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http://physics.stackexchange.com/questions/43973/applying-angular-velocity-to-a-rotation-matrix	Applying angular velocity to a rotation matrix I have a very simple question. In our project we store an object's orientation as a 3x3 matrix which holds the orthonormal base of that object's local space. For instance if the object is aligned with the axis of the world, then its orientation matrix is : ````1 0 0 0 1 0 0 0 1 ```` Now we store the angular velocity as a vector that contains values in rad/s around each of the world's axis. For example if the object should rotate around the world's y axis at 20 rad/s, then the angular velocity is ````0 20 0 ```` My question is then, say we want to apply this velocity over 1 second to the orientation matrix, how would we do this? - 2 Answers The rate of change of the rotation matrix $\boldsymbol R$ is $$\dot{\boldsymbol{R}} = \vec{\omega} \times \boldsymbol{R}$$ where $\vec\omega\times = \begin{pmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \end{pmatrix}$ is the cross product operator in 3x3 form. So if the position of a point A rotating on a frame is given by $\vec{r}_A = \vec{r} + \boldsymbol{R} \,\vec{p}$ where $\vec{p}$ is the position in local coordinates and $\vec{r}$ the location of the frame, the kinematics are $$\dot{\vec{r}}_A = \dot{\vec{r}} + \dot{\boldsymbol R}\,\vec{p} \\ \vec{v}_A = \vec{v} + \vec{\omega} \times (\boldsymbol{R}\,\vec{p})$$ $$\dot{\vec{v}}_A = \dot{\vec{v}} +\dot{\vec{\omega}} \times (\boldsymbol{R}\,\vec{p}) + \vec \omega \times (\dot{\boldsymbol{R}}\,\vec{p}) \\ \vec{a}_A = \vec{a} + \vec{\alpha}\times(\boldsymbol{R}\,\vec{p}) + \vec\omega\times\vec\omega\times(\boldsymbol{R}\,\vec{p})$$ - From Goldstein, chapter 4 eqn 4-92', for a finite rotation the change $\boldsymbol{\Delta r}$ caused by rotating a vector $\boldsymbol{r}$ through an angle $\Phi$ about a direction defined by a unit vector $\boldsymbol{n}$ ($\Phi$ positive for a counter-clockwise rotation), to a final position $\boldsymbol{r'}$ is given by: $$\boldsymbol{\Delta r} = \boldsymbol{r'-r} = [\boldsymbol{n} (\boldsymbol{n \cdot r} ) - \boldsymbol{r} ] [ 1 - \cos \Phi ] + (\boldsymbol{n \times r}) \sin \Phi$$ Relating this equation to your situation, with an angular momentum vector $\boldsymbol{\omega}$ applied for a time $t$: $$\boldsymbol{n} = \frac{\boldsymbol{\omega}}{\|\boldsymbol{\omega}\|} , \Phi = \|\boldsymbol{\omega}\| \, t$$ Since your angular velocity vector $\boldsymbol{\omega}$ is given with respect to the fixed world coordinate system $\boldsymbol{x_1, x_2, x_3}$, it's straightforward to compute components in that basis: $$\Delta r'_i = \left[ n_i \sum_{j=1}^3\left(n_j r_j \right) - r_i \right][1-\cos \Phi] + \sum_{j,k=1}^3 \left( \epsilon_{ijk} n_j r_k \right) \sin \Phi$$ In particular, these formulas apply to the basis vectors of your object's local space, so you can use them to evolve your storage matrix. For the specific example you mention, $\boldsymbol{n = x_2}$ and $\Phi =20$ rad/s $* 1$s $= 20$ rad. Plugging in the 3 object basis vectors $\boldsymbol{x'_1, x'_2, x'_3}$ (which in this case are just equal to the world basis vectors $\boldsymbol{x_1, x_2, x_3}$ respectively) for $\boldsymbol{r}$, you can see that: • The $\boldsymbol{x'_2}$ body axis is unchanged (as expected for a vector parallel to the angular velocity) • $\boldsymbol{x'_1}$ and $\boldsymbol{x'_3}$ rotate around a unit circle (again as expected). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "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": 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": 0.889476478099823, "perplexity_flag": "head"}
http://mathoverflow.net/questions/17140/why-is-the-harmonic-oscillator-so-important-pure-viewpoint-sought-how-to-moti	## Why is the harmonic oscillator so important? (pure viewpoint sought). How to motivate its role in Getzler’s work on Atiyah-Singer? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm in the process of understanding the heat equation proof of the Atiyah-Singer Index Theorem for Dirac Operators on a spin manifold using Getzler scaling. I'm attending a masters-level course on it and using Berline, Getzler Vergne. While I think I could bash my way through the details of the scaling trick known as `Getzler scaling', I have little to no intuition for it. As I understand it, one is computing the trace of the heat kernel of the ("generalized") Laplacian associated to a Dirac operator. The scaling trick reduces the problem to one about the ("supersymmetric" or "generalized") harmonic oscillator, whose heat kernel is given by Mehler's formula. I am repeatedly assured that the harmonic oscillator is a very natural and fundamental object in physics, but, being a `pure' analyst, I still can't sleep at night. What reasons are there for describing the harmonic oscillator as being so important in physics? Why/how might Getzler have thought of his trick? (Perhaps the answer to this lies in the older proofs?) Is there a good way I could motivate an attempt to reduce to the harmonic oscillator from a pure perspective? (i.e. "It's a common method from physics" is no good). I'm looking for: "Oh it's simplest operator one could hope to reduce down to such that crucial property X still holds since Y,Z"...or..."It's just like the method of continuity in PDE but a bit different because..." Thanks. - ## 3 Answers What reasons are there for describing the harmonic oscillator as being so important in physics? The harmonic oscillator tends to show up when you're expanding a potential function around non-degenerate critical points. The simplest example is a physical system described by a map $t \mapsto \phi(t) \in \mathbb{R}$. If the energy function for this system has the form $E(\phi) = \frac{1}{2}\|\dot{\phi}\|^2 + V(\phi)$, with $V$ bounded below, then the lowest energy states are going to be of the form $\phi_0(t) = \phi_0$, where the constant $\phi_0$ is a minimum of $V$, hence a critical point. So, if your map $\phi$ never deviates too much from $\phi_0$ and $\phi_0$ is a non-degenerage critical point, you can approximate the energy function by $E(\phi) = \|\frac{1}{2}\dot{\phi}\|^2 + V(\phi_0)+\frac{1}{2}V''(\phi_0)(\phi-\phi_0)^2$. In other words, the harmonic oscillator potential describes small disturbances around "generic" minima of an energy function. This situation comes up all the time in physics. For example: it shows up in Witten's Supersymmetry & Morse Theory paper, which I think would have been well-known to people working on topology and analysis in the 1980s. - 3 More generally, I think the word "harmonic oscillator" essentially means "pure non-zero quadratic function". And these are the first approximation for most systems of interest, or at least for any system where we have any chance of making an approximation. – Theo Johnson-Freyd Mar 5 2010 at 0:02 ### 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. I too am but a mere graduate student trying to sort through some of these same issues, but I might have some helpful insight. I'll let you be the judge. The basic idea behind the heat equation proof of the index theorem is to extract the right term in the asymptotic expansion for the heat kernel and then appeal to the McKean - Singer formula. According to my understanding the original strategy for doing this was to realize that the index is a cobordism invariant and thus it would suffice to do enough explicit calculations on generators for the cobordism group until all free parameters are fixed; as it turns out the complex projective spaces are a good choice. That's exactly what was done. I believe - and I really hope someone will correct me if I'm wrong since I haven't gotten my hands dirty myself - that the required calculation really boils down to dealing with the quantum mechanical harmonic oscillator when you work it out for $CP_2$. If this is correct, then the first hint that the quantum mechanical harmonic oscillator is important came from a very fundamental example. But I think a more analytic answer is also possible. Let's say instead of working with a Dirac operator acting on smooth sections of the spinor bundle you instead just consider the usual scalar Laplacian acting on functions. What happens if you imitate the heat kernel proof in this much less subtle context? You wind up reproving Weyl's asymptotic formula for the eigenvalues of the Laplacian. In essence this calculation amounts to rescaling the spacial variable so that your operator is deformed into the constant coefficient operator obtained by freezing coefficients. The basic idea of the Getzler calculus is to rescale both the spacial variable and the Riemannian metric in a compatible way - this rescaling deforms the Clifford algebra into the exterior algebra (thereby making Clifford multiplication act like an order one operator) and hence the Dirac operator into a polynomial coefficient operator. What polynomial coefficient operator is it? We have reached the limit of my ability to motivate things any further, but the answer is the quantum mechanical harmonic oscillator operator. I of course have no idea whether or not the physical significance of this operator can be accounted for according to a similar rescaling argument. I should also mention that the quantum mechanical harmonic oscillator makes no obvious appearance in the original global proofs of the index theorem. It does, however, make a non-obvious appearance via Bott Periodicity which can be proven essentially using Mehler's formula. Nigel Higson and Eric Guenter wrote a very readable paper explaining most of the details of this proof entitled something like "K-Theory and Group C* Algebras". You can find it on Nigel's website, www.math.psu.edu/higson. The last thing I will say is that I found Getzler, Berligne, and Verne to be a pretty tough way to penetrate this material. The style pays off in some of the later material, but I think I would have had a lot of trouble learning the heat kernel proof of the index theorem for the first time from that book. You might try John Roe's book "Elliptic Operators, Topology, and Asymptotic Methods" instead of or as a supplement. I hope this has been helpful! - Here is a fairly physicsy point of view. You are calculating the supertrace of the heat operator, and because this is magically time independent, you can calculate it in the small $t$ limit. The heat operator is the time-evolution operator (propagator) for a certain theory (supersymmetric quantum mechanics) and the supertrace amounts to considering it only on loops (I am skipping some details here!). The small $t$ approximation means that you can consider loops that are nearly constant. The small $t$ limit is similar to the semiclassical (small $\hbar$) limit, so we will borrow our intuition from there. In the semiclassical approximation you consider small perturbations from a classical solution and expand the action in powers of the perturbation. Because classical solutions are critical points of the action, you get no linear term, so the first interesting term in quadratic in the perturbation. So in the small $t$ limit you can ignore everything but the quadratic part of the action. A quadratic action is basically just a harmonic oscillator. I believe (and here I am expressing an intuition I have no concrete argument for!) that Getzler's rescaling amounts to converting between the small $t$ limit and the small $\hbar$ limit, since these are not quite the same thing. I also think that you are actually getting a magnetic term, not a quadratic potential (which is the harmonic oscillator), but there is a standard physics trick for going from one to the other. More generally, if you are in any physics situation where you have a minimum/critical point of a potential/action, it is probably going to make sense to expand it in powers of distance from that point, and the first term of the expansion will be quadratic and therefore a harmonic oscillator. So you can always model systems sufficiently close to their equilibrium by a system of harmonic oscillators. In quantum theory it is only a little bit of a lie to say that the only things we can solve are quadratic theories (harmonic oscillator) so nearly all understanding starts from there and builds out. The magic of the Index Theorem, and most places where QFT give you wonderful results in mathematics, is that this approximation is somehow exact. -	{"extraction_info": {"found_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": 0.9385585784912109, "perplexity_flag": "head"}
http://nrich.maths.org/6575	### Flexi Quads A quadrilateral changes shape with the edge lengths constant. Show the scalar product of the diagonals is constant. If the diagonals are perpendicular in one position are they always perpendicular? ### A Knight's Journey This article looks at knight's moves on a chess board and introduces you to the idea of vectors and vector addition. ### 8 Methods for Three by One This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different? Which do you like best? # Fix Me or Crush Me ##### Stage: 5 Challenge Level: Imagine that you have a pair of vectors ${\bf F}$ and ${\bf Z}$ {\bf F}=\pmatrix{1\cr 1 \cr 0}\quad {\bf Z}=\pmatrix{0\cr 1 \cr 1} Can you construct an example of a matrix $M$, other than the identity, which leaves ${\bf F}$ fixed, in that $M{\bf F}={\bf F}$? How many such matrices can you find? Which is the simplest? Which is the most complicated? Can you construct an example of a matrix $N$, other than the zero matrix, which crushes ${\bf Z}$ to the zero vector ${\bf 0}$, in that $N{\bf Z}={\bf 0}$? How many such matrices can you find? Which is the simplest? Which is the most complicated? Can you find a matrix which leaves ${\bf F}$ fixed and also crushes ${\bf Z}$? Can you find any (many?) vectors fixed or crushed by the following matrices? Give examples or convincing arguments if no such vectors exist. M = \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{pmatrix}, \begin{pmatrix} 1&2&3\\ 2&3&4\\ 3&4&5\\ \end{pmatrix}, \begin{pmatrix} 1&-2&1\\ 1&1&0\\ -2&1&-2\\ \end{pmatrix} Very hard extension: Imagine that you are given a vector ${\bf F}$ and a vector ${\bf Z}$. Investigate whether you will be able to make a matrix $M$ which both fixes ${\bf F}$ and crushes ${\bf Z}$. NOTES AND BACKGROUND Matrices are used to represent transformations of vectors; vectors and matrices are usually studied together as an inseparable pair. Although matrices and the rules of matrix multiplication might seem abstract upon first encounter, they are actually very natural and encode in an entirely meaningful way notions of symmetry and transformation. This problem allows you to explore the effects matrix multiplication has on various vectors. The eigenvectors of a matrix are those vectors whose direction is unchanged by the action of the matrix. The kernel of a matrix is the set of vectors which are squashed to zero. Both concepts are of fundamental importance in higher-level algebra and its applications to science. 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": 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": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9211958646774292, "perplexity_flag": "middle"}
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Infinity	All Science Fair Projects Science Fair Project Encyclopedia for Schools! Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. Infinity For the automobile brand, see Infiniti. Infinity is a word carrying a number of different meanings in mathematics, philosophy, theology and everyday life. Popular usage is often not in accordance with the term's more technical meanings. In theology, for instance in the work of Duns Scotus, the infinity of God carries the sense not so much of quantity (leading to the question, quantity of what?) but of unconstrainedness. In philosophy, infinity can be attributed to space and time, as for instance in Kant's first antinomy. In popular culture, we find Buzz Lightyear's rallying cry, "To infinity -- and beyond!" which might also be called the rallying cry of set theorists considering large cardinals, which are quantitative infinities, defining the number of things in a collection, so large that they cannot be proven to exist in the ordinary mathematics of Zermelo-Fraenkel plus Choice (ZFC), and which might be so large they embody a contradiction. In mathematics, some articles relevant to the subject can be found at limit (mathematics), aleph number, class (set theory), Dedekind infinite, large cardinal, Russell's paradox, hyperreal numbers, projective geometry, extended real number and absolute infinite. In philosophy and theology, one can investigate the Ultimate, the Absolute, God, and Zeno's paradoxes. For a discussion about infinity and the physical universe, see Universe. Contents History Ancient view of infinity The traditional view derives from Aristotle: "... it is always possible to think of a larger number: for the number of times a magnitude can be bisected is infinite. Hence the infinite is potential, never actual; the number of parts that can be taken always surpasses any assigned number." [Physics 207b8] This is often called "potential" infinity; however there are two ideas mixed up with this. One is that it is always possible to find a number of things that surpasses any given number, even if there are not actually such things. The other is that we may quantify over finite numbers without restriction. For example "For any integer n, there exists an integer m > n such that P(m)". The second view is found in a clearer form by medieval writers such as William of Ockham: "Sed omne continuum est actualiter existens. Igitur quaelibet pars sua est vere existens in rerum natura. Sed partes continui sunt infinitae quia non tot quin plures, igitur partes infinitae sunt actualiter existentes." (But every continuum is actually existent. Therefore any of its parts is really existent in nature. But the parts of the continuum are infinite because there are not so many that there are not more, and therefore the infinite parts are actually existent.) The parts are actually there, in some sense. However, on this view, no infinite magnitude can have a number, for whatever number we can imagine, there is always a larger one: "there are not so many (in number) that there are no more". Aquinas also argued against the idea that infinity could be in any sense complete, or a totality [reference]. Views from the Renaissance to modern times Galileo (during his long house arrest in Siena after his condemnation by the Inquisition) was the first to notice that we can place an infinite set into one-to-one correspondence with one of its proper subsets (any part of the set, that is not the whole). For example, we can match up the "set" of even numbers {2, 4, 6, 8 ...} with the natural numbers {1, 2, 3, 4 ...} as follows: 1, 2, 3, 4, ... 2, 4, 6, 8, ... It appeared, by this reasoning, as though a set which is naturally smaller than the set of which it is a part (since it does not contain all the members of that set) is in some sense the same size. He thought this was one of the difficulties which arise when we try, "with our finite minds", to comprehend the infinite. "So far as I see we can only infer that the totality of all numbers is infinite, that the number of squares is infinite, and that the number of their roots is infinite; neither is the number of squares less than the totality of all numbers, nor the latter greater than the former; and finally the attributes "equal", "greater", and "less", are not applicable to infinite, but only to finite, quantities." [On two New Sciences, 1638] The idea that size can be measured by one-to-one correspondence is today known as Hume's principle, although Hume, like Galileo, believed the principle could not be applied to infinite sets. Locke, in common with most of the empiricist philosophers, also believed that we can have no proper idea of the infinite. They believed all our ideas were derived from sense data or "impressions", and since all sensory impressions are inherently finite, so too are our thoughts and ideas. Our idea of infinity is merely negative or privative. "Whatever positive ideas we have in our minds of any space, duration, or number, let them be never so great, they are still finite; but when we suppose an inexhaustible remainder, from which we remove all bounds, and wherein we allow the mind an endless progression of thought, without ever completing the idea, there we have our idea of infinity ... yet when we would frame in our minds the idea of an infinite space or duration, that idea is very obscure and confused, because it is made up of two parts very different, if not inconsistent. For let a man frame in his mind an idea of any space or number, as great as he will, it is plain the mind rests and terminates in that idea; which is contrary to the idea of infinity, which consists in a supposed endless progression." (Essay, II. xvii. 7., author's emphasis) Famously, the ultra-empiricist Hobbes tried to defend the idea of a potential infinity in the light of the discovery by Evangelista Torricelli, of a figure (Gabriel's horn) whose surface area is infinite, but whose volume is finite. Not reported, this motivation of Hobbes came too late as curves having infinite length yet bounding finite areas were known much before. Such seeming paradoxes are resolved by taking any finite figure and stretching its content infinitely in one direction; the magnitude of its content is unchanged as its divisions drop off geometrically but the magnitude of its bounds increases to infinity by necessity. Potentiality lies in the definitions of this operation, as well-defined and interconsistent mathematical axioms. A potential infinity is allowed by letting an infinitely-large quantity be cancelled out by an infinitely-small quantity. Modern philosophical views Modern discussion of the infinite is now regarded as part of set theory and mathematics, and generally avoided by philosophers. An exception was Wittgenstein, who made an impassioned attack upon axiomatic set theory, and upon the idea of the actual infinite, during his "middle period". (see also Logic of antinomies) "Does the relation m = 2n correlate the class of all numbers with one of its subclasses? No. It correlates any arbitrary number with another, and in that way we arrive at infinitely many pairs of classes, of which one is correlated with the other, but which are never related as class and subclass. Neither is this infinite process itself in some sense or other such a pair of classes ... In the superstition that m = 2n correlates a class with its subclass, we merely have yet another case of ambiguous grammar." (Philosophical Remarks § 141, cf Philosophical Grammar p. 465) Unlike the traditional empiricists, he thought that the infinite was in some way given to sense experience. "... I can see in space the possibility of any finite experience ... we recognise [the] essential infinity of space in its smallest part." "[Time] is infinite in the same sense as the three-dimensional space of sight and movement is infinite, even if in fact I can only see as far as the walls of my room." "... what is infinite about endlessness is only the endlessness itself." Mathematical infinity Infinity in real analysis In real analysis, the symbol $\infty$, called "infinity", denotes an unbounded limit. $x \rightarrow \infty$ means that x grows beyond any assigned value, and $x \rightarrow -\infty$ means x is eventually less than any assigned value. Points labeled $\infty$ and $-\infty$ can be added to the real numbers as a topological space, producing the two-point compactification of the real numbers. Adding algebraic properties to this gives us the extended real numbers. We can also treat $\infty$ and $-\infty$ as the same, leading to the one-point compactification of the real numbers, which is the real projective line. Projective geometry also introduces a line at infinity in plane geometry, and so forth for higher dimensions. Infinity is represented by "∞". Infinity is often used not only to define a limit but as if it were a value in the extended real numbers in real analysis; if f(t) ≥ 0 then • $\int_{0}^{1} \, f(t) dt \ = \infty$ means that f(t) does not bound a finite area from 0 to 1 • $\int_{0}^{\infty} \, f(t) dt \ = \infty$ means that the area under f(t) increases without bound as its upper bound increases • $\int_{0}^{\infty} \, f(t) dt \ = 1$ means that the area under f(t) approaches 1, though its upper bound increases limitlessly. Infinity symbol It is unclear what the exact origins of the infinity symbol are, but the most commonly cited explanation says that it is derived from the shape of a Möbius strip twisted to look like $\infty$ since if one were to stand on a the surface of a Möbius strip, one could walk along it forever. In addition, the lemniscate curve looks like the infinity symbol, and its name is derived from the Latin lemniscus, meaning "ribbon," which is what a Möbius strip can be made of. The symbol itself is also sometimes referred to as the lemniscate. This explanation may not be correct however since the symbol had been in use to represent infinity even before August Ferdinand Möbius had discovered the Möbius strip. John Wallis is often credited for introducing this symbol through his book Arithmetica Infinitorum, which was published more than a century before Möbius was born. Conjectures of why Wallis chose this symbol say that he derived it from the Etruscan numeral for 1000, which looked somewhat like CIƆ and is sometimes used to mean "many," or that he derived it from the Greek letter ω (omega), the last letter in the Greek alphabet. The Dustbin of History by Paul JJ Payack recounts how the infinity symbol came into widespread use in mathematics. Infinity in set theory A different type of "infinity" are the ordinal and cardinal infinities of set theory. Georg Cantor developed a system of transfinite numbers, in which the first transfinite cardinal is aleph-null ($\aleph_0$), the cardinality of the set of natural numbers. This modern mathematical conception of the quantitative infinite developed in the late nineteenth century from work by Cantor, Gottlob Frege, Richard Dedekind and others, using the idea of collections, or sets. Dedekind's approach was essentially to adopt the idea of one-to-one correspondence as a standard for comparing the size of sets, and to reject the view of Galileo (which derived from Euclid) that the whole cannot be the same size as the part. An infinite set can simply be defined as one having the same size as at least one of its "proper" parts; this notion of infinity is called Dedekind infinite. Cantor defined two kinds of infinite numbers, the ordinal numbers and the cardinal numbers. Ordinal numbers may be identified with well-ordered sets, or counting carried on to any stopping point, including points after an infinite number have already been counted. Generalizing finite and the ordinary infinite sequences which are maps from the positive integers leads to mappings from ordinal numbers, and transfinite sequences. Cardinal numbers define the size of sets, meaning how many members they contain, and can be standardized by choosing the first ordinal number of a certain size to represent the cardinal number of that size. The smallest ordinal infinity is that of the positive integers, and any set which has the cardinality of the integers is countably infinite. If a set is too large to be put in one to one correspondence with the positive integers, it is called uncountable. Cantor's views prevailed and modern mathematics accepts actual infinity. Certain extended number systems, such as the hyperreal numbers, incorporate the ordinary (finite) numbers and infinite numbers of different sizes. Our intuition gained from finite sets breaks down when dealing with infinite sets. One example of this is Hilbert's paradox of the Grand Hotel. Mathematics without infinity Leopold Kronecker rejected the notion of infinity and began a school of thought in the philosophy of mathematics called finitism, which led to the philosophical and mathematical school of mathematical constructivism. Use of infinity in common speech In common parlance, infinity is often used in a hyperbolic sense. For example, "The movie was infinitely boring, but we had to wait forever to get tickets." In video games, "infinite lives" and "infinite ammo" usually mean a truly never-ending supply of lives and ammunition. Another accurate usage is an infinite loop in computer programming, a conditional loop construction whose condition always evaluates to true. As long as there is no external interaction (such as switching the computer off, or the heat death of the universe), the loop will continue to run for all time. In practice however, most programming loops considered as infinite will halt by exceeding the (finite) number range of one of its variables. See halting problem. The number Infinity plus 1 is also used sometimes in common speech. Physical infinity In physics, approximations of real numbers are used for continuous measurements and natural numbers are used for discrete measurements (i.e. counting). It is therefore assumed by physicists that no measurable quantity could have an infinite value, for instance by taking an infinite value in an extended real number system (see also: hyperreal number), or by requiring the counting of an infinite number of events. It is for example presumed impossible for any body to have infinite mass or infinite energy. There exists the concept of infinite entities (such as an infinite plane wave) but there are no means to generate such things. Likewise, perpetual motion machines theoretically generate infinite energy by attaining 100% efficiency or greater, and emulate every conceivable open system; the impossible problem follows of knowing that the output is actually infinite when the source or mechanism exceeds any known and understood system. This point of view does not mean that infinity cannot be used in physics. For convenience sake, calculations, equations, theories and approximations, often use infinite series, unbounded functions, etc., and may involve infinite quantities. Physicists however require that the end result be physically meaningful. In quantum field theory infinities arise which need to be interpreted in such a way as to lead to a physically meaningful result, a process called renormalization. Infinity in cosmology An intriguing question is whether actual infinity exists in our physical universe: Are there infinitely many stars? Does the universe have infinite volume? Does space "go on forever"? This is an important open question of cosmology. Note that the question of being infinite is logically separate from the question of having boundaries. The two-dimensional surface of the Earth, for example, is finite, yet has no edge. By walking/sailing/driving straight long enough, you'll return to the exact spot you started from. The universe, at least in principle, might have a similar topology; if you fly your space ship straight ahead long enough, perhaps you would eventually revisit your starting point. Infinity in science fiction The Hitchhiker's Guide to the Galaxy contains the following definition of infinity: "Bigger than the biggest thing ever and then some, much bigger than that, in fact really amazingly immense, a totally stunning size, real 'Wow, thats big!' time. Infinity is just so big that by comparison, bigness itself looks really titchy. Gigantic multiplied by colossal multiplied by staggeringly huge is the sort of concept we are trying to get across here." External links • A Crash Course in the Mathematics of Infinite Sets, by Peter Suber. From the St. John's Review, XLIV, 2 (1998) 1-59. The stand-alone appendix to Infinite Reflections, below. A concise introduction to Cantor's mathematics of infinite sets. • Infinite Reflections, by Peter Suber. How Cantor's mathematics of the infinite solves a handful of ancient philosophical problems of the infinite. From the St. John's Review, XLIV, 2 (1998) 1-59. 03-10-2013 05:06:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 12, "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": 0.9419181942939758, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/60158/derham-cohomology	## DeRham cohomology ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The Poincare lemma fails in positive characteristic, since pth powers vanish under differention. My question is : is there still some kind of resolution of the local system k by considering some twisted DeRham complex. - 2 Have you heard about crystalline cohomology? (see en.wikipedia.org/wiki/Crystalline_cohomology) It is not really a resolution of the local system $k$, though... – Leo Alonso Mar 31 2011 at 8:27 1 It fails even in char.0, because the Poincare lemma works only formally locally and the de Rham complex is not exact in positive degrees. On the other hand, the cohomology of any constant sheaf vanishes on an irreducible topological space, so anyway taking resolutions of $k$ would not lead no anything of interest. – Piotr Achinger Mar 31 2011 at 13:05 in char o the deRham is a resolution of C , which implies $H(X,C) \cong H_{Dr}(X)$ – chemaida Mar 31 2011 at 15:16 1 chemaida, to be clear, the Poincar\'e lemma holds for holomorphic or $C^\infty$ forms, but it can fail for the algebraic de Rham complex even in characteristic $0$ (which is what Piotr was referring to). Nevertheless the hypercohomology of the algebraic de Rham complex does give the correct answer, although the reasons are more subtle. – Donu Arapura Mar 31 2011 at 16:00 Ok..thank you for your responses..I would be interested in the subtle reason that Donu alludes to or even a reference – chemaida Apr 1 2011 at 7:25 show 1 more comment	{"extraction_info": {"found_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": 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": 0.8976932764053345, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/115877-changing-least-rapidly.html	# Thread: 1. ## Changing least rapidly? At (2,1), I have to find the direction where $f(x,y)=x^2+3xy-y^2$ is increasing fastest. So I did: $\triangledown f(2,1)= \langle 2(2)+3, 3(2)-2 \rangle = \langle 7,4 \rangle$ Next I have to find the rate of growth in this direction, so I did: $\|\|\triangledown f(2,1)\|\| = \sqrt{7^2+4^2} = \sqrt{65}$ Finally I have to find the direction a this point where f is changing "least rapidly." Not sure what to do here... 2. Originally Posted by MathSucker At (2,1), I have to find the direction where $f(x,y)=x^2+3xy-y^2$ is increasing fastest. So I did: $\triangledown f(2,1)= \langle 2(2)+3, 3(2)-2 \rangle = \langle 7,4 \rangle$ Next I have to find the rate of growth in this direction, so I did: $\|\|\triangledown f(2,1)\|\| = \sqrt{7^2+4^2} = \sqrt{65}$ Finally I have to find the direction a this point where f is changing "least rapidly." Not sure what to do here... What do you mean by "changing least rapidly"? A function increases most rapidly in the direction of the gradient, which you calculated, decreases most rapidly in the opposite direction, and has 0 rate of change at right angles to the gradient. Does "has 0 rate of change" qualify as "changes least rapidly"? 3. I'm not quite sure what it means, hence the name of the thread. I assume the magnitude of when it is decreasing fastest is the same as when it is increasing fastest. I was thinking "zero because it isn't moving sideways" but that seems to easy. Or maybe that's the point. I dunno. #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"extraction_info": {"found_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": 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": 0.9663515090942383, "perplexity_flag": "middle"}
http://www.reference.com/browse/cuspidal	Definitions # Cuspidal representation In number theory, cuspidal representations are certain representations of algebraic groups that occur discretely in $L^2$ spaces. The term cuspidal is derived, at a certain distance, from the cusp forms of classical modular form theory. In the contemporary formulation of automorphic representations, representations take the place of holomorphic functions; these representations may be of adelic algebraic groups. When the group is the general linear group $operatorname\left\{GL\right\}_2$, the cuspidal representations are directly related to cusp forms and Maass forms. For the case of cusp forms, each Hecke eigenform (newform) corresponds to a cuspidal representation. In detail, let $G$ be a connected reductive algebraic group over an algebraic number field $F$. Let $mathbb\left\{A\right\}$ be the adele of $F$. Let $Z$ be the center of $G$. Let $omega$ be a Hecke character. The character can be regarded as a character over $Z\left(mathbb\left\{A\right\}\right)$. Let $L^2\left(Z\left(mathbb\left\{A\right\}\right) G\left(F\right) backslash G\left(mathbb\left\{A\right\}\right), omega\right)$ be the space of measurable functions on $G\left(F\right) backslash G\left(mathbb\left\{A\right\}\right)$ satisfying $varphi\left(zx\right) = omega\left(z\right) varphi\left(x\right), z in Z\left(mathbb\left\{A\right\}\right), x in G\left(mathbb\left\{A\right\}\right)$ and $int_\left\{Z\left(mathbb\left\{A\right\}\right) G\left(F\right) backslash G\left(mathbb\left\{A\right\}\right)\right\} \|varphi\left(x\right)\|^2 dx < infty.$ There is a representation $R$ of $G\left(mathbb\left\{A\right\}\right)$ (called right regular representation) on the space $L^2\left(Z\left(mathbb\left\{A\right\}\right) G\left(F\right) backslash G\left(mathbb\left\{A\right\}\right), omega\right)$. The representation is defined by $R\left(g\right)varphi\left(x\right) = varphi\left(xg\right)$. A function in $L^2\left(Z\left(mathbb\left\{A\right\}\right) G\left(F\right) backslash G\left(mathbb\left\{A\right\}\right), omega\right)$ is a cuspidal function if for every parabolic subgroup $P$ of $G$ $$ int_{N(F) backslash N(mathbb{A})} f(nx) dn = 0. for almost every $x$, where $N$ is the unipotent radical of $P$. Denote the space of cuspidal functions by $L^2_0\left(omega\right)$. It is easily shown that the space is stable under the right regular representation. It can be shown that the subspace of cuspidal functions can be decomposed discretely, i.e. it can be written as a direct sum of represntations of $G\left(mathbb\left\{A\right\}\right)$. Those irreducible subrepresentations of $L^2_0\left(omega\right)$ are called cuspidal representations.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 28, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.8738362193107605, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?s=4f969aff1a2fb5ac91b89862dd4e041f&p=4020401	Physics Forums ## Inverse ratio and proportion question I understood that for a direct proportion such as "if 6 mangoes cost $12, then how many dollars worth are 12 mangoes?", we have to first divide 12 by 6 to get the no. of divisions of 6 in 12 i.e. precisely 2 and then as 1 division is worth$12 we multiply the two divisions with $12 and that's how we get$24. What I don't understand is that for an inverse proportion such as "there are a certain no. of pills and for 10 patients the pills will last for 14 days, for 35 patients how long will the pills last?", the no. of days will obviously decrease and hence more the patients less the no. of days, but to calculate the answer we divide and not multiply as we do in direct proportion question such as the one above (we multiplied 2 with 12). Why do we divide to find the answer in inverse proportion? Why can't it be subtract or add or any of those things. What I mean is that what is the logic behind the last division? Thanks! Hey physio and welcome to the forums. The simple answer is how you define the ratio and the units. In the example above we have 6 mango's for 12 dollars. If we want to find mangos per dollar we divide 6:12 by 6 giving us 1:2 or, 1 mango per $2 dollars. Now the key thing in a rate is that you have two quantities: the first one is the independent and the second is dependent. In the above we have 1 mango every two dollars. If we have to standardize our dependent variable, we set it equal to 1 which means give 1:2 (one mango for every two dollars) we divide both parts by 2 which gives 0.5:1 or half a mango per dollar. Now if we have to find dollars per mango we simply invert the mangos per dollar and set the dollar figure to one: specifically we have 0.5:1 for mangos per dollar so we have 1:0.5 dollars per mango and multiplying both sides by two gives us 2:1 or two dollars per mango. The 1:0.5 is best visualized as a fraction like $$\frac{1}{0.5}$$ The logic behind this is that if we want something per something else we get a ratio that corresponds to how many a per b (in this case 6 mangos per$12 dollars) and then we obtain the right side of the ratio to be 1. If we want the opposite rate, we invert the rate (i.e. turn a:b into b:a) and then make the right hand side of the ratio to be 1. It's exactly the same sort of thing when you are trying to convert say kilometres per hour to miles per hour or metres per second: we get our rate in the right form and then make the denominator of the fraction 1 to get "x things per unit". So lets say we have 60 kilometres per hour: we write this as 60:1. Now lets say we want to find it per minute. We know that 60 minutes = 1 hour so we change this to 60:60 in terms of kilometres per minute. Now to find in terms of minutes we standardize the denominator to 1 which means dividing both numbers by 60 giving us 1:1 or one kilometre per minute. Same sort of thing can be done for say turning a kilometre into a metre: we have 60:1 kilometres per hour to 60*1000:1 metres per hour since 1 km = 1000 m. This gives us 60,000:1 or 60,000 metres per hour. Thread Tools \| \| \| \| \|------------------------------------------------------------\|----------------------------------\|---------\| \| Similar Threads for: Inverse ratio and proportion question \| \| \| \| Thread \| Forum \| Replies \| \| \| Precalculus Mathematics Homework \| 2 \| \| \| Calculus & Beyond Homework \| 5 \| \| \| Precalculus Mathematics Homework \| 1 \| \| \| General Math \| 3 \| \| \| Precalculus Mathematics Homework \| 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": 3, "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": 0.8965306878089905, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/2685/what-is-information/3289	# What is information? We're all familiar with basic tenants such as "information cannot be transmitted faster than light" and ideas such as information conservation in scenarios like Hawking radiation (and in general, obviously). The Holographic Principle says, loosely, that information about a volume of space is encoded on its two-dimensional surface in Plank-sized bits. In all these contexts, I can take "information" to mean predictive or postdictive capability, i.e. information is what enables us to state what the outcome of a measurement was or will be (locally). But what is information, exactly? Do we have any kind of microscopic description of it? Is it just a concept and, if so, how can we talk about transmitting it? I suspect this is probably as unanswerable as what constitutes an observer/measurement for wave function collapse, but I'd love to know if we have any formulation of what information is made of, so to speak. If I'm talking nonsense, as I suspect I may be, feel free to point this out. - 2 Actually, you are talking sense and this is a good question ;-) +1 – Marek Jan 10 '11 at 21:02 ## 4 Answers Information is a purely mathematical concept, usually a a characteristic of uncertainty (of a probability distribution function), but can be interpreted in different ways. In the simplest form it is introduced in information theory as a difference between uncertainties of two distributions, with uncertainty being the logarithm of a number of possible equally-probable states of a discrete random variable. For continuous distribution, it can be introduced as a logarithm of an integral. Sometimes introduced proper information - a quantity which differs from negative entropy only by a constant independent of the distribution (this constant can be taken as zero). Thus information is a difference of proper information (difference of negative entropy) of two states. The states are represented by probability distribution functions, thus information is a formal operator of two functions. For continuous distributions (of which discrete case is a variant) proper information of distribution $w$ is $$I[w]=-H(w)=-\int_{-\infty}^{+\infty}w(x)\log(w(x))dx$$ and relative information of $w_2$ compared to $w_1$ is $$I[w_2,w_1]=H(w_1)-H(w_2)=I(w_2)-I(w_1)$$ or $$I[w_2,w_1]=\int_{-\infty}^{+\infty}\log \left(\frac{w_1(x)^{w_1(x)}}{w_2(x)^{w_2(x)}}\right)$$ This operator is not much different from norm or angle in vector spaces. It is just one measure, attributed to members of the space. Compare this with the definition of norm: $$\|\|w\|\|=\sqrt{\int_{-\infty}^{+\infty}w(x)^2dx}$$ distance $$D[w_1,w_2]=\|\|w_1-w_2\|\|=\sqrt{\int_{-\infty}^{+\infty}(w_1(x)-w_2(x))^2dx}$$ angle $$\Phi[w_1,w_2]=\arccos \frac{\int_{-\infty}^{+\infty}w_1(x)w_2(x)dx}{\sqrt{\int_{-\infty}^{+\infty}w_1(x)^2dx}\sqrt{\int_{-\infty}^{+\infty}w_2(x)^2dx}}$$ So think about information as of a mathematical quantity similar to angle. - 2 Any talk of information from a mathematical perspective really needs to mention Shannon entropy. – Noldorin Jan 10 '11 at 21:50 2 To say in simple words, norm answers the question how big something is, angle answers the question how oriented something is and entropy/information answers the question how complex something is. – Anixx Jan 10 '11 at 22:39 2 Welcome to physics.se @anixx. This is quite a remarkable answer. The viewpoint that information is a "mathematical quantity similar to angle" is also at the heart of quantum mechanics. In fact in 1981 Wootters ("Statistical distance and hilbert space", PRD) showed that the "statistical distance" between two sets of observations coincides with the angle between rays of a Hilbert space. Of course none of this would come as a surprise to R. A. Fisher ;) – user346 Jan 10 '11 at 22:46 2 Ah ok. I missed it it seems. You should definitely call it "Shannon entropy" or "Information entropy" to distinguish it from thermodynamic entropy, given this is a physics site. – Noldorin Jan 10 '11 at 22:59 2 @Anixx @space_cadet -to be abit pedantic, you should call what you wrote differential entropy. In Shannon entropy, the random variable is discrete. Differential entropy extends Shannon entropy using probability density functions which are continuous, but these can be tricky and can have values greater than 1. I agree, I like the answer. +1 – Gordon Feb 18 '11 at 4:59 show 6 more comments Since there are already outstanding technical answers to this question, I think we should add some better philosophical underpinnings for you to explore that might help with gaining a better intuitive feel for what information is. Warren Weaver provided an excellent discussion on information theory in 1949 in his paper entitled "Recent Contributions to The Mathematical Theory of Communication". In the paper he breaks down communications problems into three main categories: technical, semantic and effectiveness. He further explains that the concept of information is purely derived to address the technical problem in communications theory. A simple definition of information, provided by Weaver, is that "information is a measure of one's freedom of choice when one selects a message"; or more correctly, the logarithm of that freedom of choice. Information is thus more clearly understood as a the number of combinations of component parts that are available to be chosen arbitrarily. In this sense, on can view it as a measure of randomness associated with a string of letters. A great example is wheel of fortune. When Pat Sajak shows you a the board with the white and green blocks, he as already provided you a lot of information by placing spaces between the white blocks, because he has drastically reduced the number of possible combinations that might be possible to fill in the white blocks. The maximum information (or entropy) of the board with 52 boxes or "trilons" and using 26 letters is $26^{52} = 3.8\times 10^{73}$ combinations or between $244$ and $245$ bits of information in binary. However, if there were only 11 boxes illuminated white, then the actual information of the board has suddenly dropped to $26^{11} = 3.6\times 10^{15}$ combinations giving an actual information content (or entropy) or $51$ to $52$ bits. The relative information is $\dfrac{51}{244} = 0.21$ or 21%. The redundancy is then given by $1 - 0.21 = 0.79$ or 79%. As Vanna flips boxes, she is decreasing the relative entropy and increasing the redundancy to a point where the probability of solving the puzzle becomes very high. So in this sense, information, like entropy, is a measure of uncertainty about the system. Now there are different types of uncertainty, one is the uncertainty associated with the freedom of choice of message, and the other is noise. The uncertainty discussed in wheel of fortune example is due to the freedom of choice. In a noiseless situation, we would expect the word or phrase that Vanna unveils to be exactly the one chosen before the show. In a noisy environment, for instance, one where there is some probability of a crewmember mispelling the word while setting up the blocks, then it is possible that the final word shown is not the one chosen before the show. That uncertainty, or noise, is called equivocation, and is brought in by the environment itself. The distinction between a noisy and noiseless environment is very important. William Tuller in 1949 published a paper "THEORETICAL LIMITATIONS ON THE RATE OF TRANSMISSION OF INFORMATION" that proved that there was no limit in the amount of information that could be transmitted in a noiseless channel. This was why Shannon's paper "Communication in the Presence of Noise" was critical to communication theory in that it properly quantified what noise actually was, and how it affected communication and the transfer of information. Now, before finishing, it should be noted that Hartley in his 1928 paper "Transmission of Information" was the first to really give a modern definition of information and give it quantitative measure. I would recommend reviewing that paper as a starting point. Other significant contributions are made from other scientists, such as Wiener which is best captured in Cybernetics. On a closing note, it is refreshing that the significance of quantum noise is beginning to be discussed, and I hope it continues in the future. - 2 absolutely brilliant – Unassuminglymeek May 22 '11 at 13:09 In short: information contained in a physical system = the number of yes/no questions you need to get answered to fully specify the system. - 1 a very nice and succinct answer +1 – user346 Jan 14 '11 at 5:30 3 +1 but could be improved into "the minimum number of..." IMHO – Tobias Kienzler Apr 4 '11 at 12:35 the information is a dimensionless (unitless) - and, in this sense, "purely mathematical" - quantity measuring how much one has to learn to know something relatively to the point when he doesn't know it, expressed in particular units. Operationally speaking, it is the amount of RAM chips (or their parts) one needs to have so ehat they can remember some knowledge. Of course, by using the word "knowledge", I am just avoiding the word "information", and it is impossible to define any of these terms without any "circular references" because one has to know at least something to be able to define as elementary concepts as knowledge. One bit of information is the knowledge needed to know whether a number that can be 0 or 1 with the same probability turned out (or will turn out) to be 0 or 1. In mathematics, a more natural unit than one bit is one "e-bit" which is such that 1 bit is ln(2) "e-bits". The information needed to distinguish among "N" equally likely alternatives is ln(N) "e-bits". The natural logarithm is always more natural than other logarithms - that's why it's called natural. For example, its derivative equals 1/x, without complicated constants. The formulae for the "Shannon" dimensionless information, assuming any probabilistic distribution, are given above. In physics, every physical system with some degrees of freedom may carry some information. In quantum information, the "alternatives" are usually associated with basis vector of the allowed Hilbert space of states. But in that context, one "bit" of information is usually referred to as "qubit" or "quantum bit" which means that in the real world, the alternatives may also be combined into arbitrary complex linear superpositions, like the postulates of quantum mechanics dictate. In discussions about causality, we mean that the spatially separated objects can't really influence each other. This is guaranteed by the Lorentz symmetry. In field theory, the condition is equivalent to the constraint that the space-like separated fields $\phi(x)$ and $\phi(y)$ commute with each other (or anticommute if both of them are fermionic). Best wishes Lubos - ## protected by Qmechanic♦Mar 10 at 8:26 This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "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": 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": 0.9364728331565857, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/7638/proving-operator-identities-quantum-physics	# Proving Operator identities (Quantum Physics) How would I go about showing: $$\hat{A}^{\dagger} + \hat{B}^{\dagger} = \left( \hat{A} + \hat{B} \right) ^{\dagger}$$ - You wouldn't. Its an axiom. Or perhaps there is some deeper meaning in terms of higher-category-monoidal-bridget-fonda-F theory. – user346 Mar 27 '11 at 8:31 1 It is not an axiom. First you should remind yourself of the definition of the $\dagger$ operation. – Qmechanic♦ Mar 27 '11 at 9:28 1 I figured somebody would have a problem with that statement. I'm just finished fighting one battle and am too tired to start another. So please please, if you disagree by all means throw poop at my statements. I'm just too sleep-deprived to throw any back at present ;) – user346 Mar 27 '11 at 9:41 2 Agree with Qmechanic. It's a property of adjoints. But I am not sure whether or not we should answer this kind (i.e. pure math) of questions here. – Marek Mar 27 '11 at 9:45 Dear @Deepak Vaid. Please be sure that my above comment was mainly aimed at user2789, not you. – Qmechanic♦ Mar 30 '11 at 17:18 show 1 more comment ## 1 Answer $$\left\langle \chi \| \left( A + B \right)^\dagger \| \psi \right\rangle = \left\langle \psi \| (A +B) \| \chi \right\rangle^* =$$ $$=\left\langle \psi \| A \| \chi \right\rangle^* + \left\langle \psi \| B \| \chi \right\rangle^* = \left\langle \chi \| A^\dagger \| \psi \right\rangle + \left\langle \chi \| B^\dagger \| \psi \right\rangle = \left\langle \chi \| (A^\dagger +B^\dagger) \| \psi \right\rangle$$ for all $\|\chi\rangle$, $\|\psi\rangle$. - 3 -1 because bra-ket notation totally confuses the matters here (it's not clear when the operators act to the left and when to the right). – Marek Mar 27 '11 at 10:04 +1, I think it is a totally legitimate answer to this trivial question. An operator is fully determined by its matrix elements with respect to any pair of vectors, here called $\chi$ and $\psi$, and the Hermitian conjugate operator's matrix elements are just given by those of the original one in the opposite order, with $$. I just divided the long equation into two lines and added the missing complex conjugation signs $$. The bracket notation is OK for any space with linear operators, and by convention, the primary action of operators is on the kets - but it also works on the bras! – Luboš Motl Mar 27 '11 at 10:12 1 At any rate, the point of the proof is that the Hermitian conjugation is a linear operation (more precisely antilinear) acting on the space of operators, so if the coefficients in the linear combinations are real, and they're $1,1$ in this case, the terms may be conjugated one-by-one. – Luboš Motl Mar 27 '11 at 10:16 3 @Luboš: the reason bra-ket notation "is OK" for any space with linear operators is because it satisfies lots of useful properties of which this one (i.e. linearity of adjoints) is one case. So the usage of bra-ket notation is kind of circular here. The correct way is to work from the $\left<x, Ay\right> = \left<A^{\dagger}x, y\right>$ where everything is crystal clear. – Marek Mar 27 '11 at 11:17	{"extraction_info": {"found_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": 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": 0.9367566704750061, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/87876/fractional-chromatic-number-of-fullerenes?answertab=oldest	Fractional chromatic number of fullerenes Computations of fractional chromatic numbers this week tell me that for Fullerene Graphs the value is $5/2$. I have computed $100$ of these or more. Is there any theorem that would say this? Any information on formulas for fractional chromatic numbers of families of graphs would be welcome. I am aware that the Kneser graphs $K(a,b)$ have fractional chromatic number $a/b$. For the definition of a Fullerene graph see this MathWorld link. - If you don't get an answer here, you might try MathOverflow (but be sure to mention at each site that you have posted to the other). – Gerry Myerson Dec 3 '11 at 8:50 Are all your fullerenes on 60 vertices? – Chris Godsil Dec 4 '11 at 16:29 The ones I checked are up to 50 vertices. I am told that perhaps things don't get interesting until they are much larger, so maybe there is nothing to this "conjecture" about 5/2. – stan wagon Dec 5 '11 at 1:07 – J. M. Dec 20 '11 at 6:59 1 Answer Any fullerene contains pentagons. Each of the 5 vertices on a pentagon touches 2 others. At most 2 vertices on any pentagon can share a colour since adjacent vertices have different colours. Therefore the chromatic number of a pentagon is 5/2. If some of the vertices of a pentagon are coloured then, if for example a vertex is red and blue, the 2 opposite vertices must be coloured so that one is red and the other is blue. A fullerene can be coloured with 5/2 colours by first colouring a pentagon, then all the adjacent pentagons, then any adjacent to the adjacent pentagons... then any adjacent hexagons, then another pentagon and its adjacent pentagons, then adjacent hexagons etc. - Checking over my database I found that 3 of the 100-200 fullerenes I examined did not have 5/2... So this is intriguing. If your proof, Angela, holds water, then there will be something funny about these 3 cases. I double-checked with others that my fractional chi is correct on these 3. Example: – stan wagon Dec 9 '11 at 16:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.9179272651672363, "perplexity_flag": "middle"}
http://mathhelpforum.com/trigonometry/2398-trig.html	# Thread: 1. ## Trig Stuck on this one... The perimeter of a parallelogram is 44cm and the length of the shorter diagonal is 14cm. Find the lengths of the sides, given that they contain an angle of 60 degrees. And this... THe angles of elevation of the top of a tower from the top of a building 100m high are 50 and 75 respectively. FInd the height of the tower. I drew diagrams but it still wont work out! 2. Originally Posted by classicstrings Stuck on this one... The perimeter of a parallelogram is 44cm and the length of the shorter diagonal is 14cm. Find the lengths of the sides, given that they contain an angle of 60 degrees. And this... THe angles of elevation of the top of a tower from the top of a building 100m high are 50 and 75 respectively. FInd the height of the tower. I drew diagrams but it still wont work out! Hello, I've attached a drawing. 1. two neighbored sides form a triangle together with the short diagonal. To calculate the sides you have to use the Cosine Rule: $14^2=x^2+(22-x)^2-2 \cdot x \cdot (22-x) \cdot \cos(60^{\circ})$ with $\cos(60\circ})=\frac{1}{2}$ you'llget a quadratic equation: $3x^2-66x+288=0$ and the solutions for x= 16 or x= 6. Next problem will follow. Now I'm in a hurry. Greetings EB Attached Thumbnails 3. Originally Posted by classicstrings Stuck on this one... ... THe angles of elevation of the top of a tower from the top of a building 100m high are 50 and 75 respectively. FInd the height of the tower. I drew diagrams but it still wont work out! Hello, I'm awfully sorry, but are you sure that this is the complete problem? There must be at least one value (or another point) to calculate the height.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.905458390712738, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/18839?sort=oldest	## Infinity groupoid objects ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I was wondering if there is a model-theoretic way of defining the infinity category of infinity-groupoid objects in a category $C$ (more generally, if $C$ is an infinity category itself, but, right now a 1-category is enough). Is there a model structure on $C^{\Delta^{op}}$ such that those objects which are fibrant and cofibrant correspond to "internal Kan-complexes" in the correct way? So e.g. I want the C-enriched nerve of an actual groupoid object of C to be fibrant and cofibrant in this model structure. If you don't know the answer in general, for now I am mostly interested in the case that $C$ is the 1-category of topological spaces (here I DO NOT want to think of $C$ as being the same thing as infinity-groupoids or simplicial sets, I actually care about the topology). More generally, if $C$ is an infinity-category associated to a model category $D$, does this correspond to the Reedy-model structure on $D^{\Delta^{op}}$? EDIT: I should really be asking for a SIMPLICIAL model category structure. - ## 3 Answers I don't have an answer which includes the category of topological spaces, but the two remarkable papers "Sheafifiable Homotopy Model Categories I and II" by Tibor Beke together give you a big class of categories with good model structures on their simplicial objects. They are available on his homepage. I think your requirement for internal groupoids is satisfied for these, but better check it out yourself. - Thanks, I'll check it out. However, I am really interested in an answer which would include C=Top, and C=manifolds. – David Carchedi Mar 23 2010 at 23:52 ### 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. Did you check math.AT/0603563 (integrating $L_\infty$ algebras" by André Henriques)? I am not an expert, but I think he does a construction close to what you are looking for in the category of smooth manifolds. - There is not a model category on simplicial smooth manifolds which gives the data you're looking for, but there is a structure of category of fibrant objects such that nerves of groupoids are fibrant. The fibrations are Kan maps (a la Henriques, but with the additional requirement that the maps on vertices be submersions), and the weak equivalences are maps which induce isomorphisms on all simplicial homotopy groups (again, as Henriques defines them). Nerves of groupoids are fibrant in this sense, hypercovers are trivial fibrations, and morphisms qua principal bibundles are equivalent to morphisms via spans where the source leg is a hypercover. This category of fibrant objects structure doesn't extend to a model structure on simplicial manifolds because there aren't any positive dimensional cofibrant objects (e.g. hypercovers are trivial fibrations, but given any positive dimensional manifold, you can build a hypercover on it which doesn't admit a section). If you still want to work in a model category capturing this, the natural thing is to use Yoneda and pass to the local model structure on simplicial presheaves. You can show that this is a fully faithful embedding on the level of homotopy categories. - Jesse, how does Thm 10.3 of DHI say that this is a faithful embedding? – Thomas Nikolaus Apr 6 2012 at 14:55 Thomas, you're right to ask. I was thinking about it today, and I'm no longer convinced that you can deduce that from their theorem. On the other hand, you can show fairly directly that Yoneda gives a functor of categories of fibrant objects from Kan manifolds to locally fibrant simplicial presheaves (in particular, it preserves fibrations, weak equivalences and path objects). Then, with a little work you can show that any trivial fibration over a representable representable base can be refined to a representable trivial fibration. From this, it follows quickly that it induces a – Jesse Wolfson Apr 6 2012 at 22:21 that it induces a fully faithful embedding at the level of homotopy categories. Finally, using a result of Jardine (Prop. 2.8 of Simplicial Presheaves), we can conclude the statement I made above. – Jesse Wolfson Apr 6 2012 at 22:22 Hm, I still do not see how you show that a trivial fibration over a representable base can be refined to a representable fibration. To be honest I doubt that it is true. Can you maybe give me some more details? – Thomas Nikolaus Apr 6 2012 at 22:39 This isn't so bad. The argument is similar to Lemma 8.6 of Artin and Mazur's Etale Homotopy. The idea is that you can induct up the skeleton so that at the $n^{th}$ stage the $n$-skeleton is representable. The possibility of this refinement comes from the characterization Jardine gives of trivial fibrations in terms of maps of simplicial sheaves such that the maps from the $n$-simplices to the $n^{th}$ matching objects are generalized covers. – Jesse Wolfson Apr 7 2012 at 0:10 show 3 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "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": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9316145777702332, "perplexity_flag": "head"}
http://mathoverflow.net/questions/121095/why-do-we-use-the-less-simple-convention-for-the-definition-of-a-vector-bundle-co	## Why do we use the less simple convention for the definition of a vector bundle connection? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) For a (smooth) vector bundle $F$ over a manifold $M$, one normally defines a connection to be a linear map $$\nabla:\Gamma^{\infty}(V) \to \Omega^1(M) \otimes \Gamma^{\infty}(V),$$ satisfying $\nabla(fv) =$d$f \otimes v + f \nabla(v)$. One can then extend this to a map $$\nabla:\Omega^k(M) \otimes_{C^{\infty}(M)} \Gamma^{\infty}(V) \to \Omega^{k+1}(M) \otimes_{C^{\infty}(M)} \Gamma^{\infty}(V)$$ by defining $$\nabla(\omega \otimes v) = \text{d}\omega \otimes v + (-1)^k\omega \wedge \nabla(v).$$ The factor of $(-1)^k$ ensures that the definition is well-defined over the tensor product. Alternatively, one can use the equivalent definition of a connection as a linear map $$\nabla:\Gamma^{\infty}(V) \to \Gamma^{\infty}(V) \otimes \Omega^1(M),$$ satisfying $\nabla(vf) = v \otimes$d$f + \nabla(v)f$. However, this has the simpler extension to a map $$\nabla:\Gamma^{\infty}(V) \otimes_{C^{\infty}(M)} \Omega^k(M) \to \Gamma^{\infty}(V) \otimes_{C^{\infty}(M)}\Omega^{k+1}(M)$$ defined by $$\nabla(v \otimes \omega) = v \otimes \text{d}\omega + \nabla(v) \wedge \omega.$$ I.E. there is no $(-1)^k$ factor. My question is why isn't this simpler formulation the one that is normally used? - 1 Probably because people often skip parentheses, whereupon $\nabla \nu \wedge \omega$ is ambiguous as opposed to $\omega \wedge \nabla \nu$ which is not. – Andrew Stacey Feb 7 at 18:18 The way I survived, heuristically, so far, is as follows. Notice that the factors in the tensor products are permuted in the two approaches. If from the second approach, you want to get to the first, you have to "pass" the $\omega$ (on the right) above the $\nabla(v)$ (which is on the left in the 2nd approach). This second factor is a bundle valuated one form, ie locally a one form tensor a section of $V$, hence you reobtain the $(-1)^k$. It's a bit sketchy but I've been ok with it. – Amin Feb 7 at 21:37 Oups sorry I just looked at A. Stacey's comment; sorry if this is what it was saying... – Amin Feb 7 at 21:39 2 I am inclined to disagree with your claim that begins with "one normally defines". It is a convention that depends on the author, and varies across the literature. – S. Carnahan♦ Feb 8 at 0:32	{"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": 14, "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": 0.9464472532272339, "perplexity_flag": "head"}
http://nrich.maths.org/1926/note	### Route to Root A sequence of numbers x1, x2, x3, ... starts with x1 = 2, and, if you know any term xn, you can find the next term xn+1 using the formula: xn+1 = (xn + 3/xn)/2 . Calculate the first six terms of this sequence. What do you notice? Calculate a few more terms and find the squares of the terms. Can you prove that the special property you notice about this sequence will apply to all the later terms of the sequence? Write down a formula to give an approximation to the cube root of a number and test it for the cube root of 3 and the cube root of 8. How many terms of the sequence do you have to take before you get the cube root of 8 correct to as many decimal places as your calculator will give? What happens when you try this method for fourth roots or fifth roots etc.? ### Rain or Shine Predict future weather using the probability that tomorrow is wet given today is wet and the probability that tomorrow is wet given that today is dry. ### Climbing Powers $2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$? # Dalmatians ### Why do this problem? This problem gives scope for investigation, spotting patterns, working systematically to cover all cases and making and proving conjectures. It provides an example of the mathematics of dynamical systems. This is an important subject in higher mathematics and, in this problem, learners can work with whole numbers in a simple discrete system to discover for themselves the important concepts of cycles and fixed points. ### Possible approach Ensure that the learners understand how the mapping works then suggest that they choose their own starting numbers and work out their own sequences individually, making notes of anything interesting that they observe. They might need to spend time developing a sensible recording system to prevent confusion with the numbers at each step. After about 10 minutes ask the learners to work in pairs and explain to each other what they have discovered. Then later have a class discusion to compare findings from the whole class. ### Key questions What happens to the sequences? Will they go on for ever? Why? What patterns do you notice? Can you explain them? Do sequences have the same behaviour for ALL 2 digit starting numbers? Why? ### Possible extension Investigate the problem for sequences starting with negative numbers or 3-digit numbers or bigger numbers. In what circumstances might fixed points arise? Can students invent similar systems for themselves? ### Possible support Suggest that students start off with the concrete cases $a=b$ for 2 and 3. Then ask what they expect to happen for 88 and 99. Then try it out. Were they correct? Happy Numbers is a similar problem which lends itself to investigation using spreadsheets. For a full discussion of some simple discrete dynamical systems see: Whole Number Dynamics I Whole Number Dynamics II Whole Number Dynamics III Whole Number Dynamics IV Whole Number Dynamics V. 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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9255496859550476, "perplexity_flag": "middle"}
