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http://mathoverflow.net/questions/59020/where-can-i-find-a-modern-write-up-of-heegners-solution-of-gauss-class-number-1/59030	## Where can I find a modern write-up of Heegner’s solution of Gauss' class number 1 problem? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In a recent MO question someone mentioned Heegner's solution of the Gauss "class number 1" problem which takes the following form: • When the class number of an imaginary quadratic form is 1 an elliptic curve is defined over $\mathbb{Q}$ and a modular function takes on integer values at certain quadratic irrationalities which leads to a collection of Diophantine equations: The solution of which finishes the theorem. I sadly can't read Heegner's original work (since I cannot read German) but also I don't think it's necessarily the best thing to read for this proof due to an alleged gap. So if anyone recognizes this proof sketch sketch and knows where I could read this in detail that would be wonderful! Thanks. - 5 The "gap" in Heegner's proof is filled in/shown to be non-existent by the article of Birch I mentioned in my answer to the version of this question that you asked on Math.SE. – Emerton Mar 21 2011 at 1:33 @quanta: Years ago I lectured on a more algebraic version of Heegner's proof, with elliptic curve arguments replacing the use of the Weber functions. I still have handwritten notes from my talk; if I can decipher them, and if you send me a mailing address, I'll put something in the mail to you. – paul Monsky Mar 21 2011 at 19:23 1 In her recent paper "Normalizers of non-split Cartan subgroups, modular curves, and the class number one problem," Burcu Baran discusses Heegner's technique in clean modern language. The key observation is that (by the theory of CM), each quadratic number field of class number 1 yields integral points on many modular curves. Generally those curves should have only finitely many integral points, and when the genus is small, one can compute them all. In her paper, after explaining the technique, Baran analyzes (the last) three modular curves where this approach will work. – james-parson Mar 21 2011 at 21:21 Though slightly off-topic, I want to remark that one can also find a proof of the Stark-Heegner theorem in Elkie's exposition of the Klein quartic (the proof originates with Kenku): library.msri.org/books/Book35/files/elkies.pdf – Lennart Meier Feb 22 at 1:18 ## 1 Answer In his article On the "gap'' in a theorem of Heegner, Stark does a pretty thorough job of explaining where people thought the purported gap came from, to what extent it actually was a gap, and what you would need to fix such a thing if it existed. I'm paraphrasing, but he basically argues that the confusion stemmed from some errors (typos?) in some analytic results of Weber that Heegner had heavily used. So in a literal sense, Heegner had not proved it because he had cited faulty results, but Stark shows that he deserved credit for the theorem since using Heegner's argument with the correct versions of Weber results (which were indeed known to Weber), the job gets done. Here's the mathscinet review of the article: http://www.ams.org/mathscinet-getitem?mr=241384 -	{"extraction_info": {"found_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": 0.9399668574333191, "perplexity_flag": "middle"}
http://amathew.wordpress.com/tag/fac/	# Climbing Mount Bourbaki Thoughts on mathematics August 25, 2011 ## Analytic spaces Posted by Akhil Mathew under functional analysis \| Tags: analytic spaces, coherent sheaves, FAC, Oka's theorem \| Consider the space ${\mathbb{C}^n}$ and the sheaf ${\mathcal{O}}$ of holomorphic functions on it. One should think of this as the analog of complex affine space ${\mathbb{C}^n}$, with the Zariski topology, and with the sheaf ${\mathcal{O}_{reg}}$ of regular functions. In algebraic geometry, if ${I \subset \mathbb{C}[x_1, \dots, x_n]}$ is an ideal, or if ${\mathcal{I} \subset \mathcal{O}_{reg}}$ is a coherent sheaf of ideals, then we can define a closed subset of ${\mathbb{C}[x_1,\dots, x_n]}$ corresponding to the roots of the polynomials in ${I}$. This construction gives the notion of an affine variety, and by gluing these one gets general varieties. More precisely, here is what an affine variety is. If ${\mathcal{I} \subset \mathcal{O}_{reg}}$ is a coherent sheaf of ideals, then we define a ringed space ${(\mathrm{supp} \mathcal{O}_{reg}/\mathcal{I}, \mathcal{O}_{reg}/\mathcal{I})}$; this gives the associated affine variety. Here the “support” corresponds to taking the common zero locus of the functions in ${\mathcal{I}}$. In this way an affine variety is not just a subset of ${\mathbb{C}^n}$, but a locally ringed space. Now we want to repeat this construction in the holomorphic category. If ${\mathcal{I} \subset \mathcal{O}}$ is a finitely generated ideal—that is, an ideal which is locally finitely generated—in the sheaf of holomorphic functions on ${\mathbb{C}^n}$, then we define the space cut out by ${\mathcal{I}}$ to be ${(\mathrm{supp} \mathcal{O}/\mathcal{I}, \mathcal{O}_{reg}/\mathcal{I})}$. We can think of these as “affine analytic spaces.”	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 16, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.8892959356307983, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=1440733	Physics Forums ## why solvable is solvable When a Lie algebra is solvable, does it have something to do with actually solving something? 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 Recognitions: Gold Member Science Advisor Staff Emeritus A group, G, (whether "Lie" or not) is said to be solvable if and only if there exist a sequence of subgroups, G1, G2, ..., Gn of G, such that: G1 is the subgroup consisting only of the group identity, Gn is G itself, Each Gi is a normal subgroup of Gi+1, and Gi+1/Gi is commutative. The name, and indeed the whole definition, is from Galois's answer to the question of whether or not there could exist a "formula" for polynomials of degree 5 or higher, in terms of radicals. The point is that a polynomial equation, with integer coefficients, is "solvable by radicals" if and only if its Galois group is a "solvable" group. I think (I'm entering unsure ground here) that Galois also showed that, for n> 4, there exist a polynomial of degree n whose Galois group is all of Sn (the group of permutations on n objects) and that Sn, for n> 4, is not a "solvable" group. Recognitions: Science Advisor In his 1831 paper, Galois (age 20) defined a group for any polynomial with rational coefficients, a permutation group acting on the roots, which we now call the galois group of the polynomial. He showed (in essence) that this group acts transitively iff the polynomial is irreducible, and if so, it is solvable iff the roots of the polynomial can be expressed in closed form in terms of arithmetic operations plus the extraction of roots (i.e. the problem of finding the roots can be reduced to finding the roots of monomials). In such cases, the roots are expressed as nested radicals, in a manner which mirrors the descending subnormal chain of subgroups (the composition series exhibiting the fact that the group is indeed solvable--- the composition series can be refined so that the quotients mentioned by Halls are in fact prime order cyclic groups, corresponding to monomials of prime degree). In modern terms, he exhibited a galois duality between the subgroups of the galois group and the intermediate extension fields of the extension field of the rationals defined by adjoining the roots of our polynomial to the field of rational numbers. Galois also showed that while "special" polynomials of high degree can have solvable galois groups, the group of a "generic" n-degree irreducible polynomial is the symmetric group on n letters, $S_n$, which is not solvable for n > 4 (indeed its index two normal subgroup $A_n$ is not solvable for n > 4). Thus, there can be no general formula analogous to the quadratic formula for the quintic or higher degrees. In order to do all this, since groups, rings, and fields (and group actions, and finite projective planes...) did not yet exist, he had to invent them. Indeed, most would agree that he invented modern algebra. All this in a dozen pages, yet his paper clearly contains all the essential ideas, albeit in sketchy and sometimes delphic form. This paper is often regarded as one of the single most profound advances in mathematical thought, because Galois was the first to clearly see that algebraic objects more complicated than numbers are worthy of recognition and study. In a sense, he took the first step down the path of categorification. As we recently discussed in some other PF threads, he also introduced one of the great themes of mathematics, the notion of symmetry (and its relation to the notion of information). His remarkable achievement in completely resolving a problem which had remained unsolved for millenia, in an utterly original and completely unexpected manner, inspired many of the greatest nineteenth century mathematicians, including Sophus Lie (whose dream of doing for differential equations what Galois had done for polynomials led to the development of Lie theory as required background for Lie's theory of the symmetries of differential equations). And Galois continues to inspire mathematicians today (one might mention Grothendieck as a more recent example). For the simplest example of nested radicals, see Cardano's formula for the roots of a cubic: http://gowers.wordpress.com/2007/09/...for-the-cubic/ Thread Tools \| \| \| \| \|-----------------------------------------------\|----------------------------\|---------\| \| Similar Threads for: why solvable is solvable \| \| \| \| Thread \| Forum \| Replies \| \| \| General Math \| 17 \| \| \| Calculus & Beyond Homework \| 1 \| \| \| Quantum Physics \| 0 \| \| \| Linear & Abstract Algebra \| 3 \| \| \| Calculus \| 2 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9467911124229431, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/1491/can-i-parameterize-the-state-of-a-quantum-system-given-reduced-density-matrices?answertab=votes	# Can I parameterize the state of a quantum system given reduced density matrices describing its subparts? As the simplest example, consider a set of two qubits where the reduced density matrix of each qubit is known. If the two qubits are not entangled, the overall state would be given by the tensor product of the one qubit states. More generally, I could write a set of contraints on the elements of a two-qubit density matrix to guarantee the appropriate reduced description. Is there is a way to do this more elegantly and systematically for arbitrary bi-partite quantum systems? I'm particularly interested in systems where one of the Hilbert spaces is infinite dimensional, such as a spin 1/2 particle in a harmonic oscillator. - ## 1 Answer Density matrices often admit an interesting geometric interpretations when you map them to the space of generalized Bloch vectors, see for example the book I. Bengtsson, K. Życzkowski, Geometry of quantum states, 2006. I won't be surprised if it turns out that the result has something to do with the coset space $SU(2N)/[SU(N)\times SU(N)]$, where N is the dimensionality of the Hilbert space of a single qubit. -	{"extraction_info": {"found_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": 0.8949117064476013, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/49394/list	2 LaTeX # classificationClassification of real forms byupto inner automorphisms I hope to know the classification of real forms of complex simple Lie algebras of types A, D, E $A$, $D$, $E$ up to inner automorphisms. Let g_1 $\mathfrak{g}_1$ and g_2 $\mathfrak{g}_2$ be real forms of a complex simple Lie algebra g. $\mathfrak{g}$. We say that they are equivalent if there is an isomorphism g_1 ---> g_2 $\mathfrak{g}_1 \to \mathfrak{g}_2$ which extends to an inner automorphism of g. This $\mathfrak{g}$. However, there may result in be real forms which are isomorphic but not equivalent. Where can we find the classification of such equivalence classes of real forms? 1 # classification of real forms by inner automorphisms I hope to know the classification of real forms of complex simple Lie algebras of types A, D, E up to inner automorphisms. Let g_1 and g_2 be real forms of a complex simple Lie algebra g. We say that they are equivalent if there is an isomorphism g_1 ---> g_2 which extends to an inner automorphism of g. This may result in real forms which are isomorphic but not equivalent. Where can we find the classification of such equivalence classes of real forms?	{"extraction_info": {"found_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": 0.8913487195968628, "perplexity_flag": "head"}
http://www.ugcs.caltech.edu/~srbecker/wiki/BP	# Category:Problems ### From SparseSolver (Redirected from BP) ## Sparse approximation problems ### Best subset selection This problem is in general NP-hard. Many of the algorithms presented on this wiki are attempts to solve this problem approximately, or to solve it in certain special cases. We denote the $\ell_0$ quasi-norm to be the number of non-zeros of a vector. $\min_x \\|x\\|_0 \quad\text{subject to}\quad Ax=b$. There are also variants that account for noise, or that flip the objective and constraint: $\min_x \frac{1}{2}\\|Ax-b\\|_2^2 \quad\text{subject to}\quad \\|x\\|_0 \le k$. See pages that link to this problem ### Basis Pursuit (BP) This is one of the best-known sparse recovery problems. The matrix A is usually underdetermined or ill-conditioned. This problem is useful for finding a solution x that is sparse. $\min_x \\|x\\|_1 \quad\text{subject to}\quad Ax=b$ See pages that link to this problem #### Synthesis variant Instead of looking for a sparse solution x, we instead look for a sparse representation α in a basis or overcomplete dictionary Ψ. The problem is $\min_\alpha \\|\alpha\\|_1 \quad\text{subject to}\quad A\Psi\alpha=b$ and then x is recovered via x = Ψα. See pages that link to this problem #### Analysis variant Again, we expect x to have a sparse representation in a dictionary Ψ, but this time the problem uses the analysis operator $\Psi^\dagger$. If Ψ is a tight frame, then this is just the adjoint/transpose: $\Psi^\dagger = \Psi^$: $\min_x \\|\Psi^\dagger x\\|_1 \quad\text{subject to}\quad Ax=b$ This is equivalent to synthesis when Ψ is a basis (i.e. when its matrix form is invertible), but otherwise it is not equivalent. See Rémi Gribonval's talk on synthesis vs analysis from SPARS 2011. See pages that link to this problem ### Basis Pursuit Denoising (BPDN) This is a variant of BP that is used when measurements are inexact or noisy in some way: $\min_x \\|x\\|_1 \quad\text{subject to}\quad \\|Ax-b\\|_2 \le \epsilon.$ There are synthesis and analysis variants of BPDN in the analogous fashion to BP. See pages that link to this problem ### Lagrangian form of BPDN (LAG), aka the LASSO This is an equivalent form to BPDN for the correct choice of λ = λ(ε). It is sometimes algorithmically easier to solve than BPDN, but the precise relation between λ and ε is in general unknown (unless you have already solved the problem). $\min_x \\|x\\|_1 + \frac{\lambda}{2} \\|Ax-b\\|_2^2.$ Again, there are analysis and synthesis variants. Since this form is somewhat easier to solve than BPDN, it is the most popular, and when people refer to "the LASSO", the usually mean LAG. See pages that link to this problem ### Original LASSO formulation of BPDN The term "LASSO" is used differently by different people, so we avoid it on this wiki. The "original LASSO" or "Tibshirani's LASSO" (TibLASSO) refers to the form proposed by Tibshirani in his classic paper from the mid '90s. It is equivalent to BPDN and LAG for some choice of parameter τ, but this choice of parameter is in general unknown (unless you know a solution to the problem). $\min_x \frac{1}{2}\\|Ax-b\\|_2^2 \quad\text{subject to}\quad \\|x\\|_1 \le \tau.$ See pages that link to this problem ### Elastic Net The Elastic Net was proposed in a statistical framework to overcome limitations to the LASSO. Below, we will keep to our earlier terminology (variable $x \in \mathbb{R}^N$, matrix $A \in \mathbb{R}^{M \times N}$) and convert the standard statistics notation to our notation. Some possible deficiencies of the LASSO in statistical regression: • The LASSO selects a sparse solution in general, and it has at most M non-zeros. This can sometimes be limiting. • If a group of variables are highly correlated pairwise, the LASSO often picks just one of them, and sets the other to zero. For interpreting results, it would be more useful to select each member of the group. • In the over-determined case, if A has high correlations in the columns, the LASSO performs worse than standard ridge regression, according to Tibshirani's original '96 LASSO paper. To overcome these perceived problems, Zou and Hastie [1] propose the Elastic Net. They start by defining the "naive" elastic net: $x_\text{naive Elastic Net} = \text{argmin} \\|Ax-b\\|_2^2 + \lambda_1 \\|x\\|_1 + \lambda_2 \\|x\\|_2^2.$ Assuming that one knows the values of λ1 and λ2, then this problem can be solved by using a standard LASSO solver, since the $\\|x\\|_2^2$ term can be absorbed into the $\\|Ax-b\\|_2^2$ term by augmenting the A matrix to become [A;τI] where I is the $N \times N$ identity matrix and τ is an appropriate scalar, and by similarly augmenting b to become [b;zeros(n,1)]. If good values of λ1 and λ2 are not known, then other solution techniques that generate solutions for a range of λ may be preferred. Zou and Hastie suggest that the naive estimator produces values that are too small, so the true Elastic Net is a scaled version of the naive estimator: $x_\text{Elastic Net} = (1+\lambda_2) x_\text{naive Elastic Net}.\,$ See pages that link to this problem ### Fused Lasso Before defining the Fused Lasso, consider the original Lasso (TibLASSO) in Tibshirani's original form: $\min_x \frac{1}{2}\\| Ax - b \\|_2^2 \quad\text{such that}\quad \\|x\\|_1 \le \tau.$ where $A \in \mathbb{R}^{m \times n}$ and, as usual, $\\|x\\|_1 = \sum_{i=1}^N \|x_i\|$. The Fused Lasso modifies this by adding a penalty on the difference of the coefficients of x. The problem is: $\min_x \frac{1}{2}\\| Ax - b \\|_2^2 \quad\text{such that}\quad \\|x\\|_1 \le \tau_1 \quad\text{and}\quad \sum_{i=2}^N \|x_i-x_{i-1}\| \le \tau_2.$ The effect is to not only encourage sparsity in the coefficients, but also sparsity in the difference of neighboring coefficients, and thus coefficients are more likely to clump together (either several coefficients in a row are zero, or they are nearly the same value). The Fused Lasso was first proposed in [2] See pages that link to this problem ### Dantzig selector (DS) Proposed by Candès and Tao: $\min_x \\|x\\|_1 \quad\text{subject to}\quad \\|A^(Ax-b)\\|_\infty \le \delta.$ If there is no noise, then take δ = 0, in which case it is the same as BP whenever A has full row rank (which it usually does for practical problems). The different form of the constraint is motivated by the optimality conditions of the dual vector. See pages that link to this problem ### Total Variation (TV) problems These come in many variants (depending on the constraints), but they all have in common the total-variation norm. This can be seen as a weighted norm of complex numbers. For a TV problem, we have a vector $x \in \mathbb{R}^n$ but it will help to think of this as a matrix (i.e. an image) $X \in \mathbb{R}^{n_1 \times n_2 }$ (with n = n1n2 ). Replace $\\|x\\|_1$ in the above problems with $\\|x\\|_{TV} = \sum_{i=1}^{n_1-1} \sum_{j=1}^{n_2-1} \sqrt{ (X(i+1,j)-X(i,j))^2 + (X(i,j+1)-X(i,j))^2 }$ Sometimes this is referred to as the "isotropic TV norm", as opposed to the "anisotropic TV norm", defined as: $\\|x\\|_{ATV} = \sum_{i=1}^{n_1-1} \sum_{j=1}^{n_2-1} \sqrt{ (X(i+1,j)-X(i,j))^2 }+ \sqrt{(X(i,j+1)-X(i,j))^2 }$ The isotropic TV norm is also sometimes regularized by adding a small nonzero offset under the square root, in order to make it differentiable everywhere. When we refer to just "TV", we mean the non-smoothed, isotropic TV norm, unless otherwise specified. The classic algorithm for solving TV regularized denoising problems is "Chambolle's Algorithm": it is basically the proximity function for TV. There is no known closed-form solution for the TV proximity function. There are other variants of TV that are sometimes easier to work with; for example, smoothing the norm in different fashions (corresponding to a different finite-difference stencil), or assuming periodic boundary conditions (which makes it amenable to analysis with the FFT). The PPXA paper, by Pustelnik, Chaux and Pesquet, discusses some TV variants using filter such as the Roberts filter, a first-order finite difference filter that is different than the one above, Prewitt filters, and Sobel filters. These filters allow one to decompose the TV function into a sum of a few auxiliary functions; these auxiliary functions admit a closed-form proximity operator, although the sum of the functions still does not. See pages that link to this problem ### Variants These variations can apply to most of the problems above. #### Weighted l1 norm Weighted $\ell_1$ norm: replace $\\|x\\|_1$ with $\\|Wx\\|_1$ for some weight matrix $W \in \mathbb{R}^{p \times n}$. Almost all of the algorithms can handle (or be modified to handle) this case when W is square and non-singular, or especially when it is orthogonal, so we don't point out this case. But we do note when algorithms can handle this variant for arbitrary W, e.g. when $p \gg n$ as in when W is the analysis operator for a frame. #### Block l1 norm Block norms, for the case of recovering multiple signals simultaneously: replace $\\|x\\|_1$ with $\\|X\\|_{q,1} = \sum_{i=1}^n \\|X(i,:)\\|_q$ where $X \in \mathbb{R}^{n \times d }$, using Matlab notation. Typically q = 2 or $q = \infty$. Also known as "multiple measurement vectors (MMV)" and used with fusion frames. The constraint norm becomes the matrix Frobenius norm instead of the $\ell_2$ norm. Most algorithms that work with $\ell_1$ will also work with the block norms (some may require you to modify them manually though). See pages that link to this problem ## Low-rank problems ### Matrix completion (MC) Let $X \in \mathbb{R}^{n_1 \times n_2}$ be a matrix, and Xi,j denote the (i,j) entries. The problem of matrix completion is to estimate the matrix X given an incomplete set of entries. The incomplete set is denoted Ω, so $(i,j) \in \Omega$ means that the Xi,j entry is observed. Recovering a matrix from incomplete entries is impossible unless there is further restrictions on the matrix. Usually the restriction is in the form of an assumption about the rank of the matrix. It is generally considered desirable to solve this problem: $\min_X \,\text{rank}(X) \quad\text{subject to}\quad X_{i,j} = Y_{i,j} \;\forall (i,j) \in \Omega$ where Y contains the partial observations. In general, this problem is not computationally feasible. When people refer to Matrix Completion, it sometimes refers to any heuristic method to solve the rank minimization problem, but it can also refer to the following nuclear-norm minimization problem: $\min_X \,\\|X\\|_* \quad\text{subject to}\quad X_{i,j} = Y_{i,j} \;\forall (i,j) \in \Omega$ The nuclear norm (aka trace norm aka Schatten norm with p = 1) is the sum of the singular values. For convenience, let $\mathcal{A}_\Omega(X)$ be the sampling operator that picks out the entries in Ω, so that the constraint can be written $\mathcal{A}_\Omega(X) = \mathcal{A}_\Omega(Y).$ Some algorithms can deal with general linear operators $\mathcal{A}$. If $\mathcal{A}$ satisfies a generalization of the RIP, then most compressed sensing results apply (see Guaranteed Minimum-Rank Solutions of Linear Matrix Equations via Nuclear Norm Minimization by Recht, Fazel and Parrilo arXiv:0706.4138 ). The subsampling operator $\mathcal{A}_\Omega(X)$ does not satisfy RIP bounds, so proof techniques are quite different, and reconstruction is often more difficult as well. See pages that link to this problem (some might also be under this MC alias ). #### Noisy matrix completion (NMC) This is a slight variant of the nuclear-norm minimization problem that allows for noisy data. To account for noise, the constraints are relaxed from equality constraints: $\min_X \,\\|X\\|_* \quad\text{subject to}\quad \\|\mathcal{A}_\Omega(X) - \mathcal{A}_\Omega(Y)\\|_F \le \epsilon$ It is common to use the Frobenius norm for the constraints since this is the maximum likelihood term for iid white noise, though other choices are possible. pages that link to this problem #### Lagrangian form of noisy matrix completion (LAG-MC) Similar to noisy matrix completion, but with the constraints put into an objective term: $\min_X \,\\|X\\|_* + \frac{\lambda}{2} \\|\mathcal{A}_\Omega(X) - \mathcal{A}_\Omega(Y)\\|_F^2$ For some choice of λ and ε, this is equivalent to NMC, but the exactly equivalent parameters are generally not known until after one has solved either problem. pages that link to this problem ### Robust PCA The classic PCA technique finds a low-dimensional representation of a dataset, but it is sensitive to outliers in the data. There have been many variants of PCA that attempt to make it more robust to outliers, so there is no consensus on what the robust PCA (RPCA) technique means. However, in the field of matrix completion, RPCA usually refers to the following problem: $\min_{S,L}\, \\|L\\|_* + \alpha \\|S\\|_1 \quad\text{subject to}\quad S+L = Y$ where $\\|L\\|_$ is the nuclear norm of L and $\\|S\\|_1$ is the sum of absolute values of the entries of S (that is, think of S as a vector). The nuclear norm will often induce the rank of L to be low, and the l1 norm will induce the number of nonzeros of S to be small. The idea is that L captures the predictable, low-rank structure of the data, and S contains outliers. For an idea of what RPCA can be used for, see the demo using TFOCS. pages that link to this problem ### Low-rank representation (LRR) This is another generalization of PCA. We seek a low-rank representation of a data matrix, and not only allow outliers (like the RPCA described above) but also allow simple transformations. The problem is: $\min_{S,L}\, \\|L\\|_ + \alpha \\|S\\|_{2,1} \quad\text{subject to}\quad S+AL = Y$ where A is some known transformation (if it is not known, then it can be estimated at a separate step, and generally one would iterate estimating A and solving the LRR problem, much as in dictionary learning). For background, see Robust Recovery of Subspace Structures by Low-Rank Representation by Liu, Lin, Yan, Sun, Yu and Ma. Yi Ma's group has worked on many similar variants, such as the ones in TILT: Transform Invariant Low-rank Textures (Zhang, Ganesh, Liang, Ma) and RASL: Robust Alignment by Sparse and Low-rank Decomposition for Linearly Correlated Images (Peng, Ganesh, Wright, Xu, Ma). pages that link to this problem ## References 1. ↑ Zou and Hastie, Regularization and variable selection via the elastic net, J. R. Statist. Soc. B (2005) 67, Part 2, pp. 301--320 link 2. ↑ Tibshirani, Saunders, Rossets, Zhu and Knight, Sparsity and smoothness via the fused lasso, J. Royal Stat. Soc. Ser. B 67, 2005, 91--108. link. ## Below is a list of pages in this category This category currently contains no pages or media.	{"extraction_info": {"found_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": 54, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9142094254493713, "perplexity_flag": "middle"}
http://mathhelpforum.com/statistics/129665-expected-value.html	# Thread: 1. ## Expected Value No idea where to start. The owner of a small firm has just purchased a personal computer, which she expects will serve her for the next two years. The owner has been told that she "must" buy a surge suppressor to provide protection for her new hardware against possible surges or variations in the electrical current, which have the capacity to damage the computer. The amount of damage to the computer depends on the strength of the surge. It has been estimated that there is a 2% chance of incurring 450 dollars damage, 6% chance of incurring 150 dollars damage, and 11% chance of 100 dollars damage. An inexpensive suppressor, which would provide protection for only one surge, can be purchased. How much should the owner be willing to pay if she makes decisions on the basis of expected value? 2. Originally Posted by tootiebee No idea where to start. The owner of a small firm has just purchased a personal computer, which she expects will serve her for the next two years. The owner has been told that she "must" buy a surge suppressor to provide protection for her new hardware against possible surges or variations in the electrical current, which have the capacity to damage the computer. The amount of damage to the computer depends on the strength of the surge. It has been estimated that there is a 2% chance of incurring 450 dollars damage, 6% chance of incurring 150 dollars damage, and 11% chance of 100 dollars damage. An inexpensive suppressor, which would provide protection for only one surge, can be purchased. How much should the owner be willing to pay if she makes decisions on the basis of expected value? it is always good to define what random variable you want to compute hte xpected value for. In this case, youre interested in the expected value of the random variable X, which is the amount of \$ spent on repairing the computer from surges. With the given info, we can compute $<br /> E[X] = .02450+.06150+.11100+(1-.02-.06-.11)0$ Since the owner would only want protection against surges if that protection cost less than the amount she would pay on average to repair the computer from surges, she would be willing to pay E[X]. 3. ## Expected Value #2 To examine the effectiveness of its four annual advertising promotions, a mail order company has sent a questionnaire to each of its customers, asking how many of the previous year's promotions prompted orders that would not have otherwise been made. The accompanying table lists the probabilities that were derived from the questionnaire, where X is the random variable representing the number of promotions that prompted orders. If we assume that overall customer behavior next year will be the same as last year, what is the expected number of promotions that each customer will take advantage of next year by ordering goods that otherwise would not be purchased? X01234P(X)0.0840.2240.3130.1960.183 Expected value = A previous analysis of historical records found that the mean value of orders for promotional goods is 23 dollars, with the company earning a gross profit of 28% on each order. Calculate the expected value of the profit contribution next year. Expected value = The fixed cost of conducting the four promotions is estimated to be 10000 dollars with a variable cost of 5.25 dollars per customer for mailing and handling costs. What is the minimum number of customers required by the company in order to cover the cost of promotions? (Round your answer up to the next highest integer.) Break even point =	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9623715281486511, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/55384?sort=oldest	## Primes represented by two-variable quadratic polynomials ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm looking over a paper, "Primes represented by quadratic polynomials in two variables" [1] which attempts to characterize the density of the primes in two-variable quadratic polynomials. Its apparent improvement over previous works is that it covers not just quadratic forms plus a constant but quadratic forms plus linear forms plus a constant. I would like to know what the current state of knowledge is for this sort of problem. 1. This paper covers only the case of polynomials that "depend essentially on two variables". This has surely not been improved, else the H-L Conjecture E, F, etc. would have been solved. 2. The paper allows only integers, not half-integers, as coefficients, and so does not give the number of primes, e.g., in A117112. 3. It does not give the constant (or prove that a constant exists!) for the densities it finds. The third is my main interest at this point. References would be appreciated, whether to research papers, survey papers, or standard texts. I also have some interest in the cubic case, if anything is known. It seems that even characterizing which two-variable cubics represent an infinite number of primes is open...? Certainly the conditions Iwaniec gives for two-variable quadratics do not suffice to give an infinity of primes. For example, with the polynomial $P(x,y)=ax^2+bxy+cy^2+ex+fy+g$, setting $\Delta=b^2-4ac$ and $D=af^2-bef+ce^2+g\Delta$ and $C(x)=\sum_{p=P(x,y)\le x}$1, [1] shows that $$\frac{x}{(\log x)^{1.5}}\ll C(x)\ll\frac{x}{(\log x)^{1.5}}$$ for all $P$ with $D\neq0$ and $\Delta$ not a square, but is it known that $C(x)\sim kx/(\log x)^{1.5}$ for some $k$? Are values of $e$ or $E$ known such that $$e<\liminf\frac{C(x)(\log x)^{1.5}}{x}$$ or $$\limsup\frac{C(x)(\log x)^{1.5}}{x}< E$$ ? Similarly, if $D=0$ or $\Delta$ is a square do we know when $$\ell=\lim\frac{C(x)\log x}{x}$$ exists and what its value is? Related question: Who should I cite for these results? I'm never sure of when there might have been simultaneous discoveries or other reasons for priority issues. [1] Iwaniec, H. (1974). Primes represented by quadratic polynomials in two variables. Acta Arithmetica 24, pp. 435–459. [2] Hardy, G. H., & Littlewood, J. E. (1923). Some problems of ‘Partitio numerorum’; III: On the expression of a number as a sum of primes. Acta Mathematica 44:1, pp. 1-70. - I'm finding a number of (slightly) older results by Peter Pleasants which may address the issue. I'll give more details as I find them. – Charles Feb 14 2011 at 15:41 1 Regarding 2., I am sure that retracing the arguments of [1] will allow half-integer constants with almost no trouble. Regarding 3., this is subtler; it's easy to write down what the asymptotic density ought to be, but the combinatorial sieve techniques employed by Iwaniec and other sieve theorists often (as in [1]) exploit positivity of various gnarly terms in order to discard them completely, which inevitably produces a lower bound rather than a precise asymptotic. For the most up-to-date material, you might try Friedlander and Iwaniec's book "Opera de Cribro". – David Hansen Feb 14 2011 at 17:44 OK, my library has Opera de Cribro, I guess I'll check that out and see what it has. – Charles Feb 14 2011 at 19:47 I'm having trouble with your "For example," partly because you define but never use $\Delta$ and $D$, partly because of confusion over $X$ and $x$, and mostly because $P(x,y)=x^2+y^2$ represents all primes congruent 1 mod 4, which should be asymptotic to $x/(2\log x)$. – Gerry Myerson Feb 14 2011 at 22:40 I defined D and Delta so I could give the conditions under which the result holds. Unfortunately I neglected to give it! Let me edit. – Charles Feb 15 2011 at 2:18 ## 1 Answer The modern reference work on the subject seems to be [1], but it spends only a page and a half on the subject of primes in multivariate quadratic polynomials (pp. 396-397). More than half this space is devoted to Iwaniec's 1974 result. The balance mentions Sarnak's application to the Problem of Apollonius and a result of "J. Cho and H. Kim" on counting primes in $\mathbb{Q}[\sqrt{-2}].$ So nothing there. Pleasants [2] shows that, subject to a Davenport-Lewis [3] condition on the $h^$ (a complexity measure on the cubic form part), multivariate cubic polynomials have the expected number of primes. Unfortunately this condition requires (as a necessary but insufficient condition) that there be at least 8 variables. Further, it double-counts repeated primes. Goldoni [4] recently wrote a thesis on this general topic. His new results (Chapter 5) on the $h$ and $h^$ invariants make it easier to use the results of Pleasants but do not extend them to cubic polynomials with fewer than 8 variables. Of course I would be remiss in failing to mention the groundbreaking work of Heath-Brown [5], building on Friedlander & Iwaniec [6]. These results will no doubt clear the way for broader research, but so far have not been generalized. So in short it appears that: • Nothing further is known about primes represented by quadratic polynomials. • Apart from $x^3+2y^3$, almost nothing is known about which primes are represented by cubic polynomials, though some results are known for how often such polynomials take on prime values provided $h^*$ and hence the number of variables is large enough. On the historical side, of course Fermat is responsible for the proof of the case $x^2+y^2$. I have references that say that Weber [7] and Schering [8] handled the case of (primitive) binary quadratic forms with nonsquare discriminants, but I haven't read the papers. Motohashi [9] proved that there are $\gg n/\log^2 n$ primes of the form $x^2+y^2+1$ up to $n$, apparently (?) the first such result with a constant term. He conjectured that the true number was $$\frac{n}{(\log n)^{3/2}}\cdot\frac32\prod_{p\equiv3(4)}\left(1-\frac{1}{p^2}\right)^{-1/2}\left(1-\frac{1}{p(p-1)}\right)$$ but as far as I know the constant still has not been proved even for this special form. Edit: Apparently Bredihin [10] proved the infinitude of primes of the form $x^2+y^2+1$ some years before Motohashi. He only gave a slight upper-bound on their density, though: $O(n/(\log n)^{1.042}).$ (Motohashi improved the exponent to 1.5 in a later paper.) [1] Friedlander, J. and Iwaniec, H. (2010). Opera de Cribro. AMS. [2] Pleasants, P. (1966). The representation of primes by cubic polynomials, Acta Arithmetica 12, pp. 23-44. [3] Davenport, H. and Lewis, D. J. (1964). "Non-homogeneous cubic equations". Journal of the London Mathematical Society 39, pp. 657-671. [4] Goldoni, L. (2010). Prime Numbers and Polynomials. Doctoral thesis, Università degli Studi di Trento. [5] Heath-Brown, D. R. (2001). Primes represented by $x^3 + 2y^3$. Acta Mathematica 186, pp. 1-84. [6] Friedlander, J. and Iwaniec, H. (1997). Using a parity-sensitive sieve to count prime values of a polynomial. Proceedings of the National Academy of Sciences 94, pp. 1054-1058. [7] Weber, H. (1882). "Beweis des Satzes, dass, usw". Mathematische Annalen 20, pp. 301-329. [8] Schering, E. (1909). "Beweis des Dirichletschen Satzes". Gesammelte mathematische Werke, Bd. 2, pp. 357-365. [9] Motohashi, Y. (1969). On the distribution of prime numbers which are of the form $x^2+y^2+1$. Acta Arithmetica 16, pp. 351-364. [10] Bredihin, B. M. (1963). Binary additive problems of indeterminate type II. Analogue of the problem of Hardy and Littlewood (Russian). Izv. Akad. Nauk. SSSR. Ser. Mat. 27, pp. 577-612. - Almost surely, Heath-Brown's result on $x^3+2y^3$ will generalize readily to $ax^3+by^3$. – Greg Martin Sep 16 2011 at 21:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9059343934059143, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/120010/list	## Return to Answer 2 added 4 characters in body I don't know how much about areas you want to prove, and how developed a background the audience is supposed to have, but here is a definition of area of a bounded planar region. Take such a region $D$ in $\mathbb{R}^2$, $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\ve}{{\varepsilon}}$ consider the $\varepsilon$-grid $(\varepsilon\bZ)^2$ and denote by $N_\ve(D)$ the number of $\ve$-pixels that touch $D$. (A pixel is one of the $\ve\times\ve$-tiny squarers of the grid.) We then declare $D$ to be measurable (i.e. to have area) if the limit $\newcommand{\eA}{\mathscr{A}}$ $$\eA(D)= \lim_{\ve\to0}\ve^2 N_\ve(D). N_\ve(D)$$ exists. If this is the case, then we define the area to be the limit $\eA(D)$, and we set $\eA_\ve(D)=\ve^2N_\ve(D)$. The first step is to prove $\newcommand{\bR}{\mathbb{R}}$ that if $L,U: [a, b]\to \bR$ are Riemann integrable functions and $D(U,L)$ is the region $$D_f= \bigl\lbrace\; (x,y)\in [a,b]\times \bR;\;\;L(x) \leq y\leq U(x)\;\bigr\rbrace,$$ then $D(U,L)$ is measurable $$\eA(D(U,L))=\int_a^b \bigl(\, bigl(\; U(x)-L(x)\;\bigr) dx.$$ The next thing to prove is a weak form of the inclusion exclusion inclusion-exclusion principle: if $D_1$, $D_2$ are measurable regions that intersect along the grapf of a $C^1$-function, then $D_1\cup D_2$ is measurable and $$\eA(D_1\cup D_2)=\eA_(D_1)+\eA(D_2)D_2)=\eA(D_1)+\eA(D_2).$$ 1 I don't know how much about areas you want to prove, and how developed a background the audience is supposed to have, but here is a definition of area of a bounded planar region. Take such a region $D$ in $\mathbb{R}^2$, $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\ve}{{\varepsilon}}$ consider the $\varepsilon$-grid $(\varepsilon\bZ)^2$ and denote by $N_\ve(D)$ the number of $\ve$-pixels that touch $D$. (A pixel is one of the $\ve\times\ve$-tiny squarers of the grid.) We then declare $D$ to be measurable (i.e. to have area) if the limit $\newcommand{\eA}{\mathscr{A}}$ $$\eA(D)= \lim_{\ve\to0}\ve^2 N_\ve(D).$$ exists. If this is the case we define the area to be the limit $\eA(D)$, and we set $\eA_\ve(D)=\ve^2N_\ve(D)$. The first step is to prove $\newcommand{\bR}{\mathbb{R}}$ that if $L,U: [a, b]\to \bR$ are Riemann integrable functions and $D(U,L)$ is the region $$D_f= \bigl\lbrace\; (x,y)\in [a,b]\times \bR;\;\;L(x) \leq y\leq U(x)\;\bigr\rbrace,$$ then $D(U,L)$ is measurable $$\eA(D(U,L))=\int_a^b \bigl(\, U(x)-L(x)\;\bigr) dx.$$ The next thing to prove is a weak form of the inclusion exclusion principle: if $D_1$, $D_2$ are measurable regions that intersect along the grapf of a $C^1$-function, then $D_1\cup D_2$ is measurable and $$\eA(D_1\cup D_2)=\eA_(D_1)+\eA(D_2).$$	{"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": 44, "mathjax_display_tex": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.938568651676178, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/304769/a-tough-series-sum-k-1-infty-frac-zeta2kk12k1-need-help	# A tough series: $\sum_{k=1}^{\infty}\frac{\zeta(2k)}{(k+1)(2k+1)}$, need help I was doing a integral which ends up with a tough series part: $$\sum_{k=1}^{\infty}\frac{\zeta(2k)}{(k+1)(2k+1)}$$ Mathematica says $$\frac12$$ Which agrees with the anwer...Anyone know how to evaluate this? - Isn't there an exact expression for the zeta function of even integers? – Ron Gordon Feb 15 at 12:20 @rlgordonma, I m not familiar with zeta :(. – Ryan Feb 15 at 12:21 1 May first idea is partial fraction decomposition – Dominic Michaelis Feb 15 at 12:32 $$\frac{\sin(\pi x)}{\pi x}=\prod_{k=1}^\infty \left(1-\frac{x^2}{k^2} \right)$$ $$\begin{aligned}\log \left[ \frac{\sin(\pi x)}{\pi x}\right] &=\sum_{k=1}^\infty\log\left(1-\frac{x^2}{k^2}\right) \\ &= \sum_{k=1}^\infty \left( -\sum_{n=1}^\infty \frac{x^{2n}}{k^{2n}n}\right) \\ &= - \sum_{n=1}^\infty \frac{x^{2n}}{n}\sum_{k=1}^\infty \frac{1}{k^{2n}} \\ &= -\sum_{n=1}^\infty \frac{x^{2n}}{n}\zeta(2n)\end{aligned}$$ $$\displaystyle \ln(\sin \pi x)=\ln(\pi x)-\sum_{n=1}^\infty \frac{x^{2n}}{n}\zeta(2n)$$ We can try differentiating or integrating this equation to bring it into the required form. – Integral Feb 15 at 13:24 @ShobhitBhatnagar That's quite a voluminous comment. Maybe this would be better in the answer zone. – julien Feb 15 at 13:30 show 1 more comment ## 2 Answers Using the definition of $\zeta$ and exchanging the order of summation since everything is positive, one sees that the sum $S$ to be computed is $$S=\sum_{n=1}^{+\infty}\left(n^2u(\tfrac1n)-1\right),\quad\text{with}\quad u(t)=\sum\limits_{k=0}^{+\infty}\frac{2t^{2k+2}}{(2k+1)(2k+2)}.$$ From here, only elementary analysis is required. To compute $u$, note that $u(0)=u'(0)=0$ and $$u''(t)=\sum\limits_{k=0}^{+\infty}2t^{2k}=\frac2{1-t^2}=\frac1{1-t}+\frac1{1+t},$$ hence, for every $\|t\|\leqslant1$, $$u(t)=(1+t)\log(1+t)+(1-t)\log(1-t).$$ This yields, for each integer $n\geqslant1$, $$n^2u(\tfrac1n)=n(n+1)\log(n+1)+n(n-1)\log(n-1)-2n^2\log n,$$ hence, for every $N\geqslant1$, $$\sum_{n=1}^Nn^2u(\tfrac1n)=\sum_{n=1}^{N+1}n(n-1)\log(n)+\sum_{n=1}^{N-1}n(n+1)\log(n)-\sum_{n=1}^N2n^2\log n,$$ that is, $$\sum_{n=1}^Nn^2u(\tfrac1n)=N(N+1)\log(1+\tfrac1N).$$ Since $\log(1+x)=x-\frac12x^2+o(x^2)$ when $x\to0$, when $N\to\infty$, the RHS is $$N(N+1)\left(\tfrac1N-\tfrac12\tfrac1{N^2}+o\left(\tfrac1{N^2}\right)\right)=N+\tfrac12+o(1).$$ This proves (rigorously) that $$S=\tfrac12.$$ - Cool, nice to have something rigorous! – achille hui Feb 15 at 14:18 Interesting, here is my attempt. Nothing is rigorous. Start with $$\sum_{k=1}^{\infty} \frac{x^{2k-1}}{\Gamma(2k)(k+1)(2k+1)} = \sum_{k=1}^{\infty} \frac{4 k x^{2k-1}}{(2k+2)!} = 2\frac{d}{dx}\left[\sum_{k=1}^{\infty}\frac{x^{2k}}{(2k+2)!}\right] = 2\frac{d}{dx}\left[\frac{\cosh x - 1 - \frac{x^2}{2}}{x^2}\right] = \frac{e^x - e^{-x}}{x^2} - 2 \frac{e^x + e^{-x} - 2}{x^3}$$ Divide both sides by $e^x - 1$, integrate and use the identity for $\Re(s) > 1$: $$\zeta(s)\Gamma(s) = \int_0^{\infty}\frac{x^{s-1}}{e^x-1} dx$$ We get: $$\sum_{k=1}^{\infty} \frac{\zeta(2k)}{(k+1)(2k+1)} = \int_0^{\infty} \left[\frac{1+e^{-x}}{x^2} - 2 \frac{1-e^{-x}}{x^3}\right] dx = \int_0^{\infty} \frac{d}{dx}\left[-\frac{1}{x} + \frac{1-e^{-x}}{x^2}\right]dx\\ = -\lim_{x\to 0} \left[-\frac{1}{x} + \frac{1-e^{-x}}{x^2}\right] = \frac12$$ -	{"extraction_info": {"found_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": 16, "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": 0.9198834300041199, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/30343/help-with-summing-a-power-series/30407	# Help with summing a power series I'd like to determine the function corresponding to the following power series: $$x + \sum_{n=1}^\infty (-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n} \frac{x^{2n+1}}{2n+1},$$ where $\|x\|<1$. - 12 @bimol: Please do not ask questions in the imperative. Instead, explain what you have tried and what you need help with. I have posted a complete answer, this series corresponds to a well known function. But I have deleted this answer, and will not repost it until you explain what you have tried, and ask this question properly. – Eric♦ Apr 1 '11 at 16:40 5 @Eric @Did I think the above, and the downvotes are uncalled for (and rude). Please be more polite with new users. – Gone Apr 1 '11 at 18:39 8 It looks like answers are being downvoted as well for, it would seem, no good reason. If you think the questioner is being lazy, fine, but downvoting an answer because you don't think the question deserves an answer is extremely rude. If you downvote, at least provide a reasonable critique. – Brian Vandenberg Apr 1 '11 at 18:52 4 @Bill: Look at the original question in the edit history. It is very poor. You are right though, I was definitely being rude. But the problem is I don't know what to do when the question is this poor, other than vote to close. (which others have done) I still think that he should try to tell us a little about what he did before a complete solution is given. Perhaps I should of just asked for this in a nicer way. I don't think that it is uncalled for. Based on the originally proposed question, I don't think the downvotes for the question are uncalled for either. – Eric♦ Apr 1 '11 at 18:57 4 @user02138: You put words in his mouth for him. This new question is not his. I am not sure if it is standard to rewrite someone else's question to fit how you believe it should be asked. I am actually curious, I have wanted to do this many times. The homework tag, and the sentences you added may or may not be what the OP was thinking. – Eric♦ Apr 1 '11 at 19:20 show 11 more comments ## 4 Answers The following is an attempt to actually do some mathematics with the question. Thus I try to avoid ready-made formulas drawn out of the hat and I want to describe step by step a path which leads to the solution as routinely as possible. My first reaction when I read the formula to be explained is that I am not too happy with the denominators $2n+1$. The series is typeset with some ratios $\displaystyle\frac{x^{2n+1}}{2n+1}$, I do not know if this is meant as a hint or not but these ratios are obvious primitives, so I feel like differentiating the series $A(x)$ the OP wants to compute. Note that I have no real justification for such a move at this point but one has to start somewhere, hasn't one? Thinking hard about the terms of degree $0$ and $1$ in $A(x)$, I finally convince myself that $A(0)=0$ and $$A'(x)=\sum_{n=0}^\infty (-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n} x^{2n}.$$ Now, what is this strange looking coefficient of $x^{2n}$? This looks like a ratio of factorials, but not quite... After some fiddling around, I notice that I could use the even integers in the denominator to fill in the gaps between the odd integers in the numerator and that the resulting numerator would then be a perfect factorial. Seems like a good idea, but first I must write this rigorously, getting $$\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}=\frac{1\cdot2\cdot3\cdots(2n)}{2^2\cdot4^2\cdot6^2\cdots(2n)^2}=\frac{(2n)!}{2^{2n}(n!)^2}=\frac1{2^{2n}}{2n\choose n}.$$ My next step is to change variable. Using the variable $z=-\displaystyle\frac{x^2}4$ instead of $x$ can only simplify things, right? So now, my aim is to understand the series $$A'(x)=\sum_{n=0}^\infty{2n\choose n}z^n.$$ Ha! Binomial coefficients! That is easy, I just have to find a binomial somewhere. The expansion of $(1+u)^{2n}$ involves all the $2n$-choose-something, so I am after the term $u^n$ in $(1+u)^{2n}$. How could I keep this term and erase all the others? Well, a guy named Joseph Fourier already knew how to do that, so I will try to imitate him. Old Joseph knew that if you integrate the function $s\mapsto\mathrm{e}^{2\mathrm{i}\pi ns}\mathrm{e}^{-2\mathrm{i}\pi ks}$ on $[0,1]$ for integers $n$ and $k$, you get zero except in one case: if $k=n$, and then you get $1$. So if I replace $u$ by $\mathrm{e}^{2\mathrm{i}\pi s}$ in $(1+u)^{2n}$ and if I integrate everything multiplied by $\mathrm{e}^{-2\mathrm{i}\pi ns}$ on $[0,1]$, I will collect the coefficient of $u^n$ alone. In other words, $${2n\choose n}=\int_0^1(1+\mathrm{e}^{2\mathrm{i}\pi s})^{2n}\mathrm{e}^{-2\mathrm{i}\pi ns}\mathrm{d}s.$$ Now I multiply this by $z^n$, I sum the results over $n$ and I assume that $\|z\|$ is small enough to allow me to interchange the order of the summation and the integral (I see that $\|z\|<\frac14$ will do). The result is $$A'(x)=\int_0^1\sum_{n=0}^\infty z^n(1+\mathrm{e}^{2\mathrm{i}\pi s})^{2n}\mathrm{e}^{-2\mathrm{i}\pi ns}\mathrm{d}s.$$ The function I integrate is nothing but a geometric series, right? And the modulus of its ratio is at most $4\|z\|<1$, right? Here I am in known territory because I know how to sum a geometric series $$G(r)=\sum_{n=0}^{+\infty}r^n,$$ for every complex number $r$ such that $\|r\|<1$ and I know this because somebody told me once how another guy named Euclid did it: he (basically) wrote the series as $$G(r)=1+r+r^2+r^3+\cdots=1+r(1+r+r^2+\cdots),$$ and he noticed that the parenthesis was $G(r)$ again! In other words, $G(r)=1+rG(r)$, that is, $$G(r)=\frac1{1-r}.$$ So I can compute the series inside the integral and this gets me a simpler expression of $A'(x)$, namely $$A'(x)=\int_0^1\frac{\mathrm{d}s}{1-z(1+\mathrm{e}^{2\mathrm{i}\pi s})^{2}\mathrm{e}^{-2\mathrm{i}\pi s}}.$$ This is an integral of a rational function of sines and cosines, in this case $$A'(x)=\int_0^1\frac{\mathrm{d}s}{1-4z\cos^2(\pi s)}=\frac2\pi\int_0^{\pi/2}\frac{\mathrm{d}s}{1-4z\cos^2(s)}=\frac2\pi\int_0^{\pi/2}\frac{\mathrm{d}s}{1+x^2\cos^2(s)},$$ where in the last equality I finally decided to come back to the $x$ variable. Since $\cos^2(s)$ and $\mathrm{d}s$ are invariant by the transformation $s\to s+\pi$, I am pretty sure the change of variables $t=\tan(s)$ will be successful. Let me verify this: $\mathrm{d}t=(1+t^2)\mathrm{d}s$ and $\cos^2(s)=1/(1+t^2)$, hence $$A'(x)=\frac2\pi\int_0^{+\infty}\frac{\mathrm{d}t}{1+t^2+x^2} =\frac2\pi\int_0^{+\infty}\frac{\mathrm{d}t}{\sqrt{1+x^2}(1+t^2)},$$ that is $$A'(x) =\frac2\pi\frac1{\sqrt{1+x^2}}\left[\arctan(t)\right]_0^{+\infty} =\frac1{\sqrt{1+x^2}}.$$ Almost done! I remember that $A(0)=0$, hence $$A(x)=\int_0^x\frac{\mathrm{d}v}{\sqrt{1+v^2}}=\left[\mathrm{arcsinh}(v)\right]_0^x=\mathrm{arcsinh}(x)=\log(x+\sqrt{1+x^2}).$$ Done. - 7 Very nice! This was one of the more joyful readings on this site so far. Whoever voted this answer down must have a very strange reason... – t.b. Apr 2 '11 at 8:27 3 +1 for this wonderfully motivated derivation, Didier. I wonder how on earth you stumbled on Fourier: where do you teach, by the way :-) – Georges Elencwajg Apr 2 '11 at 8:34 @elgeorges Ha! You noticed that... :-) – Did Apr 2 '11 at 8:44 2 What an excellent integral to represent the binomial coefficients!! Thank you! I was actually looking for a bunch of these. The best one I found myself was $$\frac{2^{m+n}(-1)^{n}}{\pi i{}^{m+n}}\int_{-\infty}^{\infty}\frac{1}{(x-i)^{n+1}(x+i)^{m+1}}dx=\binom{m+n}{‌n}$$ but it is not as pretty. (My solution to this used a combinatorics identity to find the binomial power series, because the above identity is not the greatest for explaining, especially since the only way I know to prove it is contour integration. But I personally prefer integrals and swapping the order any day!!) – Eric♦ Apr 3 '11 at 13:59 @Eric Thanks. You are right that contour integration techniques are not far away from "my" formula for the binomial coefficient. – Did Apr 3 '11 at 15:34 show 2 more comments Consider the following series, \begin{align} \arcsin x = \sum_{n \geq 0} \frac{(2n)!}{2^{2n} (n!)^{2}} \frac{x^{2n+1}}{2n+1} = \sum_{n \geq 0} \frac{(2n-1)!!}{(2n)!!} \frac{x^{2n+1}}{2n+1}, \end{align} which converges for $\|x\| \leq 1$ and where $\frac{(2n-1)!!}{(2n)!!} = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots 2n}$. This can be derived by integrating the series expansion of $(1 -x^2)^{-1/2}$ termwise and paying careful attention to questions of convergence. Observe that \begin{align} -i \arcsin i x = x + \sum_{n \geq 1} (-1)^n \frac{(2n-1)!!}{(2n)!!} \frac{ x^{2n + 1}}{2n+1}, \end{align} which is the series in question. - 4 -1: This seems uncalled for, or rather, unuseful...as much fun as it is to answer a question flat out, it's obvious that the OP is in homework mode. Please consider 'deleting' this answer like Eric did for his, and undelete if the OP fixes things. You're doing their homework for them which is uncool. – Mitch Apr 1 '11 at 18:51 4 I'm surprised at people that downvoted a perfectly reasonable answer to this question. I'm just helping the OP along. – user02138 Apr 1 '11 at 18:58 5 @Mitch - If you don't like the question, downvote it. If you think the answer given is badly written or contains errors, downvote it. But this is just not classy. – Brian Vandenberg Apr 1 '11 at 19:04 1 @Brian: Thanks. I second your comment. – user02138 Apr 1 '11 at 19:05 1 @Mitch - I suppose that's one interpretation. I see the stackexchange-style sites as a far more global resource. While the answer may only serve the OP by doing his homework for him, this isn't an IRC chat-room or newsgroup post where a small audience will get something out of the response. – Brian Vandenberg Apr 1 '11 at 21:46 show 8 more comments Hint Consider the binomial series expansion for $(1+x^2)^{-\frac{1}{2}}$. You should arrive at something very similar to that series. - – Brian Vandenberg Apr 1 '11 at 21:53 Can you give a tiny bit more? What does ${- \frac{1}{2} \choose 2k}$ even mean? – Mitch Apr 1 '11 at 21:55 @Mitch - ${-n \choose k} = (-1)^{k}{{k - (n + 1)} \choose k}$, for positive $n$. The $(-1)^{k}$ coefficient keeps the value of the quantity positive (provided $n > k$?). Ignoring sign, this is the number of ways to arrange $\|k - (n + 1)\|$ items taken $k$ at a time. The $\frac{1}{2}$ I don't have a great intuitive description for, sorry. – Brian Vandenberg Apr 1 '11 at 22:13 3 ${\alpha \choose k} = \frac{\alpha (\alpha-1) \ldots (\alpha-k+1)}{k!}$. I call that the "generalized" binomial coefficient (I don't know if it is the standard term). You can find more here en.wikipedia.org/wiki/…. And here is some info on the binomial series en.wikipedia.org/wiki/Binomial_series. – alejopelaez Apr 1 '11 at 22:58 Nice...that's where the odd numbers come in the numerator, and then the $2^k k!$ in the denominator. – Mitch Apr 2 '11 at 2:33 I am going to use a fairly well known generating series. Its proof will be left until the end. Lemma We have that $$\frac{1}{\sqrt{1-4x}}=\sum_{n=0}^\infty \binom{2n}{n} x^n$$ where $\binom{2n}{n}$ refers to the central binomial coefficient. Now to solve your problem: Define $f(x)$ by $$f(x)=x+\sum_{n=1}^{\infty}(-1)^{n}\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\frac{x^{2n+1}}{2n+1}.$$ Notice that $$f(x)=x+\sum_{n=1}^{\infty}(-1)^{n}\frac{(2n)!}{n!n!}\frac{1}{4^{n}}\frac{x^{2n+1}}{2n+1}$$ so that $$f^{'}(x)=1+\sum_{n=1}^{\infty}(-1)^{n}\frac{(2n)!}{n!n!}\frac{x^{2n}}{4^{n}}=\sum_{n=0}^{\infty}\binom{2n}{n}\left(\frac{-x^2}{4}\right)^{n}.$$ By the lemma, we see that $$f^{'}(x)=\frac{1}{\sqrt{1+x^{2}}}.$$ To solve $f(x)=\int_0^x \frac{1}{\sqrt{1+x^{2}}}$, use the identity $\sinh^2(x)+1=\cosh^2 (x)$ and make the substitution $x=\sinh (u)$. (Notice the constant of integration must be zero from the original definition of $f(x)$.) Then we find $$f(x)=\sinh^{-1}(x).$$ Proof of Lemma: We have the combinatorial identity $$\sum_{i=0}^{n}\binom{2i}{i}\binom{2(n-i)}{n-i}=4^{n}$$ which follows from finding two different ways to count the number of possible ways to choose a team from a group of size $2n$. Then $$\left(\sum_{n=0}^\infty \binom{2n}{n} x^n\right)^2=\sum_{n=0}^\infty (4x)^n=\frac{1}{1-4x}.$$ -	{"extraction_info": {"found_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": 23, "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": 0.9497841000556946, "perplexity_flag": "head"}
http://mathoverflow.net/questions/95408/elliptic-curves-over-rings/95409	## Elliptic Curves over Rings? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) So an elliptic curve $E$ over a field $K$ is a smooth projective nonsingular curve of genus $1$ together with a point $O \in E$. I was reading Silverman's "Arithmetic of Elliptic Curves" and it seems that most of its treatment is over fields. My question is, does it make sense to define an elliptic curve over a ring (eg: a noncommutative ring)? If not, why not (where would the "construction" fail)? Is it simply not an object of much interest? Edit: Apparently the question of elliptic curves over noncommutative rings is considered to some extent in this. http://user.math.uzh.ch/fontein/diplom-fontein.pdf - I would say that perhaps 'explored' is saying a bit too much. He solves an equation in a noncommutative ring. If you want to see this as constructing an.elliptic curve over a noncommutative ring that's ok, but as long as there's no group law or any of the other features that do appear over commutative rings, you really don't have anything more than just any old equation. (Btw, after 8 edits your question becomes community-wiki, which voids any reputation points that were awarded to any of us. FYI.) – René Pannekoek Apr 29 2012 at 7:45 Was cross-posted on MSE after a month. I'm including a link here as notification. math.stackexchange.com/questions/154965/… – Eugene Jun 7 at 8:29 ## 2 Answers A commutative ring, yes. This is treated to some extent in Silverman's second book; for the more general story of "abelian schemes," which is what you're really after, I might look at Milne's articles in the volume Arithmetic Geometry edited by Cornell and Silverman. As for noncommutative rings, I'm afraid I have no idea -- I'm not even sure what "construction" would be in a position to fail! - So, I googled "noncommutative torus" "abelian variety" and sure enough there were results, including a paper by Manin. – Rob Harron Apr 28 2012 at 4:32 3 @Rob: "noncommutative torus," as I understand it, doesn't refer to an elliptic curve over a noncommutative ring, but to a noncommutative deformation of the ring of functions on a torus. – Qiaochu Yuan Apr 28 2012 at 16:48 2 I agree with Qiaochu here. As I recall, Manin's noncommutative tori are meant (e.g.) to stand in relation to real quadratic fields as CM elliptic curves do to imaginary quadratic fields. – JSE Apr 29 2012 at 4:28 I guess the more fundamental question would be is it even possible to define curves over noncommutative rings? – Eugene May 12 2012 at 0:35 ### 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. Elliptic curves can be defined over arbitrary base schemes $S$. In particular, for every (commutative!) ring $R$ one can talk about elliptic curves over (the spectrum of) $R$. Loosely speaking, what one gets is a family $E$ of elliptic curves parametrized by the points of $S$. One then proves the existence of the group law ($E$ can be given the structure of an $S$-group scheme), and goes from there. E.g., locally over $S$, $E$ can be put into Weierstrass form. In the book Arithmetic Moduli of Elliptic Curves by Katz and Mazur, an elliptic curve over $S$ is defined as a proper smooth morphism $f : E \rightarrow S$ of finite presentation, with a section $0 : S \rightarrow E$, such that all geometric fibers of $f$ are integral (equivalently, connected) curves of genus one. What can be done for noncommutative $R$ I don't know. It seems to me that you have to say what you mean by an elliptic curve over a noncommutative ring. One can't simply replace 'field' in 'elliptic curve over a field' by the name of some other algebraic structure and expect it to make sense, I guess. - What you are referring to in Deligne–Rapoport are generalized elliptic curves, not elliptic curves. – Rob Harron Apr 28 2012 at 14:20 Yes you're right. My mistake, I'll make an edit. – René Pannekoek Apr 28 2012 at 14:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9554303288459778, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/6811/can-peps-explain-the-holographic-principle-in-quantum-gravity	Can PEPS explain the holographic principle in quantum gravity? Condensed matter physicists have shown using quantum information that in many condensed matter systems, entanglement entropy only scales as the area of the boundary, and not the volume. This is the basis for the density matrix renormalization group and Projected Entangled Pair States (PEPS). Does this also explain the holographic principle in quantum gravity? - 1 What do you mean by explain? Most meanings of that word don't make any sense here. – Marek Mar 13 '11 at 9:48 2 I don't understand what you are asking at all. The area law scaling of entanglement entropy in the quantum many body systems that you are mentioning is the basis for the DMRG, MPS, PEPS etc, but I don't see how these variational constructions for the quantum state of these systems could "explain" the holographic principle in QG. I think that the question could be better posed as: Can the holographic principle in QG give some theoretical support to variational methods in quantum many body systems such as PEPS? – xavimol Mar 13 '11 at 12:31 PEPS == Pair Entangled Product States, correct? I edited the question to make this clear. – user346 Mar 13 '11 at 14:32 @Deepak: isn't it projected entangled pair states? Today is the first time I heard the term PEPS but google (and many articles it provides) suggests this is the most likely meaning :) – Marek Mar 13 '11 at 15:48 @Deepak and @Marek: Yes it is true, the correct term is Projected Entangled Pairs (PEPS) states that is a generalization of Matrix Product states (MPS,that is directly related with density matrix renormalization group). – xavimol Mar 13 '11 at 16:26 1 Answer Nope, the very fact that the entanglement entropy is - naturally - proportional to the surface area does not explain the holographic principle because the holographic principle, reduced to the corresponding entropy bound, implies that the total entropy of one of the systems can't exceed $A/4G$ where $A$ is the surface. The entanglement entropy is just one tiny term of the entropy, a degree of correlation between two subsystems, so its being proportional to the area is a much weaker and less surprising statement than the holographic principle. Quite generally, non-gravitational physical systems are simply not holographic in the space (bulk) that they occupy. Only (quantum) gravitational systems obey the holographic principle. In the context of AdS/CFT or AdS/anything, the holographic principle is encoded in the description of the CFT or anything by its having a secret extra (fifth) dimension of spacetime. - Nice answer @Lubos. Some people are trying to reinterpret quantum information based renormalization group procedures in quantum many body physics (an example is PEPS) in terms of holography. These procedures support the area law scaling of entanglement entropy in low dimensional systems. However, despite the interesting or even the usefulness (in terms of being an effective tool to compute things) that this interpretation can be, it is questionable how these facts could "explain" something that, at this time, has the status of principle. – xavimol Mar 14 '11 at 13:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9230684638023376, "perplexity_flag": "middle"}
