keirp/tiny_math · Datasets at Hugging Face

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http://math.stackexchange.com/questions/44139/how-many-solutions-are-there-to-fn-m-n2nmm2-q/	# How many solutions are there to $F(n,m)=n^2+nm+m^2 = Q$? Let $n,m$ be two positive integers, we consider: $$F(n,m)=n^2+nm+m^2$$ Let $Q$ be one value reach by $F(n,m)$. How many different pairs $(n,m)$ verify $F(n,m)=Q$? - 6 The trick in this instance is that $n^2+nm+m^2$ is the norm in the ring $\mathbb{Z}[\omega]$ where $\omega=\frac{1+\sqrt{-3}}2$. Then $N(n+m\omega)=n^2+nm+m^2$. Since $\mathbb{Z}[\omega]$ is a unique factorization domain, you get results like the result for the sum of two squares. – Thomas Andrews Jun 8 '11 at 17:46 @Eric and Ray: Sorry about my erroneous answer; I was too quick to "jump the gun" and as I pursued the problem after posting, I realized I overlooked the restriction to positive m, n...and further, I realized there are far more solutions than my erroneous post revealed. – amWhy Jun 8 '11 at 17:47 @Thomas Andrews: That is an excellent point! I wish had seen that comment before typing my answer. – Eric♦ Jun 8 '11 at 18:35 @Amwhy: It happens to all of us, no worries. (I have jumped the gun like this many times!) – Eric♦ Jun 8 '11 at 18:42 ## 1 Answer As it turns out, we can give a complete answer to this question. The exact number of solutions depends on the prime factorization of $Q$ Specifically, it is a function of the exponents of the prime factors which are congruent to $1$ mod $3$, with the condition that all the factors congruent to $2$ modulo $3$ have their prime factors appear with even multiplicity. Note: I have not yet provided a proof, the result that primes of the form $1+3k$ can be represented is a theorem of Jacobi. Let $$Q=\prod_i q_i^{\alpha_i} \prod_i p_i^{\beta_i}$$ where the $q_i$ are $1$ mod $3$ and the $p_i$ are $2$ mod $3$. Our equation $n^2+nm+m^2$ has solutions if and only if each $\beta_i$ is even. Proof: Take the equation modulo $3$. By case analysis for $n,m$ the right hand side cannot be congruent to $2$, and hence the statement follows. Remark: Notice the following similarity to the sum of squares problem (points on a circle). The Answer: Suppose that as before $$Q=\prod_i q_i^{\alpha_i} \prod_i p_i^{\beta_i}$$ where the $q_i$ are $1$ mod $3$ and the $p_i$ are $2$ mod $3$. Suppose as well that all of the $\beta_i$ are even. (Otherwise we can have no solutions) Let $$B=(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_n+1).$$ Then the number of non-negative integer solutions to $$m^2+mn+n^2=Q$$ is exactly $$\left\lceil\frac{B}{2}\right\rceil.$$ (Again notice the similarity to the sum of squares function) In particular, if $l(n)$ is the number of representations where $n,m$ are any integers, (that is positive or negative) then $$l(n)=2B.$$ Hope that helps. - 3 This is a very nice answer. One comment: I'm not sure why you're diagonalizing -- you don't seem to use it anywhere. More fundamentally, the given quadratic form is the norm form of the maximal order of $\mathbb{Q}(\sqrt{-3})$, whereas $x^2 + 3y^2$ is the norm form of the nonmaximal order $\mathbb{Z}[\sqrt{-3}]$, which is not a UFD. So in fact if you wanted to study the form $x^2 + 3y^2$ you would do it by comparing to $x^2 + xy + y^2$, not the other way around. Note also that even without the change of variables you are looking for lattice points on an ellipse. – Pete L. Clark Jun 8 '11 at 18:45 2 – Pete L. Clark Jun 8 '11 at 19:11 @Pete: Ok, thanks for the note! I don't know why I thought it wasn't. – Eric♦ Jun 8 '11 at 19:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 42, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9421892166137695, "perplexity_flag": "head"}
http://nrich.maths.org/6484/index?nomenu=1	Imagine that I plot the pdfs for two normal distributions on the same axes. Could I choose the parameters so that the curves intersect $0$, $1$ or $2$ times? Imagine that I plot the cdfs for two normal distributions on the same axes. Could I choose the parameters so that the curves intersect $0$, $1$ or $2$ times?	{"extraction_info": {"found_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": 0.7794747352600098, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/tagged/elgamal-signature+forgery	# Tagged Questions 0answers 77 views ### Why do we need to hash both the message and the $h$ value in ElGamal signature? The professor left us a question on ElGamal signatures: Given the hash function $H$ and message $M$, choose a random $r$ and compute $h=g^r$ and $H(M\|\|h)$. Show that, if $H(M)$ is used instead of ... 0answers 37 views ### ElGamal signature: Forging a signature of a specific form I have a question I can't solve from one of the courses I'm currently taking: Show that given a legitimate ElGamal signature $(S,R)$ on a given message $m$, an attacker can compute a signature ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8702031373977661, "perplexity_flag": "middle"}
http://www.nag.com/numeric/CL/nagdoc_cl23/html/F12/f12fdc.html	NAG Library Function Documentnag_real_symm_sparse_eigensystem_option (f12fdc) Note: this function uses optional arguments to define choices in the problem specification. If you wish to use default . 1 Purpose nag_real_symm_sparse_eigensystem_option (f12fdc) is an option setting function in a suite of functions consisting of nag_real_symm_sparse_eigensystem_init (f12fac), nag_real_symm_sparse_eigensystem_iter (f12fbc), nag_real_symm_sparse_eigensystem_sol (f12fcc), nag_real_symm_sparse_eigensystem_option (f12fdc) and nag_real_symm_sparse_eigensystem_monit (f12fec), and may be used to supply individual optional arguments to nag_real_symm_sparse_eigensystem_iter (f12fbc) and nag_real_symm_sparse_eigensystem_sol (f12fcc). The initialization function nag_real_symm_sparse_eigensystem_init (f12fac) must have been called prior to calling nag_real_symm_sparse_eigensystem_option (f12fdc). 2 Specification #include <nag.h> #include <nagf12.h> void nag_real_symm_sparse_eigensystem_option (const char str, Integer icomm[], double comm[], NagError fail) 3 Description nag_real_symm_sparse_eigensystem_option (f12fdc) may be used to supply values for optional arguments to nag_real_symm_sparse_eigensystem_iter (f12fbc) and nag_real_symm_sparse_eigensystem_sol (f12fcc). It is only necessary to call nag_real_symm_sparse_eigensystem_option (f12fdc) for those arguments whose values are to be different from their default values. One call to nag_real_symm_sparse_eigensystem_option (f12fdc) sets one argument value. Each optional argument is defined by a single character string consisting of one or more items. The items associated with a given option must be separated by spaces, or equals signs $\left[=\right]$. Alphabetic characters may be upper or lower case. The string ```'Iteration Limit = 500' ``` is an example of a string used to set an optional argument. For each option the string contains one or more of the following items: – a mandatory keyword; – a phrase that qualifies the keyword; – a number that specifies an Integer or double value. Such numbers may be up to $16$ contiguous characters in C's d or g format. nag_real_symm_sparse_eigensystem_option (f12fdc) does not have an equivalent function from the ARPACK package which passes options by directly setting values to scalar arguments or to specific elements of array arguments. nag_real_symm_sparse_eigensystem_option (f12fdc) is intended to make the passing of options more transparent and follows the same principle as the single option setting functions in Chapter e04. The setup function nag_real_symm_sparse_eigensystem_init (f12fac) must be called prior to the first call to nag_real_symm_sparse_eigensystem_option (f12fdc) and all calls to nag_real_symm_sparse_eigensystem_option (f12fdc) must precede the first call to nag_real_symm_sparse_eigensystem_iter (f12fbc), the reverse communication iterative solver. A complete list of optional arguments, their abbreviations, synonyms and default values is given in Section 10. 4 References Lehoucq R B (2001) Implicitly restarted Arnoldi methods and subspace iteration SIAM Journal on Matrix Analysis and Applications 23 551–562 Lehoucq R B and Scott J A (1996) An evaluation of software for computing eigenvalues of sparse nonsymmetric matrices Preprint MCS-P547-1195 Argonne National Laboratory Lehoucq R B and Sorensen D C (1996) Deflation techniques for an implicitly restarted Arnoldi iteration SIAM Journal on Matrix Analysis and Applications 17 789–821 Lehoucq R B, Sorensen D C and Yang C (1998) ARPACK Users' Guide: Solution of Large-scale Eigenvalue Problems with Implicitly Restarted Arnoldi Methods SIAM, Philidelphia 5 Arguments 1: str – const char Input On entry: a single valid option string (as described in Section 3 and Section 10). 2: icomm[$\mathit{dim}$] – IntegerCommunication Array Note: the dimension, dim, of the array icomm must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{licomm}}\right)$ (see nag_real_symm_sparse_eigensystem_init (f12fac)). On initial entry: must remain unchanged following a call to the setup function nag_real_symm_sparse_eigensystem_init (f12fac). On exit: contains data on the current options set. 3: comm[$\mathit{dim}$] – doubleCommunication Array Note: the dimension, dim, of the array comm must be at least $60$. On initial entry: must remain unchanged following a call to the setup function nag_real_symm_sparse_eigensystem_init (f12fac). On exit: contains data on the current options set. 4: fail – NagError Input/Output The NAG error argument (see Section 3.6 in the Essential Introduction). 6 Error Indicators and Warnings NE_BAD_PARAM On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value. NE_INTERNAL_ERROR An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance. NE_INVALID_OPTION Ambiguous keyword: $〈\mathit{\text{value}}〉$ Keyword not recognized: $〈\mathit{\text{value}}〉$ Second keyword not recognized: $〈\mathit{\text{value}}〉$ Not applicable. None. 9 Example This example solves $Ax=\lambda Bx$ in ${\mathbf{Shifted Inverse}}$ mode, where $A$ and $B$ are obtained from the standard central difference discretization of the one-dimensional Laplacian operator $\frac{{\partial }^{2}u}{\partial {x}^{2}}$ on $\left[0,1\right]$, with zero Dirichlet boundary conditions. Data is passed to and from the reverse communication function nag_real_symm_sparse_eigensystem_iter (f12fbc) using pointers to the communication array. 9.1 Program Text Program Text (f12fdce.c) 9.2 Program Data Program Data (f12fdce.d) 9.3 Program Results Program Results (f12fdce.r) 10 Optional Arguments Several optional arguments for the computational functions nag_real_symm_sparse_eigensystem_iter (f12fbc) and nag_real_symm_sparse_eigensystem_sol (f12fcc) define choices in the problem specification or the algorithm logic. In order to reduce the number of formal arguments of nag_real_symm_sparse_eigensystem_iter (f12fbc) and nag_real_symm_sparse_eigensystem_sol (f12fcc) these optional arguments have associated default values that are appropriate for most problems. Therefore, you need only specify those optional arguments whose values are to be different from their default values. The remainder of this section can be skipped if you wish to use the default values for all optional arguments. The following is a list of the optional arguments available and a full description of each optional argument is provided in Section 10.1. Optional arguments may be specified by calling nag_real_symm_sparse_eigensystem_option (f12fdc) before a call to nag_real_symm_sparse_eigensystem_iter (f12fbc), but after a call to nag_real_symm_sparse_eigensystem_init (f12fac). One call is necessary for each optional argument. All optional arguments you do not specify are set to their default values. Optional arguments you do specify are unaltered by nag_real_symm_sparse_eigensystem_iter (f12fbc) and nag_real_symm_sparse_eigensystem_sol (f12fcc) (unless they define invalid values) and so remain in effect for subsequent calls unless you alter them. 10.1 Description of the Optional Arguments For each option, we give a summary line, a description of the optional argument and details of constraints. The summary line contains: • the keywords, where the minimum abbreviation of each keyword is underlined; • a parameter value, where the letters $a$, $i\text{ and }r$ denote options that take character, integer and real values respectively; • the default value, where the symbol $\epsilon $ is a generic notation for machine precision (see nag_machine_precision (X02AJC)). Keywords and character values are case and white space insensitive. Optional arguments used to specify files (e.g., ${\mathbf{Advisory}}$ and ${\mathbf{Monitoring}}$) have type Nag_FileID. This ID value must either be set to $0$ (the default value) in which case there will be no output, or will be as returned by a call of nag_open_file (x04acc). Advisory Default $\text{}=0$ (See Section 3.2.1.1 in the Essential Introduction for further information on NAG data types.) Advisory messages are output to Nag_FileID ${\mathbf{Advisory}}$ during the solution of the problem. Defaults This special keyword may be used to reset all optional arguments to their default values. Exact Shifts Default Supplied Shifts During the Lanczos iterative process, shifts are applied internally as part of the implicit restarting scheme. The shift strategy used by default and selected by the ${\mathbf{Exact Shifts}}$ is strongly recommended over the alternative ${\mathbf{Supplied Shifts}}$ (see Lehoucq et al. (1998) for details of shift strategies). If ${\mathbf{Exact Shifts}}$ are used then these are computed internally by the algorithm in the implicit restarting scheme. If ${\mathbf{Supplied Shifts}}$ are used then, during the Lanczos iterative process, you must supply shifts through array arguments of nag_real_symm_sparse_eigensystem_iter (f12fbc) when nag_real_symm_sparse_eigensystem_iter (f12fbc) returns with ${\mathbf{irevcm}}=3$; the real and imaginary parts of the shifts are supplied in y and mx respectively. This option should only be used if you are an experienced user since this requires some algorithmic knowledge and because more operations are usually required than for the implicit shift scheme. Details on the use of explicit shifts and further references on shift strategies are available in Lehoucq et al. (1998). Iteration Limit $i$ Default $\text{}=300$ The limit on the number of Lanczos iterations that can be performed before nag_real_symm_sparse_eigensystem_iter (f12fbc) exits. If not all requested eigenvalues have converged to within ${\mathbf{Tolerance}}$ and the number of Lanczos iterations has reached this limit then nag_real_symm_sparse_eigensystem_iter (f12fbc) exits with an error; nag_real_symm_sparse_eigensystem_sol (f12fcc) can still be called subsequently to return the number of converged eigenvalues, the converged eigenvalues and, if requested, the corresponding eigenvectors. Largest Magnitude Default Both Ends Largest Algebraic Smallest Algebraic Smallest Magnitude The Lanczos iterative method converges on a number of eigenvalues with given properties. The default is for nag_real_symm_sparse_eigensystem_iter (f12fbc) to compute the eigenvalues of largest magnitude using ${\mathbf{Largest Magnitude}}$. Alternatively, eigenvalues may be chosen which have ${\mathbf{Largest Algebraic}}$ part, ${\mathbf{Smallest Magnitude}}$, or ${\mathbf{Smallest Algebraic}}$ part; or eigenvalues which are from ${\mathbf{Both Ends}}$ of the algebraic spectrum. Note that these options select the eigenvalue properties for eigenvalues of $\mathrm{OP}$ (and $B$ for ${\mathbf{Generalized}}$ problems), the linear operator determined by the computational mode and problem type. Nolist Default List Normally each optional argument specification is not printed to ${\mathbf{Advisory}}$ as it is supplied. Optional argument ${\mathbf{List}}$ may be used to enable printing and optional argument ${\mathbf{Nolist}}$ may be used to suppress the printing. Monitoring Default $\text{}=-1$ (See Section 3.2.1.1 in the Essential Introduction for further information on NAG data types.) Unless ${\mathbf{Monitoring}}$ is set to $-1$ (the default), monitoring information is output to Nag_FileID ${\mathbf{Monitoring}}$ during the solution of each problem; this may be the same as ${\mathbf{Advisory}}$. The type of information produced is dependent on the value of ${\mathbf{Print Level}}$, see the description of the optional argument ${\mathbf{Print Level}}$ in this section for details of the information produced. Please see nag_open_file (x04acc) to associate a file with a given Nag_FileID. Print Level $i$ Default $\text{}=0$ This controls the amount of printing produced by nag_real_symm_sparse_eigensystem_option (f12fdc) as follows. $=0$ No output except error messages. If you want to suppress all output, set ${\mathbf{Print Level}}=0$. $>0$ The set of selected options. $=2$ Problem and timing statistics on final exit from nag_real_symm_sparse_eigensystem_iter (f12fbc). $\ge 5$ A single line of summary output at each Lanczos iteration. $\ge 10$ If ${\mathbf{Monitoring}}$ is set, then at each iteration, the length and additional steps of the current Lanczos factorization and the number of converged Ritz values; during re-orthogonalisation, the norm of initial/restarted starting vector; on a final Lanczos iteration, the number of update iterations taken, the number of converged eigenvalues, the converged eigenvalues and their Ritz estimates. $\ge 20$ Problem and timing statistics on final exit from nag_real_symm_sparse_eigensystem_iter (f12fbc). If ${\mathbf{Monitoring}}$ is set, then at each iteration, the number of shifts being applied, the eigenvalues and estimates of the symmetric tridiagonal matrix $H$, the size of the Lanczos basis, the wanted Ritz values and associated Ritz estimates and the shifts applied; vector norms prior to and following re-orthogonalisation. $\ge 30$ If ${\mathbf{Monitoring}}$ is set, then on final iteration, the norm of the residual; when computing the Schur form, the eigenvalues and Ritz estimates both before and after sorting; for each iteration, the norm of residual for compressed factorization and the symmetric tridiagonal matrix $H$; during re-orthogonalisation, the initial/restarted starting vector; during the Lanczos iteration loop, a restart is flagged and the number of the residual requiring iterative refinement; while applying shifts, some indices. $\ge 40$ If ${\mathbf{Monitoring}}$ is set, then during the Lanczos iteration loop, the Lanczos vector number and norm of the current residual; while applying shifts, key measures of progress and the order of $H$; while computing eigenvalues of $H$, the last rows of the Schur and eigenvector matrices; when computing implicit shifts, the eigenvalues and Ritz estimates of $H$. $\ge 50$ If ${\mathbf{Monitoring}}$ is set, then during Lanczos iteration loop: norms of key components and the active column of $H$, norms of residuals during iterative refinement, the final symmetric tridiagonal matrix $H$; while applying shifts: number of shifts, shift values, block indices, updated tridiagonal matrix $H$; while computing eigenvalues of $H$: the diagonals of $H$, the computed eigenvalues and Ritz estimates. Note that setting ${\mathbf{Print Level}}\ge 30$ can result in very lengthy ${\mathbf{Monitoring}}$ output. Random Residual Default Initial Residual To begin the Lanczos iterative process, nag_real_symm_sparse_eigensystem_iter (f12fbc) requires an initial residual vector. By default nag_real_symm_sparse_eigensystem_iter (f12fbc) provides its own random initial residual vector; this option can also be set using optional argument ${\mathbf{Random Residual}}$. Alternatively, you can supply an initial residual vector (perhaps from a previous computation) to nag_real_symm_sparse_eigensystem_iter (f12fbc) through the array argument resid; this option can be set using optional argument ${\mathbf{Initial Residual}}$. Regular Default Regular Inverse Shifted Inverse Buckling Cayley These options define the computational mode which in turn defines the form of operation $\mathrm{OP}\left(x\right)$ to be performed when nag_real_symm_sparse_eigensystem_iter (f12fbc) returns with ${\mathbf{irevcm}}=-1$ or $1$ and the matrix-vector product $Bx$ when nag_real_symm_sparse_eigensystem_iter (f12fbc) returns with ${\mathbf{irevcm}}=2$. Given a ${\mathbf{Standard}}$ eigenvalue problem in the form $Ax=\lambda x$ then the following modes are available with the appropriate operator $\mathrm{OP}\left(x\right)$. ${\mathbf{Regular}}$ $\mathrm{OP}=A$ ${\mathbf{Shifted Inverse}}$ $\mathrm{OP}={\left(A-\sigma I\right)}^{-1}$ where $\sigma $ is real Given a ${\mathbf{Generalized}}$ eigenvalue problem in the form $Ax=\lambda Bx$ then the following modes are available with the appropriate operator $\mathrm{OP}\left(x\right)$. ${\mathbf{Regular Inverse}}$ $\mathrm{OP}={B}^{-1}A$ ${\mathbf{Shifted Inverse}}$ $\mathrm{OP}={\left(A-\sigma B\right)}^{-1}B$, where $\sigma $ is real ${\mathbf{Buckling}}$ $\mathrm{OP}={\left(B-\sigma A\right)}^{-1}A$, where $\sigma $ is real ${\mathbf{Cayley}}$ $\mathrm{OP}={\left(A-\sigma B\right)}^{-1}\left(A+\sigma B\right)$, where $\sigma $ is real Standard Default Generalized The problem to be solved is either a standard eigenvalue problem, $Ax=\lambda x$, or a generalized eigenvalue problem, $Ax=\lambda Bx$. The optional argument ${\mathbf{Standard}}$ should be used when a standard eigenvalue problem is being solved and the optional argument ${\mathbf{Generalized}}$ should be used when a generalized eigenvalue problem is being solved. Tolerance $r$ Default $\text{}=\epsilon $ An approximate eigenvalue has deemed to have converged when the corresponding Ritz estimate is within ${\mathbf{Tolerance}}$ relative to the magnitude of the eigenvalue. Vectors Default $\text{}=\text{RITZ}$ The function nag_real_symm_sparse_eigensystem_sol (f12fcc) can optionally compute the Schur vectors and/or the eigenvectors corresponding to the converged eigenvalues. To turn off computation of any vectors the option ${\mathbf{Vectors}}=\mathrm{NONE}$ should be set. To compute only the Schur vectors (at very little extra cost), the option ${\mathbf{Vectors}}=\mathrm{SCHUR}$ should be set and these will be returned in the array argument v of nag_real_symm_sparse_eigensystem_sol (f12fcc). To compute the eigenvectors (Ritz vectors) corresponding to the eigenvalue estimates, the option ${\mathbf{Vectors}}=\mathrm{RITZ}$ should be set and these will be returned in the array argument z of nag_real_symm_sparse_eigensystem_sol (f12fcc), if z is set equal to v (as in Section 9) then the Schur vectors in v are overwritten by the eigenvectors computed by nag_real_symm_sparse_eigensystem_sol (f12fcc).	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 119, "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": 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": 0.6098495721817017, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/107030/relationships-between-properties-of-model-categories	## relationships between properties of model categories ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I've recently found myself running up against all sorts of adjectives that can describe a model category: cofibrantly generated, combinatorial, tractable, stable, locally (finitely) presentable, (left and/or right) proper, simplicial, admits Bousfield localizations, et al. I'm hoping for a reference whose explicit goal is to give an intuitive explanation of these concepts; presumably, such a reference would also tell me what each of these adjectives can buy me. (As a bonus, this reference might even have a diagram analogous to the one on pp. 576-7 of Gortz & Wedhorn's book Algebraic Geometry, which summarizes the relationships between different properties that a morphism of schemes may satisfy. But that'd just be for fun.) - In the case of algebraic geometry there's something to prove for many of the indicated relations. For model categories, and for the properties you mention, one only has to look at the definition and see whether one property is required in the definition of another property. This is why I think this question is somewhat empty as currently stated. – Fernando Muro Sep 13 at 0:52 I've never come across such a list, and I agree with Fernando that it wouldn't be very useful for the properties that you've listed. However, a few things are known. If you add "admits left Bousfield Localizations" to your list then left proper and combinatorial implies this. So does left proper and cellular. If you replace simplicial by "quillen equivalent to a simplicial model category" then Dan Dugger has a paper saying which of the other properties imply this: pages.uoregon.edu/ddugger/smod.html. There's also "permits small object argument" $\Rightarrow$ cofibrantly generated. – David White Sep 13 at 1:17 Thanks for the feedback. The reason I asked is that I've seen these sorts of adjectives running around before, but only recently have I begun to actually wrestle with them. I wasn't sure a diagram would be helpful, but the one in Gortz-Wedhorn popped into my head when I was thinking about it. I suppose the more important part is the last sentence. Certainly nLab has been a good resource, but I think I'd benefit from something a little more narrative. I'll edit. – Aaron Mazel-Gee Sep 13 at 11:33 Well, the classical reference is Hovey, but he doesn't cover a lot of the adjectives on your list. Hirschhorn covers more and is very thorough, so that's where I would (and did) start. To my knowledge those are the only books on model categories, though there are large sections of Lurie's DAG which cover the subject and you could do a CTRL+F search on those for the adjectives of interest (esp tractable, finitely presentable, etc). See also Chorny's recent work on class-combinatorial & class accessible categories. Otherwise I recommend Dugger. His writing is very clear and down to earth. – David White Sep 13 at 13:52 Ah, so there's no magic pill that I can take and suddenly understand all this stuff? ;o) I'll give this another day or so, and if nobody else responds then I'll accept your comment as an answer. Already Hirschhorn's introduction (as well as his introductions to the various Parts and Chapters) has been quite helpful. – Aaron Mazel-Gee Sep 14 at 22:08 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": 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": 0.9558046460151672, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2009/11/02/parallelepipeds-and-volumes-i/?like=1&source=post_flair&_wpnonce=8149a33672	# The Unapologetic Mathematician ## Parallelepipeds and Volumes I And we’re back with more of what Mr. Martinez of Harvard’s Medical School assures me is onanism of the highest caliber. I’m sure he, too, blames me for not curing cancer. Coming up in our study of calculus in higher dimensions we’ll need to understand parallelepipeds, and in particular their volumes. First of all, what is a parallelepiped? Or, more specifically, what is a $k$-dimensional parallelepiped in $n$-dimensional space? It’s a collection of points in space that we can describe as follows. Take a point $p$ and $k$ vectors $\left\{v_i\right\}_{i=1}^k$ in $\mathbb{R}^n$. The parallelepiped is the collection of points reachable by moving from $p$ by some fraction of each of the vectors $v_i$. That is, we pick $k$ values $t^i$, each in the interval $\left[0,1\right]$, and use them to specify the point $p+t^iv_i$. The collection of all such points is the parallelepiped with corner $p$ and sides $v_i$. One possible objection is that these sides may not be linearly independent. If the sides are linearly independent, then they span a $k$-dimensional subspace of the ambient space, justifying our calling it $k$-dimensional. But if they’re not, then the subspace they span has a lower dimension. We’ll deal with this by calling such a parallelepiped “degenerate”, and the nice ones with linearly independent sides “nondegenerate”. Trust me, things will be more elegant in the long run if we just deal with them both on the same footing. Now we want to consider the volume of a parallelepiped. The first observation is that the volume doesn’t depend on the corner point $p$. Indeed, we should be able to slide the corner around to any point in space as long as we bring the same displacement vectors along with us. So the volume should be a function only of the sides. The second observation is that as a function of the sides, the volume function should commute with scalar multiplication in each variable separately. That is, if we multiply $v_i$ by a non-negative factor of $\lambda$, then we multiply the whole volume of the parallelepiped by $\lambda$ as well. But what about negative scaling factors? What if we reflect the side (and thus the whole parallelepiped) to point the other way? One answer might be that we get the same volume, but it’s going to be easier (and again more elegant) if we say that the new parallelepiped has the negative of the original one’s volume. Negative volume? What could that mean? Well, we’re going to move away from the usual notion of volume just a little. Instead, we’re going to think of “signed” volume, which includes the possibility of being positive or negative. By itself, this sign will be less than clear at first, but we’ll get a better understanding as we go. As a first step we’ll say that two parallelepipeds related by a reflection have opposite signs. This won’t only cover the above behavior under scaling sides, but also what happens when we exchange the order of two sides. For example, the parallelogram with sides $v_1=a$ and $v_2=b$ and the parallelogram with sides $v_1=b$ and $v_2=a$ have the same areas with opposite signs. Similarly, swapping the order of two sides in a given parallelepiped will flip its sign. The third observation is that the volume function should be additive in each variable. One way to see this is that the $k$-dimensional volume of the parallelepiped with sides $v_1$ through $v_k$ should be the product of the $k-1$-dimensional volume of the parallelepiped with sides $v_1$ through $v_{k-1}$ and the length of the component of $v_k$ perpendicular to all the other sides, and this length is a linear function of $v_k$. Since there’s nothing special here about the last side, we could repeat the argument with the other sides. The other way to see this fact is to consider the following diagram, helpfully supplied by Kate from over at f(t): The side of one parallelogram is the (vector) sum of the sides of the other two, and we can see that the area of the one parallelogram is the sum of the areas of the other two. This justifies the assertion that for parallelograms in the plane, the area is additive as a function of one side (and, similarly, of the other). Similar diagrams should be apparent to justify the assertion for higher-dimensional parallelepipeds in higher-dimensional spaces. Putting all these together, we find that the $k$-dimensional volume of a parallelepiped with $k$ sides is an alternating multilinear functional, with the $k$ sides as variables, and so it lives somewhere in the exterior algebra $\Lambda(V^*)$. We’ll have to work out which particular functional gives us a good notion of volume as we continue. ### Like this: Posted by John Armstrong \| Analytic Geometry, Geometry ## 5 Comments » 1. [...] and Volumes II Yesterday we established that the -dimensional volume of a parallelepiped with sides should be an [...] Pingback by \| November 3, 2009 \| Reply 2. [...] An Example of a Parallelogram Today I want to run through an example of how we use our new tools to read geometric information out of a parallelogram. [...] Pingback by \| November 5, 2009 \| Reply 3. [...] Now that we’ve used exterior algebras to come to terms with parallelepipeds and their transformations, let’s come back to apply these ideas to the [...] Pingback by \| November 11, 2009 \| Reply 4. [...] to come back down again, covering the same region twice with opposite signs. This is related to the signed volumes we talked about, where (in one dimension) an interval can be traversed (integrated over) from left [...] Pingback by \| January 5, 2010 \| Reply 5. [...] because of this correspondence. In fact, it’s not hard to see that they’re related to signed volumes. This is the starting point from which all integration on manifolds emerges, and everything will [...] Pingback by \| August 2, 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": 36, "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": 0.9073722958564758, "perplexity_flag": "head"}
http://mathoverflow.net/questions/82058/is-sln-mathbbz-a-cat0-group/82065	## Is $SL(n,\mathbb{Z})$ a CAT(0) group? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is it possible to find a CAT(0) space on which the matrix group $SL(n,\mathbb{Z})$ acts properly discontinuously and cocompactly? Note: when the cocompactness is dropped , it is possible. - ## 1 Answer If n=2, yes: it acts on its Bass-Serre tree. If n>2, no: your group contains distorted elements, i.e. elements conjugated to a proper power of themselves (look at unipotent matrices). Such an element will have zero displacement length, which is impossible for an infinite order element in a group acting discretely cocompactly. For this and much more, see the monograph of Bridson and Haefliger. For even more restrictions on SL_n actions, see Theorem 1.14 in: Caprace-Monod, Isometry groups of non-positively curved spaces: structure theory Journal of Topology 2 No. 4 (2009), 661–700 -	{"extraction_info": {"found_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": 0.3, "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": 0.8571606278419495, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/64064/list	Return to Question 2 added 231 characters in body A $G$-structure $\pi : B_G \rightarrow M$ is said to be of $finite$ $type$ if $\mathfrak{g}^{(k)} = 0$ for some $k \in \mathbb{N}$, where $\mathfrak{g}^{(k)}$ denotes the $k$th prolongation of the Lie algebra $\mathfrak{g} = T_{e}G$. For finite type $G$-structures, let us call the first $k$ for which $\mathfrak{g}^{(k)} = 0$ the $order$ of the $G$-structure. For example, : • $O(n)$-structures (Riemannian metrics) are of finite type and order $1$, because $\mathfrak{o} (n)^{(1)} = 0$, but . • But $Sp(n)$-structures (symplectic structures) are not of finite type because the group of symplectomorphisms is infinite dimensional. • It can also be shown that $CO(n)$-structures (conformal structures) are of finite type and order $2$ (except if the dimension is $n > 2$)2\$, in which case it is not of finite type). Are there any finite type $G$-structures of order greater than $2$? More generally, are there $G$-structures of any order? Thanks. 1 $G$-structures of finite type. A $G$-structure $\pi : B_G \rightarrow M$ is said to be of $finite$ $type$ if $\mathfrak{g}^{(k)} = 0$ for some $k \in \mathbb{N}$, where $\mathfrak{g}^{(k)}$ denotes the $k$th prolongation of the Lie algebra $\mathfrak{g} = T_{e}G$. For finite type $G$-structures, let us call the first $k$ for which $\mathfrak{g}^{(k)} = 0$ the $order$ of the $G$-structure. For example, $O(n)$-structures (Riemannian metrics) are of finite type and order $1$, because $\mathfrak{o} (n)^{(1)} = 0$, but $Sp(n)$-structures are not of finite type. It can also be shown that $CO(n)$-structures are of finite type and order $2$ (if $n > 2$). Are there any $G$-structures of order greater than $2$? More generally, are there $G$-structures of any order? Thanks.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.8875663876533508, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=246195	Physics Forums Thread Closed Page 1 of 2 1 2 > ## Axiom of Choice Can someone give me a list of problems which at first sight require the axiom of choice, but do not? 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 Quote by Dragonfall Can someone give me a list of problems which at first sight require the axiom of choice, but do not? Here are two problems. One will fit on your list, the other won't. Preliminaries: R denotes set of real numbers N denotes set of natural numbers 1) f:R->N arbitrary surjection. Show there exists an injection g:N->R s.t. f(g(b)) = b for every b in N. 2) f:R->R arbitrary surjection, continuous and non-decreasing. Show there exists an injection g:R->R s.t. f(g(b)) = b for every b in R. Well for 2, since continuous functions map compact sets to compact sets, and the intervals where f is constant are compact, g(x) can be defined as the minimal element of f inverse of x. 1 is not possible without AC. ## Axiom of Choice Here's one where 'experience' may tell you to use the axiom of choice. Suppose that S and T are infinite sets. Let f:S-->T be a surjection, and let g:T-->S be another surjection. Write down a bijection between S and T. That requires the axiom of choice. Now change the word surjection to injection. Does the proof require AC, now? I think this is quite interesting, from wikipedia, about a result of Banach & Tarski which they believed required the axiom of choice. Vitali's and Hausdorff's constructions depend on Zermelo's axiom of choice ("AC"), which is also crucial to the Banach–Tarski paper, both for proving their paradox and for the proof of another result: Two Euclidean polygons, one of which strictly contains the other, are not equidecomposable. They remark: Le rôle que joue cet axiome dans nos raisonnements nous semble mériter l'attention (The role this axiom plays in our reasoning seems, to us, to deserve attention) and point out that while the second result fully agrees with our geometric intuition, its proof uses AC in even more substantial way than the proof of the paradox. Thus Banach and Tarski imply that AC should not be rejected simply because it produces a paradoxical decomposition. Indeed, such an argument would also reject some geometrically intuitive statements! Ironically, in 1949 A.P.Morse showed that the statement about Euclidean polygons can be proved in ZF set theory and thus does not require the axiom of choice. Quote by n_bourbaki Now change the word surjection to injection. Does the proof require AC, now? Nope, since you can define h:T->S by $h(t)=f^{-1}(t)$ if $t\in f(S)$ and [/itex]h(t)=g(t)[/itex] otherwise. 1) f is invertible is it? 2) How do you know that h is an injection? In fact your h cannot be an injection, unless f is already a bijection (but you assumed that by writing down it's inverse). If we assume that you're taking a one sided inverse, then the map from h from f(S) to S is surjective, thus if there is any element in t in T\f(S), then necessarily g(t)=h(t') for some t' in f(S). You certainly don't need the axiom of choice, though not for the reasons you wrote. I thought you might like an example of two superficially similar statements one using and one not using the axiom of choice. I seem to recall Conway having a similar thing for 'dividing a set into 3'. Quote by Dragonfall Well for 2, since continuous functions map compact sets to compact sets, and the intervals where f is constant are compact, g(x) can be defined as the minimal element of f inverse of x. 1 is not possible without AC. If by "the minimal element of f inverse of x" you mean "the min of the pre-image of the unit set {x} under f", then OK. Of course, I don't what else you could have possibly meant. "1 is not possible without AC." Fine. Can't say I'd disagree with you. With a slight generalization it can be shown to be equivalent to AC. Quote by n_bourbaki Here's one where 'experience' may tell you to use the axiom of choice. Suppose that S and T are infinite sets. Let f:S-->T be a surjection, and let g:T-->S be another surjection. Write down a bijection between S and T. That requires the axiom of choice. Now change the word surjection to injection. Does the proof require AC, now? If I understand you correctly, we now have injections, and you ask for a bijection? Schroder-Bernstein. The classical proof of this theorem is a constructive-existence proof (no AC). But contrary to what was suggested in another post, the construction of the bijection is non-trivial (my opinion). Here's one you might think about: f:A->B arbitrary injection (A,B arbitrary sets). Show there exists a surjection g:B->A s.t. g(f(a)) = a for every a in A. Can we get by without AC? Recognitions: Homework Help Science Advisor There exists a subset of $\mathbb{R}$ which is not Lebesgue measurable. Quote by fopc If I understand you correctly, we now have injections, and you ask for a bijection? Yes. Schroder-Bernstein. The classical proof of this theorem is a constructive-existence proof (no AC). Yes, and the result for surjections does (I believe) need the axiom of choice. This is why I included it. Quote by morphism There exists a subset of $\mathbb{R}$ which is not Lebesgue measurable. I always thought that did require the axiom of choice, something to do with viewing R as a vector space over Q? What's the non AC method? Quote by fopc Schroder-Bernstein Ah yes, now I remember. Back when I took set theory I wrote that "if there are injections from S to T and from T to S, then S and T are in bijection" is "obvious" in an exercise. I lost many, many points. Quote by n_bourbaki I always thought that did require the axiom of choice, something to do with viewing R as a vector space over Q? What's the non AC method? Actually, that does require the Axiom of Choice (or something similar). Maybe he meant that there exists a subset of R which is not Borel measurable. That doesn't require AC. Here is the Conway paper I was thinking of. http://citeseer.ist.psu.edu/cache/pa...n-by-three.pdf it also gives a discussion of what is entailed in avoiding AC, and why one might wish to do it without just going 'ugh, it's false'. Recognitions: Homework Help Science Advisor Quote by gel Actually, that does require the Axiom of Choice (or something similar). Maybe he meant that there exists a subset of R which is not Borel measurable. That doesn't require AC. No, I meant Lebesgue measurable. Apparently all you need to construct such a set is the Hahn-Banach theorem (which is strictly weaker than choice); see this paper by Foreman and Wehrung. Quote by morphism No, I meant Lebesgue measurable. Apparently all you need to construct such a set is the Hahn-Banach theorem (which is strictly weaker than choice); see this paper by Foreman and Wehrung. ok, well that shows that it isn't equivalent to AC, but I assumed that the OP wanted things that can be proved using standard ZF axioms. Otherwise you could just say the Hahn-Banach theorem doesn't require AC as another example. In the wikipedia link I posted above, it mentions that the "ultrafilter lemma" is enough to prove the Banach-Tarski paradoxical decomposition, which would also give non Lebesgue measurable sets (which I why I added the disclaimer ...or something similar... to my prev post). Thread Closed Page 1 of 2 1 2 > Thread Tools \| \| \| \| \|--------------------------------------\|--------------------------------------------\|---------\| \| Similar Threads for: Axiom of Choice \| \| \| \| Thread \| Forum \| Replies \| \| \| General Math \| 9 \| \| \| Set Theory, Logic, Probability, Statistics \| 2 \| \| \| Set Theory, Logic, Probability, Statistics \| 3 \| \| \| General Math \| 9 \| \| \| Set Theory, Logic, Probability, Statistics \| 9 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9311724305152893, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/184301/is-there-a-non-contradictory-non-trivial-axiomatic-system-in-which-godels-theor/184307	# Is there a non-contradictory non-trivial axiomatic system in which Gödel's theorem is undecidable? Gödel's Incompleteness Theorem states that in any non-contradictory non-trivial axiomatic system there are certain statements or theorems whom cannot be proved in that system, For example in ZFC, assuming it is non-contradictory, there is the famed Continuum Hypothesis and Whitehead's problem. My question is, is there any such system (but different from ZFC) in which Gödel's Incompleteness Theorem itself is an unprovable statement? - 2 Are you sure that the Riemann Hypothesis is unprovable in ZFC (if ZFC is consistent)? You should write that up for publication -- you'll be famous! – Henning Makholm Aug 19 '12 at 14:51 This is not really about set theory as much as it is about logic and incompleteness. I retagged it appropriately. (And it's not very soft either) – Asaf Karagila Aug 19 '12 at 14:52 @HenningMakholm Ah yes you're right. Then it means that I probably read about another famous theorem that was proved unprovable within ZFC. And Asaf, yes, you're right, thanks, and I originally tagged it as soft as I did not expect a proof or anything very concrete, given the question asked. – andreas.vitikan Aug 19 '12 at 14:59 @andreas.vitikan: The most famous statements that are proved independent of ZFC are the Continuum Hypothesis and the Generalized Continuum Hypothesis. Since those have "hypothesis" in their names, they might be what you're thinking about. – Henning Makholm Aug 19 '12 at 15:01 @Henning Ah yes, that was it, thanks for spotting it! – andreas.vitikan Aug 19 '12 at 15:03 ## 2 Answers Robinson arithmetic is non-trivial enough that the incompleteness theorem applies to it, but as far as I know not strong enough to prove the incompleteness theorem itself. - Just a footnote to Henning Makholm's answer. You indeed need to look at very weak systems for examples of the phenomenon you want. For a start, as soon as you add a little bit of induction to Robinson Arithmetic (officially, induction for $\Sigma_1$ wffs) you get a theory that, for any recursively axiomatized $T$ which includes a bit of arithmetic and any sensible system of Gödel-coding, can show that $Con_T \to \neg Prov_T(\overline{\ulcorner G_T\urcorner)}$, i.e. it can prove the formalized version of (half of) Gödel's incompleteness theorem for $T$. (It was rather important to Gödel that the incompleteness theorem is elementary in this sort of way.) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9566431045532227, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/15895-algebra-polynomial.html	Thread: 1. Algebra polynomial The degree three polynomial f(x) with real coefficients and leading coefficient 1, has 4 and 3 + i among its roots. Express f(x) as a product of linear and quadratic polynomials with real coefficients. choices f(x) = (x+4)(x^2+6x+10) f(x) = (x-4)(x^2-6x-9 f(x) = (x-4)x^2-6x+10) f(x) = (x-4)(x^2-6x+9) 2. Originally Posted by wvmcanelly@cableone.net The degree three polynomial f(x) with real coefficients and leading coefficient 1, has 4 and 3 + i among its roots. Express f(x) as a product of linear and quadratic polynomials with real coefficients. All such non-zero polynomials difference by a constant multiple. So we will find one which has 1 as its leading coefficient (called monic). Now if f(x) is a polynomial in R[x] (meaning the real numbers) and if a+bi is a zero then a-bi is also a zero. So if 3+i is a zero that means 3-i is a zero. Thus, $f(x) = (x-4)(x-(3+i))(x-(3-i))$ Multiply the last two factors together, $f(x)=(x-4)(x^2 - 6x + 10)$ 3. thanks, good not have figured that one out without you.	{"extraction_info": {"found_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": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9428328275680542, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/23770/why-are-normal-crossing-divisors-nice/23776	## Why are normal crossing divisors nice? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This question is going to be extremely vague. It seems that wherever I go (especially about Grothendieck's circle of ideas) the higher-dimensional analogue of a curve minus a finite number of points is a scheme minus a normal crossing divisor. Why is that? What's so special about a normal crossing divisor that it simulates a curve minus a finite number of points better? - 1 A normal crossing divisor looks like a bunch of coordinate planes, and has monomial equations. What could be better?! :) – Mariano Suárez-Alvarez Oct 8 2010 at 0:44 ## 2 Answers It mostly has to do with finding nice compactifications. Compactifications of varieties are a good thing as they allow us to control what happens at "infinity". If the variety itself is smooth it seems a good idea (and it is!) to demand that the compactification also be smooth. However, you need the situation to be nice at infinity in order to make the study of asymptotic behaviour at infinity to be as easy as possible. The best behaviour at infinity would be if the complement were smooth but that is in general not possible. What is always possible is to demand that the complement be a divisor with normal crossings. In practice it works essentially as well as having a smooth complement: You have a bunch of smooth varieties intersecting in as nice a manner as possible. - "What is always possible is to demand that the complement be a divisor with normal crossings." - can you elaborate on that? – Makhalan Duff May 6 2010 at 20:03 13 Hironaka proved that a smooth variety can always be realized as an open subvariety of a smooth projective variety in such a way that the boundary is a divisor with normal crossings. – JS Milne May 6 2010 at 20:43 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. A related point is that if say $X\subset Y$ and you want an embedded resolution of $X$, then you can of course ask that you want a birational morphism $\pi:Z\to Y$ such that the strict transform of $X$ is smooth, but a better thing to ask is the the entire pre-image of $X$ is as nice as possible. Unfortunately (in general) you cannot make the preimage of $X$ smooth as the exceptional set will add additional components and where they meet is going to be a singular point. So, you can ask for the next best thing: normal crossings. You could even say that normal crossings is the reducible analogue of smooth. Anyway, this is the result of Hironaka, JS Milne referred to above: for any $X\subset Y$ (plus some reasonable assumptions) there exists a projective birational $\pi$ such that $Z$ is smooth and $\pi^{-1}X$ is a normal crossing divisor. If $Y\setminus X$ is smooth, then you may even require that $\pi$ is an isomorphism outside $X$. The compactification result is a simple consequence of this: if $U$ is open (say quasi-projective), pick a projective compactification $Y$ and let $X=Y\setminus U$. Perform Hironaka's embedded resolution of singularities and you get $U\subset Z$ with the complement being a normal crossing divisor. One, slightly independent word on normal crossings. There seems to be some confusion in the literature about what normal crossings mean. Or rather what the difference is between normal crossings and simple normal crossings. Well, the point is that (nowadays) the latter is understood in the Zariski topology while the former in the analytic or formal topology. In other words, simple normal crossings mean that each irreducible component is smooth and they meet transversally, while normal crossings allows for a component to meet itself transversally. In particular, a nodal curve has normal crossings but not simple normal crossings. In the above discussion and in the other answers before this one, you can always put simple normal crossings in place of normal crossings and the statements remain true. It is possible that back when Hironaka proved his famous theorem, this distinction had not been made so in older texts the meaning might be different. At the same time, according to Miles Reid, it was the Japanese who invented the term simple normal crossings. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9513657689094543, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/179105-trig-substitution-print.html	# Trig substitution Printable View • April 30th 2011, 09:05 PM Hardwork Trig substitution How can we find the integral $\int_{0}^{\pi}\cos{x}\sin{x}\,{dx}$ with the substitution $u = sinx$? I know how to find the integral, but the limits become the same sin(0) = sin(pi) = 0. What should I do in general when a given substitution makes upper bound = lower bound? • April 30th 2011, 09:21 PM TheChaz Quote: Originally Posted by Hardwork How can we find the integral $\int_{0}^{\pi}\cos{x}\sin{x}\,{dx}$ with the substitution $u = sinx$? I know how to find the integral, but the limits become the same sin(0) = sin(pi) = 0. What should I do in general when a given substitution makes upper bound = lower bound? Using a different property, you can analyse the integrand as a function with period "pi", the (signed) area of which equals zero after each period. • April 30th 2011, 09:24 PM chisigma Quote: Originally Posted by Hardwork How can we find the integral $\int_{0}^{\pi}\cos{x}\sin{x}\,{dx}$ with the substitution $u = sinx$? I know how to find the integral, but the limits become the same sin(0) = sin(pi) = 0. What should I do in general when a given substitution makes upper bound = lower bound? Is... http://quicklatex.com/cache3/ql_0a43...4bcb647_l3.png http://quicklatex.com/cache3/ql_5801...1eb2dab_l3.png ... so that, no matter which 'substitution' You do, the integral is 0... Kind regards $\chi$ $\sigma$ • May 2nd 2011, 01:32 AM Hardwork Okay. I do see how the integral is zero. I just want to know what to do in general when a substitution makes the limits the same. All times are GMT -8. The time now is 05:54 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": 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": 0.18, "absolute_threshold": 10, "end_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": 0.8513757586479187, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/90522?sort=newest	## Is there a really big ring of differential operators in characteristic p? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) $k$ is a field of characteristic $p$. $k[t]$ has canonical first-order differential operator $\partial$ As an endomorphism of $k[t]$, $\partial^p=0$. First way to fix it: Use the divided power differential operator $\partial^{[p]}$. Shortfall: As an endomorphism of $k[t]$, ${\partial^{[p]}}^p=0$ Second way to fix it: Use crystalline differential operators. Shortfall: No higher order operators on $k[t]$. Question: Is there a really big ring of differential operators which contains the divided powers $\partial^[n]$ for all $n$ and which also has a natural evaluation map to $End_k(k[t])$? - 2 There is the ring of diferential operators à la Grothendieck, which is infinite dimensional and has elements of all orders. It is an increasing union of matrix rings in the case of the line, though, so it is a bit strange. – Mariano Suárez-Alvarez Mar 8 2012 at 0:28 The comment of Mariano and resulting answer by Lars are good. However, you may be interested to look here: mathoverflow.net/questions/56860/… – B. Bischof Mar 8 2012 at 18:10 ## 1 Answer Let me give some more details on Mariano's comment: The ring of differential operators a la EGA4 in this particular case will be a free $k[t]$-algebra generated by the following operators: We write $$\partial_t^{(n)}$$ for the operator which is defined by $$\partial_t^{(n)}(t^m)={m\choose n}t^{m-n}.$$ Because of this, sometimes the notation $$\partial_t^{(n)}=\frac{1}{n!}\frac{\partial^n}{\partial t^n}$$ is used. Actually, to generate the ring, the operators $\partial_t^{(p^n)}$ suffice. Now this ring is not noetherian, but it is an increasing union of noetherian subalgebras, lets denote them by $D^{(m)}$, which are the subalgebras generated by operators of degree $\leq p^m$. Using partially divided powers, Berthelot abstractly defines rings $\mathcal{D}^{(m)}$ such that the full ring of differential operators $\mathcal{D}$ is the direct limit of the $\mathcal{D}^{(m)}$. The image of $\mathcal{D}^{(m)}$ in $\mathcal{D}$ is then precisely the $D^{(m)}$ that I defined ad-hoc above. The crystalline operators that you defined in the question correspond to Berthelot's $\mathcal{D}^{(0)}$. - YoungMathematician asks (in a deleted answer) whether ${\partial^{[p]}}^p=0$ holds in Berthelot's ring. – S. Carnahan♦ Mar 9 2012 at 7:44 Ah, I wonder which of Bertholot's rings he means, and what precisely is $\delta^{[p]}$. In general, if $\delta$ is an operator of order 1, then $\delta^p=0$ in `$\mathcal{D}^{(n)}$` for $n>1$, but not necessarily `$\mathcal{D}^{(0)}$`. Modules over`$\mathcal{D}^{(0)}$` are connections. Now if by $\delta^{[p]}$` you mean what I wrote as `$\delta^{(p}_t$,then that′s an operator of order $p$, and cannot be considered as an elementof `$\mathcal{D}^{(0)}$`, but if I remember correctly, the same reasoning applies. – Lars Mar 9 2012 at 14:13	{"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": 32, "mathjax_display_tex": 3, "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": 0.917069673538208, "perplexity_flag": "head"}
http://mathforum.org/mathimages/index.php?title=Fibonacci_Numbers&oldid=13375	Fibonacci Numbers From Math Images Revision as of 10:21, 28 June 2010 by Iris (Talk \| contribs) Fibonacci Spiral Field: Number Theory Created By: [[Author:\| ]] Website: [ ] Fibonacci Spiral The spiral curve of the Nautilus sea shell follows the pattern of the spiral drawn in a Fibonacci rectangle, a collection of squares with sides that have the length of Fibonacci numbers. Basic Description The Fibonacci sequence is the sequence $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots,$ where the first two numbers are 1s and every later number is the sum of the two previous numbers. So, given two $1$'s as the first two terms, the next terms of the sequence follows as : $1+1=2, 1+2=3, 2+3=5, 3+5=8, \dots$ Image 1 The Fibonacci numbers can be discovered in nature, such as the spiral of the Nautilus sea shell, the petals of the flowers, the seed head of a sunflower, and many other parts of the nature. The seeds at the head of the sunflower, for instance, are arranged so that one can find a collection of spirals in both clockwise and counterclockwise ways. The number of spirals differs depending on whether one counts in a clockwise or a counterclockwise way because different patterns of spirals are formed depending on the counting direction, as shown by Image 1. The two numbers of spirals are always consecutive numbers in the Fibonacci sequence. Nature prefers this way of arranging the seeds because it seems to allow the seeds to be uniformly distributed. For more information about Fibonacci patterns in nature, see Fibonacci Numbers in Nature Origin The Fibonacci sequence was studied by Leonardo of Pisa, or Fibonacci (1770-1240). In his work Liber Abacci, he introduced a problem involving the growth of the rabbit population. The assumptions were • there is one pair of baby rabbits placed in an enclosed place on the first day of January • this pair will grow for one month before reproducing and produce a new pair of baby rabbits on the first day of March • each new pair will mature for one month and produce a new pair of rabbits on the first day of their third month • the rabbits never die, so after they mature, the rabbits produce a new pair of baby rabbits every month. The problem was to find out how many pairs of rabbits there will be after one year. Image 2 On January 1st, there is only 1 pair. On February 1st, the baby rabbits matured to be grown up rabbits, but they have not reproduced, so there will only be the original pair present. Now look at any later month. June is a good example. As you can see in Image 2, all 5 pairs of rabbits that were alive in May continue to be alive in June. Furthermore, all 3 pairs of rabbits that were also alive on April 1st, which all became or were adult rabbit pairs on May 1st, reproduce, creating 3 new pairs of rabbits born in June. This means that on June 1st, there are 8 pairs of rabbits. This is equal to the 5 pairs from May 1st plus the 3 new pairs, which is the number of pairs from April 1st. This same reasoning can be applied to any month, March or later, so the number of rabbits pairs at a certain point is the same as the sum of the number of rabbit pairs in the two previous months. This is exactly the rule that defines the Fibonacci sequence. As you can see in the image, the population by month begins: 1, 1, 2, 3, 5, 8, ..., which is the same as the beginning of the Fibonacci sequence. The population continues to match the Fibonacci sequence no matter how many months out you go. An interesting fact is that this problem of rabbit population was not intended to explain the Fibonacci numbers. This problem was originally intended to introduce the Hindu-Arabic numerals to Western Europe, where people were still using Roman numerals, and to help people practice addition. It was coincidence that the number of rabbits followed a certain pattern which people later named as the Fibonacci sequence. Fibonacci Numbers in Nature Template:Hideshow A More Mathematical Explanation [Click to view A More Mathematical Explanation] Symbolic Definition of Fibonacci Sequence The Fibonacci sequence is the sequence UNIQ76e5473510 [...] [Click to hide A More Mathematical Explanation] Symbolic Definition of Fibonacci Sequence The Fibonacci sequence is the sequence $F_1, F_2, F_3, \ldots, F_n, \ldots$ where $F_n = F_{n-1} + F_{n-2} \quad \hbox{ for } n>2$, and $F_1 = 1,\ F_2 = 1$. The Fibonacci sequence is recursively definedA recursively defined sequence is one in which each term is defined by preceding terms in the sequence. For instance, $x_n=(x_{n-1})^2-3x_{n-2}$ is recursively defined. because each term is defined in terms of its two immediately preceding terms. Identities and Properties Idenitities There are some interesting identities, including formula for the sum of first $n$ Fibonacci numbers, the sum of Fibonacci numbers with odd indices and sum of Fibonacci number with even indices. Note that all the identities and properties in this section can be proven in a more rigorous way through mathematical induction. Sum of first $n$ Fibonacci numbers The sum of first$n$ Fibonacci numbers is one less than the value of the ${(n+2)}^{\rm th}$ Fibonacci number: $F_1+F_2+\dots+F_n=F_{n+2}-F_2=F_{n+2}-1$ For example, the sum of first $5$ Fibonacci number is : $F_1+F_2+F_3+F_4+F_5= 1 + 1 + 2 + 3 +5=F_7-1=12$ The example is demonstrated below. The total length of red bars that each correspond to $F_1, F_2, F_3, F_4, F_5$ is one unit less than the length of $F_7$. Image 11 To see the proof, click below [Click to see more.] $F_1=F_3-F_2$ $F_2=F_4-F_3$ $F_3=F_5-F_4$ UNIQ76e547351052 [...] $F_1=F_3-F_2$ $F_2=F_4-F_3$ $F_3=F_5-F_4$ $\dots$ $F_{n-1}=F_{n+1}-F_n$ $F_n=F_{n+2}-F_{n+1}$ Adding up all the equations, we get : $F_1+F_2+\dots+F_n=F_{n+2}-F_2$. Sum of Fibonacci numbers with odd indices The sum of first $n$ Fibonacci numbers with odd indices is equal to the ${(2n)}^{\rm th}$ Fibonacci number: $F_1+F_3+F_5+\dots+F_{2n-1}=F_{2n}$ For instance, the sum of first $4$ Fibonacci numbers with odd indices is: $F_1+F_3+F_5+F_7=1+2+5+13=21=F_8$ This example is shown below. Image 12 To see the proof, click below [Click to see more.] $F_1=F_2$ $F_3=F_4-F_2$ $F_5=F_6-F_4$ UNIQ76e547351052 [...] $F_1=F_2$ $F_3=F_4-F_2$ $F_5=F_6-F_4$ $\dots$ $F_{2n-1}=F_{2n}-F_{2n-2}$ Adding all the equations, we get : $F_1+F_3+F_5+\dots+F_{2n-1}=(F_2-F_2)+(F_4-F_4)+(F_6-F_6)+\dots+(F_{2n-2}-F_{2n-2})+F_{2n}$ $=F_{2n}$ Except for $F_{2n}$, all the terms on the right side of the equation disappear because each term is canceled out by another term that has the opposite sign and the same magnitude. Sum of Fibonacci numbers with even indices The sum of first $n$ Fibonacci numbers with even indices is one less than the ${(2n+1)}^{\rm th}$ Fibonacci number: $F_2+F_4+\dots+F_{2n}=F_{2n+1}-1$ For example, the sum of first $4$ Fibonacci numbers with even indices is : $F_2+F_4+F_6= 1+3+8=F_7-1=13-1=12$ This example is shown below. Image 13 To see the proof, click below [Click to see more.] Subtracting , the sum of Fibonacci numbers with odd indices [...] Subtracting Eq. (2), the sum of Fibonacci numbers with odd indices, from Eq. (1), the sum of the first $n$ Fibonacci numbers, we get the identity of the sum of Fibonacci numbers with even indices. Sum of the squares of Fibonacci numbers The sum of the squares of the first $n$ Fibonacci numbers is the product of the $n^{\rm th}$ and the ${(n+1)}^{\rm th}$ Fibonacci numbers. Image 14 $\sum_{i=1}^n {F_i}^2=F_n F_{n+1}$ This identity can be proved by studying the area of the rectangles in Image 14. To see the proof, click below [Click to see more.] The rectangle is called a Fibonacci rectangle, which is further described in [[Fibonacci_Numbers#Fibonacci_Numbers_in_Nature\| Fibonacci Numbers in Na [...] The rectangle is called a Fibonacci rectangle, which is further described in Fibonacci Numbers in Nature. The numbers inside each square indicate the length of one side of the square. Notice that the lengths of the squares are all Fibonacci numbers. Any rectangle in the picture is composed of squares with lengths that are Fibonacci numbers. In fact, any rectangle is composed of every square with side lengths $F_1$ through $F_n$. Moreover, the dimension of this rectangle is $F_n$ by $F_{n+1}$. We can prove this identity by computing the area of the rectangle in two different ways. The first way of finding the area is to add the area of each squares. That is, the area of the rectangle will be : ${F_1}^2+{F_2}^2+{F_3}^2+\dots+{F_n}^2$. Another way of computing the area is by multiplying the width by the height. Using this method, the area will be : $F_n F_{n+1}$. Because we are computing the area of the same rectangle, the two methods should give the same results. Thus, ${F_1}^2+{F_2}^2+{F_3}^2+\dots+{F_n}^2=F_n F_{n+1}$. For example, for the red rectangle, the width is $5$ and the height is $8$. Since $5$ is the $5^{\rm th}$ Fibonacci number and $8$ is the $6^{\rm th}$ Fibonacci number, let $n=5$. The area of the rectangle is : $1^2+1^2+2^2+3^2+5^2={F_1}^2+{F_2}^2+{F_3}^2+{F_4}^2+{F_5}^2=\sum_{i=1}^5 {F_i}^2=40$, or $5 * 8 = F_5 F_{5+1} = F_5 F_6 = 40$. Thus, $\sum_{i=1}^5 {F_i}^2=F_5 F_{5+1}$. Properties Greatest Common Divisor The greatest common divisor of two Fibonacci numbers is the Fibonacci number whose index is the greatest common divisor of the indices of the original two Fibonacci numbers. In other words, $\gcd(F_n,F_m) = F_{\gcd(n,m)}$. For instance, $\gcd(F_9,F_6)=\gcd(34,8)=2=F_3=F_{\gcd(9,6)}$. In a special case where $F_n$ and $F_m$ are consecutive Fibonacci numbers, this property says that $\gcd(F_n, F_{n+1})=F_{\gcd(n,n+1)}=F_1=1$. That is, $F_n$ and $F_{n+1}$are always relatively prime. To see the proof for this special case, click below [Click to see more.] Assume that $F_n$ and $F_{n+1}$ have some integer UNIQ76e547351052273a-math-000 [...] Assume that $F_n$ and $F_{n+1}$ have some integer $k$ as their common divisor. Then, both $F_{n+1}$ and $F_n$ are each multiples of $k$: $F_{n+1}=ka$ $F_n=kb$ Subtracting Eq. (4) from Eq. (3), we get : $F_{n-1}=k(a-b)$, which means that if two consecutive Fibonacci numbers, $F_n$ and $F_{n+1}$, have $k$ as their common divisor, then the previous Fibonacci number, $F_{n-1}$ must also be a multiple of $k$. In that case, $F_{n-1}$ and $F_n$, which are also two consecutive Fibonacci numbers, will have $k$ as a common divisor. Then, it follows that $F_{n-2}$ must also be a multiple of $k$. Repeating the subtraction of consecutive Fibonacci numbers, we can conclude that the very first Fibonacci number, $F_1 = 1$ must also be a multiple of $k$. So $k=1$, and the only common divisor between two consecutive Fibonacci numbers is 1. Thus, two consecutive Fibonacci numbers are relatively prime. Finite Difference of Fibonacci Numbers One of the interesting properties of Fibonacci numbers is that the sequence of differences between consecutive Fibonacci numbers also forms a Fibonacci sequence, as shown in the table below. For more information about the difference table, click Difference Tables. Because the first sequence of differences of the Fibonacci sequence also includes a Fibonacci sequence, the second difference also includes a Fibonacci sequence. The Fibonacci sequence is thus reproduced in every sequence of differences. We can see that the sequence of differences is composed of Fibonacci numbers by looking at the definition of Fibonacci numbers : $F_n = F_{n-1} + F_{n-2}$. The difference between two consecutive Fibonacci numbers is : $F_n - F_{n-1} = F_{n-2}$. Thus, the difference between two consecutive Fibonacci numbers, $F_n$ and $F_{n-1}$, is equal to the value of the previous Fibonacci number, $F_{n-2}$. Golden Ratio Image 9 The golden ratio appears in paintings, architecture, and in various forms of nature. Two numbers are said to be in the golden ratio if the ratio of the smaller number to the larger number is equal to the ratio of the larger number to the sum of the two numbers. In Image 9, the width of A and B are in the golden ratio if$a : b = (a+b) : a$. The golden ratio is represented by the Greek lowercase phi ,$\varphi$, and the exact value is $\varphi=\frac{1 + \sqrt{5}}{2} \approx 1.61803\,39887\dots\,$ This value can be found from the definition of the golden ratio. To see an algebraic derivation of the exact value of the golden ratio, go to Golden Ratio : An Algebraic Representation. An interesting fact about golden ratio is that the ratio of two consecutive Fibonacci numbers approaches the golden ratio as the numbers get larger, as shown by the table below. $\frac{F_{n+1}}{F_n}$ $\frac{1}{1}$=1 $\frac{2}{1}$=2 $\frac{3}{2}$=1.5 $\frac{5}{3}$=1.66667 $\frac{8}{5}$=1.6 $\frac{13}{8}$=1.625 $\frac{21}{13}$=1.61538 $\frac{34}{21}$=1.61904 $\frac{55}{34}$=1.61765 $\frac{89}{55}$=1.61818 Lets assume that the ratio of two consecutive Fibonacci numbers have a limit and verify that this limit is, in fact, the golden ratio. Let $r_n$ denote the ratio of two consecutive Fibonacci numbers, that is, $r_n=\frac{F_{n+1}}{F_n}$. Then, $r_{n-1}=\frac{F_n}{F_{n-1}}$. $r_n$ and $r_{n-1}$ are related by : $r_n=\frac{F_{n+1}}{F_n}=\frac{F_n+F_{n-1}}{F_n}=1+\frac{F_{n-1}}{F_n}=1+\cfrac{1}{{F_n}/{F_{n-1}}}=1+\frac{1}{r_{n-1}}$. Assuming that the ratio $r_n$ has a limit, let $r$ be that limit: $\lim_{n \to \infty} r_n=\lim_{n \to \infty}\frac{F_{n+1}}{F_n}=r$. Then, $\lim_{n \to \infty} r_n = \lim_{n \to \infty} r_{n-1} = r$. Taking the limit of $r_n=1+\frac{1}{r_{n-1}}$ we get : $r=1+\frac{1}{r}$ Multiplying both sides by $r$, we get ${r}^2=r+1$ which can be written as: $r^2 - r - 1 = 0$. Applying the quadratic formula An equation, $\frac{-b \pm \sqrt {b^2-4ac}}{2a}$, which produces the solutions for equations of form $ax^2+bx+c=0$ , we get $r = \frac{1 \pm \sqrt{5}} {2}$. Because the ratio has to be a positive value, $r=\frac{1 + \sqrt{5}}{2}$ which is the golden ratio. Thus, if $r_n$ has a limit, then this limit is the golden ration. That is, as we go farther out in the sequence, the ratio of two consecutive Fibonacci numbers approaches the golden ratio. In fact, it can be proved that $r_n$ does have a limit; one way is to use Binet's formula in the next section. For a different proof using infinite continued fraction A continued fraction is a fraction in which the denominator is composed of a whole number and a fraction. An infinite continued fraction of the golden ratio has the form : $\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1} {1 + \ddots\,}}}}$go to Continued Fraction Representation and Fibonacci Sequences Image 10: Vitruvian man Many people find the golden ratio in various parts of nature, art, architecture, and even music. However, there are some people who criticize this viewpoint. They claim that many mathematicians are wishfully trying to make the connection between the golden ratio and other parts of the world even though there is no real connection. One example of the golden ratio that mathematicians found in nature is the human body. According to many, an ideal human body have proportions that show the golden ratio, such as: • distance between the foot and navel : distance between the navel and the head • distance between the finger tip and the elbow : distance between the wrist and the elbow • distance between the shoulder line and top of the head : length of the head. Leonardo da Vinci's drawing Vitruvian man shown in Image 10 emphasizes the proportion of human body. This drawing shows the proportions of an ideal human body that was studied by a Roman architect Vitruvius in his book De Architectura. In the drawing, a man is simultaneously inscribed in a circle and a square. The ratio of the square side to the radius of the circle in the drawing reflects the golden ratio, although the drawing deviates from the real value of the golden ratio by 1.7 percent. The proportions of the body of the man is also known to show the golden ratio. Although people later found the golden ratio in the painting, there is no evidence whether Leonardo da Vinci was trying to use the golden ratio in his painting or not. For more information about the golden ratio, go to Golden Ratio Binet's Formula for Fibonacci Numbers Binet's Formula gives a formula for the $n^{\rm th}$ Fibonacci number as : $F_n=\frac{{\varphi}^n-{\bar{\varphi}}^n}{\sqrt5}$, where $\varphi$ and $\bar{\varphi}$ are the two roots of Eq. (5), that is, $\varphi=\frac{1 + \sqrt{5}}{2},\quad \bar{\varphi}=\frac{1-\sqrt{5}}{2}$. Here is one way of verifying Binet's formula through mathematical induction, but it gives no clue about how to discover the formula. Let $F_n=\frac{{\varphi}^n-{\bar{\varphi}}^n}{\sqrt5}$ as defined above. We want to verify Binet's formula by showing that the definition of Fibonacci numbers holds true even when we use Binet's formula. First, we will show through inductive stepAn inductive step is one of the two parts of mathematical induction (base case and inductive step) where one shows that if a statement holds true for some $n$, then the statement also holds true for $n+1$ that: $F_n=F_{n-1}+F_{n-2}\quad\hbox{ for } n>2$ and then we will show the base caseA base case is one of the two parts of mathematical induction (base case and inductive step) where one shows that a statement holds true for the lowest value of $n$, usually $n=0$ or $n=1$, depending on situation. that: $F_1=1,\quad F_2=1$. First, according to Binet's fromula, $F_{n-1}+F_{n-2} = \frac{{\varphi}^{n-1}-{\bar{\varphi}}^{n-1}}{\sqrt5}+ \frac{{\varphi}^{n-2}-{\bar{\varphi}}^{n-2}}{\sqrt5}$ $=\frac{({\varphi}^{n-1}+{\varphi}^{n-2})-({\bar{\varphi}}^{n-1}+{\bar{\varphi}}^{n-2})}{\sqrt5}$ $=\frac{({\varphi}+1){\varphi}^{n-2}-(\bar{\varphi}+1){\bar{\varphi}}^{n-2}}{\sqrt5}$. Because $\varphi$ and $\bar{\varphi}$ are the two roots of Eq. (1), the above equation becomes : $F_{n-1}+F_{n-2}=\frac{{{\varphi}^2}{\varphi}^{n-2}-{{\bar{\varphi}}^2}{\bar{\varphi}}^{n-2}}{\sqrt5}$ $=\frac{{\varphi}^n-{\bar{\varphi}}^n}{\sqrt5}$ $=F_n$, as desired. Now, because $\varphi=\frac{1 + \sqrt{5}}{2},\quad \bar{\varphi}=\frac{1-\sqrt{5}}{2}$, $F_1=\frac{\varphi-\bar{\varphi}}{\sqrt5}=\frac{1}{\sqrt5}\left (\frac{1 + \sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}\right)=\frac{1}{\sqrt5} {\sqrt5} = 1$ $F_2=\frac{{\varphi}^2-{\bar{\varphi}}^2}{\sqrt5}=\frac{(\varphi+\bar{\varphi})(\varphi-\bar{\varphi})}{\sqrt5}=\frac{1*{\sqrt5}}{\sqrt5}=1$. Binet's formula thus is a correct formula of Fibonacci numbers. Fibonacci Numbers and Fractals Fibonacci Numbers and the Mandelbrot Set Image 15 The Mandelbrot set is a set of points in which the boundary forms a fractal. It is a set of all complex numbers $c$ for which the sequence :$z_{n+1}=(z_n)^2+c \quad \hbox{ for } n=0,1,2,\dots$ does not go to infinity, starting with $z_0=0$. For instance, $c=0$ is included in the Mandelbrot set because $z_1=(z_0)^2+c=0^2+0 = 0$ $z_2=(z_0)^2+c=0^2+0=0$ $\dots$ $z_n=0^2+0=0$ for any $n$. Thus, the sequence defined by $c=0$ is bounded and $0$ is included in the Mandelbrot set. On the other hand, when we test$c=1$, Image 16 $z_1=(z_0)^2+c=1$ $z_2=(z_1)^2+c=2$ $z_3=(z_2)^2+c=5$ $\dots$ The terms of this sequence will increase to infinity. Thus, $c=1$ is not included in the Mandelbrot set. Image 17 People have been drawn to study the Mandelbrot set because of its aesthetic beauty. It is surprising to many people how a simple formula like Eq. (6) can generate a complex structure of the Mandelbrot set. The Fibonacci sequence is related to the Mandelbrot set through the period of the main cardioid and some large primary bulbs. For each bulb, there are many antennas, and the largest antenna is called the main antenna. The number of spokes in the main antenna is the period of the bulb. The period of the main cardioid is considered to be 1. In Image 17, the main antenna has five spokes, including the one connecting the primary bulb and the junction point of the antenna. The period of this bulb is five. Now, we will consider the period of the largest primary bulbs that are attached to the main cardioid and are in between two larger bulbs. In Image 18, the largest bulb between the bulb of period 1 and the bulb of period 2 is the bulb of period 3, and this bulb was found by looking for the largest bulb on the periphery of the main cardioid. The largest bulb between the bulb of period 2 and period 3 is the bulb of period 5, and the one between bulb of period 3 and period 5 is the bulb of period 8. The sequence generated in this way proceeds as 1, 2, 3, 5, 8, 13, ..., following the pattern of Fibonacci sequence. Image 18 Teaching Materials There are currently no teaching materials for this page. Add teaching materials. References http://www.worldproutassembly.org/archives/2007/08/the_mathematica.html http://www.jimloy.com/algebra/fibo.htm http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibnat.html http://milan.milanovic.org/math/english/division/division.html http://dougerino.blogspot.com/2010/03/fibonacci-phi-and-kepler.html http://www.mathematicianspictures.com/FIBONACCI/Fibonacci.htm http://www.chabad.org/library/article_cdo/aid/463900/jewish/Deciphering-Natures-Code.htm http://hiddenlighthouse.wordpress.com/2010/04/07/nature-fibonacci/ http://en.wikipedia.org/wiki/Mandelbrot_set http://plus.maths.org/issue40/features/devaney/ } Future Directions for this Page Things to add(possible ideas for future) • Fibonacci numbers and Pascal's triangle • A helper page for recursively defined sequence • A section describing the Fibonacci numbers with negative subscripts. Things to 'not' add • A derivation of the exact value of the golden ratio. The derivation is redundant with the information in the golden ratio page. Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page. 