k}=k$ for $1\le k\le10$, or is there any counter example if it is not true? Thanks. - 2 same question was posted and was answered here: math.stackexchange.com/questions/95546/… – Paul Jan 1 2012 at 11:29 ## 1 Answer Here is a slightly more general proof. Let $x$ be any vector in $R^n$. Let $x^\downarrow$ denote the vector obtained from $x$ by sorting its entries in decreasing order, so that $x_1^\downarrow \ge x_2^\downarrow \ge \cdots \ge x_n^\downarrow$. Now, let $x, z \in R^n$, and consider $x+z$. Clearly, if we apply the same permutation to $x$ and $z$ separately, the entire sum $x+z$ is also permuted the same way. Hence, we may assume wlog that $x=x^\downarrow$. Now, let $x$ be as in your question above, and let $z=-y$. Recall now the concept of majorization. A quick calculation (also see Theorem II.4.2 of Matrix Analysis by R. Bhatia) shows that $$x^\downarrow + z^\uparrow \prec x + z$$ (since we assumed wlog that $x=x^\downarrow$); also since $y=-z$ we obtain $$x^\downarrow - y^\downarrow \prec x - y.$$ But this majorization implies that for any symmetric gauge function $G$, we have $$G(x^\downarrow-y^\downarrow) \le G(x-y).$$ This then implies the original minimization claim if we choose $G=\\|\cdot\\|_1$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 26, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9400469660758972, "perplexity_flag": "head"}
http://mathoverflow.net/questions/81728/probability-of-overlapping-of-repetitive-events/81744	## Probability of overlapping of repetitive events ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The question is to compute or estimate the following probabilty. Suppose that you have $N$ (e.g. $30$) tasks, each of which repeats every $t$ min (e.g. $30$ min) and lasts $l$ min (e.g. $5$ min). If the tasks started at uniformly random point in time yesterday, what is the probability that there is a time today at which at least $m$ (e.g. $10$) of the tasks run. - ## 4 Answers Consider a circle of length $t+\ell$. Then I think your problem is asking if I drop $N$ points uniformly at random onto that circle, what is the probability that at least $m$ of them are in an interval of length $\ell$. When $m/(N-m) >> \ell / t$, so that the tail event that there are $m$ points in a fixed interval of size $\ell$ is exponentially small, you can use the union bound $$N \sum_{j\ge m} (t/(t+\ell))^{N-j} (\ell/(t+\ell))^j \binom{N}{j}$$. You could also consider using the Brownian bridge (from $(0,0)$ to $(1,1))$ approximation of the partial sum $\sum_{j=1}^k X_j$ where $X_j$ is the distance between the $j$th and $j+1$st points in say the clockwise direction, with a particular chosen first point $p$. Then the question roughly becomes what is the chance that there is some $s \in [0,1]$ such that $B_{s + \ell/(\ell + t)} - B_s$ exceeds $m/N$. So for instance if $m/N \le \ell / (t + \ell) + o(\sqrt{\ell / (t + \ell)})$ then this probability should be very close to $1$. - Thank you, this is a good start. Unfortunately, the interesting part for me is when $l/t$ is not small... – Petar Ivanov Nov 24 2011 at 6:37 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Each block of $t$ minutes today will have the same pattern of tasks. Think of the middle point of a task as a uniform random variable on the unit circle. Then, $m$ tasks overlap if their corresponding random variables fall within a ball of radius $l/t$. This is a continuous generalization of the birthday problem. If instead of the unit circle, the variable fell on the unit interval $[0,1]$, the case $m=2$ has a simple solution in Feller Vol II. (p. 42): ```$$p(\text{At least }2\text{ tasks coincide}) = \begin{cases} 1-\left[1-(N-1)\frac{l}{t}\right]^N & \text{if } \frac{l}{t} < \frac{1}{N-1} \\ 1 & \text{if } \frac{l}{t} \geq \frac{1}{N-1}. \end{cases} $$``` If $l/t$ is small, you can probably approximate the probability you are looking for with this probability. In this case, you could also find approximations using the discrete birthday problem (see Wolf Schwarz, Comparing Continuous and Discrete Birthday Coincidences: “Same-Day” versus “Within 24 Hours”, The American Statistician, 2010). This may allow you to treat the case $m>2$. - Thank you, this is a good start. Unfortunately, the interesting part for me is when l/t is not small... – Petar Ivanov Nov 24 2011 at 6:37 Prob that a task is occuring at any given point in time in interval t is l/t. Prob that a task is not occurring at any given point in time in interval t = (t-l)/t Prob that at least m of N events occur at any given point in interval is sum of binomial probabilities : (N!/i!(N-i)!) (l/t)^i*((t-l)/t)^(N-i) summed over i=m to N. Then you have to integrate 1-this sum over the range (0,t]. Then you have to take 1- this integral. This should give probability for a given interval of length t. Then you have to figure out how many intervals there are of length t in the day (suppose there are T) and using the probability previously calculated as p do another binomial calculation for 1 minus the probability that none of the T intervals have an instant in time where m or more tasks are running. - Probability that a task Is operating at a given instant = l/t. Probability that a task isn’t operating at a given instant = (t-l)/t. Probability that at least m of N tasks are operating at a given instant is C(N,i)[ (l/t)^i][(t-l)/t]^(N-i) summed for i=m to N where C(N,i) is the combination of N objects taken i at a time. Integrate the sum over (0,t] . Result is (((t-l)^N)/t^(N-1))∑i=m to N [C(N,i)(l/(t-l))^i] which further simplifies to (((t-l)^N)/t^(N-1))[(t/(t-l))^N – ((t/(t-l))^m)] This further simplifies to t[1-((t-l)/t)^(N-m)] We only have to compute the probability for one interval of length t since the start times for each task has fixed period t and task lasts for same interval l. So if the event of m tasks happening in an instant over an interval of length t doesn’t happen once, it never happens. - Please avoid double posting where possible; you should be able to edit your existing answer. – Yemon Choi Apr 27 2012 at 19:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9219125509262085, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/kaluza-klein+geometry	# Tagged Questions 3answers 290 views ### Question about associative 3-cycles on G2 manifolds Let $X$ be a manifold with $G_2$ holonomy and $\Phi$ be the fundamental associative 3-form on $X$. Let $*\Phi$ be the dual co-associative 4-form on $X$. Now consider a particular associative 3-cycle ... 6answers 1k views ### Experimental evidence of a fourth spatial dimension? As human beings, we observe the world in which we live in three dimensions. However, it is certainly theoretically possible that more dimensions exist. Is there any direct or indirect evidence ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8732873797416687, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/75798/probability-distribution-with-infinite-variance-but-finite-mean/75808	## probability distribution with infinite variance but finite mean [closed] ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Does such a distribution exist? The Cauchy distribution has infinite variance but its mean is also undefined. - 1 Look up "power law": en.wikipedia.org/wiki/Power_law . – Joseph O'Rourke Sep 19 2011 at 0:19 This is more suitable for math.stackexchange.com than here. – Deane Yang Sep 19 2011 at 0:57 Math questions posted here are supposed to be research questions, whereas this is more the sort of thing found in textbooks. I suspect you'll be back with more questions of the former kind. – Michael Hardy Sep 19 2011 at 5:49 ## 2 Answers Consider the density function $f(x)= (3/2) x^{-5/2}$ on the interval from $1$ to infinity. More generally, Google for the term "Pareto distribution". - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. In addition to the Pareto distribution for suitable parameter values, the t-distribution has a mean of 0 if the number of degrees of freedom is more than 1, but has infinite variance if it is not more than 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": 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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8987189531326294, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/42709-word-problem-about-chemical-half-lives.html	# Thread: 1. ## Word problem about Chemical Half-Lives Dunno how to write this one, so I used a screenshot. http://img111.imageshack.us/img111/8039/mathwordux7.jpg The thing that trips me up is the 0 seconds. <(I hate Word Problems!) 2. Originally Posted by mankvill Dunno how to write this one, so I used a screenshot. http://img111.imageshack.us/img111/8039/mathwordux7.jpg The thing that trips me up is the 0 seconds. <(I hate Word Problems!) the half-life is 38 seconds. thus, if you start with $A_0$, after 38 seconds, you will have $\frac 12 A_0$. thus, the following equation must hold: $\frac 12 A_0 = A_0a^{38}$ now just solve for $a$ 3. a = 38? 4. Originally Posted by mankvill a = 38? ok, so lets say a = 38, that would mean $\frac 12 A_0 = A_0 (38)^{38} \implies \frac 12 = (38)^{38}$. obviously that makes no sense . how did you come up with that answer? do you have problems solving exponential equations? 5. Originally Posted by Jhevon ok, so lets say a = 38, that would mean $\frac 12 A_0 = A_0 (38)^{38} \implies \frac 12 = (38)^{38}$. obviously that makes no sense . how did you come up with that answer? do you have problems solving exponential equations? I thought I understood it. D: ...a = 19? Or 2? 6. Originally Posted by mankvill I thought I understood it. D: ...a = 19? Or 2? haha, stop guessing, and solve it. a would have to be less than 1, of course, otherwise, if you raise it to the 38th power, you would get a number greater than or equal to 1, which is bad, since we have 1/2 on the left hand side $\frac 12A_0 = A_0a^{38}$ $\Rightarrow \frac 12 = a^{38}$ Now, how would you solve for $a$ here. what must you do to get rid of that 38th power, and be left with just $a$ on the right side? 7. Originally Posted by Jhevon haha, stop guessing, and solve it. a would have to be less than 1, of course, otherwise, if you raise it to the 38th power, you would get a number greater than or equal to 1, which is bad, since we have 1/2 on the left hand side $\frac 12A_0 = A_0a^{38}$ $\Rightarrow \frac 12 = a^{38}$ Now, how would you solve for $a$ here. what must you do to get rid of that 38th power, and be left with just $a$ on the right side? Take the 38th root of $\frac 12$? 8. Originally Posted by mankvill Take the 38th root of $\frac 12$? yes! and your answer is? (to 6 decimal places) 9. Originally Posted by Jhevon yes! and your answer is? (to 6 decimal places) I got: .981925 after rounding. 10. Originally Posted by mankvill I got: .981925 after rounding. yup 11. 2nd part: How much of a 4 gram sample would be left after 10 seconds? 12. Originally Posted by mankvill 2nd part: How much of a 4 gram sample would be left after 10 seconds? Plug in $A_0 = 4$ and $t = 10$ and solve for A(t) (remember, you just found the value for a) 13. Originally Posted by Jhevon Plug in $A_0 = 4$ and $t = 10$ and solve for A(t) (remember, you just found the value for a) To set it up, would that be: 1/2 (4) = (4)a^10 ? 14. Originally Posted by mankvill To set it up, would that be: 1/2 (4) = (4)a^10 ? no, your original equation is $A(t) = A_0a^t$ now plug in the values you got for $A_0, ~a, \mbox{ and }t$ 15. Originally Posted by Jhevon no, your original equation is $A(t) = A_0a^t$ now plug in the values you got for $A_0, ~a, \mbox{ and }t$ So A(10) = 4(a)^10?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 24, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9582164883613586, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/67815/a-quantitative-orthogonality-argument	# A quantitative orthogonality argument Let $S={\rm span}(v_1,\ldots,v_k)$ be a subspace of $R^n$, and let $P_S$ be the orthogonal projection matrix onto $S$. If for some $x$ we have $x^T v_i = 0$ for all $i=1,\ldots,k$, then we can conclude that $P_S x=0$. I would like a quantitative version of this statement which I'll now describe. Suppose that instead we just know that $\|x^T v_i\| \leq \epsilon$ for all $i=1,\ldots,k$; and moreover $\|\|x\|\|_2=1$, $\|\|v_i\|\|_2=1$ for all $i=1,\ldots,k$. I'd like an argument that concludes $\|\|P_S x\|\|_2 \leq f(\epsilon, v_1, \ldots, v_n)$ where $f(\epsilon, v_1, \ldots, v_n)$ is some function which approaches $0$ as $\epsilon \rightarrow 0$ and the other arguments are kept fixed. Naturally, the faster rate of decay as $\epsilon \rightarrow 0$, the better. - There are many projection matrices onto a given subspace. Is $P_S$ the orthogonal projection onto $S$? – robjohn♦ Sep 27 '11 at 2:28 Yes; thanks - I edited this into the question. – robinson Sep 27 '11 at 3:08 2 You're going to need additional assumptions on how close the $v_i$ are allowed to be to each other. Otherwise, you could have $v_1 = (1,0,0)$, $v_2 = (\sqrt{1-\epsilon^2},\epsilon,0)$, and $x = (0,1,0)$ with $\lVert P_S x\rVert_2 = 1$. – Rahul Narain Sep 27 '11 at 4:00 Good point; I've reworded the question to allow the function $f$ to depend on the $v_i$. – robinson Sep 27 '11 at 5:30 ## 1 Answer Intuitively, the statement is obvious. Here is a proof. Let $A=(v_1,v_2,...,v_k)\in\mathbb{R}^{n\times k}$. Then the orthogonal projection matrix $P_S$ is $$P_S=A(A^TA)^{-1}A^T\in\mathbb{R}^{n\times n}$$ Denote $y=A^Tx\in\mathbb{R}^k$. Then $\|v_i^Tx\|<\epsilon$ implies $\|y_i\|<\epsilon$ and $$\\|y\\|^2<k\epsilon^2$$ Moreover, $$\\|P_Sx\\|^2=x^TP_Sx=y^T(A^TA)^{-1}y$$ Let $(A^TA)^{-1}=U^T\Sigma U$ be an SVD of $(A^TA)^{-1}$ with $\Sigma=\mathrm{diag}(\sigma_1,...,\sigma_k)$. Denote $z=Uy$. Then $$\\|P_Sx\\|^2=y^T(A^TA)^{-1}y=z^T\Sigma z=\sum_{i=1}^k \sigma_i z_i^2\le \sigma_1\sum_{i=1}^k z_i^2=\sigma_1\\|z\\|^2=\sigma_1\\|Uy\\|^2=\sigma_1\\|y\\|^2<\sigma_1k\epsilon^2$$ where $\sigma_1$ is the largest singular value of $(A^TA)^{-1}$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 38, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9340890049934387, "perplexity_flag": "head"}
http://mathoverflow.net/questions/64688?sort=newest	Non-vanishing of group cohomology in sufficiently high degree Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Atiyah in his famous paper , Characters and cohomology of finite groups, after proving completion of representation ring in augmentation ideal is the same as $K(BG)$, gives bunch of corollaries of this main theorem. One of them that catches my interest is: For any finite non-trivial group $G$ there exists arbitrary large integer $n$ such that $H^n(G,\mathbb{Z})\neq 0$. I just wonder if anyone can prove this without this powerful theorem. - Could you clarify a qualifier in your statement. Is the theorem that for any non-trivial finite group there are infinitely many $n$ such that $H^n(G,\mathbb Z)$ is non-trivial? Or is there only one $n$, and it depends on the group $G$? – Ryan Budney May 11 2011 at 20:29 No, I meant $H^n(G,\mathbb{Z})$ cannot vanish for all sufficiently large $n$. – Sam Nariman May 11 2011 at 20:45 BTW: Swan also showed the stronger result: If $H^n(G,\mathbb{Z}) = 0$ for all sufficiently large odd $n$ then $G$ is trivial or cyclic. – Ralph May 11 2011 at 23:57 2 @Ralph: I do not believe this is literally true. The cohomology of the binary icosahedral group is concentrated in degrees divisible by $4$. – Johannes Ebert May 12 2011 at 8:23 3 @Johannes: You're right. The correct statement is that the p-Sylow subgroups are cyclic or (generalized) quaternion. – Ralph May 12 2011 at 18:33 4 Answers The first purely algebraic proof of this fact seems to be from Leonard Evens: • A Generalization of the Transfer Map in the Cohomology of Groups, Trans. Amer. Math. Soc. 108(1963), 54-65 [Theorem 3] where he proves the result with help of his norm map. After having established the basic properties of the norm map, the proof is rather elementary: Let $C$ be a cyclic subgroup of prime order of $G$ and let $x$ be a generator of $H^2(C,\mathbb{Z})$. Then the powers of $y = N^G_C(x)$ yield non-trivial cohomology classes of $H^(G,\mathbb{Z})$ in degrees divisible by $(G:C)$. From a historical point of view the norm map already occured in disguise in Evens´ paper • The Cohomology Ring of a Finite Group, Trans. Amer. Math. Soc. 101(1961), 224-239 where he proves finte generation of the cohomology ring. - You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Yes, Richard Swan is the first to prove it, The Nontriviality of the Restriction Map in the Cohomology of Groups (1959): The restriction $res^G_H:H^i(G)\rightarrow H^i(H)$ is nonzero for an infinite number of values of $i>0$. As a corollary, for any prime $p$ dividing $\|G\|$, the $p$-primary component $H^i(G)_{(p)}$ is nonzero for an infinite number of values of $i>0$. He actually proves this in a more general case, where $G$ is a compact Lie group and $H$ is a closed subgroup (defining group cohomology with the classifying spaces $BG$ and $BH$). The proof uses basic cohomological and Lie group principles. - I did not know that paper. Swan's argument begins similarly as the argument in my answer, but then it becomes more involved since a Lie group does not have a regular representation. Swan has to use a spectral sequence. – Johannes Ebert May 12 2011 at 18:01 Yea; both arguments are awesome. – Chris Gerig Jan 10 2012 at 23:01 Let $G \to GL_N (C)$ be the regular representation. Pick a nontrivial subgroup $Z/p \subset G$ for a prime $p$ and consider the composition $Z/p \to GL_N(C)$, inducing $BZ/p \to BGL_N (C)$. If we can show that this map is nonzero in arbitrarily high cohomological degree, the theorem is proven. Let $L_k$ be the $1$-dim representation of $Z/p$ with the generator acting by $e^{2 \pi i k/ p}$. The restriction of the regular representation of $G$ to $Z/p$ is a multiple (say $m$ times) of the sum $$L_0 \oplus L_1 \ldots \oplus L_{p-1}.$$ Let $x \in H^2 (Z/p)$ be the first Chern class of $L_1$; this is a generator. Since $L_i$ is the $i$th tensor power of $L_1$, the total Chern class of $L_i$ is $1+ix$. Therefore, the total Chern class of the regular representation on $Z/p$ is $$((1+x)(1+2x) \ldots 1+(p-1)x))^m.$$ In particular, the $m(p-1)$st Chern class is $$z=\prod_{k=1}^{p-1} k^m x^{m(p-1)} \neq 0,$$ the latter because $p$ is a prime. Because $H^{\ast} (Z/p; Z) = Z[x]/(px)$, all powers of $z$ are nonzero. Therefore: write $\|G\|=pm$, $p$ prime. Then $H^{2 m (p-1)k } (G) \neq 0$ for all $k\geq 1$. - Is this really saying that $H^(G)$ is nonzero in all degrees higher than some integer $i$ ? I did not know of such a result. – Chris Gerig May 11 2011 at 22:03 And should that first term in the total Chern class be (1+x) ? – Chris Gerig May 11 2011 at 22:12 Ah nevermind to my first comment, wasn't thinking clearly. – Chris Gerig May 12 2011 at 4:24 thanks for spotting the typo. – Johannes Ebert May 12 2011 at 8:19 Could you, Johannes, elaborate more? How did you compute chern class of $L_i$? – Sam Nariman May 14 2011 at 20:00 One way is to use another powerful theorem :) For example, one knows from a result of Dan Quillen that the Krull dimension of the cohomology ring with coefficients in a field with bad characteristic is the same thing as the $p$-rank of the group, that is, the maximal rank of an elementary abelian $p$-subgroup. This is nicely explained in Dave Benson's Representations and cohomology, vol. II, and is proved in [Quillen, Daniel. The spectrum of an equivariant cohomology ring. I, II. Ann. of Math. (2) 94 (1971), 549--572; ibid. (2) 94 (1971), 573--602. MR0298694 (45 #7743)] It follows that for some prime $p$, the Krull dimension of $H^\bullet(G,\mathbb F_p)$ is positive, so it cannot vanish for $\bullet\gg0$, and then the same thing holds with integer coefficients. Indeed, this shows there are infinitely many non-zero degrees, answering Ryan's question in the comments to the main question. - I realized we need not to use that powerful theorem of Quillen to show that $H^(G,\mathbb{Z}_p)$ is nonzero for arbitrary large dimension. Actually by embedding $G$ in $U(n)$ for some $n$ we can show that $H^(G,\mathbb{Z}_p)$ is finitely generated $\mathbb{Z}_p$-algebra. Using this observation, it is easy to see $H^(\mathbb{Z}_p,\mathbb{Z}_p)$ is finitely generated $H^(G,\mathbb{Z}_p)$-module. So we are done. – Sam Nariman Mar 10 2012 at 2:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 76, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9087461829185486, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/12546/software-package-to-solve-l-infinity-norm-linear-regression/12564	# Software package to solve L-infinity norm linear regression Is there any software package to solve the linear regression with the objective of minimizing the L-infinity norm. - Well, any linear-programming package would work. That leaves you with a lot of options. :) – cardinal Jul 1 '11 at 15:25 1 @Cardinal How would you recast this as a linear program? It's not evident how to do it even in trivial cases (such as two data points and one parameter): there are no constraints and the objective function is nonlinear. – whuber♦ Jul 1 '11 at 15:57 Key phrase: Chebyshev approximation. (More to follow. The idea is to introduce an extra variable then turn objective into the constraints.) – cardinal Jul 1 '11 at 17:39 – Fan Zhang Jul 1 '11 at 18:18 Well, it's a bit related, but not germane to this problem. Your problem can be solved with a simple LP. As soon as I can get to a computer, I'll post an answer. – cardinal Jul 1 '11 at 18:29 show 1 more comment ## 3 Answers Short answer: Your problem can be formulated as a linear program (LP), leaving you to choose your favorite LP solver for the task. To see how to write the problem as an LP, read on. This minimization problem is often referred to as Chebyshev approximation. Let $\newcommand{\y}{\mathbf{y}}\newcommand{\X}{\mathbf{X}}\newcommand{\x}{\mathbf{x}}\newcommand{\b}{\mathbf{\beta}}\newcommand{\reals}{\mathbb{R}}\newcommand{\ones}{\mathbf{1}_n} \y = (y_i) \in \reals^n$, $\X \in \reals^{n \times p}$ with row $i$ denoted by $\x_i$ and $\b \in \reals^p$. Then we seek to minimize the function $f(\b) = \\|\y - \X \b\\|_\infty$ with respect to $\b$. Denote the optimal value by $$f^\star = f(\b^\star) = \inf \{f(\b): \b \in \reals^p \} \>.$$ The key to recasting this as an LP is to rewrite the problem in epigraph form. It is not difficult to convince oneself that, in fact, $$f^\star = \inf\{t: f(\b) \leq t, \;t \in \reals, \;\b \in \reals^p \} \> .$$ Now, using the definition of the function $f$, we can rewrite the right-hand side above as $$f^\star = \inf\{t: -t \leq y_i - \x_i \b \leq t, \;t \in \reals, \;\b \in \reals^p,\; 1 \leq i \leq n \} \>,$$ and so we see that minimizing the $\ell_\infty$ norm in a regression setting is equivalent to the LP $$\begin{array}{ll} \text{minimize} & t \\ \text{subject to} & \y-\X \b \leq t\ones \\ & \y - \X \b \geq - t \ones \>, \\ \end{array}$$ where the optimization is done over $(\b, t)$, and $\ones$ denotes a vector of ones of length $n$. I leave it as an (easy) exercise for the reader to recast the above LP in standard form. Relationship to the $\ell_1$ (total variation) version of linear regression It is interesting to note that something very similar can be done with the $\ell_1$ norm. Let $g(\b) = \\|\y - \X \b \\|_1$. Then, similar arguments lead one to conclude that $$\newcommand{\t}{\mathbf{t}} g^\star = \inf\{\t^T \ones : -t_i \leq y_i - \x_i \b \leq t_i, \;\t = (t_i) \in \reals^n, \;\b \in \reals^p,\; 1 \leq i \leq n \} \>,$$ so that the corresponding LP is $$\begin{array}{ll} \text{minimize} & \t^T \ones \\ \text{subject to} & \y-\X \b \leq \t \\ & \y - \X \b \geq - \t \>. \\ \end{array}$$ Note here that $\t$ is now a vector of length $n$ instead of a scalar, as it was in the $\ell_\infty$ case. The similarity in these two problems and the fact that they can both be cast as LPs is, of course, no accident. The two norms are related in that that they are the dual norms of each other. - +1 Nice exposition. Thanks. – whuber♦ Jul 2 '11 at 18:10 +1 Great Solution @cardinal. – suncoolsu Jul 5 '11 at 4:39 @cardinal's answer is well-stated and has been accepted, but, for the sake of closing this thread completely I'll offer the following: The IMSL Numerical Libraries contain a routine for performing L-infinity norm regression. The routine is available in Fortran, C, Java, C# and Python. I have used the C and Python versions for which the method is call lnorm_regression, which also supports general $L_p$-norm regression, $p >= 1$. Note that these are commercial libraries but the Python versions are free (as in beer) for non-commercial use. - Malab can do it, using cvx. to get cvx (free): ````cvx_begin variable x(n); minimize( norm(A*x-b,Inf) ); cvx_end ````	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9417501091957092, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/27549/modular-invariance-for-higher-genus	# Modular invariance for higher genus As far as I understand, there are roughly 2 "common" kinds of 2D conformal field theories: 1. Theories that are defined only on the plane, more precisely, on any surface of vanishing genus. Such a theory can be mathematically described by a vertex operator algebra, or by "usual" axiomatic QFT in 2D with the added requirement of conformal invariance 2. Theories defined on arbitrary surfaces. Such a theory can be described e.g. by Segal's axioms In order for a theory of the 1st category to become a theory of the 2nd category, it must pass the test of modular invariance. In my experience, this term usually means that the theory has to be defined in genus 1, i.e. that the would-be torus partition function $$Tr(q^{L_0-\frac{c}{24}}\bar{q}^{\bar{L_0}-\frac{\bar{c}}{24}})$$ is invariant under the modular group $SL(2,\mathbb{Z})$ What about higher genus? Is the torus condition sufficient to make the theory well defined there? Or does it impose additional conditions? If so, can they be cast into an anologous elegant algebraic form? Have these conditions been proved for CFTs used in string theory? - This is a very good question. Put it in another way: Can representation of modular transformation for torus completely determine a (chiral) CFT? – Xiao-Gang Wen May 27 '12 at 11:53 ## 3 Answers The modular group $SL(2,{\mathbb Z})$ is generated to $Sp(2h,{\mathbb Z})$, for genus $h$. That's the group exchanging the 1-cycles of the Riemann surface while preserving the intersection numbers (an antisymmetric tensor). Recall that the moduli space of Riemann surfaces is higher-dimensional, namely $(6h-6)$-dimensional (real dimensions) for $h>1$. Quite generally, the modular invariance at $h=1$ guarantees the modular invariance at all finite $h$. - 1 I know about the higher genus moduli group, but how do you show h=1 modular invariance implies h>1 modular invariance? – Squark Jan 15 '12 at 19:02 Hi Squark, see e.g. sciencedirect.com/science/article/pii/0370269387909464 - scholar.google.com/… - sorry for not reviewing the papers here – Luboš Motl Jan 18 '12 at 12:22 A higher genus surface can be considered as a connected sum of tori, and the linking cylinder in the connected sum can be replaced by a sum over all the propagating particles. If the theory is modular invariant on the torus, you know that you can do modular transformations on each torus separately, and everything is consistent. It is intuitive that every large diffeomorphism of a high genus surface can be generated using generators each in the individual SL(2,Z)s of the different tori of the connected sum. So by knowing torus modular invariance of the theory, correct for all tori and insertions on the tori, you learn that the large diffeomorphisms are ok. A sketch of a proof: you cut up the torus along loops into pieces homeomorphic to triangles, then consider the image of the loops under diffeomorphism. Restricting to each torus, there is a nonzero intersection number of the image of the loops with the old loops, and you perform a large diffeomorphism whenever this can reduce the intersection number. continue on each torus until you can't go reduce the intersection numbers any further. At this point, the intersection number must be as low as possible, and this means that the curves are isotopic to their original position, so that the diffeomorphism is now continuously connected to the identity. I did not check this in detail (I should), but the main lemma you need--- that you can always reduce a non-minimal intersection number by isotoping the curves to go near one genus, and doing an SL(2,Z) is very plausible. There might also be an easier proof. - A review reference for the case of superstrings in NSR formalism: http://arxiv.org/abs/0804.3167 I have been told that pure-spinor formalism has gone slightly further on the genus count. - Thx, but I'm not sure we are talking about the same thing. You are talking about performing integration over the moduli space in higher genus while I am talking about having a CFT well defined in higher genus which is a weaker requirement – Squark Jan 15 '12 at 19:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9183623790740967, "perplexity_flag": "head"}