http://nrich.maths.org/6680/solution	nrich enriching mathematicsSkip over navigation ### Epidemic Modelling Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths. ### Very Old Man Is the age of this very old man statistically believable? ### bioNRICH bioNRICH is the area of the stemNRICH site devoted to the mathematics underlying the study of the biological sciences, designed to help develop the mathematics required to get the most from your study of biology at A-level and university. # Is Your DNA Unique? ##### Stage: 5 Challenge Level: This problem makes heavy use of combinatorics: i) We are asked the probability of a single adenine among 10 bases. If the adenine were in the the first base in the sequence, the 9 following bases could be any of the other three types. Thus the probability of this is: $$p(ANNNNNNNNNN) = \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^9 = 0.0188$$ However, it is also possible that the Adenine could have been in the any of the other positions instead. Thus the probability is increased tenfold. We can express this possibility of placing the adenine in multiple places by using the Combinations notation: $^{10}C_1$ indicates that we wish to place 1 adenine among 10 bases. Thus, overall the probability we require is: $$p(one\ adenine) = ^{10}C_1\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^9 = 0.188$$ ii) A 30% cytosine content implies the need for 45 cytosines from among the 150 bases. Thus, $$p(45C) = ^{150}C_{45}\left(\frac{1}{4}\right)^{45}\left(\frac{3}{4}\right)^{105} = 0.0272$$ iii) We are asked for the probability that there is at least one chain of at least 5 Thymines among 1000 bases. To tackle this, we must realise that a group of 5 Thymines has 996 possible locations within 1000 bases, and that the remaining 995 bases can be of any sort. Thus, $$p = ^{996}C_{1}\left({1}{4}\right)^5 = 0.973$$ iv) The probability of an individual having the same genetic composition as me implies that their every base must be identical in type and placement as mine. Therefore: $$p(same) = \left(\frac{1}{4}\right)^{6,000,000,000} = \text{exceptionally small!}$$ v) The probability of a random 6 base sequence of DNA forming GGATCC is $\left(\frac{1}{4}\right)^6$. If we simplistically say that the 6 billion base-pair human genome is composed of 1 billion different possible sites, then the number of expected sites with the correct restriction sequence is: $$\left(\frac{1}{4}\right)^6\times 1,000,000 = 2.44 \times 10^5$$ vi) If only ever 1000 bases vary across a population, then there are only 6 million variable sites in the genome. Thus, the probability of an individual being identical to me is: $$\left(\frac{1}{4}\right)^{6,000,000} = \text{very small}$$ vii) We wish to find the number of sites necessary for it to be possible to match an individual to a 99.99% probability to a piece of DNA. Thus, we want the possibility of the two samples of DNA being the same by chance as 0.01%. $$p = \left(\frac{1}{4}\right)^n = \frac{0.01}{100}$$ $$n = \frac{ln(10,000)}{ln(4)} = 6.62$$ Therefore, at least 7 of the variable sites should be investigated. viii) As before, a misidentification occurs when the two DNA samples are the same purely by chance. We want the probability of this happening to be less than 1 in 1,000,000. However, since the same variable sites are present in the same place on homologous chromosomes, the probability of two individuals being identical at both these loci is $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$. $$\therefore \left(\frac{1}{16}\right)^n = \frac{1}{1,000,000}$$ $$n = \frac{ln(1,000,000)}{ln(16)} = 4.98$$ Therefore, at least 5 sites should be investigated. 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": 3, "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": 0.3, "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": 0.9298163652420044, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/53936?sort=newest	## Simple but serious problems for the edification of non-mathematicians ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) When people graduate with honors from prestigious universities thinking everything in math is already known and the field consists of memorizing algorithms, then the educational system has failed in one of its major endeavors. If members of the next freshman class will take just one one-semester math course before becoming the aforementioned graduates, here's what I think I might do (and this posting is indeed a question, as you will see). I would not have a fixed syllabus of topics that the course must cover by the end of the semester. I would assign very simple but serious problems that I would not tell the students how to do. A few simple examples: • $3 \times 5 = 5 + 5 + 5$ and $5 \times 3 = 3 + 3 + 3 + 3 + 3$. Why must this operation thus defined be commutative? • A water lily has a single leaf floating on the surface of a pond. The leaf doubles in size every day. After 16 days it covers the whole pond. How long will it take two such leaves to cover the whole pond. (Here lots of students say "8 days". I might warn them against that. This is the very hardest problem assigned in an algebra course that I taught, according to most of the students.) • Here is a square circumscribing a circle. [Illustration here.] Here is how you use this to see that $\pi<4$. [Explanation here.] Now figure out how to prove that $\pi > 3$ by a similarly simple argument. • Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, ...... Multiples of 18 are 18, 36, 54, 72, 90, ..... The smallest one that they have in common is 36. Multiples of 63 are 63, 126, 189, 252, 315, 378,..... Multiples of 77 are 77, 154, 231, 308, 385,.... Could this sequence go on forever without any number appearing in both lists? (Usual answer: Yes. It will. Because 63 and 77 have nothing in common.) Is it the case that no matter which pair of numbers you start with, eventually some number will appear in both lists? I said simple but serious, the latter meaning they will actually learn something worth learning about mathematics or about how to think about mathematics. Not all need be as elementary as these. With some of the less elementary problems I might sketch a solution or write out a solution in detail and then ask questions about the solution. I would not fix in advance the date at which problems were to be turned in, but would set deadlines after discussion reveals that serious difficulties are overcome. I might also do some "teasing" concerning various math topics not covered. HERE'S THE QUESTION: Which published books of problems can participants in this forum recommend for this purpose? Why those ones? - 8 What do you mean exactly by a "serious problem"? Also, is this question significantly different than your previous question? mathoverflow.net/questions/28695/… – Jeremy West Feb 1 2011 at 1:51 3 @Jeremy: What I mean by "serious" is stated explicitly in my posting. How is my proposed statement of the meaning deficient? This posting OBVIOUSLY (but only if you read the whole thing) differs from that earlier posting in the content that follows after the words "HERE'S THE QUESTION", set in boldface type (Does it fail to appear in bold on your browser?). – Michael Hardy Feb 1 2011 at 2:13 2 Why are so MANY up-votes given to comments that prove that the commenter did not read the question? I've seen this with a number of other questions here. – Michael Hardy Feb 1 2011 at 3:58 15 I've hit this question with the Wiki-hammer. Please make an effort to communicate in a way that is less likely to appear condescending or sarcastic. – S. Carnahan♦ Feb 1 2011 at 3:59 4 @Michael Also, lest I seem antagonistic, I thought the original question was excellent, which is why I remembered when I read this one. – Jeremy West Feb 1 2011 at 5:06 show 34 more comments ## 5 Answers How about The Theory of Remainders by Andrea Rothbart. I remember back in the day I was struggling with the concept of modular arithmetic and randomly came across the book above. It's really well written in an unorthodox way as a dialogue between two people talking about modular arithmetic. The book introduces basic concepts of abstract algebra and has plenty of "simple, but serious" exercises. If I recall correctly, it did a really good job of motivating the concept of fields. Above anything, it was written with a high school audience in mind, so incoming freshmen should not be deterred by the level of difficulty. I also found the style of the book engaging. I dare say I was bitten by the number theory bug shortly after reading it. - +1 I believe this book is where I read a discussion concerning which scores are possible outcomes in American football, which is a really fun question. – R Hahn Feb 1 2011 at 10:17 ### 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. Your pedagogical approach sounds suspiciously like the one in many Math Circles. The National Association of Math Circles has problem lists on their website. Here's another problem set which looks interesting. If your intention is to leave these students with the sense that there is fabulous ongoing research in mathematics, I'd recommend the Five Golden Rules: Great 20-Century Mathematics and Why They Matter. It's quite accessible because it focuses on the general ideas. - One good option is "The Magic of Numbers" by Gross and Harris (not to be confused with a book of the same title by ET Bell), which was written for the eponymous class Gross used to teach at Harvard. The problems include some stuff on, say, Catalan numbers, and some reasonably serious modular arithmetic (e.g. RSA encryption) with a minimum of baggage, which should recommend the book to non-mathematicians. The Art of Problem Solving series (here) is also quite good. I learned a lot from some of those books when I was in high school--they have lots of exercises, ranging from very easy to problems I, at least, found quite difficult. And there is a lot of discussion of technique, which I think non-mathematicians often find lacking in other textbooks. And Martin Gardner's entire oeuvre is great, and I think that recommendation probably doesn't require any explanation. - Thank you. I've requested Gross & Harris via interlibrary loan. – Michael Hardy Feb 4 2011 at 0:56 No problem, enjoy! – Daniel Litt Feb 4 2011 at 5:15 It showed up today. It looks excellent so far. But there's something really weird on pages 114--115. For the proof of the infinitude of primes, it says on page 114 "we will have to argue by contradiction". That is of course nonsense, and then the usual proof by contradiction, which is pointlessly complicated, is given on page 115. Then it says "Another way to phrase this argument would be the following. Suppose we have any finite collection of primes" etc..... It shows that if you multiply them and add 1 and then factor the result, you get primes not in the set you started with. This – Michael Hardy Feb 11 2011 at 22:55 1 ....other way to phrase this is not a proof by contradiction. Yet it had just said "we have to argue by contradiction". This is a place where the book could be improved by omitting something. The proof by contradiction (which, contrary to the assertions of many universally respected authors, is not in the works of Euclid) should just be omitted. The "other way to phrase this argument", which proves that we do not "have to argue by contradiction" is in fact the one in Euclid's Elements. – Michael Hardy Feb 11 2011 at 22:57 "Heard on the Street" by Timothy Falcon Crack is a collection of brainteasers that were supposedly put to interview candidates for Wall Street jobs. The book has many more questions like the ones you asked - most of them can be solved without any heavy mathematical machinery, but they all require a little ingenuity. Sadly, the book has become famous enough that recent graduates hoping to get banking jobs often memorise all the problems, rendering the whole process useless. - But most students aren't seeking banking jobs. – Michael Hardy Feb 1 2011 at 4:05 @Michael: true, most students aren’t seeking banking jobs, but nevertheless this book contains many problems along the lines of what you asked for; plus the fact that these skills by at least some non-academics may be a helpful motivator for some students. (Even if they don’t feel “This will help me, personally, get a job”, it can at least help them stop feeling “Ugh, after college no-one even cares whether you can do math; it’s so pointless!”) – Peter LeFanu Lumsdaine Feb 1 2011 at 4:53 When I said most students aren't seeking banking jobs, I was responding to your "sadly" comment. In other words, it's not as bad as that comment might suggest. – Michael Hardy Feb 4 2011 at 1:15 Published books of problems generally mean to nudge serious, dedicated and perhaps talented mathematics students towards a research-oriented frame of mind. If you really mean a bound problem collection for general education, I expect you will have to write such a book yourself. But I would probably recommend against publishing such a book - on the grounds: don't teach until you see the whites of their eyes. A mathematics problem that will work with one cohort might variously and unpredictably either defeat or insult the intelligence of another. And, as the the response to your other question might indicate, teachers of mathematics will have very diverse views of what constitute a value partial knowledge of mathematics. That said, if I had an audience of highly intelligent but not especially mathematically oriented students, I might focus their "last look" at mathematics on Lawvere and Schanuel's Conceptual Mathematics (which has many good problems). The authors show themselves as both wise and smart. While the book could save the soul of a stray mathematician, it does not harbor any hidden agenda that means ignoring the needs of the broader audience. And while it might accidentally remediate some high school induces confusions, even the best trained students will find most of what the authors say both very new and very fundamental. Serge Lang's Math Talks for Undergraduate also attracts me, but where Lawvere and Schanuel help a student think about the larger world in a more mathematical way, Lang wants non-mathematicians to understand more about what mathematicians do. Research mathematician/teachers generally hold as a sacred shibboleth the dictum that "mathematics in not a spectator sport." In the case of a class of general education students seeing mathematics in the classroom for the last time, and and the risk of blasphemy, I question this, I question whether having these students primarily trying to solve problems for themselves necessarily constitutes the best use of their time. I believe that the mathematics community has neglected developing of literature of what one might call proof-oriented spectator mathematics. But still I might recommend a book: I taught a course recently out of Ross Honsberger's Episodes in 19th and 20th Century Euclidean Geometry where I focused on close readings of complicated but elementary proofs of concrete and yet often spectacularly counter-intuitive facts, all material most mathematics majors will never see on the grounds that it isn't sufficiently modern. But as a toy model of what mathematicians do it worked very well for my students. - 1 In the realm of what you call proof-oriented spectator mathematics, there is a very beautiful book by Stanley Ogilvy called Excursions in Geometry, that a 15-year-old who knows next to nothing can read and enjoy. (I read it when I was 14 or 15.) Non-mathematically inclined undergraduates intensely hate that book. – Michael Hardy Feb 1 2011 at 4:05 1 @David Feldman: While I agree with the sentiments of almost everything you write, and while I haven't read "Conceptual Mathematics," I notice that on Amazon its subtitle is "A First Introduction to Categories." I think that the language of categories, as much as I value it, is unlikely to do anything other than annoy the average (even quite intelligent) English major--do you disagree? – Daniel Litt Feb 1 2011 at 4:09 5 @Daniel All I can say is that I'd probably agree with you...if I'd never seen the book. But actually, I think categories do have a lot to say to an English major (think about characters and plots, etc.) for roughly the same reason they have a lot to say to computer scientists interested in the semantics of programming languages. But developing that point would probably better be done over lunch than in a MO comment. :) – David Feldman Feb 1 2011 at 4:39 Fair enough--I guess I'll read the book! – Daniel Litt Feb 1 2011 at 4:41 1 @ Michael The Ogilvy and the Honsberger are fundamentally different books. Ogilvy emphasizes theories (inversive geometry, projective geometry) from which theorems drop out - in that sense it seems "modern" and pre-professional. Honsberger just develops these wonderful, elegant, but seemingly ad hoc results. Ogilvy:Honsberger:: Chewable vitamins : exotic desserts. – David Feldman Feb 1 2011 at 5:06	{"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": 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": 0.959309995174408, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2007/12/02/another-carnival/?like=1&source=post_flair&_wpnonce=1c947633de	# The Unapologetic Mathematician ## Another Carnival The Secret Blogging Seminar is hosting the latest installment of the Carnival of Mathematics. They seem to have done a fair job of collecting links, with more (and more varied) entries than I remember seeing before. Enjoy. ### Like this: Posted by John Armstrong \| Uncategorized ## 3 Comments » 1. John, I have appreciated your help in the past and I would like to get your insight into yet another question I have concerning linear algebra. What is the relationship between the set of all linear transformations on a vector space over a field F of dimension n and the set of all nxn matrices with entries from the same field F. Clearly, it is not a one-to-one relation. There may be many, possibly an infinite number of matrices that can represent the same linear transformation with respect to different basis of the space. What exactly is the relation? Jonathan Greenberg Comment by Jonathan Grennberg \| December 11, 2007 \| Reply 2. When you’re dealing with $R$-linear functions between free $R$-modules (all vector spaces — modules over a field — are free), you can pick a basis for each of the domain and range as I discuss here. Once you do, then any such linear transformation can be written down uniquely as a matrix with respect to those bases, as I discuss in that link. Also, given a choice of bases any matrix corresponds to exactly one linear transformation. But as you notice, there’s that choice of basis. Changing the basis of a free module is the same as applying an invertible linear transformation from that module to itself. That is, $\hom(R^m,R^n)$ is the space of linear transformations from a rank $m$ module to a rank $n$ module. Then there is an action of the group $\mathrm{GL}(m,R)$ (invertible $m\times m$ matrices with entries in $R$) on the one side, and an action of $\mathrm{GL}(n,R)$ on the other. What you’re really asking about is what the set of transformations looks like once we quotient out by these group actions. And that’s where the story gets complicated. Over an algebraically closed field (like the complex numbers), the answer is contained in the Jordan normal form, which is a standard part of any good senior-level linear algebra course. It gives a canonical representative of each orbit of the group action. Beyond that it’s a huge mess. Sorry I can’t be more accurate. Comment by \| December 11, 2007 \| Reply 3. John, thank you for your insight into my question. I will have to think about your answer as I study linear algebra in more depth. JG Comment by Jonathan Greenberg \| December 11, 2007 \| 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. • ## Feedback Got something to say? Anonymous questions, comments, and suggestions at Formspring.me! %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": 9, "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": 0.923723578453064, "perplexity_flag": "head"}
http://mathhelpforum.com/pre-calculus/78358-solved-write-parametric-symmetric-equations-z-axis.html	# Thread: 1. ## [SOLVED] Write parametric and symmetric equations for the z-axis. This question just seems somewhat unclear. Asking for symmetric and parametric equations for the z axis means to have x and y equaling zero? 2. Hello, TJ2988! Write the symmetric and parametric equations for the z axis. You are expected to know this. . . A line through $P(x_1,y_1,z_1)$ with direction vector $\vec v \:=\:\langle a,b,c\rangle$ has: . . Symmetric equation: . $\frac{x-x_1}{a} \:=\:\frac{y-y_1}{b} \:=\:\frac{z-z_1}{c}$ . . Parametric equations: . $\begin{Bmatrix}x &=& x_1+at \\ y &=& y_1 + bt \\ z &=& z_1 + ct\end{Bmatrix}$ We have: . $P(0,0,0)$ and $\vec v \:=\:\langle0,0,1\rangle$ . . Go for it! 3. Okay, so then the equations would be as follows: Symmetric: (x-0)/0 = (y-0)/0 = (z-0)/1 Parametric: x=0 y=0 z=t Although the symmetric equations of x & y are ÷ 0 which is not allowed. So how would do you change it so they are not? Multiply all by zero to give x=y=0? Thank you very much for your help! 4. Originally Posted by TJ2988 Okay, so then the equations would be as follows: Symmetric: (x-0)/0 = (y-0)/0 = (z-0)/1 parametric: x=0 y=0 z=t Although the symmetric equations of x & y are ÷ 0 which is not allowed. So how would do you change it so they are not? Multiply all by zero to give x=y=0? Yes, that is exactly right. Thank you 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": 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": 0.3, "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": 0.9326466917991638, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2007/03/26/more-sketches-and-why-we-care/?like=1&source=post_flair&_wpnonce=c1ce40f6e1	# The Unapologetic Mathematician ## More sketches, and why we care Dr. Adams just sent me a link to an explanation of the technical details for mathematicians in other fields, but it’s still somewhat readable. I also have been reading the slides for Dr. Vogan’s talk, The Character Table for E8, or How We Wrote Down a 453,060 x 453,060 Matrix and Found Happiness. There’s also an audio recording available (7MB mp3). Incidentally, I’d have gone for The Split Real Form of E8, or How We Learned to Stop Worrying and Love the Character Table, but it’s all good. This talk actually manages to be very generally accessible, and includes all sorts of pretty pictures. Those of you who wanted more visuals than I provided in my rough overview might like to check that one out. Together, these two are my core that, together with some input from Dr. Zuckerman I’ll be trying to break down into smaller chunks. I highly advise reading at least Vogan’s slides and preferably also Adams’ notes. I also want to respond to a comment basically asking, “so why the heck should we care about this?” It’s an excellent question, and yet another one the newspaper reports really glossed over without taking seriously. I’ll admit that I glossed it over at first too, since I think this stuff is just too elegant not to love. Still, I’ve mulled this over not just as applies to these calculations, but with regard to a lot of mathematics at this level (thus qualifying the “why we care” as a rant). This sort of question from a non-mathematician almost always is looking for an engineering response. “What’s it good for?” means, “what can we build with it?” Honestly I have to say “not much”. Representations of Lie groups do have their uses, though, and I can point out a few things they have already been good for. As indicated in Dr. Vogan’s slides, representations of the one-dimensional Lie groups are concerned with change through time, particularly periodic changes. This means that they’re exceptionally good at talking about periodic phenomena, like waves. Sound waves, light waves, electrical circuits, vibrating strings — they’re all one-dimensional waves. So what? So every time you use the graphic equalizer on your stereo the electronics are taking the signal and performing a fast Fourier transform on it. This turns a function on the line (Lie group) into a function on the space of all representations of the group; that’s the “unitary dual” that Dr. Adams refers to. Then you can adjust the periodic components and reconstruct a new function with much fatter bass, or whatever your tastes are. The same sorts of things can be done in higher dimensions. Similar techniques revealed that you can’t hear the shape of a drum — there are differently-shaped membranes that have the same vibrational characteristics. What are “orbitals” of electrons around an atomic nucleus (hazy memories of chemistry)? They’re representations of the Lie group $SO(3,\mathbb{R})$! So what can we do with $E_8$? Nothing right now, but there’s plenty we can do (and have done) with representation theory in general. There’s another reason (beyond the intrinsic beauty of the ideas) to work out the Atlas: more data means more patterns, and more patterns means more interrelationships between seemingly-distinct fields. Quite a few of the greatest theorems in recent years have been saying that this field of mathematics over here and that one over there are “really” the same thing. Everyone knows that Andrew Wiles solved Fermat’s Last Theorem, but what he really did was show that some things in algebraic geometry (the study of solution sets of polynomials) called “elliptic curves” are deeply related to functions with a certain sort of periodicity called “modular forms”. If, as David Corfield asserts, mathematics proceeds by “telling stories”, then each field’s stories become allegories for the other. Hard questions in one area might be translated into questions we know how to solve in the other. So how does having a lot of data like the Atlas around help out? Because we discover a lot of these relationships from similar patterns in the data, and in many cases (though I hate to admit it) through the same numbers showing up over and over. As just one example, I present the Monstrous Moonshine conjecture. The Monster is a finite, simple group — no normal subgroups, so it can’t be broken down into even a semidirect product of smaller groups — of order (brace yourself) $808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,000,000,000$ That’s $8\times10^{53}$ elements being juggled around in an intricate symmetry. People sat down and calculated its character table, very much a similar project to the current one about $E_8$. And then there’s a certain special modular form called $j$ that just happens to be related to it. How so? John McKay happened to see the $j$-function written out like this: $j(\tau) = \frac{1}{q} + 744 + 196884q + 21483670q^2 + ...$ So? So he’d also seen the dimensions of representations of the Monster, which start with $1$, $196883$, $21296876$, and continue. Every single coefficient in the function came from dimensions of representations of the Monster! And it was conjectured that the pattern continued. In fact it did. Twenty-some years ago, Frenkel, Lepowsky, and Meurman constructed a representation of the Monster that made it clear, and their results are still echoing. One of my colleagues graduated last year and went on to Harvard by studying exactly the same sorts of connections. And how did it start? By recognizing patterns in a mountain of raw data about representations. What unsolved problems might be translatable into representation theory by reflections found through the Atlas data? Maybe the Navier-Stokes equations, which would give a better understanding of fluid flows and aerodynamics. Maybe the Riemann hypothesis, which would lead to a better understanding of the distributions of prime numbers, which would have an impact on modern cryptography. Who knows? Oh, and one more thing. How did someone find the Monster in the first place? Well it turns out to be a group of symmetries of a certain collection of points tiling eight-dimensional space. What collection of points? The “Leech lattice”. And you’ve already seen it: that picture of the $E_8$ root system in all the news reports is the basic cell, just like a square is the basic cell of a checkerboard tiling of the plane. And it all comes back around again. [EDIT: I've found out I was wrong about how the Monster relates to $E_8$. More info in the link.] ### Like this: Posted by John Armstrong \| Atlas of Lie Groups, rants ## 5 Comments » 1. The slide show is pretty good. In fact, most explanations (yours, Baez) are good. The problem I had with the NY Times article & others was the B.S. about the T.O.E., etc. Beyond that — and beyond the fact that the math is interesting and constructing the algorithm solid work — there is the problem of going at physics too abstractly. It is true that you can look at special relativity as a representation of a group but too often that leads to forgetting that, as Einstein said, time is measured with a clock and space with a ruler. Comment by Steve Myers \| March 26, 2007 \| Reply 2. [...] I now also have another post trying to answer the “what’s it good for?” question. That response starts at the [...] Pingback by \| March 27, 2007 \| Reply 3. [...] in my little added remarks about the Monster group in my “Why We Care” post, I was oversimplifying. First of all, the lattice is not the Leech lattice. The Leech lattice [...] Pingback by \| March 31, 2007 \| Reply 4. [...] overview for complete neophytes of what, exactly, had been calculated; and an attempt to explain why we should care. All with the promise of more information [...] Pingback by \| January 11, 2010 \| Reply 5. [...] overview for complete neophytes of what, exactly, had been calculated; and an attempt to explain why we should care. All with the promise of more information [...] Pingback by \| August 28, 2010 \| 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": 13, "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": 0.948992133140564, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/289035/how-prove-that-fx-ge2n-n1-under-these-conditions	# how prove that $\|f(x)\|\ge2^n (n+1)$ under these conditions? Assume $f:[0,1]\mapsto\mathbb{R}$ is continuous and satisfies 1. $\int_0^1x^kf(x) \, dx=0 \quad\forall k\in\{0,1,2,\ldots,n-1\}$, 2. $\int_0^1x^n f(x) \, dx=1$. How do you prove that $\exists x\in[0 ,1]$ such that $\|f(x)\|\ge2^n (n+1)$? - This a puzzler. :) If no one solves it here, but you, would appreciate if you post a solution as i failed fir an hour (except for the tremendous insight it doesn't hold for $n=0$...) – gnometorule Jan 28 at 19:40 It doesn't hold for $n=0$? Why not? – user7530 Jan 28 at 19:43 @User7530: my bad. – gnometorule Jan 28 at 19:57 ## 1 Answer Let $\tilde P_n$ be the $n$-th shifted Legendre polynomial. Then $\tilde P_n$ has degree $n$, leading coefficient equal to $\binom{2n}n$ and $$\int_0^1\tilde P_n(x)\tilde P_m(x)\,dx=\frac{\delta_{mn}}{2n+1}\ (\delta\ \text{denotes the Kronecker delta}\ )\,.$$ The hypotheses on $f$ imply that $\int_0^1\tilde P_n(x)f(x)\,dx=\binom{2n}n$. If $a_n=(2n+1)\binom{2n}n$, then by the ortogonality of the polynomials $\tilde P_n$ we have $$0\leq\int_0^1\bigl(f(x)-a_n\tilde P_n(x)\bigr)^2\,dx=\int_0^1f(x)^2\,dx-2a_n\binom{2n}n+\frac{a_n^2}{2n+1}$$ $$=\int_0^1f(x)^2\,dx-(2n+1)\binom{2n}n^2\,.$$ Since $f$ is continuous, then for some $x_0\in[0,1]$ we have $f(x_0)^2\geq(2n+1)\binom{2n}n^2\geq4^n(n+1)^2$ (this inequality can be easily proved by induction on $\boldsymbol{n\geq2}$) , and so $\|f(x_0)\|\geq2^n(n+1)$, as desired. - You probably meant $f(x_0)^2\geq (2n+1)\binom{2n}{n}^2$. – julien Jan 28 at 19:49 @julien I mean equality, by invoking mean value theorem for integrals, but yes, you can think it as an inequality. – Matemáticos Chibchas Jan 28 at 19:53 Can you tell me how the mean value theorem for integrals gives you an equality here? – julien Jan 28 at 19:58 @julien Apply mean value theorem to $F(x)=\int_0^x f(t)^2\,dt$, together with fundamental theorem of calculus. – Matemáticos Chibchas Jan 28 at 19:59 If you mean $F(1)-F(0)=F'(x_0)(1-0)=f^2(x_0)$, I still only see $f(x_0)^2\geq (2n+1)\binom{2n}{n}^2$. But maybe you meant otherwise. – julien Jan 28 at 20:06 show 2 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8769862055778503, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/4241/examples-of-fields-of-characteristic-0/4259	# Examples of fields of characteristic 0? I was preparing for an area exam in analysis and came across a problem in the book Real Analysis by Haaser & Sullivan. From p.34 Q 2.4.3, If the field F is isomorphic to the subset S' of F', show that S' is a subfield of F'. I would appreciate any hints on how to solve this problem as I'm stuck, but that's not my actual question. I understand that for finite fields this implies that two sets of the same cardinality must have the same field structure, if any exists. The classification of finite fields answers the above question in a constructive manner. What got me curious is the infinite case. Even in the finite case it's surprising to me that the field axioms are so "restrictive", in a sense, that alternate field structures are simply not possible on sets of equal cardinality. I then started looking for examples of fields with characteristic zero while thinking about this problem. I didn't find many. So far, I listed the rationals, algebraic numbers, real numbers, complex numbers and the p-adic fields. What are other examples? Is there an analogous classification for fields of characteristic zero? - ## 4 Answers Is there an analogous classification for fields of characteristic zero? Yes, but it is somewhat useless and nobody would call it a classification. Every field of characteristic zero has the form $Quot(\mathbb{Q}[X]/S)$, where $X$ is a set of variables and $S$ is a set of polynomials in $\mathbb{Q}[X]$ (which you may replace by the ideal generated by $S$, which must be prime). This may be improved by the existence of transcendence bases: Every field of characteristic zero has the form $Quot(\mathbb{Q}[X])[T]/S$, where $X$ and $T$ are sets of variables and $S$ consists of polynomials, which have each only one variable of $T$. - 2 In particular, C is such a field. Now try writing down the variables and the polynomials! – Qiaochu Yuan Sep 8 '10 at 19:58 Yuval just posted the function field construction, which is an important example for your question. You mentioned the algebraic numbers, but just to make it explicit, there are many algebraic number fields (often called just "number fields"). We have $\bar Q$, the field of all algebraic numbers, but we also have $Q(\sqrt{2})$, $Q(\sqrt[5]{7})$, $Q(i,\sqrt{2},\sqrt[5]{7})$, and any other field you wish to generate over $Q$ by some tower of finite extensions. (If we take $Q(\pi)$, we get a field isomorphic to the function field $Q(t)$, since $\pi$ is not algebraic.) You can also take infinitely many different algebraic extensions of the p-adic fields, like $Q_3(i)$ or $Q_3(\sqrt[4]{2})$. If you like big constructions, you can take the algebraic closure $\bar {Q_p}$of a p-adic field (which turns out not to be p-adically complete), then take the completion $\hat { \bar {Q_p}}$ of that to get a p-adcially complete and algebraically complete field which is isomorphic to the field of complex numbers (but not in any canonical way). - If $F$ is any field, the rational functions over it form a field $F(t)$ with the same characteristic (and cardinality, if $F$ is infinite). This field consists of all rational functions $P(t)/Q(t)$ (considered as equivalence classes, i.e. if $P_1(t) Q_2(t) = P_2(t) Q_1(t)$ then $P_1(t)/Q_1(t)$ and $P_2(t)/Q_2(t)$ are identified). You can also replace polynomials with formal power series to get a different field. And you can iterate the construction or just consider rational functions in several variables. - Yes, you are right in saying that field axioms are restrictive. Some other examples of the restriction [for finite fields] are: 1) You can't have field of arbitrary order. Only of the order $p^n$ are possible. 2) Non-zero elements of a field form a multiplicative group. When the field is finite, this group is cyclic [it's not straightforward to prove this starting with field axioms but you should try it] 3) Any finite division ring is a field [Wedderburn's theorem]. This is a very surprising result as a division ring need not by commutative. But finiteness imposes it. For the question that you mentioned [from that book], you should note that when some set with some algebraic structure [like group, ring, field] is isomorphic [as that algebraic structure] to some other set, then that under that isomorphism, we are giving an algebraic structure to that other set. Hence in your case S' is a field. - The specific issue I'm having trouble with is this: consider the finite field of order 7. It is isomorphic to a subset of $Q$. But the first is not a subfield of the second. I think I'm missing something simple here. – dls Sep 8 '10 at 16:08 All finite fields of the same order are isomorphic. That's quite restrictive. – Yuval Filmus Sep 8 '10 at 19:56 In response to dls's comment: the source of confusion is a shift in meaning in the term isomorphism. There are many bijections from the finite field $F$ of order $7$ to subsets of $Q$. But none of these are field homomorphisms, or even homomorphisms of the additive groups of the fields. Group homomorphisms must send elements of finite order to elements of finite order. So if $\phi: (F, +) \to (Q, +)$ is a group homomorphism, every element of $\phi(F)$ has finite additive order. Every nonzero element of $(Q,+)$ has infinite order. So $\phi$ must be the zero map, and not a bijection. – leslie townes Oct 23 '11 at 22:44	{"extraction_info": {"found_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": 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": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.939087450504303, "perplexity_flag": "head"}
http://en.wikibooks.org/wiki/Problems_in_Mathematics/To_be_added	# Problems in Mathematics/To be added 2 Exercise Suppose $f$ is infinitely differentiable. Suppose, furthermore, that for every $x$, there is $n$ such that $f^{(n)}(x) = 0$. Then $f$ is a polynomial. (Hint: Baire's category theorem.) Exercise $e$ and $\pi$ are irrational numbers. Moreover, $e$ is neither an algebraic number nor p-adic number, yet $e^p$ is a p-adic number for all p except for 2. Exercise There exists a nonempty perfect subset of $\mathbf R$ that contains no rational numbers. (Hint: Use the proof that e is irrational.) Exercise Construct a sequence $a_n$ of positive numbers such that $\sum_{n \ge 1} a_n$ converges, yet $\lim_{n \to \infty} {a_{n+1} \over a_n}$ does not exist. Exercise Let $a_n$ be a sequence of positive numbers. If $\lim_{n \to \infty} n \left({a_n \over a_{n+1}} - 1 \right) > 1$, then $\sum_{n=1}^\infty a_n$ converges. Exercise Prove that a convex function is continuous (Recall that a function $f: (a,b) \rightarrow \mathbb{R}$ is a convex function if for all $x,y \in (a,b)$ and all $s,t \in [0,1]$ with $s+t = 1$, $f(sx+ty) \leq sf(x)+tf(y)$) Exercise Prove that every continuous function f which maps [0,1] into itself has at least one fixed point, that is $\exists p \in [0,1]$ such that $f(p) = p$ Proof: Let $g(x) = x - f(x)$. Then Exercise Prove that the space of continuous functions on an interval has the cardinality of $\mathbb{R}$ Exercise Let $f:[a,b] \rightarrow \mathbb{R}$ be a monotone function, i.e. $\forall x,y \in [a,b]; x \leq y \Rightarrow f(x) \leq f(y)$. Prove that $f$ has countably many points of discontinuity. Exercise Suppose $f$ is defined on the set of positive real numbers and has the property: $f(xy) = f(x) + f(y)$. Then $f$ is unique and is a logarithm.	{"extraction_info": {"found_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": 31, "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": 0.9417007565498352, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/9833/constructing-95-confidence-interval-based-on-profile-likelihood	# Constructing 95% confidence interval based on profile likelihood In my elementary statistics course, I learnt how to construct 95% confidence interval such as population mean, mu based on asymptotic normality for "large" sample size. Apart from resampling methods, there is another approach based on "profile likelihood". Could someone elucidate this approach? Under what situations, the constructed 95% CI based on asymptotic normality and profile likelihood are comparable? I could not find any references on this topic, any suggested references, please? Why isn't it more widely used? - ## 1 Answer In general, the confidence interval based on the standard error strongly depends on the assumption of normality for the estimator. The "profile likelihood confidence interval" provides an alternative. I am pretty sure you can find documentation for this. For instance, here and references therein. Here is a brief overview. Let us say the data depend upon two (vectors of) parameters, $\theta$ and $\delta$, where $\theta$ is of interest and $\delta$ is a nuisance parameter. The profile likelihood of $\theta$ is defined by $L_p(\theta) = \max_{\delta} L(\theta, \delta)$ where $L(\theta, \delta)$ is the 'complete likelihood'. $L_p(\theta)$ does no longer depend on $\delta$ since it has been profiled out. Let a null hypothesis be $H_0 : \theta = \theta_0$ and the likelihood ratio statistic be $LR = 2 (\log L_p(\hat{\theta}) - \log L_p(\theta_0))$ where $\hat{\theta}$ is the value of $\theta$ that maximises the profile likelihood $L_p(\theta)$. A "profile likelihood confidence interval" for $\theta$ consists of those values $\theta_0$ for which the test is not significant. - @ocram- Thanks for the clarification. It seems the method requires some intensive computations, maximizing the profile likelihood. Just wondering why not simply resort to the bootstrap method if the estimator is not normally distributed. – user2264 Apr 21 '11 at 17:46 default	{"extraction_info": {"found_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": 0.18, "absolute_threshold": 10, "end_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": 0.9026100039482117, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/112504/complex-manifold-with-non-finitely-generated-canonical-ring	## Complex manifold with non-finitely generated canonical ring ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) P.M.H. Wilson has an example of a compact non-Kahler manifold whose canonical ring is not finitely generated; see his article and this MO question. I'm trying to understand his construction and have trouble understanding how he conludes that his ring is not finitely generated. I'll briefly describe his construction. First, Wilson fabricates a projective surface $\widetilde{\mathbb{P}}$ with a divisor $D$ such that the ring $$R(\widetilde{\mathbb{P}} ,D) := \bigoplus_{m \geq 0} H^0(X,mD)$$ is not finitely generated by following ideas of Zariski: Let $C \subset \mathbb P^2$ be an elliptic curve and $H$ a line. Blow up 12 points in general position on $C$ and a point outside of $C$ to get $\widetilde{\mathbb{P}}$. Let $C'$ be the proper transform of $C$, $E$ the exceptional divisor of the point outside of $C$ and $H' = f^H - E$. If $D := C' + H'$, then $R(\widetilde{\mathbb{P}},D)$ is not finitely generated. Second, Wilson makes a double cover $S$ of $\widetilde{\mathbb{P}}$ ramified over a general element of $\|6C' + 6H'\|$ and desingularizes to get $\alpha : \tilde S \to \widetilde{\mathbb{P}}$. Then $K_{\tilde S} \sim \alpha^(2C' + 3H' + E)$. Third, he makes a nontrivial torus bundle $\pi : W \to \widetilde{\mathbb{P}}$, that is therefore non-Kahler. The fibers are of dimension 2, so $W$ is a fourfold. One remarks that $K_{W / \widetilde{\mathbb{P}}} = \pi^\mathcal O(-2H')$. Fourth, take the fibered product $V = W \times_{\widetilde{\mathbb{P}}} \tilde S$ and let $g : V \to \widetilde{\mathbb{P}}$ be the induced morphism. As $\alpha : \tilde S \to \widetilde{\mathbb{P}}$ is finite of degree 2, this is a compact non-Kahler fourfold. We see that $K_V = g^ \mathcal O(2C' + H' + E)$. Until now, I'm mostly fine with his construction and can either see or am willing to take on faith how each step is necessary. But here Wilson claims that $R(V,K_V)$ is not finitely generated, and I don't see how. I assume there's some link between $R(V,K_V)$ and $R(\widetilde{\mathbb{P}},D)$, but it has escaped me so far. How do we see 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": 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": 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": 0.9537668228149414, "perplexity_flag": "head"}
http://mathhelpforum.com/geometry/119877-help-pythagorean-theorem-proof.html	# Thread: 1. ## Help! - Pythagorean Theorem Proof In need of help. I've searched the internet and can't seem to find a pythag proof that makes since to me. Looking at the diagram below and using the following theorem, how do you prove the pythagorean theorem. Theorem: The power of a point is well defined; that is, the same value is obtained regardless of which line l is used in the definition as long as the line has at least one point of intersection with the circle. I'd even settle for just proving the pythagorean theorem without using the above theorem. Prove: (AD)(DC) = (AE)(EB) + (BF)(FC) 2. Originally Posted by ReneePatt In need of help. I've searched the internet and can't seem to find a pythag proof that makes since to me. Looking at the diagram below and using the following theorem, how do you prove the pythagorean theorem. Theorem: The power of a point is well defined; that is, the same value is obtained regardless of which line l is used in the definition as long as the line has at least one point of intersection with the circle. I'd even settle for just proving the pythagorean theorem without using the above theorem. Prove: (AD)(DC) = (AE)(EB) + (BF)(FC) HI Take a look at the diagram i attached . $\triangle ABC$ with right angle at B . Then construct BD which is perpendicular to AC . We are trying to prove that $AC^2=AB^2+BC^2$ so here $\triangle ADB$ , $\triangle ABC$ , $\triangle BDC$ are similar . $\frac{AD}{AB}=\frac{AB}{AC}\Rightarrow AB^2=AD\cdot AC$ $\frac{BD}{DC}=\frac{AC}{BC}\Rightarrow BC^2=AC\cdot DC$ $\therefore AB^2+BC^2=AD\cdot AC+AC\cdot DC=AC(AD+DC)=AC\cdot AC$ hence proved . Attached Thumbnails 3. Originally Posted by mathaddict so here $\triangle ADB$ , $\triangle ABC$ , $\triangle BDC$ are similar . Hi, You say that $\triangle ADB$ , $\triangle ABC$ , $\triangle BDC$ are similar but there isn't a $\triangle BDC$. Did you mean $\triangle ADC$? 4. Originally Posted by ReneePatt Hi, You say that $\triangle ADB$ , $\triangle ABC$ , $\triangle BDC$ are similar but there isn't a $\triangle BDC$. Did you mean $\triangle ADC$? hey the diagram i attached previously is incorrect . This is the correct one . I accidentally swapped the A and B . Attached Thumbnails 5. Originally Posted by mathaddict hey the diagram i attached previously is incorrect . This is the correct one . I accidentally swapped the A and B . Thank you very much!! This makes so much sense.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 21, "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": 0.9583190083503723, "perplexity_flag": "head"}
http://openfoamwiki.net/index.php/HowTo_Adding_a_new_transport_equation	# HowTo Adding a new transport equation From OpenFOAMWiki Suppose you want to solve and additional scalar transport equation for the scalar ψ by adding it to an existing solver. The equation has the form $\frac{\partial}{\partial t} \left(\rho \psi \right) + \nabla \cdot \phi \psi - \nabla \cdot \Gamma \nabla \psi = S_{\psi}$ where ρ and Γ are defined as dimensionedScalar and are retrieved from a dictionary using the lookup member function. Sφ is the source term. To implement the equation, just add the line: `dimensionedScalarField psi;` to the createFields.H file of the solver, Then, solve the equation by adding the following lines in the point of the solver where you want the equation to be solved. ``` solve ( fvm::ddt(rho, psi) + fvm::div(phi, psi) - fvm::laplacian(gamma, psi) == S_psi );``` In this example, the source term is treated explicitly. To manage it implicitly, OpenFOAM provides the Sp and the SuSp functions. Suppose we can write it as Sk = Κψ. The code lines to solve the same equation with an implicit treatment of the source term are: ``` solve ( fvm::ddt(rho, psi) + fvm::div(phi, psi) - fvm::laplacian(gamma, psi) == fvm::Sp(kappa, psi) );``` SuSp(kappa, psi) can be used to discretise the source term implicitly or explicitly according to the sign of kappa.	{"extraction_info": {"found_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": 0.9223951101303101, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/28480/finite-tor-dimension	## finite tor dimension ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi. Can, every one, give me an example of finite surjective morphism of finite tor dimension (but not flat!) between reduced schemes or complex analytic spaces... Thank you. - ## 1 Answer Consider a smooth surface $Y$ with a point $p\in Y$. Let $X$ be obtained by gluing two copies of $Y$ at $p$, with the obvious morphism $X \to Y$. This is surjective and finite, and has finite Tor-dimension (because $Y$ is regular, hence every morphism to $Y$ has finite Tor dimension). However, it is not flat (for example, because $X$ is not Cohen-Macaulay). - thank you very much Angelo. In fact, all flat and surjective morphims with no Cohen-Macaulay fibers gives, by finite projection or Noether quasi-normalization, finite tor dimension morphism... – kaddar Jun 18 2010 at 7:17	{"extraction_info": {"found_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": 0.3, "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": 0.9057613611221313, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/73228/gauss-sum-with-sign-through-algebra/73229	## Gauss sum (with sign) through algebra ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $p$ be an odd prime, and $\zeta$ a primitive $p$-th root of unity over a field of characteristic $0$. Let $G = \sum\limits_{j=0}^{p-1} \zeta^{j\left(j-1\right)/2}$ be the standard Gauss sum for $p$. (An alternative definition for $G$ is $G = \sum\limits_{j=1}^{p-1}\left(\frac{j}{p}\right)\zeta^j$, where the bracketed fraction denotes the Legendre symbol.) Denote by $k$ the element of $\left\lbrace 0,1,...,p-1\right\rbrace$ satisfying $16k\equiv -1\mod p$. Then, $\prod\limits_{i=1}^{\left(p-1\right) /2}\left(1-\zeta^i\right) = \zeta^k G$. Question: Can we prove this identity purely algebraically, with no recourse to geometry and analysis? If we can, then we obtain an easy algebraic proof for the value - including the sign - of the Gauss sum $G$, since both the modulus and the argument of $\prod\limits_{i=1}^{\left(p-1\right) /2}\left(1-\zeta^i\right)$ are easy to find (mainly the argument - it's a matter of elementary geometry). Note that my "algebraically" allows combinatorics, but I am somewhat skeptical in how far combinatorics alone can solve this. Of course, we can formulate the question so that it asks for the number of subsets of $\left\lbrace 1,2,...,\frac{p-1}{2}\right\rbrace$ whose sum has a particular residue $\mod p$, but whether this will bring us far... On the other hand, $q$-binomial identities might be of help, since $\prod\limits_{i=1}^{\left(p-1\right) /2}\left(1-\zeta^i\right)$ is a $\zeta$-factorial. I am pretty sure things like this must have been done some 100 years ago. - Just a pedantic point. When $\zeta$ is an arbitrary primitive $p$-th root of $1$, there is no sense in talking about the sign of $G(\zeta)=\sum_{x=1}^{p-1}\left(x\over p\right)\zeta^x$, for the sign of $G(\zeta)$ changes if we replace $\zeta$ by $\zeta^u$, whenever $u$ is prime to $p$ and such that $\left(u\over p\right)=-1$. You can talk about the sign only when $\zeta$ is a specific primitive $p$-th root of $1$, such as $e^{2i\pi\over p}$, where $i$ is a 4-th root of $1$ in $\bf C$. – Chandan Singh Dalawat Sep 21 at 2:14 Right, thank you. My $G$ is not the standard Gauss sum for $p$ but one of its two versions. – darij grinberg Sep 21 at 3:30 ## 3 Answers Have you seen this blog post of mine? Summary: Use Geoff's argument to prove this up to a sign; then use $p$-adic arguments to nail down the sign. - 1 Thanks - a brilliant answer! (And no, for some reason I have not been following sbseminar; I believe it is because the last times I checked it, it was too much about analysis.) This really ought to be a PDF rather than a blog post (think of the people who want to print this out; also think of the ugliness of Wordpress's latex.php generated formulas). Also, I have posted a comment with some minor errors you might want to fix. – darij grinberg Aug 19 2011 at 18:00 ### 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. (restored slightly simplified version of earlier post). How about: $1- \zeta^{p-1} = -\zeta^{-1}(1-\zeta)$. Doing likewise for $\zeta^{2},\ldots,\zeta^{\frac{p-1}{2}}$, and setting $\alpha = \prod_{i=1}^{\frac{p-1}{2}} (1- \zeta^{i}),$ we see that $\bar{\alpha} = (-1)^{\frac{p-1}{2}} \bar{\zeta}^{\frac{p^2-1}{8}} \alpha.$ Hence we have `$$p = \prod_{i=1}^{p-1}(1- \zeta^i) = (-1)^{\frac{p-1}{2}} \bar{\zeta}^{\frac{p^2-1}{8}}\prod_{i=1}^{\frac{p-1}{2}} (1- \zeta^{i})^{2}.$$` - Now we have to rule out the "false" square roots... – darij grinberg Aug 19 2011 at 16:10 1 I would have called your post redundant if David Speyer had repeated your calculation. David did not do that, he referred to it. Now that you have removed the calculation, I ask that you restore it so that David's (and your) answer has more impact. Gerhard "Ask Me About System Design" Paseman, 2011.08.19 – Gerhard Paseman Aug 19 2011 at 21:52 1 OK, Gerhard, I will do that. The calculation is repeated in the blog linked to, but I agree there is a slight non-sequitur. – Geoff Robinson Aug 19 2011 at 22:21 Thank you Geoff. I like it enough that I will give it my fourth vote. Gerhard "Ask Me About System Design" Paseman, 2011.08.19 – Gerhard Paseman Aug 19 2011 at 22:35 A few historical remarks about algebraic determinations of the sign of the quadratic gaussian sum might not be out of order. The proof in David's post was first given by Kronecker, according to Hasse's Vorlesungen. The only analytic ingredient is the determination of the sign of the sin function. This proof is reproduced in Fröhlich and Taylor, Algebraic Number Theory, pp. 228--231. A different algebraic proof, using the same analytic ingredient, was given by Schur and can be found in Borevich and Shafarevich, Number Theory, pp. 349--353. Hasse's Vorlesungen also contain a proof by Mordell in which the analytic ingredient is replaced by the fact that if a polynomidal $f\in{\mathbf Z}[T]$ has opposite signs at $a,b\in{\mathbf R}$, then it has a root between $a$ and $b$. This can be proved using the purely algebraic theory of Artin and Schreier. If you are looking for a proof using more analysis, not less, see Rohrlich's survey on Root Numbers in Arithmetic of L-functions, pp. 353--448. Addendum. A nice (if somewhat dated) survey on The determination of Gauss sums can be found in the BAMS 5 (1981), 107-129. I learnt there that the proof attributed by Hasse to Kronecker actually goes back to Cauchy. New proofs are still being given; see for example Gurevich, Hadani, and Howe, Quadratic reciprocity and the sign of the Gauss sum via the finite Weil representation. Int. Math. Res. Not. IMRN 2010, no. 19, 3729–3745, available 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": 47, "mathjax_display_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": 0.9208301901817322, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-equations/166915-notation-question.html	# Thread: 1. ## Notation Question What is meant by this notation $u(x,y;\lambda)\mbox{?}$ Thanks. 2. It usually means there's some parameter $\lambda$ in the solution $u(x,y)$ (if it's in the context of the solution of a differential equations which I think has been the focus of some of your lasts posts). 3. If you wouldn't mind, can you show me an example of this? It can be as simple as possible. 4. Surely a family of functions $\{u_{\lambda}:A\subset\mathbb{R}^2\rightarrow \mathbb{R}\}$ Example: $u_{\lambda}(x,y)=\lambda^2\sin (x^3y)$ Fernando Revilla Edited: Sorry, I didn't see Danny's post. 5. Originally Posted by dwsmith If you wouldn't mind, can you show me an example of this? It can be as simple as possible. Do you know the Lagrange Mulitplier method for optimisation? 6. Like when you have an equation, constraint, and your goal is to maximize or minimize? 7. yep. 8. I don't see how they related though. I am not saying they aren't I just don't see it. Well, besides that lambda is used in the Lagrange Multiplier.	{"extraction_info": {"found_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": 5, "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": 0.95791095495224, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/3114/what-is-the-correct-value-for-certainty-in-rsa-key-pair-generation/3116	# What is the correct value for “certainty” in RSA key pair generation? I'm creating an RSA key pair in Bouncy Castle and need to specify an int value for certainty. This Stack Overflow answer says it is a relative test for how prime the values are. There is another answer that says this value should be adjusted relative to the key length. Question • What are the correct values for certainty relative to key length (how did you determine this?) • What does it mean to say "certainty of x bits" of a number? (If it's possible to sub-divide a number and certify bits, which bits are being certified?) - ## 1 Answer Certainty of $x$ bits means that the probability that something (in this case $p$ being prime) not being true is smaller than $2^{-x}$. This is the same probability as guessing a random $x$-bit value correctly on the first try, hence the name. How to select $x$? We want the probability of $p$ (and $q$) not being prime to be small enough that a failure probability in this point is not larger than other ways the system could be broken - like guessing a symmetric key, factoring the modulus etc. So here a correspondence table of symmetric and asymmetric key sizes should help. Pick the same prime certainty as you would pick an symmetric key size accompanying your public key usage. - Also, your algorithmic probability of failure is physically bounded by the failure probability of your hardware, so for instance, a $2^{-512}$ probability of failure is quite overkill. – Thomas Jul 2 '12 at 0:36 As a confirmation, the certainty in the question is traceable to that in a paragraph just above this, reading: "$\mathtt{certainty }$ - a measure of the uncertainty that the caller is willing to tolerate. The probability that the new BigInteger represents a prime number will exceed $(1-1/{2^{\mathtt{certainty}}})$. The execution time of this constructor is proportional to the value of this parameter." – fgrieu Jul 2 '12 at 7:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9164390563964844, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/63651/how-can-i-prove-formally-that-this-set-is-not-path-connected	# how can I prove formally that this set is not path connected Let $\mathbb{R}^2$ with its usual topology, let $D$ the set of all the lines that pass through the origin, with rational slope. And add to $D$ some point that does not lie in any of the lines ( call this new set $X$). For example $(e,e 2^{1/2})$ It is easy to show that this set is connected since, all the lines are connected and the intersection of these is not empty, then the union is also connected. And any set that lies between $D$ and its closure will also be connected, the closure is all $\mathbb{R}^2$ so $X$ is also connected. The question is if this set it´s also path connected, but I don´t think so , only for geometrically reasons. But in general it´s difficult to me, that this assertion )=. If someone can help me )= - 1 "it's" is a contraction; it stands for "it is". It is not the possessive (the possessive is "its"), and it is not a synonym for "is". – Arturo Magidin Sep 11 '11 at 19:51 sorry I do not speak english )= – Daniel Sep 11 '11 at 19:53 If this is homework, you should add the "homework" tag. – Nate Eldredge Sep 11 '11 at 19:57 2 @Daniel: And I'm giving you a tip on how to write it better in the future. – Arturo Magidin Sep 11 '11 at 19:57 1 Hint: Argue by contradiction. Suppose there is a path connecting the new point $z$ to some point of $D$. Call the path $(x(t), y(t))$ for $0 \le t \le 1$. What can you say about the slopes $y(t)/x(t)$? – Nate Eldredge Sep 11 '11 at 19:58 show 1 more comment ## 1 Answer Let $f$ be a path starting at the new point, with coordinate functions $f_1$ and $f_2$. Consider the function $\frac{f_2}{f_1}$, which starts out irrational, and must become rational if the path ever reaches an element of $D$. Between any irrational and rational number, there is an irrational number, and you can apply the intermediate value theorem to $\frac{f_2}{f_1}$. - This is not a homework )= – Daniel Sep 11 '11 at 20:32 @Daniel: Oh. Did you mean to make that comment on your question? – Jonas Meyer Sep 11 '11 at 20:35 good Idea, thanks! – Daniel Sep 12 '11 at 1:15	{"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": 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": 0.9359270334243774, "perplexity_flag": "head"}