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http://www.haskell.org/haskellwiki/index.php?title=User:Michiexile/MATH198/Lecture_2&diff=30193&oldid=30183	# User:Michiexile/MATH198/Lecture 2 ### From HaskellWiki (Difference between revisions) \| \| \| \| \| \|----------\|------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\|----------\|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\| \| \| \| \| \| \| Line 40: \| \| Line 40: \| \| \| \| We call a left cancellable arrow in a category a ''monomorphism''. \| \| We call a left cancellable arrow in a category a ''monomorphism''. \| \| \| \| \| \| \| - \| ====In concrete categories==== \| + \| ====In Set==== \| \| \| \| \| \| \| \| Left cancellability means that if, when we do first <math>g_1</math> and then <math>f</math> we get the same as when we do first <math>g_2</math> and then <math>f</math>, then we had equality already before we followed with <math>f</math>. \| \| Left cancellability means that if, when we do first <math>g_1</math> and then <math>f</math> we get the same as when we do first <math>g_2</math> and then <math>f</math>, then we had equality already before we followed with <math>f</math>. \| \| Line 49: \| \| Line 49: \| \| \| \| \| \| \| \| \| ====Subobjects==== \| \| ====Subobjects==== \| \| \| \| + \| \| \| \| \| + \| Consider the subset <math>\{1,2\}\subset\{1,2,3\}</math>. This is the image of an accordingly chosen injective map from any 2-element set into <math>\{1,2,3\}</math>. Thus, if we want to translate the idea of a subset into categorical language, it is not enough talking about monomorphisms, though the fact that inclusion is an injection indicates that we are on the right track. \| \| \| \| + \| \| \| \| \| + \| The trouble that remains is that we do not want to view <math>\{1,2\}</math> when it occurs as the image of <math>\{1,2\}</math> as a different subset from <math>\{5,6\}</math> mapping to <math>\{1,2\}</math>. So we need some way of figuring out how to catch these situations and parry for them. \| \| \| \| + \| \| \| \| \| + \| We'll say that a morphism <math>f</math> ''factors through'' a morphism <math>g</math> if there is some morphism <math>h</math> such that <math>f=gh</math>. \| \| \| \| + \| \| \| \| \| + \| We can also talk about a morphism <math>f:A\to C</math> factoring through an ''object'' <math>B</math> by requiring the existence of morphisms <math>g:A\to B, h:B\to C</math> that compose to <math>f</math>. \| \| \| \| + \| \| \| \| \| + \| Now, we can form an equivalence relation on monomorphisms into an object <math>A</math>, by saying <math>f\sim g</math> if <math>f</math> factors through <math>g</math> and <math>g</math> factors through <math>f</math>. The arrows implied by the factoring are inverse to each other, and the source objects of equivalent arrows are isomorphic. \| \| \| \| + \| \| \| \| \| + \| Equipped with this equivalence relation, we define a ''subobject'' of an object <math>A</math> to be an equivalence class of monomorphisms. \| \| \| \| \| \| \| \| ===Epimorphisms=== \| \| ===Epimorphisms=== \| \| Line 58: \| \| Line 70: \| \| \| \| A right cancellable arrow in a category is an ''epimorphism''. \| \| A right cancellable arrow in a category is an ''epimorphism''. \| \| \| \| \| \| \| - \| ====In concrete categories==== \| + \| ====In Set==== \| \| \| \| \| \| \| \| For epimorphims the interpretation in set functions is that whatever <math>f</math> does, it doesn't hide any part of the things <math>g_1</math> and <math>g_2</math> do. So applying <math>f</math> first doesn't influence the total available scope <mamth>g_1</math> and <math>g_2</math> have. \| \| For epimorphims the interpretation in set functions is that whatever <math>f</math> does, it doesn't hide any part of the things <math>g_1</math> and <math>g_2</math> do. So applying <math>f</math> first doesn't influence the total available scope <mamth>g_1</math> and <math>g_2</math> have. \| \| \| \| \| \| \| - \| ===Terminal objects=== \| + \| ===More on factoring=== \| \| \| \| \| \| \| - \| ====Pointless sets and global constants==== \| + \| In Set, and in many other categories, any morphism can be expressed by a factorization of the form <math>f=ip</math> where <math>i</math> is a monomorphism and <math>p</math> is an epimorphism. For instance, in Set, we know that a function is surjective onto its image, which in turn is a subset of the domain, giving a factorization into an epimorphism - the projection onto the image - followed by a monomorphism - the inclusion of the image into the domain. \| \| \| \| \| \| \| - \| ===Initial objects=== \| + \| ---- \| \| \| \| + \| \| \| \| \| + \| Note that in Set, every morphisms that is both a mono and an epi is immediately an isomorphism. We shall see in the homework that the converse does not necessarily hold. \| \| \| \| + \| \| \| \| \| + \| ===Initial and Terminal objects=== \| \| \| \| + \| \| \| \| \| + \| An object <math>0</math> is ''initial'' if for every other object <math>C</math>, there is a unique morphism <math>0\to C</math>. Dually, an object <math>1</math> is ''terminal'' if there is a unique morphism <math>C\to 1</math>. \| \| \| \| + \| \| \| \| \| + \| First off, we note that the uniqueness above makes initial and terminal objects unique up to isomorphism whenever they exist: we shall perform the proof for one of the cases, the other is almost identical. \| \| \| \| + \| \| \| \| \| + \| '''Proposition''' Initial (terminal) objects are unique up to isomorphism. \| \| \| \| + \| \| \| \| \| + \| '''Proof''': Suppose <math>C</math> and <math>C'</math> are both initial (terminal). Then there is a unique arrow <math>C\to C'</math> and a unique arrow <math>C'\to C</math>. The compositions of these arrows are all endoarrows of one or the other. Since ''all'' arrows from (to) an initial (terminal) objects are unique, these compositions have to be the identity arrows. Hence the arrows we found between the two objects are isomorphisms. QED. \| \| \| \| + \| \| \| \| \| + \| * In Sets, the empty set is initial, and any singleton set is terminal. \| \| \| \| + \| * In the category of Vector spaces, the single element vector space 0 is both initial and terminal. \| \| \| \| \| \| \| \| ===Zero objects=== \| \| ===Zero objects=== \| \| \| \| \| \| \| - \| ===Internal and external hom=== \| + \| This last example is worth taking up in higher detail. We call an object in a category a ''zero object'' if it is simultaneously initial and terminal. \| \| \| \| \| \| \| - \| * Isomorphisms and existence of inverses. \| \| \| \| - \| * Epi- and mono-morphisms and cancellability. \| \| \| \| - \| Examples in concrete categories. \| \| \| \| - \| Monomorphisms and subobjects: \| \| \| \| - \| * Factoring through. Equivalence relation by mutual factoring. \| \| \| \| - \| * Subobjects as equivalence classes of monomorphisms. \| \| \| \| - \| ** Splitting and the existence of inverses. \| \| \| \| - \| * Terminal and initial objects. \| \| \| \| - \| ** Constants. Pointless sets. \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| + \| ====Pointless sets and generalized elements==== \| \| \| \| + \| \| \| \| \| + \| Arrows to initial objects and from terminal objects are interesting too - and as opposed to the arrows from initial and to the terminals, there is no guarantee for these arrows to be uniquely determined. Let us start with arrows <math>A\to 0</math> into initial objects. \| \| \| \| + \| \| \| \| \| + \| In the category of sets, such an arrow only exists if <math>A</math> is already the empty set. \| \| \| \| + \| \| \| \| \| + \| In the category of all monoids, with monoid homomorphisms, we have a zero object, so such an arrow is uniquely determined. \| \| \| \| + \| \| \| \| \| + \| For arrows <math>1\to A</math>, however, the situation is significantly more interesting. Let us start with the situation in Set. <math>1</math> is some singleton set, hence a function from <math>1</math> picks out one element as its image. Thus, at least in Set, we get an isomorphism of sets <math>A = Hom(1,A)</math>. \| \| \| \| + \| \| \| \| \| + \| As with so much else here, we build up a general definition by analogy to what we see happening in the category of sets. Thus, we shall say that a ''global element'', or a ''point'', or a ''constant'' of an object <math>A</math> in a category with terminal objects is a morphism <math>x:1\to A</math>. \| \| \| \| + \| \| \| \| \| + \| This allows us to talk about elements without requiring our objects to even be sets to begin with, and thus reduces everything to a matter of just morphisms. This approach is fruitful both in topology and in Haskell, and is sometimes called ''pointless''. \| \| \| \| + \| \| \| \| \| + \| The important point here is that we can replace ''function application'' <math>f(x)</math> by the already existing and studied ''function composition''. If a constant <math>x</math> is just a morphism <math>x:1\to A</math>, then the value <math>f(x)</math> is just the composition <math>f\circ x:1\to A\to B</math>. Note, also, that since <math>1</math> is terminal, it has exactly one point. \| \| \| \| + \| \| \| \| \| + \| In the idealized Haskell category, we have the same phenomenon for constants, but slightly disguised: a global constant is 0-ary function. Thus the type declaration \| \| \| \| + \| <haskell> \| \| \| \| + \| x :: a \| \| \| \| + \| </haskell> \| \| \| \| + \| can be understood as syntactic sugar for the type declaration \| \| \| \| + \| <haskell> \| \| \| \| + \| x :: () -> a \| \| \| \| + \| </haskell> \| \| \| \| + \| thus reducing everything to function types. \| \| \| \| + \| \| \| \| \| + \| ---- \| \| \| \| + \| \| \| \| \| + \| Similarly to the ''global'' elements, it may be useful to talk about ''variable elements'', by which we mean non-specified arrows <math>f:T\to A</math>. Allowing <math>T</math> to range over all objects, and <math>f</math> to range over all morphisms into <math>A</math>, we are able to recover some of the element-centered styles of arguments we are used to. We say that <math>f</math> is ''parametried over <math>T</math>''. \| \| \| \| + \| \| \| \| \| + \| Using this, it turns out that <math>f</math> is a monomorphism if for any variable elements <math>x,y:T\to A</math>, if <math>x\neq y</math> then <math>f\circ x\neq f\circ y</math>. \| \| \| \| + \| \| \| \| \| + \| ===Internal and external hom=== \| \| \| \| + \| \| \| \| \| + \| If <math>f:B\to C</math>, then <math>f</math> induces a set function <math>Hom(A,f):Hom(A,B)\to Hom(A,C)</math> through <math>Hom(A,f)(g) = f\circ g</math>. Similarly, it induces a set function <math>Hom(f,A):Hom(C,A)\to Hom(C,B)</math> through <math>Hom(f,A)(g) = g\circ f</math>. \| \| \| \| \| \| \| \| \| + \| Using this, we have an occasionally enlightening \| \| \| \| \| \| \| \| \| + \| '''Proposition''' An arrow <math>f:B\to C</math> is \| \| \| \| + \| # a monomorphism if and only if <math>Hom(A,f)</math> is injective for every object <math>A</math>. \| \| \| \| + \| # an epimorphism if and only if <math>Hom(f,A)</math> is injective for every object <math>A</math>. \| \| \| \| + \| # a split monomorphism if and only if <math>Hom(f,A)</math> is surjective for every object <math>A</math>. \| \| \| \| + \| # a split epimorphism if and only if <math>Hom(A,f)</math> is surjective for every object <math>A</math>. \| \| \| \| + \| # an isomorphism if and only if any one of the following equivalent conditions hold: \| \| \| \| + \| ## it is both a split epi and a mono. \| \| \| \| + \| ## it is both an epi and a split mono. \| \| \| \| + \| ## <math>Hom(A,f)</math> is bijective for every <math>A</math>. \| \| \| \| + \| ## <math>Hom(f,A)</math> is bijective for every <math>A</math>. \| \| \| \| \| \| \| \| \| + \| ---- \| \| \| \| \| \| \| - \| ==Morphisms== \| + \| For any <math>A,B</math> in a category, the homset is a ''set'' of morphisms between the objects. For many categories, though, homsets may end up being objects of that category as well. \| \| - \| The arrows of a category are called ''morphisms''. This is derived from ''homomorphisms''. \| + \| \| \| \| \| \| \| \| - \| Some arrows have special properties that make them extra helpful; and we'll name them: \| + \| As an example, the set of all linear maps between two fixed vector spaces is itself a vector space. \| \| \| \| \| \| \| - \| ;Endomorphism:A morphism with the same object as source and target. \| + \| Alternatively, the function type <hask>a -> b</hask> is an actual Haskell type, and captures the morphisms of the idealized Haskell category. \| \| - \| ;Monomorphism:A morphism that is left-cancellable. Corresponds to injective functions. We say that ''f'' is a monomorphism if for any <math>g_1,g_2</math>, the equation <math>fg_1 = fg_2</math> implies <math>g_1=g_2</math>. In other words, with a concrete perspective, ''f'' doesn't introduce additional relations when applied. \| + \| \| \| - \| ;Epimorphism:A morphism that is right-cancellable. Corresponds to surjective functions. We say that ''f'' is an epimorphism if for any <math>g_1,g_2</math>, the equation <math>g_1f = g_2f</math> implies <math>g_1=g_2</math>. \| + \| \| \| - \| Note, by the way, that cancellability does not imply the existence of an inverse. Epi's and mono's that have inverses realizing their cancellability are called ''split''. \| + \| \| \| - \| ;Isomorphism:A morphism is an isomorphism if it has an inverse. Split epi and split mono imply isomorphism. Specifically, <math>f:v\to w</math> is an isomorphism if there is a <math>g:w\to v</math> such that <math>fg=1_w</math> and <math>g=1_v</math>. \| + \| \| \| - \| ;Automorphism:An automorphism is an endomorphism that is an isomorphism. \| + \| \| \| \| \| \| \| \| - \| ==Objects== \| + \| We shall return to this situation later, when we are better equipped to give a formal scaffolding to the idea of having elements in objects in a category act as morphisms. For now, we shall introduce the notations <math>[A\to B]</math> or <math>B^A</math> to denote the ''internal'' hom - where the morphisms between two objects live as an object of the category. This distinguishes <math>B^A</math> from <math>Hom(A,B)</math>. \| \| - \| In a category, we use a different name for the vertices: ''objects''. This comes from the roots in describing concrete categories - thus while objects may be actual mathematical objects, but they may just as well be completely different. \| + \| \| \| \| \| \| \| \| - \| Just as with the morphisms, there are objects special enough to be named. An object ''v'' is \| + \| To gain a better understanding of the choice of notation, it is worth noting that <math>\|Hom_{\textsf{Set}}(A,B)\|=\|B\|^\|A\|</math>. \| \| - \| ;''Initial'':if ''[v,w]'' has exactly one element for all other objects ''w''. \| + \| \| \| - \| ;''Terminal'': if ''[w,v]'' has exactly one element for all other objects ''w''. \| + \| \| \| - \| ;''A Zero object'': if it is both initial and terminal. \| + \| \| \| \| \| \| \| \| - \| All initial objects are isomorphic. If <math>i_1,i_2</math> are both initial, then there is exactly one map <math>i_1\to i_2</math> and exactly one map <math>i_2\to i_1</math>. The two possible compositions are maps <math>i_1\to i_1</math> and <math>i_2\to i_2</math>. However, the initiality condition holds even for the morphism set ''[v,v]'', so in these, the only existing morphism is <math>1_{i_1}</math> and <math>1_{i_2}</math> respectively. Hence, the compositions have to be this morphism, which proves the statement. \| + \| ===Homework=== \| \| \| \| \| \| \| - \| ==Dual category== \| + \| Passing mark requires at least 6 of 11. \| \| \| \| \| \| \| - \| The same proof carries over, word by word, to the terminal case. This is an illustration of a very commonly occurring phenomenon - dualization. \| + \| # Suppose <math>g,h</math> are two-sided inverses to <math>f</math>. Prove that <math>g=h</math>. \| \| \| \| + \| # (requires some familiarity with analysis) There is a category with object <math>\mathbb R</math> (or even all smooth manifolds) and with morphisms smooth (infinitely differentiable) functions. Prove that being a bijection does not imply being an isomorphisms. Hint: What about <math>x\mapsto x^3?</math>. \| \| \| \| + \| # (do this if you don't do 2) In the category of posets, with order-preserving maps as morphisms, show that not all bijective homomorphisms are isomorphisms. \| \| \| \| + \| # Consider the partially ordered set <math>P</math> as a category. Prove: every arrow is both monic and epic. Is every arrow thus an isomorphism? \| \| \| \| + \| # What are the terminal and initial objects in a poset? Give an example of a poset that has both, either and none. Give an example of a poset that has a zero object. \| \| \| \| + \| # What are the terminal and initial objects in the category with objects graphs and morphisms graph homomorphisms? \| \| \| \| + \| # Prove that if a category has one zero object, then all initial and all terminal objects are all isomorphic and all zero objects. \| \| \| \| + \| # Prove that the composition of two monomorphisms is a monomorphism and that the composition of two epimorphisms is an epimorphism. If <math>g\circ f</math> is monic, do any of <math>g,f</math> have to be monic? If the composition is epic, do any of the factors have to be epic? \| \| \| \| + \| # Verify that the equivalence relation used in defining subobjects really is an equivalence relation. Further verify that this fixes the motivating problem. \| \| \| \| + \| # Describe a representative subcategory each of: \| \| \| \| + \| #* The category of vectorspaces over the reals. \| \| \| \| + \| #* The category of finite sets. \| \| \| \| + \| #* The category formed by the poset of the integers <math>\mathbb Z</math> and the order relation <math>a\leq b</math> if <math>a\|b</math>. \| \| \| \| + \| # * An arrow <math>f:A\to A</math> in a category <math>C</math> is an ''idempotent'' if <math>f\circ f = f</math>. We say that <math>f</math> is a ''split idempotent'' if there is some <math>g:A\to B, h:B\to A</math> such that <math>h\circ g=f</math> and <math>g\circ h=1_B</math>. Show that in Set, <math>f</math> is idempotent if and only if its image equals its set of fixed points. Show that every idempotent in Set is split. Give an example of a category with a non-split idempotent. \| ## Revision as of 20:01, 23 September 2009 IMPORTANT NOTE: THESE NOTES ARE STILL UNDER DEVELOPMENT. PLEASE WAIT UNTIL AFTER THE LECTURE WITH HANDING ANYTHING IN, OR TREATING THE NOTES AS READY TO READ. ## Contents ### 1 Morphisms and objects Some morphisms and some objects are special enough to garner special names that we will use regularly. In morphisms, the important properties are • cancellability - the categorical notion corresponding to properties we use when solving, e.g., equations over $\mathbb N$: $3x = 3y \Rightarrow x = y$ • existence of inverses - which is stronger than cancellability. If there are inverses around, this implies cancellability, by applying the inverse to remove the common factor. Cancellability, however, does not imply that inverses exist: we can cancel the 3 above, but this does not imply the existence of $1/3\in\mathbb N$. Thus, we'll talk about isomorphisms - which have two-sided inverses, monomorphisms and epimorphisms - which have cancellability properties, and split morphisms - which are mono's and epi's with correspodning one-sided inverses. We'll talk about what these concepts - defined in terms of equationsolving with arrows - apply to more familiar situations. And we'll talk about how the semantics of some of the more wellknown ideas in mathematics are captured by these notions. For objects, the properties are interesting in what happens to homsets with the special object as source or target. An empty homset is pretty boring, and a large homset is pretty boring. The real power, we find, is when all homsets with the specific source or target are singleton sets. This allows us to formulate the idea of a 0 in categorical terms, as well as capturing the roles of the empty set and of elements of sets - all using only arrows. ### 2 Isomorphisms An arrow $f:A\to B$ in a category C is an isomorphism if it has a twosided inverse g. In other words, we require the existence of a $g:B\to A$ such that fg = 1B and gf = 1A. #### 2.1 In concrete categories In a category of sets with structure with morphisms given by functions that respect the set structure, isomorphism are bijections respecting the structure. In the category of sets, the isomorphisms are bijections. #### 2.2 Representative subcategories Very many mathematical properties and invariants are interesting because they hold for objects regardless of how, exactly, the object is built. As an example, most set theoretical properties are concerned with how large the set is, but not what the elements really are. If all we care about are our objects up to isomorphisms, and how they relate to each other - we might as well restrict ourselves to one object for each isomorphism class of objects. Doing this, we get a representative subcategory: a subcategory such that every object of the supercategory is isomorphic to some object in the subcategory. #### 2.3 Groupoids A groupoid is a category where all morphisms are isomorphisms. The name originates in that a groupoid with one object is a bona fide group; so that groupoids are the closest equivalent, in one sense, of groups as categories. ### 3 Monomorphisms We say that an arrow f is left cancellable if for any arrows g1,g2 we can show $fg_1 = fg_2 \Rightarrow g_1=g_2$. In other words, it is left cancellable, if we can remove it from the far left of any equation involving arrows. We call a left cancellable arrow in a category a monomorphism. #### 3.1 In Set Left cancellability means that if, when we do first g1 and then f we get the same as when we do first g2 and then f, then we had equality already before we followed with f. In other words, when we work with functions on sets, f doesn't introduce relations that weren't already there. Anything non-equal before we apply f remains non-equal in the image. This, translated to formulae gives us the well-known form for injectivity: $x\neq y\Rightarrow f(x)\neq f(y)$ or moving out the negations, $f(x)=f(y) \Rightarrow x=y$. #### 3.2 Subobjects Consider the subset $\{1,2\}\subset\{1,2,3\}$. This is the image of an accordingly chosen injective map from any 2-element set into {1,2,3}. Thus, if we want to translate the idea of a subset into categorical language, it is not enough talking about monomorphisms, though the fact that inclusion is an injection indicates that we are on the right track. The trouble that remains is that we do not want to view {1,2} when it occurs as the image of {1,2} as a different subset from {5,6} mapping to {1,2}. So we need some way of figuring out how to catch these situations and parry for them. We'll say that a morphism f factors through a morphism g if there is some morphism h such that f = gh. We can also talk about a morphism $f:A\to C$ factoring through an object B by requiring the existence of morphisms $g:A\to B, h:B\to C$ that compose to f. Now, we can form an equivalence relation on monomorphisms into an object A, by saying f˜g if f factors through g and g factors through f. The arrows implied by the factoring are inverse to each other, and the source objects of equivalent arrows are isomorphic. Equipped with this equivalence relation, we define a subobject of an object A to be an equivalence class of monomorphisms. ### 4 Epimorphisms Right cancellability, by analogy, is the implication $g_1f = g_2f \Rightarrow g_1 = g_2$ The name, here comes from that we can remove the right cancellable f from the right of any equation it is involved in. A right cancellable arrow in a category is an epimorphism. #### 4.1 In Set For epimorphims the interpretation in set functions is that whatever f does, it doesn't hide any part of the things g1 and g2 do. So applying f first doesn't influence the total available scope <mamth>g_1</math> and g2 have. ### 5 More on factoring In Set, and in many other categories, any morphism can be expressed by a factorization of the form f = ip where i is a monomorphism and p is an epimorphism. For instance, in Set, we know that a function is surjective onto its image, which in turn is a subset of the domain, giving a factorization into an epimorphism - the projection onto the image - followed by a monomorphism - the inclusion of the image into the domain. Note that in Set, every morphisms that is both a mono and an epi is immediately an isomorphism. We shall see in the homework that the converse does not necessarily hold. ### 6 Initial and Terminal objects An object 0 is initial if for every other object C, there is a unique morphism $0\to C$. Dually, an object 1 is terminal if there is a unique morphism $C\to 1$. First off, we note that the uniqueness above makes initial and terminal objects unique up to isomorphism whenever they exist: we shall perform the proof for one of the cases, the other is almost identical. Proposition Initial (terminal) objects are unique up to isomorphism. Proof: Suppose C and C' are both initial (terminal). Then there is a unique arrow $C\to C'$ and a unique arrow $C'\to C$. The compositions of these arrows are all endoarrows of one or the other. Since all arrows from (to) an initial (terminal) objects are unique, these compositions have to be the identity arrows. Hence the arrows we found between the two objects are isomorphisms. QED. • In Sets, the empty set is initial, and any singleton set is terminal. • In the category of Vector spaces, the single element vector space 0 is both initial and terminal. ### 7 Zero objects This last example is worth taking up in higher detail. We call an object in a category a zero object if it is simultaneously initial and terminal. #### 7.1 Pointless sets and generalized elements Arrows to initial objects and from terminal objects are interesting too - and as opposed to the arrows from initial and to the terminals, there is no guarantee for these arrows to be uniquely determined. Let us start with arrows $A\to 0$ into initial objects. In the category of sets, such an arrow only exists if A is already the empty set. In the category of all monoids, with monoid homomorphisms, we have a zero object, so such an arrow is uniquely determined. For arrows $1\to A$, however, the situation is significantly more interesting. Let us start with the situation in Set. 1 is some singleton set, hence a function from 1 picks out one element as its image. Thus, at least in Set, we get an isomorphism of sets A = Hom(1,A). As with so much else here, we build up a general definition by analogy to what we see happening in the category of sets. Thus, we shall say that a global element, or a point, or a constant of an object A in a category with terminal objects is a morphism $x:1\to A$. This allows us to talk about elements without requiring our objects to even be sets to begin with, and thus reduces everything to a matter of just morphisms. This approach is fruitful both in topology and in Haskell, and is sometimes called pointless. The important point here is that we can replace function application f(x) by the already existing and studied function composition. If a constant x is just a morphism $x:1\to A$, then the value f(x) is just the composition $f\circ x:1\to A\to B$. Note, also, that since 1 is terminal, it has exactly one point. In the idealized Haskell category, we have the same phenomenon for constants, but slightly disguised: a global constant is 0-ary function. Thus the type declaration `x :: a` can be understood as syntactic sugar for the type declaration `x :: () -> a` thus reducing everything to function types. Similarly to the global elements, it may be useful to talk about variable elements, by which we mean non-specified arrows $f:T\to A$. Allowing T to range over all objects, and f to range over all morphisms into A, we are able to recover some of the element-centered styles of arguments we are used to. We say that f is parametried over T. Using this, it turns out that f is a monomorphism if for any variable elements $x,y:T\to A$, if $x\neq y$ then $f\circ x\neq f\circ y$. ### 8 Internal and external hom If $f:B\to C$, then f induces a set function $Hom(A,f):Hom(A,B)\to Hom(A,C)$ through $Hom(A,f)(g) = f\circ g$. Similarly, it induces a set function $Hom(f,A):Hom(C,A)\to Hom(C,B)$ through $Hom(f,A)(g) = g\circ f$. Using this, we have an occasionally enlightening Proposition An arrow $f:B\to C$ is 1. a monomorphism if and only if Hom(A,f) is injective for every object A. 2. an epimorphism if and only if Hom(f,A) is injective for every object A. 3. a split monomorphism if and only if Hom(f,A) is surjective for every object A. 4. a split epimorphism if and only if Hom(A,f) is surjective for every object A. 5. an isomorphism if and only if any one of the following equivalent conditions hold: 1. it is both a split epi and a mono. 2. it is both an epi and a split mono. 3. Hom(A,f) is bijective for every A. 4. Hom(f,A) is bijective for every A. For any A,B in a category, the homset is a set of morphisms between the objects. For many categories, though, homsets may end up being objects of that category as well. As an example, the set of all linear maps between two fixed vector spaces is itself a vector space. Alternatively, the function type a -> b is an actual Haskell type, and captures the morphisms of the idealized Haskell category. We shall return to this situation later, when we are better equipped to give a formal scaffolding to the idea of having elements in objects in a category act as morphisms. For now, we shall introduce the notations $[A\to B]$ or BA to denote the internal hom - where the morphisms between two objects live as an object of the category. This distinguishes BA from Hom(A,B). To gain a better understanding of the choice of notation, it is worth noting that Failed to parse (unknown function\textsf): \|Hom_{\textsf{Set}}(A,B)\|=\|B\|^\|A\| . ### 9 Homework Passing mark requires at least 6 of 11. 1. Suppose g,h are two-sided inverses to f. Prove that g = h. 2. (requires some familiarity with analysis) There is a category with object $\mathbb R$ (or even all smooth manifolds) and with morphisms smooth (infinitely differentiable) functions. Prove that being a bijection does not imply being an isomorphisms. Hint: What about $x\mapsto x^3?$. 3. (do this if you don't do 2) In the category of posets, with order-preserving maps as morphisms, show that not all bijective homomorphisms are isomorphisms. 4. Consider the partially ordered set P as a category. Prove: every arrow is both monic and epic. Is every arrow thus an isomorphism? 5. What are the terminal and initial objects in a poset? Give an example of a poset that has both, either and none. Give an example of a poset that has a zero object. 6. What are the terminal and initial objects in the category with objects graphs and morphisms graph homomorphisms? 7. Prove that if a category has one zero object, then all initial and all terminal objects are all isomorphic and all zero objects. 8. Prove that the composition of two monomorphisms is a monomorphism and that the composition of two epimorphisms is an epimorphism. If $g\circ f$ is monic, do any of g,f have to be monic? If the composition is epic, do any of the factors have to be epic? 9. Verify that the equivalence relation used in defining subobjects really is an equivalence relation. Further verify that this fixes the motivating problem. 10. Describe a representative subcategory each of: • The category of vectorspaces over the reals. • The category of finite sets. • The category formed by the poset of the integers $\mathbb Z$ and the order relation $a\leq b$ if a \| b. 11. * An arrow $f:A\to A$ in a category C is an idempotent if $f\circ f = f$. We say that f is a split idempotent if there is some $g:A\to B, h:B\to A$ such that $h\circ g=f$ and $g\circ h=1_B$. Show that in Set, f is idempotent if and only if its image equals its set of fixed points. Show that every idempotent in Set is split. 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http://mathoverflow.net/questions/2390/why-does-non-abelian-group-cohomology-exist/15970	## Why does non-abelian group cohomology exist? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If K is a non-abelian group on which a group G acts via automorphisms, we can define 1-cocycles and 1-coboundaries by mimicking the explicit formulas coming from the bar resolution in ordinary group cohomology, and thus we have a reasonable notion of H^1(G, K). It turns out we have a part of the expected long exact sequence, until this construction breaks down for building H^i when i > 1, where the long exact sequence stops. There are other analogues to ordinary group cohomology as well. The only proof I've ever seen of any of this is by hand. Is there some deeper explanation of why non-abelian group cohomology exists (and then ceases to exist)? - ## 6 Answers Topologically, you could say that this is true because K(A,1) exists for nonabelian groups A. When the action of G on A is trivial, at least, H^1(G,A) should be homotopy classes of maps from K(G,1) to K(A,1) (the same way H^n(G,A)=H^n(K(G,1);A)=homotopy classes of maps K(G,1) \to K(A,n) for A abelian). In a similar way, H^0 is defined with coefficients in any pointed set. - 7 And the reason, in turn, that K(A,1) exists for nonabelian groups but not K(A,n) for higher n is the Eckmann-Hilton argument showing that pi_n is abelian for n>1. Likewise, K(A,0) exists for any pointed set, but not K(A,n) for n>0 since pi_n is a group for n>0. (-: – Mike Shulman Oct 25 2009 at 2:45 ### 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. To elaborate on Eric's answer, I believe that H1-n(G, A) is πn of the homotopy fixed point space K(A, 1)hG. That exact sequence which ends at H1--which is only a set, while H0 is a group--is actually the long exact sequence of the fibration K(A, 1)hG -> K(B, 1)hG -> K(C, 1)hG in disguise. That reindexing 1-n is related to the 1 in K(A, 1); if A is abelian, then we can replace all the occurrences of 1 by r for any r ≥ 0, giving arbitrarily long segments of the long exact sequence. (Or you can use the language of spectra: H-n(G, A) = πn((HA)hG). (HA)hG has nonzero homotopy groups only in nonpositive dimension.) K(A, 1)hG is a groupoid (it only has homotopy in dimensions 0 and 1) and I expect it should be the groupoid of some kind of extensions of G by A (where morphisms are isomorphisms of extensions which are the identity on G and A). H1(G, A) would then classify those extensions, and H0(G, A) = AG would be the automorphism group of the basepoint, which I guess is the semidirect product (this is easy to check). - I don't know if this is a "deep" or "shallow" explanation, but if anyone is still reading this thread, here is a different explanation. I'll start with the preliminary comment that the cohomology of a group $K$ is a special case of cohomology of topological spaces. In topology in general, you get the same phenomenon that $H^k(X,G)$ is well-defined either when $G$ is abelian or $k=1$. Consider the definition of simplicial cohomology for locally finite simplicial complexes. Or, more generally, CW cohomology for locally finite, regular complexes — regular means mainly that each attaching map is embedded. You can define a $k$-cochain with coefficients in a group $G$ (or even in any set) as a function from the $k$-cells to $G$. In attempting to define the coboundary of a cochain $c$ on a $k+1$-cell $e$, you should multiply together the values of $c$ on the facets of $e$. The obvious problem is that if $G$ is non-abelian, the product is order-dependent. However, if $k=1$, geometry gives you a gift: The facets are cyclically ordered, and what you mainly wanted to know is whether the product is trivial. The criterion of whether a cyclic word is trivial is well-defined in any group, not just abelian groups. A similar but simpler phenomenon occurs for the notion of a coboundary: If $e$ is an oriented edge and $c$ is a 0-cochain, there is a non-abelian version of modifying a 1-cochain by $c$ because the vertices of $e$ are an ordered pair. So far this is just a more geometric version of Eric Wofsey's answer. It is very close to the fact that $\pi_1$ is non-abelian while higher homotopy groups are abelian — and therefore non-commutative classifying spaces exist only for $K(G,1)$. However, in this version of the explanation, something extra appears when $X=M$ is a 3-manifold. If $M$ is a 3-manifold, then not only are the edges of a face cyclically ordered, the faces incident to an edge are also cyclically ordered. It turns out that, at least at the level of computing the cardinality of $H^1(M,G)$, you can let $G$ be both non-commutative and non-cocommutative. In other words, $G$ can be replaced by a finite-dimensional Hopf algebra $H$ which does not need to be commutative or cocommutative. Finiteness is necessary because it is a counting invariant. The resulting invariant $\#(M,H)$ was a topic of my PhD thesis and is explained here and here. Although the motivation is original, the invariant is a special case of more standard quantum invariants defined by other people. (The same construction was also later found by three physicists, but I can't remember their names at all right now.) Many 3-manifolds are also classifying spaces of groups, so for these groups there is the same notion of noncommutative, non-cocommutative group cohomology. - The topological viewpoint of Eric's answer applies to cohomology with nontrivial actions, too. If the action of G on A is not trivial, then the group cohomology H(G;A) can be identified with the topological cohomology of K(G,1) with local coefficients in the coefficient system (= locally constant sheaf) which is A with its action of π1(K(G,1)) = G. So the real question is then, why is the sheaf cohomology H1(X;A) defined with coefficients in a sheaf of nonabelian groups, but not Hn for n>1? This then essentially follows from the same argument (K(A,1) exists for any group A, but other K(A,n)s only for A abelian) but applied in the category (or (∞,1)-category, or model category, or whatever) of sheaves of spaces over X. There is also a sort of "higher nonabelian cohomology." For a nonabelian group A, you can't make K(A,2) but you can make B(hAut(K(A,1))) where hAut denotes the topological monoid of self-homotopy-equivalences, and you can think about homotopy classes of maps from a space X (such as K(G,1)) into B(hAut(K(A,1))) as a sort of "nonabelian H2." If A happens to be abelian, this construction contains the usual abelian H2 via the map K(A,2) = B(K(A,1)) --> B(hAut(K(A,1))) given by letting K(A,1) act on itself via left translation (since it is a topological group whenever A is abelian). But even in this case, the "nonabelian H2" contains much more than the usual abelian H2, so it's a little misleading to call it "nonabelian H2." - 2 Yes, I was viewing group cohomology as the derived version -^hG of the fixed point functor -^G, but the "global sections over K(G,1)" functor is probably more fundamental in a sense. – Reid Barton Oct 25 2009 at 3:33 A concrete and arithmetically useful way to interpret it without appeal to explicit cocycle formulas is to express everything in the language of torsors. More specifically, for arithmetic purposes if the group G is Gal(F'/F) for a Galois extension F'/F and if the group K is H(F') for an F-group scheme H of finite type and K is equipped with the evident left G-action then ${\rm{H}}^1(G,K)$ is the set of isomorphism classes of H-torsors over F which split over F' (i.e.,, admit an F'-rational point). The low-degree exact sequence can then be expressed entirely in such terms, using pushouts and pullbacks with torsors. (Implicit in the argument is effectivity of Galois descent for H-torsors, which uses that H is quasi-projective over F.) This is useful in settings as varied as H an abelian variety and H a linear algebraic group, and even the non-smooth case. In fact, when using non-smooth H it is rather restrictive to use Galois cohomology (but not unnatural if studying Tate-Shafarevich sets with coefficients in an Aut-functor, such as for a projective variety), and in such cases the "right" variant that is often more useful is to work with torsors for the fppf topology over F. The torsor viewpoint also gives a useful perspective when working over richer base rings than fields, such as rings of S-integers in a global field, even in the case of a smooth coefficient group (over the ring of S-integers), for which the etale topology is "enough". See section 5.3 in Chapter I of Serre's book on Galois cohomology for the Galois case, Milne's "Etale cohomology" book for generalization with flat and \'etale topologies, and Appendix B in my paper on "Finiteness theorems for algebraic groups over function fields" for a concrete fleshing out of the dictionary between the torsor and Galois languages (where I work with affine group schemes, due to the context of that paper). Some papers of Mazur and Grothendieck (not as co-authors...) on abelian varieties make creative use of the torsor viewpoint when working with Tate-Shafarevich groups. The exact sequence for Brauer groups in global class field theory also has a useful interpretation via torsors; see Grothendieck's papers on Brauer groups for more in that direction (and somewhere in there he also discusses Tate-Shafarevich). Beware that when the base is not a field (or even when it is a field but we relax "quasi-projective" to "locally finite type" on $H$, such as for Aut-schemes of projective or proper varieties) then effectivity of descent for torsors is not at all clear, even with quasi-projective hypotheses, and so the torsors often need to be understood to be taken in the category of algebraic spaces (for which fppf descent is always effective). In the N\'eron Models book they have a discussion (somewhere in Chapter 6, I think) on effectivity of descent for torsors if one wishes to avoid algebraic spaces (under suitable hypotheses on the "coefficient group"), but working with algebraic spaces isn't so bad once one gets used to them and it is a more natural setting due to their better general behavior with respect to descent. - Extensions exist for non-abelian groups too. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 30, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.9289658665657043, "perplexity_flag": "head"}
http://mathoverflow.net/questions/100469?sort=newest	Opinions on the Multiplication of Measures Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) A few questions, hopefully to spark some discussion. How can one define a product of measures? 1. We could use Colombeau products by embedding the measures into the distributions? I'm not sure why this approach is frowned on? 2. Since measures are map from sets into numbers, can we not define the product as the pointwise product on sets? I would like to see the counter example here. 3. Another product I can think of is to take the product measure and then use the fact that there exists a bijection from R to R^2. I guess that there exist no measurable bijection, which is why that might not work. Thanks, Daniel - 6 Before tackling this general problem, take a very special case: both measures are the Dirac delta at zero in the real line. Now "define" the product. I think you can find that question asked here more than once already... – Gerald Edgar Jun 23 at 19:44 2 Answers Counterexample for suggestion 2: Take a measure on a 2-point set, assigning to each singleton measure 1 and therefore assigning to the whole set measure 2. The pointwise product of this measure with itself still gives the singletons measure 1 but gives the whole space measure 4, so it's not a measure. For suggestion 3: There are measurable bijections from $R^2$ to $R$ when $R$ is the real line, but under the "natural" such bijections (e.g., interleaving binary expansions) the image of the product measure (Lebesgue measure times itself, on the plane) would just be Lebesgue measure on the line. By using other, specially constructed bijections, you could get the image to be any of lots of other measures on the line. - Thank you for your answer. Any opinion on whether there is any sort of product? – dcs24 Jun 23 at 16:22 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. Not a definitive answer, just some thoughts that I hope you might find useful. Firstly, there is a celebrated example of Schwartz that shows that you can't do this globally in the sense of getting a ring structure on the space of measures which extends that on the space of continuous functions. It can, however, often be fruitful to try a local approach, i.e., find pairs of measures which {\it can} be multiplied in a sensible way. For example, the product of two dirac measures is only a problem if their singularities coincide. One can, trivially, always multiply a continuous function and a measure (and with a little bit of integration theory this can be extended to more general functions). One way to deal with less trivial situations is to say that if we can cover $I$ (for the sake of simplicity, I will assume that the measures are defined on the unit interval $I$) with a finite family of relatively open subintervals so that on each interval at least one of the the measures is a continuous function (which one depending on the interval, of course), then we can define the product. This covers the above case of the product of dirac measures with distinct singularities. One would have to know more about the kind of measures that you want to multiply in order to decide if this is of much help. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9537918567657471, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/197549-proof-parametrized-curve-orthogonal-certain-vector-v.html	1Thanks • 1 Post By HallsofIvy # Thread: 1. ## Proof that a parametrized curve orthogonal to a certain vector v Let $\alpha\colon I\rightarrow R^3$ be a parametrized curve and let $\mathbf{v} \in R^3$ be a fixed vector. Assume that $\alpha'(t)$ is orthogonal to $\mathbf{v}$ for all $t \in I$ and that $\alpha(0)$ is also orthogonal to $\mathbf{v}$ Prove that $\alpha(t)$ is orthogonal to $\mathbf{v}$ for all $t \in I$ We are given that $\alpha'(t)$ and $\alpha(0)$ is orthogonal to $\mathbf{v}$ Let $\mathbf{v} = (v_1, v_2, v_3)$ and $\alpha(t) = (x(t), y(t), z(t))$ $\therefore \alpha'(t) = (x'(t), y'(t), z'(t))$ So we have: $\alpha'(t) \cdot\, \mathbf{v} = (x'(t) * \, v_1 +\,\, y'(t) * v_2 + \,\, z'(t) * v_3) = 0 \text{ [Given]}$ $\alpha(0) \cdot\, \mathbf{v} = (x(0) \, v_1 +\,\, y(0) v_2 +\,\, z(0) * v_3) = 0 \text{ [Given]}$ Now how do I use this info and prove that $\alpha(t)$ is orthogonal to $\mathbf{v}$ for all $t \in I$? Is it possible for anyone to kindly give some hints on how to prove this? 2. ## Re: Proof that a parametrized curve orthogonal to a certain vector v I think I solved it. For those who are interested: Since $\alpha'(t) \cdot\, \mathbf{v} = (x'(t) \, v_1)+\,\,( y'(t) v_2) +\,\, (z'(t) * v_3) = 0 \text{ [Given]}$ Particularly: $(x'(t) \, v_1)+\,\, (y'(t) v_2) +\,\, (z'(t) * v_3) = 0$ Taking integration of both sides: $(v_1 * x(t)) + C_1 +\,\,( v_2 * (y(t)) + C_2 +\,\, (v_3 * z(t)) + C_3 = C$ By taking all constants to the right side we get: $(v_1 * x(t)) + (v_2 * y(t)) + (v_3 * z(t)) = C - C_1 - C_2 - C_3 \text{ ................(1)}$ We also know that: $\alpha(0) \cdot\, \mathbf{v} = (x(0) \, v_1) +\,\, (y(0) v_2) +\,\, (z(0) * v_3) = 0 \text{ [Given]}$ So pluging in $t = 0$ into eq(1) we get: $C - C_1 - C_2 - C_3 = 0$ therefore: $(v_1 * x(t)) + (v_2 * y(t)) + (v_3 * z(t)) = 0$ But this means: $\alpha(t) \cdot\,\, \mathbf{v} = 0$ So $\alpha(t)$ is orthogonal to $\mathbf{v}$.[As was to be shown] 3. ## Re: Proof that a parametrized curve orthogonal to a certain vector v Looks good but I think I would go the "other way"- that is, differentiate $\alpha$ rather than integrate $\alpha'$. Let $u(t)= \alpha(t)\cdot v$. Then, since v is a constant vector, $u'= \alpha\cdot v= 0$. Since the derivative of u is 0, u is a constant. And since $u(0)= \alpha(0)\cdot v= 0$ it follows that $u(t)= \alpha(t)\cdot v= 0$ for all t. 4. ## Re: Proof that a parametrized curve orthogonal to a certain vector v Originally Posted by HallsofIvy Looks good but I think I would go the "other way"- that is, differentiate $\alpha$ rather than integrate $\alpha'$. Let $u(t)= \alpha(t)\cdot v$. Then, since v is a constant vector, $u'= \alpha\cdot v= 0$. Since the derivative of u is 0, u is a constant. And since $u(0)= \alpha(0)\cdot v= 0$ it follows that $u(t)= \alpha(t)\cdot v= 0$ for all t. Thank you for showing another way of solving this problem. I believe your way is the correct and better method. Again thanks.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 44, "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": 0.9439458250999451, "perplexity_flag": "head"}
http://mathoverflow.net/questions/105293/proofs-of-ergodicity-of-sl2-z-action-on-r2-without-using-duality/108744	## proofs of ergodicity of Sl(2, Z) action on R^2 without using duality ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The group $G=SL(2, R)$ acts linearly on $\mathbb R^2$. The Lebesgue measure of $\mathbb R^2$ is invariant and ergodic for $G$. There is a proof using duality theorem: Let $U$ be the upper triangular unipotent subgroup of $G$ and let $\Gamma=SL(2, \mathbb Z)$. Then $U$ acts ergodically on $G/\Gamma$. Therefore duality tells us that $\Gamma$ acts ergodically on $G/U\cong \mathbb R^2\backslash 0$. This method is good but it can not be extended to the group action on non homogeneous manifolds. Are there any other proofs? It is better that I can see different proofs to learn methods. - The most basic proof of the ergodicity result mentionned is by Fourier series but of course it does not generalize so it is not a good answer ... – unknown (google) Aug 26 at 11:50 ## 2 Answers What you call "duality" (that for two commuting actions under reasonable conditions ergodic properties of either action on the space of orbits of the other one are the same) goes back to the work of Furstenberg in the 60s. This idea is especially useful when dealing with the so-called geometric flows: the horocycle and the geodesic ones. However, usually one goes in the direction opposite to the one you mention, namely first studies the action on the space of $\Gamma$-orbits and then deduces the corresponding properties of geometric flows. You are referring to the situation with the horocycle flow: its ergodicity on the quotient of the hyperbolic plane $\mathbf H^2$ by a discrete subgroup $\Gamma$ is equivalent to ergodicity of the action of $\Gamma$ on the space of horocycles in $\mathbf H^2$ (the latter one being isomorphic to the linear action of $\Gamma$ on $\mathbb R^2\setminus{0}$). In the case of the geodesic flow the space of its orbits, i.e., the space of geodesics in the hyperbolic plane, is isomorphic to the square of the boundary circle with removed diagonal $\partial\mathbf H^2\times\partial\mathbf H^2\setminus \text{diag}$. In particular, invariant measures of the geodesic flow are in natural one-to-one correspondence with the so-called geodesic currents, i.e., invariant Radon measures on $\partial\mathbf H^2\times\partial\mathbf H^2\setminus \text{diag}$. For geodesic flows the case of "big" surfaces and manifolds (with infinite volume) was being considered from the very beginning of the theory in the 30s (and in the constant curvature case finalized in the so-called Hopf-Tsuji-Sullivan theorem in the late 70s), whereas for the horocycle flow it wasn't really considered (and settled) until the late 90s. Generally speaking, proving ergodicity of the horocycle flow on a general surface consists of two components: (1) establishing ergodicity of the boundary action, (2) proving ergodicty of the so-called Busemann cocycle (which determines the extension from the boundary circle to the space of horocycles). For more details see the works of Babillot - Ledrappier MR1699356, Kaimanovich MR1738739, MR1919405, MR2731695, Pollicott MR1739598, Coudene MR1836430 , Solomyak MR1860491, Ledrappier MR1871151, Hamenstadt MR1926280, Ledrappier - Pollicott MR1953296, Babillot MR2087786, Ledrappier - Sarig MR2226490, Sarig MR2827866. - ### 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 recommand the work of Thomas Roblin "Ergodicité et équidistribution en courbure négative" Mémoires de la SMF 95 (2003), in addition to the references mentioned in the first answer. In particular, he does not need any compactness or finite volume assumption, he works with variable negative curvature, and even on CAT(-1) spaces $X$, not only on manifolds. In this framework ergodic properties of the horospherical foliation are related (by duality) to ergodic properties of the action of $\Gamma$ on the space of horospheres $\partial\tilde{X}\times \mathbb{R}$ (which identifies with $\mathbb{R}^2\setminus {0 }/\pm$ in the case of hyperbolic surfaces). These ergodic properties are strongly related to the mixing property of the geodesic flow. As mentioned in the beginning of the first answer, in the much earlier approach of Furstenberg, harmonic analysis is used on $\mathbb{R}^2$ to get unique ergodicity of the action of a cocompact fuchsian group $\Gamma$ on $\mathbb{R}^2\setminus{0}$, and the unique ergodicity of the horocyclic flow on $SL(2,\mathbb{R})/\Gamma$ is deduced by duality. The modern approach (as in Roblin, or Coudène, or others) through mixing is more dynamical. -	{"extraction_info": {"found_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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9065430164337158, "perplexity_flag": "head"}
http://alanrendall.wordpress.com/2010/11/	Hydrobates A mathematician thinks aloud Archive for November, 2010 Me on TV November 26, 2010 Recently I was interviewed by TV journalists for a documentary of the channel 3Sat called “Rätsel Dunkle Materie” [The riddle of dark matter]. It was broadcast yesterday. Before I say more about my experience with this let me do a flashback to the only other time in my life I appeared on TV. On that occasion the BBC visited our school. I guess I was perhaps twelve at the time although I do not know for sure. I was filmed reading a poem which I had written myself. I was seen sitting in a window of the Bishops’ Palace in Kirkwall, looking out. I suppose only my silhouette was visible. I no longer have the text of the poem. All I know is that the first line was ‘Björn, adventuring at last’ and that later on there was some stuff about ravens. At that time I was keen on Vikings. The poem was no doubt very heroic, so that the pose looking out the window was appropriate. Coming back to yesterday, the documentary consisted of three main elements. There was a studio discussion with three guests – the only one I know personally is Simon White. There were some clips illustrating certain ideas. Thirdly there were short sequences from interviews with some other people. I was one of these people. They showed a few short extracts of the interview with me and I was quite happy with the selection they made. This means conversely that they nicely cut out things which I might not have liked so much. I was answering questions posed by one of the journalists and which were not heard on TV. They told me in advance that this would be the case. They told me that for this reason I should not refer to the question during my answers. I found this difficult to do and I think I would need some practice to do it effectively. Fortunately it seems that they efficiently cut out these imperfections. I did not know the questions in advance of the filming and this led to some hesitant starts in my answers. This also did not come through too much in what was shown. Summing up, it was an interesting experience and I would do it again if I had the chance. Of course being a studio guest would be even more interesting … I found the documentary itself not so bad. I could have done without the part about religion at the end. Perhaps the inclusion of this is connected with the fact that the presenter of the series, Gert Scobel, studied theology and also has a doctorate in hermeneutics. (I had to look up that word to have an idea what it meant.) An aspect of the presentation which was a bit off track was that it gave the impression that the idea of a theory unifying general relativity and quantum theory was solely due to Stephen Hawking. Before ending this post I should perhaps say something about my own point of view on dark matter and dark energy. Of course they are symptoms of serious blemishes in our understanding of reality. I believe that dark matter and dark energy are better approaches to explaining the existing observational anomalies than any other alternative which is presently available. In the past I have done some work related to dark energy myself. The one thing that I do not like about a lot of the research in this area is that while people are very keen on proposing new ‘theories’ (which are often just more or less vague ideas for models) there is much less enthusiasm for working out these ideas to obtain a logically sound proposal. Of course that would be more difficult. A case study in this direction was carried out in the diploma thesis of Nikolaus Berndt which was done under my supervision. The theme was to what extent the so-called Cardassian models (do not) deserve to be called a theory. We later produced a joint publication on this. It has not received much attention in the research community and as far as I know has only been cited once. Posted in Uncategorized \| 3 Comments » Twisting Gowdy November 11, 2010 I have previously written about the Gowdy solutions as the simplest model case of spatially inhomogeneous solutions of the Einstein vacuum equations and about the generalization of this to the Einstein-Maxwell equations. I have also written about spatially homogeneous solutions (Bianchi models). I have just posted a paper where I explore connections between these two classes of solutions (Gowdy and Bianchi) in the hope of learning more about both. The basic idea is simple. A Bianchi model has Killing vectors which form a three-dimensional Lie algebra. Most of these (all except types VIII and IX) have a two-dimensional Abelian subalgebra. When this is the case we can simply forget about one Killing vector and end up with a spacetime with two commuting Killing vectors. This looks very like a Gowdy spacetime. There are, however, two potential problems. The first is that Gowdy spacetimes have a certain discrete symmetry in addition to their two Killing vectors. Almost all Bianchi types do satisfy this condition, the exception being type ${\rm VI}{}_{\frac19}$. In what follows I ignore this exceptional type. The other problem is that a Gowdy spacetime is supposed to have a compact Cauchy surface and the Killing vectors are assumed to exist globally. If these conditions are weakened then more alternatives are allowed. Let me call this situation a locally Gowdy spacetime. (This is not standard terminology.) Assume that the Killing vector fields have no zeroes. Then in the Gowdy case the Cauchy surface has the topology of a three-torus $T^3$. In the locally Gowdy case the existence of a compact Cauchy surface is closely linked to those Bianchi types belonging to Bianchi Class A. I now restrict to that class. Then the additional Bianchi types which are possible are II, ${\rm VI}{}_0$ and ${\rm VII}{}_0$. In Gowdy spacetimes there are two of the Einstein equations for functions $P(t,\theta)$ and $Q(t,\theta)$ which are of central importance. I call them the Gowdy equations. In order to describe Gowdy spacetimes the functions $P$ and $Q$ should be periodic in $\theta$. To accomodate Bianchi models of types II and ${\rm VI}{}_0$ more complicated boundary conditions are required. The twisted Gowdy solutions are solutions of the Gowdy equations which satisfy these boundary conditions but need not be homogeneous. They represent finite inhomogeneous perturbations of the corresponding Bianchi models. Periodic boundary conditions can accommodate type ${\rm VII}{}_0$, which then corresponds to what are called circular loop spacetimes. In the twisted cases the topology of a compact Cauchy surface is more complicated than that of $T^3$. For type II and type ${\rm VI}{}_0$ it corresponds to a manifold admitting geometric structures of type Nil and Sol in the sense of Thurston, respectively. These are twisted topologies which are torus bundles over a circle. A lot is understood about the global dynamics of Gowdy solutions, mainly due to the work of Hans Ringström. By uncovering certain connections between different classes of solutions I have been able to transfer some of these results to the twisted Gowdy case. Unfortunately I was not able to obtain a general analysis of the late-time behaviour of twisted Gowdy solutions. The analytical techniques which were successful in the untwisted case do not seem to adapt well. This leaves a challenge for the future.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 13, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9789594411849976, "perplexity_flag": "head"}
http://medlibrary.org/medwiki/Clifford_algebras	# Clifford algebras Welcome to MedLibrary.org. For best results, we recommend beginning with the navigation links at the top of the page, which can guide you through our collection of over 14,000 medication labels and package inserts. For additional information on other topics which are not covered by our database of medications, just enter your topic in the search box below: In mathematics, Clifford algebras are a type of associative algebra. As K-algebras, they generalize the real numbers, complex numbers, quaternions and several other hypercomplex number systems.[1][2] The theory of Clifford algebras is intimately connected with the theory of quadratic forms and orthogonal transformations. Clifford algebras have important applications in a variety of fields including geometry, theoretical physics and digital image processing. They are named after the English geometer William Kingdon Clifford. The most familiar Clifford algebra, or orthogonal Clifford algebra, is also referred to as Riemannian Clifford algebra.[3] ## Introduction and basic properties Specifically, a Clifford algebra is a unital associative algebra which contains and is generated by a vector space V equipped with a quadratic form Q. The Clifford algebra Cℓ(V, Q) is the "freest" algebra generated by V subject to the condition[4] $v^2 = Q(v)1\ \text{ for all } v\in V,$ where the product on the left is that of the algebra, and the 1 is its multiplicative identity. The definition of a Clifford algebra endows it with more structure than a "bare" K-algebra: specifically it has a designated or privileged subspace that is isomorphic to V. Such a subspace cannot in general be uniquely determined given only a K-algebra isomorphic to the Clifford algebra. If the characteristic of the ground field K is not 2, then one can rewrite this fundamental identity in the form $uv + vu = 2\lang u, v\rang \text{ for all }u,v \in V,$ where ⟨u, v⟩ = (Q(u + v) − Q(u) − Q(v))/2 is the symmetric bilinear form associated with Q, via the polarization identity. The idea of being the "freest" or "most general" algebra subject to this identity can be formally expressed through the notion of a universal property, as done below. Quadratic forms and Clifford algebras in characteristic 2 form an exceptional case. In particular, if char(K) = 2 it is not true that a quadratic form determines a symmetric bilinear form, or that every quadratic form admits an orthogonal basis. Many of the statements in this article include the condition that the characteristic is not 2, and are false if this condition is removed. ### As a quantization of the exterior algebra Clifford algebras are closely related to exterior algebras. In fact, if Q = 0 then the Clifford algebra Cℓ(V, Q) is just the exterior algebra Λ(V). For nonzero Q there exists a canonical linear isomorphism between Λ(V) and Cℓ(V, Q) whenever the ground field K does not have characteristic two. That is, they are naturally isomorphic as vector spaces, but with different multiplications (in the case of characteristic two, they are still isomorphic as vector spaces, just not naturally). Clifford multiplication together with the privileged subspace is strictly richer than the exterior product since it makes use of the extra information provided by Q. More precisely, Clifford algebras may be thought of as quantizations (cf. quantization (physics), Quantum group) of the exterior algebra, in the same way that the Weyl algebra is a quantization of the symmetric algebra. Weyl algebras and Clifford algebras admit a further structure of a -algebra, and can be unified as even and odd terms of a superalgebra, as discussed in CCR and CAR algebras. ## Universal property and construction Let V be a vector space over a field K, and let Q: V → K be a quadratic form on V. In most cases of interest the field K is either R, C or a finite field. A Clifford algebra Cℓ(V, Q) is a unital associative algebra over K together with a linear map i : V → Cℓ(V, Q) satisfying i(v)2 = Q(v)1 for all v ∈ V, defined by the following universal property: given any associative algebra A over K and any linear map j : V → A such that j(v)2 = Q(v)1A for all v ∈ V (where 1A denotes the multiplicative identity of A), there is a unique algebra homomorphism f : Cℓ(V, Q) → A such that the following diagram commutes (i.e. such that f ∘ i = j): Working with a symmetric bilinear form ⟨·,·⟩ instead of Q (in characteristic not 2), the requirement on j is j(v)j(w) + j(w)j(v) = 2⟨v, w⟩ for all v, w ∈ V. A Clifford algebra as described above always exists and can be constructed as follows: start with the most general algebra that contains V, namely the tensor algebra T(V), and then enforce the fundamental identity by taking a suitable quotient. In our case we want to take the two-sided ideal IQ in T(V) generated by all elements of the form $v\otimes v - Q(v)1$ for all $v\in V$ and define Cℓ(V, Q) as the quotient algebra Cℓ(V, Q) = T(V)/IQ. The ring product inherited by this quotient is sometimes referred to as the Clifford product[5] to differentiate it from the inner and outer products. It is then straightforward to show that Cℓ(V, Q) contains V and satisfies the above universal property, so that Cℓ is unique up to a unique isomorphism; thus one speaks of "the" Clifford algebra Cℓ(V, Q). It also follows from this construction that i is injective. One usually drops the i and considers V as a linear subspace of Cℓ(V, Q). The universal characterization of the Clifford algebra shows that the construction of Cℓ(V, Q) is functorial in nature. Namely, Cℓ can be considered as a functor from the category of vector spaces with quadratic forms (whose morphisms are linear maps preserving the quadratic form) to the category of associative algebras. The universal property guarantees that linear maps between vector spaces (preserving the quadratic form) extend uniquely to algebra homomorphisms between the associated Clifford algebras. ## Basis and dimension If the dimension of V is n and {e1, …, en} is a basis of V, then the set $\{e_{i_1}e_{i_2}\cdots e_{i_k} \mid 1\le i_1 < i_2 < \cdots < i_k \le n\mbox{ and } 0\le k\le n\}$ is a basis for Cℓ(V, Q). The empty product (k = 0) is defined as the multiplicative identity element. For each value of k there are n choose k basis elements, so the total dimension of the Clifford algebra is $\dim C\ell(V,Q) = \sum_{k=0}^n\begin{pmatrix}n\\ k\end{pmatrix} = 2^n.$ Since V comes equipped with a quadratic form, there is a set of privileged bases for V: the orthogonal ones. An orthogonal basis is one such that $\langle e_i, e_j \rangle = 0 \qquad i\neq j. \,$ where ⟨·,·⟩ is the symmetric bilinear form associated to Q. The fundamental Clifford identity implies that for an orthogonal basis $e_ie_j = -e_je_i \qquad i\neq j. \,$ This makes manipulation of orthogonal basis vectors quite simple. Given a product $e_{i_1}e_{i_2}\cdots e_{i_k}$ of distinct orthogonal basis vectors of V, one can put them into standard order while including an overall sign determined by the number of pairwise swaps needed to do so (i.e. the signature of the ordering permutation). ## Examples: real and complex Clifford algebras The most important Clifford algebras are those over real and complex vector spaces equipped with nondegenerate quadratic forms. It turns out that every one of the algebras Cℓp,q(R) and Cℓn(C) are isomorphic to A or A⊕A, where A is a full matrix ring with entries from R, C, or H. For a complete classification of these algebras see classification of Clifford algebras. ### Real numbers Main article: Geometric algebra The geometric interpretation of real Clifford algebras is known as geometric algebra. Every nondegenerate quadratic form on a finite-dimensional real vector space is equivalent to the standard diagonal form: $Q(v) = v_1^2 + \cdots + v_p^2 - v_{p+1}^2 - \cdots - v_{p+q}^2$ where n = p + q is the dimension of the vector space. The pair of integers (p, q) is called the signature of the quadratic form. The real vector space with this quadratic form is often denoted Rp, q. The Clifford algebra on Rp, q is denoted Cℓp, q(R). The symbol Cℓn(R) means either Cℓn,0(R) or Cℓ0,n(R) depending on whether the author prefers positive definite or negative definite spaces. A standard orthonormal basis {ei} for Rp,q consists of n = p + q mutually orthogonal vectors, p of which have norm +1 and q of which have norm −1. The algebra Cℓp,q(R) will therefore have p vectors that square to +1 and q vectors that square to −1. Note that Cℓ0,0(R) is naturally isomorphic to R since there are no nonzero vectors. Cℓ0,1(R) is a two-dimensional algebra generated by a single vector e1 that squares to −1, and therefore is isomorphic to C, the field of complex numbers. The algebra Cℓ0,2(R) is a four-dimensional algebra spanned by {1, e1, e2, e1e2}. The latter three elements square to −1 and all anticommute, and so the algebra is isomorphic to the quaternions H. Cℓ0,3(R) is an 8-dimensional algebra isomorphic to the direct sum H ⊕ H called split-biquaternions. ### Complex numbers One can also study Clifford algebras on complex vector spaces. Every nondegenerate quadratic form on a complex vector space is equivalent to the standard diagonal form $Q(z) = z_1^2 + z_2^2 + \cdots + z_n^2$ where n = dim V, up to isomorphism so there is only one nondegenerate Clifford algebra for each dimension n. We will denote the Clifford algebra on Cn with the standard quadratic form by Cℓn(C). The first few cases are not hard to compute. One finds that Cℓ0(C) ≅ C, the complex numbers Cℓ1(C) ≅ C ⊕ C, the bicomplex numbers Cℓ2(C) ≅ M(2, C), the biquaternions where M(n, C) denotes the algebra of n×n matrices over C. ## Examples: constructing quaternions and dual quaternions ### Quaternions In this section, Hamilton's quaternions are constructed as the even sub algebra of the Clifford algebra Cℓ0,3(R). Let the vector space V be real three dimensional space R3, and the quadratic form Q be derived from the usual Euclidean metric. Then, for v, w in R3 we have the quadratic form, or dot product, $\mathbf{v}\cdot\mathbf{w}= v_1w_1 + v_2w_2 + v_3w_3.$ Now introduce the Clifford product of vectors v and w given by $\mathbf{v}\mathbf{w} + \mathbf{w}\mathbf{v} = -2 (\mathbf{v}\cdot \mathbf{w}).\!$ This formulation uses the negative sign so the correspondence with quaternions is easily shown. Denote a set of orthogonal unit vectors of R3 as e1, e2, and e3, then the Clifford product yields the relations $\mathbf{e}_2 \mathbf{e}_3 = -\mathbf{e}_3 \mathbf{e}_2, \,\,\, \mathbf{e}_3 \mathbf{e}_1 = -\mathbf{e}_1 \mathbf{e}_3,\,\,\, \mathbf{e}_1 \mathbf{e}_2 = -\mathbf{e}_2 \mathbf{e}_1,\!$ and $\mathbf{e}_1 ^2 = \mathbf{e}_2^2 =\mathbf{e}_3^2 = -1. \!$ The general element of the Clifford algebra Cℓ0,3(R) is given by $A = a_0 + a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 + a_3 \mathbf{e}_3 + a_4 \mathbf{e}_2 \mathbf{e}_3 + a_5 \mathbf{e}_3 \mathbf{e}_1 + a_6 \mathbf{e}_1 \mathbf{e}_2 + a_7 \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3.\!$ The linear combination of the even rank elements of Cℓ0,3(R) defines the even sub algebra Cℓ00,3(R) with the general element $Q = q_0 + q_1 \mathbf{e}_2 \mathbf{e}_3 + q_2 \mathbf{e}_3 \mathbf{e}_1 + q_3 \mathbf{e}_1 \mathbf{e}_2. \!$ The basis elements can be identified with the quaternion basis elements i, j, k as $i= \mathbf{e}_2 \mathbf{e}_3, j= \mathbf{e}_3 \mathbf{e}_1, k = \mathbf{e}_1 \mathbf{e}_2,$ which shows that the even sub algebra Cℓ00,3(R) is Hamilton's real quaternion algebra. To see this, compute $i^2 = (\mathbf{e}_2 \mathbf{e}_3)^2 = \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_2 \mathbf{e}_3 = - \mathbf{e}_2 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_3 = -1,\!$ and $ij = \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_3 \mathbf{e}_1 = -\mathbf{e}_2 \mathbf{e}_1 = \mathbf{e}_1 \mathbf{e}_2 = k.\!$ Finally, $ijk = \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_3 \mathbf{e}_1 \mathbf{e}_1 \mathbf{e}_2 = -1.\!$ ### Dual quaternions In this section, dual quaternions are constructed as the even Clifford algebra of real four dimensional space with a degenerate quadratic form.[6][7] Let the vector space V be real four dimensional space R4, and let the quadratic form Q be a degenerate form derived from the Euclidean metric on R3. For v, w in R4 introduce the degenerate bilinear form $d(\mathbf{v}, \mathbf{w})= v_1w_1 + v_2w_2 + v_3w_3.$ This degenerate scalar product projects distance measurements in R4 onto the R3 hyperplane. The Clifford product of vectors v and w is given by $\mathbf{v}\mathbf{w} + \mathbf{w}\mathbf{v} = -2 \,d(\mathbf{v}, \mathbf{w}).\!$ Note the negative sign is introduced to simplify the correspondence with quaternions. Denote a set of orthogonal unit vectors of R4 as e1, e2, e3 and e4, then the Clifford product yields the relations $\mathbf{e}_m \mathbf{e}_n = -\mathbf{e}_n \mathbf{e}_m, \,\,\, m \ne n,\!$ and $\mathbf{e}_1 ^2 = \mathbf{e}_2^2 =\mathbf{e}_3^2 = -1, \,\, \mathbf{e}_4^2 =0.\!$ The general element of the Clifford algebra Cℓ(R4,d) has 16 components. The linear combination of the even ranked elements defines the even sub algebra Cℓ0(R4,d) with the general element $H = h_0 + h_1 \mathbf{e}_2 \mathbf{e}_3 + h_2 \mathbf{e}_3 \mathbf{e}_1 + h_3 \mathbf{e}_1 \mathbf{e}_2 + h_4 \mathbf{e}_4 \mathbf{e}_1 + h_5 \mathbf{e}_4 \mathbf{e}_2 + h_6 \mathbf{e}_4 \mathbf{e}_3 + h_7 \mathbf{e}_1 \mathbf{e}_2\mathbf{e}_3 \mathbf{e}_4. \!$ The basis elements can be identified with the quaternion basis elements i, j, k and the dual unit ε as $i=\mathbf{e}_2 \mathbf{e}_3, j=\mathbf{e}_3 \mathbf{e}_1, k = \mathbf{e}_1 \mathbf{e}_2, \,\, \varepsilon = \mathbf{e}_1 \mathbf{e}_2\mathbf{e}_3 \mathbf{e}_4. \!$ This provides the correspondence of Cℓ00,3,1(R) with dual quaternion algebra. To see this, compute $\varepsilon ^2 = (\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4)^2 = \mathbf{e}_1 \mathbf{e}_2\mathbf{e}_3 \mathbf{e}_4 \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4 = -\mathbf{e}_1 \mathbf{e}_2\mathbf{e}_3 (\mathbf{e}_4 \mathbf{e}_4 ) \mathbf{e}_1 \mathbf{e}_2\mathbf{e}_3 = 0,\!$ and $\varepsilon i = (\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4) \mathbf{e}_2 \mathbf{e}_3 = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4 \mathbf{e}_2 \mathbf{e}_3 = \mathbf{e}_2\mathbf{e}_3 (\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4) = i\varepsilon.\!$ The exchanges of e1 and e4 alternate signs an even number of times, and show the dual unit ε commutes with the quaternion basis elements i, j, and k. ## Properties ### Relation to the exterior algebra Given a vector space V one can construct the exterior algebra Λ(V), whose definition is independent of any quadratic form on V. It turns out that if K does not have characteristic 2 then there is a natural isomorphism between Λ(V) and Cℓ(V, Q) considered as vector spaces (and there exists an isomorphism in characteristic two, which may not be natural). This is an algebra isomorphism if and only if Q = 0. One can thus consider the Clifford algebra Cℓ(V, Q) as an enrichment (or more precisely, a quantization, cf. the Introduction) of the exterior algebra on V with a multiplication that depends on Q (one can still define the exterior product independent of Q). The easiest way to establish the isomorphism is to choose an orthogonal basis {ei} for V and extend it to a basis for Cℓ(V, Q) as described above. The map Cℓ(V, Q) → Λ(V) is determined by $e_{i_1}e_{i_2}\cdots e_{i_k} \mapsto e_{i_1}\wedge e_{i_2}\wedge \cdots \wedge e_{i_k}.$ Note that this only works if the basis {ei} is orthogonal. One can show that this map is independent of the choice of orthogonal basis and so gives a natural isomorphism. If the characteristic of K is 0, one can also establish the isomorphism by antisymmetrizing. Define functions fk: V × … × V → Cℓ(V, Q) by $f_k(v_1, \cdots, v_k) = \frac{1}{k!}\sum_{\sigma\in S_k}{\rm sgn}(\sigma)\, v_{\sigma(1)}\cdots v_{\sigma(k)}$ where the sum is taken over the symmetric group on k elements. Since fk is alternating it induces a unique linear map Λk(V) → Cℓ(V, Q). The direct sum of these maps gives a linear map between Λ(V) and Cℓ(V, Q). This map can be shown to be a linear isomorphism, and it is natural. A more sophisticated way to view the relationship is to construct a filtration on Cℓ(V, Q). Recall that the tensor algebra T(V) has a natural filtration: F0 ⊂ F1 ⊂ F2 ⊂ … where Fk contains sums of tensors with rank ≤ k. Projecting this down to the Clifford algebra gives a filtration on Cℓ(V, Q). The associated graded algebra $Gr_F C\ell(V,Q) = \bigoplus_k F^k/F^{k-1}$ is naturally isomorphic to the exterior algebra Λ(V). Since the associated graded algebra of a filtered algebra is always isomorphic to the filtered algebra as filtered vector spaces (by choosing complements of Fk in Fk+1 for all k), this provides an isomorphism (although not a natural one) in any characteristic, even two. ### Grading In the following, assume that the characteristic is not 2.[8] Clifford algebras are Z2-graded algebras (also known as superalgebras). Indeed, the linear map on V defined by v ↦ −v (reflection through the origin) preserves the quadratic form Q and so by the universal property of Clifford algebras extends to an algebra automorphism α: Cℓ(V, Q) → Cℓ(V, Q). Since α is an involution (i.e. it squares to the identity) one can decompose Cℓ(V, Q) into positive and negative eigenspaces of α $C\ell(V,Q) = C\ell^0(V,Q) \oplus C\ell^1(V,Q)$ where Cℓi(V, Q) = {x ∈ Cℓ(V, Q) \| α(x) = (−1)ix}. Since α is an automorphism it follows that $C\ell^{\,i}(V,Q)C\ell^{\,j}(V,Q) = C\ell^{\,i+j}(V,Q)$ where the superscripts are read modulo 2. This gives Cℓ(V, Q) the structure of a Z2-graded algebra. The subspace Cℓ0(V, Q) forms a subalgebra of Cℓ(V, Q), called the even subalgebra. The subspace Cℓ1(V, Q) is called the odd part of Cℓ(V, Q) (it is not a subalgebra). This Z2-grading plays an important role in the analysis and application of Clifford algebras. The automorphism α is called the main involution or grade involution. Elements that are pure in this Z2-grading are simply said to be even or odd. Remark. In characteristic not 2 the underlying vector space of Cℓ(V, Q) inherits an N-grading and a Z-grading from the canonical isomorphism with the underlying vector space of the exterior algebra Λ(V).[9] It is important to note, however, that this is a vector space grading only. That is, Clifford multiplication does not respect the N-grading or Z-grading, only the Z2-grading: for instance if Q(v) ≠ 0, then v ∈ Cℓ1(V, Q), but v2 ∈ Cℓ0(V, Q), not in Cℓ2(V, Q). Happily, the gradings are related in the natural way: Z2 ≅N/2N≅ Z/2Z. Further, the Clifford algebra is Z-filtered: Cℓ≤i(V, Q) ⋅ Cℓ≤j(V, Q) ⊂ Cℓ≤i+j(V, Q). The degree of a Clifford number usually refers to the degree in the N-grading. The even subalgebra Cℓ0(V, Q) of a Clifford algebra is itself isomorphic to a Clifford algebra.