http://jeremykun.com/2012/08/26/k-nearest-neighbors-and-handwritten-digit-classification/	# K-Nearest-Neighbors and Handwritten Digit Classification Posted on August 26, 2012 by ## The Recipe for Classification One important task in machine learning is to classify data into one of a fixed number of classes. For instance, one might want to discriminate between useful email and unsolicited spam. Or one might wish to determine the species of a beetle based on its physical attributes, such as weight, color, and mandible length. These “attributes” are often called “features” in the world of machine learning, and they often correspond to dimensions when interpreted in the framework of linear algebra. As an interesting warm-up question for the reader, what would be the features for an email message? There are certainly many correct answers. The typical way of having a program classify things goes by the name of supervised learning. Specifically, we provide a set of already-classified data as input to a training algorithm, the training algorithm produces an internal representation of the problem (a model, as statisticians like to say), and a separate classification algorithm uses that internal representation to classify new data. The training phase is usually complex and the classification algorithm simple, although that won’t be true for the method we explore in this post. More often than not, the input data for the training algorithm are converted in some reasonable way to a numerical representation. This is not as easy as it sounds. We’ll investigate one pitfall of the conversion process in this post, but in doing this we separate the data from the application domain in a way that permits mathematical analysis. We may focus our questions on the data and not on the problem. Indeed, this is the basic recipe of applied mathematics: extract from a problem the essence of the question you wish to answer, answer the question in the pure world of mathematics, and then interpret the results. We’ve investigated data-oriented questions on this blog before, such as, “is the data linearly separable?” In our post on the perceptron algorithm, we derived an algorithm for finding a line which separates all of the points in one class from the points in the other, assuming one exists. In this post, however, we make a different structural assumption. Namely, we assume that data points which are in the same class are also close together with respect to an appropriate metric. Since this is such a key point, it bears repetition and elevation in the typical mathematical fashion. The reader should note the following is not standard terminology, and it is simply a mathematical restatement of what we’ve already said. The Axiom of Neighborliness: Let $(X, d)$ be a metric space and let $S \subset X$ be a finite set whose elements are classified by some function $f : S \to \left \{ 1, 2, \dots, m \right \}$. We say that $S$ satisfies the axiom of neighborliness if for every point $x \in S$, if $y$ is the closest point to $x$, then $f(x) = f(y)$. That is, $y$ shares the same class as $x$ if $y$ is the nearest neighbor to $x$. For a more in-depth discussion of metrics, the reader should refer to this blog’s primer on the topic. For the purpose of this post and all foreseeable posts, $X$ will always be $\mathbb{R}^n$ for some $n$, while the metric $d$ will vary. This axiom is actually a very strong assumption which is certainly not true of every data set. In particular, it highly depends on the problem setup. Having the wrong kinds or the wrong number of features, doing an improper conversion, or using the wrong metric can all invalidate the assumption even if the problem inherently has the needed structure. Luckily, for real-world applications we only need the data to adhere to the axiom of neighborliness in approximation (indeed, in practice the axiom is only verifiable in approximation). Of course, what we mean by “approximation” also depends on the problem and the user’s tolerance for error. Such is the nature of applied mathematics. Once we understand the axiom, the machine learning “algorithm” is essentially obvious. For training, store a number of data points whose classes are known and fix a metric. To determine the class of an unknown data point, simply use the most common class of its nearest neighbors. As one may vary (as a global parameter) the number of neighbors one considers, this method is intuitively called k-nearest-neighbors. ## The Most Basic Way to Learn: Copy Your Neighbors Let’s iron out the details with a program and test it on some dummy data. Let’s construct a set of points in $\mathbb{R}^2$ which manifestly satisfies the axiom of neighborliness. To do this, we’ll use Python’s random library to make a dataset sampled from two independent normal distributions. ```import random def gaussCluster(center, stdDev, count=50): return [(random.gauss(center[0], stdDev), random.gauss(center[1], stdDev)) for _ in range(count)] def makeDummyData(): return gaussCluster((-4,0), 1) + gaussCluster((4,0), 1)``` The first function simply returns a cluster of points drawn from the specified normal distribution. For simplicity we equalize the covariance of the two random variables. The second function simply combines two clusters into a data set. To give the dummy data class “labels,” we’ll simply have a second list that we keep alongside the data. The index of a data point in the first list corresponds to the index of its class label in the second. There are likely more elegant ways to organize this, but it suffices for now. Implementing a metric is similarly straightforward. For now, we’ll use the standard Euclidean metric. That is, we simply take the sum of the squared differences of the coordinates of the given two points. ```import math def euclideanDistance(x,y): return math.sqrt(sum([(a-b)*2 for (a,b) in zip(x,y)])) ``` To actually implement the classifier, we create a function which itself returns a function. ```import heapq def makeKNNClassifier(data, labels, k, distance): def classify(x): closestPoints = heapq.nsmallest(k, enumerate(data), key=lambda y: distance(x, y[1])) closestLabels = [labels[i] for (i, pt) in closestPoints] return max(set(closestLabels), key=closestLabels.count) return classify ``` There are a few tricky things going on in this function that deserve discussion. First and foremost, we are defining a function within another function, and returning the created function. The important technical point here is that the created function retains all local variables which are in scope even after the function ends. Specifically, you can call “makeKNNClassifier” multiple times with different arguments, and the returned functions won’t interfere with each other. One is said to close over the values in the environment, and so this programming language feature is called a function closure or just a closure, for short. It allows us, for instance, to keep important data visible while hiding any low-level data it depends on, but which we don’t access directly. From a high level, the decision function entirely represents the logic of the program, and so this view is justified. Second, we are using some relatively Pythonic constructions. The first line of “classify” uses of heapq to pick the $k$ smallest elements of the data list, but in addition we use “enumerate” to preserve the index of the returned elements, and a custom key to have the judgement of “smallest” be determined by the custom distance function. Note that the indexed “y[1]” in the lambda function uses the point represented by “y” and not the saved index. The second line simply extracts a list of the labels corresponding each of the closest points returned by the call to “nsmallest.” Finally, the third line returns the maximum of the given labels, where a label’s weight (determined by the poorly named “key” lambda) is its frequency in the “closestLabels” list. Using these functions is quite simple: ```trainingPoints = makeDummyData() # has 50 points from each class trainingLabels = [1] 50 + [2] * 50 # an arbitrary choice of labeling f = makeKNNClassifier(trainingPoints, trainingLabels, 8, euclideanDistance) print f((-3,0)) ``` The reader may fiddle around with this example as desired, but we will not pursue it further. As usual, all code used in this post is available on this blog’s Google code page. Let’s move on to something more difficult. ## Handwritten Digits One of the most classic examples in the classification literature is in recognizing handwritten digits. This originally showed up (as the legend goes) in the context of the United States Postal Service for the purpose of automatically sorting mail by the zip code of the destination. Although this author has no quantitative proof, the successful implementation of a scheme would likely save an enormous amount of labor and money. According to the Postal Facts site, there are 31,509 postal offices in the U.S. and, assuming each one processes mail, there is at least one employee at each office who would spend some time sorting by zip code. Given that the USPS processes 23 million pieces of mail per hour, a conservative estimate puts each office spending two hours of labor per day on sorting mail by zip code (resulting in a very rapid pace of 146.52 pieces of mail sorted per minute per worker). At a lower bound of \$18/hr this amounts to a cost of \$1,134,324 per day, or over 400 million dollars per year. Put in perspective, in one year the amount of money saved equals the entire two-year tuition of Moraine Valley Community College for 68,000 students (twice the current enrollment). In short, the problem of sorting mail (and of classifying handwritten digits) begs to be automated, and indeed it has been to some degree for about four decades. Let’s see how k-nearest-neighbors fares in this realm. We obtain our data from the UCI machine learning repository, and with a few minor modifications, we present it on this blog’s Google Code page (along with the rest of the code used in this post). A single line of the data file represents a handwritten digit and its label. The digit is a 256-element vector obtained by flattening a 16×16 binary-valued image in row-major order; the label is an integer representing the number in the picture. The data file contains 1593 instances with about 160 instances per digit. In other words, our metric space is $\left \{ 0,1 \right \}^{256}$, and we choose the Euclidean metric for simplicity. With the line wrapping to better display the “image,” one line from the data file looks like: ```0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 1 1 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0, 6``` After reading in the data appropriately, we randomly split the data set into two pieces, train on one piece, and test on the other. The following function does this, returning the success rate of the classification algorithm on the testing piece. ```import knn import random def column(A, j): return [row[j] for row in A] def test(data, k): random.shuffle(data) pts, labels = column(data, 0), column(data, 1) trainingData = pts[:800] trainingLabels = labels[:800] testData = pts[800:] testLabels = labels[800:] f = knn.makeKNNClassifier(trainingData, trainingLabels, k, knn.euclideanDistance) correct = 0 total = len(testLabels) for (point, label) in zip(testData, testLabels): if f(point) == label: correct += 1 return float(correct) / total ``` A run with $k=1$ gives a surprisingly good 89% success rate. Varying $k$, we see this is about as good as it gets without any modifications to the algorithm or the metric. Indeed, the graph below shows that the handwritten digits data set agrees with the axiom of neighborliness to a fair approximation. A graph of classification accuracy against k for values of k between 1 and 50. The graph clearly shows a downward trend as k increases, but all values k < 10 are comparably good. Of course, there are many improvements we could make to this naive algorithm. But considering that it utilizes no domain knowledge and doesn’t manipulate the input data in any way, it’s not too shabby. As a side note, it would be fun to get some tablet software and have it use this method to recognize numbers as one writes it. Alas, we have little time for these sorts of applications. ## Advantages, Enhancements, and Problems One reason k-nearest-neighbors is such a common and widely-known algorithm is its ease of implementation. Indeed, we implemented the core algorithm in a mere three lines of Python. On top of that, k-nearest-neighbors is pleasingly parallel, and inherently flexible. Unlike the Perceptron algorithm, which relies on linear separability, k-nearest-neighbors and the axiom of neighborliness allow for datasets with many different geometric structures. These lecture notes give a good example, as shown below, and the reader can surely conjure many more. k-nearest-neighbors applied to a data set organized in concentric circles. And of course, the flexibility is even greater by virtue of being able to use any metric for distance computations. One may, for instance, use the Manhattan metric if the points in question are locations in a city. Or if the data is sequential, one could use the dynamic time warping distance (which isn’t truly a metric, but is still useful). The possibilities are only limited by the discovery of new and useful metrics. With such popularity, k-nearest-neighbors often comes with a number of modifications and enhancements. One enhancement is to heuristically remove certain points close to the decision boundary. This technique is called edited k-nearest-neighbors. Another is to weight certain features heavier in the distance computations, which requires one to programmatically determine which features help less with classification. This is getting close to the realm of a decision tree, and so we’ll leave this as an exercise to the reader. The next improvement has to do with runtime. Given $n$ training points and $d$ features (d for dimension), one point requires $O(nd)$ to classify. This is particularly expensive, because most of the distance computations performed are between points that are far away, and as $k$ is usually small, they won’t influence in the classification. One way to alleviate this is to store the data points in a data structure called a k-d tree. The k-d tree originated in computational geometry in the problem of point location. It partitions space into pieces based on the number of points in each resulting piece, and organizes the partitions into a tree. In other words, it will partition tightly where the points are dense, and loosely where the points are sparse. At each step of traversing the tree, one can check to see which sub-partition the unclassified point lies in, and descend appropriately. With certain guarantees, this reduces the computation to $O(\log(n)d)$. Unfortunately, there are issues with large-dimensional spaces that are beyond the scope of this post. We plan to investigate k-d trees further in a future series on computational geometry. The last issue we consider is in data scaling. Specifically, one needs to be very careful when converting the real world data into numerical data. We can think of each of the features as a random variable, and we want all of these random variables to have comparable variation. The reason is simply because we’re using spheres. One can describe k-nearest-neighbors as finding the smallest (filled-in) sphere centered at the unlabeled point containing $k$ labeled data points, and using the most common of those labels to classify the new point. Of course, one can talk about “spheres” in any metric space; it’s just the set of all points within some fixed distance from the center (and this definition doesn’t depend on the dimension of the space). The important point is that a sphere has uniform length along every axis. If the data is scaled improperly, then the geometry of the sphere won’t mirror the geometry of the data, and the algorithm will flounder. So now we’ve seen a smattering of topics about k-nearest-neighbors. We’d love to continue the discussion of modifications in the comments. Next time we’ll explore decision trees, and work with another data set. Until then! ### Like this: This entry was posted in Algorithms, Discrete, Geometry, Linear Algebra and tagged dimension, machine learning, mathematics, metric, programming by j2kun. Bookmark the permalink. ## 7 thoughts on “K-Nearest-Neighbors and Handwritten Digit Classification” 1. on August 26, 2012 at 9:11 pm said: Amazing! Great post 2. on August 27, 2012 at 12:37 pm said: You might want to try k-center on the same data. See http://sites.stat.psu.edu/~jiali/course/stat597e/notes2/kcenter.pdf. Cheers, Hein (artent.net) • on August 27, 2012 at 2:04 pm said: Noted, but k-center (as with k-means and other centroid-type algorithms) is clustering, not classification. 3. on August 28, 2012 at 3:44 am said: Thanks, it was very useful. Reminded me I should try data scaling to see if I can improve performance on my problem. A cool application (and modification) of KNN is Semi-supervised condensed nearest neighbor[0] which achieves the best performance on the standard part-of-speech tagging dataset. [0] http://cst.dk/anders/scnn/ 4. on September 2, 2012 at 3:41 am said: Very nice posts! I thought post offices initially considered using neural networks for that (there’s a nice discussion about this at http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.50.2067&rep=rep1&type=ps), but I’m impressed at the accuracy of this method! • They definitely used (and I’m sure they still use) neural networks for this. Neural networks are the reason that people consider optical character recognition to be a “solved” problem. 5. Uzair on October 7, 2012 at 10:48 am said: Just fantastic – you covered so much so fluidly and compactly! 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": 27, "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": 0.9165964722633362, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/9555/laplaces-equation	# Laplace's equation I have got some mathematical difficulties in the following exercise : Calculate the potential of the polarized sphere along the z-axis. There are no free charges. For this, we need to solve Laplace's equation, by using the method of separation of variables. $\phi (r, \theta, \phi) = R(r) \Theta(\theta)$ We obtain a partial differential equation on $r$ and $\theta$ : $\frac{1}{R} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r} R = l(l+1)$ $\frac{1}{\Theta} \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \sin \theta \frac{\partial}{\partial \theta} \Theta = - l(l+1)$ My question is : "Why is $l$ to be a positive integer?" The solution I got in some course takes it for granted, but I do not know why. Thanks, Isaac - ## 2 Answers Dear Isaac, with all my respect, I don't think that Marek is answering your question. You're primarily asking why it's integer, right? First, $l$ can be both positive or negative integer. If $l=-10$, for example, $l(l+1)=90$ is clearly the same number as $l(l+1)$ for $l=9$. If you replace $l$ by $-1-l$, you get the same value of $l(l+1)$. The value $l=-1$ is equivalent to $l=0$, $l=-2$ is equivalent to $l=+1$, and so on. These eigenvalues of $l(l+1)$ are equal so the corresponding solutions shouldn't be double-counted. Now, why $l$ is an integer. Confusingly, it's because how the Legendre polynomials work: http://en.wikipedia.org/wiki/Legendre_polynomials It just happens that for integer values of $l$, the function $\Theta$ is non-singular for $\sin\theta=\pm 1$. Surprisingly, the same condition - integrality of $l$ - guarantees that the function $\Theta$ reduces to a polynomial. Clearly, this doesn't explain anything because I would have to use some maths to prove the propositions in the previous paragraph. There is a very transparent, more physical explanation why it's so. The operator acting on your $\Theta$ function - on the left-hand side - is $-\vec L\cdot \vec L$, usually denoted as $L^2$, and this operator has eigenvalues $l(l+1)$ for integer values of $l$. Why? It's because one may write $$L^2 = L_- L_+ + L_z^2 + L_z$$ where $L_{\pm} = L_x\pm i L_y$. The last term, $L_z$, has to be added because $L_+ L_-$ was only written in one order and $L_x,L_y$ (or, equivalently, $L_+,L_-$) don't commute with each other - their commutator is $i L_z$ (in $\hbar=1$ units). Now, $L\pm$ commute with $L^2$, so the state $L^+ \|\psi\rangle$ has the same eigenvalue of $L^2$ as $\|\psi\rangle$ itself. However, the eigenvalue of $L_z$ of the former state - with the action of $L_+$ - is greater by one than the $L_z$ eigenvalue of $\|\psi\rangle$ itself. Act with $L^+$ as many times as you need to get zero. You eventually have to get zero because the eigenvalue of $L_z^2$ can't be greater than the eigenvalue of $L^2$. When you get to the state with the maximum eigenvalue of $L_z$ which is non-vanishing, you will have $L_+\|\psi\rangle = 0$, so the first term in the formula for $L^2$ above will vanish. The remaining terms will give you $l(l+1)$ where $l$ is the maximum value of $L_z$ that you can get from the original state by actions of $L_+$. That's called the orbital angular momentum. In a similar way, I can act with $L_-$ to get to the minimum value of $L_z$. This minimum value of $L_z$ has to be the minus the maximum value, by rotational symmetry or by the $l\to -1-l$ symmetry above. So the difference between the maximum and minimum values of $L_z$ has to be integer - because $L_\pm$ was changing $L_z$ by steps equal to one. It follows that $l$ has to be either half-integer or integer. However, only integer values are possible for orbital angular momentum because $L_z$ has to be integer because the $\phi$-dependent wave function $\exp(im\phi)$ is only single-valued on the circle for integer values of $m$. That's why for orbital wave functions all eigenvalues $L_z$ are integer, and because $l$ is defined as the maximum $L_z$ in the "multiplet" of states that can be obtained from each other by the action of $L_\pm$, it follows that $l$ is integer, too. The formula for $L^2$ above guarantees that its eigenvalue is $l(l+1)$ for this integer $l$. - Hm, true, the question can also be interpreted this way (and is a lot more interesting, too). I wonder what OP really meant though. – Marek May 7 '11 at 16:26 It need not be positive but can also be zero. So let's suppose you're asking why it has to be non-negative. The answer to this question is that Laplace's operator $\Delta$ splits into radial and angular parts (these are precisely the operators that can be found in your equations, if you discard the $\phi$ dependence). The angular part is then simply $L^2$ where $\mathbf L$ is an angular momentum operator. Now because $L^2$ is a non-negative operator (convince yourself that a square of an operator must be non-negative) it must have only non-negative eigenvalues. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 81, "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": 0.9478182196617126, "perplexity_flag": "head"}
http://mathhelpforum.com/math-topics/205697-question-circles.html	Thread: 1. Question on circles So, I've attached ye question and have completed the square to solve it. However, what do I do after doing so , what can I do to find the answer?<br> <br> Also for part b, how do I find the point of intersection? Attached Thumbnails 2. Re: Question on circles You have the right line of thought when completing the square, however you have added to the equation without subtracting the equivalent amount to keep the equation the same. I would write: $x^2-6x+y^2-4y-12=0$ $x^2-6x+9+y^2-4y+4=12+9+4$ $(x-3)^2+(y-2)^2=5^2$ Now, it's easy to see the circle is centered at (3,2) and has radius 5. For part b) use $y=1$ and solve for $x$. You will have a quadratic in $x$ so there will be two roots, giving you 2 points, both with $y$-coordinate of 1.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "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": 0.9628006815910339, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/34713/fundamental-group-of-a-finite-set-with-discrete-topology/34715	# Fundamental Group of a finite set with discrete topology let S be a finite set with say n elements. Give it the discrete topology, Now what can we say about its fundamental group? Atleast can we determine the fundamental group of a set with two elements? Thanks - ## 2 Answers Such a space is not path-connected, so you have to keep track of the basepoint. Each path-connected component is trivially contractible (it consists only of one point). The fundamental group is trivial for each component. You don't need the finiteness of $S$. - 1 Yes, I missed out the fact that image of a connected domain is connected for a continuous map. Hence every loop in S should be a constant map. From which it turns out that the fundamental group is trivial. Thanks! – Dinesh Apr 23 '11 at 17:52 You need to choose a base point both in $S^1$ and $S$. Since $S^1$ is connected, it must be mapped to this base point entirely. Hence there is only one base-point preserving loop in $S$, and thus the fundamental group is trivial. This applies to any set with the discrete topology, no matter if it is finite or not. - @Theo: what's $S^1$? The unit circle? – Matt N. Apr 25 '11 at 8:57 @Matt: Yes. ${}$ – t.b. Apr 25 '11 at 8:58 @Theo: I assume you are using $S^1$ instead of $[0,1]$ because the question is about closed paths. Why do you have to choose a base point in the domain of the paths? Shouldn't you say "...since $S^1$ is connected, it must be mapped to this base point entirely....", where that base point is the one in $S$? – Matt N. Apr 25 '11 at 9:09 1 @Matt: I'm just using $S^1 \cong [0,1]/\{0 \sim 1\}$. The chosen base point $\ast$ on $S^1$ corresponds to the image of the identified end points of the interval. In other words, there is a bijection between loops $\gamma: [0,1] \to S$ (i.e. maps s.t. $\gamma(0) = \gamma(1) = s_0$) and base-point preserving maps $\gamma: (S^1, \ast) \to (S,s_0)$. – t.b. Apr 25 '11 at 9:14 @Theo: thank you, now I understand. I didn't understand why it was necessary to pick base points when one can show that every point in $S$ can only have the constant loop. It's necessary because that is how the fundamental group is defined. – Matt N. Apr 25 '11 at 10:43 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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9263039231300354, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/15640/what-are-electron-holes-in-semiconductors	# What are “electron holes” in semiconductors? I'm tutoring senior high school students. So far I've explained them the concepts of atomic structure (Bohr's model & Quantum mechanical model) very clearly. Now the next topic to be taught is semiconductors. I myself am not conviced with the concept of electron holes. If there is no electron then there is no electron. How can it be a hole. We define a hole when there is some thing every where except at a place. But inside an atom how can we define a HOLE. Kindly explain it with the help of Bohr's model. What was the need of introducing such abstract concept in semi conductors? - 1 Electrical conduction is like water moving through a hose; when you put water in at the spigot water comes out the other end but it is not the same water. In order for conduction to occur, the electrons have to be able to move along the material. Electron holes are like spaces that the electrons can jump to, or move through. When we talk about electron holes moving, it's like how the available space moves in a game of Chinese checkers. – AdamRedwine Oct 12 '11 at 19:41 @Adam: That's a good explanation. I would only add that metals have "loose" electrons in their valence shells, and whenever an electron wanders away from its "home" atom, the place left behind is a positively charged "hole". So the places where electrons are not can carry charge just as well as the electrons that aren't there. – Mike Dunlavey Oct 12 '11 at 21:04 @Mike, Yes, this is also the case for n-type doped semiconductors as well, but it is less obvious what the "hole" is in that case. I agree with OP that the whole concept is very difficult to grasp, which is why I responded as a comment rather than a full answer. – AdamRedwine Oct 12 '11 at 21:17 @Adam: actually that's even more reason to post what you did as a full answer, IMO, because for a difficult-to-grasp concept we're correspondingly less likely to get additional good answers. – David Zaslavsky♦ Oct 12 '11 at 21:20 – Larry Harson Oct 13 '11 at 12:36 show 1 more comment ## 6 Answers The notion of a particle in nonrelativistic quantum mechanics is very general: anything that can have a wavefunction, a probability amplitude for being at different locations, is a particle. In a metal, electrons and their associated elastic lattice deformation clouds travel as a particle. These effective electron-like negative carriers are electron quasiparticles, and these quasiparticles have a negative charge, which can be seen by measuring the Hall conductivity. Their velocity gives rise to a potential difference transverse to a wire in an external magnetic field which reveals the sign of the carriers. But in a semiconductor, the objects which carry the charge can be positively charged, which is physically accurate--- a current in such a material will give an opposite sign Hall effect voltage. To understand this, you must understand that the electron eigenstates in a periodic lattice potential are defined by bands, and these bands have gaps. When you have an insulating material, the band is fully filled, so that there is an energy gap for getting electrons to move. The energy gap generically means that an electron with wavenumber k will have energy: $$E= A + B k^2$$ Where A is the band gap, and B is the (reciprocal of twice the) effective mass. This form is generic, because electrons just above the gap have a minimum energy, and the energy goes up quadratically from a minimum. This quadratic energy dependence is the same as for a free nonrelativistic particle, and so the motion of the quasiparticles is described by the same Schrodinger equation as a free nonrelativistic particle, even though they are complicated tunneling excitations of electrons bound to many atoms. Now if you dope the material, you add a few extra electrons, which fill up these states. These electrons fill up k up to a certain amount, just like a free electron Fermi-gas and electrons with the maximum energy can be easily made to carry charge, just by jumping to a slightly higher k, and this is again just like a normal electron Fermi gas, except with a different mass, the effective mass. This is a semiconductor with a negative current carrier. But the energy of the electrons in the previous band has a maximum, so that their energy is generically $$E = -Bk^2$$ Since the zero of energy is defined by the location of the band, and as you vary k, the energy goes down. These electrons have a negative nonrelativistic effective mass, and their motion is crazy--- if you apply a force to these electrons, they move in the opposite direction! But this is silly--- these electron states are fully occupied, so the electrons don't move at all in response to an external force, because all the states are filled, they have nowhere to move to. So in order to get these electrons to move, you need to remove some of them, to allow electrons to fill these gaps. When you do, you produce a sea of holes up to some wavenumber k. The important point is that these holes, unlike the electrons, have a positive mass, and obey the usual Schroedinger equation for fermions. So you get effective positively charged positive effective mass carrier. These are the holes. The whole situation is caused by the generic shape of the energy as a function of k in the viscinity of a maximum/minimum, as produced by a band-gap. ### Bohr model holes You can see a kind of electron hole already in the Bohr model when you consider Moseley's law, but these holes are not the physical holes of a semiconductor. If you knock out an electron from a K-shell of an atom, the object you have has a missing electron in the 1s state. This missing electron continues to orbit the nucleus, and it is pretty stable, in that the decay takes several orbits to happen. The many-electron system with one missing electron can be thought of as a single-particle hole orbiting the nucleus. This single particle hole has a positive charge, so it is repelled by the nucleus, but it has a negative mass, because we are not near a band-gap, it's energy as a function of k is the negative of a free electron's energy. This negative-mass hole can be thought of as orbiting the nucleus, held in place by its repulsion to the nucleus (remember that the negative mass means that the force is in the opposite direction as the acceleration). This crazy system decays as the hole moves down in energy by moving out from the nucleus to higher Bohr orbits. This type of hole-description does not appear in the literature for Moseley's law, but it is a very simple approximation which is useful, because it gives a single particle model for the effect. The approximation is obviously wrong for small atoms, but it should be exact in the limit of large atoms. There are unexplained regularities in Moseley's law that might be explained by the single-hole picture, although again, this "hole" is a negative mass hole, unlike the holes in a positive doped semiconductor. - ## Did you find this question interesting? Try our newsletter Sign up for our newsletter and get our top new questions delivered to your inbox (see an example). email address The convention is that current flows in an electrical circuit from positive to negative. This was decided before electrons were discovered and before they were discovered to be negatively charged. You can identically consider the flow of electrical charge as either the movement of a negative electron from left to right, or the movement of the empty place an electron would be from right to left. If you think in terms of the movements of the gaps (holes) then you have a positive current flowing which matches the normal definition of electrical current. - This is not quite right, because it is a question of what are the actual physical carriers--- there is a difference between a P type and N type semiconductor which is not just a definition. – Ron Maimon Oct 13 '11 at 5:45 -1: Sorry, this is wrong, the mass of the carriers is not just a definition--- it's the mass of the carriers. – Ron Maimon Jul 19 '12 at 16:58 It is impossible to answer in terms of Bohr model. The important part of the concept which is out of the scope of the school approach is the degenerate electron gas. Below Fermi level all states at zero temperature are filled. If you take an electron out of this filled place you get a hole. Which behaves exactly as an "anti-electron". The key is to realize that a lot of electrons in the degenerate gas below Fermi level behave as there is nothing there. If you really intend to give this idea at the school-level, you could try to bring an analogy with bubbles in the water. They behave exactly (well, more or less actually) as particles with negative mass. If you want to study the situation when you have a lot of water and relatively small amount of bubbles, it is constructive not to solve equations for the whole mass of water, but replace them with the equations for the bubbles. It is more or less why people use hole language. And, technically, these holes are not "in atoms". Valence electrons which conduct in semiconductors and metals are collective. Thus, "missing valence electron" also does not belong to any particular atom. - This is true, but you can make a hole description of Moseley's law, so long as you remember that the holes have negative mass. The reason is that the region near the nucleus is a very deep potential well, and it looks the same as a filled Fermi gas near there. The orbits of the hole don't leave this region for 1 or 2 shell holes in heavy atoms. – Ron Maimon Oct 13 '11 at 6:56 @Ron Maimon, with all respect you really have no idea what you are talking about. Valence electrons are different from core electrons and do not see the potential near nucleus. Hole may be more or less localized, but first of it all, by definition it originates from the delocalized states. – Misha Oct 13 '11 at 7:40 @Misha, the ability to re-write you theory in terms of holes arises whenever there is a gap that that limits the electron energiy spectrum from above. The you cahnge sigs and "top" becomes the "bottom". Localization does not matter. – Slaviks Oct 13 '11 at 10:37 @Slaviks, localization is irrelevant. However, we are talking about semiconductors. Where holes are charge carriers which move freely and formed by valence electrons. For holes which are localized in atoms the story would be a bit different: they would not "move freely", had no momentum, etc. – Misha Oct 13 '11 at 10:57 @Misha, ok, but... an acceptor-doped semiconductor is still better described in terms of holes, even if they become localized at low T because of the Anderson-Mott metal-to-insulator transition. Localized holes in this regime would be just ionized donors – Slaviks Oct 13 '11 at 11:15 show 11 more comments What the hell are "electron HOLES" in semiconductors? Doping. I found this on youtube. For every reaction there is an equal and opposite reaction. This takes me back to my A Level's, Maths, Physics and Chemistry all rolled into one. - What was the need of introducing such abstract concept in semi conductors? Electrical physics and electrical engineering have been around much longer than knowledge of the electron has. For most things in electrical engineering, the circuits they design work exactly the same whether the charge-carrier is a negative-particle moving in one direction, or a positive-particle moving the other way! Since they didn't know whether the charge-carrier was positive or negative at the time, and it didn't really make a difference to them, electrical engineers chose to make all their electrical diagrams and symbols under the assumption that the charge-carrier was positive. Of course, it turns out they were wrong. However, by the time they discovered this, their notation had been pretty much set in stone - circuit diagrams even today are drawn as though the charge-carrier is positive, and even some elementary- and middle-school students are taught that electricity flows "from the positive terminal to the negative terminal" in a battery. If electrons are negatively-charged, what is this "positively charged particle" that flows in a circuit? It is not a real particle - there is no positively charged particle flowing in the circuit (the protons stay relatively still). Rather, the "positive charge" is really a lack of negative-charge - an empty space with higher electric-potential, which the electrons are necessarily attracted towards. (Electrons moving to the right necessarily cause holes to flow to the left) Electron holes are not a real "thing" - they are just a concept used to help explain why it doesn't matter (to electrical engineers) whether it is a positive- or negative-charge flowing, and to help them cope with the fact that all their diagrams use the "wrong" charge-carrier :) See this blog post for more info on this (and other good electricity-related questions). - 1 This is incorrect. Electron holes are a "real thing" in that they give a positive carrier in a positively doped semiconductor. The Hall voltage is opposite. It is not a matter of convention whether the charge carrier is negative of positive--- there are real physical effects when the carriers switch signs. – Ron Maimon Oct 14 '11 at 21:42 -1: Gotta downvote, this is wrong. – Ron Maimon Jul 19 '12 at 16:58 When the elctron is free in outermost orbit that empty place is called is holes.. - Technically true, but this really doesn't say anything useful... – David Zaslavsky♦ Jul 19 '12 at 4:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9507310390472412, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/24418/ordinary-cohomology-of-stacks/24428	## Ordinary cohomology of stacks ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $\mathbf{X}$ be a stack over $Top$ (a lax sheaf of groupoids, or some such thing). If it admits a surjective representable map $F \to \mathbf{X}$ then one can form the iterated fibre product to get a simplicial space $F_\bullet$, and its realisation $X$ is the homotopy type of $\mathbf{X}$, which comes with a (homotopy class of?) map $X \to \mathbf{X}$. It seems that the reasonable thing to call the singular cohomology of $\mathbf{X}$ is the singular cohomology of its homotopy type, because the homotopy type is sufficiently unique for this to be well-defined. On the other hand, consider the composition of functors $$\mathbf{h}^i : Top \overset{H^i(-;\mathbb{Z})}\to AbGp \to Set \overset{inc}\to Gpd$$ where the middle functor is forgetful, and the last one is the inclusion of sets as groupoids with no non-identity morphisms. The stack $\mathbf{X}$ gives another functor $Top \to Gpd$, and one may consider the set $H^i$ of natural transformations $\eta : \mathbf{X} \to \mathbf{h}^i$. This set has the structure of an abelian group as one may add pointwise (as $\mathbf{h}^i$ factors through abelian groups). Given any such $\eta$, we may apply it to the unique homotopy class of maps $X \to \mathbf{X}$ to obtain $\eta(X \to \mathbf{X}) \in H^i(X;\mathbf{Z})$ a cohomology class on the homotopy type. Any $Y \to \mathbf{X}$ factors up to homotopy through $X$, and so $\eta(Y \to \mathbf{X})$ is obtained by pullback from $\eta(X \to \mathbf{X})$. Thus it seems to me that the group $H^i$ is naturally isomorphic to $H^i(X;\mathbf{Z})$. My first question, at last, is: is the group $H^i$ the correct notion of the cohomology of the stack $\mathbf{X}$, which happens to coincide with the cohomology of its homotopy type? Would it still be a reasonable definition on some terrible stack that does not admit an atlas? Secondly: how does one do homology like this? Thirdly: one can do the above over $Diff$ instead of $Top$, and use de Rham cohomology in the definition of $\mathbf{h}^i$. Then it seems the group one produces, call it $H^i_{dR}$ now, still makes sense, but the homotopy type of a stack over $Diff$ is not necessarily itself a manifold and does not necessarily have a de Rham theory. Is this a reasonable thing to do? Why is it not done this way (for example by Behrend)? Fourthly: if, as I suspect, all this is known, where can I find it? - Just to make this clear to everyone, $F$ is assumed to be a topological space. Such a stack \mathbf{X} is equivalent to the stack of torsors of the topological groupoid object $X \times_{\mathbf{X}} X \rightrightarrows X$. – David Carchedi May 12 2010 at 18:43 ## 2 Answers Let me address those questions which you made that I can: 1.) is the group $\mathbf{H^{i}}$ the correct notion of the cohomology of the stack $\mathbf{X}$? Yes, I believe so, at least when the stack is nice enough. However, you should be a little careful with your proof: Behrang Noohi's develops a homotopy theory for topological stacks. In particular, every topological stack $\mathbf{X}$ admits an atlas $X \to \mathbf{X}$ which is a universal homotopy equivalence. (To be more precise, to any given any atlas $F \to \mathbf{X}$, there exists such a universal $X$- it is the classifying space of the topological groupoid associated to this atlas). Being a universal homotopy equivalence means that for any $Y \to \mathbf{X}$, the map $Y \times_{\mathbf{X}} X \to Y$ is a weak equivalence. This is the sense that any map $Y \to \mathbf{X}$ "factors through $X$ up to homotopy". Combining this observation with the naturality square for $\eta:\mathbf{X} \to h^{i}$ gives a precise proof of your statement. It should be noted that there is a wider class of stacks than just topological ones which admit such a universal weak equivalence from a space. For example, paratopological and pseudotopological stacks (see: Homotopy Theory of Topological stacks by Behrang Noohi). Morevoer, in Gepner and Henrique's "Homotopy Theory of Orbispaces", they remark that any stack over $Top$ has a weak homotopy type, not just ones with atlases. So, we get a unique up to homotopy map from a CW-complex $X$ to $\mathbf{X}$ which is homotopy terminal amongst maps from a CW-complex. I do not know whether or not this is a universal weak equivalence in any way though. However, there might be a way of using this to extend the proof to show your definition agrees with the cohomology groups of this weak homotopy type. Just for some extra knowledge, in the above paper, they also show that any map $Y \to \mathbf{X}$ factors DIRECTLY through $X \to \mathbf{X}$ (up to 2-iso) whenever $Y$ is paracompact Hausdorff. This is a really cool result. 2.) Well, you can always do homology by using this universal weak equivalence, but, this is probably not the kind of answer you are looking for. 3.) For differentiable stacks, if by differentiable stack you mean one coming from an atlas, you can always define De Rham cohomology via a double complex associated to the associated simplicial manifold arising as the nerve of any Lie groupoid representing you. See for instance "Differentiable Stacks and Gerbes" by Ping Xu. (There are a few other places I've seen this as well, I can search if you need me to). 4.) This is certainly the first time I myself have seen the cohomology defined this way, but, perhaps others have seen this. Anyhow, here is I would have a look at the three papers I've mentioned. All three are on the Arxiv. - ### 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. Does this work for de Rham cohomology? Sure. While the homotopy type of a stack in $Diff$ doesn't exactly have a de Rham complex, it does have a couple of things that work perfectly as substitutes. Behrend uses the Cech-de Rham double complex from a particular choice of atlas. The homotopy type of the stack is the geometric realisation of a simplicial manifold, and the Cech-de Rham double complex is made from the simplicial object in cochain complexes by taking the de Rham complex levelwise, so it really is the de Rham model for the simplicial manifold. Alternatively, the homotopy type of the stack has a real homotopy type (which can be defined, for example, by taking Sullivan's PL forms with real coefficients). On manifolds the cohomology of the real homotopy type is of course naturally isomorphic to the de Rham cohomology. Thus on stacks over $Diff$ the Cech-de Rham complex is a model for the real homotopy type. This wasn't one of your questions, but as long as we are varying the cohomology theory, it's worth mentioning this. The K-theory of a stack really is different from the K-theory of its homotopy type. The simplest example is the stack $*/G$. The K-theory of the stack is the G-equivariant K-theory of a point, but K-theory of the homotopy type is K(BG), which is of course the completion at the augmentation ideal by the Atiyah-Segal theorem. I think that defining K-theory in terms of natural transformations $hom(-,\textbf{X}) \to K(-)$ just gives you the K-theory of the homotopy type. - Hi Jeff. Does this double complex exist for a general stack on Diff? I'd suspect you'd need an atlas- I'm not sure how to get a simplicial manifold otherwise... – David Carchedi May 13 2010 at 2:54 I think you are right - one needs an atlas to make a cech-de rham complex. If you don't have an atlas then you could try doing something along the lines of how you might construct a homotopy type for a stack without an atlas: consider the category of all spaces mapping to the stack. This gives a diagram of simplicial manifolds, and the homotopy type should be the hocolim I think. To get the analogue of the Cech-de Rham complex, you should probably take the (ho)lim of the corresponding diagram of complexes. – Jeffrey Giansiracusa May 13 2010 at 6:30 I'm curious, do you have a good reference showing how equivariant K-theory is the correct notion of K-theory on a stack? And does this extend to stacks coming from groupoids? – David Carchedi May 14 2010 at 13:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 52, "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": 0.9296804666519165, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/267863/for-measurable-f-n-x-to-0-infty-show-that-sum-n-1-infty-f-n-inf/281734	# For measurable $f_n : X \to [0, \infty)$ show that $\sum_{n=1}^\infty f_n < \infty$ almost everywhere Let $f_n : X \to [0 \infty)$ be a sequence of measurable functions on the measure space $(X, \mathcal{F}, \mu)$. Suppose there is an $M > 0$ such that the functions $g_n = f_n\chi_{\{f_n \le M\}}$ satisfy $\|\|g_n\|\|_1 \le An^{-\frac{4}{3}}$ and for which $\mu\{f_n > M\} \le Bn^{-\frac{5}{3}}$. Here, $A$ and $B$ are positive constants independent of $n$. Prove that $h(x) = \displaystyle \sum_{n=1}^\infty f_n(x) < \infty$ for almost all $x \in X$. - ## 2 Answers As the sequence $\{\sum_{n=1}^Nf_n\chi_{\{f_n\leqslant M\}}\}$ is convergent in $L^1$, we extract an almost everywhere convergent sequence. As the concerned terms are non-negative, we actually have that $\sum_{n=1}^{+\infty}f_n\chi_{\{f_n\leqslant M\}}$ is convergent for almost everywhere $x$. By a Borel-Cantelli like argument, $\mu(\limsup_{n\to+\infty}\{f_n>M\})=0$, so for almost every $x$, we can find an integer $N(x)$ such that if $n\geqslant N(x)$ then $f_n(x)\leqslant M$. - The Borel-Cantelli Lemma and the bound on each $\mu\{f_n > M\}$ ensure that $\mu\{x \in X : f_n(x) > M \mbox{ infinitely often}\} = 0.$ Thus, for almost all $x \in X$, there is an $N(x) \in \mathbb{N}$ so that $\displaystyle \sum_{n=N(x)}^\infty f_n(x) =\sum_{n=N(x)}^\infty f_n(x)\chi_{\{f_n > M\}}(x)$. But $\displaystyle \sum_{n=1}^\infty f_n(x)\chi_{\{f_n > M\}}(x) < \infty$ for almost all $x \in X$, since $\displaystyle \int_X\sum_{n=1}^\infty f_n\chi_{\{f_n > M\}}d\mu = \sum_{n=1}^\infty\int_X f_n\chi_{\{f_n > M\}}d\mu \le \sum_{n=1}^\infty An^{-\frac{4}{3}} < \infty$ (the equality here is a consequence of the Monotone Convergence Theorem). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 30, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9392692446708679, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/106666-prove-if-g-abelian-group-then-s-order-least-6-a.html	# Thread: 1. ## Prove that if G is an abelian group then it's order is at least 6 I need to show that all groups of smaller orders are abelian but i'm not sure how to do this. 2. Originally Posted by Louise I need to show that all groups of smaller orders are abelian but i'm not sure how to do this. I'm almost sure you meant "prove that if a group is NON-abelian then its order is > 5, which of course is true ==> you need to prove all the groups of order <= 5 are abelian. Now, 1,2,3,5 are immediate (right? prime order groups are cyclic...), so the only "problem" is order 4. Supose then that G = {e,x,y,z} is a non-cyclic group (because cyclic groups we already know are abelian) with 4 elments and play with the elements: for example, what can xy, yx be so that teh axioms of group theory will be fulfilled? Of course, e = the group's unity Tonio 3. Originally Posted by Louise I need to show that all groups of smaller orders are abelian but i'm not sure how to do this. Are you allowed to use the result that a group of prime order is cyclic? And that a cyclic group is always abelian? If so, then that takes care of 2, 3 and 5. So you only have to consider the order 4 group. Consider the elements e, a, b, c (arbitrary names for an arbitrary group, e is the identity). Suppose the group is not abelian. Then (wlog) suppose $ab \ne ba$. I'll leave you to finish it off by showing that there need to be at least two other elements in the group apart from $a, b, e$. 4. Groups of order 1, 2, 3, and 5 are cyclic and so Abelian. There are only two non-isomorphic groups of order 4, the cyclic group and the "Klein four group" and they are both Abelian. Proving that those are the only two groups of order four is tedious but not too difficult.	{"extraction_info": {"found_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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9550545811653137, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/50890/mean-of-log-of-cdf	# Mean of log of cdf Let $CDF$ be the cumulative distribution function for the standard normal distribution. Let $Z$ be a standard normal random variable. Then $CDF(Z)$ is uniformly distributed on the unit interval, so by integration we can show that $E(ln(CDF(Z))) = -1$. My question: there an easy way to compute $E[\ln(CDF(Z + c))]$ for constant $c$? What I'm really looking for is $E[\ln(CDF(Z + c))]$ on the $(-\infty, 0]$ interval and the $(0, \infty)$ interval. - 1 Perhaps you can tell us in just a tad more detail what integration you are doing to come up with $\ln(\operatorname{cdf}(Z)) = -1$? $\operatorname{cdf}(Z)$ is a random variable taking on values in $(0,1)$, and its natural logarithm $\ln$ is also a random variable with values in $(-\infty,0)$. So, what does it mean when you say that $\ln(\operatorname{cdf}(Z)) = -1$?? – Dilip Sarwate Feb 26 at 21:14 Dilip, sorry for the conclusion, I meant E(ln(CDF(Z)))=−1. Or equivalently, if U is a uniformly distributed RV on the unit interval, E(ln(U)) = -1. – garyrob Feb 26 at 21:19 1 E(ln(CDF(Z))) is equivalent to E(ln(U)) where U is uniformly distributed on (0, 1]. That can be found by integrating ln(), which is x(ln(x))-x + C, which is -1 if C is 0 x is 1. – garyrob Feb 26 at 21:31 ## 2 Answers No,you'll have to do the integration numerically for each $c$ value. Let $\Phi(x)$ be the Gaussian CDF function. You want to evaluate $$I=\int \Phi'(x-c) \ln \Phi(x) dx$$ Idea 1: this looks like a convolution; the Fourier transform of $\Phi'$ is easy enough, but that of $\ln \Phi$ is not; and even if you could take the transform of the second factor, inverting the resulting expression is unlikely to be feasible. Idea 2: do it by parts, this works out to $$\Phi(x-c) \ln \Phi(x) \vert^{\infty}_{\infty} - \int_{\infty}^{\infty} \frac{ \Phi(x-c)}{\Phi(x)} \Phi'(x) dx$$ the first term is conveneiently zero, but the second is no better off. Idea 3: expand the $\ln \Phi$ term. Define $U=1-\Phi$ so that : $$I=-\int U'(x-c) \ln [ 1-U(x)] dx$$ and then expand the logarithm as though $U$ were small. You end up with terms like $U'(x-c) U^n(x)$, which are still not easily integrable due to the shift in the argument. At this point, it seems that the simplifications that arise in $E[ \ln \Phi(x)]$ aren't panning out, so, numerical integration is a feasible solution. - plugging the relevant integrand into Wolfram Alpha yielded no analytic results. – Dave Feb 27 at 0:23 Wolfram Alpha is only a crude test--there is plenty it cannot do. Your exposition here is much more interesting and convincing than that! (+1) – whuber♦ Feb 27 at 14:12 – garyrob Feb 27 at 17:44 In short there is no "easy" way to do this, and Dave has tried a few sensible approaches. One approach perhaps worth a try is to taylor expand since your function is smooth and analytic, and we have a simple form for the moments of the standard normal: Notice that $L^* = \ln(\Phi(X))$ $L = \frac{\phi(x)}{\Phi(x)}$ $L' = \frac{\phi(x)^2}{\Phi(x)^2} - x \frac{\phi(x)}{\Phi(x)} = L(x)^2 - xL(x)$ $L'' = 2LL'-L-xL' = 2L^3 - 3xL^2 +(x^2-1) L$ $L''' = 6L^2L' - 6xLL'-3L^2+2xL+(x^2-1)L' = 6L^4-6xL^3-6xL^3+6x^2L^2-3L^2 + 2xL+(x^2-1)(L^2-xL)$ $=6L^4-12xL^3+(7x^2-4)L^2 + (3x-x^3)L$ I had hoped we could spot a pattern there incolving Hermite polynomials or some-such, but I can't see one... however, because of the recurrence it is basic calculus and algebra that a computer could chunk through to arbitrary depth. Armed with these we can now notice that we can define $X = Z + c$ so that $X \sim N(c,1)$ and then Taylor expand $L^$ around the point c $E[\ln(\Phi(X))] = E[\ln(\Phi(c)) + L(c) Z + L'(c)\frac{Z^2}{2} + L''(c)\frac{Z^3}{3!}+\cdots]$ Now remember that $E[Z^p] = (p-1)!!$ for even $p$ and $0$ otherwise, so we only keep every other term: $E[\ln(\Phi(X))] = \ln(\Phi(c)) + L'(c)\frac{1}{2} + L'''(c)\frac{3!!}{4!}+ L^{(5)}(c)\frac{5!!}{6!} \cdots$ $E[\ln(\Phi(X))] = \ln(\Phi(c)) + L'(c)\frac{1}{2} + L'''(c)\frac{1}{8}+ L^{(5)}(c)\frac{1}{48} + L^{(7)}(c) \frac{1}{384} + \cdots + L^{(2k-1)}(c) \frac{1}{2^k k!} + \cdots$ (where we use that $(2k-1)!! = \frac{(2k)!}{2^k k!}$ ) So all that remains is to plug in the derivatives above and work out the total. If anyone spots a pattern then this should work exactly. Incidentally, if you hadn't asked about $\ln(\Phi(x))$ but simply $\Phi(x)$ then we have a much simpler expression. Carrying on from the expression above but using $\Phi$ in place of $L^$ we have: $E[\Phi(X)]=\Phi(c) + \phi(c) \sum_{k=1}^\infty \frac{H_{2k-1}(c)}{2^k k!}$ Where $H_n$ is the nth Hermite polynomial, and we have used the fact that $\frac{d^n \phi(x)}{dx^n}=H_n(x)\phi(x)$ Notice that inside the sum we have a polynomial over an exponential, so this is going to converge nicely. Notice also that since $H_{2k-1}(0) = 0$ for all k then in the case where $c=0$ this colapses back to 0.5 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": 49, "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": 0.9418045878410339, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/32268/suppose-phi-is-a-weak-solution-of-delta-phi-f-in-mathcalh1-then	# Suppose $\phi$ is a weak solution of $\Delta \phi = f \in \mathcal{H}^1$. Then $\phi\in W^{2,1}$ I'm trying to prove the statement in the title in as simple a way as possible. It is Theorem 3.2.9 in Helein's book "Harmonic maps, conservation laws, and moving frames", although it is not proved there. The statement is as follows. Suppose $\phi\in\mathbb{R}^m$ is a weak solution of $\Delta \phi = f \in \mathcal{H}^1$, where $\mathcal{H}^1$ is the standard Hardy space on $\mathbb{R}^m$. Then $$\Big\lVert\frac{\partial^2\phi}{\partial x^\alpha \partial x^\beta}\Big\rVert_{L^1(\mathbb{R}^m)} \le C\lVert f \rVert_{\mathcal{H}^1(\mathbb{R}^m)}.$$ My idea is to use convolution with the kernel of the Laplacian, and then differentiate, estimate in $L^1$ and somehow interpolate between the $\mathcal H^1$ and $BMO$ norms. Then since the kernel of the Laplacian is in $BMO$, I am finished. However there are two problems with my proof: I don't know how to prove that one can interpolate a convolution between $\mathcal H^1$ and $BMO$ (atomic decomposition?) and I don't know how to prove that the kernel of the Laplacian is in $BMO$. Does anyone have either a better way to prove this theorem, or a way to fix up my proof? Thanks! - Glen: I just wanted to point out that you can bump your question to the front page by editing it (of course, you should do this with reasonable restraint). I think this would get noticed more than advertising it on the chat (where there are at most 3 users a day). Also, if you have a heavy-handed solution, could you give some indication of what it is? I'd certainly be interested in seeing it. – t.b. Apr 12 '11 at 9:23 @Theo Buehler Yeah, I am aware of that method of 'bumping' questions, but is this not considered poor behaviour? I wouldn't want to go against site policy. My solution is just to flesh out what I wrote in the question, and solve the two problems. The first is easy. The second is the reason I see it as heavy handed: the interpolation of a convolution between $\mathcal{H}^1$ and $BMO$ is a result from Stein, "Harmonic Analysis", which I do not understand at the moment. Atomic decomposition is the right idea, but the theorem is just a black box for me at the moment. – Glen Wheeler Apr 12 '11 at 9:46 Maybe when I understand it a little better I will be able to provide a good answer. – Glen Wheeler Apr 12 '11 at 9:46 Thanks for the info, I appreciate it. In fact, if you add actual content (such as your first comment), I don't think anyone would construe this as poor behavior. I think you're reasonable enough to avoid exaggerating. Moreover, there is of course the bounty system. – t.b. Apr 12 '11 at 9:54 ## 1 Answer Let me take a try: Recall that the Riesz transform is bounded from $\mathcal H^1$ to $\mathcal H^1$ (and from $L^2$ to $L^2$). We have the inequality $$\\|\partial_i \partial_j u \\|_{L^2} \leq \\| \Delta u \\|_{L^2}.$$ This is because $R_i R_j \Delta u = \partial_i \partial_j u$ where $R_i$ is the $i$-th Riesz transform. Now because the Riesz transform is also $\mathcal H^1$ bounded we have $$\\|\partial_i \partial_j u \\|_{\mathcal H^1} \leq \\| \Delta u \\|_{\mathcal H^1}.$$ But the $L^1$ norm is dominated by the $\mathcal H^1$ norm so $$\\|\partial_i \partial_j u \\|_{L^1} \leq \\| \Delta u \\|_{\mathcal H^1}.$$ Further note that $$t f'(s) = f(t + s) - f(s) - \int_0^1 f''(s+r)(t - r) \, dr$$ and a similar statement holds for partial derivatives. This implies that we can control the first derivative by the second and the function itself. This should give the result as asked in the title. Alternatively we could use Mikhlin's multiplier theorem for $\mathcal H^1$ for the second part. - Hi Jonas, thanks for the answer. It looks like you've perhaps made a typo in the second inequality? Should that not be the Hardy space there? Actually, I'm confused about the second inequality. Could you please write a bit more clarification there? – Glen Wheeler Apr 12 '11 at 15:19 @Glen: I have added some information. Please tell me if it is enough or if you think it is wrong. – Jonas Teuwen Apr 12 '11 at 15:30 Oh sigh, I read the $H^1$ as the Sobolev space $W^{1,2}$ before the edit. Yes! Its correct, and it is much shorter than what I originally proposed. Thanks again for the answer. – Glen Wheeler Apr 12 '11 at 15:33 You're welcome. – Jonas Teuwen Apr 12 '11 at 15:34 @Jonas: Very nice! (I voted this up before but I was distracted by a phone call, that's why I write only now) – t.b. Apr 12 '11 at 16:17 show 4 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": 29, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9682623147964478, "perplexity_flag": "head"}