http://www.reference.com/browse/wiki/Monte_Carlo_method	Definitions # Monte Carlo method Monte Carlo methods are a class of computational algorithms that rely on repeated random sampling to compute their results. Monte Carlo methods are often used when simulating physical and mathematical systems. Because of their reliance on repeated computation and random or pseudo-random numbers, Monte Carlo methods are most suited to calculation by a computer. Monte Carlo methods tend to be used when it is infeasible or impossible to compute an exact result with a deterministic algorithm. The term Monte Carlo method was coined in the 1940s by physicists working on nuclear weapon projects in the Los Alamos National Laboratory. ## Overview There is no single Monte Carlo method; instead, the term describes a large and widely-used class of approaches. However, these approaches tend to follow a particular pattern: 1. Define a domain of possible inputs. 2. Generate inputs randomly from the domain, and perform a deterministic computation on them. 3. Aggregate the results of the individual computations into the final result. For example, the value of π can be approximated using a Monte Carlo method. Draw a square of unit area on the ground, then inscribe a circle within it. Now, scatter some small objects (for example, grains of rice or sand) throughout the square. If the objects are scattered uniformly, then the proportion of objects within the circle vs objects within the square should be approximately π/4, which is the ratio of the circle's area to the square's area. Thus, if we count the number of objects in the circle, multiply by four, and divide by the total number of objects in the square (including those in the circle), we get an approximation to π. Notice how the π approximation follows the general pattern of Monte Carlo algorithms. First, we define a domain of inputs: in this case, it's the square which circumscribes our circle. Next, we generate inputs randomly (scatter individual grains within the square), then perform a computation on each input (test whether it falls within the circle). At the end, we aggregate the results into our final result, the approximation of π. Note, also, two other common properties of Monte Carlo methods: the computation's reliance on good random numbers, and its slow convergence to a better approximation as more data points are sampled. If grains are purposefully dropped into only, for example, the center of the circle, they will not be uniformly distributed, and so our approximation will be poor. An approximation will also be poor if only a few grains are randomly dropped into the whole square. Thus, the approximation of π will become more accurate both as the grains are dropped more uniformly and as more are dropped. ## History The name "Monte Carlo" was popularized by physics researchers Stanislaw Ulam, Enrico Fermi, John von Neumann, and Nicholas Metropolis, among others; the name is a reference to a famous casino in Monaco where Ulam's uncle would borrow money to gamble. The use of randomness and the repetitive nature of the process are analogous to the activities conducted at a casino. Random methods of computation and experimentation (generally considered forms of stochastic simulation) can be arguably traced back to the earliest pioneers of probability theory (see, e.g., Buffon's needle, and the work on small samples by William Gosset), but are more specifically traced to the pre-electronic computing era. The general difference usually described about a Monte Carlo form of simulation is that it systematically "inverts" the typical mode of simulation, treating deterministic problems by first finding a probabilistic analog (see Simulated annealing). Previous methods of simulation and statistical sampling generally did the opposite: using simulation to test a previously understood deterministic problem. Though examples of an "inverted" approach do exist historically, they were not considered a general method until the popularity of the Monte Carlo method spread. Perhaps the most famous early use was by Enrico Fermi in 1930, when he used a random method to calculate the properties of the newly-discovered neutron. Monte Carlo methods were central to the simulations required for the Manhattan Project, though were severely limited by the computational tools at the time. Therefore, it was only after electronic computers were first built (from 1945 on) that Monte Carlo methods began to be studied in depth. In the 1950s they were used at Los Alamos for early work relating to the development of the hydrogen bomb, and became popularized in the fields of physics, physical chemistry, and operations research. The Rand Corporation and the U.S. Air Force were two of the major organizations responsible for funding and disseminating information on Monte Carlo methods during this time, and they began to find a wide application in many different fields. Uses of Monte Carlo methods require large amounts of random numbers, and it was their use that spurred the development of pseudorandom number generators, which were far quicker to use than the tables of random numbers which had been previously used for statistical sampling. ## Applications Monte Carlo simulation methods are especially useful in studying systems with a large number of coupled degrees of freedom, such as fluids, disordered materials, strongly coupled solids, and cellular structures (see cellular Potts model). More broadly, Monte Carlo methods are useful for modeling phenomena with significant uncertainty in inputs, such as the calculation of risk in business (for its use in the insurance industry, see stochastic modelling). A classic use is for the evaluation of definite integrals, particularly multidimensional integrals with complicated boundary conditions. Monte Carlo methods in finance are often used to calculate the value of companies, to evaluate investments in projects at corporate level or to evaluate financial derivatives. The Monte Carlo method is intended for financial analysts who want to construct stochastic or probabilistic financial models as opposed to the traditional static and deterministic models. Monte Carlo methods are very important in computational physics, physical chemistry, and related applied fields, and have diverse applications from complicated quantum chromodynamics calculations to designing heat shields and aerodynamic forms. Monte Carlo methods have also proven efficient in solving coupled integral differential equations of radiation fields and energy transport, and thus these methods have been used in global illumination computations which produce photorealistic images of virtual 3D models, with applications in video games, architecture, design, computer generated films, special effects in cinema, business, economics and other fields. Monte Carlo methods are useful in many areas of computational mathematics, where a lucky choice can find the correct result. A classic example is Rabin's algorithm for primality testing: for any n which is not prime, a random x has at least a 75% chance of proving that n is not prime. Hence, if n is not prime, but x says that it might be, we have observed at most a 1-in-4 event. If 10 different random x say that "n is probably prime" when it is not, we have observed a one-in-a-million event. In general a Monte Carlo algorithm of this kind produces one correct answer with a guarantee n is composite, and x proves it so, but another one without, but with a guarantee of not getting this answer when it is wrong too often — in this case at most 25% of the time. See also Las Vegas algorithm for a related, but different, idea. ## Monte Carlo Simulation versus “What If” Scenarios The opposite of Monte Carlo simulation might be considered deterministic modelling using single-point estimates. Each uncertain variable within a model is assigned a “best guess” estimate. Various combinations of each input variable are manually chosen (such as best case, worst case, and most likely case), and the results recorded for each so-called “what if” scenario. [citation: David Vose: “Risk Analysis, A Quantitative Guide,” Second Edition, p. 13, John Wiley & Sons, 2000.] By contrast, Monte Carlo simulation considers random sampling of probability distribution functions as model inputs to produce hundreds or thousands of possible outcomes instead of a few discrete scenarios. The results provide probabilities of different outcomes occurring. [citation: Ibid, p. 16] For example, a comparison of a spreadsheet cost construction model run using traditional “what if” scenarios, and then run again with Monte Carlo simulation and Triangular probability distributions shows that the Monte Carlo analysis has a narrower range than the “what if” analysis. This is because the “what if” analysis gives equal weight to all scenarios. [citation: Ibid, p. 17, showing graph] ### Application areas Areas of application include: • Graphics, particularly for ray tracing; a version of the Metropolis-Hastings algorithm is also used for ray tracing where it is known as Metropolis light transport • Modeling light transport in biological tissue • Monte Carlo methods in finance • Reliability engineering • In simulated annealing for protein structure prediction • In semiconductor device research, to model the transport of current carriers • Environmental science, dealing with contaminant behavior • Monte Carlo method in statistical physics; in particular, Monte Carlo molecular modeling as an alternative for computational molecular dynamics. • Search And Rescue and Counter-Pollution. Models used to predict the drift of a life raft or movement of an oil slick at sea. • In Probabilistic design for simulating and understanding the effects of variability • In Physical chemistry, particularly for simulations involving atomic clusters • In computer science • Modeling the movement of impurity atoms (or ions) in plasmas in existing and tokamaks (e.g.: DIVIMP). • In experimental particle physics, for designing detectors, understanding their behavior and comparing experimental data to theory • Nuclear and particle physics codes using the Monte Carlo method: • GEANT - CERN's simulation of high energy particles interacting with a detector. • CompHEP, PYTHIA - Monte-Carlo generators of particle collisions • MCNP(X) - LANL's radiation transport codes • MCU - universal computer code for simulation of particle transport (neutrons, photons, electrons) in three-dimensional systems by means of the Monte Carlo method • EGS - Stanford's simulation code for coupled transport of electrons and photons • PEREGRINE - LLNL's Monte Carlo tool for radiation therapy dose calculations • BEAMnrc - Monte Carlo code system for modeling radiotherapy sources (LINAC's) • PENELOPE - Monte Carlo for coupled transport of photons and electrons, with applications in radiotherapy • MONK - Serco Assurance's code for the calculation of k-effective of nuclear systems • Modelling of foam and cellular structures • Modeling of tissue morphogenesis • Computation of holograms ### Other methods employing Monte Carlo • Assorted random models, e.g. self-organised criticality • Direct simulation Monte Carlo • Dynamic Monte Carlo method • Kinetic Monte Carlo • Quantum Monte Carlo • Quasi-Monte Carlo method using low-discrepancy sequences and self avoiding walks • Semiconductor charge transport and the like • Electron microscopy beam-sample interactions • Stochastic optimization • Cellular Potts model • Markov chain Monte Carlo • Cross-Entropy Method • Applied information economics • Monte Carlo localization ## Use in mathematics In general, Monte Carlo methods are used in mathematics to solve various problems by generating suitable random numbers and observing that fraction of the numbers obeying some property or properties. The method is useful for obtaining numerical solutions to problems which are too complicated to solve analytically. The most common application of the Monte Carlo method is Monte Carlo integration. ### Integration Deterministic methods of numerical integration operate by taking a number of evenly spaced samples from a function. In general, this works very well for functions of one variable. However, for functions of vectors, deterministic quadrature methods can be very inefficient. To numerically integrate a function of a two-dimensional vector, equally spaced grid points over a two-dimensional surface are required. For instance a 10x10 grid requires 100 points. If the vector has 100 dimensions, the same spacing on the grid would require 10100 points—far too many to be computed. 100 dimensions is by no means unreasonable, since in many physical problems, a "dimension" is equivalent to a degree of freedom. (See Curse of dimensionality.) Monte Carlo methods provide a way out of this exponential time-increase. As long as the function in question is reasonably well-behaved, it can be estimated by randomly selecting points in 100-dimensional space, and taking some kind of average of the function values at these points. By the law of large numbers, this method will display $1/sqrt\left\{N\right\}$ convergence—i.e. quadrupling the number of sampled points will halve the error, regardless of the number of dimensions. A refinement of this method is to somehow make the points random, but more likely to come from regions of high contribution to the integral than from regions of low contribution. In other words, the points should be drawn from a distribution similar in form to the integrand. Understandably, doing this precisely is just as difficult as solving the integral in the first place, but there are approximate methods available: from simply making up an integrable function thought to be similar, to one of the adaptive routines discussed in the topics listed below. A similar approach involves using low-discrepancy sequences instead—the quasi-Monte Carlo method. Quasi-Monte Carlo methods can often be more efficient at numerical integration because the sequence "fills" the area better in a sense and samples more of the most important points that can make the simulation converge to the desired solution more quickly. #### Integration methods • Direct sampling methods • Random walk Monte Carlo including Markov chains • Gibbs sampling ### Optimization Another powerful and very popular application for random numbers in numerical simulation is in numerical optimization. These problems use functions of some often large-dimensional vector that are to be minimized (or maximized). Many problems can be phrased in this way: for example a computer chess program could be seen as trying to find the optimal set of, say, 10 moves which produces the best evaluation function at the end. The traveling salesman problem is another optimization problem. There are also applications to engineering design, such as multidisciplinary design optimization. Most Monte Carlo optimization methods are based on random walks. Essentially, the program will move around a marker in multi-dimensional space, tending to move in directions which lead to a lower function, but sometimes moving against the gradient. ### Inverse problems Probabilistic formulation of inverse problems leads to the definition of a probability distribution in the model space. This probability distribution combines a priori information with new information obtained by measuring some observable parameters (data). As, in the general case, the theory linking data with model parameters is nonlinear, the a posteriori probability in the model space may not be easy to describe (it may be multimodal, some moments may not be defined, etc.). When analyzing an inverse problem, obtaining a maximum likelihood model is usually not sufficient, as we normally also wish to have information on the resolution power of the data. In the general case we may have a large number of model parameters, and an inspection of the marginal probability densities of interest may be impractical, or even useless. But it is possible to pseudorandomly generate a large collection of models according to the posterior probability distribution and to analyze and display the models in such a way that information on the relative likelihoods of model properties is conveyed to the spectator. This can be accomplished by means of an efficient Monte Carlo method, even in cases where no explicit formula for the a priori distribution is available. The best-known importance sampling method, the Metropolis algorithm, can be generalized, and this gives a method that allows analysis of (possibly highly nonlinear) inverse problems with complex a priori information and data with an arbitrary noise distribution. For details, see Mosegaard and Tarantola (1995) , or Tarantola (2005) . ## Monte Carlo and random numbers Interestingly, Monte Carlo simulation methods do not generally require truly random numbers to be useful - for other applications, such as primality testing, unpredictability is vital (see Davenport (1995)). Many of the most useful techniques use deterministic, pseudo-random sequences, making it easy to test and re-run simulations. The only quality usually necessary to make good simulations is for the pseudo-random sequence to appear "random enough" in a certain sense. What this means depends on the application, but typically they should pass a series of statistical tests. Testing that the numbers are uniformly distributed or follow another desired distribution when a large enough number of elements of the sequence are considered is one of the simplest, and most common ones. ## References • Bernd A. Berg, Markov Chain Monte Carlo Simulations and Their Statistical Analysis (With Web-Based Fortran Code), World Scientific 2004, ISBN 981-238-935-0. • Arnaud Doucet, Nando de Freitas and Neil Gordon, Sequential Monte Carlo methods in practice, 2001, ISBN 0-387-95146-6. • P. Kevin MacKeown, Stochastic Simulation in Physics, 1997, ISBN 981-3083-26-3 • Harvey Gould & Jan Tobochnik, An Introduction to Computer Simulation Methods, Part 2, Applications to Physical Systems, 1988, ISBN 0-201-16504-X • C.P. Robert and G. Casella. "Monte Carlo Statistical Methods" (second edition). New York: Springer-Verlag, 2004, ISBN 0-387-21239-6 • R.Y. Rubinstein and D.P. Kroese (2007). "Simulation and the Monte Carlo Method" (second edition). New York: John Wiley & Sons, ISBN 978-0-470-17793-8. • Mosegaard, Klaus., and Tarantola, Albert, 1995. Monte Carlo sampling of solutions to inverse problems. J. Geophys. Res., 100, B7, 12431-12447. • Tarantola, Albert, Inverse Problem Theory (free PDF version), Society for Industrial and Applied Mathematics, 2005. ISBN 0-89871-572-5 • Nicholas Metropolis, Arianna W. Rosenbluth, Marshall N. Rosenbluth, Augusta H. Teller and Edward Teller, "Equation of State Calculations by Fast Computing Machines", Journal of Chemical Physics, volume 21, p. 1087 (1953) () • N. Metropolis and S. Ulam, "The Monte Carlo Method", Journal of the American Statistical Association, volume 44, number 247, pp. 335–341 (1949) () • Fishman, G.S., (1995) Monte Carlo: Concepts, Algorithms, and Applications, Springer Verlag, New York. • Judgement under Uncertainty: Heuristics and Biases, ed. D. Kahneman and A. Tversky,(Cambridge University Press, 1982) • R. E. Caflisch, Monte Carlo and quasi-Monte Carlo methods, Acta Numerica vol. 7, Cambridge University Press, 1998, pp. 1-49. ## External links • Overview and reference list, Mathworld • Introduction to Monte Carlo Methods, Computational Science Education Project • Overview of formulas used in Monte Carlo simulation, the Quant Equation Archive, at sitmo.com • The Basics of Monte Carlo Simulations, University of Nebraska-Lincoln • Introduction to Monte Carlo simulation (for Excel), Wayne L. Winston • Monte Carlo Simulation, Prof. Don M. Chance • Monte Carlo Methods - Overview and Concept, brighton-webs.co.uk • Monte Carlo Simulation - Introduction, solver.com • Molecular Monte Carlo Intro, Cooper Union • Monte Carlo techniques applied to finance, Simon Leger • Monte Carlo techniques applied in physics • MonteCarlo Simulation in Finance, global-derivatives.com • Approximation of π with the Monte Carlo Method • Risk Analysis in Investment Appraisal, The Application of Monte Carlo Methodology in Project Appraisal, Savvakis C. Savvides • Primality Testing RevisitedProc. ISSAC 1992, James H. Davenport • Example of Calculating Pi using the Monte Carlo method, C++ • Probabilistic Assessment of Structures using the Monte Carlo method, Wikiuniversity paper for students of Structural Engineering • Statistical Analysis of Simulations, by Professor Wolfhard Janke, Universität Leipzig, Germany • A very intuitive and comprehensive introduction to Quasi-Monte Carlo methods ### Software • SimulAr - Free Monte Carlo Simulation Excel Add-In • The BUGS project (including WinBUGS and OpenBUGS) • Monte Carlo Simulation Tool for Excel by Oracle (formerly Decisioneering) • Monte Carlo Simulation Tool for Excel by Palisade • Monte Carlo Simulation Tool for Excel by Lumenaut • Monte Carlo Simulation for Business, Engineered, and Environmental Systems by GoldSim • Monte Carlo Simulation, Resampling, Bootstrap tool • YASAI: Yet Another Simulation Add-In - Free Monte Carlo Simulation Add-In for Excel created by Rutgers University • Easy Monte Carlo Tool - Free Monte Carlo Tool created for UK Every Child Matters programme Wikipedia, the free encyclopedia © 2001-2006 Wikipedia contributors (Disclaimer) This article is licensed under the GNU Free Documentation License. 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All rights reserved. • Please Login or Sign Up to use the Favorites feature • Please Login or Sign Up to use the Recent Searches feature FAVORITES RECENT	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.8673558831214905, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/15033/whats-the-correct-distribution-for-page-reading-time/15037	# What's the correct distribution for page reading time? I want to simulate a user reading a book. The amount of time spent on a page will be random, but subject to a particular distribution, I just don't know which one. A normal distribution has the problem that it can produce negative values, which is clearly not right. What's the correct distribution for this kind of random variable? - ## 2 Answers It helps to have data. Jakob Nielsen has measured reading times for web pages ("How Little Do Users Read", 2008), which gives some strong hints: • The data show that the variation in reading time is directly proportional to the number of words on a page. • The variation therefore should be expressed as a time relative to the page length, not as a fixed amount. • The log of this ratio has an approximately normal distribution with a standard deviation of around 12%. This scatterplot from Nielsen's report is valuable for revealing the amount of relative variation in reading times, even though only a small portion (about 18%) of each page is actually read. Notice how the absolute variation increases with word length (page size); this complication is handled by using the log of the relative variation. Don't forget that book pages have varying amounts of words, too. The variation will depend on accumulated tiny differences related to word length, page width, paragraph length, amount of dialog, and so on. For a book with uniform looking pages we can therefore expect that variation to be normal, except for the ends and beginnings of chapters. The end pages will have approximately a uniform distribution of word lengths. The beginning pages will have approximately a normal distribution with a smaller mean than the typical (full) page, depending on the page design. This gives a complex distribution but it's relatively easy to simulate. The parameters should include 1. The mean number of words per (full) page, $w$. You can easily measure this for actual books you are trying to simulate. 2. The standard deviation in words per (full) page, $s$. Again, this is readily measured. 3. The mean number of words per start page, $u$, also easily measured. 4. The log variation in reading times, $\sigma$. Use a value around 12% to start, based on Nielsen's study; consider looking at other studies for other realistic values. 5. The user's reading speed, $v$, as words per minute, say. Values around 200-250 wpm are often used, depending on the kind of reader. 6. The mean number of pages per chapter, $n$, also easily measured. 7. The mean page turning time, $t$. You could do your own little study of readers, perhaps by spending an hour with a stopwatch in a library :-). Don't be too fussy about this number--it will depend on the book size, page material, and the reader--but it could contribute enough time to be of interest in the simulation. The simulation should comprise an entire chapter, simulated as a sequence consisting of one start page, $n-2$ normal pages, and one end page. Simulate the number of words, $m$, as $$m = (n-2)w + u + zw + r$$ where $z$ has a uniform $(0,1)$ distribution and $r$ has a normal distribution with mean $0$ and standard deviation $s \sqrt{n}$. Draw a value $x$ from a normal distribution with mean $0$ and standard deviation $\sigma$. Multiply $m$ by $w \exp(x) / v$ to simulate the reading time. Add $n t$ minutes for the page turns. This process will capture the most important influences on reading time for a book with homogeneous text, uniform reading difficulty, and no illustrations. For more complex books, such as collections of readings, math or science, books with lots of dialog, illustrated books, and so on, the model may need to be more complex in order to be realistic. ### Edit It turns out we may be able to justify and flesh out the suggestion offered by @Jason, because it happens that this complex but realistic simulation can be extremely well approximated by a version of a Gamma distribution in most cases. We have to rescale and shift the Gamma, in addition to selecting its shape parameter. Here is a (typical) example based on $100,000$ iterations with $w=300$ words per page, $s=15$ words (SD per page), $u=100$ words per start page, $\sigma=0.12$, $v=250$ words per minute, $n=8$ pages per chapter, and $t = 0.04$ minutes per page turn. The histogram gives the distribution of results while the solid red curve is the PDF for a Gamma distribution with shape parameter $27.416$, scale parameter $0.2043$, offset by $2.98$ minutes. This approximation breaks down only for extremely short chapter lengths, but is still decent even when $n=3$: The potential advantage of this observation is that you can avoid estimating many of the parameters needed to model reading a simple, homogeneous book if you are willing to specify three independent parameters of the distribution, such as its mean, standard deviation, and skewness. For instance, if you have actual data about chapter reading times you could use the first three sample moments to fit a three-parameter Gamma distribution to the data, then perform the simulation via draws from that Gamma. Furthermore, if you assume the times to read the book chapters are independent, it is easy to add these Gammas (one per chapter) to obtain a distribution for the length of time to read the entire book (because the shape parameter for the sum of Gamma distributions having a common scale factor is the sum of their shape parameters). Even with minimal data (such as used here) you could run some simulations for a single chapter, fit a Gamma to those simulation results, and proceed to deduce (rather than simulate) the total book reading times. For instance, in this case the reading times for a book of $16$ chapters should follow a Gamma distribution with shape parameter $16 \times 27.4164$, scale parameter $0.2043$, offset by $16 \times 2.98$ minutes. For many books (having many chapters), the resulting distribution will be Normal for all practical purposes. The chance assigned to negative values by this distribution would be so astronomically small that it doesn't matter. The blue curve shows the distribution of book reading times. The dashed red curve superimposed on it is a Normal approximation. Neither distribution assigns any appreciable probability to times less than 240 minutes. - You could use a gamma distribution. Check it out on Wikipedia. Gamma distributions are commonly used to model waiting times, like the situation you have here. - How is reading a page like a waiting time? – whuber♦ Aug 31 '11 at 18:12 ... waiting until someone is finished reading? – Jason Aug 31 '11 at 18:23 1 The fact that it's an elapsed time doesn't mean it acts like the "waiting time" of queuing theory! – whuber♦ Aug 31 '11 at 18:40 It's a model, not reality. Do you have a better idea for a distribution to model reading time? – Jason Aug 31 '11 at 18:54 1 @Macro How do you estimate parameters without data? – whuber♦ Sep 1 '11 at 13:47 show 2 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "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": 0.3, "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": 0.9095410704612732, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/122800?sort=oldest	Products of Boolean algebras and probability measures thereon Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) These are really two questions, but the second presupposes the first. First, let $( B_i )_{i\in I}$ be an arbitrary family of Boolean algebras. I want to directly form a product of them that is like the product topology on the product of their Stone spaces, ideally without using AC (or BPI, pun not intended). I haven't worked with Boolean algebras as such before--no doubt this will show in what I say--and my little literature search finds at least one author doing products of Boolean algebras by passing to Stone spaces, which I don't want to do. I am thinking something along the lines of the space of finite formal joins of finite formal meets. If I can't find anything, I guess what I'll do will be something like define the "cylinder elements" as finite sets $\{(i_1,a_1),...,(i_n,a_n)\}$ of index-element pairs, with $a_j \in B_{i_j}$ and the $i_j$ all distinct, and then take the space of finite sets of cylinder elements, modulo an appropriate equivalence relation. But I don't want to reinvent the wheel or multiply nonstandard notation. I'd much rather be able to cite something and use whatever notation is standard. A good feature from my point of view of the above partially-described construction is that if $I\subset J$, then the product of $(B_i)_{i\in I}$ is a subalgebra of the product of $(B_i)_{i\in J}$--I'd ideally like a construction that does that. The second thing is that I need a result showing that if I have a (finitely-additive) probability measure $P_i$ on each $B_i$ (satisfying the obvious Boolean algebra analogues of the finitely-additive probability axioms), then I can get the natural measure on the product measure. (In the above notation, we will want $P(\{\{(i_1,a_1),...,(i_n,a_n)\}\}) = \Pi_{j=1}^n P_{i_j}(a_{i_j})$.) Again, this can't be hard to prove, but I don't want to spend space writing up a proof of this if I can just find a citation. - 3 Answers This is called the "direct sum" or "internal sum" of Boolean algebras. See Introduction to Boolean Algebras by Givant and Halmos, p. 427 for the abstract definition and p. 432 for the concrete description you want (the elements of the internal sum are finite joins of finite meets of elements and complements of elements from the disjoint union of the summands). - 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 the opposite of the category of Stone spaces is equivalent to the category of Boolean algebras, the product of Stone spaces would have to correspond to the coproduct of Boolean algebras. Thus what you want is to construct the coproduct of a family of Boolean algebras directly, but without invoking Stone duality. This is just general (categorical) universal algebra. The theory of Boolean algebras is a finitary algebraic theory, so on general grounds you can construct the coproduct of a family $\{B_i\}_{i \in I}$ by taking a filtered colimit over the system of finite coproducts. Usually (and more generally for commutative rings) this is called an (infinite) tensor product: `$$\bigotimes_{i \in I} B_i = \mathrm{colim}_{\mathrm{finite}\; F \subset I} \otimes_{i \in F} B_i$$` More concretely, this colimit is simply a set-theoretic union of finite tensor products of Boolean algebras, construed as commutative algebras over the ring $\mathbb{Z}/(2)$. This description matches that in Nik's answer. For the moment I'll stop here, although I might come back to address the second query on probability measures. - 2 Hm, for some reason someone downvoted this. If I said something wrong, I would like to have it explained to me. – Todd Trimble Feb 25 at 1:55 I gave the downvotes that everyone is crying about. You did not explain how to extend measures on Boolean algebras to measures on the free products. Also, even though the category of Boolean algebras is isomorphic to the category of Boolean rings, it makes little sense to consider Boolean algebras in terms of rings. The Handbook of Boolean Algebras does not mention Boolean rings after the first section of the first chapter. – Joseph Van Name Feb 27 at 2:07 1 Crying about, Joseph? Anyway, I like to consider Boolean algebras in terms of Boolean rings. Notice for example that the Stone space can be understood in terms of the spectrum (in the algebraic geometry sense) of the Boolean ring; it is pertinent to remark here that Stone began his work by establishing the equivalence between Boolean algebras and Boolean rings and working with the space of maximal ideals thereof. It makes little sense to me that you think this POV makes little sense! – Todd Trimble Feb 27 at 3:12 First of all, I would like to say that I do not see any good reason not to use Stone spaces in order to describe the coproduct of Boolean algebras. The product of Stone spaces is the most natural way to represent free products of Boolean algebras. However, besides Stone spaces, there are a couple other ways to represent the free product of Boolean algebras. First of all, if you represent Boolean algebras in terms of algebras of sets, then the free product is easy to describe. Assume that `$(X_{i},\mathcal{A}_{i})$` is an algebra of sets for $i\in I$. Let `$X=\prod_{i\in I}X_{i}$` and let `$\pi_{i}:X\rightarrow X_{i}$` be the projection mappings. Let `$\mathcal{A}$` be the algebra of sets over `$X$` generated by sets of the form `$\pi_{i}^{-1}[R]$` for $i\in I$. Then `$\mathcal{A}$` is the free product of the Boolean algebras `$\mathcal{A}_{i}$`. Furthermore, every element in `$\mathcal{A}$` can be written as a finite disjoint union of sets of the form `$\pi_{i_{1}}^{-1}[R_{1}]\cap...\cap\pi_{i_{n}}^{-1}[R_{n}]$` for `$R_{1}\in\mathcal{A}_{1},...,R_{n}\in\mathcal{A}_{n}$` (This is because the sets of the form `$\pi_{i_{1}}^{-1}[R_{1}]\cap...\cap\pi_{i_{n}}^{-1}[R_{n}]$` form a semialgebra of sets). Also, the free product of two Boolean algebras is the Boolean power of those two Boolean algebras (i.e. `$A\ast B\simeq A^{B}$` where `$A^{B}$` denotes the Boolean power). The Boolean power $A^{B}$ is the set of all continuous functions $f:S(B)\rightarrow A$ where $A$ is given the discrete topology and $S(B)$ is the Stone space of $B$. The Boolean power $A^{B}$ can also be described as the direct limit `$^{\lim}_{\longrightarrow}A^{\|p\|}$` where $p$ ranges over the set `$\mathbb{P}_{\omega}(B)$` of finite partitions of $B$ ordered by reverse refinement. In other words, $p\leq q$ if for each $a\in q$ there is a $b\in p$ with $a\leq b$. The transitional mappings `$A^{\|p\|}\rightarrow A^{\|q\|}$` are the natural surjections whenever $p\leq q$. See the book A Course in Universal Algebra by Stanley Burris and H.P. Sankappanaver for more details on the Boolean power. Now, in order to effortlessly put a finitely additive measure on the product of Stone spaces, we will need to use Stone duality and a couple results from measure theory. In essense we use the product of countably additive measures on $\sigma$-algebras to give us the product of finitely additive measures on Boolean algebras. Assume that for $i\in I$, $B_{i}$ is a Boolean algebra and $\mu_{i}$ is a finitely additive measure on $B_{i}$. Then by the Caratheodory Extension Theorem, there is a Baire measure $\nu_{i}$ on the Stone space $S(B_{i})$ such that `$\nu_{i}(\{\mathcal{U}\in S(B_{i})\|a\in\mathcal{U}\})=\mu_{i}(a)$` for `$a\in B_{i}$`. Let $\nu$ be the product measure of the measures `$\nu_{i}$` on the product space `$\prod_{i\in I}S(B_{i})$`. Then we get a measure on the free product `$\ast_{i\in I}B_{i}$` simply by restricting $\nu$ to the clopen sets of `$\prod_{i\in I}S(B_{i})$`. Also, the free product of Boolean algebras is completely described by the following property. Assume that $B$ is a Boolean algebra and `$B_{i}\subseteq B$` is a subalgebra for $i\in I$. The system of subalgebras `$(B_{i})_{i\in I}$` is said to be independent if whenever `$i_{1},....,i_{n}\in I$` are distinct elements and `$b_{i}\in B_{i},b_{i}>0$` for `$i\in\{i_{1},...,i_{n}\}$`, then `$b_{i_{1}}\wedge...\wedge b_{i_{n}}>0$`. The Boolean algebra `$B$` is the free product of the Boolean algebras `$(B_{i})_{i\in I}$` if and only if the Boolean algebras `$B_{i}$` are independent and $B$ is generated by `$\bigcup_{i\in I}B_{i}$`. - Right, but aren't you using AC when you represent Boolean algebras in terms of algebras of sets? – Alexander Pruss Feb 24 at 21:28 That is affirmative. I will need some form of choice to represent all Boolean algebras as algebras of sets. The axiom of choice usually makes our lives much easier. – Joseph Van Name Feb 25 at 0:34 You're also using the axiom of choice to take the product $\prod X_i$ (and have it be nontrivial). – Nik Weaver Feb 25 at 17:45 I always try to use the axiom of choice as much as possible. – Joseph Van Name Feb 25 at 18:00 Actually, in the application I wanted, the $B_i$ would all be the same, so the nontriviality of $\Pi X_i$ wouldn't need Choice. – Alexander Pruss Feb 25 at 18:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 42, "mathjax_display_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": 0.9384218454360962, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=240492	Physics Forums ## Sturm-Liouville (can someone tell me if this is right?) Hi, I solved a simple Sturm-Liouville problem and not sure if its right or not because the answer is odd. I was hoping someone could tell me if I did it right. [tex] \frac{d^{2}u}{dt^{2}}=\lambda u, 0<x<L, \frac{du}{dx}(0) = 0, u(0)=L [/tex] $$\lambda > 0$$ [tex] u = c_{1}cosh(\sqrt{\lambda}x)+c_{2}sinh(\sqrt{\lambda}x) [/tex] [tex] \frac{du}{dx} = c_{1}\sqrt{\lambda}sinh(\sqrt{\lambda}x)+c_{2}\sqrt{\lambda}cosh(\sqrt{ \lambda}x) [/tex] [tex] \frac{du}{dx}(0) = c_{2}=0; u(0)=u(L)\Rightarrow c_{1}cosh(\sqrt{\lambda}x)=c_{1} [/tex] It appears I cant find c1 unless i did something wrong... $$\lambda = 0$$ $$u = c_{1}+c_{2}x$$ $$\frac{du}{dx}(0) = 0\Rightarrow c_{1}=0$$ $$u(L) = u(0) \Rightarrow c_{2}=0$$ $$\lambda < 0$$ $$u = c_{1}cos(\sqrt{\lambda}x)+c_{2}sin(\sqrt{\lambda}x)$$ $$\frac{du}{dx}(0) = \sqrt{lambda}c_{2}cos(\sqrt{\lambda}0) \Rightarrow c_{2}=0$$ $$u(0) = u(L) \Rightarrow c_{1}=c_{1}cos(\sqrt{\lambda}L)$$ $$\lambda = \left(\frac{arccos(1)}{L}\right)^{2}$$ $$u = cos\left(\frac{arccos(1)x}{L}\right)$$ is that right? i've never seen an answer like that... Any help is appreciated! PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Hey Engage, Before I dive too much further into helping you out, we need to clarify some things about your initial problem: You have written: $$\frac{d^{2}u}{dt^{2}}=\lambda u$$ However, your boundary conditions are based on x, not t. If you are to solve this ODE, you'll find a solution in terms of t, of which you've continued in terms of x. Perhaps this is just a variable mix up? Perhaps it is the case that you wish to start out with the Laplacian of u? Either way, it serves to clarify the situation. Thanks, Coto Recognitions: Gold Member Science Advisor Staff Emeritus Quote by EngageEngage is that right? You should be able to answer for yourself! You're trying to solve an equation, aren't you? Is your answer actually a solution to it? ## Sturm-Liouville (can someone tell me if this is right?) yeah, I messed up the initial problem, the independent variable is x, not t. [tex] \frac{d^{2}u}{dx^{2}}=\lambda u [/tex] $$\lambda < 0$$ makes the diff eq: $$\frac{d^{2}u}{dx^{2}}=-\lambda u$$ $$\frac{d}{dx}\left(-sin(arcos(1)\frac{x}{L})(\frac{arcos(1)}{L})\right)= -cos(arcos(1)\frac{x}{L})\left(\frac{arcos(1)}{L}\right)^2=-\lambda u$$ ... So i guess it is right... Thanks for the help! So does this mean there is only one eigenfunction then? Hey Engage, I suggest you take a look at your last step, where you solve for lambda. The idea is there, but you've overstepped what you should have done. My suggestion is to divide out the c1's such that you have: $$c_{1}=c_{1}cos(\sqrt{\lambda}L) \rightarrow 1 = cos(\sqrt{\lambda}L)$$ So then what does the argument of cosine have to be such that cos of that is always one. This is in general the way you solve these problems. You want to find the eigenvalues lambda such that this argument always holds. For finite boundary conditions, lambda is discrete. Hopefully this helps. Coto PS. It seems you do not have enough information to solve for the constant. And is it u(0) = u(L) or u(0) = L ? its u(0) = u(L). Thanks for the help. Also, i'm not quite sure what you mean Coto in your post #6. I though thtis is what I did to find lambda. did i mess that up? Thank you all for the help! Ahh, i see so arcos(1) will be 2pin where n is any integer including zero, is that correct? so $$\lambda_{n} = \left(\frac{2n\pi}{L}\right)^{2}$$ $$u_{n} = cos\left(\frac{2n\pi x}{L}\right)$$ $$u =\sum^{\infty}_{n=0}cos\left(\frac{2n\pi x}{L}\right)$$ Thread Tools \| \| \| \| \|------------------------------------------------------------------------------\|----------------------------\|---------\| \| Similar Threads for: Sturm-Liouville (can someone tell me if this is right?) \| \| \| \| Thread \| Forum \| Replies \| \| \| Calculus & Beyond Homework \| 2 \| \| \| Calculus & Beyond Homework \| 2 \| \| \| Calculus & Beyond Homework \| 1 \| \| \| Calculus & Beyond Homework \| 0 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 19, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9210147261619568, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/160390/question-on-sequences	# Question on sequences Exercise 37 in Apostol $10.20$ asks to find all complex $z$ such that $$\sum_{n=1}^{\infty} \frac{(-1)^n}{z+n}$$ converges. I suspect this requires the use of either Abel's Test or Dirichlet's Test. My attempt so far has been to set $\{b_n\}=\frac{1}{n}$, which is a decreasing sequence of real numbers that converges to $0$. Then, I set $\{a_n\}=(-1)^n\frac{n}{z+n}$. As $n\to\infty$, $$\{a_n\} \to (-1)^ne^{i\arg(\frac{n}{z+n})}$$ since $$\|\frac{n}{z+n}\|=\frac{n}{\|z+n\|}\to 1$$ In order to prove that this converges for all complex $z\not=-1,-2,\dots$, I must show that $A_n=\sum_{k=1}^{n} a_n$ is a bounded sequence (not necessarily that it converges). This would satisfy the hypotheses for Dirichlet's test. - 2 $\frac{1}{z+n}\to0$ for $n\to\infty$. so the oscillating series converges. – akkkk Jun 19 '12 at 17:26 1 @Auke: Alternating series test requires terms to (eventually) decrease to 0, not just converge. You can easily construct a counterexamples. – Erick Wong Jun 19 '12 at 17:34 ## 3 Answers $$c_n:=\frac{1}{z+n}=\frac{x+n}{\|z+n\|^2}-i\frac{y}{\|z+n\|^2}=:a_n-ib_n \quad \forall \ z=x+iy \ne -n.$$ For every $z \in \mathbb{C}\setminus(-\mathbb{N})$ the sequences $a_n, \ b_n$ decrease to $0$, therefore the series $\sum_{n=1}^\infty(-1)^nc_n$ converges for every $z \in \mathbb{C}\setminus(-\mathbb{N})$. - Clear and concise! – Andrew Salmon Jun 19 '12 at 22:55 Very nice answer! – albmiz-mth Feb 9 at 3:25 Use Cauchy property of convergent sequences and write $$\sum_{k=m}^n \frac{(-1)^k}{z+k}$$ Then write (suppose $m$ even) $$\frac{1}{z+m} - \frac{1}{z+m+1} = \frac{1}{(z+m)(z+m+1)} \sim m^{-2}$$ This should persuade you that the series converges for $z$ that makes any denominator different from zero. - We can use Abel transform. Let $s_n:=\sum_{j=0}^n(-1)^j$. We have for $n$ and $m$ integers, and $z\notin -\Bbb N$, \begin{align} \sum_{j=n+1}^{n+m}\frac{(-1)^j}{z+j}&=\sum_{j=n+1}^{n+m}\frac{s_j-s_{j-1}}{z+j}\\ &=\sum_{k=n+1}^{n+m}\frac{s_k}{z+k}-\sum_{k=n}^{m+n-1}\frac{s_k}{z+k+1}\\ &=-\frac{s_n}{z+n+1}+\frac{s_{n+m}}{z+m+n}+\sum_{k=n+1}^{n+m}\frac{s_k}{(z+k)(z+k+1)}. \end{align} This gives, for $n,m> \|z\|$, that $$\left\|\sum_{j=n+1}^{n+m}\frac{(-1)^j}{z+j}\right\|\leq \frac 1{n+1-\|z\|}+\frac 1{m+n-\|z\|}+\sum_{k=n+1}^{n+m}\frac 1{(k-\|z\|)(k+1-\|z\|)}.$$ Since the series $\sum_{k=1}^{+\infty}\frac 1{(k-\|z\|)(k+1-\|z\|)}$ is convergent, we get that the sequence $\{\sum_{k=1}^n\frac{(-1)^k}{z+k}\}$ is Cauchy if $z\neq -k$ for each integer $k$. -	{"extraction_info": {"found_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": 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": 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": 0.832169234752655, "perplexity_flag": "head"}
http://mathoverflow.net/questions/84848?sort=oldest	## volume of compact simple Lie groups under the natural Euclidean embedding ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am looking for a quick reference for the volume formula for all the compact simple Lie groups embedded as matrix groups in the natural way. The one I care most for are the real orthogonal groups. I don't see how to deduce them from the Weyl integration formula. I believe they are of the order diameter raised to the power of the dimension. So for instance for $SO(n)$ the volume should be of order $n^{\Theta(n^2)}$. - 6 For the orthogonal group $SO(n)$, isn't it just the product of the volumes of the $k$-spheres in their natural embeddings, as $k$ ranges from 1 to $n{-}1$? Similarly, for $U(n)$, it should be the product of the $2k{-}1$-spheres as $k$ ranges from $1$ to $n$, and so on. – Robert Bryant Jan 4 2012 at 1:26 Oh you are right. Thanks for the observation! – John Jiang Jan 4 2012 at 1:29 4 This is not quite true: $SO(2)$ as a circle in the vector space of $2\times 2$ matrices has radius $\sqrt 2$, in the metric that I am thinking of. – Tom Goodwillie Jan 4 2012 at 1:38 4 @Tom: You are quite right. I forgot about the factor of $(\sqrt{2})^{k-1}$ that you have to put in at each level because the natural map $\pi:SO(k)\to S^{k-1}$ is not a Riemannian submersion, but is a Riemannian submersion scaled by a factor of $1/\sqrt{2}$, i.e., each horizontal vector for this bundle is shrunk by a factor of $\sqrt{2}$ by the differential of $\pi$. Thus, the overall factor you need to multiply the answer I gave by is $(\sqrt{2})^{n(n-1)/2}$. The recipe I gave for $U(n)$ is not right either, as the map $\pi:U(k)\to S^{2k-1}$ shrinks horizontal volumes by $2^{k-1}$. Sorry. – Robert Bryant Jan 4 2012 at 2:21 I guess in general even when $\pi$ is not a Riemannian submersion one could integrate the inverse of the generalized jacobian map along the preimage of $\pi$, and then integrate that over the base manifold? – John Jiang Jan 4 2012 at 2:54 show 1 more comment ## 1 Answer Explicit formulas for the volumes of compact Lie groups, with respect to their Haar measures, given in terms of their root data are given in: M. S. Marinov: Invariant volumes of J. Phys A: Math. Gen. 13 (1980) 3357-3376. Availabe in Prof. Marinov's memorial site. The final formula is given in equation 19, which is also tabulated for the various simple types. The article compares the general formula with the sphere based computations for some of the classical groups. The method of computation in the article is based on the Weyl's integration formula, however, the volume of the flag manifold (called the orbit space in the article) was read from the evolution operator of the Laplacian on the group manifold. Prof. Marinov's interest in this subject was due to his work in quantum mechanics and path integrals on group manifolds. - I have actually run into this article before but the formula is a bit hard for me to parse. My interest comes from analyzing the Kac random walk on $SO(n)$. – John Jiang Jan 4 2012 at 8:57	{"extraction_info": {"found_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": 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": 0.9385359287261963, "perplexity_flag": "head"}
http://mathdl.maa.org/mathDL/61/?pa=newCollection&sa=viewResourceEntry&resourceId=1330	Search ## Search Course Communities: Keyword Advanced Search # SticiGui Law of Large Numbers Demonstration Course Topic(s): Probability \| Convergence Theorems, law of large numbers The Law of Large Numbers says that in repeated, independent trials with the same probability $$p$$ of success in each trial, the chance that the percentage of successes differs from the probability $$p$$ by more than a fixed positive amount, $$\epsilon > 0$$, converges to zero as the number of trials $$n$$ goes to infinity, for every positive $$\epsilon$$. Note two things: The difference between the number of successes and the number of trials times the chance of success in each trial (the expected number of successes) tends to grow as the number of trials increases. (In fact, this difference tends to grow like the square-root of the number of trials.) Although the chance of a large difference between the percentage of successes and the chance of success gets smaller and smaller as $$n$$ grows, nothing prevents the difference from being large in some sequences of trials. The assumption that this difference always tends to zero, as opposed to this difference having a large probability of being arbitrarily close to zero, is the difference between the Law of Large Numbers, which is a mathematical theorem, and the Empirical Law of Averages, which is an assumption about how the world works that lies at the base of the Frequency Theory of probability. Resource URL: http://statistics.berkeley.edu/~stark/Java/Html/lln.htm ### To rate this resource on a 1-5 scheme, click on the appropriate icosahedron: \| \| \| \| \| \| \| \| \|----\|----\|----\|----\|----\|---------------------\|----------------------\| \| \| \| \| \| \| Current rating: 2.9 \| number of votes: 205 \| \| 1 \| 2 \| 3 \| 4 \| 5 \| \| \| Subject classification(s): Univariate Distributions \| Probability \| Statistics and Probability Creator(s): Phil Stark Contributor(s): Phil Stark This resource was cataloged by Ivo Dinov Publisher: SticiGui This review was published on September 20, 2012 ### Comments Report a problem with this resource.	{"extraction_info": {"found_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": 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": 10, "end_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": 0.9340851902961731, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/18057/simpler-solution-to-this-geometry-trig-problem/18058	# Simpler solution to this geometry/trig problem? i had a geometry/trignometry problem come up at work today, and i've been out of school too long: i've lost my tools. i'm starting with a rectangle of known width (`w`) and height (`h`). For graphical simplification i can convert it into a right-angle triangle: i'm trying to find the coordinates of that point above which is perpendicular to the origin: i've labelled the opposite angle `t1` (i.e. `theta1`, but Microsoft Paint cannot easily do greek and subscripts), and i deduce that the two triangles are similar (i.e. they have the same shape): Now we come to my problem. Given `w` and `h`, find `x` and `y`. Now things get very difficult to keep drawing graphically, to explain my attempts so far. But if i call the length of the line segment common to both triangles `M`: then: ````M = w∙sin(t1) ```` Now i can focus on the other triangle, which i'll call `O`-`x`-`M`: and use trig to break it down, giving: ````x = M∙sin(t1) = w∙sin(t1)∙sin(t1) y = M∙cos(t1) = w∙sin(t1)∙cos(t1) ```` with ````t1 = atan(h/w) ```` Now this all works (i think, i've not actually tested it yet), and i'll be giving it to a computer, so speed isn't horribly important. But my god, there must have been an easier way to get there. i feel like i'm missing something. By the way, what this will be used for is drawing a linear gradient in along that perpendicular: - I think your approach is very straightforward ... it's certainly correct! If you're concerned that your final formulas look overly-complicated, it's because you stopped too soon. To finish up, you should eliminate the references to $t_1$. If you happen to know how to simplify "the sine (or cosine) of the arctan of $t_1$", great. If not, you may notice that the sine and cosine appear in your picture. Writing $s$ for the length of the hypotenuse, you have $\sin t_1 = h/s$ and $\cos t_1 = w/s$. Thus, your formulas reduce to $x=w h^2/s^2$ and $y=hw^2/s^2$. Then, since $s^2 = x^2 + y^2$ ... – Blue Jan 19 '11 at 11:49 Re: Eliminating $t_1$ ... Notice that, with your formulas, you'd compute $t_1$ via an inverse-trig function, only to toss the result into trig functions. Inverse-trigs turn "length info" into "angle info", while trigs turn "angle info" into "length info". Your solution passes through "angle info" on a journey from "length info" back to "length info". When this happens, it's a good bet that there's an algebraic expression that makes the journey directly. Here, you should suspect that $\cos {\rm atan} h/w$ is expressible algebraically in terms of $h/w$; and it is: $1/\sqrt{1+(h/w)^2}$ (or $w/s$) – Blue Jan 19 '11 at 12:18 It might looks more straightforward because i've already traveled the route, but it took me 45 minutes to get there. While i was getting there i thought, this can't be right, there has to be a better way. i had to find the size of a 3rd implied triangle, based on the similarity of two previous triangles. And `sin(cos(atan))`, "Oh boy, this is getting nasty." – Ian Boyd Jan 19 '11 at 12:49 ## 5 Answers An equation for the hypotenuse of the right triangle is $y=-\frac{h}{w}x+h$. Its slope is $-\frac{h}{w}$ (change in $y$-coordinate over change in $x$-coordinate). The slope of any line perpendicular to the hypotenuse is $\frac{w}{h}$ (perpendicular lines have slopes with product $-1$, or it is sometimes said that the slope of a perpendicular line is the "opposite reciprocal"), so an equation for the line through the origin and perpendicular to the hypotenuse is $y=\frac{w}{h}x$. The point you want is the solution to the system of equations: $$\begin{align}y&=-\frac{h}{w}x+h\\y&=\frac{w}{h}x\end{align}$$ These equations give two expressions that are both equal to $y$, so set them equal: $$-\frac{h}{w}x+h=\frac{w}{h}x$$ and solve:$$h=\left(\frac{w}{h}+\frac{h}{w}\right)x=\frac{w^2+h^2}{wh}x$$ $$x=\frac{h^2w}{w^2+h^2}$$ Use one of the original equations to find $y$: $$y=\frac{w}{h}x=\frac{hw^2}{w^2+h^2}$$ edit Working from Hans Lundmark's answer, final method, a vector from the origin to the desired point is in the same direction as $\langle\frac{1}{w},\frac{1}{h}\rangle$ or equivalently $\langle h,w\rangle$. The distance from the origin to the desired point is the length of the altitude—call this $d$. Looking at the area of the triangle, it is both $\frac{1}{2}wh$ and $\frac{1}{2}d\sqrt{w^2+h^2}$, so $d=\frac{wh}{\sqrt{w^2+h^2}}$. The desired vector is thus $$\begin{align} \frac{\langle h,w\rangle}{\|\langle h,w\rangle\|}d &=\frac{\langle h,w\rangle}{\sqrt{w^2+h^2}}\frac{wh}{\sqrt{w^2+h^2}} \\ &=\frac{\langle h,w\rangle wh}{w^2+h^2} \\ &=\left\langle\frac{h^2w}{w^2+h^2},\frac{w^2h}{w^2+h^2}\right\rangle .\end{align}$$ - I just want to make a plug for the "intercept-intercept form" of the line equation: $\frac{x}{w}+\frac{y}{h}=1$ :) – Blue Jan 19 '11 at 11:38 Of the two answers that provide final form, this is the one i understand more :) Accepted. – Ian Boyd Jan 19 '11 at 14:16 Parametrize the line from $(w,0)$ to $(0,h)$ by $(w,0) + t(-w, h)$. Then you are searching for the point $(x,y)$ on the line such that $(x,y)\cdot (-w,h) = 0$. - Is the "*" in the final equation intended to be dot-product? – Isaac Jan 19 '11 at 1:58 Yes, that is the dot product. – M.B. Jan 19 '11 at 15:12 I've edited your answer to render the ordered pairs, expression, and equation using $\LaTeX$. Feel free to re-edit or revert if you don't like the result. – Isaac Jan 19 '11 at 21:18 Another way: You have identified some angles as being equal in your last figure. This implies that several of the right triangles in the picture are similar. For example, $$\frac{h}{w} = \frac{x}{y} = \frac{y}{w-x},$$ from which it is not too difficult to solve for $x$ and $y$. And yet another: The hypotenuse lies on the line with equation $x/w+y/h=1$ (since the two points $(x,y)=(w,0)$ and $(x,y)=(0,h)$ satisfy this equation, and two points uniquely determine a line). The normal vector to a line can be read off from the coefficients of $x$ and $y$; it is $\mathbf{n}=(1/w,1/h)$ in this case. The point that you seek (let me call it $(a,b)$ here) lies on the line which goes from the origin in the direction that $\mathbf{n}$ points, so the point must be of the form $(a,b)=t \mathbf{n} = (t/w,t/h)$ for some number $t$. Substituting $(x,y)=(t/w,t/h)$ into the equation for the hypotenuse gives $t(1/w^2+1/h^2)=1$, from which we immediately find $t$ and hence also $$(a,b)=\left( \frac{1/w}{1/w^2+1/h^2}, \frac{1/h}{1/w^2+1/h^2} \right).$$ - Suppose $h = (h_1,h_2)$ and $w = (w_1, w_2)$. Then you know the slope of the line. Let's call this $m_1$. You also know that the line perpendicular to the point $(x,y)$ has slope $-\frac{1}{m_1}$. You get a system of equations. - For fun, here's another approach ... The area of your triangle is given by $\frac{1}{2} w h$, but it's also given by $\frac{1}{2}s m$, where $s$ is the length of the hypotenuse and $m$ is the length of your segment $M$. (To see why, flip the triangle over so that it's sitting on its hypotenuse; then, clearly, $s$ is the "base" and $m$ the "height".) Therefore, $$w h = s m$$ Writing $\theta$ for the angle that segment $M$ makes with the horizontal, we have that $$x = m \cos\theta = \frac{wh}{s}\cos\theta \hspace{0.5in} y = m\sin\theta = \frac{wh}{s}\sin\theta$$ By similar triangles, $\theta$ is also the angle in the top corner of your triangle, so $$\cos\theta = \frac{h}{s} \hspace{0.5in} \sin\theta = \frac{w}{s}$$ and we have $$x = \frac{wh^2}{s^2}=\frac{w h^2}{w^2+h^2} \hspace{0.5in} y = \frac{w^2 h}{s^2}=\frac{w^2h}{w^2+h^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": 66, "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": 0.9395393133163452, "perplexity_flag": "head"}
http://nrich.maths.org/1408&part=	### Calendar Capers Choose any three by three square of dates on a calendar page. Circle any number on the top row, put a line through the other numbers that are in the same row and column as your circled number. Repeat this for a number of your choice from the second row. You should now have just one number left on the bottom row, circle it. Find the total for the three numbers circled. Compare this total with the number in the centre of the square. What do you find? Can you explain why this happens? ### Adding All Nine Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself! ### Rotating Triangle What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle? # Volume of a Pyramid and a Cone ##### Stage: 3 Article by Emma McCaughan These formulae are often quoted, but rarely proved. In this article, we derive the formulae for the volumes of a square-based pyramid and a cone, using relatively simple mathematical concepts. (The ideas used in this article are also used here to find the volume of a sphere and the, impressively named, mouhefanggai shape.) ### What you need to know. You need to be familiar with the formulae for the area of a square, the area of a circle, and the volume of a cuboid. We will also use ideas of enlargement and ratio. ### The yangma. When we think of a pyramid, we usually think of one where the vertex (the top point) is above the centre of the base, a right pyramid . However in this article, we will begin with a yangma. Yangma is an ancient Chinese name for a rectangular-based pyramid whose vertex is vertically above one of the vertices of the base. We will take a yangma with a square base of side length $a$, and height also length $a$. Three of these can be fitted together to form a cube. Since we know that the volume of the cube is $a \times a \times a = a^3$, it follows that the volume of each of these yangmas is $\frac{a^3}{3}$. ### Enlargement Consider, for a moment, a cube with edges 1 unit long. You know that the volume is $1 \times 1 \times 1$. Now stretch that cube in a horizontal direction so that it measures a by 1 by 1. Its volume is now $a \times 1 \times 1$. If we imagine the cube sliced horizontally, there will be the same number of slices, but each will be a times as long. If we stretch the cube in a perpendicular direction, so that it measures a by b by 1, the volume will be $a \times b \times 1$, or $b$ times larger. If we stretch the cube in the third perpendicular direction, by a scale factor of $c$, the volume will be $a \times b \times c$, which we know as the formula for the volume of a cuboid. There are three independent directions in which we could enlarge a 3D shape. If we enlarge it by a scale factor $k$ in one of these directions, the volume will be $k$ times as large. #### Back to our yangma: Suppose we want the volume of a yangma whose height is different to the base lengths, perhaps height $h$ instead of a. Well, this is just an enlargement in the vertical direction. The scale factor? If we are turning $a$ into $h$, we have multiplied by $\frac{h}{a}$. So the volume of our new yangma is $\frac{h}{a} \times \frac{a^3}{3}$, or $\frac{a^2h}{3}$. ### Sliding the slices We now have a formula for the volume of any square-based pyramid whose vertex is above one of the vertices of the base. What about if the vertex is somewhere else - the middle, for instance? What we are going to do is to imagine the pyramid cut into lots of slices horizontally. We are going to slide these slices across, so that the top of the pyramid is above the middle of the base. (Please click here for an animation of this process, although be aware that it might take a very long time to load on some computers). If we had an infinite number of slices, our pyramid would have nice straight edges. You can probably imagine that with more slices, it would look smoother than in this illustration. Has the area of any of the slices changed? So has the volume of the shape changed? We can now see that the volume of any square-based pyramid is $\frac{a^2h}{3}$. ### Comparing a cone with a pyramid We will now look at a cone. We'll start with a right cone , whose vertex is above the centre of the base. In fact, by slicing it as in the previous section, we can show that the same formula applies for any cone. Imagine a cone whose base is a circle radius $r$, and height is $h$. This cone will fit exactly inside a square-based pyramid with base length $2r$ and height $h$. Suppose we take a slice of the pyramid with the cone inside, from some way up the pyramid. This will look like a square with a circle fitting inside. We don't know the radius of the cone at this point, so we'll call it $x$. The area of the circle is $\pi x^2$. The area of the square is $2x \times 2x=4x^2$. The ratio of the circle to the square is $\pi : 4$. The same is true for every slice we take: the area of the circle is $\frac{\pi}{4}$ of the area of the square. So, if each slice is $\frac{\pi}{4}$ the size, the volume of the cone will be $\frac{\pi}{4}$ the volume of the pyramid. The pyramid's volume is $\frac{(2r)^2h}{3} = \frac{4r^2h}{3}$ So the cone's volume is $\frac{\pi}{4} \times \frac{4r^2h}{3} = \frac{\pi r^2 h}{3}$. ### Non-square-based pyramids We can use the same principles to find the volume of any pyramid. #### Rectangular-based pyramid If we have a pyramid with rectangular base measuring $a$ by $b$, and height $h$, then this can be obtained by stretching our square-based pyramid by scale factor $\frac{b}{a}$. The new volume will be $\frac{b}{a} \times \frac{a^2 h}{3} = \frac{a b h}{3}$. #### Triangular-based pyramid If the base of the pyramid is a triangle with base $a$ and perpendicular height $b$, it will fit exactly in the rectangular pyramid above. Any slice will look like this: Although we don't know the measurements of the rectangle in this slice, the sides will still be in the ratio $a:b$ (this may take some thinking about). Let's call them $k{a}$ and $k{b}$ (where $k$ is less than 1). Area of rectangle = $k{a} \times k{b} = k^2 a{b}$ Area of triangle = $\frac{1}{2}{k}{a} \times {k}{b} = \frac{1}{2}k^2{a}{b} = \frac{1}{2} \times \mbox{area of rectangle}$. If each triangle is half the size of the rectangle, the volume of the triangular-based pyramid will be half the volume of the rectangular-based pyramid, or $a{b}h/6$. ### Generalisation The formula for the volume of any pyramid is $\frac{1}{3}\mbox{base area} \times \mbox{height}$. Verify that this works for the pyramids above (and indeed for the cone). Can you convince yourself that this is always true? 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": 47, "mathjax_display_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": 0.9255011081695557, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/246729/taylor-expansion-of-function-with-a-vector-as-variable	# taylor expansion of function with a vector as variable I know how to do a taylor expansion of a function from R to R. I dont know how to do taylor expansion of functions which have 3D vectors as variable. How can I do this? I would appreciate it if someone also worked out an example. Thank you! - The title says "vector valued" and your question mentions "vectors as variable", which is it ? – Joel Cohen Nov 28 '12 at 19:06 @JoelCohen sorry for the confusion, I am not good at the terminology. I meant a function which has a vector as a variable – Badshah Nov 28 '12 at 19:07 Do you have an example in mind? – Muphrid Nov 28 '12 at 19:14 @Muphrid f(x)=\|a-x\|^3 for example with x and a 3D vector and the vertical lines mean the length (here a is constant). – Badshah Nov 28 '12 at 19:18 So $x$ and $a$ are both vectors, just $a$ is a constant vector? – Muphrid Nov 28 '12 at 19:19 show 2 more comments ## 1 Answer Then what you're looking for is the Taylor expansion of a scalar field--a function $f$ that maps $\mathbb R^n$ to $\mathbb R$. An easy way to build up intuition about this is to do the expansion only in one direction. Let $\hat n$ be a unit vector and $t$ a scalar parameter. Let $x_0$, the point you want to expand around, be given by $x_0 = x - t \hat n$ or $x = x_0 + t \hat n$. There is only one direction connecting $x$ and $x_0$, and the magnitude can always be calculated (which fixes $t$). Then you can say $$f(x) = f(x_0 + t \hat n) = f(x_0) + \left. \frac{\partial f}{\partial t} \right\|_{x_0} t + \frac{1}{2} \left. \frac{\partial^2 f}{\partial t^2} \right\|_{x_0} t^2 + \ldots$$ Now, identify $\partial f/\partial t$ as $\hat n \cdot \nabla f$. In addition, see that $t\hat n = x - x_0$. Some clever recombining of terms gives $$f(x) = f(x_0) + (x-x_0) \cdot \nabla f\|_{x_0} + \frac{1}{2} ([x - x_0] \cdot \nabla)^2 f\|_{x_0} + \ldots$$ This is suitably general to cover any point $x$. - I dont understand why f(x)=f(x_0 +tn). could you explain? and what does de upside down delta f mean? – Badshah Nov 28 '12 at 20:28 I've added some additional language to clarify why $x = x_0 + t \hat n$. The symbol $\nabla$ (pronounced del) represents the vector derivative operator. Acting on a scalar field, it produces the gradient. – Muphrid Nov 28 '12 at 21:52	{"extraction_info": {"found_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": 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": 0.9250791668891907, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/69624/ultraproduct-of-n-dimensional-banach-spaces-and-algebras/69629	## Ultraproduct of n-dimensional Banach spaces and algebras ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi, I am interested in the following question: Fix $n$. Let $M_n$ be matrix algebra over the field of complex numbers with normalized trace $tr_n$. Let $M_n^{\omega}$ be an ultrapover of $M_n$, namely we consider the algebra of all bounded (in norm) sequences in $M_n$,say $l_{\infty}(M_n)$, and take a quotient of this space by sequences `$(a_i)_{i\in\mathbb{N}}$` with `$lim_{\omega} tr_n(a_i^*a_i)=0$`. Is $M_n^{\omega}$ is finite-dimensional? What is the structure of $M_n$? Also, if $F_i$ is $n$-dimensional Banach space, what is the dimension of the Banach space ultraproduct of ${F_i}$. - A good reference for this sort of stuff (which would also have answered your question) is Heinrich's article, now available for free: gdz.sub.uni-goettingen.de/en/dms/load/img/… – Matthew Daws Jul 6 2011 at 12:41 ## 1 Answer If the space $F$ is $k$-dimensional with basis $b_1,\ldots b_k$ then the ultraproduct with respect to the ultrafilter $U$ will also be $k$-dimensional. To see this let $x_i\in \ell_\infty(F_i)$ and let $q_{i,j}$ be scalars such that $\sum_{j=1}^kq_{i,j}b_j =x_i$. By the compactness of the ball in finite dimensional spaces the $q_{i,j}$ converge to $q_{j}$ with respect to $U$ and so $\sum_{j=1}^kq_{j}b_j = x$ modulo $U$. The same argument works of different $k$-dimensional Banach spaces as long as there is a uniform bound on the norm of the identity map from one space to the other to allow the compactness argument to be used --- in other words, there is a single ball containing the unit balls of all the $F_i$. - 1 The condition on your second paragraph is always true, as a k-dim Banach space is always at most $\sqrt{k}$ distant from $\ell^2_k$ in the Banach-Mazur distance-- see en.wikipedia.org/wiki/… – Matthew Daws Jul 6 2011 at 12:37 Thanks Matthew. I had had in mind a sequence of 2-dimensional spaces with increasingly oblong unit balls. I will have to think about why the bounded Banach-Mazur distance saves the day for such a product. – Juris Steprans Jul 6 2011 at 12:58 @Juris-- But then the "correct" unit length basis vectors also become enlongated, and so the distortions sort of cancel out... – Matthew Daws Jul 6 2011 at 15:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "mathjax_display_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": 0.8969129323959351, "perplexity_flag": "head"}