[10][11] If V is the orthogonal direct sum of a vector a of norm Q(a) and a subspace U, then Cℓ0(V, Q) is isomorphic to Cℓ(U, −Q(a)Q), where −Q(a)Q is the form Q restricted to U and multiplied by −Q(a). In particular over the reals this implies that $C\ell_{p,q}^0(\mathbf{R}) \cong C\ell_{p,q-1}(\mathbf{R})$ for q > 0, and $C\ell_{p,q}^0(\mathbf{R}) \cong C\ell_{q,p-1}(\mathbf{R})$ for p > 0. In the negative-definite case this gives an inclusion Cℓ0,n−1(R) ⊂ Cℓ0,n(R) which extends the sequence R ⊂ C ⊂ H ⊂ H⊕H ⊂ …; Likewise, in the complex case, one can show that the even subalgebra of Cℓn(C) is isomorphic to Cℓn−1(C). ### Antiautomorphisms In addition to the automorphism α, there are two antiautomorphisms which play an important role in the analysis of Clifford algebras. Recall that the tensor algebra T(V) comes with an antiautomorphism that reverses the order in all products: $v_1\otimes v_2\otimes \cdots \otimes v_k \mapsto v_k\otimes \cdots \otimes v_2\otimes v_1.$ Since the ideal IQ is invariant under this reversal, this operation descends to an antiautomorphism of Cℓ(V, Q) called the transpose or reversal operation, denoted by xt. The transpose is an antiautomorphism: (xy)t = yt xt. The transpose operation makes no use of the Z2-grading so we define a second antiautomorphism by composing α and the transpose. We call this operation Clifford conjugation denoted $\bar x$ $\bar x = \alpha(x^t) = \alpha(x)^t.$ Of the two antiautomorphisms, the transpose is the more fundamental.[12] Note that all of these operations are involutions. One can show that they act as ±1 on elements which are pure in the Z-grading. In fact, all three operations depend only on the degree modulo 4. That is, if x is pure with degree k then $\alpha(x) = \pm x \qquad x^t = \pm x \qquad \bar x = \pm x$ where the signs are given by the following table: k mod 4 0 1 2 3 $\alpha(x)\,$ + − + − (−1)k $x^t\,$ + + − − (−1)k(k−1)/2 $\bar x$ + − − + (−1)k(k+1)/2 ### Clifford scalar product When the characteristic is not 2, the quadratic form Q on V can be extended to a quadratic form on all of Cℓ(V, Q) (which we also denoted by Q). A basis independent definition of one such extension is $Q(x) = \lang x^t x\rang$ where ⟨a⟩ denotes the scalar part of a (the grade 0 part in the Z-grading). One can show that $Q(v_1v_2\cdots v_k) = Q(v_1)Q(v_2)\cdots Q(v_k)$ where the vi are elements of V – this identity is not true for arbitrary elements of Cℓ(V, Q). The associated symmetric bilinear form on Cℓ(V, Q) is given by $\lang x, y\rang = \lang x^t y\rang.$ One can check that this reduces to the original bilinear form when restricted to V. The bilinear form on all of Cℓ(V, Q) is nondegenerate if and only if it is nondegenerate on V. It is not hard to verify that the transpose is the adjoint of left/right Clifford multiplication with respect to this inner product. That is, $\lang ax, y\rang = \lang x, a^t y\rang,$ and $\lang xa, y\rang = \lang x, y a^t\rang.$ ## Structure of Clifford algebras In this section we assume that the vector space V is finite dimensional and that the bilinear form of Q is non-singular. A central simple algebra over K is a matrix algebra over a (finite dimensional) division algebra with center K. For example, the central simple algebras over the reals are matrix algebras over either the reals or the quaternions. • If V has even dimension then Cℓ(V, Q) is a central simple algebra over K. • If V has even dimension then Cℓ0(V, Q) is a central simple algebra over a quadratic extension of K or a sum of two isomorphic central simple algebras over K. • If V has odd dimension then Cℓ(V, Q) is a central simple algebra over a quadratic extension of K or a sum of two isomorphic central simple algebras over K. • If V has odd dimension then Cℓ0(V, Q) is a central simple algebra over K. The structure of Clifford algebras can be worked out explicitly using the following result. Suppose that U has even dimension and a non-singular bilinear form with discriminant d, and suppose that V is another vector space with a quadratic form. The Clifford algebra of U+V is isomorphic to the tensor product of the Clifford algebras of U and (−1)dim(U)/2dV, which is the space V with its quadratic form multiplied by (−1)dim(U)/2d. Over the reals, this implies in particular that $C\ell_{p+2,q}(\mathbf{R}) = M_2(\mathbf{R})\otimes C\ell_{q,p}(\mathbf{R})$ $C\ell_{p+1,q+1}(\mathbf{R}) = M_2(\mathbf{R})\otimes C\ell_{p,q}(\mathbf{R})$ $C\ell_{p,q+2}(\mathbf{R}) = \mathbf{H}\otimes C\ell_{q,p}(\mathbf{R}).$ These formulas can be used to find the structure of all real Clifford algebras and all complex Clifford algebras; see the classification of Clifford algebras. Notably, the Morita equivalence class of a Clifford algebra (its representation theory: the equivalence class of the category of modules over it) depends only on the signature (p − q) mod 8. This is an algebraic form of Bott periodicity. ## Clifford group In this section we assume that V is finite dimensional and the quadratic form Q is nondegenerate. The invertible elements of the Clifford algebra act on it by twisted conjugation: conjugation by x maps y ↦ xy α(x)−1. The Clifford group Γ is defined to be the set of invertible elements x that stabilize vectors, meaning that for all v in V we have: $x v \alpha(x)^{-1}\in V$ This formula also defines an action of the Clifford group on the vector space V that preserves the norm Q, and so gives a homomorphism from the Clifford group to the orthogonal group. The Clifford group contains all elements r of V of nonzero norm, and these act on V by the corresponding reflections that take v to v−2⟨v,r⟩r/Q(r) (In characteristic 2 these are called orthogonal transvections rather than reflections.) The Clifford group Γ is the disjoint union of two subsets Γ0 and Γ1, where Γi is the subset of elements of degree i. The subset Γ0 is a subgroup of index 2 in Γ. If V is a finite dimensional real vector space with positive definite (or negative definite) quadratic form then the Clifford group maps onto the orthogonal group of V with respect to the form (by the Cartan-Dieudonné theorem) and the kernel consists of the nonzero elements of the field K. This leads to exact sequences $1 \rightarrow K^ \rightarrow \Gamma \rightarrow \mbox{O}_V(K) \rightarrow 1,\,$ $1 \rightarrow K^* \rightarrow \Gamma^0 \rightarrow \mbox{SO}_V(K) \rightarrow 1.\,$ Over other fields or with indefinite forms, the map is not in general onto, and the failure is captured by the spinor norm. ### Spinor norm For more details on this topic, see Spinor_norm#Galois_cohomology_and_orthogonal_groups. In arbitrary characteristic, the spinor norm Q is defined on the Clifford group by $Q(x) = x^tx.\,$ It is a homomorphism from the Clifford group to the group K* of non-zero elements of K. It coincides with the quadratic form Q of V when V is identified with a subspace of the Clifford algebra. Several authors define the spinor norm slightly differently, so that it differs from the one here by a factor of −1, 2, or −2 on Γ1. The difference is not very important in characteristic other than 2. The nonzero elements of K have spinor norm in the group K2 of squares of nonzero elements of the field K. So when V is finite dimensional and non-singular we get an induced map from the orthogonal group of V to the group K/K2, also called the spinor norm. The spinor norm of the reflection of a vector r has image Q(r) in K/K2, and this property uniquely defines it on the orthogonal group. This gives exact sequences: $1 \to \{\pm 1\} \to \mbox{Pin}_V(K) \to \mbox{O}_V(K) \to K^/K^{2},\,$ $1 \to \{\pm 1\} \to \mbox{Spin}_V(K) \to \mbox{SO}_V(K) \to K^/K^{2}.\,$ Note that in characteristic 2 the group {±1} has just one element. From the point of view of Galois cohomology of algebraic groups, the spinor norm is a connecting homomorphism on cohomology. Writing μ2 for the algebraic group of square roots of 1 (over a field of characteristic not 2 it is roughly the same as a two-element group with trivial Galois action), the short exact sequence $1 \to \mu_2 \rightarrow \mbox{Pin}_V \rightarrow \mbox{O}_V \rightarrow 1\,$ yields a long exact sequence on cohomology, which begins $1 \to H^0(\mu_2;K) \to H^0(\mbox{Pin}_V;K) \to H^0(\mbox{O}_V;K) \to H^1(\mu_2;K).\,$ The 0th Galois cohomology group of an algebraic group with coefficients in K is just the group of K-valued points: H0(G; K) = G(K), and H1(μ2; K) ≅ K/K2, which recovers the previous sequence $1 \to \{\pm 1\} \to \mbox{Pin}_V(K) \to \mbox{O}_V(K) \to K^/K^{2},\,$ where the spinor norm is the connecting homomorphism H0(OV); K) → H1(μ2; K). ## Spin and Pin groups In this section we assume that V is finite dimensional and its bilinear form is non-singular. (If K has characteristic 2 this implies that the dimension of V is even.) The Pin group PinV(K) is the subgroup of the Clifford group Γ of elements of spinor norm 1, and similarly the Spin group SpinV(K) is the subgroup of elements of Dickson invariant 0 in PinV(K). When the characteristic is not 2, these are the elements of determinant 1. The Spin group usually has index 2 in the Pin group. Recall from the previous section that there is a homomorphism from the Clifford group onto the orthogonal group. We define the special orthogonal group to be the image of Γ0. If K does not have characteristic 2 this is just the group of elements of the orthogonal group of determinant 1. If K does have characteristic 2, then all elements of the orthogonal group have determinant 1, and the special orthogonal group is the set of elements of Dickson invariant 0. There is a homomorphism from the Pin group to the orthogonal group. The image consists of the elements of spinor norm 1 ∈ K/K*2. The kernel consists of the elements +1 and −1, and has order 2 unless K has characteristic 2. Similarly there is a homomorphism from the Spin group to the special orthogonal group of V. In the common case when V is a positive or negative definite space over the reals, the spin group maps onto the special orthogonal group, and is simply connected when V has dimension at least 3. Further the kernel of this homomorphism consists of 1 and −1. So in this case the spin group, Spin(n), is a double cover of SO(n). Please note, however, that the simple connectedness of the spin group is not true in general: if V is Rp,q for p and q both at least 2 then the spin group is not simply connected. In this case the algebraic group Spinp,q is simply connected as an algebraic group, even though its group of real valued points Spinp,q(R) is not simply connected. This is a rather subtle point, which completely confused the authors of at least one standard book about spin groups. ## Spinors Clifford algebras Cℓp,q(C), with p+q=2n even, are matrix algebras which have a complex representation of dimension 2n. By restricting to the group Pinp,q(R) we get a complex representation of the Pin group of the same dimension, called the spin representation. If we restrict this to the spin group Spinp,q(R) then it splits as the sum of two half spin representations (or Weyl representations) of dimension 2n−1. If p+q=2n+1 is odd then the Clifford algebra Cℓp,q(C) is a sum of two matrix algebras, each of which has a representation of dimension 2n, and these are also both representations of the Pin group Pinp,q(R). On restriction to the spin group Spinp,q(R) these become isomorphic, so the spin group has a complex spinor representation of dimension 2n. More generally, spinor groups and pin groups over any field have similar representations whose exact structure depends on the structure of the corresponding Clifford algebras: whenever a Clifford algebra has a factor that is a matrix algebra over some division algebra, we get a corresponding representation of the pin and spin groups over that division algebra. For examples over the reals see the article on spinors. ### Real spinors For more details on this topic, see spinor. To describe the real spin representations, one must know how the spin group sits inside its Clifford algebra. The Pin group, Pinp,q is the set of invertible elements in Cℓp, q which can be written as a product of unit vectors: ${\mbox{Pin}}_{p,q}=\{v_1v_2\dots v_r \|\,\, \forall i\, \\|v_i\\|=\pm 1\}.$ Comparing with the above concrete realizations of the Clifford algebras, the Pin group corresponds to the products of arbitrarily many reflections: it is a cover of the full orthogonal group O(p, q). The Spin group consists of those elements of Pinp, q which are products of an even number of unit vectors. Thus by the Cartan-Dieudonné theorem Spin is a cover of the group of proper rotations SO(p,q). Let α : Cℓ → Cℓ be the automorphism which is given by the mapping v ↦ −v acting on pure vectors. Then in particular, Spinp,q is the subgroup of Pinp, q whose elements are fixed by α. Let $C\ell_{p,q}^0 = \{ x\in C\ell_{p,q} \|\, \alpha(x)=x\}.$ (These are precisely the elements of even degree in Cℓp, q.) Then the spin group lies within Cℓ0p, q. The irreducible representations of Cℓp, q restrict to give representations of the pin group. Conversely, since the pin group is generated by unit vectors, all of its irreducible representation are induced in this manner. Thus the two representations coincide. For the same reasons, the irreducible representations of the spin coincide with the irreducible representations of Cℓ0p, q To classify the pin representations, one need only appeal to the classification of Clifford algebras. To find the spin representations (which are representations of the even subalgebra), one can first make use of either of the isomorphisms (see above) Cℓ0p,q ≈ Cℓp,q−1, for q > 0 Cℓ0p,q ≈ Cℓq,p−1, for p > 0 and realize a spin representation in signature (p,q) as a pin representation in either signature (p,q−1) or (q,p−1). ## Applications ### Differential geometry One of the principal applications of the exterior algebra is in differential geometry where it is used to define the bundle of differential forms on a smooth manifold. In the case of a (pseudo-)Riemannian manifold, the tangent spaces come equipped with a natural quadratic form induced by the metric. Thus, one can define a Clifford bundle in analogy with the exterior bundle. This has a number of important applications in Riemannian geometry. Perhaps more importantly is the link to a spin manifold, its associated spinor bundle and spinc manifolds. ### Physics Clifford algebras have numerous important applications in physics. Physicists usually consider a Clifford algebra to be an algebra spanned by matrices γ0,…,γ3 called Dirac matrices which have the property that $\gamma_i\gamma_j + \gamma_j\gamma_i = 2\eta_{ij}\,$ where η is the matrix of a quadratic form of signature (1,3). These are exactly the defining relations for the Clifford algebra Cℓ1,3(C) (up to an unimportant factor of 2), which by the classification of Clifford algebras is isomorphic to the algebra of 4 by 4 complex matrices. The Dirac matrices were first written down by Paul Dirac when he was trying to write a relativistic first-order wave equation for the electron, and give an explicit isomorphism from the Clifford algebra to the algebra of complex matrices. The result was used to define the Dirac equation and introduce the Dirac operator. The entire Clifford algebra shows up in quantum field theory in the form of Dirac field bilinears. The use of Clifford algebras to describe quantum theory has been advanced among others by Mario Schönberg,[13] by David Hestenes in terms of geometric calculus, by David Bohm and Basil Hiley and co-workers in form of a hierarchy of Clifford algebras, and by Elio Conte et al.[14][15] ### Computer Vision Recently, Clifford algebras have been applied in the problem of action recognition and classification in computer vision. Rodriguez et al.[16] propose a Clifford embedding to generalize traditional MACH ﬁlters to video (3D spatiotemporal volume), and vector-valued data such as optical flow. Vector-valued data is analyzed using the Clifford Fourier transform. Based on these vectors action filters are synthesized in the Clifford Fourier domain and recognition of actions is performed using Clifford Correlation. The authors demonstrate the effectiveness of the Clifford embedding by recognizing actions typically performed in classic feature ﬁlms and sports broadcast television. ## Notes 1. W. K. Clifford, "Preliminary sketch of bi-quaternions, Proc. London Math. Soc. Vol. 4 (1873) pp. 381-395 2. W. K. Clifford, Mathematical Papers, (ed. R. Tucker), London: Macmillan, 1882. 3. see for ex. Z. Oziewicz, Sz. Sitarczyk: Parallel treatment of Riemannian and symplectic Clifford algebras. In: Artibano Micali, Roger Boudet, Jacques Helmstetter (eds.): Clifford Algebras and their Applications in Mathematical Physics, Kluwer Academic Publishers, ISBN 0-7923-1623-1 [Amazon-US \| Amazon-UK], 1992, p. 83 4. Mathematicians who work with real Clifford algebras and prefer positive definite quadratic forms (especially those working in index theory) sometimes use a different choice of sign in the fundamental Clifford identity. That is, they take v2 = −Q(v). One must replace Q with −Q in going from one convention to the other. 5. Thus the group algebra K[Z/2] is semisimple and the Clifford algebra splits into eigenspaces of the main involution. 6. The Z-grading is obtained from the N grading by appending copies of the zero subspace indexed with the negative integers. 7. Technically, it does not have the full structure of a Clifford algebra without a designated vector subspace. 8. We are still assuming that the characteristic is not 2. 9. The opposite is true when using the alternate (−) sign convention for Clifford algebras: it is the conjugate which is more important. In general, the meanings of conjugation and transpose are interchanged when passing from one sign convention to the other. For example, in the convention used here the inverse of a vector is given by v−1 = vt / Q(v) while in the (−) convention it is given by v−1 = v / Q(v). 10. See the references to Schönberg's papers of 1956 and 1957 as described in section "The Grassmann–Schönberg algebra $G_n$" of:A. O. Bolivar, Classical limit of fermions in phase space, J. Math. Phys. 42, 4020 (2001) doi:10.1063/1.1386411 11. E. Conte: The solution of EPR paradox in quantum mechanics, in: Fundamental Problems of Natural Sciences and Engineering, pp. 271–204, Saint Petersburg, 2002 12. Elio Conte: On some considerations of mathematical physics: May we identify Clifford algebra as a common algebraic structure for classical diffusion and Schrödinger equations? Adv. Studies Theor. Phys., vol. 6, no. 26 (2012), pp. 1289–1307 13. Rodriguez, Mikel; Shah, M (2008). "Action MACH: A Spatio-Temporal Maximum Average Correlation Height Filter for Action Classification". Computer Vision and Pattern Recognition (CVPR). ## References • Bourbaki, Nicolas (1988), Algebra, Berlin, New York: Springer-Verlag, ISBN 978-3-540-19373-9 [Amazon-US \| Amazon-UK], section IX.9. • Carnahan, S. Borcherds Seminar Notes, Uncut. Week 5, "Spinors and Clifford Algebras". • Garling, D. J. H. (2011), Clifford algebras. An introduction, London Mathematical Society Student Texts 78, Cambridge: Cambridge University Press, ISBN 978-1-107-09638-7 [Amazon-US \| Amazon-UK], Zbl 1235.15025 • Lam, Tsit-Yuen (2005), Introduction to Quadratic Forms over Fields, Graduate Studies in Mathematics 67, American Mathematical Society, ISBN 0-8218-1095-2 [Amazon-US \| Amazon-UK], MR 2104929, Zbl 1068.11023 • Lawson, H. Blaine; Michelsohn, Marie-Louise (1989), Spin Geometry, Princeton, NJ: Princeton University Press, ISBN 978-0-691-08542-5 [Amazon-US \| Amazon-UK]. An advanced textbook on Clifford algebras and their applications to differential geometry. • Lounesto, Pertti (2001), Clifford algebras and spinors, Cambridge: Cambridge University Press, ISBN 978-0-521-00551-7 [Amazon-US \| Amazon-UK] • Porteous, Ian R. (1995), Clifford algebras and the classical groups, Cambridge: Cambridge University Press, ISBN 978-0-521-55177-9 [Amazon-US \| Amazon-UK] • Jagannathan, R., On generalized Clifford algebras and their physical applications, arXiv:1005.4300 • Sylvester, J. J., (1882), Johns Hopkins University Circulars I: 241-242; ibid II (1883) 46; ibid III (1884) 7-9. Summarized in The Collected Mathematics Papers of James Joseph Sylvester (Cambridge University Press, 1909) v III .online and further. Content in this section is authored by an open community of volunteers and is not produced by, reviewed by, or in any way affiliated with MedLibrary.org. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Clifford algebras", available in its original form here: http://en.wikipedia.org/w/index.php?title=Clifford_algebras • ## Finding More You are currently browsing the the MedLibrary.org general encyclopedia supplement. To return to our medication library, please select from the menu above or use our search box at the top of the page. In addition to our search facility, alphabetical listings and a date list can help you find every medication in our library. • ## Questions or Comments? If you have a question or comment about material specifically within the site’s encyclopedia supplement, we encourage you to read and follow the original source URL given near the bottom of each article. You may also get in touch directly with the original material provider. • ## About This site is provided for educational and informational purposes only and is not intended as a substitute for the advice of a medical doctor, nurse, nurse practitioner or other qualified health professional.	{"extraction_info": {"found_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": 64, "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": 0.8788895606994629, "perplexity_flag": "middle"}
http://en.m.wikipedia.org/wiki/Specific_detectivity	# Specific detectivity Specific detectivity, or D, for a photodetector is a figure of merit used to characterize performance, equal to the reciprocal of noise-equivalent power (NEP), normalized per unit area. Specific detectivity is given by $D^=\frac{\sqrt{A}}{NEP}$, where $A$ is the area of the photosensitive region of the detector. Its common units are $cm \cdot \sqrt{Hz}/ W$, also called the Jones in honor of R. Clark Jones who defined this magnitude.[1][2] Given that noise-equivalent power can be expressed as a function of the responsivity $\mathfrak{R}$ (in units of $A/W$ or $V/W$) and the noise spectral density $S_n$ (in units of $A/Hz^{1/2}$ or $V/Hz^{1/2}$) as $NEP=\frac{S_n}{\mathfrak{R}}$, it's common to see the specific detectivity expressed as $D^=\frac{\mathfrak{R}\cdot\sqrt{A}}{S_n}$. The unit Jones is now commonly used with the D figure of merit. It is often useful to express the specific detectivity in terms of relative noise levels present in the device. A common expression is given below. $D^* = \frac{q\lambda \eta}{hc} \left[\frac{4kT}{R_0 A}+2q^2 \eta \Phi_b\right]^{-1/2}$ With q as the electronic charge, $\lambda$ is the wavelength of interest, h is Planck's constant, c is the speed of light, k is Boltzmann's constant, T is the temperature of the detector, $R_0A$ is the zero-bias dynamic resistance area product (often measured experimentally, but also expressible in noise level assumptions), $\eta$ is the quantum efficiency of the device, and $\Phi_b$ is the total flux of the source (often a blackbody) in photons/sec/cm². ## References 1. R. C. Jones, "Quantum efficiency of photoconductors," Proc. IRIS 2, 9 (1957) 2. R. C. Jones, "Proposal of the detectivity D** for detectors limited by radiation noise," J. Opt. Soc. Am. 50, 1058 (1960), doi:10.1364/JOSA.50.001058) This article incorporates public domain material from the General Services Administration document "Federal Standard 1037C". This physics-related article is a stub. You can help Wikipedia by expanding it. ↑Jump back a section	{"extraction_info": {"found_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": 16, "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": 0.8973715305328369, "perplexity_flag": "middle"}
http://mathhelpforum.com/geometry/167244-finding-area-shaded-region.html	# Thread: 1. ## Finding the area of the shaded region. Hi I'm a college student and I'm having a bit of difficulty with my Solid Mensuration homework. Can someone help me out? I'm trying to find the area of the shaded region in the problems which I've posted pictures of in this thread. I'm having a hard time because I think the data given is insufficient, but maybe I'm wrong so I'm asking for help Attached Thumbnails 2. Originally Posted by purplesky828 Hi I'm a college student and I'm having a bit of difficulty with my Solid Mensuration homework. Can someone help me out? I've attached a picture of the problems I'm confused by. What exactly are you trying to do? Please post the whole question, not just the pictures. Edit: The OP has since edited his/her post. 3. Originally Posted by purplesky828 I'm trying to find the area of the shaded region in the problems which I've posted pictures of in this thread. I'm having a hard time because I think the data given is insufficient, but maybe I'm wrong so I'm asking for help I agree with you that technically the data is insufficient. However, a good guess on #1 would be $(85)(42.5)-\dfrac{(42.5)^2}{2}$. Do you see why? 4. Originally Posted by Plato I agree with you that technically the data is insufficient. However, a good guess on #1 would be $(85)(42.5)-\dfrac{(42.5)^2}{2}$. Do you see why? Yes I see how it would be a good guess. I was thinking of the same thing but I was unsure because the data wasn't given. Thank you very much for your help! 5. Originally Posted by purplesky828 Hi I'm a college student and I'm having a bit of difficulty with my Solid Mensuration homework. Can someone help me out? I'm trying to find the area of the shaded region in the problems which I've posted pictures of in this thread. I'm having a hard time because I think the data given is insufficient, but maybe I'm wrong so I'm asking for help	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 0.9796509146690369, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/38867/finite-neighborhood-base?answertab=active	# finite neighborhood base My question might seem obvious but still I want to sure if the result I shall state below is true: if the neighborhood base in each point of a space say X is finite. Then we have a discrete topology. Many thanks and apologies in case I just stated some insanity above :) - ## 2 Answers No, this is not true. It would imply, for instance, that every finite topological space is discrete, but this is far from the case. For a trivial example, take the trivial (or indiscrete) topology on any finite set $X$ of cardinality greater than one. Or take the Sierpinski space, i.e., the unique two-point space (up to homeomorphism) with one open point and one closed 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": 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": 0.9499073624610901, "perplexity_flag": "head"}
http://mathoverflow.net/questions/33420/norms-and-operator-algebras	## Norms and Operator Algebras ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) We know that two unital matrix algebras are isomorphic only if completely isometric. But when is a unital subspace of $N \times N$ matrices completely isometric to a matrix algebra? Can we describe this using matrix norms? - What (if anything) does Paulsen's book say? – Yemon Choi Jul 26 2010 at 18:06 3 It may also be worth giving a brief account of your motivation for this question, which sources you're familiar with, or what level you are at. This would make it easier for people to focus their answers to your question. – Yemon Choi Jul 26 2010 at 18:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.93413245677948, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/70993?sort=votes	## Proof systems and their hierarchy ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Why ZFC is placed in top of the proof system hierarchy? How it can p-simulate other systems? - 2 I really would appreciate some definition links to "proof system hierarchy" and "p-simulation". – Hans Stricker Jul 23 2011 at 15:18 2 @Hans Stricker, check the Wikipedia pages for "proof complexity" (en.wikipedia.org/wiki/Proof_complexity) and "propositional proof system" (en.wikipedia.org/wiki/Propositional_proof_system). – Kaveh Aug 2 2011 at 6:19 ## 2 Answers It is an open problem if there is an optimal propositional proof system. Therefore we don't know if ZFC as a propositional proof system is optimal either. ZFC as propositional proof system can p-simulate any propositional proof system whose soundness (if there is proof for a formula then the formula is true) as a propositional proof system is provable in ZFC. The trick (which I think is due to Steve Cook) is based on the fact that proofs are concrete finite objects, if there is a proof for a formula in a proof system, then ZFC can prove its existence (i.e. ZFC is $\Sigma_1$-complete), combining this with the provability of the soundness we derive the truth of the encoded formula in ZFC. The rest of the argument is translating this first order proof in ZFC to a propositional proof in ZFC as propositional proof system and proving the equivalence of a propositional formula with the formula itself in the proof system. This can be done in $\mathsf{TC^0}$-Frege and any system that contains it. See Logical Foundations of Proof Complexity, 2010 by Cook and Nguyen for the details. - 1 Strictly speaking, it's the function $F$ that maps ZFC-proofs of $T$ to $T$ (where $T$ is a tautology) that is the propositional proof system, not ZFC itself. Note also that the assertion that $F$ is a propositional proof system isn't provable in ZFC itself; you need to assume the soundness of ZFC. This is why people put ZFC at the top; conceivably, there could be "stronger" propositional proof systems $F$ that ZFC can't prove are actually propositional proof systems. But if you think that ZFC represents the outer limit of mathematical knowledge, you might have qualms about such $F$. – Timothy Chow Jul 22 2011 at 22:24 @Timothy, there are two equivalent definitions for a pps, one is what you described, the other one is a proof checker program (which the one I found more natural). Yes, strictly speaking, but I didn't want to go into too much details. One can add axioms to obtains a possibly stronger pps, e.g. large cardinal axioms. Note that we don't need the full soundness, we only need a very restricted form of it. In fact, as you know :), some proof complexity theorist conjecture that EF is a plausible candidate for the optimal proof system, in which case ZFC will be just equal to it. – Kaveh Aug 2 2011 at 6:13 ### 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. Iam really sorry for this delayed reply...I just digged up a bit on this topic. Here are 'my views' , Iam novice in this area, if I am wrong correct me. Frege systems ( also hilbert systems) is a better proof system than Gentzen systems as it captures "Human thinking" also. Now, if we add some rules which , say act as lemmas, they readily help us in proving a problem in the proof system. This lemma need not be proved in this problem again , we can just quote it.The Frege + rules = EXTENDED FREGE system. There is strong parallelism between Proof system and Circuit complexity proof system wherein we can simulate one problem into other.The circuit complexity class heirarchy is like Ac , NC , P/poy , TC . We concentrate on P\Poly what it means. P\poly solves the problem in polynomial time given "some polynomial time advice function". If we closely observe these advice functions, they are similar to 'lemmas' or 'rules' in extended frege system. Thus P\poly is equivalent to Extended Frege system.Is this the best proof system ? I think no because if we want to compare two outputs produced by applying different lemmas in this system we don;t have any parameter to compare. So we have Zorn's Lemma in our hand where it says that if we have upper bound in every chain, then we have maximal element in that set. This for us nothing but the "Axiom Of Choice". The proof system with the axoim of choice is ZFC. IS THIS THE ULTIMATUM ? No one knows...... - 2 As a non-expert in this particular area, I much prefer to read answers of experts than to write my own attempts. – Andrej Bauer Aug 26 2011 at 9:45 I am unable to see how this addresses the question (which, according to Kaveh's answer, seems to be based on false premises). – S. Carnahan♦ Aug 29 2011 at 2:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 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": 0.9363253712654114, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/179726-null-sequences.html	# Thread: 1. ## null sequences. Prove or disprove: If $(x_ny_n)$ is a null sequence then one or both of $(x_n), \, (y_n)$ is(are) null. My attempt: if $(x_n)\, and \, (y_n)$ both are null then certainly $(x_ny_n)$ is null, no problem with that. I assumed that $(x_n)$ is not null, $(y_n)$ is not null and as given, $(x_ny_n)$ is null. I tried to bring a contradiction since i have a feeling that the statement in the question is true. since its assumed that $(x_n)$ is not null so: there exists $\epsilon_1>0$ such that for all $N$ there exists $n \geq N$ such that $\|x_n\|>\epsilon_1$ similarly: there exists $\epsilon_2>0$ such that for all $N$ there exists $n \geq N$ such that $\|y_n\|>\epsilon_2$ and since $(x_ny_n)$ is null: there exists $N$ such that $n \geq N \Rightarrow \|x_ny_n\|<\epsilon_1 \epsilon_2$ now since i can't say that $\|x_n\|>\epsilon_1 \, and \, \|y_n\|>\epsilon_2$ occur at the same $n$ i can't find a contradiction. someone help. 2. Originally Posted by abhishekkgp Prove or disprove: If $(x_ny_n)$ is a null sequence then one or both of $(x_n), \, (y_n)$ is(are) null. Assuming both converge, suppose that neither is a null sequence. Then $(x_n)\to X\ne 0$ and $(y_n)\to Y\ne 0$. What can you say about $(x_ny_n)\to~ ?$ What does that tell you. 3. Originally Posted by Plato Assuming both converge, suppose that neither is a null sequence. Then $(x_n)\to X\ne 0$ and $(y_n)\to Y\ne 0$. What can you say about $(x_ny_n)\to~ ?$ What does that tell you. probably $(x_ny_n) \rightarrow XY \neq 0$ and probably null sequences converge to 0 and hence a contradiction. But convergence has not yet been discussed in the book i am reading(first course in real analysis- stirling K berberian) so i can only guess. 4. Originally Posted by abhishekkgp But convergence has not yet been discussed in the book i am reading(first course in real analysis- stirling K berberian) so i can only guess. How very odd. I know Berberian's book on measure theory and find it odd. Pray tell then what is a null sequence? 5. Originally Posted by Plato How very odd. I know Berberian's book on measure theory and find it odd. Pray tell then what is a null sequence? a sequence $(x_n)$ in $\mathbb{R}$ is said to be null if: for all $\epsilon>0$ there exists $N$ such that $n \geq N \Rightarrow \|x_n\|<\epsilon$ 6. Originally Posted by abhishekkgp a sequence $(x_n)$ in $\mathbb{R}$ is said to be null if: for all $\epsilon>0$ there exists $N$ such that $n \geq N \Rightarrow \|x_n\|<\epsilon$ OK go back to the OP. Let $\varepsilon = \frac{{\varepsilon _1 \varepsilon _2 }}{2}$. Make $\|x_ny_n\|<\varepsilon$ then $\varepsilon _1 \varepsilon _2\le \|x_n\|\|y_n\|=\|x_ny_n\|<\varepsilon$. That is a contradiction. 7. Originally Posted by Plato OK go back to the OP. Let $\varepsilon = \frac{{\varepsilon _1 \varepsilon _2 }}{2}$. Make $\|x_ny_n\|<\varepsilon$ then $\varepsilon _1 \varepsilon _2\le \|x_n\|\|y_n\|=\|x_ny_n\|<\varepsilon$. That is a contradiction. the problem is that i can't( or yet not able to) say that there exists an $n$ such that $\|x_n\|\|y_n\|> \epsilon_1 \epsilon_2$. I can only say that $\|x_{n_1}\|\|y_{n_2}\|>\epsilon_1 \epsilon_2$ and that's why i can't contradict anything. 8. Originally Posted by abhishekkgp the problem is that i can't( or yet not able to) say that there exists an $n$ such that $\|x_n\|\|y_n\|> \epsilon_1 \epsilon_2$. I can only say that $\|x_{n_1}\|\|y_{n_2}\|>\epsilon_1 \epsilon_2$ and that's why i can't contradict anything. I think that I misread the OP. Consider this example. $x_n = \left\{ {\begin{array}{rl} {1,} & {\text{n is even}} \\ {\frac{1}{n},} & {\text{ n is odd}} \\ \end{array} } \right.$ and $y_n = \left\{ {\begin{array}{rl} {1,} & {\text{n is odd}} \\ {\frac{1}{n},} & {\text{ n is even}} \\ \end{array} } \right.$ What is $x_ny_n~?$ 9. Originally Posted by Plato I think that I misread the OP. Consider this example. $x_n = \left\{ {\begin{array}{rl} {1,} & {\text{n is even}} \\ {\frac{1}{n},} & {\text{ n is odd}} \\ \end{array} } \right.$ and $y_n = \left\{ {\begin{array}{rl} {1,} & {\text{n is odd}} \\ {\frac{1}{n},} & {\text{ n is even}} \\ \end{array} } \right.$ What is $x_ny_n~?$ so it can happen that if we have $(x_n)$ not null, $(y_n)$ not null, we still have $(x_ny_n)$ as null!! thank you for this.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 61, "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": 0.9524378180503845, "perplexity_flag": "head"}
http://skullsinthestars.com/2011/10/09/faster-than-a-speeding-photon-precursors-test-whether-light-can-be-faster-than-light/	About these ads The intersection of physics, optics, history and pulp fiction ## Faster than a speeding photon? Precursors test whether light can be faster than light Posted on October 9, 2011 by Over the past two weeks, the biggest physics news has been the apparent observation of neutrinos (nearly undetectable subatomic particles) moving faster than the vacuum speed of light. At first glance, this would seem to violate Einstein’s special theory of relativity, which fixes the vacuum speed of light at $c = 3\times 10^8$ meters per second, and as a consequence makes it in principle impossible to travel faster than that speed. The theoretical implications are in fact a bit more subtle, but before we worry too much about those implications the experimental results will need to be checked carefully and independently verified. While we wait, it is worth noting that in June 0f 2011 a group of researchers performed an experiment to see if light itself could move faster than light! In particular, the scientists used a little known optical phenomenon known as an optical precursor to see if individual photons might travel faster than $c$ while propagating in a material. In the end, the experiment suggests that these single photons did not in fact violate Einstein’s speed limit, though the results still got a significant amount of press. The response of many physicists to the news was a collective, “Well, duh!” The prevailing attitude seems to have been: “What’s so interesting about proving something we already knew?” In this post I’d like to explore that question a little bit, and explain how some uncertainty remains about the behavior of light in materials. Along the way, we’ll introduce the fascinating phenomenon of precursors, and see how they can be used both to probe the nature of matter as well as the nature of light. To start, we need to understand a bit about how light and matter interact with one another, and how this relates to the speed of light in matter. This is a surprisingly tricky subject, which I’ve covered several times previously on this blog; we’ll review the important aspects, with a few additional details. Let’s talk about how we measure the speed of an object first. If we’re looking at the motion of a rigid object, like a speeding car or a thrown baseball, the speed $v$ can be determined simply by measuring how much time $\Delta t$ it takes for an object to travel a distance $\Delta x$. The speed is simply the distance divided by the time: $\mbox{speed} = \mbox{distance}/\mbox{time} = \Delta x/\Delta t$. There’s a small subtlety to this definition: cars and baseballs are extended objects! To accurately measure an object’s speed, we have to be consistent in how we define its position. For a car driving down the road, for instance, we should do all measurements of its position from a fixed position, such as the front bumper, to measure the speed. But what do we do when the object doesn’t have a fixed position on it? For example, what is the best way to measure the speed of a hurled bucketful of water? Water is, of course, not a rigid body, and the volume of water changes shape and spreads out as it travels. It is strange to realize, after some thought, that a blob of water like this doesn’t really have a well-defined speed. We arrive at the same problem with a pulse of light traveling in matter: in general, a light pulse has no “fixed point” upon it, and it can change shape as it travels. There isn’t a single definition for the speed of light in matter that is useful in all cases. There are three different descriptions of the speed* of light used in physics, each with its own significance and usefulness. We can introduce each of these by looking at similar definitions for our bucketful of water. We imagine that our volume of water may be broken up into a collection of well-defined droplets. The most straightforward and thorough thing to do is to measure the speed of each individual droplet: the velocity of each droplet will be referred to as its phase velocity. If there is only one or a handful or drops, this set of numbers will be a good description, but for a large number of droplets this collection of numbers will not necessarily tell us anything about the motion of the entire bulk of the water. A better option is to define the speed of the body of water by its center of mass. This definition, referred to as the group velocity, works quite well in many cases, but can also be misleading. If we’re interested, for instance, in the time it takes for water to first reach its target (say a fire), there may be droplets well ahead of the center of mass and the use of the group velocity can overestimate the time. We may also choose to define the speed of the body of water as the speed of the fastest drop in the collection; this definition is called the signal velocity. One problem with the signal velocity is the opposite of that of the group velocity: if we need to know when the bulk of a body of water reaches a fire, the signal velocity will underestimate the time. Physicists use roughly analogous definitions to characterize the speed of light in matter. Instead of “droplets”, a pulse of light can be decomposed into a collection of waves of different frequencies (colors). It can be shown that a narrow pulse of light necessarily consists of a very broad spectrum (range of colors), while a very broad, slowly varying pulse has a narrow spectrum. Also, and significant for our later discussion, a pulse with a very sharp edge necessarily has a very broad spectrum, as well. This idea is illustrated crudely below: Atoms respond differently to light of different frequencies, in a phenomenon known as dispersion. The result of this is that some frequencies of light are absorbed more than others, and in general every frequency of light travels at a different speed. The term phase velocity is used to refer to the different speeds of each frequency of light. For a pulse narrow in frequency (like the upper one above), this can sometimes serve as an approximation to the overall speed of light. However, in many circumstances this phase velocity can be faster than the vacuum speed of light $c$, and therefore does not accurately represent the speed of the pulse. Because of difficulties with the phase velocity, the group velocity is the standard method of defining the velocity of a pulse in a medium. It may be considered, in essence, the speed of the “center of mass” of the pulse. However, if the shape of the pulse is highly distorted as it propagates, the group velocity can lose all useful meaning. Also, it has been shown that the group velocity can be greater than the vacuum speed of light, a result that generated quite a bit of controversy at first. The absolute speed of light in a medium is the signal velocity, which is loosely defined as the speed at which information can be conveyed in a medium, or the fastest possible speed that the front of a light signal can travel. This is generally thought to be no faster than the vacuum speed of light $c$, in agreement with Einstein’s special theory of relativity. But is $c$ the fastest speed of light in matter? It seems likely, but a few gaps in our knowledge leave it a partially open question. The quantum-mechanical nature of light-matter interactions leaves open the possibility that some “quantum weirdness” allows Einstein’s speed limit to be broken in a subtle way. A more concrete concern involves our understanding of how light propagates in matter. It is well-known that the “causality” of a light signal — and the absolute speed of light — is built into the frequency-by-frequency response of light to matter. If one knows the speed of light, and the absorption properties of light, for every frequency, one can determine whether or not $c$ is the top speed. But we don’t know these properties for every frequency! In particular, we don’t know exactly what happens to light in matter for arbitrary high frequencies (frequency approaches infinity) or for arbitrary low frequencies (frequency approaches zero). There is always the possibility that some small violation of relativity is “hiding” in these extreme frequency ranges. This is where the idea of optical precursors becomes useful. Let’s consider the temporal and spectral properties of another pulse, with a square envelope: This pulse has a large central peak in frequency, but also has long frequency “tails” that stretch out, to some degree, to arbitrarily high frequencies (to infinity) and arbitrarily low frequencies (to zero). These tails arise due to the instantaneous start and end of the pulse in time: the rapid rise and fall that give the “square pulse” its name. At the beginning of the 20th century, physicists Arnold Sommerfeld and Léon Brillouin independently performed theoretical work to investigate what happens to such sharp-edged pulses as they propagate a long distance through matter. In 1914, they each published results showing that the main body of the pulse (traveling at the group velocity) is preceded by faster-moving waves, now known as “precursors”. The precursors are general broken into two types: the Sommerfeld precursor, which actually travels at the vacuum speed of light $c$, and the Brillouin precursor, which travels at the speed of light $c/n(0)$, where $n(0)$ is the refractive index of the medium at zero frequency. A crude illustration of the arrangement of precursors is shown below: The most remarkable thing about the precursor fields is that they are very weakly absorbed by matter, much less so than the main body of the pulse. If we send our original square pulse through a very thick piece of absorbing material, the main body will be almost completely absorbed while the precursors will be absorbed only weakly. This fact has led to some researchers suggesting that precursors could be used to “see” through normally opaque objects, like clouds. It took quite a few years for Sommerfeld and Brillouin’s theoretical work to be confirmed. The first observation of precursors was performed in 1969, using microwaves. Since then, they have been observed for a variety of wavelengths, and have been observed even in “mundane” materials such as water*. What is the origin of these precursors? I wasn’t able to find a simple explanation for them in the literature, perhaps not surprising in terms of their mathematical complexity. I would like to introduce a (hopefully accurate) description of them, in what I will refer to as the “pendulum slapping” model! We have all been taught, at some point in our education, the planetary model of the atom. In this model, an atom consists of a positively-charged nucleus surrounded by one or more orbiting negatively-charged electrons, much like the planets in the solar system orbit the Sun: With our modern understanding of quantum mechanics, we know that this model isn’t a terribly good one: electrons act much more like “clouds” of negative charge centered on the nucleus, and don’t “orbit” in a well-defined sense. However, the planetary model does get one thing right: electrons have a characteristic frequency associated with their motion, much like the planets each orbit the Sun with a characteristic frequency (the Earth orbits the Sun with a frequency of 1/365 days). In a very loose sense, we can picture atoms (and molecules) as collections of oscillating electron pendulums, each vibrating with a characteristic frequency: What happens when we apply an oscillating electric field (i.e. a light wave) to our pendulum? The electric field applies a force to the electron, driving its oscillation. The effectiveness of the electric field in moving the pendulum depends on how close its frequency is to the pendulum frequency. The pendulum will have the largest swing when the electric field matches its own frequency; this is like a child pumping his legs on a swing to increase his motion. Any other frequency of the electric field will be less effective in moving the pendulum, but pretty much every frequency will force some oscillation in it. The energy that is imparted to the pendulum is lost by the electric field, and consequently the light wave; this is the origin of the absorption of light by matter. There are two extreme limits of frequency in which no energy is transferred to the pendulum: the limit of infinite frequency and the limit of zero frequency. We can visualize what happens in these cases by imaging that we are “slapping” a pendulum with our hands on either side (or you can try it yourself, if you have a pendulum). When we slap a pendulum at a frequency much higher than its characteristic frequency, a push on the left is almost immediately canceled by a push on the right — the pendulum doesn’t move, and we’ve transferred no energy to it! In the limit of very low frequency, the pendulum moves, but doesn’t oscillate: we “lift” it to the right with our left hand, then slowly lower it back to its rest position and then “lift” it to the left with our right hand. At no point does the pendulum oscillate freely, and therefore we have transferred no energy to it. These limits, in which no energy is transferred to the pendulum, are the origins of the precursors. We have seen that a square wave has components of arbitrarily high frequency and arbitrarily low frequency, and these components are only very weakly absorbed by the medium. The high frequency components of the wave result in the Sommerfeld precursor, and the low frequency components of the wave result in the Brillouin precursor. For tests of the absolute speed of light, the Sommerfeld precursor is of particular interest, because (a) it tests the whether a violation of relativity is “hiding” at high frequencies, and (b) the speed of a precursor is theoretically supposed to be equal to $c$, making “faster than light” violations relatively easy to spot. All of this brings us back at last to the June paper in Physical Review Letters on the “Optical precursor of a single photon”, by a research group in Hong Kong. A photon is a single quantum particle of light; all of the discussion of precursors up to this point have concerned pulses consisting of many, many photons. Two questions arise when considering precursors and single photons: (1) Do precursors even exist for a single photon? One would naturally be inclined to say “yes”, but it may be that, on a quantum level, precursors inherently involve the interaction of many photons at once. (2) Can single photons travel faster than $c$? One would be inclined to say “no” in this case, but again the behavior of single photon precursors (if they exist) might be subtly different than a group of many photons. The Hong Kong researchers investigated these possibilities by producing coupled pairs of photons using the following experimental configuration (adapted and simplified from the article): The fundamental components of the system are a pair of magneto-optical traps (MOT) containing Rubidium atoms. Magneto-optical traps use light and magnetic fields to localize a group of atoms and cool them to a low temperature, forming a low-density medium. The first MOT is excited by a pair of lasers — a pump beam and a coupling beam — and through a complicated light-matter interaction pairs of photons are produced. The physics of this production is too complicated to discuss here, but the result is a higher frequency “Stokes photon” and a lower-frequency “anti-Stokes photon”. These photons are produced at the same time and are therefore correlated in time. The Stokes photon is picked up by detector 1 and sends a signal to a function generator, which triggers an electro-optic modulator that the anti-Stokes photon passes through. This modulator “chops off” the front end of the anti-Stokes photon signal, producing a sharp-edged pulse necessary for precursor generation*. This sharp-edged pulse propagates into the second MOT, in which its waveform is expected to take a precursor shape. It is essential for this experiment that a single anti-Stokes photon be measured at a time; after the second MOT, the light signal is split by a beam-splitter and sent to a pair of detectors, detectors 2 and 3. Because a single photon can only arrive at a single detector, the simultaneous tripping of both detectors implies the presence of more than one photon and the event is thrown out. The time of arrival of the single anti-Stokes photons can be tallied, producing a profile of the average behavior of the photons. The coupling laser illuminating the second MOT could modify the optical properties of the Rubidium atoms into one of two configurations. In the first configuration, the absorption properties of the medium can be completely suppressed, in a technique called electromagnetic induced transparency (EIT). With EIT, both the main body of the pulse and the precursor signal make it through the MOT and can be measured by the detectors. Furthermore, the main signal is slowed considerably in speed and is separated in time from the optical precursor. In the second configuration, the group velocity of the main signal is faster than $c$, i.e. the group velocity is “superluminal”. However, this main signal is highly absorbed. What were the results of the experiments? The researchers measured the speed of the main body of the pulse and the precursor for both MOT configurations. The precursor was clearly visible in both cases, and its speed was found to be exactly $c$, regardless of the MOT behavior. Therefore no true “faster than light” behavior was observed, even when the group velocity was greater than the vacuum speed of light. An important aspect of this result is a partial answer to a debate in quantum information theory — how fast does a single photon transmit information? As we’ve noted, there are multiple definitions of the speed of light in matter, and it has remained a bit of a mini-mystery (at least to some) which of these described the speed at which information is transmitted. The observation that a photon does have a precursor signal suggests that single photons can travel at the vacuum speed of light in matter, at least under the right conditions. The result isn’t as earth-shattering as the possible discovery of superluminal neutrinos, but it does highlight a number of unusual aspects of optics, and further strengthens our understanding of both relativity and quantum mechanics! ********************** * In physics, “velocity” is used to refer to the vectorial motion of an object: not only how fast it is going, but in what direction. ”Speed” is used to refer to the magnitude of velocity, or simply how fast the object is going. We use the terms interchangeably here in the sense of “speed”. A. Sommerfeld, Ann. Phys. (Leipzig) 349 (1914), 177. L. Brillouin, Ann. Phys. (Leipzig) 349 (1914), 203. * P. Pleshko and I. Palócz, “Experimental observation of Sommerfeld and Brillouin precursors in the microwave domain,” Phys. Rev. Lett. 22 (1969), 1201. ** S-H. Choi and U. Österberg, “Observation of optical precursors in water,” Phys. Rev. Lett. 92 (2004), 193903. * Actually, the anti-Stokes photon will already have a precursor signal due to its propagation in MOT 1! This precursor is also chopped off by the EOM. *********************** Zhang, S., Chen, J., Liu, C., Loy, M., Wong, G., & Du, S. (2011). Optical Precursor of a Single Photon Physical Review Letters, 106 (24) DOI: 10.1103/PhysRevLett.106.243602 About these ads ### Like this: Like Loading... This entry was posted in Optics, Physics, Relativity. Bookmark the permalink. ### 13 Responses to Faster than a speeding photon? Precursors test whether light can be faster than light 1. Pingback: Faster than a speeding photon? Precursors test whether light can be faster than light \| benjamin junior 2. Pingback: RB Editor’s Selections: The Expanding Universe, Triggered Earthquakes, Statistical Badness, and More Neutrinos 3. CHRIS says: Typo in last paragraph. “are” should be “our”. • skullsinthestars says: Ah, thanks! Fixed. 4. Ray Kilburn says: Previous measurements of light velocity traveling through matter is akin to the photoelectric effect for which Einstein received his Noble prize. The current description is incomplete as electrons are incorrectly described, as in this blog, as a :”pendulum” or a spring is no more accurate than the orbital model. When we correlate the wave description of the electron’s orbit with its particle description, we realize a different view of the model of the atom. Rutherford’s model of the atom, and all subsequent variations, propose an orbital model of sorts with material electrons flying around the proton nucleus. Subatomic physics and quantum theory reduced this model to electron clouds that supported the quantum uncertainty of finding the electron in any particular position around the proton. Quantum Field Theory proposes that perhaps, the electron is not orbiting the proton at all, in this sense, but that the proton is actually inside of the electron! Each subsequent electron “shell” encompasses the previous shell. When light encounters an atom, corresponding wavelengths of the photonic wave are absorbed by the electron bubble of the corresponding frequency. While the time delay between absorption and emission explains normal ionic activity, apparent increase light velocity is directly proportional to the uncertainty of the speed of light to begin with. Keep in mind that Einstein’s velocity of light is ONLY accurate in a vacuum but not accurate within a gravitation field which effect light’s true path thru space as confirmed by relativity. Therefore light could be said to NEVER travel in a straight line by any terrestrial measurement than can be made and therefore its accurate velocity can never be determinate. In absence of a quantum description of gravity, it could be said that no gravity exists within an atom as we see no gravitational effect on particles with the atom and the gravitational effect is clearly only present outside of an atomic system. Therefore, one could postulate that our measure of light velocity within matter cannot be accurately determinate. Unlike a classical particle having mass, the Photon does not emit a rotating spherically symmetrical wave system with a continuous degree of freedom as do material particles. Nor is it normally polarized in a specific directional time orientation as observed in Electrons and Protons. Instead, it appears to interact with particles and antiparticles equally in both time domains. For this reason the Photon is said to be its own antiparticle. Following this reasoning further, one could postulate that one or more of the following is true: 1) the photon exists simultaneously in both time domains, 2) it rotates back and forth between time domains, or 3) is unaffected by time domains. When a Photon is compared to a particle with mass in motion, the Photon cannot be said to have a momentum in a single vector direction, however it appears to be propagated as a directional wave packet. However the Photon appears to propagate its momentum along a three-directional vector corresponding to physical construction of the space-time lattice structure with a definitive wave structure. Along each directional vector there is an angular momentum in the direction of motion called a heliocity. But just like particles having mass, this heliocity is restricted normally to integer values. The exception to this observation is in the case of a polarized photon (neutrino/antineutrino) which has a heliocity value of half-integer values. Because virtual particles imply both the inversion of space and time, both positive and negative energy states could be said to exist. It is only a mathematical trick that requires massless particles to be anti-unitary and anti-linear because the Standard Model of classical physics does not allow negative energy states, that is, energy less than that of the so-called vacuum. Because of this inexplicable paradox, classical physics does not allow the reversal of its time components either. Both cases to be incorrect but too lengthy to describe here. 5. Сергей Морозов says: May I translate it in my blog (moogoo.ru)? 6. Pingback: Links for Tuesday « Galileo's Pendulum 7. Pingback: The Value of Experimenting With What We “Know” \| Citizen Scientists League 8. Primoz Gabrijelcic says: I don’t understand how the electro-optic modulator can react quick enough to clip the photon, which is traveling at c. That would imply that the signal that triggers the modulator travels with velocity > c. Or is the anti-Stokes photon traveling through some materil with low “speed of light”? • skullsinthestars says: One of the things I left out of my simple picture of the experimental setup is the delay line the anti-Stokes photon passes through! This photon is coupled into a long optical fiber, and with a long enough fiber the photon can be delayed as long as one likes compared to the modulator trigger — within reason! 9. Ray Kilburn says: The result of any observation of a “superluminal neutrinos” may result in the realization that neutrinos may not really be an independent particle at all as currently considered by the Standard Model and may explain the difficulty in detecting them. Consider how the neutrino was “invented” to explain differences in mass between a neutron and its resulting beta decay model which supposedly results in an electron, proton and a neutrino that apparently flies away at light speed in violation of relativity and doesn’t interact with any matter. What is generally ignored is that the additional mass measured in the neutron may be a result of the retrograde spin of the neutron (which could be considered a micro-model of a hydrogen atom (proton+electron)) and therefore may be relativistic mass and not a real mass at which explains why the neutrino is always considered massless. During beta decay, when the proton and electron are released from the neutron, the remaining relativistic mass spins off as a photon that is moving as a helicoid instead of a plane-wave photon due to the angular momentum imparted upon it from the spin of the originating neutron. Since a helicoid looks like a spring, the photon’s angular momentum tunnels the so-called “neutrino” through all mass without stopping. The is the same principal used by electrons to tunnel thru matter as well. I think that upon further examination we will find that the neutrino is really photon traveling as a helicoid and therefore unpolarized and incapable of interaction with matter. Plane-wave photons which are polarized along two planes run into everything. Real super-luminal communication of information is probably the result that the photon as a discreet entity really doesn’t travel anywhere. Consider that the photon is really at both its origin and destination at the same time, more akin to a string connecting point A to B. The phontonic wave we observe is the jiggling of the string, nothing is really moving anywhere except the wave along the string, the string doesn’t move anywhere. From an outside observer we see the wave moving along the string and measure its velocity at c. From the perspective of the photon, it is already there since zero time dilation occurs and the photon appears (from its perspective) to arrive at the same moment it departed it source. Another point to consider in the measurement of light speed in a vacuum is that true vacuums exist nowhere in reality. Even when we peer into the deepest darkest part of space, we still measure background microwave radiation in 3micron wavelength. Therefore a photon at some energy level exists at all points in space and no point in space could be said not to contain either traditional matter or a photon at some energy level. So no true vacuum could be said to exist nor can be created and therefore light true velocity will always be indeterminate in all true circumstances. Our current measurement of light’s velocity is an approximation. Theories surrounding the big bang suggests that the speed of light varies as the universes expands….the string is getting longer? 10. zephirawt says: My stance is, OPERA results are real and they should be published ASAP. IMO we can model the space-time (brane) with density gradient at the phase interface of two elastic fluids. After then a two kinds of solitons will appear: A) the one, which corresponds to photons and it spreads with slightly lower speed, than the transverse surface waves (which are serving as an analogy of light waves) B) the faster one, which corresponds the neutrinos and it will spread with slightly higher speed, than the surface ripples. The first kind of solitons results from coupling of surface ripples with longitudinal bulk waves of more dense phase, the second one from coupling of surface ripples with longitudinal waves of less dense phase. From this perspective the neutrinos would behave like the superpartners of gamma ray photons, i.e. like the lightweight photinos. 11. Anirudh Kumar Satsangi says: Dear Mr. Jug Suraiya Congratulations for your very thought provoking article ‘Einstein Won’t Mind’ (The Speaking Tree, Oct. 9, 2011, pg. 7). Einstein is great, but Newton is all time great. I would like to quote your few excellent lines from this article before I come to main theme of this article: “Faith and religious beliefs are destinations reached; science and skepticism are journeys without end”. But it’s to be modified. In Veda it is written, ‘Neti, Neti’, ‘Not this, Not this’. So in Vedic religion journeys never end. I consider Vedas the most honest scriptures which do not limit the scope of further exploration of truth. Now I come to the speed of light. This has long been proved since the time of discovery of black holes that the speed of light is not the fastest. Black holes do not allow even light to escape. It means the escape velocity at the black holes is much higher than the speed of light. Black holes are the infinitely dense ball of gravitation force. All creational forces of the universe have originated from the gravitational force field and will end up in it. The speed of light is no doubt fastest in our solar system. The source of light is Sun in our solar system. But how this light is originated? We should study the various stages involved in the formation of a star. Our Sun is also a star. The starting material for the formation of a star is mainly hydrogen gas and helium gas. If the hydrogen cloud contains a very large number of atoms, each atom feels the gravitational pull of all the atoms in the hydrogen cloud. (Here is NO LIGHT) The gas cloud becomes a permanent entity, held together by the mutual attraction of all the atoms present in it. The cloud then begins to contract under its own gravity setting off the process which will convert this huge condensed gas cloud into a star. Such a tight contracting cluster of atoms held in the grip of its own gravity, is called a protostar. The protostar is not yet a star and does NOT emit LIGHT. The temperature of this star is as low as -173 degree C. The force of gravity acting on different atoms in the protostar draws every atom towards centre. As a result, the protostar shrinks in size and its density increases. As the atoms in the protostar fall towards the centre, they pick up speed. Because of the high speed and greater density of atoms, the atoms in the gas cloud collide with one another more frequently, thereby raising its temperature from -173 degree C to about 10 ^7 degree C. At these extremely high temperatures the proton (hydrogen nuclei) at the centre of the protostar collide together and undergo a nuclear fusion to form helium nuclei. In this reaction a tremendous amount of energy is released. This further raises the temperature and pressure. The release of nuclear energy marks the birth of the star. The protostar now beings to GLOW and becomes a STAR. Here at this stage LIGHT is ORIGINATED. Thus light is NOT ETERNAL. It has a beginning and an end. So LIGHT cannot be claimed as Cosmic Constant. However, Gravitation Force is eternal. It is evident from the above description that light is latent before the birth of star. Light originates and become kinetic only after the action of gravitation force. So the speed of light can never exceed the speed of gravitation force. It cannot be ruled out that the speed of gravitation force is infinitely greater than the speed of light at black holes. 1. Gravitation Force is the Ultimate Creator 2. In Scientific Terminology Source of Gravitational Wave is God About these ads • ### Search Skulls in the Stars: • The author of Skulls in the Stars is an associate professor of physics, specializing in optical science, at UNC Charlotte. The blog covers topics in physics and optics, the history of science, classic pulp fantasy and horror fiction, and the surprising intersections between these areas. • ### Twitter Updates • RT @hitRECordJoe: RT @rupertmurdoch: "IRS scandal makes perfect case for flat tax. Nothing could be fairer...” If by "fairer" you mean bet… 39 minutes ago • RT @FreakyFrogger: It's raining diamonds on Neptune and Uranus?!?! spacedaily.com/news/carbon-99… 8 hours ago • RT @sfpelosi: Congress couldve spent the week creatig jobs or ending #sequester that cut embassy security funding. #DoYourJobGOP #WhoDoct… 19 hours ago • I think I finally understand differential geometry enough to do some productive research. Beautiful subject, really. 19 hours ago • ### Blogroll %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 17, "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": 0.9318211674690247, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/62051-uniform-distribution-random-independent-variables.html	# Thread: 1. ## uniform distribution of random independent variables okay. my exam is approaching in a few days and I have one last question. Let X1, X2, ... X7 be a set of seven independent random variables, each having the uniform distribution of values from the [3, 6] interval (this is called random independent sample of size 7 from this distribution). (a) Find the expected value and standard deviation of their sum T ≡ X1 + X2 + ... + X7. (b) What is the probability that more than three of these seven random variables will have a value smaller than 1 3? (c) Find the skewness and kurtosis of X1. answers: (a) E(x) = 7(6+3)/2 = 31.5 Var(x) = 7[(6-3)^2]/2 standard deviation = 2.2913 (b) not sure (c) would skewness be 0 since its uniform? what about kurtosis 2. Originally Posted by chrisc okay. my exam is approaching in a few days and I have one last question. Let X1, X2, ... X7 be a set of seven independent random variables, each having the uniform distribution of values from the [3, 6] interval (this is called random independent sample of size 7 from this distribution). (a) Find the expected value and standard deviation of their sum T ≡ X1 + X2 + ... + X7. (b) What is the probability that more than three of these seven random variables will have a value smaller than 1 3? Mr F says: The question is not clear to me. Is it meant to be 13, 1 or something else? (c) Find the skewness and kurtosis of X1. answers: (a) E(x) = 7(6+3)/2 = 31.5 Var(x) = 7[(6-3)^2]/2 standard deviation = 2.2913 (b) not sure (c) would skewness be 0 since its uniform? what about kurtosis (a) You're expected to know or derive that $E(X_i) = \frac{6+3}{2} = \frac{9}{2}$ and $Var(X_i) = \frac{(6-3)^2}{12} = \frac{3}{4}$. Then: $E\left(\sum_{i=1}^{7} X_i\right) = \sum_{i=1}^{7} E(X_i) = 7 \cdot \frac{9}{2} = \frac{63}{2}$. $Var\left(\sum_{i=1}^{7} X_i\right) = \sum_{i=1}^{7} Var(X_i) = 7 \cdot \frac{3}{4} = \frac{21}{4}$ (summing the variances follows because the X's are independent). (b) Is there a typo? (c) What definition/type of kurtosis are you using? 3. a) okay, confirmed b) was a typo, its 1/3, not 1 c) nothing else was specified outside of this question 4. Originally Posted by chrisc a) okay, confirmed b) was a typo, its 1/3, not 1 Mr F says: Well if that's the case, what's the probability that any of the X's can be smaller than 1/3 .....?! Look at the interval over which the pdf is not equal to zero! c) nothing else was specified outside of this question c) That's why you go to your class notes or textbook, find the definition/type that's been used and then report the answer in this thread. I had thought that's what you'd do. 5. this is the one thing we were given about for kurtosis, part of our exponential stuff (not uniform like the question asks) kurtosis: [E(X4)−4μE(X3)+6μ2E(X2)−3μ4]/σ4 = ... [24β4−4×6β4+6×2β4−3β4]/β4 = 9 does this help at all? if not, ill just hope this specific type of question doesnt make its way to the exam 6. Originally Posted by chrisc this is the one thing we were given about for kurtosis, part of our exponential stuff (not uniform like the question asks) kurtosis: [E(X4)−4μE(X3)+6μ2E(X2)−3μ4]/σ4 = ... [24β4−4×6β4+6×2β4−3β4]/β4 = 9 does this help at all? if not, ill just hope this specific type of question doesnt make its way to the exam Then calculate each of the bits in the formula: $E(X^n) = \int_3^6 x^n \cdot 1 \, dx$. $\mu = E(X_1) = \frac{9}{2}$. $\sigma = s.d.(X_1) = \frac{\sqrt{3}}{2}$. Now substitute into the formula.	{"extraction_info": {"found_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": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.931275486946106, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/90859-separation-axioms.html	# Thread: 1. ## separation axioms prove that : Every $\displaystyle \tau \scriptstyle3$ space is $\displaystyle \tau \scriptstyle 2 \frac{1}{2}$ 2. A good way to start: write out the definitions of $T_3$ and $T_{2 1/2}$ 3. Originally Posted by flower3 prove that : Every $\displaystyle \tau \scriptstyle3$ space is $\displaystyle \tau \scriptstyle 2 \frac{1}{2}$ Definition. A space X is a $T_{2\frac{1}{2}}$-space provided that for each pair x,y of distinct points of X, there exists open sets U and V with disjoint closures such that $x \in U$ and $y \in V$. Lemma 1. X is regular Hausdorff ( $T_3$) if and only if given a point x of X and a neighborhoood U of x, there is a neighborhood W of x such that $\bar{W} \subset U$. Assume X is regular Hausdorff ( $T_3$). Since X is also Hausdorff, for each pair x,y of distinct points of X, there exists disjoint open sets U and V containing x and y, respectively. By hypothesis, it follows that there is a neighborhood W of x such that $\bar{W} \subset U$ and a neighborhood Y of y such that $\bar{Y} \subset V$ by lemma 1. Since U and V are disjoint, we see that $\bar{W}$ and $\bar{Y}$ are disjoint. Now, for each pair x,y of distinct points of X, there exists open sets W and Y with disjoint closures such that $x \in W$ and $y \in Y$.Thus, X is $T_{2\frac{1}{2}}$-space. The converse ("Every $T_{2\frac{1}{2}}$-space is $T_3$") is not necessarily true. The example where a space X is $T_{2\frac{1}{2}}$-space but not $T_3$ can be found in the K-topology on $\mathbb{Re}$, where a closed set $K=\{1/n \| n \in \mathbb{Z}^+\}$ (Note that K is closed in the K-topology on $\mathbb{Re}$ ) and {0} cannot be separated by disjoint open sets.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 26, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9559143781661987, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/155409/how-do-i-solve-a-certain-characteristic-system/155494	# How do I solve a certain characteristic system? I am studying PDEs and have the following (seemingly simple) problem: Find a surface that passes through the curve $$x^2+y^2=z=1$$ and is orthogonal to the family of surfaces $$z(x+y)=c(3z+1)\qquad(c\in\Bbb R)$$ After writing down the orthogonality condition (assuming my calculations are correct $()$), this yields the following equation: $$u(3u+1)(u_x+u_y)-x-y=0$$ We usually solve such equations by using the method of characteristics, which tells us (using assumption $()$ again) to solve the following characteristic system: $$\begin{align}\dot x=&u(3u+1)\\\dot y=&u(3u+1)\\\dot u =&x+y\end{align}$$ Differentiating the last equation of this system with respect to $t$ gives us $\ddot u=\dot x+\dot y$, which using the first two equations gives us $$\ddot u = 2u(3u+1)$$ After staring at this equation for some time, I decided to ask Wolfram\|Alpha. The result seems pretty ugly, so the following questions arise: Did I make a mistake/am I missing something? Is my approach correct? How do I proceed? Thanks. - – Weltschmerz Jun 8 '12 at 7:00 ## 1 Answer Here's what I do (I'm probably wrong, but we can compare our procedures. I'm not using the comment box because there's not enough space in it). The intersection of the surfaces would be the set of points $(x,y,z)$ such that \begin{cases}u(x,y)-z=0\\z(x+y)-3cz-c=0,\end{cases} that is, those that lie both on the solution surface and the given surface (determined by $c$, and this must hold for every $c$). The surfaces are orthogonal means (I'm guessing) their respective gradients at the intersection points must be orthogonal. The gradient of the solution surface is $(u_x,u_y,-1)$, and the gradient of the given surface(s) is $(z,z,x+y-3c)$. So the condition is $zu_x+zu_y-x-y+3c=0$. Since this occurs when $z=u$ we can write it as $$uu_x+uu_y=x+y-3c.$$ We can clear $c$ from the intersection equations: $c=\frac{(x+y)u}{1+3u}$, and plugging it in the last equation gives us $$uu_x+uu_y=(x+y)(\frac{1}{1+3u}).