http://cms.math.ca/Events/summer12/abs/com	2012 CMS Summer Meeting Regina Inn and Ramada Hotels (Regina~Saskatchewan), June 2 - 4, 2012 www.cms.math.ca//Events/summer12 Combinatorics Org: Karen Meagher (Regina) and Marni Mishna (SFU) [PDF] ROBERT BAILEY, University of Regina Metric dimension of distance-regular graphs: an update [PDF] A {\em resolving set} for a graph $G$ is a collection of vertices chosen so that any vertex of $G$ is uniquely identified by the list of distances to the chosen few. The {\em metric dimension} of $G$ is the smallest size of a resolving set. At the 2009 CMS Summer Meeting in St. John's, I gave a talk entitled "Metric dimension of distance-regular graphs". Since that time, several papers on the subject have been written by myself and others. In this talk, I will give an overview of some recent results for various families of distance-regular graphs, and also some computer calculations determining the metric dimension of some small distance-regular graphs. ANDREA BURGESS, Ryerson University Cops and robbers on graphs based on designs [PDF] In the game of cops and robbers, a set of cops and a single robber occupy vertices of a graph, with cops and robber playing alternately, in each turn moving to an adjacent vertex or passing. The cops win if one of them occupies the same vertex as the robber, and the robber wins if he evades capture indefinitely. The {\em cop number} $c(G)$ is the smallest number of cops which guarantee that the cops win on the graph $G$. Meyniel's conjecture states that for a connected graph $G$ on $n$ vertices, $c(G)=O(\sqrt{n})$; families of graphs which attain the conjectured asymptotically largest cop number are known as {\em Meyniel extremal}. Known Meyniel extremal families arise from incidence graphs of projective and affine geometries. Motivated by such results, we investigate the cop number of various graphs based on combinatorial designs, such as incidence graphs, point graphs and block intersection graphs. We give bounds on the cop number of such graphs, and in some cases find new Meyniel extremal families. This is joint work with Anthony Bonato. SOPHIE BURRILL, Simon Fraser University Combinatorics in arc diagrams: crossings, nestings and topology [PDF] A variety of combinatorial objects, including matchings, set partitions and permutations, can be modelled using arc diagrams- a set of vertices on a horizontal line with arcs possibly connecting them. There are two natural ways to parameterize these diagrams: 1) by considering the number of arcs that mutually cross (crossings), and 2) by genus. We explore (also using nestings, an additional companion statistic to crossings) these two methods from an enumerative standpoint, providing links between them where appropriate. YI CAO, University of Alberta DP-complete Problems Derived from Extremal NP-complete Properties [PDF] In contrast to the extremal variants of coNP-complete problems, which are frequently DP-complete, many extremal variants of NP-complete problems are in P. We investigate the extremal variants of two NP-complete problems, the extremal colorability problem with restricted degree and the extremal unfrozen non-implicant problem, and show that both of them are DP-complete. As far as we know, no extremal variant of an NP-complete problem has been shown to be DP-complete before. This is a joint work with Dr. Joseph Culberson and Dr. Lorna Stewart. MICHAEL CAVERS, University of Calgary Graphs with large distinguishing chromatic number [PDF] The distinguishing chromatic number $\chi_D(G)$ of a graph $G$ is the minimum number of colours required to properly colour the vertices of $G$ so that the only automorphism of $G$ that preserves colours is the identity. It is known that for a graph $G$ of order $n$, the bound $1\leq\chi_D(G)\leq n$ holds, with equality in the upper bound only for complete multipartite graphs. We discuss properties of graphs with large distinguishing chromatic number and characterize the graphs $G$ of order $n$ satisfying $\chi_D(G)=n-1$ or $\chi_D(G)=n-2$. ADA CHAN, York University Trivial Nomura algebra [PDF] First introduced by Sylvester in 1867 as inverse orthogonal matrices, a type~II matrix is an invertible $n\times n$ matrix that has no zero entry, whose inverse can be easily obtained by inverting every entry, taking the transpose and multiplying by $n^{-1}$. Hadamard matrices and spin models are well known examples of type~II matrices. Nomura gave a construction of a formally dual pair of association schemes from each type~II matrix. Despite the potential of finding new dual pairs of association schemes from Nomura's construction, almost all of the type~II matrices we have examined yield the trivial association schemes. In this talk, we discuss what we can say in this disappointing situation. PETER DUKES, University of Victoria Intersections of Latin Squares [PDF] If we superimpose two latin squares on the same symbols and count the number of agreements, this is the \emph{intersection number} of the pair. Here, we consider the problem of determining which intersection numbers are possible, given the order of the squares. Early considerations solved this problem when the two squares have the same order, say $n$. Roughly speaking, the result is that every value in $[0,n^2]$ is possible, except for a few values near the top of the interval. Here, we consider and completely solve the following two parameter version of the problem. {\sl For fixed $m$ and $n$, $n<m$, what are the realizable intersection numbers for two latin squares, one of order $m$ and the other of order $n$}? This is joint work with Jared Howell. CHRIS GODSIL, University of Waterloo Graphs, Polytopes, Quadrics [PDF] Let $X$ be a graph with adjacency matrix $A$ and eigenvalue $\theta$. The projections of the standard basis vectors onto the $\theta$-eigenspace of $A$ generate a convex polytope, and properties of this polytope relate in an interesting way to properties of $X$. We can use knowledge of the facets to derive the Erd\H{o}s-Ko-Rado theorem. We can also use information about the real quadrics that contain the vertices of the polytope to constrain the structure of $X$. In my talk I will explain these connections. ANGELE HAMEL, Wilfrid Laurier University A Generalized Anonymity Metric for Probabilistic Attacks [PDF] In theory, an anonymity network allows two participants, Alice and Bob, to communicate without an adversary being aware that the message Alice sends is the same one that Bob receives. However, in practice, a statistical attack on such a system is possible through observation of the patterns of sending and receiving. How anonymous is the communication really, and how can this anonymity be measured? In 2007 Edman et al. proposed a combinatorial model for this problem. In this talk we show how, by interpreting the problem in terms of contingency tables, we can define a new measure of the anonymity provided by the system. This measure generalizes two existing measures, and provides a new framework for future results. (Joint work with Jean-Charles Grégoire). PENNY HAXELL, University of Waterloo Packing and covering in $r$-partite hypergraphs [PDF] We give the first known nontrivial upper bound (for $r>3$) on the following old problem known as Ryser's Conjecture. For an $r$-partite $r$-uniform hypergraph $H$, suppose the maximum size of a matching in $H$ is $k$. Does there exist a cover of $H$ of size at most $(r-1)k$? Here a cover is a set of vertices meeting all edges of $H$. It is immediate that $H$ has a cover of size $rk$. We show that for the cases $r=4$ and 5, there exists $x>0$ such that $H$ has a cover of size at most $(r-x)k$. (Joint work with Alex Scott) GLENN HURLBERT, Arizona State University EKR on Graphs and Lattices [PDF] The classic theorem of Erd\H os, Ko, and Rado has generated a lot of activity in recent years. One new idea explores the structure of intersecting families of maximum size under the restriction that certain pairs of elements cannot be in the same set. This corresponds to investigating the largest intersecting family of independent sets of a graph. If some such family forms a star -- some element is in every set -- the graph is said to have the EKR property. A second consideration studies intersection of other objects, such as permutations and partitions (defined by sharing the same coordinate or block, respectively), and asking the usual Erd\H os-Ko-Rado and Hilton-Milner type questions. A third notion defines the intersection of elements of a lattice by to the rank of their meet. Here, a lattice is said to be EKR if a largest intersecting family of elements forms a star; that is, it is the upset of an atom. We discuss current advances in these areas, including joint work with Bekmetjev, Brightwell, Czygrinow, Fishel, Kamat, and Meagher. SAMUEL JOHNSON, Simon Fraser University The exponential growth of restricted lattice paths [PDF] Restricted lattice paths are effective statistical mechanical models of physical and chemical phenomena. The exponential growth factor of the number of lattice paths of size $n$ corresponds to the entropy in the statistical mechanical system. We give a systematic method for proving factors for families of models sharing a common submodel with known exponential growth. This is joint work with Marni Mishna. BEN LI, University of Manitoba friendship 3-hypergraphs [PDF] Let $(X,\mathcal{B})$ be a set system in which $\mathcal{B}$ is a set of 3-subsets of $X$. Then $(X,\mathcal{B})$ is a \textit{friendship $3$-hypergraph} if it satisfies the following property: for distinct elements $u,v,w \in X$, there exists a unique fourth element $x \in X$ such that $\{u,v,x\}, \{u,w,x\}, \{v,w,x\} \in \mathcal{B}$. If a friendship $3$-hypergraph contains an element $f \in X$ such that $\{f,u,v\} \in B$ for all $u,v \in X \setminus \{f\}$, then it is called a \textit{universal friend $3$-hypergraph} and the element $f$ is called a \textit{universal friend} of the hypergraph. In this presentation, we will discuss what we know about friendship $3$-hypergraphs and universal friend $3$-hypergraphs. JESSICA MCDONALD, Simon Fraser University Average Degree in Graph Powers [PDF] The $k$th power of a simple graph $G$, denoted $G^k$, is the graph with vertex set $V(G)$ where two vertices are adjacent if they are within distance $k$ in $G$. In this talk we are interested in finding lower bounds on the average degree of $G^k$, a problem that is related to both additive number theory (via Cayley graphs) and the famous Caccetta-H\"{a}ggkvist Conjecture. Here we share essentially best possible lower bounds when $k=4$ or $k\equiv 2$ (mod 3). Joint work with M. DeVos and D. Scheide. KAREN MEAGHER, University of Regina Sperner partition systems [PDF] A Sperner set system is a set system in which no set is a subset of another. Sperner proved that the maximum size of a Sperner set system on $\{1,2,\dots,n\}$ is $\binom{n}{\lfloor \frac{n}{2}\rfloor }$ and the only systems meeting this bound are the middle levels of the poset of sets ordered by inclusion. A \textsl{Sperner partition system} is a set of partitions with the property that no class from one partition is the subset of a class from another partition. A Sperner partition system can be considered to be a resolvable Sperner set system. In 2005 Lucia Moura, Brett Stevens and I proved a bound on the size of a Sperner partition system and showed that this bound can be met if the size of the classes in the partitions are all equal. Recently, P. C. Li and I determined bounds on the sizes of some Sperner partition systems when the sizes of the classes cannot be all the same. MARNI MISHNA, Simon Fraser University Towards generating random mammalian genomes [PDF] Genome arrangements, a major mechanism of evolution, shuffle genetic material along chromosomes. Thus, a now standard approach models groups of close genomes as signed permutations. The correct data structure to study permutations in this context is the common interval tree. In this talk we will describe the process of considering tree parameters to refine the class of common interval trees (hence permutations) to those that represent mammalian genomes well. These refinements are particularly amenable to Boltzmann random generation and analytic enumerative techniques. Work in collaboration with Mathilde Bouvel, Cedric Chauve, Rosemary McCloskey, Cyril Nicaud and Carine Pivoteau. JOY MORRIS, Lethbridge Structure of circulant graphs [PDF] Some fairly recent deep results using Schur rings have provided a means of classifying circulant graphs. I will present this classification, and discuss some asymptotic and other results that follow from it. ORTRUD OELLERMANN, University of Winnipeg The Average Connectivity of Graphs and Digraphs [PDF] The connectivity between a pair $u,v$ of vertices in a graph $G$ is the maximum number of pairwise internally disjoint $u$--$v$ paths in $G$. The average connectivity of $G$ is the average connectivity between pairs of vertices of $G$ taken over all pairs. Analogous concepts can be defined for digraphs. We survey known results on this subject and present several open/partially solved problems including: (i) the problem of finding the maximum connectivity of a subgraph in a graph with a given average connectivity; (ii) the problem of finding the maximum average connectivity among all orientations of a given graph $G$; (iii) the maximum average connectivity among all graphs with a given degree sequence. ALISON PURDY, University of Regina EKR-type results using the method of Ahlswede and Khachatrian [PDF] In 1997, Ahlswede and Khachatrian published what is now commonly known as the Complete EKR Theorem. Although there have been numerous papers proving EKR-type results for objects other than sets, the techniques used by Ahlswede and Khachatrian do not appear to have received the attention that the significance of the theorem warrants. In this talk I will discuss attempts to adapt these techniques for use on other objects. This is joint work with Karen Meagher. MATEJA SAJNA, University of Ottawa On Sheehan's Conjecture for graphs with symmetry [PDF] It is known that every simple regular graph of degree $d$ that has a Hamilton cycle in fact possesses a second Hamilton cycle if $d$ is odd or $d \ge 300$. Sheehan conjectured that the statement is also true for $d=4$, which would imply that it is true for every $d > 2$. Fleischner showed that Sheehan's conjecture fails for 4-regular multigraphs. In this talk, we show that Sheehan's conjecture is true for 4-regular vertex-transitive simple graphs, and present some other recent results on the conjecture for regular simple graphs with a sufficiently large automorphism group. This is joint work with Andrew Wagner. Self-Avoiding Polygon Models of Polymer Entanglements [PDF] With the goal of understanding polymer entanglements, for over 20 years there has been interest in questions about knotting and linking of self-avoiding polygons on the simple cubic lattice. Notably, in 1988 Sumners and Whittington proved that all but exponentially few sufficiently long self-avoiding polygons are knotted. This proved the long standing Frisch-Wasserman-Delbruck conjecture that sufficiently long ring polymers will be knotted with high probability. Since then there has been progress both theoretically and numerically using lattice polygon models to investigate polymer entanglements. Much of this progress has been motivated by questions arising from the study of DNA topology. For lattice models, these questions lie at the interface between statistical mechanics, enumerative combinatorics, topology, graph theory and applied probability/Monte Carlo methods. I will review progress made and highlight new results and open problems, especially in the context of extensions of the Sumners and Whittington theoretical approach to questions about knotting and linking in self-avoiding polygon models. BRETT STEVENS, Carleton University Linear feedback shift registers and covering arrays [PDF] The set of fixed length subintervals of a linear feedback shift register form a linear code. A very nice theorem of Bose from 1961 proves that these codewords form the rows of a covering array of strength $t$ if and only if the dual linear code has minimum weight $t+1$. Munemasa observed that whenever the length of the intervals is less than a generous bound, the dual code is the Hamming code which has minimum distance 3 and this the covering array is guaranteed to have strength 2. In fact the only 3-coverage that is missing corresponds to the weight-3 multiples of the generating polynomial of the LFSR. We use this and results on difference sets over finite fields to construct a new famlily of strength 3 covering arrays which improve many best known upper bounds on covering arrays AMY WIEBE, Simon Fraser University Golay sequence and array pairs [PDF] Golay complementary sequence pairs solve several problems in digital information processing. They have a rich existence pattern which includes several infinite families over different alphabets. Many families of Golay sequence pairs can be viewed as the projection of Golay array pairs. This multi-dimensional viewpoint considerably simplifies the construction and enumeration of Golay sequence pairs. We give an unexpected application of this approach which allows a simple reinterpretation of some classical and recently-discovered Golay sequence pairs. This is joint work with Jonathan Jedwab and Matthew Parker. KAREN YEATS, Simon Fraser University Feynman graphs and a chord diagram expansion [PDF] Feynman graphs have a lot of fun combinatorial structure. Without expecting any background in physics, I will set up the situation and discuss how by playing with some recurrences we can get an expansion in terms of chord diagrams for some Green functions. ## Event Sponsors Support from these sponsors is gratefully acknowledged.	{"extraction_info": {"found_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": 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": 0.9078746438026428, "perplexity_flag": "head"}
http://mathoverflow.net/questions/55885	Why semigroups could be important? Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) There is known a lot about the use of groups -- they just really appear a lot, and appear naturally. Is there any known nice use of semigroups in Maths to sort of prove they are indeed important in Mathematics? I understand that it is a research question, but may be somebody can hint me the direction to look on so that I would see sensibility of semigroups, if you see what I mean (so some replies like look for wikipedia are not working as they are anti-answers). - 3 This is an extremely general question, given that whole books have been written about semigroups or monoids. – Jim Humphreys Feb 18 2011 at 18:21 1 @Jim: Yes, perhaps this question asks for a direction where to begin ... – Martin Brandenburg Feb 18 2011 at 18:24 1 @Victor: The wikipedia article shows that semigroups are important in applied mathematics, PDE, theoretial computer science and probability theory ... perhaps you should explain why you think that this is not evidence enough for their importance. Check also related articles such as en.wikipedia.org/wiki/C0-semigroup – Martin Brandenburg Feb 18 2011 at 18:30 2 @Martin: ok, in PDE's, in computer science (say, particularly, semi-Thue systems), the so-called Cuntz-algebras in C*-algebras etc. -- all this is where semigroups appear is like there appears something with associative operation and that's it, no further miracle of that we then go looking what is going on with those semigroups with further yielding nice results about our initial problem. And, C_0-semigroups aren't really semigroups :-) I'm not saying that semigroups are not important, I just wonder if somebody knows where semigroups can do some tricks (hence it's a good research question) – Victor Feb 18 2011 at 18:41 2 Shouldn't this be titled "Why are semigroups important?" :) – Mariano Suárez-Alvarez Feb 18 2011 at 18:55 show 4 more comments 10 Answers Semigroups provide a fundamental, algebraic tool in the analysis of regular languages and finite automata. This book chapter (pdf) by J-E Pin gives a brief overview of this area. - 1 I overlooked this answer while writing my own; Pin's book is a nice reference, thanks for mentioning it. – Michal Kotowski Feb 18 2011 at 19:06 Thanks for the book!!!!!!!!!!!!!!!!! – Victor Feb 19 2011 at 18:30 You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Am slightly surprised no one has mentioned the Galvin-Glazer proof of Hindman's theorem via the existence of semigroup structure on $\beta{\mathbb N}$, the Stone-Cech compactification of the positive integers (see, for instance, part of this note by Hindman. The relevance to the original question is that knowing that compact right topological semigroups have idempotents'' may sound recondite, but it is just what was needed to answer Galvin's original question about translation-invariant ultrafilters, which was itself motivated by a "concrete" question in additive combinatorics. On a related note, while it is in general not possible to embed a locally compact group as a dense subgroup of something compact (the map from a group to its Bohr compactification need not be injective), you can always embed it densely into various semigroups equipped with topological structure that interacts with the semigroup action: there are various of these, perhaps the most common being the WAP-compactification and the LUC-compactification. Unfortunately this often says more about the complicated behaviour of compact semitopological semigroups (and their one-sided versions) than about anything true for all locally compact groups, but the compactifications are a useful resource in some problems in analysis, and the semigroup structure gives one some extra grip on how points in this compactification behave. (Disclaimer: this is rather off my own fields of core competence.) - 1 +1: I was thinking of remarking on this. Furstenberg and Katznelson's paper on idempotents in compact semigroups (math.stanford.edu/~katznel/colorla.pdf) is amazing. – Daniel Litt Feb 26 2011 at 1:38 Victor, I don't understand your claim that $C^0$-semigroups aren't really semigroups. You are not free to decide for all the mathematical community what is a semigroup (I guess that you are interested only on discrete semigroups, aren't you ?). $C^0$-semigroups are fundamental in PDEs (in probability too as mentioned by Steinhurst). The reason is that a lot of evolution PDEs (basically all parabolic ones, like the heat equation, or Navier-Stokes) can be solved only forward but not backward. In linear PDEs, this is a consequence of the Uniform Boundedness Principle (= Banach-Steinhaus Theorem). There is a nice theory relating operators and semigroups, the former being the generator of the latter. In the linear case, a fundamental result is the Hille-Yosida Theorem. Subsequent tools are Duhamel's principle and Trotter's formula. A part of the theory extends to nonlinear semigroups. - Semigroups of bounded $L^{2}$ operators are very important in probability. They in fact provide one of the main ways show the very close connection between a self-adjoint operator and a `nice' Markov process (nice can be taken to mean strong Markov, cadlag, and quasi-left continuous.) So how does one get this semigroup from a Markov process? If $X_t$ is your process let $\mu_{t}(x,A)$ be measure with mass $\le 1$ with value $P(X_t \in A \| X_0=x)$. Then $\int f(y) \mu_t(x,dy) = T_tf(x)$ gives a semigroup of bounded $L^{2}$ operators. Why are such constructions important and natural? If $f$ is your initial distribution of something (heat for example) and $X_t$ is Brownian motion. Then $T_t$ acts by letting the heat distribution $f$ diffuse the way heat should. This then gives a nice way to connect PDE and probability theory. I'd end by offering that semigroups are important, in part, because they do arise is so many places and can bridge between disciplines. There are other reasons as well. - When people say "semigroup" in relation to functional analysis they really only mean semigroups isomorphic to the non-negative reals, right? – Qiaochu Yuan Feb 18 2011 at 20:22 1 Not necessarily, although that is the most common usage. But there is a huge literature on non-commutative dynamical systems, motivated by quantum mechanical problems. – András Bátkai Feb 18 2011 at 20:34 Thank you very much -- this is the type of answers I'm looking for!!!!!!!!!!!!!!!!!!!!!! I'll look in that direction. – Victor Feb 19 2011 at 18:28 An important application of semigroups and monoids is algebraic theory of formal languages, like regular languages of finite and infinite words or trees (one could argue this is more theoretical computer science than mathematics, but essentialy TCS is mathematics). For example, regular languages can be characterized using finite state automata, but can also be described by homomorphisms into finite monoids. The algebraic approach simplifies many proofs (like determinization of Buchi automata for infinite words or proving that FO = LTL) and gives deeper insight into the structure of languages. - Interestingly enough, N. Bourbaki describes exactly this use of the theory in his set theory book, in the appendix to chapter 1. – Harry Gindi Feb 19 2011 at 20:07 That's interesting - could you provide a reference? – Michal Kotowski Feb 19 2011 at 21:14 2 I just gave you a reference... Here, is this better: N. Bourbaki Éléments de Mathématique I: Théorie des Ensembles. – Harry Gindi Feb 20 2011 at 4:46 (Commutative) semigroups and their analysis shows up in the theory of misère combinatorial games. The "misère quotient" semigroup construction gives a natural generalization of the normal-play Sprague-Grundy theory to misere play which allows for complete analysis of (many) such games. (See http://miseregames.org/ for various papers and presentations.) - Also finitely generated commutative semigroups are equivalent to Petri nets. – Mark Sapir Feb 18 2011 at 20:09 Circuit complexity. See Straubing, Howard, Finite automata, formal logic, and circuit complexity. Progress in Theoretical Computer Science. Birkhäuser Boston, Inc., Boston, MA, 1994. If you want a research problem relating circuit complexity with (finite) semigroups, there are many in the book and papers by Straubing and others. See also Eilenberg and Schutzenberger (that is in addition to Pin's book mentioned in another answer) - about connections between finite semigroups and regular languages and automata. - Unary (1-variable) functions mapping a set X to itself under composition is a semigroup. Cayley's Theorem (one of them) says that every semigroup is isomorphic to one of this kind. - That is definitely true, but here you see semigroups appear just like they appear and nothing else. Furthermore, this sort of says that every semigroup can be realised by something. What I'd like is that something would be resolved (may in particular) by means of semigroups use. It's hard to explain what I mean. – Victor Feb 18 2011 at 18:23 Though you say that $C_0$ semigroups are not really semigroups, the structure of compact semitopological semigroups plays an important role in the investigation of their asymptotic behaviour. For example, Glicksberg-DeLeeuw type decompositions or Tauberian theorems are obtained such a way, see Engel-Nagel: One-Parameter semigroups for Linear evolution Equations, Springer, 2000, Chapter V.2. - Given a group $G$, the Block Monoid $B(G)$ consists of sequences of elements in $G$ that sum to zero. So for example, an element of $B(\mathbb{Z})$ is $(-2,-3,1,1,3)$. The monoid operation is concatenation, and the empty block is the identity element. Given a Dedekind domain, one can take its ideal class group, and consider the block monoid over that group. Note that in the obvious way, elements of the block monoid can be irreducible or not. One can study irreducible factorization in the Dedekind domain by studying irreducible factorization in the block monoid. -	{"extraction_info": {"found_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": 0.18, "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": 0.9293664693832397, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/68387-union-u-intersection-n.html	# Thread: 1. ## Union (u) and Intersection (n) the real number system ImageShack - Image Hosting :: 16012009078os8.jpg eg. Q n R = Q Find? 1)Q u H 2)J u N 3)H n J 4)H n R 5)Q u R 6)H n Q 7)J n N I cant understand how to do them ? please solve the above with explanation. 2. Originally Posted by mj.alawami the real number system ImageShack - Image Hosting :: 16012009078os8.jpg eg. Q n R = Q Find? 1)Q u H 2)J u N 3)H n J 4)H n R 5)Q u R 6)H n Q 7)J n N I cant understand how to do them ? please solve the above with explanation. W, J, H? Abusive notation. Anyway, lol... 1) It tells you that the real numbers are composed of irrational and rationals. In other words, real numbers are composed of numbers that cannot be expressed as a fraction of integers (irrational ones) and those that can be expressed as a fraction of integers (rational ones). Thus: $\mathbb{H} \cup \mathbb{Q} = \mathbb{R}$ 2) Clearly the natural numbers are a subset of the integers. Any natural number in the set {1,2,3,...} is a part of the set of integers {...,-2,-1,0,1,2,3,...}. So their union is just the integers again. Thus: $\mathbb{N} \cup \mathbb{J} = \mathbb{J}$ 3) The intersection of the irrationals and the integers is the set of all elements that BOTH irrational and integer. Can you find any such numbers? Can you find numbers that cannot be expressed as a ratio between two integers and yet are still integers? 4) The irrationals a proper subset of the reals, so if you are asking the intersection, it is the set of all numbers that both real and irrational. But by definition, all irrationals are real. So the set that you get for your answer is just the irrationals once again. In general, here are a few tips for sets: -If $X$ is a subset of $Y$ (written as $X \subseteq Y$), then their intersection is $X \cap Y = Y$. Example: the irrationals are a subset of the reals, so their intersection is just the irrationals. -If $X$ is a subset of $Y$, then their union is $X \cup Y = Y$. Example: the irrationals are a subset of the reals, so their union is just the reals. -If $X$ and $Y$ are disjoint sets (meaning that they share no elements), then their intersection is the empty set. Example: the intersection between the negative integers and the positives is empty (no such number is both negative and positive at the same time) -If $X$ and $Y$ are disjoint sets, then their union is simply a new set that contains every element of both $X$ and $Y$. Example: the union of all the nonnegative integers and the negative integer is simply the set of integers because every integer is either negative or nonnegative. Try to finish the rest yourself and show us your work. We can check it over for you. 3. Originally Posted by Last_Singularity W, J, H? Abusive notation. Anyway, lol... 1) It tells you that the real numbers are composed of irrational and rationals. In other words, real numbers are composed of numbers that cannot be expressed as a fraction of integers (irrational ones) and those that can be expressed as a fraction of integers (rational ones). Thus: $\mathbb{H} \cup \mathbb{Q} = \mathbb{R}$ 2) Clearly the natural numbers are a subset of the integers. Any natural number in the set {1,2,3,...} is a part of the set of integers {...,-2,-1,0,1,2,3,...}. So their union is just the integers again. Thus: $\mathbb{N} \cup \mathbb{J} = \mathbb{J}$ 3) The intersection of the irrationals and the integers is the set of all elements that BOTH irrational and integer. Can you find any such numbers? Can you find numbers that cannot be expressed as a ratio between two integers and yet are still integers? 4) The irrationals a proper subset of the reals, so if you are asking the intersection, it is the set of all numbers that both real and irrational. But by definition, all irrationals are real. So the set that you get for your answer is just the irrationals once again. In general, here are a few tips for sets: -If $X$ is a subset of $Y$ (written as $X \subseteq Y$), then their intersection is $X \cup Y = Y$. Example: the irrationals are a subset of the reals, so their intersection is just the irrationals. -If $X$ is a subset of $Y$, then their union is $X \cap Y = Y$. Example: the irrationals are a subset of the reals, so their union is just the reals. -If $X$ and $Y$ are disjoint sets (meaning that they share no elements), then their intersection is the empty set. Example: the intersection between the negative integers and the positives is empty (no such number is both negative and positive at the same time) -If $X$ and $Y$ are disjoint sets, then their union is simply a new set that contains every element of both $X$ and $Y$. Example: the union of all the nonnegative integers and the negative integer is simply the set of integers because every integer is either negative or nonnegative. Try to finish the rest yourself and show us your work. We can check it over for you. My attempt to answer - 3) H n J=R 4)H n R=H 5)Q u R=R 6)H n Q=R 7)J n N =N If they any one of them are wrong please tell me which on and explain it ,Thank you 4. 3) You're saying that the irrational numbers and the integers, when intersected, create the real numbers? Think about it this way... Irrational numbers are the numbers that have no fractional form, and can therefore NEVER be integers. And the set of Integers is the set of all WHOLE numbers, and can therefore never be IRRATIONAL. It is clear that these have nothing in common and are therefore disjoint sets, meaning: $\mathbb{H} \cap \mathbb{J} = \emptyset$ 4) That one is right. 5) That one is also right. 6) I see that your achilles heel is with disjoint sets. Think about this one, The set of the irrational numbers are numbers that have no fractional form, and can therefore NEVER be rational. And the set of rational numbers is the set of numbers that CAN be put into fractional form, and can therefore NEVER be irrational. It is clear that these have nothing in common and are therefore disjoint sets, meaning: $\mathbb{H} \cap \mathbb{Q} = \emptyset$ 7) That is right. 5. Originally Posted by Aryth 3) You're saying that the irrational numbers and the integers, when intersected, create the real numbers? Think about it this way... Irrational numbers are the numbers that have no fractional form, and can therefore NEVER be integers. And the set of Integers is the set of all WHOLE numbers, and can therefore never be IRRATIONAL. It is clear that these have nothing in common and are therefore disjoint sets, meaning: $\mathbb{H} \cap \mathbb{J} = \emptyset$ 4) That one is right. 5) That one is also right. 6) I see that your achilles heel is with disjoint sets. Think about this one, The set of the irrational numbers are numbers that have no fractional form, and can therefore NEVER be rational. And the set of rational numbers is the set of numbers that CAN be put into fractional form, and can therefore NEVER be irrational. It is clear that these have nothing in common and are therefore disjoint sets, meaning: $\mathbb{H} \cap \mathbb{Q} = \emptyset$ 7) That is right. Can you please check these question too for me ... N u Q =Q J n N =N Q n H = /0 (disjoint) Q u J =Q 6. Originally Posted by mj.alawami Can you please check these question too for me ... N u Q =Q J n N =N Q n H = /0 (disjoint) Q u J =Q That is correct - good job! 7. Originally Posted by Last_Singularity That is correct - good job! Thank you very much ... Can't put the rest in words #### 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": 34, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.942876935005188, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/43578/solving-time-dependent-schrodinger-equation-in-matrix-form?answertab=oldest	# Solving time dependent Schrodinger equation in matrix form If we have a Hilbert space of $\mathbb{C}^3$ so that a wave function is a 3-component column vector $$\psi_t=(\psi_1(t),\psi_2(t),\psi_3(t))$$ With Hamiltonian $H$ given by $$H=\hbar\omega \begin{pmatrix} 1 & 2 & 0 \\ 2 & 0 & 2 \\ 0 & 2 & -1 \end{pmatrix}$$ With $$\psi_t(0)=(1,0,0)^T$$ So I proceeded to find the stationary states of $H$ by finding it's eigenvectors and eigenvalues. $H$ has eigenvalues and eigenvectors: $$3\hbar\omega,0,-3\hbar\omega$$ $$\psi_+=\frac{1}{3}(2,2,1)^T,\psi_0=\frac{1}{3}(2,-1,-2)^T,\psi_-=\frac{1}{3}(1,-2,2)^T$$ Respectively. Could anyone explain to me how to go from this to a general time dependent solution, and compute probabilities of location? I have only ever encountered $\Psi=\Psi(x,y,z,t)$ before, so I am extremely confused by this matrix format. I would be extremely grateful for any help! - ## 1 Answer The general solution is $$\psi(t)=\sum_k c_k e^{-itE_k/\hbar}\psi_k$$ where the $\psi_k$ form a basis of eigenvectors with corresponding eigenvalues $E_k$, and the $c_k$ are constant. You can match arbitrary initial conditions at $t=0$ by expanding the initial state in the eigenbasis; this will determine the valued for the $c_k$. To get the statistical interpretation: The expectation of the Hermitian observable $A$ at time $t$ is given in the Schroedinger picture by $$\langle A\rangle_t:=\psi(t)^*A\psi(t).$$ Here it is assumed that $\psi(t)$ has norm 1. As the squared norm is preserved by the dynamics, this gives a well-defined expectation (i.e.,, the expectation of the identity matrix is 1 at all times). - Ok, good that's what I imagined to do, I wasn't sure. But more importantly how do you interpret this probabilistically in matrix form? – Freeman Nov 6 '12 at 20:45 @LHS: see the edited answer – Arnold Neumaier Nov 7 '12 at 8:39 Does your $c=0$ mean $c(t=0)$ ? – Nivalth Nov 7 '12 at 16:53 1 The $c_k$ are constant, and a $t$ was missing in the exponent of the first formula. $c=0$ meant $t=0$. I corrected both errors; sorry. – Arnold Neumaier Nov 8 '12 at 9:19	{"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": 19, "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": 0.9096881151199341, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/30861/isotropy-and-noise/41801	# Isotropy and Noise If you have a field which value is just Gaussian noise plus a constant do you call it isotropic? • there is no preferred direction • however it is not "the same" in all directions if "the same" means "constant" The question is about a definition of a word, thus obviously conflicting answers can arise. However what definition do you use or have seen used: "no preferred direction" or "constant wrt translation"? - I have a hard time distinguishing between your two alternatives as it is not clear what you mean by constant. Isotropy is pretty well defined from it's Greek origin, i.e. equal in all directions. In a way that implies constant, so the property cannot be a function of the direction. Though it can still vary in time or distance. – Alexander Jun 27 '12 at 16:30 How can something vary in distance and be isotropic? Central symmetry is not isotropy. And my question is whether you call something with no preferred direction isotropic even if it contains random fluctuations? – Krastanov Jun 27 '12 at 19:51 – Alexander Jun 27 '12 at 21:12 Please be clearer about the type of field: what is constant? Do you have a constant vector field plus gaussian fluctuations? I use "isotropic" as a synonym for translational invariant, and use "rotationally invariant" for "rotationally invariant". I don't know if this is universal usage. – Ron Maimon Jul 28 '12 at 5:13 There seems to be some confusion in the answers. Are you referring to a scalar field? Or a vector field? – Colin K Sep 26 '12 at 2:15 show 1 more comment ## 4 Answers The noise may have no preferred direction, but for large enough volumes the average $$\langle\mathbf{F}\rangle_V=\frac{1}{V}\int_V \mathbf{F}(x) d^3x$$ will approach the constant term. Thus the field is not isotropic and there is a preferred direction: that of averages over sufficiently large volumes. (Here, of course, specifying the direction to arbitrary precision and confidence requires sufficiently large volumes, and one has to assume this is not a problem or the question becomes ill-defined.) Although the field is not constantly in this preferred direction, this direction does exist - the field is just more likely to be in this direction. - Why is this downvoted (+1)? The OP seems to be asking about a case where there is a vector which is mostly one direction, with an additional random noise that prevents you from measuring the global average. Without more detail from OP regarding what is the vector in question (is he talking about E or the Poynting vector), it is hard to be more precise. – Ron Maimon Jul 28 '12 at 21:31 What if the average is zero? – Krastanov Oct 26 '12 at 22:31 If the average is zero then the field is indeed isotropic. – Emilio Pisanty Oct 26 '12 at 22:47 The way I would approach this is to consider blackbody radiation, where a spherical blackbody radiator is isotropic, e.g. it emits radiation equally in all directions. If we were to model the noise introduced by such a blackbody, it would be additive Gaussian white noise. In this example, it is called white noise because it has a flat power spectrum density, and it is Gaussian because the noise has Gaussian amplitude distribution. It is additive because it can be added linearly to the desired signal. In this case isotropy is a characteristic of the emitter. If I would interpret isotropy in terms of noise, I would understand it as being the same in all directions with respect to the receiver, e.g. regardless of direction, the received signal would have identical noise functions that were added to the signal. If the noise had some constant associated with it, as long is the constant was additive and the same in all directions with respect to the receiver, I would consider that isotropic as well. - You say: •there is no preferred direction but in your example this is true only at a single point i.e. at the centre of the gaussian. At every other point in your space the grad of your field is non-zero so the field at that point can't be isotropic (because the gradient will vary depending on the direction you look). An isotropic field must have zero gradient everywhere. A consequence of this is that an isotropic field is necessarily homogeneous. - I mean Gaussian noise for the field, not an amplitude that has the same functional dependence of $\exp(x^2)$ in space. I will make the text clearer. – Krastanov Jun 27 '12 at 19:48 It depends how you define your Gaussian noise. A Gaussian random field can be modelled as having a distribution which is a multi-variate Gaussian of the Fourier modes. The field is isotropic if the distribution only depends on the magnitude of the Fourier mode and the Fourier modes are uncorrelated with each other. In cosmology the primordial perturbations are thought to be such a field and current observations apear to be consistent with this, although some argue that there may be violation on the largest scales. In the comsology case the constant part is the background density and it is spatially uniform so it does not effect the isotropy. In theory one could have a constant part that varies spatially. But if your observations only consist of one realization of the field, then it is not really possible to distinguish between a spatially varying constant part and the random fluctuations. An example of an article discussing these issues is available at: http://arxiv.org/abs/astro-ph/0509301 . -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9420726895332336, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/119866/find-all-solutions-to-the-equation-ez-i/119869	Find all solutions to the equation $e^z = i$ I know that there is an equation for finding the nth roots of a complex number, which easily done once you have the modulus and argument of the complex number in question. There would be n roots. But how do I know how many solutions there are to an exponential raised to a complex number? Thanks for any help out there! - 1 Use $e^{r\theta}=r\cdot cos (\theta) + r\cdot sin(\theta)i$ – you Mar 14 '12 at 0:21 4 @you: I am not sure that is quite correct – Henry Mar 14 '12 at 0:26 oops! It's too late for me to edit it now. I mean "$re^{i\theta}$" – you Mar 14 '12 at 4:14 2 Answers $\displaystyle e^z=e^{i\frac{\pi}2+2k\pi i}$ so that $\ \displaystyle z=i\frac{\pi}2+2k\pi i\$ with $k\in \mathbb{Z}$ - Thanks for your help! – mathlearner Mar 14 '12 at 5:31 For real $a$ and $b$ you have $$\exp(a+bi) = e^a \cos(b/2\pi) + e^a \sin(b/2\pi) i$$ so now set the right hand side equal to $i$ and solve the real and imaginary parts. You get $e^a \cos(b/2\pi)=0$ but $e^a$ is positive so $b$ can only be of the form $(n+1/2)\pi$ for integer $n$, implying $\sin(b/2\pi)=\pm 1$, and you want $e^a \sin(b/2\pi)=1$ but $e^a$ is still positive so you must have $\sin(b/2\pi) = 1$ and so $e^a =1$, implying $b=(2n+1/2)\pi$ for integer $n$ and $a=0$, so the solution is $a+bi$ which is $$\left(2n+\frac{1}{2}\right)\pi i.$$ - thanks for your help! – mathlearner Mar 14 '12 at 5:31	{"extraction_info": {"found_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": 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": 0.929267406463623, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2011/04/26/tangent-spaces-and-regular-values/?like=1&source=post_flair&_wpnonce=c8d80d4293	The Unapologetic Mathematician Tangent Spaces and Regular Values If we have a smooth map $f:M^m\to N^n$ and a regular value $q\in N$ of $f$, we know that the preimage $f^{-1}(q)=A\subseteq M$ is a smooth $m-n$-dimensional submanifold. It turns out that we also have a nice decomposition of the tangent space $\mathcal{T}_pM$ for every point $p\in A$. The key observation is that the inclusion $\iota:A\to M$ induces an inclusion of each tangent space by using the derivative $\iota_{p}(\mathcal{T}_pA)\subseteq\mathcal{T}_pM$. The directions in this subspace are those “tangent to” the submanifold $A$, and so these are the directions in which $f$ doesn’t change, “to first order”. Heuristically, in any direction $v$ tangent to $A$ we can set up a curve $\gamma$ with that tangent vector which lies entirely within $A$. Along this curve, the value of $f$ is constantly $q\in N$, and so the derivative of $f\circ\gamma$ is zero. Since the derivative of $f$ in the direction $v$ only depends on $v$ and not the specific choice of curve $\gamma$, we conclude that $f_{p}(v)$ should be zero. This still feels a little handwavy. To be more precise, if $v\in\mathcal{T}_pA$ and $\phi$ is a smooth function on a neighborhood of $q\in N$, then we calculate $\displaystyle\begin{aligned}\left[f_{p}(\iota_{p}(v))\right](\phi)&=\left[\left[f\circ\iota\right]_{p}(v)\right](\phi)\\&=v\left(\phi\circ f\circ\iota\right)\\&=v\left(\phi(q)\right)\\&=0\end{aligned}$ since any tangent vector applied to a constant function is automatically zero. Thus we conclude that $\iota_{p}(\mathcal{T}_pA)\subseteq\mathrm{Ker}(f_{p})$. In fact, we can say more. The rank-nullity theorem tells us that the dimension of $\mathrm{Ker}(f_{p})$ and the dimension of $\mathrm{Im}(f_{p})$ add up to the dimension of $\mathcal{T}_pM$, which of course is $m$. But the assumption that $p$ is a regular point means that the rank of $f_{p}$ is $n=\dim(N)$, so the dimension of the kernel is $m-n$. And this is exactly the dimension of $A$, and thus of its tangent space $\mathcal{T}_pA$! Since the subspace $\mathcal{T}_pA$ has the same dimesion as $\mathrm{Ker}(f_{p})$, we conclude that they are in fact equal. What does this mean? It tells us that not only are the tangent directions to $A$ contained in the kernel of the derivative $f_$, every vector in the kernel is tangent to $A$. Thus we can break down any tangent vector in $\mathcal{T}_pM$ into a part that goes “along” $A$ and a part that goes across it. Unfortunately, this isn’t really canonical, since we don’t have a specific complementary subspace to $\mathcal{T}_pA$ in mind. Still, it’s a useful framework to keep in mind, reinforcing the idea that near the subspace $A$ the manifold $M$ “looks like” the product of $\mathbb{R}^{m-n}$ (from $A$) and $\mathbb{R}^n$, and we can even pick coordinates that reflect this “decomposition”. Like this: Posted by John Armstrong \| Differential Topology, Topology 2 Comments » 1. [...] diffeomorphism is . The tangent space at a point on the submanifold is mapped by to , and the kernel of this map is exactly the image of the inclusion . The same statements hold with and swapped appropriately, [...] Pingback by \| April 27, 2011 \| Reply 2. [...] further! Let , where is the complementary projection to . This map has maximal rank everywhere, so we know that for each the preimage is an -dimensional submanifold . The tangent space to consists of [...] Pingback by \| June 30, 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": 51, "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": 0.9309676885604858, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/116257/list	## Return to Answer 2 added 156 characters in body You can look at the short paper by Conte, Marchisio end Murre On the k-unirationality of the cubic complex (2007). It contains a proof of the unirationality of $V_6$ over a field $k$ of any characteristic $\neq 2,3$, under the assumption that $V_6$ has a $k$-rational point $p$ and that one of the two planes through $p$ on the quadric is also rational over $k$. The paper is freely available on In the webintroduction, the authors write "we follow closely Enriques construction, our only contribution being to fully explain and justify his statements". 1 You can look at the short paper by Conte, Marchisio end Murre On the k-unirationality of the cubic complex (2007). It contains a proof of the unirationality of $V_6$ over a field $k$ of any characteristic $\neq 2,3$, under the assumption that $V_6$ has a $k$-rational point $p$ and that one of the two planes through $p$ on the quadric is also rational over $k$. The paper is freely available on the web.	{"extraction_info": {"found_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": 0.9283759593963623, "perplexity_flag": "head"}
http://nrich.maths.org/472/index?nomenu=1	Sketch the members of the family of graphs given by $$y = {a^3\over (x^2+a^2)}$$ for $a=1$, $2$ and $3$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9803082346916199, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=587545&page=4	Physics Forums Page 4 of 5 < 1 2 3 4 5 > ## What exactly is an electron? In post # 47, a minimum electron radius value, (3Gm/c^2) was noted. From this radius, a fundamental mass value is defined, using a ring shape with the angular momentum (h/4pi). The charge spins at light velocity so that the effective mass times velocity times radius will equal angular momentum. m c (3Gm/c^2) = h/4pi (m)^2 = (h/4pi) (c/3G) m = (hc/12pi G)^1/2 m = (1/2) (2/3)^1/2 (Planck mass) I suggest this mass value is the fundamental value that has a specific relationship to the electron mass, the muon mass and the tau mass. The photon wavelength that has energy to produce two particles with each particle mass value equal to (hc/12pi G)^1/2 is (3pi hG/c^3)^1/2 meter. This wavelength is: wavelength = 2pi (3/2)^1/2 (Planck length) The ratio of this fundamental wavelength to the wavelength (h/2mc) is approximately 1.025x10^-22 to one. I will suggest that this is also equal to [h/(2pi)^2] divided by (2mc^2) where the m value is the electron mass. This ratio is 1.025028393x10^-22. If these ratio values are precisely correct then the true G value must be very close to 6.671745197x10^-11. Improved experiments will determine if this is correct. First off I'm no physicist but an EE. So my question might seem odd, but in light of everything that was said until now, why has nobody (except for one guy I believe) proposed string theory to try to explain what elementary particles are (electron included)? Is this because it is still an "unproven" (untested) theory? String theory seems to be acknowledged by many recognized scientists, so perhaps it is a valid one at answering the initial question: what is an electron? Btw, what a great forum this is. Just recently found it. Since then I just can't help but try to read every single posts. Waaa, I'm going crazy :) Hi kended, String Theory and Quantum Gravity are well covered in the book by Lee Smolin, titled Three Roads To Quantum Gravity. Much work is needed if string theory is to accomplish its objective. A quote from the book follows: "Modern physicists try -- to explain particles in terms of fields. But this does not eliminate all problems. Some of the most serious of these problems have to do with the fact that theory of fields is full of infinite quantities. They arize because the strength of the electric field around a charged particle increases as one gets closer to the particle. But a particle has no size, so one can get as close as one likes to it. The result is that the field approaches infinity as one appraches the particle. This is responsible for many of the infinite expressions that arize in the equations of modern physics." He suggests, we may deny that space is continuous and so it is impossible to get arbitrarily close to a particle. We may also replace particles by little loops or strings. String theory is interesting but it is not yet mature enough to explain specifics such as electron mass. Quote by DonJStevens Hi kended, String Theory and Quantum Gravity are well covered in the book by Lee Smolin, titled Three Roads To Quantum Gravity. Much work is needed if string theory is to accomplish its objective. A quote from the book follows: "Modern physicists try -- to explain particles in terms of fields. But this does not eliminate all problems. Some of the most serious of these problems have to do with the fact that theory of fields is full of infinite quantities. They arize because the strength of the electric field around a charged particle increases as one gets closer to the particle. But a particle has no size, so one can get as close as one likes to it. The result is that the field approaches infinity as one appraches the particle. This is responsible for many of the infinite expressions that arize in the equations of modern physics." He suggests, we may deny that space is continuous and so it is impossible to get arbitrarily close to a particle. We may also replace particles by little loops or strings. String theory is interesting but it is not yet mature enough to explain specifics such as electron mass. I think I may have to read some more on this subject. It seems though that the book was written in 2002. So perhaps has the subject evolved a bit since then. Now with the very recent "supposed" discovery (measurement) of the Higgs-Boson, I read that string theorists are even more excited as this would somehow fit their theory in relation to particle mass. Anyhow I just thought that instead of saying that we "don't really know what an electron is", I would rather try to explain it using string theory that many great minds do believe in and where the mathematical constructs apparently make sense that is, until proven right. Btw, your former job sounded very cool :) The charge radius of an elementary particle has nothing to do with the spatial distribution of its charge. The charge radius is a length scale characterizing a scattering cross-section. In Quantum Field Theory, elementary particles are thought as excitations of the corresponding matter field that propagate carrying energy-momentum. These excitations may be created and destroyed by the action of sources. Consequently, the energy of the field's excitation due to two sources differs from the sum of the energy of the field's excitation due to the separate presence of each source. This is interpreted as a potential energy of interaction of two sources due to the exchang of virtual particles. The sources of some kinds of particles become quantum operators themselves corresponding to a (conserved) Noether current density corresponding to a continuous symmetry of the theory. For example, the free-electron Lagrangian has a global U(1) symmetry, corresponding to the invariance of the Lagrangian with respect to an arbitrary change in phase of the "electron field". The corresponding noether current is the electric current density, that acts as a source term for the "photon field". The photon "listens" to the electric charge in its vicinity and mediates the electromagnetic interaction. Since the range of the interaction is infinite, the photons are massless, and there is only a kinetic term for the photon field. This effectively describes Quantum Electrodynamics (QED), the simplest (Abelian) gauge theory of the Standard model. So a photon that undergoes pair production does so because of because it is a perturbation energetic enough to inititiate the pair's own standing waves. But why at one point, not another? Why don't we just see a pair and the lower energy photon from, for instance, Co60, but instead see a high energy photon and the pair production photons later? Is that because this "Noether" field has to be in the correct configuration locally for the pair production to occur, or because the virtual particle must form, which isn't a given, but a statistical process? I just want to point out to the OP. You're getting down to the basic building blocks of matter. When ever you describe something, call it E, you break it down or you reduce it to it's parts or its properties, X Y and Z. I think Vanhees said it perfectly: Quote by vanhees71 To conclude: To the best of our knowledge today (i.e., in this case the standard model of particle physics) the electron is an elementary spin-1/2 Dirac particle with one negative elementary charge and a mass of about $511 \; \mathrm{keV}/c^2$. It's a lepton, i.e., participates only in the electroweak interaction (let alone gravitation, which acts universally on anything that has energy and momentum). It is so difficult to describe an electron because you're running up against the basic building blocks of the universe which cannot in principle be described. Notice that Vorhees described an electrong in terms of charge, mass, spin, and the forces with which it participates. So we have some reduction. But charge, mass, spin we cannot reduce those entities to anything else, at least not now, and they remain in principle undescribable. Despite of the great success of mathematics one should reconsider the building blocks from time to time. Dirac invented his equation to describe the properties of spin 1/2 particles. The interaction of electrons are perfectly described by this equation. Especially D. Hestenes investigated this equation in detail and found a description of the electron: The electron is circulating with speed of light which is described by the Zitter-Bewegung, generates an angular momentum - the spin, and with E = h x nu the Compton wavelength defines the circumference of the circulation. But why the charge should circulate is still open. All models in the past ignore the synchrotron radiation of the charge. Even in classical physics a circulating charge embedded in its synchrotron radiation yields the angular momentum of the particle, the Compton wavelength as the wavelength of the radiation and the classical electron radius is the result of quantum mechanic interaction with the singularity. Circulation with v = c yields mass = field energy. The spherical solution of the radiation just guides the charge onto a circular orbit and is thus the reason for the circulating charge. Details are in G. Poelz "On the Wave Character of the Electron" http://arxiv.org/abs/1206.0620 This is a very good question to pose, as quoted by many people have tried to make a modle of a electron while its basic formation is known for the most part it would be a good project to go into to try and look inside of the electron Recognitions: Gold Member Quote by Atom1 This is a very good question to pose, as quoted by many people have tried to make a modle of a electron while its basic formation is known for the most part it would be a good project to go into to try and look inside of the electron We've tried. We can't find anything inside it. And by "we" I mean thousands of people using multiple particle colliders and other experiments over the last 50 years. Recognitions: Gold Member Since MacGregor and Rivas have been mentioned perhaps it is worth mentioning a moderately priced collection of papers What is the Electron?, edited by Simulik that includes papers by each of the them as well as others. It's quirky and in print. I obtained the book, What is the Electron, yesterday. Thank you xristy for mentioning this. The Einstein question (on back cover of book) is so very significant. When he was asked what he thought about the large numbers of short lived heavy particles being produced in high-energy accelerators, Einstein pondered the question and replied, "You know, it would be sufficient to really understand the electron." At the time little attention was paid to his remark. Yet the electron remains as mysterios today as it was in Einstein's time. The electron will be less mysterious if we learn why all electrons are identical. J. A. Wheeler said "That an electron here has the same mass as an electron there is also a triviality or a miracle." (see page 1215 of book Gravitation) Recognitions: Gold Member Quote by DonJStevens I obtained the book, What is the Electron, yesterday. Thank you xristy for mentioning this. The Einstein question (on back cover of book) is so very significant. When he was asked what he thought about the large numbers of short lived heavy particles being produced in high-energy accelerators, Einstein pondered the question and replied, "You know, it would be sufficient to really understand the electron." At the time little attention was paid to his remark. Yet the electron remains as mysterios today as it was in Einstein's time. The electron will be less mysterious if we learn why all electrons are identical. J. A. Wheeler said "That an electron here has the same mass as an electron there is also a triviality or a miracle." (see page 1215 of book Gravitation) You could expand your statement to include all fundamental particles, as they are all identical to other particles of the same type. Drakkith, you are so correct, all particles of the same type are identical. This implies that nature has a specific set of requirements that must be precisely met for each particle (type). We expect that theorists will determine and define these strictly imposed requirements. The electron requirements will most probably be the first that we will understand. Recognitions: Gold Member Perhaps Don. We'll have to wait and see! We can see now what some theorists have recently written about the electron. In post # 59 a paper by G. Polz was referenced. In this paper the electron is analyzed as a toroidal ring. The author (G. Polz) also references other papers that are interesting to all who want to know more. The referenced paper by Williamson and van der Mark analyzes the electron as a photon trapped in a toroidal path. As we come closer to a correct electron model, the desire to understand becomes ever more intense. As Drakkith said: We'll have to wait and see! The book, What is the Electron? noted in post #62 is interesting. The Wave Structure of Matter is discussed (page 227 - page 250). From page 240: "Schrodinger and Clifford predicted that charge was due to wave structures in space. - - We observe this process and call it charge. But as Clifford and Schrodinger wrote, there is no charge substnce involved. It is a property of the wave structure at the center." This book allows us to see some concepts by theorists who want to help us understand the electron. Thank you xristy for noting this book. Page 4 of 5 < 1 2 3 4 5 > Tags electron Thread Tools \| \| \| \| \|---------------------------------------------------\|----------------------------------------\|---------\| \| Similar Threads for: What exactly is an electron? \| \| \| \| Thread \| Forum \| Replies \| \| \| High Energy, Nuclear, Particle Physics \| 5 \| \| \| Introductory Physics Homework \| 8 \| \| \| Quantum Physics \| 1 \| \| \| Classical Physics \| 1 \| \| \| High Energy, Nuclear, Particle Physics \| 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": 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": 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": 0.9584144949913025, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/216311/fourier-transform-and-sampling-time	# Fourier transform and sampling time Given a signal $x(t)$ and the $X(\omega)$ obtained from $x(t)$ using a FFT with a sampling time $Ts$, I get a subset of $X(\omega)$: $Y(\omega)$ obtained from $X(\omega)$ taking it between $\omega_0$ and $\omega_1$, the question is: if I make the IFFT of $Y(\omega)$, does the new signal $y(t)$ have the same sampling time $Ts$? Thanks in advance - ## 1 Answer Starting from $x(t)$, you sample it with a sampling time $T_s$, then obtain a discrete time signal $x(n T_s)$ or $x(n)$, on which you apply FFT, to obtain the discrete spectrum $X(m)$ (the notation $X(\omega)$ usually denotes the Discrete Time Fourier Transform, which is different from the Discrete Fourier Transform, a fast version of the latter is FFT). Now, if you truncate $X(m)$ by keeping its values from $m_0$ up to $m_1$ and discarding the rest, without zero padding, then you are changing the FFT resolution, which is proportional to $1/N$, where $N$ is the number of your samples. The FFT resolution is the frequency distance between adjacent FFT points and corresponds to time distance between adjacent signal points (in the time domain). So yes, you are changing the sampling rate of your original signal. My advice is to use zero padding to avoid that. - when I get $Y(\omega)$, I don't multiply with a window. In that case I would obtain a 'zero padding' of the whole signal $X(\omega)$. In my case, I get $only$ the signal between $\omega_0$ and $\omega_1$ – Riccardo.Alestra Oct 18 '12 at 14:01 I see. I will modify my answer for that case. – Manos Oct 18 '12 at 14:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 28, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "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": 0.9348449110984802, "perplexity_flag": "head"}