http://mathoverflow.net/questions/63826/minimum-norm-of-convex-hull	## Minimum norm of convex hull ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Dear all, I am currently stuck at a problem which seems too easy to be stuck at to me... ### Summary Let $H$ be the convex hull of the points $d_1,\ldots, d_n\in \mathbb{R}^d$. How can one compute $\min_{x\in H}\|\|x\|\|^2_2$ efficiently? ### Conditions one should know about When I talk about efficiency in this question, I do not talk about a polynomal time algorithm. For my purpose, $n$ is really small, $n=4$ should do in almost all cases. (Even a fast solution for the special case $n=4$ would be highly appreciated). Also, $d=2$ is the main dimension of interest, whereas it might be possible that at some time this might also be a question for the three dimensional space. This question is embedded in an algorithm that will be called thousands of times within a very short timeframe. In short, I am looking for an efficient way in terms of "Do I need 20 or 40 FLOPS"... ### First thoughts Of course, one could take this problem as a Quadratic Program by writing $D=[d_1,\ldots, d_n]$ and then minimizing $x^tD^tDx$ with respect to $\sum_{i=1}^d x= 1$ and $x_i\geq 0$. However, I just feel this might be overkill for such a simple problem. I also thought about taking a least norm approach, but here I do not really get a grip on formulating the problem the right way to throw some (to me) known technique at it. Also, one could compute the convex hull explicitly and then try to locate 0 geometrically around the hull and only compute the corresponding distance to the hull. As I made a rough sketch to the algorithm I had in mind, I realized this also gives me quite a lot of things to compute and cases to distinguish between. Your ideas would be greatly appreciated. - I think that you are trying to find the point in H that is closest to 0. If your hull is not containing 0, then the solution is one of your di points. If it includes 0, then your solution is 0 – Anadim May 4 2011 at 2:23 @Anadim: It does not have to be one of the $d_i$-points if $0\not\in H$. The searched point could also lie on any line segment between $d_i$ and another $d_j$ (assuming they both lie on the boundary of the convex hull). Also note that while finding the Point in $H$ closes to the origin solves the problem, I am only interested in the minimum distance. – Thilo Schneider May 4 2011 at 5:29 ## 3 Answers The following iterative algorithm is perhaps fast: Set $P=d_i$ where $d_i$ is of minimal norm among $d_1,\dots,d_n$. Iterate the following loop: Let $j=j(P)$ be an index such that $\alpha=\langle P,d_i-P\rangle/\sqrt{\langle d_i-P,d_i-P\rangle}, i=1,..n$ is minimal for $i=j$. If $\alpha\geq 0$ then $P$ is at minimal distance. (If $\alpha>-\epsilon$ for small $\epsilon$ you are very close to the minimum.) Otherwise replace $P$ with the point closest to the origin of the line joining (the old point) $P$ to $d_j$. End of Loop. In dimension $d=2$ and if the origin is not in the convex hull, this algorithme will ultimately only involve the two endpoints among $d_1,\dots,d_n$ of the edge realising the minimum (generically). - Thank you. Looks promising at a first glance. I will have a more in depth look tomorrow and will probably accept your answer then. – Thilo Schneider May 3 2011 at 20:20 @Roland: Consider 3 points $d_1=(1,N)$, $d_2=(1,-N)$ and $d_3=(2,0)$, where $N$ is large. I'm afraid that $P$ will not reach the segment $[d_1d_2]$ in finitely many steps, and it will converge rather slowly (like $O(N)$ iterations per digit of output). – Sergei Ivanov May 3 2011 at 21:28 @Sergei: Thank you for pointing out. Currently I assume you can easily rescue Rolands algorithm without increasing the computational effort a lot. Will give feedback once I am more certain (a.k.a once I got my idea proven). – Thilo Schneider May 4 2011 at 6:36 Sergei: You are of course absolutely right, the algorithm is iterative but does generally not stop and does not converge very quickly in your example. However, its values for $P$ oscillate between $d_1$ and $d_2$ which shows that the minimum is achieved by the segment joining $_1$ and $d_2$. Once this is known, one can use this information and drop the point $d_3$ from the initial set. The algorithm converges then in one iteration to the minimum and doing a last iteration with all three points proves that we have the minimum. – Roland Bacher May 4 2011 at 8:19 ### 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. Well, if you are in two dimensions, and have four points, your convex hull is either a triangle, or a union of two triangles. If you have three points, then the problem is a simple lagrange multiplier problem, with three equations and three unknowns, so you can easily write out the solution explicitly. If the solution has all the coefficients positive, you win. If not, do it for the edges (two equations with two unknown each). Then loop over the four possible triangles. Notice that a lot of that can be done in parallel, so while the flop count is a couple of hundred (I am guessing), it looks like 20 if done on GPU (from your phrasing, I am guessing this is a graphics hack). - Thanks for the input. I believe I found an algorithm that should work faster for my specific problem (which is not intended for the GPU - at least not yet). – Thilo Schneider May 5 2011 at 20:10 Dear all, I think I found quite a neat solution for my problem. At first I want to give humble thanks to Roland, who greatly inspired the solution I'm going to use now. Just in case somebody else might sometimes in the future be struggling with a similar problem, I'm going to give a short outline of the ideas and a rough sketch of the algorithm. After some consideration I decided to go down to only two dimensions, as this allows some neat tricks. To go along with the notation I use in my own work (and which I did not use asking the question), let $X=\{x_1,\ldots, x_n\}$ be the set that spans the convex hull $H$. First I want to reflect the main ideas, which I am certain are very obvious to most of you. 1. Let $P=\arg\min_{x\in H}\|\|x\|\|_2^2$. It is either $P\in H$ and thus $P=0$, or $P$ lies on the boundary of $H$, which means there are $x,'x\in X$ and $c\in[0,1]$ with $P=cx+(1-c)x'$. 2. Let now $\delta:=\min_{i=1,\ldots,n}\min_{j=i+1,\ldots, n}\min\left\{\|\|x\|\|_2^2: x=cx_i+(1-c)x_j, c\in[0,1]\right\}.$ Because of 1. we have $\min_{x\in H}\|\|x\|\|_2^2\in{0,\delta}$. 3. To compute $\delta$, a lot of minimum norms $d$ on line segments from $x_i$ to $x_j$ have to be computed. By usage of the scalar product and simple usage or pythagoras, one can derive the distance as $d=\begin{cases} \|\|x_i\|\|^2_2 & \text{if } \langle x_i,x_i-x_j\rangle \leq 0\\ \|\|x_j\|\|^2_2 & \text{if } \langle x_i,x_i-x_j\rangle \geq \|\|x_i-x_j\|\|_2^2\\ \|\|x_i\|\|^2_2-\frac{\langle x_i, x_i-x_j\rangle^2} {\|\|x_i-x_j\|\|^2} & \text{ otherwise}. \end{cases}$ 4. The point $P$ realizing the minimum distance in this case is $P=x_i+\frac{(x_j-x_i)\langle x_i, x_i-x_j\rangle}{\|\|x_i-x_j\|\|_2^2}$ 5. Given $\delta$ and the Point $P$ realizing $\delta$, the following holds: $0\in H$ if and only if there exists $x_k$ with $\langle P, P-x_k\rangle >0$. (Very compact proof: If one of those exists, one can reduce $\delta$ by using the method of 3. As the new minimum may not lie on the boundary as it has not been found in the first run, $0\in H$ has to hold. Otherwise assume $0\in H$ and no scalar product fulfills the given condition. Then one can write $0$ as convex combination of the $x_i$ and express $0<\|\|P\|\|_2^2=\langle{P,P-0}\rangle$ and, using linearity of the scalar product, show that this is $\leq 0$ and thus getting a contradiction). Putting all of this together one yields the following algorithm: 1. Init $d_\min = \infty$. 2. For $i\in\{0,\ldots, n-1\}$ do: 1. Compute $n_i$ = $\|\|x_i\|\|_2^2$ 2. For ($j \in \{i+1,\ldots, n-1\}$ do 1. Compute $s_{i,j}=\langle x_i, x_j\rangle$. 2. Compute `$n_j=\|\|x_j\|\|_2^2$`. 3. Decide: • If $s_{i,j}\geq n_i$ set $d=n_i, c=1$. • If $s_{i,j}\geq n_j$ set $d=n_j, c = 0$. • Else set $c = (n_i-s_{i,j})/(n_i+n_j-2s_{i,j})$ and $d=n_i- (n_i-s_{i,j})\cdot c$. 4. If `$d<d_{\min}$` set $d_{\min}= d$, $j_{\min}= j$, $i_{\min} = i$ and $c_{\min} = c$. 3. For $k\in\{0,\ldots,n\}\setminus\{i,j\}$ do: If $d > (1-c)s_{i,k}+c s_{j,k}$ return 0. 4. Return $d_\min$. This algorithm is based on the rules 1-5 above, however, the conditions are formulated a bit more efficient. Please apologize the incomplete reasoning. I found this solution appealing as it allows a lot of optimization in my special scenario. I already computed all norms of the $x_i$ somewhere else, so this is no further effort. Furthermore, the parent algorithm scans a set of values, successively replacing one $x_i$ with another $x_i'$ and then again computing the distance to the convex hull. This means that most of the distances can be saved and used later and for each iteration I only have to do very little work in updating and computing very few (as for my main problem, where $n=4$ it is 3) scalar products. Thank you for your support and the in fact very valuable input. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 99, "mathjax_display_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": 0.9454883933067322, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/169543-irreducible-polynomial.html	# Thread: 1. ## Irreducible Polynomial Let f(x) be an irreducible polynomial of degree n in Zp[x]. Show that the field E = Zp[x]/<f(x)> has exactly p^n elements. f(x) is irreducible, so it cannot be factored into a product of polynomials of lower degree. 2. Think about what the elements of $\mathbb{Z}_p[x]/\langle f(x)\rangle$ look like and what it means for them to be different. Maybe that's too vague. I mean, for example, consider the coset $a=(x^n+1)+\langle f(x)\rangle$. If you can write $x^n+1=f(x)q(x)+r(x)$, then isn't $r(x)+\langle f(x)\rangle$ the same coset as $a$? And how many different remainders can you get when you divide by $f(x)$? 3. Well, I started by letting f(x)=a_nx^n+.....+a1x+a0 I don't know if that helps me any 4. That helps to at least figure out what kind of remainders you can get when dividing by $f(x)$. That division is the key to the whole problem because when dividing by an nth degree polynomial, $f(x)$ in this case, the remainder must be of degree strictly less than n. So if you have some coset $g(x)+\langle f(x)\rangle$, you can divide $g(x)$ by $f(x)$ and represent the same coset by the remainder, which is guaranteed to be a polynomial of degree at most $n-1$. The nice thing is that you don't have to actually do any division at all (although this is one way to find inverses if you want). You just have to show that each coset can be represented by a polynomial of degree less than $n$. Now since the coefficients come from a finite field, how many different polynomials can you form? That's your answer. You do need to show that each of these different polynomials does actually represent a different coset, but you can do that by contradiction. Say that two different polynomials, each of degree at most $n-1$ represent the same coset. Write out what this means and use the fact that $f(x)$ is irreducible to get the contradiction. edit: Actually the fact that f(x) is irreducible doesn't really come into play in that last paragraph. I must have been thinking of using that fact to prove the existence of inverses, but you already know it's a field.	{"extraction_info": {"found_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": 15, "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": 0.9579214453697205, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/686/analysis-of-repeatedly-enciphered-plaintext-using-same-algorithm-key?answertab=votes	# Analysis of Repeatedly Enciphered Plaintext using Same Algorithm / Key Please forgive the impracticality of this question, but I'm curious about the behaviors of encryption algorithms applied to their own output. Suppose I have an encryption algorithm E and (using the same key(s)) I repeatedly encrypt its output for a given plaintext: `E(E(E(E(...E(plaintext)` Would a modern encryption algorithm, such as some variant of AES, display any kind of cycling with respect to its output for arbitrarily long compositions of itself? Is there some known number of compositions where the final output is itself the plaintext? - – mikeazo♦ Sep 15 '11 at 12:09 ## 2 Answers Eventually, it will cycle, but it'll almost certainly take far longer than your lifetime to do so. If that sounds too flip, here's the mathematics. Assuming AES is secure, then the probability it will revisit the original plaintext after at most r repeated encryptions is just r/2128. (This follows from a standard fact about the cycle structure of a random permutation.) For the values of r that can realistically arise in practice, this probability is negligible: less than the chances of getting hit by lightning, twice. Is there some known number r such that r repeated encryptions are guaranteed to bring you back to the plaintext you started with? Yes, for instance: r = (2128)!, i.e., 2128 factorial. Or, r = lcm(1, 2, 3, ..., 2128) also works. However, this is a stupendously, ridiculously large number. So it's completely irrelevant in practice. Bottom line. If you've got a secure block cipher, you don't need to worry about cycling. It's not an issue. It just ain't gonna happen. - A clearer way to write your third sentence would be the probability to reach the original plain text after at most r repeated encryptions is r/2^128. – Paŭlo Ebermann♦ Sep 15 '11 at 9:47 1 Or even more precisely than that: the probability to reach the original plaintext after at most r repeated encryptions is exactly r/2^128 for all r<=2^128 (assuming that the block cipher acts as a random permutation). – poncho Sep 15 '11 at 15:58 Thanks for the analysis! – Michael Petito Sep 15 '11 at 16:42 1 Thanks for all the comments, everyone. Three nitpicks: 1. The correct statement of the fact is about the probability of hitting the original plaintext after exactly r repeated encryptions, not at most r. 2. Yeah, yeah, yeah. Fine, so the lcm is smaller than the factorial. It's still a stupendously, ridiculously, impractically large number. My conclusion still holds. 3. e501, your statement about 2^64 is not correct. The birthday paradox is not relevant here. – D.W. Sep 15 '11 at 17:38 1 @D.W.: The correct statement is with at most r. – fgrieu Sep 15 '11 at 18:10 show 5 more comments Your question essentially asks, given a function $f$ that takes an input $i$ of length $k$ and produces an output $o$ of length $k$. $$i \in \{0, 1\}^k, o \in \{0, 1\}^k$$ $$f(i) = o$$ How many times on average will $f$ have to be recursively called on itself such that it collides with one of it's inputs or outputs (all previous outputs are also inputs). $$f(f(f(...f(i)...))) = i$$. Since we are assuming $f$ is a cipher, lets be generous and assume that $f$ randomly maps $i$s to $o$ (that is, lets assume $f$ is secure) such that the same $i$ always produces the same $o$. This is, $f$ is a random permutation over $2^k$ possible $i$s (remember $i$ is a binary number of $k$ digits). This question really asks: 1. how many times must a permutation on $2^k$ elements be applied to itself such that it repeats (that is the "order of a permutation") 2. and what is the average order of a random permutation on $2^k$ elements. The Landau function give us the worst possible number of steps we would need to take to get a permeation to repeat itself. This is our upper bound and the answer to question 1. Question 2 I don't have an exact answer for you but there has been a lot of mathematic work in this area (Paul Erdos worked on this a bit). There is a mathoverflow question/answer on the average order of a permutation as well. - The statements about Question 2, the birthday paradox, and the probability of a collision are not correct. This is a random permutation, not a random function. The birthday paradox does not apply. – D.W. Sep 15 '11 at 17:40 – D.W. Sep 15 '11 at 17:41	{"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": 20, "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": 0.3, "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": 0.9267190098762512, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/113968-expected-time-poisson-process.html	# Thread: 1. ## Expected time from Poisson process Hi, Let's say I have cars passing a point as a Poisson process with having stationary flow of 100 cars/hr. How would I go about finding the expected amount of time needed to observe one headway (time between cars) larger than three times the mean headway? Thanks! 2. Arrivals in a Poisson process are $\exp(\lambda)$. Now what is the expected time between cars? Easy, $\frac{1}{\lambda}$. State your question in terms of an exponential distribution (i.e. what is the prob. that an arrival takes longer than 3 times the mean), then it should be pretty clear what the answer is. Good luck.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "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": 0.9515901803970337, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/39401/is-hyperbolic-space-h3-the-best-representation-space-for-momenta-in-momentum?answertab=active	# Is Hyperbolic Space $H3$ the best representation space for momenta, in momentum-scale invariant theories? The motivation is the following: For each particle of mass m, we could write : $- (E/m)^2 + (p_x/m)^2 + (p_y/m)^2 + (p_z/m)^2 = -1$, which is nothing than a equation for a point in the hyperbolic space $\mathbb H3$, which can be seen, with Poincaré representation (Poincaré ball), as a (full) sphere (with hyperbolic distance) For photons, we may give us a very tiny mass, so photons could be considered, in the limit where this tiny mass is zero, as living at the boundary of $\mathbb H3$, which is $\mathbb CP1$ (Riemann sphere $S^2$) With these prescriptions, we cannot distinguish, photons with proportionnal momenta, but if the theory is momentum-scale invariant, it seems OK. So could we use Hyperbolic Space $H3$ as representation space for momenta, in momentum-scale invariant theories? - ## 1 Answer In English, unlike French, "impulsions" are known as "momenta". Yes, massive particles' momentum 4-vectors belong to a $H^3$. On the other hand, massless particles' momentum 4-vectors don't belong to a boundary of $H^3$. Instead, they belong to a degenerate, singular version of an $H^3$ which is the same thing as $CP^1$ (semidirectly or fibered) times the scaling $R^+$. This statement means that the overall normalization of the photon's energy or momentum always matters. If two photons are moving in the same direction, to the same point of the $CP^1$, it clearly doesn't mean that they have the same momentum or energy, does it? An X-ray flying from Paris to Lyon is something else than a radio wave photon flying from Paris to Lyon. In scale-invariant theories, particles have to be massless and physics of particles with one momentum 4-vector and its multiple is "physically equivalent" - related by a symmetry transformation. But it doesn't mean it's the same photon. The photons with differently scaled energies are still different and the ratio of energies of two photons, at least overlapping or nearby ones, may always differ from one and may be determined. - Thanks for language correction... But there is a subtelty between "physically equivalent" and "the photons are still different" that I don't understand. For instance, in calculating a scattering amplitude, if it is not scale-invariant in momenta, what does mean "physically equivalent" ? – Trimok Oct 9 '12 at 13:14 Dear Trimok, the scattering amplitude is scale-invariant, symmetric, which means that if you rescale all momenta $k$ times in it, you will get the same amplitude (up to some possible simple power-law rescaling of the amplitude by $k^\lambda$ where $\lambda$ is a simple exponent). But that doesn't mean that the actual photons don't care about the scaling of the energy. For example, if you rescale the (several) photons' momenta by different factors, you get a totally different amplitude. The symmetry only holds if you rescale everything. – Luboš Motl Oct 10 '12 at 5:06 Even if you do rescale everything - momenta of all photons - and you get the same amplitude up to a simple scaling of the amplitude, the states of the photon before and after rescaling are just not equal, much like an iPhone in Boston isn't equal to an iPhone as the same model iPhone in Marseille. – Luboš Motl Oct 10 '12 at 5:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.18, "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": 0.9248030185699463, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2011/03/17/sheaves/?like=1&source=post_flair&_wpnonce=c589d5bcbb	# The Unapologetic Mathematician ## Sheaves For the moment we will be more concerned with presheaves, but we may as well go ahead and define sheaves. These embody the way that not only can we restrict functions to localize them to smaller regions, but we can “glue together” local functions on small domains to define functions on larger domains. This time, let’s start with the fancy category-theoretic definition. For any open cover $\{U_i\}_{i\in I}$ of an open set $U$, we can set up the following diagram: $\displaystyle\mathcal{F}(U)\to\prod\limits_{i\in I}\mathcal{F}(U_i)\rightrightarrows\prod\limits_{(i,j)\in I\times I}\mathcal{F}(U_i\cap U_j)$ Let’s talk about this as if we’re dealing with a sheaf of sets, to make more sense of it. Usually our sheaves will be of sets with extra structure, anyway. The first arrow on the left just takes an element of $\mathcal{F}(U)$, restricts it to each of the $U_i$, and takes the product of all these restrictions. The upper arrow on the right takes an element of $\mathcal{F}(U_i)$ and restricts it to each intersection $U_i\cap U_j$. Doing this for each $U_i$ we get a map from the product over $i\in I$ to the product over all pairs $(i,j)$. The lower arrow is similar, but it takes an element in $\mathcal{F}(U_j)$ and restricts it to each intersection $U_i\cap U_j$. This may look the same, but the difference in whether the original set was the first or the second in the intersection makes a difference, as we shall see. Now we say that a presheaf $\mathcal{F}$ is a sheaf if and only if this diagram is an equalizer for every open cover $U_i$. For it to be an equalizer, first the arrow on the left must be a monomorphism. In terms of sets, this means that if we take two elements $s\in\mathcal{F}(U)$ and $t\in\mathcal{F}(U)$ so that $s\vert_{U_i}=t\vert_{U_i}$ for all \$latex $U_i$, then $s=t$. That is, elements over $U$ are uniquely determined by their restrictions to any open cover. The other side of the equalizer condition is that the image of the arrow on the left consists of exactly those products in the middle for which the two arrows on the right give the same answer. More explicitly, let’s say we have an $s_i\in\mathcal{F}(U_i)$ for each $U_i$, and let’s further assume that these elements agree on their restrictions. That is, we ask that $s_i\vert_{U_i\cap U_j}=s_j\vert_{U_i\cap U_j}$. If this is true for all pairs $(i,j)$, then the product $\left(s_i\right)_{i\in I}$ takes the same value under either arrow on the right. Thus it must be in the image of the arrow on the left — there must be some $s\in\mathcal{F}(U)$ so that $s\vert_{U_i}=s_i$. In other words, as long as the local elements $s_i\in\mathcal{F}(U_i)$ “agree” where their domains overlap, we can “glue them together” to give an element $s\in\mathcal{F}(U)$. Again, the example to keep in mind is that of continuous real-valued functions. If we have a continuous function $f_U:U\to\mathbb{R}$ and another continuous function $f_V:V\to\mathbb{R}$, and if $f_U(x)=f_V(x)$ for all $x\in U\cap V$, then we can define $f:U\cup V\to\mathbb{R}$ by “gluing” these functions together over their common overlap: $f(x)=f_U(x)$ if $x\in U$, $f(x)=f_V(x)$ if $x\in V$, and it doesn’t matter which we choose when $x\in U\cap V$ because both functions give the same value there. So, a sheaf is a presheaf where we can glue together elements over small domains so long as they agree when restricted to their intersections, and where this process defines a unique element over the larger, “glued-together” domain. ### Like this: Posted by John Armstrong \| Topology ## 12 Comments » 1. [...] As ever, we want our objects of study to be objects in some category, and presheaves (and sheaves) are no exception. But, luckily, this much is [...] Pingback by \| March 19, 2011 \| Reply 2. For the left-most arrow in the first diagram to work, don’t the $U_i$ have to be subsets of $U$, which isn’t necessary for an open cover of the topological space, or am I once again missing something? Comment by Avery Andrews \| March 20, 2011 \| Reply 3. oops open cover of $U$ in the topological space Comment by Avery Andrews \| March 20, 2011 \| Reply 4. It’s not required for an open cover, but since $U$ is an open subspace we can always just pass to the intersection of the covering sets with $U$ itself. We still have an open cover of $U$, but all the sets are subsets of $U$. Comment by \| March 20, 2011 \| Reply 5. I managed to think of this right after posting the question, but it’s nice to have it confirmed. Comment by Avery Andrews \| March 20, 2011 \| Reply 6. [...] Direct Image Functor So far our morphisms only let us compare presheaves and sheaves on a single topological space . In fact, we have a category of sheaves (of sets, by default) on . [...] Pingback by \| March 21, 2011 \| Reply 7. [...] that we’ve talked a bunch about presheaves and sheaves in general, let’s talk about some particular sheaves of use in differential topology. Given a [...] Pingback by \| March 23, 2011 \| Reply 8. p 106-108 of http://folli.loria.fr/cds/1999/library/pdf/barrwells.pdf is a decent piece of side-reading for this, I think. Comment by Avery D Andrews \| March 27, 2011 \| Reply 9. Nitpick: s/left/right/ in “upper arrow on the left” Comment by Chad \| April 2, 2011 \| Reply 10. [...] that we can uniquely glue together vector fields which agree on shared domains, meaning we have a sheaf of vector [...] Pingback by \| May 23, 2011 \| Reply 11. [...] for the collection of all such -forms over . It’s straightforward to see that this defines a sheaf on [...] Pingback by \| July 12, 2011 \| Reply 12. [...] of coordinates! That is, on the intersection . Since the algebras of differential forms form a sheaf , we know that we can patch these together into a unique , and this is our volume [...] Pingback by \| October 6, 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": 46, "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": 0.9316017627716064, "perplexity_flag": "head"}
http://mathoverflow.net/questions/1291/a-learning-roadmap-for-algebraic-geometry/1763	## A learning roadmap for algebraic geometry ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Unfortunately this question is relatively general, and also has a lot of sub-questions and branches associated with it; however, I suspect that other students wonder about it and thus hope it may be useful for other people too. I'm interested in learning modern Grothendieck-style algebraic geometry in depth. I have some familiarity with classical varieties, schemes, and sheaf cohomology (via Hartshorne and a fair portion of EGA I) but would like to get into some of the fancy modern things like stacks, étale cohomology, intersection theory, moduli spaces, etc. However, there is a vast amount of material to understand before one gets there, and there seems to be a big jump between each pair of sources. Bourbaki apparently didn't get anywhere near algebraic geometry. So, does anyone have any suggestions on how to tackle such a broad subject, references to read (including motivation, preferably!), or advice on which order the material should ultimately be learned--including the prerequisites? Is there ultimately an "algebraic geometry sucks" phase for every aspiring algebraic geometer, as Harrison suggested on these forums for pure algebra, that only (enormous) persistence can overcome? - 9 Akhil, to be fair to everyone else ever, I have a "foo sucks" phase for pretty much everything: I learn some shiny new facts about something, get heavily into it for a while without understanding it on more than a surface level, and then when I start to learn it "for real," there's more ugly technical details and fewer of the broad, striking generalities that attracted me to the field. So this is more a byproduct of my way of learning things than an absolute necessity. – Harrison Brown Oct 19 2009 at 23:03 Same here, incidentally. But I think the problem might be worse for algebraic geometry---after all, the "barriers to entry" (i.e. theoretical prerequisite material) are somewhat more voluminous than for analysis, no? – Akhil Mathew Oct 20 2009 at 1:04 Maybe interesting: Oort's talk on Grothendiecks mindset: staff.science.uu.nl/~oort0109/… – Thomas Riepe Nov 6 2011 at 12:41 ## 13 Answers FGA Explained. Articles by a bunch of people, most of them free online. You have Vistoli explaining what a Stack is, with Descent Theory, Nitsure constructing the Hilbert and Quot schemes, with interesting special cases examined by Fantechi and Goettsche, Illusie doing formal geometry and Kleiman talking about the Picard scheme. For intersection theory, I second Fulton's book. And for more on the Hilbert scheme (and Chow varieties, for that matter) I rather like the first chapter of Kollar's "Rational Curves on Algebraic Varieties", though he references a couple of theorems in Mumfords "Curves on Surfaces" to do the construction. And on the "algebraic geometry sucks" part, I never hit it, but then I've been just grabbing things piecemeal for awhile and not worrying too much about getting a proper, thorough grounding in any bit of technical stuff until I really need it, and when I do anything, I always just fall back to focus on varieties over C to make sure I know what's going on. EDIT: Forgot to mention, Gelfand, Kapranov, Zelevinsky "Discriminants, resultants and multidimensional determinants" covers a lot of ground, fairly concretely, including Chow varieties and some toric stuff, if I recall right (don't have it in front of me) - Thanks! So you're advising emphatically not to go the EGA-route (i.e. do nothing other than reading Grothendieck linearly for several months, but rather skip arond from different sources)? Perhaps this is the antidote for that phase. – Akhil Mathew Oct 19 2009 at 23:00 I've actually never cracked EGA open except to look up references. SGA, too, though that's more on my list. I'll probably have to eventually, but I at least have a feel for what's going on without having done so, and other people have written good high-level expositions of most of the stuff that Grothendieck did. And specifically, FGA Explained has become one of my favorite references for anything resembling moduli spaces or deformations. – Charles Siegel Oct 19 2009 at 23:03 Gelfand, Kapranov, and Zelevinsky is a book that I've always wished I could read and understand. Maybe this is a "royal road" type question, but what're some good references for a beginner to get up to that level? – jc Oct 21 2009 at 23:47 1 Well, to get a handle on discriminants, resultants and multidimensional determinants themselves, I can't recommend the two books by Cox, Little and O'Shea enough. The first one, Ideals, Varieties and Algorithms, is undergrad, and talks about discriminants and resultants very classically in elimination theory. The second, Using Algebraic Geometry, talks about multidimensional determinants. – Charles Siegel Oct 22 2009 at 1:11 ### 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. Concentrated reading on any given topic—especially one in algebraic geometry, where there is so much technique—is nearly impossible, at least for people with my impatient idiosyncracy. It's much easier to proceed as follows. 1. Ask an expert to explain a topic to you, the main ideas, that is, and the main theorems. Keep diligent notes of the conversations. 2. Try to prove the theorems in your notes or find a toy analogue that exhibits some of the main ideas of the theory and try to prove the main theorems there; you'll fail terribly, most likely. 3. Once you've failed enough, go back to the expert, and ask for a reference. 4. Open the reference at the page of the most important theorem, and start reading. 5. Every time you find a word you don't understand or a theorem you don't know about, look it up and try to understand it, but don't read too much. At this stage, it helps to have a table of contents of FGA explained-EGA-SGA where you can quickly look up unknown words. Keep diligent notes of your progress, and talk to your expert as much as possible. Then go back to step 2. An example of a topic that lends itself to this kind of independent study is abelian schemes, where some of the main topics are (with references in parentheses): 1. the rigidity lemma (Mumford, Geometric invariant theory, Chapter 6), 2. the theorem of the cube (Raynaud, Faisceaux amples sur les schémas…), 3. construction of the dual abelian scheme (Faltings-Chai, Degeneration of abelian varieties, Chapter 1), 4. questions of projectivity (Raynaud, Faisceaux amples sur les schemas…), 5. Lang-Néron theorem and $K/k$ traces (Brian Conrad's notes). 6. proof that abelian schemes assemble into an algebraic stack (Mumford, Geometric invariant theory, Chapter 7), 7. compactifications of the stack of abelian schemes (Faltings-Chai, Degeneration of abelian varieties; Olsson, Canonical compactifications…; Kato and Usui, Classidying spaces of degenerating polarized Hodge structures.) You may amuse yourself by working out the first topics above over an arbitrary base. That's enough to keep you at work for a few years! A brilliant epitome of SGA 3 and Gabriel-Demazure is Sancho de Salas, Grupos algebraicos y teoria de invariantes. It explains the general theory of algebraic groups, and the general representation theory of reductive groups using modern language: schemes, fppf descent, etc., in only 400 quatro-sized pages! - 2 I like the use of toy analogues. Personally, I don't understand anything until I've proven a toy analogue for finite graphs in one way or another. – Qiaochu Yuan Oct 20 2009 at 2:43 I need to go at once so I'll just put a link here and add some comments later. Or someone else will. The Stacks Project - nearly 1500 pages of algebraic geometry from categories to stacks. - My advice: spend a lot of time going to seminars (and conferences/workshops, if possible) and reading papers. Talk to people, read blogs, subscribe to the arxiv AG feed, etc. AG is a very large field, so look around and see what's out there in terms of current research. There's a huge variety of stuff. There's a lot of "classical" stuff, and there's also a lot of cool "modern" stuff that relates to physics and to topology (e.g. Gromov-Witten theory, derived algebraic geometry). First find something more specific that you're interested in, and then try to learn the background that's needed. - I'm only an "algebraic geometry enthusiast", so my advice should probably be taken with a grain of salt. With that said, here are some nice things to read once you've mastered Hartshorne. 1) I'm a big fan of Mumford's "Curves on an algebraic surface" as a "second" book in algebraic geometry. 2) Fulton's "Toric Varieties" is also very nice and readable, and will give access to some nice examples (lots of beginners don't seem to know enough explicit examples to work with). 3) More stuff about algebraic curves. The best book here would be "Geometry of Algebraic Curves" by Arbarello, Cornalba, Griffiths, and Harris. The next step would be to learn something about the moduli space of curves. An inspiring choice here would be "Moduli of Curves" by Harris and Morrison. 4) Intersection Theory. Fulton's book is very nice and readable. 5) Algebraic groups. I'm a big fan of Springer's book here, though it is written in the language of varieties instead of schemes. EDIT : I forgot to mention Kollar's book on resolutions of singularities. A masterpiece of exposition! - As for Fulton's "Toric Varieties" a somewhat more basic intro is in the works from Cox, Little and Schenck, and can be found on Cox's website. Also, in theory (though very conjectural) volume 2 of ACGH Geometry of Algebraic Curves, about moduli spaces and families of curves, is slated to print next year. But they said that last year...though the information on Springer's site is getting more up to date. – Charles Siegel Oct 19 2009 at 23:11 I actually possess a preprint copy of ACGH vol II, and Joe Harris promised me that it would be published soon! – Andy Putman Oct 19 2009 at 23:16 Great! I've been waiting for it for a couple of years now. Springer's been claiming the earliest possible release date and then pushing it back. – Charles Siegel Oct 19 2009 at 23:44 And now I wish I could edit my last comment, to respond to your edit: Kollar's book is great. I learned a lot from it, and haven't even gotten to the general case, curves and surface resolution are rich enough. – Charles Siegel Oct 19 2009 at 23:47 I have owned a prepub copy of ACGH vol.2 since 1979. Of course it has evolved some since then. BY now I believe it is actually (almost) shipping. – roy smith May 12 2011 at 21:04 show 1 more comment For me, I think the key was that much of my learning algebraic geometry was aimed at applying it somewhere else. Starting with a problem you know you are interested in and motivated about works very well. For me it was certain bits of geometric representation theory (which is how I ended up learning etale cohomology in the hopes understanding knots better), but for someone else it could be really wanting to understand Gromov-Witten theory, or geometric Langlands, or applications of cohomology in number theory. - I found that this article "Stacks for everybody" was a fun read (look at the title!), and provided motivation through the example of vector bundles on a space, though it doesn't go that deep: http://www.cgtp.duke.edu/~drm/PCMI2001/fantechi-stacks.pdf As for things like étale cohomology, the advice I have seen is that it is best to treat things like that as a black box (like the Lefschetz fixed point theorem and the various comparison theorems) and to learn the foundations later since otherwise one could really spend way too long on details and never get a sense of what the point is. I have certainly become a big fan of this style of learning since it can get really boring reading hundreds of pages of technical proofs. - If you want to learn stacks, its important to read Knutson's algebraic spaces first (and later Laumon and Moret-Baily's Champs Algebriques). To keep yourself motivated, also read something more concrete like Harris and Morrison's Moduli of curves and try to translate everything into the languate of stacks (e.g. as you're learning stacks work out what happens for moduli of curves). There are a lot of cool application of algebraic spaces too, like Artin's contraction theorem or the theory of Moishezon spaces, that you can learn along the way (Knutson's book mentions a bunch of applications but doesn't pursue them, mostly sticks to EGA style theorems). Another nice thing about learning about Algebraic spaces is that it teaches you to think functorially and forces you to learn about quotients and equivalence relations (and topologies, and flatness/etaleness, etc). - For a small sample of topics (concrete descent, group schemes, algebraic spaces and bunch of other odd ones) somewhere in between SGA and EGA (in both style and subject), I definitely found the book 'Néron Models' by Bosch, Lütkebohmert and Raynaud a nice read, with lots and lots of references too. - Thanks! Descent is something I've been meaning to learn about eventually and SGA looks somewhat intimidating. – Akhil Mathew Oct 20 2009 at 1:05 Underlying étale-ish things is a pretty vast generalization of Galois theory. Hendrik Lenstra has some nice notes on the Galois Theory of Schemes ( websites.math.leidenuniv.nl/algebra/GSchemes.pdf ), which is a good place to find some of this material. - Douglas Ulmer recommends: "For an introduction to schemes from many points of view, in particular that of number theory, the best reference by far is a long typescript by Mumford and Lang which was meant to be a successor to “The Red Book” (Springer Lecture Notes 1358) but which was never finished. These notes have excellent discussions of arithmetic schemes, Galois theory of schemes, the various flavors of Frobenius, flatness, various issues of inseparability and imperfection, as well as a very down to earth introduction to coherent cohomology. Some of this material was adapted by Eisenbud and Harris, including a nice discussion of the functor of points and moduli, but there is much more in the Mumford-Lang notes." Unfortunately I saw no scan on the web. The preliminary, highly recommended 'Red Book II' is online here. - Wow,Thomas-this looks terrific.I guess Lang passed away before it could be completed? Wonder what happened there. – Andrew L Jun 15 2010 at 21:47 Here is a soon-to-be-book by Behrend, Fulton, Kresch, great to learn stacks: http://www.math.uzh.ch/index.php?pr_vo_det&key1=1287&key2=580&no_cache=1 - 2 Is it really "Soon" though? The notes are missing a few chapters (in fact, over half the book according to the table of contents). – Charles Siegel Oct 22 2009 at 1:12 True, the project might be stalled, in that case one might take something else right from the beginning. Pity - I find the style great... – Peter Arndt Oct 24 2009 at 13:27 There are a few great pieces of exposition by Dieudonné that I really like. The first two together form an introduction to (or survey of) Grothendieck's EGA. The second is more of a historical survey of the long road leading up to the theory of schemes. I am sure all of these are available online, but maybe not so easy to find. • Algebraic geometry ("The Maryland Lectures", in English), MR0150140 • Fondements de la géométrie algébrique moderne (in French), MR0246883 • The historical development of algebraic geometry (available here or here) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9367238879203796, "perplexity_flag": "middle"}
http://jeremykun.com/2013/02/28/methods-of-proof-contradiction/	# Methods of Proof — Contradiction Posted on February 28, 2013 by In this post we’ll expand our toolbox of proof techniques by adding the proof by contradiction. We’ll also expand on our knowledge of functions on sets, and tackle our first nontrivial theorem: that there is more than one kind of infinity. ## Impossibility and an Example Proof by Contradiction Many of the most impressive results in all of mathematics are proofs of impossibility. We see these in lots of different fields. In number theory, plenty of numbers cannot be expressed as fractions. In geometry, certain geometric constructions are impossible with a straight-edge and compass. In computing theory, certain programs cannot be written. And in logic even certain mathematical statements can’t be proven or disproven. In some sense proofs of impossibility are hardest proofs, because it’s unclear to the layman how anyone could prove it’s not possible to do something. Perhaps this is part of human nature, that nothing is too impossible to escape the realm of possibility. But perhaps it’s more surprising that the main line of attack to prove something is impossible is to assume it’s possible, and see what follows as a result. This is precisely the method of proof by contradiction: Assume the claim you want to prove is false, and deduce that something obviously impossible must happen. There is a simple and very elegant example that I use to explain this concept to high school students in my guest lectures. Image you’re at a party of a hundred people, and any pair of people are either mutual friends or not mutual friends. Being a mathematical drama queen, you decide to count how many friends each person has at the party. You notice that there are two people with the same number of friends, and you wonder if it will always be the case for any party. And so the problem is born: prove or disprove that at every party of $n$ people, there exist two people with the same number of friends at the party. If we believe this is true, and we can’t seem to find a direct proof, then we can try a proof by contradiction. That is, we assume that there are not two people with the same number of friends. Equivalently, we can assume that everybody has a distinct number of friends. Well what could the possibilities be? On one hand, if there are $n$ people at the party, then the minimum number of friends one could have is zero (if you’re quite lonely), and the maximum is $n-1$ (if you’re friends with everybody). So there are $n$ distinct numbers, and $n$ people, and everyone has to have a different number. That means someone has to have zero friends, and someone has to be friends with everybody. But this can’t possibly be true: if you’re friends with everyone (and we’re only counting mutual friendships) then nobody can be friendless. Thus, we have arrived at a contradiction, and our original assumption must have been incorrect. That is, every party has two people with the same number of friends at the party. There are certainly other proofs of this fact (I know of a direct proof which is essentially the same proof as the one given above), and there are more mathematical ways to think about the problem. But this is a wonderful example of a proof which requires little else than the method of contradiction. ## A Reprise on Truth Tables, and More Examples Just as with our post on contrapositive implication, we can analyze proof by contradiction from the standpoint of truth tables. Recall the truth table for an implication $p \to q$: ```p q p->q T T T T F F F T T F F T``` We notice that an implication can only be false if the hypothesis $p$ is true and the consequence $q$ is false. This is the motivation for a proof by contradiction: if we show this case can’t happen, then there’s no other option: the statement $p \to q$ must be true. In other words, if supposing “p and not q” is true implies something which we know to be false, then by the very same truth table argument it must be that either “q” is true or “p” is false. In any of these cases “p implies q” is true. But all of this truth table juggling really takes us away from the heart of the method. Let’s do some proofs. First, we will prove that the square root of 2 is not a rational number. That is, we are proving the statement that if $x$ is a number such that $x^2 = 2$, then it cannot be true that $x = p/q$ where $p,q$ are integers. Suppose to the contrary (this usually marks the beginning of a proof by contradiction) that $x = p/q$ is a ratio of integers. Then we can square both sides to get $2 = x^2 = p^2 / q^2$, and rearrange to arrive at $2q^2 = p^2$. Now comes the tricky part: if a number is a divisor of $p$, then its square must divide $p^2$; this is true of all square numbers. In particular, it must be the case that the largest power of 2 dividing any square number is even (and $2^0$ counts as an even power). Now in the equation $2q^2 = p^2$ the right hand side is a square, so the largest power of two dividing it is even, and the right hand side is two times a square, so the largest power of 2 dividing it is odd (2 times an even power of 2 gives an odd power of two). But the two sides are equal! They can’t possibly have different largest powers of 2 dividing them. So we have arrived at a contradiction, and it must not be the case that $x$ is rational. This is quite a nice result, and a true understanding of the proof comes when you see why it fails for numbers which actually do have rational square roots (for instance, try it for the square root of 9 or 36/25). But the use of the method is clear: we entered a magical fairy land where the square root of 2 is a rational number, and by using nothing but logical steps, we proved that this world is a farce. It cannot exist. Often times the jump from “suppose to the contrary” to the contradiction requires a relatively small piece of insight, but in other times it is quite large. In our example above, the insight was related to divisors (or prime factorizations) of a number, and these are not at all as obvious to the layman as our “having no friends” contradiction earlier. For instance, here is another version of the square root of two proof, proved by contradiction, but this time using geometry. Another example is on tiling chessboards with dominoes (though the application of the proof by contradiction in this post is more subtle; can you pick out exactly when it’s used?). Indeed, most proofs of the fundamental theorem of algebra (these are much more advanced: [1] [2] [3] [4]) follow the same basic recipe: suppose otherwise, and find a contradiction. Instead of a obviously ridiculous statement like “1 = 0″, often times the “contradiction” at the end of a proof will contradict the original hypothesis that was assumed. This happens in a famous proof that there are infinitely many prime numbers. Indeed, if we suppose that there are finitely many prime numbers, we can write them all down: $p_1 , \dots, p_n$. That is, we are assuming that this is a list of all prime numbers. Since the list is finite, we can multiply them all together and add 1: let $q = p_1 \dots p_n + 1$. Indeed, as the reader will prove in the exercises below, every number has a prime divisor, but it is not hard to see that no $p_i$ divides $q$. This is because no matter what some number $x$ is, no number except 1 can divide both $x$ and $x-1$ (one can prove this fact by contradiction if it is not obvious), and we already know that all the $p_i$ divide $q-1$ . So $q$ must have some prime divisor which was not in the list we started with. This contradicts that we had a complete list of primes to begin with. And so there must be infinitely many primes. Here are some exercises to practice the proof by contradiction: 1. Prove that the base 2 logarithm of 3 is irrational. 2. More generally that $\log_a(b)$ is irrational if there is any prime $p$ dividing $a$ but not $b$, or vice versa. 3. Prove the fundamental theorem of arithmetic, that every natural number $n \geq 2$ is a product of primes (hint: inspect the smallest failing example). ## A Few More Words on Functions and Sets Last time we defined what it means for a function $f: X \to Y$ on sets to be injective: different things in $X$ get mapped to different things in $Y$. Indeed, there is another interesting notion called surjectivity, which says that $f$ “hits” everything in $Y$ by something in $X$. Definition: A function $f: X \to Y$ is surjective if for every element $y \in Y$ there is an $x \in X$ for which $f(x) = y$. A surjective function is called a surjection. A synonym often used in place of surjective is onto. For finite sets, we use surjections to prove something nice about the sets it involves. If $f:X \to Y$ is a surjection, then $X$ has at least as many elements as $Y$. The reader can prove this easily by contradiction. In our previous post we proved an analogous proposition for injective functions: if $f: X \to Y$ is injective, then there are at least as many elements of $Y$ as there are of $X$. If we combine the two notions, we can see that $X$ and $Y$ have exactly the same size. Definition: A function $f: X \to Y$ which is both injective and surjective is called a bijection. The adjectival form of bijection is bijective. So for finite sets, if there exists a bijection $X \to Y$, then $X$ and $Y$ have the same number of elements. The converse is also true, if two finite sets have the same size one can make a bijection between them (though a strictly formal proof of this would require induction, which we haven’t covered yet). The main benefit of thinking about size this way is that it extends to infinite sets! Definition: Two arbitrary sets $X,Y$ are said to have the same cardinality if there exists a bijection $f : X \to Y$. If there is a bijection $f: \mathbb{N} \to X$ then $X$ is said to have countably infinite cardinality, or simply countably infinite. If no such bijection exists (and $X$ is not a finite set), then $X$ is said to be uncountably infinite. So we can say two infinite sets have the same cardinality if we can construct a bijection between them. For instance, we can prove that the even natural numbers have the same cardinality as the regular natural numbers. If $X$ is the set of even natural numbers, just construct a function $\mathbb{N} \to X$ by sending $x \mapsto 2x$. This is manifestly surjective and injective (one can prove it with whatever method one wants, but it is obviously true). A quick note on notation: when mathematicians want to define a function without giving it a name, they use the “maps to” arrow $\mapsto$. The reader can think of this as the mathematician’s version of lambda expression. So the above map would be written in python: lambda x: 2x. So we have proved, as curious as it sounds to say it, that there are just as many even numbers as not even numbers. Even more impressive, one can construct a bijection between the natural numbers and the rational numbers. Mathematicians denote the latter by $\mathbb{Q}$, and typically this proof is done by first finding a bijection from $\mathbb{N} \to \mathbb{Z}$ and then from $\mathbb{Z} \to \mathbb{Q}$. We are implicitly using the fact that a composition of two bijections is a bijection. The diligent reader has already proved this for injections, so if one can also prove it for surjections, by definition it will be satisfied for bijections. ## Diagonalization, and a Non-Trivial Theorem We now turn to the last proof of this post, and our first non-trivial theorem: that there is no bijection between the set of real numbers and the set of natural numbers. Before we start, we should mention that calling this theorem ‘non-trivial’ might sound insulting to the non-mathematician; the reader has been diligently working to follow the proofs in these posts and completing exercises, and they probably all feel non-trivial. In fact, mathematicians don’t use trivial with the intent to insult (most of the time) or to say something is easy or not worth doing. Instead, ‘trivial’ is used to say that a result follows naturally, that it comes from nothing but applying the definitions and using the basic methods of proof. Of course, since we’re learning the basic methods of proof nothing can really be trivial, but if we say a theorem is non-trivial that means the opposite: there is some genuinely inspired idea that sits at the focal point of the proof, more than just direct or indirect inference. Even more, a proof is called “highly non-trivial” if there are multiple novel ideas or a menagerie of complicated details to keep track of. In any case, we have to first say what the real numbers are. Instead we won’t actually work with the entire set of real numbers, but with a “small” subset: the real numbers between zero and one. We will also view these numbers in a particular representation that should be familiar to the practicing programmer. Definition: The set of $[0,1]$ is the set of all infinite sequences of zeroes and ones, interpreted as the set of all binary decimal expansions of numbers between zero and one. If we want to be rigorous, we can define an infinite sequence to either be an infinite tuple (falling back on our definition of a tuple as a set), or we can define it to be a function $f : \mathbb{N} \to \left \{ 0, 1 \right \}$. Taking the latter view, we add one additional piece of notation: Definition: An infinite binary sequence $(b_i) = (b_1, b_2, \dots)$ is a function $b : \mathbb{N} \to \left \{ 0, 1 \right \}$ where we denote by $b_i$ the value $b(i)$. So now we can state the theorem. Theorem: The set $[0,1]$ of infinite binary sequences is uncountably infinite. That is, there is no bijection $\mathbb{N} \to [0,1]$. The proof, as we said, is non-trivial, but it starts off in a familiar way: we assume there is such a bijection. Suppose to the contrary that $f : \mathbb{N} \to [0,1]$ is a bijection. Then we can list the values of $f$ in a table. Since we want to use $b_i$ for all of the values of $f$, we will call $\displaystyle f(n) = (b_{n,i}) = b_{n,1}, b_{n,2}, \dots$ This gives us the following infinite table: $\displaystyle \begin{matrix} f(1) &=& b_{1,1}, & b_{1,2}, & \dots \\ f(2) &=& b_{2,1}, & b_{2,2}, & \dots \\ f(3) &=& b_{3,1}, & b_{3,2}, & \dots \\ \vdots & & \vdots & & \end{matrix}$ Now here is the tricky part. We are going to define a new binary sequence which we can guarantee does not show up in this list. This will be our contradiction, because we assumed at first that this list consisted of all of the binary sequences. The construction itself is not so hard. Start by taking $c_i = b_{i,i}$ for all $i$. That is, we are using all of the diagonal elements of the table above. Now take each $c_i$ and replace it with its opposite (i.e., flip each bit in the sequence, or equivalently apply $b \mapsto 1-b$ to each entry). The important fact about this new sequence is it differs from every entry in this table. By the way we constructed it, no matter which \$lateex n\$ one chooses, this number differs from the table entry $f(n)$ at digit $n$ (and perhaps elsewhere). Because of this, it can’t occur as an entry in the table. So we just proved our function $f$ isn’t surjective, contradicting our original hypothesis, and proving the theorem. The discovery of this fact was an important step forward in the history of mathematics. The particular technique though, using the diagonal entries of the table and changing each one, comes with a name of its own: the diagonalization argument. It’s quite a bit more specialized of a proof technique than, say, the contrapositive implication, but it shows up in quite a range of mathematical literature (for instance, diagonalization is by far the most common way to prove that the Halting problem is undecidable). It is worth noting diagonalization was not the first known way to prove this theorem, just the cleanest. The fact itself has interesting implications that lends itself nicely to confusing normal people. For instance, it implies not only that there is more than one kind of infinity, but that there are an infinity of infinities. Barring a full discussion of how far down the set-theory rabbit hole one can go, we look forward to next time, when we meet the final of the four basic methods of proof: proof by induction. Until then! ### Like this: This entry was posted in Number Theory, Primers, Set Theory, Teaching and tagged bijections, countability, diagonalization, mathematics, methods of proof, proof by contradiction by j2kun. Bookmark the permalink. ## 11 thoughts on “Methods of Proof — Contradiction” 1. Andrew on February 28, 2013 at 9:51 pm said: This post suffers from the usual confusion about the difference between “proof by contradiction” and “proof of negation”. Andrej Bauer has a nice post about the distinction here: http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/ The short of it is, proving a negative statement is not a proof by contradiction, in the sense that it is still a constructively valid proof. A proof by contradiction is when you prove a positive statement by assuming its negation. • This comment suffers from the usual confusion about the difference between “studying the properties of an axiomatic logical system” and “doing mathematics.” Claiming these aren’t proofs by contradiction because someone defined it as such is a vapid philosophical debate. • Andrew on February 28, 2013 at 10:48 pm said: It’s not vapid philosophical debate. The point is that they are distinct ideas, and conflating these two ideas easily leads to confused and muddled proofs. Gowers’ post that sparked Andrej’s reply was confusedly musing about a “direct” proof of the irrationality of sqrt(2). There is no such thing — irrationality is a negative property and so in the end can only be proven by the “proof of negation” technique. Clarifying this distinction gives one a better understanding of the structure of proofs and how approach proofs — and that, I believe, is the whole point of your original post. • The point is to teach people how to do proofs, and since the proof technique is identical regardless of whether you call it contradiction or “negation,” it’s actually more confusing to point it out. Everyone who actually does mathematics (and doesn’t specifically study intuitionistic logic) doesn’t care about the distinction, because double negations cancel and that’s all there is to it. Nothing about that implicit step makes a proof confused or muddled. It just shows that intuitionistic logic is too stupid to recognize an obvious equivalence of statements. This is a typical thing that happens in logic all the time: intentionally simplified models and axiomatic systems are too dumb to do natural things. It’s not a deep metamathematical insight; it’s a curiosity for logicians to tinker with. And it certainly has no place in a students’ first glimpse at proof technique. • Andrew on March 1, 2013 at 6:57 am said: Knowing this distinction simplifies the process of proving something for the simple reason that if you are proving a negative statement, you must* use “proof of negation”. However, if you are proving a positive statement, most of the time you in fact should not use proof by contradiction. In that sense, knowing the distinction is very useful to both learned mathematicians and students. • on March 1, 2013 at 10:40 am said: I don’t think you understand: the distinction only exists in a particular logical framework that nobody uses to actually do mathematics. Sure one might prefer constructive proofs of existence to relate theorems to algorithms, but there is never a time when proving a theorem by contradiction is a bad thing. 2. Greg on March 1, 2013 at 7:06 pm said: “If there is a bijection f : N -> X then X is said to have countably infinite cardinality, or simply countably infinite. If no such bijection exists, then X is said to be uncountably infinite.” Knowing that there isn’t a bijection between N and X only tells us that the two sets have different cardinalities. Don’t we still have to prove that there isn’t a surjection between N and X, before we can say that X is uncountably infinite? • on March 1, 2013 at 8:22 pm said: Yes, of course. Thanks for catching that. 3. on April 11, 2013 at 7:34 am said: Hey j2kun, thanks for this post. I think I found I typo in “Now in the equation 2q^2 = p^2 the right hand side is a square, so the largest power of two dividing it is even, and the right hand side is two times a square, so the largest power of 2 dividing it is odd” Did you mean “… and the left hand side is two times a square …”? • on April 11, 2013 at 8:34 am said: Yes of course. No matter how easy a proof is, it’s even easier to make a mistake in writing it down 4. on May 10, 2013 at 6:26 am said: Hi Jeremy, very nice primers, indeed ! Thanks for your work… Some typos/suggestions : * “there is no bijection between the set of real numbers is uncountably infinite”: looks like you changed your mind in the middle of the sentence, as how to express it * “it differs from entry f(n) precisely at digit n”: I would rather write “at least” than “precisely”, which seem to imply that it does not differ elsewhere (which we don’t know for sure, nor care about) Cancel	{"extraction_info": {"found_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": 92, "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": 0.9437165856361389, "perplexity_flag": "head"}
http://colindcarroll.com/category/math/calculus/	The integral of csc(x) [NOTE: At the end of editing this, I found that the substitution used below is famous enough to have a name, and for Spivak to have called it the "world's sneakiest substitution". Glad I'm not the only one who thought so.] In the course of working through some (very good) material on neural networks (which I may try to work through here later), I noticed that it was beneficial for a so-called “activation function” to be able to be written as the solution of an “easy” differential equation. Here by “easy” I mean something closer to “short to write” than “easy to solve”. The [activation] function sigma. In particular, two often used activation functions are $\sigma (x) := \frac{1}{1+e^{-t}}$ and $\tau (x) := \tanh{x} = \frac{e^{2x}-1}{e^{2x}+1}.$ One might observe that these satisfy the equations $\sigma' = \sigma (1-\sigma),$ and $\tau' = 1-\tau^2.$ By invoking some theorems of Picard, Lindelof, Cauchy and Lipschitz (I was only going to credit Picard until wikipedia set me right), we recall that we could start from these (separable) differential equations and fix a single point to guarantee we would end up at the functions above. In seeking to solve the second, I found after substituting cos(u) =τ that $-\int\frac{du}{\sin{u}} = x+C,$ and shortly after that, I realized I had no idea how to integrate csc(u). Obviously the internet knows (substitute v = cot(u) + csc(u) to get the integral being -log(cot(u)+csc(u))), which is a really terrible answer, since I would never have gotten there myself. Not the right approach. Instinctually, I might have tried the approach to the right, which gets you back to where we started, or by changing the numerator to cos2x+sin2x, which leads to some amount of trouble, though intuitively, this feels like the right way to do it. Indeed, eventually this might lead you to using half angles (and avoiding integrals of inverse trig functions). We find $I = \int \frac{du}{\sin{u}} = \int \frac{\cos^2{u/2} + \sin^2{u/2}}{2\cos{u/2}\sin{u/2}}.$ Avoiding the overwhelming temptation to split this integral into summands (which would leave us with a cot(u)), we instead divide the numerator and denominator by sin2(u) to find $I=\int \frac{1+\tan^2{u/2}}{2\tan{u/2}} du.$ Now substituting v = tan(u/2), we find that dv = 1/2 (1+tan2(u/2))du = 1/2(1+v2)du, so making this substitution, and then undoing all our old substitutions: $I = \int \frac{1+v^2}{v}\frac{2}{1+v^2}dv = \int \frac{dv}{v} = \log{\|v\|} + C = \log{\|\tan{\frac{u}{2}}\|}+C = \log{\|\tan{\frac{\cos^{-1}\tau}{2}}\|}+C.$ The function tau we worry so much about. Looks pretty much like sigma. Using the half angle formulae that everyone of course remembers and dropping the C (remember, there’s already a constant on the other side of this equation), this simplifies to (finally) $I = \log{\|\frac{\sqrt{1-\tau^2}}{1+\tau}\|}.$ Subbing back in and solving for $\tau(x)$ gives, as desired, $\tau(x) = \frac{e^{2x}-1}{1+e^{2x}}$. Phew. Posted in calculus, differential equations, Math Expectations II A contour plot of the function. Pretty respectable looking hills- maybe somewhere in the Ozarks- if I say so myself. As a further example of yesterday’s post, I was discussing multivariable calculus with a student who had never taken it, and mentioned the gradient. Putting our discussion into the framework of this post, here is what he wanted out of such a high dimensional analogue of the derivative of a function $f: \mathbb{R}^2 \to \mathbb{R}$ (note to impressionable readers: the function defined below is not quite the gradient): 1. Name the answer: Call the gradient D. 2. Describe the answer: D should be a function from $\mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}^3$, which takes a point in the domain, a direction in the domain, and returns the vector in the range. The idea being that if you had a map, knew where you are and in which direction you wished to travel, then the gradient should tell you what 3-dimensional direction you would head off in. Certainly there is such a function, though in some sense we are making it too complicated. As an example we have some pictures of the beautiful hills formed by the function $f(x,y) = \sin{3y} + \sin{(4x + 5y)} - x^2 - y^2 + 4.$ The (actual) gradient of this function is $\nabla f(x,y) = \left(4\cos (4x + 5y) - 2x, 3\cos(3y) - 2y + 5\cos(4x + 5y)\right)$. Plugging in a point in the plane will give a single vector, and then taking the dot product of this vector with a direction will give a rate of change for f at that point, in that direction. Specifically, if we start walking north at unit speed from the origin, the gradient will be (4,8), and I take the dot product of this with (0,1) to find that I will be climbing at 8m/s (depending on our units!) Now the correct answer from my student’s point of view would be that the answer is (0,1,8), since this is the direction in 3 dimensions that one would travel, and that the correct definition for D would have $Df(x,y) \cdot v = \left(x,y,\nabla(x,y) \cdot v \right)$. The graph of the indicated function, including the vector of the "pseudo-gradient" we discuss. Of course there are more sophisticated examples of this. Suppose a function $u: \mathbb{R}^n \to \mathbb{R}$ is harmonic. That is to say, $\Delta u := \sum_{j = 1}^n \frac{\partial^2 u}{\partial x_j^2} = 0$. Notice that in order to write down this equality, we already named our solution u. But just working from this equation, we can deduce a number of qualities that any solution must have: u is infinitely differentiable and, restricted to any compact set, attains its maximum and minimum on the boundary of that set. Such properties quickly allow us to narrow down the possibilities for solutions to problems. A picture of some hills that might be shaped like the function we're looking at. In the Ozarks of all places! Posted in Math, General, calculus, differential equations Expectations Hard thinkin' being done today. It is useful (for me!) to think about the importance of math as teaching us how to think about problems, rather than providing us with useful factoids (I’m looking at you, history class). There are a lot of problems/puzzles/patterns in the world, and the chance of seeing the same problem twice is very low (and really, I’ve never seen Batman use the Pythagorean theorem even once, so what’s the point?), so we focus on solving problems in as broad of a context as possible. In this way, I’d argue, mathematicians become very good problem solvers (“toot! toot!” <– my own horn) One method of problem solving I would like to focus on today is to name and describe your answer before you have found it. As a simple example, in order to answer the question “what number squared is equal to itself?”, we would: 1. Name the answer: Suppose x squared is equal to x. 2. Describe the answer: This is where the explicitly developed machinery comes in: We know that $x^2 = x$, so we deduct that x also has the property $x(x-1) = 0$, and conclude that either x = 0 or x = 1. A geometric way of looking at the word problem. NOT TO SCALE. As a second example, much of linear algebra is naming objects, describing them, and then realizing you accidentally completely described them. For example, suppose we wanted to identify every matrix with a number, and make sure that every singular matrix has determinant 0: 1. Name the answer: Let’s call the answer the determinant, or det() for short. 2. Describe the answer: det() should be a function from matrices to numbers, and at least satisfy the following properties: (i) det(I) = 1, so that the identity matrix is associated with the number 1 (so at least some nonsingular matrices will not have determinant zero), (ii) if the matrix A has a row of zeros, then det(A) = 0 (so that at least some singular matrices will have determinant zero, and (iii) the determinant is multilinear, which takes some motivation, will definitely respect identifying singular matrices. Well, it turns out that these three properties have already completely determined the object we are looking for! If I had been greedy and asked (iv) each nonsingular matrix is associated with a unique number, then I would have deduced that no such map exists. If I had not included property (iii), then I would have found there are many such maps. It is a fairly enjoyable exercise to deduce the other properties of determinants starting from just these three rules. More filler photos! This is from Cinqueterre in Italy, between some two of the towns. Posted in calculus, General, linear algebra, Math Another nice theorem Trying to visualize the projection map using fibers. You'll have to take my word that the lines stop before getting to the origin. Today’s Theorem of the Day (TM) I used to compute the Jacobian of a radial projection. In particular, consider the map $F: \mathbb{R}^n \to \mathbb{R}^n$ where $x \mapsto x/\|x\|$ for all $\|x\| > 1$. This projects all of n-space onto the surface of the unit ball, and leaves the interior untouched. Then we may compute the derivative $\frac{\partial F_j}{\partial x_k} = \frac{\delta_{j,k}\|x\|^2 - x_k^2}{\|x\|^3}$. To calculate the Jacobian of F means we have to calculate the determinant of that matrix. With a little figuring, we can write that last sentence as $\|JF(x)\| = \det \left(\frac{1}{\|x\|} ( I - \frac{x^Tx}{\|x\|^2} ) \right) = \frac{1}{\|x\|^n} \det \left(I-\frac{x^Tx}{\|x\|^2}\right)$. Now we apply The Theorem, which Terry Tao quoted Percy Deift as calling (half-jokingly) “the most important identity in mathematics”, and wikipedia calls, less impressively, “Sylvester’s determinant formula“. Its usefulness derives from turning the computation of a very large determinant into a much smaller determinant. At the extreme, we apply the formula to vectors u and v, and it says that $\det (I+u^Tv) = 1+v^Tu$. In our case, it yields $\|JF(x)\| = 0$. Thus we turned the problem of calculating the determinant of an n x n matrix into calculating the determinant of a 1 x 1 matrix. Pretty nifty. Posted in calculus, linear algebra, Math, Visualizing data Busy days. Somehow spring break has turned into one of the busier weeks of my year. Trying to keep up with real life work has not left a ton of time for writing anything thoughtful/reasonable, though at least for continuity I will try to keep a paragraph or so up here each day with my favorite thought of the day. This also means I can reuse some old graphics! Today I really enjoyed a particular fact about Sobolev functions. Recall that these are actually equivalence classes of functions, as they are really defined under an integral sign, which “can’t see” sets of small measure. However, the following quantifies exactly how small the bad set might have to be: If $f \in W^{1,p}(\Omega)$ for $\Omega \subset \mathbb{R}^n$, then the limit $\lim_{r \to 0} \frac{1}{\alpha(n)r^n}\int_{B(x,r)}f(y)~dy$ exists for all x outside a set E with $\mathcal{H}^{n-p+\epsilon}(E) = 0$ for all $\epsilon > 0$. Put another way, every Sobolev function may be “precisely defined” outside a set of small dimension, where the dimension gets smaller as p gets larger. I suppose a given representative may be worse, but this allows you to require that the member of the equivalence class of Sobolev functions has some nice properties. The fibers of two functions in a sequence. I was thinking the above argument might imply that the limit was not Sobolev, but the limit is precisely represented outside a set with positive 1-dimensional measure, so the result is silent on this issue. Posted in calculus, coarea, geometric measure theory, Math L’Hopital’s rule. Two photos from a recent trip up north. Major bonus points for knowing which of New England's many trails this was taken on. L’Hopital’s rule is really how every student of calculus (and I believe Leibniz, though I cannot find a reference) wishes the quotient rule worked. Specifically, that $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}.$ Of course, it can’t be that easy. We also need that f and g are differentiable in a neighborhood of a, that both function approach 0, or they both approach $\infty$, or they both approach $-\infty$ as x approaches this point a, and finally that the limit on the right hand side exists (though we all recall that if it does not work the first time, we may continue to apply L’Hopital until the limit does exist, which then justifies using L’Hopital in the first place). I was thinking of this rule in relation to generating interesting examples of limits. In particular, if we are in a situation where L’Hopital’s applies, then we can apply the rule in two ways: $\lim_{x\to a}\frac{f'(x)}{g'(x)}=\lim_{x \to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{\left(\frac{1}{g(x)}\right)'}{\left(\frac{1}{f(x)}\right)'}.$ Proceeding informally (i.e., I’m not going to keep track of hypotheses), the right hand side of this evaluates to $\lim_{x\to a}$$\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f(x)^2}{g(x)^2}\frac{g'(x)}{f'(x)}.$ This is all well and good- the right hand side looks appropriately ugly, but now the trick is picking f and g to get interesting limits. I have worked out two reasonable examples: 1. Choosing f(x) = sin(x), g(x) = x, we get $\lim_{x \to 0} \frac{\sin{x}}{x^2}\tan{x} = 1.