$$ So we agree there. As for what follows, the method of characteristics, I think you proceeded correctly. - Thanks for your thoughts. – Dejan Govc Jun 8 '12 at 10:00	{"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": 13, "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": 0.9296669960021973, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/4334/could-life-survive-a-pole-shift-caused-by-an-asteroid-collision/4424	# Could life survive a pole shift caused by an asteroid collision? Could life on earth survive a large pole shift caused by an asteroid collision? I became aware that there are people who believe that the earth's pole suddenly shifts. That is, its rotational axis changes rather than its magnetic axis. This is listed in wikipedia as follows: Cataclysmic pole shift hypothesis The cataclysmic pole shift hypothesis is the conjecture that there have been rapid shifts in the relative positions of the modern-day geographic locations of the poles and the axis of rotation of a planet. For the Earth, such a dynamic change could create calamities such as floods and tectonic events. ... The field has attracted pseudoscientific authors offering a variety of evidence, including psychic readings. ... http://en.wikipedia.org/wiki/Cataclysmic_pole_shift_hypothesis While watching a chess game at the local club a fellow member told me that he'd found the solution for where oil came from. It was long and involved. One of its features required that the earth's axis be changed by 90 degrees due to the collision of an asteroid a few million years ago. The idea was that the earth's axis used to be in the plane of the earth's orbit, but then it was knocked into its current rotation axis tilt of around 23 degrees. This changed the climate and buried a lot of stuff. The axis shift seemed to be an obvious weak link with the theory in that it involves the simplest physics. I told him that any asteroid large enough to change the earth's axis an appreciable amount (i.e. 70 degrees!) would cause so much damage as to destroy all life on the surface of the planet including its oceans. I'd have forgotten about it, but the other night, while listening to the radio, a guest appeared who had an idea that was similar. So this seems like a common target for pseudoscience. And so my question is this: Would an asteroid collision sufficient to immediately change the earth's rotation axis wipe out all life on the planet? - 4 Some people have evidently never played with a gyroscope or tried to turn a spinning bicycle wheel out of plane with their hands. What mechanism is proposed to damp the ferocious precession that would be the inevitable aftermath of twisting the axis around like that? Bah! Humbug! – dmckee♦ Feb 1 '11 at 1:24 1 What sort of asteroid could knock Earth's axis off by 70 degrees without ripping the planet to shreds, for that matter!? – Noldorin Feb 1 '11 at 2:35 @Noldorin: Yeah. I started trying to estimate the energies involved, but floundered when I realized there was no impulse-like way to get from there to here. – dmckee♦ Feb 1 '11 at 2:56 1 Some very simple calculation would answer this question. – Georg Feb 1 '11 at 10:34 1 Two votes to close because it (a) has nothing to do with physics, and (b) does not have a solution. However, I've provided a physics calculation as a solution so these objections are misplaced. – Carl Brannen Feb 3 '11 at 3:08 show 4 more comments ## 3 Answers Quite clearly this man’s theory is balderdash. A massive collision could change the tilt of the Earth’s axis, but that would be one hell of a torque. The collision would not be with an asteroid either, but with another planet that might be as big as Mars. Clearly nothing like this happened in recent geological history, though the Earth did suffer a collision of this magnitude a few 100 million years after gravitational accretion. A chunk from the collision spalled off to form the moon. This other planet probably was in some resonance condition around a Lagrange point. This would be a devastating collision, which would doubtless change this planet into something else. The solar system has not experienced such collsions since its early evolution. The solar system is now staid and respectable. Velikovskian things just do not happen. - I fully agree. If you could simply change the rotational axis via magic (with actual physics you gotta find a source of a huge amount of torque), there would be major climate change, and seas would shift around by thousands of feet, so it would be pretty tramatic, but the new earth should still support life. The real issue is that any concievable hit of that magnitude would be catastropic for nearly all life. [It is possible that simple bacteria living a couple of kilometers undergroud would outlive the heat spike (the surface would be a lava lake for years).] – Omega Centauri Feb 1 '11 at 5:09 Let's begin by choosing coordinates. Let the orbit of the earth define the x-y plane. Assume that just before the collision the earth's orbital axis is pointing in the x-direction, that is $(1,0,0)$. I'll assume that the earth absorbs the asteroid, and that after the collision the axis has been changed to $(\sin(27),0,\cos(27))$ where 27 is the degree measure of the earth's axis currently. So the relative change in angular momentum for the earth is $(1,0,0) - (\sin(27),0,\cos(27)) = (0.55,0,-.89)$ which is a vector of length 1.05 and so the angular momentum of the asteroid has to be 1.05 times the angular momentum of the earth. Make the approximation that the earth is a sphere of constant density. Then its angular momentum is given by $0.4 M_e\omega\;R_e^2$ where $M_e,R_e$ are the mass and radius of the earth and $\omega$ is its angular rate of rotation. We have $M_e = 6\times 10^{24}\;kg\;\;\;R_e = 6.4\times 10^6\;m,\;\;\;\omega = 2\pi/(24\;hours) = 2\pi/ (86400\;sec)$. I'm using sloppy approximations here; the earth is not constant density, I'm not using the sidereal day, etc. This gives $7\times 10^{33}\;kg\;m^2/s$ as the approximate angular momentum contributed by the asteroid. The formula for angular momentum is mass x velocity x radius. This is maximum when the radius is maximum; for an asteroid hitting the earth this happens when the radius is equal to the earth's radius, i.e. $6.4\times 10^6\;m$. Dividing the asteroid's angular momentum by this give its linear momentum: $P_a = 7\times 10^{33} / 6.4\times 10^6 = 10^{27}\;kg\;m/s$ The reason a relatively small asteroid can wipe out all life on earth is due to the kinetic energy it carries. Upon collision, the kinetic energy is converted to heat. If the asteroid is large and fast enough, the heat will increase the temperature of the atmosphere enough to boil off the oceans and even vaporize the salt deposits left over. Since kinetic energy is proportional to the square of velocity while momentum is only proportional to velocity, we will assume that our asteroid is moving as slowly as possible. The escape velocity of the earth is 11 km per second; to achieve a slower velocity an asteroid would have to be very lucky. So in calculating the kinetic energy of the asteroid, we'll assume a very conservative speed of 5 km/sec. Putting $10^{27} = 5000 m$ we find the mass of the asteroid as $2\times 10^{23}$ kilograms. Assuming a specific density of 5, or 5000 kg per cubic meter, this gives a radius for the asteroid of 2000 kilometers or a diameter of 5000 km. This is far more than enough to destroy all life on the planet. The asteroid's kinetic energy is $0.5\times 2\times 10^{23}\times (5000)^2 = 2.5\times 10^{30}$ Joules $= 6\times 10^{14}$ megatons. In other words, the collision would result in the release of kinetic energy equal to 600 million million hydrogen bombs going off, each with an equivalent energy of one million tons of high explosives. The earth's surface area is 500 million square kilometers or 500 million million square meters. So the energy release is equivalent to having a 1 megaton bomb going off on each square meter of the earth's surface. And since the vaporized rock from the blast is lighter than rock, this vapor will condense on the surface and apply its heat to the surface. For a video of a 500 km asteroid hitting the earth (1/1000 of the volume necessary to change the earth's axis) see:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9480854272842407, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/159054-isometric-print.html	# Isometric? Printable View • October 10th 2010, 10:23 AM Showcase_22 Isometric? Quote: Let $X=\{A,\ B,\ C, \ D\}$ with $d(A,D)=2$, but all other distances equal to 1. d is a metric. Prove that the metric space X is not isometric to any subset of $\mathbb{E}^n$ for any n. Hint: can you realise $X$ as a subset of a sphere $S^2$ of appropriate radius, with the spherical "great circle" metric? I think I have to take two points in $X$ and show that any mapping from $X$ to $\mathbb{E}^n$ would result in a different distance between them. Here's my attempt: Take a sphere of radius 1 and the points A and D. Since $d(A,D)=2$, A and D are on opposite sides of the sphere. Suppose there exists an isometry $\phi:X \rightarrow \sigma$ where $\sigma$ is is the plane used in stereographic projection (also a subset of $\mathbb{E}^n$). Then $\phi(A)=\phi(D)$. Hence $d(\phi(A),\phi(D))=0$ which is not equal to $d(A,D)=2$. Hence $\phi$ cannot be an isometry. • October 10th 2010, 11:26 AM Opalg Quote: Originally Posted by Showcase_22 I think I have to take two points in $X$ and show that any mapping from $X$ to $\mathbb{E}^n$ would result in a different distance between them. Here's my attempt: Take a sphere of radius 1 and the points A and D. Since $d(A,D)=2$, A and D are on opposite sides of the sphere. Suppose there exists an isometry $\phi:X \rightarrow \sigma$ where $\sigma$ is is the plane used in stereographic projection (also a subset of $\mathbb{E}^n$). Then $\phi(A)=\phi(D)$. Hence $d(\phi(A),\phi(D))=0$ which is not equal to $d(A,D)=2$. Hence $\phi$ cannot be an isometry. I'm not sure that the hint is very helpful, and I don't think that stereographic projection is needed. If such an isometry $\phi$ exists then the points $\phi (A),\ \phi (B),\ \phi (C)$ form an equilateral triangle with side 1, in some 2-dimensional subspace of $\mathbb{E}^n$. So do the points $\phi (B),\ \phi (C),\ \phi (D)$. Thus the images of the four points lie in some 3-dimensional subspace of $\mathbb{E}^n$, so we may as well assume that n=3. The points $\phi (A)$ and $\phi (D)$ must both lie at a distance $\sqrt3/2$ from the midpoint of the line segment joining $\phi (B)$ and $\phi (C)$. So the distance from $\phi (A)$ to $\phi (D)$ is at most $\sqrt3<2$, contradicting the condition that $\phi$ is an isometry. All times are GMT -8. The time now is 11:22 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": 41, "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": 0.9548066854476929, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2011/08/04/integrals-are-independent-of-parameterization/	# The Unapologetic Mathematician ## Integrals are Independent of Parameterization If $c:[0,1]^k\to M$ is a singular $k$-cube and $\omega$ is a $k$-form on the image of $c$, then we know how to define the integral of $\omega$ over $c$: $\displaystyle\int\limits_{c([0,1]^k)}\omega=\int\limits_{[0,1]^k}c^\omega$ On its face, this formula depends on the function $c$ used to parameterize the region of integration. But does it really? What if we have a different function $d:[0,1]^k\to M$ with the same image? For convenience we’ll only consider singular $k$-cubes that are diffeomorphisms onto their images — any singular $k$-cube can be broken into pieces for which this is true, and we’ll soon deal with how to put these together. Anyway, if $c([0,1]^k)=d([0,1]^k)$, then given our assumptions there is some diffeomorphism $f=c^{-1}\circ d:[0,1]^k\to[0,1]^k$ such that $d=c\circ f$. If $J_f$ is everywhere positive, then we say that $d$ is an “orientation-preserving reparameterization” of $c$. And I say that the integrals of $\omega$ over $c$ and $d$ are the same. Indeed, we calculate: $\displaystyle\begin{aligned}\int\limits_{c\circ f}\omega&=\int\limits_{[0,1]^k}(c\circ f)^\omega\\&=\int\limits_{[0,1]^k}f^c^\omega\\&=\int\limits_fc^\omega\\&=\int\limits_{f([0,1]^k)}\left[c^\omega\right]\left(\frac{\partial}{\partial u^1},\dots,\frac{\partial}{\partial u^k}\right)\,d(u^1,\dots,u^k)\\&=\int\limits_{[0,1]^k}\left[c^*\omega\right]\left(\frac{\partial}{\partial u^1},\dots,\frac{\partial}{\partial u^k}\right)\,d(u^1,\dots,u^k)\\&=\int\limits_c\omega\end{aligned}$ where we use our expression for the integral of $\omega$ over the image $f([0,1]^k)$ in passing from the third to the fourth line. Thus the integral is a geometric quantity, depending only on the image $c([0,1]^k)$ and the $k$-form $\omega$ rather than on any detail of the parameterization itself. About these ads ### Like this: Like Loading... Posted by John Armstrong \| Differential Topology, Topology ## 2 Comments » 1. [...] it all works out for the same reason parameterization invariance and the change of variables formula do. Passing from the boundary of the singular cube back to the [...] Pingback by \| August 20, 2011 \| Reply 2. [...] we use to reparameterize our integral. Of course, this function may not be defined on all of , but it’s defined on , [...] Pingback by \| September 5, 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. • ## RSS Feeds RSS - Posts RSS - Comments • ## Feedback Got something to say? Anonymous questions, comments, and suggestions at Formspring.me! %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 27, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8871572613716125, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/4894/why-does-symplectic-geometry-have-many-applications-in-mathematics?answertab=active	# Why does symplectic geometry have many applications in mathematics It is not quite intuitive , at least from its origin. Could any one can give me an intuitive explanation?Thank you! - ## 4 Answers Any two of (a) symplectic structure, (b) almost-complex structure, and (c) Riemannian structure determine the third. This reflects the fact that $$U(n) = O(2n) \cap GL(n,\mathbb{C}) \cap Sp(2n),$$ but the intersection of any two is already $U(n)$. This gives rise to the theory of Kähler manifolds (cf. wikipedia), which are central to mirror symmetry (and the mirror symmetry conjecture). - I suppose a question to ask here is : what kinds of applications? One of the areas where symplectic geometry is used is Hamiltonian dynamics, where the geometric framework allows one to generalize away from cotangent bundles, but this direction is directly related to the origins of the subject. Another direction comes from algebraic geometry, where the fact that complex and Riemannian structure determine a symplectic one (see Aaron's answer) makes all complex projective (and affine) algebraic varieties symplectic. Viewing them as such allows one to use differential-geometric methods, while the symplectic form provides some control (in the theory of holomorphic curves, symplectic form gives the crucial energy bounds without which things fall apart; this is why there is no Gromow-Witten theory of almost-complex manifolds). Add a bit of mirror symmetry to the mix and you have interaction with many other areas of math. There is also the fact that the wedge product is anti-symmetric, which leads the space of connections on a principal bundle over 2-d Riemannian manifold to be an (infinite dimensional) symplectic manifold (the gauge group is Hamiltonian, and curvature is the moment map, leading to flat connections being the reduction!). This leads, for G=SO(3) and the surface coming from a Heegard splitting, to Atiyah-Floer conjecture and some related stuff leads to Heegard-Floer theory. Thus connections to low-dimensional topology. The "naive" connection here is the anti-symmetry underlying symplectic form and wedge product, but probably there is a deeper reason known to wise people... - There is a simple "philosophical" answer: Symplectic forms allow you to only measure two-dimensional quantities, not one-dimensional ones (you can measure area infinitesimally). That's the basic difference with Riemannian geometry, where you can measure length infinitesimally. Furthermore symplectic geometry pops up in connection with the cotangent bundle. For instance if you have a (scalar) first order differential operator on a manifold, then the principal symbol transforms like a covector (when changing coordinates). Note that this does not require a metric. - Symplectic geometry is applicable in lot of areas such as Classical mechanics, Hamiltonian mechanics. While browsing through the web i have found a reason as to why SG is good for classical mechanics. You may read it here: http://research.microsoft.com/en-us/um/people/cohn/thoughts/symplectic.html - 3 I think the question is more about why symplectic geometry has found applications beyond classical mechanics. – Qiaochu Yuan Sep 18 '10 at 5:34 @Qiaochu Yuan: I answered the question, w.r.t my thinking. – anonymous Sep 18 '10 at 7:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9209643006324768, "perplexity_flag": "middle"}
http://mathhelpforum.com/trigonometry/39976-radians.html	# Thread: 1. ## Radians Hi, need help understanding 2 problems. Give all the solutions to the following problems, answer in radians. $\sin(3x)=\sqrt{3}/2$ and $\cos(x-pi/4)=1/\sqrt{2}$ Thanks. 2. Originally Posted by Smifsret Hi, need help understanding 2 problems. Give all the solutions to the following problems, answer in radians. $\sin(3x)=\sqrt{3}/2$ and $\cos(x-pi/4)=1/\sqrt{2}$ Thanks. You need to know what angle for sine gives the value $\frac{\sqrt{3}]{2}$. This happens to be $\frac{\pi}{3}$ Solve for $3x = \frac{\pi}{3}$ 3. Thanks, is the second one pi / 4 then? Does pi/4 count as an answer (if its right) or do they want me to write a number? 4. sin(3x)= pi/3 x1. sin(x) = pi/9 + 2 * pi * n? x2. sin(x) = pi-pi/9 + 2 * pi * n? Is that correct? 5. Ahh the question is to answer EXACTLY in radians and I was told it isn't exactly unless you write it like that. Also the formula + n * 2 * pi I was told to add because n is the number of times it has gone around the circle. Hope I explained that decently. But I understand that I shouldn't have addad sin to the solution, thank you, as always I wasn't aware that it mattered. 6. For the second part x1 = pi/2 x2 = -pi/2 Correct? 7. Originally Posted by Smifsret Ahh the question is to answer EXACTLY in radians and I was told it isn't exactly unless you write it like that. Also the formula + n * 2 * pi I was told to add because n is the number of times it has gone around the circle. Hope I explained that decently. But I understand that I shouldn't have addad sin to the solution, thank you, as always I wasn't aware that it mattered. I'm sorry, I should have realized that your x1 and x2 were solutions to the first problem, not solutions to problems 1 and 2. In that case, yes, those solutions are correct (with x replacing your sin(x).) -Dan 8. You wrote that $x - \frac{\pi}{4} = \frac{\pi}{6}$ i think that its $1/\sqrt{2}=0.70711 = 45 degrees = \frac{\pi}{4}$ Or am I completely wrong? 9. Originally Posted by Smifsret You wrote that $x - \frac{\pi}{4} = \frac{\pi}{6}$ i think that its $1/\sqrt{2}=0.70711 = 45 degrees = \frac{\pi}{4}$ Or am I completely wrong? My apologies again. You are correct. I'm going back to bed. -Dan 10. Thank you very much for the help, does that mean that my previous answers were correct?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9744579195976257, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/136215-use-cylindrical-shells.html	Thread: 1. use cylindrical shells... use cylindrical shells to find the volume of the solid generated when the region bounded by xy = 4, x + y = 5 is revolved about the x -axis. i did the work for this problem i was wondering if someone could check my work and let me know if my setup for this problem is right? thanks in advance...IMG.pdf i cant add the second page, but my answer, the final answer for the problem was volume = (58/6)pi 2. I came up with $9\pi$ but I worked it in a hurry! 3. Is this the integral? $2\pi \int_{1}^{4} y(5y - y^2 - 4) dy$ 4. um not thats not quite what i got..one of of the functions xy = 4, so one of the functions should be divided by x or y, right? 5. I'm not sure what your asking, but using the shell method and rotating the solid about the x-axis would require the equation: $V = \int_{c}^{d} p(y)h(y) dy$ right? And if so, p(y) = y, the distance from the axis of revolution to the representaive rectangle (incremental slice) and h(y) would represent the length of the representative rectangle which would be the difference of the two given functions x = 5 - y minus x = 4/y. 6. I'm sorry, the integral that I gave still had "y" in it. That was an error, it should have been: $\int_{1}^{4}(5y - y^2 -4) dy$ 7. The work you submitted is correct: However, there may be a mistake in your algebra. Check out my solution and compare. $2\pi[\frac{5y^2}{2} - \frac{y^3}{3} - 4y]_{1}^{4}$ $2\pi[(40 - \frac{64}{3} - 16) - (\frac{5}{2} - \frac{1}{3} - 4)]$ $2\pi[(\frac{120 - 64 - 48}{3} - (\frac{15 - 2 - 24}{6})]$ $2\pi(\frac{8}{3} + \frac{11}{6})<br />$ = $2\pi(\frac{27}{6})$ = $9\pi$ 8. thanks im going to go back and check the algebra on that one, thanks again for the catch..	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9505492448806763, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/208888/every-automorphism-of-a-tree-with-an-odd-number-of-vertices-has-a-fixed-point?answertab=votes	# Every automorphism of a tree with an odd number of vertices has a fixed point If $T$ is a tree, and $T$ has an odd number of vertices, then $\forall f$, where $f$ is automorphism $\Rightarrow \exists$ fixed point (vertex). What it means: Formally, an automorphism of a tree $T$ is a permutation $\sigma$ of the vertex set $V$ of tree, such that the pair of vertices $(u, v)$ form an edge if and only if the pair $(\sigma(u),\sigma(v))$ also form an edge. If $T$ is a tree with vertices ($v1,..vn$), then after applying $f$, some vertex will be always stay in place. if somebody want to illustrate this, I will do. - I removed the automorphic-forms tag since that's something totally different. – Ted Oct 7 '12 at 20:28 I don't understand the last sentence. – joriki Oct 7 '12 at 20:36 @joriki what sentence? – jofisher Oct 7 '12 at 20:41 As I said, the last one. – joriki Oct 7 '12 at 21:01 "if somebody want to illustrate this, I will do." if you don't understand the problem above, I can illustrate this for more understanding. – jofisher Oct 7 '12 at 21:14 ## 1 Answer Call the original tree $T_0$. Remove all the leaves of $T_0$ to get a tree $T_1$. Then remove all the leaves of $T_1$ to get $T_2$. After finitely many iterations, you get a tree $T_n$ which is all leaves, and therefore consists of either a single vertex, or two vertices connected by an edge. Any automorphism $\sigma$ of the original tree $T_0$ must take every $T_i$ to itself. If $T_n$ consists of a single vertex, we are done. If $T_n$ consists of 2 vertices, and $\sigma$ fixes them, we are done. Suppose $\sigma$ reverses the two vertices of $T_n$. Then $\sigma$ fixes the edge $e$ of $T_n$. Remove the edge $e$ from the original tree $T_0$. We get a forest with 2 components, and $\sigma$ must be an automorphism of this forest which reverses the 2 components. But this is impossible, because $T_0$ has an odd number of vertices and therefore the number of vertices in the two components must be different (one must be odd and the other must be even). - 1 Very handy. Thank you! – jofisher Oct 7 '12 at 21:47 I have a small question. The automorphism transfers vertex to the vertex with the same degree (in common case). But in your case transferred the vertices only at the same level from the bottom. Why, if a vertex from ${T}_{i}$ that can not be transferred to a vertex from ${T}_{j}$..? This is the only question. – jofisher Oct 9 '12 at 12:57 1 Are you asking why $\sigma$ must map $T_i$ to itself? Because we know an automorphism must map leaves of $T_0$ to other leaves of $T_0$. Therefore, it must also map the rest of $T_0$ to the rest of $T_0$. But the rest of $T_0$ is $T_1$. Therefore $\sigma$ must map $T_1$ to $T_1$, so $\sigma$ is also an automorphism of $T_1$. Now apply this argument repeatedly to see that $\sigma$ maps every $T_i$ to itself. – Ted Oct 9 '12 at 16:29 Yep, I thought the same. But scared a little bit. Thank you. – jofisher Oct 9 '12 at 17:31	{"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": 51, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9281609058380127, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Exterior_covariant_derivative	# Exterior covariant derivative In mathematics, the exterior covariant derivative, sometimes also covariant exterior derivative, is a very useful notion for calculus on manifolds, which makes it possible to simplify formulas which use a principal connection. ## Definition Let P → M be a principal G-bundle on a smooth manifold M. If ϕ is a tensorial k-form on P, then its exterior covariant derivative is defined by $D\phi(X_0,X_1,\dots,X_k)=\mathrm{d}\phi(h(X_0),h(X_1),\dots,h(X_k))$ where h denotes the projection to the horizontal subspace, Hx defined by the connection, with kernel Vx (the vertical subspace) of the tangent bundle of the total space of the fiber bundle. Here Xi are any vector fields on P. Dϕ is a tensorial (k + 1)-form on P. ## Properties Unlike the usual exterior derivative, which squares to 0 (that is d2 = 0), we have $D^2\phi=\Omega\wedge\phi$ where Ω denotes the curvature form. In particular D2 vanishes for a flat connection. ## References • Kobayashi, Shoshichi and Nomizu, Katsumi (1996 (New edition)). . Wiley-Interscience. ISBN 0-471-15733-3. This Differential geometry related article is a stub. You can help Wikipedia by expanding it.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "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": 0.8711704015731812, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/32894-sylow-p-subgroups-print.html	# Sylow p-subgroups Printable View • April 1st 2008, 09:49 PM math_science_dude Sylow p-subgroups Hello, I'm trying to show that if H is a subgroup of a finite group G, and p is a prime that the number of sylow p-subgroups of H $n_H$ is less thanor equal to the number of sylow p-subgroups of $n_G$. This is all I have: Clearly if p does not divide \|G\| then p does not divide \|H\| and so $n_H = n_G = 0$. Let $p^j$ and $p^k$ be the greatest powers of p that divides \|H\| and \|G\| respectively. Case 1: j = k. In this case any sylow p-subgroup of H is a sylow p-subgroup of G and so clearly $n_H \le n_G$. Case 2: j < k. This is where i got stuck. I am at a point where I need to show that each sylow p-subgroup of G can contain only one sylow p-subgroup of H. I do not know if this is right path, maybe I overlooked a simple theorem. Any assistance is greatly appreciated. Thank you. m_s_d All times are GMT -8. The time now is 06:34 AM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9213710427284241, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/71750/can-we-say-for-sure-that-the-function-is-increasing-to-mean-that-the-rst-de/71754	# Can we say (for sure) that "the function is increasing” to mean that the ﬁrst derivative is positive? Can we say (for sure) that "the function is increasing” to mean that the ﬁrst derivative is positive? Whenever $f'$ (the first derivative) is positive the function is increasing,but does that imply if a function is increasing the first derivative must be positive? - 2 No. First, increasing functions need not to be everywhere differentiable (for example $f\colon \mathbb R\to \mathbb R$ given by $f(x) = x$ for $x \ge 0$ and $f(x) = 2x$ for $x \le 0$), second: If $f$ is differentiable, you only need to have $f' \ge 0$, not $f' > 0$, for example: $g(x) = x^3$ has $g'(0) =0$. – martini Oct 11 '11 at 16:29 3 Increasing functions don't even have to be continuous! – barrycarter Oct 11 '11 at 16:41 ## 3 Answers No. Consider a dense countable subset $Q$ of $\mathbb R$ and a family $(a_q)_{q\in Q}$ of positive real numbers such that $\sum\limits_{q\in Q}a_q(1+\|q\|)$ converges. For every $x$, let $(x)^+=\max\{x,0\}$ denote the positive part of $x$. Then, the function $f$ defined by $$f(x)=\sum\limits_{q\in Q}a_q(x-q)^+$$ is well defined for every real number $x$. The left and right derivatives of $f$ exist everywhere, with $$f'_\ell(x)=\sum\limits_{q<x}a_q\qquad\text{and}\qquad f'_r(x)=\sum\limits_{q\leqslant x}a_q.$$ Thus, $f$ is strictly increasing and strictly convex, differentiable at every point not in $Q$, and not differentiable at every point in $Q$. To prove the existence of $f'_\ell$ and $f'_r$ at every point, one can come back to the definitions of the left and right derivatives as the limits, if these exist, of $\pm(f(x\pm h)-f(x))/h$ when $h\to0^+$. Or one can use directly the fact that each function $g_q$ defined by $g_q(x)=(x-q)^+$ has left and right derivatives $(g_q)'_\ell(x)=[x>q]$ and $(g_q)'_r(x)=[x\geqslant q]$. - No. Consider function $f(x)=x^3$. It is increasing on $\mathbb R$ but $f'(0)=0$. - Someone mentioned $f(x)=x^3$, for which the derivative at one point is $0$ but the function is everywhere increasing. It can also happen that the derivative at one point is undefined and the function is everywhere increasing. For example, let $$f(x) = \begin{cases} x & \text{if }x<0, \\ 2x & \text{if }x\ge 0. \end{cases}$$ The derivative is undefined at $x=0$. The function is everywhere increasing. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 38, "mathjax_display_tex": 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": 0.9386005401611328, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Solution_point	# Equation (Redirected from Solution point) This article may be expanded with text translated from the corresponding article in the French Wikipedia. (February 2013) Click [show] on the right to read important instructions before translating. Google's machine translation is a useful starting point for translations, but translators must revise errors as necessary and confirm that the translation is accurate, rather than simply copy-pasting machine-translated text into the English Wikipedia. Do not translate text that appears unreliable or low-quality. If possible, verify the text with references provided in the foreign-language article. After translating, `{{Translated\|fr\|Équation}}` must be added to the talk page to ensure copyright compliance. The first use of an equals sign, equivalent to 14x+15=71 in modern notation. From The Whetstone of Witte by Robert Recorde (1557). In mathematics an equation is an expression of the shape A = B, where A and B are expressions containing one or several variables called unknowns. An equation looks like an equality, but has a very different meaning: An equality is a mathematical statement that asserts that the left-hand side and the right-hand side of the equals sign (=) are the same or represent the same mathematical object; for example 2 + 2 = 4; is an equality. On the other hand, an equation is not a statement, but a problem consisting in finding the values, called solutions, that, when substituted to the unknowns, transform the equation into an equality. For example, 2 is the unique solution of the equation x + 2 = 4, in which the unknown is x.[1] Equation may also refer to a relation between some variables that is expressed by the equality of some expressions of their values. For example the equation of the unit circle is x2 + y2 = 1, which means that a point belongs to the circle if and only if its coordinates are related by this equation. Most physical laws are expressed by equations. One of the most popular ones is Einstein's equation E=mc2. The = symbol was invented by Robert Recorde (1510–1558), who considered that nothing could be more equal than parallel straight lines with the same length. Centuries ago, the word "equation" frequently meant what we now usually call "correction" or "adjustment". This meaning is still occasionally found, especially in names which were originally given long ago. The "equation of time", for example, is a correction that must be applied to the reading of a sundial in order to obtain mean time, as would be shown by a clock. ## Parameters and unknowns See also: Expression (mathematics) Equations often contain other variables than the unknowns. These other variables that are supposed to be known are usually called constants, coefficients or parameters. Usually, the unknowns are denoted by letters at the end of the alphabet, x, y, z, w, …, while coefficients are denoted by letters at the beginning, a, b, c, d, … . For example, the general quadratic equation is usually written ax2 + bx + c = 0. The process of finding the solutions, or in case of parameters, expressing the unknowns in terms of the parameters is called solving the equation. Such expressions of the solutions in terms of the parameters are also called solutions. A system of equations is a set of simultaneous equations, usually in several unknowns, for which the common solutions are sought. Thus a solution to the system is a set of one value for each unknown, which is a solution to each equation in the system. For example, the system $\begin{align} 3x+5y&=2\\ 5x+8y&=3 \end{align}$ has the unique solution x = -1, y = 1. ## Analogous illustration Illustration of a simple equation; x, y, z are real numbers, analogous to weights. The analogy often presented is a weighing scale, balance, seesaw, or the like. Each side of the balance corresponds to each side of the equation. Different quantities can be placed on each side, if they are equal the balance corresponds to an equality (equation), if not then an inequality. In the illustration, x, y and z are all different quantities (in this case real numbers) represented as circular weights, each of x, y, z has a different weight. Addition corresponds to adding weight, subtraction corresponds to removing weight from what is already placed on. The total weight on each side is the same. ## Types of equations Equations can be classified according to the types of operations and quantities involved. Important types include: • An algebraic equation or polynomial equation is an equation in which both sides are polynomials (see also system of polynomial equations). These are further classified by degree: • linear equation for degree one • quadratic equation for degree two • cubic equation for degree three • quartic equation for degree four • quintic equation for degree five • A Diophantine equation is an equation where the unknowns are required to be integers • A transcendental equation is an equation involving a transcendental function of its unknowns • A parametric equation is an equation for which the solutions are sought as functions of some other variables, called parameters appearing in the equations • A functional equation is an equation in which the unknowns are functions rather than simple quantities • A differential equation is a functional equation involving derivatives of the unknown functions • An integral equation is a functional equation involving the antiderivatives of the unknown functions • An integro-differential equation is a functional equation involving both the derivatives and the antiderivatives of the unknown functions • A difference equation is an equation where the unknown is a function f which occurs in the equation through f(x), f(x-1), ..., f(x-k), for some whole integer called the order of the equation. If x is restricted to be an integer, a difference equation is the same as a recurrence relation ## Identities Main articles: Identity (mathematics) and List of trigonometric identities An identity is a statement resembling an equation which is true for all possible values of the variable(s) it contains. Many identities are known, especially in trigonometry. Probably the best known example is: $\sin^2(\theta)+\cos^2(\theta)=1,$, which is true for all values of θ. In the process of solving an equation, it is often useful to combine it with an identity to produce an equation which is more easily soluble. For example, to solve the equation: $3\sin(\theta) \cos(\theta)= 1,$ where θ is known to be between zero and 45 degrees, use the identity: $\sin(2 \theta)=2\sin(\theta) \cos(\theta),$ so the above equation becomes: $\frac{3}{2}\sin(2 \theta) = 1$ Whence: $\theta = \frac{1}{2} \arcsin\left(\frac{2}{3}\right),$ which comes to about 20.9 degrees. ## Properties Two equations or two systems of equations are equivalent if they have the same set of solutions. The following operations transform an equation or a system into an equivalent one: • Adding or subtracting the same quantity to both sides of an equation. This shows that every equation is equivalent to an equation in which the right-hand side is zero. • Multiplying or dividing both sides of an equation by a non-zero constant. • Applying an identity to transform one side of the equation. For example, expanding a product or factoring a sum. • For a systems: adding to both sides of an equation the corresponding side of another, equation multiplied by the same quantity. If some functions is be applied to both sides of an equation, the resulting equation has the solutions of the initial equation among its solutions, but may have further solutions called extraneous solutions. If the function is not defined everywhere, (like 1/x that is not defined for x=0) some solutions may be lost. Thus caution must be exercised when applying such a transformation to an equation. For example, the equation $x=1$ has the solution $x=1.$ Raising both sides to the exponent of 2 (which means, applying the function $f(s)=s^2$ to both sides of the equation) changes our equation into $x^2=1$, which not only has the previous solution but also introduces the extraneous solution, $x=-1.$ Above transformations are the basis of most elementary methods for equation solving and some less elementary ones, like Gaussian elimination For more details on this topic, see Equation solving. ## References 1. "Equation". Dictionary.com. Dictionary.com, LLC. Retrieved 2009-11-24.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 11, "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": 0.9384692907333374, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/31321/testing-periodogram-peaks-sine-like-wave-or-ar-ma-arma-noise	# Testing periodogram “peaks”: sine-like wave or AR/MA/ARMA noise? I'm performing an harmonic fit to data I know (from physical constraints) come from a periodic source of the form $$\sum_j^M \sum_i^N a_{i,j}\sin(2\pi f_it)+b_{i,j}\cos(2\pi f_it)$$ using the Lomb-Scargle periodogram (I know this is not the optimal thing to do, but it's fast and reliable so far), where I'm also trying to estimate $N$ and $M$, the number of frequencies and the number of harmonics per frequency to fit. The number of harmonics is really not a very hard problem to solve once the number of frequencies is fixed, because I can perform methods like Ridge Regression, LASSO, etc. in order to find the best subset of harmonics, so my problem is really on estimating the optimal number of frequencies. The thing is that I know red noise is almost always present, and certain harmonic fits give AR-like spectra on the residuals (and, at least to me, it seems like I could even have ARMA-like spectra). The "classical" Lomb-Scargle periodogram, on the other hand, assumes white noise in order to test the significance of the peaks, so the significance test on the peaks doesn't seem like a good idea in order to test if there are any residual frequencies on the periodogram. I've been trying to derive significance tests assuming ARMA-like noise, but the problem of defining the order $p,q$ of the process arises: do you know of a way of differencing between correlated noise and sine wave-like spectra? This paper shows a way of doing this assuming AR(1) noise (in fact something more like a CAR-type noise?), but the problem of it is that it actually knows there is red noise. I think I can "see" how to do it: sine wave-like spectra has almost-symmetric decreasing peaks around the main peak because of the (uneven) sampling, which ARMA-like noise doesn't have (the amplitudes are random with an apparent upper limit dictated by the form of the spectra). I've also been thinking in performing fits with ARMA and sine-like waves, and perform model selection via AIC...any suggestions? - ## 1 Answer A general test for periodicity is Fisher's periodogram test. I assume you are familar with it. But in case you aren't here are some links to recent papers related to it. http://users.ics.aalto.fi/harrila/publications/RobFisherPeriodicity.pdf	{"extraction_info": {"found_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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9421370029449463, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/74311/list	## Return to Answer 2 added 257 characters in body Especially if your interested in dynamical systems, I highly recommend Abraham--Marsden--Ratiu, Manifolds, tensor analysis, and applications. For a more Riemannian-geometric/global-analytic focus, you might want to try Klingenberg, Riemannian Geometry, or Lang, Differentiable Manifolds. There is a standard way to construct a canonical topology on $C^r(M,N)$ for $M$ compact, one that turns $C^r(M, N)$ into a Banach manifold. But I don't think there is a canonical metric on $C^r(M, N)$ unless you put some additional structure on $N$. 1 Especially if your interested in dynamical systems, I highly recommend Abraham--Marsden--Ratiu, Manifolds, tensor analysis, and applications. For a more Riemannian-geometric/global-analytic focus, you might want to try Klingenberg, Riemannian Geometry, or Lang, Differentiable Manifolds.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.3, "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": 0.8721901178359985, "perplexity_flag": "middle"}
http://computationalblackbody.wordpress.com/	# Model of Atmosphere with CO2 shows Small Emissivity Posted on February 10, 2013 by Here is a another argument indicating that the effect of the atmospheric trace gas CO2 on the radiation balance of the Earth is small. Our model of blackbody radiation consists of collection of oscillators with small damping with equal oscillator internal energy T representing temperature, with oscillator resonance frequencies n varying from 1 to a cut-off set at T and each oscillator radiating • $E_n = \gamma T n^2$ where $\gamma$ is a universal constant, which is Planck’s Law. Summing over n from 1 to T, we obtain the total radiance • $E =\sum_n \gamma T n^2 = \sigma T^4$ which is Stefan-Boltzmann’s law with $\sigma = \gamma /3$. In the case of only one resonance frequency $n = T$, the radiance would be reduced to • $e = \gamma T T^2 =\gamma T^3 = 3 E/T$ with the reduction factor $1/T$. The radiance of an atmosphere which is fully opaque over the entire spectrum would radiate $E$, while an atmosphere opaque only for a specific frequency near cut-off $T$, would radiate $e\approx 3E/T$ with a reduction factor $1/T$. We conclude that the emissivity of transparent atmosphere with a trace gas like CO2 with only a few isolated resonances, would scale like 1/T and thus be small as soon as T is bigger than say 100 K. We thus find theoretical evidence from a basic model that the emissivity of the Earth’s atmosphere with the trace gas CO2 would be small, and thus that CO2 would have little effect on the Earth’s radiation balance. We thus find theoretical evidence from a basic model that the emissivity of the Earth’s atmosphere with the trace gas CO2 would be small, and thus that CO2 would have little effect on the Earth’s radiation balance. Posted in Uncategorized \| # Blackbody as Transformer of Radiation Posted on September 9, 2012 by Our blackbody model, which has a high frequency cut-off scaling with temperature, absorbs all incoming radiation (forcing), re-emits frequencies below cut-off and stores frequencies above cut-off as internal energy, which is assumed to be equally distributed over frequencies below cut-off. The result is that a blackbody subject to forcing with frequencies above cut-off, will heat up with a corresponding increase of cut-off until the highest frequency of the forcing is reached, assuming the blackbody is not somehow cooled. The effect is that the blackbody acts as a transformer of radiation from high to low frequencies by absorbing frequencies above current cut-off while emitting frequencies below cut-off. The Earth subject to forcing from the Sun acts as such a transformer, see Blackbody: Transformer of Radiation. Posted in Uncategorized \| # Kirchhoff’s Law 2 Posted on September 8, 2012 by Kirchhoff’s Law states that for a blackbody emissivity = absorptivity. In our wave equation with small damping as a model of blackbody in radiative equilibrium with an exterior forcing, this is built into the model for frequencies below high-frequency cut-off. This is because for these frequencies the outgoing radiation simply reproduces the part of the incoming radiation which is not reflected. Or put differently: In radiative equilibrium there is below cut-off no distinction between incoming radiation and outgoing radiation as explained in the previous post. However, if the incoming waves (the forcing) contain frequencies above the present cut-off of the absorbing body, then the game changes, as these frequencies contribute to the internal energy of the body increasing its temperature and are not (re-)emitted. This is a case with positive absorptivity and zero emissivity of the frequencies above cut-off. This leads to coefficients of absorptivity depending on temperature, which complicates the picture. The effect is that different bodies with different cut-off at the same distance to the Sun, can take on different equilibrium temperatures, as evidenced in e.g. Satellite Thermal Control Engineering and What Is the Temperature of Space? Posted in absorptivity, emissivity, high frequency cut-off \| # String and Soundboard Posted on September 8, 2012 by Our model of blackbody radiation has an acoustic analog in the form of a string interacting with a soundboard, with the following key components (as total or for each frequency in a spectral decomposition): • internal string energy • string amplitude • soundboard amplitude • force balance between string and soundboard = wave equation • small damping in the interaction expressed by the wave equation. The small damping is an important feature of the model since it creates an “optimal” interaction between the string and the soundboard in the sense that the tone generated by the soundboard will be both loud and have long sustain. With zero damping the interaction will small and the tone weak, and with too big damping the sustain will be short. In radiation the damping is small. The effect of the small damping is that the string velocity is out-of-phase with the forcing, which is the key to optimal interaction between string and soundboard. Radiative equilibrium corresponds to acoustic equilibrium, which can be interpreted in two different ways: 1. The vibration of the soundboard is sustained by the vibration of the string = outgoing acoustic waves. 2. The vibration of the string is sustained by the vibration of the soundboard = incoming acoustic waves. The same equilibrium can thus be interpreted as incoming or outgoing, which can be expressed as absorptivity = emissivity, which thus is built into the model. This argument applies to frequencies below cut-off with only radiative damping being active. For frequencies above cut-off the interaction between string and soundboard is different with internal heating of the string, and emissivity in general different from absorptivity. There is also non-equilibrium dynamics below cut-off with e.g. the string energy being transferred to the soundboard as the string amplitude decreases. Posted in acoustics, resonance \| # Equidistribution 1 Posted on September 7, 2012 by Equidistribution in our wave model of blackbody radiation means that the internal energy of different frequencies is the same, that is, all frequencies have the same temperature. More specifically, the internal energy $IE_\nu$ of the wave component of frequency $\nu$ with amplitude $U_\nu$ is measured by (with subindices indicating differentiation with respect to time $t$ and space $x$ and integration in space and time over a period) • $IE_\nu = \frac{1}{2}\int (U_{\nu ,t}^2+U_{\nu ,x}^2)dxdt$. Equidistribution means that $IE_\nu$ is the same for all frequencies and then defines a common temperature $T =IE_\nu$. In the piano model equidistribution corresponds to pressing all the keys with the same force to give all strings of the piano the same string energy. Planck’s Law $E(\nu ,T) =\gamma T\nu^2$ assumes equidistribution with all frequencies having the same internal energy measured by temperature $T$. With equidistribution as a characteristic of blackbody radiation the radiation intensity spectrum shows the characteristic quadratic increase with frequency. Without equidistribution the spectrum can show different dependence on frequency. Posted in equidistribution, temperature \| # Kirchhoff’s Law 1 Posted on September 7, 2012 by Kirchhoff’s Law of Thermal Radiation: 150 Years starts off by: • Kirchhoff’s law is one of the simplest and most misunderstood in thermodynamics. Let us see what we can say about Kirchhoff’s Radiation Law stating that the emissivity and absorptivity of a radiating body are equal, in the setting of the wave model with damping presented in Computational Blackbody Radiation and Mathematical Physics of Blackbody Radiation: • $U_{tt} - U_{xx} - \gamma U_{ttt} - \delta^2U_{xxt} = f,$ where the subindices indicate differentiation with respect to space $x$ and time $t$, and 1. $U_{tt} - U_{xx}$ models a vibrating material string with \$U\$ displacement 2. $- \gamma U_{ttt}$ is a dissipative term modeling outgoing radiation 3. $- \delta^2U_{xxt}$ is a dissipative term modeling internal heating by friction 4. $f$ is the amplitude of the incoming forcing, 5. $T$ is temperature with $T^2=\int\frac{1}{2}(\vert U_t^2+U_x^2)dxdt$, 6. the wave equation expresses a balance of forces, where $\gamma$ and $\delta^2$ are certain small damping coefficients defined by spectral decomposition as follows in a model case: • $\gamma = 0$ if the frequency $\nu > \frac{1}{\delta}$ • $\delta = 0$ if the frequency $\nu < \frac{1}{\delta}$, where $\delta = \frac{h}{T}$ represents a “smallest coordination length” depending on temperature $T$ and $h$ is a fixed smallest mesh size (representing some atomic dimension). This represents a switch from outgoing radiation to internal heating as the frequency $\nu$ passes the threshold $\frac{T}{h}$, with the threshold increasing linearly with $T$. The idea is that a hotter vibrating string is capable of radiating higher frequencies as coherent outgoing radiation. The switch acts as a band filter with frequencies outside the band being stored as internal heat instead of being radiated: The radiator is then muted and heats up internally instead of delivering outgoing radiating. A spectral analysis, assuming that all frequencies share a common temperature, shows an energy balance between incoming forcing $f$ measured as • $F = \int f^2(x,t)\, dxdt$ assuming periodicity in space and time and integrating over periods, and (rate of) outgoing radiation \$R\$ measured by • $R = \int \gamma U_{tt}^2\, dxdt$, and (rate of) internal energy measured by • $IE = \int \delta^2U_{xt}^2\, dxdt$, together with the oscillator energy • $OE =T^2 = \frac{1}{2}\int (U_t^2 + U_x^2)\, dxdt$ with the energy balance in stationary state with $OE$ constant taking the form • $F = \kappa (R + IE)$ with $\kappa\lessapprox 1$ is a constant independent of $T$, $\gamma$, $\delta$ and $\nu$. In other words, • incoming energy = $\kappa\times$ outgoing radiation energy for $\nu <\frac{1}{\delta}$ • incoming energy = $\kappa\times$ stored internal energy for $\nu >\frac{1}{\delta}$, which can be viewed as an expression of Kirchhoffs’ law that emissivity equals absorptivity. The equality results from the independence of the coefficient $\kappa$ of the damping coefficients $\gamma$ and $\delta^2$, and frequency. Summary: The energy of damping from outgoing radiation or internal heating is the same even if the damping terms represent different physics (emission and absorption) and have different coefficients ($\gamma$ and $\delta^2$). PS: Note that internal heat energy accumulating under (high-frequency) forcing above cut-off eventually will be transformed into low-frequency outgoing radiation, but this transformation is not part of the above model. \| # Helmholtz Reciprocity 1 Posted on September 7, 2012 by Helmholtz Reciprocity states that certain phenomena of propagation of light and interaction between light and matter, are reversible. Let us study reversibility in our model of blackbody radiation: • $U_{tt} - U_{xx} - \gamma U_{ttt} - \delta^2U_{xxt} = f$ where the subindices indicate differentiation with respect to space $x$ and time $t$, and 1. $U_{tt} - U_{xx}$ represents a vibrating string with U displacement 2. $- \gamma U_{ttt}$ is a dissipative term modeling outgoing radiation = emission 3. $- \delta^2U_{xxt}$ is a dissipative modeling internal heating = absorption 4. $f$ is incoming forcing/microwaves, where $\gamma$ and $\delta^2$ are positive constants connected to dissipative losses as outgoing radiation = emission and internal heating = absorption. We see emission represented by $-\gamma U_{ttt}$ and absorption by $-\delta^2U_{xxt}$. We now ask: 1. How is the distinction between emission and absorption expressed in this model? 2. Is Helmholtz Reciprocity valid (emission and absorption are reverse processes)? 3. Is Kirchhoff’s Radiation Law (emissivity = absorptivity) valid? Before seeking answers let us recall the basic energy balance between incoming forcing $f$ measured as • $F = \int f^2(x,t)\, dxdt$ assuming periodicity in space and time and integrating over periods, and (rate of) outgoing radiation = emission \$R\$ measured by • $R = \int \gamma U_{tt}^2\, dxdt$, the oscillator energy $OE$ measured by • $OE =\frac{1}{2}\int (U_t^2 + U_x^2)\, dxdt$ and (rate of) internal energy = absorption measured by • $IE = \int \delta^2U_{xt}^2\, dxdt$ • $F = R + IE$ • incoming energy = emission + absorption. The model has a frequency switch switching from emission to absorption as the frequency increases beyond a certain threshold proportional to temperature in accordance with Wien’s displacement law. We now return to questions 1 – 3: Both terms generate dissipative effects when multiplied with $U_t$ as $R = \int \gamma U_{tt}^2\, dxdt$ and $IE = \int \delta^2U_{xt}^2\, dxdt$, but the terms involve different derivatives with $U_{tt}$ acting only in time and $U_{xt}$ acting also in space. The absorption $U_{xt}^2$ represents a smoothing effect in space, which is irreversible and thus cannot be reversed into emission as reversed absorption. The emission $U_{tt}^2$ represents a smoothing effect in time, which is irreversible and thus cannot be reversed into absorption as reversed emission. In other words, in the model both absorption and emission are time irreversible and thus cannot be reversed into each other. We conclude that the model does not satisfy Helmholtz reciprocity. Nevertheless, the model satisfies Kirchhoff’s law as shown in a previous post. Conclusion: The space derivative in $U_{xt}$ models absorption as process of smoothing in space with irreversible transformation of high frequencies in space into low frequencies with a corresponding increase of internal energy as heat energy. Absorbed high frequencies can with increasing temperature be rebuilt through (resonance in) the wave equation into high frequency emission. Absorption and emission are not reverse processes, but my be transformed into each other through (resonance in) the wave equation and the switch. We may compare absorption with a catabolic process of destroying (space-time) structure and emission with an anabolic process of building structure, with the wave equation as a transformer. For more details see Mathematical Physics of Blackbody Radiation and The Clock and the Arrow. 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http://math.stackexchange.com/questions/90340/proving-p-to-q-to-r-to-p-to-q-to-p-to-r	# Proving $(p \to (q \to r)) \to ((p \to q) \to (p \to r))$ I'm looking for a way to prove $$(p \to (q \to r)) \to ((p \to q) \to (p \to r))$$ from the axioms $$\begin{align} & p \to (q \to p) \\ & (p \to (p \to q)) \to (p \to q) \\ & (p \to q) \to ((q \to r) \to (p \to r)) \\ & (\sim p \to \, \sim q) \to (q \to p) \\ \end{align}$$ using universal substitution and modus ponens I suspect the fourth axiom is not necessary for the proof. I have been working in Tarski's Introduction to Logic and am trying to establish the equivalence of his axiom system with the axioms used at us.metamath.org $$\begin{align} & p \to (q \to p) \\ & (p \to (q \to r)) \to ((p \to q) \to (p \to r)) \\ & ( \sim p \to \, \sim q) \to (q \to p) \\ \end{align}$$ which will allow me to connect Tarski's 4 axioms with all sorts of interesting proofs on that site. - 1 Do you have the deduction theorem available for your system? – Henning Makholm Dec 11 '11 at 2:32 @toph: I agree that the fourth axiom is not relevant here. The proposition you are trying to prove is essentially an internal form of modus ponens, which is valid intuitionitically, but the fourth axiom is essentially double negation elimination. – Zhen Lin Dec 11 '11 at 2:55 This appears to be hard -- are you sure the four axioms you've given are all there is? The first three ones are the K, W, and (almost) B axioms of the BCKW system of combinatory logic, but the C axiom is missing. That makes me doubt that the system you present is complete. – Henning Makholm Dec 11 '11 at 4:45 2 The axioms are taken from page 147 of Tarski's Introduction to Logic. I must (guiltily) add that there are three other axioms: (p <-> q) -> (p -> q), (p <-> q) -> (q -> p), (p -> q) -> ((q -> p) -> (p <-> q). I couldn't see how these axioms, which basically define equivalence, would be necessary. But following your remarks I suspect the last axiom above in particular might be crucial. Thanks for your help so far! – toph Dec 11 '11 at 13:11 3 Aha! Wikipedia's List of logic systems lists your three (original) axioms as Hilbert's second system for "Postive implicational calculus" (i.e., the implicational fragment of intuitionistic logic). Since your target sentence is intuitionistically valid, it indeed ought to be derivable. Still beats me how, though. – Henning Makholm Dec 11 '11 at 15:23 show 7 more comments ## 4 Answers As you suspected, this can be proved from the first three axioms only. I couldn't find a short proof, though – I tried brute force enumeration of the theorems deducible from the three axioms (by taking all pairs of theorems already proved and unifying one with the premise of the other), but didn't find your target in the first $80000$ theorems proved. I then found some guidance in the article on relevance logic in the Handbook of Philosophical Logic. Relevance logic focuses on the fragment of logic in which, roughly speaking, the premises are relevant to the conclusions. It doesn't include the axiom $p\to(q\to p)$, which allows us to add an irrelevant premise to a theorem already proved without that premise, and is thus strictly weaker than the system you're using, but we can nevertheless make use of the results cited in that article. I'll first describe the structure of the proof and how I found it, and then give the proof in detail. Here are the names I'll use for the axioms; the first column names the corresponding axioms of combinator logic, for comparison with the discussion in the comments under the question: $$\begin{array}{c\|l\|l} \mathbf I&\text{self-implication}&p\to p\\ \mathbf K&\text{weakening}& p \to (q \to p) \\ \hline \mathbf B&\text{prefixing}& (p \to q) \to ((r \to p) \to (r \to q)) \\ \mathbf A&\text{suffixing}& (p \to q) \to ((q \to r) \to (p \to r)) \\ \hline \mathbf W&\text{contraction}& (p \to (p \to q)) \to (p \to q) \\ \mathbf S&\text{self-distribution}&(p \to (q \to r)) \to ((p \to q) \to (p \to r))\\ \hline \mathbf C&\text{permutation}&(p\to(q\to r))\to(q\to(p\to r))\\ &\text{assertion}&p\to((p\to q)\to q) \end{array}$$ (The names are the ones used in the article, except I use "weakening" instead of "positive paradox", since it's shorter and makes more sense to me.) Theorem $1$ of the article states that, with modus ponens (and implicitly universal substitution), the axiom sets formed by self-implication and one each from the three pairs prefixing/suffixing, contraction/self-distribution and permutation/assertion lead to the same theory. What you have is weakening, suffixing and contraction. Self-implication can be deduced from weakening and contraction in a single step (by substituting $p$ for $q$ everywhere). Thus, if we can deduce assertion in your system, the theorem will tell us that we can deduce everything else, including your target, self-distribution. I did find a proof for assertion by brute force search. The article doesn't give a proof of its Theorem $1$ and only says that it can be proved by consulting a book that isn't available online and doing some "fiddling", so we still have to show how to get from self-implication, suffixing, contraction and assertion to self-distribution. I found a deduction of self-distribution online that uses prefixing and permutation. It turns out that prefixing is deducible in a single step from suffixing and permutation, so the problem remained only to deduce permutation. Again, I found a proof for this by brute force search. So here's the entire proof put together, starting with your three axioms and ending with your target. First, a high-level description similar to the actual calls in my Java code: ````assertion = t (t (weakening,suffixing),contraction); permutation = t (suffixing,m (assertion,suffixing)); prefixing = m (suffixing,permutation); target = t (m (prefixing,prefixing),t (permutation,m (contraction,prefixing))); ```` Each call to `m` is an application of modus ponens, in which the first argument is $A$, the second argument is $A\to B$ and the most general unifier that makes the $A$s coincide is used. Each call to `t` is an invocation of transitivity (i.e. deducing $A\to C$ from $A\to B$ and $B\to C$), which can be implemented as ````t (A->B,B->C) = m (B->C,m (A->B,suffixing)) ```` using suffixing, or as ````t (A->B,B->C) = m (A->B,m (B->C,prefixing)) ```` once prefixing is available. Here's the proof spelled out in $14$ steps. The first table shows the theorems used to generate the antecedents $A$ and the implications $A\to B$ for modus ponens, as well as the resulting theorems $B$: $$\begin{array}{c\|c\|c\|c\|c} &&A&A\to B&B\\\hline \text{(a)}&\text{weakening}&&&p \to (q \to p)\\ \text{(b)}&\text{suffixing}&&&(p \to q) \to ((q \to r) \to (p \to r))\\ \text{(c)}&\text{contraction}&&&(p \to (p \to q)) \to (p \to q)\\ \hline \text{(d)}&\text{}&\text{(a)}&\text{(b)}&((p \to q) \to r) \to (q \to r)\\ \text{(e)}&&\text{(b)}&\text{(d)}&p \to ((p \to q) \to (r \to q))\\ \text{(f)}&\text{}&\text{(e)}&\text{(b)}&(((p \to q) \to (r \to q)) \to s) \to (p \to s)\\ \text{(g)}&\text{assertion}&\text{(c)}&\text{(f)}&p \to ((p \to q) \to q)\\ \text{(h)}&&\text{(g)}&\text{(b)}&(((p \to q) \to q) \to r) \to (p \to r)\\ \text{(i)}&\text{}&\text{(b)}&\text{(b)}&(((p \to q) \to (r \to q)) \to s) \to ((r \to p) \to s)\\ \text{(j)}&\text{permutation}&\text{(h)}&\text{(i)}&(p \to (q \to r)) \to (q \to (p \to r))\\ \text{(k)}&\text{prefixing}&\text{(b)}&\text{(j)}&(p \to q) \to ((r \to p) \to (r \to q))\\ \text{(l)}&&\text{(k)}&\text{(k)}&(p \to (q \to r)) \to (p \to ((s \to q) \to (s \to r)))\\ \text{(m)}&&\text{(c)}&\text{(k)}&(p \to (q \to (q \to r))) \to (p \to (q \to r))\\ \text{(n)}&\text{}&\text{(j)}&\text{(b)}&((p \to (q \to r)) \to s) \to ((q \to (p \to r)) \to s)\\ \text{(o)}&&\text{(m)}&\text{(n)}&(p \to (q \to (p \to r))) \to (q \to (p \to r))\\ \text{(p)}&\text{*}&\text{(l)}&\text{(b)}&((p \to ((q \to r) \to (q \to s))) \to t) \to ((p \to (r \to s)) \to t)\\ \text{(q)}&\text{self-distribution}&\text{(o)}&\text{(p)}&(p \to (q \to r)) \to ((p \to q) \to (p \to r))\\ \end{array}$$ The asterisks mark intermediate steps in invocations of transitivity. Note that most theorems with more than three variables occur only in such intermediate steps. Substitutions are being performed as late as possible; by performing them as early as possible, the proof could be written using only theorems with at most three variables. The second table shows the substitutions used; you can also find these automatically by unification. The variables are named such that they appear in alphabetical order in the resulting theorems. $$\begin{array}{c\|l\|l} &A&A\to B\\\hline \text{(d)}& p\mapsto q,q\mapsto p& p\mapsto q,q\mapsto (p \to q),r\mapsto r\\ \text{(e)}& p\mapsto r,q\mapsto p,r\mapsto q& p\mapsto r,q\mapsto p,r\mapsto ((p \to q) \to (r \to q))\\ \text{(f)}& p\mapsto p,q\mapsto q,r\mapsto r& p\mapsto p,q\mapsto ((p \to q) \to (r \to q)),r\mapsto s\\ \text{(g)}& p\mapsto (p \to q),q\mapsto q& p\mapsto p,q\mapsto q,r\mapsto (p \to q),s\mapsto ((p \to q) \to q)\\ \text{(h)}& p\mapsto p,q\mapsto q& p\mapsto p,q\mapsto ((p \to q) \to q),r\mapsto r\\ \text{(i)}& p\mapsto r,q\mapsto p,r\mapsto q& p\mapsto (r \to p),q\mapsto ((p \to q) \to (r \to q)),r\mapsto s\\ \text{(j)}& p\mapsto q,q\mapsto r,r\mapsto (p \to r)& p\mapsto (q \to r),q\mapsto r,r\mapsto p,s\mapsto (q \to (p \to r))\\ \text{(k)}& p\mapsto r,q\mapsto p,r\mapsto q& p\mapsto (r \to p),q\mapsto (p \to q),r\mapsto (r \to q)\\ \text{(l)}& p\mapsto q,q\mapsto r,r\mapsto s& p\mapsto (q \to r),q\mapsto ((s \to q) \to (s \to r)),r\mapsto p\\ \text{(m)}& p\mapsto q,q\mapsto r& p\mapsto (q \to (q \to r)),q\mapsto (q \to r),r\mapsto p\\ \text{(n)}& p\mapsto q,q\mapsto p,r\mapsto r& p\mapsto (q \to (p \to r)),q\mapsto (p \to (q \to r)),r\mapsto s\\ \text{(o)}& p\mapsto q,q\mapsto p,r\mapsto r& p\mapsto q,q\mapsto p,r\mapsto (p \to r),s\mapsto (q \to (p \to r))\\ \text{(p)}& p\mapsto p,q\mapsto r,r\mapsto s,s\mapsto q& p\mapsto (p \to (r \to s)),q\mapsto (p \to ((q \to r) \to (q \to s))),r\mapsto t\\ \text{(q)}& p\mapsto p,q\mapsto (p \to q),r\mapsto r& p\mapsto p,q\mapsto p,r\mapsto q,s\mapsto r,t\mapsto ((p \to q) \to (p \to r))\\ \end{array}$$ - 1 Well done! ${}$ – Henning Makholm Dec 21 '11 at 13:48 Would you mind if I add a column to your first table showing the corresponding combinator names for comparison with the discussion between me and Zhen Lin in the comments? – Henning Makholm Dec 21 '11 at 13:53 @Henning: That would be great, thanks! – joriki Dec 21 '11 at 13:55 @Henning: Interesting -- so the answer actually fits well with your discussion, in that the hard part that I did by computer search was to deduce $\mathbf C$/permutation (up to step (j)), and the rest was known. – joriki Dec 21 '11 at 14:29 It seems to me that a much simpler and human readable proof is possible, unless I’m misunderstanding something. Using the Deduction Theorem, the result is relatively straightforward to prove. This motivated me to prove the Deduction Theorem for this logical system, which I found to be less straightforward, but still not particularly difficult. To make sure we’re all on the same page, the logical system in question consists of the inference rule modus ponens (MP) and the following three axiom schema: axiom 1 $\;\;\;\;\; p \; \rightarrow \; (q \rightarrow p)$ axiom 2 $\;\;\;\;\;(p \rightarrow q) \;\; \rightarrow \;\; [\; (q \rightarrow r) \rightarrow (p \rightarrow r) \; ]$ axiom 3 $\;\;\;\;\;[\; p \rightarrow (p \rightarrow q) \; ] \;\; \rightarrow \;\; (p \rightarrow q)$ We want to show that the following wff (well formed formula) is provable in this logical system: $$[p \rightarrow (q \rightarrow r)] \;\; \rightarrow \;\; [(p \rightarrow q) \rightarrow (p \rightarrow r)]$$ By 3 applications of the Deduction Theorem (proved further below), it suffices to prove $r$ under the following assumptions: $p \rightarrow (q \rightarrow r),$ $p \rightarrow q,$ and $p.$ That is, it suffices to prove $$p \rightarrow (q \rightarrow r), \; p \rightarrow q, \; p \;\vdash \; r$$ (1) $\;\;\;p \rightarrow q$ (2) $\;\;\;$(line 1) $\rightarrow [\;(q \rightarrow r) \rightarrow (p \rightarrow r)\;]$ (3) $\;\;\;(q \rightarrow r) \rightarrow (p \rightarrow r)$ (4) $\;\;\;p \rightarrow (q \rightarrow r)$ (5) $\;\;\;p$ (6) $\;\;\;q \rightarrow r$ (7) $\;\;\;p \rightarrow r$ (8) $\;\;\;r$ Reasons for the above steps (1) $\;\;\;$assumption (2) $\;\;\;$axiom 2 (3) $\;\;\;$MP (lines 1, 2) (4) $\;\;\;$assumption (5) $\;\;\;$assumption (6) $\;\;\;$MP (lines 5, 4) (7) $\;\;\;$MP (lines 6, 3) (8) $\;\;\;$MP (lines 5, 7) In trying to prove the Deduction Theorem for this logical system (i.e. $\Gamma, \;p \vdash q$ implies $\Gamma \vdash p \rightarrow q$), I simply followed the standard proof (which makes use of axiom 1 and the wff we originally wanted to prove), and noted that the standard proof only requires us to make use of the following 3 results: 1. If $q$ is an axiom or a member of $\Gamma$, then for any wff $p$ we can prove $p \rightarrow q$ in our logical system. 2. We can prove $p \rightarrow p$ in our logical system. 3. Given $p \rightarrow r$ and $p \rightarrow (r \rightarrow q)$, we can prove $p \rightarrow q$ in our logical system. proof of 1: $\;\;\;$Apply MP to $q$ and $q \rightarrow (p \rightarrow q)$ (axiom 1). proof of 2: $\;\;\;$Apply MP to $p \rightarrow (p \rightarrow p)$ (axiom 1) and axiom 3. proof of 3: $\;\;\;$This is the difficult part. Below is a proof of what’s needed, namely $$p \rightarrow r, \; p \rightarrow (r \rightarrow q) \;\vdash \; p \rightarrow q$$ (1) $\;\;\;p \rightarrow r$ (2) $\;\;\;p \rightarrow (r \rightarrow q)$ (3) $\;\;\;$(line 1) $\rightarrow \; [(r \rightarrow q) \rightarrow (p \rightarrow q)]$ (4) $\;\;\;(r \rightarrow q) \rightarrow (p \rightarrow q)$ (5) $\;\;\;$(line 2) $\;\;\rightarrow \;\; \{\;$(line 4)$\rightarrow [p \rightarrow (p \rightarrow q)] \; \}$ (6) $\;\;\;$(line 4) $\;\rightarrow \; [p \rightarrow (p \rightarrow q)]$ (7) $\;\;\;p \rightarrow (p \rightarrow q)$ (8) $\;\;\;[p \rightarrow (p \rightarrow q)] \; \rightarrow \; (p \rightarrow q)$ (9) $\;\;\;p \rightarrow q$ Reasons for the above steps (1) $\;\;\;$assumption (2) $\;\;\;$assumption (3) $\;\;\;$axiom 2 ($r$ is $q$) (4) $\;\;\;$MP (lines 1, 3) (5) $\;\;\;$axiom 2 ($r$ is $p \rightarrow q$) (6) $\;\;\;$MP (lines 2, 5) (7) $\;\;\;$MP (lines 4, 6) (8) $\;\;\;$axiom 3 (9) $\;\;\;$MP (lines 7, 8) Here is how I discovered the above proof. Working backwards, I noticed that the conclusion of axiom 3 was what I wanted, so I made note of the fact that it would be enough to obtain $p \rightarrow (p \rightarrow q).$ Then I tried working forward. First, I applied axiom 2 followed by MP to the assumption $p \rightarrow r,$ using $q$ as the introduced suffix. (Since I already had $p$ and $r$ appearing, this seemed to be a natural way to get $q$ to appear.) Then I tried applying axiom 2 followed by MP to the assumption $p \rightarrow (r \rightarrow q).$ At some point (perhaps my 3rd attempt), I used $p \rightarrow q$ as the introduced suffix, motivated by the fact that this got line 4 to show up. After this, the proof immediately fell into place, since in line 5 the conclusion of the conclusion is $p \rightarrow (p \rightarrow q),$ which I had previouly noted was sufficient. Incidentally, the logical system above is the same (in the sense of having the same set of provable wffs) as the logical system with the inference rule MP and the following two axioms: axiom 1 and the wff we originally wanted to prove. Each of these logical systems is also equal to the logical system with MP and Deductive Theorem as inference rules and no axioms (thus, one might call this system “DT Logic”). I think logicians call this the positive implicational fragment of intuitionistic propositional logic, but I like “DT Logic” better. Other axiomatizations of DT Logic can be found at the Wikipedia page “List of logic systems” under the category “Positive implicational calculus”. For completeness, here’s a proof that DT Logic can be characterized by no axioms along with the inference rules MP and DT (and also the Rule of Assumptions, I suppose). It suffices to prove, in this no-axiom logical system, axiom 1 and the wff we were proving in this thread. 1. $\;\;\;p,\; q \vdash p\;$ implies $\;p \vdash q \rightarrow p\;$ implies $\;\vdash p \rightarrow (q \rightarrow p)$ 2. $\;\;\;p \rightarrow (q \rightarrow r), \; p \rightarrow q, \; p \; \vdash \; r\;\;\;$ (MP, 3 times) implies $\;p \rightarrow (q \rightarrow r), \; p \rightarrow q \; \vdash \; p \rightarrow r\;\;\;$ (DT) implies $\;p \rightarrow (q \rightarrow r) \; \vdash \; (p \rightarrow q) \rightarrow (p \rightarrow r)\;\;\;$ (DT) implies $\;\vdash \; [p \rightarrow (q \rightarrow r)] \;\; \rightarrow \;\; [(p \rightarrow q) \rightarrow (p \rightarrow r)] \;\;\;$ (DT) - 1 Well done too. In combinator-logical notation, this amounts to using $\mathbf W(\mathbf A\;M\;(\mathbf A\;N))$ instead of $\mathbf S\;M\;N$ in the translation of an application -- which is indeed so simple that we ought to have been able to find it by hand. – Henning Makholm Dec 27 '11 at 18:10 1 Incidentally, your "MP+DT" formulation is known (at least in computer-sciency contexts) as "natural deduction" for $\to$, and corresponds exactly, via the Curry-Howard isomorphism, to the simply typed lambda calculus. – Henning Makholm Dec 27 '11 at 18:14 I only saw this now; great! – joriki May 14 '12 at 7:55 For comparison, here is Joriki's solution in the combinator language we used in the comment thread between me and Zhen Lin: $$\begin{align} \mathbf X &= \mathbf A (\mathbf A \; \mathbf K \; \mathbf A) \mathbf W \\ \mathbf C &= \mathbf A \; \mathbf A (\mathbf A \; \mathbf X) \\ \mathbf B &= \mathbf C \; \mathbf A \\ \mathbf S &= \mathbf A (\mathbf B \; \mathbf B) \; (\mathbf A \; \mathbf C (\mathbf B \; \mathbf W)) \end{align}$$ where $\mathbf X$ is an ad-hoc name for Joriki's "assertion" formula. Zhen Lin's constuction for the final line $$\mathbf{S} = \mathbf{A A} ( \mathbf{A} ( \mathbf{B W} ) ( \mathbf{A A} ) )$$ is slightly more efficient than Joriki's because it contains only one $\mathbf B$ that needs to be unfolded. This yields the final term $$\mathbf S = \mathbf A\; \mathbf A\;(\mathbf A ( \mathbf A \; \mathbf A (\mathbf A \; (\mathbf A (\mathbf A \; \mathbf K \; \mathbf A) \mathbf W) ) \mathbf A \; \mathbf W) \; (\mathbf A\; \mathbf A))$$ which encodes a Hilbert-style proof with 15 fully substituted axiom instances and 14 modus ponens steps. - Thanks very much for this! An impressive demonstration of the conciseness of combinator logic :-) – joriki Dec 21 '11 at 15:21 This isn't much at all, but I think too long for a comment. Managing parenthesis for long formulas becomes difficult for me, so I'll use Polish notation. Somewhere in the proof (which most fittingly I would think would come as the penultimate step of the proof) we have to have one of the following: 1. C CCpqCpr C CpCqr CCpqCpr (substitution instance of axiom 1). 2. CC CpCqr C CpCqr CCpqCpr C CpCqr CCpqCpr (substitution instance of axiom 2). 3. C CpCqr x C x CCpqCpr C CpCqr CCpqCpr (substitution instance of axiom 3). 4. CC NCCpqCpr NCpCqr C CpCqr CCpqCpr (substitution instance of axiom 4). Now, CCpqCpr is not a theorem in a complete classical propositional calculus. So, cross out 1. The antecedent of 2; CCpCqrCCpCqrCCpqCpr can't get proven using the rules here, since we'd have to have CpCqr as a theorem. Edit: so, we can't have 2 as the penultimate step of the proof. We might still have CCpCqrCCpCqrCCpqCpr as a theorem (if we don't have it as a theorem we know right then and there the system is incomplete with respect to the full, classical propositional calculus), but we'd have to have CCpCqrCCpCqrCCpqCpr as the consequent of some other theorem, along with the antecedent of this other theorem as a theorem also. So, either a step of one of the forms of 3, or 4 will appear in the proof somewhere. The antecedent of 3 I'd guess contingent, while the antecedent of 4 does hold as a tautology... though I'm not sure it qualifies as a theorem in this system. - 1 Please don't leave downvotes without stating the reason. From the FAQ: "Add comments indicating what, specifically, is wrong." – joriki Dec 13 '11 at 4:58 – Carl Mummert Dec 13 '11 at 19:14 2 I added a second downvote because writing formulas in Polish notation only serves to obfuscate an otherwise very nice post. – Carl Mummert Dec 13 '11 at 19:17 @Carl: Many thanks for the link. If I remember correctly this is the second time I've learned something important from you about this site. Also thanks for explaining your downvote ;-) – joriki Dec 13 '11 at 19:22 3 @CarlMummert So goes your opinion. You can find hints in the logical literature that some have disagreed at points at least, such as Rose and Rosser (unless they wanted to obfuscate things), as described by Chang jstor.org/pss/20015986. After all, "C" comes as an abbreviation for "conditional" while "->" isn't like that. So, I don't see how they can "only serve to obfuscate", though they may obfuscate to you for sure. – Doug Spoonwood Dec 14 '11 at 4:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 106, "mathjax_display_tex": 12, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9391961097717285, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/2158/division-of-factorials/2165	# Division of Factorials I have a partition of a positive integer (p). How can I prove that the factorial of p can always be divided by the product of the factorials of the parts? As a quick example 9!/(2!3!4!) = 1260 (no remainder), where 9=2+3+4. I can nearly see it by looking at factors, but I can't see a way to guarantee it. - 12 @Andrew: Then why mention it? :-) – ShreevatsaR Aug 11 '10 at 17:43 @ShreevatsaR I apologise if it was spam. It was just to show that there is yet another way of proving it, and I may return to this page after publication. – Andrew Aug 11 '10 at 17:58 2 So @Andrew, were you really intending to post your proof here? – J. M. Oct 29 '11 at 11:05 ## 8 Answers The key observation is that the product of $n$ consecutive integers is divisible by $n!$. This can be proved by induction. - A quicker proof than induction would be to note that (by definition) $\binom{k}{n}$ is an integer, hence $n!\|k\cdot (k-1)\ldots (k-n+2)\cdot (k-n+1)$. – Andrew Aug 11 '10 at 17:24 Is that because given n (or more) consecutive integers one of them will always be divisible by n? – Guillermo Phillips Aug 13 '10 at 13:05 The "high-level" way to see this is to recall that whenever a finite group $G$ has a subgroup $H$, we know that $\|H\|$ divides $\|G\|$. Then note that $S_n$ clearly contains $S_{n_1} \times ... \times S_{n_k}$ as a subgroup for any partition $n_1 + ... + n_k = n$. (This is actually the same as the combinatorial interpretation in Robin Chapman's answer, since what we are counting is the number of cosets $G/H$, and these cosets are precisely what the multinomial theorem is counting.) This basic lemma is surprisingly useful. For example, it is not hard to use it to show that $m! (n!)^m$ divides $(mn)!$. - 3 Among many nice answers, I like this one the best. – Pete L. Clark Aug 11 '10 at 20:17 Below is a sketch of a little-known purely arithmetical proof that binomial coefficients are integral. I purposely constructed the proof so that it would be comprehensible to an educated layperson. The proof gives an algorithm to rewrite a binomial coefficient as a product of fractions whose denominators are coprime to any given prime. The method of proof is best comprehended by applying the algorithm to a specific example. [Note: It may prove helpful to first read this simpler example before proceeding to the exposition below]. E.g. consider $\ \ \binom{39}{17}\: =\: \frac{39!}{22! \; 17!}\: =\: \frac{23 \cdot 24 \cdots\; 39}{1 \cdot 2 \cdots\; 17}\:.\$ When this fraction is reduced to lowest terms $\rm\:a/b\:,\$ no prime $\rm\ p > 17\$ can divide its denominator $\rm\: b\:,\:$ since $\rm\ b\:\|\:17\:!\ \:$ Hence, to show that $\rm\ a/b\$ is an integer, it suffices to show that no prime $\rm\ p \le 17\$ divides its denominator $\rm\: b\:$. E.g. we show that $2$ doesn't divide $\rm b$. The highest power of $2$ in the denominator terms is $16 < 17$. Now align the numerator and denominator terms $\rm (mod\ 16)$ by shifting the 1st numerator term so it lies above its value $\rm (mod\ 16)$, viz. $23 \equiv 7 \pmod{16}$ so right-shift the numerator terms until $23\:$ lies above $7$ $$\frac{}{1}\frac{}{2}\frac{}{3}\frac{}{4}\frac{}{5}\frac{}{6}\frac{23}{7}\frac{24}{8}\frac{25}{9}\frac{26}{10}\frac{27}{11}\frac{28}{12}\frac{29}{13}\frac{30}{14}\frac{31}{15}\frac{32}{16}\frac{33}{17}\frac{34}{}\frac{35}{}\frac{36}{}\frac{37}{}\frac{38}{}\frac{39}{}$$ Now I claim that $2$ doesn't divide the reduced denominator of each aligned fraction. Indeed $\ 24/8 = 3$, $\: 26/10 = 13/5$, $\:\ 28/12 = 7/3$, $\:\ 30/14 = 15/7$, $\:\ 32/16 = 2$. This holds because these fractions $\rm\; c/d \;$ satisfy $\rm c \equiv d\ (mod\ 16)\:$ i.e. $\rm c = d + 16\: n \;$ so $\rm 2\|d \Rightarrow 2\|c$, $\rm\; 4\|d \Rightarrow 4\|c$, $\:\cdots$, $\rm\; 16\|d \Rightarrow 16\|c$, i.e. any power of $2\:$ below $16$ that divides $\rm d$ must also divide $\rm c$, so it cancels out upon reduction. Therefore to prove that $2$ doesn't divide the reduced denominator of $\binom{39}{17}$ it suffices to prove the same for the "leftover" fraction composed of the above non-aligned terms $(34 \cdots 39)/(1 \cdots 6) = \binom{39}{6}$. Being an $\rm\binom{n}{k}$ with smaller $\rm k = 6 < 17\:,\:$ this follows by induction. Since the same proof works for any prime $\rm p$, we conclude that no prime divides the reduced denominator of $\binom{39}{17}$, therefore it is an integer.$\quad$ QED Informally, the reason that this works is because the denominator sequence starts at $1$, which is coprime to every prime $\rm p$. This ensures that it is the "greediest" possible contiguous sequence of integers, in the sense that its product contains the least power of $\rm\:p\:$ compared to other contiguous sequences of equal length. The algorithm extends to multinomials by using the simple reduction of multinomials to products of binomials mentioned in my prior post here. - Very elegant idea! – Andres Caicedo Nov 9 '10 at 5:24 These quotients are integers since they solve counting problems. For instance, how many nine-letter words are there with 2 As, 3 Bs and 4 Cs? For the full story see the multinomial theorem. - 2 While true, it's not a good answer - usually one needs to be convinced that the quotients are integers before they can understand why the results are the solution to a counting problem... – BlueRaja - Danny Pflughoeft Aug 11 '10 at 16:19 11 @BlueRaja: Not at all! Showing that some ratio is the solution to a counting problem is my favourite way of proving divisibility. You don't need to be convinced that the quotient is an integer here; once you show it's the number of [something], it has to be an integer. (You don't usually check divisibility and compare prime factors when you show that the number of ways of choosing k out of n is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.) – ShreevatsaR Aug 11 '10 at 17:37 If you believe (:-) in the two-part Newton case, then the rest is easily obtained by induction. For instance (to motivate you to write a full proof): $$\frac{9!}{2! \, 3! \,4!} = \frac{9!}{5!\, 4!} \frac{5!}{2!\, 3!}$$ - I wanted to indent the equality, but the simplest standard TeX spacing "\ \ \ \ " didn't work. – wlod Aug 11 '10 at 15:48 use two dollars for centered LaTeX, (single dollar for inline). [; \cdots ;] doesn't work on this site.... – anon Aug 11 '10 at 15:58 We use dollar signs to denote TeX equations like `$this$`. – KennyTM Aug 11 '10 at 15:58 1 In terms of multinomial coefficients: choosing (2, 3, and 4) out of 9 is the same as first choosing 4 from the original 9, then choosing 2 and 3 out of the remaining 5. Etc. – ShreevatsaR Aug 11 '10 at 17:42 Here is another proof: We use Legendre's formula for the exact power of a prime $p$ which divides $n!$ which is given by $$\sum_{k=1}^{\infty} [\frac{n}{p^k}]$$ where $[x]$ is the integer part of $x$. Coupled with $$\sum_{i=1}^{n} [x_i] \le [\sum_{i=1}^{n} x_i]$$ we get that if $\sum_{i=1}^{n} a_i = N$ then $$\sum_{i=1}^{n} [\frac{a_i}{p^k}] \le [\frac{N}{p^k}]$$ And so any prime power which divides $(a_1)! \dots (a_n)!$ divides $N!$ and so $\frac{N!}{(a_1)! \dots (a_n)!}$ is an integer. - Besides the obvious combinatorial interpretations, one also has the has the following reduction from multinomial coefficient to products of binomial coefficients. Namely for $n = i+j+k+\cdots + m$ $$\frac{n!}{i!j!k!\cdots m!} = \binom{n}{i} \frac{(n-i)!}{j!k!\cdots m!} = \binom{n}{i}\binom{n-i}{j} \frac{(n-i-j)!}{k!\cdots m!} = \;\cdots$$ - 4 This is implicitly what wlod's answer does, with an example. – ShreevatsaR Aug 13 '10 at 16:57	{"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": 69, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9339596033096313, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/155259-relations-set.html	# Thread: 1. ## Relations in a set Hello, Given is a positive integer $k$. Prove that out of every set of integers which has more than $3^k$ elements one can pick out a sub-set $S$ with $k+1$ elements and the following quality: for any two subsets $A,B\subseteq S$ the sum of all elements in A is different from the sum of all elements in B. I've been trying to play with base-system representation for a while but it leads me nowhere. Your help will be appreciated. 2. I'm not sure about this, but I do have an idea about how you could approach this. Maybe you can prove that there is a subset you can make such that all numbers are "out of reach". For example, for k = 3 and the set is all numbers up until 27, the numbers 1, 4, 9 and 27 satisfy S. Maybe if you order them and pick the $3^0$th, $3^1$th, $3^2$th, $3^3$th, etc. ? 3. Ok, but these numbers don't have to start with 1 and increment by 1, they can be any numbers, including negatives. 4. Yeah, I know. Hence the "maybe if you order them". I have no idea how to actually prove this, but this was something that sprang to mind.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9622909426689148, "perplexity_flag": "head"}