http://mathoverflow.net/questions/25933/kleisli-monad-bijection/26044	## Kleisli Monad bijection ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) For a monad $(T,\mu,\eta)$ if $T(A) = T(B)$, does this imply that $\mu_A = \mu_B$? I want to know because in the bijection between Kleisli triples and monads, given a monad, we define $f^* := T(f) ; \mu_B$ if $f : A\to T(B)$ (c.f. Prop 1.6), but this needs to be well defined even when $T$ is not an embedding. Clarification (more formal): Given a monad $(T,\mu,\eta)$, I need to define ${}^$ on the class $\{f \;\|\; \exists A,B \in \|\mathbf{C}\| . f : A \to T B \}$. If $T$ were an embedding then I could just take $B := T^{-1}(\mathrm{cod}(f))$. What do I do though when $T$ is not an embedding? All I know about the codomain of an element of this class is that it is in the image of $T$. There may be many $B$'s and the $\mu_B$'s may be different! - 2 The objects of the Kleisli category are the objects of the underlying category, not T of them, so the reconstruction of the monad from the Kleisli category is always perfectly well-defined, even though the answer to your first question is probably "no." – Mike Shulman May 26 2010 at 0:56 Maybe I did not make myself clear enough. It is a subtle point (maybe trivial?), and one that I missed until I tried to formally write down a definition in Isabelle. Given a monad $(T,\mu,\eta)$, I need to define $$ on the class $\{f\|\exists A,B \in \|\mathbf{C}\| . f : A \to T B \}$. If $T$ were an embedding then I could just take $B := T^{-1}(\mathrm{cod}(f))$. What do I do though when $T$ is not an embedding? All I know about the codomain of an element of this class is that it is in the image of $T$. There may be many $B$'s and the $\mu_B$'s may be different! – apk May 26 2010 at 10:08 You could just define the morphisms to be triples $(A, B, f : A \to TB)$. Then you would know which $B$ to choose. – Neel Krishnaswami May 26 2010 at 11:27 I was thinking about this, but that's a different category, isn't it? – apk May 26 2010 at 11:32 1 I should add that your difficulty is partly a function of the fact that you are trying to define $$ on the universal collection of all the morphisms, rather than by considering families of hom-sets indexed by domain and codomain. In a dependent type theory, I would usually try to define Hom to be a type operator dependent upon domain and codomain. Isabelle must have some standard techniques to work around the lack of dependency -- I would look at those. (My previous comment is one way of doing it, but if there's an idiomatic thing you should do that.) – Neel Krishnaswami May 26 2010 at 11:33 show 1 more comment ## 3 Answers This didn't fit in the comments, so I'm posting it as an answer. Ignoring size issues, we can define a category as a set of objects, together with a family of sets of morphisms, with one set for each domain and codomain -- ie, as a dependent record: $$\mathrm{Cat} = \sum \mathrm{Obj}:\mathrm{Set}.\;\sum \mathrm{Mor} : \mathrm{Obj} \times \mathrm{Obj} \to \mathrm{Set}.\; \ldots \mathit{category\; axioms} \ldots$$ So if you have a category $C \equiv (\mathrm{Obj}, \mathrm{Mor}, \ldots)$ and a monad $(T, \mu, \eta)$, the Kleisli category will be of the form $(\mathrm{Obj}, (\lambda AB.\;\mathrm{Mor}(A, TB)), \ldots)$. Then, the extension operator will be an operation whose type is $\prod A,B:O.\;\mathrm{Mor}(A, TB) \to \mathrm{Mor}(TA, TB)$, which can be defined in the obvious way, as $\lambda A\;B\;f.\;T(f);\mu_B$. Note that the objects for the domain and codomain come in as arguments, so there's no need to reconstruct them from the data of the function $f$. (In fact, if you spell out the definition of functor for this setup, you'll see that even $T$ will be indexed, so its action on morphisms really ought to be written $T_{A,B}(f)$. I just left them out since these arguments are obvious from context.) - Thanks for this. But I'm not talking about the Kleisli Category $C_T$, I am talking about the Kleisli triple over $C$ (sorry, this was my mistake; I'll edit the question). In every place where I have seen the Kleisli triple defined from a Monad, the underlying Category is the same with no indexing of objects. When you index the objects, surely it is a different category, however you present the theory. – apk May 26 2010 at 12:56 ### 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. I agree with Mike that you shouldn't NEED this, BUT the answer is yes. The monad $T$ arises as coming from the adjunction $U:C^T \to C:F$, where $C^T$ is its category of Eilenberg-Moore algebras. Since this adjunction is monadic, it reflects isos. Hence, if $TX=U(FX)$ is iso to $TY=U(FY)$, then $FX$ and $FY$ are iso. Now, consider $id_{UFX}$, then, since here is a bijection DEPENDING NATURALLY ON A=UFX and B=FX between $Hom(FA,B)$ and $Hom(A,UB)$, we have that $\epsilon_{FX}$ and $\epsilon_{FY}$ are the same (up to identification via isomorphisms), where $\epsilon$ represents the counit of the adjunction. Now, multiplication $\mu$ of the monad $T$ has $\mu_X=G(\epsilon_{FX})$. This completes the proof. - In fact, the Kleisli star is a partial map ${}^ : \mathrm{Mor}(\mathbf{C}) \times \mathrm{Obj}(\mathbf{C}) \times \mathrm{Obj}(\mathbf{C}) \to \mathrm{Mor}(\mathbf{C})$. So there is no problem! -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 54, "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": 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": 0.9546073079109192, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/65256/problem-about-assasin-of-module	# Problem about Assasin of module I found this problem and I don't understand the solution. I will appreciate your help. Let $A = \mathbb{Q}[X_1,...,X_n,...], a = (X_1^2,...,X_n^2,...)$ and $M = A/a$. Show that $Ass_A (M) = \emptyset$. Why isn't $Ass_A(M) = (X_1,...,X_n,...)$ ? I'm writing this because the site is giving me this error : "Oops! Your question couldn't be submitted because: ````* It does not meet our quality standards.". This feels verry strange!!! ```` - – joriki Sep 17 '11 at 12:15 3 Perhaps because the question repeatedly includes the word "ass" ? – Daniel McLaury Sep 17 '11 at 15:42 ## 2 Answers 1. $Ass_A(M)$ denotes the set of associated primes of $M$, i.e. the set of prime ideals $\mathfrak p$ of $A$ for which there is an embedding $A/\mathfrak p \hookrightarrow M$. (Note that in particular that $Ass_A(M)$ is not an ideal of $A$ --- unlike the annihilator of $M$, which is an ideal of $A$. This is why the assertion $Ass_A(M) = \emptyset$ even makes sense.) 2. A contextual remark: if $M$ is an module over a Noetherian ring, then $Ass_A(M)$ is always non-empty. The point of problem you are asking about is to show that this can be false for non-Noetherian $A$ (such as the $A$ in your question). 3. If $\mathfrak p$ is an element of $Ass_A(M)$, then (as you implicitly observe in your post) it contains the annihilator $Ann_A(M)$ --- which in your case is $(X_1^2,X_2^2,\ldots)$, and hence, being prime, it contains the ideal $(X_1, X_2, \ldots)$. Thus, to solve the problem, you need to show that there is no embedding $\mathbb Q[X_1,\ldots]/(X_1,\ldots) \hookrightarrow M$, that is, that there is no non-zero element of $M$ annihilated by $(X_1,X_2, \ldots).$ 4. If you don't see how to do this straight away, try thinking about the case when $A$ has only finitely many indeterminates, i.e. when $A = \mathbb Q[X_1, \ldots,X_n],$ and when $M = A/(X_1^2,\ldots,X_n^2)$. In this case, by remark 2 above, it must be possible to find a non-zero element of $M$ which is annihilated by $(X_1,X_2,\ldots, X_n)$. Find this element explicitly. Once you have found it, look at it, and see why you can't construct an analogous element in the case of infinitely many variables. - The annihilator of a module $M$ over a ring $A$ is the set of elements $a$ in $A$ such that $aM=0$. This is an ideal of $A$ usually denoted by $\textrm{Ann}_A(M)$ and I think this is what you mean by assasin. In this case, the annihilator of $M$ over $A$ is the zero ideal because if $X_i X_{i+1} \neq 0$ in $M$ for every $i \geq 1$. If you actually mean the set of associated primes of $M$ over $A$ I think it works as follows. The ring $A/a$ has precisely one (prime) ideal: the image of $(X_1,X_2,\ldots)$ in $A/a$. Now, the annihilator of this submodule is just zero. This is a prime ideal of $A$ and thus, the set of associated primes should be a singleton. Maybe I'm wrong? -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 49, "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": 0.958065390586853, "perplexity_flag": "head"}
http://mathoverflow.net/questions/82726?sort=newest	## The Plancherel Formula for SL2 ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) What are the important questions to start to study about Plancherel Formula for SL2? And what is the relation this topic and the Arthur's paper, "On some problems suggested by the trace formula (1984)" Can anyone suggest any source shows this relation? - ## 2 Answers Another standard reference for the Plancherel formula is the book: $SL_2(\mathbb{R})$, Addison-Wesley, 1974; of course $SL$ stands for Serge Lang. The Plancherel formula and the Selberg trace formula have in common that a trace is computed in two ways. But they deal with different representations of $G$. For $G$ a suitable locally compact group (type I, unimodular, separable), and $f\in C_c(G)$, the Plancherel formula is $f(e)=\int_{\hat{G}}Tr\pi(f)\;d\mu(\pi)$, where $\mu$ is the Plancherel measure on the dual $\hat{G}$; and the difficulty is, in concrete examples, to describe $\hat{G}$ and $\mu$ explicitly (or at least to describe explicitly the support of $\mu$, called the tempered dual). For the Selberg trace formula, I found the Wikipedia article fairly readable: http://en.wikipedia.org/wiki/Selberg_trace_formula - ### 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 can't speak to the second question, but a canonical source on Plancherel for $SL(2)$ is An Introduction to Harmonic Analysis on Semisimple Lie Groups V.S. Varadarajan -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 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": 0.9036362171173096, "perplexity_flag": "middle"}
http://mathhelpforum.com/trigonometry/172911-evaluating-particular-product-sines.html	# Thread: 1. ## Evaluating a particular product of sines I need to evaluate the product $\prod_{m=n-a}^{n-1} \sin \left( \pi m /n \right)$. I've found this identity: $\prod_{m=1}^{n-1} \sin \left( \pi m /n \right) \equiv \frac{n}{2^{n-1}}$, which looks useful but isn't quite what I need. Does anyone know of a generalisation of this identity so that the lower range is $n-a$? 2. $\displaystyle \prod^{n-1}_{n-a}(\sin(\pi m/n)) = \prod^{n-1}_{1}(\sin(\pi m/n)) \div \prod^{n-a}_{1}(\sin(\pi m/n))$ or something like that. I'm not thinking very carefully about the indices and have to run, but this is the idea. Correct it where appropriate. 3. Thanks, I'd already thought of that (in fact the second product only runs to $n-a+1$. But I'd like to express the product as a function not involving trig functions. 4. I think you mean $n - a - 1$, but as for the rest of your question, I'm not sure I understand. Call $m-1 = n - a - 1$ then $\displaystyle \prod^{n-1}_{n-a}(\sin(\pi m/n)) = \prod^{n-1}_{1}(\sin(\pi m/n)) \div \prod^{m-1}_{1}(\sin(\pi m/n)) = \frac{n}{2^{n-1}} \div \frac{m}{2^{m-1}} = \frac{n}{2^{n-1}} \cdot \frac{2^{n- a - 1}}{n - a} = \frac{n}{2^{a}(n-a)}$. If that's not what you're looking for then I don't know what is! 5. Yes, sorry I meant -1 of course. Originally Posted by ragnar Call $m-1 = n - a - 1$ then $\displaystyle \prod^{n-1}_{n-a}(\sin(\pi m/n)) = \prod^{n-1}_{1}(\sin(\pi m/n)) \div \prod^{m-1}_{1}(\sin(\pi m/n)) = \frac{n}{2^{n-1}} \div \frac{m}{2^{m-1}}$. This is confusing - it looks to me like you're using $m$ as both the product-variable and as a constant used to define the product range. What I think you're trying to do is this: Let $b=n-a$ so that $\displaystyle \prod_{m=1}^{n-a-1} \sin \left( \frac{m\pi}{n} \right) = \prod_{m=1}^{b-1} \sin \left( \frac{m\pi}{n} \right)$ If that's what you mean, then your next part is wrong, since the identity in my first post doesn't apply. Or did I misunderstand? 6. Originally Posted by wglmb I need to evaluate the product $\prod_{m=n-a}^{n-1} \sin \left( \pi m /n \right)$. I think you're out of luck here. What I mean by that is that I don't think there is a formula of the type that you are looking for. Think about the particular case of the product that arises when $a=1$. There is then only one term in the product, and it is equal to $\sin(\pi/n)$. For most values of n, there is no explicit formula for $\sin(\pi/n)$ in terms of more elementary functions of n. So even in that special case, there is no satisfactory solution to the problem. Originally Posted by wglmb I've found this identity: $\prod_{m=1}^{n-1} \sin \left( \pi m /n \right) \equiv \frac{n}{2^{n-1}}$, which looks useful but isn't quite what I need. The fact that there is a good solution in this case is due to the fact that the numbers $\sin \left( \pi m /n \right)\ (1\leqslant m\leqslant n-1)$ are the roots of a fairly easily computable polynomial of degree n–1, and the product of the roots is given by the constant term in the polynomial. But that doesn't apply if you only have a subset of the roots. 7. I see, thank you - that was very enlightening! #### 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": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9784256219863892, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/54847/ac-circuit-theory-maximum-power-transfer	# AC Circuit Theory Maximum Power Transfer I am currently hitting a barrier with regards to the basics of this question. A 50HZ supply is connected to various given impedance's, calculate the maximum power transfer. Maximum power transfer I understand is basically the Thevenin/Norton equivalent. Though THERE IS NO EMF indicated in the question. Is there a method around this problem (Calculate the Norton/Thevenin with only Impedance values)? Any help would be superbly fantastic - We need the specific structure of the circuit so that we can give you specific Hints. But the method for this question is, do the Thevenin analysis of your circuit: (1) Short circuit all emfs in the circuit and calculate the impedance at your terminals, (ii) calculate the $V_AB$ so that you have the Thevenin emf. The load therefore will have to match the internal impedance of your equivalent circuit, for maximum power transfer. Try it. – JKL Feb 23 at 13:50 ## 1 Answer The Maximum Power Transfer Theorem is not so much a means of analysis as it is an aid to system design. Simply stated, the maximum amount of power will be dissipated by a load resistance when that load resistance is equal to the Thevenin/Norton resistance of the network supplying the power. If the load resistance is lower or higher than the Thevenin/Norton resistance of the source network, its dissipated power will be less than maximum. This is essentially what is aimed for in radio transmitter design , where the antenna or transmission line “impedance” is matched to final power amplifier “impedance” for maximum radio frequency power output. Impedance, the overall opposition to AC and DC current, is very similar to resistance, and must be equal between source and load for the greatest amount of power to be transferred to the load. A load impedance that is too high will result in low power output. A load impedance that is too low will not only result in low power output, but possibly overheating of the amplifier due to the power dissipated in its internal (Thevenin or Norton) impedance. -	{"extraction_info": {"found_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": 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": 0.9156945943832397, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/216929/put-n-items-into-m-groups-such-that-the-sums-of-the-ns-in-every-m-are-as-even-a	Put N items into M groups such that the sums of the N's in every M are as even as possible I have a `multiset`, `S`, that contains `N` items that I wish to place into `M` different groups so that the sum of all `N`'s in each group `M` is as evenly distributed as possible amongst all `M`'s. For example: If my multiset were: `S = {6, 3, 5, 2, 7, 11, 2}`, and number of groups `M = 3`, I may expect a result like so (my long hand approximation): M1 = {11, 2} sum of 13 M2 = {7, 5} sum of 12 M3 = {6, 3, 2} sum of 11 Is this something that could be done formulaically or would this be better approached algorithmically, and how under either case might I go about solving this problem? EDIT: Clarification. Speaking in programming terms in which I can articulate myself more clearly, I have an `array`, `S`, that contains `N integers`, I am looking to find a method to split these distinct integers into `M groups` so that the sums of each group have as little difference between them as is possible based on the set of numbers in `S`. The example above well demonstrates how such a group would occur under the test scenario given... - When you say "the sum of all N's," presumably then these elements of $N$ are not just "items," but actual numbers? – Thomas Andrews Oct 19 '12 at 15:27 If you want the "absolute best" way, then you have to define what the "best" answer is. ""As evenly as possible" is not well-defined. On definition might be to minimize the standard deviation of the sums of each group. – Thomas Andrews Oct 19 '12 at 15:30 @ThomasAndrews apologies for any ambiguity. S = {6, 3, 5, 2, 7, 11, 2} refers an array holding 7 sub arrays, of which the count of the first array is 6, 3 the second. So by saying the sum of all N's in each group, I mean the total of a sub-collection of S, e.g. S[0] + S[1] = 9. Im just trying to split them into M even columns if that makes sense. – Gareth Harding Oct 19 '12 at 16:05 @ThomasAndrews as evenly as possible was the best way I could think of, each M can't always be equal, so I'm trying to find the solution that has the least difference between sets of M – Gareth Harding Oct 19 '12 at 16:06 1 Answer The partition problem, which asks whether a given multiset of natural numbers can be partitioned into two submultisets with equal sums, is NP-complete. Since it can easily be reduced to your problem (by finding the most even distribution and checking whether it's perfectly even), your problem is also NP-complete. (In a nutshell, that means that it would be a revolution if you could find an algorithm that solves it in time polynomial in $N$.) However, the Wikipedia article seems to contain some efficient algorithms, some of which seem adaptable to your problem. - Thank you very much for this insight. – Gareth Harding Oct 22 '12 at 9:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9613750576972961, "perplexity_flag": "middle"}
http://mathhelpforum.com/math-topics/121793-quadratic-problem-help-print.html	# Quadratic Problem Help Printable View • December 28th 2009, 09:22 AM andi01 Quadratic Problem Help Thanks for your help in advance, here's the problem: The path of a golf ball can be modeled by a quadratic equation. The path of a ball hit an angle of 10 degrees can be modeled by h=-0.002d^2+0.4d where h is the height in meters and d is the horizontal distance the ball travels in meters until it first hits the gorund. a) What is the maximum height reached by the ball? The answer is 20m, usually I would just sub in 0 for d to find h but in this case it's different. What should I do? • December 28th 2009, 12:41 PM Soroban Hello, andi01! Quote: The path of a golf ball can be modeled by a quadratic equation. The path of a ball hit an angle of 10° can be modeled by: $h\:=\:-0.002d^2+0.4d$ where $h$ is the height in meters and $d$ is the horizontal distance. a) What is the maximum height reached by the ball? The answer is 20m. Usually I would just sub in 0 for d to find h. . . Why? Think of what you said . . . You let $d = 0$. The ball has not moved yet. What is its height? . . Obviously it's still on the ground! Here's one way of looking at it: The graph of $h \:=\:-0.002d^2 + 0.4d$ is a down-opening parabola: $\cap$ Its maximum is at its vertex. The vertex is at: . $v \:=\:\frac{\text{-}b}{2a} \quad\Rightarrow\quad d \:=\:\frac{\text{-}0.4}{2(\text{-}0.002)} \:=\:100$ The maximum height is: . $h \:=\:-0.002(100^2) + 0.4(100) \:=\:20$ m. All times are GMT -8. The time now is 01:12 AM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "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": 0.9448469877243042, "perplexity_flag": "middle"}
http://mathematica.stackexchange.com/questions/5563/how-to-load-a-package-without-naming-conflicts/5567	# How to load a package without naming conflicts? This question applies to any package, but I encountered this problem while working with graphs. There are symbols in the Combinatorica package (such as Graph, IncidenceMatrix, EdgeStyle, and others) that have the same name as analogous symbols in System. If I execute Needs[Combinatorica], then I can access CombinatoricaGraph by the name Graph, but if I want to access SystemGraph, I have to write SystemGraph. I want to use the Combinatorica prefix to access all the symbols in Combinatorica, and I want to access System` symbols without using a prefix. And I don't want to have the symbols Graph, IncidenceMatrix, and so on, in red in Mathematica because of the naming conflict. Is there a way to use the Combinatorica` package without introducing naming conflicts? - ## 3 Answers Shadowing occurs only when there are two functions with the same name that are in \$ContextPath. So right after you do <<Combinatorica`, do the following: $ContextPath = Rest@$ContextPath; What this does is that it removes Combinatorica (which is the package you just loaded). Now the only Graph function that's on the path is SystemGraph and you can call it simply by the name, without the prefix. To access any functions from the package, use the prefix, as CombinatoricaGraph. If Combinatorica` was loaded a while ago and you have loaded other packages in between, Rest@... is not going to be helpful. In that case, use: $ContextPath = DeleteCases[$ContextPath, "Combinatorica`"]; - The answer of @R.M. already explains the essence of the problem. You can streamline the process of removing the Combinatorica from the \$ContextPath by loading it via Block[{\$ContextPath}, Needs["Combinatorica`"]] (or use Get intead of Needs, although Needs is a preferred way to load a package). In this way, you don't have to do anything afterwards, since, once the scope of Block is left, the \$ContextPath automatically is reset to its previous value. This is a generally useful trick to load packages without adding them to the \$ContextPath. I found it quite useful on many occasions. - 1 Here's something that's good to be aware of (see under P.S.) If using Combinatorica, there's a chance one might one to use the ToCombinatoricaGraph function as well. – Szabolcs Apr 21 at 2:34 One way is to use something like Graph=SystemGraph; and use Graph which refers to SystemGraph thereafter. You can use any other name such as g=System`Graph as well. The downside is that you have to do it for any function. Edit: This does the trick: rep = StringReplace[#, "System`" -> ""] &; Scan[ Unprotect[#]; ToExpression[# <> "=System`" <> #]; Protect[#]; &, Intersection[ rep /@ Names["Combinatorica`"], rep /@ Names["System`"] ] ] - 1 You don't have to do the rep /@ Names["Combinatorica`*"] part, you know... – J. M.♦ May 15 '12 at 3:55 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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 5, "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": 0.921532392501831, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Biorthogonal_system	Biorthogonal system In mathematics, a biorthogonal system is a pair of indexed families of vectors $\tilde v_i$ in E and $\tilde u_i$ in F such that $\langle\tilde v_i , \tilde u_j\rangle = \delta_{i,j} ,$ where E and F form a pair of topological vector spaces that are in duality, ⟨,⟩ is a bilinear mapping and $\delta_{i,j}$ is the Kronecker delta. A biorthogonal system in which E = F is an orthonormal system. In L2 [0, 2π] the functions cos(nx) and sin(nx) form a biorthogonal system. Another example is the pair of sets of respectively left and right eigenvectors of a matrix, indexed by eigenvalue.[citation needed] Projection Related to a biorthogonal system is the projection $P:= \sum_{i \in I} \tilde u_i \otimes \tilde v_i$, where $\left( u \otimes v\right) (x):= u \langle v, x\rangle$; its image is the linear span of $\{\tilde u_i: i \in I\}$, and the kernel is $\{\langle\tilde v_i, \cdot\rangle = 0: i \in I \}$. Construction Given a possibly non-orthogonal set of vectors $\mathbf{u}= (u_i)$ and $\mathbf{v}= (v_i)$ the projection related is $P= \sum_{i,j} u_i \left( \langle\mathbf{v}, \mathbf{u}\rangle^{-1}\right)_{j,i} \otimes v_j$, where $\langle\mathbf{v},\mathbf{u}\rangle$ is the matrix with entries $\left(\langle\mathbf{v},\mathbf{u}\rangle\right)_{i,j}= \langle v_i, u_j\rangle$. • $\tilde u_i:= (I-P) u_i$, and $\tilde v_i:= \left(I-P \right)^* v_i$ then is an orthogonal system. References • Jean Dieudonné, On biorthogonal systems Michigan Math. J. 2 (1953), no. 1, 7–20 [1]	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 15, "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": 0.8753294348716736, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/164989-secant-method-error-bound-between-kth-iteration-root.html	# Thread: 1. ## Secant Method - Error bound between kth iteration and root Hey guys. I know from the proof of the secant method that given the two initial values, [LaTeX ERROR: Convert failed] and [LaTeX ERROR: Convert failed] , are sufficiently close to the solution, [LaTeX ERROR: Convert failed] , then [LaTeX ERROR: Convert failed] for [LaTeX ERROR: Convert failed] where [LaTeX ERROR: Convert failed] is the solution after [LaTeX ERROR: Convert failed] solutions. This essentially says that the error bound between the actual solution and the $k$'th iterate is less than half the distance between the iterate before that and the actualy solution. How would i show using this fact (and possible other results) that [LaTeX ERROR: Convert failed] , which means that the error bound between the actual solution and the k'th iterate is less than the distance between the $k$'th iterate and the the $k-1$'th iterate. I'm very stumped on this question so if anyone has an idea i'd be extremely grateful Edit: I've been told that the Contraction Mapping Method could help me show this. As a reminder, It turns out that the error bound for this method is given by However, just like the error bound for the secant method that i gave in my question, this is not computable as the value of [LaTeX ERROR: Convert failed] is unknown. Now for the Contraction Mapping Method, the way they have manipulated the error bound is as shown, I can see the similarities between this and what i am trying to show with the Secant Method, but i don't understand exactly what they have done here, especially concerning the part under 'Now' on the above image. Can someone explain this to me and how i could use in my original problem?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9254299998283386, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/74693/on-pseudo-rational-modular-forms-of-weight-2-and-level-n	## On pseudo rational modular forms of weight 2 and level N ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) So consider the $\mathbb{Q}$-vector space $V$ of functions which satisfy the following conditions (1) $f:\mathbb{H}\rightarrow\mathbb{C}$ is holomorphic. Here $\mathbb{H}$ stands for the upper half plane. (2) $f(z+1)=f(z)$ (3) The Fourier series of $f$ at infinity has the form $\sum_{n\geq 1} a_nq^n$ where $q=e^{2\pi iz}$ where the $a_n$'s are rational numbers (4) $\frac{1}{Nz^2}f(\frac{-1}{Nz})=\pm f(z)$ Examples of non-zero elements of $V$ are given for example by $\mathbb{Q}$-linear combinations of modular forms associated to rational elliptic curves of conductor $N$ that have the same sign in their functional equation. Let us denote this sub vector space by $W$. Unless I made a mistake in my calculation, an example of an element of $V$ which is not in $W$ could be ``` $$ \sum_{d\|N} d a_d E_2(dz) $$ ``` with $\sum d a_d=0$ and $a_d=a_{N/d}$. If $N$ is sufficiently composite then we may find such $a_d$'s. Here $E_2$ is the weight $2$ Eisenstein series suitably normalized. Q: How big is $V$ and is it possible to describe it in some interesting way? - Not in general — not even after correcting the factor $1/(N^2z^2)$ to $1/Nz^2$. There are several problems here. First, not every elliptic curve of conductor $N$ works: you must impose a sign condition. Second, the modular forms of level $N$ have a rational basis but usually most of them come from abelian varieties of dimension $>1$. Finally, once $N$ is at all large the transformations $z \mapsto z+1$ and $z \mapsto -1/Nz$ generate only a tiny part of the relevant modular group, so there should be plenty of other "pseudo modular" cusp forms. – Noam D. Elkies Sep 6 2011 at 23:38 Yes I just edited. The transformation $z\mapsto \frac{-1}{Nz}$ is not in the modular group unless $N=1$ but I guess you could work in a bigger group which contains the involution of level $N$. – Hugo Chapdelaine Sep 6 2011 at 23:43 So @Noam do you think you could use Poincare's trick and average out over the group generated by $z\mapsto z+1$ and $z\mapsto \frac{-1}{Nz}$. Of course, one has to be careful about convergence issues. – Hugo Chapdelaine Sep 6 2011 at 23:45 @Hugo C.'s edit: indeed the Fricke involution $z \mapsto -1/Nz$ is not in $\Gamma_0(N)$ but is in the normalizer of $\Gamma_0(N)$ in ${\rm SL}_2({\bf R})$. However, once $N>4$ Fricke and $z \mapsto z+1$ do not generate the full normalizer. Also I don't think $E_2(z) - NE_2(Nz)$ works: that's a modular form of weight $2$ for all of $\Gamma_0(N)$ [proportional to the logarithmic differential of the modular function $\Delta(z) / \Delta(Nz)$], but it does not vanish at $q=0$. – Noam D. Elkies Sep 6 2011 at 23:47 @Hugo C.'s question: It's not clear to me whether the Poincaré trick would work well enough to produce a modular form with rational coefficients. But you can probably find a subgroup $G$ contained in $\Gamma_0(N)$ with finite index and use holomorphic differentials on the Riemann surface ${\cal H}^* / G$. – Noam D. Elkies Sep 6 2011 at 23:49 show 3 more comments ## 1 Answer To summarize some remarks in the comments. The function $f \cdot d \tau$ will be a differential on $Y_N:=\mathbf{H}/\Gamma$, where $\Gamma_N$ is the group generated by the two matrices $$\left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right) \ , \ \left( \begin{matrix} 0 & -1 \\ N & 0 \end{matrix} \right)$$ If $N > 4$, then $\Gamma_N$ will have infinite index in $\mathrm{SL}_2(\mathbf{Z})$. In particular, if $X_N$ is the complex curve obtained by filling in the puncture at $\infty$ (to account for the condition that the $q$-expansion has positive coefficients), then $X_N$ (for $N > 4$) will be an open curve, and so $H^1(X_N,\Omega^1)$ will be infinite dimensional (and not particularly interesting). It's a theorem of Hecke that $\Gamma_N$ is the normalizer of $\Gamma_0(N)$ for $N \le 4$. In this way, Hecke was able to show that an $L$-series $L(f,s) = \sum a_n n^{-s}$ which satisfied a functional equation of a certain kind (with "conductor" $N \le 4$) was exactly the Mellin transform of a modular form. Being modular is thus (for $N > 4$) a stronger condition than simply asking that the $L$-series satisfies the appropriate functional equation. Historically this was interesting, because one could conjecture that the $L$-series of elliptic curves satisfied a functional equation of a certain kind, which is (a priori) a weaker conjecture than asking that elliptic curves are modular. Weil, however, showed that if the $L$-series attached to an elliptic curve $E/\mathbf{Q}$ satisfied the right functional equation, and the same was true of all the quadratic twists of $E$, then $E$ was actually modular (the so called converse theorem, which has been vastly generalized). - Thanks Michael for your answer, but I'm wondering if there is some kind of combinatorial-arithmetic description of this vector space of 1-forms. Do you think it will be exhausted by linear combinations of modular forms attached to abelian varieties and Eisenstein series? – Hugo Chapdelaine Sep 7 2011 at 2:37 No, it will be much much bigger. A good model to think of is the space of holomorphic differentials on the disc $\|z\| < 1$, which is isomorphic to the space of power series with radius of convergence $\ge 1$ (in particular, it has uncountable dimension). – Michael Sep 7 2011 at 3:15 I agree that the space will likely be too huge to say much about, but it's not as simple as Michael's answer suggests, because Hugo C. asked for rational $q$-expansions, and it's not immediately obvious there'll be lots of non-modular forms for "$\Gamma_N$" with rational coefficients, let alone an uncountable-dimensional space of such forms. – Noam D. Elkies Sep 7 2011 at 4:09 Noam makes a good point - I actually didn't notice that you asked for $\mathbf{Q}$-coefficients. However, as soon as you have a single holomorphic function $z$ on $X_N$ with rational coefficients, then, after normalizing $z$ to have norm $\le 1$, and letting $f$ be any classical weight $2$ form with rational coefficients on $X_0(N)^{+}$, then $f \cdot B(z,1)$ is a subspace of the holomorphic weight $2$ form with rational coefficients, where $B(z,1)$ denotes the power series in $\mathbf{Q}[[z]]$ with radius of convergence at least $1$> – Michael Sep 7 2011 at 5:35 Right, right so this space will be incredibly huge! So too construct your function $z$ you might try to take the quotient of two linearly independent "rational "Poincare series of the same weight. Something like $\sum_{\gamma\in \Gamma_N} H(\gamma(z))$ for two suitable $H$'s should work. – Hugo Chapdelaine Sep 7 2011 at 11:33 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": 95, "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": 0.9462186694145203, "perplexity_flag": "head"}
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http://www.physicsforums.com/showthread.php?p=810950	Physics Forums ## Work Done "On" or "By" The System Will someone help solidy this concept of Work 'On' a system and Work 'Done' by a system. If I lift an object from some defined zero potential. Then the work BY the system is -mgh, because gravity 'g' is in the reverse direction as the displacement 'h', but work done ON the system is +mgh because it is the opposite of work BY the system. But what do I do if I don't know the work done by the system, i.e. there is not a given 'g' vector pointing down? How can I tell if the work is plus or minus? PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus Recognitions: Gold Member Homework Help Science Advisor Work is always associated with a particular force. The work done by the force is: $$W = \int \vec{F}.d\vec{r}$$ In your example of lifting an object, the work done by the gravitational force is negative, while the work done by the lifting force is positive. Mentor Blog Entries: 1 Quote by brentd49 But what do I do if I don't know the work done by the system, i.e. there is not a given 'g' vector pointing down? How can I tell if the work is plus or minus? Can you give an example of what you mean? As James R stated, work is always associated with a particular force acting through a displacement. If the displacement is in the direction of the force, the work is positive; if opposite to the force, negative. ## Work Done "On" or "By" The System It's sometimes easier to find the change in energy first (whether +ve or -ve) and decide whether it is positive or negative afterwards. energy lost ===> ΔE is negative ==> work done by system energy gained => ΔE is positive ==> work done on system A good example is the first law of thermodynamics: change in internal energy = heat added to system - work done by system; OR, equivalently: change in internal energy = heat added to system + work done on system E.g. in a four-stroke engine, if you consider a single cylinder, work is done BY the system during the power stroke (i.e. BY the gas expanding on the piston to move it outwards and thus transferring energy outsdie the system to the other cylinders) and work is done ON the system during the exhaust stroke (i.e. by the piston (the force is supplied by the power stroke of a different cylinder) ON the exhaust gas as it is pushed out of the cylinder). Mentor For me it's always easiest to draw a picture. Start with a big circle that defines the boundary of your system, then place objects in it. And if often doesn't matter whether something is placed inside or outside, as long as you are consistent. Since gravity seems like an all-encompassing force, it can get confusing, but if the system is just the box, then both you lifting and gravity acting against you are outside the system. And if you put you in the system, you need to put gravity in the system as well. Otherwise, you could accidentally create a perpetual motion machine (no inputs, only outputs). Thread Tools \| \| \| \| \|--------------------------------------------------------\|----------------------------\|---------\| \| Similar Threads for: Work Done "On" or "By" The System \| \| \| \| Thread \| Forum \| Replies \| \| \| Calculus & Beyond Homework \| 9 \| \| \| Advanced Physics Homework \| 7 \| \| \| General Math \| 13 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9302473664283752, "perplexity_flag": "middle"}
http://understandinguncertainty.org/node/622	# Quantifying the Risk of Natural Catastrophes How do companies prepare for the financial impact of natural catastrophes? How can they possibly have an idea of what the potential cost can be for events that haven't yet happened? Shane Latchman explains the way companies in the insurance industry are using catastrophe models to help make sense of a very uncertain future... ## Introduction Catastrophe modelling companies serve to measure the financial impact of natural catastrophes on infrastructure, with a view to estimating expected losses. Although natural catastrophes often have a grave humanitarian impact with regard to loss of life, this aspect is not currently included and only the event’s financial impact will be discussed below. The purpose of catastrophe modelling (known as cat modelling in the industry) is to anticipate the likelihood and severity of catastrophe events—from earthquakes and hurricanes to terrorism and crop failure—so companies (and governments) can appropriately prepare for their financial impact. In this article we will explain the three main components of a cat model, starting with an event’s magnitude and ending with the damage and financial loss it inflicts on buildings. We then discuss what metrics are used in the catastrophe modelling industry to quantify risk. ## The Three Modules A catastrophe model can be roughly divided into three modules, with each module performing a different step towards the overall calculation of the loss that results from a catastrophe. ### Hazard Module The hazard module looks at the physical characteristics of potential disasters and their frequency, whereas the vulnerability module assesses the vulnerability (or “damageability”) of buildings and their contents when subjected to natural and man-made disasters. The very first task when trying to estimate potential future losses from a peril is to create a catalogue of possible future events, which form the basis for drawing conclusions about the perils (e.g. hurricanes) that may strike, their intensity, and the likelihood that they will strike. Statistical and physical models are used to simulate a large catalogue of events. For example historical data on the frequency, location, and intensity of past hurricanes are modeled and used to predict 10,000 scenario-years of potential hurricane experience. Each of the 10,000 years should be thought of as 10,000 potential realisations of what could happen in the year 2010, for example, and not simulations—or predictions—of hurricane activity from now until the year 12010. A notional 10,000 year sample is simulated and the collection of resulting events is called an ensemble. Within the ensemble we have information on the characteristics of the event, as well as the simulated year of its occurrence, so that over the 10,000 year catalogue we might have, say, 30,000 events and in our output we may have 0 events in Year 1, 3 in Year 2, 1 in Year 3 and so on. Thus any year in the catalogue gives a snapshot of the potential hurricane experience in a single calendar year. While the event catalogue doesn't consist of events that have actually happened in the past, historical data on the frequency, location, and intensity of past hurricanes is used to generate a realistic 10,000 year simulation. Since the past is not always indicative of the future, the event catalogue therefore can include events that are more extreme than those that have occurred in the past (in accordance with their probability of occurrence). ### The vulnerability module After simulating an event of a given magnitude, the damage it does to buildings must now be computed. The extent to which a particular building will be damaged during a hurricane depends on many factors, but certain characteristics of a building serve as good indicators of its vulnerability and, in particular, of its likely damage ratio. The damage ratio is the ratio of the cost to repair a building to the cost of rebuilding it. It is, of course, quite possible for seemingly identical buildings to experience different levels of damage when impacted by the same intensity of windspeed. This is due to small differences in construction and local site-specific effects that can have a major impact on losses. To capture this variability in damage, we look not at a single value for the damage ratio, but at a whole distribution of possible values. The mean of this distribution is known as the mean damage ratio. A graph of the mean damage ratio as a function of intensity (e.g. wind speed for hurricanes) is plotted below and is called a vulnerability function. The graph below (left plot) shows that as the intensity increases, the mean damage ratio also increases, as expected. Figure 1: Graph showing relationship between Vulnerability Module (left) and Financial Module (right) ### Financial Module The damage ratio distribution for a specific event is then multiplied by the building replacement value to obtain the loss distribution (Figure 1 above, right plot). These calculations are done within the financial module which also incorporates specific insurance policy conditions that are crucial in accurately determining the insurer's loss. This module computes the combined loss distribution of all buildings through a process known as convolution. This is a means of computing all possible combinations of the loss distributions $L_i + L_j$ , and their associated probabilities, given the probability distributions of $L_i$ and $L_j$ separately. In this case, $L_i$ and $L_j$ are the loss distributions for two locations, 1 and 2 respectively, for each event. This is shown formally below, where $L$ represents the total loss for 2 locations, $P_1(L_i)$ is the probability distribution for location 1, and $P_2(L_j)$ the probability distribution for location 2. $$P(L) = \sum_{L=L_i+L_j} P_1(L_i) \times P_2(L_j)$$ This is illustrated in Figure 2 below. We can see that if we have two loss distributions for the two locations (shown in the top plots), then when these are convolved, the range of the resulting loss distributions is equal to the sum of the ranges of the individual loss distributions. We can also see, that to get a total loss of 10, this is brought about by three combinations (0,10), (10,0) and (5,5). The individual probabilities for each component of the combinations are multiplied together and the 3 products summed to give the probability of a total loss of 10. $$P(10) = P_1(0) \times P_2(10) + P_1(10) \times P_2(0) + P_1(5) \times P_2(5)$$ This process is completed for all other possible combinations and thus the convolved loss distribution for the total loss of the two locations is obtained. Figure 2: Figure showing the result of convolution on two loss distributions ## Risk Quantification ### Exceedance Probability Curves An Exceedance Probability curve (known as an EP curve) describes the probability that various levels of loss will be exceeded. For example, if we simulate 10,000 years of hurricanes (outlined in the Hazard section above), the highest causing loss will have a 0.01% chance of being exceeded. This is because there is only one event with a loss greater than or equal to that loss in 10,000 years of events (1/10,000 = 0.01%). Figure 3: Graph showing Exceedance Probability curve An EP curve is generated by running the catalogue against exposure (buildings) and obtaining losses for each event and year. The events are then grouped by year (the reader should recall at this point that each simulated event has a simulated year to which it is associated) to determine the loss-causing events for each year. The total mean loss for each year is then found by adding the mean losses for each event together. It should also be noted that the mean of the convolved loss distribution for a year is the same as the sum of the mean losses of each event in that year. The losses are then sorted in descending order and plotted to give the exceedance probability and corresponding loss at that probability. The EP curve is the basis upon which insurers estimate their likelihood of experiencing various levels of loss. Another common form of the above curve is to invert the exceedance probability to obtain the corresponding return period. For example, we can estimate the loss from a certain hurricane, find where that lies on the exceedance probability curve, and invert the exceedance probability to deduce, for example, that that hurricane has an exceedance probability of 1%, for a particular insurer’s portfolio, it is therefore a “1 in 100 year storm”. Nevertheless, although a loss of, say, \$100m has an annual loss exceedance probability of just 1% in any given year, this probability compounds over several years. For example, a 1% annual exceedance probability of a \$100m loss compounds to 9.6% (1−0.910) probability over 10 years. Consideration of the compounding of loss exceedance probabilities should also be taken into account by the insurer. As regards correlation of building losses based on geographic distance, the beauty of cat models is that the correlation of losses is built in from the ground up by virtue of the model knowing the location of each risk. Thus the EP curve already takes into account geographical correlation and so correlation does not need to be added in after assembling the EP curve. ### Percentiles One way for insurers to assess the potential payouts that could be required at various return periods is through plotting percentiles around the EP curve Figure 4: Graph showing percentiles around Exceedance Probability Curve As can be seen in Figure 4, an insurer can assess their risk at the 250-year (0.4% exceedance probability) return period by looking at the mean loss at that return period—\$10m in this example—and observe the range of losses from the 5th to 95th percentile—\$7m to \$19m. Note that there are two sets of probabilities involved; that is, if 10,000 years are run and when all the losses are computed and ranked, the 40th largest loss (corresponding to the 10,000/40 = 250 year return period) will produce a loss with a 90% confidence interval of \$7m to \\$19m. These percentiles are produced by the Vulnerability Module and are the result of similar buildings having a different response to the same level of intensity of the event; this is explained in more detail in the Vulnerability Section above. To capture this variability in damage, we thus look not at a single value for the loss, but at a whole distribution of possible values, and looking at particular percentiles is a useful way to visualise this variability. Another observation concerning percentiles is the relative narrowing of the bands as the number of locations increases. We can see below that as the number of locations increases, the relative uncertainty (also known as the coefficient of variation) decreases at each exceedance probability. The coefficient of variation is the ratio of the standard deviation of loss of an event to the mean of that event and decreases as the number of locations increases. This means that more locations leads to higher relative certainty: this is an example of the general statistical rule that larger sample sizes lead to estimates with lower relative error. Figure 5: Graph showing the narrowing of percentile bands as the number of locations increases ### Tail Value at Risk Another measure sometimes used to estimate risk is Tail Value at Risk (TVaR). This metric finds the average of all losses beyond a specified return period. For example, the 5,000 year TVaR is the average of the largest and second largest losses. Thus if an insurer would like to keep reserve capital (capital required to pay out to home and business owners after catastrophes) to withstand a 1 in 250 year loss, the insurer may decide to look at the 1 in 250 year TVaR amount, which will allow for losses greater than the 250 year return period. Figure 6: Plot of TVaR and EP curve ### Catalogue size Another means of estimating risk is for an insurer to run different sized catalogues against their exposure. This allows for an alternative view of risk, looking at a larger number of events; for example using a 50,000 or 100,000 year catalogue of events. Larger catalogues will usually contain larger maximum events but the EP curve for the 50,000 year catalogue is not necessarily bounded above the 10,000 year catalogue. ### Pricing Some insurers use the results of catastrophe models as one input into pricing policies. A policy is a contract between an insurer and the insured that requires the insurer to pay claims on the policy. The policy can consist of a group of buildings distributed over a wide or small geographic area and that are insured for individual or multiple perils (e.g. hurricanes and earthquakes). Insurers can model an individual policy and obtain metrics, such as the annual expected loss and the standard deviation of this loss and use this (after accounting for commission, expenses, profit loading and so on) to help calculate the amount they should charge the policyholder to allow for reasonable profit while minimising the risk of insolvency (bankruptcy). ### Conclusion We have seen that the hazard module gives us a way of assessing the extent to which a particular region is at risk from natural and man-made (e.g. Terrorism) catastrophes of various magnitudes. Then the vulnerability and financial modules allow us to compute the loss due to damage from these catastrophes. Finally, we saw the use of the Exceedance Probability curve in estimating risk from extreme events and using percentiles and other risk metrics to aid the interpretation of the EP curve. So while we may not be able to avert disasters, we can at least prepare for their financial impact. ## About the author Shane Latchman works as a Client Services Associate with the catastrophe modelling company AIR Worldwide Limited (www.air-worldwide.com). Shane works with the earthquake model for Turkey, the hurricane model for offshore assets in the Gulf of Mexico and the testing and quality assurance of the Financial Module. Shane received his BSc in Actuarial Science from City University and his Masters in Mathematics from the University of Cambridge, where he concentrated on Probability and Statistics. ## Comments Anonymous (not verified) Sun, 05/02/2010 - 23:44 ### Black Swans It is a pity that you do not mention/discuss the concept "Black Swan" (see Nassim Taleb's books on this topic). It is very relevant in this context! Anonymous (not verified) Fri, 07/09/2010 - 18:56 ### Indirect financial impacts Thanks for the interesting article Shane. While I was reading this, I did have a thought. Whilst the direct financial impacts associated with the interaction of a hazard (natural or man-made) and vulnerable & exposed infrastructure is definitely a key component of financial 'loss', it does not give a full picture. For example, there can be huge losses associated damaged infrastructure i.e. the cost is not just how much it will be to repair damage, but how much money is lost for every day that the infrastructure is not available. And the infrastructure has primary, secondary etc users, and each of those will have losses whilst the infrastructure is not functioning... What I mean is that despite sophisticated modelling in the 'Financial Module', it seems there are still many unknowns about what the true (financial) costs of a disaster will be. Anonymous (not verified) Wed, 08/11/2010 - 22:30	{"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": 11, "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": 0.9423229098320007, "perplexity_flag": "middle"}
http://euler.slu.edu/escher/index.php/Tessellations_by_Polygons	Tessellations by Polygons Seville, Spain. Explorations The explorations for this section include: Explorations using Geogebra For those who have access to The Geometer's Sketchpad the following explorations are available. The first part of the Introductions to GSP Exploration is heavily based on and inspired by materials developed by Mike Riedy. [1]. Some Basic Tessellations Recall that a polygon is a closed plane figure made by joining line segments. You might want to review the relevant material in Fundamental Concepts concerning polygons before reading this section. The fundamental question we will discuss in this section is: Which polygons tessellate? More precisely, which polygons can be used as the only tile in a monohedral tessellation of the plane? Before moving on, you may want to do the Tessellation Exploration: The Basics The most common and simplest tessellation uses a square. Squares easily form horizontal strips: Stacks of these strips cover a rectangular region and the pattern can clearly be extended to cover the entire plane. The same technique works with parallelograms, and so: All parallelograms tessellate. Special parallelograms such as rectangles, and rhombuses also tessellate. Looking for other tessellating polygons is a complex problem, so we will organize the question by the number of sides in the polygon. The simplest polygons have three sides, so we begin with triangles: All triangles tessellate. To see this, take an arbitrary triangle and rotate it about the midpoint of one of its sides. The resulting parallelogram tessellates: The picture works because all three corners (A, B, and C) of the triangle come together to make a 180° angle - a straight line. This property of triangles will be the foundation of our study of polygon tessellations, so we state it here: The sum of angles of any triangle is 180°. Moving up from triangles, we turn to four sided polygons, the quadrilaterals. Before continuing, try the Quadrilateral Tessellation Exploration. Tessellations by Quadrilaterals Recall that a quadrilateral is a polygon with four sides. The sum of angles in any quadrilateral is 360° To prove, divide a quadrilateral into two triangles as shown: Since the angle sum of any triangle is 180°, and there are two triangles, the angle sum of the quadrilateral is 180° + 180° = 360°. Taking a little more care with the argument, we have: $\alpha_1 + \delta_1 + \gamma = 180^\circ$ and $\alpha_2 + \delta_2 + \beta = 180^\circ$. Then $360^\circ = \alpha_1 + \delta_1 + \gamma + \alpha_2 + \delta_2 + \beta = \alpha_1 + \alpha_2 + \beta + \gamma + \delta_1 + \delta_2 = \angle A + \angle B + \angle C + \angle D$. This division into triangles does not calculate the angle sum of the quadrilateral. The point of all the letters is that the angles of the triangles make the angles of the quadrilateral, which would not work if the quadrilateral was divided as shown on the right. We now turn to the main result of this section: All quadrilaterals tessellate. Begin with an arbitrary quadrilateral ABCD. Rotate by 180° about the midpoint of one of its sides, and then repeat using the midpoints of other sides to build up a tessellation. The angles around each vertex are exactly the four angles of the original quadrilateral. Since the angle sum of the quadrilateral is 360°, the angles close up, the pattern has no gaps or overlaps, and the quadrilateral tessellates. Recall from Fundamental Concepts that a convex shape has no dents. All triangles are convex, but there are non-convex quadrilaterals. The technique for tessellating with quadrilaterals works just as well for non-convex quadrilaterals: It is worth noting that the general quadrilateral tessellation results in a wallpaper pattern with p2 symmetry group. Tessellations by Convex Polygons Every shape of triangle can be used to tessellate the plane. Every shape of quadrilateral can be used to tessellate the plane. In both cases, the angle sum of the shape plays a key role. Since triangles have angle sum 180° and quadrilaterals have angle sum 360°, copies of one tile can fill out the 360° surrounding a vertex of the tessellation. The next simplest shape after the three and four sided polygon is the five sided polygon: the pentagon. The angle sum of any pentagon is 540°, because we can divide the pentagon into three triangles: Since each triangle has angle sum 180° the angle sum of the pentagon is 180° + 180° + 180° = 540°. Rather than repeat the angle sum calculation for every possible number of sides, we look for a pattern. The angle sum of a triangle (3-gon) is 180°, the angle sum of a quadrilateral (4-gon) is 2x180°, and the angle sum of a pentagon is 3x180°. A general polygon with n sides can be cut into n − 2 triangles and so we have: The sum of the angles of an n-gon is $(n-2)\times 180^\circ$. Unlike the triangle and quadrilateral case, the pentagon's angle sum of 540° is not helpful when trying to fit a bunch of pentagons around a vertex. In fact, there are pentagons which do not tessellate the plane. For example, the regular pentagon has five equal angles summing to 540°, so each angle of the regular pentagon is $\frac{540^\circ}{5} = 108^\circ$. Attempting to fit regular polygons together leads to one of the two pictures below: Both situations have wedge shaped gaps that are too narrow to fit another regular pentagon. Thus, not every pentagon tessellates. On the other hand, some pentagons do tessellate, for example this house shaped pentagon: The house pentagon has two right angles. Because those two angles sum to 180° they can fit along a line, and the other three angles sum to 360° (= 540° - 180°) and fit around a vertex. Thus, some pentagons tessellate and some do not. The situation is the same for hexagons, but for polygons with more than six sides there is the following: No convex polygon with seven or more sides can tessellate. This remarkable fact is difficult to prove, but just within the scope of this book. However, the proof must wait until we develop a counting formula called the Euler characteristic, which will arise in our chapter on Non-Euclidean Geometry. Nobody has seriously attempted to classify non-convex polygons which tessellate, because the list is quite likely to be too long and messy to describe by hand. However, there has been quite a lot of work towards classifying convex polygons which tessellate. Because we understand triangles and quadrilaterals, and know that above six sides there is no hope, the classification of convex polygons which tessellate comes down to two questions: • Which convex pentagons tessellate? • Which convex hexagons tessellate? Question 2 was completely answered in 1918 by K. Reinhardt.[2] Reinhardt showed that there are only three types of convex hexagons which tessellate: \| \| \| \| \|-----------------------------------------------\|------------------------------------------------------\|---------------------------------------------\| \| \| \| \| \| Type 1B + C + D = 360° A + E + F = 360° a = d \| Type 2 A + B + D = 360° C + E + F = 360° a = d c = e \| Type 3 A = C = E = 120° a = a' c = c'e = e' \| Reinhardt also addressed Question 1 and gave five types of pentagon which tessellate. In 1968, R. Kershner[3] found three new types, and claimed a proof that the eight known types were the complete list. A 1975 article by Martin Gardner[4] in Scientific American popularized the topic, and led to a surprising turn of events. In fact Kershner's "proof" was incorrect. After reading the Scientific American article, a computer scientist, Richard James III, found a ninth type of convex pentagon that tessellates. Not long after that, Marjorie Rice, a San Diego homemaker with only a high school mathematics background, discovered four more types, and then a German mathematics student, Rolf Stein, discovered a fourteenth type in 1985. Since 1985, no new types have been discovered, and many mathematicians believe that the list is finally complete. However, there is no well accepted proof of the classification, so it remains possible that there is a fifteenth or even many more types of convex pentagons that tessellate. Today, question 1 is an open problem, a problem whose solution is unknown. Summary of Polygon Tessellations Sides Angle Sum Tessellates? 3 180° Yes. All triangles tessellate. 4 360° Yes. All quadrilaterals tessellate. 5 540° Sometimes. There is a list of 14 types of convex pentagons which tessellate, but nobody knows if the list is complete. 6 720° Sometimes. There is a list of 3 types of convex hexagons which tessellate, and these are the only three. 