http://nrich.maths.org/5626/solution	Quaternions and Rotations Find out how the quaternion function G(v) = qvq^-1 gives a simple algebraic method for working with rotations in 3-space. Quaternions and Reflections See how 4 dimensional quaternions involve vectors in 3-space and how the quaternion function F(v) = nvn gives a simple algebraic method of working with reflections in planes in 3-space. Two and Four Dimensional Numbers Stage: 5 Challenge Level: Do you think of complex numbers as very abstract and strange, and the idea of 4-dimensional numbers as even stranger? If you can add and multiply two by two matrices then you will see here how two by two matrices provide a model for complex numbers with a two by two matrix corresponding to i (the square root of -1). Two by two matrices also provide a model for the 4-dimensional numbers called quaternions and for the many different square roots of -1 that occur in this number system. Read on... Here is another excellent solution from Andrei (Bucharest, Romania). (1) (a) First we shall add and multiply the two matrices, obtaining: $$\pmatrix {x & -y \cr y & x} + \pmatrix {u & -v \cr v & u} = \pmatrix {x+u & -y-v)\cr y+v & x+u}$$ and $$\pmatrix {x & -y \cr y & x} \pmatrix {u & -v \cr v & u} = \pmatrix {xu-yv & -xv-yu\cr xv+yu & xu-yv}.$$ [Note the similarities here to the addition and multiplication of complex numbers.] (b) By simple calculations we observe that $\pmatrix {0 & 0 \cr 0 & 0}$ is the identity for addition and $\pmatrix {1 & 0 \cr 0 & 1}$ is the identity for multiplication. The inverses for addition and multiplication are obtained from the conditions that i) addition of the given matrix with its inverse gives the identity for addition, so the inverse of $\pmatrix {x & -y \cr y & x}$ for addition is: $\pmatrix {-x & y \cr -y & -x}$ and ii) multiplying the given matrix with its inverse gives the identity for multiplication; the inverse is ${1\over (x^2+y^2)}\pmatrix {x & y \cr -y & x}.$ We see that both identity matrices and both inverses are from the set C. (c) Here, we shall consider R as the set of matrices of the form $\pmatrix {x & 0 \cr 0 & x}$. These matrices could be written as $x\pmatrix {1 & 0 \cr 0 & 1} = xI_2$. Evidently $\pmatrix {x & 0 \cr 0 & x} \pm \pmatrix {y & 0 \cr 0 & y} = \pmatrix {x\pm y & 0 \cr 0 & x\pm y} = (x\pm y)I_2$. The identity for addition is $\pmatrix {0 & 0 \cr 0 & 0}$ and the inverse of $\pmatrix {x & 0 \cr 0 & x}$ is $\pmatrix {-x & 0 \cr 0 & -x}$, which are both from R. For multiplication $\pmatrix {x & 0 \cr 0 & x}\pmatrix {y & 0 \cr 0 & y} =x\pmatrix {1 & 0 \cr 0 & 1}\cdot y\pmatrix {1 & 0 \cr 0 & 1} = \pmatrix {xy & 0 \cr 0 & xy}$. The multiplicative identity is $\pmatrix {1 & 0 \cr 0 & 1}$ and the multiplicative inverse of $\pmatrix {x & 0 \cr 0 & x}$ is $\pmatrix {{1\over x} & 0 \cr 0 & {1\over x}}$. The distributive law of addition and multiplication is the same as that of real numbers: $$\pmatrix {x & 0 \cr 0 & x}\cdot \left(\pmatrix {y & 0 \cr 0 & y}+ \pmatrix {z & 0 \cr 0 & z}\right)= \pmatrix {x & 0 \cr 0 & x}\cdot \pmatrix {y & 0 \cr 0 & y} + \pmatrix {x & 0 \cr 0 & x}\cdot \pmatrix {z & 0 \cr 0 & z} = \pmatrix {x(y+z) & 0 \cr 0 & x(y+z)}.$$ This proves that the arithmetic of R is the same as the arithmetic of real numbers. (d) Considering 2 complex numbers, $x + iy$ and $u + iv$: $(x + iy) + (u + iv) = (x + u) + i(y+v)$ $(x + iy) (u + iv) = (xu - yv) + i(xv + yu).$ By simple computation we can show that multiplication in the set of two by two matrices C* is commutative and that the distributive law holds in C. This is so because the laws apply to every operation on the components. We observe that addition and multiplication in C are the same as addition and multiplication of complex numbers. (e) Note that $\pmatrix {0 & -1 \cr 1 & 0}^2 = \pmatrix {-1 & 0 \cr 0 & -1}$ so we see that C* contains a model for $\sqrt -1$, the complex number $i$. The matrix $\pmatrix {x & -y \cr y & x}$ corresponds to the complex number $x + iy$. Multiplying this number by $i$, we obtain $-y + ix$, i.e. the matrix $\pmatrix {-y & -x \cr x & -y}$. We shall consider that matrix $\pmatrix {a & b \cr c & d}$ corresponds to $i$: So, $\pmatrix {x & -y \cr y & x}\pmatrix {a & b \cr c & d} =\pmatrix {-y & -x \cr x & -y}$. The only solution for this equation is $\pmatrix {0 & -1 \cr 1 & 0}$. A different approach is the following: we associate the point $(x,y)$ in the complex plane with the complex number $(x + iy)$ and also with the matrix $\pmatrix {x & -y \cr y & x}$. The geometrical significance of the multiplication by $i$ of a complex number is the counterclockwise rotation of the point by $\pi /2$ so $(x,y) \to (-y,x)$. In matrix notation this corresponds to $$\pmatrix {0 & -1 \cr 1 & 0}\pmatrix {x & -y \cr y & x} =\pmatrix {-y & -x \cr x & -y}.$$ (2)(a)Working out the squares of the matrices $B= \pmatrix {i & 0\cr 0 & -i}$ , $C= \pmatrix {0 & 1\cr -1 & 0}$ and $D=\pmatrix {0 & i\cr i & 0}$ we get $B^2=C^2=D^2=\pmatrix{-1 & 0\cr 0 & -1}$. So all these three matrices are square roots of -1. (b) Similarly $BC=D=-CB$, (c)$CD=B=-DC$ (d) $DB=C=-BD$ so these matrices are models of the quaternions $i, j$ and $k$. Then $$a\pmatrix {1 & 0\cr 0 & 1}+b\pmatrix {i & 0\cr 0 & -i} +c\pmatrix {0 & 1\cr -1 & 0}+d\pmatrix {0 & i\cr i & 0}$$ provides a model for the quaternion number system. Simple calculations show that all the field axioms hold except t (3) Now, we shall calculate $i, i j, i j k, i j k i,...$ and we shall try to find a pattern. Using the relationships already established: $ij = k$, $ijk = k^2 = -1$, $ijki = -i$, $ijkij = -ij = -k$, $ijkijk = -kj = i$. This means that the sequence is periodic with period 6, namely $i, k, -1, -i, -k, 1, i, k ...$. The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found 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": 53, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9257075786590576, "perplexity_flag": "head"}
http://mathhelpforum.com/discrete-math/126851-need-help-proving-odd-number-divided-2-never-integer.html	# Thread: 1. ## Need help proving an odd number divided by 2 is never an integer I know it's true. This is actually part of a bigger proof, but I got it down to this, but I can't just say it's true. I have to somehow show it's true by relating it to an integer. $\frac{2k+1}{2}$ where 2k+1 is the definition of an odd integer. I can rearrange it and get k + 1/2, but I still need to prove that k + 1/2 is not an integer. I can't just say it. The only thing we have learned about that can identify integers is divisibility. i.e. a\|b means a = bk for some k that's an integer Anyone know how I can prove this? If it helps, the original question was: $\forall q \in \textbf{Q}, \exists r \in\textbf{Q}$ so that q + r is not an integer Which I figured I could do two cases where the rational number q is an integer, or when it isn't. So now that I'm at the case where it is an integer, I'm trying to prove that adding 1/2 (r value I picked) to it makes it so it's no longer an integer. Which got me to (2k+1)/2 2. If $\frac{k+1}{2}=j$ then $2k+1=2j$. There seems to be something wrong with that. NO? 3. Okay, so what you're saying is (2k+1)/2 = some integer j so then 2k + 1 = 2j where left is def of an odd and right is def of an even number This proves that 2k+1 is not even, but how does it prove that it's not an integer? 4. You can prove it by contradiction. If (2k+1)/2 is an integer, then it must be either even or odd. If it is even: (2k+1)/2 = 2j for some integer j 2k+1 = 4j 2k+1 = 2(2j) Since addition and multiplication are closed operations on integers, 2j is an integer, therefore 2(2j) is even (by the definition of evenness). Contradiction, since 2k+1 is odd by the definition of oddness. If it is odd: (2k+1)/2 = 2j + 1 for some integer j. 2k+1 = 2(2j + 1) Again, since addition and multiplication are closed operations on integers, 2j is an integer. Therefore 2(2j + 1) is an even number by definition. 2k + 1 is an odd number by definition => contradiction. Therefore (2k+1)/2 is not an integer. 5. Thanks! So to prove something like this, we have to break it up into two cases? And as long as you prove that its neither even nor odd, then it can't be an integer.	{"extraction_info": {"found_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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9605792164802551, "perplexity_flag": "head"}
http://mathoverflow.net/questions/44137?sort=votes	## FFT setting the boundary conditions ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am using the FFT method to solve the two dimensional PDE on a $2\pi\times 2\pi$ square. I have boundary conditions that $u(x,0,t)=0$ and $u(x,2\pi,t)=0$. How do I go about setting these boundaries? I can't just expand in different basis functions because of the way my fft function works. Thanks in advance. - Sounds like a job for the discrete sine transform. Any reference on FFT should include something on how to get a canned FFT routine to do DST. Alternatively, you might want to look at FFTW. – J. M. Oct 29 2010 at 15:04 ## 1 Answer If you don't mind wasting some resources, extend the $y$ coordinate to the interval $[0,4\pi]$. Then only look at solutions that are odd (anti-symmetric) under the flip $y\mapsto 4\pi-y$. In particular, if your PDE has source terms, they have to be anti-symmetrically mirrored from the $[0,2\pi]$ to the $[2\pi,4\pi]$ half of the extended interval. When restricted to $y\in[0,2\pi]$ a periodic solution on the extended interval will satisfy the desired boundary conditions. - This is essentially what a discrete sine transform does, only with less waste of space due to the exploitation of the antisymmetry. – J. M. Oct 31 2010 at 10:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.91043621301651, "perplexity_flag": "middle"}
http://polymathprojects.org/2009/07/28/deterministic-way-to-find-primes-discussion-thread/?like=1&source=post_flair&_wpnonce=744caefc44	# The polymath blog ## July 28, 2009 ### Deterministic way to find primes: discussion thread Filed under: discussion,finding primes — Terence Tao @ 3:09 pm The proposal “deterministic way to find primes” is not officially a polymath yet, but is beginning to acquire the features of one, as we have already had quite a bit of interesting ideas. So perhaps it is time to open up the discussion thread a little earlier than anticipated. There are a number of purposes to such a discussion thread, including but not restricted to: 1. To summarise the progress made so far, in a manner accessible to “casual” participants of the project. 2. To have “meta-discussions” about the direction of the project, and what can be done to make it run more smoothly. (Thus one can view this thread as a sort of “oversight panel” for the research thread.) 3. To ask questions about the tools and ideas used in the project (e.g. to clarify some point in analytic number theory or computational complexity of relevance to the project). Don’t be shy; “dumb” questions can in fact be very valuable in regaining some perspective. 4. (Given that this is still a proposal) To evaluate the suitability of this proposal for an actual polymath, and decide what preparations might be useful before actually launching it. To start the ball rolling, let me collect some of the observations accumulated as of July 28: 1. A number of potentially relevant conjectures in complexity theory and number theory have been identified: P=NP, P=BPP, P=promise-BPP, existence of PRG, existence of one-way functions, whether DTIME(2^n) has subexponential circuits, GRH, the Hardy-Littlewood prime tuples conjecture, the ABC conjecture, Cramer’s conjecture, discrete log in P, factoring in P. 1. The problem is solved if one has P=NP, existence of PRG, or Cramer’s conjecture, so we may assume that these statements all fail. The problem is probably also solved on P=promise-BPP, which is a bit stronger than P=BPP, but weaker than existence of PRG; we currently do not have a solution just assuming P=BPP, due to a difficulty getting enough of a gap in the success probabilities. 1. Existence of PRG is assured if DTIME(2^n) does not have subexponential circuits (Impagliazzo-Wigderson), or if one has one-way functions (is there a precise statement to this effect?) 2. Discrete log being in hard (or easy) may end up being a useless hypothesis, since one needs to find large primes before discrete logarithms even make sense. 2. If the problem is false, it implies (roughly speaking) that all large constructible numbers are composite. Assuming factoring is in P, it implies the stronger fact that all large constructible numbers are smooth. This seems unlikely (especially if one assumes ABC). 3. Besides adding various conjectures in complexity theory or number theory, we have found some other ways to make the problem easier: 1. The trivial deterministic algorithm for finding k-bit primes takes exponentially long in k in the worst case. Our goal is polynomial in k. What about a partial result, such as exp(o(k))? 1. An essentially equivalent variant: in time polynomial in k, we can find a prime with at least log k digits. Our goal is k. Can we find a prime with slightly more than log k digits? 2. The trivial probabilistic algorithm takes O(k^2) random bits; looks like we can cut this down to O(k). Our goal is O(log k) (as one can iterate through these bits in polynomial time). Can we do o(k)? 3. Rather than find primes, what about finding almost primes? Note that if factoring is in P, the two problems are basically equivalent. There may also be other number theoretically interesting sets of numbers one could try here instead of primes. 4. At the scale log n, primes are assumed to resemble a Poisson process of intensity 1/log n (this can be formalised using a suitably uniform version of the prime tuples conjecture). Cramer’s conjecture can be viewed as one extreme case of this principle. Is there some way to use this conjectured Poisson structure in a way without requiring the full strength of Cramer’s conjecture? (I believe there is also some work of Granville and Soundarajan tweaking Cramer’s prediction slightly, though only by a multiplicative constant if I recall correctly.) See also the wiki page for this project. ## 58 Comments » 1. This project is moving faster than I had thought, so I have quickly whipped up a wiki page for it: http://michaelnielsen.org/polymath1/index.php?title=Finding_primes Also, it looks like we will have to launch this particular polymath well before October… any thoughts on the pros and cons on “officially” making this a polymath (e.g. polymath4, since Gil has claimed the Hirsch conjecture for polymath3) in the near future? I can act as a moderator, though I will need some assistance in the short term as I will be travelling starting today for the next few days. Comment by — July 28, 2009 @ 3:33 pm • I’ve just announced this blog publicly on my own blog, so I expect activity on this project to increase even more in the near future… I was sort of hoping for a slow, controlled rollout of this project, but it seems that these things have a life of their own, and essentially only operate at one speed (fast). So we’ll have to play catch-up. One thing that needs to be done soon is to improve the wiki page so that people can join in more easily. For instance, this project is going to throw around a lot of conjectures in both computational complexity (e.g. P=BPP) as well as analytic number theory (e.g. prime tuples conjecture). One immediate project would be to flesh out the list of such conjectures already on the wiki with some links, either to external sites (e.g. Wikipedia) or some internal page that will explain all the jargon that is being thrown around. In the meantime, please use this thread to ask “dumb” questions (e.g. “What is BPP?”), especially if their answer is likely to be clarifying to other casual participants of this blog. The other thing is that this is likely to be my last post for four or five days as I will shortly travel overseas. Given the experience with minipolymath I know that a leadership vacuum can be damaging to a polymath project, so I hope people will take the initiative and step in to organise things if necessary. (Tim Gowers, Gil Kalai, and Michael Nielsen are also administrating this blog and may be able to help out with technical issues, e.g. editing a comment, or empowering a moderator to edit comments.) Comment by — July 28, 2009 @ 4:34 pm 2. If this does turn into a polymath project by default — which is OK by me as it’s a problem I’d very much like to see solved — can I strongly urge that the start is delayed for a week or two? I have a selfish and an altruistic reason for this. (I seem to remember having written precisely that sentence somewhere else recently.) The selfish reason is that I’ll be on an internet-free holiday for the whole of next week. The more serious reason is that I’m sure I’m not the only one who gets rapidly lost when theoretical computer scientists drop names of complexity classes all over the place. I don’t know how they do it, but they seem to know what they all are, and how they are related to each other (either actually or conjecturally). I can see myself getting rapidly left behind, or at least left behind from parts of the conversation, unless some serious work is put in first in order to get a good wiki going that explains, informally but in detail and with good examples, what all the various conjectures from complexity theory and from number theory are. Comment by — July 28, 2009 @ 4:40 pm • Fortunately, there is already such a wiki that lists and explains almost all complexity classes known to mankind: http://qwiki.stanford.edu/wiki/Complexity_Zoo As an example, on PromiseBPP is says: “Same as PromiseRP, but for BPP instead of RP.” And then on PromiseRP it says: “The class of promise problems solvable by an RP machine. I.e., the machine must accept with probability at least 1/2 for “yes” inputs, and with probability 0 for “no” inputs, but could have acceptance probability between 0 and 1/2 for inputs that do not satisfy the promise. “ Comment by Wim vam Dam — July 28, 2009 @ 4:59 pm • For senile mathematicians like me, that isn’t enough. I need to be given some examples of problems in BPP, RP, PromiseBPP, etc., together with informal explanations of why they are in those classes. Otherwise, I just don’t have any confidence that I understand what I have read. Comment by — July 28, 2009 @ 5:24 pm • [Airport has internet, yay!] Given that I’m also going to be largely internet-free next week, a delay of a week or more does sound desirable. I think what I will do is put a note on the research thread asking people not to work too hard on the problem for now, though clearly if a useful thought comes to mind and it is solid enough to be of immediate value to other participants, it is still worthwhile to record it. Meanwhile, we can focus efforts on building the wiki with some good exposition, particularly with regard to complexity classes, which seem to be particularly relevant for this project. I’m imagining taking a highly specialised subslice of the Complexity Zoo and expanding it on the wiki in a way aimed at complexity theory novices (with emphasis on examples, relationship to the prime finding problem, etc.) Comment by — July 28, 2009 @ 7:00 pm • I’ll gladly do some of that — in fact, I might start this evening. Comment by — July 28, 2009 @ 7:33 pm 3. I’d be happy to take a crack at contributing on this problem, but it seems like a bit of a “scary” one for a Polymath project. By this I mean that it’s a problem that’s been around for a while but I haven’t heard of many attacks on it. (Perhaps we should ask Manindra/Neeraj/Nitin if they spent time on it…) Contrast this with Primality, which did have a fair bit of partial progress before its dramatic solution (e.g., n^{log log log n} time algorithm, poly-time algorithm assuming RH, in NP, in RP, in co-RP, Agrawal-Biswas). Perhaps a less scary question in a similar vein is that of factoring (univariate) polynomials over finite fields. It has long been known that this has a randomized poly-time algorithm (by poly, we mean poly(n, log p) for factoring degree-n polynomials over F_p), and it is still unknown if this can be derandomized. I’m not sure what the current state of the art is, but I’m pretty sure there’s been a steady drip of work on special cases, how to use fewer than expected random bits, what you can get if you assume ERH, average-case analysis, and so forth. Might be easier for Polymath to get some traction on this problem. Comment by Ryan O'Donnell — July 29, 2009 @ 6:34 am • Ryan, I sort of agree with you, and certainly the problem you suggest sounds like a very good one to try. But in a way I’m quite interested to see what the result will be if this problem turns into a fully-fledged Polymath project. It does feel “too difficult” somehow, but for that very reason it could be a very useful experiment in understanding what Polymath can and cannot achieve. My guess is that there would be a lot of initial discussion, with many interesting ideas proposed and briefly explored, and that after a shortish time like two or three weeks it would begin to feel as though we were going round in circles. But even if that happened, I think we could aim to produce a very interesting wiki that served as a summary of the thoughts that had taken place. And there is always the chance that the prediction I’ve just made is wrong: perhaps some piece of magic would happen and a serious result (not necessarily a solution to the original problem, but something related) would get proved. Either way, we would have valuable new information about Polymath projects in general. This isn’t a knock-down argument of course — maybe there are other projects that would yield more information of this kind more efficiently. But I think this one should be at least reasonably efficient, since either my prediction is right (in which case either the project lasts a short time or it leads to something very good) or it is wrong (in which case something unexpected has happened and we’ve learned more about Polymath). I definitely think we should bear in mind that polynomial-factoring question though. Another question is whether we should keep that as an entirely separate project, or whether it should be part of the discussion about finding primes. For what it’s worth, I’d vote for keeping it separate. Comment by — July 29, 2009 @ 7:30 am • By the way, it was pointed out to me today by Sergey Yekhanin and others that the polynomial factorization problem is open even in the case of degree 2, in which case it is equivalent to the problem of deterministically finding a quadratic nonresidue in F_p in time polylog(p) — and that this is a notorious open problem in algorithmic number theory. Comment by Ryan O'Donnell — July 30, 2009 @ 7:18 am 4. [...] also the discussion thread for this proposal, which will also contain some expository summaries of the comments below, as well as the wiki page [...] Pingback by — July 31, 2009 @ 5:12 am 5. An unlikely idea, but maybe a polymath project is just the right place to throw it into the discussion: Recently Viola, Bogdanov, Lovett, et al. have made progress on unconditional PRGs that fool constant-degree polynomials over finite fields. So, if the AKS primality test could be reformulated in this model of computation, i.e. as a constant-degree d polynomially-sized polynomial AKS: {0,1}^n –> {0,1}, the problem would be solved for any fixed n. Comment by — July 31, 2009 @ 4:10 pm 6. I’ll have a crack at trying to summarise some of the progress of the last few days, which is part of what this discussion thread is for; feel free to expand upon this update. * The wiki page is slowly being updated. Please take a look at it and see if there are any quick contributions you can add (e.g. adding new papers, concepts, etc.) Don’t be shy; we can always edit the content later, but the key thing is to have the content there in the first place. * It looks like we’re not going to solve the problem just by soft complexity theory techniques; in the language of Impagliazzo, we have one way to find primes in “Algorithmica” (and maybe “Heuristica”) type worlds, in which we can solve NP type problems easily, and we have another way to find primes in “Minicrypt” or “Cryptomania” type worlds where we have good pseudorandom generators to derandomise the probabilistic algorithm, but there is this nasty middle world of “Pessiland” where we don’t have good tools. * The function field version is known (a result of Adelman-Lenstra). * If factoring is cheap, then it would suffice to find a short interval (or other small, easily explored set) which contains a non-smooth number. There is a chance that additive combinatorial techniques (applied to iterated sumsets of logarithms of primes) could be relevant here. There is also a substantial number theory literature on smooth numbers that should be looked into, although most of the focus is on getting smooth numbers rather than non-smooth numbers. (I tried to apply the ABC conjecture to rule out large consecutive strings of smooth numbers, but it looks like it is not the right tool – the smooth numbers don’t need to have repeated prime factors and so one gets little gain from ABC.) I probably missed some stuff; feel free to add, correct, or elaborate. Comment by — July 31, 2009 @ 7:51 pm 7. [...] http://polymathprojects.org/2009/07/28/deterministic-way-to-find-primes-discussion-thread/ [...] Pingback by — August 2, 2009 @ 4:39 pm 8. This could be interesting http://mathworld.wolfram.com/PrimeDiophantineEquations.html. Are there any approach based on Diophantine representation? Comment by Anonymous — August 3, 2009 @ 5:12 pm • Hmm. Well, by Matiyasevich, any recursively enumerable set (not just the primes) can be described via Diophantine equations. We know that the problem of finding primes quickly becomes impossible if one replaces the primes with other recursively enumerable sets (e.g. the set of high-complexity numbers), so it seems unlikely that one could proceed purely by Diophantine equation theory, but will have to use something specific about the primes. Comment by — August 5, 2009 @ 2:03 am 9. An update on the recent discussion: * We have a couple of negative results and heuristics that seem to rule out some types of approaches. For instance, we can’t use any soft complexity theory argument that doesn’t explicitly use some special property of the primes (besides the fact that they are dense and testable) because there are sets of “pseudoprimes” which are dense and testable (assuming an oracle) but which cannot be located deterministically. (See this page for details.) On the other hand, this objection disappears if a factoring oracle is assumed. * In a similar spirit, if we delete all the “constructible” primes from the set of primes, leaving only the “generic” primes, then the conjecture fails, while leaving all average-case results about primes (and some worst-case results too) intact. So it is unlikely that one can proceed purely by using average-case results about primes. Though again, this objection may disappear if a factoring oracle is assumed. * Using sieve theory to find non-smooth numbers also seems problematic. The problem is that sieve theory is largely insensitive to the starting point of an interval. Since the interval [1,x] contains nothing but x-smooth numbers, it is thus not possible to use sieve theory to locate non x-smooth numbers in any other interval of length x. In particular, this suggests that sieves and factoring oracles cannot be used to locate primes larger than x without searching through more than x numbers. * The smallest explicit set known which is guaranteed to contain a number with a prime factor of at least x is of size about $x^{0.535}$ (unconditionally) or $x^{1/2+o(1)}$ (assuming GRH), and is basically just an interval. Can one do better? There are some polynomial forms ($a^2+b^4$ and $a^3+2b^3$) that are known to capture their primes, but can they be made sparser? (It may become necessary to start a reading seminar on Friedlander-Iwaniec at some point.) * Tim’s proposal to try to use additive combinatorics to show that non-smooth numbers are so uniformly distributed that they lie in every short interal (comment #18 in the research thread) remains untouched so far by the above objections; hopefully when Tim gets back from his vacation we can analyse it more thoroughly. * There seems to be a fundamental barrier that in order to deterministically find an integer that avoids m different congruence relations, one has to search at least m+1 different numbers. Thus, for instance, to find a number not divisible by any prime less than $\sqrt{n}$, one needs to search about $\sqrt{n}$ numbers, which is consistent with the fact that even assuming GRH, we currently need about $\sqrt{n}$ searches to deterministically obtain a prime of size n. It would be good to find a way to substantially break this barrier. (For small moduli one can use the Chinese remainder theorem to concatenate some moduli, but if searching for numbers of size n, one can only apply this theorem to about $O(\log n)$ of the moduli.) * It may be that even the very weakest of the conjectures (e.g. beating the $\sqrt{n}$ barrier for deterministically finding a prime larger than n, assuming a factoring oracle) may be beyond the ability to solve without powerful number theory or complexity conjectures. However, it may still be instructive to try to collect reasons why the problem is difficult (i.e. add to the list of obstructions mentioned above). * The research thread is getting very long. I guess it should soon be time to officially open the polymath project and start a fresh thread, summarising some of the key insights so far. I might do this in a week or so (I know Tim wanted to hold off until he’s back from vacation). Comment by — August 5, 2009 @ 4:56 pm 10. Let me inaugurate the asking of dumb questions with a request and a comment. First, could someone make an attempt at summarizing Gowers’ combinatorial idea (related to post 18), I don’t entirely follow it. Secondly, I have a (rather obvious) comment. There seems to be several strategies of the form, “we can assume that conjectures X, Y and Z don’t hold, for otherwise we can solve the problem”. For example, we may assume that Cramer’s conjecture fails and P != NP. However, if we can succeed under these hypothesis, then it seems likely we will obtain an existence proof without an explicit algorithm. For example, we won’t have access to the supposed polynomial algorithm implied by the assumption that P=NP, although our solution would be based on running that algorithm. If we assume, say, Cramer’s conjecture then we have a very simple algorithm (we just start searching through the k-digit numbers) however if we don’t know the implied constant in expression O(ln^2(k)) then we have no idea how long it will take our algorithm to terminate. I would be very eager to even see an existence proof, but would such a proof really have any advantage over the random polynomial algorithm? Comment by Anon — August 6, 2009 @ 9:30 pm • Thanks for “breaking the ice”, as it were! Tim’s idea goes something like this. We want to find a prime with at least k digits. If we assume a factoring oracle, it’s enough to find a non-smooth number, or more precisely a number containing at least one prime factor with at least k digits. The idea is then to look at some interval [n,n+a] (with n much larger than $10^k$, and $a$ represents the “cost budget” for the search – polynomial in k for the strong conjecture, $\exp(o(k))$ for the weak conjecture) – the exact tuning of parameters is one of the things that will have to be done at some point) and factor all of them, hoping to find a large prime. The only way this will fail is if none of the numbers in [n,n+a] have prime factors with at least k digits. In particular, if one lets J be the set of all primes with k digits, then the iterated power sets $J, J^2, J^3, \ldots$ of J are going to completely miss [n,n+a]. If we let K be the logarithms of the elements in J, this means that the iterated