http://mathoverflow.net/questions/100652/efficient-algorithm-for-projection-onto-a-convex-set/100703	## Efficient Algorithm For Projection Onto A Convex Set ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given $\mathbf{x} \in \mathbb{R}^n$ and $\tau$ a scalar, I would like to solve the following Euclidean projection problem: ```$\underset{\mathbf{p}}{\mathrm{argmin}} \; \\|\mathbf{p}-\mathbf{x}\\|_2 \;\; \mathrm{s.t.} \;\; \displaystyle \sum_{i}{ \left \\| \left [ \begin{array}{c} \mathbf{f}_i^\mathrm{T} \\ \mathbf{g}_i^\mathrm{T} \end{array} \right] \cdot \mathbf{p} \right \\|_2 } \leq \tau $```, where $\mathbf{f}_i,\mathbf{g}_i \in \mathbb{R}^n$. The above is a convex function over a convex set and as such should have a unique solution. Moreover, we can find the upper bound on the summation as follows: ```$\displaystyle \sum_{i}{ \left \\| \left [ \begin{array}{c} \mathbf{f}_i^\mathrm{T} \\ \mathbf{g}_i^\mathrm{T} \end{array} \right] \cdot \mathbf{p} \right \\|_2 } \leq \\|\mathbf{p}\\|_2 \cdot \sum_i \sigma_i $```, where $\sigma_i$ is the operator norm of the $2 \times n$ matrix $[\mathbf{f}_i \;\; \mathbf{g}_i]^{\mathrm{T}}$. I have been using CVX to solve the above, but it's just too slow in its current form. I have not figured out how to make use of them, but the operator norms are easily found before-hand. Can anyone suggest a re-formulation of the above or an algorithm that is tailored to these types of problems? - Have you tried minConf? di.ens.fr/~mschmidt/Software/minConf.html – Vidit Nanda Jun 26 at 1:22 @Vel Nias: I looked at minConf but unless I have misread the page, it requires one to provide a function to compute the projection onto the convex set, which is the problem I am trying to find an efficient solution for. – AnonSubmitter85 Jun 26 at 1:28 I'm confused: aren't you trying to minimize the distance to a fixed $x \in \mathbb{R}^n$ constrained by a single convex inequality? It sounds like "distance to $x$" is the function that you should use. Have I misunderstood? – Vidit Nanda Jun 26 at 1:51 1 @Vel Nias: I think you understand things correctly. However, in order to use either minConf_SPG() or minConf_SPG() I have to supply a projection function of the form: funProj(x) = argmin_y \|\|x - y\|\|_2, subject to y is in X. Is this not the very problem I trying to solve? I am left looking at minConf thinking it is for problems for which my problem above is only a sub-problem. – AnonSubmitter85 Jun 26 at 2:04 1 Either you can solve it as an SOCP (presumably that's something you've already tried), or you can use an ADMM style method to solve it. The details are somewhat messy, but it seems that one can do a fairly good job of efficiently solving this. Some of the key ideas that you can exploit, are discussed in the paper: Barbero and Sra (2011), people.kyb.tuebingen.mpg.de/suvrit/work/papers/… – S. Sra Jun 26 at 14:29 show 3 more comments ## 3 Answers It's easier to just write $A_i = [f_i \ g_i]^T$ so then the problem is: $$\min_p \frac{1}{2}\\|x-p\\|^2 \quad \mbox{s.t. } \sum_i \\|A_ip\\| \le \tau$$ Forming the Lagrangian and minimizing shows that the optimal $p^$ satisfies: $$p^ = x - \lambda^* \sum_i A_i^T \frac{ A_i p^* }{ \\|A_i p^* \\|}$$ So one obvious thing to try is just iterating this equation. We can solve for $\lambda^$ with a 1-dimensional root-finding procedure since we know that $\sum_i \\|A_i p^\\| = \tau$ (assuming that $x$ is not already in the interior of the feasible region). In summary, we iterate: $$d_n = \sum_i A_i^T \frac{ A_i p_{n-1} }{ \\|A_i p_{n-1} \\|}$$ $$p_n = x - \lambda_n d_n, \mbox{ where } \lambda_n \mbox{ is given by } \sum_i \\|A_i(x - \lambda_n d_n)\\| = \tau$$ Of course, you need to be careful not to divide zero by zero in case $A_i p_n=0$. Unfortunately, I don't know if this procedure converges, but it might be sufficient for your purposes. - This iteration is a simple version of what is called "the method of multipliers" in optimization. Its convergence isn't robust, but can be improved by using an augmented Lagrangian to stabilize the algorithm. – Brian Borchers Jun 27 at 14:54 I don't see how we know that $sum_{i}{\\| A_i (x-\lambda_n d_n )\\|} = \tau$ has a solution. – AnonSubmitter85 Jul 21 at 20:36 ### 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. You haven't told us anything about the size of your problem instances. How many terms are there in the sum of norms? What is $n$? Your problem is an example of a "sum of norms" optimization problem. Searching with Google Scholar will lead you to published research on this class of problems. CVX is using a standard approach for solving this problem by reformulating it as a second order cone programming (SOCP) problem and then using a primal-dual interior point method (SeDuMi's or SDPT3) to solve the resulting SOCP. For small problem instances, this should be a very robust and reasonably fast approach to solving the problem, but there are faster primal-dual codes for SOCP available (both CPLEX and MOSEK can be used to solve SOCP's.) You could ask CVX to extract the SOCP problem and export it in SeDuMi format and then try to use another solver on the SOCP. You might also look at first order methods to solve the SOCP- For example, I believe that TFOCS could be used to solve the problem. - I am working with images, so each $n$ is the total number of pixels and there will be $n+p-1$ terms in the sum of norms, where $p$ will probably be less than 20 . I am currently trying to get a rough cut working, so I am using smaller images, say, $100 \times 100 \Rightarrow n = 10000$. However, I'd like to use more realistic image sizes, so $n$ could be over a million. – AnonSubmitter85 Jun 27 at 6:07 For problems of that size, you'll definitely want to look at first order methods such as the alternating direction method of multipliers. – Brian Borchers Jun 27 at 14:51 Check out Dattorro's convex optimization book, page 748. -	{"extraction_info": {"found_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": 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": 0.9429044127464294, "perplexity_flag": "head"}
http://cms.math.ca/cmb/v44/	Canadian Mathematical Society www.cms.math.ca \| \| Site map \| CMS store location: Publications → journals → CMB « 2000 (v43) 2002 (v45) » Volume 44 (2001) Page Contents 3 Alexandru, Victor; Popescu, Nicolae; Zaharescu, Alexandru The generating degree $\gdeg (A)$ of a topological commutative ring $A$ with $\Char A = 0$ is the cardinality of the smallest subset $M$ of $A$ for which the subring $\Z[M]$ is dense in $A$. For a prime number $p$, $\C_p$ denotes the topological completion of an algebraic closure of the field $\Q_p$ of $p$-adic numbers. We prove that $\gdeg (\C_p) = 1$, \ie, there exists $t$ in $\C_p$ such that $\Z[t]$ is dense in $\C_p$. We also compute $\gdeg \bigl( A(U) \bigr)$ where $A(U)$ is the ring of rigid analytic functions defined on a ball $U$ in $\C_p$. If $U$ is a closed ball then $\gdeg \bigl( A(U) \bigr) = 2$ while if $U$ is an open ball then $\gdeg \bigl( A(U) \bigr)$ is infinite. We show more generally that $\gdeg \bigl( A(U) \bigr)$ is finite for any {\it affinoid} $U$ in $\PP^1 (\C_p)$ and $\gdeg \bigl( A(U) \bigr)$ is infinite for any {\it wide open} subset $U$ of $\PP^1 (\C_p)$. 12 Anisca, Razvan; Ilie, Monica This paper is concerned with the structure of the arithmetic sum of a finite number of central Cantor sets. The technique used to study this consists of a duality between central Cantor sets and sets of subsums of certain infinite series. One consequence is that the sum of a finite number of central Cantor sets is one of the following: a finite union of closed intervals, homeomorphic to the Cantor ternary set or an $M$-Cantorval. 19 Brindza, B.; Pintér, Á.; Schmidt, W. M. In this note the multiplicities of binary recurrences over algebraic number fields are investigated under some natural assumptions. 22 Evans, Ronald Precise, elegant evaluations are given for Gauss sums of orders six and twelve. 27 Goodaire, Edgar G.; Milies, César Polcino We show that an $\RA$ loop has a torsion-free normal complement in the loop of normalized units of its integral loop ring. We also investigate whether an $\RA$ loop can be normal in its unit loop. Over fields, this can never happen. 36 Kapovich, Michael; Millson, John J. The Hamiltonian potentials of the bending deformations of $n$-gons in $\E^3$ studied in \cite{KM} and \cite{Kl} give rise to a Hamiltonian action of the Malcev Lie algebra $\p_n$ of the pure braid group $P_n$ on the moduli space $M_r$ of $n$-gon linkages with the side-lengths $r= (r_1,\dots, r_n)$ in $\E^3$. If $e\in M_r$ is a singular point we may linearize the vector fields in $\p_n$ at $e$. This linearization yields a flat connection $\nabla$ on the space $\C^n_$ of $n$ distinct points on $\C$. We show that the monodromy of $\nabla$ is the dual of a quotient of a specialized reduced Gassner representation. 61 Kats, B. A. The paper is dealing with determination of the integral $\int_{\gamma} f \,dz$ along the fractal arc $\gamma$ on the complex plane by terms of polynomial approximations of the function~$f$. We obtain inequalities for polynomials and conditions of integrability for functions from the H\"older, Besov and Slobodetskii spaces. 70 Lempert, László; Szőke, Róbert Motivated by deformation theory of holomorphic maps between almost complex manifolds we endow, in a natural way, the tangent bundle of an almost complex manifold with an almost complex structure. We describe various properties of this structure. 80 Levin, Michael Applying the Sullivan conjecture we construct compacta of certain cohomological and extensional dimensions. 87 Lieman, Daniel; Shparlinski, Igor Let $p$ be prime and let $\vartheta\in\Z^_p$ be of multiplicative order $t$ modulo $p$. We consider exponential sums of the form $$S(a) = \sum_{x =1}^{t} \exp(2\pi i a \vartheta^{x^2}/p)$$ and prove that for any $\varepsilon > 0$ $$\max_{\gcd(a,p) = 1} \|S(a)\| = O( t^{5/6 + \varepsilon}p^{1/8}) .$$ 93 Neumann, B. H. A challenge by R.~Padmanabhan to prove by group theory the commutativity of cancellative semigroups satisfying a particular law has led to the proof of more general semigroup laws being equivalent to quite simple ones. 97 Ou, Zhiming M.; Williams, Kenneth S. An asymptotic formula is obtained for the number of cyclic quartic fields over $Q$ with discriminant $\leq x$. 105 Pilipović, Stevan The singular spectrum of $u$ in a convolution equation $\mu * u = f$, where $\mu$ and $f$ are tempered ultradistributions of Beurling or Roumieau type is estimated by $$SS u \subset (\mathbf{R}^n \times \Char \mu) \cup SS f.$$ The same is done for $SS_{}u$. 115 Roy, Damien The purpose of this paper is to show the limitations of the conjectures of algebraic approximation. For this, we construct points of $\bC^m$ which do not admit good algebraic approximations of bounded degree and height, when the bounds on the degree and the height are taken from specific sequences. The coordinates of these points are Liouville numbers. 121 Wojciechowski, Michał Simple necessary conditions for weak type $(1,1)$ of invariant operators on $L(\rr^d)$ and their applications to rational Fourier multiplier are given. 126 Zeron, E. Santillan Around 1995, Professors Lupacciolu, Chirka and Stout showed that a closed subset of $\C^N$ ($N\geq 2$) is removable for holomorphic functions, if its topological dimension is less than or equal to $N-2$. Besides, they asked whether closed subsets of $\C^2$ homeomorphic to the real line (the simplest 1-dimensional sets) are removable for holomorphic functions. In this paper we propose a positive answer to that question. 129 Currás-Bosch, Carlos In this paper the germ of neighborhood of a compact leaf in a Lagrangian foliation is symplectically classified when the compact leaf is $\bT^2$, the affine structure induced by the Lagrangian foliation on the leaf is complete, and the holonomy of $\bT^2$ in the foliation linearizes. The germ of neighborhood is classified by a function, depending on one transverse coordinate, this function is related to the affine structure of the nearly compact leaves. 140 Gotay, Mark J.; Grabowski, Janusz We prove an algebraic no-go theorem'' to the effect that a nontrivial \pa\ cannot be realized as an associative algebra with the commu\-ta\-tor bracket. Using it, we show that there is an obstruction to quantizing the \pa\ of polynomials generated by a nilpotent \ba\ on a \sm. This result generalizes \gr 's famous theorem on the impossibility of quantizing the Poisson algebra of polynomials on $\r^{2n}$. Finally, we explicitly construct a polynomial quantization of a \sm\ with a solvable \ba, thereby showing that the obstruction in the nilpotent case does not extend to the solvable case. 150 Jakóbczak, Piotr Let $B_N$ be the unit ball in $\mathbb{C}^N$ and let $f$ be a function holomorphic and $L^2$-integrable in $B_N$. Denote by $E(B_N,f)$ the set of all slices of the form $\Pi =L\cap B_N$, where $L$ is a complex one-dimensional subspace of $\mathbb{C}^N$, for which $f\|_{\Pi}$ is not $L^2$-integrable (with respect to the Lebesgue measure on $L$). Call this set the exceptional set for $f$. We give a characterization of exceptional sets which are closed in the natural topology of slices. 160 Langlands, Robert P. James Arthur was awarded the Canada Gold Medal of the National Science and Engineering Research Council in 1999. This introduction to his work is an attempt to explain his methods and his goals to the mathematical community at large. 210 Leung, Man Chun We construct unbounded positive $C^2$-solutions of the equation $\Delta u + K u^{(n + 2)/(n - 2)} = 0$ in $\R^n$ (equipped with Euclidean metric $g_o$) such that $K$ is bounded between two positive numbers in $\R^n$, the conformal metric $g=u^{4/(n-2)}g_o$ is complete, and the volume growth of $g$ can be arbitrarily fast or reasonably slow according to the constructions. By imposing natural conditions on $u$, we obtain growth estimate on the $L^{2n/(n-2)}$-norm of the solution and show that it has slow decay. 223 Marshall, M. A generalization of Schm\"udgen's Positivstellensatz is given which holds for any basic closed semialgebraic set in $\mathbb{R}^n$ (compact or not). The proof is an extension of W\"ormann's proof. 231 Rosenblatt, Joseph M.; Willis, George A. Let $G$ be an infinite discrete amenable group or a non-discrete amenable group. It is shown how to construct a net $(f_\alpha)$ of positive, normalized functions in $L_1(G)$ such that the net converges weak to invariance but does not converge strongly to invariance. The solution of certain linear equations determined by colorings of the Cayley graphs of the group are central to this construction. 242 Schueller, Laura Mann The zeta function of a nonsingular pair of quadratic forms defined over a finite field, $k$, of arbitrary characteristic is calculated. A.~Weil made this computation when $\rmchar k \neq 2$. When the pair has even order, a relationship between the number of zeros of the pair and the number of places of degree one in an appropriate hyperelliptic function field is 257 Abánades, Miguel A. Let $X$ be a reduced nonsingular quasiprojective scheme over ${\mathbb R}$ such that the set of real rational points $X({\mathbb R})$ is dense in $X$ and compact. Then $X({\mathbb R})$ is a real algebraic variety. Denote by $H_k^{\alg}(X({\mathbb R}), {\mathbb Z}/2)$ the group of homology classes represented by Zariski closed $k$-dimensional subvarieties of $X({\mathbb R})$. In this note we show that $H_1^{\alg} (X({\mathbb R}), {\mathbb Z}/2)$ is a proper subgroup of $H_1(X({\mathbb R}), {\mathbb Z}/2)$ for a nonorientable hyperelliptic surface $X$. We also determine all possible groups $H_1^{\alg}(X({\mathbb R}), {\mathbb Z}/2)$ for a real ruled surface $X$ in connection with the previously known description of all possible topological configurations of $X$. 266 Cencelj, M.; Dranishnikov, A. N. We show that every compactum has cohomological dimension $1$ with respect to a finitely generated nilpotent group $G$ whenever it has cohomological dimension $1$ with respect to the abelianization of $G$. This is applied to the extension theory to obtain a cohomological dimension theory condition for a finite-dimensional compactum $X$ for extendability of every map from a closed subset of $X$ into a nilpotent $\CW$-complex $M$ with finitely generated homotopy groups over all of $X$. 270 Cheung, Wai-Shun; Li, Chi-Kwong Let $c = (c_1, \dots, c_n)$ be such that $c_1 \ge \cdots \ge c_n$. The $c$-numerical range of an $n \times n$ matrix $A$ is defined by $$W_c(A) = \Bigl\{ \sum_{j=1}^n c_j (Ax_j,x_j) : \{x_1, \dots, x_n\} \text{ an orthonormal basis for } \IC^n \Bigr\},$$ and the $c$-numerical radius of $A$ is defined by $r_c (A) = \max \{\|z\| : z \in W_c (A)\}$. We determine the structure of those linear operators $\phi$ on algebras of block triangular matrices, satisfying $$W_c \bigl( \phi(A) \bigr) = W_c (A) \text{ for all } A \quad \text{or} \quad r_c \bigl( \phi(A) \bigr) = r_c (A) \text{ for all } A.$$ 282 Lee, Min Ho; Myung, Hyo Chul Jacobi-like forms for a discrete subgroup $\G \subset \SL(2,\mbb R)$ are formal power series with coefficients in the space of functions on the Poincar\'e upper half plane satisfying a certain functional equation, and they correspond to sequences of certain modular forms. We introduce Hecke operators acting on the space of Jacobi-like forms and obtain an explicit formula for such an action in terms of modular forms. We also prove that those Hecke operator actions on Jacobi-like forms are compatible with the usual Hecke operator actions on modular forms. 292 McKay, Angela There is a theorem, usually attributed to Napoleon, which states that if one takes any triangle in the Euclidean Plane, constructs equilateral triangles on each of its sides, and connects the midpoints of the three equilateral triangles, one will obtain an equilateral triangle. We consider an analogue of this problem in the hyperbolic plane. 298 Muić, Goran In this paper the author prove that standard modules of classical groups whose Langlands quotients are generic are irreducible. This establishes a conjecture of Casselman and Shahidi for this important class of groups. 313 Reverter, Amadeu; Vila, Núria We give an explicit recipe for the determination of the images associated to the Galois action on $p$-torsion points of elliptic curves. We present a table listing the image for all the elliptic curves defined over $\QQ$ without complex multiplication with conductor less than 200 and for each prime number~$p$. 323 Schuman, Bertrand Soit $H_0 = \frac{x^2+y^2}{2}$ un hamiltonien isochrone du plan $\Rset^2$. On met en \'evidence une classe d'hamiltoniens isochrones qui sont des perturbations polynomiales de $H_0$. On obtient alors une condition n\'ecessaire d'isochronisme, et un crit\ere de choix pour les hamiltoniens isochrones. On voit ce r\'esultat comme \'etant une g\'en\'eralisation du caract\ere isochrone des perturbations hamiltoniennes homog\`enes consid\'er\'ees dans [L], [P], [S]. Let $H_0 = \frac{x^2+y^2}{2}$ be an isochronous Hamiltonian of the plane $\Rset^2$. We obtain a necessary condition for a system to be isochronous. We can think of this result as a generalization of the isochronous behaviour of the homogeneous polynomial perturbation of the Hamiltonian $H_0$ considered in [L], [P], [S]. 335 Stacey, P. J. Irrational rotation $C^*$-algebras have an inductive limit decomposition in terms of matrix algebras over the space of continuous functions on the circle and this decomposition can be chosen to be invariant under the flip automorphism. It is shown that the flip is essentially the only toral automorphism with this property. 337 Vinet, Luc; Zhedanov, Alexei We continue to study the simplest closure conditions for chains of spectral transformations of the Laurent biorthogonal polynomials ($\LBP$). It is shown that the 1-1-periodic $q$-closure condition leads to the $\LBP$ introduced by Pastro. We introduce classes of semi-classical and Laguerre-Hahn $\LBP$ associated to generic closure conditions of the chain of spectral transformations. 346 Wang, Wei In this paper, we establish the existence of positive solution of a nonlinear subelliptic equation involving the critical Sobolev exponent on the Heisenberg group, which generalizes a result of Brezis and Nirenberg in the Euclidean case. 355 Weaver, Nik We examine Hilbert bimodules which possess a (generally unbounded) involution. Topics considered include a linking algebra representation, duality, locality, and the role of these bimodules in noncommutative differential geometry 370 Weston, Anthony Motivated by a question of Per Enflo, we develop a hypercube criterion for locating linear isometric copies of $\lone$ in an arbitrary real normed space $X$. The said criterion involves finding $2^{n}$ points in $X$ that satisfy one metric equality. This contrasts nicely to the standard classical criterion wherein one seeks $n$ points that satisfy $2^{n-1}$ metric equalities. 376 Zhang, Xi 385 Ballantine, Cristina M. Let $F$ be a totally real number field and let $\GL_{n}$ be the general linear group of rank $n$ over $F$. Let $\mathfrak{p}$ be a prime ideal of $F$ and $F_{\mathfrak{p}}$ the completion of $F$ with respect to the valuation induced by $\mathfrak{p}$. We will consider a finite quotient of the affine building of the group $\GL_{n}$ over the field $F_{\mathfrak{p}}$. We will view this object as a hypergraph and find a set of commuting operators whose sum will be the usual adjacency operator of the graph underlying the hypergraph. 398 Cardon, David A.; Ram Murty, M. We find a lower bound on the number of imaginary quadratic extensions of the function field $\F_q(T)$ whose class groups have an element of a fixed order. More precisely, let $q \geq 5$ be a power of an odd prime and let $g$ be a fixed positive integer $\geq 3$. There are $\gg q^{\ell (\frac{1}{2}+\frac{1}{g})}$ polynomials $D \in \F_q[T]$ with $\deg(D) \leq \ell$ such that the class groups of the quadratic extensions $\F_q(T,\sqrt{D})$ have an element of order~$g$. 408 Falbel, E. We classify finite subgroups of $\SO(4)$ generated by anti-unitary involutions. They correspond to involutions fixing pointwise a Lagrangian plane. Explicit descriptions of the finite groups and the configurations of Lagrangian planes are obtained. 420 Gauthier, P. M.; Pouryayevali, M. R. Functions defined on closed sets are simultaneously approximated and interpolated by meromorphic functions with prescribed poles and zeros outside the set of approximation. 429 Henniger, J. P. In answer to a question posed in \cite{G}, we give sufficient conditions on a Lie nilmanifold so that any ergodic rotation of the nilmanifold is metrically conjugate to its inverse. The condition is that the Lie algebra be what we call quasi-graded, and is weaker than the property of being graded. Furthermore, the conjugating map can be chosen to be an involution. It is shown that for a special class of groups, the condition of quasi-graded is also necessary. In certain examples there is a continuum of conjugacies. 440 Hironaka, Eriko In this paper we find a formula for the Alexander polynomial $\Delta_{p_1,\dots,p_k} (x)$ of pretzel knots and links with $(p_1,\dots,p_k, \nega 1)$ twists, where $k$ is odd and $p_1,\dots,p_k$ are positive integers. The polynomial $\Delta_{2,3,7} (x)$ is the well-known Lehmer polynomial, which is conjectured to have the smallest Mahler measure among all monic integer polynomials. We confirm that $\Delta_{2,3,7} (x)$ has the smallest Mahler measure among the polynomials arising as $\Delta_{p_1,\dots,p_k} (x)$. 452 Ishihara, Hironobu Let $\ce$ be an ample vector bundle of rank $r$ on a projective variety $X$ with only log-terminal singularities. We consider the nefness of adjoint divisors $K_X + (t-r) \det \ce$ when $t \ge \dim X$ and $t>r$. As an application, we classify pairs $(X,\ce)$ with $c_r$-sectional genus zero. 459 Kahl, Thomas Nous montrons que la A-cat\'egorie d'un espace simplement connexe de type fini est inf\'erieure ou \'egale \a $n$ si et seulement si son mod\ele d'Adams-Hilton est un r\'etracte homotopique d'une alg\ebre diff\'erentielle \a $n$ \'etages. Nous en d\'eduisons que l'invariant $\Acat$ augmente au plus de 1 lors de l'attachement d'une cellule \`a un espace. We show that the A-category of a simply connected space of finite type is less than or equal to $n$ if and only if its Adams-Hilton model is a homotopy retract of an $n$-stage differential algebra. We deduce from this that the invariant $\Acat$ increases by at most 1 when a cell is attached to a space. 469 Marcoux, Laurent W. In this article it is shown that every bounded linear operator on a complex, infinite dimensional, separable Hilbert space is a sum of at most eighteen unilateral (alternatively, bilateral) weighted shifts. As well, we classify products of weighted shifts, as well as sums and limits of the resulting operators. 482 Mezo, Paul We prove an identity between weighted orbital integrals of the unit elements in the Hecke algebras of $\GL(r)$ and its $n$-fold metaplectic covering, under the assumption that $n$ is relatively prime to any proper divisor of every $1 \leq j \leq r$. 491 Wang, Weiqiang We give canonical resolutions of singularities of several cone varieties arising from invariant theory. We establish a connection between our resolutions and resolutions of singularities of closure of conjugacy classes in classical Lie algebras. 504 Zhang, Yong We show that, if a Banach algebra $\A$ is a left ideal in its second dual algebra and has a left bounded approximate identity, then the weak amenability of $\A$ implies the ($2m+1$)-weak amenability of $\A$ for all $m\geq 1$. 509 2001, for 2001 - pour No abstract.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 207, "mathjax_display_tex": 5, "mathjax_asciimath": 4, "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": 0.8812189102172852, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=4092761	Physics Forums Recognitions: Gold Member ## Dirac delta in curved space The Dirac delta function is defined as: $$\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = 1$$ Or more generally the integral is, $$\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {dx'} )dx}$$ But if the metric varies with x, then the integral becomes, $$\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx}$$ But from the wikipedia.com article we have, $$\delta (f(x)) = \frac{{\delta (x - {x_z})}}{{\left\| {f'({x_z})} \right\|}}$$ where xz is such that $f({x_z}) = 0$. And since, $$\frac{d}{{dx}}(\int_{{x_0}}^x {\sqrt {g(x')} dx'} ) = \sqrt {g(x)}$$ and $$f(x) = \int_{{x_0}}^x {\sqrt {g(x')} dx'} = 0$$ when x= x0 so that xz= x0. Then the above becomes, $$\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} = \int_{ - \infty }^{ + \infty } {\frac{{\delta (x - {x_0})}}{{\left\| {\sqrt {g({x_0})} } \right\|}}\sqrt {g(x)} dx}$$ Now obviously if g(x) is constant, we recover the original Dirac delta function. But what if g(x) is not constant, can I still recover the original Dirac delta function? Then the Dirac delta would behave the same in any space. Have I evaluated ${\left\| {\sqrt {g({x_0})} } \right\|}$ wrong? Could there be some sense is saying this should simply by ${\sqrt {g(x)} }$ so that it cancels out? Maybe x never really equals x0, but only approaches it, so that we use some other property of the Dirac delta. 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 OK, I think I got it. Starting with $$\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} = \int_{ - \infty }^{ + \infty } {\frac{{\delta (x - {x_0})}}{{\left\| {\sqrt {g({x_0})} } \right\|}}\sqrt {g(x)} dx}$$ I suppose it is safe to assume that the metric g(x) is real and always positive so the squareroot is positive, and we can drop the absolute sign. And since we can bring the denominator outside the integral, we have, $$\frac{1}{{\sqrt {g({x_0})} }}\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})\sqrt {g(x)} dx}$$ And then the sifting property lets us bring the function multiplied by the Dirac delta outside the integral with an x value of x0, to get, $$\frac{1}{{\sqrt {g({x_0})} }}\sqrt {g({x_0})} \int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx }$$ But since the squareroots of g cancel, this is just $\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx}$. So, $$\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} = \int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = 1$$ Or the Dirac delta function is invariant with changes of coordinates, metric, curvature or spaces. But I suppose if we think in higher dimensional terms, the integral inside the Dirac delta is a line integral, and the integral outside the delta is a volume integral, right? So would we still get the cancellation of squareroots of g in higher dimensions? Or would the g(x) term in the line integral not be squarerooted like the volume integral outside the delta? Well, the Dirac distribution satisifes: $$\int \delta(x-x_0) f(x) dx = f(x_0)$$ That pretty much answers your question. ## Dirac delta in curved space BTW, if I am not mistaken we should take minus the squareroot of the det of the metric in the change of variables in integration. Recognitions: Gold Member I'm not satisfied because I'm not sure how this generalizes in multidimensional space. What does $$\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx}$$ look like in 3D? The $${\int_{{x_0}}^x {\sqrt {g(x')} dx'} }$$ part would be a line integral. How does that generalize in 3D with a changing metric? The $$\int_{ - \infty }^{ + \infty } {\delta (......)\sqrt {g(x)} dx}$$ part would be a volume integral where the squareroot of the metic is commonly used. So I guess my biggest question is how to use the metric in a line integral. And since you can have more than one line from x0 to x in 3D, how do you determine which line to take, if it even matters? Would you integrate the line integral as it travels over the geodesic? Well it should be similar to the one dimensional case. I assume you know that the n-dimensional Dirac Delta distribution is made up of a product of 1-dimensional Dirac Delta distribution. So you do the same trick you did before and get something like: $$\int_{\mathbb{R}^n} \delta ^n (\bar{x}-\bar{x_0}) \frac{\sqrt{g(\bar{x})}}{\sqrt{g(\bar{x_0})}} d\bar{x}$$ Ok, in your case I think (though not certain), that th line integral from one dimensional becomes if $$\bar{x} =(x^1,x^2,...,x^n)$$ then the line integral becomes: $$\int_{x^1_0}^{x^1}...\int_{x^n_0}^{x^n} \sqrt{g(x')}dx'^1...dx'^n$$ Recognitions: Gold Member The differential line segment squared is: $$d{s^2} = {g_{ij}}(\vec x)d{x^i}d{x^j}$$ So the line integral in a general curved space is: $$\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} }$$ which is usually parameterized as: $$\int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t))\frac{{d{x^i}}}{{dt}}\frac{{d{x^j}}}{{dt}}} } \,\,\,dt$$ And when we put this inside the Dirac delta to get $$\delta (\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } )$$ I don't know how this can be put in the form $$\delta (f(\vec x)) = \frac{{\delta (\vec x - {{\vec x}_z})}}{{\left\| {f'({{\vec x}_z})} \right\|}}$$ where $f({{\vec x}_z}) = 0$. Can ${g_{ij}}(\vec x)d{x^i}d{x^j}$ be put in the form of a determinant of the metric time differentials so that we can evaluate $g(\vec x)$ at ${{\vec x}_0}$ and bring it outside the delta in the denominator where it cancels as before? Recognitions: Homework Help Quote by friend The differential line segment squared is: $$d{s^2} = {g_{ij}}(\vec x)d{x^i}d{x^j}$$ So the line integral in a general curved space is: $$\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} }$$ which is usually parameterized as: $$\int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t))\frac{{d{x^i}}}{{dt}}\frac{{d{x^j}}}{{dt}}} } \,\,\,dt$$ And when we put this inside the Dirac delta to get $$\delta (\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } )$$ I don't know how this can be put in the form $$\delta (f(\vec x)) = \frac{{\delta (\vec x - {{\vec x}_z})}}{{\left\| {f'({{\vec x}_z})} \right\|}}$$ where $f({{\vec x}_z}) = 0$. Can ${g_{ij}}(\vec x)d{x^i}d{x^j}$ be put in the form of a determinant of the metric time differentials so that we can evaluate $g(\vec x)$ at ${{\vec x}_0}$ and bring it outside the delta in the denominator where it cancels as before? Recognitions: Gold Member In one dimensional space, $\vec x = (x)$, which means ${x^i} = x$ and $d{x^i} = dx$, so that $$\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} = \sqrt {{g_{ii}}(\vec x)d{x^i}d{x^i}} = \sqrt {g(x)} dx$$ and then $$\delta (\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } ) = \delta (\int_{{x_0}}^x {\sqrt {g(x')} dx} ') = \frac{{\delta (x - {x_0})}}{{\sqrt {g({x_0})} }}$$ And this ${\sqrt {g({x_0})} }$ cancels with the one in the volume integral. And the Dirac delta is invariant in 1D. In 2D, we have $\vec x = ({x^1},{x^2})$, and $${g_{ij}}(\vec x)d{x^i}d{x^j} = {g_{11}}({x^1},{x^2})d{x^1}d{x^1} + {g_{12}}({x^1},{x^2})d{x^1}d{x^2} + {g_{22}}({x^1},{x^2})d{x^2}d{x^2} + {g_{21}}({x^1},{x^2})d{x^2}d{x^1}$$ And in flat space, ${g_{12}} = {g_{21}} = 0$, so that the above becomes, $${g_{ij}}(\vec x)d{x^i}d{x^j} = {g_{11}}d{x^1}d{x^1} + {g_{22}}d{x^2}d{x^2}$$ Then if ${g_{11}} = {g_{22}} = g$, we get $${g_{ij}}(\vec x)d{x^i}d{x^j} = g{(d{x^1})^2} + g{(d{x^2})^2}$$ And, $$\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} = \sqrt g \sqrt {{{(d{x^1})}^2} + {{(d{x^2})}^2}} = \sqrt g ds$$ So that the $\sqrt g$ cancels out to make the integral of the Dirac delta invariant with changes in rescaling of flat space. I wonder what this means. Is the Dirac delta invariant with Lorentz transformation since those are transformation from one flat spacetime to another flat spacetime? If we parameterize the line integral with $\vec x = \vec x(t)$ to get, $$\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } = \int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t'))\frac{{d{x^i}}}{{dt'}}\frac{{d{x^j}}}{{dt'}}} } \,\,\,dt'$$ Can we show that the time variable must be orthogonal to the flat space dimensions? Can we prove the necessity of Minkowski metric for an invariant integral of a parameterized Dirac delta function? Recognitions: Gold Member Quote by friend I wonder what this means. Is the Dirac delta invariant with Lorentz transformation since those are transformation from one flat spacetime to another flat spacetime? If we parameterize the line integral with $\vec x = \vec x(t)$ to get, $$\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } = \int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t'))\frac{{d{x^i}}}{{dt'}}\frac{{d{x^j}}}{{dt'}}} } \,\,\,dt'$$ Can we show that the time variable must be orthogonal to the flat space dimensions? Can we prove the necessity of Minkowski metric for an invariant integral of a parameterized Dirac delta function? Notice, for example, that in $$\int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t))\frac{{d{x^i}}}{{dt}}\frac{{d{x^j}}}{{dt}}} } \,\,\,dt$$ if we make ${g_{ij}}(\vec x) = 0$ whenever i≠j, then we get $$\int_{{t_0}}^t {\sqrt {{g_i}(\vec x(t)){{\left( {\frac{{d{x^i}}}{{dt}}} \right)}^2}} } \,\,\,dt$$ Now, in order to make $\sqrt {g(x)}$ come outside the Dirac delta and cancel as before, we need $${\left( {\frac{{d{x^i}}}{{dt}}} \right)^2} = 1$$ Or, we can put this in a different way by saying $${(ds)^2} = {(dt)^2} - {(dx)^2} = 0$$ This looks like the Minkowski metric. But I don't know what it would mean to have ds≠0. Recognitions: Gold Member Now that I think about it, when I write: $$\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = \int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {dx'} )dx} = \int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} = 1$$ I don't actually know whether the ${g(x)}$ in the volume form of $\int_{ - \infty }^{ + \infty } {\delta (...)\sqrt {g(x)} dx}$ outside the dirac delta is really the same as the ${g(x')}$ in $\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )$ in the line integral inside the dirac delta. The volume form ${g(x)}$ outside the delta is $$\sqrt {g(x)} = \det ({J^T}J)$$ where $$J = \left( {\begin{array}{*{20}{c}} {\frac{{\partial {u^1}}}{{\partial {x^1}}}}&{\frac{{\partial {u^1}}}{{\partial {x^2}}}}\\ {\frac{{\partial {u^2}}}{{\partial {x^1}}}}&{\frac{{\partial {u^2}}}{{\partial {x^2}}}} \end{array}} \right)$$ for a change of coordinates from u to x. But the ${g(x')}$ in the line integral inside the delta is $${g_{ij}}(\vec x)d{x^i}d{x^j} = {g_{11}}d{x^1}d{x^1} + {g_{12}}d{x^1}d{x^2} + {g_{22}}d{x^2}d{x^2} + {g_{21}}d{x^2}d{x^1}$$ And then when ${g_{12}} = {g_{21}} = 0$ and ${g_{11}} = {g_{22}} = g$, the ${g(x')}$ is just a constant ${g}$. Are these two g's actually the same thing... even in curved space? How does that work? Recognitions: Gold Member Previously, I considered: Quote by friend $$\delta (x - {x_0})=\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx} ') = \frac{{\delta (x - {x_0})}}{{\sqrt {g({x_0})} }}$$ But on second thought, this is probably not useful since I could do the same thing with the dirac delta at the right as I did on the left. I could iterate the process to get another squareroot of g in the denominator making it equal the dirac delta divided by the squarroot of g squared. In fact, I could iterate this an infinite number of times to get a squarroot of g to the power of infinity. This only works if g=1 for flat space, and I have not converted the flat dirac delta to a dirac delta in curved space. I will have to look for something else. Recognitions: Homework Help Perhaps this will help? http://www.fen.bilkent.edu.tr/~ercelebi/mp03.pdf It gives the form of the dirac delta in curvilinear coordinates (in 3d). I would guess that to achieve your purposes you would need to generalize to arbitrary dimension and modify the scale factors ##h_i## to account for different metrics. Recognitions: Gold Member Quote by Mute Perhaps this will help? http://www.fen.bilkent.edu.tr/~ercelebi/mp03.pdf Thank you. That's a start. My main goal is to investigate how the properties of the dirac delta function might be used to determine the properties of the background spacetime metric. For example, if the dirac delta describes the density of a mass point, is that enough to determine the properties of the metric? Or, if the dirac delta represents causality, is that enough to determine the Minkowski metric? So I consider equation such as: $$\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx}=\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {dx'} )dx}=\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx}=\int_{ - \infty }^{ + \infty } {\delta (\int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t))\frac{{d{x^i}}}{{dt}}\frac{{d{x^j}}}{{dt}}} } \,\,\,dt){\sqrt {g(x)} }dx}=1$$ Or, if I also apply the first variation to all this and set it equal to zero, is there enough information to specify the metric of Special Relativity or General Relativity? Wouldn't it be interesting if it did? Quote by friend Thank you. That's a start. My main goal is to investigate how the properties of the dirac delta function might be used to determine the properties of the background spacetime metric. For example, if the dirac delta describes the density of a mass point, is that enough to determine the properties of the metric? My gut reaction would be yes. On ℝ at least, metrics define Borel measures and every Borel measure is a Schwartz distribution. IIRC, Schwartz distributions can be broken up into sums of other singular distributions related to Dirac. Honestly, this sounds like the domain of current theory, so ask mathematician who specialise in that? Recognitions: Gold Member Quote by pwsnafu My gut reaction would be yes. On ℝ at least, metrics define Borel measures and every Borel measure is a Schwartz distribution. IIRC, Schwartz distributions can be broken up into sums of other singular distributions related to Dirac. Honestly, this sounds like the domain of current theory, so ask mathematician who specialise in that? God! I hope it's not that complicated. Recognitions: Gold Member Quote by pwsnafu My gut reaction would be yes. On ℝ at least, metrics define Borel measures and every Borel measure is a Schwartz distribution. IIRC, Schwartz distributions can be broken up into sums of other singular distributions related to Dirac. Honestly, this sounds like the domain of current theory, so ask mathematician who specialise in that? Hopefully, it is easier than this. For example, the dirac delta function is a continuous version of the Kronecker delta which is used as the metric in flat space. But the spacetime metric of relativity is also flat spacetime metric. Is it also discrete version of a delta function? Thread Tools \| \| \| \| \|--------------------------------------------------\|------------------------------\|---------\| \| Similar Threads for: Dirac delta in curved space \| \| \| \| Thread \| Forum \| Replies \| \| \| Quantum Physics \| 6 \| \| \| Calculus \| 3 \| \| \| Special & General Relativity \| 7 \| \| \| Differential Geometry \| 6 \| \| \| Advanced 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": 34, "mathjax_display_tex": 46, "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": 0.9123882055282593, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/14276/why-cant-quantum-teleportation-be-used-to-transport-information	# Why can't quantum teleportation be used to transport information? Kaku Michio says in an interview that we've teleported photons, cesium atoms and beryllium atoms. Having watched a lot of Kaku as well as way too many astrophysics documentaries in general, I know he's exaggerating. From what I understand, some state of those particles were teleported. Here's my question: what state exactly is being teleported, and why can't it be used to transfer classical information? How can we know experimentally that a teleportation occured, if the transported information can't be read classically? Most of the news articles for the general public just throw around the word "state" without explaining the idea (or flat out tell us like Kaku that atoms were teleported), and Wikipedia uses math and terminology that's way over my head. Is there a layman's explanation by any chance? - 2 +1, damn good question – Arjang Sep 2 '11 at 10:26 2 @Arjang Thanks, I was a bit nervous about posting because this is really not my field at all :) – Rei Miyasaka Sep 2 '11 at 10:53 ## 2 Answers To elaborate on Misha's answer: here is a spiel on teleportation from quantum information theory. I'm going to do some actual computations, but they will be made as simple as possible for novice readers. ### Beginning of the sketch Let us simplify the discussion by limiting ourselves to the teleportation of a single bit's worth of information, e.g. a spin ½ particle. Let us denote spin-down by the bit-value 0, and spin-up by 1. A general (pure, i.e. non-randomized) state of the spin ½ particle is given by $\|\sigma\rangle_s = \alpha_0 \|0\rangle_s + \alpha_1 \|1\rangle_s$: a superposition of the two spin-values, where $\|\alpha_0\|^2$ and $\|\alpha_1\|^2$ must sum to 1 and are otherwise unconstrained complex numbers. If you wanted to store a single bit, you could just use $\alpha_0 = 1$ or $\alpha_1 = 1$ to represent "0" or "1" respectively. (The subscript s just means that this indicates the state of a spin "s".) For teleportation between two distant locations A and B (where A is in the same general area as the spin s), you need an entangled state as a resource, such as $$\|\beta_{00}\rangle_{A,B} = \frac{1}{\sqrt 2}\Bigl[\; \|0\rangle_A\|0\rangle_B + \|1\rangle_A\|1\rangle_B \Bigr]$$ which is a superposition of two "classical" states: one where both A and B are in the state 0, and one where both are in the state 1. This is a notation which represents a vector, but that's not important for this post. • You should think of the symbols $\|u\rangle_{\!\!X}$ as being like variables that you can multiply together in whichever order you like; each one describes the state of one part of the system; and combinations of these products can be used to describe superpositions. (This "multiplication" is actually a tensor product; but that's also not important for this post.) • The factor of $1/\!\sqrt{2}\,$ is just for the sake of book-keeping, so that we can compute probabilities. We won't be doing a lot of probability calculations, but there's an interesting comparison coming below which makes it worthwhile to mention normalization. —— If you're happy with the notion of a vector, what this is doing is ensuring that the vector has norm 1; corresponding roughly to being in this entangled two-particle state with certainty. (Shorter vectors represent states which occur only with some probability: we will see an example later.) The terms $\|0\rangle_A\|0\rangle_B$ and $\|1\rangle_A\|1\rangle_B$ are also unit vectors, and are orthogonal to one another because they represent perfectly distinguishable classical states. Their sum has length $\sqrt 2$, so we divide by this to get another unit vector. States in quantum info are typically represented by unit vectors.) Another way of reading this is that if we measured this state to get bit values, we would have a probability of $(1/\!\sqrt 2)^2 = \frac12$ change of obtaining 00 for A and B, and the same chance of obtaining 11 for A and B. Now, I'm going to make a sharp right turn, and talk for a little about cryptography, because the protocoal for teleportation bears a striking resemblance to the protocol for transmitting a message secretly with a one-time pad. ### A digression into classical cryptography A one-time pad is nothing more than two copies of the same sequence of random numbers, e.g. strings of random digits. When you want to encrypt a message which is n bits long, to send from Alice (A) to Bob (B), the two sequences of random bits must have length at least n. For example, if you want to encode a single bit of information, it is enough for A and B to share a single random bit. In a manner of speaking, the random bits in their position must be in a joint state of the form $$\mathbf{Pads} = \frac{1}{2} \Bigl[ \Bigl(\langle 0\rangle_A \text{ and } \langle 0 \rangle_B\Bigr) + \Bigl(\langle 1 \rangle_A \text{ and } \langle 1 \rangle_B\Bigr) \Bigr]$$ where $\langle x \rangle_{\!A/B}\,$ just describes the scenario of Alice's/Bob's bit being x, the addition represents a sort of disjunction between the two (exclusive) events, and where the scaling of 1/2 represents the probabilities with which these two events occur. (I've written it this way deliberately; the mathematics for writing this as a classical probability distribution is almost identical to what one would use for writing quantum states such as $\|\beta_{00}\rangle$.) If Alice has a single bit s ∈ {0,1} that she wants to send to Bob, she measures — yes, measures; you could also say compute if you like — the parity of s with her bit A, $$s' = s \oplus A = \begin{cases} 0,& \text{if $s = A$} \\ 1,& \text{if $s \ne A$} \end{cases}$$ where $\oplus$ is addition modulo two [$0 \oplus 0 = 1 \oplus 1 = 0$, and $0 \oplus 1 = 1 \oplus 0 = 1$] and then sends the result s' to Bob, who either flips his own bit B according to whether s' is zero or one: $$B' = B \oplus s'\;.$$ Exercise for the reader: show that B' = s, Alice's bit prior to encryption. Of course, before obtaining s', Bob has no way of guessing what the value of B' will be: no amount of staring at his copy of the original shared bit will reveal any operations Alice has done. Back to teleportation! ### How teleportation actually works (mathematically speaking) In order to teleport $\|\sigma\rangle_s$, Alice performs a joint measurement on s and A. There are four possible outcomes, so the most information Alice can obtain out of her measurement is two bits. The measurement she performs is in the basis $$\begin{align} \|\beta_{00}\rangle_{s,A} \;&=\; \tfrac{1}{\sqrt 2}\Bigl[\; \|0\rangle_s\|0\rangle_A + \|1\rangle_s\|1\rangle_A \Bigr], \\ \|\beta_{01}\rangle_{s,A} \;&=\; \tfrac{1}{\sqrt 2}\Bigl[\; \|0\rangle_s\|1\rangle_A + \|1\rangle_s\|0\rangle_A \Bigr], \\ \|\beta_{10}\rangle_{s,A} \;&=\; \tfrac{1}{\sqrt 2}\Bigl[\; \|0\rangle_s\|0\rangle_A - \|1\rangle_s\|1\rangle_A \Bigr], \\ \|\beta_{11}\rangle_{s,A} \;&=\; \tfrac{1}{\sqrt 2}\Bigl[\; \|0\rangle_s\|1\rangle_A - \|1\rangle_s\|0\rangle_A \Bigr]. \end{align}$$ There is a reason why we choose these states in particular; the high-level summary for why we choose these states is given in the final paragraph where I complete the analogy to the one-time pad. (This is not the only possible choice of basis, but it's the simplest.) She will obtain one of the states $$\|\beta_{zx}\rangle_{s,A} = \tfrac{1}{\sqrt 2}\Bigl[\; \|0\rangle_s\|x\rangle_A + (-1)^z \|1\rangle_s\|1 \oplus x\rangle_A \Bigr]$$ as a result, which can be represented succinctly by the two bits x, z ∈ {0,1}. How do we compute the effect of this measurement? By applying the operator $$\langle \beta_{zx} \|_{s,A} = \tfrac{1}{\sqrt 2}\Bigl[\; \langle 0\|_s\langle x\|_A + (-1)^z \langle 1\|_s \langle 1 \oplus x\|_A \Bigr]$$ to the input state $\|\sigma\rangle_s \|\beta_{00}\rangle_{A,B}$, distributing the $\langle \text{whatever}\|_X$ symbols over addition like linear operators (which they are), expanding the "multiplication" of the state $\|\sigma\rangle_s$ with $\|\beta_{00}\rangle_{A,B}$, and using the combination rule $\langle u \|_{\!\!X}\, \| v \rangle_{\!\!X} = \langle u \| v\rangle_{\!\!X} = \delta_{u,v}$ for symbols with a common subscript (this is mechanical calculation, skip to the end if you prefer): $$\begin{align} \langle \beta_{xz} \|_{s,A} &\Bigl[ \|\sigma\rangle_s \| \beta_{00}\rangle_{A,B} \Bigr] \\\\\\=&\; \tfrac{1}{\sqrt 2}\langle \beta_{xz} \|_{s,A}\Bigl[\; \alpha_0 \|0\rangle_s \|0\rangle_s\|0\rangle_A + \alpha_0 \|0\rangle_s \|1\rangle_s\|1\rangle_A + \alpha_1 \|1\rangle_s \|0\rangle_s\|0\rangle_A + \alpha_1 \|1\rangle_s \|1\rangle_s\|1\rangle_A \Bigr] \\\\\\=&\; \tfrac{1}{2} \Bigl[\; \langle 0\|_s\langle x\|_A + (-1)^z \langle 1\|_s \langle 1 \oplus x\|_A \Bigr] \times \\&\qquad\qquad \Bigl[\; \alpha_0 \|0\rangle_s \|0\rangle_A\|0\rangle_B + \alpha_0 \|0\rangle_s \|1\rangle_A\|1\rangle_B + \alpha_1 \|1\rangle_s \|0\rangle_A\|0\rangle_B + \alpha_1 \|1\rangle_s \|1\rangle_A\|1\rangle_B \Bigr] \\\\\\=&\; \tfrac{1}{2} \Bigl[\; \alpha_0 \langle 0\|0\rangle_s \langle x\|0\rangle_A\|0\rangle_B + \alpha_0 \langle 0\|0\rangle_s \langle x\|1\rangle_A\|1\rangle_B \\&\qquad + \alpha_1 \langle 0\|1\rangle_s \langle x\|0\rangle_A\|0\rangle_B + \alpha_1 \langle 0\|1\rangle_s \langle x\|1\rangle_A\|1\rangle_B \\&\qquad + (-1)^z \alpha_0 \langle 1\|0\rangle_s \langle 1 \oplus x \|0\rangle_A\|0\rangle_B + (-1)^z \alpha_0 \langle 1\|0\rangle_s \langle 1 \oplus x \|1\rangle_A\|1\rangle_B \\&\qquad + (-1)^z \alpha_1 \langle 1\|1\rangle_s \langle 1 \oplus x \|0\rangle_A\|0\rangle_B + (-1)^z \alpha_1 \langle 1\|1\rangle_s \langle 1 \oplus x \|1\rangle_A\|1\rangle_B\Bigr] \\\\\\=&\; \tfrac{1}{2} \Bigl[\; \alpha_0 \langle x\|0\rangle_A\|0\rangle_B + \alpha_0 \langle x\|1\rangle_A\|1\rangle_B \\&\qquad + (-1)^z \alpha_1 \langle 1 \oplus x \|0\rangle_A\|0\rangle_B + (-1)^z \alpha_1 \langle 1 \oplus x \|1\rangle_A\|1\rangle_B\Bigr] \\\\\\=&\; \tfrac{1}{2} \delta_{x,0} \Bigl[\; \alpha_0 \|0\rangle_B + (-1)^z \alpha_1 \|1\rangle_B \Bigr] + \tfrac{1}{2} \delta_{x,1} \Bigl[\; \alpha_0 \|1\rangle_B + (-1)^z \alpha_1 \|0\rangle_B \Bigr] \\\\\\=&\; \tfrac{1}{2} \Bigl[\; \alpha_0 \|x\rangle_B + (-1)^z \alpha_1 \|1 \oplus x\rangle_B \Bigr]. \end{align}$$ This calculation is doing nothing more than computing the marginal state on B, conditioned on obtaining the state $\|\beta_{zx}\rangle_{\!s,A}$ on s and A. People still argue about the interpretation of this concept, but the mathematics itself is identical to what you would do for computing classical probability distributions on B conditioned on some property of s and A holding (such as the parity of s and A being some bit x ∈ {0,1}). The scalar factor of $\frac12$ just represents the fact that the outcome $\|\beta_{zx}\rangle$ of the measurement arises, independently of the values of x and z, with probability $(\frac12)^2 = \frac14$; the final state is some state on B which depends on the values of x and z. Without knowing what x and z are, Bob doesn't know the result of the measurement collapse — any more than he could guess Alice's message in the one-time pad protocol by staring at his own random bit. In fact, we can show that without knowledge of x and z, Bob's state looks perfectly random to him, just as it would have if he had measured B before Alice did anything, and just as the random bit in a one-time pad looks. But if Alice were to send x and z to him, he could perform operations on his state to obtain the state $\|\sigma\rangle_B = \alpha_0 \|0\rangle_B + \alpha_1 \|1\rangle_B$. Which operations he performs depends on the values of x and z, just as whether Bob flips his bit in the one-time pad protocol depends on the message Alice sends him. ### Completing the analogy to the one-time pad The measurement above, in the basis of $\|\beta_{zx}\rangle$ states, corresponds exactly to a parity measurement in the classical one-time pad scenario: the two bits x and z represent measured correlations between the spins s and A, in the standard basis and in the basis $\|+\rangle \propto \|0\rangle + \|1\rangle$ and $\|-\rangle \propto \|0\rangle - \|1\rangle$. In particular, you'll notice that whenever we obtain an outcome of $\|\beta_{zx}\rangle$ where x = 0, Bob's final state is incorrect only by the sign of the $\|1\rangle$ component; and when x = 1, then mapping each component of Bob's state from $\|0\rangle$ to $\|1\rangle$ (and vice-vera) gives him a state which is again correct up to a sign. The x bit corresponds exactly to the bit-parity measurement in a one-time pad classically; and without that bit, you could never hope to communicate classical information by teleportation. But on the plus side, although you have to send a bit of communication, whichever state s was in to begin with has been transmitted to Bob with perfect security! - Thanks for all that. I'm going to spend a while to try to learn as much of the notation and terminology as I can to see if I can fit even a small piece of this inside my head! – Rei Miyasaka Sep 3 '11 at 8:05 Of course you can "teleport" information. The problem is that to make this you still need classical channel which is limited by the speed of light. If you have no classical channel, teleportation process itself can not help to give any new information at recepient side due to its quantum nature. If you have no information of what happens on the other end of teleportation process from the other sources, results are just random for you. It is complicated, and I see no way to explain it without quantum mechanics. But the whole idea has a bit different formulation than stated in your question. Probably, you should read something about Quantum key distribution. It is more technical area and may be formulated in more technical terms. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 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": 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": 0.9527863264083862, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/167744-matrix-equation.html	# Thread: 1. ## Matrix equation Let M = { [a -b] , a,b E R } ..............[b a] Solve X^4+X^2+I=0, where X E M I attempted this question a while ago. I remember finding the eigenvalues and eigenvectors (X^4+X^2=-I and treat X^4+X^2 as a matrix itself) but it led to nowhere. I then tried to use a linear map but no luck either. I would love to hear ideas on how to solve this one. Thanks. 2. You would solve for the eigenvalues by solving the numeric equation $x^4+ x^2- 1= 0$. If we let $y= x^2$, then we have $y^2+ y- 1= 0$ which, by the quadratic formula, has roots $\frac{-1\pm i\sqrt{3}}{2}$. The eigenvalues of the original equation are the roots of those and so will be very messy! 3. $M = \begin{bmatrix}a & -b \\ b & a\end{bmatrix} = \gamma\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$ Where $\gamma = \sqrt{a^2 + b^2}$ and $\theta = \tan^{-1}(b/a)$. So M is a rotation by $\theta$ with a scaling by $\gamma$. Immediately using a geometric interpretation, we can see that a rotation by 60 degrees is a solution for X. Also a rotation by 120 degrees is another solution. There are more solutions, but hopefully this helps you get started finding roots. 4. An alternative: we can use the natural field isomorphism: $\varphi: \mathcal{M}\rightarrow{\mathbb{C}},\quad \varphi\begin{bmatrix}{a}&{-b}\\{b}&{\;\;a}\end{bmatrix}=a+bi$ so, solving $X^4+X^2+I=0$ in $\mathcal{M}$ it is equivalent to solve $z^4+z^2+1=0$ in $\mathbb{C}$. But, $z^4+z^2+1=(z^2+1)^2-z^2=(z^2+z+1)(z^2-z+1)=0$ $z=\dfrac{-1\pm\sqrt{3}i}{2},\;z=\dfrac{1\pm\sqrt{3}i}{2}$ As a consequence the roots of the given equation are the elements of the set: $\mathcal{S}=\left\{{\dfrac{1}{2}\begin{bmatrix}{-1}&{-\sqrt{3}}\\{\sqrt{3}}&{-1}\end{bmatrix},\;\dfrac{1}{2}\begin{bmatrix}{-1}&{\sqrt{3}}\\{-\sqrt{3}}&{-1}\end{bmatrix}, \;\dfrac{1}{2}\begin{bmatrix}{1}&{-\sqrt{3}}\\{\sqrt{3}}&{1}\end{bmatrix},\;\dfrac{1} {2}\begin{bmatrix}{1}&{-\sqrt{3}}\\{\sqrt{3}}&{1}\end{bmatrix}}\right\}$ Fernando Revilla 5. Thanks for your help guys. FernandoRevilla, I haven't learn anything about natural field isomorphism so I don't understand what you did there...Is that some kind of a linear map? Is there any link about it so I can have a read? I tried google but nothing really came up. EDIT: you can actually prove that its a linear map. =) Thanks that's a very pretty solution! snowtea, could you please explain how you got 60 as a solution using geometric interpretation? I didn't quite see that =S I get everything up to "So M is a rotation by with a scaling by gamma." X^4 and X^2 mean rotations by 4theta and 2theta respectively(with scalings too). But what about X^4+X^2+I? How did you see that 60 is a solution so quickly? 6. Let X be a rotation by 60 degrees (no scaling). Draw a vector v. Multiplying by I keeps it the same. Multiplying it by X^2 rotates it by 120 degrees. Multiplying it by X^4 rotates it by 240 degrees. (X^4+X^2+I) v is the sum of all the above vectors. You can show this is always 0 using some simple trigonometry. 7. Originally Posted by snowtea Let X be a rotation by 60 degrees (no scaling). Draw a vector v. Multiplying by I keeps it the same. Multiplying it by X^2 rotates it by 120 degrees. Multiplying it by X^4 rotates it by 240 degrees. (X^4+X^2+I) v is the sum of all the above vectors. You can show this is always 0 using some simple trigonometry. Thanks that makes sense 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": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9346800446510315, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/186607/does-an-orientable-subbundle-of-an-orientable-vector-bundle-always-have-a-orient	# Does an orientable subbundle of an orientable vector bundle always have a orientable complement? If I have an orientable vector bundle $E$ and a subbundle $F$ on a manifold $M$, where both the bundles are orientable, does $F$ have a complement in $E$ which is also orientable? Does it have a complement bundle at all? That is, a subbundle of $E$ that is pointwise a complement of $F$. What if $F$ is of codimension $1$? Is the complement always trivial in this case? Does anything change if $E$ is specifically the tangent bundle $TM$? - ## 2 Answers If $E=F\oplus G$, then $\Lambda^\det E=\Lambda^\det F\otimes\Lambda^\det G$. If both $\Lambda^\det E$ and $\Lambda^\det G$ are trivial, then so is $\Lambda^\det F$, because the three being line bundles, you can «divide by $\Lambda^\det G$» in that equality. - By $\Lambda^\det$ I mean the maximal exterior power here. – Mariano Suárez-Alvarez♦ Aug 25 '12 at 3:36	{"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": 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": 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": 0.9112259149551392, "perplexity_flag": "head"}
http://mathoverflow.net/questions/78278/is-su-infty-amenable	## Is $SU(\infty)$ amenable? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) We can write the finitary special unitary group $SU(\infty)$ as the direct limit $\varinjlim SU(n)$ of ordinary special unitary groups. These groups $SU(n)$ are compact, thus amenable. In other words each of them has an invariant mean (in this case, from Haar measure). Is $SU(\infty)$ amenable? Of course one then asks the same question for the full unitary group $U(\infty)$, the special orthogonal group $SO(\infty)$ and the unitary symplectic (quaternion unitary) group $Sp(\infty)$. - I think that it contains free groups, isn't it? – Valerio Capraro Oct 16 2011 at 19:44 @Valerio Capraro Certainly it does, but that only makes it non-amenable as a discrete group. – John Wiltshire-Gordon Oct 16 2011 at 21:12 5 Sorry if this is silly but what exactly do you mean by amenability for non locally compact groups such as $SU(\infty)$? I'm not at all an expert on this but one definition I guess could be that if $G$ acts on a compact space then there is an invariant probability measure. With that definition I think any direct limit of amenable (in that sense) groups is amenable by weak compactness of probability measures on a compact space. – Vitali Kapovitch Oct 16 2011 at 21:53 ## 1 Answer The answer is that $G=SU(\infty)$ (with the direct limit topology of the usual Hilbert-Schmidt topologies) is extremely amenable. This means (by definition) that every continuous action of $G$ on a compact set has a fixed point. This was proved as an application of the isoperimetric inequality by Gromov and Milman M. Gromov and V.D. Milman, A topological application of the isoperimetric inequality, Amer. J. Math. 105 (1983), 843–854. Since $SU(\infty)$ is not locally compact, various characterizations of amenability have to be adapted. One way to define amenability for such groups is to say that there exists a $G$-invariant mean on the algebra of bounded uniformly continuous real-valued functions on $G$. Extreme amenability then gives even the existence of a $G$-invariant character on this algebra. This is somewhat unintuitive, since obviously no compact group can admit such a character on the algebra of continuous functions on it. Extreme amenability is a concept which is related to phenomena of measure concentration. Gromov and Milman proved (as an application of lower bounds on the Ricci curvature) that for every sequence of measurable subsets $A_n \subset SU(n)$ with $\liminf_{n \to \infty} \mu_n(A_n) \neq 0$, one has $$\lim_{n \to \infty} \mu_n(A_{n,\epsilon}) = 1.$$ Here, $\mu_n$ denotes the normalized Haar measure and $A_{n,\varepsilon}$ the $\varepsilon$-neighborhood of $A_n$ in the unnormalized Hilbert-Schmidt metric. This concentration phenomenon can be used to prove extreme amenability of $SU(\infty)$. - I found another affirmative answer in Example 3.4 of a paper "Functional Analytic Background for a Theory of Infinite-Dimensional Reductive Lie Groups" by Daniel Beltita. He shows that if $G$ is a topological group and there is a directed system $\{G_\alpha\}_{\alpha \in A}$ of amenable topological subgroups whose union is dense in $G$ then $G$ is amenable. Here "amenable" means that there is a left-invariant mean. – Joseph Wolf Oct 25 2011 at 18:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "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": 0.9356181025505066, "perplexity_flag": "head"}