7 & up 900°, 1080°, ... No convex n-gon tessellates for $n \geq 7$ Tessellations by Regular Polygons Recall that a regular polygon is a polygon whose sides are all the same length and whose angles all have the same measure. A regular n-gon has n equal angles that sum to $(n-2)180^\circ$, so: The corner angle of a regular n-gon is $\frac{(n-2)180^\circ}{n}$. The table shows the corner angles for the first few regular polygons: \| \| \| \| \| \| \| \| \| Number of Sides \| Corner angle \| \|-----\|-----\|------\|------\|---------\|------\|------\|------\|---------------------\|------------------\| \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| 11 \| 12 \| \| 60° \| 90° \| 108° \| 120° \| ~128.6° \| 135° \| 140° \| 144° \| ~147.3° \| 150° \| Regular Tessellations Regular tessellation A tessellation using one regular polygon tile, arranged so that edges match up. Corners of the tiles need to fit together around a point, which means the corner angle of the regular polygon must evenly divide 360°. Since $6 \times 60^\circ = 360^\circ$, there is a regular tessellation using six triangles around each vertex. Since $4 \times 90^\circ = 360^\circ$, there is a regular tessellation using four squares around each vertex. And since $3 \times 120^\circ = 360^\circ$ there is a regular tessellation using three hexagons around each vertex. We have already seen that the regular pentagon does not tessellate. A regular polygon with more than six sides has a corner angle larger than 120° (which is 360°/3) and smaller than 180° (which is 360°/2) so it cannot evenly divide 360°. We conclude: There are three regular tessellations of the plane: by triangles, by squares, by hexagons. A major goal of this book is to classify all possible regular tessellations. Apparently, the list of three regular tessellations of the plane is the complete answer. However, these three regular tessellations fit nicely into a much richer picture that only appears later when we study Non-Euclidean Geometry. Tessellations using different kinds of regular polygon tiles are fascinating, and lend themselves to puzzles, games, and certainly tile flooring. Try the Pattern Block Exploration. Archimedean tessellations Floor in Seville, Spain Side walk in Seville, Spain An Archimedean tessellation (also known as a semi-regular tessellation) is a tessellation made from more that one type of regular polygon so that the same polygons surround each vertex. \| \| \| \| \| \|-----------\|-------------\|-----------\|-------------\| \| \| \| \| \| \| (3,6,3,6) \| (3,4,6,4) \| (3,12,12) \| (3,3,3,4,4) \| \| \| \| \| \| \| (4,8,8) \| (3,3,4,3,4) \| (4,6,12) \| (3,3,3,3,6) \| We can use some notation to clarify the requirement that the vertex configuration be the same at every vertex. We can list the types of polygons as they come together at the vertex. For instance in the top row we see on the left a semi-regular tessellation with at every vertex a (3,6,3,6) configuration. We see a 3-gon, a 6-gon, a 3-gon and a 6-gon. The other tessellations on the top row have a (3,4,6,4), a (3,12,12), and a (3,3,3,4,4) configuration. These configurations are unique up to cyclic reordering (and possibly reversing the order). For example (3,12,12) can also be written as (12,12,3) or (12,3,12). In the bottom row we have (4,8,8), (3,3,4,3,4), (4,6,12) and (3,3,3,3,6) configurations. Recall that creating a tessellation requires the angle sums of the polygons to add up to 360° around a vertex. We know from a previous section that the angle of a regular n-gon is $\frac{(n-2)180^\circ}{n}$. If we have a collection of n-gons: n1,n2,...,nk , then the angles of those polygons need to add up to 360°: $\frac{(n_1-2)180^\circ}{n_1}+ \frac{(n_2-2)180^\circ}{n_2} + ... + \frac{(n_k-2)180^\circ}{n_k} = 360^\circ$ Dividing both sides by 180° gives us the following equation: $\frac{n_1-2}{n_1}+ \frac{n_2-2}{n_2} + ... + \frac{n_k-2}{n_k} = 2$. For example, suppose that n1 = 3,n2 = 3,n3 = 3,n4 = 4,n5 = 5, then we have: $\frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{2}{4} + \frac{2}{4} = 2$. This means that 3 triangles and 2 squares will give us a vertex type. In this case we can arrange these polygons around the vertex in two different ways: (3,3,3,4,4) and (3,3,4,3,4). Both of these will give rise to a semi-regular tessellation. Recall that a semi-regular tessellation satisfies the following requirements: • All the tiles are regular polygons • It is a tessellation, hence no gaps or overlaps • All the vertices are of the same type. There are only 21 combinations of regular polygons that will fit around a vertex. And of these 21 there are there are only 11 that will actually extend to a tessellation. Below are the different vertex types. An asterisk indicates that this vertex type cannot be extended to a tessellation. type N=3 N=4 N=5 N=6 N=7 N=8 N=9 N=10 N=12 N=15 N=18 N=20 N=24 N=42 vertex configuration(s) 1 6 (3,3,3,3,3,3) 2 4 1 (3,3,3,3,6) 3 3 2 (3,3,3,4,4) 4 3 2 (3,3,4,3,4) 5 2 1 1 (3,3,4,12) () 6 2 1 1 (3,4,3,12) () 7 2 2 (3,3,6,6) () 8 2 2 (3,6,3,6) 9 1 2 1 (3,4,4,6) () 10 1 2 1 (3,4,6,4) 11 1 1 1 (3,7,42) () 12 1 1 1 (3,8,24) () 13 1 1 1 (3,9,18) () 14 1 1 1 (3,10,15) () 15 1 2 (3,12,12) 16 4 (4,4,4,4) 17 1 1 1 (4,5,20) () 18 1 1 1 (4,6,12) 19 1 2 (4,8,8) 20 2 1 (5,5,10) () 21 3 (6,6,6) From the table we can see that: Theorem 1: There are 3 regular tessellations of the plane. Theorem 2: There are 8 semi-regular (Archimedean) tessellations of the plane. Exercises Polygonal Tessellation Exercises Relevant examples from Escher's work • Fundamental forms of regular division of the plane, Visions of Symmetry pg. 33 • Sketch #A7 (Regular division with triangles) • Tessellation by triangles, sketch (2) from the abstract motif notebook, Visions of Symmetry pg. 83. • Sketch #131-134 (Pentagon tessellations), and Tiled Column, New Lyceum, Baarn • Tessellation by hexagons, Visions of Symmetry pg. 90 Related Sites • Java applet for quadrilateral tessellations. By Shodor Interactive. • Topics in Tiling by Eric Weisstein at Wolfram Research. Convex Pentagons • The 14 known types of convex pentagons that tessellate. By http://www.mathpuzzle.com • Pentagon tilings at Wikipedia.org. • A brief history of convex pentagon tessellations. By Doris Schattschneider. • Intriguing Tessellations. By Marjorie Rice. • Tessellations by Hexagons. By David King. Notes 1. ↑ http://forum.swarthmore.edu/sketchpad/intro/gsp.introlab1.html 2. ↑ K. Reinhardt, Über die Zerlegung der Ebene in Polygone. (Inaugural-Disstertation, Univ. Frankfurt a.M.) R. Noske, Borna and Leipzig, 1918. 3. ↑ R. Kershner, On Paving the Plane, The American Mathematical Monthly 75, October 1968, pg. 839-844 4. ↑ Martin Gardner, Time Travel and Other Mathematical Bewilderments Ch. 13. W.H. Freeman, 1988	{"extraction_info": {"found_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": 15, "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": 0.8866819143295288, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/4049/meaning-of-the-anti-commutator-term-in-the-uncertainty-principle/4077	# Meaning of the anti-commutator term in the uncertainty principle What is the meaning, mathematical or physical, of the anti-commutator term? $\langle ( \Delta A )^{2} \rangle \langle ( \Delta B )^{2} \rangle \geq \dfrac{1}{4} \vert \langle [ A,B ] \rangle \vert^{2} + \dfrac{1}{4} \vert \langle \{ \Delta A, \Delta B \} \rangle \vert^{2}$, where $\Delta A, \Delta B, A$ and $B$ are operators. The inequality is still true, and the anti-commutator term "strengthens" the inequality, but why does it appear? - ## 2 Answers Dear Rodrigo, it's an interesting stronger version of the uncertainty principle for general operators $A,B$ that I've never seen before but I just verified it holds. Just to be sure, the anticommutator is simply $$\{A,B\}\equiv AB+BA.$$ I like when the braces are only used for pairs of Grassmannian objects but people use it as a bookkeeping device to simplify $AB+BA$ in all situations. Nothing difficult about the notation. Note that the commutator and anticommutator appear totally symmetrically in the inequality, a fact we will derive. To see why the stronger inequality holds, open Wikipedia here http://en.wikipedia.org/wiki/Uncertainty_principle#Mathematical_derivations where only the simpler version of the inequality (without the squared anticommutator) is proved by combining two inequalities. The first one, $$\|\|A\psi\|\|^2 \|\|B\psi\|\|^2 \geq \|\langle A\psi\|B\psi\rangle\|^2$$ remains unchanged. However, the second inequality from the Wikipedia article may be strengthened to a full-fledged equality $$\|\langle A\psi\|B\psi\rangle\|^2 = \left\| \frac{1}{2i} \langle \psi \| AB-BA \| \psi \rangle \right\|^2 + \left\| \frac{1}{2} \langle \psi \| AB+BA \| \psi \rangle \right\|^2$$ This identity simply says that the squared absolute value of a complex number is the sum of the squared real part and the squared imaginary part (which was omitted on Wikipedia). Combining the previous two inequalities, one gets your "stronger" uncertainty principle. (Of course, the equation derived above is uselessly weak unless the expectation values of $A,B$ vanish themselves. It can be strengthened into yours by repeating the same prodedure for $\Delta A = A-\langle A\rangle$ and similarly $\Delta B = B-\langle B\rangle$ instead of $A,B$.) I wrote what the anticommutator means mathematically and why the inequality is true. Now, what does the anticommutator term mean physically? I don't know what this question mean. It's a term in an equation that I can read and explain for you again. The precise answers in physics are given by mathematics. So I guess that the answer you want to hear is that it means nothing physically, it's just pure mathematics. This fact doesn't mean that it can't be useful. Well, in normal cases, the stronger version is not "terribly" useful because the anticommutator term is only nonzero if there is a "correlation" in the distributions of $A,B$ - i.e. if the distribution is "tilted" in the $A,B$ plane rather than similar to a vertical-horizontal ellipse which is usually the case in simple wave packets etc. Maybe this is what you wanted to hear as the physical explanation of the anticommutator term - because $AB+BA$ is just twice the Hermitean part of $AB$, it measures the correlation of $A,B$ in the distribution given by the wave function - although the precise meaning of these words has to be determined by the formula. - Very nice, both of you. – Carl Brannen Jan 28 '11 at 15:58 1 This "strong equality" is used as a stepping stone in proving the more common expression for HP (in position-momentum terms for example). That's how I was taught it at least. – Noldorin Jan 28 '11 at 16:41 What do you mean by HP? – Rodrigo Thomas Jan 28 '11 at 18:15 I guess that Noldorin means the Heisenberg principle but I don't understand why the comment should be right... – Luboš Motl Jan 28 '11 at 21:11 @Luboš Motl Could you give some example of this correlation? – Rodrigo Thomas Jan 29 '11 at 18:40 show 1 more comment Dear Rodrigo, I am not aware of any direct physical meaning of the anti-commutator term, but it is useful when you want to pin down the states that saturate the inequality in the Heisenberg principle. Obviously, two conditions must be met if an equality in the usual uncertainty principle should occur: the anti-commutator term must vanish and the Cauchy-Schwarz inequality (see the comment by Luboš Motl) must be saturated. The latter happens if and only if the vectors $\Delta A\|\psi\rangle$ and $\Delta B\|\psi\rangle$ are collinear, let's say $\Delta A\|\psi\rangle=\lambda\Delta B\|\psi\rangle$. This is equivalent to $(A-\lambda B)\|\psi\rangle=(\langle A\rangle-\lambda\langle B\rangle)\|\psi\rangle$, that is, $\psi$ is an eigenvector of $A-\lambda B$. But then the expectation value of the anti-commutator becomes $\langle\{\Delta A,\Delta B\}\rangle=(\lambda+\lambda^*)\langle(\Delta B)^2\rangle$, which vanishes only if $\lambda$ is purely imaginary unless, of course, $\|\psi\rangle$ is an eigenvector of $B$ in which case the whole inequality is trivial. So in the end, the state $\|\psi\rangle$ saturates the inequality in the (usual formulation of the) uncertainty principle if and only if it is an eigenstate of $A-\lambda B$ for some purely imaginary $\lambda$. This happens for example for the coherent states of the harmonic oscillator. -	{"extraction_info": {"found_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": 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": 0.9330567717552185, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/3141/can-has-string-theory-solved-cosmological-constant-problem	# Can/has string theory solved cosmological constant problem? Can someone explain to me what the KKLT paper says, and what has and hasn't it achieved regarding the ability to construct solutions with a small positive or negative cosmological constant in string theory? - For starters, how about a link to the paper? – David Zaslavsky♦ Jan 17 '11 at 10:08 @David: AFAIK there is only one KKLT paper, so I linked to it. – Marek Jan 17 '11 at 10:35 ## 1 Answer Dear gob, the KKLT 2003 paper http://arxiv.org/abs/hep-th/0301240 is constructing a large number of stabilized vacua with a negative cosmological constant - anti de Sitter or AdS vacua - and, with a somewhat smaller certainty, many positive-cosmological-constant de Sitter or dS vacua derived from them. The precise paper is nontrivial because of the particular technical context how this is achieved; interpretation will be left to the end. It considers type IIB string theory on Calabi-Yau manifolds with a topology. This topology has typically lots of "cycles" - noncontractible submanifolds - and there can be an integer-valued generalized (NS-NS and/or R-R) magnetic flux through each "cycle". By taking combinations, one may obtain googols (or powers of googols) of different stable points in the space of configurations of the "string fields". The vacua they obtain at the beginning are AdS vacua. They're supersymmetric and the cosmological constant is negative. KKLT also show that there exist related vacua in which some "antibranes" are added. Those break supersymmetry but are metastable, with lifetimes that vastly exceed the current age of the Universe, so these vacua are potential candidates to correspond to our world. The KKLT computation of the de Sitter vacua's lifetime has come under some scrutiny and physicists differ in their opinion whether the existence of a large number of de Sitter vacua has been established; the situation is much clearer with the unbroken-supersymmetric AdS part of their construction. Interpretation These vacua are stable - that's a good thing because they don't include any exactly massless scalar fields whose value could spontaneously change, thus producing new (unobserved) long-range forces (and allowing the fundamental constants such as the fine-structure constant to change rapidly). The instabilities are nonperturbative - essentially quantum tunneling into other vacua. The precise choice which tunneling directions are favored has been updated by some newer papers, too. It remains a controversial technical question. It seems very likely that string theory's equations have many solutions - people often say $10^{500}$ although the number is not known at any accuracy - that qualitatively look like the Universe around us. By their sheer number, some of them will have a small value of the cosmological constant, comparable to the observed value. The smallness was only guaranteed by having many sufficiently (seemingly) "random" solutions to choose from. Some of them will have a small cosmological constant. Anthropic selection A highly disputed question is whether we are living in a randomly chosen Universe or whether there exists a cosmological mechanism or an equation that picks a privileged vacuum (or at least a much smaller subset). As Weinberg argued decades ago, the existence of galaxies - which seems needed for the existence of intelligent beings - may act as a selection criterion because galaxies only arise if the cosmological constant is tiny, comparable to the observed one. According to the anthropic principle, we don't need to explain why the constants of Nature don't take values incompatible with life; if they did, there would be no one who could complain that the hopes for life were doomed. The physicists defending the so-called "anthropic principle" usually say that there is no other selection criterion aside from the condition that galaxies and life may arise in the Universe. Moreover, they often assume that all the Universes with equal chances to produce life have the same "prior probability" - which means that we should live in a typical or random or average Universe among those that are compatible with life. As they usually admit, the actual probability distribution on the Universes is unknown and even in principle, we don't know an algorithm to calculate it. Even the most hardcore anthropic physicists know that they don't know whether the chances of a particular vacuum are increased by its longer lifetime, bigger volume, bigger expectation value of planets with life, and other factors. All these things are unknown and arguments favoring one answer over others are, so far, philosophical in character, not scientific. If the KKLT or related construction is right and we live in a rather random Universe selected from the string theory solutions, then string theory solves the cosmological constant problem. String theory surely admits solutions with nonzero values of the cosmological constant. And a sufficient number of candidates that produce a cosmological constant of a random size is predicted and the emergence of life is a sufficient criterion to choose a solution that agrees with the real world. In practice, it could be difficult to locate the right solution if there are googols (or powers of googol) of candidates and ours is pretty much random - and the other parameters of particle physics depend on the properties of the compactification and fluxes somewhat randomly, too. Other physicists, such as myself, are convinced that a deeper understanding of the structure of the space of vacua - the landscape - and the cosmological mechanisms that may related or comparing different places of the landscape will show that our vacuum is much more special and physics may ultimately become able to determine in which one we live. We're not there yet but it is fair to say that string theory has produced an internally consistent solution that may produce stable vacua with all the qualitative particle species and interactions we know as well as a reasonable value of the cosmological constant. Needless to say, it's the only framework known to modern physics that can offer such a realistic candidate description. We won't know for some time whether this candidate description is valid. In particular, even if string theory is valid which is almost certainly the case, we won't know for a long time whether our Universe is close to one of the type IIB compactifications or whether it can only be described by another category of string models such as heterotic string theory; heterotic M-theory; type IIA braneworlds; M-theory on $G_2$ holonomy manifolds, or others. A much more detailed comparison of the vacua with the observed reality is needed to settle such questions. Best wishes Lubos -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9403721690177917, "perplexity_flag": "middle"}
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http://mathoverflow.net/questions/15292/why-cant-there-be-a-general-theory-of-nonlinear-pde	## Why can’t there be a general theory of nonlinear PDE? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Lawrence Evans wrote in discussing the work of Lions fils that there is in truth no central core theory of nonlinear partial differential equations, nor can there be. The sources of partial differential equations are so many - physical, probabilistic, geometric etc. - that the subject is a confederation of diverse subareas, each studying different phenomena for different nonlinear partial differential equation by utterly different methods. To me the second part of Evans' quote does not necessarily imply the first. So my question is: why can't there be a core theory of nonlinear PDE? More specifically it is not clear to me is why there cannot be a mechanical procedure (I am reminded here by [very] loose analogy of the Risch algorithm) for producing estimates or good numerical schemes or algorithmically determining existence and uniqueness results for "most" PDE. (Perhaps the h-principle has something to say about a general theory of nonlinear PDE, but I don't understand it.) I realize this question is more vague than typically considered appropriate for MO, so I have made it CW in the hope that it will be speedily improved. Given the paucity of PDE questions on MO I would like to think that this can be forgiven in the meantime. - 4 Are there any Markov or Novikov type theorems for PDEs ? i.e. presumably you could encode algorithmically unsolvable problems into the language of PDEs. Meaning, knowledge of some aspect of the solution (bounded orbit, say) is equivalent to knowing the solution to an algorithmically unsolvable problem? If there were such theorems that would partially address your question. – Ryan Budney Feb 14 2010 at 23:14 9 Perhaps the kind of negative result you are looking for is the theorem of Pour-el and Richards that the 3-dimensional wave equation has non-computable solutions with computable initial conditions. This is in their book Computability in Analysis and Physics (Springer-Verlag 1989). – John Stillwell Feb 14 2010 at 23:43 @Ryan, John--Good point! (One or both of) You should put this in an answer...I seem to recall hearing something once along the lines of PDEs being Turing-universal. But perhaps there can be a general theory of PDE that correspond to a restricted model of computation? – Steve Huntsman Feb 15 2010 at 1:10 4 When I teach basic differential equations, I stress analogies with algebraic equations. While this is probably more simple-minded than you were looking for, I point out (without attempting a thorough justification) that although there is a good theory of linear (algebraic equaions) a general theory to solve all algebraic equations, no matter how irregular, is hopelessly out of reach. And we have no right to expect better of differential equations. – Mark Meckes Mar 23 2010 at 16:31 1 This also brings to mind the preface (books.google.com/…) from "Lectures on Partial Differential Equations" by Arnol'd. Unfortunately the google books version cuts out after the first page, and I can't find another English version online. You can find a Russian version by googling for "Лекции об уравнениях с частными производными". – Josh Guffin Sep 29 2010 at 18:26 show 1 more comment ## 8 Answers I find Tim Gowers' "two cultures" distinction to be relevant here. PDE does not have a general theory, but it does have a general set of principles and methods (e.g. continuity arguments, energy arguments, variational principles, etc.). Sergiu Klainerman's "PDE as a unified subject" discusses this topic fairly exhaustively. Any given system of PDE tends to have a combination of ingredients interacting with each other, such as dispersion, dissipation, ellipticity, nonlinearity, transport, surface tension, incompressibility, etc. Each one of these phenomena has a very different character. Often the main goal in analysing such a PDE is to see which of the phenomena "dominates", as this tends to determine the qualitative behaviour (e.g. blowup versus regularity, stability versus instability, integrability versus chaos, etc.) But the sheer number of ways one could combine all these different phenomena together seems to preclude any simple theory to describe it all. This is in contrast with the more rigid structures one sees in the more algebraic sides of mathematics, where there is so much symmetry in place that the range of possible behaviour is much more limited. (The theory of completely integrable systems is perhaps the one place where something analogous occurs in PDE, but the completely integrable systems are a very, very small subset of the set of all possible PDE.) p.s. The remark Qiaochu was referring to was Remark 16 of this blog post. - I wonder: can one not model Turing machines using ODEs? – Mariano Suárez-Alvarez Feb 15 2010 at 2:15 1 And even the completely integrable systems are full of surprises, such as the Camasso–Holm equation, where the solution concept needs some tweaking in order to make the Cauchy problem well posed. – Harald Hanche-Olsen Feb 15 2010 at 2:20 @Mariano: yes, as covered in your subsequent question: mathoverflow.net/questions/15309 – Steve Huntsman Feb 15 2010 at 4:26 Leave it to Terry Tao to give the most knowledgable and succinct response to a deep question.His grasp of the Big Picture and relevant publications in any field never ceases to amaze me. – Andrew L Jun 4 2010 at 21:13 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I agree with Craig Evans, but maybe it's too strong to say "never" and "impossible". Still, to date there is nothing even close to a unified approach or theory for nonlinear PDE's. And to me this is not surprising. To elaborate on what Evans says, the most interesting PDE's are those that arise from some application in another area of mathematics, science, or even outside science. In almost every case, the best way to understand and solve the PDE arises from the application itself and how it dictates the specific structure of the PDE. So if a PDE arises from, say, probability, it is not surprising that probabilistic approximations are often very useful, but, say, water wave approximations often are not. On other hand, if a PDE arises from the study of water waves, it is not surprising that oscillatory approximations (like Fourier series and transforms) are often very useful but probabilistic ones are often not. Many PDE's in many applications arise from studying the extrema or stationary points of an energy functional and can therefore be studied using techniques arising from calculus of variations. But, not surprisingly, PDE's that are not associated with an energy functional are not easily studied this way. Unlike other areas of mathematics, PDE's, as well as the techniques for studying and solving them, are much more tightly linked to their applications. There have been efforts to study linear and nonlinear PDE's more abstractly, but the payoff so far has been rather limited. - Further to my comment above, on the theorem of Pour-el and Richards: it originally appeared in Advances in Math. 39 (1981) 215-239, entitled "The wave equation with computable initial data such that its unique solution is not computable." I think it is fair to say that they get the wave to simulate a universal Turing machine, albeit with very complicated encoding. However, this may all be irrelevant to explaining why "nonlinear PDE are hard" because the wave equation is linear! - 1 Yes, I would say that there is a general theory of linear PDE, and Hörmander pretty well captures the basics. – Steve Huntsman Feb 15 2010 at 3:00 3 Yes, there is a general theory of linear PDE developed largely by Hormander, but of what use is it? In some sense, the space of all possible linear PDE's can be viewed as a singular algebraic variety, where Hormander's theory applies only to generic (smooth) points and the most interesting and heavily studied PDE's all lie in a lower-dimensional subvariety and mostly in the singular set of the variety. – Deane Yang Feb 15 2010 at 3:17 1 Also, even though you can't solve the halting problem for Turing machines, the existence, uniqueness, and computability (by definition!) of solutions to the Turing machine “equations of motion” are all utterly trivial. For PDEs, nothing could be farther from the truth. Similarly for ODEs: The local theory is easy, it's long term and global behaviour that is difficult. But for PDEs, even the local theory is fiendishly difficult. (Except for the Cauchy-Kowalevskaja theorem, which despite (or because of?) its generality also turns out to be of rather limited use.) – Harald Hanche-Olsen Feb 15 2010 at 3:31 Some more random thoughts: The closest thing I've ever seen to a "general theory of nonlinear PDE's" is Gromov's book, Partial Differential Relations. He does many things in there that I don't understand, but one application that he applied his theory to is isometric embeddings of Riemannian manifolds into Euclidean space or other higher dimensional Riemannian manifolds (the problem made famous by Nash). Moreover, in a paper by Bryant, Griffiths, and me (but in a section written by the other two and not me), it is shown that in some sense, the linearized PDE corresponding to the isometric embedding of an $n$-dimensional Riemannian manifold into $n(n+1)/2$-dimensional Euclidean space looks like a generic $n$-by-$n$ system of first order linear PDE's. I'm not aware of any other place where a "generic" system of PDE's arises naturally. The results in this paper inspired some efforts by Jonathan Goodman and me (unpublished) as well as Nakamura and Maeda (TAMS 313 (1989) 1-51) to extend Hormander's theory of linear PDE's (at least those of real principal type) to nonlinear PDE's. (It should be noted that much more interesting work in this direction was done for the 2-dimensional case, starting with the Ph.D. thesis of C. S. Lin) But maybe you really meant "the general theory of nonlinear PDE's that are elliptic, hyperbolic, or parabolic" and not really the all encompassing "general theory of nonlinear PDE's"? There's far too much junk in the latter. - As far as my very limited understanding goes, the h-principle is not really a "general theory of nonlinear PDE's" but mostly applies to underdetermined systems, which happen to arise a lot in geometric applications, but not as much in physics. – Otis Chodosh Jan 11 2012 at 3:54 Yes, Gromov's study of PDE's is pretty much limited to underdetermined systems and therefore is definitely not a study of general PDE's. But it applies to general underdetermined PDE's, and that's probably the broadest class of PDE's that anyone has been able to study using a unified approach. – Deane Yang Jan 11 2012 at 8:39 Here is a 7-page review of Partial Differential Relations by Dusa McDuff: projecteuclid.org/DPubS/Repository/1.0/… – Tom LaGatta Mar 20 at 10:44 Tom, thanks. I've certainly seen that when it first came out. – Deane Yang Mar 21 at 2:41 To elaborate on Steve Huntsman's comment, I remember reading the following on Terence Tao's blog: there exist PDE that can simulate Newtonian mechanics, and using such a PDE and the correct initial conditions it is possible, in principle, to simulate an arbitrary analog Turing machine. So a general-purpose algorithm to determine even the qualitative behavior of an arbitrary PDE cannot exist because such an algorithm could be used to solve the halting problem. - I think there is something you can call a general theory of PDEs. It started already long time ago with Meray, Riquier, Janet, Elie Cartan. There is an important survey article by Donald Spencer: Overdetermined systems of linear partial differential equations , Bull. Amer. Math. Soc. 75 (1969), 179-239. see also the recent book by Seiler: Involution:The Formal Theory of Differential Equations and its Applications in Computer Algebra, springer, 2010. This book contains lots of references to this topic. It is a bit strange why this line of research is not very well known. - In response to "It is a bit strange why this line of research is not very well known": 1) Actually, this stuff has become much better known through the work and books by Bryant, Chern, Goldschmidt, Griffiths, Ivey, and Landsberg. 2) Most PDE's that arise from other areas of mathematics and sciences are either scalar or determined systems. For such PDE's, the formal theory tells you nothing more than what the Cauchy-Kovalevski theorem says. 3) The formal theory tells you nothing about the global behavior and regularity of solutions to PDE's. – Deane Yang Jun 4 2010 at 18:47 @Deane. Your comment 2) is irrelevant for several reasons. a) Cauchy-Kovalevskaia theorem tells you nothing about the Cauchy problem for the heat equation, Navier-Stokes system or Schrödinger equation, because the order with respect to time ($=1$) is smaller than the total order ($=2$). b) Real problems are posed in domains with boundaries, and the boundary conditions can be non-homogeneous. You may need a very much elaborated theory to prove the solvability. Hyperbolic initial-boundary-value problems are notoriously difficult (see the book by S. Benzoni-Gavage and myself); C.-K. is useless. – Denis Serre Nov 18 2010 at 7:51 Denis, your statements are consistent with and provide some details that underly mine. – Deane Yang Jan 20 2011 at 4:27 this is a comment to Deane Yang, but apparently it was too long so here is a separate answer. My background is in numerical solution of PDEs 1) while I know about this, it is not at all well-known by people who numerically solve PDEs. 2) this is not true. Most computations are systems of PDEs. I think most computations are done with systems where there are no actual theory, i.e. existence and uniqueness results. Think about Navier-Stokes. Many systems are NS coupled with for example convection diffusion type systems (small amounts of material in the flow etc). Then there are liquid crystals, Maxwell, elasticity, flow coupled with elasticity etc. Of course when the computers were slower one had to simplify to get a scalar equation and then hope that it gives something reasonable. Of course Cauchy-Kovalevskaia as such is irrelevant because one wants the solutions in Sobolev spaces. But the whole formal theory started as a generalization of CK. 3) this is not true. For example there are systems which are not elliptic initially but whose involutive forms are elliptic. This gives a priori regularity results and existence results. Also one could argue that the word "determined" (and over/underdetermined) can't be defined in general without formal theory. - Here are my reactions: 1) Is the formal theory useful for numerical solutions? Could you provide references for this? 2) There are certainly systems consisting of an evolution equation that is coupled with a constraint or gauge condition. Navier-Stokes is like this. The formal theory provides no new insights for these systems, either. 3) What example do you have in mind? I know this statement as an abstract theorem, but I have never seen it used anywhere. – Deane Yang Jun 4 2010 at 20:26 I will simply quote Heisenberg. (This an approximative quote from memory.) One can say almost everything about nothing, and almost nothing about everything. - But why is the study of PDE's "everything"? In comparison to, say, the study of polynomials? – Deane Yang Jan 11 2012 at 8:37 Of course, the study of PDEs is not everything, but the point of Heisenberg's quote (and he was referring specifically to nonlinear pde-s) is within the pde Universe, the statement that something is nonlinear carries zero information. PS The study of polynomials can be viewed as a subclass of the study of PDE's (think characteristic polynomials of constant coefficients o.d.e.'s) More generally the theory of D-modules suggests that substantial chunk of algebraic geometry is closely connected to PDE's. – Liviu Nicolaescu Jan 11 2012 at 11:28	{"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": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9438585042953491, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/87143?sort=votes	## Algebraic function with extra condition, what can it be? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have an unknown function $\psi(\xi_1,\dots,\xi_n)$, such that $\psi$ satisfy an (unknown) polynomial equation with coefficients polynomials in the $\xi_i$. The function is homogeneous, that is, $\psi(t \xi_1,\dots, t \xi_n) = t^c \psi(\xi_1,\dots,\xi_n)$. I also know that $\|\psi(e^{i \theta_1},\dots,e^{i \theta_n})\|=1$ when $\sum_i \theta_i =0$, which implies that $$\psi(e^{i \theta_1},\dots,e^{i \theta_n}) = e^{i A(\theta_1,\dots,\theta_n)}$$ for some real-valued function $A$. Clearly, one option is that $\psi(\xi_1,\dots,\xi_n) = \xi_1^{p_1} \cdots \xi_n^{p_n}$ such that $p_1+\cdots+p_n = c$, but can one exclude any other form? How do one take advantage of the fact that $\psi$ is algebraic in the $\xi_i$? - ## 2 Answers So, here is my stab at a proof, which actually do not require algebraicness of $\psi$ Notice that we have $\|\psi(\xi_1, \dots \xi_n)\| = 1$ whenever $\xi_1 \cdots \xi_n = 1.$ Thus, using the homogeneity property, we may see that $$\psi(t^{1-n}\xi_1, t \xi_2, t\xi_3, \dots ,t\xi_n) = \phi(\xi_1,\dots,\xi_n) e^{i A(t)}$$ for any $\xi_1,\dots,\xi_n,$ since we may normalize the $\xi_i$:s with $(\xi_1 \xi_2 \cdots \xi_n)^{1/n}$ and move this to the other side ($\phi$). Now, changing $t$ will not change the modulus of the product of the parameters to the function, so it may only affect the argument, and hence the form above. Now, using homogeneity again yields $$t^c\psi(t^{-n}\xi_1, \xi_2, \xi_3, \dots ,\xi_n) = \phi(\xi_1,\dots,\xi_n) e^{i A(t)}$$ Differentiating both sides w.r.t. $t$ and then putting $t=1$ gives $$\psi - n \xi_1 \psi'_1 = \psi i A'(1)$$ Now, this is a differential equation, which is easy to solve, one sees that $\psi = \xi_1^\beta F(\xi_2,\dots,\xi_n)$ for some constant $\beta$ and unknown function $F$. However, this argument can be made for any of the variables, yielding the desired result. We do need that $\psi$ is differentiable, but this should be true almost everywhere for algebraic functions. - ### 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. Is your function entire? An entire algebraic function is just a polynomial function. So you know that it's a sum of terms of that type. (Unless it might have poles, in which case it's a rational function.) Fix the terms $\xi_1,...,\xi_{n-1}$ at roots of unity and let $xi_n$ vary. Then $\xi_n$ is a polynomial that takes roots of unity. Therefore it must be a constant power of $\xi_n$ times a root of unity. (by Schwartz reflection) By continuity, the power cannot depend on $\xi_1,...,\xi_{n-1}$, so the only terms must have $\xi_n$ that power. We can continue this argument for each $\xi_i$, thus proving that the expresion consists of only a single term. So, up to multiplication by a root of unity, it must be the option you gave. - If your function is allowed to have poles, this is false. Let $f$ be any rational linear transformation of the Riemann sphere that preserves the roots of unity, and let $g$ be a homogeneous polynomial of degree zero, then multiplying by $f\circ g$ will preserve all the properties given, and you can make your function arbitrarily complicated like this. – Will Sawin Feb 2 2012 at 18:20 well maybe not arbitrarily. But extremely. – Will Sawin Feb 2 2012 at 18:20 @Will Sawin: A polynomial of degree zero? That sounds very much like a constant... – Per Alexandersson Feb 2 2012 at 22:09 oh sorry. i meant a rational function of total degree zero like $\zeta_1^2\zeta_2^{-1}\zeta_3^{-1}$ – Will Sawin Feb 3 2012 at 7:22 @Will Sawin: But the rational linear transform must be homogeneous as well... – Per Alexandersson Feb 4 2012 at 11:24 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": 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": 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": 0.9195618033409119, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=3932172	Physics Forums Page 1 of 3 1 2 3 > ## Why does light move? Or how does light move? There's a source of light, for example a torch but what is it that propelled the photons into my eye from the torch? I'm new to this so I don't know, I mean I know light is a wave but it's also a photon init, so why does that photon move? PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus Although this is far from a good analogy, but what's propelling the Earth to revolve around the Sun? Ummm, as far as I understand it something moves through space at the same speed forever unless its path is altered by another force? I'm new to this I'm sorry if it's a stupid question. But as I think about it, the Earth, fundamentally moving through space because of energy is received from some other place, like...maybe it gained momentum from the gravity of something and swung off into space? But light isn't gaining momentum from anywhere there's no force that initially accelerated it is it? I'm not sure, as I say I'm a total n00b compared to all of you I was never smart enough at maths to do physics but I can generally get my head around concepts if I'm given the opportunity to ask questions. Recognitions: Gold Member ## Why does light move? The source of typical light is the transition of electrons in a nucleus from a higher energy to a lower energy orbit. When they give off energy some of it is visible light, more is electromagnetic radiation we cannot see by eye. All electromagnetic radiation moves at a constant speed, 'c', and so do the quanta of electromagnetic radiation, photons. When a photon of the right energy interacts with an electron say in silicon, a different form of eenergy is produced: an electric current. Ultimately nobody really knows 'why' things operate this way, but we have a lot of math that describes how things operate; if they did not operate very nearly as they do we do know we would not be here to observe such activity. Without electrons orbiting nuclei to form the elements and light being absorbed and emitted, life could not exist....in fact neither would stars, plants,planets nor much of anything else. Recognitions: Gold Member but what's propelling the Earth to revolve around the Sun? Thats another basic of our universe: gravity. It also progagates at the speed of light. When earth formed it did so from revovling gases...and that anguklar momentum remained as the earth solidified....nearby gases formed elsewhere and became other planets, sun, asteroid, ice moons, and so forth. Even entire galaxies have such rotations, like our own Milky Way. Quote by RichyB Ummm, as far as I understand it something moves through space at the same speed forever unless its path is altered by another force? I'm new to this I'm sorry if it's a stupid question. Correct, although this is true in non-relativistic mechanics. In relativistic mechanics, there is the possibility of the existence of massless particles, such as the photons. It may be shown that the parallel and perpendicular accelerations of a massless particle subject to a force F is: [tex] \mathbf{a}_{\\|} = \frac{c^2}{E} \, \left(\frac{m c^2}{E} \right)^{2} \, \mathbf{F}_{\\|} [/tex] [tex] \mathbf{a}_{\bot} = \frac{c^2}{E} \, \mathbf{F}_{\bot} [/tex] where $E = c \, \sqrt{p^2 + (m c)^2}$ is the relativistic energy of the particle, and m is the rest mass. It is easily seen that massless particles do not feel any parallel acceleration (since their speed is always c), but the radius of curvature of the trajectory is given by: [tex] \frac{c^2}{R} = \frac{c^2}{E} \, F_{\bot} \Rightarrow \frac{1}{R} = \frac{F_{\bot}}{E} [/tex] But, for photons, I don't know what force may act on them, since they are uncharged. You cannot use the above formulae for a gravitational field, since they are derived in special relativity. Quote by RichyB But as I think about it, the Earth, fundamentally moving through space because of energy is received from some other place, like...maybe it gained momentum from the gravity of something and swung off into space? But light isn't gaining momentum from anywhere there's no force that initially accelerated it is it? I'm not sure The photon gains momentum and energy in the process of its emission, and keeps it until absorbed. Quote by Naty1 The source of typical light is the transition of electrons in a nucleus from a higher energy to a lower energy orbit. When they give off energy some of it is visible light, more is electromagnetic radiation we cannot see by eye. All electromagnetic radiation moves at a constant speed, 'c', and so do the quanta of electromagnetic radiation, photons. When a photon of the right energy interacts with an electron say in silicon, a different form of eenergy is produced: an electric current. Ultimately nobody really knows 'why' things operate this way, but we have a lot of math that describes how things operate; if they did not operate very nearly as they do we do know we would not be here to observe such activity. Without electrons orbiting nuclei to form the elements and light being absorbed and emitted, life could not exist....in fact neither would stars, plants,planets nor much of anything else. So the movement of one electron from one orbit in the nucleus to another gives off one photon? How many different orbit levels are there for an electron to have? Cause I was once told that every electron in the universe has a different one, and once one changes energy levels, every other one in the universe has to shift instantly to another one because they can't occupy the same one? Brian Cox said that. Quote by RichyB Cause I was once told that every electron in the universe has a different one, and once one changes energy levels, every other one in the universe has to shift instantly to another one because they can't occupy the same one? Brian Cox said that. Please give a reference to this claim. "I was once told..." by no matter whom does not count as a reputable source. The source of 'typical' (!!) light is NOT the transition of electrons in a nucleus from a higher to a lower energy level. Recognitions: Gold Member Science Advisor Quote by truesearch The source of 'typical' (!!) light is NOT the transition of electrons in a nucleus from a higher to a lower energy level. ohhh really ? and where do you think the photon comes from ? Dave Recognitions: Science Advisor Quote by RichyB Cause I was once told that every electron in the universe has a different one, and once one changes energy levels, every other one in the universe has to shift instantly to another one because they can't occupy the same one? What somebody was probably trying to tell you, and I'm just guessing based on your description, is that every electron must be in different quantum state. That is true, because electrons are fermions, and no two fermions can have exactly the same state. This is called Pauli Exclusion Principle. Feel free to look it up for more details. However, two electrons in two different atoms are already inherently in different states, as location is part of the state. So transition of electron in one atom doesn't depend on state of electron in another atom. (This is simplifying things quite a bit, but it should at least clear up some confusion.) Mentor Blog Entries: 1 Quote by Naty1 The source of typical light is the transition of electrons in a nucleus from a higher energy to a lower energy orbit. Is that what you meant to say? Quote by K^2 What somebody was probably trying to tell you, and I'm just guessing based on your description, is that every electron must be in different quantum state. That is true, because electrons are fermions, and no two fermions can have exactly the same state. This is called Pauli Exclusion Principle. Feel free to look it up for more details. However, two electrons in two different atoms are already inherently in different states, as location is part of the state. So transition of electron in one atom doesn't depend on state of electron in another atom. (This is simplifying things quite a bit, but it should at least clear up some confusion.) Yeah, yeah this is what he said. It was in a talk he gave, thanks for that, it's on youtube in a video called 'A night with the stars'. I wish I could spend a week with Brian Cox I'd have so many questions. Depends what you mean by 'typical' light. If you mean em radiation detected by the eye (known as 'visible' radiation) then this comes from changes in electron energy levels of orbiting electrons in atoms. X-rays (not visible to the human eye) come from changes in electron energy levels closer to the nucleus of an atom. Gamma rays come from the nucleus. Any basic physics text book will explain this light does not move. in our particular reference frame, it appears to travel at the fixed velocity of C. however, at that speed, there is no where for it to travel from or to - time has stopped and all distance references have been reduced to zero. the photon essentially occupies all of spacetime in between the time it is emitted and the time it is absorbed. I agree with you jnorman and I love the summary that I once read (can't remember where !!!) Light takes no time to get from one point to the next and for light there is no distance between one point and the next. Wish I understood it !! Recognitions: Gold Member Quote by jnorman light does not move. in our particular reference frame, it appears to travel at the fixed velocity of C. however, at that speed, there is no where for it to travel from or to - time has stopped and all distance references have been reduced to zero. the photon essentially occupies all of spacetime in between the time it is emitted and the time it is absorbed. Light most definitely moves. How else would it get from a light bulb to your eye? The issue is that you cannot assign a reference frame to light because there is no reference frame where light is at rest. But this does not mean it does not move, as it moves relative to our frames. I've read a explanation of everyone and everything in the universe moving at c through the time dimension in their rest frames, while light simply moves at c in the space dimension, as it doesn't have a rest frame. Quote by truesearch I agree with you jnorman and I love the summary that I once read (can't remember where !!!) Light takes no time to get from one point to the next and for light there is no distance between one point and the next. Wish I understood it !! I believe you are saying that light does not experience time, as it travels at c and according to the math time should stop at that velocity. I'd say that the math simply doesn't work when you input a velocity of c into the equations, so you cannot depend on it. Page 1 of 3 1 2 3 > Thread Tools Similar Threads for: Why does light move? Thread Forum Replies General Physics 1 Introductory Physics Homework 3 Special & General Relativity 3 Special & General Relativity 6 General Physics 6	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "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": 0.9617655277252197, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/241525/generating-random-variables-from-standard-uniform-distribution-on-1-0?answertab=active	# Generating random variables from standard uniform distribution on $(1,0)$ A random number generator generates random values from the standard uniform distribution $\mathrm{Uniform}(0,1)$. Call this random variable $U$. Starting with a random value $u$ from $U$, show all the steps necessary to generate a random variable $X$ such that the probability mass function $p(x)$ of $X$ is: $p(1)=0.3$ $p(2)=0.15$ $p(3)=0.35$ $p(4)=0.2$ I'm not looking for an exact answer as much as an explanation to help with these problems in the future. - 1 Try partioning $[0,1]$ into disjoint parts with lengths given by your probabilities using the fact that $P(a\leq U\leq b)=b-a$ when $a<b$ and $a,b\in[0,1]$. – Stefan Hansen Nov 20 '12 at 19:46 ## 1 Answer Your random number generator will produce (pseudo) random numbers uniformly distributed on $(0,1)$. Suppose that we want to simulate a random variable $X$ which takes on value $3$ with probability $0.25$, value $7$ with probability $0.35$, and value $45.6$ with probability $0.40$ (of course, this is not your problem, just a similar one). Use the following idea. If the random number generator produces a number between $0$ and $0.25$, report that $X$ has taken on value $3$. If the random number generator produces a number between $0.25$ and $0.25+0.35$, that is, between $0.25$ and $0.60$, report that $X$ has taken on the value $7$. Finally, if the random number generator produces a number between $0.25+0.35$ and $1$, report that $X$ has taken on the value $45.6$. Note that the probability that the random number generator produces a number between $0$ and $0.25$ is $0.25$. So with probability $0.25$ we will be reporting that $X$ has taken on value $3$. The probability that the random number generator produces a number between $0.25$ and $0.25+0.35$ is $0.35$, so with probability $0.35$ we will be reporting that $X$ has taken on value $17$. And so on. Remarks: 1) In principle, the random number generator produces "random" reals between $0$ and $1$. So the probability it produces a particular number $u$, like $1/\pi$, is $0$. In practice, the numbers produced are say decimals, to $10$ decimal places. So our program needs to deal with the highly improbable but possible situations where the random number produced is exactly at a boundary, like $0.25$. It doesn't really matter what we do. 2) The procedure we use can also be described in terms of the cumulative distribution function of the random variable we are trying to simulate. Perhaps you are expected to do it in that style, in preparation for more complex problems when we are simulating a continuously distributed random variable $X$. If a description in terms of the cdf is required, please indicate. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 58, "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": 0.9159103035926819, "perplexity_flag": "head"}
http://mathoverflow.net/questions/110404/when-does-a-distinguished-matching-exist/111309	## When does a `distinguished matching' exist? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose I have a bipartite graph on a pair of vertex sets $X$ and $Y$. Definition: A distinguished matching is a subset $DX\subseteq X$ and a subset $DY\subseteq Y$ such that: • For all $y\in Y$, there exists at least one $x\in DX$ such that $(x,y)$ is an edge. (We say that `$DX$ covers Y'); • For all $x\in X$, there exists at most one $y\in DY$ such that $(x,y)$ is an edge. (We say that `$DY$ is distinguished'); • For all $x\in DX$ there exists a unique $y\in DY$ such that $(x,y)$ is an edge, and for all $y\in DY$ there exists a unique $x\in DX$ such that $(x,y)$ is an edge. (We say that `$DX$ and $DY$ are matched'). (Note: The third condition has been changed twice. I hope it's now correct.) Questions: 1. What are necessary and sufficient conditions for the existence of a distinguished matching? 2. In the event that such a matching exists is there an efficient algorithm to find such a matching? 3. The term `distinguished matching' is my own. Perhaps this notion has been studied by graph theorists under another name. If so, please give me some references! Applications: Suppose that $Y$ is the set of elements of some group $G$, and suppose that $X$ is the set of maximal abelian subgroups of $G$. An edge $(x,y)$ is drawn if the element $y$ is contained in the subgroup $x$. Suppose there is a distinguished matching. Then the set $DX$ is a minimally-sized cover of $G$ by abelian subgroups; the set $DY$ is a maximally-sized set of pairwise non-commuting elements. It is easy to see that a minimal cover by abelian subgroups must be at least as big as a maximal set of pairwise non-commuting elements. The extremal situation is when they're the same size and that's what a matching yields. Finite groups admitting such a matching include rank 1 groups of Lie type. Finite groups that don't admit such a matching include $Sym(n), n\geq 15$. There are other group-theoretic variations on this idea: just change the adjectives abelian and non-commuting in the set-up. Credits: These type of coverings have been studied in group theory at various times. I came across them in joint work with A. Azad and J. Britnell. I'm mainly interested in the situation where the graph is finite, but any thoughts would be welcome. - 3 How do you rule out the possibility that $DY = \{ y_1 \}$ and $DX = \{ x_1,x_2,x_3 \}$ with edges $(x_i,y_1)$ for $i=1,2,3$? I guess I must be missing something? Thanks! – Patricia Hersh Oct 23 at 13:41 5 Is there a typo or am I missing something? The first condition doesn't involve DY, and the second one is always satisfied when DY is empty or a singleton, so they can't possibly imply that \|DX\| = \|DY\|. – Johan Wästlund Oct 23 at 13:46 Patricia and Johan, thank you for your comments which are absolutely correct. I have adjusted the definition and I hope everything now makes sense. – Nick Gill Oct 24 at 8:49 1 Patricia’s example satisfies the adjusted definition as well. – Emil Jeřábek Oct 24 at 10:32 1 heck, i've not made a very good job of this - sorry! i'm going to have another go in a moment at writing down all of the required conditions. my aim was to have as few as possible, so that the definition did not become hideous, but i have overdone it... – Nick Gill Oct 24 at 12:09 show 5 more comments ## 1 Answer Answering one's own question is a bit weird, but I think I might now understand what is going on here... Define a new graph $\mathcal{H}$ whose vertices are the elements of $Y$. Connect two vertices $y_1, y_2 \in Y$ if they have a common neighbour in $X$. Observe that elements of $X$ correspond to maximal cliques in the new graph $\mathcal{H}$. The subset $DX$ corresponds to a set of maximal cliques that cover all of the vertices of $\mathcal{H}$. Thus $\|DX\|$ must be at least as big as the clique cover number of $\mathcal{H}$. Observe next that $DY$ is a set of vertices of $\mathcal{H}$ which are all pairwise disconnected. Thus $DY$ is an independent set. It is clear that the clique cover number of a graph is always at least the independence number of the graph. Now the requirement that $\|DX\|=\|DY\|$ implies that the clique cover number of $\mathcal{H}$ is the same as the size of a maximal independent set (the independence number) of $\mathcal{H}$. Thus we are in an extremal situation. What is more this situation has been discussed elsewhere on MathOverflow. The rest of this answer is a summary of the answers given in the other MO thread... The condition that our graph $\mathcal{H}$ has coinciding clique cover number and independence number is equivalent to the condition that the complement graph $\mathcal{H}'$ has coinciding chromatic and clique numbers. This latter condition seems difficult, however a definition from the literature is relevant here: a perfect graph is one in which the chromatic number of every induced subgraph equals the clique number of that subgraph. A characterization of perfect graphs exists: perfect graphs are the same as Berge graphs. The proof of this characterization is known as the Strong Perfect Graph Theorem. This theorem is a very big deal, and its tremendous difficulty is enough to make one think that characterising graphs with the weaker condition of coinciding chromatic and clique numbers is likely to be nigh on impossible. (I'll wait and see how people respond to this before I accept my own answer! Many thanks to the people who commented and forced me to get my question right in the first place.) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 53, "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": 0.9646434783935547, "perplexity_flag": "head"}
http://mathoverflow.net/questions/97005?sort=votes	Planar sets closed under intersection of circles Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $P$ be the plane with a point at infinity. By plane, I mean the Euclidian plane, and therefore it has circles. A line is also a circle, though its center is at infinity. If $A\subset P$ has cardinal $\|A\|=3$, there exists a unique circle (possibly a line) containing $A$; let me denote it $\Gamma_A$. Let me say that a subset $X$ of $P$ is circularly stable if it satisfies the following property: For every subset $A,B\subset X$ with $\|A\|=\|B\|=3$ and $\Gamma_A\ne \Gamma_B$, the intersection of $\Gamma_A$ and $\Gamma_B$ is included in $X$. Every non-colinear $X$ with $\|X\|\le4$ is circularly stable. A line or a circle is circularly stable, and $P$ itself is so too. If $X$ has a non-void interior, then $X=P$. Q: What can look like a circularly stable subset $X$ of the plane ? For instance, is it true that if $\|X\|\ge5$ and $X$ is circularly stable, then $X$ is dense in $P$, or $X$ is a line or a circle ? Motivation: In classical geometry, one may wander what are the continuous maps $f:P\rightarrow P$ which transform circles or lines into circles or lines. If the guess above is true, then $f$ is completely determined by the images of $5$ non-colinear points. - 1 Answer Send one of the points to infinity by a Mobius tranformation. Your set of points now has the property that the intersection of any two lines passing from pairs of points in the set is also in the set. Such configurations are either all collinear, dense in the plane, or one of these two exceptional cases: • A point together with a collection of points in a line • The four vertices of a parallelogram and the intersection of its diagonals. The first case cannot be circularly stable as soon as $\|X\|\geq 6$. This is because if we let $P$ be the apex and $A,B,C,D$ be collinear points in that order, then the circumcircles of $PAC$ and $PBD$ intersect in a point not in $X$. We conclude that if $\|X\|\geq 7$ and $X$ is circularly stable, then either $X$ is collinear or it is dense in the plane. To classify all circularly stable sets with $\|X\|=6$ that are not collinear we must be in the second case above. Let the points be $A,B,C,D$ as vertices of the parallelogram and $O$ the intersection of the diagonals. Now, the circumcircle of $ABC$ and the line $BD$ must intersect at $D$ so $ABCD$ is inscribed in a circle, i.e. is a rectangle. Now looking at the circumcircle of $ABO$ and the line $BC$ we must have $BC$ tangent to this circumcircle so $ABCD$ must be a square. It's easy to check that a square, its center and the point at infinity form a circularly stable set. There are no circularly stable sets with $\|X\|=5$ that are not collinear. If there were it would come from the first case. We have the apex $A$ and three collinear points $B,C,D$. The circumcircle of $ABD$ and the line $AC$ intersect in a different point, contradiction. On the other hand, as you mention in the OP, any set with $\|X\|\le 4$ is circularly stable. The result I used above is proved in "A dense planar point set from iterated line intersections" by D. Ismailescu and R. Radoicic. - @Gjerhji: I think, meant complex-projective (or Moebius) transformation (since real ones do not map circles to circles). – Misha May 15 2012 at 15:22 @Misha, thank you, I changed projective to Mobius :) – Gjergji Zaimi May 15 2012 at 15:27 @Gjergji. Isn't it a problem in the second case ? Let us denote $a,b,c,d$ the points on the rectangle ($ab$, ..., $da$ being edges), then $O$ its center. The intersection of $\Gamma_{abO}$ and $bc=\Gamma_{bc\infty}$ has an other point, not in $A$.Can we really have $\|A\|=6$ ? Even a $5$-element circularly stable $A$ is not clear to me. – Denis Serre May 16 2012 at 6:15 @Denis, I added a few more details. You're right, a rectangle and it's center is only circularly stable if it is a square. – Gjergji Zaimi May 16 2012 at 17:39 @Gjergji. Im OK with your answer now. The picture with six vertices is interesting. It raises questions. – Denis Serre May 18 2012 at 7:28 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": 63, "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": 0.9362377524375916, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/27849/branch-point-twist-fields-and-operator-insertions-on-a-riemann-manifold?answertab=votes	# Branch-point twist fields and operator insertions on a Riemann manifold I am having trouble understanding how Eq (2.6) in this paper (PDF) $$Z[\mathcal{L},\mathcal{M}_{n}]\propto\langle\Phi(u,0)\tilde{\Phi}(v,0)\rangle_{\mathcal{L}^{(n)},\mathbb{R}^{2}}$$ generalizes to Eqn (2.7) $$\langle\mathcal{O}(x,y;\mbox{ sheet i })...\rangle_{\mathcal{L},\mathcal{M}_{n}}=\frac{\langle\Phi(u,0)\tilde{\Phi}(v,0)\mathcal{O}_{i}(x,y)...\rangle_{\mathcal{L}^{(n)},\mathbb{R}^{2}}}{\langle\Phi(u,0)\tilde{\Phi}(v,0)\rangle_{\mathcal{L}^{(n)},\mathbb{R}^{2}}}$$ It's quite possible that some of you with more experience in CFTs will be able to immediately answer this for me without the context, but here it is anyway: we want to evaluate the partition function on the Riemann manifold $\mathcal{M}_{n}$ which consists of $n$ flat 2D sheets joined together at the branch cut between $u$ and $v$ in the manner shown in Fig 1 in the paper. We do this by modeling the manifold as $n$ disconnected flat sheets with twist fields inserted at the branch points. It turns out that the original partition function is proportional to the correlation function of two twist fields as in Eq. 2.6. Later on, the paper makes use of the correlation function with insertions of the stress-energy tensor, so that generalization (with the equality sign) is crucial. Your help is appreciated! In particular, how does this NOT mean that the partition function in Eq 2.6 is not actually proportional to the two-point function but is simply equal to one? (I am replacing the $\mathcal{O}$'s in (2.7) with one to make this claim) - Reading my question again, I'm slightly amused by how I instinctively use the pronoun "we" instead of the more honest "they". – dbrane May 6 '12 at 0:06 This may be out of my depth but I'll give it a look! – David Zaslavsky♦ May 8 '12 at 22:31 ## 2 Answers I think LHS of eqn 2.7 is normalized, meaning $\frac{1}{Z}\int\mathcal{D}\phi \mathcal{O} exp\left(-S_{E}\left[\phi\right]\right)$ evaluated on $\mathcal{M}_{n}$ If you put $\mathcal{O} =1$, you get 1. But $Z$ itself is proportional to the correlation function of the two primary fields ref: http://arxiv.org/abs/hep-th/0405152 sect IIIA Hope this is useful - You're right! I stupidly overlooked the fact that correlation functions aren't just path integrals with operator insertions but are normalized to the partition function, at least in the treatment by these authors. – dbrane May 10 '12 at 1:49 @dbrane: These formulas define the action of the twist fields. I didn't realize you were only confused about the normalization. – Ron Maimon May 10 '12 at 5:05 The second formula is not a generalization of the first, it is a simple consequence of the definition of the twist fields, like the first. These formulas are just defining the formal path integral for the Riemann surface, and then the "twist field" correlation functions are essentially defined by to reproduce the correlation functions. (I will be explicit about the definition, see below). The paper uses the notation $\tau$ and $\bar{\tau}$ for the twist fields, instead of $\Phi$ and $\bar{\Phi}$, I'll use that. These fields don't appear in the Lagrangian, you don't integrate over them, they don't have dynamics, they do not represent particles of any kind, they aren't anything physical like that at all. These are made-up fields. They act to change the path integral boundary conditions using some cut lines, and they are local operators only in that inserting them depends only on the position where you insert them, not on how you draw the cut lines between them. To see what they are, consider (following Cardy's explanation) n-copies of the given field theory, say a theory of a single scalar $\phi$, and then the n-copies are $\phi_i$ with i from 1 to n. Since you just duplicated the theory n-times with no interaction, you have an action which is the sum of the action of each field separately. This n-fold not-mutually interacting copies have an obvious permutation symmetry, any permutation of the fields is equivalent to any other. This cyclic symmetry is a no-brainer--- the fields are all the same, they each have the same action. Consider the subgroup of cyclic permutations which shift field $i$ to field $i+1$, and suppose we use this symmetry to define a new field theory, in which there is a special horizontal line going between two horizontally separated points A and B. When you cross this line going up, field i turns into field i+1, when you cross going down, it turns back to field i. To say that the field turns into another field is simply saying that the fields are "discontinuously" changing when crossing this line (I put discontinuously in quotes because quantum fields are always discontinuous, but the value at one point in the path integral is dependent on the values nearby, and in this case the field you depend on changes) --- the action in the path integral makes the value of the field $\phi_1$ just below the line related to the value of $\phi_2$ just above the line, and not at all related to the value of $\phi_1$ just above the line. You can imagine this in a simulation of the free field theory. In such a simulation, you pick a lattice site, and replace the value at the site with the average of the same field at the four neighbors, plus a random gaussian value of a certain fixed width (depending on the lattice spacing). Right below the magic line, the up neighbor you average to find the desired value of field 1 is of field number 2, and similarly for field 2, you use field 3 above the line in the average, and so on cyclically. This takes the n-copy theory on $R^2$ to the theory on a Riemann surface (by the definition of a Riemann surface with cuts), by definition. But now notice something nice--- the line position is completely arbitrary, so long as the endpoints stay the same. If you move the line up, so long as it has the same endpoint, you can just locally redefine the lattice values using the cyclic permutation symmetry so that the theory with this crazy shift stays exactly the same. For example, suppose we move all the positions in the middle of the line up by one lattice spacing. For all the points that were previously above, but are now below, simply move the value of $\phi_i$ in any configuration to $\phi_{i-1}$. The cut still does the same thing, the partition function is exactly the same, but the cut is in a completely different position. This is the standard thing in Riemann surfaces--- the position of the cut is arbitrary. What this means is that the theory with the cut can be thought of as a theory with two insertions, at either end of the cut. The property of these insertions is that when you take field $\phi_i$ around the left one counterclockwise, you end up at $\phi_{i+1}$, and if you do the same thing around the right one, you end up at $\phi_{i-1}$. This defines the twist fields $\tau$ and $\bar\tau$, which are acting at A and B. What they do is produce a start and end of a branch cut, and you can link the starts and ends to each other (or to infinity) and you have the same partition function. So to find the correlation function with twist insertions: • put a cut in between the $\tau$'s and $\bar\tau$'s (any which way) • simulate the theory with the cuts. • find the expected value of O. If O=1, then you just get the partition function on the Riemann surface (the partition function of the cut, which is just 1 in the simulation definition above, since in a probabilistic simulation Z=1). If you simulate a nontrivial operator O, you find the correlation function of O in the presence of the cut. This is what Cardy's two formulas mean. The fact that you can interpret the twist fields as local operators is very useful, and allows you to find OPE's and solve the Riemann surface problem. But the formulas you give are simply definitions, and any confusion is in the high level picture defining these. I hope this clears everything up, but it isn't saying anything more or different than Cardy, except in different words. Perhaps this will click better. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "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": 0.9358117580413818, "perplexity_flag": "head"}