sumsets $K, K+K, K+K+K, \ldots$ completely miss $[\log n, \log(n+a)]$. On the other hand, there are a number of results in additive combinatorics which have the general flavour that repeated sumsets tend to spread out very “uniformly”, unless they are concentrated in an arithmetic progression. If all the parameters are favourable, the hope is then that one can use the additive combinatorics machinery to show that iterated sumsets will hit $[\log n, \log (n+a)]$ unless K is concentrated on an arithmetic progression, so that J is concentrated on or near a geometric progression, which perhaps can be ruled out by number-theoretic means. It may well be that the parameters are such that this doesn’t work, but I think there is a non-zero chance of success, especially if one is willing to settle for the weak conjecture or even just to beat the current world record of $(10^k)^{1/2}$. I’m thinking of making the exploration of the feasibility of Tim’s idea one of the main focuses of the polymath project when it formally launches. As for your second comment, there is a standard observation (which I guess I should have made explicitly, but I put it on the wiki) that if a polynomial-time algorithm to achieve a task (in this case, to find a prime), then an explicit polynomial-time algorithm to achieve the same task also exists. The idea is basically to run all the possible polynomial-time algorithms in parallel. More precisely, if one enumerates all the possible algorithms in the world as $A_1, A_2, \ldots$, then what one can do is, for each i, run each of the algorithms $A_1,\ldots,A_i$ for time $k^i$, then increment i and keep going. This will terminate successfully in polynomial time if even just oen of the algorithms $A_i$ finds a prime in polynomial time. It’s sort of a sleazy trick, but this is, technically, an explicit algorithm. (My guess, though, it will be relatively easy to extract a more elegant algorithm out of any positive result that we actually end up obtaining, even if it is initially conditional in nature.) I should also point out that this problem is primarily of pure mathematical interest, given that the current probabilistic methods to find primes are quite satisfactory. See also this post by Lance Fortnow on this project. Comment by — August 6, 2009 @ 11:10 pm 11. “So, given sufficiently strong conjectures in complexity theory (I don’t think P=BPP is quite sufficient, but there are stronger hypotheses than this which would work), one could solve the problem.” Maybe I overlook something simple but what specific conjecture in complexity theory will solve the problem positively? Another question: I wonder what is the algorithmic status of other related problems regarding gaps of primes. (Even not worrying too much about randomized algorithms.) a) Finding large gaps between primes (Rankin and related results), b) finding primes with small gaps (Goldon-Pintz-Yildirim), c) finding AP of length 3 of at least k-digits primes (Green-Tao) If we allow Cramer-model based conjectures I suppose we have deterministic polynomial algorithm for all these as well as for d) find k-digits twin primes. But these are extremely strong conjectures. Also I wonder if the little advantage we gained by allowing factoring oracles disappears at once for all these variants. Comment by — August 8, 2009 @ 11:25 am • Well, as Lance further explained on his blog there is a concrete conjecture referred to as full derandomization that implies that the there is a polynomial time deterministic algorithm for finding primes http://blog.computationalcomplexity.org/2006/07/full-derandomization.html The (very strong) conjecture is stated in terms of Hardness and is related to Hardmess vs randomization theme in CC. http://cseweb.ucsd.edu/~russell/survey.ps Comment by — August 8, 2009 @ 3:59 pm • Well, with regards to (c) [finding AP of primes] the number of AP are dense enough that one can just randomly take arithmetic progressions of k-digit numbers and check them for primality and one will probably get a hit in polynomial time. Similarly, for (b) (primes with small gap), assuming GUE one has Poisson type statistics for prime gaps, and so one will get a prime gap much smaller than log n just by searching randomly. Or, one can just go for the gold and search for twin primes randomly, though of course this is conditional on the twin prime conjecture (with the expected asymptotic, or at least a comparable lower bound.) Unconditionally, I think it’s open, because the GPY results only give a very sparse set of small prime gaps. Similarly for the Rankin results in (a). I quite like the problem you proposed in the other thread, to find adjacent pairs n, n+1 of square-free numbers. Such pairs exist in abundance (since $6/\pi^2 \approx 60\%$ of all numbers are square-free) but there is no obvious way to generate them. As square-free numbers seem to be substantially more “elementary” than primes, this may well be a good toy problem (and has some of the same flavour, namely trying to avoid a large number of residue classes at once, though the number of residue classes to avoid is much sparser than that of the primes (note that $\sum_n \frac{2}{n^2}$ converges, while $\sum_p \frac{1}{p}$ does not) so this is presumably an easier problem to solve. In a few days I plan to launch the finding primes project formally as polymath4 (polymath3 being claimed by the Hirsch conjecture), in part because the current research thread is getting bloated. I think I will put the square-free problem as one of the toy problems to consider. Comment by — August 8, 2009 @ 4:48 pm • My questions regarding (a) (b) and (c) is what can be proved without assuming any conjectures… Comment by — August 8, 2009 @ 6:03 pm • Possible approach. To generate lots adjacent pairs of square-free numbers with reasonable probability, note that one of the numbers has to be even. So construct even square-free numbers by multiplying small primes including 2 (to get an even one) and test the numbers either side – depends on having a good test for square-freeness, but factoring will do this. Comment by Mark Bennet — August 8, 2009 @ 6:19 pm • We might be able to make some progress on the square-free problem from a well-known conjecture regarding Sylvester’s sequence. Sylvester’s sequence is defined by the recurrence: $S_0=2$ $s_n = 1+ \prod_{i=0}^{n-1}s_i$. It is conjectured (and there is numerical evidence to support this) that every element of this sequence is square-free. Additionally, it isn’t hard to show that the numbers $s_n$ are coprime. To see this note that for every $l$ < $m$ we have that $s_n \equiv 1 \mod s_l$. On the conjecture, we then would have that $s_n-1$ and $s_n$ are both square-free. We should also (I think) be able to quickly compute the sequence from the recurrence. One interesting caveat is that the sequence grows double-exponentially, so it wouldn't (at least directly) give a square-free number of $k$-digits for every $k$. Comment by Mark Lewko — August 9, 2009 @ 12:04 am 12. [...] which has now become rather lengthy. (Simultaneously with this research thread, we also have the discussion thread to oversee the research thread and to provide a forum for casual participants, and also the wiki [...] Pingback by — August 9, 2009 @ 4:00 am 13. [...] be explored soon. In addition to this thread, there is also the wiki page for the project, and a discussion thread aimed at more casual participants. Everyone is welcome to contribute, as always. Possibly related [...] Pingback by — August 9, 2009 @ 4:35 am 14. A few comments: 1) About the shortest interval which guarantees primes: The 0.535 can be replaced by 0.525. Proceedings of the London Mathematical Society 83, (2001), 532–562. R. C. Baker, G. Harman and J. Pintz, The difference between consecutive primes, II The authors sharpen a result of Baker and Harman (1995), showing that [x, x + x^{0.525}] contains prime numbers for large x. 2) About question 28 by Gil, 5th August: 2a) In addition to Terry’s reference to Schinzel’s Hypothesis H, saying that irreducible polynomials take infinitely many prime values, unless there is an obvious reason why not: a quantitive version of Schinzel’s conjecture is called “Bateman-Horn conjecture”, giving the density c_f/log X, where the constant c_f depends on the polynomial, and also predicting densities for k-tuples. http://en.wikipedia.org/wiki/Bateman-Horn_conjecture 2b) An elementary way to see that an arithmetic progression cannot contain too many primes is: The sum over the reciprocals of the AP: Sum_{k < X} 1/(ak+b) grows like (log X)/a, whereas the sum of the reciprocals of primes p<X grows like log log X only. So the density of the primes in the AP must be zero. Comment by — August 9, 2009 @ 9:22 am 15. Can we have examples/counterexamples, whereever possible, that illustrate how a conjecture/theorem holds/does not hold Comment by tful — August 9, 2009 @ 4:35 pm • Is there a specific result or conjecture on the wiki or elsewhere that you have in mind? I guess I’m asking for an example for where an example is needed. :-) Comment by — August 10, 2009 @ 10:27 pm 16. (I’m posting this here because after I thought of it, I could hardly not post it, but it’s incredibly silly and should probably not clutter up the research thread.) I have a simple deterministic algorithm, with a 19th-century analytic NT flavor, that essentially finds a prime of at least n bits in time roughly linear in n (assuming an oracle for primality, of course.) First, pick an arbitrary n-bit integer; there are a number of ways to do this in the literature. Then, check the next O(n) odd integers for primality. If one of them is prime, we’re done; if not, pick $3^n$; since prime powers can for the interesting purposes be considered essentially the same as the primes, we have essentially found a prime. (P.S. Yes, I realize that “prime powers can be considered essentially primes” is a criminal oversimplification, but there are situations — in particular the proof of the PNT — where it’s true and useful, and once I thought of this bad joke it wouldn’t get out of my head.) Comment by — August 11, 2009 @ 11:11 am 17. I have a question of procedure. To improve my own understanding I’ve been working on some notes about a proof of Bertrand’s postulate. I was imagining posting these to the wiki (and cross-posting to my blog) as background material. However, it occurs to me to wonder how that fits with the spirit of “small increments”. I would perhaps suggest that the small increments philosophy only need apply to the research and writeup, not necessarily to the preparation of background materials for the wiki. But I wanted to run this by others in case they see a problem or disagree. Comment by — August 11, 2009 @ 5:58 pm • Well, from previous polymath experiences I think that the “small increments” philosophy is mostly applicable to the type of work which _can_ be divided up into small increments; there are some tasks which really do need someone to go away and work for an hour, and it seems counterproductive to ask that such tasks be prohibited or that it is somehow split up into five-minute chunks. It seems that we are approaching the point where detailed computations need to be made, but as long as people aren’t spending entire days offline trying to crack the whole thing, I think we’re still adhering to the spirit of the polymath project. (But it was courteous to notify the rest of us what you are doing.) Comment by — August 11, 2009 @ 7:02 pm 18. Wow, there is so much activity in the research that it is hard to keep up! I am going to try to summarise some of the discussion, and to put it on the wiki. This may take a while though… We seem to be considering quite a few distinct strategies: 1. Studying iterated sumsets of log-primes (Gowers #I.18, Tao #3, Gowers #7, #17), which now looks like it has hit some fundamental obstructions (Kalai #36, Gowers #38); 2. Studying signed sums of prime reciprocals (Croot #10), which still looks promising but needs further analysis; 3. Computing pi(x) and using Riemann zeroes (Kowalski #28, Croot #35, PhilG #37) – I haven’t digested this thread yet, but will try to soon. 4. Demonstrating the existence of a factored number near n by repeated adjustments (Gowers #13, #16, #21). May also be subject to the obstructions to 1. 5. Finding pseudoprimes first (Wigderson #18, Kalai #22) 6. Analysing special sequences, e.g. Sylvester sequence (Lewko #2, Dyer #24) 7. Lattice strategies for the adjacent square-frees (or nearby square-frees) problem (Croot #29). May be subject to the same objections to 1. There’s also some progress in clarifying the role (or lack thereof) of P=BPP (Trevisan #34) There are now multiple indications that the square root barrier (beating $(10^k)^{1/2}$) is quite a serious one. Perhaps we should lower our sights still further and shoot for the “ultraweak” conjecture of finding a prime larger than $10^k$ in time $o(10^k)^{1/2}$. We’ve also realised that to get a non-smooth number in a small interval one cannot use average-case information about the set of all primes, because most primes will not actually divide into a number in that small interval, and if one deletes the few primes that do, then this is not detectable in the average-case statistics (unless the size of the interval exceeds the square root barrier). This type of problem seems to kill strategy 1., and may also seriously injure strategies 4 and 7. More updates later, hopefully… in the meantime I will try incorporating the recent progress into the wiki. Comment by — August 12, 2009 @ 3:35 pm 19. Incidentally, I have a post on my own blog on the AKS primality test at http://terrytao.wordpress.com/2009/08/11/the-aks-primality-test/ which may be of relevance. Somewhat more whimsically, XKCD seems to also be contributing to the project with its latest comic: http://www.xkcd.com/622/ Comment by — August 12, 2009 @ 3:55 pm 20. The research thread has passed 120 comments now… is it time to start a new thread? I realised that maybe we can use the ratings system to answer this question efficiently… one can use the ticks or crosses below this comment to vote on whether it is time to start afresh and try to summarise the progress so far (tick), or continue with the thread as is (cross). Of course, one can also make comments in the more traditional way. Comment by — August 12, 2009 @ 9:01 pm 21. [...] always, oversight of this research thread is conducted at the discussion thread, and any references and detailed computations should be placed at the wiki. Leave a [...] Pingback by — August 13, 2009 @ 5:13 pm 22. This is regarding my most recent reply to post #7 on research III. I got a bit caught up in the verifications and I didn’t realize just how long this description ended up being. This clearly belongs in the wiki and not on the research thread. Mea culpa! If possible, I would like some assistance to declutter the research thread and move this where it belongs on the wiki. I will also add a description of the systems $S^i_2$ and $T^i_2$ for context. Comment by — August 15, 2009 @ 3:25 am • Dear Francois, It does seem like a good idea to put lengthier or more technical computations on the wiki. If you do so, you can put a new comment giving the link and a summary of the contents next to your previous comments and I can then remove any obsolete comments. Comment by — August 15, 2009 @ 3:04 pm 23. The poor formatting of my post #15 with all those line-breaks in strange places makes me wonder: is there a way to set up wordpress so that it is possible to “preview” a submission before actually submitting it? Comment by Albert Atserias — August 17, 2009 @ 3:45 pm • I fixed the problem manually. I think it is possible to install a previewing feature, but we would have to use a customised installation of wordpress rather than relying on wordpress.com’s free platform as they explicitly disallow preview for security reasons, as stated here: http://en.forums.wordpress.com/topic.php?id=7517&page&replies=4 Certainly, if and when we migrate to a more optimal platform, this is one of the features we will want to have. Comment by — August 17, 2009 @ 3:51 pm 24. I’m wondering why the idea of trying to generate numbers that satisfy the AKS congruences and variants has received very little (visible) attention so far. Admittedly this looks like a very difficult problem, but it has many advantages over “pure” divisibility/density based approaches. An important advantage is that such congruences are apparently “razor-proof”. More generally, you can’t delete an arbitrary “thin-but-large” set of (pseudo-)primes and expect the resulting set to be polytime recognizable. Maybe I missed something that rules out this method? Comment by — August 18, 2009 @ 7:08 pm • Well, I tried a little to get somewhere, but even the Fermat congruences such as $2^n = 2 \hbox{ mod } 2$ seem hard to catch, let alone the stronger AKS congruences [of which the Fermat congruences are the constant term]. (There is the classical observation that if n is a pseudoprime wrt 2, then $2^n-1$ is also, but this doesn’t seem to extend to, say, give a simultaneous pseudoprime wrt 2 and 3.) Part of the problem is that the quantity $2^n \hbox{ mod } n$ seems to obey no useful properties as n varies; the exponentially large nature of the quantity which one is seeking the remainder of seems to defeat most naive approaches to the problem. The one thought I had was that it may be possible to apply the AKS algorithm to several integers n in parallel by recycling some steps, which could shave a log factor or two off of the best deterministic algorithm (if we insist on computing primality by AKS, rather than assuming a O(1)-time oracle), but this is perhaps not very exciting. Comment by — August 18, 2009 @ 8:57 pm • It does look like a very difficult problem. I thought that understanding a little about the function $G(n,d) = \mathrm{gcd}(2^n-1,\dots,d^n-1)$ might help. The only idea I got toward this is the following. Expanding $(x-1)(x-2)\cdots(x-d) = x^d - a_{d-1}x^{d-1} - \cdots - a_1x - a_0$ and looking at the solutions $(s_n)$ of the linear recurrence $s_{n+d} = a_{d-1}s_{n+d-1} + \cdots + a_0 s_n$ with $s_0 = 0$ and $s_1,\dots,s_{d-1} \in \mathbb{Z}$, then the multiples of $G(n,d)$ are precisely the set of possible values of $s_n$. I tried to use some linear recurrence technology (which I don’t know very well) to get some information about $G(n,d)$ but I didn’t get anywhere. Anyway, I just wanted to throw that out there in case somebody might find this useful… Comment by — August 19, 2009 @ 1:28 am • This is not directly relevant to the finding primes project, but I also thought about deconstructing the proof that $2^n \neq 2 \hbox{ mod } n$ implies that n is composite to see if it could give a factoring algorithm. The problem is that the proof of the implication requires poly(n) arithmetic operations (to get from the bijectivity of $x \mapsto 2x \hbox{ mod } n$ when n is odd to the fact that $(n-1)! = 2^{n-1} (n-1)!$ requires poly(n) applications of the associative law, and dividing out by (n-1)! requires a further n operations if one wants to be sure that one is not dividing by zero) so does not suggest any algorithm for factoring that is not exponentially large in the number of digits. Comment by — August 18, 2009 @ 9:02 pm • I’m glad you thought about doing that. I believe it is still an open problem whether Fermat’s Little Theorem is provable in $\mathrm{I}\Delta_0 + \Omega_1$! Comment by — August 19, 2009 @ 1:08 am 25. Let me propose the following meta-problem: What’s the “simplist” algorithm/proof of the fact that we can deterministically find a $k$ digit prime in $O(10^{k})^{1-\delta}$ time for some small $\delta>0$. That is, can we get a result of this form using anything less powerful than the work of Baker-Harmen or Friedlander-Iwaniec theorem? Comment by Mark Lewko — August 19, 2009 @ 5:13 am • According to this paper Odlyzko, Andrew M.(1-BELL) Analytic computations in number theory. (English summary) Mathematics of Computation 1943–1993: a half-century of computational mathematics (Vancouver, BC, 1993), 451–463, http://www.ams.org/mathscinet-getitem?mr=1314883 there is an elementary algorithm to compute $\pi(x)$ in $x^{2/3+o(1)}$ time, though I don’t know whether this is a deterministic algorithm (I don’t have access to the paper). If so, that would solve this problem by binary search. Comment by — August 22, 2009 @ 4:44 pm • I typed in the paper to google, and found that Odlyzko has a link on his website at: http://www.dtc.umn.edu/~odlyzko/doc/arch/analytic.comp.pdf I looked up the section where it mentions the $x^{2/3 + o(1)}$ algorithm, and apparently there is an analytic algorithm to compute $\pi(x)$ in time $x^{1/2+o(1)}$ using the zeta function (but not the explicit formula)! So it seems we were trying to solve a problem that has already been solved. Still, it would be good to have other solutions. Comment by Ernie Croot — August 22, 2009 @ 5:31 pm 26. This may not be the right place to say this, but based on some earlier discussion about finding sequences divisible by a sparse set of primes I’d like to suggest a toy problem of independent interest: given a primitive arithmetic progression, find a sequence that grows polynomially (or at least slower than $e^{ \sqrt[k]{n} }$ for every $k$) such that the only prime factors of its entries are, with finitely many exceptions, in that progression. It’s known that one can do more or less this for progressions $a \bmod n$ such that $a^2 \equiv 1 \bmod n$ by finding an actual polynomial, but it’s not necessary that the sequence actually be a polynomial. The growth condition above ensures that such a sequence has infinitely many distinct prime divisors, hence one would obtain a proof of Dirichlet’s theorem without using the machinery of zeta functions. Comment by — August 22, 2009 @ 8:21 am 27. I have a question regarding the w-trick. It has been remarked in several places that we might be able to use the square-free gap results and the w-trick to find large pairs of consecutive square-free numbers. However, it isn’t clear to me how the details of this would work. The w-trick allows us to pass to an arithmetic progression with a density of square-free numbers arbitrary close to 1. Now, assuming that we can adapt the square-free gap results to arithmetic progressions (which seems reasonable to me), this would allow us to find consecutive square-free terms in that arithmetic progression, however this won’t produce a pair of genuinely consecutive square-free integers. Comment by Mark Lewko — August 24, 2009 @ 11:12 pm • Dear Mark, one way to proceed is to first find a pair of consecutive residues b, b+1 mod W^2 that are not divisible by any square less than W^2 (this is possible by the Chinese remainder theorem). The density of square-frees in b mod W and in b+1 mod W are both close to 1, so there will be a lot of consecutive square-frees in this pair of progressions. Comment by — August 28, 2009 @ 5:13 am 28. Let me make a naive observation (based on four data points). During this project we have studied algorithms for computing the following number-theoretic quantities. (1) $\pi(n) = \sum_{p\leq n} 1$ (2) $D_{1}(n) = \sum_{x\leq n} \tau (x)$ (3) $D_{k}(n) = \sum_{x\leq n} \tau_{k}(x)$ (4) $Q(n) = \sum_{x\leq n} \|\mu(x)\|$ In the following table, the first column indicates (ignoring log terms) how fast our best algorithm computes the above quantities, the second column indicates the best known error term on these quantities, and the third column indicates the conjectured error term on each of these quantities. (1) $n^{1/2}$ $n$ $n^{1/2}$ (2) $n^{1/3}$ $n^{.314}$ $n^{1/4}$ (3) $n^{1-1/k}$ $n^{(k-1)/(k+2)}$ $n^{(k-2)/2k}$ (4) $n^{1/3}$ $n^{1/2}$ $n^{.314}$ In some cases (such as (1)) we can compute a quantity such that the order of the computation time is smaller than that of the best known error term. However, in none of these cases can we compute a quantity where the exponent on the computation time is better than the conjectured error term. This leads to the following rather vague question. Is anyone aware of any number theoretic quantity whose computation time is smaller than the conjectured error term? Comment by Mark Lewko — August 31, 2009 @ 1:17 am • Related to the comment above, I’d be interested to know if anyone is aware of any result of the following form: Give a non-trivial $\Omega$-bound for computing some non-trivial number-theoretic quantity. Comment by Mark Lewko — August 31, 2009 @ 4:13 am • Well, there are artificial counterexamples… $\sum_{m \leq n} 2^m \mu(m)$, for instance, is computable in time poly(n) but has an error term which is exponential in n. But yes, in general there seems to be a strong correlation between a quantity being easy to compute on one hand, and being easy to estimate on the other (Raphy Coifman is fond of pushing this particular philosophy). Comment by — August 31, 2009 @ 9:57 pm 29. [...] the meantime, Terence Tao started a polymath blog here, where he initiated four discussion threads (1, 2, 3 and 4) on deterministic ways to find primes (I’m not quite sure how that’s [...] Pingback by — October 26, 2009 @ 10:43 pm 30. “Find efficient deterministic algorithms for finding various types of “pseudoprimes” – numbers which obey some of the properties of being prime…” The following define two deterministic prime generating algorithms which generate some curious composites, which are essentially a small (tiny?) subset of Hardy’s round numbers. (Note that the focus is on generating the prime gaps, and not the primes themselves.) TRIM NUMBERS A number is Trim if, and only if, all its divisors are less than the Trim difference $d_{n}$, where: (i) $t_{1} = 2$, and $t_{n+1} = t_{n} + d_{n}$; (ii) $d_{1} = 1$, and $a(2, 1) = 1$; (iii) $d_{n}$ is the smallest even integer that does not occur in the n’th sequence: $\{a(n, 1), \ldots, a(n, n-1)\}$; (iv) $j_{i} \geq 0$ is selected so that: $0 \leq a(n+1, i) = ((a(n, i) - d(n) + j(i)p_{i}) < p_{i}$, where $p_{i}$ is the i'th prime. It follows that $t_{n+1}$ is a prime unless all its prime divisors are less than $d_{n}$. Question: What is the time taken to generate $t_{n}$? COMPACT NUMBERS A number is Compact if, and only if, all its divisors, except a maximum of one, are less than the Compact difference $d_{n}$, where: (i) $c_{1} = 2$, and $c_{n+1} = c_{n} + d_{n}$; (ii) $d_{1} = 1$, and $a(2, 1) = 1$; (iii) $d_{n}$ is the smallest even integer that does not occur in the n’th sequence: $\{a(n, 1), \ldots, a(n, k)\}$; (iv) $j_{i} \geq 0$ is selected so that, for all $0 < i \leq k$: $0 \leq a(n+1, i) = (a(n, i) - d_{n} + j_{i}p_{i}) < p_{i}$; (v) $k$ is selected so that: $p_{k}^{2} < c(n) \leq p_{k+1}^{2}$; (vi) if $c_{n} = p_{k+1}^{2}$, then: $a(n, k+1) = 0$. It follows that $c_{n+1}$ is either a prime, or a prime square, unless all, except a maximum of 1, prime divisors of the number are less than $d_{n}$. What is the time taken to generate $c_{n}$? The distribution of the Compact numbers suggests that the prime difference may be $O(\pi(n)^{1/2})$. The algorithms can be seen in more detail at: http://alixcomsi.com/A_Minimal_Prime.pdf Comment by — September 6, 2010 @ 6:05 am 31. I wanted to know if there is a solution to a slightly different but simple question: Give a number $n$, how to find a prime larger than $n$ in time bounded by a polynomial in $\log n$? Comment by — November 18, 2010 @ 4:35 pm RSS feed for comments on this post. TrackBack URI Theme: Customized Rubric. Blog at WordPress.com. %d bloggers like this:	{"extraction_info": {"found_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": 120, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9432585835456848, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2007/07/01/the-strictification-theorem/?like=1&_wpnonce=d19428bcb3	# The Unapologetic Mathematician ## The “Strictification” Theorem I want to get in another post in the current line before I head off to Portugal on Tuesday. This theorem is another heavy one to prove, so if the proof of the Coherence Theorem was overwhelming you might just want to skim this one and remember the result: Any monoidal category $\mathcal{M}$ is monoidally equivalent to a strict monoidal category $\mathcal{S}$ — one whose objects form an honest monoid without any “up to isomorphism” dodging about. This will justify the usual laxity with which we treat associativity and unit laws in monoidal categories, writing $A\otimes B\otimes C\otimes\cdots$ without regard to parentheses and such. When we naively write down a monoidal product like this we’re treating it as a list of objects of $\mathcal{M}$, and we take the product of two lists just by concatenation. So let’s start by defining the collection of objects of $\mathcal{S}$ to be the free monoid on the objects of $\mathcal{M}$, which is exactly this collection of lists. We’ll write a typical list as $A=[A_1,A_2,...,A_m]$. Then we have the concatenation product $\left[A_1,...A_m\right]\cdot\left[B_1,...B_n\right]=\left[A_1,...A_m,B_1,...,B_n\right]$ and the empty list $\left[\right]$ for an identity object. All the monoid rules hold “on the nose” already, so we don’t need to worry about associators and all. We don’t have any arrows in our category yet, but let’s press on a bit first to define a map $F:\mathcal{S}\rightarrow \mathcal{M}$. If this has any hope of being the object function of a monoidal functor we’ll need $F([])=\mathbf{1}$. We also need to figure out how $F$ should deal with a monoidal product. What we’re going to do is just choose to make the result have all its parentheses on the right. We define $F([A_1,A_2,...])=A_1\otimes F([A_2,...])$ to get (for example) $F([A_1,A_2,A_3,A_4])=A_1\otimes(A_2\otimes(A_3\otimes A_4))$. Now we’ll go back and add arrows to $\mathcal{S}$ in just such a way as to make $F$ into a monoidal functor. Given two lists $A$ and $B$ in $\mathcal{S}$ I’ll define $\hom_\mathcal{S}(A,B)=\hom_\mathcal{M}(F(A),F(B))$. Notice that this makes the functor $F$ fully faithful by definition, and that every object $M$ in $\mathcal{M}$ is $F([M])$, so we’re well on our way to having an equivalence of categories! What’s missing now? Well, we still don’t know quite how to take the monoidal product of arrows in $\mathcal{S}$, so it’s still not quite a monoidal category yet. We can use the isomorphisms of hom-sets we just set up to do this. Given $f:A\rightarrow C$ and $g:B\rightarrow D$ in $\mathcal{S}$ their monoidal product $f\cdot g:A\cdot B\rightarrow C\cdot D$ is an arrow in $\hom_\mathcal{M}(F(A\cdot B),F(C\cdot B))$. We define it to be the composite: $F(A\cdot B)\rightarrow F(A)\otimes F(B)\rightarrow F(C)\otimes F(D)\rightarrow F(C\cdot D)$ Here the middle arrow is the monoidal product of $f$ and $g$ in $\mathcal{M}$, and the outer two arrows are the (unique!) ways to change parentheses as needed. The uniqueness follows from the Coherence Theorem, so we have a well-defined monoidal product on $\mathcal{S}$. So now $F$ is a monoidal functor. We take $F_$ to be the identity $F([])=\mathbf{1}$, and we let $F_{(A,B)}:F(A)\otimes F(B)\rightarrow F(A\cdot B)$ be the (unique!) arrow moving all the parentheses to the right. The Coherence Theorem gives us this uniqueness, and also handles the diagrams required for a monoidal functor. Now for the other direction we’ll just go ahead and define $G:\mathcal{M}\rightarrow\mathcal{S}$ by setting $G(A)=[A]$ and $G(f)=f$. We also define $G_:[]\rightarrow[\mathbf{1}]$ by noticing that $G_*\in\hom_\mathcal{M}(\mathbf{1},\mathbf{1}$, so we can just use the identity. We define $G_{(A,B)}:[A,B]\rightarrow[A\otimes B]$ similarly. I’ll leave the cleanup to you: $G$ does satisfy coherence conditions for a monoidal functor, $F\circ G:\mathcal{M}\rightarrow\mathcal{M}$ is the identity functor, and there is a monoidal natural isomorphism from $G\circ F$ to the identity functor on $\mathcal{S}$. It’s all pretty striaghtforward, though you’ll have to appeal to the Coherence Theorem a couple times like we have above. ### Like this: Posted by John Armstrong \| Category theory ## 3 Comments » 1. Thanks for this nice explanation. In a monoidal closed category, there’s the curry isomorphism between \$A \otimes B \multimap C\$ and \$B \multimap A \multimap C\$. Are all monoidal closed categories equivalent to strict ones where this isomorphism is the identity? Comment by \| September 23, 2008 \| Reply 2. You know, I don’t know offhand. I’d be inclined to say yes, but I’ve got a calculus class coming up in a few minutes. I’ll toss it around and see if I come up with anything. Comment by \| September 23, 2008 \| Reply 3. Mike, I just saw this comment. John Baez asked me this question a while back in email, and I sent back a reply; I don’t know whether you got it. I outlined an argument for why the answer should be ‘yes’. Comment by \| December 17, 2008 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for 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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 49, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9113948345184326, "perplexity_flag": "head"}