$ Also, moderate amounts of bonus points for naming (at least) two universities in the northeast with this mascot. 2. Choosing $f(x) = e^x-1$ and $g(x) = \log{x}$, and applying (hopefully correctly!) a number of logarithm rules, we can get $\lim_{x \to 1} \frac{(e^x-1)^2}{\log{(x^{\log{(xe^x)}})}} = 0.$ What would be interesting is to find an example where it is difficult/impossible to evaluate without recognizing that it was created using this process. This second example might fit the “difficult” bill, as I would not want to take the derivative of the denominator directly, but factoring, you might recognize it as $xe^x (\log{x})^2$, and then be able to reverse engineer this process, somehow. As usual, just a thought I’ve been playing with. Posted in calculus, Math, teaching More with fibers of functions I posted earlier on a way of visualizing the fibers of certain maps from high dimensions to low dimensions. Specifically, if the range can be embedded in the domain so that f is the identity of the image of the range, then we can draw the inverse image at each point. I had some images of functions whose inverse image was a torus, but had trouble making these sorts of images for maps $f: \Omega \subset \mathbb{R}^3 \to \mathbb{R}^2$, so that the inverse image of a point is a line. Well, no more! Here are two images, one is the projection of a cube onto a square, and the other is somewhat more complicated, and is the string hyperboloid map. See the previous post for more details on these specific maps, but I just thought these were nice images! Fibers of the projection map from the cube to the square. Fibers of the "twisted cylinder", which are again straight lines. Posted in calculus, Math, Visualizing data More geometry with inverse images Yesterday’s post was on inverse images of functions as sets, and ways to visualize them. Today, I realized that despite my early series of posts on the Jacobian derailing, I probably have enough background to describe the area and coarea formulae. The two give a relationship between the “size” of the fibers and the derivative of the map. The first thing I’ll need to do is define the Jacobian for maps $f: \mathbb{R}^m \to \mathbb{R}^n$. The definition will be slightly different depending on whether m or n is larger, but if $n \geq m$, then $\|Jf(x)\| := \sqrt{\|Df(x)^T \cdot Df(x)\|}$, and if $m \geq n$, then $\|Jf(x)\| := \sqrt{\|Df(x) \cdot Df(x)^T\|}$ where Df is the n x m matrix of partial derivatives of f, and we use the absolute value bars to indicate a determinant. Notice that if m = n, then the definitions agree, and it is just the absolute value of the determinant of the matrix of partial derivatives. If n = 1 so that f is a real-valued function, then the Jacobian is the length of the gradient of f. Now then, the area formula says that for a Lipschitz $f:\mathbb{R}^m \to \mathbb{R}^n$ with $m \leq n$, and any Lebesgue measurable $U \subset \mathbb{R}^m$, $\int_U \|Jf(x)\| d\mathcal{L}^m(x) = \int_{\mathbb{R}^n} \#(f^{-1}(y) \cap U)~d\mathcal{H}^m(y),$ A hyperboloid projecting onto a circle. where, for a set S, $\#(S)$ denotes the cardinality of S, i.e., how many points are in S. We expect this number to be finite (for most functions f I think of, each inverse image has cardinality either 1 or 0). Indeed, notice that if f is a smooth embedding, then f is one-to-one, and the right hand side of the above is always 1 if f maps there. Hence the right hand side will be $\mathcal{H}^m(f(U))$, the area of the image of U under f. This explains why it is called the area formula- it agrees with the classical area of parametrized surfaces. The coarea formula (the subject of my research) keeps all the conditions above, but now f maps from high dimensions into a lower one, so $m \geq n$. We have $\int_U \|Jf(x)\| d\mathcal{L}^m(x) = \int_{\mathbb{R}^n} \mathcal{H}^{m-n}(f^{-1}(y) \cap U)~d\mathcal{H}^n(y).$ In plain English, the integral of the Jacobian of f is equal to the integral of the length of the fibers of f. [Technical sentences coming up!] One surprising fact is that this coarea formula was first proven in 1959 in Herbert Federer’s paper “Curvature Measures“, while the area formula was (is) a basic calculus fact, at least for smooth functions. The formula has since played a role in image processing, as when f is a real valued function, the quantity is usually referred to as the total variation. De Giorgi showed that the fibers of functions which minimized the left hand side are actually minimal surfaces. A string hyperboloid. I’ve included a few illustrations of how the coarea formula might relate to “projections” of hyperboloids onto the circle. The first shows such a hyperboloid, along with the fibers of the “projection”. The coarea of this map will be the surface area of the hyperboloid. Such a hyperboloid can be made with string from a cylinder, and twisting the top. See the second figure. The final .gif illustrates continuing to twist the top, and the resulting surfaces. In each case, integrating the function whose level sets are these straight lines will return the surface area of the hyperboloid Twisting hyperboloids Posted in calculus, coarea, geometric measure theory, Math Fibers of functions Inverses Something that is easy to miss in early calculus classes is that the inverse of a function is typically not a function. We go through this whole confusing notion first with the square root (because while it is true that if $x^2 = 16$ then $x = \pm 4$, we all know that we like +4 better), then with trig functions. I would argue that it is helpful to think about the inverse of a function as a set, and then point out the wonderful fact that if all the inverses of individual points have only one or zero members, then there is a function g so that g(f(x)) = x. Typically though, inverse images will have more than one point. Indeed, for a map $f: \mathbb{R}^m \to \mathbb{R}^n$, you will expect $f^{-1}(y)$ to be m-n dimensional, if m is bigger than n, and a point otherwise. Intuitively, this is because we have m equations and n unknowns, leaving us with m-n free variables. This suggests a way of visualizing functions that I have actually never seen used (references to where it has been used are welcome). What I have in mind is that, if you have a function $f: U \subset \mathbb{R}^m \to \mathbb{R}^n$, and it so happens that $f(U)$ can be isometrically embedded back into U by choosing the well from the sets $f^{-1}(y)$, then we may plot the inverse images of f on the same graph as we draw the domain of f. That last paragraph was confusing, so let me give an example right away. We will look at the function f which maps from the solid torus (donut) to the real numbers, so The map of the torus that gives the radius of a point. The line in red is the range of the map. Notice it intersects with every shell exactly once. that f(x) is the distance of x from the center of the solid torus. Hence $f^{-1}(r)$ will be the (not solid) torus of radius r. I have made the graph I describe above for this map. Notice that the image of the torus under f, a circle, is indicated in blue in the left of the graph. This picture has a nice intuition: each surface will map down to one point (so our intuition earlier holds up- f maps a three dimensional object down onto one dimension, so the inverse images are all two dimensional), so we can easily look at this and see the domain, range and action of f on the domain. Notice also that to plot this in a traditional manner it would take either 4 dimensions as a graph, or 1 overloaded dimension as a parametric plot. This particular example could* be displayed using a movie, though again we would be displaying fibers of the map. The last image of this sort is where we instead map a torus (again, non-solid) to a circle. Notice that now the map is from a 2-D surface to a 1-D curve, so we expect (and see that) the fibers to be 1 dimensional. The inverse images of the torus-radius map, as the radius goes from 0 to 1. Inverse images of a projection-of-sorts of the torus onto a circle. Posted in calculus, coarea, Math, MATLAB A note on graphs As a followup to my previous post on functions, I’d like to talk about graphs. More specifically, ways of visualizing functions. As before, we’ll go by cases, though this time we are somewhat limited by having only two dimensions to depict a function. Real valued functions of one real variable: These are what we all started with in high school, and unfortunately all most people will ever see (though maybe also some parametric graphs). When someone says “graph” this is typically what they mean. This happens to be a graph of the function $f(x) = x^2 + \sin{4x}$, displayed from x = -2 to +2. To make a graph like this by hand, one would go to x = 0, figure out that f(0) = 0, then put a single dot at the point (0,0). Then we would move to, say, x = 1, see that f(1) = 1+sin(4), and put another dot there. We repeat this process ad infinitum, and get a picture. If we wanted to make this by computer, we ask MATLAB to make a vector with, in this case, 100 elements, looking like x = (-2.00, -1.95, -1.90, … , 1.85, 1.90, 1.95, 2.00). Then, since MATLAB enjoys doing arithmetic a vector at a time, we calculate a new vector y = x.^2+sin(4*x), and then plot x and y by typing plot(x,y). The computer then plots all the points in the vectors, and connects them to create the smoothish line above. Real valued functions of two real variables: These are the objects of study in multivariable calculus, and I will stop explaining everything so much. Suffice it to say that we plot the height of a function above the point (x,y), so we can start creating things that look like surfaces: This guy looks pretty wild (if I do say so myself), but we’re just getting started. For interest’s sake, it is the graph of $f(x,y) = x^2 + y^2 + \sin(4x) + \sin(4y).$ Notice that neither function we have plotted has had any self intersections, and each has been a proper function. It passes the “vertical line test”, to borrow a phrase no one uses outside of calc 1. If we want spheres and donuts, we can’t let the vertical line test stop us. Parameterizations of a line in the plane: Above, it cost us one dimension for each variable in the domain, and one more for the range, meaning we’ll have to be clever to draw the graph of a real valued function of 3 or more variables. However, with parametric plots, we will use precisely as many dimensions as are in the range. First, a parametrization of a line in the plane can be thought of intuitively as telling a line how to sit in the plane. The above is the graph u(t) = (sin(3t),sin(8t)). As mentioned, it won’t be possible to realize the above curves as a traditional graph, but I should say that it is easy to realize any traditional graph as a parametric graph. For example, the graph of y = sin(x) is given parametrically by u(t) = (t,sin(t)). Functions of one real variable into three space: This is very useful, and gets you used to the idea that we are really just telling a line where to perch. I’ve animated this graph, rather than using shadows, to give a sense of depth. The above is the plot u(t) = (cos(t)sin(.2t),sin(t)sin(.2t),cos(.2t)). I chose this because the curve would sit on the surface of the sphere. It is late, and the post is long, so I will give just one more example for now. Functions of two variables into three space: These are much more general surfaces then we saw earlier. Again, we can think of the maps as telling the plane how to sit in three space. The above is a torus, which is generated using the parameterization $u(s,t) = \cos(s)(2+\cos(t)),\sin(s)(2+\cos(t)),\sin(t)).$ By adding sin(5s) to the z coordinate, you can get the following, rather more complicated looking graph: Hopefully sometime in the future I’ll talk about more exotic techniques of graphing. Posted in calculus, Math, MATLAB, Programming RSS Feed Blog at WordPress.com. \| Theme: Quintus by Automattic.	{"extraction_info": {"found_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": 68, "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": 0.9216828942298889, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/134813/examples-of-comma-categories	Examples of comma categories There is a basic construction in category theory which I've only just recently become acquainted with, that is the comma category. It seems to be a quite basic construction for which, however, I've seen really few "real life" examples. I know the slice, coslice and arrow categories are particular cases of a comma category. This is in MacLane, or in the wikipedia article. In that article there are also the examples of pointed sets or graphs, which are of a more concrete nature. I'm asking, then, for more examples of this construction in mathematics. Examples of (co)slice categories are also welcome. Here's one example I've come up with. The completion of a metric space $M$ consists of a pair $(\overline{M},i)$ where $\overline{M}$ is a complete metric space and $i:M\to \overline{M}$ is a uniformly continuous function which satisfies the following universal property: if $N$ is another complete metric space and $g:M\to N$ is a uniformly continuous function, then there exists a unique uniformly continuous function $h:\overline{M}\to N$ such that the following diagram commutes: I claim this completion is an initial object in a suitable comma category. Consider the functors where $\mathbf{Met_u}$ is the category of metric spaces with uniformly continuous functions and $\mathbf{CompMet_u}$ is the category of complete metric spaces with uniformly continuous functions. The functor $F$ is such that $F(\star)=M$ where $\star$ is the sole object of $\mathbf{1}$, and $U$ is the inclusion functor. Then an initial object of $(F\downarrow U)$ is exactly a completion of $M$. Bonus question: is this approach to the completion not interesting/not useful? I ask this because it seems the categorial approach to completions has nothing to do with comma categories. (I can see, though, that the universal property of this completion can also be seen as an adjunction). - I'm sorry for the "bonus question" part, I know it is not good practice to ask more than one question per question (hmm, that's slightly redundant). If it is a more interesting question than it seemed to me and there is someone willing to give a comprehensive answer that is not suitable for a comment here, then I will happily split it into another question. – Bruno Stonek Apr 21 '12 at 14:45 Have you read the proof of the general adjoint functor theorem? It constructs left adjoints via initial objects of certain comma categories... – Zhen Lin Apr 21 '12 at 16:23 @ZhenLin: I haven't (but plan on doing so relatively shortly). That's very interesting to me, and it can't be just a coincidence that both my example and the one given in the answer by Matt N. are both left adjoints... Thank you. – Bruno Stonek Apr 21 '12 at 18:07 Did you want things that weren't slice/coslice categories? I assume you know that you can think of $R\text{-}\mathbf{Alg}$ as the slice category $(R\downarrow \mathbf{CRing}$. – Alex Youcis Apr 21 '12 at 18:08 @AlexYoucis: no, I didn't mean to say that, I'm also interested in "concrete" examples of (co)slice categories! I didn't know what you mention, perhaps you could post an answer? – Bruno Stonek Apr 21 '12 at 18:11 show 2 more comments 2 Answers One example of a comma category $(F \downarrow G)$, I think of it as "simplified" comma category but I don't think that's a commonly used term, is what you get when you take $F: A \to C$ to be the selection functor (i.e. $A = \textbf{1}$). More concretely, if $G:\textbf{Group} \to \textbf{Set}$ is the forgetful functor mapping a group to its underlying set and $F: \textbf{1} \to \textbf{Set}$ is the selection functor selecting a set $S$ then you get the category $(S \downarrow G)$ where the objects are pairs $(f,Y)$ where $f:S \to Y$ is a morphism in $\textbf{Set}$ and $Y = G(X)$ is the underlying set of some group $X$. The morphisms $(f,Y) \to (f^\prime, Y^\prime)$ are induced by group homomorphisms $\varphi : X \to X^\prime$ such that $G(\varphi) \circ f = f^\prime$. You can use this category to define the free group over a set $S$, namely, it is a group $F$ such that $(f,G(F))$ is an initial object in $(S \downarrow G)$. Hope this serves as an example. As for the bonus question: I'll have to pass on that. - Yes, it is a nice example, thank you! I'm so used to thinking of the free functor as a left adjoint of the forgetful functor, that I hadn't thought of that triangle as an arrow in a comma category. This example is similar to the one I gave, where also one of the functors was a "selection functor" as you call them; and it could also be seen as an adjoint (see the nlab article). I guess seeing both these constructions as adjoints is nicer because it gets functors (i.e. free functor, completion functor) out of them, not only objects. – Bruno Stonek Apr 21 '12 at 15:04 @BrunoStonek Yes, actually I'd call it the same as what you give in your question. : ) I suspect there are many more "same" examples of this kind: all things defined in terms of universal properties, such as for example free modules. – Matt N. Apr 21 '12 at 15:09 Ok, if you are interested in slice/coslice categories then there are two obviously interesting ones: It's pretty trivial that the category $\mathbf{Top}_\ast$ of pointed topological spaces is just $(\bullet\downarrow\mathbf{Top})$. As another example let's find out what $(R\downarrow\mathbf{CRing})$ looks like, when $R$ is come commutative unital ring. We see that, almost by definition, we can think of the objects of $(R\downarrow\mathbf{CRing})$ as being commutative unital rings $A$ with a distinguished ring homomorphism $f:R\to A$. Ok, there's no more "simplification" that can be done there. So, what do the arrows in this comma category look like? Well, if we have two objects $f:R\to A$ and $g:R\to B$, we see that an arrow between them is an arrow $h:A\to B$ (of course, technically it's a pair of arrows, but when the left element of $(-\downarrow-)$ is discrete this is always the identity arrow, and so unimportant) such that $g=h\circ f$. But, this precisely the formulation for commutative algebras over $R$. In other words, we have commutative unital rings $A$ with specified $\mathbf{CRing}$-arrows $R\to A$ and the morphisms between two such objects $(R,A,f)$ and $(R,B,g)$ is just a $\mathbf{CRing}$-arrow $A\to B$ which respects $f$ and $g$. Thus, we see that $\left(R\downarrow\mathbf{CRing}\right)\cong R\text{-}\mathbf{CAlg}$. - Thank you for your elaboration. – Bruno Stonek Apr 21 '12 at 18:23	{"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": 60, "mathjax_display_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": 0.9545318484306335, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/77119/linear-homogeneous-algebraic-equations/77125	# Linear Homogeneous Algebraic Equations From the following two linear homogeneous algebraic equations: $$A \sin\left(\frac{kl}{\sqrt2}\right) = B \sin(kl)$$ $$\frac{kA}{\sqrt2}\cos\left(\frac{kl}{\sqrt2}\right) = kB\cos(kl)$$ form matrix of these 2 equations, and setting the determinant equal to zero will lead to: $$\frac1{\sqrt2}\cos\left(\frac{kl}{\sqrt2}\right)\sin(kl) - \sin\left(\frac{kl}{\sqrt2}\right)\cos(kl) = 0.$$ k is unknown. l is constant. A and B are constants. I'm trying to find a nonzero solution when the determinant of the equation system vanishes. How do I solve this? - Which of those letters are the unknowns, and which are constants? – Henning Makholm Oct 30 '11 at 7:18 @Chris: please use mathjax commands for your question. The question is not clear. – Hassan Muhammad Oct 30 '11 at 7:36 @Henning: It appears that $A$ and $B$ are the unknowns. – Brian M. Scott Oct 30 '11 at 7:39 These are not algebraic equations. Also there's very little to do with linear algebra here. – Gerry Myerson Oct 30 '11 at 9:53 ## 2 Answers $(A,B)=(0,0)$ is always a solution. If $k=0$ then both equations vanish, and any $(A,B)$ at all will be a solution. If $k\ne 0$ but $l=0$, then the first equation vanishes and the second one reduces to $A=\sqrt 2 B$. There are other special values of $kl$ where the two equations are proportional such that there are nonzero solutions, but you will have to find them numerically. Your determinant procedure looks correct and simplifies to $\frac{\tan(kl)}{\sqrt 2} = \tan(\frac{kl}{\sqrt 2})$, which probably has no nice closed-form solutions other than $kl=0$. - As previous comments indicate, it is not clear what you are trying to find. However, It is possible to solve this by getting a realation between A and B as follows: $A \sin\left(\frac{kl}{\sqrt2}\right) = B \sin(kl)$ (1) $\frac{kA}{\sqrt2}\cos\left(\frac{kl}{\sqrt2}\right) = kB\cos(kl)$ (2) which is: ${A}\cos\left(\frac{kl}{\sqrt2}\right) = {\sqrt2} B\cos(kl) , k \ne 0$ (3) Square and add both sides of (1) and (3): $A^2=B^2(1+(\cos(kl))^2)$ When I first posted this comment, I got a simpler experession but I was corrected by Pedja, I hope this can help you... - $A^2=B^2(1+(\cos(kl))^2)$ – pedja Oct 30 '11 at 8:30 1 @pedja: Thanks, you are correct of course. I have fixed that in the post. – Emmad Kareem Oct 30 '11 at 8:38 Replace <>= by $\backslash\mathtt{ne}$. – Did Oct 30 '11 at 8:43 @Didier Piau: Thanks, done. I am not very familiar with the syntax yet. – Emmad Kareem Oct 30 '11 at 8:45 1 @Emmad, right. The best way to learn is to look at the source of other questions or answers and to see how they encode such and such. – Did Oct 30 '11 at 8:48 show 1 more comment	{"extraction_info": {"found_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": 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": 0.9519175887107849, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/43681/motivating-the-de-rham-theorem	## Motivating the de Rham theorem ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In grad school I learned the isomorphism between de Rham cohomology and singular cohomology from a course that used Warner's book Foundations of Differentiable Manifolds and Lie Groups. One thing that I remember being puzzled by, and which I felt was never answered during the course even though I asked the professor about it, was what the theorem could be used for. More specifically, what I was hoping to see was an application of the de Rham theorem to proving a result that was "elementary" (meaning that it could be understood, and seen to be interesting, by someone who had not already studied the material in that course). Is there a good motivating problem of this type for the de Rham theorem? To give you a better idea of what exactly I'm asking for, here's what I consider to be a good motivating problem for the Lebesgue integral. It is Exercise 10 in Chapter 2 of Rudin's Real and Complex Analysis. If $\lbrace f_n\rbrace$ is a sequence of continuous functions on $[0,1]$ such that $0\le f_n \le 1$ and such that $f_n(x)\to 0$ as $n\to\infty$ for every $x\in[0,1]$, then $$\lim_{n\to\infty}\int_0^1 f_n(x)\thinspace dx = 0.$$ This problem makes perfect sense to someone who only knows about the Riemann integral, but is rather tricky to prove if you're not allowed to use any measure theory. If it turns out that there are lots of answers then I might make this community wiki, but I'll hold off for now. - 2 I don't think this deserves to be an answer, so I 'll leave it as a comment. I tend to think of the other way, that de Rham cohomology is the thing that one is often interested in. From this point of view de Rham's theorem gives a tool for understanding it at a deeper level, and also for computing it by bring in topological methods. – Donu Arapura Oct 26 2010 at 16:42 I've thought of it as the natural extension of the 'miracle' from algebraic topology that the various homology theories (singular, simplicial, CW) are all equal where it makes sense to compare them. Though, to be honest, the de Rham theorem is a much bigger miracle than the other ones. – Ketil Tveiten Mar 2 2011 at 13:02 ## 11 Answers Here is a really "trivial" application. Since a volume form (say from a Riemannian metric) for a compact manifold $M$ is clearly closed (it has top degree) and not exact (by Stoke's Theorem), it follows that the cohomology is non-trivial, so $M$ cannot be contractible. - What about $D^n$? – Nikita Kalinin Oct 26 2010 at 19:44 5 @Nikita: $D^n$ is a manifold-with-boundary, not a manifold (which by definition is locally-diffeo to $R^n$). Perhaps I should have said "closed manifold" to make this point clear. – Dick Palais Oct 26 2010 at 20:11 ### 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. I don't know if it is necessary to add yet another answer, but this theme is close to my heart. I'm not a historian, and I would be happy if someone corrects me here, but I have the impression that the idea of understanding a differential in terms of its periods, which would go back to Riemann at least, would have been a historical antecedent to de Rham's theorem. In other words, I don't think the theorem came out of a vacuum. To explain what I mean by periods, suppose that $X$ is a compact Riemann surface of genus $g$. Then $H_1(X,\mathbb{Z})=\mathbb{Z}^{2g}$, with a basis of loops $\gamma_i$ constructed in the usual way. De Rham's theorem gives an isomorphism of the first de Rham space $H^1(X,\mathbb{C})\cong \mathbb{C}^{2g}$ by identifying a $1$-form $\alpha$ with its period vector $(\int_{\gamma_i}\alpha)$. Of course, the 19th century people would have been more interested in the case where $\alpha$ is holomorphic. In this case, the space of holomorphic forms injects into $H^1(X,\mathbb{C})$ (Proof: $\alpha=df$ implies that $f$ is holomorphic and therefore constant). This is why they could talk about this without explicitly defining cohomology first. - 2 It would make sense if this were precisely why cohomology was considered in the first place. – Mariano Suárez-Alvarez Oct 27 2010 at 16:00 This is a very good answer. – Kevin Lin Oct 27 2010 at 16:09 There is quite a number of surprising and deep statements that can be proven using de Rham. The examples I list are not elementary in any sense, but give a glimpse at the power of the theory. They all have in common that they employ features of the de Rham theory that are not at hand in singular theory. Often de Rham theory is presented as simply being the quickest way to develop cohomology theory, but in my opinion this misses the point. First of all, whether the development of the theory is really simpler than singular theory is contestable, especially if you consider that you get a considerably weaker theory as long as you if you restrict your toolkit to the Eilenberg-Steenrod axioms. Secondly, the real power of de Rham theory becomes apparent when you study specific situations where you can apply different methods than that of standard homology theory. What are these specific situations? Well, I have three examples in mind, but certainly there are much more: 1.) Connections and curvature, i.e. Chern-Weil theory. This can be motivated by the Gauss-Bonnet formula, or, better, the Gauss-Bonnet-Chern theorem, equating the Euler number of a manifold with an integral of some differential form constructed from the curvature. Already the statement that this integral is an integer is pretty intriguing if you do not know about de Rham's theorem. 2.) Symmetry! If a compact group acts on the manifold, you can restrict to invariant forms. If the action is homogeneous, you are left with a finite-dimensional complex. So symmetry can be reduced to cut down the size of the de Rham complex, leading for example to the isomorphism $H^{\ast}(G) \cong (\Lambda \mathfrak{g}^{\ast})^G$ for compact $G$, which came as a real surprise to me when I saw it first. As far as I know, this is the simplest way to the real cohomology of Lie groups. 3.) Kähler metrics! The Hodge decomposition is of course even less elementary than the previous examples, but the statement that the dimension of the space of holomorphic 1-forms on a closed Riemann surface $S$ is precisely the genus (defined as the number of handles) is rather mysterious in the first place. - Re 2): Perhaps it's worth mentioning that this connection between the cohomology of a compact connected Lie group G and the cohomology of its Lie algebra is what led Cartan to conjecture the de Rham theorem (cf. the first paragraph of Chevalley--Eilenberg). – Faisal Oct 26 2010 at 23:01 Yes, this isomorphism lies quite at the heart of the whole story! And one can also mention that Cartan and Weil combined 1) and 2) and that this ultimately lead to the whole development of rational homotopy theory (modelling spaces by commutative d.g.a's). – Johannes Ebert Oct 26 2010 at 23:18 Dear Timothy, here is a theorem which, according to your wish, "could be understood, and seen to be interesting, by someone who had not already studied the material in that course": Brouwer's celebrated fixed point theorem! It says that every continuous function from the n-dimensional closed ball $B^n \subset \mathbb R^n$ into itself has a fixed point. Please notice that I wrote "continuous" and didn't even mention the word "differentiable"! So how does De Rham solve the problem ? Step 1 Reduce to showing that there is no continuous retraction to the inclusion $S^{n-1} \to B^n$ of the boundary sphere. This reduction needs only completely elementary vector (= "analytic") geometry. Step 2 Reduce the no-retraction statement to non-contractibility of $\mathbb R^{n}\setminus O$. Again, this is easy and requires little more than the definition of contractibility Step 3 Prove the non-contractibility of $\mathbb R^{n}\setminus O$ by showing that $H^{n-1}( \mathbb R^{n}\setminus O)\simeq\mathbb R$, whereas contractible manifolds have zero de Rham cohomolgy in positive degree. This is the step where De Rham's cohomology shines in all its splendour! In the same vein you can also prove that the n-dimensional sphere $S^n$ has a tangent everywhere non-vanishing vector field if and only if $n$ is odd. An excellent source for this material is Madsen and Tornehave's extremely well-written From Calculus to Cohomology (Cambridge University Press). - This answer is quite related to Francesco Polizzi's answer. – Kevin Lin Oct 27 2010 at 16:15 Differential forms and cohomology are somewhat less intuitive than integration (at least for me), so maybe it is no easy to find such a neat example. Anyway, let's try this one. Consider the 1-form $\omega:=\frac{xdy-ydx}{x^2+y^2}$ in $X:=\mathbb{R}^2 \setminus 0$. It provides the standard example of closed form which is not exact, and in fact it is essentially the only example on $X$, because of the following Proposition. Every 1-form on $X$ which is closed but not exact is of type $a\omega + \eta$, where $a \in \mathbb{R}$ and $\eta$ is an exact 1-form. This statement makes perfect sense to everyone who understands differential forms, and at first glance it does not seem obvious at all. On the other hand, it is an immediate consequence of De Rham theorem: in fact, since $X$ retracts on $S^1$, we have $H^1_{DR}(X)=H^1_{sing}(X, \mathbb{R})=H^1_{sing}(S^1, \mathbb{R})= \mathbb{R}$, with generator $[\omega]$. - 1 This kind of example is relevant to, and was probably first studied in the context of, classical things in physics - fluid mechanics, electricity and magnetism. – Kevin Lin Oct 26 2010 at 18:00 1 To be picky this doesn't seem to be an application of de Rham's theorem. The calculation is just as easy to do directly in de Rham cohomology. – Torsten Ekedahl Oct 26 2010 at 18:51 2 @Torsten I agree with you, but I was not claiming that the computation in De Rham cohomology is particularly difficult. I just see this as a kind of "archetipal" example of De Rham philosophy: non obvious statements about differential forms become transparent facts in topology (and sometimes conversely, see Dick Palais'answer). – Francesco Polizzi Oct 26 2010 at 19:51 $\frac{1}{4\pi} \oint_{\gamma_1}\oint_{\gamma_2} \frac{\mathbf{r}_1 - \mathbf{r}_2}{\|\mathbf{r}_1 - \mathbf{r}_2\|^3} \cdot (d\mathbf{r}_1 \times d\mathbf{r}_2)$ is an integer when $\gamma_1, \gamma_2: S^1 \to \mathbb{R}^3$ are non-intersecting differentiable curves. Seriously? This number tells you how many times $\gamma_1$ winds around $\gamma_2$ (The linking number). My wife was a math and biochem major as an undergraduate interested in applying knot theory to genomics, and she and I spent countless hours trying to make sense of this without any knowledge of cohomology. Builds character I guess. - An interesting application of De Rham's theorem is to show that certain differential manifolds are not diffeomorphic. Here are two examples. 1) For $n$ even the sphere $S^n$ and real projective space $\mathbb P^n(\mathbb R)$ are not diffeomorphic since $H^n(S^n) \simeq \mathbb R$ while $H^n(\mathbb P^n(\mathbb R))=0$. Ah, you say, but I can see that with the concepts of orientation or fundamental group $\pi_1$: I don't need your Swiss's stuff! Fair enough: these are reasonable elementary alternatives. 2) Fix $N\geq 2$ and delete $k$ points from $\mathbb R ^N$: call $X_k$ the resulting manifold. Then for $k\neq l$, the manifolds $X_k$ and $X_l$ are not diffeomorphic since $dim_{\mathbb R} H^{N-1}(X_k )=k\neq l=dim_{\mathbb R} H^{N-1}(X_l )$ . However they are both orientable, and simply connected for $N\geq 3$. So the elementary tools of example 1) do not apply. - At first I always thought about the deRham theorem in terms of vector analysis and fluid dynamics. For instance, if one has a curl-free vector field, then one might want to write it as a gradient field of a function. But if your domain has holes (of a certain kind) this will not necessarily be true. The analogous statement holds true for divergence-free vector fields that you want to write as the curl of another vector field. - I don't know if this counts as "elementary", but I think the whole story connecting the topology and Morse theory of a closed oriented surface with its deRham cohomology is quite pretty. I recommend the book "Differential Topology" by Guillemin and Pollack. Or Milnor's "Topology from a Differentiable Viewpoint". - More elementary than my previous post is this "de Rham for the punctured plane in a nutshell", which shows how differential forms capture essential topological information. Put $\omega:= \frac{1}{2\pi i} z^{-1} dz$, a closed $1$-form on $C^{\times}$. Given any smooth path $c$ in the punctured plane, the integral $\int_{c} \omega$ gives a lift of $c$ to $C$ (i.e., the logarithm of $c(1)/c(0)$). If $c$ is closed, you get an integer, call it $\langle \omega, c\rangle$, which is of course the usual path-lifting from covering space theory. But you can also view it as an integration of a specific form over cycles! It is not hard to show, using integration, that $c$ is nullhomotopic iff $\langle \omega,c\rangle=0$, which gives, by the way, a computation of $\pi_1 (C^{\times})$. But now you can vary $\omega$ instead. If $d\omega \neq 0$, examples show that $\langle \omega, c\rangle$ is not homotopy-invariant, so discard that case. If $\omega =df$, show that $\rangle \omega,c\rangle =0$. If $\langle \omega,c\rangle=0$ for all $c$, a likewise elementary argument shows that $\omega=df$, and you have proven de Rhams theorem for $C^{\times}$. You can do the same computation for $S^1=R /Z$ instead, but there you are tempted to integrate only over the fundamental class, which hides essential features of that yoga. B.t.w.:Moritas book "Geometry of differential forms" contains another application cute application of the de Rham theorem: a definition of the integral Hopf invariant of a map $f:\bS^3 \to \bS^2$ in terms of differential forms (page 133). Morita also discusses Gauss' integral formula for the linking number. - One can use the de Rham theorem to define the Lebesgue integral without ever using any notion of measure theory. More precisely, the integral can be defined as the composition of the following sequence of maps: C∞cs(Dens(M))→Hncs,dR(M,Or(M))→Hncs(M,Or(M))→H0(M)→H0(∙)=R. Here C∞cs(Dens(M)) denotes the space of all smooth densities with compact support. The space C∞cs(Dens(M)) is mapped to Hncs,dR(M,Or(M)) (the nth de Rham cohomology of M with compact support twisted by the orientation sheaf of M) by the obivous map given by the definition of de Rham cohomology. The space Hncs,dR(M,Or(M)) is isomorphic to Hncs(M,Or(M)) (the nth twisted ordinary cohomology of M with compact support) by the de Rham theorem. The space Hncs(M,Or(M)) is isomorphic to H0(M) by Poincaré duality. Finally, H0(M) can be mapped to H0(∙)=R by the usual pushforward map for homology. More details are available in this answer: http://mathoverflow.net/questions/38439/integrals-from-a-non-analytic-point-of-view/38479#38479 Here is an easy application of the above definition: The easiest version of Stokes' theorem states that ∫dω=0, where ω∈Ωn-1(M,Or(M)). Proof: ∫ factors through the map to the de Rham cohomology. The form dω is a coboundary, hence its image vanishes in the de Rham cohomology and the integral equals zero. - 2 This is obviously cheating, since you need to use integration to DEFINE the de Rham isomorphism. To clear the fancy language, let us assume $M=\bR$. What you defined is a functional on the space of compactly supported smooth functions on $\bR$, which is of course the same as the Riemann integral. So you end up precisely where you started. – Johannes Ebert Oct 26 2010 at 18:27 2 @Johannes Ebert. We can get the comparison without knowing integration theory. First prove Poincaré Lemma (i.e. de Rham cohomology of '$\mathbf{R}^n$' vanishes in degree $>0$), then, using partitions of unity, compare de Rham cohomology and sheaf cohomology with coefficients in the constant sheaf $\mathbf{R}$, which, in turn, may be compared with the simplicial-like singular cohomology. None of these steps requires any integration theory of any kind; see for instance these elementary lecture notes of Shapira: people.math.jussieu.fr/~schapira/lectnotes/… – Denis-Charles Cisinski Oct 26 2010 at 23:08 2 That does not matter; all this super-fancy stuff brings you back to the point of departure. Next question: if you define integration in this way, how exactly do you integrate an n-form over a simplex?? You only learn how to integrate it against a homology class (to speak in fancy language, you probably get a pairing of simplices and forms in some $E_{\infty}$-sense). Having the Riemann integral as a linear functional can be done without simplices, sheaves, homological algebra etc. This linear functional is where the real work begins. – Johannes Ebert Oct 26 2010 at 23:44 2 And you did not gain a single piece of insight that helps you with that real work. Of course you can define $L^1 (M)$ formally as a completion (not totally unreasonable, but there are some flaws, see the discussion in Langs ''Real and functional analysis'') and try to prove Fubini and the transformation formula with topology. I am pretty sure that the arguments quickly turn into a monstrous mess, if you try to prove e.g. regularity theory of elliptic operators with that notion of integral. – Johannes Ebert Oct 27 2010 at 0:01 2 @Dmitri, Denis: This is what I meant with "super-fancy language": you read "Koszul-complex", "coregular" and do not notice that the fundamental theorem of calculus is the real meat of the proof. – Johannes Ebert Oct 27 2010 at 17:05 show 33 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 94, "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": 0.941144585609436, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-equations/125801-eulers-method-question.html	# Thread: 1. ## Eulers Method Question We are covering Eulers method and I am unsure about one thing. After calculating a problem that states use Euler’s Method with a step size of h = 0.1 to find approximate values of the solution at t = 0.1, 0.2, 0.3, 0.4, and 0.5, I am asked to now calculate using a step size of h = .05. I still use the same t values as before correct? 2. Originally Posted by fishguts We are covering Eulers method and I am unsure about one thing. After calculating a problem that states use Euler’s Method with a step size of h = 0.1 to find approximate values of the solution at t = 0.1, 0.2, 0.3, 0.4, and 0.5, I am asked to now calculate using a step size of h = .05. I still use the same t values as before correct? You will have to find the approximate solution values at $t=0.05, 0.1, .., 0.45, 0.5.$ The time values you use are all of the form $t_0+n\times h$ where $n$ is a natural number and $t_0$ is start time. CB	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 0.885766863822937, "perplexity_flag": "head"}
http://mathoverflow.net/questions/61009/compact-elements-and-continuous-functors/61013	## compact elements and continuous functors ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi, I am interested in abstracting the Scott topology from Domains to Categories. One can find a definition of a continuous functor which is just such an abstraction: A functor F:C→D is continuous if it preserves all small (weighted) limits that exist in C, i.e. if for every small category diagram A:E→C in C there is an isomorphism F(limA)≃lim(F∘A).F(\lim A) \simeq \lim (F\circ A). I found this on the n-Lab. What I am interested in is a notion of compact element which is defined for Domains as this: if k is less than the sup of any directed subset D, then there is an element x in D such that k is less than x. Any reference would be good. My intuitions are telling me that if a category is compact, in this sense, it should have a kind of finite presentation. This would be an abstraction from a dcpo of groups where the compact elements are finitely generated. I realize this might not make sense so any help would be great. - 2 The equivalent of compact elements could be the following: an object $X$ is called finitely presentable when $\mathrm{Hom}(X,-)$ preserves directed colimits. The equivalent of a Scott domain would then be a locally finitely presentable category: one in which every object is a colimit of finitely presentable ones. The nLab page doesn't have many pointers, but Adamek and Rosicky have a nice book about the topic (albeit not oriented towards domains). – Chris Heunen Apr 8 2011 at 1:35 Hi Chris -- I was in the middle of writing my answer and didn't see your comment, which contains much of what I said. – Todd Trimble Apr 8 2011 at 1:41 Thanks very much. I might be able to start working on this a bit more now. – Ben Sprott Apr 8 2011 at 7:50 Hey, It looks like I am interested in yet a higher level of abstraction. When I asked about finitely present categories, I was being specific. I mean that categories themselves could be the elements of the domain. I apologize for the roughness with which I am speaking. I think the intuition I am working on is borrowed from the fact that the set of endomaps of a domain is also a domain. This is, itself, a path of abstraction. If the categories are the elements of the domain, then the compact objects which Todd talks aobut can be abstracted up into compact categories. – Ben Sprott Apr 11 2011 at 5:57 I also think that the n-categorical framework is at work here too. – Ben Sprott Apr 11 2011 at 5:58 ## 2 Answers Your intuitions look good. The analogue of compact element for general categories is the notion of finitely presentable object: • An object $c$ of a category $C$ is finitely presentable if the hom-functor $\hom_C(c, -): C \to Set$ preserves directed colimits (colimits of diagrams $D \to C$ where $D$ is a directed poset). These coincide with finitely presentable algebras when $C$ is a category of algebras of a finitary algebraic theory. This is part of the very beautiful theory of locally finitely presentable categories (and the more general locally presentable categories); an excellent reference for this is the book by Adamek and Rosicky, Locally Presentable and Accessible Categories. "Locally finitely presentable" means a category which is cocomplete such that every object is a directed colimit of finitely presentable objects (for example, every group can be presented as a directed colimit of finitely presentable groups). Locally finitely presentable posets are the same as algebraic lattices. Edit: Finn is right that finitely presentable objects are also called "compact objects". - ### 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. Have you tried nLab again? I'm not sure continuous functors are what you're looking for, incidentally. It seems more likely that a 'Scott-continuous functor' should be one that preserves filtered colimits. Replacing the preorders (and metric spaces) of domain theory with (enriched) categories is not a new idea. Have a look at Categories for fixpoint semantics (1978) by Daniel Lehmann, and Solving recursive domain equations with enriched categories (1994) by Kim Ritter Wagner. - Functors that preserve filtered colimits are often called "finitary". – Mike Shulman Apr 8 2011 at 6:27 Hey, It looks like I am interested in yet a higher level of abstraction. When I asked about finitely present categories, I was being specific. I mean that categories themselves could be the elements of the domain. I apologize for the roughness with which I am speaking. I think the intuition I am working on is borrowed from the fact that the set of endomaps of a domain is also a domain. This is, itself, a path of abstraction. If the categories are the elements of the domain, then the compact objects which Todd talks aobut can be abstracted up into compact categories. – Ben Sprott Apr 11 2011 at 5:56	{"extraction_info": {"found_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": 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": 0.9484431743621826, "perplexity_flag": "head"}
http://mathoverflow.net/questions/97888/generalization-of-gausss-inequality-for-not-necessarily-unimodal-distributions	## Generalization of Gauss’s inequality for not necessarily unimodal distributions? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Gauss's inequality is for unimodal distributions, concerning distance from the mode. A similar result is Vysochanskiï–Petunin inequality, which is for the distance from the mean rather than the mode. Chebyshev's inequality generalizes Vysochanskiï–Petunin inequality by concerning distance from the mean without requiring unimodality. I wonder if there is generalization of Gauss's inequality for distributions not necessarily unimodal? Thanks! - ## 1 Answer Chebyshev's inequality is crude: The probability of being more than $k$ standard deviations away from the mean is at most $1/k^2$. A similar coarse estimate applies to the distance from the mode or any other point. Let $r_x^2$ be the average of the square of the distance from $x$, which can be computed as $(x-\mu)^2 + \sigma^2$. Then the probability of being more than $kr_x$ away from $x$ is at most $1/k^2$. This is true for all $x$, including a mode. The same works if we let $x$ denote a set, including a set of modes, although computing $r_x^2$ becomes harder. - Thanks! How is $r_x^2$ defined when x is a set? – Tim May 25 at 1:01 The distance between the point $y$ and the set $X$ is $d(y,X) = \inf_{x \in X} d(y,x)$. Then $r_X^2$ can be defined as the average of the square of the distance to $X$. This is no longer a quadratic polynomial so it can't be computed as easily as $(x-\mu)^2 + \sigma^2$, but the inequality still holds. – Douglas Zare May 25 at 1:30	{"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": 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": 0.3, "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": 0.9401414394378662, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/81863/uniform-convergence-of-series-of-functions	# uniform convergence of series of functions Prove that $f(x)=\displaystyle{\sum_{n=1}^{\infty}} \dfrac{e^x \sin (n^2x)}{n^2}$ is convergent for every $x \in \mathbb{R}$ and that its sum $f(x)$ is a continuous function on $\mathbb{R}$. This is my tentative to solve the problem: $f(x)= e^x \sum {\frac{\sin (n^2x)}{n^2}}$. So, to prove that $f(x)$ is convergent, I only need to prove that $\sum {\frac{\sin(n^2x)}{n^2}}$ is convergent. since absolute value of $\frac {\sin (n^2x)}{n^2} \le \frac{1}{n^2}$ because absolute value of $\sin (n^2x)$ is $\le 1$ and the series: $\sum {\frac{1}{n^2}}$ is convergent, then by the $M$-test the series $\sum {\frac{\sin (n^2x)}{n^2}}$ is uniformly convergent and thus $f(x)$ is continuous on $\mathbb{R}$. Please let me know whether my solution is true? Also, do I have to distinguish the two cases where $x=0$ and $x \ne 0$? Do I have to prove that $f$ is continuous at $x=0$ separately? - @Srivatsan I edited LaTeX only. I think, you should also notice the spelling "beccause". – gaurav Nov 14 '11 at 3:18 ## 1 Answer You're very close. Minor notes: • No, you don't have to consider $x=0$ as a special case, because $\|\sin(0)\|\leq 1$ is still true. • $f(x)= e^x \sum(\sin(n^2x)/n^2)$ is true, but note now that the series is different. You gave an argument for why the series here converges uniformly, but note that this does not imply that the initial series defining $f(x)$ converges uniformly (and it does not). • Related to the last point, you technically never addressed why the original series converges for every $x$. This follows from the absolute convergence test, or from the work you already did and the fact that if $\sum a_n$ converges and $b\in \mathbb R$, then $\sum ba_n$ converges to $b\sum a_n$. (In other words, you could just make a little more explicit some of the work you already did. How much detail to include is largely a matter of taste.) • Other details that may be mentioned to finish this off (perhaps intentionally omitted in your question because they are straightforward): • For each $n$, $x\mapsto \sin(n^2x)/n^2$ is continuous, which is why you can conclude that the uniformly convergent series $\sum\sin(n^2 x)/n^2$ is continuous. • $x\mapsto e^x$ is continuous. • The product of 2 continuous functions is continuous. - 1 +1. "this does not imply that the initial series defining $f(x)$ converges uniformly (and it does not)" - So this gives an example where a sequence of continuous functions converges to a continuous limit, but the convergence is not uniform. In fact, this might even the hidden context of the question. – Srivatsan Nov 14 '11 at 3:15 2 Srivatsan makes a good point. Here at least there is uniform convergence on compact subsets which is also generally strong enough to conclude that the limit is continuous (and fortunately as the OP found there is a way to avoid even worrying about that). For examples where even uniform convergence on compact subsets fails but the limit is still continuous, take a sequence $f_n:[0,1]\to \mathbb R$ such that $f_n$ increases linearly from $0$ to $1$ on the interval $[0,1/n]$, decreases linearly from $1$ to $0$ on the intreval $[1/n,2/n]$, and is $0$ on $[2/n,1]$. Then $f_n\to 0$ nonuniformly. – Jonas Meyer Nov 14 '11 at 3:24 Ok. great. So this proves that the function f(x) is continuous because it is the product of two continuous functions. But, How does it follow that f(x) is convergent? and what type of convergence f is following? Is it pointwise convergence? Can anyone help please? – M.Krov Nov 14 '11 at 17:25 @Zi2018Alpha: I added a new bullet on convergence of the original series. The problem says "is convergent for every $x$," so yes it means pointwise convergence. That is, for each (fixed) $x$, the series $\displaystyle{\sum_{n=1}^{\infty}} \dfrac{e^x \sin (n^2x)}{n^2}$ converges. The absolute convergence test is one way to show this. It is also true in general that if $\sum a_n$ converges and $b\in\mathbb R$, then $\sum ba_n$ converges to $b\sum a_n$, so you don't need to directly analyze the original series (at least if you can justify my claim). – Jonas Meyer Nov 14 '11 at 18:17	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 51, "mathjax_display_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": 0.9543540477752686, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/20789/electric-field-at-a-point-inside-a-capacitor	# Electric field at a point inside a capacitor Let's say I have a parallel plate capacitor. How would I find the electric field at a certain point INSIDE the capacitor (inside the dielectric let's say). From what I understand, the flux of the electric field will be constant everywhere (even if there is more than 1 different dielectric), but the electric field varies. Is this correct? - ## 3 Answers (Im assuming effectively infinite capacitors ($\sqrt{A}>>d$), otherwise icky fringe fields come into play) Calculating the net field inside an infinite dielectric sheet is simple. If the field due to the capacitor plates is E, the net field inside the dielectric will be $\epsilon_rE$. This means that the dielectric itself has an induced field of $E(1-\epsilon_r)$ in the opposite direction (this field only exists inside the dielectric). This also applies to a multiple dielectric system. Inside each dielectric sheet, there will be a similar induced field, and we get the same value of net electric field (where $\epsilon_r$ is the dielectric constant of the sheet in question) So what's actually happening here? A dielectric contains charges that are pretty much bound in place. When you apply a field E, these charges excercise whatever little freedom they have to move, and they form tiny dipoles (I think the electron cloud of the atom/molecule polarizes, but i'm not sure). These tiny dipoles will be randomly scattered, but will all be aligned with the field. We can vector add their dipole moments to make a column of large dipoles. Since dipoles are after all pairs of charges, we can look at this as two sheets of opposite charge embedded on either face of the dielectric. These sheets create no field outside, but they create a net induced field inside (Just how a capacitor creates field only between its plates, as $E=\frac{q}{2A\epsilon_0}$ for a single plate; both Es cancel outside the capacitor) Now, when talking about the flux, we have a bit of an issue. If we are considering a perfectly infinite capacitor, the flux is infinite, so we cant discuss it. If we consider an effectively infinite capacitor, the flux should be constant (no net sources/sinks of of flux.). But, since the field is changing, we have a bit of confusion. This is due to boundary conditions. Actually, the field is diverging a bit in a lemon-like way when it reaches the dipole. Near the center, this divergence is small, but it gets large when go near the top/bottom of the capacitor. So, the electric field has reduced in value, but it has spread out as well, so net flux is constant. Draw a little diagram of the field lines, and you'll understand this clearly. Find some diagram of how field lines bend through a dielectric for reference if you want, - If you have a single dielectric then you can simply preform Gauss's law at the plate. This will entail imagining a Gaussian pill-box. Say the plates have a charge $Q$ and $-Q$, then if we draw the pill-box around the plate with $Q$ and call the down direction positive $\hat{z}$ we have $$\frac{Q}{\epsilon}=\int \vec{E}\cdot d\vec{a} =EA\implies \vec{E}=\frac{Q}{\epsilon A}\hat{z}$$ where $\epsilon$ is the permittivity of the material. You can use the Electric Displacement field $\vec{D}$ to get the same answer quicker with the boundary conditions that $(\vec{D}_2 -\vec{D}_2)\cdot\vec{n}=\sigma_{\text{free}}$, then from that you can use the relation $\vec{D}=\epsilon \vec{E}$ to get $\vec{E}$ also. If there are multiple dielectrics and you want to find the field in the middle of the capacitor, say at the boundary between two dielectrics, you will have to use the boundary conditions on dielectric media $$(\vec{D}_2 -\vec{D}_2)\cdot\vec{n}=\sigma_{\text{free}}$$ $$(\vec{E}_2 - \vec{E}_1 )\times \vec{n}=0$$ along with the fact that the potential is a solution to Laplace's equation. This should be enough for the majority of problems one comes across. I hope this helps, - If the capacitor is of the flat plate variety with area A and separation l, the potential increases linearly from one plate to the other. From the potential you can get the electric field by taking the gradient. Note that the dielectric means that there's a difference between E and D but this should be discussed in your textbook. -	{"extraction_info": {"found_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": 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": 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": 0.9276736974716187, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/29344/list	## Return to Answer 2 accent I've never heard of $K_2(R)$ having a description as a more easily/elementarily described object attached to $R$. Borel showed that $K_2(R)$ is torsion. The order of $K_2(R)$ appears in "Theorem" 31 of Soule's Soulé's notes on higher algebraic K-theory of rings of algebraic integers where $K_2(R)$ (I think that "Theorem" 31 is called the Quillen-Lichtenbaum conjecture). 1 I've never heard of $K_2(R)$ having a description as a more easily/elementarily described object attached to $R$. Borel showed that $K_2(R)$ is torsion. The order of $K_2(R)$ appears in "Theorem" 31 of Soule's notes on higher algebraic K-theory of rings of algebraic integers where $K_2(R)$ (I think that "Theorem" 31 is called the Quillen-Lichtenbaum conjecture).	{"extraction_info": {"found_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": 0.9670246839523315, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/32071/is-the-dual-notion-of-a-presheaf-useful/32162	## Is the dual notion of a presheaf useful? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It seems that there is a common theme in mathematics where, if we want to find out about a category C, then we look at $\hat{C}$ (the category of contravariant functors from $C$ to $Set$). There are all sorts of good reasons for this (Yoneda's lemma being a big one, and the fact that this is a topos). There are other versions, sometimes you look at contravariant functors into some other category, like groups (e.g. for algebraic groups). But I have never seen the dual notion: covariant functors from some other category INTO $C$. Is this notion as useful as the notion of a presheaf? I guess it's like looking at a category "over" $C$ as opposed to a category "under" it. I guess, hidden in this question is the question: Can one learn anything about a category $C$ by looking at presheaves of $C$? For example, does the difference between presheaves of sets and presheaves of ableian groups tell us anything about the differences between the category of Groups and the category of Sets? EDIT: There seems to be a bit of confusion of what I mean by "dual of a presheaf." I don't mean "copresheaves" (which I didn't know existed before I asked this), I mean what you get by reversing the arrow of the functor not taking the opposite category. So I'm looking at functors into the category of interest, as opposed to out of them. I can see how this is confusing because usually "dual" doesn't turn around functors, just arrows inside categories. So I guess I mean presheaves in $C^{op}$... (but covariant instead of contravariant?)... who knows. - 8 Sure it is: functors into C are diagrams in C. See en.wikipedia.org/wiki/Diagram_(category_theory) . – Qiaochu Yuan Jul 15 2010 at 22:31 1 (In particular, one can distinguish Ab and Set by differences in how limits and colimits of their diagrams behave. For example, the (finite) coproduct and product are identical in Ab, but different in Set. Similarly, every element of Set is a coproduct of copies of the terminal object, but in Ab the terminal and initial objects agree.) – Qiaochu Yuan Jul 15 2010 at 22:35 1 That's a fair answer, but diagrams are kind of a limited interpretation of a functor into C, don't you think? It would seem that this notion is only manageable when the category mapping into C is quite small... – Dylan Wilson Jul 15 2010 at 22:47 3 When you form the nerve of a category $\mathcal C$, the n-simplices are contravariant functors from an interval into $\mathcal C$. – Josh Shadlen Jul 15 2010 at 22:59 17 There was a quote on Tao's blog about how morphisms out of an object give you global information whereas morphisms into an object tend to give more local information. One (trivial) way of seeing this is that morphisms out of the category necessarily involve all objects, but morphisms in might only involve some small subcollection. So if you want to understand the category as a whole, morphisms out might be the way to go. – Kevin Ventullo Jul 16 2010 at 8:58 show 6 more comments ## 6 Answers Let me try and take a stab at this question. I will give not so much as an answer to your question, but more of a rambling collection of remarks. The post turned out to be much longer than I wanted, because as B. Pascal once complained, I did not have the time to write a shorter one. Expect some inconsistencies, a lot of hand-waving, etc. To put some order in my thoughts, I will try to argue the following four points. Fix a category $\mathcal{A}$, which is the category "of interest". Then: (1) Functors into $\mathcal{A}$ are interesting. (2) Functors into $\mathcal{A}$ tell you everything about $\mathcal{A}$. (3) Part of the perceived asymmetry follows from the fact that in many cases the interesting category $\mathcal{A}$ is equivalent to a presheaf category (or some localization thereof). (4) There is not really an asymmetry between functors out of and functors into $\mathcal{A}$, but more of a duality. The first point is easy to settle, as Qiaochu Yuan already gave a class of interesting examples: for a category of interest $\mathcal{A}$ and any category $\mathcal{I}$, a functor $\mathcal{I}\to \mathcal{A}$ is a diagram of shape $\mathcal{I}$ in $\mathcal{A}$, so hell yeah, functors into $\mathcal{A}$ are important. You may think this a rather pedestrian example, but below I give more examples. To explain (2), instead of working with categories, that is, in the $2$-category of categories, let us go one level down to the category of sets. A set $X$ is determined completely by its elements, that is, maps $\ast\to X$ where $\ast$ is a singleton set (any one will do: for the sake of determinacy, take the singleton set comprised of the element $\emptyset$). Now in a general category, we cannot speak of elements, but we can (and do) speak of generalized elements. Definition: Let $\mathcal{A}$ be a category and $a$ an object of $\mathcal{A}$. A generalized element of $a$ is a map $x\colon d\to a$. This is also written symbolically as $x\in_{d} a$. We do not need generalized elements to develop category theory, but personally, I found them useful to build upon the intuition gained from working with more "concrete" categories. A very nice discussion of generalized elements is in Awodey's book on category theory. Now the kicker: Yoneda's lemma tells us that if we know all the generalized elements of an object $a$ then we know everything about $a$, including of course, the arrows out of $a$ (note: and by duality, if we know all the arrows out of $a$ we will know everything about $a$). Two possible objections may be raised: 1. The original example involves a $2$-category: does not make much of a difference, as my reply to Kevin's comment (see above) still applies. And besides, there are $2$-categorial versions of Yoneda lemma. 2. Yoneda's lemma works ok, but you need the knowledge of all generalized elements and these range over a potentially proper class of objects: that is true, but in virtually every interesting category one can trim down this proper class to a (small) set, even a singleton set. I will not spell out the proper definitions; they are intimately tied with "smallness" conditions and the adjoint functor theorems (and yes, the category of (small) categories satisfies them). For (3), let $\mathcal{C}^{\mathcal{B}}$ be a functor category. For size reasons, $\mathcal{B}$ has to be small (note: if you have no scientifick problems with creation ex nihilo, you can always spawn a larger universe by invoking the axiom of universes and sidestep this particular size issue). Since functor categories inherit most of the good properties of the codomain category, you will want $\mathcal{C}$ to be as good behaved as possible (e.g. complete, cocomplete, abelian, symmetric monoidal closed, etc.). It is not true that the structure of $\mathcal{B}$ is irrelevant for the structure of $\mathcal{C}^{\mathcal{B}}$, but it is true that in general, $\mathcal{B}$ only needs a bare minimum of structure. This asymmetry between the domain and the codomain categories is reinforced by the fact that many of the interesting categories $\mathcal{A}$ are equivalent to functor categories, even presheaf categories (or some localization of them). Here are two examples. 1. Let us consider the category of groups $\operatorname{Grp}$, undoubtebly a category "of interest". There is a category $\operatorname{Th}(\operatorname{Grp})$ that has all finite products such that $\operatorname{Grp}$ is equivalent to the category of product-preserving functors $\operatorname{Th}(\operatorname{Grp})\to \operatorname{Set}$ where $\operatorname{Set}$ is the category of sets. This equivalence can be generalized to a very large class of categories of "algebraic flavor" and even some that at first sight do not bear the least resemblance to "algebraic categories". A few extra remarks about this example. First, the category $\operatorname{Th}(\operatorname{Grp})$ is a category constructed to make the equivalence work (it is the free category with products on a group object), in other words, it does not exactly fall within the class of categories "of interest". It's a similar to the example of diagram categories in (1), where the domain, a free category on a graph, is just a categorial construction to make the identification of diagrams with functors, not an interesting category by itself. Second, by replacing $\operatorname{Set}$ by another category $\mathcal{B}$ (with at least finite products), you can now speak of groups in $\mathcal{B}$. This gives another class of examples where functors into a category are interesting. 2. If $(\mathcal{C}, \mathcal{J})$ is a site (a category with a Grothendieck topology), by first taking the category of presheaves and then a suitable localization, one obtains the category of sheaves. This produces a host of geometric categories, like manifolds and schemes. Much like in example 1, the interest is not so much on $(\mathcal{C}, \mathcal{J})$ and even less in the codomain, which is usually the category of sets (other categories for codomain also work, but the categorial requirements for everything to work smoothly are fairly strong), but in the (pre)sheaf category. For my last point (4), a lot could be said, but I will just point you to two articles by F. W. Lawvere in the TAC Reprints, "Metric Spaces, Generalized Logic and Closed Categories" and "Taking categories seriously" (google for them, they are available online). In them, Lawvere makes several remarks about the duality between spaces and algebras of functions which are directly relevant to your question. To show that there is not so much an asymmetry but a duality between functors into and out of, let me give you two examples. 1. In your (that is, the OP) last post, you speak about stacks. A stack on a category $\mathcal{A}$ can be defined as a functor with values in $\mathcal{A}$ satisfying some conditions -- this is the fibered category approach to stacks. But a stack can also be defined as weak $2$-functor with values in the $2$-category of categories (satisfying some extra conditions). For reasons that I will not explain, the first approach is better, but nevertheless the point should be clear. There are actually many examples of this "duality", that identifies some category of functors into $\mathcal{A}$ with some category of functors out of $\mathcal{A}$. 2. Let me end up with an example from physics that further illustrates this duality. Quantum field theories are notoriously hard objects to define (let alone study). Several years ago, V. Turaev defined the notion of a Homotopy Quantum Field Theory, HQFT for short (check his papers in the arxiv if you are interested) which is a very simple, "toy" example of a QFT. If $X$ is a topological space, we can define a category that has for objects manifolds $M$ equipped with a homotopy class of maps $g\colon M\to X$ and a morphism $(M, g)\to (N, h)$ is a cobordism $W\colon M\to N$ with a homotppy class of maps into $X$ extending $g$ and $h$. An HQFT is a monoidal functor from this category into another monoidal category (usually, the category of finite-dimensional complex linear spaces). I am omitting lots of details, but the gist is that an HQFT gives us invariants of manifolds $M$ by mapping $M$ into some fixed background space $X$. But we can turn things around, for the category of $X$-HQFT's is a (functorial) invariant of the homotopy type of $X$, an invariant cooked up by mapping manifolds into $X$. Hope it helps, regards, G. Rodrigues - This was an even better answer than I expected possible! I especially love the last two examples (the first answer makes sense out of what i attempted to say in my answer, and the second is exactly the kind of "application" of the concept that I hoped existed!). Thank you! – Dylan Wilson Jul 18 2010 at 15:56 ### 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. Covariant functors from the category of pointed sets to the category of pointed topological spaces are sometimes called $\Gamma$-spaces, and they have been important in algebraic topology. One reason is that $\Gamma$-spaces model infinite loop spaces (and therefore connective spectra) and are very helpful for understanding stable homotopy theory. $\Gamma$-spaces also serve as a model for particularly well-behaved covariant functors from the category of pointed topological spaces to itself. Of course, these functors play an important role in topology as well. I like to think of Goodwillie's Calculus of Homotopy Functors (and also of Michael Weiss's Orthogonal Calculus) as a kind of "sheaf theory for covariant functors". In these theories, covariant functors are analogous to presheaves and linear functors are analogous to sheaves (The definition of a linear functor is essentially a homotopy-invariant version of the definition of a sheaf). The process of approximating a general functor by a linear one is analogous to sheafification, and so forth. These theories provide methods for studying certain types of functors, but of course they also tell you something about the category of spaces itself. - You can define a category whose objects are all functors $X\colon I \to \mathcal{C}$, where $I$ is any small category, and whose morphisms are given in the same way as for ind-categories: $$Hom((I,X),(J,Y)) = lim_{i \in I} colim_{J \in J} Hom(X(i),Y(j)).$$ (If you restrict attention to filtered categories $I$, this is precisely the category $Ind-\mathcal{C}$.) There is a fully faithful functor $F$ from this category, let's call it $Cat-\mathcal{C}$, to $\hat{\mathcal{C}}$: $$(I,X) \mapsto (c \mapsto colim_{i \in I} Hom(c,X(i))).$$ This functor is an equivalence of categories if one gets some set-theoretic problems out of the way. For example, if $\mathcal{C}$ is small, then any contravariant functor from $\mathcal{C}$ to sets is a colimit of representable functors, which shows that $F$ is essentially surjective. So, basically, this is the same construction as the Yoneda embedding! - With no disrespect to my fellows answerers, the examples so far are a bit exotic. If $X$ is a space/manifold, the collection of continuous/$C^\infty$ functions on open subsets with compact support forms a copresheaf (the maps are extensions by $0$). For an even simpler example, start with a presheaf of modules $M$, then the dual $U\mapsto M(U)^$ will be a copresheaf. You can replace `$()^$` with your favourite contravariant functor. Also, I seem to recall that Bredon thought the notion was useful enough to include a discussion of them in his book on sheaf theory. That said, I have no very good explanation for why (pre)sheaves are more common, and seemingly more useful, than the dual notion. I suppose one explanation, is that in some cases by dualizing you can get rid of the "co"-ness without loss of information. I've used this trick myself. - 1 I may be parsing the answers wrong, but I believe you are making the same mistake as some other posters. Precosheaves are covariant functors out of the category of interest, while the OP is interested in functors into the category of interest. For reference, yes (pre)cosheaves are interesting, as they appear in Borel-Moore cohomology (e.g. Bredon's book), as distributions in topoi (in the sense of Lawvere), etc. – G. Rodrigues Jul 16 2010 at 14:08 Yes, you're absolutely right! Although I'll leave it up, I retract my answer. I made the mistake of typing it over morning coffee before the effect was evident.... – Donu Arapura Jul 16 2010 at 14:17 Yes, Rodrigues is correct- I should make that more clear. But thank you for your otherwise interesting answer! – Dylan Wilson Jul 16 2010 at 18:30 1 Quantum fields exotic?! They're all around you! – José Figueroa-O'Farrill Jul 17 2010 at 0:03 1 Of course. Unfortunately they're exotic to me, since my physics education ended a bit too early. But if I keep reading things like your post on particles & group reps., then who knows? – Donu Arapura Jul 17 2010 at 0:46 show 1 more comment Edit I seem to have misunderstood the nature of the duality in the question. This answer is not relevant. I'll keep it in case it has any archaeological interest. In local quantum field theory, one does use the notion of a co-presheaf (or is it a pre-cosheaf?) of operator algebras. You can read about it (perhaps not with those words) in Streater and Wightman's PCT, spin-statistics, and all that and perhaps also in the links from the above wikipedia page. This is perhaps not surprising as it is the states of the system which form a presheaf. - Perhaps I can ask a naive question with no disrespect intended: In regard to the humorous exchange with Donu, isn't it the case that the cosheaf point of view would be regarded as somewhat exotic by many quantum field theorists? – Minhyong Kim Jul 18 2010 at 8:44 1 Absolutely! The word "local" in front of "quantum field theory" gives it away. (I could have used "algebraic", "axiomatic",... as synonymous.) The preface to Streater and Wightman has a funny quote about this. "The physicists who have engaged in this kind of work are sometimes dubbed the Feldverein. Cynical observers have compared them to the Shakers, a religious sect of New England who built solid barns and led celibate lives, a non-scientific equivalent of proving rigorous theorems and calculating no cross sections." :-) – José Figueroa-O'Farrill Jul 18 2010 at 16:32 Sort of an answer to my own question that I just stumbled across: A "stack over a category $S$" is a functor from some other category into S that satisfies various properties. If these properties are similar or dual to the ones that a presheaf must satisfy in order to be a sheaf... then I would say that the notion I was talking about in my question was that of a "pre-stack," and that the answer is "Yes, they are useful... they give rise to the notion of stacks!" EDIT: Considering my original question simply asks if the notion of functors into a category being useful, it doesn't really matter if the property of a stack that separates it from any old functor into $S$ is related to the property of a sheaf (that separates it from any old functor out of $S$)... What matters, I think, is that stacks are a type of functor INTO $S$ as opposed to out of it, and they are, indeed, useful. But it would be nice if someone could clean this answer by erasing this sentence and replacing it with some "big-picture" description of what (if anything) a stack over $S$ tells us about $S$... in relation to, say, a sheaf on $S$. Am I still incomprehensible? It is rather late... Of course, I may be spouting nonsense... so I'm making this answer a community wiki in the hopes that someone who knows something about stacks can verify or falsify what i've written, and hopefully expand. (Meta-note: I put this as an answer because I wasn't really sure what to do... I don't think it properly belongs as an edit to the original question... it's basically a different question "are stacks basically what I was thinking of?" but it was sufficiently related and similar to my original question that I don't think it merits a whole NEW question... so I put it here as community wiki. If this is not the correct thing to do, then I trust a moderator or someone much wiser than me will take down this post!) - I don't follow your second sentence, and am indeed not sure that what you've written is helpful as (the seed of) an answer to your original question. – Yemon Choi Jul 18 2010 at 7:24 Well... I tried to clear a few things up. But if this is still nonsense in, say, a day or two... by all means erase this answer! – Dylan Wilson Jul 18 2010 at 8:03 1 In my minuscule understanding, you can't take a stack over any old category, whereas you can take a presheaf on pretty much anything. I think you are veering close to "stacks are cool, maybe they're relevant!" which does not strike me as a good way to proceed. (I'm assuming you've already looked on wiki?) – Yemon Choi Jul 18 2010 at 9:37 I have equally miniscule understanding... but it seems to me that you can take a "pre-stack" over anything? And then turn it into a stack if it's over a site? And I really have no idea if stacks are cool, since I know so little about them. But you're right in that I'm pretty much asking "are stacks relevant to what I was asking?" as sort of a sub-question. Meanwhile, I should read the post below this because it looks like Rodrigues mentions stacks at the end. – Dylan Wilson Jul 18 2010 at 15:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 103, "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": 0.9504696130752563, "perplexity_flag": "head"}