http://mathforum.org/mathimages/index.php?title=Volume_of_Revolution&oldid=7780	# Volume of Revolution ### From Math Images Revision as of 15:14, 8 July 2009 by Lmasis1 (Talk \| contribs) Solid of revolution This image shows a solid of revolution Solid of revolution Field: Calculus Created By: Lizah Masis # Basic Description This image shows the solid formed after revolving the region bounded by $y=x^2$, $y=0$,$x=0$ and $x=1$, about the $x$-axis # A More Mathematical Explanation Note: understanding of this explanation requires: *Pre-calculus and elementary calculus [Click to view A More Mathematical Explanation] When finding the volume of revolution of solids, in many cases the problem is not with the calculus, [...] [Click to hide A More Mathematical Explanation] When finding the volume of revolution of solids, in many cases the problem is not with the calculus, but with actually visualizing the solid. To find the volume of a solid like a cylinder, usually we use the formula ${\pi} {r^2} h$. Alternatively we can imagine chopping up the cylinder into thin cylindrical plates, much like slicing up bread, computing the volume of each slice, each of which is ${\Delta x }$ units thick , then summing up the volumes of all the slices. The disc method is much like slicing up bread and computing the volume of each slice http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html This is the idea behind computing the volume of solids whose shapes are complicated. Given a function, we can graph it, then revolve the plane area we get about a fixed axis to obtain a solid called the solid of revolution.In general,the plane area can be bounded by many curves of almost any form. The volume of the solid can then be computed using the disc method. Note: There are other ways of computing the volumes of complicated solids other than the disc method. In the disc method, we imagine chopping up the solid into thin cylindrical plates, each ${\Delta x }$ units thick, calculating the volume of each plate, then summing up the volumes of all plates. For example, let's consider a region bounded by $y=x^2$, $y=0$,$x=0$ and $x=1$ <-------Plotting the graph of this area, If we revolve this area about the x axis ($y=0$), then we get the main image on the right hand side of the page This image shows a plane area being revolved to create a solid http://curvebank.calstatela.edu/volrev/volrev.htm To find the volume of the solid using the disc method: Volume of one disc = ${\pi} y^2{\Delta x}$ where $y$- which is the function- is the radius of the circular cross-section and $\Delta x$ is the thickness of each disc Volume of all dics: Volume of all discs = ${\sum}{\pi}y^2{\Delta x}$, with $X$ ranging from 0 to 1 If we make the slices infinitesmally thick, the Riemann sum becomes the same as: $\int_0^1 {\pi}y^2\,dx ={\pi}\int_0^1 (x^2)^2\, dx$ Evaluating this intergral, ${\pi}\int_0^1 x^4 dx$ =$[{{x^5\over 5} + C\|}_0^1] {\pi}$ =$[{1\over 5} + {0\over 5}] {\pi}$ =${\pi}\over 5$ volume of solid= ${\pi\over 5} units^3$ ## References Bread image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html Revolving image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.	{"extraction_info": {"found_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": 24, "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": 0.8449844121932983, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/48287/earth-moves-how-much-under-my-feet-when-i-jump?answertab=active	Earth moves how much under my feet when I jump? If I'm standing at the equator, jump, and land 1 second later, the Earth does NOT move 1000mph (or .28 miles per second) relative to me, since my velocity while jumping is also 1000mph. However, the Earth is moving in a circle (albeit a very large one), while I, while jumping, am moving in a straight line. How much do I move relative to my starting point because of this? I realize it will be a miniscule amount, and not noticeable in practise, but I'd be interested in the theoretical answer. - 2 Answers My answer: 816.126 nanometers. Someone please review my thought process before I accept my own answer + declare myself awesome. OK, I originally asked this question to save myself from doing some math, but this was clearly a bad idea, as it doubtless angered the Math Gods. I've now come up with an answer, which, of course, agrees with neither of the previous answers (one of which was deleted). I'm using a diagram, which of course makes my answer correct . Generalizing the problem a bit: if I travel distance 't' from the starting point, my new angle relative to the Earth's center is Atan[t/r], where r is the Earth's radius. I call this angle psi. The hypotenuse of SOE is $\sqrt{r^2+t^2}$, so I am hovering above the surface: $$\sqrt{r^2+t^2}-r$$ If the Earth didn't rotate, I'd have moved $\psi r$ (along the Earth's surface) from my original position. If the Earth rotates theta in this time, I've moved $(\psi-\theta)r$. Plugging in some values (assuming earth circumference exactly 40Mm), where $s$ is how long I stay in the air: $$t = 4010^6/86400 s$$ $$\psi = \arctan[t/r] = \arctan [(4010^6/86400s)/(4010^6/2/\pi)] =$$ $$\arctan[\pis/43200]$$ $$\theta = 2s\pi/86400$$ $$\psi-\theta = \arctan(\pi s/43200) - 2 s \pi/86400$$ $$d = 4010^6/2/\pi(\arctan(\pi s/43200) - 2 s \pi/86400)$$ My hovering altitude is: $$\sqrt{(4010^6/2/\pi)^2 + (4010^6/86400 s)^2} - (4010^6/2/\pi)$$ For s=1, I get d=816.126 nanometers, and hover altitude of 1.68cm (the latter seems high). As a note $d$ appears to grow as $s^3$, so the longer you hover, the more you move per second (does this explain helicopter drift?) - I would highly recommend against using numbers in formulas like these, just you $\omega$ or something for angular frequence, etc. The way you wrote in now, makes is pretty unreadible. – Bernhard Jan 6 at 18:28 While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both continuing to undergo a gravitational acceleration towards sun. However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, resulting in so-called tidal forces. The solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about $0.52$ $10^{−6} m/s^2$. You would have to specify where on the equator you are jumping relative to the sun-earth axis, but as a rough estimate, the effect during a 1 second jump will not be larger than 0.5 micrometer. In reality, it will be much less, as a significant part of the effect would be in changing the height of your jump, not in changing your landing position. Also, you have to include tidal forces due to the moon (which are of the same order of magnitude), and effects due to the rotation of the earth around it's own axis. The tidal effects due to the Milky Way can safely be ignored. Edit: The effect due to earth's rotation around it's own axis can be estimated as follows. At the equator, the circumference of the Earth is 40,000 kilometers, and the day is 86,400 seconds long, so the speed of earth's surface at the equator is roughly 460 m/s. When during the jump you spend your time on average roughly one meter above earth's surface, your velocity lags 460 m/s times 1/6.4x10^6 (the denominator corresponding to earth's radius in meters) which equates to about 70 micrometer per second. So, when jumping exactly vertically, after a second you land roughly 70 micrometer west of where you started. - I was referring to the Earth's rotation about its own axis when I said "moving in a circle", not the Earth's revolution around the Sun. – barrycarter Jan 4 at 4:40 Included the rotation effect in an edit. – Johannes Jan 4 at 13:20 Why wouldn't my velocity remain constant at 460m/s, albeit in a different direction (straight line vs circular around the Earth's center)? – barrycarter Jan 4 at 15:18 Your velocity lags in the sense that at radius 6.4x10^6 + 1 m away from earth's center, the rotational speed corresponding to a radius of 6.4x10^6 m is not enough to keep up with earth's rotation. In other words, you keep your speed, and therefore you lack angular speed. – Johannes Jan 4 at 15:28	{"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": 9, "mathjax_display_tex": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.941290557384491, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/17915/the-role-of-resistor-in-e-g-an-and-gate?answertab=active	# The role of resistor in e.g. an AND gate What is the role of the resistor in e.g. an AND gate like this one? : One often sees lots of resistors in electric circuits, but I haven't really understood their role. - This is rather more a technology question that one of physics per se. I'll ask the mods at Electronics.SE if they want it. – dmckee♦ Dec 6 '11 at 16:29 I suppose that is why I can't find the answer in my High School basic physics books. – Anders Hovgaard Dec 6 '11 at 16:38 It's a "pull-up" resistor that pulls the out terminal up to +5 volts. – John McVirgo Dec 6 '11 at 16:48 1 Pull up is lab slang, it does not explain what the resistor does when A or B or both are low: limiting the current through the diodes and acting as a voltage divider between itself and the equivalent resistance of the diodes when conducting. Funny that such oldfashioned RDL pops up again :=) AFAIK RDL was never produced as integrated circuits, integration started with RTL, but I am not absolutely shure. – Georg Dec 6 '11 at 17:55 The question can be analyzed by the methods of physics and Electronics.SE doesn't want it (too basic), so I'm not going to close it, but lets have answers with physics in them. – dmckee♦ Dec 6 '11 at 18:40 ## 2 Answers The resistor provides the logic one when both inputs are high. As far as choosing a resistor, this is a matter of two contradictory requirements. (1) You want the resistor to be low so that the circuit's propagation delay for low to high transitions is short. That is, you want it to drive a capacitive load (for example, wire plus parasitic capacitances) quickly. If the load is C and your time budget for the rise time delay is T, then you want $RC < T$ so pick $R < C/T$. (2) You want the resistor to be high so that the circuit does not dump a lot of current through the diodes when the output is low. If the maximum current your weakest input can sink (at its low voltage level) is I, then you want $IR > V - V_d$ where $V_d$ is the voltage drop of the forward biased diode, so $R> (V-V_d)/I$. This type of AND gate is somewhat primitive in that its low output voltage is a diode drop higher than its low voltage input. Therefore its output has a diode drop less noise immunity than its input (as compared to the same high voltage level). One can only do this so many times before running out of noise margin. This is why modern logic families have less primitive output stages. - Resistors are generally used to dimension electrical devices to the ranges in voltage, current, time constants, what have you, that are needed. In this specific example the resistor is used to dimension the voltage drop in case one of the inputs has low voltage (lower than $V$), so that a current flows from $V$ to the input (it can only flow in this direction because of the way the diodes are connected). Once a current $I$ flows, a drop of $V_{drop}=IR$ will drop across the resistor and lower the output voltage. This ensures the functionality as an AND gate. If both input voltages are high, no or only little current will flow, thus only a small voltage will drop keeping the output voltage high. If one of the too inputs has low voltage, the output oltage will drop. In logical terms high voltages are logical 1s and low voltages are logical 0s. To summarize $V$ is dimensioned to define what "low" and "high" voltages are. and $R$ is dimensioned to define how big the voltage drop is going to be. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9424561858177185, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/165465/do-function-with-the-following-property-have-special-name	Do function with the following property have special name? I'm writing "a structure preserving surjection" way too much when I need to refer a function of the following property: $$Y \subseteq Z, X \subseteq Z. g: Z \to A, g \text{ is some fixed function}.$$ $$\phi : X \to Y, \phi \text{ is a surjection, and } g(\phi(x)) = g(x).$$ Would a category theorist frown if I call $\phi$ a epimorphism? - 1 I expect they would object mightily, unless the functions with this property are right-cancellable. – Arturo Magidin Jul 1 '12 at 22:01 Regular people might object too ^^ – Olivier Bégassat Jul 1 '12 at 22:40 1 Answer If you don't mind making up names (obviously, make sure you define them prominently), then here are two ideas: • The preimage under $g$ of each point in $g(Y)$ contains two points, one of which is mapped to the other by $\phi$. Perhaps you could call $\phi$ a preimage map. • The image of $X\subset Z$ under $g$ is invariant under $\phi$. Perhaps you could call $\phi$ an invariance, or equivariance. (I like the second idea better than the first. In fact, I see some nonce uses of invariance in what appears to be this sense on Google Books (example).) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9244775772094727, "perplexity_flag": "middle"}
http://www.reference.com/browse/Differential+equations+of+addition	Definitions # Differential equations of addition In cryptography, differential equations of addition (DEA) are one of the most basic equations related to differential cryptanalysis that mix additions over two different groups (e.g. additions over GF$\left(2^\left\{32\right\}\right)$ and GF$\left(2\right)$) and where input and output differences are expressed as XORs. ## Examples of Differential Equations of Addition Differential equations of addition (DEA) are of the following form: $\left(x+y\right)oplus\left(\left(xoplus a\right)+\left(yoplus b\right)\right)=c$ where $x$ and $y$ are $n$-bit unknown variables and $a$, $b$ and $c$ are known variables. The symbols $+$ and $oplus$ denote addition modulo $2^n$ and bitwise exclusive-or respectively. The above equation is denoted by $\left(a, b, c\right)$. Let a set $S=\left\{\left(a_i, b_i, c_i\right)\|i$ is an integer less than $k\right\}$ denote a system of $k$ DEA where $k$ is a polynomial in $n$. It has been proved that the satisfiability of an arbitrary set of DEA is in the when a brute force search requires an exponential time. ## Usage of Differential Equations of Addition Solution to an arbitrary set of DEA (either in batch and or in adaptive query model) was due to Souradyuti Paul and Bart Preneel. The solution techniques have been used to attack the stream cipher Helix. ## References • Souradyuti Paul and Bart Preneel, Solving Systems of Differential Equations of Addition, ACISP 2005. Full version (PDF) • Souradyuti Paul and Bart Preneel, Near Optimal Algorithms for Solving Differential Equations of Addition With Batch Queries, Indocrypt 2005. Full version (PDF) • Helger Lipmaa, Johan Wallén, Philippe Dumas: On the Additive Differential Probability of Exclusive-Or. FSE 2004: 317-331.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 18, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.8957198262214661, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/32152/number-of-realizations-of-particular-triad-type?answertab=votes	# Number of realizations of particular triad type Given four types of triads (figure below) their probabilities in a random Bernoulli digraph are as follows: • $T_{003}$: $(1-p)^6$ • $T_{012}$: $6p(1-p)^5$ • $T_{102}$: $3p^2(1-p)^4$ • $T_{111D}$: $6p^3(1-p)^3$ For example, for triad $T_{012}$ there are six realizations of the asymmetric dyad and for triad $T_{102}$ there are three realizations of the mutual dyad. My question is how to properly express the number of realizations of each triad type. - ## 1 Answer If you have, for example, a large random Bernoulli digraph $G$ and are looking for the number of "copies" of a subgraph $H$, we can write something along the lines: "...the number of induced subgraphs of $G$ isomorphic to $H$ is...". This is the typical definition in the network motif literature (which seems to be the modern spin on dyads, triads, etc.). Variations on this theme are not unheard of (e.g. to account for the fact that the above definition counts overlapping "copies" of $H$ separately). Note the word "induced", which is frequently (and, from a graph theory perspective, erroneously) omitted from many publications in this area. For example, the subgraph labelled "003" would occur exactly ${n \choose 3}$ times as a subgraph in any $n$-node network, whereas, it would likely occur fewer times as an induced subgraph. Note: it's also fairly normal (at least in graph theory) to call these random graphs, Erdős-Rényi random graphs, since they are generated in the same spirit as the undirected model. This terminology is used, for example, in: B. Bollobás, O. Riordan, Mathematical results on scale-free random graphs, in Handbook of graphs and networks, 2002. Here's the default reference when using Erdős-Rényi random graphs: P. Erdos, A. Renyi, On the evolution of random graphs, Publ. Math. Inst. Hung. Acad. Sci, Vol. 5 (1960), pp. 17-61. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.9204266667366028, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/161770-solid-revolution.html	# Thread: 1. ## Solid of revolution The area under the curve y= sin x, between x =0 and x = π Show that the volume of the solid formed by rotating this area through 360 about the x-axis is exactly ½(π^2) This is what I have after doing some of the work: V = π (-cos^2x) limits π and 0 I am not sure how to take it any further to obtain the answer. Please help me out guys... 2. Originally Posted by mike789 The area under the curve y= sin x, between x =0 and x = π Show that the volume of the solid formed by rotating this area through 360 about the x-axis is exactly ½(π^2) This is what I have after doing some of the work: V = π (-cos^2x) limits π and 0 I am not sure how to take it any further to obtain the answer. Please help me out guys... your antiderivative is incorrect ... use the power reduction identity that follows. $\displaystyle V = \pi \int_0^\pi \sin^2{x} \, dx$ $\displaystyle V = \int_0^\pi \frac{1-\cos(2x)}{2} \, dx$ now try it again. 3. Originally Posted by skeeter your antiderivative is incorrect ... use the power reduction identity that follows. $\displaystyle V = \pi \int_0^\pi \sin^2{x} \, dx$ $\displaystyle V = \int_0^\pi \frac{1-\cos(2x)}{2} \, dx$ now try it again. Do i need to integrate and substitute the limits? 4. Originally Posted by mike789 Do i need to integrate and substitute the limits? if that's what you mean by evaluating the definite integral using the Fundamental Theorem of Calculus, then yes. #### 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": 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": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9399417042732239, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=621949	Physics Forums ## Tough problem involving algebra 1. The problem statement, all variables and given/known data Original problem asked me to prove $$\int_a^b \! x \, \mathrm{d} x = \frac{b^2-a^2}{2}$$ using Riemann sums. I've already seen a simpler formula using the left side of the rectangles but I'm curious as to how you would manipulate the formula below by hand to get the answer 2. Relevant equations $$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}[\frac{(b-a)i}{n}+a][\frac{b-a}{n}]$$ 3. The attempt at a solution pages and pages of algebra leading nowhere! 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 You can write that thing into $$I = \lim_{n\rightarrow\infty} \sum_{i=1}^{n} [\frac{(b-a)i}{n}+a](\frac{b-a}{n}) = a(b-a) \lim_{n\rightarrow\infty} \sum_{i=1}^{n} \frac{1}{n} + (b-a)^2 \lim_{n\rightarrow\infty} \sum_{i=1}^{n} \frac{i}{n^2}$$ Now, can you evaluate the sums and then take the limits? Could you please explain how you got that? I'm not exactly sure how to evaluate the last limit/sum but does everything else evaluate to just $$a(b-a)+(b-a)^2$$? Recognitions: Gold Member Science Advisor Staff Emeritus ## Tough problem involving algebra Quote by autodidude Could you please explain how you got that? I'm not exactly sure how to evaluate the last limit/sum but does everything else evaluate to just $$a(b-a)+(b-a)^2$$? No, it doesn't. What are $\sum\frac{1}{n}$ and $\sum\frac{1}{n^2}$? I don't know - the only way I know how to interpret the first one right now is to divide 1 into n parts but then I'm not sure what to do so it becomes an infinitely small number if n->infinity. I guess it would be the same for the second one but it gets smaller faster Quote by autodidude Could you please explain how you got that? I'm not exactly sure how to evaluate the last limit/sum but does everything else evaluate to just $$a(b-a)+(b-a)^2$$? The first sum evaluates to 1 but the second does not. http://www.americanscientist.org/iss...-of-reckoning/ might help. consider this instead....$I_{n} = \sum_{i=0}^{n} [\frac{(b-a)i}{n}+a](\frac{b-a}{n}) = \frac{a(b-a)}{n} \sum_{i=0}^{n} 1 + \frac{(b-a)^2}{n^2} \sum_{i=0}^{n} i$ now we know adding 1 n times... is n, how about adding numbers from 1 to n? know any relevant formulas for that? once you simplify the expression to only relying on n, take the limit of $I_n$ as n goes to infinity you have to remember, your sums arent over n, they are over i, so n can be take out front I'm quite sure the formula is $$\frac{n(n+1)}{2}$$ I tried subbing that in and after some algebra, I've got this: $$\frac{(b-a)^2}{2}+\frac{b^2-a^2}{2n}$$ Am I on the right track? Also, when you write the sigmas out without the limits, does n just become some arbitrary integer? And is there any difference between writing a sigma with infinity on top and writing sigma with an n but with lim(h->infinity)? Quote by autodidude I'm quite sure the formula is $$\frac{n(n+1)}{2}$$ I tried subbing that in and after some algebra, I've got this: $$\frac{(b-a)^2}{2}+\frac{b^2-a^2}{2n}$$ Am I on the right track? Yeah except you of course should have b2-a2, not (b-a)2. Then just take limit n→∞. Quote by autodidude Also, when you write the sigmas out without the limits, does n just become some arbitrary integer? And is there any difference between writing a sigma with infinity on top and writing sigma with an n but with lim(h->infinity)? Infinite sums are always defined as that limit so there's no worries there. You want to find the n:th partial sum, and then take the limit. Thread Tools \| \| \| \| \|------------------------------------------------------\|----------------------------------\|---------\| \| Similar Threads for: Tough problem involving algebra \| \| \| \| Thread \| Forum \| Replies \| \| \| Precalculus Mathematics Homework \| 8 \| \| \| General Math \| 6 \| \| \| Calculus & Beyond Homework \| 1 \| \| \| Introductory Physics Homework \| 2 \| \| \| Introductory Physics 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": 4, "mathjax_display_tex": 10, "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": 0.9286440014839172, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/22080/does-a-ski-racer-with-a-larger-mass-have-an-advantage	# Does a ski racer with a larger mass have an advantage? Does a ski racer with a greater mass have an advantage over a racer with a lesser mass? If mass of one racer is 54 kg and the mass of a more slender racer is 44 kg I know the speed at which they descend should be equal if they were to fall in a vacuum. How does friction force, air resistance and momentum play a part in determining the advantage or disadvantage of larger mass in ski racing? - ## 2 Answers Viscous drag will tend to favor the heavier person, since surface area scales slower than mass, i.e. $A \propto r^2$, $M \propto r^3$, $F_D / F_G \propto M^{-1/3}$. But I would generally expect this effect to be weak except at very high speeds. Friction would cancel out if you were skiing on ice, however in soft snow the heavier person would tend to sink deeper in the snow and that extra compaction of the snow is going to take away more energy. Unless the heavier skier has larger skis, i.e. it really would depend on the loading per area of the skis. - Not exactly. Friction makes no difference, as it is a force proportional to mass ($\mu_k mg\cos\theta$). Since $F=ma$, the $m$ cancels off and we get that the effect of friction on acceleration is constant for different masses. The same goes for gravity. Recall Galileo's famous drop-two-balls-from-the-leaning-tower experiment. There are only two forces left. Both depend upon the size and shape of the body and not the mass. Since we have to divide by mass to get acceleration, out of two people of the same shape, the heavier one gets the benefit. This can be said to be due to momentum/inertia. The two forces are buoyant force and viscous drag. Buoyant force is directly proportional to volume, and is pretty negligible. Viscous drag depends upon the size and shape and lots of tiny things. Generally, aheavier person will be broader as well, so this one can go either way. All in all, it really doesn't matter what a skier's mass is, except for overweight people (too much drag I would think). The technique matters much more. - 1 Buoyancy? That's the same as gravity. – Bernhard Mar 8 '12 at 19:09 @Bernhard well, kind of (i consider it to be caused by the gravitationally induced pressure difference). Anyways, it has a different direction and behaviour, so i'll keep it seperate. – Manishearth♦ Mar 9 '12 at 1:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9509871006011963, "perplexity_flag": "middle"}
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(the length of vars is also variable) ...	{"extraction_info": {"found_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": 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": 0.8953995704650879, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/29082/would-time-freeze-if-you-can-travel-at-the-speed-of-light	# Would time freeze if you can travel at the speed of light? I read with interest about Einstein's `Theory of Relativity` and his proposition about the speed of light being the speed limit for anything with mass. So, if I were to travel in a spacecraft at the speed of light, would I freeze and stop moving? Would the universe around me freeze and stop moving? Who would the time stop for? - 3 Since you can't move at the speed of light, you should phrase this question about travelling at less than the speed of light. – Ron Maimon May 28 '12 at 7:23 even so time would not change for you individually only if you returned to the place that you started would you realize that you have had a different experiance of time without returning to compare your time/ageing with the origin of your travels you would not experiance for instance slow motion or any such observable discrepency – Argus May 28 '12 at 18:07 Does the "place" actually have anything to do with it. I've seen this in documentaries, well the "3" I've watched, and they use this idea but isn't the idea to do with relative velocity? i.e. if you went out into space for 1 light year and then returned all at the speed of light, and I stayed in earth orbit going round and round at the speed of light for 2 years there wouldn't be any relative difference between you or I would there? Clearly u would have had the better time but age wise = same? – rism Aug 5 '12 at 4:21 ## 7 Answers Yes, I agree with David. If somehow, you were able to travel at the speed of light, it would seem that 'your time' would not have progressed in comparison to your reference time once you returned to 'normal' speeds. This can be modeled by the Lorentz time dilation equation: $$T=\frac{T_0}{\sqrt{1 - (v^2 / c^2)}}$$ When traveling at the speed of light ($v=c$), left under the radical you would have 0. This answer would be undefined or infinity if you will (let's go with infinity). The reference time ($T_0$) divided by infinity would be 0; therefore, you could infer that time is 'frozen' to an object traveling at the speed of light. - – Ignacio Vazquez-Abrams May 28 '12 at 19:43 4 I'm kind of wary about using the phrase "if you were able to travel at the speed of light," because you can't, and in my experience when you say anything less than "this is unequivocally not possible" people start to get ideas that they really shouldn't. Although this does get the point across. – David Zaslavsky♦ May 29 '12 at 17:49 You can't travel at the speed of light. So it's a meaningless question. The reason some people will say that time freezes at the speed of light is that it's possible to take two points on any path going through spacetime at less than the speed of light and calculate the amount of time that a particle would experience as it travels between those points along that path. The calculation is $$\Delta\tau^2 = \Delta t^2 - \frac{1}{c^2}(\Delta x^2 + \Delta y^2 + \Delta z^2)$$ where $\Delta\tau$ is the amount of time experienced by the traveling particle, and the other $\Delta$'s are the differences in space and time coordinates between the two points as measured by an external observer. If you take this same calculation and blindly apply it to a path which is at the speed of light, you get $\Delta\tau = 0$. - nice answer, but probably not at the level of the op – Larry Harson May 28 '12 at 16:41 Hey, nothing wrong with exposing people to a little math ;-) – David Zaslavsky♦ May 28 '12 at 17:00 Photons travel at almost fully that speed all the time, except in glass, air, not quite empty space. They seem to never get bored. – roadrunner66 Mar 12 at 7:50 It is not that time "freezes" or "resumes", but rather that an object moving at a certain speed experiences time differently. Time still proceeds at one second per second regardless, and an object moving at light speed will still take one year to cover one light-year; what changes is that for that object, time appears to pass much more slowly. - Velocity is relative, so it doesn't matter if you're "travelling" at some speed relative to something, or something is travelling at some velocity relative to you - the effects are the same. Right now you have objects in the universe travelling at a wide range of velocities relative to you. If you decided to change your speed to close to the speed of light compared to what it is now, you will find that there is still the same range of velocities of objects relative to you. That's because objects that were travelling close to c in the direction of your increase will have slowed down, and objects that were travelling in the opposite direction will have increased their velocity. However, you will also find that as objects increase their speed relative to you, the sequence of events there slow down, and that includes the running of their clocks from your view point, which approaches zero as their speed approaches the speed of light. - I'm a total novice but i watched NOVA's "Brian Greene's Fabric of the Cosmos" on YouTube and as i understand it, "your" space and time share a finite amount of energy at any given moment called Space-Time. So the finite amount energy available at any given moment can be consumed "either" by motion through space or ticks of some clock. But the more you use that energy for one aspect of SpaceTime i.e. motion or ticks, the less there is for the other... kinda like the way your computer seems to run slower the more programs you run at once i.e. a finite amount of processing power is available at any given moment to be distributed among applications. The upper limit of motion being the speed of light. So the closer you travel to the speed of light the more energy for a given moment is assigned/consumed by motion, leaving less and less energy for time to tick (relative to someone else). Therefore the slower time goes (relatively speaking). So the faster you go, the slower time "appears" to go relative to a slower observer because most of your space time energy is being consumed by motion whereas more of theirs is being used for time to tick. - You seem to have gotten a garbled message here. It is true that all inertial observed can agree on the "interval" between to points in space-time and that the interval is calculated by using space and time components with opposite signs, but the rest is a bit of a mess... – dmckee♦ Aug 5 '12 at 5:23 The time wouldn't freeze. Instead, all events in the world will happen at the same time and place (from the viewpoint of the observer travelling at the light speed). It would be better to say that the world (i.e. space & time) would collapse into single point. - Hibernation is one way to freeze time,that is to arrive at future date,the natural way to time travel. Linear or circular movement at very high speed is the scientific theory to reach future. - ## protected by Qmechanic♦May 4 at 12:58 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": 5, "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": 0.9703965783119202, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations?answertab=active	# Is there a general formula for solving 4th degree equations? There is a general formula for solving quadratic equations, namely the Quadratic Formula. For third degree equations of the form $ax^3+bx^2+cx+d=0$, there is a set of thee equations: one for each root. Is there a general formula for solving equations of the form $ax^4+bx^3+cx^2+dx+e=0$ ? How about for higher degrees? If not, why not? - Could you change the title to "Is there a general formula for solving 4th degree polynomial equations" or "Is there a general formula for solving quartic equations?" – user126 Aug 5 '10 at 23:40 9 Put `ax^4+bx^3+cx^2+dx +e = 0` in wolfram alpha (wolframalpha.com) and then sit back and watch the fireworks! – ja72 Aug 16 '11 at 2:00 ## 7 Answers There is, in fact, a general formula for solving quartic (4th degree polynomial) equations. As the cubic formula is significantly more complex than the quadratic formula, the quartic formula is significantly more complex than the cubic formula. Wikipedia's article on quartic functions has a lengthy process by which to get the solutions, but does not give an explicit formula. Beware that in the cubic and quartic formulas, depending on how the formula is expressed, the correctness of the answers likely depends on a particular choice of definition of principal roots for nonreal complex numbers and there are two different ways to define such a principal root. There cannot be explicit algebraic formulas for the general solutions to higher-degree polynomials, but proving this requires mathematics beyond precalculus (it is typically proved with Galois Theory now, though it was originally proved with other methods). This fact is known as the Abel-Ruffini theorem. Also of note, Wolfram sells a poster that discusses the solvability of polynomial equations, focusing particularly on techniques to solve a quintic (5th degree polynomial) equation. This poster gives explicit formulas for the solutions to quadratic, cubic, and quartic equations. edit: I believe that the formula given below gives the correct solutions for x to $ax^4+bx^3+cx^2+d+e=0$ for all complex a, b, c, d, and e, under the assumption that $w=\sqrt{z}$ is the complex number such that $w^2=z$ and $\arg(w)\in(-\frac{\pi}{2},\frac{\pi}{2}]$ and $w=\sqrt[3]{z}$ is the complex number such that $w^3=z$ and $\arg(w)\in(-\frac{\pi}{3},\frac{\pi}{3}]$ (these are typically how computer algebra systems and calculators define the principal roots). Some intermediate parameters $p_k$ are defined to keep the formula simple and to help in keeping the choices of roots consistent. Let: \begin{align} p_1&=2c^3-9bcd+27ad^2+27b^2e-72ace \\\\ p_2&=p_1+\sqrt{-4(c^2-3bd+12ae)^3+p_1^2} \\\\ p_3&=\frac{c^2-3bd+12ae}{3a\sqrt[3]{\frac{p_2}{2}}}+\frac{\sqrt[3]{\frac{p_2}{2}}}{3a} \end{align} $\quad\quad\quad\quad$ \begin{align} p_4&=\sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+p_3} \\\\ p_5&=\frac{b^2}{2a^2}-\frac{4c}{3a}-p_3 \\\\ p_6&=\frac{-\frac{b^3}{a^3}+\frac{4bc}{a^2}-\frac{8d}{a}}{4p_4} \end{align} Then: $$\begin{align} x&=-\frac{b}{4a}-\frac{p_4}{2}-\frac{\sqrt{p_5-p_6}}{2} \\\\ \mathrm{or\ }x&=-\frac{b}{4a}-\frac{p_4}{2}+\frac{\sqrt{p_5-p_6}}{2} \\\\ \mathrm{or\ }x&=-\frac{b}{4a}+\frac{p_4}{2}-\frac{\sqrt{p_5+p_6}}{2} \\\\ \mathrm{or\ }x&=-\frac{b}{4a}+\frac{p_4}{2}+\frac{\sqrt{p_5+p_6}}{2} \end{align}$$ (These came from having Mathematica explicitly solve the quartic, then seeing what common bits could be pulled from the horrifically-messy formula into parameters to make it readable/useable.) - You beat me to the answer by a few seconds :). – Akhil Mathew Jul 27 '10 at 18:45 +1, I like the poster a lot. thx for the link. – Chao Xu Jul 27 '10 at 18:59 The computational effort is simplified greatly if you "depress" the quartic (i.e. remove the cubic term through a change of variables) by appealing to Vieta's formulae (the mean of the roots is -b/(4a)) first. If you will notice, all the roots of the original equation have a -b/(4a) term added in to compensate for this preliminary translation. – J. M. Aug 6 '10 at 0:43 @J. Mangaldan: Absolutely. My goal (in the edit) was to create a fully-general formula that could be applied straight away; it is not at all illustrative of how to get such a formula. (Your observation is true for the nth-degree polynomial equation formula for n=2, 3, and 4: each formula has a -b/(na) term common to every solution corresponding to the depression.) – Isaac Aug 6 '10 at 0:48 The formula given above is not correct. One can try some examples: x^4-1,x^4+x^2 and x^4+5*x^2+4. – user14620 Aug 15 '11 at 15:02 show 1 more comment http://planetmath.org/encyclopedia/QuarticFormula.html This has the whole thing written out... Might take a few hours to actually input it. - Yes. As an answer I will use a shorter version of this Portuguese post of mine, where I deduce all the formulae. Suppose you have the general quartic equation (I changed the notation of the coefficients to Greek letters, for my convenience): $$\alpha x^{4}+\beta x^{3}+\gamma x^{2}+\delta x+\varepsilon =0.\tag{1}$$ If you make the substitution $x=y-\frac{\beta }{4\alpha }$, you get a reduced equation of the form $$y^{4}+Ay^{2}+By+C=0\tag{2},$$ with $$A=\frac{\gamma }{\alpha }-\frac{3\beta ^{2}}{8\alpha ^{2}},$$ $$B=\frac{\delta }{\alpha }-\frac{\beta \gamma }{2\alpha ^{2}}+\frac{\beta }{ 8\alpha },$$ $$C=\frac{\varepsilon }{\alpha }-\frac{\beta \delta }{4\alpha ^{2}}+\frac{ \beta ^{2}c}{16\alpha ^{3}}-\frac{3\beta ^{4}}{256\alpha ^{4}}.$$ After adding and subtracting $2sy^{2}+s^{2}$ to the LHS of $(2)$ and rearranging terms, we obtain the equation $$\underset{\left( y^{2}+s\right) ^{2}}{\underbrace{y^{4}+2sy^{2}+s^{2}}}-\left[ \left( 2s-A\right) y^{2}-By+s^{2}-C\right] =0. \tag{2a}$$ Then we factor the quadratic polynomial $$\left(2s-A\right) y^{2}-By+s^{2}-C=\left(2s-A\right)(y-y_+)(y-y_-)$$ and make $y_+=y_-$, which will impose a constraint on $s$ (equation $(4)$). We will get: $$\left( y^{2}+s+\sqrt{2s-A}y-\frac{B}{2\sqrt{2s-A}}\right) \left( y^{2}+s+% \sqrt{2s-A}y+\frac{B}{2\sqrt{2s-A}}\right) =0,$$ $$\tag{3}$$ where $s$ satisfies the resolvent cubic equation $$8s^{3}-4As^{2}-8Cs+\left( 4Ac-B^{2}\right) =0.\tag{4}$$ The four solutions of $(2)$ are the solutions of $(3)$: $$y_{1}=-\frac{1}{2}\sqrt{2s-A}+\frac{1}{2}\sqrt{-2s-A+\frac{2B}{\sqrt{2s-A}}}, \tag{5}$$ $$y_{2}=-\frac{1}{2}\sqrt{2s-A}-\frac{1}{2}\sqrt{-2s-A+\frac{2B}{\sqrt{2s-A}}} ,\tag{6}$$ $$y_{3}=-\frac{1}{2}\sqrt{2s-A}+\frac{1}{2}\sqrt{-2s-A-\frac{2B}{\sqrt{2s-A}}} ,\tag{7}$$ $$y_{4}=-\frac{1}{2}\sqrt{2s-A}-\frac{1}{2}\sqrt{-2s-A-\frac{2B}{\sqrt{2s-A}}} .\tag{8}$$ Thus, the original equation $(1)$ has the solutions $$x_{k}=y_{k}-\frac{\beta }{4\alpha }.\qquad k=1,2,3,4\tag{9}$$ Example: $x^{4}+2x^{3}+3x^{2}-2x-1=0$ $$y^{4}+\frac{3}{2}y^{2}-4y+\frac{9}{16}=0.$$ The resolvent cubic is $$8s^{3}-6s^{2}-\frac{9}{2}s-\frac{101}{8}=0.$$ Making the substitution $s=t+\frac{1}{4}$, we get $$t^{3}-\frac{3}{4}t-\frac{7}{4}=0.$$ One solution of the cubic is $$s_{1}=\left( -\frac{q}{2}+\frac{1}{2}\sqrt{q^{2}+\frac{4p^{3}}{27}}\right) ^{1/3}+\left( -\frac{q}{2}-\frac{1}{2}\sqrt{q^{2}+\frac{4p^{3}}{27}}\right) ^{1/3}-\frac{b}{3a},$$ where $a=8,b=-6,c=-\frac{9}{2},d=-\frac{101}{8}$ are the coefficients of the resolvent cubic and $p=-\frac{3}{4},q=-\frac{7}{4}$ are the coefficients of the reduced cubic. Numerically, we have $s_{1}\approx 1.6608$. The four solutions are : $$x_{1}=-\frac{1}{2}\sqrt{2s_{1}-A}+\frac{1}{2}\sqrt{-2s_{1}-A+\frac{2B}{% \sqrt{2s-A}}}-\frac{\beta }{4\alpha },$$ $$x_{2}=-\frac{1}{2}\sqrt{2s_{1}-A}-\frac{1}{2}\sqrt{-2s_{1}-A+\frac{2B}{% \sqrt{2s-A}}}-\frac{\beta }{4\alpha },$$ $$x_{3}=-\frac{1}{2}\sqrt{2s_{1}-A}+\frac{1}{2}\sqrt{-2s_{1}-A-\frac{2B}{% \sqrt{2s-A}}}-\frac{\beta }{4\alpha },$$ $$x_{4}=-\frac{1}{2}\sqrt{2s_{1}-A}-\frac{1}{2}\sqrt{-2s_{1}-A-\frac{2B}{% \sqrt{2s-A}}}-\frac{\beta }{4\alpha },$$ with $A=\frac{3}{2},B=-4,C=\frac{9}{16},D=\frac{9}{16}$. Numerically we have $x_{1}\approx -1.1748+1.6393i$, $x_{2}\approx -1.1748-1.6393i$, $x_{3}\approx 0.70062$, $x_{4}\approx -0.35095$. Another method is to expand the LHS of the quartic into two quadratic polynomials, and find the zeroes of each polynomial. However, this method sometimes fails. Example: $x^{4}-x-1=0$. If we factor $x^{4}-x-1$ as $x^{4}-x-1=\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right)$ expand and equate coefficients we will get two equations, one of which is $-1/c-c^{2}\left( 1+c^{2}\right) ^{2}+c=0$. This is studied in Galois theory. The general quintic is not solvable in terms of radicals, as well as equations of higher degrees. - What has not been mentioned thus far is that one can in fact use any number of "auxiliary cubics" in the solution of the quartic equation. Don Herbison-Evans, in this page (adapted from his technical report), mention five such possible auxiliary cubics. Given the quartic equation $$x^4 + ax^3 + bx^2 + cx + d = 0$$ the five possible auxiliary cubics are referred to in the document as Christianson-Brown: $$y^3 +\frac{4a^2b - 4b^2 - 4ac + 16d - \frac34a^4}{a^3 - 4ab + 8c}y^2 + \left(\frac3{16}a^2 - \frac{b}{2}\right)y + \frac{ab}{16} - \frac{c}{8} + \frac{a^3}{64} = 0$$ Descartes-Euler-Cardano: $$y^3 + \left(2b - \frac34 a^2\right)y^2 + \left(\frac3{16}a^4 - a^2b + ac + b^2 - 4d\right)y + abc - \frac{a^6}{64} + \frac{a^4b}{8} - \frac{a^3c}{4} - \frac{a^2b^2}{4} - c^2 = 0$$ Ferrari-Lagrange $$y^3 + by^2 + (ac - 4d)y + a^2d + c^2 - 4bd = 0$$ Neumark $$y^3 - 2by^2 + (ac + b^2 - 4d)y + a^2d - abc + c^2 = 0$$ Yacoub-Fraidenraich-Brown $$(a^3 - 4ab + 8c)y^3 + (a^2b - 4b^2 + 2ac + 16d)y^2 + (a^2c - 4bc + 8ad)y + a^2d - c^2 = 0$$ See the page for how to obtain the quadratics that will yield the solutions to the original quartic equation from a root of the auxiliary cubic. Let me also mention this old ACM algorithm in Algol. Netlib has a C implementation of that algorithm. - To your question, "if not, why not", I could say this (which I think goes a little further that the previous answers). A polynomial $p(x) \in \mathbb{Q}[x]$ of degree $n$ has a Galois group $G = G(p)$ attached to it. This is a subgroup of the symmetric group $S_{n},$ and this identification comes about because $G$ is a group of permutations of the roots of $p(x)$. The equation $p(x)$ is solvable by radicals if and only if $G$ is a solvable group, which is a key theorem of Galois theory. When $n \leq 4$, the symmetric group $S_n$ is itself a solvable group, so all its subgroups are solvable, and the group $G$ must be solvable. As remarked in earlier answers, when $n \geq 5,$ the group $S_n$ is never solvable. This does not mean that the polynomial $p(x)$ is never solvable by radicals, but (depending what its Galois group is), we can not be sure that it is (and there are explicit examples when it is not for each $n \geq 5$). I do make the point though that the solvability or otherwise of $p(x)$ by radicals really depends on the factorization of $p(x)$ as a product of irreducible polynomials, rather than just the degree of $p(x)$. If $p(x)$ is a product of irreducible polynomials which each have degree at most 4, then its Galois group is solvable, so $p(x)$ is solvable by radicals. - Regarding the inability to solve the quintic, this is sort-of true and sort-of false. No, there is no general solution in terms of $+$, $-$, $\times$ and $\div$, along with $\sqrt[n]{}$. However, if you allow special theta values (a new operation, not among the standard ones!) then yes, you can actually write down the solutions of arbitrary polynomials this way. Also, you can do construct lengths equal to the solutions by intersecting lower degree curves (for a quintic, you can do so with a trident and a circle.) - 2 "However, if you allow special theta values (a new operation, not among the standard ones!) then yes, you can actually write down the solutions of arbitrary polynomials this way." Do you know of a reference which expands on this point? – Zach Conn Dec 8 '10 at 20:35 1 – J. M. Aug 16 '11 at 4:47 Yes, there is a quartic formula. There is no such solution by radicals for higher degrees. This is a result of Galois theory, and follows from the fact that the symmetric group $S_5$ is not solvable. It is called Abel's theorem. In fact, there are specific fifth-degree polynomials whose roots cannot be obtained by using radicals from $\mathbb{Q}$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 62, "mathjax_display_tex": 30, "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": 0.9155659675598145, "perplexity_flag": "head"}
http://mathoverflow.net/questions/37647/the-dold-thom-theorem-for-infinity-categories/37898	## The Dold-Thom theorem for infinity categories? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $\mathcal{M}$ denote the category of finite sets and monomorphisms, and let $\mathcal T$ denote the category of based spaces. For a based space $X \in \mathcal T$, one has a canonical funtor $S_X : \mathcal M \rightarrow \mathcal T$ defined by $\{n\} \mapsto X^n$. The definition on morphisms is to insert basepoints on the factors which are not in the image of a given monomorphism. As is well know, the homotopy groups of $\mathrm{colim} S_X = SP^\infty X$ give the homology of $X$ (this is the Dold-Thom theorem), and the homotopy groups of $\mathrm{hocolim} S_X = SP^\infty_h X$ given the stable homotopy of $X$. Is there a model for $SP^\infty X$, the ordinary infinite symmetric product, as a homotopy colimit as opposed to a categorical colimit? The motivation for this question comes from thinking about $\infty$-categories. In an $\infty$-category, one does not really have a good notion (at least not one that I am aware of) of strict categorical colimits. So I'm wondering if there is, nonetheless, some easily defined functor on the $\infty$-category of spaces which will let us calculate ordinary homology. In short, is there any $\infty$-categorical analog of the Dold-Thom theorem? Update: Following up on André's remark it seems using the orbit category is heading in the right direction, at least for the $n$-th approximations. I'll just quickly sketch what I have so far: Let $\mathcal O(\Sigma_n)$ denote the orbit category. The objects are the homogeneous (discrete) spaces $\Sigma_n/H$ (with left actions) as $H$ runs over all the subgroups of $\Sigma_n$, and the morphisms are the $\Sigma_n$-equivariant maps. There is a canonical functor $$\Sigma_n \rightarrow \mathcal O(\Sigma_n)^{op}$$ where we regard $\Sigma_n$ as a category with one object as usual. Given a $\Sigma_n$ space $X$, right Kan extension along this inclusion produces a $\mathcal O(\Sigma_n)^{op}$ diagram $\tilde X$ defined by $$\tilde X(\Sigma_n/H) = X^H$$ It turns out that the above inclusion is final so that it induces an isomorphism of colimits. Hence $\mathrm{colim}_{\mathcal O(\Sigma_n)} \tilde X \cong X_{\Sigma_n}$, i.e., the coinvariants. It's also not hard to see that the undercategories are copies of $B\Sigma_n$, hence not contractible, so we don't expect an equivalence of homotopy colimits, which is good. On the other hand, I can now show that when $X$ is discrete, the canonical map $$\mathrm{hocolim} \tilde X \rightarrow \mathrm{colim} \tilde X$$ is an equivalence. My methods here do not generalize to all spaces, so if someone has a reference for why this is true in general, that would be much appreciated. (I think something like this must appear in May's book on equivariant homotopy theory if it's true, but I did not have it available this weekend.) The remaining part would be to let $n \rightarrow \infty$, but somehow this seems like it should not be too bad. (Something like: make a functor $\mathcal M \rightarrow \mathcal Cat$ by $n \mapsto \mathcal O(\Sigma_n)$. Take the Grothendieck construction. Some natural diagram on this category might give the right answer.) - 1 I only know how to go the other direction. There's a model category with a monoidal Quillen equivalence to the category of spaces so that the infinite symmetric product gives the stable homotopy groups rather than the homology. (This is forthcoming work of Sagave and Schlichtkrull.) I'm not sure what really makes symmetric products of ordinary spaces behave as nicely as they do. – Tyler Lawson Sep 3 2010 at 20:01 2 This is an interesting question, and I'd also like to know an answer! – Charles Rezk Sep 3 2010 at 23:24 1 Let's look at the n-th finite approximation of the space you care about, i.e., the n-th symmetric power of X. That's the strict coinvariants of S_n acting on X^n. If you want a homotopy invariant description of those coinvariants, you should start with a model of X^n that lives in the oo-category of S_n-spaces. By Elmendorff's thm, that category is equivalent to the oo-category of O_{S_n} - spaces (diagrams indexed y the orbit category of S_n). I forget what the homotopy invariant desrption of strict coinvariants is in terms of O_{S_n} - spaces... – André Henriques Sep 4 2010 at 3:28 @André: After mulling it over a bit last night, I came to a similar conclusion about somehow getting orbit categories into the picture, so this seems like a promising thing to look at. I'll report back if I find anything. – Eric Finster Sep 4 2010 at 9:02 @Eric: "As is well know...the homotopy groups of hocolim `$SX = SP^\inft_h X$` give the stable homotopy of X." Isn't this a rather recent theorem of Christian Schlichtkrull? (Algebr. Geom. Topol. 7 (2007), 1963--1977) – Dan Ramras Sep 5 2010 at 17:16 show 1 more comment ## 1 Answer It so happens that Emmanuel Dror Farjoun is visiting the EPFL this week. I figured I'd ask him about this problem at lunch today. What a coincidence! He proved exactly this statement using the exact same techniques. In fact, the construction of $SP^n$ as a homotopy colimit is the subject of Chapter 4 in "Cellular Spaces, Null Spaces, and Homotopy Localization," Lecture Notes in Mathematics, 1622. It turns out, the idea works more generally so that we can always replace strict colimits with homotopy colimits: define an orbit on a category $\mathcal C$ to be a functor $O : \mathcal C \rightarrow \mathcal Set$ such that $\mathrm{colim}_{\mathcal C} O \cong *$. There is a category of such functors which we call the orbit category of $\mathcal C$, denoted $\mathcal O(\mathcal C)$. The Yoneda embedding factors through $\mathcal O(\mathcal C)$, and the right Kan extension along this inclusion always results in a "free" diagram. I still want to play around with the construction a bit to see if there are any wrinkles with $n \rightarrow \infty$, and if I can use this to give easy calculations of homology in the $\infty$-category $\mathcal S$, but I think it's safe to say at this point that answer to my question is yes. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9286248087882996, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/258654/cyclic-vector-exists-for-symmetric-operator-iff-there-no-repeated-eigenvalues?answertab=oldest	# cyclic vector exists for symmetric operator iff there no repeated eigenvalues Considering a symmetric operator $A$ acting on a finite dimensional Hilbert space $H$, we say $x\in H$ is a cyclic vector for $A$ if the set of finite linear combinations of $\{A^n x:n=0,1,2,...\}$ is equal to $H$. I am looking for a proof of whether $A$ must have a cyclic vector iff $A$ has no repeated eigenvalues. Hints are welcomed. - 1 It would be helpful if you comment on what have you attempted and tried so far. – Javier Álvarez Dec 14 '12 at 12:55 Up to isomorphism you can assume that the Hilbert space is $\mathbb{C}^n$ and the operator is represented by a diagonal matrix $$\begin{bmatrix} \lambda_1 && \\ & \ddots & \\ &&\lambda_n \end{bmatrix}$$This way we are reduced to a problem in linear algebra. Have you tried this already? – Giuseppe Negro Dec 14 '12 at 13:09 1 If $H$ is of finite dimension, then "dense" is a strange term to use; $H$ itself is the only dense subspace. – Marc van Leeuwen Dec 14 '12 at 13:10 1 Is your Hilbert space real or complex? When you say that $A$ is symmetric do you mean that $A$ is self-adjoint or that some matrix representing is equal to its (regular, non-conjugate) transpose? – brom Dec 14 '12 at 15:39 ## 1 Answer I will assume that we are talking a real Hilbert space. Everything works the same in the complex case if we use "hermitian" instead of "symmetric". Note that, from your definition, the condition "$A$ has a cyclic vector $x$" is equivalent to $$\tag{1} H=\{p(A)x:\ p\in\mathbb R[t]\}.$$ If we fix an orthonormal basis for $H$, the fact that $A$ is symmetric guarantees that $A$ is diagonalizable, i.e. $A=VDV^T$, with $V$ orthogonal and $D$ diagonal containing the eigenvalues of $A$ (counting multiplicities) in its diagonal. As $p(A)=Vp(D)V^T$, condition $(1)$ translates into $$\tag{2} H=\{p(D)y:\ p\in\mathbb R[t]\},$$ where $y=V^Tx$. The key observation is that $\{p(D):\ p\in\mathbb{R}\}$ is a vector space (since $\mathbb R[t]$ is), and that $$\tag{3} \dim\{p(D):\ p\in\mathbb{R}\}=\text{the number of distinct eigenvalues of }A.$$ This can be seen from the fact that $p(D)$ is the diagonal matrix with diagonal $p(D_{11}),\ldots,p(D_{nn})$, and that given distinct numbers $\lambda_1,\ldots,\lambda_k$, the set $\{(p(\lambda_1),\ldots,p(\lambda_k)):\ p\in\mathbb R[t]\}$ equals $\mathbb R^k$ Now, if we have equality in $(2)$, then the set on the right has dimension $n$, and by $(3)$ we get that $A$ has $n$ distinct eigenvalues. Conversely, if $A$ has $n$ distinct eigenvalues, then choosing $y$ to be the vector with all its coordinates equal to one (i.e. the sum of the unit eigenvectors for $A$), we get that the set on the right-hand-side on $(2)$ has dimension $n$, and so it equals all of $H$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 44, "mathjax_display_tex": 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": 0.3, "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": 0.9204919338226318, "perplexity_flag": "head"}