http://mathoverflow.net/questions/87688?sort=newest	## Linear Algebra - Find an N-Dimensional Vector Orthogonal to A Given Vector ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm writing an eigensolver and I'm trying to generate a guess for the next iteration in the solve that is orthogonal to to all known eigenvectors calculated thus far. This means that if I have only one eigenvector, that is say 2 million entries long, I need to generate a vector orthogonal to it. I don't think Gram Schmidt works here because I don't have a set of vectors to orthogonalize. What I have is a single vector, in the first eigensolve, and I need to generate another that is orthogonal. So, in summary: given one vector, create from nothing another vector which is orthogonal to it. The method must support N-Dimensional vectors (where N could be millions). I should add that writing a generalized cross product algorithm is not appealing. I'd prefer another way. - 1 Are there any other constraints? Otherwise, let $a$ be the given vector. If $a_1=0$, then $(1,0,0,\dots)$ works; if $a_2=0$, then $(0,1,0,0,\dots)$ works; otherwise, $(a_2,-a_1,0,0,\dots)$ works. – Klaus Draeger Feb 6 2012 at 17:21 Guess (perhaps by coin flipping) the first say 7/8 of the coordinates, and then use guessing or algebra to determine the rest. This method will rarely fail, and usually because some small set of coordinates are supposed to be zero instead of nonzero. Gerhard "Ask Me About System Design" Paseman, 2012.02.06 – Gerhard Paseman Feb 6 2012 at 17:21 Very useful comments, thank you. – wavepacket Feb 6 2012 at 17:41 ## 2 Answers Suppose the vector is $(x_1,x_2,\dots)$. The following algorithm should work: 1. If $x_1=0$ then take $(1,0,0,\dots)$. 2. If $x_1$ is non-zero and $x_2=0$ then take $(0,1,0,0,\dots)$. 3. If $x_1$ and $x_2$ are non-zero then take $(-x_2,x_1,0,0,\dots)$ - This seems to be a popular answer. I'll give it a shot, thanks. – wavepacket Feb 6 2012 at 17:41 Just for the record - 2 and 3 are in fact applications of the same rule :-) – Vladimir Dotsenko Feb 6 2012 at 18:56 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Actually you might use Gram Schmidt here. Given a set of ortogonal vectors $x_1,x_2,\ldots,x_k$ you can use Gram-Shmidt algorithm for set of vectors ${x_1,x_2,...,x_k,e_i}$ adding basis vector to system of ortogonalysed vectors (note that you need use Gram Schmidt procedure only to find last vector since first k vectors are already orthogonal). Then (since vectors $e_1,e_2,\ldots,e_{k+1}$ are linearly independent) for some i between 1 and k+1 Gram Schmidt will give you non-zero vector which is ortogonal to given vectors $x_1,x_2,\ldots,x_k$ So to find a guess you simply need to use Gram Schmidt procedure several times (no more than k+1 for the first guess and no more then two times for next guesses). To simplify this procedure you can do this only with first $k+1$ coordinates of vectors, so you will find a vector of form $(y_1,y_2,\ldots,y_{k+1},0,0,\ldots)$. Answers of Kapil and Klaus are actually equivalent to using this route. - Or just choose the next vector randomly. In particular, what's wrong with doing Gram-Schmidt to a set of randomly generated vectors? – Deane Yang Feb 6 2012 at 20:10 nothing wrong I guess, don't know if probabilistys algorithm qualifies as an answer to op, so I chose deterministic one. – Ostap Chervak Feb 7 2012 at 20:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9225122928619385, "perplexity_flag": "middle"}
http://[email protected]/bookstore?fn=20&arg1=stmlseries&ikey=STML-34	New Titles \| FAQ \| Keep Informed \| Review Cart \| Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education Return to List Number Theory in the Spirit of Ramanujan Bruce C. Berndt, University of Illinois, Urbana-Champaign, IL SEARCH THIS BOOK: Student Mathematical Library 2006; 187 pp; softcover Volume: 34 ISBN-10: 0-8218-4178-5 ISBN-13: 978-0-8218-4178-5 List Price: US\$36 Member Price: US\$28.80 Order Code: STML/34 See also: Higher Arithmetic: An Algorithmic Introduction to Number Theory - Harold M Edwards Ramanujan is recognized as one of the great number theorists of the twentieth century. Here now is the first book to provide an introduction to his work in number theory. Most of Ramanujan's work in number theory arose out of $$q$$-series and theta functions. This book provides an introduction to these two important subjects and to some of the topics in number theory that are inextricably intertwined with them, including the theory of partitions, sums of squares and triangular numbers, and the Ramanujan tau function. The majority of the results discussed here are originally due to Ramanujan or were rediscovered by him. Ramanujan did not leave us proofs of the thousands of theorems he recorded in his notebooks, and so it cannot be claimed that many of the proofs given in this book are those found by Ramanujan. However, they are all in the spirit of his mathematics. The subjects examined in this book have a rich history dating back to Euler and Jacobi, and they continue to be focal points of contemporary mathematical research. Therefore, at the end of each of the seven chapters, Berndt discusses the results established in the chapter and places them in both historical and contemporary contexts. The book is suitable for advanced undergraduates and beginning graduate students interested in number theory. Readership Undergraduate and graduate students interested in number theory, including $$q$$-series and theta functions. Reviews "... undergraduates will find no better place to meet the mind behind the towering reputation." -- D. V. Feldman, University of New Hampshire for CHOICE Reviews "This is a delightful little book on selected topics in number theory. ...I highly recommend this book to all mathematicians. It is a great resource both to learn from and to teach from. Even the experts will enjoy his new perspective on these old questions." -- Journal of Approximation Theory "This slender volume is extremely well-written and contains a wealth of material. It is a lucid and accessible introduction to a rich and fascinating area of mathematics, written by the world's leading expert. For anyone with a knowledge of calculus wanting to learn about the mathematical work of Ramanujan, this book is the best place to start." -- Shaun Cooper, Massey University - New Zealand Newsletter "..."Number theory in the spirit of Ramanujan" is a gem that deserves a place in every mathematician's library." -- Zentralblatt MATH AMS Home \| Comments: [email protected] © Copyright 2012, American Mathematical Society Privacy Statement	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9084171652793884, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-equations/181881-not-getting-right-answer-reduction-order-problem.html	Thread: 1. Not getting the right answer in Reduction of Order Problem $x''=\frac{t+1}{t}x'-\frac{1}{t}x$ with $x_1(t)=e^t$ The goal is to find the other solution. my attempt $x_2=ux_1$ $x_2'=e^{t}u'+e^{t}u$ $x_2''=e^tu''+2e^tu'+e^tu$ Substituting into the original equation I get... $u''=-\frac{t-1}{t}u'$ substituting with u'' with v'... $v'=-\frac{t-1}{t}v$ I integrate to $ln v=ln t-t+c1$ $v=t\cdot e^{-t}=u'$ integrating again... $u=(-t-1)e^{-t}$ $x_2=(-t-1)e^{-t}\cdot e^t$ $x_2=-t-1$ but the answers supposed to be $t-1$ where did I go wrong? thanks 2. Why do you think you're wrong? If you substitute your answer into the original ODE you get $0 = \dfrac{t+1}{t} (-1) - \dfrac{1}{t} \left(-t-1\right)$ which I believe identically satisfies your ODE! 3. Because of this... http://www.mathhelpforum.com/math-he...wn-181831.html So there are more than two solutions? 4. If you look, there's a slight typo (on the negative sign). 5. Who has the typo? On the maple I only see that he moved the right part of the equation to the left side which seems like it's allowed... 6. Originally Posted by Carbon Who has the typo? On the maple I only see that he moved the right part of the equation to the left side which seems like it's allowed... In my finial line in the other thread I wrote $e^{t}e^{-t}(t+1)=t-1$ It should be $t+1$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9510935544967651, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/38004/slight-extension-to-classification-of-finitely-generated-abelian-groups/38013	## slight extension to classification of finitely generated abelian groups ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is there a classification theorem for "finitely generated abelian groups with a distinguished element"? If it helps, you can restrict this to the cases where the order of each element divides the order of the distinguished element. My idea would be to try to use this for a classification theorem for rings with a finitely generated additive group, where the distinguished element is 1. - Well, it would in particular be a finitely generated abelian group, so the usual structure theorem would apply; the distinguished element would, presumably, show up in the definition of homomorphisms (which would be required to respect this zero-ary operation) and of substructure (same reason, you require the substructure to be closed under the zero-ary operation). But you would still have the decomposition into direct sums of cyclic groups... just that those cyclic summands need not be "subgroups-with-distinguished-element". – Arturo Magidin Sep 7 2010 at 19:59 well I think those cyclic summands also have distinguished elements. For example $(\mathbb{Z}^2,(1,2))$ is the direct sum of $\mathbb{Z},1$ and $(\mathbb{Z},2)$. Then every such finitely generated group can be expressed uniquely as a direct sum of summands of the form $(\mathbb{Z},n)$ with $n\ge 0$ and $(\mathbb{Z}/p^n,p^k)$ with $p$ prime and $k\le n$. – HenrikRüping Sep 7 2010 at 20:15 @Henrik: but then the direct summands are not subgroups-with-distinguished-element. While you can map $(\mathbb{Z},1)$ to $(\mathbb{Z}^2,(1,2))$ by sending $1$ to $(1,2)$ (and must map it so), you cannot map $(\mathbb{Z},2)$ to $(\mathbb{Z},(1,2))$: you would have to map the distinguished element $2$ to $(1,2)$, and that means that the image of $1$ would have to satisfy $f(1)+f(1) = (1,2)$ in $\mathbb{Z}^2$. No such element exists, so the second direct summand you have is not isomorphic to any subgroup-with-distinguished-element of $(\mathbb{Z}^2,(2,1))$. This seems like a likely headache. – Arturo Magidin Sep 7 2010 at 20:28 I see thanks a lot. – HenrikRüping Sep 7 2010 at 23:02 ## 4 Answers You say that we may assume that the order of every element divides the order of the distinguished element. That's the easy case. If the distinguished element generates a finite cyclic group $C$ of order $n$ and every element has order dividing $n$, then in fact $C$ is a direct summand. ("For the ring $\mathbb Z/n$ the free module of rank one is injective.") - Oh, I see, you are thinking about rings. But the $1$ will be the least of your worries, even if you stick to commutative rings. Even when the order of $1$ is a prime $p$ you will be seeking to classify in particular all $0$-dimensional local rings over the field with $p$ elements, which sounds hard over any field. – Tom Goodwillie Sep 7 2010 at 21:59 Every possible order divides infinity. If the distinguished element has infinite order, is it necessarily a member of a cyclic direct summand? I googled "the free module of rank one is injective.", and it didn't find it (not even on this page). Where would I find a proof that C is a direct summand? – Ricky Demer Sep 7 2010 at 23:30 1 An $R$-module $I$ is injective if for modules $M\subset N$ the restriction map $Hom(N,I)\to Hom(M,I)$ is always surjective). This is true for all pairs $(N,M)$ if it is true in the case when $N=R$. In the case when $R=\mathbb Z/n$ ($n\ge 1$) it is easy to see that this condition holds for $I=R$. This observation is the basis for the usual proof(s) of the classification of finite abelian groups, isn't it? Choose a cyclic subgroup of maximal size ad split it off. Induct. – Tom Goodwillie Sep 8 2010 at 0:34 And if $1$ has infinite order in a ring then let $n$ be the largest integer such that there exists $x$ in the ring with $nx=1$. If $n\ge 2$ then $n^2x^2=1$ gives a contradiction. – Tom Goodwillie Sep 8 2010 at 0:39 That skips all the work, which is showing that there is a largest such integer. However, what it gives is only that if 1 is in a direct summand, then 1 generates the direct summand. Does 1 always generate a direct summand? – Ricky Demer Sep 8 2010 at 2:05 show 5 more comments ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This is only a partial answer assuming the "if this help" (interpreted in a certain way, hoping it is the intended one). Let G be a finite abelian group and g in G such that the order of each element of G divides the order of g. (I assume this means that the order of g is finite.) There exist n_1,...,n_r such that n_i divides n_(i+1) and and isomorphism f from G to C_(n_1) + ... + C_(n_r) such that f(g) = (0,...,0,1). If the element does not have maximal order the situation is certainly more complex. - Sorry, the first answer was not there when I started mine, thus the duplication. Feel free to delete it. – quid Sep 7 2010 at 22:06 I don't think I will, but ... How would I delete an answer? – Ricky Demer Sep 8 2010 at 5:04 I have some general comments. Summary: such a general theory would be much harder than that for abelian groups (and in fact, would contain the classification of finitely generated abelian groups as a special case), and you would lose a lot of the "good" properties of the category of abelian groups along the way. Many constructions (such as the coproduct) would be different from what we are used to. The advantage is that you can fit into the general context of Universal Algebra, so all of the familiar "abstract nonsense" theory will apply (suitably interpreted). On the other hand, if you restrict to the case you are actually interested in, in which the order of each element divides the order of the distinguished element, then you end up with a much simpler theory, but it falls outside the usual confines of Universal Algebra. You can write down an easy description, but it's not going to be well-behaved relative to homomorphisms. Some details. To consider the general theory from the point of view of Universal Algebra, you are looking at algebras of type (2,1,0,0), where the binary operation is denoted $+$ (with the usual infix notation), the unary operation is "$-$", one of the zero-ary operations is denoted "$0$", and the second unary operation is denoted "$1$". The operations are required to satisfy the obvious identities: $(a+b)+c = a+(b+c)$, $a+b=b+a$, $a+0=a$, $a+(-a)=0$, and no identity is required of the second unary operation $1$. Homomorphisms are required to respect addition, inversion, map $0$ to $0$, and map $1$ to $1$. The class of abelian groups (as usual) is a variety of algebras of this type, characterized by the identity $1=0$. So you have an embedding from abelian groups to "pointed abelian groups", and this embedding has left adjoint which corresponds to taking the quotient modulo the subgroup generated by $1$. You also have a "forgetful" functor from "pointed abelian groups" to abelian groups, obtained by forgetting about the pointed element; This class, however, is no longer an abelian category. I'm pretty sure of the following, but I could have made some mistakes: The product of two pointed abelian groups $(A,1)$ and $(B,1')$ is, as Henry suggested in the comments, the abelian group $A\times B$ with distinguished element $(1,1')$, and the structure projections are the canonical projections. The free pointed abelian group on a set $X$ will be the direct sum of $\|X\|+1$ copies of $\mathbb{Z}$, with the generator of the extra copy being the distinguished element. The coproduct of $(A,1)$ and $(B,1')$ will be the quotient of $A\oplus B\oplus\mathbb{Z}$ modulo the subgroup generated by $(1,0,1)$ and $(0,1',1)$, and the distinguished element is the image of $(0,0,1)$. The canonical immersions go are the canonical maps into the direct sum followed by the quotient map. This category includes the entire theory of abelian groups as the class of those groups in which $0=1$; so any classification theory will be much harder than the one for abelian groups. For example, the $1$ generator pointed abelian groups: every $1$-generator pointed abelian group is really a $2$-generator abelian group: it is generated by the given generator plus the distinguished element $1$. So suppose $x$ is the given generator. If $\langle 1\rangle\subseteq\langle x\rangle$, then you get an isomorphism type of the form $(C,C')$, where $C$ is a cyclic group and $C'$ is a subgroup of $C$ ($C$ corresponds to the cyclic group generated by $x$, and $C'$ to the subgroup generated by $1$). But these do not exhaust all possibilities. You also get groups in which $\langle x\rangle\cap\langle 1\rangle = \{0\}$; if the orders are coprime, you are back to the previous case, but if the orders are not coprime then the description is a bit more complicated. And then the case where they intersect nontrivially but neither contains the other, and so on. As you can see, things get complicated even in this simplest of cases... Two pointed abelian groups that are isomorphic as pointed abelian groups will be isomorphic as abelian groups, but you can have to pointed abelian groups whose forgetful image is isomorphic, but that are not isomorphic as pointed abelian groups; for example, $(\mathbb{Z},1)$ and $(\mathbb{Z},2)$ are not isomorphic as pointed abelian groups. The advantage is that you have the entire weaponry from Universal Algebra at your disposal, as well as two reasonably well-behaved functors to abelian groups: the forgetful functor, and the adjoint of the inclusion of the variety of abelian groups to the variety of pointed abelian groups. The disadvantage is that it is unfamiliar, and more complicated than the usual theory of abelian groups. (Think about how things get more difficult going form abelian groups to not-necessarily-abelian groups). Different take: For the application you want, however, you don't need such a general theory. In your application, the order of every element divides the order of the distinguished element. The disadvantage is that this cannot be expressed as an identity, so you do not have a variety of algebras. You do not have the nice constructions such as coproducts, products, free objects, etc. a priori. On the other hand, I think that the classification is pretty straightforward: every such group can be expressed as a product of cyclic groups, $C_{n_1}\times\cdots\times C_{n_k}$, with $n_1\|n_2\|\cdots\|n_k$, and with distinguished element a specific generator of the largest cyclic factor in the case where $n_k>0$, and an arbitrary nonzero element of the last cyclic factor if $n_k=0$ and a torsion element plus a nonzero element of the last cyclic factor if $n_k=0$. However, this decomposition is not decomposing the objects into something which is a direct product of pointed abelian groups (the direct summands other than the largest ones do not contain the distinguished element). The proof in the finite case can be done the usual way; for an abelian $p$-group $A$, if $x$ is an element of maximal order then $A$ is isomorphic to $A/\langle x\rangle \oplus \langle x\rangle$, so you can start by picking the distinguished element; otherwise, consider the $p$-parts and think of them as quotients of the original pointed group, with distinguished element the image of the distinguished element; then put them together in the usual way, and the distinguished element will correspond to a generator of the largest cyclic factor. In the infinite case, it gets a bit more complicated, but it should go through: take $A/A^{tor}$, with distinguished element the image of the distinguished element of $A$; this is a direct sum of cyclic groups of infinite order as an abelian group, and the subgroup generated by the distinguished element is a submodule of rank $1$. So if I remember my modules over PIDs correctly, there is a basis for $A/A^{tor}$ of the form $x_1,\ldots,x_r$, in which $1$ is a multiple of $x_r$; then deal with the torsion part as you would with a normal finite abelian group. Edit: No, the infinite case does not go through like this; see comment. You can get it so that it will be of the form $(a,(0,\ldots,0,d))$, where $a$ is in the torsion part and $(0,\ldots,0,d)$ is an element of the torsion free part with $d>0$, but you may not be able to "get rid" of the $a$. This is problematical. The advantage is that there's the description of all such groups. There are some obvious properties, such as: you can only map from $(A,1)$ to $(B,1)$ if the order of the distinguished element of $A$ is a multiple of the order of the distinguished element of $B$; the map must send the distinguished generator of the largest cyclic factor of the expression for $A$ to the distinguished generator of the largest cyclic factor of $B$. Among the disadvantages is that it is not a "decomposition" in the sense of expressing the pointed group in terms of "smaller" pointed groups, and so it may not lead to any sensible "simplification", just a "description", which is probably not what you want. A more sensible simplification would require you to identify all "product irreducible" pointed abelian groups, and this might be pretty difficult. - 3 has infinite order in Z, which is certainly maximal, but Z/<3> + <3> has an element of order 3 and Z doesn't, so Z is not isomorphic to Z/<3> + <3>. Could you give or point me to a proof of what you meant to say? Also, a "description" is what I want, since I'm just wondering what elements I would have to worry about being 1 in an invariant factor decomposition. – Ricky Demer Sep 8 2010 at 0:13 Oh, sorry; I'm misreporting/misremembering a theorem from Lang. It holds for finitely generated abelian $p$-groups, not for finitely groups. I'll fix it. – Arturo Magidin Sep 8 2010 at 0:28 ...and it seems that the last part of the penultimate paragraph, when your distinguished element is of infinite order, may not be quite right either. As I note in another comment, I don't see how to arrange it if we had $A=C_2\times C_0$ with distinguished element $(1,2)$ (where $C_2$ is cyclic group of order $2$, $C_0$ the infinite cyclic group). The quotient is cyclic of order $4$, so we cannot write it as $C_2\times C_0$ with distinguished element contained in the summand $C_0$. – Arturo Magidin Sep 8 2010 at 4:54 As far as I understand, the classification problem for finite abelian groups with a fixed element is not such a 'slight' extension. (Without any simplifying assumptions, such as divisibility of order.) It is probably an overkill for your situation, but it actually already came up on mathoverflow at least once: see the answer and comments to this question, so I thought I'd mention it: The answer and comments in the link lead to the following conjecture (which hopefully should not be too hard). First, we may assume without losing generality that G is an abelian p-group. Its isomorphism type is given by a partition. Conjecture. If g∈G is a non-zero element, the isomorphism type of (g,G) is given by a bipartition. Justification. There is an analogy between the theory of finite abelian groups and the theory of Jordan form (both classify modules over PIDs). Under this analogy, p-groups correspond to vector spaces equipped with a nilpotent operator. (And indeed, both sides of the analogy are classified by partitions.) The problem at hand is thus analogous to classifying finite dimensional vector spaces equipped with a nilpotent operator and a non-zero vector. Such triples are known to be classified by bipartitions: see Achar, Henderson: Orbit closures in an enhanced nilpotent cone. - Yeah, when I asked this I thought it would be much simpler than it turned out to be. However, how do we "assume without losing generality that G is an abelian p-group"? The unsolved case is when the distinguished element has infinite order. – Ricky Demer Sep 8 2010 at 19:01 That's why I said 'finite abelian groups'. However, the case of an element of infinite order easily reduces to the case of finite group (unless I am hallucinating). Indeed, suppose our group is $G\times T$, where G is free and T is finite. Suppose the image of $g\in G\times T$ in G is divisible by k (but not by anything larger). Then it is clear that classification problem is the same as classifying the image of g in $T/(k\cdot T)$, which is a finite group – t3suji Sep 8 2010 at 20:56 It is not clear to me how it is enough to classify the image of g in T/(k*T). – Ricky Demer Sep 9 2010 at 0:42 Suppose you have a group H that is isomorphic to $G\times T$. Let us fix such an isomorphism $\phi$, then an element $h\in H$ would correspond to a pair $(g,t)\in G\times T$. Now, what is $(g,t)$ defined up to? If we change $\phi$ by an automorphism of $G\times T$, the following things happen: $g$ gets acted upon by an automorphism of $G$, $t$ is acted upon by an automorphism of $T$, and an element of $k\cdot T$ is added to $t$ (where $k$ is the max. divisibility of $g$). All this follows because we know what automorphisms of $G\times T$ look like. Any such change of the pair $(g,t)$ occurs. – t3suji Sep 9 2010 at 2:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 149, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9419305920600891, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2009/10/07/the-chain-rule-2/?like=1&_wpnonce=84af3d0b4c	# The Unapologetic Mathematician ## The Chain Rule Since the components of the differential are given by partial derivatives, and partial derivatives (like all single-variable derivatives) are linear, it’s straightforward to see that the differential operator is linear as well. That is, if $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ and $g:\mathbb{R}^m\rightarrow\mathbb{R}^n$ are two functions, both of which are differentiable at a point $x$, and $a$ and $b$ are real constants, then the linear combination $af+bg$ is also differentiable at $x$, and the differential is given by $\displaystyle d(af+bg)=adf+bdg$ There’s not usually a product for function values in $\mathbb{R}^n$, so there’s not usually any analogue of the product rule and definitely none of the quotient rule, so we can ignore those for now. But we do have a higher-dimensional analogue for the chain rule. If we have a function $g:X\rightarrow\mathbb{R}^n$ defined on some open region $X\subseteq\mathbb{R}^m$ and another function $f:Y\rightarrow\mathbb{R}^p$ defined on a region $Y\subseteq\mathbb{R}^n$ that contains the image $g(X)$, then we can compose them to get a single function $f\circ g:X\rightarrow\mathbb{R}^p$ defined by $\left[f\circ g\right](x)=f(g(x))$. And if $g$ is differentiable at a point $x$ and $f$ is differentiable at the image point $g(x)$, then the composite function is differentiable at $x$. First of all, what should the differential be? Remember that the differential $dg(x)$ is a linear transformation that takes displacements $t\in\mathbb{R}^m$ from the point $x$ and turns them into displacements $dg(x)t\in\mathbb{R}^n$ from the point $g(x)$. Then the differential $df(y)$ is a linear transformation that takes displacements $s\in\mathbb{R}^n$ from the point $y$ and turns them into displacements $df(y)s\in\mathbb{R}^p$ from the point $f(y)$. Putting these together, we have a composite linear transformation $df(g(x))dg(x)$ that will start with a linear transformation that takes displacements $t\in\mathbb{R}^m$ from the point $x$ and turns them into displacements $df(g(x))dg(x)t$ from the point $f(g(x))$. I assert that this is composite transformation is exactly the differential of the composite function. Just as a sanity check, what happens when we look at single-variable real-valued functions? In this case, $df(y)$ and $dg(x)$ are both linear transformations from one-dimensional spaces to other one-dimensional spaces. That is, they’re represented as $1\times1$ matrices that just multiply by the single real entry. So the composite of the two transformations is given by the matrix whose single entry is the product of the two matrices’ single entries. In