http://mathematica.stackexchange.com/questions/17391/coding-mistake	# Coding mistake? [closed] I have just started using Mathematica with v9.0. I am trying to follow a computation from a book on Fourier series with the function $f(x)=x$ on the interval $-\pi < x < \pi$. Here is the code I have tried: f[x_] = x; a[0] = 1/2 Pi Integrate[f[x], {x, -Pi, Pi}] a[n] = 1/Pi Integrate[f[x] Cos (nx), {x, -Pi, Pi}] b[n] = 1/Pi Integrate[f[x] Sin (nx), {x, -Pi, Pi}] I got the same result for a[0] and a[n] as per the book; for b[n] the answer generate by Mathematica is 0, but the answer in the book is $\frac{2}{n}(-1)^{n+1}$. Can someone explain why is this the case? Note: I found a post here Fourier Coefficients in Mathematica but I can't seem to follow it. - 4 Sin(nx) is the product (or multiplication) of the symbol Sin with the symbol nx. You should be using Sin[n x]. – jVincent Jan 7 at 19:05 1 Cos(nx) will be interpreted as Cosnx try: Cos[n x] (note the space beteen n and x, otherwise it'll be interpreted as variable called nx) – ssch Jan 7 at 19:05 2 @Sandra For integer $n$, you have $\sin(n\pi)=0$ and $\cos(n\pi) = (-1)^{n}$. Include the $-$ sign too and the exponent becomes $n+1$. You can do this with Mathematica by telling it that $n\in\mathbb{Z}$ as Simplify[(-2 n Pi Cos[n Pi] + 2 Sin[n Pi])/(n^2 Pi), Assumptions :> {n ∈ Integers}] – rm -rf♦ Jan 7 at 19:26 1 Your formula for a[0] should be 1/(2 Pi) Integrate[f[x], {x, -Pi, Pi}]. (1/2 Pi is the same as Pi/2 -- an unlikely coefficient for the integral in this context.) – Michael E2 Jan 7 at 19:44 1 Sandra, welcome to Mathematica.SE, and thank you for taking the time to format your question so nicely! – Simon Woods Jan 7 at 20:27 show 4 more comments ## closed as too localized by Artes, rcollyer, Oleksandr R., whuber, Sjoerd C. de VriesJan 8 at 21:24 This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ. ## 1 Answer I appreciate very much, that you wrote up such a nicely formatted question, although this is your first post. Therefore, let's put the comments into an answer. Your first issue was that you used ( ) where you should have used [ ]. That's maybe not obvious for starters and I have seen this mistake very often. There are different types of braces and it's important to always use the right ones. Here is an overview 1. [ ] are always used when you want to call a function like Sin[x]. 2. ( ) are used change the order of evaluation like you would do in normal mathematics too. Therefore, you would use a(b+c) to add b and c before doing the multiplication. 3. { } are to construct lists, matrices and other tensors. {1,2,3} is a list or vector and yes, {x,-Pi,-Pi} is a list too. 4. [[ ]] is used to take parts of lists and vectors. {x,-Pi,Pi}[[3]] would give you the Pi. See the documentation to Part to learn what else is possible. Your second issue was that you assumed nx is the multiplication of n and x which it is not. You have to understand that when you use var it would be impossible to decide for Mathematica whether you mean the variable name var or the multiplication var. Therefore, always put explicit multiplication nx which is the most clear way or leave a space between them. A third option which is not recommended in this situation is to add parenthesis. n(x) is interpreted as nx too. Looks odd here but this notation is very common in expressions like (x + 1)(x + 2). Your third issue that the expression is the hardest one because in general it is not obvious and sometimes even not possible to force Mathematica to simplify an expression into a form you prefer. Additional to the solution to Hypnotoad using Simplify[(-2 n Pi Cos[n Pi] + 2 Sin[n Pi])/(n^2 Pi), Assumptions :> {n \[Element] Integers}] you can use Refine which gives the form of expr that would be obtained if symbols in it were replaced by explicit numerical expressions satisfying the assumptions assum. Refine[(-2 n Pi Cos[n Pi] + 2 Sin[n Pi])/(n^2 Pi), n \[Element] Integers] Your fourth issue is about the right hand side of your definition of a[n] and b[n]. When your goal is to define a general coefficient where you can change the value of n to calculate for instance b[5], then those lines will not work. Try yourself what happens if you use b[4]. To make it work, you have to use b[n_] = 1/Pi Integrate[f[x] Sin [n*x], {x, -Pi, Pi}] Note the underscore! Now it's possible to calculate b[4]. When you want to know why this is, you should start reading the tutorial about defining functions and follow the references. (Thanks to caya to point out that I had forgotten this issue) - thank you for your graceful recap of the issues and answers. – Joseph Jan 8 at 3:11 @halirutan, I think you forgot to mention the all important missing pattern n_ in LHS. – caya Jan 8 at 12:19 @caya Thanks, I added it as last point. – halirutan Jan 9 at 8:59 lang-mma	{"extraction_info": {"found_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": 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": 0.9374278783798218, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Complete_theory	# Complete theory In mathematical logic, a theory is complete if it is a maximal consistent set of sentences, i.e., if it is consistent, and none of its proper extensions is consistent. For theories in logics which contain classical propositional logic, this is equivalent to asking that for every sentence φ in the language of the theory it contains either φ itself or its negation ¬φ. Recursively axiomatizable first-order theories that are rich enough to allow general mathematical reasoning to be formulated cannot be complete, as demonstrated by Gödel's incompleteness theorem. This sense of complete is distinct from the notion of a complete logic, which asserts that for every theory that can be formulated in the logic, all semantically valid statements are provable theorems (for an appropriate sense of "semantically valid"). Gödel's completeness theorem is about this latter kind of completeness. Complete theories are closed under a number of conditions internally modelling the T-schema: • For a set $S\!$: $A \land B \in S$ if and only if $A \in S$ and $B \in S$, • For a set $S\!$: $A \lor B \in S$ if and only if $A \in S$ or $B \in S$. Maximal consistent sets are a fundamental tool in the model theory of classical logic and modal logic. Their existence in a given case is usually a straightforward consequence of Zorn's lemma, based on the idea that a contradiction involves use of only finitely many premises. In the case of modal logics, the collection of maximal consistent sets extending a theory T (closed under the necessitation rule) can be given the structure of a model of T, called the canonical model. ## Examples Some examples of complete theories are: • Presburger arithmetic • Tarski's axioms for Euclidean geometry • The theory of dense linear orders • The theory of algebraically closed fields of a given characteristic • The theory of real closed fields • Every uncountably categorical countable theory • Every countably categorical countable theory ## References • Mendelson, Elliott (1997). Introduction to Mathematical Logic (Fourth edition ed.). Chapman & Hall. p. 86. ISBN 978-0-412-80830-2. This mathematical logic-related article is a stub. You can help Wikipedia by expanding it.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "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": 0.8719308376312256, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/39051/how-can-such-a-high-exponent-arise-in-this-physics-equation?answertab=oldest	# How can such a high exponent arise in this physics equation? This site http://what-if.xkcd.com/14/ states that during a helium flash, "the reaction rate is proportional to the 40th power of the temperature". Taking for granted that this is true, how can such a large exponent arise in a physic equation? It just makes no sense to me that some physical process could give rise to this phenomenon. An explanation of how it can arise for a helium flash would be satisfactory, but the question is really more general. How can a physical process accumulate enough factors to get up to x^40? Thermodynamic processes tend to be in the x^2 or x^3's and geometry can only add a few more x^3's. Note that we are not talking about the scale of things (obviously the 40th power is not a big deal in the length or time scales), we are talking about a physics /equation/. - 1 That's a good question. High powers may occur but I still think that the explanation here is different. The dependence isn't an exact 40th power in any sense. In fact, it is exponential – the fusion depends on quantum tunneling that probes the exponentially decreasing "tails" of the distribution and at higher temperatures, one may get much further in the tail, thus speeding up the rate exponentially - and the (approximately) 40th power is arguably just some best interpolation of the intrinsically exponential function for some interval of parameters. – Luboš Motl Oct 4 '12 at 15:05 1 I would assume it's roughly the value of the power derivative $\frac{d ln(f(r))}{d ln(r)}$, which gives the exponent of a the local approximate power law. It's probably an exponential law like @LubošMotl said. I remember once I did a calculation about stellar power production that yielded a power derivative of about 22, so why not 40? – Benji Remez Oct 4 '12 at 18:32 1 What I meant is that the quantity in question is my be an exponential function of the form $y(x) = e^x$. Both quantum tunneling and Boltzmann factors are exponential, and they both come into play in stellar fusion. Try looking up the Gamow Peak. Now, if you have some power law %f(r) = r^n%, notice that its logarithmic derivative %\frac{d(log(f(r))}{d(log(r))}\$ is equal to %n%, so this yields the exponent of the law. In principle, it may be applied to any general function, and its value would give the appropriate exponent should you wish to locally approximate that function as a power law. – Benji Remez Oct 5 '12 at 6:12 1 When you have exponential behavior which goes as exp(ax - b x^2), where a is positive and b is positive, you can approximate the thing in the exponent as a logarithm, by scaling by a constant, and then the power can be whatever you like. – Ron Maimon Oct 5 '12 at 7:39 1 – Warrick Oct 5 '12 at 11:33 show 3 more comments ## 1 Answer Expanding the comments above: The quantity in question probably doesn't obey a power law in the traditional sense. It is most likely an exponential law $y(x) = e^x$, since it involves thermodynamic Boltzmann factors and quantum tunneling probabilities, and both are exponential. Having said that, it can still be locally approximated as a power law. Consider a function $f(r) = r^n$. If you take its logarithmic derivative $\dfrac{d(ln\ f(r))}{d(ln\ r)}$, you recover the exponent $n$. You may also apply this derivative to any general function. The value you obtain (as a function of $r$) is then the appropriate exponent should you wish to use a power-law local approximation. Check out this document; it both shows the exponential calculations and the use of this derivative. -	{"extraction_info": {"found_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": 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": 0.938818097114563, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/191672-need-help-differentiating-expression-answer-given-just-unsure-steps.html	# Thread: 1. ## Need help differentiating an expression. Answer is given just unsure of the steps. Hello, this problem states to differentiate the expression with respect to q. I am not sure of the steps taken to get the final result. This is an economics problem differentiating the variable cost function to get the marginal cost function. Attached Thumbnails 2. ## Re: Need help differentiating an expression. Answer is given just unsure of the steps So you want to differentiate: $6(\frac{q}{10k^{0.4}})^{1.67}$ with respect to q? In which case, we have: $6\frac{q^{1.67}}{(10k^{0.4})^{1.67}}$ = $\frac{6q^{1.67}}{(10k^{0.4})^{1.67}}$ But the $\frac{6}{(10k^{0.4})^{1.67}}$(everything except the $q^{1.67}$) is just constant. It's a stable value, like 4 or 7 - it isn't a variable. For the sake of simplicity, I'm going to let $\frac{6}{(10k^{0.4})^{1.67}}=c$ When we differentiate $4x^n$ with respect to $x$, we get: $4nx^{n-1}$. When we differentiate $ax^n$ with respect to $x$, we get: $anx^{n-1}$ We have: $c\times~q^{1.67}$ So, differentiating, we get: $1.67c\times~q^{0.67}$ Then, slipping c back in as $\frac{6}{(10k^{0.4})^{1.67}}$, we have: $\frac{1.67\times~6\times~q^{0.67}}{(10k^{0.4})^{1. 67}}$ and $1.67\times~6=10.02$ which is where that number arises. #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.8985183835029602, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/70320/for-mathfrak-g-a-lie-algebra-of-type-e-7-mathfrak-h-a-cartan-subalgeb/106607	## For $\mathfrak g$ A Lie algebra of type $E_7$, $\mathfrak h$ a Cartan subalgebra and $\Delta$ the resulting root system, does $Aut(\mathfrak g,\mathfrak h)\rightarrow Aut(\Delta)$ split over the Weyl group? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given a complex simple Lie algebra $\mathfrak g$ of type $E_7$, Cartan subalgebra $\mathfrak h$ and simple roots $\alpha_1,…\alpha_n$, suppose $\pi$ is an involution of the extended Dynkin diagram. I would like to know whether $\pi$ must be induced from an involution of $\mathfrak g$. Writing $\Delta$ for the root system and W for the Weyl group, since $Aut(\Delta)/W$ is isomorphic to the group of automorphisms of the Dynkin diagram and this is trivial for $E_7$, the answer is "yes" if the map $Aut(\mathfrak g,\mathfrak h)\rightarrow Aut(\Delta)$ splits over the Weyl group. That is, it would suffice if there is a subgroup of the automorphism group of $Aut(\mathfrak g, \mathfrak h)$ which is isomorphic to the Weyl group under this mapping. Is this true? Or do you otherwise know whether every involution of the extended Dynkin diagram for $E_7$ must arise from an involution of $\mathfrak g$? Thanks very much! - 2 Obviously what one would like is a functorial way of constructing Lie algebras from root systems, and one can't have it because the Weyl group acts on the root system and doesn't act on the group. Which is basically your splitting question. The closest I've seen to a functorial construction is in Jacob Lurie's undergrad thesis math.harvard.edu/~lurie/papers/thesis.pdf and a construction Richard Borcherds explained to me that doubles the root system in order to let the normalizer of the Z-torus act: math.berkeley.edu/~allenk/courses/spr02/261b/… – Allen Knutson Jul 22 2011 at 6:17 1 Thank you for this comment Allen, it was very helpful in answering a question I hadn't been able to ask. – B. Bischof Sep 7 at 14:47 ## 3 Answers There is a related old paper by Tits on normalisers of tori, but my copy is long gone and I'm not sure whether the splitting issues had been addressed there. In the case of $E_7$, the sequence does not split over $W$ if the argument below is correct. Indeed suppose that it does. Then $W$ is a subgroup of $G={\rm Aut}(g)$ and for every $\alpha\in \Delta$ we have an involution $\theta_\alpha\in G$ corresponding to the reflection $s_\alpha\in W$. We look at the fixed point algebra $g^{\theta_\alpha}$. If $s_\alpha(\beta)=\beta$ for $\beta\in\Delta$ then ($\theta_\alpha$ being an involution) for any $e_\beta\in g_\beta$ we have that $\theta_\alpha(e_\beta)=c_\alpha(\beta)e_\beta$, where $c_\alpha(\beta)=\pm 1$. Now the centraliser $C_W(s_\alpha)$ contains a reflection subgroup of type $A_1 + D_6$ which acts ttransitively on the set $\Delta_\alpha$ of all roots orthogonal to $\alpha$. From this it is immediate that $c_\alpha(\beta)=c_\alpha(\gamma)$ for all $\beta,\gamma\in\Delta_\alpha$. As we can find $\beta,\gamma\in\Delta_\alpha$ with $\beta+\gamma\in\Delta_\alpha$ and $c_\alpha(\beta+\gamma)=1$, we deduce that $c_\alpha(\beta)=1$ for all $\beta\in\Delta_\alpha$. Now we can compute $\dim\ g^{\theta_\alpha}$. Each pair of distinct roots $(\beta, s_\alpha(\beta))$ contributes $1$ to $\dim\ g^{\theta_\alpha}$. As $\|\Delta_\alpha\|=60$ and $\|\Delta\|=126$, the number of such pairs is $33$ and the above discussion shows that $\dim\ g^{\theta_\alpha}=6+60+33=99$. But it is well known (and goes back to E. Cartan's work on simple Lie algebras over $\mathbb{R}$) that there are three conjugacy classes of involutions in $G$ whose fixed point algebras have dimensions $63$, $69$ and $79$. So the extension does not split over $W$. As for the second part of the question, it seems likely to me that the involution of the extended Dynkin diagram of $\Delta$ can be lifted to an involution of $G$. It has the form $w_0w_1$ where $w_0$ is the longest element of $W$ and $w_1$ is that of the Weyl group of the $E_6$-subdiagram. Let $l$ be the corresponding Levi subalgebra of $g$. One just needs to find an involution in $N_G(l)$ which acts as the nontrivial graph automorphism on $[l,l]$ and as $-1$ on the centre of $l$. - ### 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. Sasha Premet has provided a concrete direct approach to the question, which at first sight looks convincing. But I'd like to follow up with more detail about one aspect of the question. First, the paper by Tits which he mentions is Normalisateurs de tores I, J. Algebra 4 (1966), 96-116 (alas there is no part II). Basically Tits is investigating the relationship between the normalizer of a maximal torus, in a split reductive group over some field, and the resulting Weyl group. But he doesn't quite state explicitly here in which simple groups the extension splits (in other words, when is the Weyl group naturally a subgroup of the algebraic group). Later some topologists at Rice (Morton Curtis, Alan Wiederhold, Bruce Williams) wrote a couple of joint papers, the first one Normalizers of maximal tori appearing in Springer Lect. Notes in Math. 418 (1974), 31-47. Here their framework involves compact connected semisimple Lie groups, so a standard translation into the language of complex Lie algebras or corresponding algebraic groups is needed. But they do show directly that two such Lie groups are isomorphic if and only if the respective normalizers are isomorphic (Theorem 1). Along with this, they work out explicitly a table (Theorem 2) showing which compact simple Lie groups (simply connected or taken modulo center) have a natural subgroup isomorphic to the Weyl group. Here only type `$G_2$` among the exceptional groups has such a splitting of the normalizer. The method is reasonable but too long to outline here. Amusingly, they write at first: It seems strange that Theorems 1 and 2 do not seem to have been known, because their proofs use no techniques not known for many years. But then an added footnote states that they learned recently that in unpublished work Tits had earlier proved Theorem 2 in a more comprehensive way. One other comment is that automorphisms of finite order of semisimple complex Lie algebras play a major role in the work of Victor Kac and are discussed in his book on infinite dimensional Lie algebras as well as in section X.5 of the classic book by Helgason Differential Geometry, Lie Groups, and Symmetric Spaces. But these sources don't seem to address the specific question asked. - Jim, many thanks for pointing out the reference to the paper by topologists. Regarding Theorem 2, did the authors prove that the extension is non-split outside type $G_2$? The argument I gave can be carried out for other types as well (with some modfications). – Alexander Premet Sep 8 at 8:52 @Sasha: Yes. Theorem 2 covers each simple type including `$E_7$`. The authors consider only simply connected or adjoint groups (which is enough for this example), but in their footnote they mention that Tits has also treated the various intermediate groups in his unpublished work. I can't immediately compare their approach with yours. – Jim Humphreys Sep 8 at 18:30 Thanks very much all for these most helpful answers, much appreciated! Regarding the 2nd (and easier) part of the question, as you suspected this is true. In the preprint http://front.math.ucdavis.edu/1111.4028 Katharine Turner and I were needing this for some work on harmonic maps, and the form of the statement we prove there is below. It seems like something that would be known to folks working in this area, but we couldn't find a reference. Every involution of the extended Dynkin diagram for a simple complex Lie algebra $\mathfrak {g} ^\mathbb {C}$ is induced by a Cartan involution of a real form of $\mathfrak {g} ^\mathbb {C}$. More precisely, let $\mathfrak {g}^\mathbb {C}$ be a simple complex Lie algebra with Cartan subalgebra $\mathfrak {t} ^\mathbb {C}$ and choose simple roots $\alpha_1,\ldots,\alpha_N$ for the root system $\Delta (\mathfrak {g} ^\mathbb {C},\mathfrak {t} ^\mathbb {C})$. Given an involution $\pi$ of the extended Dynkin diagram for $\Delta$, there exists a real form $\mathfrak {g}$ of $\mathfrak {g} ^\mathbb {C}$ and a Cartan involution $\theta$ of $\mathfrak {g}$ preserving $\mathfrak {t} =\mathfrak {g}\cap\mathfrak {t} ^\mathbb {C}$ such that $\theta$ induces $\pi$ and $\mathfrak {t}$ is a real form of $\mathfrak {t} ^\mathbb {C}$. The Coxeter automorphism $\sigma$ determined by $\alpha_1,\ldots,\alpha_N$ preserves the real form $\mathfrak {g}$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 92, "mathjax_display_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": 0.947953999042511, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/149430-taylor-series-expansion.html	# Thread: 1. ## Taylor series expansion Find the Taylor series for $%5Csqrt%7Bx%7D$ centered around $x=1$. I've gotten pretty close but I don't know how to put it back into summation format. This is my working: $f%28x%29%20=%20%5Csum_%7Bn=0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7Bf%5En%281%29%7D%7Bn%21%7D%28x-1%29%5En$ $f%5E1%281%29%20=%20%5Cfrac%7B1%7D%7B2%7D$ $f%5E2%281%29%20=%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Cleft%28-%5Cfrac%7B1%7D%7B2%7D%5Cright%29$ $f%5E3%281%29%20=%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Cleft%28-%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Cleft%28-%5Cfrac%7B3%7D%7B2%7D%5Cright%29$ $f%5E4%281%29%20=%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Cleft%28-%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Cleft%28-%5Cfrac%7B3%7D%7B2%7D%5Cright%29%5Cleft%28-%5Cfrac%7B5%7D%7B2%7D%5Cright%29$ So we have: $\sqrt{x} = 1 + \frac{\frac{1}{2}}{1!}(x-1) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(x-1)^2+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}(x-1)^3 + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{4!}(x-1)^4 + \cdots$ So I know the summation form must have $1+%5Csum_%7Bn=1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%28-1%29%5E%7Bn+1%7D%5Cfrac%7B1%7D%7B2%5En%7D%5Ctext%7Bsomething%7D%7D%7Bn%21%7D%28x-1%29%5En$ but I can't find the pattern for $%281%29,%20%281%29%28-1%29,%20%281%29%28-1%29%28-3%29,%20%281%29%281%29%28-3%29%28-5%29$ 2. $something=\frac{(2n)!}{n!2^n}$ 3. Originally Posted by Also sprach Zarathustra $something=\frac{(2n)!}{n!2^n}$ Hi how is it (2n)! ? Isn't that even? 4. Hello! You need to find some kind of formula to: $1\cdot 3 \cdot 5 \cdot 7 \cdot ... \cdot (2n-1)$ but, the above equals to: $\frac{(2n)!}{n!2^n}$ Check this formula for $n=4$. Try to prove it! It is easy! By the way: ${(2n)!}=1\cdot 2 \cdot 3 \cdot 4 \cdot ... \cdot (2n)$ 5. yup that was exactly the formula for the multiplication of odd numbers however how do you prove it? in other words how did you get that 'formula'? Also: can't I have something like this: $1+%5Cfrac%7B1%7D%7B2%7D%28x-1%29%20+%20%5Cfrac%7B1%7D%7B2%7D%5Csum_%7Bn=1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%28-1%29%5En%5Cfrac%7B%281%29%283%29%5Ccdots%282n-1%29%7D%7B2%5En%7D%7D%7B%28n+1%29%21%7D%28x-1%29%5E%7Bn+1%7D$ 6. A product of even numbers, 2468...(2n) can be written as (21)(22)(23)(24)...(2n). Now factor out those "2"s: (2222...2)(1234...n)= $2^n n!$. To simplify a product of odd numbers, 1357...(2n-1), multiply and divide by the "missing" even numbers: $\frac{12345678...(2n-1)(2n)}{2468*...(2n)}$ The numerator is now (2n)! and the denominator is $2^n n!$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 10}, "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": 0.9331405162811279, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/86906/complexity-of-matching-red-and-blue-points-in-the-plane/86922	## Complexity of matching red and blue points in the plane. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm just asking because I'm curious. I was seeking references on the following problem, that a friend exposed to me last holidays : ## Problem Given $n$ red points and $n$ blue points in the plane in general position (no 3 of them are aligned), find a pairing of the red points with the blue points such that the segments it draws are all disjoint. This problem is always solvable, and admits several proof. A proof I know goes like this : Start with an arbitrary pairing, and look for intersections of the segments it defines, if there are none you're done. If you found one, do the following operation : ````r r r r \ / \| \| X => \| \| / \ \| \| b b b b ```` (uncross the crossing you have found), you may create new crossings with this operation. If you repeat this operation, you cannot cycle, because the triangle inequality shows that the sum of the length of the segments is strictly decreasing. So you will eventually get stuck at a configuration with no crossings. ## Questions 1. What is the complexity of the algorithm described in the proof ? 2. What is the best known algorithm to solve this problem ? I wouldn't be surprised to learn that this problem is a classic in computational geometry, however googling didn't give me references. Since some computational geometers are active on MO, I thought I could get interesting answers here. - You might gain insight by viewing each move as a transposition applied to a permutation, and looking at a directed graph of all permutations joined by edges when two permutations differ by a transposition and the direction goes from larger sum of lengths to smaller. This problem of traversing such a graph has likely been studied in combinatorics; I would be surprised to find an example that requires worse than O(n^2) moves. Gerhard "Ask Me About System Design" Paseman, 2012.01.28 – Gerhard Paseman Jan 28 2012 at 18:22 Try adding "ghostbusters" to your searches. – Zsbán Ambrus Jan 28 2012 at 19:50 Note that you could also solve a minimal weight matching problem on a weighted bipartite graph where the edge weights are the distances of red points from blue points. This has $O(n^3)$ runtime, still polynomial but asymptotically slower than the algorithms suggested in the answers. – Zsbán Ambrus Jan 28 2012 at 21:57 Thanks to all for your replies and comments. I knew how to prove the existence with the ham sandwich theorem but wasn't sure how efficient it was as an algorithm. – Thomas Richard Jan 29 2012 at 9:59 Could we tag this with [co.combinatorics]? I think the part of the question that's still unanswered, that is, whether if you naively uncross edges it may take more than polynomial time to arrive at a non-crossing matching, is a combinatorial question. – Zsbán Ambrus Feb 15 2012 at 10:16 ## 3 Answers The Ghosts and Ghostbusters problem can be solved in $O(n\log n)$ time, which is considerably faster than the $O(n^2\log n)$-time algorithm suggested by CLRS. The ham sandwich theorem implies that there is a line $L$ that splits both the ghosts and the ghostbusters exactly in half. (If the number of ghosts and ghostbusters is odd, the line passes through one of each; if the number is even, the line passes through neither.) Lo, Matoušek, and Steiger [Discrete Comput. Geom. 1994] describe an algorithm to compute a ham-sandwich line in $O(n)$ time; their algorithm is also sketched here. Now recursively solve the problem on both sides of $L$; the recursion stops when all subsets are empty. The total running time obeys the mergesort recurrence $T(n) = O(n) + 2T(n/2)$ and thus is $O(n\log n)$. This algorithm is optimal in the algebraic decision tree and algebraic computation tree models of computation, because you need $\Omega(n\log n)$ time in those models just to decide whether two sets of $n$ points are equal. - ### 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. Check out the classic Cormen, Leiserson, Rivest, Stein, ''Introduction to algorithms'', second edition. In chapter 33 (Computational Geometry). See the exercises at the end of the whole chapter: exercise 33-3 is your problem. The full solution isn't described, but you will find a hint for a polynomial time algorithm. I have no idea whether that's the fastest algorithm known. - In the paper Geometry Helps in Matching, by P. Vaidya, he shows that the minimum matching (which is what you are finding here) can be found in $O(n^2 \log^3(n))$ time. - The abstract of that paper claims that they can get $O(n^2 log^3 n)$ time only in the case of $L_1$ and $L_\infty$ metrics (those two are the same in 2 dimensions of course). They give a somewhat slower algorithm for the $L_2$ case. Does a minimal total length in $L_\infty$ metric guarantee no intersections between the segments? – Zsbán Ambrus Jan 29 2012 at 11:05 The triangle inequality in those metrics is Minkowski's inequality, so certainly $L_1$ should work... – Igor Rivin Jan 29 2012 at 16:11 No way. The triangle inequality proof only works with L_2 because you can break up the crossing segments to two smaller segments at their intersection points. In fact, consider the two red points a = (0, 0), b = (1, 0), and the two black points c = (1, 2), d = (2, 2). Then the matching a-c, b-d has total $L_\infty$ length 3 + 3 = 6; but the matching a-d, b-c which has segments crossing each other has total $L_\infty$ length 4 + 2 = 6, so a matching shortest in $L_\infty$ need not be non-crossing. – Zsbán Ambrus Jan 30 2012 at 9:05	{"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": 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": 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": 0.9400143027305603, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/153302-name-sets-numbers-each-number-belongs.html	# Thread: 1. ## Name the sets of numbers to which each number belongs? Well yeah I just started Algebra II this year. I don't really understand this yet. Can someone do a couple of these? and then explain? I just scanned it into my computer. Here's what I can use: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 2. 1) 6/7 is a rational number because it can be expressed as a ratio of two integers. Since all rational numbers are real numbers, 6/7 is also a real number. 3) 0 is a whole number and an integer. It is also a rational number (you could write 0 as 0/1) and a real number. 8) 26.1 is a rational number (you could write it as 261/10). Rational numbers can also be decimals that terminate, like this one. 14) $\sqrt{42}$ is an irrational number. It's decimal equivalent would be a non-repeating, non-terminating decimal. All square roots of positive integers that are not perfect squares are irrational numbers. The numbers $\pi$ and e are also irrational numbers. Since all irrational numbers are also real numbers, $\sqrt{42}$ is also a real number. 19) 894,000 is a natural number, a whole number, an integer, a rational number, and a real number. A couple of other points: - A number cannot be BOTH rational and irrational. It is one or the other. - Note the hierarchy of these sets of numbers. Example: all integers are rational numbers, but not all rational numbers are integers.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "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": 0.9306051135063171, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/90996-solved-limits-question-tuffy.html	# Thread: 1. ## [SOLVED] Limits question tuffy any ideas how $Lim_{n\rightarrow\infty}\frac{((n+1)!)^2(2n)!}{(2n +2)!(n!)^2}=Lim_{n\rightarrow\infty}\frac{(n+1)^2} {(2n+2)(2n+1)}=\frac{1}{4}$ I can only get as far as $\frac{(n+1)^2(2n)!}{(2n+2)!}$ ?? thnx 2. Originally Posted by i_zz_y_ill any ideas how $Lim_{n\rightarrow\infty}\frac{((n+1)!)^2(2n)!}{(2n +2)!(n!)^2}=Lim_{n\rightarrow\infty}\frac{(n+1)^2} {(2n+2)(2n+1)}=\frac{1}{4}$ I can only get as far as $\frac{(n+1)^2(2n)!}{(2n+2)!}$ ?? thnx So $\lim_{n \to \infty} \frac{[(n+1)!]^{2}(2n)!}{(2n+2)!(n!)^{2}} = \lim_{n \to \infty} \frac{[n! \cdot (n+1)]^{2} \cdot (2n)!}{(2n+1)! \cdot (2n+2) \cdot (n!)^{2}} = \lim_{n \to \infty} \frac{(n+1)^{2}}{(2n+2)(2n+1)}$ $= \lim_{n \to \infty} \frac{n^2+2n+1}{(2n+2)(2n+1)}$. Note that the highest degree term in the numerator is $n^2$ and the highest degree term in the denominator is $4n^2$. So as $n \to \infty$, $\frac{n^2}{4n^2} \to \frac{1}{4}$. 3. Try to use these: $\begin{gathered}<br /> \left( {\left( {n + 1} \right)!} \right)^2 = \left( {\left( {n + 1} \right)n!} \right)^2 = \left( {n + 1} \right)^2 \left( {n!} \right)^2 \hfill \\<br /> \left( {2n + 2} \right)! = \left( {2n + 2} \right)\left( {2n + 1} \right)\left( {2n} \right)! \hfill \\ <br /> \end{gathered}$	{"extraction_info": {"found_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": 11, "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": 0.952163577079773, "perplexity_flag": "middle"}
http://www.thefullwiki.org/Tessellation	# Tessellation: Wikis Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles. ### Related top topics Categories: Tiling > Mosaic > Symmetry # Encyclopedia Updated live from Wikipedia, last check: May 14, 2013 04:29 UTC (40 seconds ago) ### From Wikipedia, the free encyclopedia A tessellation of pavement A Honeycomb is an example of a tessellated natural structure A tessellation or tiling of the plane is a collection of plane figures that fills the plane with no overlaps and no gaps. One may also speak of tessellations of the parts of the plane or of other surfaces. Generalizations to higher dimensions are also possible. Tessellations frequently appeared in the art of M. C. Escher. Tessellations are seen throughout art history, from ancient architecture to modern art. In Latin, tessella is a small cubical piece of clay, stone or glass used to make mosaics.[1] The word "tessella" means "small square" (from "tessera", square, which in its turn is from the Greek word for "four"). It corresponds with the everyday term tiling which refers to applications of tessellations, often made of glazed clay. ## Wallpaper groups Tilings with translational symmetry can be categorized by wallpaper group, of which 17 exist. All seventeen of these patterns are known to exist in the Alhambra palace in Granada, Spain. Of the three regular tilings two are in the category p6m and one is in p4m. ## Tessellations and colour If this parallelogram pattern is colored before tiling it over a plane, seven colors are required to ensure each complete parallelogram has a consistent color that is distinct from that of adjacent areas. (This tiling can be compared to the surface of a torus.) Tiling before coloring, only four colors are needed. When discussing a tiling that is displayed in colors, to avoid ambiguity one needs to specify whether the colors are part of the tiling or just part of its illustration. See also symmetry. The four color theorem states that for every tessellation of a normal Euclidean plane, with a set of four available colors, each tile can be colored in one color such that no tiles of equal color meet at a curve of positive length. Note that the coloring guaranteed by the four-color theorem will not in general respect the symmetries of the tessellation. To produce a coloring which does, as many as seven colors may be needed, as in the picture at right. ## Tessellations with quadrilaterals Copies of an arbitrary quadrilateral can form a tessellation with 2-fold rotational centers at the midpoints of all sides, and translational symmetry whose basis vectors are the diagonal of the quadrilateral or, equivalently, one of these and the sum or difference of the two. For an asymmetric quadrilateral this tiling belongs to wallpaper group p2. As fundamental domain we have the quadrilateral. Equivalently, we can construct a parallelogram subtended by a minimal set of translation vectors, starting from a rotational center. We can divide this by one diagonal, and take one half (a triangle) as fundamental domain. Such a triangle has the same area as the quadrilateral and can be constructed from it by cutting and pasting. ## Regular and irregular tessellations Hexagonal tessellation of a floor A regular tessellation is a highly symmetric tessellation made up of congruent regular polygons. Only three regular tessellations exist: those made up of equilateral triangles, squares, or hexagons. A semiregular tessellation uses a variety of regular polygons; there are eight of these. The arrangement of polygons at every vertex point is identical. An edge-to-edge tessellation is even less regular: the only requirement is that adjacent tiles only share full sides, i.e. no tile shares a partial side with any other tile. Other types of tessellations exist, depending on types of figures and types of pattern. There are regular versus irregular, periodic versus aperiodic, symmetric versus asymmetric, and fractal tessellations, as well as other classifications. Penrose tilings using two different polygons are the most famous example of tessellations that create aperiodic patterns. They belong to a general class of aperiodic tilings that can be constructed out of self-replicating sets of polygons by using recursion. A monohedral tiling is a tessellation in which all tiles are congruent. Spiral monohedral tilings include the Voderberg tiling discovered by Hans Voderberg in 1936, whose unit tile is a nonconvex enneagon; and the Hirschhorn tiling discovered by Michael Hirschhorn in the 1970s, whose unit tile is an irregular pentagon. ## Self-dual tessellations Tilings and honeycombs can also be self-dual. All n-dimensional hypercubic honeycombs with Schlafli symbols {4,3n−2,4}, are self-dual. A ]] ## Tessellations and computer models A tessellation of a disk used to solve a finite element problem. These rectangular bricks are connected in a tessellation, which if considered an edge-to-edge tiling, topologically identical to a hexagonal tiling, with each hexagon flattened into a rectangle with the long edges divided into two edges by the neighboring bricks. This basketweave tiling is topologically identical to the Cairo pentagonal tiling, with one side of each rectangle counted as two edges, divided by a vertex on the two neighboring rectangles. In the subject of computer graphics, tessellation techniques are often used to manage datasets of polygons and divide them into suitable structures for rendering. Normally, at least for real-time rendering, the data is tessellated into triangles, which is sometimes referred to as triangulation. Tessellation is a staple feature of DirectX 11 and OpenGL.[2] In computer-aided design the constructed design is represented by a boundary representation topological model, where analytical 3D surfaces and curves, limited to faces and edges constitute a continuous boundary of a 3D body. Arbitrary 3D bodies are often too complicated to analyze directly. So they are approximated (tessellated) with a mesh of small, easy-to-analyze pieces of 3D volume — usually either irregular tetrahedrons, or irregular hexahedrons. The mesh is used for finite element analysis. Mesh of a surface is usually generated per individual faces and edges (approximated to polylines) so that original limit vertices are included into mesh. To ensure approximation of the original surface suits needs of the further processing 3 basic parameters are usually defined for the surface mesh generator: • Maximum allowed distance between planar approximation polygon and the surface (aka "sag"). This parameter ensures that mesh is similar enough to the original analytical surface (or the polyline is similar to the original curve). • Maximum allowed size of the approximation polygon (for triangulations it can be maximum allowed length of triangle sides). This parameter ensures enough detail for further analysis. • Maximum allowed angle between two adjacent approximation polygons (on the same face). This parameter ensures that even very small humps or hollows that can have significant effect to analysis will not disappear in mesh. Algorithm generating mesh is driven by the parameters (example: CATIA V5 tessellation library). Some computer analyses require adaptive mesh. Mesh is being locally enhanced (using stronger parameters) in areas where it is needed during the analysis. Some geodesic domes are designed by tessellating the sphere with triangles that are as close to equilateral triangles as possible. ## Tessellations in nature Basaltic lava flows often display columnar jointing as a result of contraction forces causing cracks as the lava cools. The extensive crack networks that develop often produce hexagonal columns of lava. One example of such an array of columns is the Giant's Causeway in Northern Ireland. The Tessellated pavement in Tasmania is a rare sedimentary rock formation where the rock has fractured into rectangular blocks. ## Number of sides of a polygon versus number of sides at a vertex For an infinite tiling, let a be the average number of sides of a polygon, and b the average number of sides meeting at a vertex. Then (a − 2)(b − 2) = 4. For example, we have the combinations $(3, 6), (3 \tfrac{1}{3},5), (3 \tfrac{3}{4},4 \tfrac{2}{7}), (4, 4), (6, 3)$, for the tilings in the article Tilings of regular polygons. A continuation of a side in a straight line beyond a vertex is counted as a separate side. For example, the bricks in the picture are considered hexagons, and we have combination (6, 3). Similarly, for the basketweave tiling often found on bathroom floors, we have $(5, 3\tfrac13)$. For a tiling which repeats itself, one can take the averages over the repeating part. In the general case the averages are taken as the limits for a region expanding to the whole plane. In cases like an infinite row of tiles, or tiles getting smaller and smaller outwardly, the outside is not negligible and should also be counted as a tile while taking the limit. In extreme cases the limits may not exist, or depend on how the region is expanded to infinity. For finite tessellations and polyhedra we have $( a - 2 ) ( b - 2 ) = 4 ( 1 - \frac{\chi}{F} ) ( 1 - \frac{\chi}{V} )$ where F is the number of faces and V the number of vertices, and χ is the Euler characteristic (for the plane and for a polyhedron without holes: 2), and, again, in the plane the outside counts as a face. The formula follows observing that the number of sides of a face, summed over all faces, gives twice the total number of sides in the entire tessellation, which can be expressed in terms of the number of faces and the number of vertices. Similarly the number of sides at a vertex, summed over all vertices, also gives twice the total number of sides. From the two results the formula readily follows. In most cases the number of sides of a face is the same as the number of vertices of a face, and the number of sides meeting at a vertex is the same as the number of faces meeting at a vertex. However, in a case like two square faces touching at a corner, the number of sides of the outer face is 8, so if the number of vertices is counted the common corner has to be counted twice. Similarly the number of sides meeting at that corner is 4, so if the number of faces at that corner is counted the face meeting the corner twice has to be counted twice. A tile with a hole, filled with one or more other tiles, is not permissible, because the network of all sides inside and outside is disconnected. However it is allowed with a cut so that the tile with the hole touches itself. For counting the number of sides of this tile, the cut should be counted twice. For the Platonic solids we get round numbers, because we take the average over equal numbers: for (a − 2)(b − 2) we get 1, 2, and 3. From the formula for a finite polyhedron we see that in the case that while expanding to an infinite polyhedron the number of holes (each contributing −2 to the Euler characteristic) grows proportionally with the number of faces and the number of vertices, the limit of (a − 2)(b − 2) is larger than 4. For example, consider one layer of cubes, extending in two directions, with one of every 2 × 2 cubes removed. This has combination (4, 5), with (a − 2)(b − 2) = 6 = 4(1 + 2 / 10)(1 + 2 / 8), corresponding to having 10 faces and 8 vertices per hole. Note that the result does not depend on the edges being line segments and the faces being parts of planes: mathematical rigor to deal with pathological cases aside, they can also be curves and curved surfaces. ## Tessellations of other spaces An example tessellation of the surface of a sphere by a truncated icosidodecahedron. A torus can be tiled by a repeating matrix of isogonal quadrilaterals. M.C.Escher, Circle Limit III (1959) As well as tessellating the 2-dimensional Euclidean plane, it is also possible to tessellate other n-dimensional spaces by filling them with n-dimensional polytopes. Tessellations of other spaces are often referred to as honeycombs. Examples of tessellations of other spaces include: • Tessellations of n-dimensional Euclidean space. For example, filling 3-dimensional Euclidean space with cubes to create a cubic honeycomb. • Tessellations of n-dimensional elliptic space. For example, projecting the edges of a regular dodecahedron onto its circumsphere creates a tessellation of the 2-dimensional sphere with regular spherical pentagons. • Tessellations of n-dimensional hyperbolic space. For example, M. C. Escher's Circle Limit III depicts a tessellation of the hyperbolic plane using the Poincaré disk model with congruent fish-like shapes. The hyperbolic plane admits a tessellation with regular p-gons meeting in q's whenever $\tfrac{1}{p}+\tfrac{1}{q} < \tfrac{1}{2}$; Circle Limit III may be understood as a tiling of octagons meeting in threes, with all sides replaced with jagged lines and each octagon then cut into four fish. ## See also • Convex uniform honeycomb - The 28 uniform 3-dimensional tessellations, a parallel construction to this plane set • Coxeter groups - algebraeic groups that can be used to find tesselations • Four color theorem • Girih tiles • Honeycomb (geometry) • Jig-saw puzzle • List of regular polytopes • List of uniform tilings • Mathematics and fiber arts • Mosaic • Nikolas Schiller - Artist using tessellations of aerial photographs • Penrose tilings • Pinwheel tiling • Polyiamond - tilings with equilateral triangles • Polyomino • Quilting • Self-replication • Tile • Tiling puzzle • Tiling, Aperiodic • Tilings of regular polygons • Trianglepoint - needlepoint with polyiamonds (equilateral triangles) • Triangulation • Uniform tessellation • Uniform tiling • Uniform tilings in hyperbolic plane • Voronoi tessellation • Wallpaper group - seventeen types of two-dimensional repetitive patterns • Wang tiles ## Notes 1. ^ tessellate, Merriam-Webster Online 2. ^ http://developer.amd.com/gpu_assets/AMD_vertex_shader_tessellator.pdf ## References • Grunbaum, Branko and G. C. Shephard. Tilings and Patterns. New York: W. H. Freeman & Co., 1987. ISBN 0-7167-1193-1. • Coxeter, H.S.M.. Regular Polytopes, Section IV : Tessellations and Honeycombs. Dover, 1973. ISBN 0-486-61480-8. Categories: \| \| # Related links Up to date as of November 16, 2009 # Related topics Up to date as of August 19, 2010 Got something to say? Make a comment.	{"extraction_info": {"found_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": 4, "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": 0.9041978120803833, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/238980/exponents-in-odd-and-even-functions?answertab=oldest	# Exponents in Odd and Even Functions I was hoping someone could show or explain why it is that a function of the form $f(x) = ax^d + bx^e + cx^g+\cdots$ going on for some arbitrary length will be an odd function assuming $d, e, g$ and so on are all odd numbers, and likewise why it will be even if $d, e, g$ and so on are all even numbers. Furthermore, why is it if say, $d$ and $e$ are even but $g$ is odd that $f(x)$ will then become neither even nor odd? Thanks. - ## 3 Answers An even function is one for which $f(-x) = f(x)$ for all values of $x$ (e.g. evaluating at -6 is the same as evaluating at 6). If $n$ is an even exponent, then $(-x)^n = (-1)^nx^n = x^n$, since an even number of negative signs will cancel out. If all the exponents are even, then this happens for every term in the polynomial, so the result is the same as the original polynomial. An odd function is one for which $f(-x) = -f(x)$ for all values of $x$ (i.e. the minus sign factors out). If $n$ is an odd exponent, then $(-x)^n = (-1)^nx^n = -x^n$, since an odd number of negative signs leaves just one negative sign remaining. If all the exponents are odd, then we get: $$f(-x) = ax^d + bx^e + cx^g + \cdots = -ax^d - bx^e - cx^g - \cdots = -(ax^d + bx^e + cx^g + \cdots) = -f(x).$$ If there is a mixture of odd an even exponents, then neither of these nice properties will hold, so the function will be neither even nor odd. - Thanks, that's exactly what I was looking for. – user48226 Nov 17 '12 at 1:41 If the exponents are all odd, then $f(x)$ is the sum of odd functions, and hence is odd. If the exponents are all even, then $f(x)$ is the sum of even functions, and hence is even. As far as your last question, the sum of an odd function and even function is neither even nor odd. ## Proof: Sum of Odd Functions is Odd: Given two odd functions $f$ and $g$. Since they are odd functions $f(-x) = -f(x)$, and $g(-x) = -g(x)$. Hence: \begin{align} f(-x) + g(-x) &= -f(x) - g(x) \\ &= -(f+g)(x) \\ \implies (f+g) & \text{ is odd if $f$ and $g$ are odd.} \end{align} ## Proof: Sum of Even Functions is Even: Given two even functions $f$ and $g$, then $f(-x) = f(x)$, and $g(-x) = g(x)$. Hence: \begin{align} f(-x) + g(-x) &= f(x) + g(x) \\ \implies (f+g) &\text{ is even if $f$ and $g$ are even.} \end{align} ## Proof: Sum of an odd function and even function is neither odd or even. If $f$ is odd, and $g$ is even \begin{align} f(-x) + g(-x) &= -f(x) + g(x) \\ &= -(f-g)(x) \\ \implies (f+g)&\text{ is neither odd or even if $f$ is even, $g$ is odd.} \end{align} - Thanks, that's a cool way to prove it. – user48226 Nov 17 '12 at 2:05 It’s just a matter of checking the definitions of even function and odd function. Let’s look at a simple case that displays all of the possible behaviors of the more general case: consider $f(x)=ax^m+bx^n$. • Suppose that $m$ and $n$ are even, say $m=2k$ and $n=2\ell$; then $$(-x)^m=(-x)^{2k}=\left((-x)^2\right)^k=\left(x^2\right)^k=x^{2k}=x^m\;,$$ and similarly $(-x)^n=x^n$, so $$f(-x)=a(-x)^m+b(-x)^n=ax^m+bx^n=f(x)\;,$$ and by definition $f$ is an even function. • Now suppose that $m$ and $n$ are odd, say $m=2k+1$ and $n=2\ell+1$. Then $$(-x)^m=(-x)^{2k+1}=(-x)^{2k}(-x)=x^{2k}(-x)=-x^{2k+1}=-x^m\;,$$ and similarly $(-x)^n=-x^n$, so $$f(-x)=a(-x)^m+b(-x)^n=-ax^m-bx^n=-\left(ax^m+bx^n\right)=-f(x)\;,$$ and by definition $f$ is an odd function. • Finally, suppose that $m$ is odd and $n$ is even. (The argument is the same if $m$ is even and $n$ is odd, just reversing the rôles of $m$ and $n$.) Then by what we’ve already seen we know that $$f(-x)=a(-x)^m+b(-x)^n=-ax^m+bx^n\;,$$ which is neither $f(x)$ nor $f(-x)$. Thus, $f$ is neither even nor odd. Nothing essentially different happens if there are more than two terms. In fact the names even function and odd function come from the behavior of even and odd powers, respectively. - Thanks, also a solid explanation. – user48226 Nov 17 '12 at 1:49 @ascii: You’re welcome. – Brian M. Scott Nov 17 '12 at 1:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 55, "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": 10, "end_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": 0.9119294285774231, "perplexity_flag": "head"}
http://mathoverflow.net/questions/123760/topological-characterization-of-the-closed-interval-0-1	## Topological characterization of the closed interval $[0,1]$. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This question is related to question 92206 "What properties make $[0, 1]$ a good candidate for defining fundamental groups?" but is not exactly equivalent in my opinion. It is even suggested in one of the answers to 92206 that "there is nothing fundamental about the unit interval," but i would like to know what is fundamental about the unit interval. I have learned some answers from the answers to 92206, but i wonder if there is more. I am mostly interested in its significance for the study of Hausdorff compacts (metrizability, Urysohn's lemma). I have some ideas and have asked already on Math.StackExchange, but decided to duplicate here. One purely topological way to define $[0, 1]$ up to homeomorphism would be to define path connectedness first: $x_1$ and $x_2$ are connected by a path in a topological space $X$ if for every Hausdorff compact $C$ and $a,b\in C$, there is a continuous $f\colon C\to X$ such that $f(a)=x_1$ and $f(b)=x_2$. Then it can be said that in every Hausdorff space $X$ with $x_1$ and $x_2$ connected by a path, every minimal subspace in which $x_1$ and $x_2$ are still connected by a path is homeomorphic to $[0, 1]$. Maybe in some sense it can be said that $[0, 1]$ is the "minimal" Hausdorff space $X$ such that every Hausdorff compact embeds into $X^N$ for some $N$, but i do not know if this can be made precise. In one of the answers to 92206 it was stated that $([0, 1], 0, 1)$ is a terminal object in the category of bipointed spaces equipped with the operation of "concatenation." This is the kind of answers i am interested in. As 92206 was concerned mostly with the fundamental group and tagged only with [at.algebraic-topology] and [homotopy-theory], i am asking this general topological question separately. Update. Related question: "Topological Characterisation of the real line". - mathoverflow.net/questions/76134/… – Gjergji Zaimi Mar 6 at 14:41 Thanks for the link. – Alexey Muranov Mar 6 at 14:56 1 This has already been discussed various times on mathoverflow. You should make precise why you are not satisfied with the answers, or/and in how far your question is different. – Martin Brandenburg Mar 6 at 15:51 @Martin can you give links please? I have looked for this question, but haven't found. It does not exist on MathOverflow in this form. I will edit it if i find a better way to state it, but if i just keep adding precisions, i am afraid it will become unreadable. – Alexey Muranov Mar 6 at 16:12 I wouldn't mind changing it into wiki. – Alexey Muranov Mar 6 at 16:16 show 6 more comments ## 4 Answers In this edit, a PS is added at the end. Not too long ago (2005) Harvey Friedman announced an attractive, novel characterization of the unit interval that seems to be little known, and might be the kind of answer you are looking for: Up to isomorphism, the unit interval is the only complete totally ordered set (with end points) that has a continuous "betweenness function". [Since arithmetical operations are continuous, it is clear that there are lots of continuous betweenness functions on the unit interval]. Here is the official characterization: Theorem (H. Friedman). Let $X$ be a linearly ordered set with left and right endpoints and the least upper bound property. The following two statements are equivalent: (a) $X$ is isomorphic to the usual closed unit interval. (b) There is some $f:X^{2}\rightarrow X$ that is continuous relative to the order-topology on $X$, and $x < f(x,y) < y$ whenever $x < y$. Friedman's proof appears in this FOM posting. PS: Friedman's proof, when coupled with usual techniques of imposing an order on a continuum with at most two non-cut points (the "separation order") yields the following purely topological characterization [the main new idea being: in the classical characterization of the unit interval as the unique second countable continuum with exactly two non-cut points, "second countable" can be swapped with "supports a continuous betweenness function relative to the separation order"]. Theorem. Up to homeomorphism, the unit interval is the only continuum $X$ (Hausdorff, connected, and compact) in which all but two points of $X$ are cuts points, and which additionally has the property that $X^2$ supports a continuous "betweenness" function (relative to the separation order). - Thanks, this looks interesting, i will think about it. – Alexey Muranov Mar 6 at 18:39 1 Is this a topological characterization? – Martin Brandenburg Mar 6 at 22:00 @Martin: admittedly this is not a purely topological characterization, but clearly topology plays a pivotal role in its formulation (continuity of $f$ with respect to the product topology on $X^2$ induced by the order topology on $X$). – Ali Enayat Mar 7 at 0:28 @Martin/Ali: I guess one could use the van Dalen-Wattel topological characterization of ordered spaces to get rid of $<$. – Ramiro de la Vega Mar 7 at 15:07 @Ramiro: thanks for the pointer; I noticed another way to couch the result in terms of purely topological notions (as in the edit). – Ali Enayat Mar 7 at 21:16 show 1 more comment ### 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 following topological characterization is close to that of the real line which is indicated in another MO thread, but does not seem to have been pointed out here. Let $X$ be a topological space, let $a,b$ be points of $X$ such that $a\neq b$. Assume that $X$ is compact, connected and separable. The following conditions are equivalent: 1. There exists a homeomorphism $f\colon [0,1]\to X$ such that $f(0)=a$ and $f(1)=b$. 2. Every connected subset of $X$ containing $\lbrace a,b\rbrace$ is equal to $X$. 3. The space $X$ is locally connected, and any connected and compact subset of $X$ which contains $\lbrace a,b\rbrace$ is equal to $X$. 4. For every $x\in X\setminus\lbrace a,b\rbrace$, the space $X\setminus\lbrace x\rbrace$ is not connected. - 1 What definition of separable do you use (i have heard two)? In any case, i prefer definitions without cardinality assumptions. – Alexey Muranov Mar 6 at 18:52 Alexey, I'm perplexed: are the two definitions you of separable have heard not equivalent? – François G. Dorais♦ Mar 6 at 22:22 @Alexey: separable = having a countable dense subset. What other definition have you heard? – ACL Mar 6 at 22:52 P.R. Halmos calls a space separable if it has a countable base of open sets. I think it is now called "second countable". (Also Halmos uses a different definition of Borel sets that is now common: for him, they are elements of the sigma-ring generated by compact sets in a locally compact space.) My first impression is that his definitions could be more meaningful than the ones i've heard elsewhere – Alexey Muranov Mar 7 at 6:17 A space which is homeomorphic to the closed unit interval is called a simple arc in the monograph "Dynamic topology" by Whyburn and Duda and there is a characterisation of it on p. 70 of this book. This assumes that the given space is a metric space, a condition which can be avoided by using the Urysohn metrization theorem. On request, the characterisation is as follows: a space is a simple arc if and only if it is a non-degenerate (i.e., with more than one point) compact, connected set which is second countable and such that each point (with the exception of two specified ones---the endpoints) is a cut point. (A point in a connected space is a cut point, if its complement is disconnected). - 3 Can you add this characterization to your answer? – Martin Brandenburg Mar 6 at 18:01 This is only a guess, I don't know whether is true, but I find it plausible. Consider all the triplets $(X, x_0, x_1)$ where $X$ is a compact, second-countable, connected Hausdorff space, and $x_0,x_1$ are distinct points in $X$. We say that the triplet satisfies the property $P$ if the following hold. • For any $x\in X$, $x\neq x_0,x_1$ the complement $X\setminus \lbrace x\rbrace$ has exactly two connected components, each containing exactly one of the points $x_0,x_1$. Denote by $C_i$ the component containing $x_i$, $i=0,1$. • Set $$\bar{C}_i:= C_i\cup\lbrace x\rbrace$$ Then each of the triplets $(\bar{C}_i, x_i,x)$ is homeomorphic to the triplet $(X, x_i, x_{1-i})$. Here is my claim: a triplet $(X.x_0,x_1)$ satisfies property $P$ if and only if it is homeomorphic to the triplet $(\; [0,1],0,1\;)$ Acknowledgments. I want to thank all the commenters for the useful explanations. (I've added second-countability to my claim which turned out to be an old result.) Here's an odd simple observation which I find intriguing, and it may or may not be useful. Observe that on the space $\newcommand{\eT}{\mathscr{T}}$ $\eT$ of (homeomorphisms types of) second countable, compact, connected, Hausdorff triplets $(X,x_0,x_1)$ there is a structure of associative semigroup $$(X, x_0,x_1)* (Y,y_0,y_1)= (Z,x_0,y_1),$$ where $Z$ is the space obtained by gluing $X$ to $Y$ by identifying $x_1$ with $y_0$. This operation is clearly involved in defining the fundamental group. Denote by $\newcommand{\be}{\boldsymbol{e}}$ $\be$ the triplet $([0,1],0,1)$. Observe that $\be$ is an idempotent $\be\ast\be=\be$. This fact alone makes possible the definition of the fundamental group. The results described in the comments show that $\be$ is characterized by the property $$\be =t_1\ast t_2,\;\;t_1,t_2\in \eT \Longleftrightarrow t_1=t_2=\be.$$ Here is an amusing question. Is it true that $\be$ is the unique idempotent of the semigroup $(\eT,\ast)$? I'm inclined to believe that the answer is positive. - 2 The long line as I know it doesn’t, as one of the components $C_i$ will be a short line. Anyway, one could add “second-countable” to the list to be safe against similar examples. – Emil Jeřábek Mar 6 at 15:21 1 Compactness of $X$ guarantees 2nd countability. By the way, what is "the long line"? I have never encountered this term. – Liviu Nicolaescu Mar 6 at 15:31 2 Compactness implies 2nd countability for metric spaces; there are plenty of compact spaces that aren't second-countable. A compact, connected, homogeneous Suslin line would give a nice counterexample, if there happens to be such a thing. – François G. Dorais♦ Mar 6 at 16:20 1 Actually, according to math.stackexchange.com/a/322519/22772, $([0,1],0,1)$ is characterized as the unique compact, Hausdorff, second-countable space $X$ with points $x_0\ne x_1$ such that $X\smallsetminus\{x_i\}$ is connected for $i=0,1$, and $X\smallsetminus\{x\}$ is disconnected for every $x\ne x_0,x_1$. This seems to imply that Liviu’s conditions with second countability added also characterize $[0,1]$. – Emil Jeřábek Mar 6 at 16:56 1 @Liviu: Maybe you want to add "connected" to the definition of $eT$. Otherwise the Cantor space with any two points gives another idempotent. – Ramiro de la Vega Mar 7 at 15:23 show 5 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 96, "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": 0.9362525939941406, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/44389/relative-velocity-and-momentum?answertab=oldest	# Relative Velocity and Momentum The question is, "A $45.5~kg$ girl is standing on a $140~kg$ plank. Both originally at rest on a frozen lake that constitutes a friction-less, flat surface. The girl begins to walk along the plank at a constant velocity of $1.47~m/s$ relative to the plank." (a) What is the velocity of the plank relative to the ice surface? (b) What is the girl's velocity relative to the ice surface? In a commentary given on this problem, they are essentially that, when the velocity of the girl relative to the plank is $\vec{v}_{gp}$, and the velocity of the plank relative to the ice is $\vec{v}_{gp}$, the velocity of the girl relative to the ice is simply $\vec{v}_{gi}= \vec{v}_{gp} + \vec{v}_{gp}$. I just can't conceive why finding the velocity of the girl relative to the earth is a simply sum. Could someone please elucidate this for me? Also, they treat this situation as if momentum is conserved, that is, $\Delta \vec{P}=0$; but isn't it necessary that friction be present, between the plank and the girl's shoes, for there to be any motion? Wouldn't this force cause a change in momentum? - ## 1 Answer I will try to attempt an answer here. You have two questions: One on relative velocity and the other on momentum. For the first, I believe you're confused on the notion of relative velocity. So let me intuitively explain the idea: When we say what is the velocity of B with respect to A, what we mean is that if A were a person, with which velocity would he see B moving? Please note that A thinks that he is stationary. So if A is riding a bicycle, he would realize that he is moving towards the street lamps in front of him. Of course, this is because he knows that he is the one who is actually moving. If he instead thought that he is the one who is stationary then he would think that the street lamps are instead moving towards him, and the velocity with which they move towards A is the relative velocity of the lamps with respect to A. Okay now let's say there are two people, A and B and they start moving towards each other with velocities $5$ and $7~m/s$ respectively. What is the relative velocity of B with respect to A? Every second B moves $7~m$ towards A but A in the same amount of time covers $5~m$ towards B. So if A thinks he is stationary then he would think that B just moved $7~m + 5~m = 12~m$ toward him in one second, hence the answer is $12~m/s$. I hope you can now relate this to the case where both of them are moving in the same direction. Now for your example. Let's say the magnitude of the velocity of the plank and the girl is $v_p$ and $v_g$. By magnitude of the velocity I mean the magnitude of the velocity of the plank with respect to the ice, so don't get confused. Also these two variables are simply the magnitudes, so both of them are positive numbers with units $m/s$. Now if you think about this in terms of the analogy given above, it should make sense that relative to the plank, the girl is moving away from it with a speed of $v_p + v_g$, the value of which is given to be $1.47~m/s$. So if $v_{gp}$ is the relative velocity of the girl with respect to the plank then, $v_{gp} = v_p + v_g$. You may find the equation that I have given here is different to the one given in the commentary in your book, that's because my variables do not include the negative sign of the direction in the variables themselves, it is written explicitly in the equation. The normal force from the ice and the weight of the girl and the plank combined cancel each other, so there is no external force on the girl-plank system. Hence the momentum is conserved. Since the initial momentum was zero, the final momentum must also be zero. Therefore, $m_gv_g + (-m_pv_p) = 0$ where $m_g$ and $m_p$ is the mass of the girl and the plank respectively. Note that the momentum of the plank is negative since the velocity is in the opposite direction and I haven't included the negative sign in the velocity variable of the plank. You can now solve the two equations to give you the solution. The frictional force between the girl and the plank is within the system i.e. it is an internal force. This force is what causes the change in momentum of the girl and the plank from zero to whatever they have now. But the momentum of them combined is still the same i.e. zero. I think I may have not explained the frictional force part clearly, so if you have doubts, feel free to ask. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_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": 0.9616391658782959, "perplexity_flag": "head"}