http://mathhelpforum.com/math-topics/52799-circles-please-help.html	# Thread: 1. ## Circles - Please Help Two circles touch internally at a point A and the smaller of the two circles passes through O, the centre of the larger circle. AB is any chord of the larger circle, cutting the smaller circle S. The tangents to the larger circle at A and B meet at a point T. Prove: i) AB is bisected at S. ii) O, S and T are collinear. 2. Hello, xwrathbringerx! Two circles touch internally at a point $A$ and the smaller of the two circles passes through $O$, the centre of the larger circle. $AB$ is any chord of the larger circle, cutting the smaller circle $S.$ The tangents to the larger circle at $A$ and $B$ meet at a point $T.$ Prove: a) $AB$ is bisected at $S.$ b) $O, S\text{ and }T$ are collinear. Draw $OT.$ Draw radii $OA = OB = r$ $TA = TB$ Tangents to a circle from an extrenal point are equal. Points $O$ and $T$ are equidistant from points $A$ and $B.$ . . Hence, $OT$ is the perpendicular bisector of $AB.$ $OA$ is a diameter of the small circle. $\angle OSA$ is inscribed in a semicircle: . $\angle OSA = 90^o$ Hence, $S$ lies on $OT.$ And the two proofs follow . . . 3. Can I simply say OA is a diameter of the small circle?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 24, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.8869450092315674, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/194865/inequality-3-leftia2ib2ic2-right-geq-ab2bc2ca2	# Inequality. $3\left(IA^2+IB^2+IC^2\right) \geq AB^2+BC^2+CA^2$ (Korea 1998) Let $I$ be the incenter of a triangle $ABC$.Prove that: $$3\left(IA^2+IB^2+IC^2\right) \geq AB^2+BC^2+CA^2.$$ Thanks :) - We prefer to avoid subjective titles. – Peter Tamaroff Sep 12 '12 at 19:26 Presumably you're looking at this in order to prepare for an upcoming competition. What use is it, then, if others solve the problem for you? – Yuval Filmus Sep 12 '12 at 23:43 @YuvalFilmus I won't participate to a competition. It is only my pleasure to solve inequalities. If you want you can give me an idea. – Iuli Sep 13 '12 at 6:46 ## 1 Answer First note that $IA^2=(p-a)bc/p$ and similarly for $IB, IC$. Substitute to get$$3abc(\frac{p-a}{ap}+\frac{p-b}{bp}+\frac{p-c}{cp})-a^2-b^2-c^2>=0$$ Replace $p$ with $(a+b+c)/2$ and get rid of the fractions to arrive at:$$-a^3 + 2 a^2 b + 2 a b^2 - b^3 + 2 a^2 c - 9 a b c + 2 b^2 c + 2 a c^2 + 2 b c^2 - c^3>=0$$ Substitute $a=x+y, b=x+z, c=y+z$ because they are triangle sides and simplify to get:$$x^3 - x^2 y - x y^2 + y^3 - x^2 z + 3 x y z - y^2 z - x z^2 - y z^2 + z^3>=0$$ This I think you know how to prove. - How do you know that : $$IA^2=\frac{(p-a) \cdot bc}{p} ?$$ Thanks :) – Iuli Sep 13 '12 at 8:01 2 – ivan Sep 13 '12 at 8:18	{"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": 8, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.9096247553825378, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/14083/modular-forms-and-the-riemann-hypothesis/14161	## Modular forms and the Riemann Hypothesis ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is there any statement directly about modular forms that is equivalent to the Riemann Hypothesis for L-functions? What I'm thinking of is this: under the Mellin transform, the Riemann zeta function $\zeta(s)$ corresponds to a modular form $f$ (of weight 1/2). The functional equation of $\zeta(s)$ follows from the transform equation of $f$. So what is the property of $f$ that would be equivalent to the (conjectural) property that all the non-trivial zeros of $\zeta(s)$ lie on the critical line? Or perhaps is there any statement about some family of modular forms that would imply RH for $\zeta(s)$? - why is $\zeta(s)$ the Mellin transform of a modular form? – natura Feb 4 2010 at 8:06 There is a discussion of this fact in Milne's "Modular Functions And Modular Forms" notes, page 95-96. – Anonymous Feb 4 2010 at 8:19 16 The zeta function is the Mellin transform of the Jacobi theta function, a weight 1/2 modular form. This fact, observed and exploited by Riemann, is at the root of all later developments relating modular/automorphic forms and L-functions. Specifically regarding the zeroes of the zeta-function, Hardy used it in his proof thathere are infinitely many zeroes on the critical line. I think the question is far from rubbish. In general, it wouldn't hurt to give the questioner the benefit of the doubt, especially if you are not expert in the field yourself. – Emerton Feb 4 2010 at 16:40 2 To continue: the idea of using families of modular forms to study RH is an important one, employed by Sarnak and Iwaniec among others, and is inspired in part by Deligne's use of the monodromy of families in his proof of RH in char. p (Weil's Riemann Hypothesis). This is how I interpret the last line of the question. Even if it is not what the questionner had in mind, it is concrete mathematics in the direction they intimated, and would be more useful as an answer than "rubbish". – Emerton Feb 4 2010 at 16:44 1 @Emerton. I realize from your comment that I was wrong and I have removed the comments implying that the question was "rubbish". – Anweshi Feb 4 2010 at 18:56 show 1 more comment ## 5 Answers It can be hard to translate some properties between modular forms and L-functions. As far as I know, there is no simple property of a modular form that is equivalent to the Riemann Hypothesis for its corresponding L-function. Here is a baby problem to think about. The exponential function and the gamma function form a Mellin pair. How would you detect the periodicity of $e^{ix}$ (to pick a well-known property off the top of my head) from its integral representation in terms of the gamma function? In this paper, Conrey describes an approach of Iwaniec to RH using a family of elliptic curves. See page 12 for Iwaniec's method, as well as the conclusion with some comments on families. - ### 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 RH for an L-series L(s) is equivalent to an assertion about the locations of the poles of the logarithmic derivative (log L(s))' of L(s). In this way one can relate square-root savings in estimates for log-weighted partial sums of the traces of the Hecke operators of a given form to the RH for its L-series -- for the weight 1/2 theta series you've mentioned (whose Mellin transform is zeta(2s)), this corresponds to an improved error term in the prime number theorem. As David Hansen points out above, modular forms live in linear spaces but RH is not linearly robust, so any answer to your question will have to take some additional structure of those spaces (e.g. the action of the Hecke algebra) into account. As for your second question, there are statements about families of modular forms that imply RH for zeta; see for instance MR0633666 (and papers that reference it). - I know two statements about modular forms that are Riemann Hypothesis-ish. First, note that the constant term of the level-one non-holomorphic Eisenstein series $E_s$ is $y^s+c(s)y^{-s}$, and that the poles of $c(s)$ are the same as the poles of $E_s$. We can directly calculate that $c(s)={\Lambda(s)\over\Lambda(1+s)}$ (this depends on your precise normalization of the Eisenstein series), where $\Lambda$ is the completed zeta function. We can actually say something about the location of the poles of $E_s$ (using the spectral theory of automorphic forms). Unfortunately, we only know how to control poles for ${\rm Re}(s)\ge 0$. This does give an alternate proof of the nonvanishing of $\zeta(s)$ at the edge of the critical strip (from the lack of poles of ${\Lambda(it)\over\Lambda(1+it)}$), but it doesn't seem possible to go further to the left (though it does generalize to other $L$-functions appearing as the constant term of cuspidal-data Eisenstein series). Second, the values of modular forms at certain (Heegner) points in the upper-half plane can be related to zeta functions. For example, $E_s(i)={\Lambda_{{\mathbb Q}(i)}(s)\over \Lambda_{\mathbb Q}(2s)}$. The general statement is simple to express adelically. Take a quadratic extension $k_1$ of $k$, and let $H$ denote $k_1^\times$ as a $k$-group and $E_s$ the standard level-one Eisenstein series on $G=GL_2(k)$. Take a character $\chi$ on $Z_{\mathbb A}H_k\backslash H_{\mathbb A}$ then $$\int_{Z_{\mathbb A}H_k\backslash H_{\mathbb A}}E_s(h)\chi(h)\ dh={\Lambda_{k_1}(s,\chi)\over \Lambda_k(2s)}$$ where $Z$ denotes the center of $G$, and we have normalized the measure on the quotient space to be 1. Note that since $H$ is a non-split torus in $G$, the quotient is compact, so the integral is finite. In fact, the integrand is invariant (on the right) under a compact open subgroup $K$ of $H_{\mathbb A}$, so the integral is actually over the double coset space $Z_{\mathbb A}H_k\backslash H_{\mathbb A}/K$, which is actually a finite group. In order to get the Riemann zeta function in the numerator on the right-hand-side, you would need to integrate over a split torus, which is precisely the Mellin transform, and you would have convergence issues. Note that if it did converge, the Mellin transform of $E_s$ would be $$\int_{Z_{\mathbb A}M_k\backslash M_{\mathbb A}} E_s(a)\|a\|^v\ da={\Lambda(v+s)\Lambda(v+1-s)\over\Lambda(2s)}$$ The second idea is more commonly discussed in the context of subconvexity problems for general $L$-functions. (See Iwaniec's Spectral Methods of Automorphic Forms, especially Chp 13.) A class of subconvexity results is the Lindelof Hypothesis, which is one of the stronger implications of the Riemann Hypothesis. - No. Let's remember what the basic properties of a modular form are. A modular form is either a section of some high tensor power of a line bundle over $\Gamma \backslash \mathfrak{H}$ for some discrete cofinite $\Gamma \subset \mathrm{SL}_2(\mathbb{R})$ (holomorphic modular forms) or an element of $L^2(\Gamma \backslash \mathfrak{H})$ (Maass forms). A modular form gives rise to a Dirichlet series via Mellin transform, as you say. Now, if $\Gamma$ is a congruence subgroup and the modular form in question is an eigenform of all the Hecke operators, then it is expected that its corresponding L-function satisfies a Riemann hypothesis. However, this property is very sensitive: if $f$ and $g$ are modular eigenforms, then the modular form $f+10^{-10}g$ (say) will not be an eigenform, but it will look a lot like $f$. In particular, its Dirichlet series will look a lot like the Dirichlet series for $f$, but its RH will certainly be destroyed. There is no good criterion solely in terms of the modular form... However, people expect that any dirichlet series with an Euler product and a functional equation (and some other mild properties) will satisfy RH. You can look up the "Selberg class" of Dirichlet series for more information on this. Of course, modular eigenforms are a primary source of Dirichlet series with Euler product... - Perhaps the questioner meant eigenform''; that's certainly how I understood the question. – Emerton Feb 4 2010 at 16:35 @Hansen and @Anonymous: your answers are appreciated. I want to know why people almost never discuss this question, so even the answer that the question is not a good one is appreciated, provided it also gives a reason, like you did. As Emerton suggested, I want to know whether RH could be stated for eigenforms directly, instead of the L-functions. I'm no expert in this field, but it seems to me that analytic properties of modular forms are easier to understand (than those of L-functions), so why not expressing RH in the space of modular forms and working with them? @Anonymous: do you know of any readily accessible source for statements about families of modular forms that imply RH for zeta? I don't have access to MathSciNet. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 48, "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": 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": 0.9370462894439697, "perplexity_flag": "head"}
http://mathoverflow.net/questions/103288?sort=votes	## A combinatorial question ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This possibly easy question is related to this one. Let $s_1,...,s_n$ be a sequence of natural numbers (some of them may be equal to 0). Consider the following sequence of multisets of 2-vectors (each vector is counted with its multiplicity) of natural numbers $A_1=\{ (s_1,s_2),...(s_n,s_1)\}$, $A_2=\{(s_1+s_2,s_3),...,(s_n+s_1,s_2)\}$, ...,$A_n=\{(s_1+...+s_n,0)\}$. Note that each multiset is closed under taking cyclic shift on the set of indices $1,2,...,n$. Question Does this sequence of multisets determine the sequence $s_1,...,s_n$ up to a cyclic shift? In particular, is the Prouhet-Morse-Thue sequence reconstructible (up to a cyclic shift)? Example If we encode every pair $(i,j)$ by a monomial $a^ib^j$ in commuting variables $a,b$, and interpret the multiplicity as a coefficient, we encode every multiset as a polynomial in $a,b$ over $\mathbb{Z}$. Here are the first 7 polynomials corresponding to the Prouhet-Morse-Thue word of length $8$ p_8="10010110": $$3a+1+3b+ab\\ 3a+b+3ab+{a}^{2}\\ 2ab+2a+2{a}^{2}b+2{a}^{2}\\ 2{a}^{2}+2\,ab+2{a}^{2}b+2\,{a}^{3}\\ 3{a}^{2}b+3{a}^{3}+{a}^{3}b+{a}^{2}\\ 3{a}^{3}b+3{a}^{3}+{a}^{4}+{a}^{2}b\\ 4{a}^{4}+4{a}^{3}b$$ Does this sequence of polynomials determine the word $p_8$ up to a cyclic shift? Update Answer for $p_8$ is "yes" (computed using Maple). - Is $A_n$ correct? It isn't a multiset of pairs like the others. – Brendan McKay Jul 27 at 10:11 @Brendan, you are right, I have made it a pair. In fact $A_n$ is redundant because the sum $s_1+...+s_n$ is already encoded in $A_1$. – Mark Sapir Jul 27 at 10:16 How is this question related to the previous one? – Mikael de la Salle Jul 27 at 15:24 @Mikhael: Consider a word $w(A,B)=A^{s_1}BA^{s_2}B...A^{s_n}b$ where some $s_i$ may be 0. Let $A$ be the diagonal matrix $diag(a,b,c)$, and $B$ be a generic matrix as in the previous question. Monomials in $f_w$ correspond to sequences of numbers $1,2,3$: for every sequence $i_1...i_n$ we get a monomial $a^pb^qc^rB[i_1,i_2]B[i_2,i_3]...B[i_n,i_1]$ where $p,q,r$ are easy to define. Now if you collect all terms containing the same product $B[i_1,i_2]...B[i_n,i_1]$, its coefficient is a polynomial in $a,b,c$. For sequences $1,2,3,...,3$, $1,1,2,3,...,3$,... you get polynomias as here. – Mark Sapir Jul 27 at 15:41 It means that if $w$ was determined by these multisets (=polynomials), then the answer to the previous question would be "yes". Now that we know that $w$ is not determined, one needs to consider different sequences of $1,2,3$. In the prev. answer the sequences were $(1,2,3,...,3)$, $(1,1,2,3,...,3)$, and so on. It did not come up nicely in LaTeX. – Mark Sapir Jul 27 at 15:45 show 5 more comments ## 1 Answer Unless I made mistake a counterexample is formed by the binary $m$-sequence of length 7 $s=(1,0,0,1,0,1,1)$ and its reversal $\tilde{s}=(1,1,0,1,0,0,1)$ that is not a cyclic shift of $s$. Both lead to the sequence of generating functions $1+2a+2b+2ab$, $2a+b+a^2+2ab+a^2b$, $a+a^2+2ab+a^3+2a^2b$, $a^2+ab+2a^3+2a^2b+a^3b$, $2a^3+2a^2b+a^4+2a^3b$ and $3a^4+4a^3b$. The $m$-sequences are examples of de Bruijn -sequences. That is binary sequences of length $2^n-1$ such that every sequence of $n$ bits (with the exception of $n$ zeros) occurs exactly once in the cycle. This is, of course, then a natural source for an eventual counterexample as the condition is automatically satisfied for $A_j, j\lt n.$ Length 7 is the shortest, where not all de Bruijn sequences are cyclic shifts of each other. An $m$-sequence is generated by a linear feedback shift register that has a primitive polynomial of degree $n$ from the ring $\mathbb{F}_2[x]$ as a feedback polynomial. If you decimate an $m$-sequence with a decimation exponent $d$, $\gcd(d, 2^n-1)=1,$ by cyclically taking every $d^{th}$ member, you get another $m$-sequence. Therefore for larger $n$ the number of non-cyclically equivalent $m$-sequences increases. For example, the sequence $s$ is generated by the feedback polynomial $x^3+x^2+1$, or equivalently by the given first 3 bits and the recurrence relation $$s_n=s_{n-3}+s_{n-2}\pmod2$$ with subscript arithmetic done modulo seven. The reversed sequence $\tilde{s}$ is similarly generated by the reciprocal polynomial $x^3+x+1$. I dare not guess yet, whether all $m$-sequences of a given length give rise to the same sequence of multisets. - Yes, it is correct. I also found that the Prouhet-Morse-Thue word $p_{16}$ is not recovered. Thank you! – Mark Sapir Jul 27 at 15:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 68, "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": 0.921979546546936, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/163792/convex-programming-when-the-problem-has-an-underlying-combinatorial-structure-th?answertab=active	# Convex programming when the problem has an underlying combinatorial structure that's a DAG I have a nonlinear convex objective function to minimize. The function is defined on a set of variables: $\{ x_1,x_2, \ldots ,x_p \},$ where each $x_i$ is a number associated with a path in the DAG. I'm wondering if there's previous work available for such convex programming problems when the problem has an underlying combinatorial structure such as a graph? - ## 1 Answer If $x_i$ would be associated not with paths but nodes of undirected graph it would looks similar to result of "functional lifting" method of convexificaton, for example Tom Goldstein et al "Global minimization of markov random fields with applications to optical flow" -	{"extraction_info": {"found_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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9384527206420898, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/189401/proving-that-a-n-and-b-n-relatively-prime-implies-ab-n-relatively-prime/189403	# Proving that $a,n$ and $b, n$ relatively prime implies $ab,n$ relatively prime Question: Suppose $a,b \in \Bbb N$, $\gcd (a,n) = \gcd(b,n) = 1$. The question is to prove or give a counterexample: $\gcd(ab,n) = 1$. My Work: This is what I have so far (for $\alpha, \beta, \gamma, \delta \in \Bbb Z$): \begin{align} \gcd(a,n) = 1 \ &\Rightarrow 1 = \alpha a + \beta n\\ \gcd(b,n) = 1 \ &\Rightarrow 1 = \gamma b + \delta n \end{align} Multiplying the top equation by $b$, and the bottom by $a$, I have $$b + a = (\alpha + \gamma)ab + (\beta b + \delta a)n$$ Here is where I am stuck. I now know that you can write a linear combination of $ab, n$ in this form, where all coefficients are integers, but I think I may have gone down the wrong road in this proof in multiplying by $a,b$. Hints would be appreciated. - 6 You question does not match your title. Should the question end $gcd(ab,n)=1$? – Kris Williams Aug 31 '12 at 20:18 One of the points of this question is to stress that a number $g$ being the gcd of a set of numbers is usually a stronger requirement than being the gcd of any proper subset of them. – Kirk Boyer Aug 31 '12 at 20:23 1 @jmi4 The first comment is correct. Does your question concern the statement in the title, or the different statement in the body? – Gone Aug 31 '12 at 21:11 Sorry everyone! The body is a typo. I mean the $\gcd(ab,n) = 1$ – JJR Aug 31 '12 at 22:16 ## 6 Answers Let $a=2, b=4, n=5$ shows statement is false. - You could have $n=3$ too ... – Mark Bennet Aug 31 '12 at 20:14 Well...this is embarrassing. Thanks! – JJR Aug 31 '12 at 20:15 @MarkBennet Or any odd. – i. m. soloveichik Aug 31 '12 at 20:15 Indeed - I had your example first, and then I thought 3 was a lower number - and I put up a higher one in my answer simply as an invitation to think a bit about what the question meant. – Mark Bennet Aug 31 '12 at 20:19 Am I having a moment of stupidity or isnt gcd$(8,5) = 1$, supporting the claim? – fretty Sep 1 '12 at 18:31 The following is an answer to the question in the title, using the technique that you tried to use. Of course the answer has nothing to do with the question in the body of the post. Because $a$ and $n$ are relatively prime, there exist integers $q$ and $r$ such that $qa+rn=1$. Similarly, there exist integers $s$ and $t$ such that $sb+tn=1$. Rewrite these equations as $qa=1-rn$ and $sb=1-tn$. Multiply. We obtain $$(qa)(sb)=(1-rn)(1-tn).$$ Expand and rearrange a bit. We get $$(qs)ab+(r+t-rtn)n=1,$$ which shows that $ab$ and $n$ are relatively prime. - Suppose $n=17$, I think you can try $a$ and $b$ less than 17 and coprime to 17, in fact because 17 is a prime ... - Hint $\$ By Euclid's Lemma, the naturals coprime to $\rm\:n\:$ are closed under multiplication. But sets $\rm\:S\ne \{1\}$ of positive naturals closed under multiplication always have non-coprime elements, e.g. $\rm\:1\ne a\in S\:\Rightarrow\:a^2\in S,\:$ so $\rm\:(a,a^2) = a \ne 1.$ As for Euclid's Lemma (the question in the title), by Bezout's Lemma, the elements coprime to $\rm\,n\,$ are exactly the invertible elements mod $\rm\,n,\,$ and a product of invertibles is invertible, thus $$\rm (a,n) = 1 = (b,n)\:\Rightarrow\:\exists\, \bar a,\bar b:\ a\bar a\equiv 1 \equiv b\bar b\,\ (mod\ n)\: \Rightarrow\ ab\ \bar a\bar b\equiv 1\,\ (mod\ n)\:\Rightarrow\: (ab,n) = 1$$ - Here is another simple solution for the question in the title. Suppose by contradiction that $ab$ and $n$ are not relatively prime. Then they have a common prime divisor $p$. Then $p$ divides $n$, and it also divides $ab$, hence either $a$ or $b$. Contradiction. - Think in terms of the prime factorizations of $a,b$, and $n$. (It doesn’t matter whether you’ve proved unique factorization yet; this is just a way to approach the problem.) Let $A$ be the set of prime factors of $a$, $b$ the set of prime factors of $b$, and $N$ the set of prime factors of $n$. Your hypothesis is basically that $A\cap N=\varnothing=B\cap N$, and the question is whether this imlies that $A\cap B=\varnothing$. When you put the question this way, the answer is pretty easy to see: the fact that $A$ and $B$ are both disjoint from $N$ says nothing at all about whether $A$ and $B$ overlap. Indeed, if $A=B\ne\varnothing$ it’s automatic that $A$ and $B$ will overlap. This immediately points you towards a counterexample like i.m. soloveichik’s. It actually gives you an even simpler one: just let $a=b=2$ and $n=1$! -	{"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": 68, "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": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9382261037826538, "perplexity_flag": "head"}
http://mathoverflow.net/questions/112015?sort=newest	## The history of the geometrization of closed surfaces ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) 1. Who first recognized that the torus supports a flat structure? 2. Who first characterized the moduli space of flat structures on the torus? 3. Who first recognized that the closed, orientable genus 2 supports a hyperbolic structure? 4. Who first thought of a geometrized surface in terms of the property that for any two points $A$ and $B$ there exists an isometry of the surface that takes $A$ to $B$? [Reposted from Math Stack Exchange.] - I don't understand property 4. For instance, think of a square torus and take A=C. There's a whole circle of points of (small) constant distance from A, but the group of isometries of the torus that fix A is finite. – HW Nov 10 at 19:43 I suppose you are right. I will fix this. – Jonah Sinick Nov 10 at 20:18 2 Isn't property 4 false for all surfaces of genus $g \geq 2$? The isometry group is certainly contained in the automorphism group of the complex structure, which is finite by Hurwitz's theorem. – Will Sawin Nov 10 at 20:22 2 Probably what is meant by property 4 is that there is a local isometry which takes A to B. Perhaps one even wants to say that there is a local isometry from A to B inducing any given isometry of their tangent spaces. – unknown (google) Nov 10 at 20:43 ## 2 Answers I believe that on the first two questions, there is no answer (so they are not well-posed:-) The reason is that the flat structure on a torus and the moduli spaces of tori where very well understood long before the notions of "flat", "structure", "moduli space" and even "tori" were introduced. Mathematicians just did not think in these terms. Once these notions were introduced, the tori became a simple example. Teichmuller was probably a person who made the notion of "moduli space" completely rigorous, though it was considered for the first time by Riemann. And Teichmuller considers tori at a great length to explain what he means. But certainly he did not find anything new about tori. This was just the simplest example. (Many people think, perhaps correctly, that Abel and Jacobi already knew everyhting about tori, though they did not use this word). Modular group was known to Gauss at about the same time. It is true, these things were developed for another 70 years or so, culminating in the Fricke-Klein and enormous French "treateases". But I know people who still prefer Jacobi's exposition, and even study Latin to read it:-) The metric of constant negative curvature on surfaces of genus >1 is another story. This was introduced by Poincare. Even the Poincare metric on the unit disc was apparently Poincare's invention. (Schwarz lemma was stated and proved much later, and not by Schwarz:-) About the last question, I suppose Klein has to be credited. It was Klein who in his "Erlangen program" explained explicitly "what geometry is", and how it is connected with a group action of certain kind. Klein was in close collaboration with Lee at that time. There is an excellent "primary source" on how all these things really developed: Klein's Lectures on the history of Math in XIX century. - Dear Alexandre, Thank you for the thoughtful answer. – Jonah Sinick Nov 11 at 1:55 1 Poincare disc model (as well as Klein model) was invented by Beltrami. – Anton Petrunin Nov 11 at 4:10 Anton: that's interesting. Poincare does not mention Beltrami. Can you give a reference? – Alexandre Eremenko Nov 11 at 5:04 Poincare did mention Beltrami, he was only popularizing his construction. Eugenio Beltrami, Teoria fondamentale degli spazii di curvatura costante, Annali. di Mat., ser II 2 (1868), 232-255 – Anton Petrunin Nov 11 at 5:10 ... which were both first invented by Bolyai in 1832. – Daniel Moskovich Nov 11 at 12:04 show 4 more comments ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. See this Wikipedia article, and the references therein. Jeremy Gray's article is particularly enlightening. - Thank you Igor. – Jonah Sinick Nov 11 at 1:56	{"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": 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": 0.9575414061546326, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/95714/reduced-cyclic-homology-of-a-free-product-of-unital-algebras	## (Reduced) cyclic homology of a free product of unital algebras ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Rather embarrassingly, the question is on how to prove something I used to know how to do... but I thought that the collective wisdom of MO might be able to quickly sketch how the argument should go, or direct me to a reference. I hope this is still a legitimate use of MO, though if you disagree please open a meta thread and let me know. Throughout all algebras are over a field $k$ of characteristic zero (in fact ${\mathbb C})$ and have identity elements. $\newcommand{\rhc}{\overline{HC}}$ Given such $A$ and $B$, let $A\ast B$ be their free product (coproduct in the category of unital algebras). I seem to remember convincing myself, or reading, that $$\rhc_n (A\ast B) \cong \rhc_n(A) \oplus \rhc_n(B)$$ where $\rhc$ denotes reduced cyclic homology (so $\rhc_\bullet(k)=0$ rather than having non-zero contributions in even degrees). Unfortunately I can't remember how the proof goes. My vague recollection is that it goes via the identification of $\rhc$ as the derived functor associated to the comonad `${\rm Vect}_k \leftrightarrow {\rm Alg}_k$` (which follows from the fact that for any vector space $V$, the reduced cyclic homology of the tensor algebra $T(V)$ vanishes in all positive degrees). But how to proceed? It should be some kind of acyclic complex argument, related to the fact that $T(V_1\oplus V_2) \cong T(V_1)\ast T(V_2)$, but I can no longer recall how these kinds of argument work in detail. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 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": 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": 0.9402703046798706, "perplexity_flag": "head"}
http://all-science-fair-projects.com/science_fair_projects_encyclopedia/Complex_number	# 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. # Complex number The complex numbers are an extension of the real numbers, in which all non-constant polynomials have roots. The complex numbers contain the imaginary unit i, with $\mathrm{i} = \sqrt{-1}$. Every complex number can be represented in the form a + ib, where a and b are real numbers called the real part and the imaginary part of the complex number, respectively. The sum and product of two complex numbers are: $( a + b\mathrm{i} ) + ( c + d\mathrm{i} ) = ( a + c ) + \mathrm{i} ( b + d ) \,$ $( a + b\mathrm{i} ) \cdot ( c + d\mathrm{i} ) = ( ac - bd ) + \mathrm{i} ( bc + ad ). \,$ Complex numbers were first introduced in connection with explicit formulas for the roots of cubic polynomials. In mathematics, the term "complex" when used as an adjective means that the field of complex numbers is the underlying number field considered, for example complex matrix, complex polynomial and complex Lie algebra. Contents ## History of complex numbers The earliest fleeting reference to square roots of negative numbers occurred in the work of the Greek mathematician and inventor Heron of Alexandria in the 1st century AD, when he considered the volume of an impossible frustum of a pyramid. They became more prominent when in the 16th century closed formulas for the roots of third and fourth degree polynomials were discovered by Italian mathematicians (see Niccolo Fontana Tartaglia, Gerolamo Cardano). It was soon realized that these formulas, even if one was only interested in real solutions, sometimes required the manipulation of square roots of negative numbers. This was doubly unsettling since not even negative numbers were considered to be on firm ground at the time. The term "imaginary" for these quantities was coined by René Descartes in the 17th century and was meant to be derogatory. (See imaginary number for a discussion of the "reality" of complex numbers.) The 18th century saw the labors of Abraham de Moivre and Leonhard Euler. To De Moivre is due (1730) the well-known formula which bears his name, de Moivre's formula: $(\cos \theta + \mathrm{i} \sin \theta)^{n} = \cos n \theta + \mathrm{i} \sin n \theta \,$ and to Euler (1748) Euler's formula of complex analysis: $\cos \theta + \mathrm{i} \sin \theta = e ^{\mathrm{i}\theta }. \,$ The existence of complex numbers was not completely accepted until the geometrical interpretation (see below) had been described by Caspar Wessel in 1799; it was rediscovered several years later and popularized by Carl Friedrich Gauss, and as a result the theory of complex numbers received a notable expansion. The idea of the graphic representation of complex numbers had appeared, however, as early as 1685, in Wallis's De Algebra tractatus. Wessel's memoir appeared in the Proceedings of the Copenhagen Academy for 1799, and is exceedingly clear and complete, even in comparison with modern works. He also considers the sphere, and gives a quaternion theory from which he develops a complete spherical trigonometry. In 1804 the Abbé Buée independently came upon the same idea which Wallis had suggested, that $\pm\sqrt{-1}$ should represent a unit line, and its negative, perpendicular to the real axis. Buée 's paper was not published until 1806, in which year Jean-Robert Argand also issued a pamphlet on the same subject. It is to Argand's essay that the scientific foundation for the graphic representation of complex numbers is now generally referred. Nevertheless, in 1831 Gauss found the theory quite unknown, and in 1832 published his chief memoir on the subject, thus bringing it prominently before the mathematical world. Mention should also be made of an excellent little treatise by Mourey (1828), in which the foundations for the theory of directional numbers are scientifically laid. The general acceptance of the theory is not a little due to the labors of Augustin Louis Cauchy and Niels Henrik Abel, and especially the latter, who was the first to boldly use complex numbers with a success that is well known. The common terms used in the theory are chiefly due to the founders. Argand called cosφ + i * sinφ the direction factor, and $r = \sqrt{a^2+b^2}$ the modulus; Cauchy (1828) called cosφ + i * sinφ the reduced form (l'expression réduite); Gauss used i for $\sqrt{-1}$, introduced the term complex number for a + bi, and called a2 + b2 the norm. The expression direction coefficient, often used for cosφ + i * sinφ, is due to Hankel (1867), and absolute value, for modulus, is due to Weierstrass. Following Cauchy and Gauss have come a number of contributors of high rank, of whom the following may be especially mentioned: Kummer (1844), Leopold Kronecker (1845), Scheffler (1845, 1851, 1880), Bellavitis (1835, 1852), Peacock (1845), and De Morgan (1849). Möbius must also be mentioned for his numerous memoirs on the geometric applications of complex numbers, and Dirichlet for the expansion of the theory to include primes, congruences, reciprocity, etc., as in the case of real numbers. Other types have been studied, besides the familiar a + bi, in which i is the root of x2 + 1 = 0. Thus Ferdinand Eisenstein has studied the type a + bj, j being a complex root of x3 - 1 = 0. Similarly, complex types have been derived from xk - 1 = 0 (k prime). This generalization is largely due to Kummer, to whom is also due the theory of ideal numbers, which has recently been simplified by Felix Klein (1893) from the point of view of geometry. A further complex theory is due to Evariste Galois, the basis being the imaginary roots of an irreducible congruence, $F(x) \equiv 0$ (mod p, a prime). The late writers (from 1884) on the general theory include Weierstrass, Schwarz , Richard Dedekind, Otto Hölder, Berloty , Henri Poincaré, Eduard Study, and Macfarlane. The formally correct definition using pairs of real numbers was given in the 19th century. ## Definition Formally we may define complex numbers as ordered pairs of real numbers (a, b) together with the operations: • $( a , b ) + ( c , d ) = ( a + c , b + d ) \,$ • $( a , b ) \cdot ( c , d ) = ( ac - bd , bc + ad ). \,$ So defined, the complex numbers form a field, the complex number field, denoted by C (or $\mathbb{C}$ in blackboard bold). We identify the real number a with the complex number (a, 0), and in this way the field of real numbers R becomes a subfield of C. The imaginary unit i is the complex number (0,1). In C, we have: • additive identity ("zero"): (0,0) • multiplicative identity ("one"): (1,0) • additive inverse of (a,b): (−a,−b) • multiplicative inverse of non-zero (a,b): $\left({a\over a^2+b^2},{-b\over a^2+b^2}\right).$ C could also be defined as the topological closure of algebraic numbers and the algebraic closure of R. ### Geometry A complex number can also be viewed as a point or a position vector on the two dimensional Cartesian coordinate system. This representation is called the complex plane or Argand diagram. In the figure, we have $z = x + \mathrm{i}y = r (\cos \phi + \mathrm{i}\sin \phi ). \,$ The latter expression is sometimes shorthanded as r cis φ, where r = \|z\| is called the absolute value of z and φ = arg(z) is called the complex argument of z. However, Euler's formula states that ei φ = cisφ. The exponential form gives us a better insight than the shorthand rcisφ, which is almost never used in serious mathematical articles. By simple trigonometric identities, we see that $r_1 e^{\mathrm{i} \phi_1} \cdot r_2 e^{\mathrm{i} \phi_2} = r_1 r_2 e^{\mathrm{i} (\phi_1 + \phi_2)} \,$ and that $\frac{r_1 e^{\mathrm{i} \phi_1}} {r_2 e^{\mathrm{i} \phi_2}} = \frac{r_1}{r_2} e^{\mathrm{i} (\phi_1 - \phi_2)}. \,$ Now the addition of two complex numbers is just the vector addition of two vectors, and the multiplication with a fixed complex number can be seen as a simultaneous rotation and stretching. Multiplication with i corresponds to a counter clockwise rotation by 90 degrees. The geometric content of the equation i2 = -1 is that a sequence of two 90 degree rotation results in a 180 degree rotation. Even the fact (-1) · (-1) = +1 from arithmetic can be understood geometrically as the combination of two 180 degree turns. ### Absolute value, conjugation and distance Recall that the absolute value (or modulus or magnitude) of a complex number z = r eiφ is defined as \|z\| = r. Algebraically, if z = a + ib, then \|z\| = √(a² + b² ). One can check readily that the absolute value has three important properties: $\| z + w \| \leq \| z \| + \| w \| \,$ $\| z w \| = \| z \| \; \| w \| \,$ $\| z / w \| = \| z \| / \| w \| \,$ for all complex numbers z and w. By defining the distance function d(z, w) = \|z - w\| we turn the complex numbers into a metric space and we can therefore talk about limits and continuity. The addition, subtraction, multiplication and division of complex numbers are then continuous operations. Unless anything else is said, this is always the metric being used on the complex numbers. The complex conjugate of the complex number z = a + ib is defined to be a - ib, written as $\bar{z}$ or z. As seen in the figure, $\bar{z}$ is the "reflection" of z about the real axis. The following can be checked: $\overline{z+w} = \bar{z} + \bar{w}$ $\overline{zw} = \bar{z}\bar{w}$ $\overline{(z/w)} = \bar{z}/\bar{w}$ $\bar{\bar{z}}=z$ $\bar{z}=z$ if and only if z is real $\|z\|=\|\bar{z}\|$ $\|z\|^2 = z\bar{z}$ $z^{-1} = \bar{z}\|z\|^{-2}$ if z is non-zero. The latter formula is the method of choice to compute the inverse of a complex number if it is given in rectangular coordinates. That conjugate commutes with all the algebraic operations (and many functions; e.g. $\sin\bar z=\overline{\sin z}$) is rooted in the ambiguity in choice of i (-1 has two square roots); note, however, that conjugate is not differentiable (see holomorphic). The complex argument of z=reiφ is φ. Note that the complex argument is unique modulo 2π, that is, any two values of the complex argument differ by an integer multiple of 2π. ### Complex number division Given a complex number (a + ib) which is to be divided by another complex number (c + id) whose magnitude is non-zero, there are two ways to do this. The first way has already been implied: to convert both complex numbers into exponential form, from which their quotient is easy to derive. The second way is to express the division as a fraction, then to multiply both numerator and denominator by the complex conjugate of the denominator. This causes the denominator to simplify into a real number: ${a + ib \over c + id} = {(a + i b) (c - i d) \over (c + i d) (c - i d)} = {(a c + b d) + i (b c - a d) \over c^2 + d^2}$ $= \left({a c + b d \over c^2 + d^2}\right) + i \left( {b c - a d \over c^2 + d^2} \right).$ ### Matrix representation of complex numbers While usually not useful, alternative representations of complex fields can give some insight into their nature. One particularly elegant representation interprets every complex number as 2×2 matrix with real entries which stretches and rotates the points of the plane. Every such matrix has the form $\begin{pmatrix} a & -b \\ b & \;\; a \end{pmatrix}$ with real numbers a and b. The sum and product of two such matrices is again of this form. Every non-zero such matrix is invertible, and its inverse is again of this form. Therefore, the matrices of this form are a field. In fact, this is exactly the field of complex numbers. Every such matrix can be written as $\begin{pmatrix} a & -b \\ b & \;\; a \end{pmatrix} = a \begin{pmatrix} 1 & \;\; 0 \\ 0 & \;\; 1 \end{pmatrix} + b \begin{pmatrix} 0 & -1 \\ 1 & \;\; 0 \end{pmatrix}$ which suggests that we should identify the real number 1 with the matrix $\begin{pmatrix} 1 & \;\; 0 \\ 0 & \;\; 1 \end{pmatrix}$ and the imaginary unit i with $\begin{pmatrix} 0 & -1 \\ 1 & \;\; 0 \end{pmatrix}$ a counter-clockwise rotation by 90 degrees. Note that the square of this latter matrix is indeed equal to -1. The absolute value of a complex number expressed as a matrix is equal to the square root of the determinant of that matrix. If the matrix is viewed as a transformation of a plane, then the transformation rotates points through an angle equal to the argument of the complex number and scales by a factor equal to the complex number's absolute value. The conjugate of the complex number z corresponds to the transformation which rotates through the same angle as z but in the opposite direction, and scales in the same manner as z; this can be described by the transpose of the matrix corresponding to z. If the matrix elements are themselves complex numbers, then the resulting algebra is that of the quaternions. In this way, the matrix representation can be seen as a way of expressing the Cayley-Dickson construction of algebras. ## Some properties ### Real vector space C is a two-dimensional real vector space. Unlike the reals, complex numbers cannot be ordered in any way that is compatible with its arithmetic operations: C cannot be turned into an ordered field. ### Solutions of polynomial equations A root of the polynomial p is a complex number z such that p(z) = 0. A most striking result is that all polynomials of degree n with real or complex coefficients have exactly n complex roots (counting multiple roots according to their multiplicity). This is known as the Fundamental Theorem of Algebra, and shows that the complex numbers are an algebraically closed field. Indeed, the complex number field is the algebraic closure of the real number field. It can be identified as the quotient ring of the polynomial ring R[X] by the ideal generated by the polynomial X2 + 1: $\mathbb{C} = \mathbb{R}[ X ] / ( X^2 + 1). \,$ This is indeed a field because X2 + 1 is irreducible. The image of X in this quotient ring becomes the imaginary unit i. ### Algebraic characterization The field C is (up to field isomorphism) characterized by the following three facts: • its characteristic is 0 • its transcendence degree over the prime field is the cardinality of the continuum • it is algebraically closed Consequently, C contains many proper subfields which are isomorphic to C. Another consequence of this characterization is that the Galois group of C over the rational numbers is enormous, with cardinality equal to the power set of the continuum. ### Characterization as a topological field As noted above, the algebraic characterization of C fails to capture some of its most important properties. These properties, which underpin the foundations of complex analysis, arise from the topology of C. The following properties characterize C as a topological field: • C is a field. • C contains a subset P of nonzero elements satisfying: • P is closed under addition, multiplication and taking inverses. • If x and y are distinct elements of P, then either x-y or y-x is in P • If S is any nonempty subset of P, then S+P=x+P for some x in C. • C has a nontrivial involutive automorphism x->x, fixing P and such that xx* is in P for any nonzero x in C. Given these properties, one can then define a topology on C by taking the sets • $B(x,p) = \{y \| p - (y-x)(y-x)^*\in P\}$ as a base, where x ranges over C, and p ranges over P. To see that these properties characterize C as a topological field, one notes that P ∪ {0} ∪ -P is an ordered Dedekind-complete field and thus can be identified with the real numbers R by a unique field isomorphism. The last property is easily seen to imply that the Galois group over the real numbers is of order two, completing the characterization. has shown that the only connected locally compact topological fields are R and C. This gives another characterization of C as a topological field, since C can be distinguished from R by noting the nonzero complex numbers are connected whereas the nonzero real numbers are not. ## Complex analysis The study of functions of a complex variable is known as complex analysis and has enormous practical use in applied mathematics as well as in other branches of mathematics. Often, the most natural proofs for statements in real analysis or even number theory employ techniques from complex analysis (see prime number theorem for an example). Unlike real functions which are commonly represented as two dimensional graphs, complex functions have four dimensional graphs and may usefully be illustrated by color coding a three dimensional graph to suggest four dimensions, or by animating the complex function's dynamic transformation of the complex plane. ## Applications ### Control theory In control theory, systems are often transformed from the time domain to the frequency domain using the Laplace transform. The system's poles and zeros are then analyzed in the complex plane. The root locus, Nyquist plot, and Nichols plot techniques all make use of the complex plane. In the root locus method, it is especially important whether the poles and zeros are in the left or right half planes, i.e. have real part greater than or less than zero. If a system has poles that are • in the right half plane, it will be unstable, • all in the left half plane, it will be stable, • on the imaginary axis, it will be marginally stable . If a system has zeros in the right half plane, it is a nonminimum phase system. ### Signal analysis Complex numbers are used in signal analysis and other fields as a convenient description for periodically varying signals. The absolute value \|z\| is interpreted as the amplitude and the argument arg(z) as the phase of a sine wave of given frequency. If Fourier analysis is employed to write a given real-valued signal as a sum of periodic functions, these periodic functions are often written as the real part of complex valued functions of the form $f ( t ) = z e^{\mathrm{i}\omega t} \,$ where ω represents the angular frequency and the complex number z encodes the phase and amplitude as explained above. In electrical engineering, this is done for varying voltages and currents. The treatment of resistors, capacitors and inductors can then be unified by introducing imaginary frequency-dependent resistances for the latter two and combining all three in a single complex number called the impedance. (Electrical engineers and some physicists use the letter j for the imaginary unit since i is typically reserved for varying currents and may come into conflict with i.) ### Improper integrals In applied fields, the use of complex analysis is often used to compute certain real-valued improper integrals, by means of complex-valued functions. Several methods exist to do this, see methods of contour integration. ### Quantum mechanics The complex number field is also of utmost importance in quantum mechanics since the underlying theory is built on (infinite dimensional) Hilbert spaces over C. ### Relativity In special and general relativity, some formulas for the metric on spacetime become simpler if one takes the time variable to be imaginary. ### Applied mathematics In differential equations, it is common to first find all complex roots r of the characteristic equation of a linear differential equation and then attempt to solve the system in terms of base functions of the form f(t) = ert. ### Fluid dynamics In fluid dynamics, complex functions are used to describe potential flow in 2d. ### Fractals Certain fractals are plotted in the complex plane e.g. Mandelbrot set and Julia set. ## Further reading • An Imaginary Tale, by Paul J. Nahin; Princeton University Press; ISBN 0691027951 (hardcover, 1998). A gentle introduction to the history of complex numbers and the beginnings of complex analysis. ## External links 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": 39, "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": 0.9257316589355469, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/95206-setup-help-print.html	# Setup help Printable View • July 15th 2009, 11:58 AM Danneedshelp Setup help Q: Find all subsets of the set that forms a basis for $\mathbb{R}^{3}$. $S=\{(1,3,-1),(-4,1,1),(-2,7,-3),(2,1,1)\}$ A: well, I did a similar one of these and got the same answer as book, but with a bunch of negative 1's unlike their strictly positive solution. I figured it was ok, because subspaces are closed under scalar multiplication (then again, we are just talking about a subset). I don't know if that’s alright. My approach is to set up a homogenous linear combination and solve for the constants. Then I set up my new linear combination and then solve for each vector in the set S. Then that gives me all my subspaces. I have to hope that S is linearly dependent for this to work. Is that the correct method?? • July 15th 2009, 12:06 PM Gamma Well there are 4C3=4 choices of 3 elements from that set of 4 elements (recall the ordering of the basis does not affect spanning or linear independence). So pick three of them and make each vector a column of a 3x3 matrix. Take the determinant, if it is 0 then they are not a basis since they are not linearly independent. If the determinant is not zero then they are linearly independent and there are three of them, so they span $\mathbb{R}^3$ a three dimensional vector space. Do this for each of the four possibilities. • July 15th 2009, 02:35 PM Danneedshelp I see, so something like this... $S=\{(1,3,-2),(-4,1,1),(-2,7,-3),(2,1,1)\}$ $\det{\begin{pmatrix}1&-4&-2\\3&1&7\\-2&1&-3\end{pmatrix}}=0$ $\det{\begin{pmatrix}1&-2&2\\3&7&1\\-2&-3&1\end{pmatrix}}=30$ $\det{\begin{pmatrix}1&-4&2\\3&1&1\\-2&1&1\end{pmatrix}}=30$ $\det{\begin{pmatrix}-4&-2&2\\1&7&1\\1&-3&1\end{pmatrix}}=-60$ Therefore, we have three subsets that form a basis for $\mathbb{R}^{3}$. They are... $S_{1}=\{(1,3,-2),(-2,7,-3),(2,1,1)\}$, $S_{2}=\{(1,3,-2),(-4,1,1),(2,1,1)\}$, $S_{3}=\{(-4,1,1),(-2,7,-3),(2,1,1)\}$. • July 15th 2009, 09:20 PM Gamma Right on As long as those determinants are correct your solution is correct. All times are GMT -8. The time now is 05:09 PM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "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": 0.9524245262145996, "perplexity_flag": "head"}
http://shreevatsa.wordpress.com/2012/10/	# The Lumber Room "Consign them to dust and damp by way of preserving them" ## The potions puzzle by Snape in Harry Potter and the Philosopher’s Stone with one comment Sometime this week, I reread the first Harry Potter book (after at least 10 years… wow, has it been that long?), just for contrast after reading Rowling’s adult novel The Casual Vacancy (on which more later). Anyway, in the penultimate chapter there is a puzzle: He pulled open the next door, both of them hardly daring to look at what came next — but there was nothing very frightening in here, just a table with seven differently shaped bottles standing on it in a line. “Snape’s,” said Harry. “What do we have to do?” They stepped over the threshold, and immediately a fire sprang up behind them in the doorway. It wasn’t ordinary fire either; it was purple. At the same instant, black flames shot up in the doorway leading onward. They were trapped. “Look!” Hermione seized a roll of paper lying next to the bottles. Harry looked over her shoulder to read it: Danger lies before you, while safety lies behind, One among us seven will let you move ahead, Another will transport the drinker back instead, Two among our number hold only nettle wine, Three of us are killers, waiting hidden in line. Choose, unless you wish to stay here forevermore, First, however slyly the poison tries to hide You will always find some on nettle wine’s left side; Second, different are those who stand at either end, But if you would move onward, neither is your friend; Third, as you see clearly, all are different size, Neither dwarf nor giant holds death in their insides; Fourth, the second left and the second on the right Are twins once you taste them, though different at first sight. I became curious about whether this is just a ditty Rowling made up, or the puzzle actually makes sense and is consistent. It turns out she has constructed it well. Let’s take a look. This investigation can be carried out by hand, but we’ll be lazy and use a computer, specifically Python. The code examples below are all to be typed in an interactive Python shell, the one that you get by typing “python” in a terminal. So what we have are seven bottles, of which one will take you forward (F), one will let you go back (B), two are just nettle wine (N), and three are poison (P). ```>>> bottles = ['F', 'B', 'N', 'N', 'P', 'P', 'P'] ``` The actual ordering of these 7 bottles is some ordering (permutation) of them. As 7 is a very small number, we can afford to be inefficient and resort to brute-force enumeration. ```>>> import itertools >>> perms = [''.join(s) for s in set(itertools.permutations(bottles))] >>> len(perms) 420 ``` The `set` is needed to remove duplicates, because otherwise `itertools.permutations` will print 7! “permutations”. So already the number of all possible orderings is rather small (it is $\frac{7!}{2!3!} = 420$). We can look at a sample to check whether things look fine. ```>>> perms[:10] ['PNFNPBP', 'NPPBNFP', 'FNNPBPP', 'PPPFNBN', 'NPPNBFP', 'PFNNBPP', 'NPBPPFN', 'NBNPPFP', 'NPPFBNP', 'BNPFNPP'] ``` Now let us try to solve the puzzle. We can start with the first clue, which says that wherever a nettle-wine bottle occurs, on its left is always a poison bottle (and in particular therefore, a nettle-wine bottle cannot be in the leftmost position). So we must restrict the orderings to just those that satisfy this condition. ```>>> def clue1(s): return all(i > 0 and s[i-1] == 'P' for i in range(len(s)) if s[i]=='N') ... >>> len([s for s in perms if clue1(s)]) 60 ``` (In the code, the 7 positions are 0 to 6, as array indices in code generally start at 0.) Then the second clue says that the bottles at the end are different, and neither contains the potion that lets you go forward. ```>>> def clue2(s): return s[0] != s[6] and s[0] != 'F' and s[6] != 'F' ... >>> len([s for s in perms if clue1(s) and clue2(s)]) 30 ``` The third clue says that the smallest and largest bottles don’t contain poison, and this would be of help to Harry and Hermione who can see the sizes of the bottles. But as we readers are not told the sizes of the bottles, this doesn’t seem of any help to us; let us return to this later. The fourth clue says that the second-left and second-right bottles have the same contents. ```>>> def clue4(s): return s[1] == s[5] ... >>> len([s for s in perms if clue1(s) and clue2(s) and clue4(s)]) 8 ``` There are now just 8 possibilities, finally small enough to print them all. ```>>> [s for s in perms if clue1(s) and clue2(s) and clue4(s)] ['PPNBFPN', 'BPNPFPN', 'BPFPNPN', 'BPPNFPN', 'PNPFPNB', 'BPNFPPN', 'PPNFBPN', 'PNFPPNB'] ``` Alas, without knowing which the “dwarf” and “giant” bottles are, we cannot use the third clue, and this seems as far as we can go. We seem to have exhausted all the information available… Almost. It is reasonable to assume that the puzzle is meant to have a solution. So even without knowing where exactly the “dwarf” and “giant” bottles are, we can say that they are in some pair of locations that ensure a unique solution. ```>>> def clue3(d, g, s): return s[d]!='P' and s[g]!='P' ... >>> for d in range(7): ... for g in range(7): ... if d == g: continue ... poss = [s for s in perms if clue1(s) and clue2(s) and clue4(s) and clue3(d,g,s)] ... if len(poss) == 1: ... print d, g, poss[0] ... 1 2 PNFPPNB 1 3 PNPFPNB 2 1 PNFPPNB 2 5 PNFPPNB 3 1 PNPFPNB 3 5 PNPFPNB 5 2 PNFPPNB 5 3 PNPFPNB ``` Aha! If you look at the possible orderings closely, you will see that we are down to just two possibilities for the ordering of the bottles. Actually there is some scope for quibbling in what we did above: perhaps we cannot say that there is a unique solution determining the entire configuration; perhaps all we can say is that the puzzle should let us uniquely determine the positions of just the two useful bottles. Fortunately, that gives exactly the same set of possibilities, so this distinction happens to be inconsequential. ```>>> for d in range(7): ... for g in range(7): ... if d == g: continue ... poss = [(s.index('F'),s.index('B')) for s in perms if clue1(s) and clue2(s) and clue4(s) and clue3(d,g,s)] ... if len(set(poss)) == 1: ... print d, g, [s for s in perms if clue1(s) and clue2(s) and clue4(s) and clue3(d,g,s)][0] ... 1 2 PNFPPNB 1 3 PNPFPNB 2 1 PNFPPNB 2 5 PNFPPNB 3 1 PNPFPNB 3 5 PNPFPNB 5 2 PNFPPNB 5 3 PNPFPNB ``` Good. Note that there are only two configurations above. So with only the clues in the poem, and the assumption that the puzzle can be solved, we can narrow down the possibilities to two configurations, and be sure of the contents of all the bottles except the third and fourth. We know that the potion that lets us go forward is in either the third or the fourth bottle. In particular we see that the last bottle lets us go back, and indeed this is confirmed by the story later: “Which one will get you back through the purple flames?” Hermione pointed at a rounded bottle at the right end of the line. […] She took a long drink from the round bottle at the end, and shuddered. But we don’t know which of the two it is, as we can’t reconstruct all the relevant details of the configuration. Perhaps we can reconstruct something with the remaining piece of information from the story? “Got it,” she said. “The smallest bottle will get us through the black fire — toward the Stone.” Harry looked at the tiny bottle. […] Harry took a deep breath and picked up the smallest bottle. So we know that the bottle that lets one move forward is in fact in the smallest one, the “dwarf”. ```>>> for d in range(7): ... for g in range(7): ... poss = [s for s in perms if clue1(s) and clue2(s) and clue4(s) and clue3(d,g,s)] ... if len(poss) == 1 and poss[0][d] == 'F': ... print d, g, poss[0] ... 2 1 PNFPPNB 2 5 PNFPPNB 3 1 PNPFPNB 3 5 PNPFPNB ``` This narrows the possible positions of the smallest and largest bottles (note that it says the largest bottle is one that contains nettle wine), but still leaves the same two possibilities for the complete configuration. So we can stop here. Conclusion What we can conclude is the following: apart from the clues mentioned in the poem, the “dwarf” (the smallest bottle) was in either position 2 (third from the left) or 3 (fourth from the left). The biggest bottle was in either position 1 (second from the left) or 5 (sixth from the left). With this information about the location of the smallest bottle (and without necessarily assuming the puzzle has a unique solution!), Hermione could determine the contents of all the bottles. In particular she could determine the location of the two useful bottles: namely that the bottle that lets you go back was the last one, and that the one that lets you go forward was the smallest bottle. ```>>> for (d,g) in [(2,1), (2,5), (3,1), (3,5)]: ... poss = [s for s in perms if clue1(s) and clue2(s) and clue4(s) and clue3(d, g, s)] ... assert len(poss) == 1 ... s = poss[0] ... assert s.index('B') == 6 ... assert s.index('F') == d ... print (d,g), s ... (2, 1) PNFPPNB (2, 5) PNFPPNB (3, 1) PNPFPNB (3, 5) PNPFPNB ``` It is not clear why she went to the effort to create a meaningful puzzle, then withheld details that would let the reader solve it fully. Perhaps some details were removed during editing. As far as making up stuff for the sake of a story goes, though, this is nothing; consider for instance the language created for Avatar which viewers have learned. Added: See also http://www.zhasea.com/logic/snape.html which does it by hand, and has a perfectly clear exposition (it doesn’t try the trick of guessing that solution is unique before reaching for the additional information from the story). Written by S Mon, 2012-10-22 at 02:00:12 +05:30 Posted in entertainment, mathematics Tagged with harry potter, programming, puzzle	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9565847516059875, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/205280/sobolev-approximation-theorem-example	# Sobolev Approximation Theorem Example The approximation theorem states that if $U$ is bounded, $u \in W^{1,p}(U)$ for some $1 \leq p < \infty$ then there are functions $u_m \in C^\infty(U) \cap W^{k,p}(U)$, such that $u_m \rightarrow u$ in $W^{k,p}(U)$. This is probably very simple, but I'm trying to follow along with an example in a book I am reading. I'm looking at the set $\Omega = (-1,0) \cup (0,1)$ and the function $u : \Omega \rightarrow \mathbb{R}$ defined by $u(x) = \left\{ \begin{array}{ll} 1, & x > 0 \\ 0, & x < 0 \end{array} \right.$. It's clear to me that this function is in $W^{1,p}(\Omega)$ for $1 \leq p \leq \Omega$; however, I'm having trouble proving that it cannot be approximated by functions in $C^\infty(\overline{\Omega})$. It seems that any sequence approaching $u$ would approach the heaviside function which I know is not in $L^p$ and therefore not in $W^{1,p}$; however, actually proving this rigorously has eluded me for a while now. I also had a question about a global form of the approximation theorem. If we also add that $\partial U$ is $C^1$ to the hypothesis, then we have functions $u_m \in C^\infty(\overline{U})$ such that $u_m \rightarrow u$ in $W^{k,p}(U)$. I know that $W^{k,p}$ is Banach so I was wondering if in this case we are able to assume that $\lim\limits_{m \rightarrow \infty} u_m$ exists in $W^{1,p}$. I'm just not sure if the $C^\infty$ functions intersected with the $W^{k,p}$ functions are complete. Anyway, thanks a lot for looking. I appreciate any help you guys can provide. - BTW, the Heaviside function is in $L^p$ (on bounded sets). The problem does not come from the fact that the limit function is not integrable but from the missing regularity of the limit function on the closure of the set. – Dirk Oct 1 '12 at 6:23 ## 1 Answer For the first question: in one dimension $W^{1,1}(U)$ embeds into $C_b(U)$ (bounded continuous functions with supremum norm). Therefore convergence in $W^{1,1}$ implies uniform convergence, which is clearly impossible. Also, on bounded domains the larger values of $p$ correspond to smaller spaces, so the case $p=1$ is in fact the general case. I don't really understand the second question. Convergence in $W^{k,p }$ for $k\ge1$ certainly implies convergence in $W^{1,p}$. The intersection of $C^{\infty }$ with any Sobolev space is never complete; in most cases (namely finite $p$) it is dense. -	{"extraction_info": {"found_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": 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": 0.9638782143592834, "perplexity_flag": "head"}
http://mathoverflow.net/questions/37682?sort=oldest	## Computing equivalent vector of random variables from covarience matrix ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given a covariance matrix, how can I construct a vector of expressions of randomly distributed variables whose covariance matrix is equal to the given one? EDIT: All variables are normally distributed. I have an algorithm that gets the covariances correct, but not the variances on the diagonal: ````a = [0]*len(r) for x, row in enumerate(cov_matrix(r)): for y, item in enumerate(row): if x > y: continue v = noise(math.sqrt(abs(item))) a[x] += v if item > 0: a[y] += v else: a[y] -= v ```` I feel like this should be simple ... - Meta-question: Why did you write that as a comment instead of an answer? – Forrest Sep 4 2010 at 14:59 Good point. I was originally planning to leave a short comment, but I ended up giving more detail than I had thought I would. I deleted the comment and reposted it as an answer (and added a paragraph generalizing the construction slightly). – Darsh Ranjan Sep 4 2010 at 23:41 ## 2 Answers If $A$ is your target covariance matrix and $LL^T = A$, and $x = (x_1, \ldots, x_n)$ is a vector of independent random variables with mean zero and variance 1, then $y = Lx$ has the required covariance. Here $L$ is a matrix and $L^T$ is its transpose. $L$ can just be the Cholesky factor of $A$. ((Check: $\mathrm{cov}(y) = E[yy^T] = E[(Lx)(Lx)^T] = E[Lxx^TL^T] = LE[xx^T]L^T$ (by linearity of expectation) $= L\mathrm{cov}(x)L^T = LIL^T = LL^T = A$. $\mathrm{cov}(y) = E[yy^T]$ because $y$ has mean 0, and likewise for $\mathrm{cov}(x)$.) That's not too far from a "complete" solution, actually. If you start with a vector $y$ of random variables with mean zero and covariance matrix $A$, then if $A = LL^T$ and $x = L^{-1}y$, then $\mathrm{cov}(x) = I$. That doesn't necessarily imply that the components of $x$ are independent; it means they are uncorrelated. So the most general construction is to begin with a vector $x$ of uncorrelated random variables with mean zero and variance 1 and let $y=Lx$. (I only mean that every example can theoretically be obtained that way, not that it's necessarily the best or most computationally efficient way to do it.) - It depends on what you are doing. Say you want uniformly distributed random variables with a particular correlation structure. You can't just generate uncorrelated uniform random variables and apply the Cholesky decomposition because a sum of uniform random variables is no longer uniform. It will have the correct correlation (as you show) but the marginal distributions will not be uniform. – Robby McKilliam Sep 4 2010 at 23:47 The way I understood the problem, the only goal is to produce a random vector with a prescribed covariance matrix. If more information is prescribed, like the marginal distributions of the components, then I don't know if my method can be extended. – Darsh Ranjan Sep 5 2010 at 1:32 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This question is perhaps more suited to stats exhange. Darsh suggested using the Cholesky decomposition, but this only works if the distribution of the random variables you want to generate is Gaussian. Otherwise there are two techniques that I know of, the Iman-Conover method and the methods based on Copulas. - "Using [...] Cholesky decomposition [...] only works if the distribution of the random variables you want to generate is Gaussian": No (re-read Darsh's comment). – Didier Piau Sep 4 2010 at 6:29 @Didier: See my comment on Darsh's answer. – Robby McKilliam Sep 5 2010 at 2:06	{"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": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9100992679595947, "perplexity_flag": "head"}