other words, in one variable the differentials looks like single real numbers $f'(y)$ and $g'(x)$, and their composite is given by multiplication: $f'(g(x))g'(x)$. This is exactly the one-variable chain rule. To understand multiple variables we have to move from products of real numbers to compositions of linear transformations, which will be products of real matrices. Okay, so let’s verify that $d\left[f\circ g\right](x)=df(g(x))dg(x)$ does indeed act as a differential for $f\circ g$. It’s clearly a linear transformation between the appropriate two spaces of displacements. What we need to verify is that it gives a good approximation. That is, for every $\epsilon>0$ there is a $\delta>0$ so that if $\delta>\lVert t\rVert>0$ we have $\displaystyle\left\lVert\left[f(g(x+t))-f(g(x))\right]-df(g(x))dg(x)t\right\rVert<\epsilon\lVert t\rVert$ First of all, since $f$ is differentiable at $g(x)$, given $\tilde{\epsilon}>0$ there is a $\tilde{\delta}$ so that if $\tilde{\delta}>\lVert s\rVert>0$ we have $\displaystyle\left\lVert\left[f(g(x)+s)-f(g(x))\right]-df(g(x))s\right\rVert<\tilde{\epsilon}\lVert s\rVert$ Now since $g$ is differentiable it satisfies a Lipschitz condition. We showed that this works for real-valued functions, but extending the result is very straightforward. That is, there is some radius $r_1$ and a constant $M>0$ so that if $r_1>t>0$ we have the inequality $\lVert g(x+t)-g(x)\rVert<M\lVert t\rVert$. That is, $g$ cannot stretch displacements by more than a factor of $M$ as long as the displacements are small enough. Now $r_1$ may be smaller than $\frac{\tilde{\delta}}{M}$ already, but just in case let’s shrink it until it is. Then we know that $\displaystyle\lVert g(x+t)-g(x)\rVert<M\lVert t\rVert<Mr_1<M\frac{\tilde{\delta}}{M}=\tilde{\delta}$. so we can use this difference as a displacement $s$ from $g(x)$. We find $\displaystyle\begin{aligned}\left\lVert\left[f(g(x+t))-f(g(x))\right]-df(g(x))\left(g(x+t)-g(x)\right)\right\rVert&<\tilde{\epsilon}\lVert g(x+t)-g(x)\rVert\\&<\tilde{\epsilon}M\lVert t\rVert\end{aligned}$ Now we’re going to find a constant $N\geq0$ and a radius $r_2$ so that $\displaystyle\left\lVert df(g(x))\left(g(x+t)-g(x)\right)-df(g(x))dg(x)t\right\rVert\leq\tilde{\epsilon}nN\lVert t\rVert$ whenever $r_2>\lVert t\rVert>0$. Once this is established, we are done. Given an $\epsilon>0$ we can set $\tilde{\epsilon}=\frac{\epsilon}{M+nN}$ and let $\delta$ be the smaller of the two resulting radii $r_1$ and $r_2$. Within this smaller radius, the desired inequality will hold. To get this result, we choose orthonormal coordinates on the space $\mathbb{R}^n$. We can then use these coordinates to write $\displaystyle df(g(x))\left(\left[g(x+t)-g(x)\right]-dg(x)t\right)=\left[D_if\right](g(x))\left(\left[g^i(x+t)-g^i(x)\right]-dg^i(x)t\right)$ But since each of the several $g^i$ is differentiable we can pick our radius $r_2$ so that all of the inequalities $\displaystyle\left\lvert\left[g^i(x+t)-g^i(x)\right]-dg^i(x)t\right\rvert<\tilde{\epsilon}\lVert t\rVert$ hold for $r_2>\lVert t\rVert>0$. Then we let $N$ be the magnitude of the largest of the component partial derivatives $\left\lVert\left[D_if\right](g(x))\right\rVert$, and we’re done. Thus when $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then the composite $f\circ g$ is differentiable at $x$, and the differential of the composite function is given by $\displaystyle d\left[f\circ g\right](x)=df(g(x))dg(x)$ the composite of the differentials, considered as linear transformations. ### Like this: Posted by John Armstrong \| Analysis, Calculus ## 20 Comments » 1. Another crisp, fine exposition. My wife is still annoyed by a time over 20 years ago when we went to a Pasadena restaurant with a younger alumnus friend of mine, whom I’d guided into Mathematical Music Theory (where he got his double degree at UC Santa Cruz and published in IEEE and other venues later), and played in a rock band together. he had a girlfriend with him who, once she found that my wife was a professor, kept interrupting to demand an explanation of the Chain Rule. My wife has taught that, but this was a social event, not a colloquium. Where was I? Oh, right, nicely done! I still think that you should find a coauthor to make this into a textbook. Comment by \| October 7, 2009 \| Reply • The textbook idea has merit, but wouldn’t it have to be rather large? Also the blog format has the advantage that people can ask questions and get answers. Btw John: has anyone come thru with the syntactic reasoners article. If not, send me your email addie and I can provide. Comment by Avery Andrews \| October 8, 2009 \| Reply • Charles sent along a copy, thanks. As for the book, I’m more inclined towards a lay-audience book, at least at first. I’ve thought about the subject, but again I need to find someone with a bit more of a background in the history and philosophy of mathematics to help keep me honest. I’ll be more comfortable searching for someone once I get more settled in terms of an actual job. Unemployment isn’t conducive to the peace of mind I’d need to get that ball rolling. Comment by \| October 8, 2009 \| Reply • Will your lay-audience book include the category theory you’ve discussed here? I think that could be one of the distinguishing features– there exist plenty of calculus and linear algebra books, but not many lay-audience books talking about category theory. Actually, a nowhere-dense introduction to category theory by itself would, I think, make the book worthwhile. Comment by \| October 10, 2009 \| Reply • Probably not. As I’ve mentioned before, the one I’m thinking of seeks to answer the more philosophical question, “What Is A Number?” Comment by \| October 10, 2009 \| Reply • Well but it seems to me that categorifying the integers is part of the answer to that. & if a book is ‘nowhere dense’, it is likely to be so heavy that just looking at it will make your wrists hurt Comment by Avery Andrews \| October 11, 2009 \| Reply • I mean more like an expansion on the standard content in the first couple days of an advanced calculus class, from Peano axioms through the continuum. Comment by \| October 11, 2009 \| Reply • I sometime fantasize about math ebooks that would initially give you a fairly crisp statement of what is true and why, but would then expand to more discursive text when you clicked on a ‘please explain’ icon. Perhaps also some kind of rating/reward system to encourage people to try to figure out as much as possible for themselves. Comment by Avery Andrews \| October 11, 2009 \| Reply 2. A while ago you were nice enough to give props out to my podcast Combinations and Permutations and I am now back(I have been back a few times to read of course but this is my first comment) to let you know that I have a 2nd podcast where I am interviewing mathematicians. The first episode where I interview Gary Chartrand is now up at Strongly Connected Components. Of course I would like you to listen but much more I would love to now have you as a guest on my new podcast. If this is interesting to you please email me at [email protected] Comment by \| October 8, 2009 \| Reply 3. [...] Invariant Rule An immediate corollary of the chain rule is another piece of “syntactic [...] Pingback by \| October 8, 2009 \| Reply 4. Re: philosophical question, “What Is A Number?” Excerpt from GENE515 (by Jonathan Vos Post) This partly answers a question asked of me when I was 21 years old, by my doctoral thesis advisor Oliver G. Selfridge [Father of Machine Perception]. When I say “Man” I am echoing and older text, and not excluding Woman. Excerpt from GENE515 What is Man, that he may know Number? What is Number that it may be known by Man? As we are mathematicians, we are in the image of our creator, The Mathematician, who has other attributes beyond our comprehension, and is Transfinite. He freely gives us this world, and the cosmos beyond, and the flora and fauna over which to be stewards, and our fellow human beings to love, which is in the image of His love, which is transfinite. We have free will, and for those of use who choose to be mathematicians, he gives us the integers as toys, in which is His book coded. We play with those toys, some of us in solitude, some of us playing together. And when we put aside childish things, behold, we still have the gift of Number, and they are more than first we knew. Eureka!, and Aha!, and knowing what Mozart meant when he said that he did not write music, but it was already there and he plucked it from thin air as it blew past. And what Ramanujan said was given him by a Goddess, And what Gauss could see as a child, and Riemann in the looking glass of Primes, and Galois by candlelight in the brief hours before his fatal duel. Euclid, alone, has looked on beauty bare. But we mathematicians today are not alone, far from it, cradled in the same Web woven of Number, binary and octal and hex, decimal and alphanumeric, vector and raster, and more in cables, trunks, and as wifi in the By knowing Number more deeply, we more deeply know ourselves, and our Creator. Every word begins and ends with the empty word; the empty word begins and ends with itself. Comment by \| October 10, 2009 \| Reply 5. [...] Differential Operators Because of the chain rule and Cauchy’s invariant rule, we know that we can transform differentials along with [...] Pingback by \| October 12, 2009 \| Reply 6. [...] with the function . This function is clearly differentiable, with constant derivative . And so the chain rule tells us [...] Pingback by \| October 20, 2009 \| Reply 7. [...] like we said when discussing the chain rule, the differential at the point defines a linear transformation from the -dimensional space of [...] Pingback by \| November 11, 2009 \| Reply 8. [...] differentiable functions on two open regions and in , with , and let be their composite. Then the chain rule tells us [...] Pingback by \| November 12, 2009 \| Reply 9. [...] derivative is identically zero as well. But since the are composite functions we can also use the chain rule to evaluate these partial derivatives. We [...] Pingback by \| November 27, 2009 \| Reply 10. [...] Then we can use the chain rule: [...] Pingback by \| January 14, 2010 \| Reply 11. [...] the multivariable chain rule. We [...] Pingback by \| March 31, 2011 \| Reply 12. [...] in the ball the whole segment for is contained within the ball. We define a function and use the chain rule to [...] Pingback by \| May 4, 2011 \| Reply 13. [...] that , , and . The chain rule lets us then [...] Pingback by \| June 27, 2011 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for 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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 91, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9497509598731995, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/28210/formal-logic-proof-of-absolute-value-formula/28215	# Formal logic proof of absolute value formula I've been trying to prove this for a long time now, anyone willing to offer some help or get me pointed in the right direction? $(x>0 \implies z = x) \wedge (x < 0 \implies z = -x) \implies z \ge 0$ - Re-tagged as "logic" instead of "higher-order-logic", as this is first-order logic. – Alex Becker Mar 21 '11 at 0:13 ## 2 Answers Hint: What happens if x = 0? Does that say anything about z? - As Robert's answer indicates, the statement is false. However, changing the first $>$ to $\geq$ gives a true statement which I would use a proof by contradiction, that is assume $z < 0$ and $(x \geq 0 \implies z = x) \wedge (x < 0 \implies z = -x)$ and derive a contradiction. This can be done as follows: $(z < 0) \wedge ((x>0 \implies z = x) \wedge (x < 0 \implies z = -x))$ $\implies (((x \geq 0 \implies z = x) \wedge z < 0) \wedge ((x < 0 \implies z = -x) \wedge z < 0))$ $\implies ((x \geq 0 \implies x < 0) \wedge (x < 0 \implies -x < 0))$ $\implies ((F \wedge (x < 0 \implies -x < 0)) \vee ((x \geq 0 \implies x < 0) \wedge F))$ $\implies (F \vee F)$ $\implies F$ $\therefore \neg((z < 0) \wedge ((x \geq 0 \implies z = x) \wedge (x < 0 \implies z = -x)))$ $\therefore (x \geq 0 \implies z = x) \wedge (x < 0 \implies z = -x) \implies z \ge 0$ - As @Robert's answer indicates, the statement is actually false, since $x$ need not be either $<0$ or $>0$; it could be $0$ (if it's implicitly assumed to be a real number; else it could also be $2+3\mathrm{i}$ or something else entirely). It follows that this proof must be flawed. The error is in replacing $(x>0\implies x<0)$ by $F$, since this implication is actually true if $x>0$ is false. – joriki Mar 24 '11 at 6:13 @Joriki: Thanks, I missed that. Fixed. – Alex Becker Mar 24 '11 at 6:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9447028040885925, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/1028/questions-about-kl-divergence?answertab=votes	Questions about KL divergence? I am comparing two distributions with KL divergence which returns me a non-standardized number that, according to what I read about this measure, is the amount of information that is required to transform one hypothesis into the other. I have two questions: a) Is there a way to quantify a KL divergence so that it has a more meaningful interpretation, e.g. like an effect size or a R^2? Any form of standardization? b) In R, when using KLdiv (flexmix package) one can set the 'esp' value (standard esp=1e-4) that sets all points smaller than esp to some standard in order to provide numerical stability. I have been playing with different esp values and, for my data set, I am getting an increasingly larger KL divergence the smaller a number I pick. What is going on? I would expect that the smaller the esp, the more reliable the results should be since they let more 'real values' become part of the statistic. No? I have to change the esp since it otherwise does not calculate the statistic but simply shows up as NA in the result table... Thanks in advance, Ampleforth - 3 Answers The KL(p,q) divergence between distributions p(.) and q(.) has an intuitive information theoretic interpretation which you may find useful. Suppose we observe data x generated by some probability distribution p(.). A lower bound on the average codelength in bits required to state the data generated by p(.) is given by the entropy of p(.). Now, since we don't know p(.) we choose another distribution, say, q(.) to encode (or describe, state) the data. The average codelength of data generated by p(.) and encoded using q(.) will necessarily be longer than if the true distribution p(.) was used for the coding. The KL divergence tells us about the inefficiencies of this alternative code. In other words, the KL divergence between p(.) and q(.) is the average number of extra bits required to encode data generated by p(.) using coding distribution q(.). The KL divergence is non-negative and equal to zero iff the actual data generating distribution is used to encode the data. - KL has a deep meaning when you visualize a set of dentities as a manifold within the fisher metric tensor, it gives the geodesic distance between two "close" distributions. Formally: $ds^2=2KL(p(x, \theta ),p(x,\theta + d \theta))$ The following lines are here to explain with details what is meant by this las mathematical formulae. Definition of the Fisher metric. Consider a parametrized family of probability distributions $D=(f(x, \theta ))$ (given by densities in $R^n$), where $x$ is a random variable and theta is a parameter in $R^p$. You may all knnow that the fisher information matrix $F=(F_{ij})$ is $F_{ij}=E[d(\log f(x,\theta))/d \theta_i d(\log f(x,\theta))/d \theta_j]$ With this notation $D$ is a riemannian manifold and $F(\theta)$ is a Riemannian metric tensor. (The interest of this metric is given by cramer Rao lower bound theorem) You may say ... OK mathematical abstraction but where is KL ? It is not mathematical abstraction, if $p=1$ you can really imagine your parametrized density as a curve (instead of a subset of a space of infinite dimension) and $F_{11}$ is connected to the curvature of that curve... (see the seminal paper of Bradley Efron http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aos/1176343282) The geometric answer to part of point a/ in your question : the squared distance $ds^2$ between two (close) distributions $p(x,\theta)$ and $p(x,\theta+d \theta)$ on the manifold (think of geodesic distance on earth of two points that are close, it is related to the curvature of the earth) is given by the quadratic form: $ds^2= \sum F_{ij} d \theta^i d \theta^j$ and it is known to be twice the Kullback Leibler Divergence: $ds^2=2KL(p(x, \theta ),p(x,\theta + d \theta))$ If you want to learn more about that I suggest reading the paper from Amari http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aos/1176345779 (I think there is also a book from Amari about riemannian geometry in statistic but I don't remember the name) - – Rob Hyndman Jul 30 '10 at 13:10 Since I am not a mathematician nor a statistician, I would like to restate what you were saying to make sure I did not mis-understand. So, you are saying that taking ds^2 (twice the KL) would have a similar meaning as R^2 (in a regression model) for a general distribution. And that this could actually be used to quantify distances geometrically? Does ds^2 have a name so I can do more reading about this. Is there a paper that directly describes this metric and shows applications and examples? – Ampleforth Aug 3 '10 at 22:22 – robin girard Aug 4 '10 at 17:25 This seems to be a characterization of directional derivative of KL rather than of KL itself, and it doesn't seem possible to get KL divergence out of it because unlike the derivative, KL-divergence doesn't depend on the geometry of the manifold – Yaroslav Bulatov Aug 11 '10 at 23:37 Suppose you are given n IID samples generated by either p or by q. You want to identify which distribution generated them. Take as null hypothesis that they were generated by q. Let a indicate probability of Type I error, mistakenly rejecting the null hypothesis, and b indicate probability of Type II error. Then for large n, probability of Type I error is at least $\exp(-n \text{KL}(p,q))$ In other words, for an "optimal" decision procedure, probability of Type I falls at most by a factor of exp(KL(p,q)) with each datapoint. Type II error falls by factor of $\exp(\text{KL}(q,p))$ at most. For arbitrary n, a and b are related as follows $b \log \frac{b}{1-a}+(1-b)\log \frac{1-b}{a} \le n \text{KL}(p,q)$ and $a \log \frac{a}{1-b}+(1-a)\log \frac{1-a}{b} \le n \text{KL}(q,p)$ If we express the bound above as the lower bound on a in terms of b and KL and decrease b to 0, result seems to approach the "exp(-n KL(q,p))" bound even for small n More details on page 10 here, and pages 74-77 of Kullback's "Information Theory and Statistics" (1978). As a side note, this interpretation can be used to motivate Fisher Information metric, since for any pair of distributions p,q at Fisher's distance k from each other (small k) you need the same number of observations to to tell them apart - +1 I like this interpretation! could you clarify " p below e "? why do you take small e ? you say "the probability of making the opposite mistake is" it is an upper bound or exact probability? If I remember, this type of approach is due to Chernoff, do you have the references (I find your first reference is not clarifying the point :) ) ? – robin girard Aug 12 '10 at 11:08 Why do I take small e...hmm...that's what Balasubramanian's paper did, but now, going back to Kullback, it seems his bound holds for any e, and he also gives bound for finite n, let me update the answer – Yaroslav Bulatov Aug 12 '10 at 20:07 ok, we don't need small e (now called b, Type II error) to be small for bound to hold, but b=0 is the value for which the simplified (exp(-n KL(p,q)) bound matches the more complicated bound above. Curiously enough, lower bound for Type I error given 0 Type II error is <1, I wonder if <1 Type II error rate is actually achievable – Yaroslav Bulatov Aug 12 '10 at 22:54 Actually a much easier to understand reference for this is Cover's "Elements of Information Theory", page 309, 12.8 "Stein's Lemma" – Yaroslav Bulatov Aug 13 '10 at 21:25	{"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": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9248917698860168, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/181219-determinant-matrix.html	# Thread: 1. ## Determinant of matrix Uploaded with ImageShack.us Is this a 3 by 4 matrix? I know how to find the determinant of a 3 by 3 matrix. How is the determinant calculated for this matrix? 2. You would be trying to evaluate $\displaystyle \left\|\begin{matrix}1 & 2 & \phantom{a}-1\\ a & 3 & \phantom{a -}1\\ 2 & 1 & a - 1\end{matrix}\right\|$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9338977336883545, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/yang-mills%20lagrangian-formalism	# Tagged Questions 2answers 145 views ### Kugo and Ojima's Canonical Formulation of Yang-Mills using BRST I am trying to study the canonical formulation of Yang-Mills theories so that I have direct access to the $n$-particle of the theory (i.e. the Hilbert Space). To that end, I am following Kugo and ... 2answers 199 views ### Can auxiliary fields be thought of as Lagrange multipliers? In the BRST formalism of gauge theories, the Lautrup-Nakanishi field $B^a(x)$ appears as an auxiliary variable \mathcal{L}_\text{BRST}=-\frac{1}{4}F_{\mu\nu}^a F^{a\,\mu\nu}+\frac{1}{2}\xi B^a B^a + ... 1answer 209 views ### Spontaneous symmetry breaking and 't Hooft and Polyakov monopoles What is spontaneous symmetry breaking from a classical point of view. Could you give some examples, using classical systems.I am studying about the 't Hooft and Polyakov magnetic monopoles solutions, ... 1answer 246 views ### Gauge-invariant field strength term in Yang-Mills Lagrangian I am reading the chapter of non-abelian gauge invariance from Peskin and Schroeder. Why is the term $-\frac{1}{4}(L_{\mu\nu}^i)^{2}$ gauge invariant? 2answers 325 views ### The Faddeev-Popov Lagrangian This is a non-abelian continuation of this QED question. The Lagrangian for a non-abelian gauge theory with gauge group $G$, and with fermion fields and ghost fields included is given by ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8486555814743042, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/34432/list	## Return to Answer 2 Added a few LaTeX tags. It is usually extraordinarily difficult to prove uniqueness of a simple group given its order, or even given its order and complete character table. In particular one of the last and hardest steps in the classification of finite simple groups was proving uniqueness of the Ree groups of type 2G2 $^2G_2$ of order q^3(q^3+1)(q-1), $q^3(q^3+1)(q-1)$, (for q $q$ of the form 3^(2n+1)) $3^{2n+1}$) which was finally solved in a series of notoriously difficult papers by Thompson and Bombieri. Although they were trying to prove the group was unique, proving that there were at most 2 would have been no easier. Another example is given in the paper by Higman in the book "finite simple groups" where he tries to characterize Janko's first group given not just its order 175560, but its entire character table. Even this takes several pages of complicated arguments. In other words, there is no easy way to bound the number of simple groups of given order, unless a lot of very smart people have overlooked something easy. 1 It is usually extraordinarily difficult to prove uniqueness of a simple group given its order, or even given its order and complete character table. In particular one of the last and hardest steps in the classification of finite simple groups was proving uniqueness of the Ree groups of type 2G2 of order q^3(q^3+1)(q-1), (for q of the form 3^(2n+1)) which was finally solved in a series of notoriously difficult papers by Thompson and Bombieri. Although they were trying to prove the group was unique, proving that there were at most 2 would have been no easier. Another example is given in the paper by Higman in the book "finite simple groups" where he tries to characterize Janko's first group given not just its order 175560, but its entire character table. Even this takes several pages of complicated arguments. In other words, there is no easy way to bound the number of simple groups of given order, unless a lot of very smart people have overlooked something easy.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9731239676475525, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/20201/how-are-the-pauli-matrices-for-the-electron-spin-derived	# How are the Pauli matrices for the electron spin derived? Could you explain how to derive the Pauli matrices? $$\sigma_1 = \sigma_x = \begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix}\,, \qquad \sigma_2 = \sigma_y = \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix}\,, \qquad \sigma_3 = \sigma_z = \begin{pmatrix} 1&0\\0&-1 \end{pmatrix}$$ Maybe you can also link to an easy to follow tutorial ? - ## 4 Answers That certainly depends on what exactly you mean. I take your question as "how do you see that the (non-relativistic) electron spin (or more generally, Spin-1/2) is described by the Pauli matrices?" Well, to start, we know that measuring the electron spin can only result in one of two values. From this we see that we need matrices of at least dimension 2. The simplest choice is then of course exactly dimension 2. Moreover the spin is an angular momentum, and thus described by three operators obeying the angular momentum algebra: $[L_i, L_j] = \mathrm{i} \hbar\epsilon_{ijk} L_k$. This together with matrix dimension 2 basically restricts the choice to sets of three matrices which are equivalent to $\hbar/2$ times the Pauli matrices (the freedom of choice of those matrices corresponds to the freedom to use three arbitrary orthogonal directions as $x$, $y$ and $z$ direction). So now why choose from those equivalent choices exactly the Pauli matrices? Well, there's always one measurement direction which is represented by a diagonal matrix; this makes calculations much easier. Of course it makes sense to choose the matrices in a way that this direction is one of the coordinate directions. By convention, the $z$ direction is chosen. This ultimately fixes the matrices. - The Hilbert space for spin 1/2 is two-dimensional - there are two possible values spin can take: $\hbar/2$ or $-\hbar/2$ (this is taken from experiment). Now, in two-dimensional Hilbert space spin operator has to be self-adjoined (this comes from foundations of QM). Furthermore, sum of its eigenvalues has to be 0 - because sum of eigenvalues is just a sum of possible results of measurements - in this case $\hbar/2-\hbar/2 = 0$. Therefore the operator corresponding to the measurement of spin in a given direction has to be 2x2 complex hermitan traceless matrix (no commutation relations used so far!). This family of matrices is a three-parameter one (one constant on diagonal, two off-diagonal for real and imaginary part, rest is determined by hermicity and tracelessness). Furthermore, every such an operator has to be diagonal in its eigenbasis - which corresponds to the measurement in a given direction. Let us name this direction $z$ direction. We see, that the only possible traceless hermitean diagonal matrix is a multiplicity of Pauli matrix $\sigma_z$. Now, we write the remaining two operators as $L_x$ and $L_y$. Next, we have to use commutation relations of angular momentum: $[L_y,\sigma_z]=L_x$, $[L_x,\sigma_z]=-L_y$ and $[L_x, L_y] = \sigma_z$. Remembering that $L_y$ as well as $L_x$ are determined by 3 real constants each, we need 6 linear equations to solve for them all at once. The above commutation relations give exactly 6 equations (3 matrix equations, two-dimensional matrices) so we can solve for $L_x$ and $L_y$ which are exactly the remaining Pauli matrices. - Yet another way is from almost purely experimental considerations. Consider a Stern-Gerlach set up; the identity is $$\mathbb{I}=\|\uparrow\rangle \langle \uparrow \|+\|\downarrow\rangle \langle \downarrow \|$$ Now experimentally we can only get two different numbers, $\hbar/2$ and $-\hbar/2$, and regardless of how we are oriented, in the natural basis corresponding to our orientation the spin operator must take the form $$S=\frac{\hbar}{2}(\|\uparrow\rangle \langle \uparrow \|-\|\downarrow\rangle \langle \downarrow \|$$ So now lets fix a basis, say the z, and hit the identity from one side with a spin vector in a different basis, say the x $$\|\uparrow\rangle_{x}=\|\uparrow\rangle\langle \uparrow \|\uparrow\rangle_{x}+\|\downarrow\rangle \langle \downarrow \|\uparrow\rangle_{x}$$ Through various arguments, the two inner-products can be evaluated to give $\pm1/\sqrt{2}$ to get $$\|\uparrow\rangle_{x}=\frac{1}{\sqrt{2}}\|\uparrow\rangle-\frac{1}{\sqrt{2}}\|\downarrow\rangle$$ to see the arguments see Ballentine, or Sakurai, or Schwinger. Now to get $S_x$ we use the expression for $S$ from above $$S_x=\frac{\hbar}{2}[(\|\uparrow\rangle \langle \uparrow \|)_x-(\|\downarrow\rangle \langle \downarrow \|)_x]$$ where my subscript x means in the x basis, but we know how to express those vectors in the "known" basis, the z basis. Substitution yeilds $$S_x=\frac{\hbar}{2}(\|\uparrow\rangle\langle\downarrow \|+\|\downarrow\rangle\langle\uparrow \|)$$ Consider the following equal signs, begotten with entirely using the identity symbol $$S_x=\sum_{n=1}^{2}\sum_{m=1}^{2}\|n\rangle\langle n\|S_x\|m\rangle\langle m\|$$ The $n$ and $m$ run from 1 to 2 standing for spin up and down. Now evaluate the guy in the middle. this is a two by two matrix whos elements are $$\langle \uparrow \|S_x\|\uparrow\rangle\quad\langle \uparrow\|S_x\|\downarrow\rangle\quad\langle\downarrow\|S_x\|\uparrow\rangle\quad\langle\downarrow\|S_x\|\downarrow\rangle$$ to see what these equal just hit the $S_x$ buy above with the two up/down vectors in the z basis and use the orthogonality condition to make them either 1 or zero. The result is one of the Pauli matrices. Hope this helps. - Spin is an angular momentum, so in the rest frame it is a 3-dimensional vector, or 4-dimensional vector with zero time component: $\vec{v} = (v_1,v_2,v_3)$ Each 3D vector can be associated with a 2x2 matrix by the following rule: $V = \begin{vmatrix} v_3 & v_1-iv_2 \\ v_1+iv_2 & -v_3 \end{vmatrix}$ In particular, if you chosse $\vec{v}$ as a basis vector: $\vec{v} = (1,0,0)$, it is associated with matrix $H_1=\begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix}$ Similarly, for $\vec{v} = (0,1,0)$ you will get $H_2=\begin{vmatrix} 0 & -i \\ i & 0 \end{vmatrix}$ and for $\vec{v} = (0,0,1)$ you will get $H_3=\begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix}$ These are Pauli matrices. Now arbitrary vector corresponds to linear combination of $H_1, H_2$ and $H_3$. For given $\vec{v} = (v_1,v_2,v_3)$ you will get $V = v_1 H_1 + v_2 H_2 + v_3 H_3$ Further you can use matrices to act on special objects with 2 complex components. These objects are called "spinors". They are used to construct wave functions of fermions with 1/2 spin. For instance, we can choose spinor of the form $s_3=\begin{vmatrix} 1 \\ 0 \end{vmatrix}$ and act on it with matrix $H_3$. We will obtain: $H_3 s_3=\begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix} \begin{vmatrix} 1 \\ 0 \end{vmatrix} = \begin{vmatrix} 1 \\ 0 \end{vmatrix} = s_3$ As you can see, $s_3$ is an "eigenspinor" of $H_3$ with eigenvalue +1. There is also another "eigenspinor" of $H_3$ with eigenvalue -1: $\begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix} \begin{vmatrix} 0 \\ 1 \end{vmatrix} = -\begin{vmatrix} 0 \\ 1 \end{vmatrix}$ Why do we use spinors instead of scalars, vectors and tensors? Because: 1. We can construct vectors out of spinors. These vectors are always isotropic, but they have spatial direction. 