http://mathoverflow.net/questions/4994?sort=oldest	## Fundamental Examples ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is not unusual that a single example or a very few shape an entire mathematical discipline. Can you give examples for such examples? (One example, or few, per post, please) I'd love to learn about further basic or central examples and I think such examples serve as good invitations to various areas. (Which is why a bounty was offered.) Related MO questions: What-are-your-favorite-instructional-counterexamples, Cannonical examples of algebraic structures, Counterexamples-in-algebra, individual-mathematical-objects-whose-study-amounts-to-a-subdiscipline, most-intricate-and-most-beautiful-structures-in-mathematics, counterexamples-in-algebraic-topology, algebraic-geometry-examples, what-could-be-some-potentially-useful-mathematical-databases, what-is-your-favorite-strange-function To make this question and the various examples a more useful source there is a designated answer to point out connections between the various examples we collected. In order to make it a more useful source, I list all the answers in categories, and added (for most) a date and (for 2/5) a link to the answer which often offers more details. (~year means approximate year, year means a year when an older example becomes central in view of some discovery, year? means that I am not sure if this is the correct year and ? means that I do not know the date. Please edit and correct.) Of course, if you see some important example missing, add it! Logic and foundations: $\aleph_\omega$ (~1890), Russell's paradox (1901), Halting problem (1936), Goedel constructible universe L (1938), McKinsey formula in modal logic (~1941), 3SAT (1970), The theory of Algebraically closed fields (ACF) (?), Physics: Brachistochrone problem (1696), Ising model (1925), The harmonic oscillator,(?) Dirac's delta function (1927), Heisenberg model of 1-D chain of spin 1/2 atoms, (~1928), Feynman path integral (1948), Real and Complex Analysis: Harmonic series (14th Cen.) {and Riemann zeta function (1859)}, the Gamma function (1720), li(x), The elliptic integral that launched Riemann surfaces (1854?), Chebyshev polynomials (?1854) punctured open set in C^n (Hartog's theorem 1906 ?) Partial differential equations: Laplace equation (1773), the heat equation, wave equation, Navier-Stokes equation (1822),KdV equations (1877), Functional analysis: Unilateral shift, The spaces $\ell_p$, $L_p$ and $C(k)$, Tsirelson spaces (1974), Cuntz algebra, Algebra: Polynomials (ancient?), Z (ancient?) and Z/6Z (Middle Ages?), symmetric and alternating groups (1832), Gaussian integers ($Z[\sqrt -1]$) (1832), $Z[\sqrt(-5)]$,$su_3$ ($su_2)$, full matrix ring over a ring, $\operatorname{SL}_2(\mathbb{Z})$ and SU(2), quaternions (1843), p-adic numbers (1897), Young tableaux (1900) and Schur polynomials, cyclotomic fields, Hopf algebras (1941) Fischer-Griess monster (1973), Heisenberg group, ADE-classification (and Dynkin diagrams), Prufer p-groups, Number Theory: conics and pythagorean triples (ancient), Fermat equation (1637), Riemann zeta function (1859) eliptic curves, transendental numbers, Fermat hypersurfaces, Probability: Normal distribution (1733), Brownian motion (1827), The percolation model (1957), The Gaussian Orthogonal Ensemble, the Gaussian Unitary Ensemble, and the Gaussian Symplectic Ensemble, SLE (1999), Dynamics: Logistic map (1845?), Smale's horseshoe map(1960). Mandelbrot set (1978/80) (Julia set), cat map, (Anosov diffeomorphism) Geometry: Platonic solids (ancient), the Euclidean ball (ancient), The configuration of 27 lines on a cubic surface, The configurations of Desrague and Pappus, construction of regular heptadecagon (1796), Hyperbolic geometry (1830), Reuleaux triangle (19th century), Fano plane (early 20th century ??), cyclic polytopes (1902), Delaunay triangulation (1934) Leech lattice (1965), Penrose tiling (1974), noncommutative torus, cone of positive semidefinite matrices, the associahedron (1961) Topology: Spheres, Figure-eight knot (ancient), trefoil knot (ancient?) (Borromean rings (ancient?)), the torus (ancient?), Mobius strip (1858), Cantor set (1883), Projective spaces (complex, real, quanterionic..), Poincare dodecahedral sphere (1904), Homotopy group of spheres, Alexander polynomial (1923), Hopf fibration (1931), The standard embedding of the torus in R^3 (1934 in Morse theory), pseudo-arcs (1948), Discrete metric spaces, Sorgenfrey line, Complex projective space, the cotangent bundle (?), The Grassmannian variety,homotopy group of spheres (1951), Milnor exotic spheres (1965) Graph theory: The seven bridges of Koenigsberg (1735), Petersen Graph (1886), two edge-colorings of K_6 (Ramsey's theorem 1930), K_33 and K_5 (Kuratowski's theorem 1930), Tutte graph (1946), Margulis's expanders (1973) and Ramanujan graphs (1986), Combinatorics: tic-tac-toe (ancient Egypt(?)) (The game of nim (ancient China(?))), Pascal's triangle (China and Europe 17th), Catalan numbers (mid 19th century), (Fibonacci sequence (12th century; probably ancient), Kirkman's schoolgirl problem (1850), surreal numbers (1969), alternating sign matrices (1982) Algorithms and Computer Science: Newton Raphson method (17th century), Turing machine (1937), RSA (1977), universal quantum computer (1985) Social Science: Prisoner's dilemma (1950) (and also the chicken game, chain store game, and centipede game), the model of exchange economy, second price auction (1961) - 15 I think that this should be community wiki. – Andrew Stacey Nov 11 2009 at 7:55 12 @Jose: Hard to say exactly. My instinct is that the kind of answers that this question will garner are those that didn't involve much actual thought, and the votes up or down will be more an assessment of whether the voter liked the example rather than whether the voter liked the answer (which, ideally, should contain an explanation of why that example shaped the discipline); both of these indicate that the answerers should not gain reputation for their answers, hence community wiki. – Andrew Stacey Nov 11 2009 at 9:50 4 I've hit this with the wiki hammer. – Scott Morrison♦ Nov 11 2009 at 19:34 7 I can't imagine a counterexample to the following rule: Any question whose purpose is to produce a sorted list of resources (i.e. the question includes, or should include, "one per post please") should be community wiki. – Anton Geraschenko♦ Nov 12 2009 at 8:03 8 Why does this question have a bounty anyway? – Kevin Lin Nov 21 2009 at 17:33 show 17 more comments ## 129 Answers The harmonic oscillator is a fundamental example in both classical and quantum mechanics. - 1 Of course, for this question and many others there is no meaning to "correct answer". In fact I liked all the answers to the question and I hope more answers will come along. Jose was the most valuable partner to this endeavor and he contributed several great answers both before and after the boundy was announced. – Gil Kalai Nov 26 2009 at 12:39 1 @Gil, now that the bounty has been delivered you might "unaccept" this answer. – Scott Morrison♦ Nov 30 2009 at 16:20 1 Is it possible? Is it moral? – Gil Kalai Dec 1 2009 at 14:53 ### 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. There's always the venerable normal distribution (for probability theory). - The complex projective space is the fundamental example in toric geometry, symplectic and GIT quotients,... - The cotangent bundle is the fundamental example of symplectic manifold/phase space. - It has often been said that if you understand $su_3$ you understand all simple Lie algebras, so that should make it the fundamental example. (Personally I think that it suffices to understand $su_2$!) - 3 On the same line, SU(2) (compared to U(1), which is abelian) already shows a lot of features of the compact non abelian groups (Peter-Weil theorem, irreducible representations of (every) dimension greater than 1, ...). – Gian Maria Dall'Ara Nov 11 2009 at 9:11 1 I guess you could say that the heat equation "shaped" functional analysis since so many of the early tools were developed to study just that equation (more so if you say fourier analysis). Similarly (as has been noted) the spheres are currently shaping algebraic topology since so many of the tools are developed to get at the stable homotopy of spheres. I didn't put my comment on your "harmonic oscillator" example because that has played a role in shaping quantum mechanics. Part of it is purely timing: good examples have a chance to shape a subject if they are encountered in its infancy. – Andrew Stacey Nov 11 2009 at 10:33 show 2 more comments I do think Milnor's exotic sphere distinguish differential topology from general topology, but i don't know if this is an example of the kind you want. Answered by: Yuhao Huang - In the theory of holomorphic functions of several variables, Hartogs's theorem that any holomorphic function on a punctured open set of ( ) can holomorphically be continued through the deleted point really started the subject and is still the most spectacular divide between and [ where it is completely false : on look at 1/z or, worse, exp(1/z): these functions clearly can't be continued holomorphically through zero] - 1 Dear Gil and Jonas:thanks for the comments and, yes, you are both right. Hartogs's example-theorem shows that there exist in C^n domains which are not regions of holomorphy,a phenomenon impossible in dimension one.This launched the notion of pseudoconvex domains (which exclude Hartogs-type extensions of holomorphic functions) and ultimately led to the concept of Stein manifolds, central in complex geometry (the analogues of affine varieties in algebraic geometry) – Georges Elencwajg Nov 11 2009 at 10:28 show 3 more comments The full matrix rings of any order over another ring (and their direct union of row-finite, column-finite matrices) are a fundamental example for Noncommutative Ring Theory: they are simple enough to be easily understood "in a glimpse" but complex enough to highlight many interesting concepts of the theory. - Laplace's equation is the fundamental example of a PDE. If I could broaden the question to allow a triumvirate of examples, I'd say Laplace's equation, the heat equation, and the wave equation are the canonical examples of PDEs, representing elliptic, parabolic, and hyperbolic equations respectively. - The torus is THE example in many branches of math. Algebraic Topology: it is the example where you can compute explicitly its fundamental group as well as its covering spaces, universal cover and everything else. Algebraic geometry: provided it is smooth, is a cubic which is not rational. It is a compact Lie group (and then a non so trivial example of a trivial tangent bundle). It is compact in and has zero curvature (Riemannian geometry). It is a Riemann Surface. Actually one can compute very explicitly the moduli space of such Riemann surfaces as well as (very explicitly) the mapping class group (which is a beautiful example). It is an elliptic curve, an abelian variety... and it has all those properties I don't know about...meaning, an endless amount of properties. - 3 But in what way has the torus shaped any of these subjects? It's certainly a good example demonstrating many of the features, but I don't see (from your answer) that it has played a significant role in shaping them. – Andrew Stacey Nov 11 2009 at 9:51 5 "Algebraic Topology: it is the example where you can compute explicitly its fundamental group as well as its covering spaces, universal cover and everything else." I don't see how this distinguishes the torus from any other closed surface, from the algebro-topological point of view. Indeed, if you're interested in the fundamental group, the fact that it's abelian in this case makes it highly unrepresentative. – HW Nov 11 2009 at 17:36 The Fano plane in finite geometry http://en.wikipedia.org/wiki/Fano_plane - The Prisoner's Dilemma in Game Theory. - 2 Prisoner's Dilemma is certainly a fundamental example but there are other games which I think have been as rich in encouraging interesting research: Chicken en.wikipedia.org/wiki/Chicken_game Chain store en.wikipedia.org/wiki/Chainstore_paradox centipede en.wikipedia.org/wiki/Centipede_game One reason these are interesting games is that help one try to understand what is "rational" behavior and help distinguish between what people do in practice as compared with some "abstract model" of rationality. – Joseph Malkevitch Dec 25 2009 at 15:10 show 2 more comments A lot of algebraic topology was developed with computing the higher homotopy groups of spheres in mind. - - Ravi Vakil gives interesting examples in algebraic geometry: "The existence of some of these pathologies is common knowledge'', but I had never known what they were.". - The Noncommutative Torus in non-commutative geometry. (Maybe it has just shaped the subject because it is about the only thing one can handle explicitly) - Answered by Agol: The figure eight knot (complement) is the starting point for much of hyperbolic geometry. Although other hyperbolic manifolds were discovered before it, the figure eight knot complement has one of the simplest hyperbolic structures to analyze. Thurston first proved his hyperbolic Dehn surgery theorem for the figure eight knot complement - after understanding the proof in this case, the general case is not much harder to understand. It is the simplest knot for which every 3-manifold is a branched cover over it. It was one of the first (non-torus) knots for which the knot-complement problem was proven. It has the most number of non-hyperbolic Dehn-fillings over any one-cusped hyperbolic 3-manifold. It is the smallest volume orientable hyperbolic manifold with one cusp. It was the first knot proven that all non-trivial Dehn fillings have a finite-sheeted cover with positive first betti number. It was the first knot for which the volume conjecture has been verified. (see also this Wikipedia article.) - The KdV equation in integrable systems. It was through a numerical study of KdV that the word soliton was coined. This numerical study lead to much analytical work, including the development of Lax Pairs. (Answer by Aaron Hoffman) - The ring ${Z}[\sqrt{-5}]$ is a fundamental example of non-unique factorization in rings in algebraic integers. Perhaps of more historical relevance is the example of p=37 that shows that Lamé's "proof" of Fermat's Last Theorem fails. I'm not sure Kummer used p=37 though. In any case, examples of non-unique factorization in rings in algebraic integers lead to the whole theory of ideal numbers and later Dedekind domains ans set the tone for algebraic number theory. - 1 Unique prime factorization fails for the ring of integers underlying the p = 23 case of FLT. Kummer's method of proof works for p = 23, but not p = 37 (as Kummer was well aware of). – Jonah Sinick Nov 16 2009 at 19:35 show 1 more comment For a picture that launched a thousand papers, I'd nominate the bifurcation diagram of the logistic map. (image via Wikipedia). Answered by Martin M. W. - 1 If pictures are allowed then the Mandelbrot set and its Julia sets are also quite fundamental as prime examples of what lies hidden in dynamical systems. Of course, as the Wikipedia page on the Mandelbrot set says, there is a close relation between the Mandelbrot set and the bifurcation diagram of the logistic map. – lhf Nov 11 2009 at 18:37 3 Roughly speaking, the bifurcation diagram is what you get by going through the Mandelbrot set along the real axis. – Lasse Rempe-Gillen Mar 3 2010 at 10:54 show 1 more comment The Petersen Graph in graph theory - The Fermat Equation xn + yn - zn = 0. This has truly been much more than an example in both algebra and number theory: it was one of the main motivations to develop the theory of unique factorization domains, Dedekind domains, class numbers, regular primes, etc. in the 19th century. In the late 20th century it provided a motivation for Wiles to work on modularity of elliptic curves. In the 21st century, the equation c1 xa + c2 yb - c3 zc = 0 is similarly motivational for things like Q-curves, Galois representations, hypergeometric abelian varieties... Answer by Peter L. Clark: - The integral $\int \frac{dx}{\sqrt{x(x-1)(x-\lambda)}}$ for $\lambda\neq 0,1$ essentially launched both complex analysis and algebraic geometry, via Riemann's discovery of the Riemann surface that is the natural domain of a function, leading to both analytic theory of Riemann surfaces and to the study of algebraic curves, leading to...complex analysis including in several variables and complex algebraic geometry, as we now know it. - 9 This great answer have earned Charles the first gold badge in mathoverflow history. Congratulations, Charles! – Gil Kalai Nov 27 2009 at 12:38 In convex geometry, the Euclidean ball. In fact (as I think Gil knows but many other readers here probably don't) a huge portion of (high-dimensional) convex geometry consists of results that show that arbitrary high-dimensional convex bodies behave like the Euclidean ball in various ways. And if I may be permitted to add another complementary example or two, the simplex and cube are for many purposes the least "Euclidean ball-like" convex bodies, so they are useful for understanding the limitations of the Euclidean ball as a prototype for arbitrary bodies. - The cyclic polytope in the study of convex polytopes in high dimensions. It is the convex hull of n points on the moment curve (t,t^2,t^3,...,t^d). It is simplicial and has the property that every [d/2] points form a face. (So, for example, in 4 dimension every two vertices form an edge.) - Recursion theory has the halting problem. From it, you obtain the first example of an interesting Turing degree, and generalizing it, you get the Turing jump operator. The halting problem also plays an important role in (the boundary of) computational complexity theory, since it's one of the main tools for demonstrating the insolubility of computational problems. - In geometry, group theory and other areas: The Leech Lattice: - In complex dynamics: The Mandelbrot set - In modal logic there is a particularly simple formula, called McKinsey formula: ◻⋄p→⋄◻p. It is so simple, yet it defines a frame property which cannot be expressed in first-order logic. Also, with the right selection of other formulas, it gives rise to frame incompleteness examples (logics that are consistent, but are not logics of any class of frames whatsoever). - Relevant to many areas (but mostly topology) is the Cantor set. It is an example of a set with properties too numerous to list here. To name a few: it is uncountable, compact, nowhere dense and it has Lebesgue measure 0. - 5 Another good reason: Every compact Hausdorff space is a quotient of the Cantor set!! See Tom Leinster's answer here mathoverflow.net/questions/5357/… It's just incredible - I still can't get over it. – Peter Arndt Nov 20 2009 at 15:30 5 @Peter Arndt: Every compact Hausdorff metrizable (equivalently, second-countable) space is a quotient of the Cantor set. There are compact Hausdorff spaces of greater than continuum cardinality, and these evidently are not quotients of the Cantor set. – Pete L. Clark Jan 15 2010 at 10:30	{"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": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9072248339653015, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/190146/what-is-the-inverse-function-of-fx-192x-16x2	# What is the Inverse function of $f(x)=192x-16x^2$? What is the inverse function of: $$f(x)=192x-16x^{2}$$ I have been finding myself going in a circle in trying to complete this problem which otherwise looks simple, but for some reason I am at a block. Could someone help explaining how to go about this problem? - This function is not monotone since $f'(12)=0$ hence it is not $1-1$ thus not invertible – Belgi Sep 2 '12 at 19:14 1 @Belgi Even though your four pieces of fact are correct, the "hence"-part is wrong. There are loads of non-monotone functions which are 1-1 and perfectly invertible, even assuming everywhere-differentiability. $x^3$ is the easiest example. – Arthur Sep 2 '12 at 19:26 @Arthur $x^3$ is monotone. What it fails is having a nonvanishing derivative. If you want a non-monotonic injective function, by IVT it will have to be discontinuous. – Logan Maingi Sep 2 '12 at 19:31 @Arthur: $f(x)=x^3$ is monotone. It’s even strictly monotone. – Brian M. Scott Sep 2 '12 at 19:32 I'm sorry, it's the "since"-part that is wrong, then. – Arthur Sep 2 '12 at 19:36 show 2 more comments ## 4 Answers Instead of completing the square, as in lab bhattacharjee’s answer, you can use the quadratic formula. You have $y=192x-16x^{2}$; rewrite it as $16x^2-192x+y=0$, and treat $y$ as the constant term to get $$x=\frac{192\pm\sqrt{192^2-64y}}{32}=6\pm\frac{\sqrt{64(576-y)}}{32}=6\pm\frac{\sqrt{576-y}}4\;.$$ This gives you two functions, both defined for $y\le 576$: $$x=6+\frac{\sqrt{576-y}}4\;,$$ and $$x=6-\frac{\sqrt{576-y}}4\;.$$ The first corresponds to the righthand side of the parabola $y=192x-16x^2$, and the second to its lefthand side. - The answers of @nayrb and @lab bhatacharjee are correct as far as they go, but as a (retired) teacher, I feel that I must address a fault of the math education system, here in the States as well as, perhaps, elsewhere. To describe a function, you need to mention the domain and the target space, and that is particularly important here. Sketch the graph! You see that there’s a maximum at $(6,576)$, and that on either side of the vertical line $x=6$, there are points on the graph at equal height. So the function fails the “horizontal line test” unless you restrict the domain. Let’s restrict to the interval $\langle-\infty,6]$, i.e. the closed half-line to the left of $6$. But what about our inverse function? It’s not defined for $y>576$, and that means that its domain has to be no bigger than $\langle-\infty,576]$. It’s only now, once we’ve restricted both the domain and the target space of our original function, that we really have an inverse function, and its formula is indeed the one given by @lab, with the minus sign of course. - Hi Jonathan! This originated in my calculus class here at UConn. Thanks for the consultation. – Jeremy Teitelbaum Sep 2 '12 at 19:43 A function is an object with three parts: a domain, a range (allowable set out outputs) and a rule tying each element of the domain to some element of the range. Failure to specify these is failure to completely specify a function. If you are asking if a function is 1-1 or onto, this is especially important. – ncmathsadist Sep 2 '12 at 20:18 @ncmathsadist, thanks for the confirmation. Only thing is, in high-schools nowadays, it seems that the word “range” is used for what I learned to call the image, and no word at all is used for what I have just called the target space. So I think that the word “range” can be misunderstood by many. – Lubin Sep 3 '12 at 20:49 People fail to make this important distinction. The range is the set of allowable outputs. The image is the set of realized outputs. When the image and the range coincide, the function is onto. I have taught HS math and computer science. I always have a discussion about the importance of specifying domain and range in functions in math and CS classes. We call them "pure functions" in a CS class. – ncmathsadist Sep 3 '12 at 20:51 Sorry to rejoin this discussion late. @ncmathsadist, you and I are on the same side, but I don’t think we have seen the same phenomena. In the high-school where I volunteer, the “range” is the set of values taken on. And I have to say that more than 50 years ago, George Mackey, a far greater mathematician then either of us, routinely said in class, “Let $f$ be a function with domain $S$, and with range in $\mathbb{R}$.” To me, this says clearly that for him, the range was the set of values taken on. For this reason, I sedulously avoid the use of the word. – Lubin Sep 5 '12 at 0:59 show 1 more comment Let $f(x)=y=192x-16x^2\implies -y=16(x^2-12x)=16(x-6)^2-16\cdot 36$ $\implies 576-y=16(x-6)^2$ $\implies x=f^{-1}(y)=6±\frac{\sqrt{576-y}}{4}$ Clearly $y ≤576$ to make $f(y)$ invertible in real numbers. - you lost me at 16(x-6)^2. where does that come from? I appreciate the help – Carl Strum Sep 2 '12 at 19:20 @CarlStrum $-y=16(x^2-12x)= 16(x^2 -12x +36 -36) = 16((x-6)^2 -36) =16(x-6)^2-16\cdot 36$ – Arthur Sep 2 '12 at 19:22 Do you think you could use words to explain what you just did? How did you just know to choose 36 to add/subtract to that equation. (i mean i know it doesn't change the problem, but how did you know to use the specific number 36?) – Carl Strum Sep 2 '12 at 19:27 $x^2-12x=(x)^2-2(x)(6)+(6)^2-(6)^2=(x-6)^2-36$, is it ok? – lab bhattacharjee Sep 2 '12 at 19:29 I appreciate all the help but I feel like I'm being misunderstood. I am interested in how you knew to use whatever technique it was to get those answers. I.E. I am a little behind on the concept of "completing the square" if that is what's being used to solve the problem. – Carl Strum Sep 2 '12 at 19:33 show 1 more comment The function will only have an inverse if you restrict it to a region where it is one-to-one. You can look at this graph: and see that you need to restrict your function to one of the regions where $6\le x\le \infty$ and $-\infty<y\le 576$ or $-\infty<x\le 6$ and the same $y$-region. The quadratic formula (as indicated above) can give you the proper formula. Wolfram Alpha works it out nicely here: Wolfram Alpha Knows the Quadratic Formula You can "show steps" to see how Wolfram Alpha does this; it's essentially completing the square. The two solutions correspond to the two branches of the inverse function. - Looks like you’re relatively new here. Welcome! - jdL – Lubin Sep 2 '12 at 19:47 I'm trying to see if I can get my Calc students to use this site as a resource. – Jeremy Teitelbaum Sep 2 '12 at 19:51 It’s more than a resource, ’cause they’ll see mathematics at many different levels. Should be an eye-opener! – Lubin Sep 2 '12 at 19:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9337806105613708, "perplexity_flag": "middle"}
http://mathhelpforum.com/trigonometry/127744-proof-required.html	# Thread: 1. ## proof required... if cosA+cosB+cosC=0=sinA+sinB+sinC prove that cos^2 A + cos^2 B + cos^2 C = sin^2 A + sin^2 B + sin^2 C = 3/2 I have no idea how to equate the two. This was part of the questions in the chapter on complex numbers, so perhaps I should have included it in number theory. But any proof is welcome. 2. Originally Posted by nahduma if cosA+cosB+cosC=0=sinA+sinB+sinC prove that cos^2 A + cos^2 B + cos^2 C = sin^2 A + sin^2 B + sin^2 C = 3/2 I have no idea how to equate the two. This was part of the questions in the chapter on complex numbers, so perhaps I should have included it in number theory. But any proof is welcome. Let $u = e^{iA}\ (=\cos A + i\sin A)$, and similarly $v = e^{iB}$ and $w = e^{iC}$. Then $u+v+w = 0$. Also, u, v and w all have absolute value 1, so their inverses are equal to their complex conjugates: $u^{-1} = \overline{u}$, and similarly for v and w. Then $uv+vw+wu = uvw\bigl(u^{-1} + v^{-1} + w ^{-1}\bigr) = uvw\bigl(\overline{u+v+w}\bigr) = 0$. Thus $0 = (u+v+w)^2 = u^2+v^2+w^2 + 2(uv+vw+wu)$, and it follows that $u^2+v^2+w^2 = 0$. Take the real part of that to see that $\cos(2A) + \cos(2B) + \cos(2C) = 0$. The results then follow from the facts that $\cos^2A = \tfrac12\bigl(1+\cos(2A)\bigr)$ and $\sin^2A = \tfrac12\bigl(1-\cos(2A)\bigr)$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9443057775497437, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/19218/certain-material-heating-water-in-a-recipient/19234	# Certain material heating water in a recipient I don't know how to resolve a problem, but I don't want the answer since I'm almost going to have it resolved. What the problem says is we have 85 liters of water at 7ºC in an iron pot of 29kg. We want the water to be at 86ºC. The temperature of the iron pot is 12ºC. The water is heated by fire wood (65% of the thermal energy is wasted in combustion), and has a heating value of 12 MJ/kg. We need to determine the amount of wood needed to heat the water. I would apply the normal formula of energy transfer: m1c1(t2-t1)=m2c2(t2-t1). But since we are using fire wood there, I don't know how to resolve this problem, probably because I'm missing something there... and I don't know what should I do. Which formula or principle would I need to use when having the material and the heating value of the material used? - 1 Hi Anne - it sounds like you're talking about a homework problem, right? I added the tag for it. As for your question, it's not bad, but I think you're omitting some information that we need in order to help you out. In fact, you come across as if you're trying to avoid revealing the true nature of the problem. Unless there really is some reason to keep your work super-secret, it's better if you don't do that. Give us everything, including the full text of the problem. The more information we have to work with, the better we can help you. – David Zaslavsky♦ Jan 7 '12 at 6:06 1 (cont.) For example, what values do you have that you can plug into the equation you mentioned? Once you've figured that out, what values do you still need, and why aren't you able to get them? – David Zaslavsky♦ Jan 7 '12 at 6:08 Hi David! Well, it's not actual homework but let's take it like that. I didn't posted the full problem since I didn't wanted an answer. What the problem says is we have 85 liters of water at 7ºC in an iron pot of 29kg. We want the water to be at 86ºC. The temperature of the iron pot is 12ºC. The water is heated by fire wood (65% of the thermal energy is wasted in combustion), and has a heating value of 12 MJ/kg. We need to determine the amount of wood needed to heat the water. And I don't know how I should understand this with the formula I posted... – Anne Jan 7 '12 at 6:12 1 OK, well, the homework tag is for homework-like questions so it probably applies even if this isn't actually homework. Also, we're not going to give you a full answer even if you do post the whole problem - it's right there in our homework FAQ that giving complete answers to homework-like questions is against the site policy. So don't worry about that! I'd suggest just editing the text of the problem (basically what you just put in your comment) into your question. – David Zaslavsky♦ Jan 7 '12 at 6:14 1 ... and it's done! Thanks David. I felt guilty last time because of that and didn't want to make the same mistake again. I'm trying to learn, that's why I don't want the final answer, but how to continue resolving my problem, since I'm unable to plug all the data in the formula - so probably I'm missing something as I said. – Anne Jan 7 '12 at 6:18 show 5 more comments ## 3 Answers If you know the specific heat of water and iron you can work out how much energy is needed to bring the water and the pot to 86°C. You need to work out how many kg of wood you need to burn to produce that amount of energy. The question even tells you how much of the energy produced by burning the wood is wasted and doesn't go towards heating the pot and water. Whoever set the question is probably looking for the simple answer, but there are lots of extra bits you could have fun with. For example you're told the volume of the water not it's mass, but specific heat is normally quoted per unit mass. That means you need to know the density of water at 7°C to work out it's mass, and the density isn't exactly 1g/cm$^3$. Also the specific heat of both water and iron are functions of temperature. - Which formula or principle would I need to use when having the material and the heating value of the material used? I would suggest starting with the conservation of energy. Have you tried drawing a block diagram showing where the energy (heat) is going from/to within the limits of the setup? Lastly keep an eye on the first law of thermodynamics... - I would apply the normal formula of energy transfer: m1c1(t2-t1)=m2c2(t2-t1). That's not the "normal formula of energy transfer". That's just the formula used when you mix liquids. A more generalized formula for energy transfer would be "heat lost=heat gained", or the first law of thermodynamics $$\Delta Q=\Delta U+\Delta W$$ Where $\Delta Q$ is heat supplied, $\Delta U$ is change in internal energy, and $\Delta W$ is work done by system. In the case of your problem, as there is no work being done, just balance the heats. See how much heat goes into the system, and then compare this with the total heat required (sum of all $mc(T_2-T_1)$) to raise the temperature of the system. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9721000790596008, "perplexity_flag": "head"}
http://terrytao.wordpress.com/tag/convex-geometry/	What’s new Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao # Tag Archive You are currently browsing the tag archive for the ‘convex geometry’ tag. ## (Gil Kalai) The weak epsilon-net problem 22 April, 2007 in guest blog, math.MG, question \| Tags: convex geometry, entropy, epsilon-net, Gil Kalai \| by Gil Kalai \| 13 comments [This post is authored by Gil Kalai, who has kindly “guest blogged” this week’s “open problem of the week”. - T.] This is a problem in discrete and convex geometry. It seeks to quantify the intuitively obvious fact that large convex bodies are so “fat” that they cannot avoid “detection” by a small number of observation points. More precisely, we fix a dimension d and make the following definition (introduced by Haussler and Welzl): • Definition: Let $X \subset {\Bbb R}^d$ be a finite set of points, and let $0 < \epsilon < 1$. We say that a finite set $Y \subset {\Bbb R}^d$ is a weak $\epsilon$-net for X (with respect to convex bodies) if, whenever B is a convex body which is large in the sense that $\|B \cap X\| > \epsilon \|X\|$, then B contains at least one point of Y. (If Y is contained in X, we say that Y is a strong $\epsilon$-net for X with respect to convex bodies.) For example, in one dimension, if $X = \{1,\ldots,N\}$, and $Y = \{ \epsilon N, 2 \epsilon N, \ldots, k \epsilon N \}$ where k is the integer part of $1/\epsilon$, then Y is a weak $\epsilon$-net for X with respect to convex bodies. Thus we see that even when the original set X is very large, one can create a $\epsilon$-net of size as small as $O(1/\epsilon)$. Strong $\epsilon$-nets are of importance in computational learning theory, and are fairly well understood via Vapnik-Chervonenkis (or VC) theory; however, the theory of weak $\epsilon$-nets is still not completely satisfactory. One can ask what happens in higher dimensions, for instance when X is a discrete cube $X = \{1,\ldots,N\}^d$. It is not too hard to cook up $\epsilon$-nets of size $O_d(1/\epsilon^d)$ (by using tools such as Minkowski’s theorem), but in fact one can create $\epsilon$-nets of size as small as $O( \frac{1}{\epsilon} \log \frac{1}{\epsilon} )$ simply by taking a random subset of X of this cardinality and observing that “up to errors of $\epsilon$“, the total number of essentially different ways a convex body can meet X grows at most polynomially in $1/\epsilon$. (This is a very typical application of the probabilistic method.) On the other hand, since X can contain roughly $1/\epsilon$ disjoint convex bodies, each of which contains at least $\epsilon$ of the points in X, we see that no $\epsilon$-net can have size much smaller than $1/\epsilon$. Now consider the situation in which X is now an arbitrary finite set, rather than a discrete cube. More precisely, let $f(\epsilon,d)$ be the least number such that every finite set X possesses at least one weak $\epsilon$-net for X with respect to convex bodies of cardinality at most $f(\epsilon,d)$. (One can also replace the finite set X with an arbitrary probability measure; the two formulations are equivalent.) Informally, f is the least number of “guards” one needs to place to prevent a convex body from covering more than $\epsilon$ of any given territory. • Problem 1: For fixed d, what is the correct rate of growth of f as $\epsilon \to 0$? Read the rest of this entry »	{"extraction_info": {"found_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": 30, "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": 0.9487913250923157, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2007/06/18/cones-and-cocones/?like=1&source=post_flair&_wpnonce=97257dee4e	# The Unapologetic Mathematician ## Cones and cocones There are a few auxiliary concepts we’ll need before the next major topic. Let’s start with two categories $\mathcal{J}$ and $\mathcal{C}$, and the category of functors $\mathcal{C}^\mathcal{J}$. If the following seems very complicated, consider $\mathcal{J}$ to be any particular toy category you’d like, so this category of functors is a diagram category. Now, for every object $C\in\mathcal{C}$ there is a constant functor that sends every object of $\mathcal{J}$ to $C$, and every morphism to the identity on $C$. Actually, this assignment of the constant functor to an object of $C$ is a functor from $\mathcal{C}$ to $\mathcal{C}^\mathcal{J}$. Indeed, given a morphism $f:C\rightarrow D$ we get a natural transformation from the one constant functor to the other, whose component at each object of $\mathcal{J}$ is $f$. We call this the “diagonal” functor $\Delta:\mathcal{C}\rightarrow\mathcal{C}^\mathcal{J}$. That is, $\Delta(C)$ is the constant functor with value $C$. Let $F:\mathcal{J}\rightarrow\mathcal{C}$ be some particular functor in $\mathcal{C}^\mathcal{J}$. A cone is an object in the comma category $(\Delta\downarrow F)$. Let’s unpack this definition. Since $\Delta$ has $\mathcal{C}$ as its source, and $F$ is a fixed object — the same thing as a functor from the category $\mathbf{1}$ to $\mathcal{C}^\mathcal{J}$, we know what objects of this category look like. An object of the category $(\Delta\downarrow F)$ consists of an object $C$ of $\mathcal{C}$ and an arrow from $\Delta(C)$ to $F$. But an arrow in $\mathcal{C}^\mathcal{J}$ is a natural transformation of functors. That is, for each object $J$ of $\mathcal{J}$ we need an arrow in $\mathcal{C}$ from $\left[\Delta(C)\right](J)$ to $F(J)$. But $\left[\Delta(C)\right](J)=C$ for all objects $J$. So we just need an object $C\in\mathcal{C}$ and an arrow $C\rightarrow F(J)$ for each $J\in\mathcal{J}$. Of course, there’s also naturality conditions to be concerned with. If $j:J_1\rightarrow J_2$ is an arrow in $\mathcal{J}$, and $c_1:C\rightarrow F(J_1)$ and $c_2:C\rightarrow F(J_2)$ are the required arrows from $C$, then naturality requires that $c_2=F(j)\circ c_1$. So we need a collection of arrows from $C$ to the objects in the image of $F$ that are compatible with the arrows from $\mathcal{J}$. Such a collection defines an object in the comma category $(\Delta\downarrow F)$ — a cone on $F$. Cocones are defined similarly. A cocone on $F$ is an object in the comma category $(F\downarrow\Delta)$. That is, it’s an object $C\in\mathcal{C}$ and a collection of arrows from the objects in the image of $F$ that are compatible with the arrows from $\mathcal{J}$. The description may seem a little odd, but try writing it out for some very simple categories $\mathcal{J}$. For example, let $\mathcal{J}$ be a set. Then try letting it be an ordinal, or another preorder. After you write down the definition of a cone and a cocone on some simple categories the general idea should seem to make more sense. ### Like this: Posted by John Armstrong \| Category theory ## 4 Comments » 1. [...] a functor the limit of (if it exists) is a couniversal cone on — a terminal object in the comma category . That is, it consists of an object and arrows [...] Pingback by \| June 19, 2007 \| Reply 2. [...] are Adjoints When considering limits, we started by talking about the diagonal functor . This assigns to an object the “constant” functor that sends each object of to and [...] Pingback by \| July 20, 2007 \| Reply 3. 2 ‘formula does not parse’ latex errors here Comment by Avery \| January 5, 2008 \| Reply 4. Yeah, those have been creeping in. WordPress changed something in their parser and now brackets give trouble unless they’re given as \left[ \right] pairs. I don’t know which pages have the problems, but when I stumble across them I fix them. Thanks for pointing this example out to me. Comment by \| January 6, 2008 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 59, "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": 0.9189394116401672, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Multiset	# Multiset In mathematics, the notion of multiset (or bag) is a generalization of the notion of set in which members are allowed to appear more than once. For example, there is a unique set that contains the elements a and b and no others, but there are many multisets with this property, such as the multiset that contains two copies of a and one of b or the multiset that contains three copies of both a and b. The term "multiset" was coined by Nicolaas Govert de Bruijn in the 1970s.[1] The use of multisets in mathematics and beyond predates the name "multiset" by many centuries: Knuth (1998) attributes the first study of multisets to the Indian mathematician Bhascara Acharya (circa 1150), who described permutations of multisets. ## Overview The number of times an element belongs to the multiset is the multiplicity of that member. The total number of elements in a multiset, including repeated memberships, is the cardinality of the multiset. For example, in the multiset {a, a, b, b, b, c} the multiplicities of the members a, b, and c are respectively 2, 3, and 1, and the cardinality of the multiset is 6. To distinguish between sets and multisets, a notation that incorporates brackets is sometimes used: the multiset {2,2,3} can be represented as [2,2,3].[2] In multisets, as in sets and in contrast to tuples, the order of elements is irrelevant: The multisets {a, a, b} and {a, b, a} are equal. ## History Blizard (1989) traced multisets back to the very origin of numbers, arguing that “in ancient times, the number n was often represented by a collection of n strokes, tally marks, or units.” These and similar collections of objects are multisets as strokes, tally marks, or units are considered indistinguishable. This shows that people implicitly used multisets even before mathematics emerged. This shows that necessity in this structure has been always so urgent that multisets have been several times rediscovered and appeared in literature under different names (Blizard, 1991). For instance, they are called bags by Peterson (1981), who attributed this term to Cerf, Gostelow, Estrin, and Volanski (1971). A multiset has been also called an aggregate, heap, bunch, sample, weighted set, occurrence set, and fireset (finitely repeated element set) (cf. (Blizard, 1991; Singh, et al, 2007)). Although multisets were implicitly utilized from ancient times, their explicit exploration happened much later. The first known study of multisets is attributed to the Indian mathematician Bhascara Acharya (circa 1150), who described permutations of multisets (Knuth, 1998). Angelelli (1965) discovered another early reference to the concept of multiset in the work of Marius Nizolius (1498-1576). Much later Kircher (1650) found the number of multiset permutations when one element can be repeated. Prestet (1675) published a general rule for multiset permutations. Wallis (1685) explains this rule in more detail. In the explicit form, multisets appeared in the work of Richard Dedekind "Was sind und was sollen die Zahlen?" (1888). However, mathematicians formalized multisets and began to study them as a precise mathematical object (structure) only in the 20th century. ## Formal definition Within set theory, a multiset may be formally defined as a 2-tuple $\left(A, m\right)$ where $A$ is some set and $m \colon A \to \mathbb{N}_{\geq 1}$ is a function from $A$ to the set $\mathbb{N}_{\geq 1} = \left\{1, 2, 3, \dots\right\}$ of positive natural numbers. The set $A$ is called the underlying set of elements. For each $a$ in $A$ the multiplicity (that is, number of occurrences) of $a$ is the number $m\!\left(a\right)$. If a universe $U$ in which the elements of $A$ must live is specified, the definition can be simplified to just a multiplicity function $m_U\colon U \to \mathbb{N}$ from $U$ to the set $\mathbb{N} = \left\{0, 1, 2, 3, \dots\right\}$ of natural numbers, obtained by extending $m$ to $U$ with values 0 for each element not in $A$. This extended multiplicity function is the multiplicity function called $1_A$ below. Like any function, the function $m$ may be defined as its graph: the set of ordered pairs $\left\{\left(a, m\left(a\right)\right) : a \in A\right\}$. With these definitions the multiset written as $\left\{a, a, b\right\}$ is defined as $\left(\left\{a, b\right\}, \left\{\left(a, 2\right), \left(b, 1\right)\right\}\right)$, and the multiset $\left\{a, b\right\}$ is defined as $\left(\left\{a, b\right\}, \left\{\left(a, 1\right), \left(b, 1\right)\right\}\right)$. The concept of a multiset is a generalization of the concept of a set. A multiset corresponds to an ordinary set if the multiplicity of every element is one (as opposed to some larger natural number). An indexed family, $\left(a_i\right)$, where $i$ is in some index-set, may define a multiset, sometimes written $\left\{a_i\right\}$, in which the multiplicity of any element $x$ is the number of indices $i$ such that $a_i = x$. The condition for this to be possible is that no element occurs infinitely many times in the family: even in an infinite multiset, the multiplicities must be finite numbers. It is possible to extend the definition of a multiset by allowing multiplicities of individual elements to be infinite cardinals instead of natural numbers. Not all properties carry over to this generalization however, and this article uses the definition above, with finite multiplicities. ## Multiplicity function The set indicator function of a normal set is generalized to the multiplicity function for multisets. The set indicator function of a subset A of a set X is the function $\mathbf{1}_A : X \to \lbrace 0,1 \rbrace \,$ defined by $\mathbf{1}_A(x) = \begin{cases} 1 &\text{if }x \in A, \\ 0 &\text{if }x \notin A. \end{cases}$ The set indicator function of the intersection of sets is the minimum function of the indicator functions $\mathbf{1}_{A\cap B}(x) = \min\{\mathbf{1}_A(x),\mathbf{1}_B(x)\}.$ The set indicator function of the union of sets is the maximum function of the indicator functions $\mathbf{1}_{A\cup B}(x) = \max\{{\mathbf{1}_A(x),\mathbf{1}_B(x)}\}.$ The set indicator function of a subset is smaller than or equal to that of the superset $A\subseteq B \Leftrightarrow \forall x \mathbf{1}_{A}(x) \le \mathbf{1}_{B}(x).$ The set indicator function of a cartesian product is the product of the indicator functions of the cartesian factors $\mathbf{1}_{A \times B}(x,y) = \mathbf{1}_A(x) \cdot\mathbf{1}_B(y).$ The cardinality of a (finite) set is the sum of the indicator function values $\|A\|=\sum_{x\in X} \mathbf{1}_{A}(x).$ Now generalize the concept of set indicator function by releasing the constraint that the values are 0 and 1 only and allow the values 2, 3, 4 and so on. The resulting function is called a multiplicity function and such a function defines a multiset. The concepts of intersection, union, subset, cartesian product and cardinality of multisets are defined by the above formulas. The multiplicity function of a multiset sum, is the sum of the multiplicity functions $\mathbf{1}_{A \uplus B}(x) = \mathbf{1}_A(x) + \mathbf{1}_B(x).$ The multiplicity function of a multiset difference is the zero-truncated subtraction of the multiplicity functions $\mathbf{1}_{A \setminus B}(x) = \max(0, \mathbf{1}_A(x) - \mathbf{1}_B(x)).$ The scalar multiplication of a multiset by a natural number n may be defined as: $\mathbf{1}_{n \otimes A}(x) = n \times \mathbf{1}_A(x). \,$ A small finite multiset, A, is represented by a list where each element, x, occurs as many times as the multiplicity, 1A(x), indicates. $\{1,1,1,3\} \cap \{1,1,2\} = \{1,1\} \,$ $\{1,1\} \cup \{1,2\} = \{1,1,2\}\,$ $\{1,1\} \subseteq \{1,1,1,2\}\,$ $\left\|\{1,1\}\right\|=2 \,$ $\{1,1\} \times \{1,2\} = \{(1,1), (1,1), (1,2), (1,2)\} \,$ $\{1,1\} \uplus \{1,2\} = \{1,1,1,2\} \,$ ## Examples One of the simplest and most natural examples is the multiset of prime factors of a number n. Here the underlying set of elements is the set of prime divisors of n. For example the number 120 has the prime factorization $120 = 2^3 3^1 5^1\,$ which gives the multiset {2, 2, 2, 3, 5}. A related example is the multiset of solutions of an algebraic equation. A quadratic equation, for example, has two solutions. However, in some cases they are both the same number. Thus the multiset of solutions of the equation could be { 3, 5 }, or it could be { 4, 4 }. In the latter case it has a solution of multiplicity 2. A special case of the above are the eigenvalues of a matrix, if these are defined as the multiset of roots of its characteristic polynomial. However a choice is made here: the (usual) definition of eigenvalues does not refer to the characteristic polynomial, and other possibilities give rise to different notions of multiplicity, and therefore to different multisets. The geometric multiplicity of λ as eigenvalue of a matrix A is the dimension of the kernel of A - λI, which is often smaller than its multiplicity as root of the characteristic polynomial (the algebraic multiplicity) when the latter is larger than 1. The set of eigenvalues of A is also the set of roots of its minimal polynomial, but the multiset of those roots need not be the same either as the one defined using algebraic multiplicity, or using the geometric multiplicity. ## Free commutative monoids The free commutative monoid on a set X (see free object) can be taken to be the set of finite multisets with elements drawn from X, with the monoid operation being multiset sum and the empty multiset as identity element. Such monoids are also known as (finite) formal sums of elements of X with natural coefficents. The free commutative semigroup is the subset of the free commutative monoid which contains all multisets with elements drawn from X except the empty multiset. Free abelian groups are formal sums (i.e. linear combinations) of elements of X with integer coefficients. Equivalently, they may be seen as signed finite multisets with elements drawn from X. ## Counting multisets Bijection between 3-subsets of a 7-set (left) and 3-multisets with elements from a 5-set (right) So this illustrates that $\textstyle {7 \choose 3} = \left(\!\!{5 \choose 3}\!\!\right)$. See also: Stars and bars (combinatorics) The number of multisets of cardinality k, with elements taken from a finite set of cardinality n, is called the multiset coefficient or multiset number. This number is written by some authors as $\textstyle\left(\!\!{n\choose k}\!\!\right)$, a notation that is meant to resemble that of binomial coefficients; it is used for instance in (Stanley, 1997), and could be pronounced "n multichoose k" to resemble "n choose k" for $\tbinom nk$. Unlike for binomial coefficients, there is no "multiset theorem" in which multiset coefficients would occur, and they should not be confused with the unrelated multinomial coefficients that occur in the multinomial theorem. The value of multiset coefficients can be given explicitly as $\left(\!\!{n\choose k}\!\!\right) = {n+k-1 \choose k} = \frac{(n+k-1)!}{k!\,(n-1)!} = {n(n+1)(n+2)\cdots(n+k-1)\over k!},$ where the second expression is as a binomial coefficient; many authors in fact avoid separate notation and just write binomial coefficients. So, the number of such multisets is the same as the number of subsets of cardinality k in a set of cardinality n + k − 1. The analogy with binomial coefficients can be stressed by writing the numerator in the above expression as a rising factorial power $\left(\!\!{n\choose k}\!\!\right) = {n^{\overline{k}}\over k!},$ to match the expression of binomial coefficients using a falling factorial power: ${n\choose k} = {n^{\underline{k}}\over k!}.$ There are for example 4 multisets of cardinality 3 with elements taken from the set {1,2} of cardinality 2 (n=2, k=3), namely : {1,1,1}, {1,1,2}, {1,2,2}, {2,2,2}. And there are also 4 subsets of cardinality 3 in the set {1,2,3,4} of cardinality 4 (n+k-1 = 4), namely : {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}. One simple way to prove the equality of multiset coefficients and binomial coefficients given above, involves representing multisets in the following way. First, consider the notation for multisets that would represent { a, a, a, a, a, a, b, b, c, c, c, d, d, d, d, d, d, d } (6 as, 2 bs, 3 cs, 7 ds) in this form: $\bullet \bullet \bullet \bullet \bullet \bullet \mid \bullet \bullet \mid \bullet \bullet \bullet \mid \bullet \bullet \bullet \bullet \bullet \bullet \bullet$ This is a multiset of cardinality k=18 made of elements of a set of cardinality n=4. The number of characters including both dots and vertical lines used in this notation is 18 + 4 − 1. The number of vertical lines is 4 − 1. The number of multisets of cardinality 18 is then the number of ways to arrange the 4 − 1 vertical lines among the 18 + 4 − 1 characters, and is thus the number of subsets of cardinality 4 − 1 in a set of cardinality 18 + 4 − 1. Equivalently, it is the number of ways to arrange the 18 dots among the 18 + 4 − 1 characters, which is the number of subsets of cardinality 18 of a set of cardinality 18 + 4 − 1. This is ${4+18-1 \choose 4-1}={4+18-1 \choose 18} = 1330,$ thus is the value of the multiset coefficient and its equivalencies: $\left(\!\!{4\choose18}\!\!\right)={21\choose18}=\frac{21!}{18!\,3!}={21\choose3}=\left(\!\!{19\choose3}\!\!\right),$ $=\frac{{\color{Periwinkle}\mathfrak{ 4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15\cdot16\cdot17\cdot18}} \mathbf{\cdot19\cdot20\cdot21}}{\mathbf{1\cdot2\cdot3}{\color{Periwinkle}\mathfrak{\cdot 4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15\cdot16\cdot17\cdot18}}},$ $=\frac{ 1\cdot2\cdot3\cdot4\cdot5\cdots16\cdot17\cdot18\;\mathbf{\cdot\;19\cdot20\cdot21}} {\,1\cdot2\cdot3\cdot4\cdot5\cdots 16\cdot17\cdot18\,\times\,\mathbf{1\cdot2\cdot3\quad}},$ $=\frac{19\cdot20\cdot21}{1\cdot2\cdot3}.$ One may define a generalized binomial coefficient ${n \choose k}={n(n-1)(n-2)\cdots(n-k+1) \over k!}$ in which n is not required to be a nonnegative integer, but may be negative or a non-integer, or a non-real complex number. (If k = 0, then the value of this coefficient is 1 because it is the empty product.) Then the number of multisets of cardinality k in a set of cardinality n is $\left(\!\!{n\choose k}\!\!\right)=(-1)^k{-n \choose k}.$ This fact led Gian-Carlo Rota to ask "Why are negative sets multisets?".[citation needed] He considered that question worthy of the attention of philosophers of mathematics. ### Recurrence relation A recurrence relation for multiset coefficients may be given as $\left(\!\!{n\choose k}\!\!\right) = \left(\!\!{n\choose k - 1}\!\!\right) + \left(\!\!{n-1\choose k}\!\!\right) \quad \mbox{for } n,k>0$ with $\left(\!\!{n \choose 0}\!\!\right) = 1,\quad n\in\N, \quad\mbox{and}\quad \left(\!\!{0 \choose k}\!\!\right) = 0,\quad k>0.$ The above recurrence may be interpreted as follows. Let [n] := {1, ..., n} be the source set. There is always exactly one (empty) multiset of size 0, and if n = 0 there are no larger multisets, which gives the initial conditions. Now, consider the case in which n,k > 0. A multiset of cardinality k with elements from [n] might or might not contain any instance of the final element n. If it does appear, then by removing n once, one is left with a multiset of cardinality k − 1 of elements from [n], and every such multiset can arise, which gives a total of $\left(\!\!{n\choose k - 1}\!\!\right)$ possibilities. If n does not appear, then our original multiset is equal to a multiset of cardinality k with elements from [n − 1], of which there are $\left(\!\!{n-1\choose k}\!\!\right).$ Thus, $\left(\!\!{n\choose k}\!\!\right) = \left(\!\!{n\choose k - 1}\!\!\right) + \left(\!\!{n-1\choose k}\!\!\right).$ ## Polynomial notation The set {x} may be represented by the monomial x. The set of subsets, { {}, {x} }, is represented by the binomial 1 + x. The set {x,y} may be represented by the monomial x·y. The set of subsets, { {}, {x}, {y}, {x,y} }, is represented by the polynomial (1 + x)·(1 + y) = 1 + x + y + x·y. The multiset {x,x} may be represented by the monomial x·x = x2. The multiset of submultisets, { {}, {x}, {x}, {x,x} }, is represented by the polynomial (1 + x)2 = 1 + x + x + x·x = 1 + 2·x + x2. The multiset of submultisets of the multiset xn is $(1+x)^n=\sum_{k=0}^n{n \choose k}\cdot x^k.$ That is why the binomial coefficient counts the number of k-combinations of an n-set. The multiset xK·yN−K, containing N elements, K of which are hits, is called a statistical population, and a submultiset is called a statistical sample. The set of samples is (1 + x)K·(1 + y)N−K which by the binomial theorem equals $\sum_{n=0}^N\sum_{k=0}^K{K \choose k}\cdot{N-K \choose n-k}\cdot x^k\cdot y^{n-k}.$ So the number of n-samples with k hits is ${K \choose k}\cdot{N-K \choose n-k}.$ See hypergeometric distribution and inferential statistics for further on the distribution of hits. The infinite set of finite multisets of elements taken from the set {x}: { {}, {x}, {x,x}, {x,x,x}, ... } may be represented by the formal power series S = 1 + x + x2 + x3 + ... = 1 + xS . The formal solution, S = (1 − x)−1, makes sense as a set of multisets, but the intermediate result, 1−x, does not make sense as a set of multisets. The infinite set of finite multisets of elements taken from the set x·y is (1 − x)−1·(1 − y)−1 = 1 + (x + y) + (x2 + x·y + y2) + ... The special case y=x : The infinite multiset of finite multisets of elements taken from the multiset x2 is (1 − x)−2 = 1 + 2·x + 3·x2 + ... The general case: The infinite multiset of finite multisets of elements taken from the multiset xn is $(1-x)^{-n}=\sum_{k=0}^\infty{-n \choose k} \cdot(-x)^k$ . This explains why "multisets are negative sets". The negative binomial coefficients count the number of k-multisets of elements taken from an n-set. ## Cumulant generating function A non-negative integer, n, can be represented by the monomial xn . A finite multiset of non-negative integers, say {2, 2, 2, 3, 5}, can likewise be represented by a polynomial f(x), say f(x) = 3·x2 + x3 + x5 . It is convenient to consider the cumulant generating function g(t) = log(f(et)), say g(t) = log(3·e2·t + e3·t + e5·t) . • The cardinality of the multiset is eg(0) = f(1), say 3 + 1 + 1 = 5. • The derivative g is g '(t) = f(et)−1·f '(et)·et, say g '(t) = (3·e2·t + e3·t + e5·t)−1·(6·e2·t + 3·e3·t + 5·e5·t) . • The mean value of the multiset is μ = g '(0) = f(1)−1·f '(1), say μ = (3+1+1)−1·(6+3+5) = 2.8 . • The variance of the multiset is σ2 = g ' '(0) . The numbers ( μ, σ2, ··· ) = ( g '(0), g ' '(0), ··· ) are called cumulants. The infinite set of non-negative integers {0, 1, 2, ···} is represented by the formal power series 1 + x + x2 + ··· = (1 − x)−1. The mean value and standard deviation are undefined. Nevertheless it has a cumulant-generating function, g(t) = −log(1−et). The derivative of this cumulant-generating function is g '(t) = (e−t−1)−1. A finite multiset of real numbers, A = { Ai }, is represented by the cumulant generating function $g_A(t) = \log \left(\sum_i e^{t \cdot A_i}\right) .$ This representation is unique: different multisets have different cumulant generating functions. See partition function (statistical mechanics) for the case where the numbers in question are the energy levels of a physical system. The cumulant-generating function of a multiset of n real numbers having mean μ and standard deviation σ is: g(t) = log(n) + μ·t + 2−1·(σ·t)2 + ··· , and the derivative is simply: g '(t) = μ + σ2·t + ··· The cumulant-generating function of set, {k}, consisting of a single real number, k, is g(t) = k·t , and the derivative is the number itself: g '(t) = k . So the concept of the derivative of the cumulant generating function of a multiset of real numbers is a generalization of the concept of a real number. The cumulant-generating function of a constant multiset, {k, k, k, k, ··· , k} of n elements all equal to the same real number k, is g(t) = log(n)+k·t , and the derivative is the number itself: g '(t) = k , irrespective of n. The cumulant-generating function of the multiset of sums of elements of two multisets of numbers is the sum of the two cumulant-generating functions: $\begin{align} g_{A+B}(t) & = \log \left(\sum_i \sum_j e^{t\cdot(A_i+B_j)}\right) = \log \left(\sum_i\sum_j e^{t\cdot A_i}\cdot e^{t\cdot B_j}\right) \\ & = \log \left(\sum_i e^{t\cdot A_i}\cdot \sum_j e^{t\cdot B_j}\right) = \log \left(\sum_i e^{t\cdot A_i}\right)+ \log\left(\sum_j e^{t\cdot B_j}\right) \\ & = g_A(t) + g_B(t). \end{align}$ There is unfortunately no general formula for computing the cumulant generating function of a product $g_{A\cdot B}(t) = \log \left(\sum_i\sum_j e^{t\cdot A_i\cdot B_j}\right) \,$ but the special case of a constant times a multiset of numbers is: $g_{k\cdot A}(t) = \log \left(\sum_i e^{t\cdot (k\cdot A_i)}\right) = \log \left(\sum_i e^{(t\cdot k)\cdot A_i}\right) = g_A\left(k\cdot t\right).$ The multiset 2·A = {2·Ai} is not the same multiset as 2×A =A+A = {Ai+Aj}. For example, 2·{+1,−1} = {+2,−2} while 2×{+1,−1} = {+1,−1} + {+1,−1} = {+1+1,+1−1,−1+1,−1−1} = {+2,0,0,−2}. $g_{k\times A}(t) = k\cdot g_{A}(t).$ The standard normal distribution is like a limit of big multisets of numbers. $\lim_{k \rarr \infty}k^{-1}\cdot (k^2\times \{+1,-1\}).$ This limit does not make sense as a multiset of numbers, but the derivative of the cumulant generating functions of the multisets in question makes sense, and the limit is well defined. $\begin{align} \lim_{k \rarr \infty} g'_{k^{-1}\cdot (k^2\times \{+1,-1\})}(t) & = \lim_{k \rarr \infty} \frac{d(k^2\cdot \log(e^{+t\cdot k^{-1}}+e^{-t\cdot k^{-1}}))}{dt} \\ & = \lim_{k \rarr \infty} \frac{d(k^2\cdot \log(2)+2^{-1}\cdot t^2+\cdots)}{dt}=t. \end{align}$ The constant term k2·log(2) vanishes by differentiation. The terms ··· vanish in the limit. So for the standard normal distribution, having mean 0 and standard deviation 1, the derivative of the cumulant generating function is simply g '(t) = t . For the normal distribution having mean μ and standard deviation σ, the derivative of the cumulant generating function is g '(t) = μ + σ2·t . See also random variable. ## Applications Multisets have various applications (Singh, et al, 2007). They are becoming the main structure of combinatorics because in its search for higher rigorousness, modern combinatorics has been developed not for sets but for multisets (Aigner, 1979; Anderson, 1987; Stanley, 1999). Now multisets has become an important tool in databases (Libkin and Wong, 1994; 1995; Grumbach and Milo, 1996). For instance, multisets are often used to implement relations in database systems. Multisets also play an imperative role in computer science (Knuth, 1997; 1998). There are also other applications. For instance, Rado (1975) uses multisets as a device to investigate the properties of families of sets. He writes, "The notion of a set takes no account of multiple occurrence of any one of its members, and yet it is just this kind of information which is frequently of importance. We need only think of the set of roots of a polynomial f(x) or the spectrum of a linear operator." ## Generalizations Different generalizations of multisets have been introduced, studied and applied to solving problems. Yager (1986) introduced fuzzy multisets under the name fuzzy bags. Grzymala-Busse (1987) introduced the concept of a rough multiset, using multirelations. Blizard (1989) generalized multisets to real valued multisets, i.e., multisets in which multiplicity of an element can be any non-negative real number. Blizard (1990) also introduced multisets with negative multiplicity. Loeb (1992) introduced hybrid sets as a generalization of multisets in which multiplicity of an element is any integer number. Miyamoto (2001) further generalized multisets introducing multisets in which multiplicity of an element is a any real-valued step function. Alkhazaleh, Salleh and Hassan (2011) introduced soft multisets, while Alkhazaleh and Salleh (2012) introduced fuzzy soft multisets. Burgin (1990; 1992; 2004; 2011) unified multisets and all their generalizations by the concept of a named set, which also comprises all other generalizations of sets. ## Notes 1. Knuth, Donald E. (1998). (3rd edition ed.). Addison Wesley. p. 694. ISBN 0-201-89684-2. Knuth also lists other names that were proposed for multisets, such as list, bunch, bag, heap, sample, weighted set, collection, and suite. 2. Hein, James L. (2003). Discrete mathematics. Jones & Bartlett Publishers. pp. 29–30. ISBN 0-7637-2210-3. ## References 1. Aigner, M. Combinatorial Theory, Springer Verlag, New York/Berlin, 1979 2. Alkhazaleh, S. and Salleh, A.R. Fuzzy Soft Multiset Theory, Abstract and Applied Analysis, 2012, article ID 350600, 20 p. 3. Alkhazaleh, S., Salleh, A.R. and Hassan, N. Soft Multisets Theory, Applied Mathematical Sciences, v. 5, No. 72, 2011, pp. 3561–3573 4. Anderson, I. Combinatorics of Finite Sets, Clarendon Press, Oxford, 1987 5. Angelelli, I. (1965) Leibniz's misunderstanding of Nizolius' notion of 'multitudo', Notre Dame Journal of Formal Logic, v. 6, pp. 319–322 6. Blizard, Wayne D. (1989) "Multiset theory," Notre Dame Journal of Formal Logic, Volume 30, Number 1, Winter: pp. 36–66. doi:10.1305/ndjfl/1093634995 http://projecteuclid.org/euclid.ndjfl/1093634995 MR990203 0668.03027 7. Blizard, W.D. (1989) Real-valued Multisets and Fuzzy Sets, Fuzzy Sets and Systems, v. 33, pp. 77–97 8. Blizard, W. (1990) Negative Membership, Notre Dame Journal of Formal Logic, v. 31, No. 1, pp. 346–368 9. Blizard, W.D. (1991) The Development of Multiset Theory, Modern Logic, v. 1, No.4, pp. 319–352 10. Bogart, Kenneth P. (2000). Introductory combinatorics, 3rd. ed. San Diego CA: Harcourt. 11. Burgin M. Theory of Named Sets as a Foundational Basis for Mathematics, In: Structures in Mathematical Theories, San Sebastian, 1990, pp. 417–420 (http://www.blogg.org/blog-30140-date-2005-10-26.html) 12. Burgin, M. (1992) On the concept of a multiset in cybernetics, Cybernetics and System Analysis, No. 3, pp. 165–167 13. Burgin, M. Unified Foundations of Mathematics, Preprint Mathematics LO/0403186, 2004, 39 p. (electronic edition: http://arXiv.org) 14. Burgin, M. (2011), Theory of Named Sets, Mathematics Research Developments, Nova Science Pub Inc, ISBN 978-1-61122-788-8, books.google.com 15. Dedekind R. Was sind und was sollen die Zahlen?, Vieweg, Braunschweig, 1888 16. Gessel, Ira M., and Stanley, Richard P. (1995) "Algebraic enumeration" in Graham, R. L., Grötschel, M., & Lovász, L., eds., Handbook of combinatorics, Vol. 2. Elsevier: 1021–1061. ISBN 0-444-82351-4, 0-444-88002-X, 0-262-07171-1, 0-262-07169-X. 17. Grzymala-Busse, J. Learning from examples based on rough multisets, in Proceedings of the 2nd International Symposium on Methodologies for Intelligent Systems, Charlotte, NC, USA, 1987, pp. 325–332 18. Grumbach, S. and Milo, T. (1996) Towards tractable algebras for bags, Journal of Computer and System Sciences, v. 52 , No. 3, pp. 570–588 19. Hickman, J. L. (1980) "A note on the concept of multiset," Bulletin of the Australian Mathematical Society 22: 211–17. 20. Kircher, A. Musurgia Universalis, Corbelletti, Rome, 1650 21. Knuth, Donald E. (1998). The Art of Computer Programming – Vol. 2: Seminumerical Algorithms Addison Wesley. p. 694. ISBN 0-201-89684-2. 22. Knuth, D. The Art of Computer Programming, v.3: Sorting and Searching, Addison-Wesley, Reading, Mass., 1998 23. Libkin, L. and Wong, L. (1994) Some properties of query languages for bags, in Proceedings of the Workshop on Database Programming Languages, Springer Verlag, pp. 97–114 24. Libkin, L. and Wong, L. (1995) On representation and querying incomplete information in databases with bags, Information Processing Letters, v. 56, No. 4, pp. 209–214 25. Loeb, D. (1992) Sets with a negative numbers of elements, Advances in Mathematics, v. 91, pp. 64–74 26. Miyamoto, S. Fuzzy Multisets and their Generalizations, in ‘Multiset Processing’, LNCS 2235, pp. 225–235, 2001 27. Prestet, J. Elemens des Mathematiques, André Pralard, Paris, 1675 28. Singh, D. (1994) A Note on the Development of Multiset Theory, Modern Logic, v. 4, pp. 405–406 29. Singh, D., Ibrahim, A. M., Yohanna, T. and Singh, J. N. (2007) An overview of the applications of multisets, Novi Sad Journ. Math., v. 37, No. 2, pp. 73–92 30. Stanley, Richard P. (1997, 1999) Enumerative Combinatorics, Vols. 1 and 2. Cambridge University Press. ISBN 0-521-55309-1, 0-521-56069-1. 31. Syropoulos, Apostolos (2001) "Mathematics of Multisets" in C. S. Calude et al., eds., Multiset processing: Mathematical, computer science, and molecular computing points of view, LNCS 2235. Springer-Verlag: 347–358 32. Yager, R. R. (1986) On the Theory of Bags, International Journal of General Systems, v. 13, pp. 23–37 33. Wallis, A treatise of algebra, John Playford, London, 1685	{"extraction_info": {"found_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": 78, "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": 0.8335877060890198, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=4209731	Physics Forums Is Energy proptional to frequency(standing wave) Here is an extract from my Physics textbook. According to classical physics, each of these standing waves carries energy, and as their frequency increases, so does their energy . I don't understand how this is true . I think that the classical wave theory says that the energy is proportional to the intensity while on the other hand it is the Quantum theory which predicts : E = $h$ f So, is there an error in the book regarding this issue ? 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 According to classical physics, each of these standing waves carries energy, and as their frequency increases, so does their energy . See http://www.egglescliffe.org.uk/physi...ody/bbody.html on how the standing wave analysis of standing waves in a cavity led to an incorrect result of the radiation emmitted by a blackbody. Yes, in classical physics way back then it was known that the energy of an electromagnetic wave increases with frequency. http://galileo.phys.virginia.edu/cla...radiation.html gives some more history on the subject. Quote by 256bits See http://www.egglescliffe.org.uk/physi...ody/bbody.html on how the standing wave analysis of standing waves in a cavity led to an incorrect result of the radiation emmitted by a blackbody. Yes, in classical physics way back then it was known that the energy of an electromagnetic wave increases with frequency. http://galileo.phys.virginia.edu/cla...radiation.html gives some more history on the subject. I don't think you get my point. I want to know how is it possible that classical physics which was formulated 400 years ago predicted what is now known to be true . We now know that the energy of a photon of E.M wave is equal to E = h f In other words it is dependent upon frequency. From classical physics, i think we all know that for a wave, its energy is propotional to its intensity. What exactly did classical physics say ? (frequency or intensity) Recognitions: Homework Help Science Advisor Is Energy proptional to frequency(standing wave) It is not the energy of radiation which increases with frequency. It is the photon energy - the step size of possible energies. This quantization is not classical - you do not get it in classical physics. I read this link http://www.egglescliffe.org.uk/physi...ody/bbody.html I don't really understand how the introduction of quantized (special and discrete) frequency of oscillation of electrons solved the Ultra Violet Catastrophe According to classical physics, the energy emitted by the black body would increase to infinity as the wavelength decreases. (how is this problem solved by quantized energy levels) Also, I am facing another problem : If the wavelength decreases ie frequency increases(to infinity); according to quantum theory, wouldn't the energy of each photon tend to infinity as well ? and hence give the exact same result as the classical theory ? Mentor Quote by hms.tech What exactly did classical physics say ? (frequency or intensity) It's not either-or. A classical wave has energy proportional to its frequency and also proportional to the square of its amplitude (sometimes called intensity). Recognitions: Homework Help I agree with Vanadium. In particular, for the example of standing waves on a string, the frequency is quantised, so it is highly analogous to quantum physics. But what they didn't know back then was that light is only emitted in quanta. That is a more recent development (along with the rest of quantum mechanics). Recognitions: Homework Help Science Advisor @hms.tech: In classical physics, radiation is not quantized. A black body could emit x-rays (or any other frequency) without the requirement of a minimal amount of energy. As every frequency should receive the same intensity and the frequency range is not limited, the total radiation should be infinite. Quantum mechanics solves this issue with the photon energy: The minimal energy required to emit radiation of a specific frequency. For x-rays, that energy threshold is so high that materials at room temperature do not reach it (or so extremely rare that it does not matter), and the used frequency range is finite. If the wavelength decreases ie frequency increases(to infinity); according to quantum theory, wouldn't the energy of each photon tend to infinity as well ? It does, but the number of photons drops quicker. Quote by hms.tech I read this link http://www.egglescliffe.org.uk/physi...ody/bbody.html I don't really understand how the introduction of quantized (special and discrete) frequency of oscillation of electrons solved the Ultra Violet Catastrophe According to classical physics, the energy emitted by the black body would increase to infinity as the wavelength decreases. (how is this problem solved by quantized energy levels) Also, I am facing another problem : If the wavelength decreases ie frequency increases(to infinity); according to quantum theory, wouldn't the energy of each photon tend to infinity as well ? and hence give the exact same result as the classical theory ? the problem you are facing was first solved exactly by Max plank. Recognitions: Homework Help Quote by mfb Quantum mechanics solves this issue with the photon energy: The minimal energy required to emit radiation of a specific frequency. For x-rays, that energy threshold is so high that materials at room temperature do not reach it (or so extremely rare that it does not matter), and the used frequency range is finite. Yeah, the way I think of it, there is a bunch of molecules flying around, also with rotational degrees of freedom. And for a molecule to go from one state to the other, it can absorb or emit a photon. So to get a high frequency photon, you would need a molecule to go from a very high energy state to a much lower energy state. So this requires a molecule in a high energy state, but this doesn't happen very often, therefore high energy photons do not happen very often. Of course, this is a very qualitative, hand-wavey description, but I think it is roughly a good idea of what is going on. According to Wikipedia : (on Ultraviolet Catastrophe) Max Planck solved the problem by postulating that electromagnetic energy did not follow the classical description, but could only be emitted in discrete packets of energy proportional to the frequency, as given by Planck's law. This has the effect of reducing the number of possible excited modes with a given frequency in the cavity described above, and thus the average energy at those frequencies Can someone explain the bolded part of the extract from this article The probability of a very high frequency mode with very high energy (according to quantum theory) getting a big fraction of the available energy, at the expense of the huge number of less greedy modes, is very low. Hence the non-occurrence of the u-v catastrophe. [How hand-waving can you get?] The word you want is 'emboldened', by the way! Thread Tools Similar Threads for: Is Energy proptional to frequency(standing wave) Thread Forum Replies Introductory Physics Homework 15 Advanced Physics Homework 1 Classical Physics 7 Introductory Physics Homework 4 Introductory Physics Homework 4	{"extraction_info": {"found_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": 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": 0.9396889805793762, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/54720?sort=newest	## Verifying coefficients of modular forms ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi, I was wondering about good techniques that one can use to show that given certain coefficients, they are the Fourier coefficients of a cusp form, assuming we know the desired weight and level. I am aware of Weil's "converse theorem", but am not aware of any examples of it being used to prove something is a modular form. So, even a pointer to an example of that would be useful. I'm curious if there are other more direct methods as well. Thanks! - ## 5 Answers Maybe I can add something: 1. Wiles famous proof of Fermat's Last Theorem (and the later proof of modularity of all elliptic curves over $\mathbf{Q}$ by Breuil, Conrad, Diamond, Taylor) is a sort of ultimate example in which "given certain coefficients [associated to an elliptic curve by point count], they are [proved to be] the Fourier coefficients of a cusp form". This theorem also produces a general strategy for showing that a list of numbers are the coefficients of a cusp form in some cases -- relate those numbers to the point counts of an elliptic curve. Similar remarks apply for abelian varieties these days, due to Ribet's theorem that Serre's conjecture implies modularity of all abelian varieties over $\mathbf{Q}$ of ${\rm GL}_2$-type. 2. There is something called the Faltings-Serre method, which is an alternative approach to proving modularity statements in particular cases. A google search for "Faltings Serre" will yield many hits. Here's a particular example: http://www.math.wisc.edu/~boston/ell.pdf . There is also a paper by Edray Goins that explains how to use the Faltings-Serre method to get new results in particular cases that go beyond (1), and it has a good list of references: http://homepage.mac.com/ehgoins/papers/serre-faltings.pdf 3. Another example is http://wstein.org/papers/artin/, which is a paper Kevin Buzzard and I wrote, in which we show that lists of numbers (traces of Frobenius) associated to certain A_5 Galois representations are the coefficients of weight 1 modular forms. One can get the same result by "pure thought" these days as an application of Khare-Wintenberger's proof of Serre's conjecture. But still, there is some value in the computational techniques introduced in that paper. There are similar ideas in Joe Buhler's Ph.D. thesis and also in Springer Lecture Notes in Math Volume 1585. 4. There is a Theorem of Sturm (see, e.g., Theorem 9.18 of my modular forms book http://wstein.org/books/modform/stein-modform.pdf and the references) that tends to be very relevant to your question in practice. Sturm's result basically tells you that if the first "few" coefficients of two $q$-expansions of modular forms agree modulo a prime, then all terms agree. This list could go on and on. Also, there are similar ideas and techniques for Hilbert modular forms. - How sharp is Sturm's result? – Dror Speiser Feb 8 2011 at 20:12 1 @Dror: Sturm's result in general says one has to check about $2d$ coefficients, where $d$ is the dimension. For two arbitrary cusp forms, that's pretty good (massively better than "effective Chebotarev", say), but for eigenforms it is far from optimal. Under some hypothesis, Sturm reduces the number of coefficients to roughly $2d/2^v$, for some $v$ (this $v$ is $\leq$ the number of primes dividing the level). There is a result in the Buzzard-Stein paper I mentioned that improves on Sturm's result in the case of eigenforms with nontrivial character. – William Stein Feb 9 2011 at 7:56 ### 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. ( tried posting this earlier today, apologies if it appears twice) For an example of a direct application of Weil's converse theorem, see Shimura's original formulation of the Shimura lift (Annals 1973). He constructs a map from modular forms of half-integral weight to modular forms of integral weight by defining an L-series built out of the Fourier coefficients of the input form, and spends most of the paper proving the analytic properties needed for the converse theorem. - If you assume that the modular form is a newform with prescribed weight and level (EDIT : you also need to know the eigenvalue of the Atkin-Lehner involution), then its Fourier coefficients can sometimes be guessed using the fact that the associated L-series satisfies a functional equation. The point is that using the functional equation, the L-series can be computed by a rapidly convergent series, so it is only necessary to have few Fourier coefficients at hand to compute an approximation of it. This even gives a very pratical and powerful method for finding modular forms, at least when the level is small. You should look at the following example written by Tim Dokchitser : http://magma.maths.usyd.edu.au/magma/handbook/text/1392#15270 Dokchitser's algorithm for computing L-functions has been implemented in Sage and Magma. Note that the method outlined here is (to my knowledge) purely experimental, in the sense that it doesn't prove that the resulting q-expansion is modular. - 1 This "approximate functional equation" technique is surprisingly useful in practice. For instance: sometimes you have the L-function of a variety at all primes except, say, 2, because the special fiber at 2 is horrible. But the method described here allows you to guess the Euler factor at 2 with moral certainty if you know (a long list of) the other Euler factors. – JSE Feb 8 2011 at 14:36 @JSE : in the situation you describe, isn't the method even perfectly rigourous (because Fourier coefficients are integers, it suffices to find approximate them within $1/2$). Ah, maybe the problem is that in general the functional equation of the L-function is only conjectural ? – François Brunault Feb 8 2011 at 15:20 Right, that was the case in the applications I seem to remember (though of course so many more Galois representations are provably modular these days...) – JSE Feb 9 2011 at 5:57 I think Ramsey's answer hits the important points, but I wanted to comment on the converse theorem. The converse theorem takes as an input a Dirichlet series (more generally, an $L$-function) whose analytic continuation (plus quite a bit more) is already known and tells you that it actually came from an automorphic form. Unfortunately, the only methods that I know for proving analytic continuation require knowing that it already comes from an automorphic form. So it certainly sounds circular, and I've only seen it applied to proving global functoriality (for quasi-split classical groups, small symmetric powers, etc). Let me quickly(!) explain the argument: Basically (eliding the hard details to the point where I may say things that are literally, but hopefully not morally, incorrect) say you start out with an automorphic representation $\pi$ on some classical group $G$, living inside $GL(N)$. The Langlands-Shahidi method proves that the twisted $L$-functions $L(s,\pi\times \tau)$ are nice (entire, functional equation, etc), with $\tau$ an automorphic representation of $GL(m)$ with $m < N$. Local functoriality gives lifts from $\pi_\nu$ on $G_\nu$ to $\Pi_\nu$ on $GL(N)$ such that the local $L$-factors coincide, in that $L_\nu(s,\pi_\nu\times\tau_\nu)=L_\nu(s,\Pi_\nu\times\tau_\nu)$. Since $L(s,\Pi\times\tau)=\prod_\nu L_\nu(s,\Pi_\nu\times\tau_\nu)=L(s,\pi\times\tau)$, it is nice for all $\tau$, so the converse theorem applies and $\Pi$ is an automorphic representation of $GL(N)$. Modularity / potential automorphy methods (e.g. Serre's Conjecture, Modularity Theorem) exist, too. In the weight 2 case, given enough information about the coefficients and a sufficient amount of pluck, it is not inconceivable that you could show that the coefficients actually come from an elliptic curve (or several elliptic curves if your modular form is not an eigenform), then use modularity to deduce that they come from a modular form. - It depends on what you mean by "certain coefficients" and how they're given. This sort of sounds like you have a (finite) handful of coefficients in a $q$-expansion and want to try to "fill them out" to a modular form of prescribed weight and level. If this is the case, it is pretty easy. This is because the space of such forms is finite-dimensional, and a basis of the space is readily computed (using, for example, the SAGE software). This reduces the problem to basic linear algebra. If, on the other hand, you have an entire $q$-expansion, it's not so clear. You can truncate and employ the above procedure. I'll just mention three things: 1) If your expansion isn't a modular form, then this will eventually (by taking larger chunks) tell you so. 2) If it is a modular form (but you have no a priori knowledge of this), then it just gives you a lot of evidence that your expansion is a modular form, but doesn't prove it. Of course, if you know more about your coefficients you might be able to go back and prove that it is the form evidenced above. 3) If you do happen to know that your expansion is a modular form of given weight and level and just want to know "what it is," then this method will eventually provably determine what it is (in terms of the basis) by taking large enough chunks. As for the converse theorem... you'd certainly need all the coefficients, but I have a hard time imagining that method to be very practical (but I'm not very analytically minded). - Regarding your point 2, you can run into undecidability problems if you don't restrict how the coefficients are presented to you (but I haven't heard of this being a problem very often in practice). – S. Carnahan♦ Feb 8 2011 at 5:17 @Ramsey: Point 1) seems vague and only practical if there is something that limits the order of the expansion that needs to be considered. Otherwise one eventually will run into computer resource limitations - which are in fact quite severe in SAGE even at modest weights and levels. – Laie Feb 8 2011 at 17:33 2 @Laie: It is vague, and relies on the fact that if a collection of $q$-expansions are linearly independent, then so are their truncations to the first $N$ terms for large $N$. Perhaps the $N$ required can be huge, I'm not sure. I'd be curious (for several reasons related to these answers) to see what the OP actually has. – Ramsey Feb 8 2011 at 19:28	{"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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9341574311256409, "perplexity_flag": "head"}