http://mathoverflow.net/questions/50366/logdet-maximization-under-logdet-inequalitie-constraint	## logdet maximization under logdet inequalitie constraint ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Consider the following convex problem $\max_{\mathbf{A}} \log \|\mathbf{I} + \mathbf{CA} \|$ subject to $\log \|\mathbf{I} + \mathbf{FA} \| \leq \xi$ where $\mathbf{A}$ is a semidefinite positive matrix and $\mathbf{I}$ is the identity matrix. Does exist any closed form solution to $\mathbf{A}$? If not, does exist efficient algorithms to solve this problem? Thanks in advance, Mikitov - 1 I am curious as to whether the problem is really convex. The determinant is log-concave for positive definite arguments, and maximization of a concave function is equivalent to the minimization of a convex function -- so that's ok; however, your constraint is concave. Can you tell us anything about $\mathbf{C}$ and $\mathbf{F}$? – Gilead Dec 25 2010 at 19:19 I'm sorry, I meant your constraint may be concave, not is concave. – Gilead Dec 25 2010 at 19:34 Also, I think it is possible that the problem will not have a simple closed-form solution. If you consider the KKT system for this problem, the inequality constraint will induce complementarity constraints ($\mu \cdot (\log \|\mathbf{I} + \mathbf{FA} \| - \xi) =0$), which need to be resolved algorithmically. – Gilead Dec 26 2010 at 2:32 $\mathbf{F}$ and $\mathbf{C}$ are meant to be herminitian. Yes, being rigourous the optimization problem is $\min_{A} \log \| \mathbf{I} + \mathbf{FA}\|$ subject to $\log \| \mathbf{I} + \mathbf{CA}\| \geq \epsilon$ and also $\mathbf{A}$ being semidefinite positive. Can this problem be casted as a convex problem? Ok about the closed form solution. Thanks, Mikitov – mikitov Dec 26 2010 at 10:31 1 Off-hand, it seems to me that the problem is convex only if $\mathbf{I}+\mathbf{CA}\succ\mathbf{0}$ and $\mathbf{I}+\mathbf{FA}\prec\mathbf{0}$. If that is the case, you may be able to pose this as some kind of semidefinite program -- but that takes me out of my province. – Gilead Dec 26 2010 at 22:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "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": 0.9324663281440735, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/56988/set-theory-proving-statements-about-sets/56990	Set Theory: Proving Statements About Sets Let A, B, and C be arbitrary sets taken from the positive integers. I have to prove or disprove that: A ∩ B ∩ C = ∅, then (A ⊆ ~B) or (A ⊆ ~C) Here is my disproof using a counterexample: If A = {} the empty set, B = {2, 3}, C = {4, 5} With these sets defined for A, B, and C, the intersection includes the disjoint set, and then that would lead to A being a subset of B or A being a subset of C which counteracts that if A ∩ B ∩ C = ∅, then (A ⊆ ~B) or (A ⊆ ~C) Is this a sufficient proof? - 1 The empty set is a subset of any set so you do not arrive at a contradiction. I'm not sure if the "~" is significant in your notation but ignoring it, the claim is clearly false. Just take A,B, and C to be the singletons {1}, {2}, and {3} respectively. The intersection of the three is empty and none of these set s is a subset of the others. – user3180 Aug 11 '11 at 23:29 1 the ~ means complement, so it is important – Krysten Aug 11 '11 at 23:31 1 Ok, so ignore my first counterexample. But the empty set is still a subset of every set. How about this. Take A,B, and C to be {1,2}, {2,3}, and {1,3}. If I understand the question correctly, this should work. – user3180 Aug 11 '11 at 23:34 Right, so why can't I use that to disprove this – Krysten Aug 11 '11 at 23:36 1 Sorry, I may still be confused about the question but to produce a counterexample (at least as I understand the question) you need three sets which do not all contain a common element, and the first set, A, cannot be a subset of ~B or ~C. Since the empty set is a subset of everything, it is a subset of ~B and ~C which is actually consistent with the result. – user3180 Aug 11 '11 at 23:42 show 5 more comments 2 Answers Let $A=\{2,3\}$, $B=\{1,3\}$, and $C=\{1,2\}$. The intersection of the three sets is empty. But none of them is a subset of the complement of another. By symmetry, it is sufficient for example to show that $A$ is not a subset of the complement $B^c$ of $B$. Note that $B^c$ consists of all integers except $1$ and $3$. Since $A$ does contain $3$, $A$ is not a subset of $B^c$. Comment: As has been pointed out in a comment by @Joe, the empty set is a subset of every set, so setting $A=\emptyset$ cannot give you a counterexample, whatever be the choice of $B$ and $C$. - Isn't the intersection of A and B 3 and the intersection of A and C is 2? – Krysten Aug 11 '11 at 23:38 Ahh, now I see how it works. Very Clever. – Krysten Aug 11 '11 at 23:46 Yes, but $A\cap B\cap C$ is the intersection of the three sets, and there is no number which is in all three sets. – André Nicolas Aug 11 '11 at 23:48 How did you know to pick those element for A, B, and C? Was it just trial and error? – Krysten Aug 11 '11 at 23:52 It was clear that we would have a counterexample if $A\cap B$ and $A\cap C$ are each non-empty. So I wanted to arrange for that, and have $A \cap B \cap C$ empty. I like symmetry, hence the example. – André Nicolas Aug 12 '11 at 0:00 show 2 more comments As another counterexample, let $A=\mathbb{N}$, and $B$ and $C$ be disjoint sets with at least one element each. Elaborating: On the one hand, since $B$ and $C$ are disjoint, $$A \cap B \cap C = \mathbb{N} \cap B \cap C = B \cap C = \emptyset.$$ On the other hand, $A = \mathbb{N}$ is not contained in the complement of $B$, since $B$ is not empty, and the same for $C$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.9389934539794922, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/57061/will-we-feel-right-up-inside-the-airplane-if-the-airplane-accelerates-toward-ear/57076	# Will we feel right up inside the airplane if the airplane accelerates toward earth at 20m/s^2? Suppose you are in the airplane and the airplane falls toward the earth with acceleration of $20m/s^2$(double of gravitational acceleration $g$). This double acceleration by airplane will cancel the gravitational acceleration and create gravitational acceleration upward from the earth. If this happens, you will be pushed to the ceiling of the airplane if there is such a airplane that falls vertically from its rest position. While falling, the environment in the airplane will be the same as on the earth because in both case you will be pushed in the direction of the acceleration. My question is if this happens, will we feel right up while while being pushed to the ceiling of airplane? Will we feel just standing on earth? - 1 – Bernhard Mar 16 at 21:40 ## 2 Answers If the plane accelerated down at 2g (ie 20m/s^2) then yes you would feel 1 gravity upward. If the plane was also upside down (!) you would feel that you were sitting in your seat normally - With respect to the aeroplane, the acceleration will be g upward(against the earth's gravitational pull). However a person in the plane will feel weightless (till he strikes the ceiling) as there is no normal reaction of the aeroplane acting on him. There is only the gravitational force exerted on him by the earth. He won't feel as if he were standing on the earth. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9500147700309753, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/5402/secure-mac-implies-that-probability-of-same-tags-on-different-messages-is-neglig?answertab=active	Secure MAC implies that probability of same tags on different messages is negligible So let any secure MAC (message authentication code) be given. Intuitively, I think it is clear that the probability of getting the same tag on two different messages is very small, i.e. negligible. I want to prove this statement mathematically. How can we (a) formulate this in an exact manner (because, negligible in what? I guess negligible in $\|k\|$, where $k$ is the chosen key - I am not sure about this) (b) (once we have a precise formulation) prove this statement? Supposing that this probability is not negligible, we'll have to construct a PPT adversary $\mathcal A$ that, somehow, is able to "beat" the system (which would be in contradiction with the fact that the MAC is secure). - If you'll have to construct a PPT adversary, then you'll only be able to show this for $\hspace{1.25 in}$ PPT-computable sequences of pairs of messages. $\:$ – Ricky Demer Nov 18 '12 at 6:06 The term neglible usually refers to some security parameter (which often is the key length). The same term is used in the definition of "secure" for your MAC, which will give you the point of attack. (Also, welcome to Cryptography Stack Exchange.) – Paŭlo Ebermann♦ Nov 19 '12 at 14:37 1 Answer Is this a homework problem? Here's a hint: Suppose that the probability of getting the same tag on two different messages is $p$. Show how to construct an adversary that breaks the MAC (i.e., forges a valid tag), and that has success probability $p$. Assuming the MAC is secure, what can you conclude about the possible range of values for $p$? - It's not a homework problem. By the way, I don't see the relevance of such a comment/question. – Dorothy Nov 18 '12 at 22:12 1 – D.W. Nov 18 '12 at 22:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9182525873184204, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/171161-conjugate-prior-gamma-distribution.html	# Thread: 1. ## Conjugate prior for gamma distribution Conjugate prior for the gamma distribution (with parameters $\alpha,\beta$) is supposed to be: $\frac{1}{Z}\frac{p^{\alpha -1}e^{-\beta q}}{{\gamma(\alpha)}^r \beta^{-\alpha s}}$ with the hyperparameters, p, q, r, s when both shape parameter and the scale parameter are known. Here I have a gamma distributed variable x, with parameters $\alpha$ and $\beta$. And I have a conjugate prior with hyperparemeters p,q,r and s. Can anybody help me solving the integration over the parameters $\alpha,\beta$? $\int_{a,b} \frac{1}{\gamma(\alpha)\beta^\alpha} x^{\alpha-1} e^{-x/\beta} \frac{1}{Z}\frac{p^{\alpha -1}e^{-\beta q}}{{\gamma(\alpha)}^r \beta^{-\alpha s}} d{a,b}$ It has been a long time since I took a calculus class in the university, so I don't remember much about integration solving. Any assistance will be appreciated!	{"extraction_info": {"found_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": 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": 0.8707534074783325, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/150004/evaluate-int-bar-z-dz?answertab=active	# Evaluate $\int \bar z dz$ How would I evaluate $\int \bar z dz$, with 1: the contour $\gamma$ being the straight line segment from $0$ to $1+i$ 2: the contour $\sigma$ being the straight line segment from $0$ to $1$, followed by the straight line segment from $1$ to $1+i$. For the first one, I tried working it from the definition $\int_\gamma f(z) dz =\int_a^bf(\gamma(t))\gamma'(t) dt$. The only question I have is, is defining $\gamma(t)=t+it$, $t\in [0,1]$ correct? For the second one, how would I define the contour $\sigma$ and thus evaluate $\int \bar z dz$? I can't seem to be able to define $\sigma$. Alternatively, is there any other way to evaluate $\int \bar z dz$ apart from going straight from the definition? - ## 1 Answer Your $\gamma$ for 1) is fine. For 2), just choose a path which connects $0$ and $1$ on, say, $[0,1]$ and $1$ and $1+i$ on $[1,2]$ such that path is continuous and piecewise differentiable on $[0,2]$. You could, for example, define $\sigma(t) := t$ (which is the same as $t +0i$) for $0 \le t \le 1$ and $\sigma(t) := 1+ (t-1)i$ for $1 \le t \le 2$ - Thanks @Thomas , for the path connecting $0$ and $1$, I should choose $t$? But how would I choose a path for $1$ and $1+i$. Is it $it$? – Derrick May 26 '12 at 11:34 1 Yes to both questions, of course with $\,0\leq t \leq 1$ – DonAntonio May 26 '12 at 11:42 1 added a possible parametrization of the path to the answer. – user20266 May 26 '12 at 11:43	{"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": 37, "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": 0.9507097601890564, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/92100-max-min-help-cant-find-critical-points.html	# Thread: 1. ## Max and Min, Help Cant find the critical points? Hi I would really appreciate if anyone could help me with the following exercises k, l, m, n. I cant seem to solve the derivatives to find the critical points. THANK VERY MUCH in advance. Attached Thumbnails 2. You need to use partial differenitation for these problems, then solve for $\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = \b{0}$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9393545389175415, "perplexity_flag": "middle"}
http://nrich.maths.org/5749	### Plane to See P is the midpoint of an edge of a cube and Q divides another edge in the ratio 1 to 4. Find the ratio of the volumes of the two pieces of the cube cut by a plane through PQ and a vertex. ### Four Points on a Cube What is the surface area of the tetrahedron with one vertex at O the vertex of a unit cube and the other vertices at the centres of the faces of the cube not containing O? ### Instant Insanity We have a set of four very innocent-looking cubes - each face coloured red, blue, green or white - and they have to be arranged in a row so that all of the four colours appear on each of the four long sides of the resulting cuboid. # Cheese Cutting ##### Stage: 5 Challenge Level: Visualisation of 3D objects is a very useful real-life skill used in a range of everyday situations, from packing suitcases and boxes into the back of a car to travel to university through to parking the car in a tight space upon arrival. We can practise and develop our 3D skills and intuition through the study of the following sorts of mathematical problems. Spatial problems are mathematically quite different to algebraic problems and you may find one type of problem easier or more difficult depending on how your brain works! I have some cubes of cheese and cut them into pieces using straight cuts from a very sharp cheese wire. In between cuts I do not move the pieces from the original cube shape. For example, with just one cut I will obviously get two smaller pieces of cheese, with two cuts I can get up to 4 pieces of cheese and with three cuts I can get up to $8$ pieces of cheese, as shown in the picture: Suppose I now make a fourth cut. How many individual pieces of cheese can I make? Suppose now that I am allowed more generally to cut the block $N$ times. Can you say anything about the maximum or minimum number of pieces of chesse that you will be able to create? Although you will not be able to determine the theoretical maximum number of pieces of cheese for $N$ cuts, you can always create a systematic cutting system which will generate a pre-detemined number of pieces (for example, making $N$ parallel cuts will always result in $N+1$ pieces of cheese). Investigate developing better cutting algorithms which will provide larger numbers of pieces. Using your algorithm what is the largest number of pieces of cheese you can make for $10$, $50$ and $100$ cuts? 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": 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": 0.9394541382789612, "perplexity_flag": "head"}
http://mathoverflow.net/questions/15444/the-phenomena-of-eventual-counterexamples/27550	## The phenomena of eventual counterexamples ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Define an "eventual counterexample" to be • $P(a) = T$ for $a < n$ • $P(n) = F$ • $n$ is sufficiently large for $P(n) = T\ \ \forall n \in \mathbb{N}$ to be a 'reasonable' conjecture to make. where 'reasonable' is open to interpretation, and similar statements for rational, real, or more abstractly ordered sets for $n$ to belong to are acceptable answers. What are some examples of eventual counterexamples, famous or otherwise, and do different eventual counterexamples share any common features? Could we build an 'early warning system' set of heuristics for seemingly plausible theorems? edit: The Polya conjecture is a good example of what I was trying to get at, but answers are not restricted to number theory or any one area. - 5 Your question seems interesting. Could you put in at least one elementary example to explain your formal definition? – Colin Tan Feb 16 2010 at 13:18 show 9 more comments ## 35 Answers It was once conjectured that factors of $x^n-1$ over the rationals had no coefficient exceeding 1 in absolute value. The first counterexample comes at $n=105$. - 23 And in fact these coefficients (eventually!!) grow exponentially fast. See cecm.sfu.ca/~ada26/cyclotomic for a nice compendium of cyclotomic polynomials with enormous coefficients. – Jacques Carette Feb 17 2010 at 3:53 1 I often heard this, but I've never seen a citation. Who conjectured that? – Kevin O'Bryant Jun 9 2010 at 2:13 2 @Kevin, I don't know. I thought I once came across a reference to someone who computed up to $n=100$ in the year 1940 or so, stopped there and made the conjecture, but I haven't had any luck tracking it down. Noticing the breakdown at 105 is attributed to Migotti, 1883, and a proof that the coefficients can be arbitrarily large is due to Schur, published by Emma Lehmer in 1936, so if I'm right about the computations in 1940 then they were done by someone who was out of the loop and perhaps it's best not to embarrass any descendants by dredging up the reference. – Gerry Myerson Jun 9 2010 at 3:32 show 2 more comments ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The least positive integer for which the equality \left\lceil \frac{2}{2^{1/n}-1}\right\rceil = \left\lfloor \frac{2n}{\log 2} \right\rfloor. fails is $n=777,451,915,729,368$. See http://oeis.org/A129935. Another example that I like is the number $f(n)$ of inequivalent differentiable structures on $\mathbb{R}^n$. We have $f(n)=1$ if $n\neq 4$, while $f(4)=c$, the cardinality of the continuum. - Strong Law of Small Numbers by Guy. Steve - 3 Item 13: D.H. & Emma Lehmer discovered that $2^n\equiv 3 (\mod n)$ for $n=4700063497,$ but for no smaller $n>1.$ – Victor Protsak Jun 13 2010 at 1:15 2 Do you mean $n > 2$? Last I checked we have $2^2 \equiv 3 \pmod{2}$. – David Mandell Freeman Jul 29 2010 at 20:28 7 @David Mandell Freeman: Huh? Isn't the left side even and the right side odd? – Sridhar Ramesh Jul 3 2011 at 20:36 show 2 more comments I'm trying to reconstruct an example I saw somewhere some years back. It goes something like this: $\gcd(n^5-5,(n+1)^5-5)=1$ is true for $n=1,2,\dots,1435389$ but fails for $n=1435390$ (when the gcd is 1968751). - 9 I found a place which has this example, and it has many more examples: math.niu.edu/~rusin/known-math/96/smallnums – Gerry Myerson Feb 17 2010 at 1:16 1 The resultant of $x^{17}+9$ and $(x+1)^{17}+9$ is some (large) integer, D. So there are polynomials $a(x)$ and $b(x)$ with integer coefficients and degree at most 16 such that $a(x)(x^{17)+9)-b(x)((x+1)^{17}+9)=D$. Now reduce modulo a prime p dividing D to get the equation $a(x)(x^{17)+9)=b(x)((x+1)^{17}+9)$ in Z_p[x]. Now $x^{17}+9$ has 17 distinct zeros in Z_p, and they can't all be zeros of a(x), so at least one of them is a zero of $(x+1)^{17}+9$, and you're done. – Gerry Myerson Feb 17 2010 at 2:23 3 I don't understand why sometimes I get to see a math preview and sometimes not. I didn't see one when I made the comment above and it appears that I left out some closing braces, so some formulas are missing. I don't know how to edit my comment to put those braces in, but it doesn't matter, since my argument was more complicated than necessary anyway. If the resultant of two polynomials is divisible by some prime p, then the two polynomials have a common factor over the integers modulo p. These polynomials either split completely or are irreducible, so they must have a common linear factor. – Gerry Myerson Feb 17 2010 at 5:17 show 1 more comment The essence of the phenomenon of eventual counterexamples is that a certain pattern that holds among small numbers, turns out not to be universal. In the very best examples, such as the examples provided in the other answers, which I have enjoyed very much, what we have is an easily described property $P(n)$, whose first failing instance is very large in comparison. Indeed, the quality of answer might be measured by the difference between the size of the description of the property and the size of the first failing instance of it. When an easily described property holds for a very long time and then suddenly fails at some very large number, we are surprised. Therefore, to my mind the phenomenon of eventual counterexamples is intimately wrapped up with the possibility of providing very short descriptions of enormous numbers. Surely we are all able easily to provide short descriptions of some very large numbers, such as $2^{100}$ or $2^{2^{100!}}$. In order to go beyond exponentiation and factorials, we might make use of other easily described functions exhibiting even more enormous growth. The Ackermann function, for example, defined by a simple one-line recursion, has diagonal values 1, 3, 7, 61, $2^{2^{2^{65536}}}$, with the next value $A(5)$ mind-bogglingly huge. All such examples, short descriptions of large numbers, can be systematically transformed into instances of eventual counterexamples. For if $d$ is a short description of an enormous number $N$, then the property $P(k)=$"$k$ does not exhibit $d$" is easily described and holds for all values $k$ below $N$, but not of $N$ itself. Thus, it does very well by the quality measure I mentioned above. So to my mind, the real issue is: what are the largest numbers that you can describe by a very short description? This question can be made precise by requiring the description to be expressible in a particular formal language. Once the language is rich enough, however, this problem will certainly wade into interesting foundational waters, for the question of whether a given description actually succeeds in describing a number---for example, "the length of the shortest proof of a contradiction in ZFC"---may be independent of our basic axioms, even if it is enormous. - 4 This is a great perspective – Q.Q.J. Jun 15 2010 at 14:32 3 Yes, but it seems that one has to take into account also the difficulty of generating the underlying sequence. For example, the polynomial x^2−x+41 gives primes up to x=40, and 40 is not a big number by "absolute" measure, it is big compared to say other polynomials in generating primes. – timur Oct 10 2010 at 3:15 2 2^2^2^65536 isn't `mind-bogglingly huge'?! – Bob Durrant May 20 2011 at 9:26 show 2 more comments In reference to the Prime Number Theorem (then Conjecture) both Gauss and Riemann further conjectured that $\pi(n) < Li(n)$ (where $\pi(n)$ is the number of primes from $1$ to $n$ and $Li(n)$ is the logarithmic integral, $\int_2^n \frac{1}{ln(t)}dt$). Although it has been proven that this does not hold (Littlewood), that there exists some $n$ such that $\pi(n) \geq Li(n)$, the first $n$ where this takes place is so huge no-one has worked it out yet (allegedly). The number is known as Skewes' Number. It is known to be between $10^{14}$ and $1.39822\times 10^{316}$, and strongly believed to be about $1.397162914\times 10^{316}$. (References at the foregoing link.) - Freeman Dyson observed in my presence that the sequence with initial condition $a_0=3,a_1=0,a_2=2$, and recurrence $a_{n+3}=a_{n+1}+a_{n}$ almost has the property that $n\mid a_n$ if and only if $n$ is prime or 1, except that it doesn't. He challenged us (grad students) to explain this near-phenomenon'', as it seems too close to being too good to be true to be coincidence. I've never seen an explanation. Since this is Math Overflow, I'll give the spoiler, the first counterexample is $n=271441$. - 7 a_n is the sum of the nth powers of the roots of x^3 = x + 1, so the divisibility follows from the fact that the Frobenius map permutes the roots of a polynomial. Are you asking for an explanation of the failure of the converse? I see no reason to expect the converse to be true. – Qiaochu Yuan Jun 9 2010 at 2:44 2 I guess if anything needs an explanation it's why does it take so long for a counterexample to turn up. These numbers are (I think) the "Perrin pseudoprimes," see research.att.com/~njas/sequences/A013998 – Gerry Myerson Jun 9 2010 at 3:40 1 Suppose for example that n = pq for p, q distinct primes and let a, b, c be the roots of x^3 = x + 1. In order for a^n + b^n + c^n to be divisible by n we require that a^q + b^q + c^q be divisible by p and a^p + b^p + c^p be divisible by q. This is just highly unlikely; one might expect that a^p, b^p, c^p and a^q, b^q, c^q are just the roots of some random irreducible cubic polynomial mod q and mod p, respectively. Replacing x^3 = x + 1 by an irreducible polynomial of higher degree might conceivably lead to even larger pseudoprimes. – Qiaochu Yuan Jun 9 2010 at 3:56 3 Here is a related perspective. a_n counts the number of closed walks of length n on a certain graph G on 3 vertices; the cyclic group Z/nZ acts on these walks in the obvious way and the residue of a_n mod n is the number of walks lying in an orbit which is not of full size. When n is prime, orbits can either have size p or size 1 and the latter can't occur if there are no loops in G, which there aren't. When n is composite, the situation is much more complicated and it would be very surprising if the number of walks in non-full orbits was still divisible by n. – Qiaochu Yuan Jun 9 2010 at 4:02 12 I think the spirit of the observation was akin to observing that ``$e^{\pi \sqrt{163}}$ is an integer, except that it isn't.'' Or that ``the image of $0,1,\dots$, under $x\mapsto x^2-x+41$ is always prime, except that it isn't.'' Now, nobody would expect these criteria hold, but it is shocking that such simple expressions can come so close. And ultimately, there is deeper meaning to the observations. In the current phenomenon, no informed number theorist would suspect that the sequence detects primes perfectly, but it is shocking (to me, at least) that so simple a sequence comes so close. – Kevin O'Bryant Jun 9 2010 at 15:52 show 3 more comments A famous example is the isomorphism problem for integral group rings: suppose $G$ and $H$ are two finite groups of order $n$ such that $\mathbb{Z}G \cong \mathbb{Z}H$ does it mean that $G \cong H$? It was proved to be true for many cases and for many $n$'s and I think it was believed to be true in all cases. Nonetheless, eventually a counter example was found, see http://www.jstor.org/pss/3062112. - Any finite loop space has the rational cohomology of a Lie group -- up to rank 65. From then on, there are counterexamples in every dimension. The smallest known dimension of a counterexampe is 1250, but whatever the actual smallest dimension is, counterexamples will occur in every dimension after that. - 3 And, for the record, a finite loop space is a finite CW-complex $X$ that is homotopy equivalent to $\Omega Y$ for some space $Y$. – André Henriques May 1 2011 at 21:53 The Mertens conjecture. - 3 I think that Littlewood's result on the difference between the number of primes <x and li(x) was a surprise to many. – paul Monsky Jun 9 2010 at 2:41 show 1 more comment The numbers 12, 121, 1211, 12111, 121111, etc., are all composite - until you get to the one with 138 digits, that's a prime. Saw this in a talk Lenny Jones gave at the New Orleans meeting earlier this month. - 3 If you take a random sequence that grows like 12*10^n, the prime number theorem says you have something like a 13% chance of making it to 137 digits without seeing a prime. So, even if you've seen that the first 137 numbers of the form 12111...11 are composite, is the conjecture that all such numbers are composite really a reasonable one to make? – Vectornaut Apr 21 2011 at 2:26 5 @Vectornaut, while I think your point is valid, it needs to be adjusted slightly, because the sequence is far from random. For example, in a pattern like that you won't get any primes unless the final digits are odd, and that increases the chance that any individual term is prime, which in turn decreases by quite a bit the chance that 137 terms are composite. – gowers Apr 21 2011 at 11:10 Smallest counterexample to "There is no positive integer $n$ such that the concatenation of (the decimal representation of) $n$ with itself is a square" is $n=13223140496$, according to http://www.research.att.com/~njas/sequences/A102567; $1322314049613223140496 = 36363636364^2$. - 2 Are all those 3s and 6s on the RHS an accident? – David Mandell Freeman Jul 29 2010 at 20:33 2 Yes - and no. If you look at research.att.com/~njas/sequences/A106497 which is the sequence of right sides, they are all highly patterned numbers, related to the decimal expansions of $a/11$ and $a/7$ for various $a$. Whether they must be of this form, I do not know. – Gerry Myerson Jul 29 2010 at 23:24 In answering another MathOverflow question on Graham's number, I quoted from Harvey Friedman's Enormous Numbers in Real Life. Perhaps eventual counterexamples bear some relation to proof strength in certain systems of logic? Anyway, that example there could be rephrased to fit the current question. Suppose I look at strings on three symbols, and given a word w of length n I look at subwords of the form (forgive the AWK notation) spc[i] = substr(w,i,i+1), i.e. those substrings starting at the ith character going for length i+1 characters. So spc[1] gets the first two characters of w, spc[2] == w[2]w[3]w[4], and so on. I manage to find, for every n that I can compute, a string w_n that I use for w above such that for 0 < i < j < = n/2, spc[i] is not a subsequence of spc[j]. Others find such examples for even larger values of n. It would be reasonable for me to believe I could find arbitrarily long strings with this property. Enter Harvey Friedman: "THEOREM 8.1. Let k >= 1. There is a longest finite sequence x1,...,xn from {1,...,k} such that for no i < j <= n/2 is xi,...,x2i a subsequence of xj,...,x2j. For k >= 1, let n(k) be the length of this longest finite sequence. Paul Sally runs a program for gifted high school students at the University of Chicago. He asked them to find n(1), n(2), n(3). They all got n(1) = 3. One got n(2) = 11. Nobody reported much on n(3). I then started to ask several mathematicians to give an estimate on n(3), some of them very famous. I got guesses like this: 60, 100, 150, 200, 300. They were not in combinatorics. Recently I asked Lovasz, telling him about these five guesses. He guessed 20,000. THEOREM 8.2. n(3) > A(7,184). Lovasz wins, as his guess is closer to A(7,184) than the other guesses. Recall the discussion about A(5,5) being incomprehensibly large. With the help of computer investigations (with R. Dougherty), I got: THEOREM 8.3. n(3) > A(7198, 158386). A good upper bound for n(3) is work in progress. Crude result: A(n,n) where n = A(5,5). " Here A(n,n) is defined earlier in Friedman's paper as an Ackermann-like sequence. I suspect n(3) squishes Graham's number quite unlike a galactic black hole absorbing a prion or even a quark. EDIT: I have been corrected; in the squishing hierarchy, n(4) squishes Graham's number, which squishes n(3). Again, unlike any physical realization I can imagine. END EDIT The moral here is: "Don't jump to conclusions without a sufficiently strong proof system as back up". Gerhard "Ask Me About System Design" Paseman, 2010.02.17 - The Borwein Integrals are integrals of products of the sinc function. They exhibit certain "apparent patterns" which, while eventually breaking down, are actually indicative of something larger at work. (The example given on the Wikipedia page is a good one.) - Here's another one, maybe mostly of historical interest. Fermat once conjectured that all numbers of the form $$p=2^{2^n}+1$$ are prime, which he had the means to verify up to $n=4$. It took more than 100 years until Euler showed that this fails at $n=5$. Today we still don't know if there are any other Fermat primes, so quite possibly Fermat's conjecture fails in the worst possible way. - - Shapiro inequality: Let $x_1,x_2\dots x_n,x_{n+1},x_{n+2}$ be positive real numbers with $x_{n+1}=x_1$ and $x_{n+2}=x_2$. Now the inequality $\sum_{i=1}^{n} \frac{x_i}{x_{i+1}+x_{i+2}} \geq \frac{n}{2}$ must be true if $n<14$ or if $n\leq 23$ and $n$ is odd. So $n=14$ is the first $n$ where a counterexample can be found. I know that 14 is not that large a number, but remember that for each n we have a problem with a lot of freedom. - One of my favourite examples in this context is the following: Define a sequence $(s_n)$ by $s_1=8$, $s_2=55$ and for $n\geq3$ $s_n$ the smallest integer such that $s_n/s_{n-1}>s_{n-1}/s_{n-2}$ so that $s_3=379$ as $379/55>55/8$. Then we have $s_n=6s_{n-1}+7s_{n-2}-5s_{n-3}-6s_{n-4}$ for $5\leq n\leq11056$ but not for $n=11057$ (I have lost track of the name of the person to whom this is due, but it is, nowadays, easily verified on a computer). - 3 This may have come out of David Boyd's research on PV and Salem numbers. – Gerry Myerson Jun 12 2010 at 23:53 3 I found the source; David W Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances In Number Theory 333-340, Oxford University Press, 1993, MR 96i:11017. Boyd had several earlier papers on Pisot sequences, and this example may also be given in one of the earlier papers. – Gerry Myerson Jun 15 2010 at 3:51 show 1 more comment Let $a_1=1$, $a_{n+1}=(1+a_1^2+a_2^2+\dots+a_n^2)/n$. Are all terms integer? No, the first non-integer is $a_{44}$. I do not know neither reference (my source is private communication by Dmitry Rostovsky, and he does not remember where is it from), nor deep reason (if they exist) why first 43 terms are integer. - 1 This is discussed in E15 of Guy, Unsolved Problems In Number Theory. He says F Gobel found the recursion yielded many integers, but Hendrik Lenstra found that first counterexample. Guy gives generalizations and many references. – Gerry Myerson Oct 11 2010 at 3:11 1 Following up some of those references, I found a claim that $a_1=11$, $a_{n+1}=a_n(a_n^2+n)/(n+1)$ gives integers up to (but not beyond) $n=600$ or so. – Gerry Myerson Oct 11 2010 at 5:57 - In this thread search down for the answer by sigfpe . - 2 Direct link: mathoverflow.net/questions/11517/… – JBL Sep 3 2010 at 13:20 The De Giorgi conjecture is true for dimensions $\leq 8$. I guess this doesn't really count because De Giorgi himself only conjectured it for those dimensions based on the fact that Bernstein Theorem of minimal graphs is only true in dimensions $\leq 8$... (To stay within the realm of geometry, if someone finds a counterexample to the positive mass theorem in high dimensions, that would be an example too.) - - R. M. Grassl and A. P. Mullhaupt, "Hook and Shifted Hook Numbers", Discrete Mathematics, Volume 79, Number 2, January (1990) pp. 153-167 "An infinite number of counter examples is provided for the conjecture that a shifted tableau shape is uniquely determined by its multiset of shifted hook numbers. Nevertheless, the previous conjecture of the first author that there was only one example of nonuniqueness is discussed and it is shown that it is «almost» true, based on computer search." There were about five million examples before the counterexample, and approximately 1 mole of examples before the next counterexample is thought to occur. - 1 The way I read the abstract (at sciencedirect.com/…), the 2nd counterexample came up somewhere in the first five million cases, and it was the 3rd counterexample that was expected to be a mole away. – Gerry Myerson Feb 18 2010 at 23:16 show 1 more comment D. H. Lehmer showed that the first prime value of the Ramanujan tau-function, defined by $$\sum_{n=1}^\infty \tau(n) q^n = q \prod_{n=1}^\infty (1-q^n)^{24} = q - 24q^2 + 252q^3 - 1472q^4 + \dots,$$ occurs at the 63001st term. This is slightly less surprising when one knows that prime values can only occur for odd square inputs. - 1 When does the first zero value occur? :P – Victor Protsak Jun 12 2010 at 22:50 show 1 more comment From the Wikipedia category of disproved conjectures: • Borsuk's conjecture • The Chinese hypothesis • Euler's sum of powers conjecture - This is a bit tongue-in-cheek, but what about Special Relativity? In this case let property $P(x), x\in \mathbb{R}$ be the property that a given velocity $x$ is attainable. After all, Galilean Transforms allow one to change to a frame moving at an arbitrary velocity. Only Einstein's interpretation of the discoveries of Lorenz and Poincaré allowed for us to realize that property $P$ is only true if $x \in [-3 \times 10^8, 3 \times 10^8]$ - The first counterexample to the second Hardy-Littlewood conjecture is expected to occur somewhere between $10^{174}$ and $10^{1199}$ (at least, according to the references from the Wikipedia page), though it has not yet been definitively established that such a counterexample exists. - 1 I fixed the link (hopefully...) – Julien Melleray Mar 30 2012 at 6:48 show 1 more comment I'm surprised no one has mentioned Graeco-Latin Squares http://en.wikipedia.org/wiki/Graeco-Latin_square Euler showed these exist for $n$ odd, or any multiple of 4. As none exist for $n=2$ or $6$, he conjectured that none exist for any $n\equiv 2 (mod 4)$. As it happens, such exist for any $n\geq 3$ except $6$. This is quite a famous example, if small. - Robert Baillie has a paper on arxiv today (http://arxiv.org/abs/1105.3943) which shows how in principle one can construct examples of formulae which hold for $N=0,1,2,\ldots,k$, for arbirtrarily large $k$, then fail for all larger $N$. His largest example holds with $k\approx \exp(\exp(\exp(\exp(\exp(\exp(e))))))$. - 1 That's the same paper Seva linked to in his answer, posted about 3 hours earlier than this one. – Gerry Myerson May 20 2011 at 10:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 127, "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": 0.3, "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": 0.9328598380088806, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/52812-q-ary-code-proof-using-spheres.html	# Thread: 1. ## q-ary code proof using spheres Show that there is no q-ary code of length 5 and distance 3 with more than $\frac{q^5}{5q-4}$ codewords. You should prove this from first principles using spheres and without using known bounds on the number of codewords. This is my working so far: First prove that there is a q-ary code of length 5 and distance 3 with exactly $\frac{q^5}{5q-4}$ codewords. consider the spheres of radius 1 centred on codewords. Each such sphere contains 1 + 5 = 6 words. Since the code has distance 3, no two such spheres intersect. Otehrwise we would have a pair of codewords at distance 2 or less. Since there are $\frac{q^5}{5q-4}$ codewords, there are $\frac{6q^5}{5q-4}$ distinct words in these spheres. However there are only $q^5$ distince q-ary words of length 5. therefore we have to show that $\frac{6q^5}{5q-4}$ = $q^5$ for some q that is a natural non-zero number. i worked out q = 2 therefore such a code exists. but now im having problems proving that no q-ary code with MORE THAN $\frac{q^5}{5q-4}$ codewords exists. also, with q = 2 the no of codewords = $\frac{16}{3}$ can anyone tell me why this isnt a whole number? 2. There is some mistake in your argument. If you look at a sphere with radius one, then in order a word has distance 1 to the center if they differ in one place. Since it is of length 5 then we can choose one of this place. BUT for each place we have a (q-1) way to change it . So the number of codes in this sphere is 1+5(q-1)=5q-4. As you said all of this sphere are disjoint. If there are N codes then we need the total word to be less or equal to q^5. Then we need $(5q-4)N\leq q^5$. There fore $N\leq \frac{q^5}{5q-4}$ as desired. 3. Originally Posted by watchmath There is some mistake in your argument. If you look at a sphere with radius one, then in order a word has distance 1 to the center if they differ in one place. Since it is of length 5 then we can choose one of this place. BUT for each place we have a (q-1) way to change it . So the number of codes in this sphere is 1+5(q-1)=5q-4. As you said all of this sphere are disjoint. If there are N codes then we need the total word to be less or equal to q^5. Then we need $(5q-4)N\leq q^5$. There fore $N\leq \frac{q^5}{5q-4}$ as desired. thanks i sort of understand what ur saying, but how do i solve $N\leq \frac{q^5}{5q-4}$? or how do i prove that there is no such answer $N$ when i dont even know $q$? 4. Maybe my wording is not quite nice. Another attempt. Let N be the number of codewords. As I proved earlier in each ball there are 1+5(q-1)=5q-4 q-ary words. Since we have N balls (because the center is the codewords) and the union of all of these balls is a subset of the set of all q-ary (some q-ary words can be outside the balls) then N(5q-4) (the number of all q-ary words in the balls) can not be bigger than the number of all q-ary, that is $N(5q-4)\leq q^5$. From this we conclude that $N\leq \frac{q^5}{5q-4}$. Therefore N, the number of N words is not more than $\frac{q^5}{5q-4}$. 5. Thanks so much!! Proof by contradiction i love 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": 0, "codecogs_latex": 19, "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": 0.9373009204864502, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/27111/i-have-a-statistic-how-do-i-calculate-its-distribution	# I have a statistic, how do I calculate its distribution? I am comparing images for correlation. The images are all correlated, but I would like to determine when one pair is much more highly correlated, relative to another pair. I am using as a statistic the difference in the logs of p-values produced by a Spearman rank correlation test. If this is difference is small, there is a normal degree of correlation. If this difference is big, there is a higher degree of correlation in one pair than the other pair. Let $A$, $B$, and $C$ be images. I would like to determine a distribution and calculate a p-value for $x = \vert \log(\mbox{SpearmanRankCorr}(A,B)) - \log(\mbox{SpearmanRankCorr}(A,C)) \vert$ - ## 1 Answer You might want to consult the theory about transformations of random variables, since a statistic is basically a function of a random variable which is the data. http://math.arizona.edu/~jwatkins/f-transform.pdf Also the best source is 'Statistical Inference' by George Casella. - Is this the only way to go about determining a distribution from a statistic? If I try to perform, $$f_{Y}(y) = f_{X}(g^{-1}(y)) \bigl\vert \dfrac{d}{dy}g^{-1}(y) \bigr\vert,$$ any statistic $Y = g(X)$ of the data, $X$, involving the Spearman rank correlation test will not be one-to-one. Assuming I had a statistic, $g(X)$, that was invertible, the underlying data, $X$, is image data that doesn't follow any simple distributions. (I guess I could model it as a complicated mixture of Gaussians, right?) Is there any other way to describe a distribution of an arbitrary statistic? – cjohnson318 Apr 25 '12 at 22:09 as far as i know this is the only way, unless there is a known approximation or result that i'm not aware of about your specific problem. Also about the fact that it is not one to one, you can use the following result from casella, which says that in the formula move you have the sum of the right hand side above for g_{1}, g_{2}, etc, where the g's correspond to each interval where the distribution is defined. for example in the case of the absolute value you have two components, one for (-\infty,0) plus other for (0,\infty). – raygozag Apr 26 '12 at 2:31 Whammy. I was hoping that there was some kind of boot-strapping solution. Thank you for pointing out the piece-wise inversion result, I did not know about that. – cjohnson318 Apr 26 '12 at 14:11	{"extraction_info": {"found_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": 0.9376311898231506, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/270752/lipschitz-continuity-result	# Lipschitz continuity result Hi: I'm reading a proof in a convex optimization book and I don't understand something. The author states the following: Let $Q$ be a subset of $R^{n}$. We denote by $C_{L}^{k,p}(Q)$, the class of functions with the following properties: A) any $f \in C_{L}^{k,p}(Q)$ is $k$ times differentiable on $Q$. B) Its $p$-th derivative is Lipschitz continuous on $Q$ with the constant $L$. i.e: $\\|f^{p}(x) - f^{p}(y)\\| \le L\\|x - y\\|$. This is fine. But then later, inside the proof, he states the following: if $f \in C_{L}^{2,1}(R^{n})$, then for any $s \in R^{n}$ and $\alpha > 0$, we have $$\left\\|\left(\int_{0}^{\alpha}f''(x + \tau ~ s)d ~ \tau\right) s\right\\| = \\|f^{\prime}(x + \alpha s) - f^{\prime}(x) \\| \le \alpha L \\|s\\|\;.$$ I don't understand how this is implied by the definition of Lipschitz continuity defined previously. Thanks a lot. - ## 1 Answer Just putting in the definition of $f\in C^{2,1}_L(\mathbb{R}^n)$ gives $$\\|f^{(1)}(x+\alpha s)-f^{(1)}(x)\\|\le L\\|x+\alpha s -x\\|=L\\|\alpha s\\| =L\alpha \\|s\\|$$ where I used positive homogeneity of the norm $\\|\cdot\\|$ in the last equality. - Hi Pavel: The book I am reading is titled "Introductory Lectures on Convex Optimization" by Yuri Nesterov. The guy is a god in the field so I'm pretty sure the statement is correct. Hi n.c.: Thank you for your help. I'll print it out and see if I follow it. – mark leeds Jan 5 at 15:06 No need to; I misread the definition. – user53153 Jan 5 at 15:57	{"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": 15, "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": 0.9057154655456543, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/31326/how-should-one-interpret-the-comparison-of-means-from-different-sample-sizes	# How should one interpret the comparison of means from different sample sizes? Take the case of book ratings on a website. Book A is rated by 10,000 people with an average rating of 4.25 and the variance $\sigma = 0.5$. Similarly Book B is rated by 100 people and has a rating of 4.5 with $\sigma = 0.25$. Now because of the large sample size of Book A the 'mean stabilized' to 4.25. Now for 100 people, it may be that if more people read Book B the mean rating may fall to 4 or 4.25. • how should one interpret the comparison of means from different samples and what are the best conclusions one can/should draw? For example - can we really say Book B is better than Book A. - Are you specifically interested in the rating context? – Jeromy Anglim Jun 29 '12 at 5:21 @JeromyAnglim - Hmmm...probably. Not sure. That's the most common example. What did you have in mind? – PhD Jun 29 '12 at 6:27 2 See my answer regarding Bayesian rating systems below. Applied rating contexts typically have hundreds or thousands of objects being rated, and the aim is often to form the best estimate of the rating for the object given the available information. This is very different to a simple two group comparison as you might find say in an medical experiment with two groups. – Jeromy Anglim Jun 29 '12 at 6:31 ## 2 Answers You can use a t-test to assess if there are differences in the means. The different sample sizes don't cause a problem for the t-test, and don't require the results to be interpreted with any extra care. Ultimately, you can even compare a single observation to an infinite population with a known distribution and mean and SD; for example someone with an IQ of 130 is smarter than 97.7% of people. One thing to note though, is that for a given $N$ (i.e., total sample size), power is maximized if the group $n$'s are equal; with highly unequal group sizes, you don't get as much additional resolution with each additional observation. To clarify my point about power, here is a very simple simulation written for R: ````set.seed(9) # this makes the simulation exactly reproducible power5050 = vector(length=10000) # these will store the p-values from each power7525 = vector(length=10000) # simulated test to keep track of how many power9010 = vector(length=10000) # are 'significant' for(i in 1:10000){ # I run the following procedure 10k times n1a = rnorm(50, mean=0, sd=1) # I'm drawing 2 samples of size 50 from 2 normal n2a = rnorm(50, mean=.5, sd=1) # distributions w/ dif means, but = SDs n1b = rnorm(75, mean=0, sd=1) # this version has group sizes of 75 & 25 n2b = rnorm(15, mean=.5, sd=1) n1c = rnorm(90, mean=0, sd=1) # this one has 90 & 10 n2c = rnorm(10, mean=.5, sd=1) power5050[i] = t.test(n1a, n2a, var.equal=T)$p.value # here t-tests are run & power7525[i] = t.test(n1b, n2b, var.equal=T)$p.value # the p-values are stored power9010[i] = t.test(n1c, n2c, var.equal=T)$p.value # for each version } sum(power5050<.05)/10000 # this code counts how many of the p-values for [1] 0.6888 # each of the versions are less than .05 & sum(power7525<.05)/10000 # divides the number by 10k to compute the % [1] 0.4219 # of times the results were 'significant'. That sum(power9010<.05)/10000 # gives an estimate of the power [1] 0.3206 ```` Notice that in all cases $N=100$, but that in the first case $n_1=50$ & $n_2=50$, in the second case $n_1=75$ & $n_2=25$, and in the last case $n_1=90$ and $n_2=10$. Note further that the standardized mean difference / data generating process was the same in all cases. However, whereas the test was 'significant' 69% of the time for the 50-50 sample, power was 42% with 75-25 and only 32% when the group sizes were 90-10. I think of this by analogy. If you want to know the area of a rectangle, and the perimeter is fixed, then the area will be maximized if the length and width are equal (i.e., if the rectangle is a square). On the other hand, as the length and width diverge (as the rectangle becomes elongated), the area shrinks. - power is maximized?? I'm not quite sure I understand. Could you please provide an example if possible? – PhD Jun 29 '12 at 6:33 2 The reason the t test can handle unequal sample sizes is that it takes account of the standard error of the estimates of the means for each group. That is the standard deviation of the group's distribution divided by the square root of the group's sample size. The goup with the much larger sample size will have the smaller standard error if the population standard deviations are bith equal or nearly so. – Michael Chernick Jun 29 '12 at 16:03 @gung - I'm not sure I really know which 'language' this simulation is written. I'm guessing 'R'? and I'm still trying to decipher it :) – PhD Jun 29 '12 at 18:53 The code is for R. I have commented it to make it easier to follow. You can just copy & paste it into R and run it yourself, if you have R; the `set.seed()` function will insure you get identical output. Let me know if it's still too difficult to follow. – gung Jun 29 '12 at 19:41	{"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": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9308846592903137, "perplexity_flag": "middle"}
http://www.wikiwaves.org/Strip_Theory_Of_Ship_Motions._Heave_%26_Pitch	# Strip Theory Of Ship Motions. Heave & Pitch From WikiWaves Wave and Wave Body Interactions Current Chapter Strip Theory Of Ship Motions. Heave & Pitch Next Chapter Ship Roll-Sway-Yaw Motions Previous Chapter Forward-Speed Ship Wave Flows ## Ship motions with forward speed Ship advances in the positive x-direction with constant speed $U\,$. Regular waves with absolute frequency $\omega_0\,$ and direction $\beta\,$ are incident upon the ship. The ship undergoes oscillatory motions in all six degrees of freedom $\xi_j(t), \quad j=1,\cdots,6\,$. The ship seakeeping problem will be treated in the frequency domain under the assumption of linearity. ## Ship-hull boundary condition The relevant rigid body velocities for a ship translating with a constant forward velocity and heaving and pitching in head waves are: $\vec{U}=U\vec{i}; \quad \mbox{Forward speed} \,$ $\vec{\dot{\xi}}_3(t)=\dot{\xi}_3(t)\vec{k}; \quad \mbox{Heave velocity (positive up)} \,$ $\vec{\dot{\xi}}_5(t)=\dot{\xi}_5(t)\vec{j}; \quad \mbox{Pitch angular velocity (positive,bow down)} \,$ Let $\vec{n}(t)\,$ the time-dependent unit normal vector to the instantaneous position of the ship hull. Let $\vec{n_0}\,$ be its value when the ship is at rest. • The body boundary condition will be derived for small heave & pitch motions and will be stated on the mean position of the ship at rest. The nonlinear boundary condition on the exact position of the ship hull is: $\frac{\partial\Phi}{\partial n}=\overrightarrow{V} \cdot \vec{n} \,$ where, $\Phi=\Phi_3 + \Phi_5 \,$ $\overrightarrow{V} \cong U \vec{i} + \vec{K} \left( \dot{\xi}_3 - X \dot{\xi}_5 \right) \,$ • The unsteady velocity potential $\Phi\,$ has been written as the linear superposition of the heave and pitch components. The total rigid-body velocity is the sum of vertical velocity due to the ship heave and pitch. • The normal vector$\vec{n}(t)\,$ is defined by $\vec{n}(t) = \vec{n}_0 + \vec{\xi}_5 \times \vec{n}_0 \,$ where $\vec{\xi}_5(t)\,$ is the ship pitch rotation angle, and: $\vec{n}_0 \equiv (n_1,n_2,n_3) \,$ Keeping only the unsteady components: $\vec{V} \cdot \vec{n} = U \xi_5 n_3 + \left( \dot{\xi}_3 - X \dot{\xi_5} \right) n_3 = \frac{\partial\Phi}{\partial n}$ • Note that the steady component $U n_1 \,$ has already been accounted for in the statement of the steady flow. Let $\Phi=\mathrm{Re}\left(\phi e^{i\omega t} \right), \quad \phi=\Pi_3 X_3 + \Pi_5 X_5$ where $\phi_3\,$ and $\phi_5\,$ are the unit-amplitude heave & pitch complex velocity potentials in time-harmonic flow. Also: $\xi_3(t) = \mathrm{Re} \left\{ \Pi_3(\omega) e^{i\omega t} \right\}, \quad \xi_5(t) = \mathrm{Re} \left\{ \Pi_5(\omega) e^{i\omega t} \right\}$ It follows that: $\frac{\partial X_3}{\partial n} = i \omega n_3; \quad \mbox{on} \ \bar{S}_B \,$ $\frac{\partial X_5}{\partial n} = - i \omega X n_3 + U n_3; \quad \mbox{on} \ \bar{S}_B \,$ • Note the forward-speed effect in the pitch boundary condition and no such effect in heave. • Diffraction problem: $\phi=\phi_7 \,$ $\frac{\partial\phi_7}{\partial n} = -\frac{\partial\phi_I}{\partial n}, \quad \frac{\partial}{\partial n} \equiv \vec{n} \cdot \nabla \,$ where: $\phi_I = \frac{i g A}{\omega_0} e^{KZ-iKX\cos\beta - iKY\sin\beta} \,$ • Radiation problem: $\phi = \phi_3 + \phi_5 \,$ We will consider the special but important case of heave & pitch which are coupled and important modes to study in the ship seakeeping problem. • Heave and pitch are the only modes of interest in head waves ($\beta = 180^\circ \,$) when all other modes of motion (Roll-Sway-Yaw) are identically zero for a ship symmetric port-starboard. Surge is nonzero but generally small for slender ships and in ambient waves of small steepness. • Bernoulli equation The linear hydrodynamic disturbance pressure due to unsteady flow disturbances is given relative to the ship frame: $P = - \rho \left(\frac{\partial}{\partial t} - U \frac{\partial}{\partial X} \right) \Phi \,$ where $\Phi\,$ is the respective real potential. ## Radiation problem $\Phi = \mathrm{Re} \left\{ \phi e^{i\omega t} \right\} \,$ $P = \mathrm{Re} \left\{ \mathbb{P} e^{i\omega t} \right\} \,$ $\mathbb{P} = - \rho \left( i\omega - U \frac{\partial}{\partial X} \right) \left( \Pi_3 \phi_3 + \Pi_5 \phi_5 \right)$ where the complex velocity potentials satisfy the 2D boundary value problems derived earlier for slender ships. • The hydrodynamic pressure will be integrated over the ship hull to obtain the added-mass and damping coefficients next: $F_i(t) = \iint_{\bar{S}_B} P n_i \mathrm{d}S, \quad i = 3,5 \,$ The expressions derived below extend almost trivially to all other modes of motion. In general, $n_i \equiv \vec{n} = \left(n_1,n_2,n_3\right);\quad i=1,2,3$ $n_i \equiv \vec{X} \times \vec{n} = \left(n_4,n_5,n_6 \right); \quad i=4,5,6$ So for heave: $n_i = n_3\,$ And pitch: $n_i = n_5 = - X n_3 + X n_1 \simeq - X n_3 \left(n_1 \ll n_3 \right) \,$ Expressing $F_i \,$ in terms of the heave & pitch added mass & damping coefficients when the ship is forced to oscillate in calm water, we obtain when accounting for cross-coupling effects: $F_i^H (t) = - \sum_{j=3,5} \left[ A_{ij} \frac{\mathrm{d}^2\xi_j}{\mathrm{d}t^2} + B_{ij} \frac{\mathrm{d}\xi_j}{\mathrm{d}t} \right]$ Hydrostatic restoring effects are understood to be added to $F_i^H\,$. Introducing complex notation: $\mathbb{F}_i^H(t) = \mathrm{Re} \left\{ \mathbb{F}_i e^{i\omega t} \right\} \,$ $\mathbb{F}_i(\omega) = \left[ \omega^2 A_{ij} (\omega) - i\omega B_{ij} (\omega) \right] \Pi_j$ where the summation notation over $i\,$ is understood hereafter. Introducing the definition of $\mathbb{F}_i\,$ in terms of the hydrodynamic pressure we obtain after some simple algebra: $\mathbb{P} = \mathbb{P}_j \Pi_j \equiv \mathbb{P}_3 \Pi_3 + \mathbb{P} \Pi_5$ $\mathbb{F}_i = \left( \iint_{\bar{S_B}} \mathbb{P}_j n_j \mathrm{d}S \right) \Pi_j$ And: $\omega^2 A_{ij} (\omega) - i\omega B_{ij}(\omega) = \iint_{\bar{S_B}} \mathbb{P}_i n_j \mathrm{d}S$ where from Bernoulli: $\mathbb{P}_3 = \left( i\omega - U \frac{\partial}{\partial X} \right) \phi_3 = \left( i\omega - U \frac{\partial}{\partial X} \right) \left( i\omega X_3 \right)$ $\mathbb{P}_5 = \left( i\omega - U \frac{\partial}{\partial X} \right) \phi_5 = \left( i\omega - U \frac{\partial}{\partial X} \right) \left( -i\omega X_3 + U X_3 \right)$ ## Strip theory • Strip theory is a popular approximation of the 3-D Neumann-Kelvin formulation for ships which are slender as is most often the case when vessels are expected to cruise at significant forward speeds. The principal assumption is: $\frac{B}{L}, \ \frac{T}{L} = O(\varepsilon), \quad \varepsilon \ll 1 \,$ where $B: \mbox{Ship maximum beam} \,$ $T: \mbox{Ship maximum draft} \,$ $L: \mbox{Ship water line length} \,$ • The principal assumption of strip theory is that certain components of the radiation and diffraction potentials vary slowly along the ship length leading to a simplification of the n-K formulation. • In head or bow waves where heave and pitch attain their maximum values, the encounter frequency $\omega\,$ is usually high. ### Radiation problem The ship is forced to oscillate in heave & pitch in calm water while advancing at a speed $U\,$. • Due to slenderness the variation of the flow in the x-direction is more gradual than its variation around a ship section. So $\frac{\partial}{\partial X} \Phi \ll \frac{\partial\Phi}{\partial Y}, \ \frac{\partial\Phi}{\partial Z} \,$ where $\Phi = \Phi_3 + \Phi_5 \,$. Thus the 3D Laplace equation simplifies into a 2D form for the heave & pitch potentials. (The same argument applies to Roll-Sway-Yaw). Thus $\left( \frac{\partial^2}{\partial Y^2} + \frac{\partial^2}{\partial X^2} \right) \Phi_j \cong 0, \ \mbox{in fluid} \ j=2, \cdots, 6$ • The 2D equation applies for each "strip" location at station-X. • The ship-hull condition at station-X for the heave & pitch potentials remains the same: $\Phi = \mathrm{Re} \left\{ \phi e^{i\omega t} \right\} \,$ $\phi = \Pi_3 X_3 + \Pi_5 X_5 \,$ $\frac{\partial X_3}{\partial n} = i \omega n_3; \quad \mbox{on} \ \bar{S_B}$ $\frac{\partial X_5}{\partial n} = - i \omega X n_3 + U n_3; \quad \mbox{on} \ \bar{S_B}$ where now: $\left( \frac{\partial^2}{\partial Y^2} + \frac{\partial^2}{\partial Z^2} \right) X_j = 0, \quad \mbox{in fluid domain} \,$ • Define the normalized potential $\psi_3\,$: $\frac{\partial \psi_3}{\partial n} = n_3; \quad \mbox{on} \ \bar{S_B}$ $\left( \frac{\partial^2}{\partial Y^2} + \frac{\partial^2}{\partial Z^2} \right) \psi_3 = 0, \quad \ \mbox{in fluid}$ • There remains to simplify the 3D n-K free-surface condition. • The solution of the 2D BVP for $\psi_3\,$ along 30 - 40 stations used to describe the hull form of most ships can be carried out very efficiently by standard 2D panel method. • In terms of $\psi_3 (Y,Z; X) \,$ the heave & pitch potentials follow in the form: $\phi = \Pi_3 X_3 + \Pi_5 X_5 \,$ $X_3 = i \omega \psi_3 \,$ $X_5 = ( - i \omega X + U ) \psi_3 \,$ $\Phi = \mathrm{Re} \left\{ \phi e^{i\omega t} \right\} = \Phi_3 +\Phi_5$ • The 2D heave added-mass and damping coefficients due to a section oscillation vertically are defined by the familiar expressions $\mbox{Station-X}: \quad a_{33} (\omega) - \frac{i}{\omega} b_{33} (\omega) = \rho \int_{C_(X)} \psi_3 n_3 \mathrm{d}l$ where $\omega = \left\| \omega_0 - U \frac{\omega_0^2}{g} \cos\beta \right\| \,$ is the encounter frequency. $\left( i\omega - U \frac{\partial}{\partial X} \right)^2 \phi + g \phi_Z = 0, \quad Z=0$ • The ship slenderness and the claim that $\omega\,$ is usually large in head or bow waves is used to simplify the above equation as follows $- \omega^2 \phi + g \phi_Z = 0, \quad Z=0 \,$ by assuming that $\omega \gg \left\|U \frac{\partial}{\partial X} \right\|$. A formal proof is lengthy and technical. It follows that the normalized potential $\psi_3\,$ also satisfies the above 2D FS condition and is thus the solution of the 2D boundary value problem stated below at station-X: $\mbox{As} \ Y \to \infty; \quad \psi_3 \sim \frac{i g A \pm}{\omega} e^{KZ\mp iKy+i\omega t} \,$ . Upon integration along the ship length and over each cross section at station-X the 3D added-mass and damping coefficients for heave & pitch take the form: $A_{33} = \int_{L} \mathrm{d}X a_{33} (X) \,$ $B_{33} = \int_{L} \mathrm{d}X b_{33} (X) \,$ $A_{35} = - \int_{L} \mathrm{d}X a_{33} - \frac{U}{\omega^2} B_{33} \,$ $A_{53} = - \int_{L} \mathrm{d}X a_{33} + \frac{U}{\omega^2} B_{33} \,$ $B_{35} = - \int_{L} \mathrm{d}X b_{33} + \frac{U}{\omega^2} A_{33} \,$ $B_{53} = - \int_{L} \mathrm{d}X b_{33} - \frac{U}{\omega^2} A_{33} \,$ $A_{55} = \int_{L} \mathrm{d}X X^2 a_{33} + \frac{U^2}{\omega^2} A_{33} \,$ $B_{35} = \int_{L} \mathrm{d}X X^2 b_{33} - \frac{U^2}{\omega^2} B_{33} \,$ where the 2D added-mass and damping coefficients were defined above: $a_{33} - \frac{i}{\omega} b_{33} = \rho \int_{C(X)} \psi_3 n_3 \mathrm{d}l$ ### Diffraction problem • We will consider heave & pitch in oblique waves. Note that in oblique waves the ship also undergoes Roll-Sway-Yaw motions which for a symmetric vessel and according to linear theory are decoupled from heave and pitch. • Relative to the ship frame the total potential is: $\Phi = \Phi_I + \Phi_D = \mathrm{Re} \left\{ \left( \phi_I + \phi_D \right) e^{i\omega t} \right\}$ where $\omega\,$ is the encounter frequency, and: $\phi_I = \frac{i g A}{\omega_0} e^{KZ-iKX\cos\beta-iKY\sin\beta} \quad K=\frac{\omega_0^2}{g}$ Define the diffraction potential as follows: $\phi_0 = \frac{i g A}{\omega_0} e^{-iKX\cos\beta} \psi_7 (Y,Z;X)$ In words, factor-out the oscillatory variation $e^{-iKX\cos\beta} \,$ out of the scattering potential. • The ship slenderness approximation now justifies that: $\frac{\partial\psi_7}{\partial X} \ll \frac{\partial\psi_7}{\partial Y}, \ \frac{\partial\psi_7}{\partial Z} \,$ Note that this is not an accurate approximation for $\phi_D\,$ when $K=\frac{2\pi}{\lambda}\,$ is a large quantity or when the ambient wavelength $\lambda\,$ is small. • Substituting in the 3D laplace equation and ignoring the $\frac{\partial\psi_7}{\partial X}, \ \frac{\partial^2\psi_7}{\partial X^2} \,$ terms we obtain $\left(\frac{\partial^2}{\partial Y^2} + \frac{\partial^2}{\partial Z^2} - K^2 \cos^2 \beta \right) \psi_7 \cong 0 \,$ • This is the modified 2D Helmholtz equation in most cases the $K^2\cos^\beta\,$ term is not important for reasons to be discussed and the 2D Laplace equation follows. • The body boundary condition simply states: $\frac{\partial\phi_D}{\partial n} = - \frac{\partial\phi_I}{\partial n} \,$ with: $\frac{\partial}{\partial n} \equiv n_1 \frac{\partial}{\partial X} + n_2 \frac{\partial}{\partial Y} + n_3 \frac{\partial}{\partial Z}$ Due to slenderness: $n_1 \ll n_2, n_3 \,$ Invoking the slenderness approximations for the $\frac{\partial}{\partial X} \,$ derivatives of $\psi_7\,$, the body boundary condition simplifies to: $\frac{\partial\psi_7}{\partial N} = - \frac{\partial}{\partial N} \left( e^{KZ-iKY\sin\beta} \right); \ \mbox{on} \ C(X)$ with: $\frac{\partial}{\partial N} = N_2 \frac{\partial}{\partial Y} + N_3 \frac{\partial}{\partial Z}$ The potential $\psi_7\,$ satisfies the 2D Laplace equation, after dropping the $K^2 \cos^2\beta \,$ term, or $\left(\frac{\partial^2}{\partial Y^2} + \frac{\partial^2}{\partial Z^2} \right) \psi_7 = 0 \,$ And the free-surface condition derived next: The 3D linear free surface condition for $\phi_D\,$ is: $\left( i\omega - U \frac{\partial}{\partial X} \right)^2 \phi_D + g \frac{\partial \phi_D}{\partial Z} = 0; \quad Z=0$ The same condition is satisfied by $\phi_I\,$: $\left( i\omega - U \frac{\partial}{\partial X} \right)^2 \phi_I + g \frac{\partial\phi_I}{\partial Z} = 0; \quad Z=0$ Verify identically for $\phi_I\,$ that: $i \omega - U \frac{\partial}{\partial X} \equiv i \omega_0 \,$ So the forward-speed effect disappears from the free surface condition of $\phi_I\,$. The same will be shown to be approximately true for the diffraction potential $\phi_D\,$: $\phi_D = \frac{i g A}{\omega_0} e^{- i K X \cos \beta} \psi_7(Y,Z; X)$ $\frac{\partial\phi_D}{\partial X} = \frac{i g A}{\omega_0} \left\{ (- i K \cos \beta) \psi_7 (Y,Z; X) + \frac{\partial \psi_7}{\partial X} \right\} e^{-i K X \cos \beta}$ Due to slenderness: $\left\| \frac{\partial\psi_7}{\partial X} \right\| \ll \|K\cos\beta\| \|\psi_7\|,$ Therefore: $\frac{\partial\phi_D}{\partial X} \simeq - i K \cos \beta \phi_D$ And from the free-surface condition $\left( i\omega - U \frac{\partial}{\partial X} \right) \phi_D \simeq i \omega_0 \phi_D$ where: $\omega = \omega_0 - U K \cos \beta \,$ So the free-surface condition for $\phi_D\,$ becomes to leading order in slenderness: $- \omega_0^2 \phi_D + g \frac{\partial\phi_D}{\partial Z} = 0, \quad Z=0$ • So in the statement of the diffraction problem forward-speed effects are absent! The same is not true in the radiation problem even for slender ship. • This "independence" of the diffraction problem on forward speed effects is verified very well in experiments and in fully three-dimensional solutions. In summary, the boundary-value problem satisfied by the diffraction potential around a slender ship takes the form: $\phi_D = \frac{ i g A}{\omega_0} e^{-iKX\cos\beta} \psi_7 (Y,Z;X)$ $\left( \frac{\partial^2}{\partial Y^2} + \frac{\partial^2}{\partial Z^2} \right) \psi_7 = 0, \quad \mbox{in fluid} \,$ $- \omega_0^2 \psi_7 +9 g \frac{\partial\psi_7}{\partial Z} = 0, \quad \mbox{on} \ Z=0$ $\frac{\partial\psi_7}{\partial N} = - \frac{\partial}{\partial N} \left( e^{KZ-iKY\sin\beta} \right)$ Or in graphical form at station-X: • As $Y \to \pm \infty \qquad \psi_7 \sim \frac{ i g A_7^\pm}{\omega_0} e^{KZ \mp i K Y}$. The hydrodynamic pressure in the diffraction problem takes the form: $P = \mathrm{Re} \left\{ \mathbb{P} e^{i\omega t} \right\} \,$ $\mathbb{P} = - \rho \left( i\omega-U\frac{\partial}{\partial X} \right) \frac{i g A}{\omega_0} e^{-iKX\cos\beta} \left( \psi_I + \psi_7 \right)$ $\cong \rho g A e^{-iKX\cos\beta} \left( \psi_I + \psi_7 \right)$ where $\frac{\partial\psi_7}{\partial X} \,$ derivatives are dropped in favor of $K\cos\beta\,$, due to slenderness. The heave and pitch exciting forces follow simply by pressure integration $X_i(t) = \mathrm{Re} \left\{ \mathbb{X}_i e^{i\omega t} \right\}; \quad i=3,5$ $\mathbb{X}_3 = \rho g A \int_L \mathrm{d}X e^{-iKX\cos\beta} \int_{C(X)} \left( \psi_I + \psi_7 \right) n_3 \mathrm{d}l$ $\mathbb{X}_5 = \rho g A \int_L \mathrm{d}X (-X) e^{-iKX\cos\beta} \int_{C(X)} \left( \psi_I + \psi_7 \right) n_3 \mathrm{d}l$ where the approximation $n_5 = -X n_3 \,$ was used. • In summary, a version of strip theory was derived for slenderships which has been found to be the most rational and accurate relative to other alternatives. This is particularly true for the treatment of the diffraction problem. • The coupled heave & pitch equation of motion take the familiar form $\left[ - \omega^2 \left( A_{ij} + M_{ij} \right) + i\omega B_{ij} + X_{ij} \right] \pi_j = \mathbb{X}_i \left(\omega_0\right) \quad i,j = 3,5$ where in practice: $\omega = \left\| \omega_0 - U \frac{\omega_0^2}{g} \cos\beta \right\|$ And the $A_{ij}, \ B_{ij} \,$ are functions of $\omega\,$ defined above as integrals of their 2D counterparts $a_{33}\,$ & $b_{33}(\omega)\,$. • The heave & pitch exciting force amplitudes $\mathbb{X}_i(\omega_0)\,$ are functions of $\omega_0\,$, hence not dependent on $U\,$. • All hydrodynamic effects oscillate at $\omega\,$! Ocean Wave Interaction with Ships and Offshore Energy Systems	{"extraction_info": {"found_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": 154, "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": 0.9110390543937683, "perplexity_flag": "middle"}

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