2. We can use Pauli matrices and their linear combinations to rotate vectors constructed from spinors. The rotation rule is very simple: if $s$ is initial spinor, then the rotated spinor is $\bar{s}=(\exp{iV})s$. If we construct a vector from $\bar{s}$, it will be equivalent to "ordinary" rotation of the vector constructed from $s$ around the axis parallel to $\vec{v}$ to the angle $\|\vec{v}\|$. Now if, for instance, $s$ is an "eigenspinor" of $H_3$, then direction of the vector constructed from $s$ will be invariant w.r.t. rotations around z axis. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 57, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9229525923728943, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/29655/why-does-spin-have-a-discrete-spectrum/29672	# Why does spin have a discrete spectrum? Why is it that unlike other quantum properties such as momentum and velocity, which usually are given through (probabilistic) continuous values, spin has a (probabilistic) discrete spectrum? - @user758556: in our current consensus model of the universe, position and momentum are not quantised. There are speculative theories which says otherwise. – genneth Jun 7 '12 at 3:51 5 @Farhad: mathematically the difference between position and angle is that of compactness. This directly implies the quantisation of the conjugate variables. The deeper question of "why angles" is perhaps a little beyond the scope of physics. – genneth Jun 7 '12 at 3:53 1 spin is not necessarily assigned a "definite quantized" value. Non-eigenstates of an operator represent non definite (quantum) states and you come across "probabilistic values" - even for spin. Stern-Gerlach experiment illustrates this: an eigenstate of $S_y$ is equally likely to be found with $z$-spin up or down. @Farhad Perhaps you are asking about the discrete nature of the observable spin VS the continuum nature of other observables, e.g. position, are you? – Jorge Campos Jun 7 '12 at 5:22 2 I think that the question should be phrased differently. You don't mean to ask why spin is quantized, everything is. What you mean is why spin has discrete spectrum. – MBN Jun 7 '12 at 7:22 Well, just to be clear, momentum can be quantized depending on the system and the boundary conditions e.g. particle in a box. The comments below seem to indicate spin is sort of like a box with periodic boundary conditions. – DJBunk Jun 7 '12 at 13:10 show 1 more comment ## 3 Answers I recently was writing about this on wikipedia. The most intuitive way to see why an operator like $S_z$ has discrete values is based on its relation to rotation operators: $R_{internal}(\hat{z},\phi) = \exp(-i\phi S_z / \hbar)$ where the left side means rotation of angle $\phi$ about the $z$-axis, but only rotating the "internal state" of particles not its spatial position (see wikipedia article for details). Since a rotation of $\phi=720^\circ$ [see below] is the same as no rotation at all (i.e. the identity operator), you conclude that the eigenvalues of $S_z$ can only be integers or half-integers. ...much like how a standing wave on a circular string has to have an integer number of wavelengths. -- Wait, why did I say $720^\circ$ not $360^\circ$?? Well, there are two mathematical groups that could plausibly correspond to rotation in the real world: $SO(3)$ and $SU(2)$. In $SO(3)$ but not $SU(2)$, rotating $360^\circ$ is the same as not rotating at all. In BOTH of them, rotating $720^\circ$ is the same as not rotating at all. So we can be totally sure that the $720^\circ$ rotation operator is the identity operator, whereas for $360^\circ$ it would just be a guess based on extrapolating from classical physics intuition. As long as there are fermions present, the guess is wrong! Rotating a fermion by $360^\circ$ corresponds to flipping the sign of its wavefunction. - Nice answer, but maybe you should add a paragraph for why the momentum of a free particle is not quantized, for completeness in the answer. – anna v Jun 8 '12 at 11:50 The deeper reason is that the components of the spin (angular momentum) vector generate the group of rotations. This group is compact, which means that a rotation perpendicular to an arbitrary direction necessarily closes. This implies for mathematical reasons (valid for every compact Lie group) that its representations as operators in a Hilbert space come in discrete batches only, and the eigenvalues of any component, in general functions of the label of the representation, must in the compact case be discrete. In contrast, position and momentum generate the noncompact Weyl group (a central extension of the phase space translations), and a translation laong an arbitrary phase space direction never closes. This implies that the eigenvalues vary continuously. - I will take a hand waving guess at this. Nature is quantum mechanical , i.e. it is ruled by quantum mechanical equations that define motion etc. The classical Lagrangians are a limiting case mostly for large dimensions. Quantization appears when the variables are constrained , for example in the confines of a potential well. One finds that only quantized values are allowed, so in a confining potential also momentum will be quantized as long as there are discrete energy levels. So the question can only be correct if one considers the unconfined particles and becomes "why unconfined elementary particles have a quantized spin in contrast to momentum or energy etc" . My intuitive answer is: probably because spin is a rotation and rotations are limited by the 0 to 2pi confinement of the values of phi, a finite constraint, in contrast to momentum which can go from zero to infinity. Constraints are conditions for quantization. As a help in intuition look at section 14 of Schiff's quantum mechanics, separation in spherical coordinates of the schroedinger equation for spherically symmetric potentials. . The angular equations have no dependence on the potential and their solutions are quantized. - – anna v Jun 7 '12 at 8:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9229182004928589, "perplexity_flag": "head"}
http://mathoverflow.net/questions/96338/integration-under-functional-sign/96354	## Integration under functional sign ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $f(x,y)$ be some bounded with its derivatives continuous function on $\Omega \times \overline{\Omega}$, where $\Omega$ is a domain in $\mathbb{R}^n$. Let $f(\cdot,y) \in \mathcal{E}(\Omega)$ for any fixed $y$. Let $L_{x} \in \mathcal{E}'(\Omega)$. Is it true that $$\int\limits_{\overline{\Omega}} L_{x}f(x,y) \, \mu(dy) = L_{x} \int\limits_{\overline{\Omega}} f(x,y) \, \mu(dy)$$ holds for any probability measure $\mu$ in $\overline{\Omega}$? If it is true, how to show it? If $f(x,y) \in \mathcal{E}(\Omega \times \Omega')$ where domain $\Omega'$ is such that $\overline{\Omega} \subseteq \Omega'$ then the equality holds by virtue of the tensor product of distributions theorem. - Do you need that the derivatives of $f(\cdot,y)$ to be bounded, in other to ensure us that $x\mapsto \int_{\overline \Omega}f(x,y)\mu(dy)$ is infinitely many times differentiable? – Davide Giraudo May 8 2012 at 15:08 Thanks, I've corrected the question. – Nimza May 8 2012 at 17:46 ## 2 Answers Suppose first that $L_x\in C^\infty_0(\Omega)$. Then the equality you ask about is Fubini's theorem. Suppose now that $L_x$ is not necessarily smooth. Choose a sequence $\newcommand{\ve}{\varepsilon}$ $L_{\ve,x}\in \mathscr{E}'(\Omega)$ that converges to $L_x$ in the weak sense. Then one needs to prove that $$\lim_{\ve\to 0} L_{\ve,x}\int_\Omega f(x,y)d\mu(y)=L_x\int_\Omega f(x,y)d\mu(y), \tag{A}$$ $$\lim_{\ve\to 0}\int_\Omega (L_{\ve,x}-L_x)f(x,y)d\mu(y)=0. \tag{B}$$ The equality (A) is an immediate consequence of the weak convergence. The equality (B) requires an additional assumption on $f$. Denote by $K$ a compact set containing the support of $L_x$ and $L_{\ve, x}$, $\ve$ sufficiently small. If we assume that for any multi-index $\alpha$ we have $$\sup_{x\in K, y\in \Omega} \partial^\alpha_x f(x,y) <\infty, \tag{C}$$ then (B) follows by invoking the uniform boundedness principle for $\mathscr{E}'(\Omega)$ which states that if a sequence $u_n \in \mathscr{E}'(\Omega)$ converges weakly to $0$, then $u_n(\phi)\to 0$ uniformly for $\phi$ in a bounded subset of $\mathscr{E}(\Omega)$. I recall that a subset $\Phi\subset \mathscr{E}(\Omega)$ is bounded if for any compact $K\subset \Omega$ and any multi-index $\alpha$ we have $$\sup_{x\in K, \phi\in \Phi} \partial^\alpha_x\phi(x) <\infty.$$ Update. Let me set $\phi_y:=f(x,y)$. To insure the integrability of $y\mapsto L(\phi_y)$ for any $L\in\mathscr{E}'(\Omega)$ it suffices to assume that the map $\Omega\ni y\mapsto \phi_y\in\mathscr{E}(\Omega)$ is continuous, i.e., for any $y_0\in \Omega$, any $\ve>0$, any compact $K\subset \Omega$ and any multi-index $\alpha$ there exists $\delta>0$ such that $$\|y-y_0\|<\delta \Rightarrow \sup_{x\in K}\left\|\partial^\alpha_x\bigl(\; \phi_y(x)-\phi_{y_0}(x)\;\bigr )\right\| <\ve.$$ - Great thanks! Can you give me please some reference on such version of uniform boundness principle? – Nimza May 8 2012 at 15:34 2 The most complete reference would be Francois Treve, Topological Vector Sapeces, Distributions and Kernels, now in Dover. However this is a bit harder to digest given its generality. A more readable and helpful source would be vol. 2 of Gelfand and Shilov's treatise on generalized functions. There they investigate complete countably normed spaces, or Frechet spaces and they prove among other things the uniform boundedness principle. The space $\mathscr{E}(\Omega)$ is such a space. – Liviu Nicolaescu May 8 2012 at 16:00 Thanks for update and for references. – Nimza May 8 2012 at 17:49 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Davide's right. Neither integral makes sense because the function was only continuous. If you assume $f$ is a smooth test function, then a priori it's only clear when $\mu$ is a finite linear combination of point masses. Thus, you cannot avoid using a Riemann sums trick: approximate the more general measure with a sequence of finitely supported measures $\mu_n$. My impression is that the general theory of distributions cannot start without taking Riemann-type sums at some point and that any argument probably has this maneuver underlying it somewhere. For the right hand side, you need to prove that $\int f(x,y) d\mu_n \to \int f(x,y) d\mu(y)$ in some $C^k$ topology as functions of $x$ over the support of $L_x$. For the left hand side, check that $\mu_n \rightharpoonup \mu$ weakly and $L_x f(x,y)$ is continuous in $y$. This step again uses that $L_x$ is continuous with respect to $C^k$ convergence for some $k$. Note, if you establish this identity when $\mu$ is, say, an absolutely continuous measure with a smooth density function, then you can pass to the limit for a general finite measure by using a mollifying kernel (analogous to taking $L_{\epsilon, x}$ in Liviu's argument, but this is a mollification in the $y$ variable, and is just measure theoretic). - @ Phil The collection of functions $\Phi:=(\; f(-,y)\;)_{y\in \Omega}$ is bounded in $\mathscr{E}(\Omega)$ due to (C). Set $$s(\varepsilon):=\sup_{y\in \Omega} \|(L_{\varepsilon, x}-L_x)f(x,y)\|$$ The uniform boundedness principle implies $s(\varepsilon)\to 0$. Hence $$\left\|\;\int_\Omega(L_{\varepsilon, x}-L_x)f(x,y)d\mu(y)\;\right\| \leq \int_Omega s(\varepsilon) d\mu(y)=s(\varepsilon)\to 0.$$ – Liviu Nicolaescu May 8 2012 at 16:23 (It looks like there is a coding error in the comment.) I think I'm starting to see the relationship between these two arguments. On the one hand, your argument appeals to Fubini to get started; the argument that I suggested actually avoids Fubini's theorem by taking Riemann sums. I actually thought that you couldn't get away without this move (for the reason in my post), but it looks like you can just get by with the continuity in the $y$ variable? Thoughts? – Phil Isett May 8 2012 at 20:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 64, "mathjax_display_tex": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9336673617362976, "perplexity_flag": "head"}
http://mathoverflow.net/questions/70679/cohomology-of-springer-resolution	## Cohomology of Springer resolution ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This question is elementary. Let $G$ be a simple algebraic group over $\mathbb{C}$, and let $B$ be a choice of Borel subgroup, with unipotent radical $U$ with Lie algebra $\mathfrak{n}$. Then the Springer resolution of the nilpotent cone is $Z = G \times_B \mathfrak{n}$; it is identified with the cotangent bundle of $G/B$. In "Cohomology and the resolution of the nilpotent variety", Math. Ann. 1976, Hesselink proves that for each $p > 0$, $H^p(Z, \mathcal{O}_Z) = 0$. My question is as follows: this group is identified with $\oplus_{l \geq 0} H^p(G/B, \text{Sym}^l \mathfrak{n}^\vee)$, so it is equivalent to show the vanishing of each of these summands. Why does the following simple argument not work? For each $l \geq 0$, $\text{Sym}^l \mathfrak{n}^\vee$ (as a representation of $B$) has a filtration with 1-dimensional graded pieces where $T$ acts with anti-dominant weights (where the positive roots are with respect to $B$). The cohomology of the line bundles associated to these graded pieces vanishes by Kempf's vanishing theorem, so the long exact sequence in sheaf cohomology should imply that the higher cohomology of $\text{Sym}^l \mathfrak{n}^\vee$ also vanishes. - The typography in the last paragraph got mixed up a little. Aside from that, it's good to keep in mind that most of the machinery makes sense over an algebraically closed field of arbitrary characteristic. This especially includes Kempf's vanishing theorem. Also, many people assume that `$B$` corresponds to negative roots in order to avoid some awkwardness about dominant weights. – Jim Humphreys Jul 18 2011 at 22:03 ## 1 Answer The reason your argument doesn't work is because it's not true that $\text{Sym}^l \mathfrak n^\vee$ has a filtration with 1-dimensional graded pieces where $T$ acts with anti-dominant weights. In fact, this is false even in the case $l = 1$. I'll assume, as you do, that $B$ corresponds to the positive roots. Then the weights of $\mathfrak n^\vee$ correspond to the negative roots, and most of the negative roots are NOT anti-dominant. (In fact, even the negative simple roots are not antidominant, unless all components of $G$ are of type $A_1$). This shows why something subtle is going on here: $H^i( G/B, \mathfrak n^\vee ) = 0$ for all $i > 0$, but it is not the case in general that $H^i( G/B, \; \textrm{gr} \; \mathfrak n^\vee ) = 0$ for all $i > 0$. - I see. Thank you for explaining this to me! – JT Jul 19 2011 at 20:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9409776329994202, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Photoacoustic_Doppler_effect	# Photoacoustic Doppler effect From Wikipedia, the free encyclopedia Jump to: navigation, search The photoacoustic Doppler effect, as its name implies, is one specific kind of Doppler effect, which occurs when an intensity modulated light wave induces a photoacoustic wave on moving particles with a specific frequency. The observed frequency shift is a good indicator of the velocity of the illuminated moving particles. A potential biomedical application is measuring blood flow. Specifically, when an intensity modulated light wave is exerted on a localized medium, the resulting heat can induce an alternating and localized pressure change. This periodic pressure change generates an acoustic wave with a specific frequency. Among various factors that determine this frequency, the velocity of the heated area and thus the moving particles in this area can induce a frequency shift proportional to the relative motion. Thus, from the perspective of an observer, the observed frequency shift can be used to derive the velocity of illuminated moving particles.[1] ## Theory To be simple, consider a clear medium firstly. The medium contains small optical absorbers moving with velocity vector $\vec{v}$. The absorbers are irradiated by a laser with intensity modulated at frequency $f_{0}$. Thus, the intensity of the laser could be described by: $I={I }_{0 } \left[ 1+cos \left ( 2 \pi f_{0}t \right ) \right ] /2$[2] Figure 1: Overview of PAD Effect[2] When $\vec{v}$ is zero, an acoustic wave with the same frequency $f_{0}$ as the light intensity wave is induced. Otherwise, there is a frequency shift in the induced acoustic wave. The magnitude of the frequency shift depends on the relative velocity $\vec{v}$, the angle $\alpha$ between the velocity and the photon density wave propagation direction, and the angle $\theta$ between the velocity and the ultrasonic wave propagation direction. The frequency shift is given by: $f_{PAD}=-f_{0} \frac{ v}{ c_{0}} cos \alpha +f_{0} \frac{v}{c_{a}} cos \theta$[2] Where $c_{0}$ is the speed of light in the medium and $c_{a}$ is the speed of sound. The first term on the right side of the expression represents the frequency shift in the photon density wave observed by the absorber acting as a moving receiver. The second term represents the frequency shift in the photoacoustic wave due to the motion of the absorbers observed by the ultrasonic transducer.[2] In practice, since $\frac{c_{0}}{c_{a}}\sim 10^{5}$ and $v \ll c_{a}$, only the second term is detectable. Therefore, the above equation reduces to: $f_{PAD}= f_{0} \frac{v}{c_{a}} cos \theta = \frac{v}{\lambda} cos \theta$[2][3] In this approximation, the frequency shift is not affected by the direction of the optical radiation. It is only affected by the magnitude of velocity and the angle between the velocity and the acoustic wave propagation direction.[2] This equation also holds for a scattering medium. In this case, the photon density wave becomes diffusive due to light scattering. Although the diffusive photon density wave has a slower phase velocity than the speed of light, its wavelength is still much longer than the acoustic wave.[3] ## Experiment Figure 2: Average Photoacoustic Doppler Shift vs. Velocity for a Scattering Medium [3] In the first demonstration of the Photoacoustic Doppler effect, a continuous wave diode laser was used in a photoacoustic microscopy setup with an ultrasonic transducer as the detector. The sample was a solution of absorbing particles moving through a tube. The tube was in a water bath containing scattering particles[2] Figure 2 shows a relationship between average flow velocity and the experimental photoacoustic Doppler frequency shift. In a scattering medium, such as the experimental phantom, fewer photons reach the absorbers than in an optically clear medium. This affects the signal intensity but not the magnitude of the frequency shift. Another demonstrated feature of this technique is that it is capable of measuring flow direction relative to the detector based on the sign of the frequency shift.[2] The reported minimum detected flow rate is 0.027 mm/s in the scattering medium.[3] ## Application One promising application is the non-invasive measurement of flow. This is related to an important problem in medicine: the measurement of blood flow through arteries, capillaries, and veins.[3] Measuring blood velocity in capillaries is an important component to clinically determining how much oxygen is delivered to tissues and is potentially important to the diagnosis of a variety of diseases including diabetes and cancer.However, a particular difficulty of measuring flow velocity in capillaries is caused by the low blood flow rate and micrometre-scale diameter.Photoacoustic Doppler effect based imaging is a promising method for blood flow measurement in capillaries. ### Existing techniques Based on either ultrasound or light there are several techniques currently being used to measure blood velocity in a clinical setting or other types of flow velocities. #### Doppler ultrasound The Doppler ultrasound technique uses Doppler frequency shifts in ultrasound wave. This technique is currently used in biomedicine to measure blood flow in arteries and veins. It is limited to high flow rates ($>1$cm/s) generally found in large vessels due to the high background ultrasound signal from biological tissue.[3] #### Laser doppler flowmetry Laser Doppler Flowmetry utilizes light instead of ultrasound to detect flow velocity. The much shorter optical wavelength means this technology is able to detect low flow velocities out of the range of Doppler ultrasound. But this technique is limited by high background noise and low signal due to multiple scattering. Laser Doppler flowmetry can measure only the averaged blood speed within 1mm$^{3}$ without information about flow direction.[3] #### Doppler optical coherence tomography Doppler Optical coherence tomography is an optical flow measurement technique that improves on the spatial resolution of laser Doppler flowmetry by rejecting multiple scattering light with coherent gating. This technique is able to detect flow velocity as low as $100 \mu$m/s with the spatial resolution of $5\times 5\times 15\mu$m$^{3}$. The detection depth is usually limited by the high optical scattering coefficient of biological tissue to $<1$mm.[3] ### Photoacoustic doppler flowmetry Photoacoustic Doppler effect can be used to measure the blood flow velocity with the advantages of Photoacoustic imaging. Photoacoustic imaging combines the spatial resolution of ultrasound imaging with the contrast of optical absorption in deep biological tissue.[1] Ultrasound has good spatial resolution in deep biological tissue since ultrasonic scattering is much weaker than optical scattering, but it is insensitive to biochemical properties. Conversely, optical imaging is able to achieve high contrast in biological tissue via high sensitivity to small molecular optical absorbers, such as hemoglobin found in red blood cells, but its spatial resolution is compromised by the strong scattering of light in biological tissue. By combining the optical imaging with ultrasound, it is possible to achieve both high contrast and spatial resolution.[1] The photoacoustic Doppler flowmetry could use the power of photoacoustics to measure flow velocities that are usually inaccessible to pure light-based or ultrasound techniques. The high spatial resolution could make it possible to pinpoint only a few absorbing particles localized to a single capillary. High contrast from the strong optical absorbers make it possible to clearly resolve the signal from the absorbers over the background. ## References 1. ^ a b c LV Wang and HI Wu (2007). Biomedical Optics: Principles and Imaging. Wiley. ISBN 9780471743940 Check `\|isbn=` value (help). 2. H. Fang, K. Maslov, L.V. Wang. "Photoacoustic Doppler Effect from Flowing Small Light-Absorbing Particles." Physical Review Letters 99, 184501 (2007) 3. H. Fang, K. Maslov, L.V. Wang. "Photoacoustic Doppler flow measurement in optically scattering media." Applied Physics Letters 91 (2007) 264103	{"extraction_info": {"found_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": 20, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9072128534317017, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/138151-definition-stationary-markov-chain.html	# Thread: 1. ## Definition of Stationary markov chain hello Is a staionary markov chain a chain with a stationary distribution (not necessarily unique)? or the stationary distribution needs to be unique (so the chain is irreducible and positive recurrent) to be a stationary markov chain? or is something totally different? Ive read something about being P-invariant under "time translations" but i can not find this definition formally. any help would be great thank you! 2. Originally Posted by mabruka hello Is a staionary markov chain a chain with a stationary distribution (not necessarily unique)? or the stationary distribution needs to be unique (so the chain is irreducible and positive recurrent) to be a stationary markov chain? or is something totally different? Ive read something about being P-invariant under "time translations" but i can not find this definition formally. any help would be great thank you! From my knowledge, stationary refers to time homogeneity which is $\mathbb{P}(X_n \| X_{n-1}=x)=\mathbb{P}(X_{n-1}\|X_{n-2}=x)$. A context may help in determining if this is what you are after. 3. Originally Posted by mabruka hello Is a staionary markov chain a chain with a stationary distribution (not necessarily unique)? or the stationary distribution needs to be unique (so the chain is irreducible and positive recurrent) to be a stationary markov chain? or is something totally different? Ive read something about being P-invariant under "time translations" but i can not find this definition formally. any help would be great thank you! In general: Stationary Markov chain $=$ Markov chain with transition probabilities that represent the "long run" fraction in state $i.$ The stationary fraction in state $i$ is given by the solution $\pi (i) \Rightarrow$ uniqueness. For definiteness: Consider a Markov chain whose stationary distribution is given by $\vec\pi = [1/4 \ \ 2/4 \ \ 1/4]<br />$ $\Rightarrow$ Stationary Markov chain = $\lim_{n\rightarrow \infty} p^n (i, j)= \left( \begin{array}{ccc} <br /> 1/4 & 2/4 & 1/4 \\ 1/4 & 2/4 & 1/4 \\ 1/4 & 2/4 & 1/4 \end{array} \right)$ In plain english: If we have a Markov chain whose stationary distribution exists, then the stationary Markov chain has the solutions $\pi (i)$ written in every entry of it's $i^{th}$ column. Say in the above example we had sample space $S=\{1 \ \ 2 \ \ 3\}$ where $1 =$ lower class, $2 =$ middle class $3 =$ upper class. Then the stationary Markov chain represents the fraction of people who will be in the lower, upper and middle classes as $t\rightarrow \infty.$ $e.g.$ in the long run $1/4$ of the population will be in the lower class. Nahwatimsayin? 4. Thank you both for your answers. Actually what Focus said is what im interested in. Context: We have $\{\xi_t\}_{t\in \mathbb R}$ an irreducible, recurrent and stationary Markov process with finite space state S and invariant distribution $\Pi=\{\pi_i\}_{i\in S}$ on $(\Omega,\mathbb F, \mathbb P)$. We consider $\Omega$to be the canonical space given by Kolmogorov's Extension theorem for $\xi$. My question about stationarity comes after introducing the concept of reversibility: $\xi$ is reversible if $(\xi_{t_1},\ldots,\xi_{t_k})$ has the same distribution as $(\xi_{T-t_1},\ldots,\xi_{T-t_k})$ for all $k\geq1$, $t_1<\ldots,<t_k$ and $T\in \mathbb R$. A measurable transformation is introduced (time shift) : $\theta_t:\Omega\longrightarrow \Omega$ $\theta_t \omega(s) =\omega(s+t)$ Then it is stated: Since $\xi$ is stationary, $\mathbb P$ is $\theta_t$ -invariant, i.e. $\theta_t \mathbb P =\mathbb P$. My question: What role does stationarity play so it makes our $\mathbb P$ $\theta_t$ - invariant and how does this follows from the definition of stationarity?? That is why i first asked about the formal definition of stationarity. 5. Originally Posted by mabruka Thank you both for your answers. Actually what Focus said is what im interested in. Okay... I assume you realize that the definition of stationary Markov chain that I gave directly implies time homogeneity, or $\mathbb{P}(X_n \| X_{n-1}=x)=\mathbb{P}(X_{n-1}\|X_{n-2}=x).$ I was simply trying to paint a vivid picture... 6. So stationarity and time homogeneiety are two different definitions, but stationarity implies time homogeneiety right?