http://mathhelpforum.com/math-topics/128508-calculating-speed.html	# Thread: 1. ## Calculating speed. I currently have a machine that can calculate the time it takes for an object to travel one meter. (two lasers first one starts a timer second one stops it, The lasers are positioned exactly 1 meter apart.) So if an object was going 10 M/s (Meters per second) it would take 0.1 of a second (10 nanoseconds) to travel one meter. If it was going 1 meter per second it would take 1 second, obviously. What would the equation needed to turn the time taken into the speed (In Meters per second preferably) For those interested the program used is called Algodoo 2. Originally Posted by Mystery I currently have a machine that can calculate the time it takes for an object to travel one meter. (two lasers first one starts a timer second one stops it, The lasers are positioned exactly 1 meter apart.) So if an object was going 10 M/s (Meters per second) it would take 0.1 of a second (10 nanoseconds) to travel one meter. If it was going 1 meter per second it would take 1 second, obviously. What would the equation needed to turn the time taken into the speed (In Meters per second preferably) For those interested the program used is called Algodoo $v = \frac{\Delta x}{\Delta t}$ also ... 0.1 sec does not equal 10 nanoseconds. 10 nanoseconds = $10^{-8}$ sec 0.1 sec = 100 milliseconds 3. So basically Velocity is the Distance Divided by the time taken. thanks. 4. However, it doesn't matter in this particular problem, because the laser travels one straight path. But generally in these problems you want the average speed. The formula $v = \frac{\Delta x}{\Delta t}$ gives you the "average velocity" of the object, $\Delta x$ is the object's displacement (change in position given by $\Delta x = x_f - x_i$). So the formula is not always the same thing as $\frac{d}{\Delta t}$ which gives you the "average speed"! Because in the latter formula, "d" means the total distance traveled which isn't always the same as displacement. Here's an example: If a runner runs a distance d of 10 km and yet ends up at his starting point, his displacement is zero, so his average velocity is zero! But his speed is clearly NOT zero!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9563674330711365, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/109181/what-can-i-do-with-what-i-know	# What can I do with what I know? [closed] This might be difficult for me to put into words, but bear with me because I think it's an important question. Among the many people who study math, I am one of them. I'm not particularly advanced but I enjoy it. I'm working through calculus 2 in my free time. I do it for fun. It occurred to me: I have a certain amount of ideas in my head: calculus, trig, geometry. These are all tools that required hundreds of years of development. I've developed my ability to reason as best as I can so far. However, I find that I'm not capable of very much, other than I can work through a textbook, I can solve the problems, maybe even sometimes in clever ways that surprise me. But what can I do? If I am capable of more, I don't know it. The question, more specifically, is this: if someone has studied up to, say, calc 2, what should they be capable of? Mathematicians and physicists over the years have achieved much more with much less. So what I'm saying is, with what I know, I feel like I should be capable of more and I'm sure there must be students out there that feel the same. I'd like to think that with knowledge comes more ability than just regurgitating their studies or passing contrived tests. Insight, intuition should lead them to discover new-found abilities. Well I hope I've made my question clear and, if I have not, feel free to delete it. Hopefully if I've left any gaps, then the reader can read between the lines and understand what I'm getting at. Okay think of it this way. Let's say someone has a powerful computer and they're using it as a paperweight. How could they use this resource to the fullest? I'm expecting question of a format like, "well, if you have trig under your belt, I'd expect you to be able to give me a good estimate of the size of the Earth. You have the tools for it." Thanks for reading, it's a bit long-winded. - 7 I feel this is a bit too soft a question, but: what you can do is really up to you. You encounter some problem/application that interests you in the course of reading different things, and you see if you have the tools necessary for tackling. If you don't, you prepare; if you do, then you can start experimenting. With luck, your experiments might turn out to be something special. – J. M. Feb 14 '12 at 5:42 9 One unexpected and very important side effect of having some background in more advanced mathematics is that you get true mastery of the elementary stuff. Many real world problems require only complete control of very basic mathematics. But not all that many people, even among those who "did well" in high school mathematics but did not go on, have enough control over that material to be able to solve real problems with confidence. – André Nicolas Feb 14 '12 at 5:54 3 What can you do with your math knowledge? You mean like doing even more math, or real-world jobs, or taking over the world, or what? This is way too broad in my opinion. – anon Feb 14 '12 at 6:00 2 I don’t think that it’s a bad question to ask, but it’s almost impossible to answer. Ideally you should be able to spot problems that can be solved by the tools that you’ve studied, but in practice @André’s observation is absolutely correct. Typically you acquire real mastery of a set of tools only when you start using them routinely in some more advanced context. – Brian M. Scott Feb 14 '12 at 6:50 3 One thing that's fun to do is write your own physics simulations. – Rahul Narain Feb 14 '12 at 8:14 show 4 more comments ## closed as not constructive by Austin Mohr, anon, lhf, Asaf Karagila, Zev Chonoles♦Feb 15 '12 at 1:37 As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or specific expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, see the FAQ for guidance. ## 1 Answer In my opinion, Calculus (SV, MV, Diff. Eqns) and Linear Algebra forms the basis of most things in Engineering (I mean basic engineering. Not high end stuff, you can get there later though) However, mathematics acts as hygiene and not a motivator. Let me explain what I mean: Pick up any field of engineering. Let us assume you chose Mechanical Engineering. The major components of ME are Heat Transfer/Fluid Mechanics and maybe newer topics like Robotics, CAD CAM or whatever. Among all of these, the aforementioned mathematical topics suffice to understand all that is being said. But effort is required to link an equation with the "real world". In you multivariable/differential equation course, you probably learnt how to solve $x\frac{dy}{dx} = 0$. You may be answers but they won't have any physical intuition. This same equation might show up in heat transfer where $y$ would signify the temperature at a given point. If you know math, you are 1 good book away from knowing the basics of an engineering field. The absence of math knowledge (no matter how good your engineering intuition is) is going to cause problems in the long run. To cite another example: Say you pick up "rocket science", the prerequisites for basic astrodynamics is Calculus with a little physics (Thermodynamics). There will be people who point out that Calculus and LA are not enough and that one needs other subjects to succeed in Engineering. This is not entirely true. You might need a few basic concepts from Probability/Statistics/Topology etc. but they can be learnt on an ad-hoc basic. Note: While the theme of SE is not to spurt out opinions but facts, this question invited one so I provided. -	{"extraction_info": {"found_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": 0.9623880982398987, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/29202?sort=newest	## Does a “Chern character” exist for any generalized cohomology theory? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The Chern character is a ring homomorphism from the complex K-theory to the usual cohomology. 1) I wonder if there are "Chern character"-like ring homomorphisms from other generalized cohomology theory to the usual cohomlogy. Are they related with Atiyah-Hirzebruch? 2) And if there are such nice homomorphisms, what is the "Todd genus" in these cases, making the generalization of that famous diagram in Grothendieck–Hirzebruch–Riemann–Roch commute? When I think about it, I cannot even recall seeing anything like this in real K-theory, but that is probably because I dont really know real K-theory at all. Thank you very much. - 1 I think, your question is related to the orientability of cohomology theories. There are many multiplicative transformations between bordism theories and other (co)homology theories such as oriented bordism to singular homology, spin bordism to K-theory, string bordism to TMF. As you see, the natural home for such a morphism is not always singular cohomology, but eg in the case of spin bordism you can, of course, compose the morphism to K-theory with the chern character. – Lennart Meier Jun 23 2010 at 8:39 ## 2 Answers In Oscar Randall-Williams' answer "connective" is unnecessary. Also, there is no need to choose that isomorphism (from a rational theory to the ordinary theory with the same coefficient groups); it is canonical. The generalization of the Todd genus or Todd class arises when the multiplicative theory $E$ has a "complex orientation": a multiplicatively well-behaved way of producing Thom isomorphisms in $E^$-theory for all complex vector bundles. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. For any (connective) spectrum $E$ one may rationalise it to get a rational spectrum $E_\mathbb{Q}$, and a map $E \to E_\mathbb{Q}$. Now rational spectra split as wedges of Eilenberg-Mac Lane spectra, so one may choose an isomorphism `$E_\mathbb{Q}^(X) \simeq H^(X;\pi_{-}(E)\otimes \mathbb{Q})$`, and the rationalisation gives a map `$$ch_E : E^(X) \longrightarrow H^(X;\pi_{-}(E)\otimes \mathbb{Q}).$$` For complex K-theory this gives the Chern character, and for real K-theory it gives the Pontrjagin character. Of course, if $E$ is a ring spectrum so is $E_\mathbb{Q}$, and one must identify the induced ring structure on `$H^(X;\pi_{-*}(E)\otimes \mathbb{Q})$`. -	{"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": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8998364806175232, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/11800/can-energy-be-taken-out-of-the-qft-vacuum/31173	# Can energy be taken out of the QFT vacuum? There have been recent questions about the vacuum. In my simplified knowledge the vacuum is like a ground state energy level, and also that there might even exist other lower energy levels than the vacuum we find ourselves in. The sea is a soup of created and annihilated pairs of virtual particles with virtual energies and momenta. In a normal sea on earth, which also represents a ground state of water, energy can be taken out of random waves by the clever construction of valves that allow only one way motion of water. Is it conceivable that a gadget of similar function could be found for the vacuum sea, or is it forbidden by conservation laws? My intuition tells me that it might be possible if GR is taken into account, but my physics knowledge does not stretch to support this. Edit : An explanation of why I am asking this question: Let me expand on the example of the sea. The energy from the waves comes from either tides, i.e. gravitational forces, or wind (temperature differentials). If these were missing the oceans would be like glass representing a unique ground state of the gravitational well of the earth. In an analogy, a gravitational wave going through the vacuum would be supplying energy to the vacuum sea. I have been thinking of this analogy ever since cold fusion surfaced and refuses to die out, the most recent one being discussed here too. Approached from nuclear physics orders of magnitude the claims seem preposterous. There are people though who believe they have results of extra energy over input energy, much more than chemical reactions could supply. This set me thinking on vacuum energy and the analogy with getting energy from the sea. A crystal is a prime candidate for any exploration of such concepts and in all cold fusion "successful" results crystals have been used. Now if the effect depended on the vacuum and how much distorted it was by a passage of a gravitational wave at the time of the experiment, or the exact orientation of the crystal, or the type of impurities in the crystal ( F centers etc) one would expect to get haphazard results, and non repeatable by other experimenters. Of course this would be the first experimental evidence of gravitational waves :). Edit 20/7/12 Maybe I should clarify that an acceptable answer in the negative would be one based on conservation laws. I believe that data trumps theory, and next in line are conservation laws,because they are the distillation of an enormous amount of data. Some people seem to think that theoretical definitions can substitute for proof in physics, but physical theories change, solid data do not, and this is physics, not axiomatic mathematics. - 1 Wouldn't that be very similar to Maxwell's demon? – Willie Wong Jul 2 '11 at 14:44 If the vacuum is the ground state, what does it mean to have a lower state. Analogy would be a value smaller than the minimum. Also in the particle interpretation, what would it correspond to, negative number of particles? – MBN Jul 2 '11 at 16:32 – anna v Jul 2 '11 at 19:01 I looked at it but it doesn't help my confusion. What is the vacuum state? I thought it is the state killed by all the anihilation operators. What would a lower than that mean? – MBN Jul 3 '11 at 0:08 1 @Marek The question is not about pumping energy from different vacua, although it might be related, if, for example, vacua had an "uncertainty principle" type of existence. The question relates to the marrying of QFT and gravity. Whether a gravitational wave would change the vacuum energy level as it is propagating in time so that energy might be captured of its passage, as in the sea wave analogy. – anna v Jul 4 '11 at 3:58 show 7 more comments ## 7 Answers Well, let us be honest here. This question was not supposed to be answered from the very beggining. First of all we don't know how to mix quantum mechanics and gravity. There is no good consistent theory for that. Another thing is that today "the Dirac sea" analogy is not considered to be a very good thing. It is a pre-QFT naiive picture. Finally we are supposed to talk about "waves" in this "sea"... While the "sea" itself is an obsolete analogy... And all that is in a context of non-existing theory... Come on... Now. There is actually a formal way to answer the question, because the question is about the "QFT vaccuum". And the QFT vacuum have a precise definition. Which basically says that it is "something you cannot take energy from". Actually, we start from that definition to build QFT. So the answer is: "you cannot take energy out of the QFT vacuum by definition". Maybe we are wrong to start from this definition. Maybe for quantum gravity we need another starting point. But then it wouldn't be a QFT -- it would be a new theory which will have QFT as a limiting case. And in the range of validity of QFT that "formal" answer will hold. - 1 Thanks for regurgitating my answer.............. – Chris Gerig Jul 19 '12 at 22:12 See my comment to @ChrisGerig . There are theories mooted which do have several vacua, QFT would be working on one of those, at the moment the one we find ourselves in. I believe my simplistic analogy of the tides should give an intuition. – anna v Jul 20 '12 at 3:09 "it would be a new theory which will have QFT as a limiting case." This is what I am exploring with this question. "And in the range of validity of QFT that "formal" answer will hold." Do you have a strong argument that you know how that limiting case is approached? For example, superconductivity , quantum mechanics on kilometer scales, shows that it is not dimension that defines absolutely the limit between QM and classical.There is the added ingredient of complexity of bulk matter and its different responses. – anna v Jul 21 '12 at 4:04 @Kostya , if you expand a bit your last paragraph I would accept your answer. I am in disagreement with the answer chosen by the bounty setter. It is not the answer of a physicist, but of a mathematician. A theory can be beautiful and fully consistent in its definitions. The results may agree with data to the available limit of measurements, but that does not mean that the definitions define nature, as progress in physics has shown us again and again. It is at the limits of the validity of a theory that it may be tested against nature with surprising results, for the theory. – anna v Jul 25 '12 at 3:10 @annav I've tried to expand my last paragraph. But I can't. I just really have nothing else to say on the point: whatever I try, I either already said that or don't know what I'm talking about. I should note that I'm a bit formal here as well -- I'm answering the question. I'm not participating in any discussion or "exploration" or whatever. Let me finally stress that I'm against this distinction you draw between "mathematicians" and "physicists". Intuition is great. But if rationality tells that intuition is wrong -- have courage to leave it. – Kostya Jul 25 '12 at 14:29 show 1 more comment No you cannot take energy out of the vacuum, BY DEFINITION. Using this analogy with a "sea" is nonsense, because the statement "vacuum is a sea of virtual particles" is also an ambiguous statement not to be taken literally. The definition of vacuum is ground-state energy, and so if you could take energy away from it, then the energy of the vacuum would be lowered, contradicting the fact that it's already in the lowest possible state. This is also the reason why spontaneous absorption does not occur with electrons/atoms (whereas spontaneous emission can occur). I want to point out why this question is not closed yet: Anna's train of thought is "The answer does not satisfy me because it is dependent on definitions which change as theories change." BUT, the question itself is dependent on definitions... you can't ask about apples and then say 'well maybe they can be oranges'. If you can pose a question about 'energy' and 'vacuum' then they have to be defined. I now vote to close! - The ground state energy is the energy of the quantum mechanical system whose potentials one knows and has solved. It is different for different initial set ups. I am talking of something like the Theory of Everything, which in some speculative theories does have several ground states. – anna v Jul 20 '12 at 3:05 p.s. BTW, nature does not follow definitions, it is observers who make them, and new definitions can supersede or embed old ones. – anna v Jul 20 '12 at 3:20 3 Well, usually the physics of tomorrow hides in the metaphysics of today, though I do not think of this as metaphysics. Rather at the frontier of physics. Your answer does not satisfy me, because it says: these are the definitions, do not bother me with facts. We would have never gotten quantum mechanics if questioning were stop by current definitions. – anna v Jul 20 '12 at 5:57 1 The answer does not satisfy me because it is dependent on definitions which change as theories change. Are you saying that gravitational waves do not carry energy by "definition" and it is futile to try and discover them? – anna v Jul 20 '12 at 6:30 2 For you to even begin talking to anyone, you have to specify what a vacuum is and what energy is, otherwise your question doesn't even make sense, and no logical statements can be made! Therefore, once you bring in the words "energy" and "vacuum", then a priori they have a definition which your question must adhere to. And this question has an answer... the end. – Chris Gerig Jul 20 '12 at 6:38 show 5 more comments Peter Milonni has devoted a lot of work to this area, i will recommend this book: The Quantum Vacuum, it basically gives a overview on lot of physical systems where vacuum field effects are dominant, and how certain boundary conditions can effect them to produce unexpected effects the expansion of fields around the vacuum in harmonic modes with quantized particle population of the modes is only valid when the physics can be fairly approximated in terms of the vacuum eigenmodes (which are the regular field wave functions). Of course, when there are special boundary conditions, this doesn't change too much since we still have eigenmodes (which correspond to the boundary geometry) but still can talk about particle populations when the system (meaning, the boundary conditions) are dynamic (moving mirrors, or superconductor walls moving over the superconductivity phase) i don't think there is an authoritative answer about the validity of these approximations. So in short, there could be interesting things to say about the vacuum dynamics when a more friendly framework exists to make computations in highly dynamical regimes I made a somewhat relevant question a while ago about casimir walls that melted and froze back with an oscillatory transversal magnetic field. i'm sorry, i'm not addressing your larger question, as to what relevant things change when taking GR in consideration. Hopefully someone more knowledgeable on the matter will. - -1: This is nonsense. You can't take energy out of the vacuum, and this is conservation laws. – Ron Maimon Sep 4 '11 at 23:22 1 @Ron, there is no such thing as an energy conservation law in GR. – lurscher Sep 5 '11 at 13:53 you aren't doing GR. – Ron Maimon Sep 5 '11 at 13:56 GR effects should manifest locally as out-of-equilibrium boundary conditions, which is why my answer tried to make those points. – lurscher Sep 5 '11 at 14:17 GR effects do not manifest locally--- the lack of conservation in GR can be attributed to new material entering a region on cosmological scales, or leaving it. The conservation of energy in QFT vacuum is a fact, and it is impossible to pump energy out unless the vacuum is unstable and you then wreck the whole universe by tunneling to a lower energy state. – Ron Maimon Sep 5 '11 at 14:25 show 3 more comments The ground state (=vacuum) is an eigenstate of the system Hamiltonian. The wave function is unique, no energy uncertainty exist. In this sense there is no fluctuation of energy to harvest. "Fluctuations" exist for non commuting with the Hamiltonian variables but it cannot be used to get some energy for the reason given above. In GR the energy is not conserved and this is not connected to the vacuum. Edit: in heat machines one tries to increase the temperature difference $T_{hot}-T_{cold}$ to increase the machine efficiency. Vacuum is the coldest body ever known. So the vacuum "cleaner" efficiency cannot exceed zero. - 2 Your last sentence is the crux of the matter. Why wouldn't general relativity create fluctuations in the vacuum, in a TOE? – anna v Jul 2 '11 at 18:40 Because GR is not a unique theory of gravity. There are theories compatible with a flat space-time with the energy/momentum conservation laws. – Vladimir Kalitvianski Jul 2 '11 at 18:59 It certainly is not a heat machine. @Vladimir , the system of comments does not accept your name in the beginning of the comment!! – anna v Jul 3 '11 at 19:18 In theory, QFT vacuum state is isotropic and invariant for all observers, as the base of all ladder operator algebras. It does not have any features other than the fluctuations. Hence, you cannot extract any more energy from it. In practice, extracting energy out of the vacuum is not only possible, but it has been already achieved in laboratory using optical parametric amplifiers (NOPA); they are used for squeezing input light, but when there is nothing in the input, the output is a field that has lower average standard deviation of the energy than the normal vacuum in the range of frequencies where the amplifier is active. If we assign to the normal vacuum zero energy, then this "squeezed vacuum" field must have negative energy. Of course, let's make perfectly clear that this mechanism cannot be used to extract useful energy, since the energy spent pumping the amplifiers greatly exceed anything that you could extract. “In theory, theory and practice are the same. In practice, they are not.” ― Albert Einstein - The concept of squeezing does NOT truly beat the "standard quantum limit", because although it squeezes one variable, the other canonical-variable increases. – Chris Gerig Jul 19 '12 at 23:02 @ChrisGerig, "it squeezes one variable, the other canonical-variable increases" that is correct. But do not forget you can choose what canonical variables you use. In most experimental setups, the variables chosen are amplitude versus phase. In this case, amplitude-squeezed vacuum has smaller amplitude fluctuations, which implies smaller average energy – diffeomorphism Jul 20 '12 at 16:15 @ChrisGerig, there is however, a valid critisism, if we accept that some form of quantum inequalities must hold, then the negative energy regions need to be overcompensed by regions with extra positive energy, so even the time-averaged energy density is nonnegative or the space-averaged energy density is nonnegative, but not both – diffeomorphism Jul 20 '12 at 16:17 There has been a proposal to use Casimir force and meta materials to build a vacuum energy extractor. Casimir force applied on parallel plates made of "normal" material pushes the plates outwardly and inwardly for metamaterials. I do not know if the idea has yet been tested. http://physicsworld.com/cws/article/news/2007/may/02/casimir-force-could-drive-tiny-ratchets - The vacuum state has a property named passivity, which means that any local operation onto the vacuum state does not extract but inject energy to the system. This implies that energy cannot be taken out of the vacuum only by local operations. However, if we adopt local operations and classical communication (LOCC), a part of zero-point energy of the vacuum state can be extracted. The scheme is called quantum energy teleportation. More information is available in wikipedia and a review article by Hotta, who first proposed the concept: http://www.tuhep.phys.tohoku.ac.jp/~hotta/extended-version-qet-review.pdf . - ## protected by Qmechanic♦Jan 28 at 23:03 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": 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": 0.9452539682388306, "perplexity_flag": "middle"}
http://nrich.maths.org/7302/solution	### Chocolate There are three tables in a room with blocks of chocolate on each. Where would be the best place for each child in the class to sit if they came in one at a time? ### Plants Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this? ### Gran, How Old Are You? When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is? # Count the Digits ##### Stage: 1 and 2 Challenge Level: We had quite a few responses from pupils who had found out things by exploring this situation. Calum, Christopher and Matthew from St. Andrews in Scotland wrote to say: We found out that using the digits $1, 2$ and $3$ making a four-digit number you would start doubling eventually when you reach line $10$ every time no matter what order you put the digits in. We found out that using the digits $1, 2$ and $4$ making a four-digit number you would start doubling eventually when you reach line $7$ every time no matter what order you put the digits in. We found out that using the digits $1, 3$ and $4$ making a four-digit number you would also start doubling eventually when you reach line $7$ every time no matter what order you put the digits in. We found out that using the digits $2, 3$ and $4$ making a four-digit number you would start doubling eventually when you reach line $6$ every time no matter what order you put the digits in. Sayeed from St. Michael's London also sent in a well thought out reponse:- This is my answer when starting with $4$ digits: $3243, 122314, 21221314, 31321314, 31123314, 31123314$ It will continue as the same number forever. I notice: Each row has two more digits then the previous row until the rows have the maximum amount of digits ($8$) possible. In each row with the maximum digits, the second digit always has a $1$, the fourth digit always has a $2$, and the sixth digit always has a $3$ and so on. Every row ends in $14$ excluding the starting row. This is my answer when starting with $5$ digits: $22411, 212214, 213214, 21221314, 31321314, 31123314, 31123314$ It will continue as the same number forever. I notice the same thing that happens with the starting $4$ digit row except the rows getting two more digits each time until there are maximum digits. I also notice that with the starting four digit row it takes four counts till the number continues as the same number forever. But for the starting five digit row it takes five counts till the number continues as the same number forever! Thomas from Colet Court School said the following and attached his numbers. When you start with $4$ digits the series converges to $21322314$ whatever the first $4$ digits. When you start with $5$ digits it gives the same result. Finally, Miss Stanley's Numeracy group from Greystoke Leicester wrote:- We liked this challenge and worked very hard. Trying other numbers using the same rules we found that we could continue until the numbers were the same, because that number would keep repeating. We discovered some have shorter sequences and some have longer sequences until the same number repeats, but we're not sure why yet. Caitlin and Millie found that some numbers ($4122$) didn't seem to have an end because we spotted the pattern that it kept repeating itself, so we decided to stop. Some of us even moved onto extending this challenge to $5, 6$,and even $7$ digits. Many of us spotted that the larger the number of digits in the starting number, the shorter the sequence was to get to the end. Thank you, we enjoyed this challenge. Well done to all the contributors, it sounds as if you really enjoyed this. (A late arrival came from Adam at Cypress School who noted something special about $1$ & $4$. I am wondering if that was because it was not a usual thing you'd find going on in many Mathematics lessons? On the last day of the month we recieved this excellent presentation of Oscar from Spain. You have to take out one digit of the $1,2,3,4$ which are the possible digits to make the starting number. If you take out a number and want to get a $4$ digit number, you have to repeat one of the other $3$ numbers. If you take out $1$, you have possible starting numbers $2234, 2334, 2344$ and other possible numbers that you get changing the order of the digits in each of those $3$ numbers. As the order does not affect digit counting, those give the same counting sequence. The counting is: $A 2 B 3 C 4$ and A-B-C have to be $2-1-1$ (for $2234$) or $1-2-1$ (for $2334$) or $1-1-2$ (for $2344$). The next counting in all cases is $2 1 2 2 1 3 1 4$ and sequence is: $3 1 3 2 1 3 1 4$ $3 1 1 2 3 3 1 4$ and this last number stays the same if you count the digits. If you take out $2$ you get the number:$2 1 3 2 2 3 1 4$ If you take out $3$ you get the same as if you take out $1$, and if you take out $4$ you get the same as if you take out $2$. The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 53, "mathjax_display_tex": 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": 0.9445727467536926, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/33373/list	Return to Answer 2 deleted 2 characters in body When property P is universal ($\forall ...$) it is likely to correspond to closed sets, and thus be preserved under intersection. Examples: axioms of a group, ring, field, directed graph; having symmetry under a given group. However, if P is existential ($\exists ...$) it corresponds to open sets and is more likely to be preserved by unions (or coproducts)products), not intersections. Examples: being algebraically closed, having at least 53 elements. (Well, algebraic closure is $\forall \exists$ so of course it is even more complicated. But falling out of the pure $\forall$ class it fails the intersection property.) The first situation is possibly more common because we want structures to satisfy some, well, structural properties. Properties expressed by equations usually correspond to closed sets. To some extent this is formalized in Birkhoff's theorem On equational presentations. Any book on Universal Algebra will discuss it. Also, the sample of concepts is biased, because definitions that become standard are often selected for their useful formal properties. Concepts not having stability under intersection (or union, or inheritance by sub- or super-structures) are less likely to be used. 1 When property P is universal ($\forall ...$) it is likely to correspond to closed sets, and thus be preserved under intersection. Examples: axioms of a group, ring, field, directed graph; having symmetry under a given group. However, if P is existential ($\exists ...$) it corresponds to open sets and is more likely to be preserved by unions (or coproducts), not intersections. Examples: being algebraically closed, having at least 53 elements. (Well, algebraic closure is $\forall \exists$ so of course it is even more complicated. But falling out of the pure $\forall$ class it fails the intersection property.) The first situation is possibly more common because we want structures to satisfy some, well, structural properties. Properties expressed by equations usually correspond to closed sets. To some extent this is formalized in Birkhoff's theorem On equational presentations. Any book on Universal Algebra will discuss it. Also, the sample of concepts is biased, because definitions that become standard are often selected for their useful formal properties. Concepts not having stability under intersection (or union, or inheritance by sub- or super-structures) are less likely to be used.	{"extraction_info": {"found_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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9580432176589966, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Third_derivative	# Third derivative Calculus Definitions Concepts Rules and identities Integral calculus Definitions Integration by Formalisms Definitions Specialized calculi In calculus, a branch of mathematics, the third derivative is the rate at which the second derivative, or the rate of change of the rate of change is changing. The third derivative of a function y=f(x) can be denoted by $\frac{d^3y}{dx^3},\quad f'''(x),\quad\text{or }\frac{d^3}{dx^3}[f(x)].$ Other notations can be used, but the above are the most common. ## Mathematical definitions Let $f(x)=x^4$. Then $f'(x)=4x^3$, and $f''(x)=12x^2$. Therefore, the third derivative of f(x) is, in this case, $f'''(x)=24x$ or, using Leibniz notation, $\frac{d^3}{dx^3}[x^4]=24x$ Now for a more general definition. Let $f(x)$ be any function of x. Then the third derivative of $f(x)$ is given by the following: $\frac{d^3}{dx^3}[f(x)]=\frac{d}{dx}[f''(x)]$ The third derivative is the rate at which the second derivative (f''(x)) is changing. ## Applications in physics Main article: Jerk (physics) In physics, particularly kinematics, jerk is defined as the third derivative of the position function of an object. It is basically the rate at which acceleration is changing. In mathematical terms: $\bold{j}(t)=\frac{d^3\bold{r}}{dt^3}$ where j(t) is the jerk function with respect to time, and r(t) is the position function of the object with respect to time. ## Economic example President Richard Nixon, when campaigning for a second term in office announced that the rate of increase of inflation was decreasing, which has been noted as "the first time a sitting president used the third derivative to advance his case for reelection."[1] Since inflation is itself the negation of a derivative—the rate at which the purchasing power of money decreases—then the rate of increase of inflation is the derivative of inflation, or the negation of the second derivative of the function of purchasing power of money with respect to time. Stating that a function is decreasing is equivalent to stating that its derivative is negative, so Nixon's statement is that the second derivative of inflation—or the negation of the third derivative of purchasing power—is negative. Thus, Nixon's statement is that the third derivative of the purchasing power is positive. Nixon's statement allowed for the rate of inflation to increase, however, so his statement was not as indicative of economic growth as it sounds. ## References 1. Rossi, Hugo (October 1996). "Mathematics Is an Edifice, Not a Toolbox". Notices of the American Mathematical Society 43 (10): 1108. Retrieved 13 November 2012.	{"extraction_info": {"found_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": 10, "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": 0.9369078874588013, "perplexity_flag": "head"}
http://mathoverflow.net/questions/24658/construct-scheme-from-quivers/25023	## construct scheme from quivers? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I heard from some guys working in noncommutative geometry talking about the idea that one can construct the noncommutative space from quivers. I feel it is rather interesting. However, I can not image how one can construct a non-affine scheme or space. For affine case, one might consider the path algebra of a quiver $Q$, say, $KQ$. It is a hereditary algebra(or quasi-free algebra, formally smooth algebra). How can one construct a non-affine space from quiver? Is it possible to construct a noncommutative space using different interesting quivers as "local data"? If it is possible, the algebraic geometry machine working on this space will be helpful to study the representations of quiver? Thanks - ## 3 Answers It is possible to construct more general schemes (even 'projective' schemes) starting from quiver-representations. The idea is to consider a stability structure and consider the corresponding moduli spaces of semi-stable quiver representations as introduced by Alastair King (Moduli of representations of finite dimensional algebras, Quat. J. Math. Oxford 45 (1994) 515-530.) If one starts with a quiver without oriented cycles, these moduli spaces are projective and so it makes sense to take the collection of all these moduli spaces (or multiples of a fixed dimension vector compatible with the stability structure) as a noncommutative projective space. This point of view was advocated in 'Noncommutative compact manifolds constructed from quivers'. More generally, allowing cycles, the collection of moduli spaces (or multiples of fixed dim vector) for all dim vectors compatible with the stabilitu structure can be viewed as a noncommutative scheme. Here the main idea is that polynomial semi-invariants of quivers are known to be generated by so called determinental semi-invarinats (a result by Aidan Schofield and Michel Van den Bergh 'Semi-invariants of quivers for arbitrary dimension vectors'. What this says is that these collections of moduli spaces have an affine cover determined by universal localizations of the path algebra. This point of view is expressed here. EDIT : Here's an example to illustrate all of this : noncommutative projective n-space. Consider the quiver with 2 vertices and n+1 arrows, all from the first to the second vertex, Take stability structure (-1,1), then we consider only representations of dimension vector (m,m). A semi-stable representation is one given by the n+1 mxm matrices X0,...,Xn such that there are numbers ai with det(a0X0+a1X1+...+anXn) is non-zero. For example, consider the open piece where Xi is invertible, then one can use Xi to identify the two vertex-spaces and the other arrows now become loops in that common vertex. That is, this open piece gives all representations of the free algebra in n variables. Hence, we can cover the moduli space by affine pieces, all iso to the reps of a one-vertex quiver with n loops. Clearly one can generalize this procedure to glue certain representations of two very different quivers. Formally invert one arrow in your quiver, this comes down to taking a stability structure with 0's at all other vertices and (-1,1) at start and end-point of your arrow. Then the 'Zariski open piece' of repQ consists the representations of a new quiver Q' in which one identifies the two vertices, and turns all other arrows between these two vertices into loops at the new vertex. In this way to quivers can be 'glued' to form more general schemes provided they are isomorphic after doing this 'invert one arrow' trick on each of them a finite number of times. Universal localizations corresponding to 'determinental semi-invariants' are a generalization of this idea, allowing paths (or even collections of linear combinations of paths) to be 'inverted'. (End EDIT) A recent incarnation of this strain of ideas (restricting to stable rather than semi-stable representations) is taken by Markus Reineke in his study of 'noncommutative Hilbert schemes' which have been used by Markus recently in 'Cohomology of quiver moduli, functional equations, and integrality of Donaldson-Thomas type invariants'. As to the local structure of these noncommutative schemes. Their 'local rings' are indeed again path algebras of 'local quivers' which can be determined from the stable components of the semi-stable representation. The details are here. If you want to extend all of this to general formally smooth algebras (rather than just path algebras) a place to start is with 'One quiver to rule them all'. Finally, Ive tried to collect all of this in a book as Daniel mentioned in his answer (one might replace 'blablabla' by 'noncommutative geometry'...). The latest version of it can be downloaded here. - Excellent! – Victor Protsak May 17 2010 at 19:12 ### 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. As far as I know a "non-commutative space" is for many people just an $A_\infty$-category. Every scheme determines a non-commutative space: its derived category of coherent sheaves (with an $A_\infty$ enrichment). Also to an (non-commuative) algebra $A$ you can associate a non-commutative space, the derived category of modules over $A$. In some cases these two types of non-commutative space are isomorphic. The classical example (due to Belinson) is projective space whose derived category is equivalent to the derived category of modules over $End(O \oplus O(1) \oplus \dots O \oplus O(n) )$. This algebra has a description as path algebra over a quiver with relations. E.g. for $P^1$ you get the Kronecker quiver (two vertices two arrows in the same direction). - I think you should start looking at the webpage of Lieven Le Bruyn. He most definitely has some notes that will quell your thirst in this respect. For one he has written a book that is freely available on his webpage ("Noncommutative Geometry and Cayley-smooth Orders"). Another good person whose papers and webpage it's reasonable to pay a visit to if you're interested in this area is Markus Reineke. Victor Ginzburg's notes on Noncommutative geometry arXiv:math/0506603 can maybe also be of interest even though I don't know if there are any quivers lurking there. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9199309349060059, "perplexity_flag": "middle"}
http://mathhelpforum.com/math-topics/201411-velocity-time-curve-problem.html	# Thread: 1. ## velocity-time curve problem A train start to move from rest and then travel its first part with constant acceleration of $f_1$then travel the second part of the constant velocity of V then travel the third part of the constant deceleration of $f_2$ then it comes to rest.So draw a velocity - time curve and it travel constant velocity with the $\frac{4}{3}t$ of the total time and the average speed of the total travelling is KV.so find out the value of k? so please give me the ideas to do that. 2. ## Re: velocity-time curve problem Originally Posted by srirahulan A train start to move from rest and then travel its first part with constant acceleration of $f_1$then travel the second part of the constant velocity of V then travel the third part of the constant deceleration of $f_2$ then it comes to rest.So draw a velocity - time curve and it travel constant velocity with the $\frac{4}{3}t$ of the total time and the average speed of the total travelling is KV.so find out the value of k? so please give me the ideas to do that. At constant acceleration $f_1$, after time t, the speed will be $f_1t$. You say "the first part" but you don't say what the time is or if "the first part" is the same length of the time as the other "parts". The graph of that is a straight line with slope $f_1$. Calling the length of time for the first part " $T_1$", the velocity at the end of the first part will be $f_1T_1$ which, presumably, is the constant speed of the "second part". Taking the length of time for the "second part" " $T_2$, the second part of the graph is a horizontal straight line with height V, from $T_1$ to $T_2$. The third part of the graph will be a straight line with slope $-f_1$. Of course, the train will stop after time interval $T_2$ where $V- f_3T_3- T_2)= 0$ I don't uderstand what the "it travel constant velocity with the $\frac{4}{3}t$ of the total time" means. You did not mention "t" before and 4/3 is larger than 1. 3. ## Re: velocity-time curve problem I am really sorry i assume the total time for travelling is t.when it travel in the constant velocity of V with 3/4t of the total time (sorry for mention 4/3)	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "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": 0.9187273979187012, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-applied-math/198820-partitions-proof.html	# Thread: 1. ## Partitions proof Let the number of partitions of n into parts each less than or equal to m be denoted P(n,m). Prove P(n,m) = P(n, m-1) + P(n-m,m) 2. ## Re: Partitions proof Let the number of partitions of n into parts each less than or equal to m be denoted P(n,m). Prove P(n,m) = P(n, m-1) + P(n-m,m) I have written on this question. So I tell you whoever wrote your question has left out some cases. What happens if $m=1~?$ What happens if $m=n~?$ Here is the MathCad way of doing it. [ATTACH=CONFIG]23869[/ATTACH] Attached Thumbnails 3. ## Re: Partitions proof what is mathcad? >.> ... i'm not really following that equation I'm a little confused because there is a formula in the book about partitioning n into exactly r parts where order counts = C(n-1,r-1) but then there's another one that says "the number of ways of writing n as the sum of r or fewer integers where the order of the summands matters is C(r+n-1,n)" but i didn't think they applied because order shouldn't count, should it? I was also trying to play with the concept that the number of partitions of n into r or less parts = the number of partitions of n into parts that are r or less, but have gotten nowhere so far :\ Originally Posted by Plato $m=1~?$ What happens if $m=n~?$ if m = 1 there should only be one partition, n itself, right? and if m = n, then all the partitions = 1 4. ## Re: Partitions proof could you be more specific please?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9308232069015503, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/118515?sort=oldest	## Derivative indicator function ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am wondering what is the derivative of the following function with respect to $x(t)$ in sense of distributions. $$I\left(\int_0^t x(\tau)d\tau \leq c\right)$$ where $I$ is the indicator function and $c$ is a constant. - What is the indicator function? – Liviu Nicolaescu Jan 10 at 11:01 Distributions on a space of functions? Even measures there are notoriously difficult (cylindrical measures of L. Schwartz). – Peter Michor Jan 10 at 11:36 Could you add a bit more details? For example, it is important what you mean by indicator function -- the standard definition I know is extended-real-valued, and hence has no derivative in the sense of classical analysis. There are other derivative concepts that are applicable here, but it would help to know the context of this question. – Christian Clason Jan 10 at 13:01 ## 2 Answers Let me try. Your mapping is: $1 - H(x-c)$ , where $H$ is the Heaviside function, derivative is $-\delta(x-c)$. composed with $\int_0^t : C^\infty(\mathbb R) \to \mathbb R$, which is bounded and linear, surjective. No way with the chain rule, in fact, there is no chain rule in this situation. Simpler approach, but it might not answer your question: Suppose $s\mapsto x(\tau,s)$ is a smooth variation of $x(\tau)$, i.e., $x\in C^\infty(\mathbb R^2)$. Then $$s\mapsto 1-H\Big(\int_0^t x(\tau,s)d\tau -c\Big)$$ can be viewed as a distribution in $s$. Take a test function $f(s)$ and consider $$\int \Big(1-H\Big(\int_0^t x(\tau,s)d\tau -c\Big)\Big)(-f'(s))ds$$ Now try formal partial integration with respect to $s$. Since the left hand integrand is 1 or 0, by the fundamental theorem it will be the difference of the values of $f$ at those $s$ where this jumps. If you fix $x(\tau,s)$ you can work this out. - ### 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. You are asking for the derivative of a non-linear function on an infinite dimensional space. (You do not specify the latter but the space locally integrable functions seems a natural candidate). The derivative can only exist in a very weak sense so it is natural to go for the directional derivative which has two inputs---the point $x$ where the derivative is computed and the direction $y$ in which the rate of change takes place. A back of the envelope calculation suggests that the derivative should then be $$-\sum \frac {Y(a)}{x(a)}\delta_a$$ where we use capitals to indicate the primitives which occur in the formulation and the sum is taken over the $a$ in the pre-image of $c$ under $X$. Some general remarks: the question was posed in such a vague manner that it is not really possible to give a precise, rigorous answer. Presumably you have some concrete application in view and I suggest that you check the above formula there to see if it leads to the expected result (ones I looked at were the functions $t$---with primitive $t^2$---and $\sin t$ for the function $x$). As is evident from the above formula, the derivative is in a much weaker sense than the usual concepts of functional analysis---one requires special conditions on $x$ (beyond just smoothness) for the above expression to make sense and the limit of the difference quotients used to define the derivative does not take place in the underlying function space but in a larger space (of distributions---hence the Dirac functions in the formula. I presume that this is the reason for the reference to distributions in your question). - 1 am having TeX problems in editing this answer. the denominator in the formula should be encased in absolute value symbols. – jbc Feb 10 at 6:30	{"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": 26, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9281629920005798, "perplexity_flag": "head"}
http://cms.math.ca/Competitions/MOCP/2002/prob_jun.mml	Canadian Mathematical Society www.cms.math.ca \| \| \| \| \| \| \|----------\|----\|-----------\|----\|----------\| \| \| \| \| \| \| \| Français \| \| Site map \| \| \| CMS store \| \| \| location: PROBLEMS FOR JUNE Valeria Pandelieva 641 Kirkwood Avenue Ottawa, ON K1Z 5X5 no later than July 21, 2002. 151. Prove that, for any natural number $n$, the equation $x\left(x+1\right)\left(x+2\right)\dots \left(x+2n-1\right)+\left(x+2n+1\right)\left(x+2n+2\right)\dots \left(x+4n\right)=0$ does not have real solutions. 152. Andrew and Brenda are playing the following game. Taking turns, they write in a sequence, from left to right, the numbers 0 or 1 until each of them has written 2002 numbers (to produce a 4004-digit number). Brenda is the winner if the sequence of zeros and ones, considered as a binary number (i.e., written to base 2), can be written as the sum of two integer squares. Otherwise, the winner is Andrew. Prove that the second player, Brenda, can always win the game, and explain her winning strategy (i.e., how she must play to ensure winning every game). 153. (a) Prove that, among any 39 consecutive natural numbers, there is one the sum of whose digits (in base 10) is divisible by 11. (b) Present some generalizations of this problem. 154. (a) Give as neat a proof as you can that, for any natural number $n$, the sum of the squares of the numbers $1,2,\dots ,n$ is equal to $n\left(n+1\right)\left(2n+1\right)/6$. (b) Find the least natural number $n$ exceeding 1 for which $\left({1}^{2}+{2}^{2}+\dots +{n}^{2}\right)/n$ is the square of a natural number. 155. Find all real numbers $x$ that satisfy the equation ${3}^{\left[\left(1/2\right)+\mathrm{log}{}_{3}\left(\mathrm{cos}x+\mathrm{sin}x\right)\right]}-{2}^{\mathrm{log}{}_{2}\left(\mathrm{cos}x-\mathrm{sin}x\right)}=\sqrt{2} .$ [The logarithms are taken to bases 3 and 2 respectively.] 156. In the triangle $\mathrm{ABC}$, the point $M$ is from the inside of the angle $\mathrm{BAC}$ such that $\angle \mathrm{MAB}=\angle \mathrm{MCA}$ and $\angle \mathrm{MAC}=\angle \mathrm{MBA}$. Similarly, the point $N$ is from the inside of the angle $\mathrm{ABC}$ such that $\angle \mathrm{NBA}=\angle \mathrm{NCB}$ and $\angle \mathrm{NBC}=\angle \mathrm{NAB}$. Also, the point $P$ is from the inside of the angle $\mathrm{ACB}$ such that $\angle \mathrm{PCA}=\angle \mathrm{PBC}$ and $\angle \mathrm{PCB}=\angle \mathrm{PAC}$. (The points $M$, $N$ and $P$ each could be inside or outside of the triangle.) Prove that the lines $\mathrm{AM}$, $\mathrm{BN}$ and $\mathrm{CP}$ are concurrent and that their intersection point belongs to the circumcircle of the triangle $\mathrm{MNP}$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 29, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.946619987487793, "perplexity_flag": "head"}
http://mathoverflow.net/questions/108903/imaginary-quadratic-field-contained-in-hecke-orbit-field/108950	## Imaginary quadratic field contained in Hecke orbit field? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $\tau$ in the upper half plane lie in an imaginary quadratic field $K$. Then is $K \subset \mathbb{Q}({j(g \tau) \ \| \ g \in GL_2^+(\mathbb{Q}) })$? (here $j$ is the modular $j$-function, and $GL_2^+(\mathbb{Q})$ means positive determinant - i.e. we adjoined all $j$ values of elliptic curves with CM by an order in $K$). If the above isn't true, then is $K$ contained in $\mathbb{Q}(S)$ where $S$ is the set of $j$ values of all CM elliptic curves? I'm also interested in whether the appropriate statement is true for a general Shimura variety. - ## 1 Answer "No" for both questions about CM elliptic curves and "I'm not even sure I know what the question would be" about general Shimura varieties. Basic idea: If $j$ is the $j$-invariant of a CM elliptic curve, then there is some imaginary quadratic discriminant $\Delta \equiv 0$ or $1 \bmod 4$ such that $\mathbf{Q}(j) \cong \mathbf{Q}[X]/H_\Delta(X)$ where $H_\Delta(X) \in \mathbf{Z}[X]$ is the Hilbert Class Polynomial of discriminant $\Delta$, whose roots are the $j$-invariants of elliptic curves over the complex numbers (actually $\overline{\mathbf{Q}}$ is enough) with CM by $\mathbf{Z}\left[\dfrac{ \Delta + \sqrt \Delta}{2}\right]$. The point is that now we can see that there is an embedding $\mathbf{Q}(j) \hookrightarrow \mathbf{R}$, for all possible $j$. Therefore there is an embedding $\mathbf{Q}(S) \hookrightarrow \mathbf{R}$. To see this, it's enough to note that for any two CM $j$-invariants $j_1$ and $j_2$ that there exists an embedding $\mathbf{Q}(j_1,j_2)\hookrightarrow \mathbf{R}$. Let $J_1$ and $J_2$ be the canonical image of $j_1$ and $j_2$ in the real numbers. Then $\mathbf{Q}(j_1)$ embeds into the real numbers as $\mathbf{Q} + \mathbf{Q}J_1 + \dots + \mathbf{Q}J_1^{h_1 -1}$ and $\mathbf{Q}(j_2)$ embeds into the real numbers as $\mathbf{Q} + \mathbf{Q}J_2 + \dots + \mathbf{Q}J_2^{h_2 -1}$. Therefore $\mathbf{Q} + \mathbf{Q}J_1 + \mathbf{Q}J_2 + \dots + \mathbf{Q}J_1^{h_1 -1}J_2^{h_2 -1}$ is a copy of $\mathbf{Q}(j_1,j_2)$ inside of $\mathbf{R}$. Notice that I didn't use direct sums because $\mathbf{Q}(j_1)$ and $\mathbf{Q}(j_2)$ might not be linearly disjoint over $\mathbf{Q}$! This is also the reason I didn't use a tensor product argument. In any case, this inductive step allows us to work with direct limits and embed $\mathbf{Q}(S)$ into $\mathbf{R}$. Therefore if we assume that there is an embedding $K\hookrightarrow \mathbf{Q}(S)$ then there must be an embedding $K\hookrightarrow\mathbf{Q}(S) \hookrightarrow \mathbf{R}$, which is absurd. Therefore, there is no embedding $K\hookrightarrow \mathbf{Q}(S)$. To show that we have an embedding $\mathbf{Q}(j)\hookrightarrow\mathbf{R}$, consider that there is some $\tau \in \mathcal{H}$ of the form $\dfrac{1}{2}\sqrt\Delta$ or $\dfrac{ 1 + \sqrt \Delta}{2}$ such that $j(\tau)$ is a root of $H_\Delta(X)$. But then $j(\tau)$ is real, because the inverse image of the reals under the $j$-function contains the lines $\lbrace iy : y \ge 1\rbrace$ and $\lbrace 1/2 + iy : y \ge (1/2)\sqrt 3\rbrace$. Therefore we have our embedding $\mathbf{Q}(j) \hookrightarrow \mathbf{R}$. - @stankewicz:Fair enough, you have an embedding $\mathbb{Q}(j)$ into $\mathbb{R}$. But could you please give more details on how you embed $\mathbb{Q}(S)$ into $\mathbb{R}$ (surely $H_{\Delta}(X)$ could have some complex roots)?. – Adam Harris Oct 8 at 13:25 @stankewicz: Isn't it true that if the class of $\mathfrak{a}$ in the ideal class group has order greater than 2 then $j(\mathfrak{a})$ isn't real? – Adam Harris Oct 8 at 13:33 I edited the answer to be more explicit. It is very possible for Hilbert Class Polynomials to have complex roots and it's easy to find examples by quickly playing around with MAGMA or sage. It just doesn't matter, because as soon as there is ONE real embedding, any subfield has to also have a real embedding. – stankewicz Oct 8 at 15:13 @stankewicz: Thanks, but I'm still confused: What is this `canonical image' of $j_i$ in the reals? If it is $j(\mathcal{O})$ where $\mathcal{O}$ is your order then what if the canonical images are equal i.e. $J_1=J_2$? i.e. how do you embed the splitting field of $H_D(X)$ in the reals if it has a complex root? – Adam Harris Oct 8 at 20:08 I don't embed the splitting field of $H_D(X)$ into the reals because you can't. Is your Hecke orbit field the union of $\mathbf{Q}(j(E))$ over the set of elliptic curves with CM by an order inside $K$, or is it the union of the Galois closures? If you mean Galois closures, the answer becomes yes and in fact the Galois closure for $E$ with CM by ANY order in $K$ will work. – stankewicz Oct 8 at 21:25 show 9 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": 76, "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": 0.9445396661758423, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2008/09/27/analytic-functions/?like=1&_wpnonce=9d779cb1c0	# The Unapologetic Mathematician ## Analytic Functions Okay, we know that power series define functions, and that the functions so defined have derivatives, which have power series expansions. And thus these derivatives have derivatives themselves, and so on. Thus a function defined by a power series in a given disk is actually infinitely differentiable within that same disk. What about the other way? Say if we have a function $f$ with arbitrarily high derivatives at the point $z_0$. We know that if this function has a power series about $z_0$, then the only possible sequence of coefficients is given by the formula $\displaystyle a_k=\frac{f^{(k)}(z_0)}{k!}$ But does this sequence actually give a power series expansion of $f$? That is, does the (in)famous “Taylor series” $\displaystyle\sum\limits_{k=0}^\infty\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k$ converge to the function $f$ in any neighborhood of $z_0$? If so, we’ll call the function “analytic” at $z_0$. So, are all infinitely-differentiable functions analytic? Are all functions for which the Taylor series at $z_0$ actually the limit of said Taylor series near $z_0$? Well, the fact that we have a special name should give a hint that the answer isn’t always “yes”. We’ve been working with complex power series, but let’s specialize now to real power series. That is, all the coefficients are real, we center them around real points, and they converge within a real disk — an interval — of a given radius. Now in this context we can consider the function defined by $f(x)=e^{-x^{-2}}$ away from $x=0$. It’s straightforward to calculate $\displaystyle\lim\limits_{x\rightarrow0}e^{-x^{-2}}=0$ And if we define $f(0)=0$ it turns out to even be differentiable there. The derivative turns out to be $\left(2x^{-3}\right)e^{-x^{-2}}$. And we can also calculate $\displaystyle\lim\limits_{x\rightarrow0}2x^{-3}e^{-x^{-2}}=0$ And so on. The $n$th-derivative will be $P_n(x^{-1})e^{-x^{-2}}$, where $P_n$ is a polynomial. We can calculate $\frac{d}{dx}P_n(x^{-1})e^{-x^{-2}}=\left(P'_n(x^{-1})(-x^{-2})+P_n(x^{-1})(2x^{-3})\right)e^{-x^{-2}}$ That is, we can set $P_0=1$ and $P_{n+1}=2X^3P_n-X^2P_n'$, and thus recursively define a sequence of polynomials. Thus for each degree we have a polynomial in $x^{-1}$ multiplied by $e^{-x^{-2}}$, and in the limit the latter clearly wins the race. For each derivative we can fill in the “gap” at $x=0$ by defining $f^{(n)}(0)=0$. But now when we set up the Taylor series around $x_0=0$ what happens? The series is $\displaystyle\sum\limits_{k=0}^\infty\frac{f^{(n)}(0)}{k!}x^k=\sum\limits_{k=0}^\infty\frac{0}{k!}x^k$ Which clearly converges to the constant function ${0}$. That is, the Taylor series of this function at $x_0=0$ converges to nothing like the function itself. This function is infinitely differentiable at ${0}$, but it is not analytic there. There are a lot of theorems about what conditions on an infinitely-differentiable function make it analytic, but I’m going to leave them alone for now. ### Like this: Posted by John Armstrong \| Analysis, Calculus ## 5 Comments » 1. [...] I’ve decided I really do need one convergence result for Taylor series. In the form we’ll consider today, it’s an extension of the ideas in the Fundamental [...] Pingback by \| September 30, 2008 \| Reply 2. [...] we can construct the Taylor series at . The coefficient formula tells [...] Pingback by \| October 7, 2008 \| Reply 3. [...] algebra. Instead, I set off on power series and how power series expansions can be used to express analytic functions. Then I showed how power series can be used to solve certain differential equations, which led us [...] Pingback by \| October 16, 2008 \| Reply 4. [...] it might seem like a really weird digression, but let’s look at the Taylor series for the function . Yes, that was purely a product of our work on analysis over , but let’s [...] Pingback by \| March 10, 2009 \| Reply 5. [...] this the end? Not quite. Just like in one variable we have analytic functions. Once we’re in and we have all higher derivatives we can use [...] Pingback by \| October 21, 2009 \| 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": 32, "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": 0.9256283044815063, "perplexity_flag": "head"}
http://mathhelpforum.com/discrete-math/137064-graph-theory-proof.html	# Thread: 1. ## graph theory proof Prove that if G is a graph with no cycles of even length, then every cycle in G is an induced subgraph of G. any help would be great thanks. 2. Suppose G has no cycles of even length. Let $H=\{ h_1,h_2,\ldots,h_n\}$ be a cycle of G (of odd length) and suppose it is not an induced subgraph of G (assume wlog that $h_i E_H h_{i+1}$ and $h_n E_H h_1$. Then there is some i, j (wlog i<j) such that $h_i E_G h_j$ but $\neg(h_i E_H h_j)$. Consider the two cycles $h_1\rightarrow \cdots\rightarrow h_i\rightarrow h_j\rightarrow h_{j+1}\rightarrow \cdots \rightarrow h_n\rightarrow h_1$ and $h_i\rightarrow h_{i+1}\rightarrow \cdots\rightarrow h_{j-1}\rightarrow h_j\rightarrow h_i$ A moment shows (and drawing a picture will help) that one of these has even length. Contradiction. 3. thanks for the reply wgunther, im kind of new to the world of graph theory, what do you mean by wlog?? and i can see quite clearly that this statement is obviously true and can show it with a diagram but how to show proof of it in mathematical terms is where i get abit lost. 4. Wlog means without loss of generality. When I say E sub H I mean the edge relation on the graph H. It's pretty simple to prove from here. Count the number of vertices in each. One has n+j-i and the other is i-j. They sum to n. n is odd, thus one of the two is even	{"extraction_info": {"found_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": 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": 0.9690490365028381, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=3810906	Physics Forums ## How to normalize wave functions in QFT? such as \lambda \phi 4 theory? In quantum mechanics, most wave functions are normalized with \int \|\phi\|^2 dx^3 =1. But I did not see any field in the quantum field theory is normalized. I understand they maybe just plain waves and does not need to be normalized. But in some cases, if we do not expand the field as plain wave, how to normalize them? For instance, in the \lambda \phi 4 theory, the field \phi has the dimension of GeV. Should we use \int \|\phi\|^4 dx^4 =1 instead of \int \|\phi\|^2 dx^3 =1 or \int \|\phi\|^4 dx^3 =1? 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 The wave function in quantum mechanics represents the state of the system. The field in quantum field theory is not the state, so why do you expect that it be normalized? On the other hand the states in QFT are assumed normalized. If we want to solve the \lambda \phi 4 equation in the same way as solve the Dirac equation, I think we should normalize the result. So we can say that the trivial expresion of <initial \| final> = <\phi\|\phi>=1 means, when no perturbation, the amplitude from \phi to its own is 1. For instance, the free particle solution of the Dirac equation is normalized to 1 by a factor \sqrt{m/E}. Blog Entries: 19 Recognitions: Science Advisor ## How to normalize wave functions in QFT? such as \lambda \phi 4 theory? Normalization of a field is needed only when that field ITSELF is interpreted as a probability density amplitude. The field \phi in \phi^4 is usually not interpreted in that way. Recognitions: Science Advisor In QFT the normalization of the field operators are defined via the equal-time commutation relations, $[\phi(t,\vec{x}),\Pi(t,\vec{y})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y}).$ Here $\Pi$ is the canonical field momentum for $\phi$. In $\phi^4$ theory, it's $\Pi(x)=\frac{\partial \mathcal{L}}{\partial \dot{\phi}(x)}=\dot{\phi}(x).$ For the asymptotically free states, symbolized by external legs in Feynman diagrams, the states are to be normalized in the usual way to $\delta$ distributions (supposed you have taken account of wave-function renormalization in the external legs, i.e., left out all self-energy insertions in them, see Weinberg, QT of Fields, vol. 1). Thread Tools \| \| \| \| \|---------------------------------------------------------------------------------------------\|---------------------------\|---------\| \| Similar Threads for: How to normalize wave functions in QFT? such as \lambda \phi 4 theory? \| \| \| \| Thread \| Forum \| Replies \| \| \| Advanced Physics Homework \| 17 \| \| \| Advanced Physics Homework \| 5 \| \| \| Advanced Physics Homework \| 4 \| \| \| Advanced Physics Homework \| 8 \|	{"extraction_info": {"found_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": 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": 0.8952093720436096, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/closed-form+exponentiation	# Tagged Questions 0answers 37 views ### can´t solve this kind of exponential equations I have no clue how can i solve this kind of exponential equation in closed form: $a^n - b^n \le c$ where $a > 1$ and $-1 < b < 0$ thank you very much for your help 0answers 84 views ### How to solve these exponential equations for D? I'm curious if this is even possible to solve for D. D is the only variable, x, y, z, and w are all constants, and e is the mathematical constant e. (\frac{x+yD^{2}}{zD})^{\sqrt{2D}} = ... 3answers 324 views ### How do I find if $\frac{e^x}{x^3} = 2x + 1$ has an algebraic solution? Is there some way of solving $$\frac{e^x}{x^3} = 2x + 1$$ non-numerically? How would I go about proving if there exists a closed form solution? Similarly how would I go about proving if there exists ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9344586133956909, "perplexity_flag": "head"}
http://mathhelpforum.com/trigonometry/78556-sine-rule.html	# Thread: 1. ## sine rule (Give non-exact answers to 3 significant figures.) In a triangle, the largest side has length 2 cm and one of the other sides has length √2 cm. Given that the area of the triangle is 1cm2, show that the triangle is right-angled and isosceles. Am not sure how to state off, can anyone just start me off . thanks. 2. Hi Using cross product properties we know that the area of the triangle is half the product of 2 lengths of the triangle multiplied by the absolute value of sine of the angle formed by the two sides $A = \frac12\:\|\|\overrightarrow{AB}\wedge\overrightarro w{AC}\|\| = \frac12\:AB\:AC\:\|sin(BAC)\|$ With this formula you can compute the angle BAC, the length of the other side and conclude 3. $1 = \frac{1}{2} \times 2 \times \sqrt{2} \times sin\theta$ $\theta = 45$ where do I go from here? how do I show it is a right angled triangle and isosceles.? 4. Hello, Tweety In a triangle, the largest side has length 2 cm and one of the other sides has length √2 cm. Given that the area of the triangle is 1 cm², show that the triangle is right-angled and isosceles. Code: ``` * * * _ * * √2 * * * * * θ * * * * * * * * 2``` Formula: . $A \;=\;\frac{1}{2}\,bc\sin A$ The area is one-half the product of two sides and the sine of the included angle. We have: . $\frac{1}{2}\left(\sqrt{2}\right)(2)\sin\theta \:=\:1 \quad\Rightarrow\quad \sin\theta \:=\:\frac{1}{\sqrt{2}} \quad\Rightarrow\quad \theta \:=\:45^o$ Hence, the triangle looks like this: The right half is identical to the left half. Code: ``` * * \| * _ * \| * _ √2 * \|1 * √2 * \| * * 45° \| 45° * * * * * * * * * * 1 1``` Therefore, the triangle is an isosceles right triangle.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.8952499628067017, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/67814/inverse-of-the-curl	# Inverse of the curl Given $\nabla \cdot u = 0$ and $w = \nabla \times u$ equation 9 of http://projecteuclid.org/euclid.cmp/1103941230 has the identity $u = - \nabla \times (\nabla^{-1} w)$. What is $\nabla^{-1}$ exactly and how was this identity derived? Thanks a lot. - 3 @QED: it would be helpful if you told us where you got this from... – Mariano Suárez-Alvarez♦ Sep 27 '11 at 2:39 QED is working through a 1984 article in Comm. Math. Phys. by Kato, Majda, and Beale, link under my answer. – Will Jagy Sep 27 '11 at 4:14 @Will: either your knowledge of the literature is simpy outstanding or you have great changes of getting the site's Mind-Reader of the Year Award. :) – Mariano Suárez-Alvarez♦ Sep 27 '11 at 4:21 Mariano, I am a pretty good mind-reader, but my first indication was the OP putting a comment with a link to the reference under my answer. – Will Jagy Sep 27 '11 at 4:29 Oh. Well, you'll have to wait a bit more for the award, then! – Mariano Suárez-Alvarez♦ Sep 27 '11 at 4:41 show 3 more comments ## 2 Answers The fact that $u$ is divergence free does mean that $u$ is the curl of something, locally at least. The fact that we have, for some $v,$ that $u = \nabla \times v$ amounts to little more than the fact that mixed partial derivatives commute, and in general is called Poincare's Lemma. Furthermore, one can replace any such $v$ by $v + \nabla \cdot f$ for some function $f.$ As in the comments, there is therefore little reason to talk about an operator $\nabla^{-1},$ such a thing is not going to be well defined. If you want to give an exact reference and convince us that a responsible person wrote the material you are quoting, things might be different. - 2 – user16697 Sep 27 '11 at 3:11 There is no such thing as $\nabla^{-1}$. What is true is that $\nabla \times (\nabla \times u) = \nabla(\nabla \cdot u) - \nabla^2 u$. - 2 QED is working his way through a 1984 article in Comm. Math. Phys. by Kato, Majda, and Beale, link under my answer. Yesterday's question was about equation (7), today equation (9) – Will Jagy Sep 27 '11 at 3:37 2 So presumably this was a misprint and they meant $(\nabla^2)^{-1} w$, i.e. some vector field $f$ such that $\nabla^2 f = w$, instead of $\nabla^{-1} w$. This is nonunique because you could add any vector field whose components are harmonic functions. – Robert Israel Sep 28 '11 at 22:21 @Robert, That could be it but if so how do I prove $u = - \nabla \times f$? I tried to show the divergence and curl of $- \nabla \times f$ match that of $u$ but could only do divergence.. and I'm not sure if both those are enough to prove it. – user16697 Sep 28 '11 at 22:47 – user16697 Sep 28 '11 at 22:59 1 – Hans Lundmark Sep 29 '11 at 8:57 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": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9384549856185913, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/partitions?page=5&sort=newest&pagesize=15	# Tagged Questions Questions related to the different ways of expressing an integer as a sum of integers; or, questions related to the subdivision of a set into smaller disjoint sets; questions related to the subdivision of an interval into smaller intervals that intersect only at the endpoints. 1answer 70 views ### HAKMEM 18(B): Cubic Partitions Taken from HAKMEM 18. Quoting... A partition of $N$ is a finite string of non-increasing integers that add up to $N$. Thus 7 3 3 2 1 1 1 is a partition of 18. Sometimes an infinite string of ... 2answers 248 views ### A question on partitions of n Let $P$ be the set of partitions of n. Let $\lambda$ denote the shape of a particular partition. Let $f_\lambda(i)$ be the frequency of $i$ in $\lambda$ and let \$a_\lambda(i) := \# \lbrace j : ... 3answers 396 views ### An Inequality Involving Bell Numbers: $B_n^2 \leq B_{n-1}B_{n+1}$ The following inequality came up while trying to resolve a conjecture about a certain class of partitions (the context is not particularly enlightening): $$B_n^2 \leq B_{n-1}B_{n+1}$$ for \$n \geq ... 2answers 292 views ### How can I reduce a number? I'm trying to work on a program and I think I've hit a math problem (if it's not, please let me know, sorry). Basically what I'm doing is taking a number and using a universe of numbers and I'm ... 1answer 296 views ### Number of solutions of Frobenius equation I have one problem which needs to count the number of solution of the equation $$2x+7y+11z=42$$ where $x,y,z \in \{0,1,2,3,4,5,\dots\}$. My attempt: I noticed that that maximum value of $z$ could ... 1answer 261 views ### Recurrence for the partition numbers I'm reading Analytic Combinatorics [PDF] book by Flajolet and Sedgewick, and I can't figure out one of the steps in the derivation of the $P_n$ — number of partitions of size $n$ (or coefficients in ... 2answers 200 views ### Two combinatorics problems. I'm not 100% confident in my answers These are two problems from my combinatorics assignment that I'm not quite confident in my answer. Am I thinking of these the right way? Problem 1: On rolling 16 dice. How many of the $6^{16}$ ... 2answers 298 views ### Learning about partitions and modular forms I'm interested in learning about partitions and modular forms. I already know algebra and analysis (complex and real). Can any one suggest me books or other materials from where I can learn these ... 0answers 121 views ### Jordan Measure, Semi-closed sets and Partitions From earlier question here. Consider $$\mu \left( (0,1) \cap \mathbb Q \right) = 0$$ where $$(0,1) = (0,\frac{1}{3})\cup [\frac{1}{3}, \frac{2}{3}] \cup (\frac{2}{3}, 1)$$ so \mu ((0,1) \cap ... 2answers 248 views ### Jordan Measures, Open sets, Closed sets and Semi-closed sets I cannot understand: $$\bar{\mu} ( \{Q \cap (0,1) \} ) = 1$$ and (cannot understand this one particularly) $$\underline{\mu} ( \{Q \cap (0,1) \} ) = 0$$ where $Q$ is rational numbers, why? I ... 1answer 744 views ### Ellipse 3-partition: same area and perimeter Inspired by the question, "How to partition area of an ellipse into odd number of regions?," I ask for a partition an ellipse into three convex pieces, each of which has the same area and the same ... 1answer 971 views ### The number of partitions of $n$ into distinct parts equals the number of partitions of $n$ into odd parts This seems to be a common result. I've been trying to follow the bijective proof of it, which can be found easily online, but the explanations go over my head. It would be wonderful if you could give ... 1answer 369 views ### Partition number problem Denote by $I_m=\{0,1,2,…m\}$, by $N_s=\{1,2,…,s\}$ , by $\overline s$ least common multiple of elements of set $N_s$ and by $p(k,N_s)$ the number of partitions of natural number $k$ in parts used ... 1answer 152 views ### Number of ways to sum square numbers to yield a given number I would like to know how many choices of $x_i$ there are such that $$\sum_{i=1}^{n}x_i^2=m$$ where $n$, $m$ are given. The $x_i$ can be any nonnegative integer and need not be unique and the order is ... 2answers 438 views ### Partition Problem, verifying solution in polynomial time I add a look at the partition problem, this problem is know as the Easiest hard problem since it is NP-complete and seems pretty easy. From wikipedia on NP-complete: In computational complexity ... 1answer 712 views ### Hardy Ramanujan Asymptotic Formula for the Partition Number I am needing to use the asymptotic formula for the partition number, $p(n)$ (see here for details about partitions). The asymptotic formula always seems to be written as, \$ p(n) \sim ... 1answer 93 views ### Something basic in “l-adic properties of the partition function” paper I am trying to understand the basic result in this paper: http://www.aimath.org/news/partition/folsom-kent-ono.pdf My problem is with the example at the end of page 2. I understand it's supposed to ... 1answer 198 views ### Closed-form Expression of the Partition Function $p(n)$ I feel like I have seen news that a paper was recently published, at most a few months ago, that solved the well-known problem of finding a closed-form expression for the partition function $p(n)$ ... 2answers 81 views ### List all 3 part compositions of 5 I am looking at a past exam written by a student. There was a question I believed he got correct but received only 1/4. The marker wrote down "4 more compositions, order matters". This is the ... 2answers 150 views ### Trouble understanding proof that the unit interval cannot be partitioned in a certain way From the book "Putnam and Beyond." The problem: Show that the interval [0, 1] cannot be partitioned into two disjoint sets A and B such that B = A + a for some real number a. Proof: Assume ... 0answers 45 views ### Prove that single machine can reduce from 2-partition problem I'm given the following problem: Prove that single machine \sum U_i (number of tardy job) with release date constraints problem can reduce from 2-partition problem ... 2answers 97 views ### Generating sequences of numeric partitions Definition: A tuple $\lambda = (\lambda_1, \ldots, \lambda_k)$ of natural numbers is called a numeric partition of $n$ if $1 \leq \lambda_1 \leq \cdots \leq \lambda_k$ and \$\lambda_1 + \cdots + ... 1answer 293 views ### identity proof for partitions of natural numbers Definition: A tuple $\lambda = (\lambda_1, \cdots, \lambda_k)$ of Natural Numbers is called a numeric partition of n if $1 \leq \lambda_1 \leq \cdots \leq \lambda_k$ and \$\lambda_1 + \cdots + ... 1answer 295 views ### How to find the coefficient of a term in this expression How to determine the coefficient of z3q100 in I stumbled upon this problem while trying to solve this type of partition problem: Find the number of integer solutions to x + y + z = 100 such that 3 ... 1answer 326 views ### Number of permutations with a given partition of cycle sizes Part of my overly complicated attempt at the Google CodeJam GoroSort problem involved computing the number of permutations with a given partition of cycle sizes. Or equivalently, the probability of a ... 1answer 153 views ### Why is there a derivative in this formula? This is a very simple question. Why is Rademacher's formula presented with d/dx in it? Why not just "do" the derivative? Then replace x with n? Is it so there is only one transcendental ... 1answer 427 views ### Feeding real or even complex numbers to the integer partition function $p(n)$? Like most people, when I first encountered $n!$ in grade school, I graphed it, then connected the dots with a smooth curve and reasoned that there must be some meaning to $\left(\frac43\right)!$ — ... 1answer 131 views ### What is the standard way of writing this partition function? [closed] What is the standard way of writing the number of partitions of n where each number is less than or equal to k. Is it $p(n,k)$ or $p(k,n)$? So far in my reading, have seen both. 3answers 261 views ### Partition an integer $n$ by limitation on size of the partition According to my previous question, is there any idea about how I can count those decompositions with exactly $i$ members? for example there are $\lfloor \frac{n}{2} \rfloor$ for decompositions of $n$ ... 3answers 166 views ### Decomposition by subtraction In how many ways one can decompose an integer $n$ to smaller integers at least 3? for example 13 has the following decompositions: \begin{gather*} 13\\ 3,10\\ 4,9\\ 5,8\\ 6,7\\ 3,3,7\\ 3,4,6\\ ... 1answer 143 views ### Seeking some details about what is denoted by the partition function $P(n,k)$ Quoting from Wolfram MathWorld, "$P(n,k)$ denotes the number of ways of writing $n$ as a sum of exactly $k$ terms or, equivalently, the number of partitions into parts of which the largest is exactly ... 2answers 180 views ### Asymptotics for partitions of $n$ with largest part at most $k$ (or into at most $k$ parts) Let $\bar p_k(n)$ be the number of partitions of $n$ with largest part at most $k$ (equivalently, into at most $k$ parts). Is there an elementary formula for the asymptotic behavior of $\bar p_k(n)$ ... 1answer 175 views ### Seeking a textbook proof of a formula for the number of set partitions whose parts induce a given integer partition Let $t \geq 1$ and $\pi$ be an integer partition of $t$. Then the number of set partitions $Q$ of $\{1,2,\ldots,t\}$ for which the multiset $\{\|q\|:q \in Q\}=\pi$ is given by \[\frac{t!}{\prod_{i \geq ... 1answer 88 views ### Validity of a q-series theorem Define the $q$-analog $(a;q)_n = \prod_{k=0}^n \left(1 - aq^k\right)$. I want to prove the identity $\frac{(q^2;q^2)_\infty}{(q;q)_\infty}=\frac{1}{(q;q^2)_\infty}$. I viewed the LHS this way: ... 2answers 263 views ### Natural set to express any natural number as sum of two in the set Any natural number can be expressed as the sum of three triangular numbers, or as four square numbers. The natural analog for expressing numbers as the sum of two others would apparently be the sum ... 1answer 179 views ### What is the number of ways to choose x groups from y items? (partitions with x cells of a multiset) Where a group can consist of 1 or more items, groups don't have to be equally sized and items can be duplicates. Example - Choose 3 groups: Items: 1 2 2 3 Groups: (1) (2 2) (3) (1 2) (2) (3) (3 ... 2answers 233 views ### Graph coloring problem (possibly related to partitions) Given an undirected graph I'd like to color each node either black or red such that at most half of every node's neighbors have the same color as the node itself. As a first step, I'd like to show ... 1answer 436 views ### Upper bound on integer partitions of n into k parts Recent news piqued my interest in integer partitions again. I'm working my way back through an old text and I'm completely hung up on this problem: Recall that $p_k(n)$ is the number of partitions ... 1answer 46 views ### Notation for “duplicating” partitions I'm using Macdonald's "Symmetric Functions and Hall Polynomials" as a reference and did not find what I was looking for -- apologies if I only missed it. As an example, let us consider the partition ... 1answer 772 views ### On problems of coins totaling to a given amount I don't know the proper terms to type into Google, so please pardon me for asking here first. While jingling around a few coins, I realized that one nice puzzle might be to figure out which $n$ or so ... 0answers 173 views ### Visualizing the Partition numbers (suggestions for visualization techniques) So Ken Ono says that the partition numbers behave like fractals, in which case I'd like to try to find an appropriately illuminating way of visualizing them. But I'm sort of stuck at the moment, so ... 4answers 2k views ### Algorithm for generating integer partitions I'm looking for a fast algorithm for generating all the partitions of an integer up to a certain maximum length; ideally, I don't want to have to generate all of them and then discard the ones that ... 6answers 2k views ### Making Change for a Dollar (and other number partitioning problems) I was trying to solve a problem similar to the "how many ways are there to make change for a dollar" problem. I ran across a site that said I could use a generating function similar to the one quoted ... 1answer 108 views ### Number of distributions leaving none of $n$ cells empty The solution for the number of distributions leaving none of the $n$ cells empty (with unlike cells and $r$ unlike objects) is given by ... 3answers 284 views ### “Converting” equivalence relations to partitions There is a direct relationship between equivalence relations and partitions. Is there a way to simply use an equivalence relation's definition to get the matching partition? And what about the other ... 1answer 323 views ### Matrix representation of a partition Is there a natural way to represent all the partitions of an integer set {1,2,3,...,n} as a matrix in the similar way permutations can be mapped to group of matrices?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 71, "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": 0.9233819246292114, "perplexity_flag": "head"}