http://nanohub.org/topics/RoleofNormalProcessesinThermalConductivityofSilicon?version=7	Search Need Help? # Support ## Support Options • ### Knowledge Base Find information on common issues. • ### Ask the Community Ask questions and find answers from other users. • ### Wish List Suggest a new site feature or improvement. • ### Support Tickets Check on status of your tickets. ## Submit a Support Ticket Home › Topics › Role of Normal Processes in Thermal Conductivity of Silicon ## Role of Normal Processes in Thermal Conductivity of Silicon James Loy, Prabhakar Marepalli ME 503 Final Project Report Abstract In the past decade, with the miniaturization of electronic circuits and growing interest in microscale heat transfer, Boltzmann transport equation (BTE) emerged as an important model for predicting the thermal behavior at small scales. Recent advancements in numerical methods and computational power enabled to solve this equation rigorously by relaxing several assumptions. But the scattering term of BTE is incredibly complex, still requiring some simplifications to make the solution possible while preserving relevant physics. One approximation is the single mode relaxation time (SMRT) approximation. This approximation assumes that each phonon scatters only to a lattice equilibrium energy. For many materials near equilibrium and at high enough temperatures, this works well. However, for low temperatures, and for low dimensional materials like graphene, this model fails to include phonon scattering due to normal processes, which play an indirect role in thermal resistance. In this project, we include normal phonon scattering in the BTE through the use of a shifted equilibrium distribution proposed by Callaway. We solve the non-gray BTE using finite volume method (FVM) with coupled ordinates method (COMET) to compute the thermal conductivity and temperature distribution of silicon at different temperatures and length scales. We implement full phonon dispersion for all polarizations under isotropic assumption. The effect of including normal processes on the thermal conductivity predictions is rigorously analyzed. Our results show that ignoring the normal process overpredicts the thermal conductivity – which is a physically intuitive result. We also observe that the thermal conductivity increases and trends towards an asymptotic value as the length scale increases. By analyzing the temperature distribution, we also show that inclusion of normal processes diffuses the energy across the material – which is an expected result. ## 1. Introduction With increasing miniaturization of integrated electronic circuits (ICs) following the Moore’s law [1] several challenges pop up in trying to keep up with the trend. One of the major bottlenecks is the high density localized heat generation in ICs that impair the device performance. In order to better understand the thermal behavior at those scales it is essential to create robust models that can accurately predict the device failure. As the device size in ICs these days are less than 32nm where the ballistic behavior dominates, Fourier law cannot be used to make accurate predictions. One of the widely used alternatives to use the Boltzmann Transport Equation (BTE) which models the phonon distribution – the major heat carrier in semiconductors – for a given macroscopic device conditions. Phonon is a quantum of thermal vibration which is considered as particle in BTE. Several researchers have developed simplified versions of BTE making intuitive assumptions to make it analytically or numerically solvable. Using these assumptions they were able to predict the thermal behavior of various materials. One of the most celebrated applications of BTE developed in early 1950s is to compute the thermal conductivity of a given material as a function of temperature, composition, and geometry, etc. [2], [3]. These analyses consider various phonon scattering mechanisms responsible for the thermal conductivity of a given material and models the effect of these scattering rates on the phonon distribution. Widely considered scattering mechanisms are isotopic scattering, boundary scattering, and three-phonon scattering (Umklapp (U) and Normal (N) scattering). Isotopic scattering is caused due to the fact that any given material, by nature, has various isotopic compositions in it. Boundary scattering is dominant at low temperatures and length scales where the phonons hit the physical boundary of the material thereby causing resistance to heat flow. Three-phonon scattering takes place when three phonons interact and results in frequency modification. This process is also called intrinsic or inelastic scattering which occurs due to the anharmonic nature of interatomic potential. These mechanisms explain why even perfectly pure crystals do not have infinite thermal conductivity (In most of the modeling procedures the interatomic potential is assumed harmonic which fail to show the three-phonon scattering). Over the last decade, with the improvement in computational power and numerical methods, many assumptions are relaxed and the BTE is solved more rigorously using entire phonon dispersion and relaxation time approximation [4]. Common methods to obtain relaxation rates include fitting the rate expressions to experimental values or perturbation theory [2], [5]. While the influence of all scattering terms on overall thermal conductivity at different temperatures and geometries has been well analyzed, the three-phonon N-processes have been neglected in most of the computations. The reason for neglecting normal processes is the premise that they conserve phonon momentum and hence do not offer thermal resistance. While the above fact is partially true, N-processes populate phonons in that region that can participate in U-processes thereby indirectly contributing to the overall thermal conductivity. Neglecting N-processes still provided reasonable comparison with experiment for materials like Si and Ge because these are 3D materials and the number of phonons that participate in U-processes are comparably higher than N-process phonons. But in case of lower dimensional (2D) materials like graphene, the population of phonons in large-wavevector region is very small and neglecting N-process provides false prediction of diverging thermal conductivity. This is because of the fact that the scarce phonon population makes them travel ballistically without many collisions. On the other hand, it was shown in [6] that on including N-processes the conductivity asymptotes to a constant value. In this paper, we simulate the thermal conductivity of Silicon by including N-processes with full BTE model using relaxation time approximation and full phonon dispersion. N-process scattering formulation developed by Callaway [2] is used by strictly enforcing momentum conservation for N-process phonons and energy conservation for N, and U process phonons. We only consider isotopic and three-phonon scattering mechanisms in this project. The simulation is performed on Silicon owing to easy availability of its dispersion curves and relaxation rates. Isotropic assumption is made in k-space which is reasonable for Si. In our simulation, we solve the energy form of non-gray BTE [7] simultaneously with overall energy conservation (N+U processes) to extract energy distribution and temperatures respectively in a coupled fashion. This provides quick convergence compared to sequential solution of these equations (non-gray BTE and energy conservation). Then we perform a detailed study of the effect of temperature, geometry, and mainly N-process scattering on the overall bulk thermal conductivity of Si. We organize the rest of the paper as follows. In Section 2, we explain the physics of Normal scattering processes and discuss the situations when their effect would be significant. In Section 3, we provide a literature review of different models and assumptions used to simulate and analyze the effect of N-processes. In Section 4, we use Callaway’s thermal conductivity model to make a first order prediction of thermal conductivity of silicon. Here we briefly discuss the assumptions made in the model and their implications. In Section 5, we discuss the numerical method and solution procedure we used for simulating the thermal conductivity. Here we provide a detailed description of equations involved and the formulation used. In Section 6, we present and discuss the results of our simulations. We conclude in Section 7 by providing a direction for possible future work. ## 2. Normal process scattering Three-phonon scattering A three phonon scattering process results in frequency modification of the resultant phonons. They are also called as inelastic scattering events. These processes can be described by the energy and momentum relations shown in Figure 1 [8]. As shown in the figure, these processes can be classified into Normal and Umklapp processes. A Normal process conserves energy and momentum whereas Umklapp process only conserves energy. Another illustration in Figure 2 shows why U-processes do not conserve momentum. The Brillouin zone of the given material is shown in gray. Incoming phonons of wave vector k1 and k2 combine to form a single phonon of wave vector k3. The left part of the figure shows N-process in which the resultant phonon lies inside the Brillouin zone; whereas the resultant phonon for U-process (figure on the right side) has such a high wave vector that it is knocked out of the Brillouin zone. By mapping it back into Brillouin zone using reciprocal lattice vector G, we can see that the resultant phonon of wave vector k3 is in the direction opposite to that of k1 and k2. This explains why U-processes impede phonon momentum and thereby the heat flow. On the other hand, as N-processes do not impede phonon momentum they do not impede the heat flow directly. But they contribute indirectly by redistributing overall phonon population which can further participate in U-processes. Figure 1: Three-phonon scattering events [8] Figure 2: Illustration of momentum conservation by N and U processes [Wikipedia] Importance of N-process As discussed earlier, given the nature of N-processes it would be worthwhile to think of situations when N-process contribution is indeed significant. First, we consider how low-frequency modes interact with high frequency-modes near the Brillouin zone boundary. Considering the selection rules (see Figure 1) of U-processes, only a mode of some minimum frequency ωi can participate in them [9]. This prohibits the interaction of low-frequency modes with that of Brillouin zone boundary. But intuitively we know that these modes should somehow contribute to thermal resistance. This can be explained by the premise that N-processes that involve these low-frequency modes generate the modes this minimum frequency ωi, which can then participate in U-processes, thereby providing thermal resistance. One of the other observations on importance of N-processes is discussed in [6] where it is shown that, with N-processes, graphene’s thermal conductivity diverges with increasing flake diameter thereby providing length dependence. But by including N-processes they showed that the conductivity asymptotes to a constant value. 3. Literature review of Normal processes In this section, we provide a brief literature review of the N-process analysis and their findings. We begin with Callaway’s phenomenological model for lattice thermal conductivity at low temperatures [2]. In this work, Callaway uses a relaxation time approximation for the scattering term of BTE, and assumes that all momentum destroying processes (isotopic, boundary, and Umklapp scattering) tend toward an equilibrium Planck distribution, whereas N processes lead to a displaced Planck distribution. Using this approximation, the scattering terms can be written as: ${{\left( \frac{\partial N}{\partial t} \right)}_{c}}=\frac{N(\lambda )-N}{{\tau }_{N}}+\frac{{N}_{0}-N}{{\tau }_{u}}$ – (1) where is the distribution function, is the relaxation time for all normal processes and is the relaxation time for all other momentum destroying processes, is the equilibrium Planck’s distribution and is displaced Planck’s distribution defined as $N(\lambda )={{\left[ \exp \left( \frac{\hbar \omega -\lambda .\mathsf{k}}{{k}_{B}T} \right) \right]}^{-1}}$ – (2) The term is along the direction of temperature gradient and defines the amount of energy redistributed by N-processes. He computed by enforcing momentum conservation for all N-process phonons using $\int{{\left( \frac{\partial N}{\partial t} \right)_{N}}\text{k}}{{d}^{3}}k=\int{\frac{N(\lambda )-N}{{\tau }_{N}}\text{k}{{d}^{3}}k=0}$ – (3) Using the assumptions that only acoustic phonons contribute to thermal conductivity all acoustic modes can be averaged using a single group velocity, and the relaxation rates for all processes can be expressed as a function of frequency and temperature he computed a simplified expression for thermal conductivity as $k=\frac{{k}_{B}}{2{\pi }^{2}c}({I}_{1}+\beta {I}_{2})$ ${{I}_{1}}=\int_{0}^{{{11k}_{B}}\Theta /\hbar }{{{\tau }_{c}}\frac{{{\hbar }^{2}}{{\omega }^{2}}}{{{k}_{B}}^{2}{{T}^{2}}}\frac{{{e}^{\hbar \omega /{{k}_{B}}T}}}{{{\left( {{e}^{\hbar \omega /{{k}_{B}}T}}-1 \right)}^{2}}}{{\omega }^{2}}d\omega }$ ${{I}_{2}}=\int_{0}^{{{k}_{B}}\Theta /\hbar }{\frac{{{\tau }_{c}}}{{{\tau }_{N}}}\frac{{{\hbar }^{2}}{{\omega }^{2}}}{{{k}_{B}}^{2}{{T}^{2}}}\frac{{{e}^{\hbar \omega /{{k}_{B}}T}}}{{{\left( {{e}^{\hbar \omega /{{k}_{B}}T}}-1 \right)}^{2}}}{{\omega }^{2}}d\omega }$ where is the combined relaxation time given by , is the frequency, is the temperature, and is the normalized Plancks constant. Using the above expression, he computed the thermal conductivity of germanium and compared it to various experimental observations to obtain fitting constants for relaxation rates. Once the constants are obtained, he found a striking match between theory and experiments. These constants are then used to examine the effects of individual processes. The second term in the thermal conductivity expression is commonly referred as correction term that accounts for correction due to N-processes. He computed that the value of this term is 10% of the overall conductivity in case of a pure germanium sample. This model for thermal conductivity has been rigorously used to compute thermal conductivities for various materials and found to provide good match at low temperatures [3], [9]. One of the papers that made improvements to Callaway’s model is Armstrong’s paper using two-fluid model [9]. In this work Armstrong divides phonons into two groups – propagating and reservoir modes. He assumes that both low and high-frequency phonons participate in N-processes. High-frequency phonons account for the fact that they can also participate in N-processes by splitting into phonons of lower frequencies. Instead of Callaway’s displaced Planck’s distribution, he used two displaced distribution functions for N-process phonons. This is because the high frequency phonons participating in N-process cannot equilibriate to very low value of distribution function that low-frequency phonons equilibriate to. He also considers the effect of different polarizations. Using this assumptions he solves the modified BTE called Boltzmann-Peierls equation for phonons written as $-\frac{{{N}_{q}}-N(\beta )}{{{\tau }_{NN}}}-\frac{{{N}_{q}}-{{N}_{0}}}{{{\tau }_{R}}}={{c}_{q}}.\nabla T\frac{d{{N}_{q}}}{dT}$ – (5) where N0 is the equilibrium distribution, Nq is the phonon occupation number, N(β) is the displaced distribution, cq is the group velocity of a given polarization, τNN and τR are the characteristic relaxation times,$\nabla T$ is the temperature gradient across the material. Recently, several rigorous computational simulations have been performed by relaxing many assumptions that were made to simplify the equation. For example, Yunfei Chen et al., [10] used Monte Carlo (MC) simulation to solve the BTE to compute the thermal conductivity of silicon nanowire. In this work, N-processes are included using genetic algorithm to generate the phonons that satisfy momentum conservation and tend towards displaced equilibrium distribution defined by Callaway. 4. Thermal conductivity of Silicon using Callaway’s analysis As a first approximation to our simulation, we start with applying Callaway’s model to compute the thermal conductivity of silicon. Both the terms of thermal conductivity expression are retained. We use the same relaxation rate expressions used by Callaway given as ${{\tau }_{u}}^{-1}=A{{\omega }^{4}}+{{B}_{1}}{{T}^{3}}{{\omega }^{2}}+c/LF$ – (6) where Aω4 represents isotopic scattering; B1T3ω2 includes the Umklapp processes with B1 containing the exponential temperature factor e − Θ / aT where Θ is the Debye temperature; and c / LF represents the boundary scattering with F being the correction factor due to both smoothness of the surface and the finite length to thickness ratio of the sample [3]. For N-processes we use ${{\tau }_{N}}^{-1}={{B}_{2}}{{T}^{3}}{{\omega }^{2}}$. The combined relaxation rate is then defined as ${{\tau }_{c}}^{-1}=A{{\omega }^{4}}+({{B}_{1}}+{{B}_{2}}){{T}^{3}}{{\omega }^{2}}+c/LF$ – (7) The above relaxation rates are used in the thermal conductivity expression to fit the constants A, (B1 + B2), and F. Though B1 implicitly has temperature dependence, we neglect that dependence while fitting. Using the experimental results shown in Table 1 [11], [12] we obtain the following values of fitting constants: A = .22e-44 sec3 B1+B2 = 2.9e-24 sec/deg3 F = .8 \| \| \| \| \| \|------\|-----------\|--------\|-----------\| \| T(K) \| k (W/cmK) \| \|T (K) \| K (W/cmK) \| \| 2 \| .44 \| 200 \| 2.66 \| \| 4 \| 3.11 \| 300 \| 1.56 \| \| 6 \| 8.99 \| 400 \| 1.05 \| \| 8 \| 16.4 \| 500 \| 0.80 \| \| 10 \| 24.0 \| 600 \| 0.64 \| \| 20 \| 47.7 \| 700 \| 0.52 \| \| 30 \| 44.2 \| 800 \| 0.43 \| \| 40 \| 36.6 \| 900 \| 0.36 \| \| 50 \| 28.0 \| 1000 \| 0.31 \| \| 100 \| 9.13 \| 1100 \| 0.28 \| \| 150 \| 4.10 \| \| \| Table1: Thermal conductivity measurements of Silicon [11], [12]. We compare the thermal conductivity obtained by substituting fitted constants into equation (1.4) and observe a good comparison with experiments at low temperatures, see Figure 3. The fitted equation is really helpful to quickly investigate the effects of individual scattering events. For example, we can fix all the constants and just vary the constant to examine the change of thermal conductivity with various isotope scattering events. The value of A for a given isotopic concentration can be found from Klemens expression [13] for isotopic scattering. This is how several researchers exploit this equation to better understand the thermal conductivity behavior of different materials. Figure 3: Callaway’s model for thermal conductivity. The constants of relaxation rates are fitted to match the model with experiment at low temperatures. While the above means of using Callaway’s expression would be extremely helpful, our focus in this project is to better understand the model itself. A small discussion on correction term ( ) and the justification for neglecting it might shed some light in this direction. As we discussed earlier, Callaway used a combined relaxation rate ${{\tau }_{c}}^{-1}={{\tau }_{N}}^{-1}+{{\tau }_{u}}^{-1}$ in his derivation. But as N-processes do not actually contribute to thermal resistance, the τc term under predicts the thermal conductivity. Most of the times the under prediction is very low and hence can be safely neglected. But in some cases, it might the prediction might differ by around 20%. An example is when isotopic scattering is absent. In this case, the predicted thermal conductivity with and without the correction term is more than 20% of its overall value. This can be qualitatively seen in Figure 4. (Note that the y-axis here is logarithmic). So it can be explained that the correction term compensates for the fact that using τc provides more resistance to the conductivity. Figure 4: Meaning of correction term in Callaway’s model for thermal conductivity. The correction term compensates the fact that the overall relaxation rate used in Callaway’s model underpredicts the thermal conductivity 5. Numerical Method All of the governing equations are solved using the Finite Volume method [14]. The BTE is a partial differential equation involving two vector spaces (physical space and wave vector space). We therefore must discretize both spaces accordingly. With this, we will arrive at a linear system with at least one equation per discretized physical space and per discretized wave vector space. The physical domain is discretized into several arbitrary convex polyhedral. Here we have chosen a square domain which can easily be discretized using a non-uniform structured grid, shown below: Figure 5: 80×80 grid used as the physical domain. The side length is unity and is scaled to the appropriate domain size. To discretize the BTE, we integrate over a finite control volume in physical space and wave vector space then apply the divergence theorem to the convective operator as shown below: $\int\limits_{\Delta \mathsf{K}}{\int\limits_{\Delta V}{\nabla \cdot \left( \mathsf{v}e'' \right)dV{{d}^{3}}\mathsf{K}}}=\int\limits_{\Delta \mathsf{K}}{\int\limits_{\Delta \mathsf{A}}{e''\mathsf{v}\cdot d\mathsf{A}{{d}^{3}}\mathsf{K}=}}\int\limits_{\Delta \mathsf{K}}{\int\limits_{\Delta V}{\frac{{{e}^{0}}-e''}{{{\tau }_{U}}}+\frac{e_{\mathsf{\lambda }}^{0}-e''}{{{\tau }_{N}}}dV{{d}^{3}}\mathsf{K}}}$ – (8) We now apply our discretization. In the wave vector space, a central difference approximation is made on both sides of the equation, thus the finite volume of k-space appears on both sides and is dropped. We arrive at the following discrete equation set: $\sum\limits_{f}{{{e}_{f}}''}v\cdot \Delta {{\mathsf{A}}_{f}}+e''\Delta V\left( \tau _{N}^{-1}+\tau _{U}^{-1} \right)-{{e}^{0}}\frac{\Delta V}{{{\tau }_{U}}}-e_{\mathsf{\lambda }}^{0}\frac{\Delta V}{{{\tau }_{N}}}=0$ – (9) The energy conservation equation can be rearranged as follows: $\int{\frac{{{e}^{0}}}{{{\tau }_{U}}}{{d}^{3}}\mathsf{K}=}\int{\frac{e''}{{{\tau }_{U}}}{{d}^{3}}\mathsf{K}}-\int{\frac{e_{\mathsf{\lambda }}^{0}-e''}{{{\tau }_{N}}}{{d}^{3}}\mathsf{K}}$ – (10) When we apply a second order finite volume discretization we arrive at the following: $\sum{{{e}^{0}}\frac{\Delta \mathsf{K}}{{{\tau }_{U}}}}=\sum{e''\left( \tau _{N}^{-1}+\tau _{U}^{-1} \right)\Delta \mathsf{K}}-\sum{e_{\mathsf{\lambda }}^{0}\frac{\Delta \mathsf{K}}{{{\tau }_{U}}}}$ – (11) Because of the tight inter-equation coupling, caused by the in-scattering term of the BTE, it is advantageous to visit each physical cell and solve all points in wave vector space and the energy conservation equation in a coupled fashion. This procedure is very similar to that shown in [15], thus we have adopted the same name, the Coupled Ordinates METhod (COMET). In COMET, the success hinges on the ability to solve all the BTE’s and the energy conservation equation. To do this, we must determine the common variable which couples all said equations. The lattice temperature shows up in the function for the equilibrium distribution function, albeit nonlinear. To extract the lattice temperature equation, we simply linearize the equilibrium function using a Taylor series expansion: ${{e}^{0,new}}={{e}^{0,old}}+\left( {{T}^{new}}-{{T}^{old}} \right){{\left( \frac{\partial {{e}^{0}}}{\partial T} \right)}^{old}}$ – (12) The derivative used is a very familiar value, the specific heat of the specific phonon frequency at the previous iterations temperature. For this procedure, we will construct a matrix which solves for the correction to the previous values of e'' and T. In this way, the residual of the previous iteration act as a source for the correction equation. With this, the BTE and the energy conservation equations become: $\begin{align} & \sum\limits_{f}{\Delta {{e}_{f}}''}\mathsf{v}\cdot \Delta {{\mathsf{A}}_{f}}+\Delta e''\Delta V\left( \tau _{N}^{-1}+\tau _{U}^{-1} \right)-\Delta T\frac{\Delta V}{{{\tau }_{U}}}{{\left( \frac{\partial {{e}^{0}}}{\partial T} \right)}^{old}}={{R}_{BTE}} \\ & -\Upsilon \sum{\Delta e''}\left( \tau _{N}^{-1}+\tau _{U}^{-1} \right)\Delta \mathsf{K}+\Delta T={{R}_{energy}} \\ & {{\Upsilon }^{-1}}={{\sum{\left( \frac{\partial {{e}^{0}}}{\partial T} \right)}}^{old}}\frac{\Delta \mathsf{K}}{{{\tau }_{U}}} \\ \end{align}$ – (13) where Δe is enew − eold with old and new representing the current and previous iteration values of simulation. RBTE is the residual for the BTE while Renergy is the residual for the energy equation. These equations form a linear system of order NK + 1, where NK is the number if points in wave vector space. The shape of the matrix is especially convenient, forming an arrowhead pointing to the lower right (all diagonals populated, last row populated, last column populated) which can be solved directly in O(N) operations. ### Legal nanoHUB.org, a resource for nanoscience and nanotechnology, is supported by the National Science Foundation and other funding agencies.	{"extraction_info": {"found_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": 18, "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": 0.9014769196510315, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics/40371	## Why is the Gaussian so pervasive in mathematics? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This is a heuristic question that I think was once asked by Serge Lang. The gaussian: $e^{-x^2}$ appears as the fixed point to the Fourier transform, in the punchline to the central limit theorem, as the solution to the heat equation, in a very nice proof of the Atiyah-Singer index theorem etc. Is this an artifact of the techniques (such as the Fourier Transform) that people like use to deal with certain problems or is this the tip of some deeper platonic iceberg? - 2 I think the appearances you mention can be sensibly grouped in two broad classes: Fourier-analytic (including the CLT, see e.g. terrytao.wordpress.com/2010/01/05/…) and heat-kernelish (incl. Atiyah-Singer). Perhaps microlocal analysis is the best bridge between these. – Steve Huntsman Sep 28 2010 at 5:35 8 I just want to point out that the appearance of the Gaussian in Atiyah-Singer isn't so mysterious. In the McKean-Singer formula - $Index(D) = Tr_s (e^{-tD^2})$ - the Gaussian can be replaced with any smooth function which rapidly decays at infinity. The Gaussian is used because $e^{-tD^2}$ is the solution operator for a very well understood differential equation, namely the heat equation, and thus we can look up the relevant asymptotic analysis in old PDE textbooks. – Paul Siegel Sep 28 2010 at 12:01 7 There is a book "The Normal Distribution: Characterizations with Applications" by Bryc which could be taken as a large collection of probabilistic examples to add to your list. One of my favorites: the normal distribution in $\mathbb{R}^n$ is the unique (up to scaling) rotation-invariant probability measure with independent components. – Mark Meckes Sep 28 2010 at 12:11 4 The central limit theorem and the heat equation make no mention of the gaussian in their basic setups; it turns out to be the answer to very natural questions. So I think those two (on the surface completely unrelated) examples show that this phenomenon is definitely not just an artifact of techniques. In particular, there are other methods of proving the CLT or solving the heat equation besides Fourier transforms. – Mark Meckes Sep 28 2010 at 13:01 3 The factoring that appears in that volume calculation happens precisely because the standard normal distribution has independent components; the reason you can relate the normal distribution to the sphere is its rotation invariance. So that use of the Gaussian is an application of my favorite property which I mentioned above. As for the CLT and the heat equation, there certainly are deep relationships, but I don't know about "obvious" or "less heavy handed". – Mark Meckes Sep 28 2010 at 19:34 show 5 more comments ## 6 Answers Quadratic (or bilinear) forms appear naturally throughout mathematics, for instance via inner product structures, or via dualisation of a linear transformation, or via Taylor expansion around the linearisation of a nonlinear operator. The Laplace-Beltrami operator and similar second-order operators can be viewed as differential quadratic forms, for instance. A Gaussian is basically the multiplicative or exponentiated version of a quadratic form, so it is quite natural that it comes up in multiplicative contexts, especially on spaces (such as Euclidean space) in which a natural bilinear or quadratic structure is already present. Perhaps the one minor miracle, though, is that the Fourier transform of a Gaussian is again a Gaussian, although once one realises that the Fourier kernel is also an exponentiated bilinear form, this is not so surprising. But it does amplify the previous paragraph: thanks to Fourier duality, Gaussians not only come up in the context of spatial multiplication, but also frequency multiplication (e.g. convolutions, and hence CLT, or heat kernels). One can also take an adelic viewpoint. When studying non-archimedean fields such as the p-adics $Q_p$, compact subgroups such as $Z_p$ play a pivotal role. On the reals, it seems the natural analogue of these compact subgroups are the Gaussians (cf. Tate's thesis). One can sort of justify the existence and central role of Gaussians on the grounds that the real number system "needs" something like the compact subgroups that its non-archimedean siblings enjoy, though this doesn't fully explain why Gaussians would then be exponentiated quadratic in nature. - ### 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. I recently came across a strange and beautiful connection between the Gaussian $e^{-x^2}$ and the method of least squares. It turns out that the square in $e^{-x^2}$ and the square in least squares'' is the same square. Let $(x_i,y_i)$ (with $1\leq i \leq n$) be the data set, and assume that for each $x$, the $y$'s are normally distributed with mean $\mu(x)=\alpha x+\beta$ and variance $\sigma^2$. Then, the likelihood of generating our data (assuming that the data points are independent) is $$\prod_{i=1}^n \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{(y_i-\mu(x_i))^2}{2\sigma^2}\right) =\left(\frac{1}{\sqrt{2\pi \sigma^2}}\right)^n \exp\left( \frac{-1}{2\sigma^2} \sum_{i=1}^n (y_i - \alpha x+\beta)^2 \right)$$ We would obviously want to choose the parameters $\alpha,\beta$ so that the likelihood is maximized, and this is accomplished by minimizing $$\sum_{i=1}^n (y_i - \alpha x+\beta)^2.$$ In other words, the least squares approximation is the one that makes the data set most likely to happen. - The analogous thing happens for the Laplace distribution and absolute loss, etc. There is certainly a relationship between loss criteria and distributions under which that loss minimizer is the MLE, but I'm not sure it goes any deeper. – R Hahn Sep 28 2010 at 19:35 This is yet another manifestation of my favorite property, mentioned in the comments... – Mark Meckes Sep 28 2010 at 19:36 @R Hahn: See my comment below the question, (2nd from top). The relation does indeed go deeper. – S. Sra Sep 29 2010 at 13:30 @Suvrit Yeah, the geometry of the exponential family is quite natural and on the real line within this family the Gaussian is also very natural (symmetric, etc), the same way Euclidean spaces are. But there are also lots of other interesting spaces :-) – R Hahn Sep 29 2010 at 16:13 Isn't this connection the reason why the Gaussian is called Gaussian? If I remember correctly, Gauss introduced "least squares" as a regression technique not because its derivative gives rise to a linear problem but because the square fits the "normal error model", i.e. corresponds to the Gaussian distribution. – Dirk Nov 14 at 12:51 This is just a minor amplification of one of Terry Tao's points. For any prime $p$, the ring `$\mathbb{Z}_p$` of $p$-adic integers forms an open compact additive subgroup of `$\mathbb{Q}_p$`, the completion of $\mathbb{Q}$ under the p-adic metric, and its characteristic function should be viewed as a p-adic analogue of the Gaussian. It displays many analogues of the nice properties we see from the Gaussian: 1. It is smooth (in the sense that the smooth functions on totally disconnected spaces are defined as the locally constant functions, but this isn't completely tautological, since this class of functions turns out to be useful). 2. It is taken to itself under the $p$-adic Fourier transform when normalizations are chosen appropriately. 3. It obeys something like a central limit theorem. For example, if you flip lots of coins, and ask for the number of heads mod $p^n$, you will, for sufficiently long trials, get a distribution that is close to uniform. It sounds like there could be a way to interpret this sort of convolution in terms of heat flowing, but I don't know a precise statement. The situation with the real line is more complicated because it is connected but not compact, and therefore has no open compact subgroups. There is a maximal compact multiplicative monoid (occasionally called the "ring of integers of $\mathbb{R}$" informally), given by the closed interval $[-1,1]$. You can think of the Gaussian as a smoothing of the uniform distribution on $[-1,1]$, but it is not clear to me that this particular analogy is very fruitful. - I'm not an expert, but I believe Stein's method gives a more satisfying connection between the CLT and the heat equation, in particular one that does not involve the Fourier transform. Stein's characterization of the normal distribution and convergence to normality involves an operator closely related to the generator of the Ornstein-Uhlenbeck process. On the other hand, the fact that latter has the Gaussian as its invariant measure can be obtained by a trivial transformation of the fundamental solution of the heat equation. - The sum of two independent normal random variables is again normal, i.e., the shape of the distribution is unchanged under addition except for stretching and scaling. Moreover, the normal distribution is unique among distributions with finite variance in having this property. Many phenomena in nature come from adding together various independent or almost independent terms. Therefore, we would expect the normal distribution to show up a lot in nature-inspired mathematics. - 10 The normal distribution is not unique in this property. The same is true of all stable distributions. – Mark Meckes Sep 28 2010 at 12:06 1 @Mark: Thanks, I have added the assumption of finite variance. – Bjørn Kjos-Hanssen Sep 28 2010 at 17:12 One of the reasons for the ubiquity of the Gaussian is displayed in what is probably the most electrifying half page of scientific prose ever written---Maxwell's argument that the distribution of the velocities of molecules in the ideal gas is Gaussian (now known as the Maxwell-Boltzmann distribution). The only physical assumptions used are that the density function depends only on the absolute value of the velocity (and not the direction) and that the components in the directions of the coordinate axes are statistically independent. Mathematically, ths means that the only functions in $3$-space which depend only on the distance $r$ from the origin and which split as the product of three functions of one variable are those of the form $ae^{br^2}$. Maxwell does this by inspection but it is easy to give a rigorous proof (under very weak smoothness conditions) and the result holds, of course, in any dimension greater then or equal to $2$. Maxwell's reasoning can be found in his collected papers, or, more accesssibly, in Hawking's anthology "On the Shoulders of Giants". - And this is yet another manifestation of the property I mentioned in the comments on the OP... – Mark Meckes Nov 14 at 14:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9365675449371338, "perplexity_flag": "head"}
http://mathoverflow.net/questions/107013/birth-death-process-associated-with-orthogonal-polynomials	## Birth-Death Process associated with Orthogonal Polynomials ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have read in various places the following objects are related: • orthogonal polynomials • birth-death processes • Lattice paths • continued fractions After a lot of searching online, I found sketches about the various relationships • the moments of orthogonal polynomials count weighted lattice paths • the Kolmogoroff forward/backward equations of birth-death process can be solved using orthogonal polynomials of a given weight. However the papers have been hard to decipher, such as this one by Karlin and McGregor, or the first few pages in this paper by Fabrice Gullemin. The birth-death process on $$1 \rightleftharpoons 2 \rightleftharpoons 3 \rightleftharpoons 4 \rightleftharpoons \dots$$ where $\mathrm{Rate}(n \to n+1)=\lambda_n, \mathrm{Rate}(n+1 \to n)=\mu_n, \mathrm{Rate}(n \circlearrowleft)=-(\mu_n + \lambda_n)$ The Karlin and McGregor showed you can find orthogonal polynomials $Q_n(x)$ with respect to a measure $\mu(dx)$ that solve the transition probabilities (the odds of starting at m and winding up at n at time t) $$\mathbb{P}(\Lambda(t) = n \| \Lambda(0) = m) = \pi_n \int_0^\infty e^{-tx} Q_n(x)Q_m(x) \mu(dx)$$ They even figured out a way to construct the polynomials $$Q_{-1}(x)=0, Q_0(x)=1, \lambda_n Q_{n+1}(x) +(x - \lambda_n - \mu_n) Q_n(x) + \mu_n Q_{n-1}(x) = 0$$ And then one notices the 3-term relation is the "magic box" recurrence formula for continued fractions. I first read about these relationships in regards to the Rogers-Ramanujan continued fraction. $r(q) = \cfrac{1}{1 + \cfrac{q}{1 + \cfrac{q^2}{1 + \dots}}} \quad\quad\quad\quad = 1 - q + q^2 - q^4 + \dots$ However the birth and death process is rather complicated. What is the birth-death process associated to the Hermite polynomials or the Legendre polynomials? or maybe in general What do we reconstruct the birth-death processes correspond to the classical orthogonal polynomials? - 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": 4, "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": 0.9191819429397583, "perplexity_flag": "middle"}
http://math.stackexchange.com/users/5191/zhen-lin?tab=activity&sort=all&page=151	# Zhen Lin reputation 12265 bio website location age member for 2 years, 4 months seen 9 hours ago profile views 4,519 bio visits 20,701 reputation website member for 2 years, 4 months 12265 badges location seen 9 hours ago # 3,494 Actions Aug3 comment subset relations among Sobolev spaces and their duals@user14178: A bijection is not the same thing as equality. The set of all even numbers is in bijection with the set of all numbers, but is also a proper subset. (This is one of the possible definitions for an infinite set.) Aug3 revised Proof of non-existence of a certain metric on $S^2$deleted 100 characters in body Aug3 comment Proof of non-existence of a certain metric on $S^2$@Jacob: Good point. I was thinking of a snappier / more general form of the argument the OP used when I wrote that. Aug3 comment What are the rules for basic algebra when modulo real numbers are involvedYou can't do modular arithmetic with real numbers. (Actually, you can't even do modular arithmetic with rational numbers.) All you're doing is taking the remainder after some division operation. Aug3 answered Proof of non-existence of a certain metric on $S^2$ Aug3 comment Did Zariski really define the Zariski topology on the prime spectrum of a ring?@Pierre: It seems to me that you are collecting references and examples of (mis)attribution of concepts in mathematics. May I ask what it is for? If it's for an upcoming article I would certainly like to read it. Aug3 comment Isometric actions@Chu: It might be useful if you explain why you want such an example? Aug3 comment Sheaf cohomology: what is it and where can I learn it?@Akhil: Thanks for the reference. To think that one textbook would go from simple groups, rings, and modules all the way up to homological algebra...! I will have a look at it, and Iversen's book, sometime when I can obtain it. Aug3 comment A question in complex analysis about analytic functionIt's a simple continuity argument. Can you see that $f$ cannot vanish on the domain? Aug3 comment Why bother with Mathematics, if Gödel's Incompleteness Theorem is true?@Adrian, re your edit: One does not need to think about other universes having different mathematical axioms. Right here on Earth there are mathematicians who study the possibility of mathematics under different axiom systems. Perhaps if history turned out differently we would have the axiom of determinacy instead of the axiom of choice. Aug3 comment Reference request: group theoryI remember reading it and not learning anything about the representation theory of Lie algebras, which is a deficiency in my mind given its use in modern theoretical physics. (And I still don't know anything about it...) Could you suggest an introductory text which covers representations? Aug3 comment countably infinite union of countably infinite sets is countable@gary: As Asaf says, in the absence of the axiom of choice, it is possible for a countable union of countable sets to be uncountable. For example, there is a model of ZF where $\mathbb{R}$ is a countable union of countable sets. Aug2 comment Sheaf cohomology: what is it and where can I learn it?(A reference for the claim that soft sheaves are acyclic would also be much appreciated.) Aug2 comment Sheaf cohomology: what is it and where can I learn it?@Akhil: I've just had a chance to think about your comments carefully. I can see that the fact that the cohomology of the constant sheaf agrees with de Rham cohomology follows from the fact that acyclic resolutions give the same answer as injective resolutions. (I was at first puzzled about how the de Rham complex could even be exact, let alone be an acyclic resolution, but then I realised I needed to look at stalks rather than sections.) What's still mysterious to me is why acyclic resolutions should give the same answer. Could you provide a reference or an informal explanation? Aug2 comment Cofusing partial order “implies”, on logic and that on setsYou seem to be confusing the fact that $\implies$ (or, to be more precise, $\vdash$) is a preorder on the set of propositions with the idea of reasoning about partially ordered sets. Aug2 comment understanding of a proof of the invariance of 3-D laplacian?@Jack: It's the bit starting at ‘substituting the previous equation in’. It is literally just the chain rule. Aug1 answered understanding of a proof of the invariance of 3-D laplacian? Aug1 comment understanding of a proof of the invariance of 3-D laplacian?The first equality is the chain rule. (If this is not obvious, perhaps the notation needs to be improved.) Aug1 comment Why bother with Mathematics, if Gödel's Incompleteness Theorem is true?A not-so-well-kept secret of mathematics is that mathematicians are not that bothered about rigour. Mathematicians have been working with things not well-defined since the dawn of time and will continue to do so. What changes is the standard of ‘well-definedness’, and from time to time there will be a crisis over the lack of foundations. People have been doing arithmetic for aeons before Peano wrote down his axioms, and for the most part, no one worried about it. Perhaps a better question to ask is, what is mathematics? Aug1 answered Injective Holomorphic Functions that are not Conformal?	{"extraction_info": {"found_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": 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": 0.9491919279098511, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/22207/ropes-and-pulleys-really-unintuitive-answer?answertab=active	# Ropes and Pulleys - Really unintuitive answer I usually don't want to do this, but please go to this link, the solution is too big to post it here http://engineering.union.edu/~curreyj/MER-201_files/Exam2_2_26_09_Solution.pdf And go to page 5 of the pdf. Briefly, the problem say Determine the velocity of the 60-lb block A if the two blocks are released from rest and the 40-lb block B moves 2 ft up the incline. The coefficient of kinetic friction between both blocks and the inclined planes is $\mu_k$ = 0.10. Things I am confused with the solution 1.First of all, I seriously thought lb was mass not force. after some googling, it turns out they are used interchangeably... 2.Where did they even get $2s_a + s_b = 0$ from? Why did they determine the change in distance this way? My first assumption was that if block B moved up 2ft, then block A should move down 2ft (the rope must "move" 2ft too right?). Then I wasn't sure, so I did a few triangles and found that the angle made a difference 3.Where did $2v_A = -v_B$ come from? 4 The FBD for block A is confusing, why is the friction force $F_A$ in the direction of the ropes? I thought it was block B that is going down? Am I the only one who had trouble deducting that the pulley and block A are the same object? 5.Look at the final answer, how could $v_b$ be negative? The problem says block B goes UP. If you are wondering, this is not homework. I am just interested in this problem and it is out of curiosity and very confused with the concepts. I've had at most introductory physics experience, but I think I should have been able to solve this still. Thank you for reading - ## 2 Answers "2.Where did they even get 2sa+sb=0 from?" From the assumption that the length of the rope does not change. "Why did they determine the change in distance this way? My first assumption was that if block B moved up 2ft, then block A should move down 2ft (the rope must "move" 2ft too right?)." No, there is a difference between a movable pulley and a fixed pulley, so A moves down 1 ft. "3.Where did 2vA=−vB come from?" From the same assumption as above. However, the direction of vA is shown incorrectly (or, alternatively, the signs in the formula are wrong). "4 The FBD for block A is confusing, why is the friction force FA in the direction of the ropes? I thought it was block B that is going down?" It is writen in the statement of the problem that block B went up. " Am I the only one who had trouble deducting that the pulley and block A are the same object?" I don't quite understand what this phrase means exactly and how it is relevant. "5.Look at the final answer, how could vb be negative? The problem says block B goes UP." See the answer to your item 3. - @Akhmetli, for 3) (and 5), you mean in the key right? That's what I am confused too. Forget 4), block A is going down, so it make sense for the friction to be in the direction of the ropes – jak Mar 11 '12 at 4:18 @jak: I mean in FBD for block A - I guess this is the only place where they show the direction of vA. – akhmeteli Mar 11 '12 at 4:34 Just one last thing, so the numerical answers is right, but just the signs are wrong? Sorry for being repetitive – jak Mar 11 '12 at 20:53 @jak: Sorry, don't have time to check the solution. – akhmeteli Mar 11 '12 at 21:32 Oh it's fine. Is it possible to solve this without knowing what is "mechanical advantage" – jak Mar 13 '12 at 1:29 show 1 more comment 2 - the block 'A' is on a pulley, when it moves 1 ft the rope moves 2ft. It has a 2:1 mechanical advantage, normally you would use this to move block 'A' half the distance you pull the rope and with twice the force - the point of a pulley is to use a smaller force for a longer distance. - sorry, I'll look up this "mechanical advantage" stuff on my own. Don't know why I expected you to explain everything to me. – jak Mar 11 '12 at 4:27 Do you mind if I bug you once more? Without knowing "mechanical advantage" could I have solved this? – jak Mar 13 '12 at 1:28 @jak - you don't have to know the word, but you do have to knowhow pulleyswork – Martin Beckett Mar 13 '12 at 2:58 I never learn the concept, but is it just something you could see "oh B moves 2ft up, so A must go down 2ft!". I have difficulties wrapping my heads around this – jak Mar 14 '12 at 18:52 – Martin Beckett Mar 14 '12 at 18:55 show 1 more comment	{"extraction_info": {"found_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": 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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9557629227638245, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/98516/partitioning-the-vertices-of-an-n-cube-with-random-hyperplane-cuts/98528	## Partitioning the vertices of an n-cube with random hyperplane cuts ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) An evolutionary biologist asked me a question which boils down, at least in part, to what seems to me an interesting question of combinatorial/probabilistic geometry. It is an old chestnut of a problem to ask: into how many pieces can an n-sphere be cut by k hyperplanes? (Here I want the hyperplanes to be honest linear subspaces, not affine ones as in the classical "lazy caterer" problem, but the flavor is much the same.) Now suppose that instead of the sphere, I have the 2^n vertices of the n-cube, i.e. the set {-1,1}^n. I cut this set with k random hyperplanes. Now I have a partition of 2^n. What do I expect this partition to look like? E.G. how many blocks are there? How big is the largest blocks? Are the biggest blocks "close to each other" in the sense that you can pass from one to the other without crossing very many of the hyperplanes? (To formalize this, one might say that the structure one is studying isn't just a partition, but a partition in which each block is identified with an element of (Z/2Z)^k, thus providing a notion of Hamming distance between block.) I have asked this question in a rather vague way by not specifying what range of k relative to n is in play. This should actually be considered part of the question: what are the threshold curves in the (n,k) plane, if any, where the partition sharply changes its expected nature? My biologist friend is certainly interested mostly in the case k > n; I think he's most interested in the case where k is bounded between n and a constant multiple of n, but I'm not sure. I expect he would be interested to know, for instance, how big k needs to be before all blocks of the partition are singletons almost surely. Further remarks: though I don't think this is relevant to MBF, one could certainly pass from a discrete to a continuous setting and ask about the statistics of the partition of the volume of the unit (n-1)-sphere by the k hyperplane cuts, which would also be interesting. Or, instead of letting the cuts be chosen randomly from a continuous distribution, you could let them be chosen from the vertices of a cube in the dual R^n; in other words, you could choose at random from hyperplanes of the form x_1 +- x_2 +- ... +-x_n. This last version is probably closest to what MBF is actually thinking about. Update: A couple of people asked about the biological context. Here's the original paper. http://www.ploscompbiol.org/article/info%3Adoi%2F10.1371%2Fjournal.pcbi.1000202 - Just to make sure, your notion of hyperplane means that the partition has a certain symmetry. In particular, the number of pieces with t points in the piece is always even. Is that what you want? Gerhard "Ask Me About System Design" Paseman, 2012.05.31 – Gerhard Paseman May 31 at 20:52 1 Yeah, that seemed the cleanest way to set it up. Fans of odd numbers are welcome to divide everything by two. – JSE May 31 at 21:13 For those of us with broader interests, can you (edit the post to or give a couple of comments to) provide a brief summary of the original question? (Some of us might want to try a different reformulation.) Gerhard "Acquiring Minds Want To Know" Paseman, 2012.06.01 – Gerhard Paseman Jun 1 at 15:42 1 I'd like to know the connection with evolutionary biology. There is a natural question of how many crossovers it takes to separate genes which are physically close, which might correspond to a nonuniform distribution of hyperplanes which is not symmetric in the coordinates. – Douglas Zare Jun 1 at 17:42 I've added a link to the relevant biology paper in the post. – JSE Jun 1 at 23:16 ## 1 Answer I corrected an earlier version which assumed that only adjacent vertices need to be separated to guarantee the parts are singletons, which fedja pointed out was incorrect. Thanks. Here is a partial answer to the question of how many cuts it takes before the nonempty pieces are singletons with high probability when the hyperplanes are chosen uniformly. Consider the roughly $4^n$ pairs of vertices of the cube. Each pair is separated by a hyperplane iff the partition separates all vertices. The expected number of pairs which intersect no hyperplane is at most $4^n$ times the probability that a particular edge is missed, since adjacent vertices are the least likely to be separated. The probability that none of the $k$ hyperplanes intersects an edge is $p^k$, where $p$ is the probability that each hyperplane misses the edge. What does it take for a hyperplane to separate two adjacent vertices? These points determine a great circle, and are at angle $\arccos \frac{n-2}{n} \approx \frac{2}{\sqrt n}$. The hyperplane almost surely intersects this circle in two antipodal points. If these intersect the arc of about $\frac{2}{\sqrt n}$ radians, then the hyperplane intersects the edge. So, the probability that a random hyperplane intersects an edge is about $\frac{2}{\pi\sqrt n}$. The probability the edge is missed is the complement, about $1- \frac{2}{\pi\sqrt n}$. The probability all $k$ hyperplanes miss this edge is about $(1- \frac{2}{\pi\sqrt n})^k \approx \exp(-\frac{2k}{\pi\sqrt n})$. The expected number of pairs not separated by any hyperplane is at most about $4^n \exp(-\frac{2k}{\pi\sqrt n})$. If you choose $k\approx c n^{3/2}$ then the expected number of pairs not separated by any hyperplane is at most $1$. For much larger $k$ the expected number of pairs of vertices in the same part, hence the probability that two vertices are in the same part, becomes small. - 2 Each edge intersects a hyperplane iff the partition separates all vertices. Really? I'm not that sure. Take a few planes in $R^3$ parallel to the vector $1,1,0$. You can cross any edge you want with such a hyperplane. However it is quite hard to separate $(1,1,1)$ and $(-1,-1,1)$ by any such plane... – fedja Jun 1 at 1:47 Hmm, good point. – Douglas Zare Jun 1 at 1:53 3 Somewhat annoyingly, it seems like the usual second moment method to get a matching lower bound may not quite work here, or at least not easily. The trouble is that if a hyperplane intersects a given edge, it is significantly more likely to intersect each other edge in the same direction (roughly speaking, if the edges are Hamming distance $d \approx n/2$ apart, hitting the first edge increases the probability the second edge is hit by a factor of $2$). This ends up making the probability that both edges are missed by all $n^{3/2}$ edges much larger than the square of the prob. one is missed – Kevin P. Costello Jun 1 at 21:34	{"extraction_info": {"found_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": 0.9413127899169922, "perplexity_flag": "head"}
http://mathoverflow.net/questions/47079?sort=oldest	## Line bundles in abelian $\otimes$-categories ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) By an abelian $\otimes$-category I mean a symmetric monoidal category $(\mathcal{A},\otimes,\mathcal{O})$, such that $\mathcal{A}$ also is an abelian category and for every $M \in \mathcal{A}$ the functor $M \otimes -$ is cocontinuous (i.e. right exact and preserves coproducts; in particular additive). A line bundle is defined as an object $\mathcal{L}$ of $\mathcal{A}$ such that there is some object $\mathcal{L}'$ such that $\mathcal{L} \otimes \mathcal{L}' \cong \mathcal{O}$. An example is the category of (quasi-coherent) modules on a locally ringed space. The line bundles then coincide with the modules which are locally free of rank $1$ (see here). Now I want to show that in general these line bundles have similar properties as in the case of the module category. For example it's not hard to show that if $\mathcal{L}$ is a line bundle, then it is flat in the sense that $\mathcal{L} \otimes -$ is exact (it is even an automorphism of $\mathcal{A}$ with inverse $\mathcal{L}^{-1} \otimes -$). The isomorphism classes of line bundles yield a group, which may be denoted as $\text{Pic}(\mathcal{A})$. Question 1. Is there any literature about these abelian $\otimes$-categories which treats them systematically? Perhaps the "usual" definition differs a little from mine, this does not matter. Question 2. Let $\mathcal{L}$ be a line bundle and $\phi : \mathcal{L} \to \mathcal{L}$ an epimorphism. Does it follow that $\phi$ is an isomorphism? Question 3. Let $\mathcal{L}$ be a line bundle and assume $\phi : \mathcal{L} \to \mathcal{L}$ is an epimorphism. Does it follow that there is an epimorphism $\psi : \mathcal{L}^{-1} \to \mathcal{L}^{-1}$ such that $\phi \otimes \psi$ corresponds to the identity of $\mathcal{O}$ under the isomorphism $\mathcal{L} \otimes \mathcal{L}^{-1} \cong \mathcal{O}$? Question 4. Assume we also have a $\lambda$-structure on $\mathcal{A}$ which is compatible with the given data. Is it possible to give a reasonable definition of a locally free object of rank $n$? See also this question. - 3 For questions 1, 2, and 3, it does not seem necessary to require that the monoidal product be symmetric. – André Henriques Nov 23 2010 at 19:41 What about doing algebraic geometry over $\mathcal{A}$ in the sense of Toën and Vaquié arXiv:math/0509684 (published in J. K-theory 3 (2009), 437-500)? This way, you will be able to speak of locally free objects as usual (this might answer questions 1 and 5). Also, the comment of Theo Johnson-Freyd below suggests that the answer to Question 2 is yes: apply his argument to the smallest tensor abelian category generated by $\mathcal{L}$ (which is small). – Denis-Charles Cisinski Nov 27 2010 at 23:36 Could you elaborate what the paper has to do with my question? I don't want to enrich my data with topologies, it should stay homological. Otherwise it's just a trivial translation. Also I'm not sure if the Theorem cited by Theo applies here, does it really give a full faithful exact tensor functor? – Martin Brandenburg Nov 28 2010 at 15:14 1 The point of Toën and Vaquié is precisely that the monoidal structure itself defines canonically a Zariski topology, as well as a faithully flat topology (so, without any extra data, if you think of your $\mathcal{A}$ as an abstract category of modules over a commutative ring, you will have a lot of the tools you are used to with usual commutative algebra). This idea of doing algebraic geometry over an abelian tensor category was also used by Deligne to study tannakian categories; see the Grothendieck Festschrift vol II, p. 111-195. With Toën and Vaquié, you don't need the abelian structure. – Denis-Charles Cisinski Nov 28 2010 at 22:38 Q2: No, take Z[x]-modules on which x acts injective. Q3: No, since this is Q2 by Steve's answer. Q4: Yes, this will appear in my thesis ... – Martin Brandenburg Feb 1 at 13:19 ## 1 Answer Question 2: this does not really depend on line bundles. If $\phi:\mathcal {L\to L}$ is a non-invertible epimorphism, then $\phi\otimes 1:\mathcal{ L\otimes L'\to L\otimes L'}$ is epi, since $-\otimes\mathcal L'$ is exact, and non-invertible, since otherwise $\phi\otimes 1\otimes 1:\mathcal{L\otimes L'\otimes L\to L\otimes L'\otimes L}$ would be invertible. Thus there is a non-invertible epimorphism $\mathcal {O\to O}$. Question 3: this has the same answer as Question 2. If the answer to Q2 is yes, then we can take $\psi$ to be the identity and get a positive answer to Q3. If the answer to Q2 is no, then (as above) if $\phi:\mathcal{ L\to L}$ is a non-invertible epimorphism, also $\phi\otimes 1:\mathcal{ L\otimes L'\to L\otimes L'}$ is a non-invertible epimorphism. But now if $\psi:\mathcal {L'\to L'}$ is any map, then $\phi\otimes\psi=(1\otimes\psi)\circ(\phi\otimes 1)$ and if this is invertible then $\phi\otimes 1$ is split monic and so invertible (since it is already known to be epi). - Q2: Ok this reduces it to the case of maps $\mathcal{O} \to \mathcal{O}$. Are there examples? Q3: If $\phi$ is an isomorphism, then $\phi \otimes L'$ does not have to be the identity! – Martin Brandenburg Nov 23 2010 at 23:21 Q3: you're right, I should have used $\mathcal{(\theta^{-1}\otimes 1_{L'})(1\otimes\phi^{-1}\otimes 1)(\theta\otimes_{L'}}$ for $\psi$, where $\theta:\mathcal{\O\cong L\otimes L'}$. – Steve Lack Nov 24 2010 at 0:05 sorry, try again: $(\theta^{-1}\otimes 1_{L'})(1_{L'}\otimes \phi^{-1}\otimes 1_{L'})(\theta\otimes 1_{L'})$ where $\theta:O\cong L'\otimes L$. (Hope this works, it's tricky without preview facility) – Steve Lack Nov 24 2010 at 0:08 If you are going to find a counterexample to Q2 (i.e. a $\otimes$ category when 1 has a non-invertible epi), you'll have to look pretty far. Most such categories (say, the small ones) embed in R-mod-R for some ring R, and the endomorphisms of 1 in R-mod-R are the center of R, and the only epis are invertible. – Theo Johnson-Freyd Nov 25 2010 at 19:53 @Theo: You quote arxiv.org/abs/math/0004160, right? But there we only have a right exact embedding. – Martin Brandenburg Dec 3 2010 at 14:48 show 1 more comment	{"extraction_info": {"found_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": 0.9199950695037842, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/30437/why-is-there-only-one-inertial-frame-that-ct-and-x-are-orthogonal/30458	# why is there only one inertial frame that $ct$ and $x$ are orthogonal? It is very long time ago that I took a physics lesson, so I want to refresh my memory. I think I learned that there is only one inertial frame in Minkowski spacetime (or special relativity time) that $ct$ and $x$ are orthogonal. (inertial frames are assumed as Lorentz-invariant, and we assume one space axis, $x$.) So, why is it? ($ct$ is in vertical axis, $x$ is in horizontal axis.) To avoid confusion: I think I found a question on my old textbook :) Edit: orthogonality in Euclidean term is assumed. Show that the $S'$ axes, $x'$ and $ct'$, are nonorthogonal in a spacetime diagram. Assume that the $S'$ frame moves at the speed of $v$ relative to the $S$ frame ($S'$ is moving away from $S$ to the side of $+x$) and that $t = t' = 0$ when $x = x' = 0$. (The $S$ axes, $x$ axis and $ct$ axis, are defined orthogonal in a spacetime diagram.) - 1 Your textbook uses misleading notation. The direction of the $S'$ axes in an $S$ spacetime diagram look non-orthogonal in the euclidean sense, which is the wrong way to look at it: the geometry of spacetime uses the Lorentz inner product, with respect to which both the $S$ and $S'$ are indeed orthogonal. – Emilio Pisanty Jun 20 '12 at 14:07 @EmilioPisanty Right.. let us just assume Euclidean sense. – Bill Fritzgard Jun 20 '12 at 14:21 That's just the thing. To your euclidean eyes, the axes don't look orthogonal, but as far as spacetime goes then they're perfectly fine. You might wish to call Lorentz-inner-product orthogonality something different, say, "good-angle-nality", but then the result you are asking about simply says that for every particle worldline there exists a unique spatial axis that is good-angle-nal to it. After a while, you'll get tired, and realize that it's easier to call it orthogonal while keeping in mind that it means something different from the (euclidean!) usual. – Emilio Pisanty Jun 20 '12 at 15:11 @EmilioPisanty Thanks. – Bill Fritzgard Jun 20 '12 at 15:15 ## 2 Answers This is false. Each inertial frame has spatial and temporal axes that are orthogonal to each other (in the Lorentz inner product, of course). Of course, different frames have different axes: temporal axes differ (to reflect relatively-moving origins, which also happens in galilean relativity) and spatial axes also differ (to reflect relativity of simultaneity, which does not happen in galilean relativity). You may be thinking of the fact that if we concentrate on some inertially-moving particle, then there is a unique spatial axis that is orthogonal to the particle's worldline. This axis is of course the spatial axis of the rest frame of the particle: the Lorentz frame that has its temporal axis along the particle's worldline. - You're right. Sorry for my comment. – Bill Fritzgard Jun 20 '12 at 14:19 There is a way to formulate your question that it has a nontrivial answer-- suppose you have both a euclidean and Lorentzian notion of perpendicularity. So that the dot product of two vectors has two meanings: $$A\cdot_E B = A_t B_t + A_x B_x$$ $$A\cdot_L B = A_t B_t - A_x B_x$$ Then if both $A\cdot_E B = 0$ and $A \cdot_L B = 0$, then by adding and subtracting the equations you learn that $A_t B_t = A_x B_x=0$, so that exactly one of the vectors has a component along the t axis, and exactly one of the vectors have a component along the x-axis, and this proves the result. The reason is that when you do a Lorentz rotation (a boost), you tilt the time axis toward the velocity vector, and you tilt the space axis toward the velocity vector too, so that they hug the 45-degree line of light propagation more closely, so the result is obvious from the picture--- the only time they are Euclidean perpendicular is when you are at rest. -	{"extraction_info": {"found_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": 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": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9320351481437683, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/62254/best-way-to-integrate-int-0-infty-frace-at-e-btt-textdt	# Best way to integrate $\int_0^\infty \frac{e^{-at} - e^{-bt}}{t} \text{d}t$ Today I had an exam and I mixed up the integration by parts formula. The question was to integrate $$\int_0^\infty \frac{e^{-at} - e^{-bt}}{t} \text{d}t$$ I will try solve this again with the right formula when I arrive home. I would appreciate if somebody could tell me the solution so I can double check and maybe give a hint to another way of solving this instead of integration by parts (if possible). - 2 Hint: Write the integrand as an integral over [a,b], then switch the order of integration. – Ragib Zaman Sep 6 '11 at 10:21 – Gerben Sep 6 '11 at 12:45 I guess this is a duplicate then, but @Ragib's answer is different (from robjohn's) and also nice. – Srivatsan Sep 6 '11 at 12:46 ## 2 Answers I may as well give an answer: Note $$\frac{ e^{-at} - e^{-bt} }{t} = \int^b_a e^{-xt} dx$$ so our integral is $$\int^{\infty}_0 \int^b_a e^{-xt} dx dt = \int^b_a \int^{\infty}_0 e^{-xt} dt dx$$ $$= \int^b_a \frac{1}{x} dx = \log(b/a)$$ This is a general method, and often this whole process is compressed into a well known integral called Frullani's Integral. - nice solution you gave! – Chris's wise sister Aug 5 '12 at 14:45 By differentiating under the integral sign. Fix $a$, and let $$g(b) = \int_{0}^\infty \frac{e^{-at} - e^{-bt}}{t} dt.$$ Differentiating w.r.t. $b$, we get $$g'(b) = \frac{d}{db}\int_{0}^\infty \frac{e^{-at} - e^{-bt}}{t} dt = \int_{0}^{\infty} \frac{\partial}{\partial b} \frac{e^{-at} - e^{-bt}}{t} dt.$$ Doing the differentiation, $$g'(b) = \int_{0}^{\infty} e^{-bt} dt = - \left.\frac{e^{-bt}}{b} \right\|_{0}^{\infty} = \frac{1}{b}.$$ Thus we must have $g(b) = \ln b + C$ for some constant $C$. To determine $C$, plug in $a$, and use the fact that $g(a) = 0$. Note. I am yet to convince myself that all the steps of the proof are rigorous. I will edit my answer later if additional argument is necessary. - @Ragib Sure. I will delete my comments now (and this one in a little while). – Srivatsan Oct 3 '11 at 17:22	{"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": 8, "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": 0.9429055452346802, "perplexity_flag": "head"}