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 37, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9079594016075134, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/52328/fast-question-about-lagrangian/52330	# Fast question about Lagrangian I've seen some problems solved in a weird way, I just want to be sure: the whole kinetic energy has to be in the lagrangian, right? For example, if we have a particle fixed in a plane with spherical coordinates $(r,\theta,\phi)=(r_0,\theta_0,\phi)$, and that plane is rotating with a constant angular velocity $\omega=(0,0,\omega_0)$, so that the $\phi$ coordinate of the particle is $\phi=\omega t$, then that term of the kinetic energy: in spherical coordinates: $\frac{1}{2}mr^2\omega^2\sin^2(\theta)$, that term has to go in the lagrangian, am I right? And does it have to go in the expression of the total mechanical energy? - Could you clarify what exactly is rotating - the particle or the plane? How is the angle $\theta$ defined? – Joe Jan 27 at 19:25 @Joe I edited it to try to explain it better. – MyUserIsThis Jan 27 at 19:29 ## 1 Answer If you can form the problem in such a way as the is a contribution to the kinetic energy that can not change during the time modeled then you can drop that term. Why? Because the Euler-Lagrange equation is only concerned with differentials and a added or subtracted constant does not affect them. The simplest example of this is writing the problem in a pair of reference frames where the system is at rest in one and in constant linear motion in another. The KE of the center of mass does not affect the outcome. - I think I got it, thanks – MyUserIsThis Jan 27 at 19:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9424154758453369, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/68488/a-wrong-proof-about-dedekind-domains	# A wrong proof about Dedekind domains I "proved" that a Dedekind domain is a PID, but as we know this is wrong (for example $\mathbb{Z}[\sqrt{-5}]$). I do not know what is wrong in my proof: Suppose $R$ is a Dedekind domain, $I$ is any nonzero ideal of $R$, $\mathfrak p_i$ all the primes of $I$. If $r_i$ is the $\mathfrak p_i$-adic value of $I$, we can find an element $x\in R$ with its $\mathfrak p_i$-adic value $r_i$, so $I=(x)$ is principal. Who can tell me what is wrong in my "proof"? - If there are infinitely many primes in $R$, then are you sure you can specify the valuation of $x$ at each one? – Keenan Kidwell Sep 29 '11 at 14:26 ## 1 Answer First, your presentation seems slightly muddled. I think you are arguing the following: Let $I$ be a nonzero ideal in a Dedekind domain $R$. Then $I$ admits a factorization into primes $I = \mathfrak{p}_1^{a_1} \cdots \mathfrak{p}_r^{a_r}$. Now you observe that (for instance by the Chinese Remainder Theorem, or Artin-Whaples approximation) there exists an element $x \in R$ such that for all $1 \leq i \leq r$, $\operatorname{ord}_{\mathfrak{p}_i}(x) = a_i$. Finally you want to claim that $I = (x)$. (I hope I am understanding correctly.) The problem is with the very last step: it need not be the case that $I = (x)$. You have enforced that for all $1 \leq i \leq r$, $\operatorname{ord}_{\mathfrak{p}_i}(x) = \operatorname{ord}_{\mathfrak{p}_i}(I) = a_i$. But what about all the other prime ideals of $R$? The Chinese Remainder Theorem does not say that you can find such an $x$ which is divisible by the given set of primes to prescribed multiplicities and is not divisible by any other prime ideals of $R$: indeed, as you have noticed, this is necessarily false in any Dedekind domain which is not a PID. Nevertheless the argument above does prove something. In fact it proves several things: 1) ("Moving Lemma") Given any nonzero ideal $I$ in a Dedekind domain and any finite set $S$ of nonzero prime ideals of $R$, there exists $0 \neq x$ in the fraction field $K$ such that the fractional ideal $(x) I$ is not divisible by any $\mathfrak{p} \in S$. This implies: 2) If $R$ has only finitely many prime ideals, then it is a PID. In fact it also implies: 3) If all but finitely many prime ideals of $R$ are principal, then all prime ideals of $R$ are principal and thus $R$ is a PID. Fact 2) above is a standard exercise in this subject. For some reason 3) is not: it seems to have first been recorded by Luther Claborn in the 1960's: see Corollary 1.6 here. I "rediscovered it" a few years ago when teaching a course on algebraic number theory. To better appreciate the result, note that the ring $\mathbb{Z}[\sqrt{-3}]$ has infinitely many prime ideals, exactly one of which is not principal. But it is not a Dedekind domain, since it is not integrally closed in its fraction field. - Yes,that is my meaning and you find the reason,thanks.I lose other primes,their adic values maybe positive,Chinese Remainder Theorem can not go to infinite,so we just can get that there is some idea J with IJ=(a). – Strongart Oct 1 '11 at 10:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9503650665283203, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=186987&page=6	Physics Forums Thread Closed Page 6 of 8 « First < 3 4 5 6 7 8 > Mentor ## Why do we rotate along with the earth's rotation? Quote by singh94 u cant force me to stop writing in short forms Hint: "Mentor" = "Moderator" here. Mentors have the power to ban people. If you insist on using text-speak, you are welcome to find another physics forum that allows you to do so. Blog Entries: 8 Recognitions: Gold Member Quote by singh94 the earth is rotating the center may not move from its position but it is also rotating if u take the earth and shrink it to the size of the center of the earth then it would still rotate right? so the position of a point on earths surface to the center remains same. The further out you move away from the centre of the earth, the greater the distance you travel and the higher your velocity becomes. A good example is a geostationary satellite. It is constantly above one point on the surface of the earth, but it's distance travelled and speed is many times greater. Mentor Quote by singh94 i thought u people would be and mature enough to understand that im just saying that do i need to exert a force which gives me a speed of 1005 miles so that i can run against the direction of the earth.also i know what r the units of force velocity etc. which concern physics thank u very much. I know that this has already been answered, but I felt that I should respond since it was addressed to me: 1. You mean that you thought that we would be psychic enough that we would somehow be able to make (correct) assumptions about your level of physics knowledge, in spite of the fact that you failed to communicate that level clearly? Well, NO, we can't read your mind. All we have to go on when it comes to gauging your physics knowledge are the things that you write. EDIT: and besides, there is no such thing as "a force that gives you a speed of 1005 mph." Any force CAN give you that speed provided you apply it for long enough time. Once again, based on what you have typed, we have no choice but to assume that you don't understand the impulse-momentum theorem. 2. You DON'T have to "run against the direction of the Earth." It's NOT like being on a treadmill, because you are MOVING WITH THE EARTH. In other words, if you don't move, it means you are stationary with respect to the Earth's surface. Even when you are standing still, it appears to an outside observer (who is not rotating with the Earth) that both the Earth and you (on it) are rotating at the same rate. Do you get it now? Also, read the rest of the thread. You do NOT need to invoke friction at any point in history (not even going back to Earth's formation) to understand why everything that makes up part of the Earth is rotating. Only conservation of angular momentum need be applied. Mentor Quote by cepheid You do NOT need to invoke friction at any point in history (not even going back to Earth's formation) to understand why everything that makes up part of the Earth is rotating. Only conservation of angular momentum need be applied. That's going a bit too far. In the long term, you cannot apply conservation of angular momentum for the simple reason that the Earth's angular momentum has not been constant. the Earth's rotation rate was considerably higher (4-6 times higher!) shortly after the Moon formed compared to its current rate. The atmosphere and oceans are not moving around the Earth at 4-6 times Earth rotation rate precisely because of friction. Mentor Quote by D H That's going a bit too far. In the long term, you cannot apply conservation of angular momentum for the simple reason that the Earth's angular momentum has not been constant. the Earth's rotation rate was considerably higher (4-6 times higher!) shortly after the Moon formed compared to its current rate. The atmosphere and oceans are not moving around the Earth at 4-6 times Earth rotation rate precisely because of friction. Thanks. Yeah, that is a fair point. I failed to consider tidal effects. Edit: but weren't the atmosphere and oceans being torqued on as well? Quote by cepheid You do NOT need to invoke friction at any point in history (not even going back to Earth's formation) to understand why everything that makes up part of the Earth is rotating. Only conservation of angular momentum need be applied. I thought conservation of angular momentum was applicable only to a rigid body rotating on its own axis. In that case we'd have to be glued to earth's surface from time immemorial. (But that's what the gravitational force and friction do. Gluing us to the surface) So, the answer is a complex interplay of conservative and non-conservative forces, and conservation of motion. Recognitions: Gold Member I agree with asdofindia. It is a result of several forces over a long period of time. But in the short term, we don't need friction to keep us rotating with the earth. Imagine a magnetic sphere. Rotating. Imagine a sticky man on it with iron legs. Let's say he already has acquired the velocity of the point right under his legs. At every point, the man's velocity wants him to go tangential to the surface, to be thrown away from the sphere. But the magnetic force pulls him onto the sphere. (Centrifugal and centripetal forces) Agreed till now? The centrifugal force and centripetal forces are exactly opposite and cancel each other. But they've got no component along the surface of the sphere!! (So there's got to be friction???!!!???!!!) I don't think I've understood my answer. Recognitions: Gold Member The centrifugal force and centripetal forces are exactly opposite and cancel each other. But they've got no component along the surface of the sphere!! What do you mean by this? See, if we draw a free body diagram. We'd draw an arrow from the man to the centre, calling it centripetal force. And another opposite to it calling it centrifugal force, right? I thought they'd cancel, but I don't think I've clearly finished that thought process, I made a quick reply... And of course they wouldn't have any component tangential to the surface. But a body already moving with a velocity tangential do not need a force to keep it moving along the tangent. But that's along the tangent... Oh... I think I'm confused. Let me think for a while... Recognitions: Gold Member The force holding the man to your sphere is the magnetic force between his legs and the sphere. This force is greater than the upward force trying to push him away. If it wasn't any small simply movement by the man would send him moving upwards and away. Because of this, the man is constantly being pulled down towards the sphere while also moving...tangently??...through space around the sphere. The magnetic force on him is similar to the gravitational force on a satellite in orbit. The satellite is constantly falling towards the earth, but also moving, resulting in an orbit. That makes me wonder whether or not to subtract mv2/r from GMm/r2 when calculating the weight of a body Edit:http://en.wikipedia.org/wiki/Centrif...tious%29#Earth Appears like there are effects due to centrifugal force Recognitions: Gold Member There are most definately effects due to the centrifigal force. They are usually so small as to be irrevelent in day to day stuff. However, for space launches we do calculate that effect as well as the velocity of the earths rotation from the launch site in order to put something into orbit correctly. Centripetal force, Fg=GMm/r2 Centrifugal force, Fr=mv2/r So, when v is > vmax where vmax> $$\sqrt{GM/r}$$ we do fly away!! If it's $$\sqrt{2}$$ times the vmax in the above equation, it'd escape earth's gravity too (because that'd be 11.2 km/s) If v/vmax is between 1 and $$\sqrt{2}$$, it'd fly away and fall down back. (But luckily on earth, v is approximately 1/17 times vmax) That's why we're not flying away from earth. Now, let me do something I've never done before. Since the centrifugal force on earth is cancelled out by the gravity (289 times stronger), we can safely assume the body to be at rest with respect to the reference frame of earth. (I don't know if that's right, because as I said, I've never done this before) (There'd be no longer any effects that'd be observed due to rotational motion of earth with respect to the space around it) So, since earth and the body are both at rest (in that inertial frame) there wouldn't be the need of any frictional force. We have to view this in the inertial frame of earth. (I have understood my point) Quote by singh94 i mean a person who is in space. if everything was not moving(not at respect to eachother but in actual) then he would be at rest with respect to them. now do u get it I'm not sure what you are asking, but when you are in orbit in space then we can confidently say that you are NOT rotating with the earth. You are moving much, much faster than the earth is spinning. Now what happens when you re-enter the atmosphere? The atmosphere WILL slow you the fuk back down to it's own speed, and it will use extreme force to do so. The force is so great in fact, that whatever is re-entering must use some kind of heat shield, otherwise it will burn up with a brightness of the sun. So you see what happens whenever something is not rotating with the earth. The earth WILL slow or speed it up to match its own rotation, and violently so if need be. If you don't believe me go jump out of a moving car, and you'll see first hand how it feels when something is NOT rotating with the earth. Blog Entries: 8 Recognitions: Gold Member asdofindia centripetal force is not gravity. Your equation above defines gravity not centripetal force. Note, your equation is virtually constant wherever you are on the planet, which is gravity, but the centripetal force is almost zero at the poles and maximum at the equator. Quote by jarednjames asdofindia centripetal force is not gravity. Your equation above defines gravity not centripetal force. At the poles there is virtually no centrifugal forces compared to at the equator. So the centripetal force is......? Thread Closed Page 6 of 8 « First < 3 4 5 6 7 8 > Thread Tools \| \| \| \| \|------------------------------------------------------------------------\|----------------------------------\|---------\| \| Similar Threads for: Why do we rotate along with the earth's rotation? \| \| \| \| Thread \| Forum \| Replies \| \| \| Introductory Physics Homework \| 2 \| \| \| Precalculus Mathematics Homework \| 5 \| \| \| Earth \| 21 \| \| \| Classical Physics \| 1 \| \| \| Classical Physics \| 5 \|	{"extraction_info": {"found_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": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9532005190849304, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/34142?sort=newest	## Projection of Borel set from $R^2$ to $R^1$ ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hello This should be easy to prove but i have no idea how to do it: If $X \subseteq \mathbb{R}^2$ is borel then $f(X)$ is borel where $f(x,y) = x$ Thanks Tobias - 2 It's so easy that it's wrong, one would say. :) – Mark Schwarzmann Aug 2 2010 at 21:49 True if you replace "Borel" by "$F_\sigma$", since compact sets will project to compact sets. (But only because $\mathbb R$ is $\sigma$-compact.) – Goldstern May 23 2011 at 15:32 ## 4 Answers This is false; take a look at http://en.wikipedia.org/wiki/Analytic_set for a quick introduction. For details, look at Kechris's book on Classical Descriptive Set Theory. There you will find also some information on the history of this result, how it was originally thought to be true, and how the discovery of counterexamples led to the creation of descriptive set theory. - 2 In fact the general analytic subset of $\mathbb{R}$ is the projection of some $G_\delta$ set in $\mathbb{R} \times \mathbb{R}$. – Gerald Edgar Aug 1 2010 at 22:30 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Since this error of Lebesgue has come up on MO a few times, it may be of interest to see his mistaken argument in its entirety. It is on pp. 191--192 of "Sur les fonctions representables analytiquement" in J. de math. pures et appl. (1905). Lebesgue calls Borel sets "B-measurable sets", and he builds them from closed intervals (or their cartesian products, in higher dimensions) by what he calls operations I and II, which are countable union and intersection. In my translation it reads: I wish to prove that, if $E$ is B-measurable, then so is its projection. This is evident when $E$ is an interval, because then $e$ [the projection] is one too. But every B-measurable set comes from intervals by applications of operations I and II, which are preserved by projection, so the proposition is established. That's it. Evidently he just didn't think about the projection of an intersection. - Great specimen for "Common false beliefs" collection! – Victor Protsak Aug 3 2010 at 0:09 I once tracked down the exact mistake Lebesgue made in his published "proof" that the projection of a Borel set in the plane is a Borel set in the line. It came down to his claim that if `$\{A_n\}$` is a decreasing sequence of subsets in the plane with intersection $A$, the the projected sets in the line intersect to the projection of $A$. Of course this is nonsense. Lebesgue knew projection didn't commute with countable intersections, but apparently thought that by requiring the sets to be decreasing this would work. (My answer is a slight expansion of Edgar's comment alluded to above.) - 1 Thanks for your explanation of that. I think everyone in the area knows about this "mistake" but I've never heard any sort of explanation of the cause before. – Carl Mummert Aug 2 2010 at 22:02 This was a famous mistake made by Lebesgue (see also Gerald Edgar's answer to this MO question). Suslin showed that a plane Borel set exists whose projection is not a Borel set. See the references to the original article by Suslin and related works 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": 16, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9547387361526489, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/90669-limit-epsilon-delta-proof-print.html	# Limit epsilon/delta proof Printable View • May 26th 2009, 11:13 PM Roam Limit epsilon/delta proof http://img40.imageshack.us/img40/39/20688555.gif Show that the limit as (x,y) -> 0 doesn't exist. Attempt: To prove it, I must use the definition of limit (the epsilon and delta thing). Well, if the limit exists, and is 1, given any $\epsilon >0$ we can find a $\delta >0$ such that $\sqrt{x^2 + y^2} < \delta$ if $\|f(x,y) -1\| < \epsilon$ perhaps we can ty $\epsilon = 1/2$ I'm not sure how to complete this proof to conclude that $\lim_{(x,y) \to (0,0)} f(x,y)$ doesn't exist... some help with this is very much appreciated. • May 27th 2009, 02:56 AM Roam Am I on the right track? ... I need some help finishing this proof. All times are GMT -8. The time now is 02:55 AM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9265873432159424, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/88184-antidifferentiation.html	# Thread: 1. ## Antidifferentiation I was working on some new homework and I'm having a hard time antidifferentiating problems like these: $\int_2^5 (3-(6/x))^2dx$ For this one, I tried multiplying it out and got ${\int _2^5 (9-36/x+36/x^2)}dx$ and tried to use that to find the integral, but I ran into the problem of having to differentiate $36/x^2$ Another problem that I can't do is $x={\int_0^y\sqrt{9t^2+6t}dt}$ for when y is greater than or equal to 1 and less than or equal to 5. On this problem, I have to find the length of the curve. I know the process, but I can't do it if I can't antidifferentiate that equation. 2. Do you mean you were having a problem finding the antiderivative of $\frac{36}{x^{2}}$? $\int 36x^{-2}\,\,dx= \frac{36x^{-2+1}}{-1} = -\frac{36}{x}$. With your second problem, complete the square and use trig substitution.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9731478691101074, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/39922-related-rates-word-problem.html	1Thanks • 1 Post By topsquark # Thread: 1. ## related rates word problem A street light is at the top of a 17 foot tall pole. A woman 6 ft tall walks away from the pole with a speed of 8 ft/s along a straight path. How fast is the tip of her shadow moving when she is 45 feet from the base of the pole? 2. I found a very similar problem on the internet that I know will help. It's here and it's Example 6. 3. Originally Posted by keemariee A street light is at the top of a 17 foot tall pole. A woman 6 ft tall walks away from the pole with a speed of 8 ft/s along a straight path. How fast is the tip of her shadow moving when she is 45 feet from the base of the pole? Always draw a diagram!! Now we can use similar triangles to get $\frac{x+y}{17}=\frac{y}{6}$ Now lets multiply th equation by 17 to get $x+y=\frac{17}{6}y \iff x=\frac{11}{6}y \iff y=\frac{6}{11}x$ Now we know that the tip of the shawdow will be moving at the rate $\frac{dx}{dt}+\frac{dy}{dt}$ Take the derivative of $y=\frac{6}{11}x$ with respect to time gives $\frac{dy}{dt}=\frac{6}{11}\frac{dx}{dt}$ You should be able to finish from here. 4. Hello, keemariee! With these "streetlight" problems, we must be very careful. Sometimes they ask how fast the tip of the shadow is moving. Sometimes they ask how fast how fast the shadow is growing. A street light is at the top of a 17 foot tall pole. A woman 6 ft tall walks away from the pole with a speed of 8 ft/s along a straight path. How fast is the tip of her shadow moving when she is 45 feet from the base of the pole? Code: ``` A * \| * \| * \| * C 17 \| * \| \| * \| 6\| * \| \| * B ------------------------------* E : y D x-y : : - - - - - - x - - - - - - - - :``` The streetlight is: $AB = 17$ The woman is: $CD = 6$ Let $y = BD.\quad\frac{dy}{dt} \,= \,8\text{ ft/sec}$ Let $x \:=\:BE\qquad\text{ then: }DE \:=\: x-y$ We have: . $\Delta CDE \sim \Delta ABE\quad\Rightarrow\quad \frac{x-y}{6} \:=\:\frac{x}{17}\quad\Rightarrow\quad x \:=\:\frac{17}{11}y$ Differentiate with respect to time: . $\frac{dx}{dt} \:=\:\frac{17}{11}\!\cdot\!\frac{dy}{dt}$ Since $\frac{dy}{dt} \,=\,8\!:\;\;\frac{dx}{dt}\:=\:\frac{17}{11}(8) \:=\:\frac{136}{11} \:=\:12\frac{4}{11}\text{ ft/sec}$ 5. ## Re: related rates word problem Originally Posted by Soroban Hello, keemariee! With these "streetlight" problems, we must be very careful. Sometimes they ask how fast the tip of the shadow is moving. Sometimes they ask how fast how fast the shadow is growing. Code: ``` A * \| * \| * \| * C 17 \| * \| \| * \| 6\| * \| \| * B ------------------------------* E : y D x-y : : - - - - - - x - - - - - - - - :``` The streetlight is: $AB = 17$ The woman is: $CD = 6$ Let $y = BD.\quad\frac{dy}{dt} \,= \,8\text{ ft/sec}$ Let $x \:=\:BE\qquad\text{ then: }DE \:=\: x-y$ We have: . $\Delta CDE \sim \Delta ABE\quad\Rightarrow\quad \frac{x-y}{6} \:=\:\frac{x}{17}\quad\Rightarrow\quad x \:=\:\frac{17}{11}y$ Differentiate with respect to time: . $\frac{dx}{dt} \:=\:\frac{17}{11}\!\cdot\!\frac{dy}{dt}$ Since $\frac{dy}{dt} \,=\,8\!:\;\;\frac{dx}{dt}\:=\:\frac{17}{11}(8) \:=\:\frac{136}{11} \:=\:12\frac{4}{11}\text{ ft/sec}$ Hi all, I know it's an old problem (since 2008...) but can someone show me how did you get this $\Delta CDE \sim \Delta ABE\quad\Rightarrow\quad \frac{x-y}{6} \:=\:\frac{x}{17}\quad\Rightarrow\quad x \:=\:\frac{17}{11}y$ I know it should be something really simple but I can't see it, please 6. ## Re: related rates word problem Originally Posted by dokrbb Hi all, I know it's an old problem (since 2008...) but can someone show me how did you get this $\Delta CDE \sim \Delta ABE\quad\Rightarrow\quad \frac{x-y}{6} \:=\:\frac{x}{17}\quad\Rightarrow\quad x \:=\:\frac{17}{11}y$ I know it should be something really simple but I can't see it, please You don't know how to prove they are similar triangles, or how to get the equation? I'll assume the equation. The two similar triangles are CDE and ABE. So if we take the ratio of any two sides of CDE you can write a "similar" expression for ABE. For example, look at the sides x - y and 6 in CDE. These are the horizontal "leg" and vertical leg of CDE. So we know that the ratio we want from ABE will involve the horizontal (x) and vertical legs. Thus $\frac{\text{horizontal leg}}{\text{vertical leg}} = \frac{x - y}{6} = \frac{x}{17}$ -Dan 7. ## Re: related rates word problem Originally Posted by topsquark You don't know how to prove they are similar triangles, or how to get the equation? I'll assume the equation. The two similar triangles are CDE and ABE. So if we take the ratio of any two sides of CDE you can write a "similar" expression for ABE. For example, look at the sides x - y and 6 in CDE These are the horizontal "leg" and vertical leg of CDE. So we know that the ratio we want from ABE will involve the horizontal (x) and vertical legs. Thus $\frac{\text{horizontal leg}}{\text{vertical leg}} = \frac{x - y}{6} = \frac{x}{17}$ -Dan No, I actually got it, the ratio we use thanks to the fact that we have two similar triangles, my question is at the 12th grade algebra level , how we actually get $x = \frac{17}{11}y$ from $\frac{x - y}{6} = \frac{x}{17}$ 8. ## Re: related rates word problem Originally Posted by dokrbb No, I actually got it, the ratio we use thanks to the fact that we have two similar triangles, my question is at the 12th grade algebra level , how we actually get $x = \frac{17}{11}y$ from $\frac{x - y}{6} = \frac{x}{17}$ sorry for the stupid question - I figured it out	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 27, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9239094853401184, "perplexity_flag": "middle"}
http://www.physicsforums.com/showpost.php?p=3732680&postcount=2	View Single Post Quote by qwirky64 I'm stuck on what I'm sure is a very simple problem. I'm trying to calculate predicted output for my share trading model. For example, for a strategy the probability of it winning on each trade may be 0.4 and hence a loss is 0.6. But for each trade it wins, my account balance is increased by 5% and each loss it loses 2%. How do I calculate how much I would have won or lost after n amount of trades? Can you break it down to on average each trade its a win or loss of x% so I can just use a simple compound interest formula to calculate expected profit after n amount of trades? Your net gain/loss n trades is $\delta_G np - \delta_L n(1-p)$ so for 10 trades you have: 0.05 (10)0.4 -0.02(10)0.6= 0.08 This is a net gain per trade of 0.008 (0.8%). The formula for compound interest is $N(t)=N_{0}e^{rt}$ where r is the rate and t is the time in units over which the rate is calculated.	{"extraction_info": {"found_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}, "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}, "english_confidence": 0.9536528587341309, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/82665-solved-inequality-if-x-real-why-does-mean-equation-has-real-roots.html	# Thread: 1. ## [SOLVED] Inequality. If x is real, why does it mean this equation has real roots? Hiya. There's an example in a book that I can follow the algebra of, but I don't quite understand the reasoning of one of the steps. The underlined part is the one and only part I don't understand. It would be great if someone could explain it some more. Here's what the book has: Q: If x is real, find the possible values of the function: $\frac{x^2}{x+1}$ A: Use $y = \frac{x^2}{x+1} \Rightarrow x^2-yx-y=0$ Since x is real, the roots of this equation are real, so $b^2-4ac \ge 0$ Then there's a bit of algebra to get: $y\le-4$ or $y\ge0$ The algebra's fine and I get that if the roots are real, then $b^2-4ac \ge 0$, but I don't see why x being real means that that equation must have real roots. If anyone could help me get some more understanding of this, intuitive or otherwise, it'd be greatly appreciated. Thanks. 2. By roots I assume they mean "values of y". Since x is real there exist integers m and n such that x=n/m. $y = \frac{x^2}{x+1} = \frac{\frac{n^2}{m^2}}{\frac n m + 1}=\frac{n^2}{nm+m^2}=\frac a b$ where $a=n^2$ and $b=nm+m^2$ and a,b are both integers. Therefore, y must be real. 3. Originally Posted by Kiwi_Dave Since x is real there exist integers m and n such that x=n/m. I'm not sure but I think that would only be true if the statement was "Since x is rational...". But x being real doesn't necessarily mean it can be represented by a fraction. Because x could be irrational at the same time as being real. If someone could confirm or deny that, it'd be appreciated. Thanks for the input either way, it's interesting to see people's approach to the problem. Would love to see more. 4. You're absolutely right. 5. Originally Posted by tleave2000 Q: If x is real, find the possible values of the function: $\frac{x^2}{x+1}$ A: Use $y = \frac{x^2}{x+1} \Rightarrow x^2-yx-y=0$ Since x is real, the roots of this equation are real, so $b^2-4ac \ge 0$.... You have defined x to be a real number. To solve the quadratic equation, you will apply the Quadratic Formula. This will give you an expression for x (in terms of y). Since x is defined as being real, then you cannot have a negative inside the square root in the Formula. Therefore the "b^2 - 4ac" part has to be non-negative. 6. Thanks that's good. Although I think it might have been a restatement of the part I had a problem with rather than quite explaining it, it might have helped me to specify my question better: Why does x being real force the discriminant of $x^2-yx-y=0$ to be positive? The domain (x) has been real for all quadratic equations that I've dealt with so far, but I've found plenty of ones with no real roots. So why does this one have to have real roots, just because x is real? 7. Originally Posted by tleave2000 Why does x being real force the discriminant of $x^2-yx-y=0$ to be positive? If the discriminant is negative, will the square root evaluate to a real number? Since x is real, and since $x\,=\, \frac{-b\, \pm\, \sqrt{b^2\, -\, 4ac}}{2a}$, can you have a value in the square root that is negative? 8. Originally Posted by tleave2000 Why does x being real force the discriminant of $x^2-yx-y=0$ to be positive? The domain (x) has been real for all quadratic equations that I've dealt with so far, but I've found plenty of ones with no real roots. So why does this one have to have real roots, just because x is real? $x$ is a root of the equation and we have defined $x$ to be real. So any roots of the equation must be real. We do not consider values of $y$ that produce non-real roots, for then $x$ would not be real. Consequently, we take $b^2-4ac\geq0.$ 9. Originally Posted by Reckoner $x$ is a root of the equation and we have defined $x$ to be real. Ok I think I get it now, thanks : ).	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 23, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9634426832199097, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/43146/list	## Return to Answer 2 added 182 characters in body In the notation of Time scale calculus, indefinite the ordinary calculus derivative df/dt and the forward difference operator $\Delta f$ are both written as $f^\Delta$. Indefinite sums and indefinite integrals are both written as $\int{f(t)\Delta t}$ and called indefinite integrals. The context would say $\mathbb{T}=\mathbb{Z}, \mathbb{T}=\mathbb{R}$ or other $\mathbb{T}\subset\mathbb{R}$. 1 [made Community Wiki] In the notation of Time scale calculus, indefinite sums and indefinite integrals are both written as $\int{f(t)\Delta t}$ and called indefinite integrals. The context would say $\mathbb{T}=\mathbb{Z}, \mathbb{T}=\mathbb{R}$ or other $\mathbb{T}\subset\mathbb{R}$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9169594049453735, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/11729?sort=newest	## Are there any nonlinear solutions to f(x+1) - f(x) = f'(x)? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) (Asked by bcross at math.iuiui.edu on the Q&A board at JMM) Are there any nonlinear solutions to $f(x+1) - f(x) = f'(x)$? - Shouldn't that be iupui instead of iuiui? – Jonas Meyer Jan 14 2010 at 7:55 There's a thread about this somewhere on the Art of Problem Solving forums, and I remember a tricky solution involving an exponential a^x satisfying a - 1 = log a, but I'm forgetting some important detail. – Qiaochu Yuan Jan 14 2010 at 13:44 If you look for exponential solutions, you get that equation. However, $a-1 \gt= ln a$ with equality at $a=1$, which doesn't make for an interesting $a^x$. – Douglas Zare Jan 14 2010 at 14:09 What I remember someone doing in the thread I referred to is that they proved the existence of a complex value of a and did something like taking the real part. Again, I think I'm forgetting an important detail. – Qiaochu Yuan Jan 14 2010 at 15:05 check also mathoverflow.net/questions/40275/… and mathoverflow.net/questions/39853/… – Pietro Majer Nov 15 2011 at 14:47 ## 3 Answers This is an elaboration of Qiaochu Yuan's prior comment: there are complex solutions (in fact, infinitely many) to $e^t-1 = t$, and then $e^{tx}$ is a solution. - 4 And if you don't want complex solutions to the original problem, just take real and imaginary parts... they are also solutions. – Gerald Edgar Jan 14 2010 at 15:13 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Yes, there exist nonlinear solutions. Multiplying by $e^{x+1}$ and setting $g(x):=e^x f(x)$ transforms the question into finding a solution to $g(x+1)=eg'(x)$ not of the form $e^x(ax+b)$. Start with any $C^\infty$ function on $\mathbb{R}$ whose Taylor series centered at $0$ and $1$ are identically $0$, but which is nonzero somewhere inside $(0,1)$. Restrict it to $[0,1]$. Let $g(x)$ on $[0,1]$ be this. Using $g(x+1):=eg'(x)$ for $x \in [0,1]$ extends $g(x)$ to a $C^\infty$ function $g(x)$ on $[0,2]$, which can then be extended to $[0,3]$, and so on. In the other direction, use $g(x) := \int_0^x e^{-1} g(t+1) dt$ to define $g(x)$ for $x \in [-1,0]$, and then for $x \in [-2,-1]$, and so on. These piece together to give a $C^\infty$ function $g(x)$ on all of $\mathbb{R}$. The corresponding $f(x)$ satisfies $f(0)=0$ and $f(1)=0$ but is not identically $0$, so it is not linear. - Perhaps I'm confused, but if g is all nonnegative on [0,1], it seems to me that you are getting a discontinuity at 0. – Douglas Zare Jan 14 2010 at 9:08 Douglas Zare, how is that? The limit from the left at 0 is clearly 0=g(0), whether or not g is nonnegative on [0,1] – Jonas Meyer Jan 14 2010 at 9:36 The limit from the left is the average value of g on $[0,1]$ divided by $e$, no? – Douglas Zare Jan 14 2010 at 9:39 No. For z a a very small positive number, g(-z) is $- e^{-1} \int_{1-z}^1 g(u) du$, which approaches zero as z goes to 0. (I've just rewritten Bjorn's formula to not have any hidden negatives.) – David Speyer Jan 14 2010 at 14:27 Ok, now I understand. Thanks. I misread the integral, and incorrectly assumed that these functions would be 0 at -1. – Douglas Zare Jan 14 2010 at 14:56 Theorem 1 in [Sugiyama, Shohei. On the existence and uniqueness theorems of difference-differential equations. Kōdai Math. Sem. Rep. 12 1960 179--190. MR0121552] (which you can probably get from here) gives an existence and uniqueness theorem which provides non-linear solutions on finite intervals. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 40, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9204773306846619, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/163646-real-valued-random-varaible-please-chech-my-definition.html	Thread: 1. Real valued Random Varaible - Please chech my definition I am trying to understand Random Variables. Please do let me know if my definiton is correct - or I have got things mixed up. 1. Let $\Omega$ be the sample space i.e. set of all possible outcomes of an experiment; This can be countable or uncountable. 2. $\omega$ be the sample point i.e. a specific outcome of the experiment A real valued random variable is a Function, $X:\Omega \rightarrow \mathbb{R}$ And that is it !! Informally for every $\omega \in \Omega$, there is a unique $X(\omega) \in \mathbb{R}$ Is this definition correct / exact ? Reason of my confusion is that some books define random variable as "a funciton which maps event in the given sample space to a real number." - I think this definition is wrong. Any comments? I am trying to understand two concepts 1. E(X) - Expected value of X 2. E(X\|A) - Conditional expectaion under event A I think in 2 above, we have not changed the way X was defined on the sample space, only thing that changed is probabiltiy measure - Is this understanding correct? 2. Hello, It's ok for the definition of $\Omega$. For the definition of a real valued random variable, it's correct for the first part : it's a function from $\Omega$ to $\mathbb R$. But I don't agree with the second part : for $\omega$, there is a unique $X(\omega)$. It's just a function. If you want to go further, you'll have to talk about measurability of the random variable. I don't know if you know this though. I am trying to understand two concepts 1. E(X) - Expected value of X 2. E(X\|A) - Conditional expectaion under event A I think in 2 above, we have not changed the way X was defined on the sample space, only thing that changed is probabiltiy measure - Is this understanding correct? No, in general, A will be another random variable, not an event. At least that's how conditional expectation works. E(X\|A) will be another random variable. You can keep the probability measure, because the sample space isn't much modified. In a non-formal way The thing that changes is that in the "information" of X, we only keep the parts that are "related" to A. E(X\|A) is viewed as the orthogonal projection of X in $L^2(A)$ I guess this is a bit abstract, but since I don't know what your background is, I just hope you will find your answer among all this stuff 3. Oops. I didn't follow your "orthogonal projection of X" part. Guess need to do some reading. Also I always thought conditional expectation is always wrt to an event. I have basic idea of analysis and am trying to learn these things on my own. Any good reference you can suggest plz - formal but not too technical. For e.g. I do not know measure theory - so can I still pick this stuff? 4. If you can please answer this questions - Consider a probability space - $(\Omega, F, P)$ Question - so is something like E(X\|A) defined? Or this is something which is not defined. To be precise - 1. By X, I mean a real random variable. 2. By A, I mean an event i.e. an element of sigma-algebra, F 5. Okay, I'll try to answer more precisely.. But I'm not a teacher so I don't have all the experience of explaining this stuff Originally Posted by aman_cc Oops. I didn't follow your "orthogonal projection of X" part. Guess need to do some reading. It doesn't matter then, it was sort of an extra. Also I always thought conditional expectation is always wrt to an event. That's for conditional probability. It is possible to define a conditional expectation wrt an event, but in advanced probability, it is "generalized" to conditional expectation wrt a random variable (more precisely, the sigma-algebra generated by the random variable) (quote]I have basic idea of analysis and am trying to learn these things on my own. Any good reference you can suggest plz - formal but not too technical. For e.g. I do not know measure theory - so can I still pick this stuff?[/QUOTE] Well there is a time when you have to know some basics of measure theory. I don't know books, especially in English since English is not my native tongue. I've heard of Durrett's Probability : Theory and examples. But I can't say if it's the best. 6. Originally Posted by aman_cc If you can please answer this questions - Consider a probability space - $(\Omega, F, P)$ Okay, so you have the F in your definitions. Then I'll add some things to my first post. What is a real valued random variable X ? Yes it's a function, but a function from the probability space $(\Omega,\mathcal F,P)$ to $(\mathbb R,\mathcal E$ $\mathcal F$ is a sigma-algebra related to the space $\Omega$, same for $\mathcal E$, a sigma-algebra related to $\mathbb R$. X will map from $\Omega$ to $\mathbb R$, but it is also measurable. Which means that for any $E\in\mathcal E$, the set $X^{-1}(E)=\{\omega\in\Omega,X(\omega)\in E\}$ will belong to $\mathcal F$. When you do exercises, you don't need this though ! But I'm giving you the complete definition of a real valued random variable. Question - so is something like E(X\|A) defined? Or this is something which is not defined. To be precise - 1. By X, I mean a real random variable. 2. By A, I mean an event i.e. an element of sigma-algebra, F Yes it is defined. It is defined as $\displaystyle \frac{1}{P(A)}\cdot \int_A X dP=\frac{1}{P(A)} \cdot E[X\mathbf{1}_{X\in A}]$. 7. Thanks a lot. I have followed a few things - but guess I need to do a bit more reading before I can move ahead on this subject. Thanks very much for your help !	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 25, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9468768835067749, "perplexity_flag": "head"}
http://mathoverflow.net/questions/42207/maximum-average-value-within-a-rectangular-bounding-box/42256	## Maximum average value within a rectangular bounding box ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The goal is to expedite detection using the sliding window approach. In other words, an object classifier is known and I need to find where the possible locations of this object are in an image. This is a general problem in object detection. We are given an intensity map (positive values - could be detection scores) on a rectangular grid (e.g. MxN intensity image). The goal is to find the bounding box (i.e. a rectangle of size mxn, where m and n are known and greater than 1) where the average intensity in the bounding box is maximum among all boxes. The brute-force algorithm would be to evaluate this value for all boxes (i.e. linear filtering) and take the maximum. Are there any more efficient ways to do this? how about approximate algorithms? This question was for one choice of m and n. But now there's a finite set of m's and n's that I needed to find optimal locations for in the image. Do I rerun the previous algorithm for each choice of m and n independently or can I do something more efficiently? Thanks - 2 I might not understand the question. Why isn't a 1x1 box around the maximum value always the answer? – Greg Muller Oct 14 2010 at 20:57 1 This question seems more appropriate for stackoverflow. Even if resubmitted there, you should probably give more details about the problem. For example: are m and n given? If not, you could just take the spot in the grid with the largest intensity and call that a 1x1 box with the largest average intensity. If they are given, what else have you tried? Dynamic programming somehow seems natural. – Noah Stein Oct 14 2010 at 21:09 -1: vote to close; seems like homework to me that the student needs to think about and solve on their own. Or of course the simple way to do it if $m$ and $n$ are given is to perform a $2$-dimensional convolution of the given $M$ by $N$ matrix with the $m \times n$ sized matrix consisting of all ones. The resulting convolution, called it $X$, has a maximal entry or entries identifying the positionning of the $m$ by $n$ matrix. This looks like homework for an image processing type of class. I'd vote to close it if I had closing vote power number of magic points. – sleepless in beantown Oct 14 2010 at 21:15 What algorithms have you tried? What's your motivation behind this problem? Is it homework? Please take a look at the FAQ's and consider that a different forum might be more appropriate for this question, and that even on a different forum you might need to clarify and explain the problem more explicitly. – sleepless in beantown Oct 14 2010 at 21:23 i updated the question, so any extra comments would be appreciated. Thanks! – Bernard Oct 15 2010 at 3:35 ## 4 Answers Given your intensity map $M_{ij}$, create an array $S_{ij}$ with $S_{ij} = \sum_{k \le i} \sum_{l \le j} M_{kl}$ Now you just need to look for a maximum value of $S_{ij} - S_{i-m,j} - S_{i,j-n} + S_{i-m,j-n}$ and you can use the same array $S$ for any values of $m,n$. - There was a similar problem in the IOI - the programming version of IMO. The solution above was suggested as the optimal solution. – Per Alexandersson Dec 24 2010 at 14:05 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. If m=n=1 you must use brute force (i.e. look at each cell once). If m=M-1 and n=N-1 then you only need to look at the outer boundary (the cells not forced to be in the window) That would locate the optimal window although its average value would require looking inside, In general you can first compute the differences a(i+m,j)-a(i,j) and also a(i,j+n)-a(i,j) and then with m (or n) additions of these find the effect of moving the window down (or left) by a step. I haven't thought about trying to find the optimal of each size. (As commented, the optimal among all the sizes would be 1x1) - i updated the question, so any extra comments would be appreciated. Thanks! – Bernard Oct 15 2010 at 3:35 While your question as stated seems ill-formed (as commenters point out, the optimal solution is a 1x1 box around the maximum value), let's assume that there's some other constraint (MxN is too large for example, or there's a lower bound on the size of the query box) that makes this approach infeasible. In that case, at least for approximations, and if MxN is very large, an approach based on $\epsilon$-approximations might work for you. Roughly speaking, you're trying to do range querying over a set of ranges that are "well structured" (formally, have low VC-dimension) and you'd like to extract a sample of the input so that range queries on this sample approximate range queries on the real input. It turns out that a random sample of the input of size roughly $O(1/\epsilon^2 \log 1/\epsilon)$ would suffice to estimate the ranges within error $\epsilon$. Algorithmically, then you merely implement your expensive procedure on this small sample, and presumably that would be more efficient. - The question is really asking about cross-correlation using matched filters. The known classifiers for objects in the set $S={s_1,s_2,... , s_x}$ are $m \times n$ matrices or subimages of sizes $m_i \times n_i$ for each object $s_i$. So now, $m$ and $n$ are no longer necessarily the same for the different objects being classified. The matched filters can be convolved in 2-dimensions one at a time with the image $M$ and peaks ascertained in order to find candidate locations in which the target objects $s_i$ occur. Once you have the cross-correlation, you look for local maxima in order to select the candidate locations. If there is any redundancy between the targets to be classified (e.g. $s_a$ and $s_b$ are very similar looking), then you might be able to get around having to do cross-correlation for each target with the entirety of the image. But if there are no similarities between the different targets to be detected, it is very likely that there is no better solution than doing full convolutions and cross-correlations for each object classifer $s_i$ over the image $M$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 30, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9464617371559143, "perplexity_flag": "middle"}
http://www.haskell.org/haskellwiki/Simple_to_complex	# Simple to complex ### From HaskellWiki It is generally a good idea to construct complex functions from simpler ones rather than making simple functions special cases of complex functions. Obviously the latter strategy does not work alone, thus using it means mixing it with the former one. That leads no longer to a clear hierarchy of functions, but to an entangled graph of dependencies. ## 1 Functions The lazy evaluation feature of Haskell tempts you to ignore the principle of building complex functions on simpler ones. See "Eleven reasons to use Haskell as mathematician" where it is presented as an advantage that lazy evaluation automatically simplifies the computation of a cross product of two 3D vectors if only a single component of the result is requested. However, computing a single component of a cross product means computing the determinant of a $2\times2$-matrix, which is certainly useful of its own. So instead of using laziness for reducing the cross product to a determinant, the better concept is clearly to write a function for computing the $2\times2$-determinant and invoke this three times in the implementation of the cross product. ## 2 Types Another bad example is the numerical linear algebra package MatLab. Its type hierarchy starts at complex valued matrices from which you can build more complex types. That is, there are no simpler types, no real valued matrices, complex numbers, real numbers, integers nor booleans. They have to be represented by complex valued matrices. This is not very natural since some operations like transcendent powers are not easily ported to matrices. That is many operations must check at run-time, whether the input values have appropriate properties, e.g. being $1\times1$-matrices. Actually, some kinds of integers and booleans (logical values) have been added later, but they interact weirdly with MatLab's base type. The mistake, that the language designers of MatLab made, is the following: They thought MatLab would remain a special purpose language for numerical linear algebra forever, so they decided to implement only the most complex type that would ever be encountered in this domain. As MatLab grew, they extended the program to fit new needs, image import and export, GUI programming and so on, but the initial decision for a universal type didn't scale well. It's not a good idea to mimic this in Haskell. Start with simple types and build complexer ones out of it. Make sure that you use fancy type constructs not in the core of a library, if at all. Move them as far as possible to leaf modules. ## 3 Type class methods There are many methods in Data.List that can be generalised to other data structures, like map , filter , foldr , (++) . Some people wonder why they aren't replaced by their generalized counterparts, namely the methods in the type classes. First of all, there are didactic reasons. Higher order functions like foldr are hard to grasp for newbies. That's why they often stick to manual encoding of recursion. (See Avoid explicit recursion) Now imagine, that Data.List.foldl is no longer exposed, but only its generalization, namely the method Data.Foldable.foldl where not only the element types and the working function are generalized away, but there is even no longer a particular data structure you are working on. Nothing concrete any longer, only abstract things. The best thing to explain this abstract function, is to use an example. E.g. sum xs = List.foldl (+) 0 xs . Now, this requires to have a function List.foldl . Close to didactic reasons there are reasons of code comprehensibility. If you know you are working on lists, map and filter tell the reader, that you are working on lists. The reader of a program doesn't need to start human type inference to deduce this. The function return may mean Just , (:[]) or Right depending on the monad. If you are working on data of type IO (Maybe a) , the program reader must deduce whether return means the one of IO or the one of Maybe . If you only want to process a specific type, tell it the reader of your program. Last but not least, there are the reasons of type safety and communication with the compiler. If the compiler knows, that you are programming for a specific type, it will check types stronger and will give more precise error messages. Imagine you generalize Data.List.filter to ```filter :: (MonadPlus m) => (a -> Bool) -> m a -> m a filter p m = m >>= \a -> if p a then return a else mzero``` This is a nice thing, but it should not replace Data.List.filter . The type inference of the compiler will fail, if you write too general code. Say, you are in GHCi, have the above definition of filter and you write `Prelude> filter Char.isUpper (return 'a')` To what monad this shall be specialised? Maybe ? IO ? List ? Rely on type defaulting? You will certainly have to add a type annotation. These problems are not only of interest for the standard libraries, but should be taken into account, when designing custom type classes. See the notes on Slim instance declarations.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9085736870765686, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/93691/elementary-proof-of-mathbbq-zeta-n-cap-mathbbq-zeta-m-mathbbq-whe/93694	# Elementary proof of $\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m)=\mathbb{Q}$ when $\gcd(n,m)=1$. In an answer to another question I used the fact that $\mathbb{Q}(\zeta_m)\subseteq \mathbb{Q}(\zeta_n)$ if and only if $m$ divides $n$ (here $\zeta_n$ stands for a primitive $n$th root of unity, Edit: and neither $m$ nor $n$ is twice an odd number; see KCd comments below). More generally, one can show that $\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m)=\mathbb{Q}$ when $\gcd(n,m)=1$. The only proof of this fact that comes to mind uses facts about discriminants of cyclotomic extensions, and the fact that every non-trivial number field extension over $\mathbb{Q}$ ramifies at least at one prime (see, for instance, Washington, "Introduction to Cyclotomic Fields", Chapter 2, Proposition 2.4). Since the original question that I was trying to answer was somewhat elementary, I was left wondering if there are more elementary proofs of the fact $$\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m)=\mathbb{Q}, \text{ when } \gcd(n,m)=1.$$ By "elementary proof" here I mean some proof that does not involve algebraic number theory results about discriminants, or ramification of primes in rings of integers of number fields. Can anyone think of an elementary proof? Thanks! - The fact stated in the first sentence is incorrect due to problems at 2: try $m=2$ and $n=1$. You need to use the standard indexing convention that $m$ and $n$ are not two times an odd number. – KCd Mar 16 '12 at 1:27 1 A more general formula is ${\mathbf Q}(\zeta_m) \cap {\mathbf Q}(\zeta_n) = {\mathbf Q}(\zeta_{(m,n)})$. This is a lot more subtle than the corresponding composite field formula ${\mathbf Q}(\zeta_{m}){\mathbf Q}(\zeta_n) = {\mathbf Q}(\zeta_{mn})$ because the second formula is actually true over any base field in place of ${\mathbf Q}$ (well, one over which those primitive roots of unity exist) but the first formula is false: $F(\zeta_m) \cap F(\zeta_n)$ need not be $F(\zeta_{(m,n)})$. For example, ${\mathbf F}_3(\zeta_5) \cap {\mathbf F}_3(\zeta_7)$ is ${\mathbf F}_9$, not ${\mathbf F}_3$. – KCd Mar 16 '12 at 1:31 That composite field formula in my previous comment should have been ${\mathbf Q}(\zeta_m){\mathbf Q}(\zeta_n) = {\mathbf Q}(\zeta_{[m,n]})$. – KCd Mar 16 '12 at 1:39 @KCd, thanks! I've changed the first paragraph accordingly. – Álvaro Lozano-Robledo Mar 16 '12 at 2:05 The examples I gave using finite fields, to show $F(\zeta_5) \cap F(\zeta_7)$ need not equal $F$, can be bootstrapped to characteristic 0 using $p$-adics: let $F$ be ${\mathbf Q}_3$ instead of ${\mathbf F}_3$. – KCd Mar 16 '12 at 3:16 ## 2 Answers $\newcommand{\Q}{\mathbf Q}$This answer assumes that we're willing to use $[\Q(\zeta_n) : \Q] = \varphi(n)$, which is not obvious. A freely available reference is Milne's notes, Lemma 5.9 and Theorem 5.10. Since $n$ and $m$ are coprime, the proof you gave shows that $\mathbf Q(\zeta_{nm}) = \mathbf Q(\zeta_n, \zeta_m)$, and the totient function satisfies $\varphi(nm) = \varphi(n)\varphi(m)$. Now $$[\mathbf Q(\zeta_{nm}) : \mathbf Q] = [\mathbf Q(\zeta_n, \zeta_m) : \mathbf Q(\zeta_n)][\mathbf Q(\zeta_n) : \mathbf Q].$$ If we had $\mathbf Q(\zeta_n) \cap \mathbf Q(\zeta_m) \neq \mathbf Q$ then the degree of $\zeta_m$ over $\mathbf Q(\zeta_n)$ would be less than $\varphi(m)$. - This is very nice, but it is worth pointing out that here you are using the fact that the degree of $\mathbb{Q}(\zeta_n)$ is $\varphi(n)$, which is not trivial by any means. – Álvaro Lozano-Robledo Dec 23 '11 at 16:02 @ÁlvaroLozano-Robledo True! But that fact seems necessary to talk about these fields at all. I will add a reference. – Dylan Moreland Dec 23 '11 at 16:13 I agree, but the proofs of $\mathbb{Q}(\zeta_{mn})=\mathbb{Q}(\zeta_n,\zeta_m)$ and $[\mathbb{Q}(\zeta_n):\mathbb{Q}]=\varphi(n)$ are at two completely different levels. In my original question, I guess I was wondering if there is a proof at the very elementary level of $\mathbb{Q}(\zeta_{mn})=\mathbb{Q}(\zeta_n,\zeta_m)$. The answer is probably no, and probably, at the very least, one needs to use the fact about the degree? – Álvaro Lozano-Robledo Dec 23 '11 at 16:21 @ÁlvaroLozano-Robledo I wish I knew! My only argument for this calculation being necessary is that I think I would have read about an alternative if it existed, which isn't a very good argument at all. Sorry to have misinterpreted your question :-/ – Dylan Moreland Dec 23 '11 at 19:05 @ÁlvaroLozano-Robledo: Dear Alvaro, I don't think it is unreasonable to use the fact about degrees to prove this. As KCd points out in his comment above, the fact you are asking about is not true if you replace $\mathbb Q$ by other fields. He gives counterexamples involving finite fields and $p$-adic fields, but here is another counterexample involving a number field: If $F = \mathbb Q(\sqrt{3})$, then $F(\zeta_4) = F(\zeta_3) \neq F$. If we tried to apply Dylan's argument, it would break down at the point where we had to compute $[F(\zeta_{12}):F]$; we have that ... – Matt E Mar 16 '12 at 3:55 show 2 more comments This is a complement to Dylan's answer: we prove that $\mathbb{Q}(\zeta_{mn})=\mathbb{Q}(\zeta_m,\zeta_n)$. We have $(\zeta_m)^{mn}=1^n=1$ and $(\zeta_n)^{nm}=1^m=1$, and so the inclusion $\mathbb{Q}(\zeta_{mn})\supseteq\mathbb{Q}(\zeta_m,\zeta_n)$ is clear. Conversely, $\zeta_{mn}^m$ is a primitive $n$-th root of unity, and $\zeta_{mn}^n$ is a primitive $m$-th root of unity, which gives the reverse inclusion. I believe this is as elementary as it gets. - Yes, I agree that his is elementary. The simplicity of this step is what got me wondering whether there is also such an elementary proof of $\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m)=\mathbb{Q}$, when $\gcd(n,m)=1$. – Álvaro Lozano-Robledo Dec 23 '11 at 19:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 61, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9472336769104004, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/3323/why-isnt-the-nelson-siegel-model-arbitrage-free/3360	# Why isn't the Nelson-Siegel model arbitrage-free? Assume $X_t$ is a multivariate Ornstein-Uhlenbeck process, i.e. $$dX_t=\sigma dB_t-AX_tdt$$ and the spot interest rate evolves by the following equation: $$r_t=a+b\cdot X_t.$$ After solving for $X_t$ using $e^{tA}X_t$ and Ito and looking at $\int_0^T{r_s\;ds}$, it turns out that $$\int_0^T{r_s\;ds} \sim \mathcal{N}(aT+b^{T}(I-e^{-TA})A^{-1}X_0,b^{T}V_Tb)$$ where $V_t$ is the covariance matrix of $\int_0^T(I-e^{-(T-u)A})A^{-1}\sigma dB_u$. This gives us the yield curve $$y(t)=a+\frac{b^{T}(I-e^{-tA})A^{-1}X_0}{t}+\frac{b^{T}V_tb}{2t}$$ and by plugging in $A= \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \\ \end{pmatrix}$ we finally arrive at $$y(t)=a+\frac{1-e^{-\lambda t}}{\lambda t}C_0+e^{-\lambda t}C_1+\frac{b^{T}V_tb}{2t}.$$ The formula above without $\frac{b^{T}V_tb}{2t}$ is known as the Nelson-Siegel yield curve model. Could somebody clarify why neglecting $\frac{b^{T}V_tb}{2t}$ leads to arbitrage opportunities? So I am essentially asking the following question: Why is the above model (with $\frac{b^{T}V_tb}{2t}$) arbitrage free? - Where did you get this derivation of the NS model? I'm only familiar with Filipovic (1999) (onlinelibrary.wiley.com/doi/10.1111/1467-9965.00073/abstract) showing that arbitrage is possible in the standard NS formulation and the solution of Christensen et al. (2009) (frbsf.org/publications/economics/papers/2007/wp07-20bk.pdf) who add a yield adjustment term. – Bob Jansen May 6 '12 at 11:53 – Artiom Fiodorov May 6 '12 at 12:07 ## 2 Answers The original Nelson Siegel paper describes a parsimonious model of the term structure using only four or three (if $\lambda_t$ is fixed). Filipovic (1999) proves that this model can never be used in a arbitrage free context, paraphrasing the abstract: We introduce the class of consistent state space processes, which have the property to provide an arbitrage-free interest rate model when representing the parameters of the Nelson–Siegel (NS) family. (We show that) there exists no nontrivial interest rate model driven by a consistent state space Itō process. This problem is solved by Christensen et al. (2009). They provide some ODE's which must hold for an AFNS and write that the "key difference between Dynamic NS and AFNS is the maturity dependent yield-adjustment term" and show how to solve for this term. They show that the yield adjustment term is empirically small and that their model fares well in out-of-sample prediction, consistently outperforming, for example, the canonical $A_0(3)$ model (of Duffee 2002). - Let $P(t,T)$ be the time-$t$ price of the zero-coupon bond expiring at $T$. The no-arbitrage condition forces: $$e^{-\int_0^tr_sds}P(t,T)=\mathbb{E}[e^{-\int_0^Tr_sds}\|\mathcal{F_t}],$$ where $\mathcal{F_t}$ is the filtration of the Brownian motion up to time $t$. Note that the expression on the right is a martingale by the tower property of expectations, so by the First Theorem of Asset pricing, there is no arbitrage. It immediately follows that $$P(t,T)=\mathbb{E}[e^{-\int_t^Tr_sds}],$$ which will further result in the yield curve specified as above (with $V_t$ term). Therefore neglecting the covariance term could result in arbitrage. In fact, I was told that there is a proof showing it is not indeed arbitrage-free, but I am not going to go into that. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9107919335365295, "perplexity_flag": "middle"}

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.