http://www.haskell.org/haskellwiki/index.php?title=Simple_to_complex&diff=15414&oldid=8607	# Simple to complex ### From HaskellWiki (Difference between revisions) \| \| \| \| \| \|----------\|---------------------------------------------------------\|----------------------\|------------------------------------------------------------------------------------------------------------------------\| \| (init) \| \| (type class methods) \| \| \| Line 5: \| \| Line 5: \| \| \| \| That leads no longer to a clear hierarchy of functions, \| \| That leads no longer to a clear hierarchy of functions, \| \| \| but to an entangled graph of dependencies. \| \| but to an entangled graph of dependencies. \| \| \| \| + \| \| \| \| \| \| \| \| \| == Functions == \| \| == Functions == \| \| Line 56: \| \| Line 57: \| \| \| \| Move them as far as possible to leaf modules. \| \| Move them as far as possible to leaf modules. \| \| \| \| \| \| \| \| \| + \| \| \| \| \| + \| == Type class methods == \| \| \| \| + \| \| \| \| \| + \| There are many methods in <hask>Data.List</hask> that can be generalised to other data structures, \| \| \| \| + \| 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 <hask>foldr</hask> are hard to grasp for newbies. \| \| \| \| + \| That's why they often stick to manual encoding of recursion. \| \| \| \| + \| (See [[Things_to_avoid#Avoid_explicit_recursion]]) \| \| \| \| + \| Now imagine, that <hask>Data.List.foldl</hask> is no longer exposed, \| \| \| \| + \| but only its generalization, namely the method <hask>Data.Foldable.foldl</hask> \| \| \| \| + \| 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 [http://research.microsoft.com/~simonpj/Papers/giving-a-talk/giving-a-talk-html.html use an example]. \| \| \| \| + \| E.g. <hask>sum xs = List.foldl (+) 0 xs</hask>. \| \| \| \| + \| Now, this requires to have a function <hask>List.foldl</hask>. \| \| \| \| + \| \| \| \| \| + \| <!-- \| \| \| \| + \| Modularity reason: \| \| \| \| + \| It is sometimes not possible to generalize all the functionality that belongs together. \| \| \| \| + \| Example? \| \| \| \| + \| On the other hand: The generaliation can guide your decision, what belongs together, \| \| \| \| + \| or vice versa, what you think, belongs together, should be generlized together. \| \| \| \| + \| --> \| \| \| \| + \| \| \| \| \| + \| <!-- \| \| \| \| + \| Close to didactic reasons there are reasons of comprehensible program code. \| \| \| \| + \| > And a lot of standard list functions can be \| \| \| \| + \| > generalized to MonadPlus, for example you can define \| \| \| \| + \| > \| \| \| \| + \| > filter :: (MonadPlus m) => (a -> Bool) -> m a -> m a \| \| \| \| + \| \| \| \| \| + \| Always using the most generalized form is not a good idea. If you know you are working on lists, 'map' \| \| \| \| + \| and 'filter' tell the reader, that they are working on lists. The reader of a program doesn't need to \| \| \| \| + \| start human type inference to deduce this. Also the type inference of the compiler will fail, if you \| \| \| \| + \| write too general. Say, you are in GHCi, have your definition of 'filter' and you write \| \| \| \| + \| \| \| \| \| + \| Prelude> filter Char.isUpper (return 'a') \| \| \| \| + \| \| \| \| \| + \| To what monad this shall be specialised? Rely on type defaulting? Ok, you can use type signatures. \| \| \| \| + \| --> \| \| \| \| + \| \| \| \| \| + \| These problems should be taken account, when designing custom type classes. \| \| \| \| + \| See the notes on [[Slim instance declarations]]. \| \| \| \| + \| \| \| \| \| + \| \| \| \| \| + \| See also \| \| \| \| \| \| \| \| [[Category:Style]] \| \| [[Category:Style]] \| ## Revision as of 06:32, 5 September 2007 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 Things_to_avoid#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 . These problems should be taken account, when designing custom type classes. See the notes on Slim instance declarations. See also	{"extraction_info": {"found_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": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.902523398399353, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/42507/a-path-between-subsets-of-mathbb-rn	# A path between subsets of $\mathbb R^n$ Let $S=\{a_i\}$ be a countable set of bounded,connected and closed subsets of $\mathbb R^n$, each of nonzero area, such that any two $a_i$ and $a_j$ may only intersect at their boundary and such that the union of all a_i cover $\mathbb R^n$ (i.e. a tiling of $\mathbb R^n$ by a countable set of tiles). Let A be the statement that for all unions $T_k$ of any collection of $a_i$'s, there is no $a_j$ such that $a_j$'s intersection with $T_k$ is the completey boundary of $T_k$.(i.e. no tile completely enclose another set of tiles). Is A sufficient for the existence of an ordering (a sequence) of the $a_i$'s such that each $a_i$ in S appear exactly once, and for any two consecutive $a_i$,$a_j$ in the sequence they share a part of their boundary? Does such a sequence exist if all the tiles are equal upto rotations and translations? - 1 I am not sure that this question fits [graph-theory], I'm not sure that it doesn't either too. – Asaf Karagila Jun 1 '11 at 5:52 ## 1 Answer I don't think so. If $a$ has only two neighbors, $x$ and $y$, then the ordering must contain $x,a,y$ or $y,a,x$ or start with $a$. Now it's easy to get three (or more) sets $a,b,c$ such that each has only the two neighbors $x,y$ and now no ordering can have all 5 sets. - I think you need four such sets? Otherwise you'd have the ordering $axbyc$. (As you say, it's also easy to get four, by considering isolated chunks on the boundary between $x$ and $y$.) – joriki Jun 1 '11 at 6:10 @joriki, sure, you can do $axbyc$, but you haven't done all of ${\bf R}^n$ that way (the sets are all bounded), and you can't leave $c$ to go anywhere else. – Gerry Myerson Jun 1 '11 at 6:15 I see, sorry, I forgot about the condition that they cover $\mathbb R^n$. – joriki Jun 1 '11 at 6:18	{"extraction_info": {"found_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": 0.18, "absolute_threshold": 10, "end_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": 0.9643752574920654, "perplexity_flag": "head"}
http://mathhelpforum.com/geometry/45779-locus-orthocenters.html	# Thread: 1. ## locus of orthocenters Determine the locus of the point $O$ of intersection of the altitudes (orthocenter) of a triangle $ABC$, if the locus of the vertex $A$ is a line parallel to $BC$. 2. Originally Posted by scipa Determine the locus of the point $O$ of intersection of the altitudes (orthocenter) of a triangle $ABC$, if the locus of the vertex $A$ is a line parallel to $BC$. The locus of the orthocenter is a parabola passing through B and C. Let $\alpha$ denote the interior angle at A, then 1. the parabola cuts the parallel to BC through A in 2 points if $\alpha > 90^\circ$; 2. the parallel to BC through A is tangent to the parabola if $\alpha = 90^\circ$; 3. the vertex of the parabola is located between the parallel to Bc and the line BC if $\alpha < 90^\circ$; Attached Thumbnails 3. Originally Posted by scipa Determine the locus of the point $O$ of intersection of the altitudes (orthocenter) of a triangle $ABC$, if the locus of the vertex $A$ is a line parallel to $BC$. Second attempt: Place the base BC of the triangle on the x-axis. Let d denote the distance between the 2 parallels. B(0,0) and C(c,0) and A(x,d). The slope of AC is $m_{AC}=-\frac d{c-x}$. Then the height through B (perpendicular to AC) has the slope: $m_{h_B}=\frac{c-x}d$ Then the orthocenter is $H \left( x,\frac{(c-x)x}d\right)$ That means all orthocenters lie on the curve of $y=-\frac1dx^2+\frac cdx$ I've choosen d = 5, C(8,0) Attached Thumbnails #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"extraction_info": {"found_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": 20, "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": 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": 0.833477795124054, "perplexity_flag": "middle"}
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I have $p\cdot x=Et-\bf p\cdot x$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "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": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9018620848655701, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/11061/how-is-shown-that-photon-speed-is-constant-using-qed?answertab=oldest	# how is shown that photon speed is constant using QED? In Feynman's simple QED book he talks about the probability amplitude P(A to B) ,where A and B are events in spacetime, and he says that it depends of the spacetime interval but he didn't put the expression. I would like to know what expression was he referring to. He said then that the amplitudes for a photon to go faster or slower than c are very smaller than the contribution from speed c. How is that this contributions cancel out for long distances? I would like also that you explain how it is related to the path integral of the form exp(iKL) of the question How is the path integral for light explained, or how does it arise? I think this path integral is very strange because it doesn't have spacetime fixed points but space fixed points. Its paths doesn't have the same time if we assume speed constant so they wont reach the same event. - 2 I have to run (literally) so just a note: $K \cdot L$ is an invariant. If you transform to a different frame you both blueshift $K$ and contract $L$ and these effects compensate. As for the main answer: as always, for nice theories (meaning weakly interating) classical limit of a path integral recovers the classical theory with variational principle. This is a general fact in any quantum mechanical theory and QED is no exception. So in this case you'll recover the motion of light along null geodesics. – Marek Jun 13 '11 at 16:26 ## 1 Answer This is a very good question, but it is really two completely different questions in one. ### Feynman's propagator The probability amplitude for a photon to go from x to y can be written in many ways, depending on the choice of gauge for the electromagnetic field. They all give the same answer for scattering questions, or for invariant questions involving events transmitted to a macroscopic measuring device, but they have different forms for the detailed microscopic particle propagation. Feynman's gauge gives a photon propagator of: $P(k) = {g_{\mu\nu} \over w^2 - k^2 + i\epsilon}$ And it's Fourier transform is $2\pi^2 P(x,t) = {g_{\mu\nu} \over {t^2 - x^2 + i\epsilon}}$ This is the propgation function he is talking about. It is singular on the light cone, because the denominator blows up, and it is only this singularity which you can see as propagating photons for long distances. For short distances, you see a $1/s^2$ propagation where s is the interval or proper time, between source and sink. To show that you recover only physical light modes propagating, the easiest way is to pass to Dirac gauage. In this gauge, electrostatic forces are instantaneous, but photons travel exactly at the speed of light. It is not a covariant gauge, meaning it picks a particular frame to define instantaneous. The issues with the Feynman gauge is that the propagator is not 100% physical, because of the sign of the pole on the time-time component of the photon propagator. You have to use the fact that charge is conserved to see that non-physical negative-coefficient-pole states are not real propagating particles. This takes thinking in the Feynman picture, but is not a problem in the Dirac picture. The equivalence between the two is a path integral exercise in most modern quantum field theory books. ### Fermat's principle and Lagrangians Fermat's principle, as you noted, is not a usual action principle because it doesn't operate at fixed times. The analog of the Fermat principle in mechanics is called the principle of Maupertuis. This says that the classical trajectory is the one which minimizes $$J = \int p dx = \int \sqrt{2m(E-V(x))} dx$$ between the endpoints. This principle is also timeless, and it can be used to construct an approximate form for the time Fourier transform of the propagator, and this is called the Gutzwiler trace formula. the Gutzwiller trace formula is the closest thing we have to a proper quantum analog of the Maupertuis principle at this time. ### Lagrangian for light The analog of the Lagrangian principle for light is just the principle of that light travels along paths that minimize proper time, with the additional constraint that these proper times are zero. The Lagrangian is $m\int ds = m\int \sqrt{1-v^2} dt$ but this is useless for massless particles. The proper transformation which gives a massless particle propagator is worked out in the early parts of Polyakov's "Gauge Fields and Strings" as a warm-up to the analogous problem for string theory. The answer is: $S= \int {\dot{x}^2\over 2} + m^2 ds$ The equivalence between this form and the previous one is actually sort of obvious in Euclidean space, because of the central limit theorem you must get falling Gaussians with a steady decay rate. Polyakov works it out carefully because the anlogous manipulations in string theory are not obvious at all. The second form is not singular as m goes to zero, and gives the proper massless propagator. Transitioning between the two introduces an "einbein" along the path, a metric tensor in one dimension. -	{"extraction_info": {"found_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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9372928142547607, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/27262/mermin-wagner-theorem-in-the-presence-of-hard-core-interactions	# Mermin-Wagner theorem in the presence of hard-core interactions It seems quite common in the theoretical physics literature to see applications of the "Mermin-Wagner theorem" (see wikipedia or scholarpedia for some limited background) to systems with hard-core interactions; for example to conclude that genuine crystal phases for a system of hard disks (with possible additional interactions) cannot exist in 2 dimensions. If this particular claim has been proved rigorously a few years ago (see this paper), it is known in general that the presence of hard-core interactions can lead to phases with broken continuous symmetry (a specific example is given below). To keep things simple, let me focus on nearest-neighbor spin models on the square lattice, with spins taking values in the unit circle. So let us consider a formal Hamiltonian of the form $$\sum_{i\sim j} V(S_i,S_j),$$ with $V$ is a continuous function, assumed to be invariant under the action of $SO(2)$: $V(r_\theta S_i, r_\theta S_j) = V(S_i,S_j)$, where $r_\theta$ rotates the spin by an angle $\theta$. In that case, it is known that all pure phases of the model are invariant under the action of $SO(2)$. (Note that we do not even require $V$ to be smooth, so that the usual, both heuristic and rigorous, arguments relying on a Taylor expansion and a spin-wave argument do not apply immediately; that one can do so was proved here). Substantially more general results are actually known, but this will suffice for my question. What I am interested in is what happens for models of this type in the presence of hard-core interactions. No general results are known, and the situation is proved to be subtle. Indeed, consider for example the Patrascioiu-Seiler model, in which $$V(S_i,S_j) = \begin{cases} -S_i\cdot S_j & \text{if }\|\delta(S_i,S_j)\|\leq\delta_0,\\ +\infty & \text{otherwise,} \end{cases}$$ where $\delta(S_i,S_j)$ denotes the angle between the spins $S_i$ and $S_j$, and $\delta_0>0$ is some parameter. In other words, this model coincides with the classical XY model apart from the additional constraint that neighboring spins cannot differ too much. For this model, it is proved here that, when $\delta_0<\pi/4$, there exist (non-degenerate) phases in which rotation invariance is broken. Nevertheless, one expects that phases obtained, say, by taking the thermodynamic limit along square boxes with free, periodic or constant boundary conditions should be rotation invariant. So, now, here is my question: Are there any heuristic physical arguments supporting the validity of a version of the "Mermin-Wagner theorem" in such situations? All the heuristic arguments I know of fail in such a context. Having good heuristic arguments might help a lot in extending the rigorous proofs to cover such situations. Edit: Let me precise my question, as the (quite long) discussion with Ron Maimon below shows that I haven't stated it in a clear enough way. I am not interested in a discussion of why the counter-example given above leads to a violation of MW theorem and whether it is physically realistic (as far as I am concerned, its main relevance was to show that one has to make some assumptions on the interaction $V$ in order to have rotation invariance of all infinite-volume Gibbs states, and this is exactly what this example does). What I am really interested in is the following: does there exist heuristic (but formulated in a mathematical way, not just some vague remarks) arguments with which physicists can derive the MW theorem in the presence of hard-core interactions? I would even be interested in arguments that apply in the absence of hard-core interactions, but when $V$ is not differentiable (even though this case is treated rigorously in the reference given above). - The question was 100% clear, and the answer I gave you is the correct one--- there is no difference in the heuristic argument for Mermin Wagner in the presence of hard-core interactions or in their absence, because your counterexample is nonsense. The argument is the usual establishing of averaged spin-wave states, and finding that the free energy is the free-field, which has only log divergent configurations. The only way this fails is if you have frozen out all fluctuations, so that there are no spin-waves at all, and this is your example. This is not essentially to do with hard-walls. – Ron Maimon Oct 11 '11 at 2:16 The reason your methods do not work for the hard-potential case is not because the argument of Mermin and Wagner is failing, because your methods are not in the spirit of Mermin-Wagner. The reason they fail is because you are not defining an averaged long-distance spin-wave field, but trying to do things with the bare configurations, which is hopeless. The way to define a long-distance theory is renormalization (in the Wilson sense, but simpler because the limit is free). – Ron Maimon Oct 11 '11 at 2:20 – Yvan Velenik Oct 11 '11 at 6:59 @RonMaimon: I think that you're somewhat unfair (or misinformed) when you say "your methods are not in the spirit of Mermin-Wagner". The methods we used (which were introduced in the late 1970s by Dobrushin&Shlosman and by Pfister) seem to me very much in the spirit of MW: roughly speaking, what they do is compare the (finite-volume) Gibbs measure before and after deforming the configurations by a global spin wave, and show that the effect of this deformation becomes negligible when the system size gets big (and thus the wave-length large). [to be continued] – Yvan Velenik Oct 11 '11 at 14:22 [continuation] The difficulty when hard-core is present is that most configurations become forbidden (have infinite energy) after deformation by the spin wave. One can then try to resort to a configuration dependent deformation (this is what is done by Richthammer in his treatment of the hard discs model), but this is technically quite difficult, unfortunately. – Yvan Velenik Oct 11 '11 at 14:26 show 5 more comments ## 1 Answer First, I will translate the relevant passages in your paper from mathematese. ### The argument in your reference You are studying an X-Y model with the constraint that neighboring spins have to always be within a certain angle of each other. You define the collection of statistical-mechanics Gibbs distributions using a given boundary condition at infinity, as the boundaries get further and further away. Then you note that if the field at the boundary makes the spin turn around from top to bottom the maximum possible amount, then the spins are locked in place--- they can't move, because they need to make a certain winding, and they unless they are at the maximum possible angle, they can't make the winding. Using these boundary conditions, there is no free energy, there is no thermodynamics, there is no spin-wave limit, and the Mermin Wagner theorem fails. You also claim that the theorem fails with a translation invariant measure, which is just given by averaging the same thing over different centers. You attempt to make the thing more physical by allowing the boundary condition to fluctuate around the mean by a little bit $\delta$. But in order to keep the boundary winding condition tight, as the size of the box $N$ goes to infinity, $\delta$ must shrink as $1\over N$, and the resulting Free energy of your configuration will always be subextensive in the infinite system limit. If $\delta$ does not shrink, the configurations will always randomize their angles, as the Mermin-Wagner theorem says. The failures of the Mermin-Wagner theorem are all coming from this physically impossible boundary situation, not really from the singular potentials. By forcing the number of allowed configurations to be exactly 1 for all intents and purposes, you are creating a situation where each different average value of the angle has a completely disjoint representative in the thermodynamic limit. This makes the energy as a function of the average angle discontinuous (actually, the energy is infinite except for near one configuration), and makes it impossible to set up spin waves. This type of argument has a 1d analog, where the analog to Mermin-Wagner is much easier to prove. ### 1-dimensional mechanical analogy To see that this result isn't Mermin-Wagner's fault, consider the much easier one-dimensional theorem--- there can be no 1d solid (long range translational order). If you make a potential between points which is infinite at a certain distance D, you can break this theorem too. What you do is you impose the condition that there are N particles, and the N-th particle is at a distance ND from the first. Then the particles are forced to be right on the edge of the infinite well, and you get the same violation: you form a 1d crystal only by imposing boundary conditions on a translation invariant potential. The argument in 1d that there can be no crystal order comes from noting that a local defect will shift the average position arbitrarily far out, so as you add more defects, you will wash out the positional order. ### Mermin-Wagner is not affected The standard arguments for the Mermin-Wagner theorem do not need modification. They are assuming that there is an actual thermoodynamic system, with a nonzero extensive free energy, an entropy proportional to the volume, and this is violated by your example. The case of exactly zero temperature is also somewhat analogous--- it has no extensive entropy, and at exactly zero temperature, you do break the symmetry. If you have an extensive entropy, there is a marvelous overlap property which is central to how physicists demonstrate the smoothness of the macroscopic free-energy. The Gibbs distribution at two angles infinitesimally separated sum over almost the same exact configurations (in the sense that for a small enough angle, you can't tell locally that it changed, because the local fluctuations swamp the average, so the local configurations don't notice) The enormous, nearly complete, overlap between the configurations at neighboring angles demonstrates that the thermodynamic average potentials are much much smoother than the possibly singular potentials that enter into the microscopic description. You always get a quadratic spin-wave density, including in the case of the model you mention, whenever you have an extensive free energy. Once you have a quadratic spin-wave energy, the Mermin Wagner theorem follows. ### Quick answer the Gibbs distributions for orientation $\theta$ and the Gibbs distributions for orientation $\theta'$ always include locally overlapping configurations as $\theta$ approaches $\theta'$. This assumption fails in your example, because even an infinitesimal change in angle for the boundary condition changes the configurations completely, because they do not have extensive entropy, and are locked to within a $\delta$, shrinking with system size, of an unphysically constrained configuration. - Thanks for your answer. I must confess that I don't get all your points, though, and shall probably have to translate everything back to mathematese, as you say. Nevertheless, if I agree that our example is quite contrived and unrealistic (it was given to explain why we do need some regularity on the interaction in our paper, namely continuity or the more general condition (26)), I don't think that the resulting Gibbs state is just a mixture of measures with all spins frozen in a given direction (even though this would be a Gibbs measure, albeit a pathological one). [to be continued] – Yvan Velenik Oct 7 '11 at 9:41 [continuation] But in any case, my question was about a heuristic argument that would explain how to implement a spin-wave argument in the presence of hard-core interactions. you seem to give one, but I can't follow it, as it is too imprecise. Could you add some more math? Actually, even expliciting your argument when the interaction is continuous (but not differentiable) would be already nice (even though we do know how to prove that rigorously, and a nice physical argument can be extracted from the proof). – Yvan Velenik Oct 7 '11 at 9:53 Also, how would you argue (heuristically but, please, with some math details) that hard discs in the plane cannot form a crystalline phase (i.e., translation invariance cannot be broken)? To keep things as simple as possible, assume that there is no other interactions beside hard-core. A rigorous proof is known (see the ref in my question) but it is quite intricate. Of course, here it's density waves rather than spin waves, but that's rather irrelevant. Note that here the enery of a configuration is either $0$ or $+\infty$, which makes smooth deformation of configurations tricky. – Yvan Velenik Oct 7 '11 at 15:37 Although that's not the main point, let me come back to your comment about $\delta$ in your answer. Sure, $\delta$ has to go to $0$ with $N$. But that does not imply that the resulting Gibbs measure has not an extensive entropy, because the spins far away from the boundary have room to wiggle. That's only spins at a finite distance from the origin that you see in the thermodynamic limit. So I am pretty sure (although it would some time to come up with a formal proof) that under the measure we "construct", the angle between two given vertices (say $(0,0)$ and $(1,0)$) has a non-zero variance. – Yvan Velenik Oct 7 '11 at 15:46 1 Still I think that you misunderstand my very question: I don't care about the precise counter-example we gave in the paper (and which was just a remark in the paper). What I care about is a (not necessarily rigorous) mathematical argument that derives absence of continuous symmetry breaking in situations where you have a hard-core interactions. You have not given me that, unfortunately. As I said above, even a nice heuristic (but mathematical!) argument in the case of non-differentiable interactions would be interesting (even though we treated those in the paper). – Yvan Velenik Oct 10 '11 at 7:08 show 17 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "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": 0.9454402923583984, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2011/03/16/presheaves/?like=1&_wpnonce=401286028f	# The Unapologetic Mathematician ## Presheaves Strictly speaking, sheaves are not just about differential topology. And it’s also possible to get away without talking about them explicitly in differential topology. But it’s in differential topology that they really start to make their presence felt. The best description of a sheaf I ever got doesn’t really translate to text, unfortunately. I forget the originator, as told to me by the professor explaining it — and if he remembers and can fill in the gap this can be recorded for posterity — but it really does capture the essence. A sheaf is a thing where the topology goes this way (moving one’s open hand around in circles at about chest level) and the algebra goes this way (moving one’s hand up and down across chest level). To be a bit more explicit, sheaves are about taking algebraic structures and localizing them to open sets in a topological space. I’ll start with the simplest versions, which are sheaves of sets. Further, I’ll actually start with something simpler: presheaves. So, a presheaf $\mathcal{F}$ of sets on a topological space $X$ is a choice of a set $\mathcal{F}(U)$ for each open set $U\subseteq X$. We call the members of $\mathcal{F}(U)$ the elements — or the “sections” — of the presheaf “over $U$“. But there’s an important condition on this choice: if $V\subseteq U$ is a smaller open subset of $X$, then we should be able to “restrict” our element $f\in\mathcal{F}(U)$ to an element $f\vert^U_V\in\mathcal{F}(V)$. Thus, if we have an inclusion $V\subseteq U$, then we have a restriction map $\cdot\vert^U_V:\mathcal{F}(U)\to\mathcal{F}(V)$. These restriction maps are subject to a couple conditions. First of all, if we restrict from $U$ to itself, then we shouldn’t change anything. That is, the restriction map $\cdot\vert^U_U$ is always the identity map. Secondly, if we restrict from $U$ to $V$, and then from $V$ to $W$, we should get the same result as if we just restricted directly from $U$ to $W$. That is, we have the equation $\cdot\vert^V_W\circ\cdot\vert^U_V=\cdot\vert^U_W$. Because of this, we’ll often just write the restriction map as $\cdot\vert_V$, since which subset it came from doesn’t really matter. We can express this definition more succinctly if we remember that containment of open subsets of a topological space constitutes a partial order, and thus defines a category $\mathrm{Subset}(X)$. The objects are the open sets themselves, and there is a unique arrow from $V$ to $U$ if $V\subseteq U$. If we look closely, we’ll find that what we’ve defined as a presheaf is actually a contravariant functor from this category to the category of sets! For every arrow $V\mapsto U$ we have an arrow $\cdot\vert_U:\mathcal{F}(U)\to\mathcal{F}(V)$ — in the “opposite direction”, since the functor is contravariant. The conditions we impose on the restriction maps just say that they preserve identity arrows and compositions. Now, there’s nothing inherently special about sets here. We can set up exactly the same construction with any target category to define, say, a presheaf of rings to be a contravariant functor $\mathcal{F}:\mathrm{Subset}(X)\to\mathbf{Ring}$. This assigns a ring $\mathcal{F}(U)$ to every open set $U$, and the restriction maps have to be ring homomorphisms. In the same way we get presheaves of groups, of abelian groups, or of vector spaces over a given field. The one possibly confusing case is when we talk about a presheaf of modules over a presheaf of rings. In this case, say we have a presheaf $\mathcal{F}$ of rings on a topological space $X$. A presheaf of modules $\mathcal{M}$ over $\mathcal{F}$ assigns an abelian group $\mathcal{M}(U)$ to every open set $U\subseteq X$ — it’s a presheaf of abelian groups — in such a way that $\mathcal{M}(U)$ is a module over $\mathcal{F}(U)$. The restriction map has to work with the restriction map of $\mathcal{F}$, so we have $f(m)\vert_V=f\vert_V\left(m\vert_V\right)$. The canonical example to keep in mind is continuous, real-valued functions on a topological space. This is a sheaf of real algebras that associates to the open set $U\subseteq X$ the algebra $\mathcal{F}(U)$ of real-valued functions that are defined and continuous there. Clearly we can restrict such a function $f$ to whatever open subset $V$ we want — and, in fact, we have. The nice thing is that this gives us a way of talking about and dealing with functions on our space that may not be defined or continuous everywhere. Just work within a suitable open set where the function does play nice! If you need to work with two functions defined over different open sets, just restrict them both to their common intersection and work there. Many structures we run into in differential geometry will be naturally expressible in terms of presheaves, just like this. ### Like this: Posted by John Armstrong \| Topology ## 18 Comments » 1. I’m having a bit of trouble figuring out what $f\in\mathcal F$ is in paragraph 3. A member of the set that is the value of the presheaf at $U$? Or something different? Comment by Avery Andrews \| March 17, 2011 \| Reply 2. I mean $f\in\mathcal F(U)$. Comment by Avery Andrews \| March 17, 2011 \| Reply 3. Exactly. If $\mathcal{F}(U)$ is a set, then $f\in\mathcal{F}(U)$ is an element of that set. I chose $f$ because of the example listed in the last paragraph. Maybe it would help to read that example and then go back through the definition to see how it fits. Comment by \| March 17, 2011 \| Reply 4. [...] the moment we will be more concerned with presheaves, but we may as well go ahead and define sheaves. These embody the way that not only can we restrict [...] Pingback by \| March 17, 2011 \| Reply 5. Got it. I wouldn’t have had a problem if “these are” was replaced by “the members of $\mathcal{F}(U)$ are”. Comment by Avery D Andrews \| March 17, 2011 \| Reply 6. In the 4th paragraph, I do not understand why the equation is not written the other way around, with restriction from U to V written on the right of restriction from V to W Comment by Francis \| March 18, 2011 \| Reply • You’re right, it probably should be written the other way. Thanks. Comment by \| March 18, 2011 \| Reply 7. John, I heard the “this way – that way” description of sheaves from Bill Massey. You might have been there at the time, or you might have gotten it from me later, since I certainly worked that description into my repertoire. I still use it whenever I talk about sheaves to a broad audience. Comment by Greg Friedman \| March 19, 2011 \| Reply 8. By the way, in the penultimate paragraph, it might be less confusing to write \$(fm)\|_V=f\|_V m\|_V\$, both because on the left hand side you’re restricting all of \$fm\$ and also because \$f(m)\$ looks like you’re evaluating some function at \$m\$. Comment by Greg Friedman \| March 19, 2011 \| Reply 9. In the last paragraph, shouldn’t it be ‘presheaf’ not ‘sheaf’ (a sheaf seems to need an extra restriction). Comment by Avery D Andrews \| March 19, 2011 \| Reply 10. Avery, it turns out that the continuous real-valued functions do form a sheaf. But since a sheaf is a presheaf, it’s an example of the latter as well. Comment by \| March 19, 2011 \| Reply 11. [...] Between Presheaves As ever, we want our objects of study to be objects in some category, and presheaves (and sheaves) are no exception. But, luckily, this much is [...] Pingback by \| March 19, 2011 \| Reply 12. Regarding the bit about doing topology horizontally and algebra vertically: somewhere I read a riff on this as follows. ‘Sheaf’ in ordinary English can mean something like a bundle of stalks, the stalks being aligned vertically. Or, it can mean a sheaf of papers, each leaf lying horizontally (like a section of a sheaf). The word play is supposed to be even better with the French equivalent faisceau. I wish I could remember where I read this; I’m pretty sure it was in some categorical discussion about sheaves. Comment by \| March 20, 2011 \| Reply 13. [...] Direct Image Functor So far our morphisms only let us compare presheaves and sheaves on a single topological space . In fact, we have a category of sheaves (of sets, by [...] Pingback by \| March 21, 2011 \| Reply 14. [...] more construction we’ll be interested in is finding the “stalk” of a presheaf over a point . We want to talk about how a presheaf behaves at a single point, but a single point [...] Pingback by \| March 22, 2011 \| Reply 15. [...] of Functions on Manifolds Now that we’ve talked a bunch about presheaves and sheaves in general, let’s talk about some particular sheaves of use in differential [...] Pingback by \| March 23, 2011 \| Reply 16. [...] subspace we can restrict a vector field on to one on , which means we’re talking about a presheaf. In fact, it’s not hard to see that we can uniquely glue together vector fields which agree [...] Pingback by \| May 23, 2011 \| Reply 17. Iterated integration of multi-integrals seems to constitute a presheaf where the topological space is \$\{ 1, …, n \}\$ and \$F(U)\$ is the set of integrable functions \$f(x_k \| k \in U)\$ (e.g. \$f(x_2, x_3, x_5)\$ for \$U = \{ 2, 3, 5 \}\$). The restriction map \$U \rightarrow V\$ is integration over \$(x_k \| k \in U \ V)\$. Comment by \| November 7, 2012 \| 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": 52, "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": 0.9407278299331665, "perplexity_flag": "head"}
http://nrich.maths.org/6140	# Big and Small Numbers in Biology ##### Stage: 4 Challenge Level: Biology makes good use of numbers both small and large. Try these questions involving big and small numbers. You might need to use standard biological data not given in the question. Of course, as these questions involve estimation there are no definitive 'correct' answers. Just try to make your answers to each part as accurate as seems appropriate in the context of the question. 1. Estimate how many of the smallest viruses would fit inside the largest bacterium. What assumptions do you make in the calculation? 2. Why might it be misleading to say that the size of a bacterium is 200 microns? How might you provide a more accurate description of the size? 3. It has been said that 1g of fertile soil may contain as many as 2500 million bacteria. Do you think that this is a high density of bacteria? Estimate the percentage, by weight, of the soil that comprises bacteria. If the bacteria were evenly spread out, estimate the distance between the bacteria and compare this to the size of the bacteria. Does this surprise you? 4. The shape of the earth may be approximated closely by a sphere of radius 6 x10$^{6}$m. In June 2008, its human population, according to the US census bureau , was thought to be 6,673,031,923. If everyone spread out evenly on the surface of the earth, what area of the planet would they each have? 5. Humans typically live on land. Readjust your answer to the previous question making use of the fact that about 70% of the surface of the earth is water. 6. Compare the previous three parts of the question. Are humans more densely packed than the bacteria in the soil? (given that humans live on the surface of the earth and bacteria live inside a volume of soil, you might want to consider how best to measure the 'density') 7. Suppose that fertile land on earth extends, on average, down to about 10cm. Estimate how many cubic mm of fertile soil the earth contains. Estimate the number of bacteria living in the fertile land on earth. 8. Question the assumption concerning the average depth of fertile land in the previous question. Would you say that it should be smaller, larger or is about right? You might want to use these suggested data (from here ) that the percentages of earth's land surface can be divided into different types: 20% snow covered, 20% mountains, 20% dry/desert, 30% good land that can be farmed, 10% land with no topsoil. What other data might you need to make a more accurate assessment? 9. There are about 300000 platelets in a cubic mm of human blood. How many platelets might you expect to find in a healthy adult male? 10. There are about 4 - 6 million erythrocytes and 1000 - 4500 lymphocytes in a cubic mm of blood. A sample of blood on a slide is 2 microns thick. Would you expect many erythrocytes or lymphocytes to overlap on the microscope image? Extension: In mathematics, a bound for a measurement gives two numbers between which we know for certain that the real measurement must lie. For example, a (not very good) bound on the height of the members of a class would be 1m < heights < 2m. In the previous questions can you find bounds on the quantities? First suggest a really rough bound which you would know to be true and then see if you can sensibly improve on it. 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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.946002185344696, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/56534/surface-automorphisms-and-conformal-automorphisms/56538	## Surface automorphisms and conformal automorphisms ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given a closed orientable surface $S$ and a topological automorphism $\sigma$ of $S$, it is not in general possible to find a conformal structure $\Sigma$ on $S$ so that $\sigma$ is isotopic to a conformal automorphism of the Riemann surface $(S,\Sigma)$. For example by the theorem of Hurwitz that the conformal automorphism group is finite, while $\sigma$ on the other hand may be of infinite order in the mapping class group. But by a theorem of Colin Maclachlan in "Modulus space is simply connected", Proc. Amer. Math. Soc. 29 (1971), 85–86, every surface automorphism is isotopic to the composition of finitely many conformal automorphisms (for varying complex structures on $S$). For being isotopic to a conformal automorphism is equivalent to being isotopic to an topological automorphism of finite order (one direction by Hurwitz, the other by averaging a metric). Maclachlan proved that the mapping class group is generated by elements of finite order. I am interested in the minimal number $m(\sigma)$ of conformal structures required, especially for the torus, where the mapping classes have a nice explicit description. Unfortunately when I tried to use this explicit description, it translated into some obscure number theory with a Diophantine flavor. I could not even show that for the torus in general arbitrarily many conformal structures would be needed, i.e. that $m(\sigma)$ is unbounded for $T^2$. This is my question. An upper bound for $m(\sigma)$ for $T^2$ in terms of the explicit description of $\sigma$ by a 2 by 2 integer matrix would also be interesting. Perhaps the higher genus case could be worth looking at after the torus case. I got stuck on $T^2$ and gave up quite a long time ago. But now that Math Overflow is here, I can ask this as a question. CLARIFICATION: For the case of a topological torus I am concretely asking how many conformal automorphisms (relative to various complex structures on the topological torus) I need to compose together to have enough freedom to represent a topological automorphism up to isotopy. I tend to agree with Sam Nead that for every positive integer $K$ there will be some topological automorphism $\sigma$ that will require at least $K$ conformal automorphisms in order to be so represented. But I don't know how to prove this, though Sam Nead's comment on proving it seems reasonable. - 1 @Marius - Your question is not clearly stated, as evidenced by the fact that Andy and Scott are telling you that various mapping class groups are generated by torsion while I have read something very different (probably due to you mentioning "Diophantine"). Would you care to clarify your question? – Sam Nead Feb 24 2011 at 17:38 So, in conclusion, I think the question is `Do the torsion elements of SL(2,Z) boundedly generate?' Quite a lot of work has been done on bounded generation, I believe. – HW Feb 24 2011 at 19:09 The answer is "no". This follows from work of Bestvina and Fujiwara. See below. – Sam Nead Feb 24 2011 at 19:30 Your question is also posed in the final paragraph of Brendle and Farb's paper "Every mapping class group is generated by 6 involutions" and is explicitly answered in papers of Korkmaz and also Kotschick -- see the second footnote of [BF]. – Sam Nead Feb 24 2011 at 19:55 ## 2 Answers Complete edit, after talking to a colleague - Suppose that $G$ is the mapping class group of a surface $S$. Then you are asking: Is there a number $K$, depending only on $S$, with the following property? For every $\sigma \in G$ there are torsion elements $\tau_i \in G$ so that $\sigma = \Pi_{i = 1}^K \tau_i$. As Henry Wilton puts it - you are asking if the mapping class group is boundedly generated by torsion. The answer is "No". This follows from a paper of Bestvina and Fujiwara "Bounded cohomology of subgroups of mapping class groups". They show that the group $G$ admits unbounded quasi-homomorphisms. See the first five pages of their paper. Edit - to give a few details. A $D$-quasi-homomorphism is a map $\phi$ from $G$ to the reals so that for all $g,f \in G$ we have $\|\phi(gf) - \phi(g) - \phi(f)\| < D$. It is an exercise to show that if $g$ is torsion then $\|\phi(g)\| < D$. Thus, if $G$ was boundedly generated, say with constant $K$, then we would have, for all $g \in G$, that $\|\phi(g)\| < 2KD$. This is a contradiction. - Yes, this is the tangible version of my question for the case of torus. I agree with you that the answer is likely to be "no". – Marius Overholt Feb 24 2011 at 17:47 Remark - Marius is referring to the previous version of this answer. – Sam Nead Feb 25 2011 at 20:21 ### 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. It was proven in J. MacCarthy and A. Papadopoulos. Involutions in surface mapping class groups. Enseign. Math. 33 (1987), 275–290. that the mapping class group is generated by the conjugates of a single involution (for $g$ large, I think at least $2$). Call this involution $\tau$ and let $S$ be the Riemann surface that $\tau$ acts on. Then the conjugate $\tau^f$ by $f$ in the mapping class group acts on the Riemann surface $f_{\ast} (S)$, which is isomorphic to $S$. In other words, up to isomorphism you only need one conformal structure. - @Andy - I think you misunderstood my question. Unfortunately I did not formulate the question precisely enough, so that there are now several different answers. It looks like the problem with your answer, from my point of view, is that the conjugating map $f$ may not be of finite order. If it were, your answer probably would give $m(\sigma) \leq 3$, though I am not quite sure of this. – Marius Overholt Feb 24 2011 at 17:46 @Sam -- Scott and I gave answers to a reasonable interpretation of the question. It isn't fair to downvote us for not reading the OP's mind correctly. – Andy Putman Feb 24 2011 at 17:52 @Andy - I've removed the downvote and I apologize for doing so. A vote, up or down, is a single bit and thus is a limited form of communication. Let me instead ask you directly: why is your answer on-point, given Marius' edit of his post and then his comment just above? – Sam Nead Feb 25 2011 at 20:20 @Marius : It doesn't matter whether $f$ is of finite order or not. If $S$ is a Riemann surface and $\phi : S' \rightarrow S$ is a diffeomorphism, then we can pull back the Riemann surface structure on $S$ to get a Riemann surface structure on $S'$. With this Riemann surface structure, $\phi : S' \rightarrow S$ will be a biholomorphism. Moreover, if $\tau : S \rightarrow S$ is a biholomorphic map, then $\tau \circ \phi \circ \tau^{-1} : S' \rightarrow S'$ will be a biholomorphic map. I'm just applying this fact when $S' = S$. – Andy Putman Feb 25 2011 at 20:34 (continued) The following might make this less confusing. In my answer, $f : S \rightarrow S$ is a biholomorphic map between two different Riemann surfaces on the same topological surface $S$ (they are isomorphic Riemann surfaces, but the surfaces are "wearing" their Riemann surfaces structures on different ways). This is exactly what happens when the whole mapping class group acts on Teichmuller space. Two points in the same mapping class group orbit on Teichmuller space are isomorphic Riemann surfaces, but they are not the same Riemann surface! – Andy Putman Feb 25 2011 at 20:37 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 60, "mathjax_display_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": 0.938133955001831, "perplexity_flag": "head"}
http://mathhelpforum.com/geometry/199196-area-enclosed-semi-circle-quarter-circle.html	Thread: 1. Area enclosed by a semi-circle and a quarter circle Find the shaded area, which is enclosed by a square, semi-circle and a quarter of circle. Attached Thumbnails 2. Re: Area enclosed by a semi-circle and a quarter circle Draw the common chord of the two circles.Area of shaded region is the sum of the two segments.Central angles of these are readily calculable 3. Re: Area enclosed by a semi-circle and a quarter circle Got it. Thanks a lot, bjhopper! 4. Re: Area enclosed by a semi-circle and a quarter circle Could you explain this in a little bit more depth? I am not seeing how to get this. Thank you. 6. Re: Area enclosed by a semi-circle and a quarter circle An alternative is to use calculus. The quarter circle has radius $\displaystyle \begin{align} 7\,\textrm{cm} \end{align}$ and is centred at $\displaystyle \begin{align} (0, 0) \end{align}$, so has the equation $\displaystyle \begin{align} x^2 + y^2 = 49 \end{align}$. The semicircle has radius $\displaystyle \begin{align} \frac{7}{2}\,\textrm{cm} \end{align}$ and is centred at $\displaystyle \begin{align} \left(7, \frac{7}{2}\right) \end{align}$, so has the equation $\displaystyle \begin{align} \left(x - 7\right)^2 + \left(y - \frac{7}{2}\right)^2 = \frac{49}{4} \end{align}$. So if we are trying to find the area of the enclosed region, we use $\displaystyle \begin{align} \int{\int_R{}\,dA} \end{align}$, where $\displaystyle \begin{align} A \end{align}$ represents an infinitesimal element of area. Since you are dealing with circular bounds, it will be easiest to convert to polars, so each element of area can be written as $\displaystyle \begin{align} r\,dr\,d\theta \end{align}$. Now as for the bounds, it can clearly be seen that since the semicircle has a radius of $\displaystyle \begin{align} \frac{7}{2}\,\textrm{cm} \end{align}$ and the quarter circle has a radius of $\displaystyle \begin{align} 7\,\textrm{cm} \end{align}$, the region of integration extends over $\displaystyle \begin{align} \frac{7}{2} \leq r \leq 7 \end{align}$. If we solve the two equations simultanously, we can see that they intersect at $\displaystyle \begin{align} (x, y) = \left(\frac{21}{5}, \frac{28}{5}\right) \end{align}$. This means at the point of intersection: $\displaystyle \begin{align} x &= r\cos{\theta} \\ \frac{21}{5} &= 7\cos{\theta} \\ \frac{3}{5} &= \cos{\theta} \\ \theta &= \arccos{\frac{3}{5}} \end{align}$ So the region of integration is swept out over $\displaystyle \begin{align} 0 \leq \theta \leq \arccos{\frac{3}{5}} \end{align}$. Therefore our double integral is $\displaystyle \begin{align} A &= \int_0^{\arccos{\frac{3}{5}}}{\int_{\frac{7}{2}}^7 {r\,dr}\,d\theta} \\ &= \int_0^{\arccos{\frac{3}{5}}}{\left[\frac{r^2}{2}\right]_{\frac{7}{2}}^7\,d\theta} \\ &= \int_0^{\arccos{\frac{3}{5}}}{\frac{7^2}{2} - \frac{\left(\frac{7}{2}\right)^2}{2}\,d\theta} \\ &= \int_0^{\arccos{\frac{3}{5}}}{ \frac{147}{8} \, d\theta } \\ &= \left[\frac{147\,\theta}{8}\right]_0^{\arccos{\frac{3}{5}}} \\ &= \frac{147\arccos{\frac{3}{5}}}{8} - \frac{147 \cdot 0}{8} \\ &= \frac{147\arccos{\frac{3}{5}}}{8} \\ &\approx 17.039\,\textrm{cm}^2 \end{align}$ 7. Re: Area enclosed by a semi-circle and a quarter circle That's two different answers, (11.781 and 17.039). When I first saw this problem I automatically thought calculus. On reflection, after bjhopper's response, I think I prefer the geometric method. However, I already had a result, (using Cartesian co-ordinates rather than polar). I'd changed the orientation so that the quarter circle had its centre at (0,7) and the half circle had its centre at (7/2,0) so as to have simple upper and lower boundary curves. That leads to the integral $\int ^{28/5}_{0} \sqrt{7x-x^{2}}-7+\sqrt{49-x^{2}}\quad dx$ which evaluates to approximately 11.7813. 8. Re: Area enclosed by a semi-circle and a quarter circle Nice work Yeoky Thanks. 10. Re: Area enclosed by a semi-circle and a quarter circle May I know how you integrate? Thanks. 11. Re: Area enclosed by a semi-circle and a quarter circle Thank a lot, Prove it. May I know the answer should be 17.039 or 11.781?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9387797713279724, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/3111/solving-equations-with-multiple-absolute-values?answertab=votes	# Solving equations with multiple absolute values I know the method of solving the equation like this $\|2x+1\|=\|3x+9\|$ but the problem is if the same equation would be like this $\| 2x+1 \| = x \| 3x+9 \|$, how can I solve this? - 1 Again, the point that has been raised to you in your past few questions: treat the absolute value as a function with two cases. $\|2x+1\|=\mathrm{something}$ can mean either of $2x+1=\mathrm{something}$ or $-(2x+1)=\mathrm{something}$. You will have four "cases" since you have two absolute values; work from that. – J. M. Aug 23 '10 at 12:28 actuall we can solve the first example as 2x+1=3x+9 or 2x+1=-(3x+1) we have only two cases to solve , now i want to discover that how we can solve the second example in this way – Zia ur Rahman Aug 23 '10 at 12:35 1 The same technique still applies, Zia. – J. M. Aug 23 '10 at 12:37 – J. M. Aug 23 '10 at 12:47 1 Eleven exclamation marks look like someone's shouting. What shall I picture for myself in the face of eleven interrogation marks? – Rasmus Aug 23 '10 at 12:54 show 3 more comments ## locked by mixedmath♦Sep 4 '12 at 4:16 This question exists because it has historical significance, but it is not considered a good, on-topic question for this site, so please do not use it as evidence that you can ask similar questions here. This question and its answers are frozen and cannot be changed. More info: FAQ. ## 2 Answers To solve $\|f(x)\|=g(x)\|h(x)\|$ in general, find the sets $P=\{x:f(x)g(x)\ge 0\}$ and $N=\{x:f(x)g(x)\lt 0\}$. Then solve the two equations 1. $f(x)=g(x)h(x)$ - but discard solutions that are not in $P$. 2. $f(x)=-g(x)h(x)$ - but discard solutions that are not in $N$. You will be left with the solutions of $\|f(x)\|=g(x)\|h(x)\|$. - Both equations you mention can be solved in the same way. As others have said in the comments on your question here and the comments and answers on this question, $\|ax-b\|=\begin{cases} ax-b & \text{ if } x\ge \frac{b}{a} \\ b-ax & \text{ if } x< \frac{b}{a} \end{cases}$, where I'll call $\frac{b}{a}$ a "turning point" value. So, for any equation an absolute value, there is at least one turning point value of the variable for each absolute value. In both of your examples, the turning point values are $-\frac{1}{2}$ and $-3$. The $n$ (2 in your examples) turning point values divide the number line into $n+1$ (3 in your examples) regions where, within each region, each absolute value can be replaced by either the expression inside the absolute value or its opposite. Each resulting equation can then be solved, restricting solutions to the corresponding region on the number line. Let's consider a different example: $x\|4x-1\|=\|3x-2\|$. The turning points are $\frac{1}{4}$ and $\frac{2}{3}$. $$\begin{matrix} \leftarrow & \frac{1}{4} & \text{---} & \frac{2}{3} & \rightarrow \\ x(1-4x)=(2-3x) & \| & x(4x-1)=(2-3x) & \| & x(4x-1)=(3x-2) \\ x-4x^2=2-3x & \| & 4x^2-x=2-3x & \| & 4x^2-x=3x-2 \\ 0=4x^2-4x+2 & \| & 4x^2+2x-2=0 & \| & 4x^2-4x+2=0 \\ \text{no }x\in\mathbb{R} & \| & x=-1\text{ or }x=\frac{1}{2} & \| & \text{no }x\in\mathbb{R} \end{matrix}$$ (edit: Note that between any two adjacent regions, the only change in the equations at the tops of the regions is the one term with turning point at the boundary between the two regions.) Now, the solutions for each region-specific equation apply only to that region, so the solutions -1 and $\frac{1}{2}$ in the middle region only apply to the interval $[\frac{1}{4},\frac{2}{3}]$, so -1 is not a solution (it's not in the interval) but $x=\frac{1}{2}$ is a solution (and it's the only solution to $x\|4x-1\|=\|3x-2\|$). Regardless of the algebraic method of solution, graphing both sides of the equation and looking for points of intersection is a good way to check your solution. -	{"extraction_info": {"found_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": 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": 0.9268468618392944, "perplexity_flag": "head"}
http://cms.math.ca/10.4153/CJM-2010-068-1	Canadian Mathematical Society www.cms.math.ca \| \| \| \| \| \|----------\|----\|-----------\|----\| \| \| \| \| \| \| \| \| Site map \| \| \| CMS store \| \| location: Publications → journals → CJM Abstract view Holomorphic variations of minimal disks with boundary on a Lagrangian surface Read article [PDF: 166KB] http://dx.doi.org/10.4153/CJM-2010-068-1 Canad. J. Math. 62(2010), 1264-1275 Published:2010-08-18 Printed: Dec 2010 • Jingyi Chen, Department of Mathematics, The University of British Columbia, Vancouver, BC, V6T 1Z2 • Ailana Fraser, Department of Mathematics, The University of British Columbia, Vancouver, BC, V6T 1Z2 Features coming soon: Citations (via CrossRef) Tools: Search Google Scholar: Format: HTML LaTeX MathJax PDF Abstract Let $L$ be an oriented Lagrangian submanifold in an $n$-dimensional Kähler manifold~$M$. Let $u \colon D \to M$ be a minimal immersion from a disk $D$ with $u(\partial D) \subset L$ such that $u(D)$ meets $L$ orthogonally along $u(\partial D)$. Then the real dimension of the space of admissible holomorphic variations is at least $n+\mu(E,F)$, where $\mu(E,F)$ is a boundary Maslov index; the minimal disk is holomorphic if there exist $n$ admissible holomorphic variations that are linearly independent over $\mathbb{R}$ at some point $p \in \partial D$; if $M = \mathbb{C}P^n$ and $u$ intersects $L$ positively, then $u$ is holomorphic if it is stable, and its Morse index is at least $n+\mu(E,F)$ if $u$ is unstable. MSC Classifications: 58E12 - Applications to minimal surfaces (problems in two independent variables) [See also 49Q05] 53C21 - Methods of Riemannian geometry, including PDE methods; curvature restrictions [See also 58J60] 53C26 - Hyper-Kahler and quaternionic Kahler geometry, special'' geometry © Canadian Mathematical Society, 2013 © Canadian Mathematical Society, 2013 : http://www.cms.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": 20, "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": 0.3, "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": 0.7702418565750122, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/108045-subspace.html	# Thread: 1. ## subspace hi we have $\mathbb{E}=\left \{ (x,y)\in \mathbb{R}^{2};xy=0 \right \}$,i want to know if $\mathbb{E}$ is stable for addition.i also want to know some informations about stability in sets it's new for me. thanks. 2. Originally Posted by Raoh hi we have $\mathbb{E}=\left \{ (x,y)\in \mathbb{R}^{2};xy=0 \right \}$,i want to know if $\mathbb{E}$ is stable for addition.i also want to know some informations about stability in sets it's new for me. thanks. What does being "stable for addition" for a set mean, anyway? Does it mean it is closed wrt addition? If so then it's simple: check what happens with (1,0) and (0,1). Tonio 3. well for $X=(1,0)$ and $Y=(0,1)$ elements of $\mathbb{E}$,we have X+Y=(1,1) not element of $\mathbb{E}$,does this equivalent of saying that $\mathbb{E}$ is not stable for addition ? 4. Originally Posted by Raoh well for $X=(1,0)$ and $Y=(0,1)$ elements of $\mathbb{E}$,we have X+Y=(1,1) not element of $\mathbb{E}$,does this equivalent of saying that $\mathbb{E}$ is not stable for addition ? Ok, what part of "what does being stable for addition mean" you didn't understand?? I DO NOT KNOW what does being stable for addition mean: I said that if it is what ALLLL the rest of the world calls "closed under addition" then...etc.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.971991240978241, "perplexity_flag": "middle"}
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Therefore, the 2D Jacobi iterative solver is ... 4answers 255 views ### Some good reading on polygon algorithms What are some good resources (books, articles, sites) about polygon intersection and union algorithms? 0answers 147 views ### Why is my lower convex hull extraction algorithm not working? Recently, I wrote an algorithm to obtain a delaunay triangulation of a random point set in $I=[-10,10]$x$[-10,10] \subset R^2$ by projecting these points onto the 3 dimensional paraboloid $z=x^2+y^2$, ... 1answer 301 views ### Sort a cloud of points with respect to an unstructured mesh of hexahedral cells Question How would you sort a cloud of points with respect to an unstructured mesh of hexahedral cells? Each cell has a centre and a unique label to represent it. There are two cloud points ... 1answer 103 views ### Given values on a mesh, what algorithm can I use to construct efficiently level set contours? 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(open-source preferred)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9130470752716064, "perplexity_flag": "middle"}
http://www.mathplanet.com/education/algebra-1/factoring-and-polynomials/factor-polynomials-on-the-form-of-ax%5E2-plus-bx-plusc	# Factor polynomials on the form of ax^2 + bx +c In the same way as you could factor trinomials on the form of $\\ x^{2}+bx+c \\$ You can factor polynomials on the form of $\\ ax^{2}+bx+c \\$ If a is positive then you just proceed in the same way as you did previously except now $\\ ax^{2}+bx+c=\left ( x+m \right )\left ( ax+n \right ) \\\\where\: \: c=mn,\: \: ac=pq\: \: \\and\: \: b=p+q=am+n \\$ Example: $\\ 3x^{2}-2x-8 \\$ We can see that c (-8) is negative which means that m and n does not have the same sign. We now want to find m and n and we know that the product of m and n is -8 and the sum of m and n multiplied by a (3) is b (-2) which means that we're looking for two factors of -24 whose sum is -2 and we also know that one of them is positive and of them is negative. $\\ \begin{matrix} Factors\: of\: -24 \: \: \: & Sum\: of\: factors\\ -1, 24 & 23\\ 1,-24&-23 \\ -2,12&10 \\ 2,-12&-10 \\ -3,8&5 \\ 3,-8& -5\\ -4,6&2 \\ {\color{green} 4,-6}&{\color{green} -2} \end{matrix} \\$ This means that: $\\ 3x^{2}-2x-8=\\=3x^{2}+\left ( 4-6 \right )x-8 =\\=3x^{2}+4x-6x-8 \\$ We can then group those terms that have a common monomial factor. The first two terms have x together and the second two -2 and then factor the two groups. $\\ =\begin{pmatrix} 3x^{2}+4x \end{pmatrix}+\begin{pmatrix} -6x-8 \end{pmatrix}= \\\\=x\begin{pmatrix} 3x+4 \end{pmatrix}-2\begin{pmatrix} 3x+4 \end{pmatrix} \\$ Notice that both remaining parenthesis are the same. This means that we can rewrite this using the distributive propertyit as: $\\ =\begin{pmatrix} x-2 \end{pmatrix}\begin{pmatrix} 3x+4 \end{pmatrix}=3x^{2}-2x-8 \\$ This method is called factor by grouping. A polynomial is said to be factored completely if the polynomial is written as a product of unfactorable polynomials with integer coefficients. Video lesson: Factor the following polynomial $\\2x^2+10+12\\$ Next Class: Quadratic equations, The graph of y = ax^2 + bx + c • Pre-Algebra • Algebra 1 • Algebra 2 • Geometry • Sat • Act	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 9, "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": 0.9663882851600647, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/166399-x-u-v-y-u-v-u-x-y-v-x-y.html	# Thread: 1. ## x(u,v),y(u,v) to u(x,y),v(x,y) Can anybody help with this: I have $x=u+sin v, y=v+cos u$. How to extract u,v as functions from x,y, i.e. to get u(x,y) and v(x,y)? 2. Essentially, you want to solve the two non-linear equations x= u+ sin v, y= v+ cos u for u and v in terms of x and y. I doubt that there will be any elementary solution for that. 3. yes, i believe in that, so i am looking for some particular step which could make this task easier. 4. Does a problem specifically ask for u(x,y) and v(x,y)? Because if it is only asking for derivatives at certain points, you don't actually need the explicit formulas. 5. No, the task actually is: $x=u+sin(v)$ $y=v+cos(u)$ $z(u,v)=uv^2$ Find $\frac{\partial z}{\partial x} + \frac {\partial z}{\partial y}$ at point $(\frac {\pi +3}{3} , \frac {\pi +1}{3}).$ So I must find $\frac{\partial z}{\partial u} \frac{\partial u}{\partial x}$, $\frac{\partial z}{\partial u} \frac{\partial u}{\partial y}$, $\frac{\partial z}{\partial v} \frac{\partial v}{\partial x}$, $\frac{\partial z}{\partial v} \frac{\partial v}{\partial y}$. 6. Right, so you don't actually need the explicit formula. For example, to find $\frac{\partial u}{\partial x}$ and $\frac{\partial v}{\partial x}$. Take the partials of the equations directly (independent variables are $x,y$): $<br /> \frac{\partial x}{\partial x} = \frac{\partial u}{\partial x} + \cos(v)\frac{\partial v}{\partial x}<br />$ $<br /> \frac{\partial y}{\partial x} = \frac{\partial v}{\partial x} - \sin(u)\frac{\partial u}{\partial x}<br />$ Note that $\frac{\partial x}{\partial x} = 1$ and $\frac{\partial y}{\partial x}=0$. Solve this system of linear equations at the given point for $\frac{\partial v}{\partial x}, \frac{\partial u}{\partial x}$. Similar process for other derivatives. Note you do need to solve for u and v at the point $(\frac {\pi +3}{3} , \frac {\pi +1}{3})$, and you use these values when evaluating things like sin(u). 7. Originally Posted by waytogo No, the task actually is: $x=u+sin(v)$ $y=v+cos(u)$ $z(u,v)=uv^2$ Find $\frac{\partial z}{\partial x} + \frac {\partial z}{\partial y}$ at point $(\frac {\pi +3}{3} , \frac {\pi +1}{3}).$ So I must find $\frac{\partial z}{\partial u} \frac{\partial u}{\partial x}$, $\frac{\partial z}{\partial u} \frac{\partial u}{\partial y}$, $\frac{\partial z}{\partial v} \frac{\partial v}{\partial x}$, $\frac{\partial z}{\partial v} \frac{\partial v}{\partial y}$. Do you know about Jacobians? I would suggest you check the point. It looks a little dubious. 8. Is $\left(\frac{\pi+ 3}{3}, \frac{\pi+ 1}{3}\right)$ (x, y) or (u, v)? 9. the point was incorrect, the actual point is $(\frac{\pi+3}{3},\frac{\pi+1}{2})$. If I have caught your idea, I should use theorem about the inverse Jacobian? 10. The point now makes sense. Yes, inverse Jacobians is the way I would do 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": 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": 0.3, "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": 0.9049712419509888, "perplexity_flag": "middle"}

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