http://mathoverflow.net/questions/104134?sort=votes	## Is a colimit of fibration still a fibration? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $\mathcal{C}$ a category. Let $Fib(\mathcal{C})$ the category of fibrations (on $\mathcal{C}$) with morphisms the cartesian functors $T: (\mathcal{A}, P)\to (\mathcal{B}, Q)$ i.e. $T: \mathcal{A}\to \mathcal{B}$ with $Q\circ T=P$ (where $P: \mathcal{A}\to \mathcal{C}$, $P: \mathcal{B}\to \mathcal{C}$ are fibrations on $\mathcal{C}$). From literature (es. "Categorical logic and Type THeory" B.Jacobs) the pullbak of two fibration (i.e. the product in $Fib(\mathcal{C})$) is still a fibration, and this has a easy generalization to a multi-pullback i.e. a the inclusion $Fib(\mathcal{C})\subset CAT\downarrow \mathcal{C}$ create (small) product, the some is true for the kernel of a couple, then the inclusion above create all (small) limits. THe some is easly true for coproducts. I ask: do the inclusion $Fib(\mathcal{C})\subset CAT\downarrow \mathcal{C}$ create also cokernels? Counterexamples?. Observation $Fib(\mathcal{C})$ is equivalent to the category of pseudo-functors $P: \mathcal{C}^{op}\to CAT$ with pseudo-transformations as morphisms, and the punctual limits or colimits of pseudofunctors is still a pseudofuntors (I seems yes, anyway this is true for funtors and natural tranformations considering the category $sFib(\mathcal{C})$ of fibrations with split clevages and clevage preserving functors ), then $Fib(\mathcal{C})$ (or at least $sFib(\mathcal{C})$) is complete and cocomplete. - ## 1 Answer $\mathrm{Fib}(\mathcal{C})$ is both monadic and comonadic over $\mathrm{Cat}/\mathcal{C}$, in the pseudo sense. That is, there are a monad $T$ and a comonad $G$ on $\mathrm{Cat}/\mathcal{C}$ such that $\mathrm{Fib}(\mathcal{C})$ is equivalent to the 2-category of strict $T$-algebras and pseudo $T$-morphisms, and also to the 2-category of strict $G$-coalgebras and pseudo $G$-morphisms. By definition $T$ is the span-composite with $\mathcal{C} \leftarrow \mathcal{C}^{\mathbf{2}} \to \mathcal{C}$, while $G$ is its right adjoint. Therefore, by a theorem of Blackwell-Kelly-Power and its dual, the forgetful functor creates strict PIE-limits and strict PIE-colimits, hence also (weak) 2-limits and 2-colimits. - Thank you very much. I guess you first answere is from the R. Street article on Sydney category seminar (LNm 420). WHere I can find references about Blackwell-Kelly-Power theorem? – Buschi Sergio Aug 6 at 21:13 The first observation is made in multiple places, another one is Street's "Fibrations in bicategories". The Blackwell-Kelly-Power paper is "Two-dimensional monad theory". – Mike Shulman Aug 7 at 6:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 30, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.8914625644683838, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/76594/throwing-dice-a-randomly-determined-number-of-times	# Throwing dice a randomly determined number of times Let $X$ be the number that shows up when rolling a die. Now throw another dice $X$ times $(Y_1, ..., Y_X)$ and calculate the sum $Z = \sum_{k=1}^X Y_k$. What kind of Stochastic Process is this? How do you calculate mean value and variance of $Z$? - ## 3 Answers Look up Wald's Lemma. This lemma is applicable here as $X$ and $Y_k$ are independent. Variance also can be found using this lemma and using the fact that $Z^2=\sum_{k=1}^{X}Y_k^2+\sum_{i<j}Y_i Y_j$ where we have $X(X-1)/2$ terms in the second sum. - Does this help find the variance? – opt Oct 28 '11 at 6:42 Yes, so long as $X$ and $Y_k$ are independent which is actually the case here. – Ashok Oct 28 '11 at 7:49 @Ashok Thank you for the answer. However, I wonder why the Wikipedia article does not mention that independence of X and Y_k is necessary. – artistoex Oct 28 '11 at 15:26 Condition (3) is essentially saying about the amount of dependence is allowed. See the discussion of assumptions below the statement in the Wikipedia page. – Ashok Oct 29 '11 at 3:03 It is a compound probability distribution. You can marginalize by taking the average of the six pmfs, giving a distribution that looks like $p(1) = 1/6^2,\dots, p(36) = 1/6^7$ and then you can find the variance of this finite marginal distribution directly. For the mean you can just take the mean of the six distribution means. - Your random variable is also a branching process $(Z_n)$ observed at time $n=2$, where the offspring distribution is given by a single roll of the die. You want to substitute $n=2$ below, where $\mu$ and $\sigma^2$ are the mean and variance for a single die roll $$\mathbb{E}(Z_n)=\mu^n\quad\mbox{ and }\quad\mathbb{V}(Z_n)={\sigma^2\mu^n(\mu^n-1)\over\mu^2-\mu}.$$ - Glad to know this interpretation! – Ashok Oct 28 '11 at 13:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "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": 0.9139391183853149, "perplexity_flag": "head"}
http://www.mpi-inf.mpg.de/~fietzke/benford.html	# An illustration of Benford's Law Benford's law, also known as the Newcomb-Benford law or the first digit law, states that in sets of numbers from many real-life sources of data, the first digits follow a logarithmic distribution. More specifically, the probability that the first digit is $$n$$ is given by $$\log_{10}\left( 1 + \frac{1}{n} \right).$$ Such distributions can arise for different reasons, including exponential growth processes, or as properties of finite sets of rational numbers representing results of physical measurements. Here we focus on exponential growth. In the application below, you can define a sequence of numbers by giving initial value(s) and a rule. The diagram on the left shows the distribution of first digits in the sequence. The Benford distribution is shown in the background. The diagram on the right shows a plot of the sequence. The dark areas correspond to numbers starting with 1. Click Run to see the distribution of first digits evolve. Observe how, as the exponential curve gets steeper and steeper, it spends more time in lower first-digit areas than in higher first-digit areas, at any order of magnitude. Some sequences to try: • Simple exponential growth: $$f(0)=1, f(n)=f(n-1) * 1.1$$ • Fibonacci numbers: $$f(0)=0, f(1)=1, f(n)=f(n-1)+f(n-2)$$ • Linear growth: $$f(0)=0, f(n)=f(n-1) + 1$$ Linearly growing sequences don't follow Benford's law, but lower first digits are still more probable than higher first digits "most of the time". This explains why lower first digits can also be more frequent in data like street numbers. ## Try it out! Initial value(s): $$f(0), f(1), \dots$$ Rule: $$f(n) =$$ $$f($$$$)=$$ First-digit distribution 50% 100% 1 2 3 4 5 6 7 8 9 Plot y scale: 1.0 Author: Arnaud Fietzke	{"extraction_info": {"found_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": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.8906566500663757, "perplexity_flag": "middle"}
http://www.mashpedia.com/Partition_of_a_set	# Partition of a set Language A partition of a set into 6 parts: an Euler diagram representation The 52 partitions of a set with 5 elements The traditional Japanese symbols for the chapters of the Tale of Genji are based on the 52 ways of partitioning five elements. In mathematics, a partition of a set X is a division of X into non-overlapping and non-empty "parts" or "blocks" or "cells" that cover all of X. More formally, these "cells" are both collectively exhaustive and mutually exclusive with respect to the set being partitioned. ## Definition A partition of a set X is a set of nonempty subsets of X such that every element x in X is in exactly one of these subsets. Equivalently, a set P is a partition of X if and only if all of the following conditions hold: 1. It does not contain the empty set. 2. The union of the elements of P is equal to X. (The elements of P are said to cover X.) 3. The intersection of any two distinct elements of P is empty. (We say the elements of P are pairwise disjoint.) In mathematical notation, these conditions can be represented as 1. $\varnothing \notin P$ 2. $\bigcup P = X$ 3. $(A \in P \and B\in P \and A \neq B) \Rightarrow A \cap B = \varnothing$ where $\varnothing$ is the empty set. The elements of P are called the blocks, parts or cells of the partition.[1] The rank of P is \|X\| − \|P\|, if X is finite. ## Examples • Every singleton set {x} has exactly one partition, namely { {x} }. • For any nonempty set X, P = {X} is a partition of X, called the trivial partition. • For any non-empty proper subset A of a set U, the set A together with its complement form a partition of U, namely, {A, U−A}. • The set { 1, 2, 3 } has these five partitions: • { {1}, {2}, {3} }, sometimes written 1\|2\|3. • { {1, 2}, {3} }, or 12\|3. • { {1, 3}, {2} }, or 13\|2. • { {1}, {2, 3} }, or 1\|23. • { {1, 2, 3} }, or 123 (in contexts where there will be no confusion with the number). • The following are not partitions of { 1, 2, 3 }: • { {}, {1, 3}, {2} } is not a partition (of any set) because one of its elements is the empty set. • { {1, 2}, {2, 3} } is not a partition (of any set) because the element 2 is contained in more than one block. • { {1}, {2} } is not a partition of {1, 2, 3} because none of its blocks contains 3; however, it is a partition of {1, 2}. ## Partitions and equivalence relations For any equivalence relation on a set X, the set of its equivalence classes is a partition of X. Conversely, from any partition P of X, we can define an equivalence relation on X by setting x ~ y precisely when x and y are in the same part in P. Thus the notions of equivalence relation and partition are essentially equivalent.[2] The axiom of choice guarantees for any partition of a set X the existence of a subset of X containing exactly one element from each part of the partition. This implies that given an equivalence relation on a set one can select a canonical representative element from every equivalence class. ## Refinement of partitions Partitions of a 4-set ordered by refinement A partition α of a set X is a refinement of a partition ρ of X—and we say that α is finer than ρ and that ρ is coarser than α—if every element of α is a subset of some element of ρ. Informally, this means that α is a further fragmentation of ρ. In that case, it is written that α ≤ ρ. This finer-than relation on the set of partitions of X is a partial order (so the notation "≤" is appropriate). Each set of elements has a least upper bound and a greatest lower bound, so that it forms a lattice, and more specifically (for partitions of a finite set) it is a geometric lattice.[3] The partition lattice of a 4-element set has 15 elements and is depicted in the Hasse diagram on the left. Based on the cryptomorphism between geometric lattices and matroids, this lattice of partitions of a finite set corresponds to a matroid in which the base set of the matroid consists of the atoms of the lattice, the partitions with $n-2$ singleton sets and one two-element set. These atomic partitions correspond one-for-one with the edges of a complete graph. The matroid closure of a set of atomic partitions is the finest common coarsening of them all; in graph-theoretic terms, it is the partition of the vertices of the complete graph into the connected components of the subgraph formed by the given set of edges. In this way, the lattice of partitions corresponds to the graphic matroid of the complete graph. Another example illustrates the refining of partitions from the perspective of equivalence relations. If D is the set of cards in a standard 52-card deck, the same-color-as relation on D – which can be denoted ~C – has two equivalence classes: the sets {red cards} and {black cards}. The 2-part partition corresponding to ~C has a refinement that yields the same-suit-as relation ~S, which has the four equivalence classes {spades}, {diamonds}, {hearts}, and {clubs}. ## Noncrossing partitions A partition of the set N = {1, 2, ..., n} with corresponding equivalence relation ~ is noncrossing provided that for any two 'cells' C1 and C2, either all the elements in C1 are < than all the elements in C2 or they are all > than all the elements in C2. In other words: given distinct numbers a, b, c in N, with a < b < c, if a ~ c (they both are in a cell called C), it follows that also a ~ b and b ~ c, that is b is also in C. The lattice of noncrossing partitions of a finite set has recently taken on importance because of its role in free probability theory. These form a subset of the lattice of all partitions, but not a sublattice, since the join operations of the two lattices do not agree. ## Counting partitions The total number of partitions of an n-element set is the Bell number Bn. The first several Bell numbers are B0 = 1, B1 = 1, B2 = 2, B3 = 5, B4 = 15, B5 = 52, and B6 = 203. Bell numbers satisfy the recursion $B_{n+1}=\sum_{k=0}^n {n\choose k}B_k$ and have the exponential generating function $\sum_{n=0}^\infty\frac{B_n}{n!}z^n=e^{e^z-1}.$ The number of partitions of an n-element set into exactly k nonempty parts is the Stirling number of the second kind S(n, k). The number of noncrossing partitions of an n-element set is the Catalan number Cn, given by $C_n={1 \over n+1}{2n \choose n}.$ ## Notes 1. Brualdi, pp. 44–45 2. Schechter, p. 54 3. Birkhoff, Garrett (1995), Lattice Theory, Colloquium Publications 25 (3rd ed.), American Mathematical Society, p. 95, ISBN 9780821810255 . ## References • Brualdi, Richard A. (2004). Introductory Combinatorics (4th edition ed.). Pearson Prentice Hall. ISBN 0-13-100119-1. • Schechter, Eric (1997). Handbook of Analysis and Its Foundations. Academic Press. 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Images Source: Flickr. Images licensed under the Creative Commons CC-BY-SA	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8685792684555054, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/194236/problem-with-a-circumference/194243	# Problem with a circumference I have the following equation for a circumference: $$9 X^2 + 25 Y^2 - 36 X - 50 Y = 154.$$ So far I only used this general equation: $X^2 + Y^2 + A X + B Y + C = 0$, but now $X^2$ and $Y^2$ are not alone and are being multiplied by an integer. I would like to know how to find out the center and the radius of this equation - What is your question? – Clive Newstead Sep 11 '12 at 16:59 Get the center and the radius of this equation – Pacha Sep 11 '12 at 16:59 This is an ellipse, not a circle, so does not have a 'radius'; but it does have major and minor axes. I'll elaborate in an answer. – Clive Newstead Sep 11 '12 at 17:01 you might be right! – Pacha Sep 11 '12 at 17:16 ## 2 Answers First of all, you ask for the center and radius, but you have an ellipse and not a circle. You can talk about a center but not really a radius. If what you want is to know the general shape of what you have, here is what you do. Start by completing the square for both $X$ and $Y$. This will give you something of the form $$9(X - H)^2 + 25(Y - K)^2 = S$$ Now, divide both sides by $S$ to get something of the form $$\frac{(X - H)^2}{A^2} + \frac{(Y - K)^2}{B^2} = 1$$ This is a general form of the equation of an ellipse with center $(H, K)$ (not the most general form, by the way). Then $A$ represents the distance from the center to the edge of the ellipse, if traveling only left or right. And $B$ represents the distance from the center to the edge if traveling up or down. So, the $A$ and $B$ represent something similar to a radius, but not a real radius. - Your equation is that of an ellipse, not a circle. However, if you substitute $x=3X$ and $y=5Y$, then your equation describes a circle in the $(x,y)$-plane. You can then find the properties (centre, major and minor axes) of the ellipse in the $(X,Y)$-plane by considering the properties of the circle, and applying graph transformations $(x,y) \mapsto (X,Y)$. For instance, if the centre of the circle in the $(x,y)$-plane is $(a,b)$, then the centre of the ellipse in the $(X,Y)$-plane is $(\frac{a}{3}, \frac{b}{5})$, since $X=\dfrac{x}{3}$ and $Y=\dfrac{y}{5}$. You can do something similar to find the major and minor axes if the ellipse. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 22, "mathjax_display_tex": 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": 0.9218121767044067, "perplexity_flag": "head"}
http://scicomp.stackexchange.com/questions/1909/fem-singularity-of-the-stiffness-matrix	# FEM: singularity of the stiffness matrix I'm solving the differential equation $$\left( \sigma^{2}(x) u ''(x) \right)'' = f(x), \;\;\; 0 \leqslant x \leqslant 1$$ with initial conditions $u(0) = u(1) = 0$, $u''(0) = u''(1) = 0$. Here $\sigma(x) \geqslant \sigma_{0} > 0$ is parameter. In operator form we can rewrite the differential equation as $Au = f$, where operator $A$ is positive definite. Following FEM scheme, I reduce my problem to an optimisation problem $$J(u) = (Au,u) - 2(f,u) \to \min_{u}$$ I introduce finite elements $h_{k}(x)$ as $$v_{k}(x) = \left\{ \begin{array}{rl} 1 - \left( \frac{x-x_{k}}{h} \right)^2, & x \in [x_{k-1},x_{k+1}] \\ 0, & \text{otherwise} \end{array} \right.$$ for any $k = 1,\ldots,n-1$, where $x_{k} = hk$, $h = \frac{1}{n}$. Finite elements $v_{0}(x)$ and $v_{n}(x)$ are introduced similarly. I try to find numericaly the vector $\alpha$ such that $u(x) = \sum_{k=0}^{n} \alpha_{k} v_{k}(x)$ solves the optimisation problem. We have $$J(u) = \sum\limits_{i=0}^{n} \sum\limits_{j=0}^{n} \alpha_{i} \alpha_{j} (Av_{i},v_{j}) - \sum\limits_{i=0}^{n} 2\alpha_{i} (v_{i},f) = \alpha^{T} V \alpha - 2\alpha^{T} b \to \min\limits_{\alpha},$$ where $b_{i} = (f,v_{i})$ and $V_{i,j} = (Av_{i},v_{j})$. After differentiation with respect to $\alpha$ I receive $$V\alpha = b,$$ but here the stiffness matrix $V$ is singular. So what I have to do? Maybe I have to choose other finite elements? - Hi, Nimza, do you have a test problem that you know the exact solution? If yes, try solving $V^T V \alpha = V^T b$ first to test if your basis is correct inside the domain, if everything looks correct, then maybe it is the incorrectly posed BC makes the matrix singular. But the BC seems OK to me though. – Shuhao Cao Apr 12 '12 at 14:49 ## 2 Answers In decreasing order of likelihood 1. Incorrect basis. From you description, it appears that you have exactly two quadratic functions with support on each element. That space is not a partition of unity and is not $C^1$ (continuous first derivatives). To discretize your fourth order problem directly (instead of reducing it to a system of second order equations, for example), you will need a $C^1$ basis. Note that the $C^1$ basis should be able to exactly reproduce all linear functions. 2. Insufficient boundary conditions. This will be blatantly obvious if you compute and plot the null space. 3. Incorrect assembly. Check the map from elements to assembled ordering to confirm that it is what you expected, for example that it isn't reversing the orientation of elements. 4. Incorrect local assembly. In 1D, you can analytically compute what the element stiffness matrix looks like (perhaps for a simplified case) and check that the code reproduces it. - Thank you. 1. I think that I will need a $C^2$ basis because $(Au,v) = \int_{0}^{1} \sigma^{2}(x) u''(x) v''(x) dx$. Then, if I consider only functions that satisfy boundary conditions then $\ker A = \{ 0 \}$. – Nimza Apr 11 '12 at 21:11 1 – Jed Brown Apr 11 '12 at 23:25 Ok. I corrected my basis: now $v_{k}(x) = \cos^2 \left(\frac{\pi}{2h}(x-x_{i}) \right)$ on $[x_{i-1},x_{i+1}]$ and $i = 1,\ldots,n-1$. Now it is $C^1$. But method still doesn't work. – Nimza Apr 12 '12 at 8:58 The $C^1$ basis should be able to reproduce linear functions, but this cannot. Once you fix that, check that integrals are being performed correctly, then check boundary conditions. – Jed Brown Apr 12 '12 at 13:30 Clearly the problem has an ODD order derivative. More specifically for larger Péclet numbers, the stiffness matrix might not maintain 'fine' shape, which creates zeros during assembly and hence gets singular or sometimes very small determinant that are noticeable by the oscillations in solution plot. The solution to these kind of problem is the use of penalty, among other methods. More specifically this is called Petrov-Galerkin method. Sorry for my bad English comprehension. -	{"extraction_info": {"found_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": 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": 0.9299185276031494, "perplexity_flag": "middle"}
http://mathhelpforum.com/number-theory/9183-ph-others-so-inclined.html	# Thread: 1. ## for PH and others so inclined "Show that $\frac{x^{5}}{5}+\frac{x^{3}}{3}+\frac{7x}{15}$ is always an integer for integral values of x". 2. It is relatively easy (but messy) to show that for each positive integer N $3N^5 + 5N^3 + 7N$ is a multiple of 15. an other 3. That was a fast response, Plato . I noticed, since the semesters have come to a close, there hasn't been much posting. Anyway, maybe I am off base, but $\frac{x^{5}}{5}+\frac{x^{3}}{3}+\frac{7x}{15}=\fra c{x^{5}-5}{5}+\frac{x^{3}-3}{3}+x$ Now, use Fermat's theorem: $x^{5}-x\equiv{0} \;\ mod(5)$ and $x^{3}-x\equiv{0} \;\ mod (3)$ Therefore, it's an integer. Seems rather logical. But, I am not that much of a number theorist. Have been learning it, though. 4. Originally Posted by Plato It is relatively easy (but messy) to show that for each positive integer N $3N^5 + 5N^3 + 7N$ is a multiple of 15. an other Hello Plato, use mathematical induction to prove this property: Prove that $3n^5+5n^3+7n=15 \cdot k\ , \ k \in \mathbb{Z}$ $step\ 1:\ n=1\ :\ 3+5+7=15=15 \cdot 1 \text{ is true}$ $step \ 2:\ assumption: 3n^5+5n^3+7n=15 \cdot k \text{ is true}$ $step\ 3:\ conclusion: 3(n+1)^5+5(n+1)^3+7(n+1)=$ $3n^5+15n^4+30n^3+30n^2+15n+3+5n^3+15n^2+15n+5+7n+7 =$ $\underbrace{3n^5+5n^3+7n}_{k\cdot 15}+15(n^4+2n^3+3n^2+2n+1)+3+5+7=$ The 1rst summand is a multiple of 15 according assumption, the 2nd summand is a multiple of 15 because it contains the factor 15 and the last 3 summands add up to 15. Thus $3(n+1)^5+5(n+1)^3+7(n+1)=$ is multiple of 15 and therefore the asumption is true for all $n \in \mathbb{N}$ EB Merry Christmas and a happy New Year.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9204093813896179, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/25512/determining-the-field-from-the-topology-of-the-affine-space-over-it	## Determining the field from the topology of the affine space over it. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The inspiration is from another question in which it is remarked that the passage to Zariski topology is a very interesting functor from Commutative Rings to Topological spaces. I am trying to understand how "faithful" this functor is. Any two fields have the same topology. But, with polynomial rings, we have more hope. So. Let $K$, $L$ be two fields. Then if $Spec\ K[X]$ and $Spec\ L[X]$ are homeomorphic, does it follow that $K$ and $L$ are isomorphic? Similarly: Let $K$, $L$ be two fields. Then if $Spec\ K[X_1, \ldots , X_n]$ and $Spec\ L[X_1, \ldots , X_n]$ are homeomorphic, does it follow that $K$ and $L$ are isomorphic? - 1 The affine line over a field carries the cofinite topology, so its homeomorphism type depends only on its cardinality. – Qiaochu Yuan May 21 2010 at 17:52 Ahh, posted as an answer as you posted your comment! – Charles Siegel May 21 2010 at 17:53 Surely the answer is no (by which I mean I do not have a proof). At the very least, if $K$ is the algebraic closure of $\mathbb{Q}$ and $L$ is the algebraic closure of $\mathbb{Q}(t)$, then $\operatorname{Spec} K[t_1,\ldots,t_n]$ and $\operatorname{Spec} L[t_1,\ldots,t_n]$ should be homeomorphic for all $n$. It seems reasonable to conjecture that the only invariant of $K$ that these spectra see is the cardinality of $K$. – Pete L. Clark May 21 2010 at 17:59 ## 2 Answers The following paper of Hrushovski-Zilber shows that if we restrict our attention to algebraically closed fields $F$, then $F$ is uniquely determined up to isomorphism by its "Zariski geometry". Presumably an examination of the proof will show that an algebraically closed field $F$ is determined by $Spec(F[x_{1}, \cdots F[x_{n}])$ for sufficiently large $n$. Hrushovski, Ehud; Zilber, Boris (1996). "Zariski Geometries". Journal of the American Mathematical Society 9: 1–56. Another "reference": http://en.wikipedia.org/wiki/Zariski_geometry - Interesting. The relevant result seems to be Proposition 1.1. If you take $C$ to be the affine line over an algebraically closed field $F$ and $C'$ to be the affine line over an algebraically closed field $F'$, then the result says that an isomorphism of Zariski geometries from $C$ to $C'$ induces an isomorphism from $F$ to $F'$. It also says that a morphism of Zariski geometries is an isomorphism if it induces a homeomorphism on $C^n$ for all $n$.... – Pete L. Clark May 21 2010 at 20:17 ...But what I wasn't immediately able to see was whether a morphism of Z.G.'s from $C$ to $C'$ involves more data than just a collection of such homeomorphisms -- e.g. whether various geometric compatibility conditions need to be satisfied. I would be very interested to see these details worked out. – Pete L. Clark May 21 2010 at 20:19 The experts in this area (such as Dave Marker) would know all about this. Hopefully one of them will stumble across this question and save us the trouble of working out the details ourselves! – Simon Thomas May 21 2010 at 20:32 In this type of argument you usually reconstruct the field from a 2-dimensional family of curves on $C\times C$. So probably you can get by knowing that the homeomorphism between $C$ and $C_1$ lifts to homeomorphisms of the Zariski topologies of $C^n$ and $C_1^n$ for $n=2,3,4$. – Dave Marker May 24 2010 at 6:56 1 A simple version of this idea would be if $K$ and $L$ are algebraically closed fields and we had a homeomorphism between $P_K^2\timesP_K^2$' and $P_L^2\times P_L^2$. Let $V\subset P^2_K \times P^2_K$ be the incidence variety for lines in $P_2$. Desargues ideas allow us to reconstruct $K$ from $V$. A homeomorphism would give us an isomorphic variety in $P_L^2\times P_L^2$. This would allow us to interpret $K$ in $L$. But by a result of Poizat then $K$ and $L$ would be isomorphic. – Dave Marker May 24 2010 at 12:53 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This is false. Let $K=\bar{\mathbb{Q}}$ and $L=\bar{\mathbb{F}}_2$. These are clearly both algebraically closed, of different characteristics, so $K\not\cong L$. However, if we ONLY look at the topology, $\mathrm{Spec}(K[x])$ and $\mathrm{Spec}(L[x])$ will be be countable sets with the finite complement topology on the closed points, with a single generic point, so they're homeomorphic. For algebraically closed fields, the TOPOLOGY on the affine line over the field is determined by the cardinality. For higher dimensions, it's less clear to me, because you might be able to recover characteristic (I'm a char 0 kind of person, so I don't know) from how the various curves/hypersurfaces sit inside it. - @CS: I don't know for sure, but I am skeptical that you can recover the characteristic of an infinite field from the topologies on affine spaces over it. – Pete L. Clark May 21 2010 at 18:01 I'm also skeptical, but my positive characteristic intuition is dismal enough that I'm not willing to come out and say it. – Charles Siegel May 21 2010 at 18:11 @Charles: um, well, in that case, maybe you should...never mind, I'm sure we'll turn out to be right. :) – Pete L. Clark May 21 2010 at 18:56	{"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": 62, "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": 0.9528334736824036, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/1881/realize-a-mac-using-a-pseudo-random-function/1883	Realize a MAC using a Pseudo-random function? Given a pseudo-random function and assuming that we do not have any other tools, How can we construct a MAC? I believe this can be done. Would like to know if there is more than one way of doing this. - 3 Answers If you have a PRF (with larger input than output), you can use it as compression function in a Merkle-Damgård structure, yielding a hash function which you can subsequently turn into a MAC with HMAC. Indeed, the security proof of HMAC relies on indistinguishability of the compression function from a PRF. There are still an awful lot of details, though. And the PRF assumption is quite strong. - Just to quibble over minor points, I think NMAC makes more sense than HMAC here (since you're starting with a compression function, rather than a hash function), and I believe the requirement is that the output has to be at least as long as the PRF key (since you'd be chaining through the key input). – Seth Feb 19 '12 at 10:03 If you have a keyed PRF, where the key is sufficient to turn the function into an independent PRF, you can turn it into a keyed PRP using a Luby Rackoff construct. If you got a keyed PRP, you can use it in CBC-MAC mode. - I am not sure if $\mathsf{PRF}$ alone can be used to construct; however there is a natural way to construct $\mathsf{MAC}$ using $\mathsf{PRF}$ and $\mathsf{UHF}$. You can see the following papers for more detail: 1. LFSR-based hashing and authentication. In CRYPTO 1994. 2. Bucket hashing and its application to fast message authentication. In CRYPTO 1995. 3. On fast and provably secure message authentication based on universal hashing. In CRYPTO 1996. I just found a very easy way to construct $\mathsf{MAC}$ from $\mathsf{PRF}$ in the lecture notes of Barak. More concisely, it says that if $\{ f_k \}_{k \in \mathcal{K}}$ is a family of $\mathsf{PRF}$, then the following is a $\mathsf{MAC}$: $$\mathsf{SIGN}_k(x) = f_k(x) \qquad \mathsf{VERIFY}_k(x,s)=1 \text{ iff } f_k(x)=s.$$ I think this is pretty neat. - I think the exact details depend from the used definition of pseudo-random function. If you have one that allows input of any size (like a hash function and the $f_k$ used here), it is a lot easier than if you have one with fixed input size. – Paŭlo Ebermann♦ Feb 20 '12 at 20:17	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "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": 0.9214656949043274, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/26607/what-does-the-semiring-of-ideals-of-a-ring-r-tell-us-about-r/26668	## What does the semiring of ideals of a ring R tell us about R? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Here is something I've wondered about since I was an undergraduate. Let $R$ be a ring (commutative, let's say, although the generalization to noncommutative rings is obvious). Ideals of $R$ can be multiplied and can be added (the ideal $I+J$ is the ideal generated by $I$ and $J$), and multiplication distributes over addition. Therefore we can consider the semiring $S$ of ideals of $R$. The question is, does the structure of $S$ tell us anything interesting about the structure of $R$? Or vice versa? I asked this question on sci.math.research last year and got a few replies but nothing very substantive. http://mathforum.org/kb/thread.jspa?messageID=6599151 For a more concrete question: Give an interesting sufficient condition for $S$ to be finitely generated. Conversely, if $S$ is finitely generated, does that imply anything interesting about $R$? - 1 Just one example which comes into my mind: If R is a direct product of n fields, then the semiring of ideals is the power set of $1,...,n$ with union as sum and intersection as product (zero element is the empty subset, unit element is the whole set). You cannot recover the fields. – Martin Brandenburg Jun 1 2010 at 0:05 So basically you want to know which properties can be covered? – Martin Brandenburg Jun 1 2010 at 0:08 Sorry you didn't like my answers... – Pete L. Clark Jun 1 2010 at 2:31 @Pete: I appreciated your answers, but am still hoping for something more substantive. – Timothy Chow Jun 1 2010 at 4:35 1 To multiply ideals shouldn't they be two-sided ideals? There are interesting (noncommutative) rings with no nontrivial two-sided ideals (for example, the first Weyl algebra, which is differential operators in 1 variable with polynomial coefficients), so in this generality I believe not very much can be said. – Peter Samuelson Jul 29 2010 at 17:57 ## 4 Answers Here are a few observations. None of them require our ring to be commutative. First, notice that one can recover the natural partial ordering of the ideals via addition, because for any two ideals $I$ and $J$ of $R$, $I\subseteq J$ if and only if $I+J=J$. (More generally, $I+J$ is the smallest ideal containing both $I$ and $J$.) Second, this allows us to recover the prime ideals of $R$. This is because an ideal $P$ of $R$ is prime if and only if, for any ideals $I$ and $J$ of $R$, $IJ\subseteq P$ implies $I\subseteq P$ or $J\subseteq P$. (The same can be said for the semiprime ideals of $R$, which are the radical ideals of $R$ in case $R$ is commutative.) Third, we can recover the Zariski topology on the prime spectrum of $R$ because it is defined using the natural partial ordering on the ideals of $R$. - The Zariski topology doesn't require the ring to be commutative? – Pete L. Clark Jun 1 2010 at 3:48 No, the Zariski topology makes sense for any ring! This is because, for ideals $I$ and $J$, $V(I)\cap V(J) = V(IJ)$ and $\bigcup_j V(I_j) = V(\sum I_j)$. What you lose, however, is the fact that ring homomorhpisms determine continuous functions between prime spectra. – Manny Reyes Jun 1 2010 at 4:13 For a noncommutative ring and in the case of $\textit{primitive}$ ideals, this is known as "Jacobson topology". (Primitive ideals are proper analogues of maximal ideals in the noncommutative setting.) It's a terminological issue, I simply cannot remember whether the same term applies to the whole prime spectrum, but certainly, "Zariski topology" implies that the ring is commutative. – Victor Protsak Jun 1 2010 at 5:15 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This is sort of a sideways answer, but: in many ways the monoid $\operatorname{Prin}(R)$ of principal ideals carries more information. If $R$ is a domain $\operatorname{Prin}(R)$ is a cancellative monoid so injects into its group completion, the group of divisibility $K^{\times}/R^{\times}$ of $R$. Many of the factorization properties of $R$ can be gracefully rephrased in terms of $\operatorname{Prin}(R)$ and/or $K^{\times}/R^{\times}$. See for instance Section 4.1 of http://www.math.uga.edu/~pete/factorization2010.pdf for more on this point of view. - I've been interested in this lately. Hopefully you have seen this? Golan, Jonathan S.(IL-HAIF) Semirings for the ring theorist. Rev. Roumaine Math. Pures Appl. 35 (1990), no. 6, 531–540. Golan cautions that treating semirings as 'poor man's rings' is not always good. They can really be different animals altogether. I think someone noted above that the semiring of ideals is additively idempotent. In a sense, this is as far as you can get from an additive group. The compensation for the loss of the additive group is the complete lattice structure compatible with the multiplication. That is, if A\leq B, then AC\leq BC and CA\leq CB. Ring theorists have been saying things about rings via the lattice of one sided ideals for years! The lattices of onesided ideals are almost as nice, except you lose compatibility of multiplication with the order, and there is no longer a twosided identity for the semiring. These are called quantales in some places. - This is pretty simple too, but here it goes. I take R to be commutative and have 1. If S is generated by just one ideal P, then all the ideals of R are of the form P^k. Thus R is local. If P^2 is not all of P then any p in P\P^2 generates P, and each P^k is generated by p^k. Hence the only prime ideal is P, and it is exactly the ideal of nilpotents (since these are the intersection of all prime ideals). It follows that some P^k = 0. If P^2 is all of P then again there is just the one prime ideal P, but now P = P^k = 0, so R is a field. So either R is a field or there is a nilpotent p in R s.t. all x in R are of the form u*p^k for some unit u and non-negative integer k. (Just consider the biggest k for which x is a multiple of p^k.) Sometimes, but not always (see below), we can identify R with a quotient of the polynomial algebra (R/P)[X] (note that R/P is a field), namely (R/P)[X]/(X^k) where k is the smallest s.t. P^k = 0. Conversely, any quotient F[X]/(X^k), F a field, has its semiring of ideals generated by (X). - The identification of $R$ with a quotient of $(R/P)[X]$ is a little shaky. How does $\mathbb{Z}/9\mathbb{Z}$ fit into your classification scheme? About the best you can say, I think, is that if $S$ is one-generated, then $R \cong V/t^k$ for some discrete valuation ring $(V,t)$. – Graham Leuschke Jun 1 2010 at 19:00 You're right, of course. I've edited accordingly. I wasn't sure myself but I wanted it to be true so badly! – Pietro KC Jun 1 2010 at 20:31 From an ideal-theoretic point of view, DVRs are just as good as F[X]/(X^k) -- in fact better, since you don't have to worry about silly arithmetic issues with the field. Might as well just say that $S$ is one-generated iff $R$ is a non-trivial image of a DVR. Cleaner that way. – Graham Leuschke Jun 1 2010 at 22:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 56, "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": 0.9424570798873901, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/38471/strategies-for-introducing-advanced-statistics-to-various-audiences	# Strategies for introducing advanced statistics to various audiences I work mainly with non-statisticians in fields such as medicine, social sciences and education. Whether I am consulting with graduate students, helping researchers with articles or reviewing articles for journals, I often have the problem that someone (client, author, dissertation committee, journal editor) wants to use some relatively well-known technique when it is either entirely inappropriate or when better but lesser-known methods exist. Often, I will explain the alternative technique but then be told "everybody does it the other way". I'd be interested in how others deal with this sort of difficulty. ADDITIONS @MichaelChernick suggested I could share some stories, so I will Currently I am working with one person who is duplicating a previous paper and adding one independent variable to see if it helps. The previous paper is, frankly, terrible. It treats dependent data as if they were independent; it is tremendously overfit and there are other problems too. Yet he (my client) submitted an earlier version as a dissertation and not only got his degree but was widely praised for the research. Many times I have tried to convince people not to dichotomize variables. This comes up very often in medicine. I patiently point out that dicohotomizing (say) birthweight into low and normal (usually at 2,500 g) means treating a 2,499 g baby as just like a 1,400 g one; but treating the 2,501 gram baby quite differently. The clinician agrees with me that this is silly. Then says to do it that way. I had a graduate student client long ago whose committee insisted on a cluster analysis. The student did not understand the method, the method did not answer useful questions, but that's what the committee wanted, so that's what they got. The entire field of statistical graphics is one where, for many, "this is how grandpa did it" is enough. Then there are people who seem to just push buttons. I remember one presentation (not by someone I helped!) who had taken an entire questionnaire and factor analyzed it. One of the variables she included was ID number! Oy. - 6 Peter, Meta is for questions about this site. I suspect you may have been thinking about "Community Wiki" status, which is for useful, interesting questions unlikely to have an objectively best answer (or which are likely to need collaborative efforts to answer at all). Accordingly, I have interpreted your suggestion as a CW request and implemented that. – whuber♦ Oct 2 '12 at 16:36 Thanks @whuber that is what I meant – Peter Flom Oct 2 '12 at 20:32 1 A quick note on dichotomisation: I actually think this is a misplaced intuition from people who know "a little bit" of stats. In the fields you speak of I would imagine a lot of analysis would be geared around decision making (eg, should I start treatment for disease X or Y?). This is dichotomus - and often a useful dichotomy. If you only have a small number of variables, then a proper decision analysis or hypothesis test may well mimic this - we may have "if x>10 take option 1, otherwise take option 2". – probabilityislogic Oct 3 '12 at 7:44 1 Also an analysis based on dichotomised variables is very easy to remember if you don't have access to a computer. – probabilityislogic Oct 3 '12 at 7:46 1 – Chris Beeley Oct 4 '12 at 19:29 ## 5 Answers This is a tricky question! First, some thoughts on why this happens. I work in an area which does (or at least should) make extensive use of statistics, but where most practitioners are not statistical experts. Consequently one sees a lot of "I put a vector into excel's t-test function and this number fell out. Therefore my paper is supported by statistics." The main reason I see for this happening is that lack of statistics knowledge starts at the top. If your reviewers and thesis committee don't keep up to date on statistical techniques, then you need to justify use of anything that is "unconventional". For example, in a thesis, I opted to use violin plots instead of box plots to show the shape of a distribution. The use of this technique required extensive documentation in the thesis, as well as a prolonged discussion in my defense where all of the committee members wanted to know what this strange plot meant, despite both the descriptions in the text and the references to the source material. Had I just used a box plot (which shows strictly less information in this case, and can easily deceive the viewer about the shape of a distribution if it is multi-modal) no one would have said anything, and my defense would have been easier. The point is, in non-stats fields practitioners face a difficult choice: We can read about and then use the correct methods, which entails a bunch of work that none of our higher ups are interested in; or we can just go with the flow, get the rubber stamp on our papers and theses, and keep using incorrect but conventional methods. I think a good approach is to emphasize the consequences of failing to use correct techniques. This might entail: • Giving a real world example of how someone in their field experienced the consequences of poor inference. This is easier in some fields than others. Examples where careers were damaged are especially good. • Explaining that doing incorrect analysis can leave you in a situation where your results are very unlikely to transfer to the real world, which could cause harms (e.g. In my field, if your A.I. system prototype appears statistically better than the competition, but in fact is the same, then spending the next 6 months building a full implementation is a really bad idea. • Pick techniques which will save the users lots of time. Enough time so that they can spend what they save explaining the techniques to the higher ups. - 1 Good discussion and good answer +1. – Michael Chernick Oct 2 '12 at 18:39 Good points @John – Peter Flom Oct 2 '12 at 20:33 1 +1 for pointing out the consequences. It can do wonders for getting people to switch to better methods. – Leo Oct 3 '12 at 6:35 Speaking from the perspective of a psychologist with only slight statistical sophistication: When you introduce the method, also introduce the tools. If you tell most researchers in my field a long story about a great new method, they're going to spend the whole time worried that the punchline is "and all you have to do is brush up on your differential calculus and then take a two week training course!" (or "and buy a \$2000 stats package!" or "and adapt 5000 lines of Python and R code!"). Whereas if there's an implementation of the method available in the stats package they already use, or in a piece of free software with a comprehensible GUI, and they can get up to speed on it in a day or two, they might be willing to give it a try. I'm aware that this approach can seem venal and unscientific, but it's easy for people to fall into when they're worried about grants and publications, and don't see learning huge amounts of math as likely to help them keep their jobs. - 2 @octem Couldn't the investigator trust the statistician to do that part of the collaboration. Why does the investigator need the tool. I liken this with the physician to reversing the tables. How would he feel if I said give me a quick tutorial on how to do this surgery and I will go ahead and do it for you. I think shock and dsimay and it is illegal for me to practice mewdicine without a license. That is probably a good thing. But doesn't the statistician deserve equal respect. Why expect that I can just give him the tool and let him hack away with his lack of training. – Michael Chernick Oct 2 '12 at 18:37 1 In social psychology, it's not usual to have a statistician (because it's not usual to have enough funding to pay a statistician). Now I'm in public health / health psychology. Big grants usually include salary for a statistician, but a lot of our work is done on stingy little pilot grants where we can't even afford salary for the PI. That's the perspective I'm coming from... if you're in a field where most projects include a trained statistician, then I agree, this kind of resistance would not be reasonable. – octern Oct 2 '12 at 18:45 1 @octem Thanks for the response. Just to follow-up with the analogy. There was a time when abortion was illegal in the US and some women went to other countries or had it done in a back room illegally without sterile conditions and great health risks. It may not seem like a great analogy but not being able to afford a statistician justifies doing an inferior job? I know medicine is more of a life or death thing but bad science has bad consequences too? Misuse of data could lead to practicing bad medicine because an unsafe drug gets used when it shouldn't be. – Michael Chernick Oct 2 '12 at 19:11 @MichaelChernick Look, I'm not happy about the state of statistical analysis in the social sciences either. But the question was how to get researchers to adopt a new statistical approach, and I'm giving an answer that's appropriate to a large subpopulation of researchers -- whether we like it or not. – octern Oct 2 '12 at 19:33 1 @octem Fair enough and I think I understood that. I gave you an upvote before I asked the question. The attitude of the investigators to think that the cheap way is the way to go is what I am questioning and not the fact that you recognize the problem exists. It does and I agree with you there. But somehow in the long run I think we just need to get more respect and recognition that our job is not trivial. – Michael Chernick Oct 2 '12 at 19:39 Thanks for this nice question Peter. I work at a medical research institution and deal with physicians who do research and publish in the medical journals. Often they are more interested in getting their paper published than "doing the statistics completely right". So when I propose an unfamilar technique they will point to a similar paper and say "look they did it this way and got their results published." There is a problem I think when the published paper is really bad and has mistakes. It is difficult to argue even though I have a great reputation. Some docs have big egos and think they can learn almost anything. So they think they understand the statistics when they don't and can be insistent. It can get frustrating. When it is a t test and Wilcoxon is more appropriate I get them to do a Wilk Shapiro test and if normality is rejected we include both methods and explain why Wilcoxon is better. I sometimes can convince them and often they depend on me for statistics, so I have a little more clout then a general consultant might have. I also ran into a situation where I did Kaplan-Meier curves for them and we used the log rank test but Wilcoxon gave a different result. It was hard for me to decide and in such situations I think it is best to present both methods and explain why they differ. The same goes for using Peto vs Greenwood confidence intervals for the survival curve. Explaining the Cox proportion hazard assumption can be difficult and they often misinterpret odds ratios and relative risk. There is no simple answer. I had a boss here who was a top medical researcher in cardiology and he sometimes referees for journals. He was looking at a paper that dealt with diagnosis and used AUC as a measure. He had never seen an AUC curve before and came to me to see if I thought it was valid. He had doubts. It turned out to be appropriate and I explained it to him as best I could. I have tried to lecture on biostatistics to physicians and have taught biostatistics in public health schools. i try to do it better than others have and produced a book for health science majors introductory course in 2002 with an epidemiologist as coauthor. Wiley wants me to do a second edition now. In 2011 I published a more concise book that I tried to cover just the essentials so that busy MDs might take the time to reasd it and reference it. That is how I deal with it. Maybe you can share your stories with us. - These are good points @Michael. I will add some stories – Peter Flom Oct 2 '12 at 20:35 1 @PeterFlom I think we might have some very similar experiences. I alos think you are getting some other very good responses in other people's answers. – Michael Chernick Oct 2 '12 at 20:38 There are some nice comments already made here, but I'll throw in my 2 cents. I'll preface this all by saying that I'm assuming we're talking about a situation where using the traditional "canned" techniques will damage the substantive conclusions reached by the analysis. If that's not the case, then I think that sometimes doing an overly simplistic analysis is excusable both for brevity and for ease of understanding when the target audience are laymen. Is it really such a crime to assume independence when the intraclass correlation is .02 or to assume linearity when the truth is $\log(x); \ x \in (1,2)? \$ I'd say no. In my career I do a lot of interdisciplinary research and has lead me to work with closely with substance abuse researchers, epidemiologists, biologists, criminologists and physicians at various times. This typically involved analysis of data where the usual "canned" approaches would fail for various reason (e.g. some combination of biased sampling and clustered, longitudinally and/or spatially indexed data). I also spent a couple years consulting part time in graduate school, where I worked with people from a large variety of fields. So, I've had to think about this a lot. My experience is that the most important thing is to explain why the usual canned approaches are inappropriate and appeal to the person's desire to do "good science". No respectable researcher wants to publish something that is blatantly misleading in its conclusions because of inappropriate statistical analysis. I've never encountered someone who says something along the lines of "I don't care whether the analysis is correct or not, I just want to get this published" although I'm sure such people exist - my response there would be to end the professional relationship if at all possible. As the statistician, it's my reputation that could be damaged if someone who actually knows what they're talking about happens to read the paper. I admit that it can be challenging to convince someone that a particular analysis is inappropriate, but I think that as statisticians we should (a) have the knowledge necessary to know exactly what can go wrong with the "canned" approach and (b) have the ability to explain it is a reasonably comprehensible way. Unless you're working as a statistics or math professor, a part of your job is going to be to work with non-statisticians (and even sometimes if you are a stat/math prof). Regarding (a), if the statistician doesn't have this knowledge, why would they be discouraging the canned approach? If the statistician is saying "use a random effects models" but can't explain why assuming independence is a problem, then aren't they guilty of giving in to dogma in the same way the client is? Any reviewer, statistician or not, can make pedantic critiques of a statistical modeling approach because, let's face it - all models are wrong. But, it requires expertise to know exactly what could go wrong. Regarding (b), I've found that graphical depictions of what could go wrong typically "hit home" the most. Examples: • In the example given by Peter about categorizing continuous data, the best way to show why this is a bad idea is to graph the data in its continuous form and compare it with its categorical form. For example, if you're making your response variable binary then plot the continuous variable vs. $x$, and, if it doesn't look an awful lot like a step function, then you know the discretization lost valuable information. If this difference isn't drastic or resulting in any changes in the substantive conclusions, you could also see this from the plot. • When the proposed "form" of the model (e.g. linear) is inappropriate. For example, if the regression function "plateaus" like $y = x$ for $x \in (0,1)$ but $y = 1$ for $x > 1$ then a linear fit's slope will be too shallow and, depending on the data, this could push the $p$-value below significance despite there being an obvious relationship between $x$ and $y$. • Another common situation (also mentioned by Peter) is explaining why assuming independence is a bad idea. For example, you can show with a plot that positive autocorrelation will typically produce data that is more "clustered" and the variance will be underestimate for that reason, giving some intuition of why the naive standard errors tend to be too small. Or, you could also plot the data with the fitted curve that assumes independence and one can visually see how the clusters influence the fit (effectively lowering the sample size) in a way that is not present in independent data. There are a million other examples but I'm working with space/time constraints here :) When pictures simply won't do for whatever reason (e.g. showing why one approach is underpowered) then simulation examples are also an option that I've employed from time to time. - Some random thoughts because this is a complex issue... I feel that a big problem is the lack of math education in a variety of professional disciplines and graduated programs. Without a mathematical understanding of statistics, it becomes a bunch of formulas to be applied according the case. Also, for getting a real understanding of the matter, professors should talk about the original problems that the original authors were facing at the time they published their approaches. One can learn more from that than from reading thousands books on the subject. Statistics is a toolbox for solving problems, but it is also an art and faces the same issues than any other art. One can learn how to make sounds with an instrument. But by being able of "playing" an instrument one does not become a musician. However, is not uncommon to find people that see themselves as musicians without having studied a single concept of rhythm, melody and harmony. In the same line, for getting papers published, most people don't need to know nor understand the concepts behind a formula... nowadays scientists just need to know what key they have to press and when it has to be pressed, period. So this has nothing to do with the "ego" of MDs. This is a subcultural problem, a problem more related with education, customs and values of the scientific community. What one can expect in an era in which there are thousands and thousands and thousands of useless papers and books being published for fulfilling some academic requisites/policies? In an era in which the amount of papers one publishes is more important than the quality of them? Mainstream scientists are not worried about the good science anymore. They are slaves of numbers. They are affected (or infected) by the administrative bug of our era... So, from my perspective, a good course in statistics should include the mathematical, historical and philosophical basis of the approach being studied, always highlighting the several paths one can take for solving a single problem. Finally, if I were a professor in statistics/probability my first lecture(s) would be dedicated to problems like shuffling cards or tossing a coin. That will put the audience in the right position for listening... probably. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9674004316329956, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/106070/if-a2-i-identity-matrix-then-a-pm-i?answertab=active	# If $A^2 = I$ (Identity Matrix) then $A = \pm I$ So I'm studying linear algebra and one of the self-study exercises has a set of true or false questions. One of the question is this: If $A^2 = I$ (Identity Matrix) Then $A = \pm I$ ? I'm pretty sure it is true but the answer say it's false. How can this be false (maybe its a typography error on the book)?. Thanks. - 15 Try $$A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$ – Dylan Moreland Feb 5 '12 at 20:13 1 I'd point out that it is true if you're working with $1$-by-$1$ matrices (over $\mathbb C$, or any other integral domain). But for $n \geq 2$ the ring of $n$-by-$n$ matrices over any non-trivial ring is not an integral domain: this means that $(A+I)(A-I) = 0$ doesn't necessarily imply that $A + I = 0$ or $A - I = 0$. – Matt Feb 5 '12 at 20:30 – Jonas Meyer Feb 5 '12 at 20:56 There's an entire family of so-called involutory matrices. Look up Householder reflectors, for instance. – J. M. Feb 6 '12 at 5:11 ## 5 Answers A simple counterexample is $$A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$$ We have $A \neq \pm I$, but $A^{2} = I$. - I know $2·\mathbb C^2$ many counterexamples, namely $$A=c_1\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}+c_2\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}\pm\sqrt{c_1^2+c_2^2\pm1}\begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix},$$ see Pauli Matrices $\sigma_i$. These are all such matrices and can be written as $A=\vec e· \vec \sigma$, where $\vec e^2=\pm1$. - "Most" (read: diagonalizable) matrices can be viewed simply as a list of numbers -- its eigenvalues -- in the right basis. When doing arithmetic with just this matrix (or with other matrices that diagonalize in the same basis), you just do arithmetic on the eigenvalues. So, to find diagonalizable solutions to $A^2 = I$, we just need to write down a matrix whose eigenvalues satisfy $\lambda^2 = 1$ -- and any such matrix will do. When thinking about matrices in this way -- as a list of independent numbers -- it makes it easy to think your way through problems like this. - Every matrix satisfying $A^2=I$ is diagonalizable, because either it is $\pm I$ or its minimal polynomial is $(x-1)(x+1)$. The general solution is obtained by taking all diagonal matrices with entries $\pm 1$ on the diagonal and conjugating by invertible matrices. – Jonas Meyer Feb 6 '12 at 5:03 1 Jonas Meyer, this is only true if $char F \ne 2$. Otherwise, there are such matrices which are not diagonalizable, – anonymous Feb 6 '12 at 8:18 1 @Jonas: That's a good point to mention as an appendix, but dealing properly with non-diagonalizable matrices in this fashion is somewhat more sophisticated. The only reason I mentioned the word was so that I didn't mislead Randolf into thinking this method works (unmodified) for all matrices; e.g. that the argument I gave isn't sufficient to tell us that this (or any!) equation has only diagonalizable solutions. – Hurkyl Feb 6 '12 at 9:53 @anonymous: Good point, e.g. $\begin{bmatrix}1&1\\ 0&1\end{bmatrix}$. @@Hurkyl: I agree, it is best as an appendix. I appreciate your caution, but wanted to point out that your method does lead to the general solution (in the characteristic $0$ case that the OP is probably working in). – Jonas Meyer Feb 6 '12 at 15:49 The following matrix is a conterexample $A = \left( {\begin{array}{cc} -1 & 0 \\ 0 & 1 \\ \end{array} } \right)$ - In dimension $\geq 2$ take the matrix that exchanges two basis vectors ("a transposition") - If you want to exchange the (standard) basis vectors $e_{i}$ and $e_{j}$ ($1 \leq i,j \leq n$), then use the matrix $A = [m_{ij}]$ with $m_{kk} = 1, k\neq i,j$, $m_{ij} = m_{ji} = 1$ and $m_{kl} = 0$ for all other values of $k$ and $l$. For example, if you want $e_2$ and $e_3$ exhanged in $\mathbb{R}^{3}$, take $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$ It is clear that such a matrix always satisfies $A^2 = I$, since applying it twice always gets you back to where you started. – Martin Wanvik Feb 5 '12 at 21:01 Thank you @Martin Wanvik, pretty clear explanation. – Randolf R-F Feb 5 '12 at 21:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 43, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.905007004737854, "perplexity_flag": "head"}
http://mathoverflow.net/questions/5659/abstract-nonsense-versions-of-combinatorial-group-theory-questions/14229	## Abstract nonsense versions of “combinatorial” group theory questions ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In particular, I'm just curious whether there's a version of the Sylow theorems (which are very combinatorially-flavored) which allows horizontal and/or vertical categorification? Or at least can be stated in more category-theoretic terms? - I was wondering this just last week. I have prelims to start worrying about, and categorical statements always stick best. – alekzander Nov 16 2009 at 0:49 alexzander, that's my justification too. I never did like abstract algebra 'til I learned to start thinking of it in categorical terms. – Harrison Brown Nov 16 2009 at 3:06 Although in my case it's not prelims so much as undergraduate-level abstract algebra test, so I'm probably overthinking it. Still curious about the question on its own merits, though. – Harrison Brown Nov 16 2009 at 3:15 ## 5 Answers Sylow subgroups are an example of a type of object satisfying a sort of universal property. Exploring other objects with similar properties gave birth to the modern theory of finite soluble groups in the 1960s. If X is a class of finite groups and G is a finite group, then P is an X-covering subgroup of G if P is in X, and whenever P ≤ H ≤ G, N ⊴ H, and H/N in X, then PN=H. In other words, P covers every X-factor of G. If X is the class of finite p-groups, then X-covering subgroups of G and Sylow p-subgroups of G are the same thing. Indeed, if P is an X-covering group, and if H is a Sylow p-subgroup containing P and N=1, then we must have H=P. If P is a Sylow p-subgroup and H contains P with N ⊴ H and [H:N] a power of p, then [H:NP] is a divisor of [H:N] and [H:P], so must be 1. Notice how the "containment" part of the Sylow theorems is replaced with a "covering" condition that behaves better with the normal structure of the group. If G is a finite solvable group and X is the class of nilpotent groups, then there is a sort of "Sylow nilpotent subgroup", the X-covering groups or Carter subgroups. They were studied by R.W. Carter who described them as self-normalizing nilpotent subgroups. Like Sylow p-subgroups, there is exactly one conjugacy class of Carter subgroups, and they have some reasonable arithmetic properties. People tried to determine which classes X of groups are such that X-covering groups exist and are unique up to conjugacy. Roughly speaking, this was the dawn of the modern theory of finite soluble groups, with Gaschütz's (et al.) classification of such X as "saturated formations". This shifts focus away from the subgroup P to the class X. If X is sufficiently nice, then there will be a nicely embedded X-subgroup for any finite group. Sylow p-subgroups also satisfy a dual condition, they are also X-injectors for the class X of finite p-groups. If X is a class of finite groups, and G is a finite group, then P is an X-injector of G if for every subnormal subgroup N of G, P∩N is a maximal X-subgroup of N. The dual definition of covering group (for X a saturated formation) is that P is an X-covering group iff PN/N is a maximal X-subgroup of G/N for every N ⊴ G. If X is the class of finite nilpotent groups, then X-injectors are called Fischer subgroups and again form a single, well-behaved conjugacy class of subgroups. A Fischer subgroup of a finite soluble group is a nilpotent subgroup that contains every nilpotent subgroup that it normalizes. This is similar to the idea that a Sylow p-subgroup contains every p-group that it normalizes. X such that X-injectors form a unique conjugacy class are called Fitting classes, due to their similarity to Fitting's lemma on subnormal nilpotent subgroups. A very approachable introduction to these ideas is B.F. Wehrfritz's tiny textbook for a Second Course on Group Theory. Some of these ideas are described in Robinson's textbook for a Course in the Theory of Groups, but I believe it spends very little time on general formations. The standard textbook source for formations, especially in the soluble universe, is K. Doerk and T. Hawkes's book Finite Soluble Groups. Doerk&Hawkes explains several of Gaschütz's arithmetically defined Xs, which you might find a good contrast. - ### 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 may be splitting hairs, but I think the Sylow theorems are more arithmetic than combinatorial. Depending on your standards of good categorification, I think this makes it difficult to encode precise congruence behavior such as the index of the normalizer being congruent to 1 mod p, and so on. There is a homotopy-theoretic way to look at the Sylow theorems involving fusion systems. These are categories that encode the conjugation data between the p-Sylow subgroups of a finite group, and let you study the homotopy type of the p-completed classifying space of a finite or compact group. Bob Oliver has some papers on this subject. I don't know if this level of sophistication will help you learn the basic theorems, though. When I was studying Sylow, I personally preferred exercises, like showing that there are no simple groups of order 56. - I learned the Sylow theorems precisely like that. Homework problem in my abstract algebra class "Show that there are no nonabelian simple groups of order less than 60." – Charles Siegel Nov 16 2009 at 6:16 Yeah, okay, good point; it's probably too much to hope for there to be something easy enough to apply in day-to-day situations. But there's no obvious reason (to me, anyway) that there shouldn't be an extension even to finite groupoids. – Harrison Brown Nov 16 2009 at 16:09 1 J.L. Alperin's "Sylow Intersections and Fusion" (MR0215913; DOI: 10.1016/0021-8693(67)90005-1) is a nice group theoretic introduction to some of the key ideas behind fusion. It is a more precise version of Sylow's conjugacy theorem. Linckelmann's introduction (MR2336638) is one of the easiest ways to read the category theoretic version of fusion. Alperin's Local Representation Theory textbook introduces the Brauer subpairs theory nicely; that is the other main source of fusion systems. – Jack Schmidt Feb 5 2010 at 4:21 Philosophically I think of the Sylow theorems as a statement about "localization of groups" at a prime. For abelian groups one can make this precise, since abelian groups, as Z-modules, can be realized as sheaves over Spec Z and the stalk over p is precisely the Sylow p-subgroup (although for finitely generated abelian groups these are trivial to describe). The hard part would be trying to carry this picture over to general groups. - You might find the theory of fusion categories interesting (this is, unfortunately for the terminology, disjoint from Scott's mention of "fusion systems"). A fusion category is a type of tensor category which generalizes both Rep(G) and Vec_G, the latter being the category of vector spaces graded by a group G (so it's Groethendieck ring is the group algebra kG). These are tensor categories which are semi-simple and have finitely many isomorphism classes of simple objects. Each simple object has a notion of dimension, called Frobenius Perron dimension, and the categories themselves have a dimension associated to them, such that in the two cases Rep(G) and Vec_G, the dimension of the category is \|G\|. While I don't know of Sylow theorems for fusion categories per se, there are many familiar theorems, saying e.g. that every fusion category of order p^k is nilpotent, and classifying groups of small orders p, pq, pqr, p^k, etc. There is also a version of Burnside's theorem stating that every category of dimension p^aq^b is solvable. It's quite possible that there are some analogs of Sylow's theorems in this direction. I'd recommend "On Fusion Categories" http://arxiv.org/abs/math/0203060, by Etingof Nikshych and Ostrik, which introduces many of these ideas. More extensively, there are course notes from a class Etingof taught on this, which can be found on his webpage, http://math.mit.edu/~etingof. Papers by various subsets of (these three, union Gelaki) introduce the concepts and claims I mentioned above. As Scott mentioned, this kind of abstract nonsense won't help much with prelims questions, most likely, but might make for a fun distraction =]. - One of the simplest consequences of Sylow's first theorem is the Cauchy theorem, saying that a group of order divisible by a prime $p$ contains an element of order $p$. I'd like to point out that there is an analog of Cauchy's theorem for semisimple Hopf algebras, arXiv:math/0311199, which has been generalized to quasi-Hopf algebras (i.e., fusion categories with integer dimensions) in arXiv:math/0601012. -	{"extraction_info": {"found_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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9356942176818848, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/84705/is-the-moduli-space-of-curves-defined-over-the-field-with-one-element/84785	## Is the moduli space of curves defined over the field with one element? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) There are various frameworks around which enlarge the category of rings to include more exotic objects such as the 'field with one element,' `$\mathbb{F}_1$`. While these frameworks differ in their details, there are certain things this should be true of any object that deserves to be called `$\mathbb{F}_1$`. For example, The algebraic K-theory of `$\mathbb{F}_1$` should be sphere spectrum, and the theory of toric varieties should be defined over `$\mathbb{F}_1$`. Question 1: Is there a moral reason why the moduli space of curves should (or should not) be defined over Spec `$\mathbb{F}_1$`? EDIT: For anyone who would like to be more concrete, I'm happy to take the Toen-Vaquie definition of schemes over `$\mathbb{F}_1$`. (see arXiv:math/0509684). In this setup (and most of the other frameworks I know) an affine scheme over `$\mathbb{F}_1$` is just a commutative monoid $M$. After base change to $\mathbb{Z}$ this becomes the monoid ring $\mathbb{Z}[M]$. So here is a more precise question: Question 2: Does the moduli space of curves $\mathcal{M}_{g,n}$ (over $\mathbb{Z}$, say) admit a covering by affine charts of the form spec $\mathbb{Z}[M_i]$ for commutative monoids $M_i$? If so, can this covering be chosen so that (as in the case of toric varieties) the gluing is entirely determined by maps of monoids? - 3 I think what you mean is that spec $\mathbb{F}_1$ should be a final object, so everything maps to it. This is different from being defined over it. E.g. $\mathbb{Z}$ is initial among rings and so any scheme maps to spec $\mathbb{Z}$, but there are certainly things that are not defined over $\mathbb{Z}$ - i.e., there exists schemes over spec $k$ that are not the pullback (along spec $k$ $\to$ spec $\mathbb{Z}$) of a scheme over $\mathbb{Z}$. – Jeffrey Giansiracusa Jan 2 2012 at 12:28 2 My opinion is that this is not a valid question unless you specify at least one definition of the field with one element. Could you please give a reference to the definition you are using? – Jason Starr Jan 2 2012 at 13:33 8 Did you look at mathoverflow.net/questions/8190/…? Outside of $M_{0,n}$ (moduli of curves of genus zero with $n$ marked points), it seems doubtful to me that any of the other $M_{g,n}$ (except a few small g,n cases?) would be defined over $F_1$, whatever that means. For example, don't the L-functions of schemes over $F_1$ (after base change to schemes over $Z$) coincide with L-functions of mixed Tate motives? Maybe $M_{1,1}$ would be interesting, since the orbifold points could be definable over $F_1$ (or at least something "below" $Z$). – Marty Jan 2 2012 at 18:18 4 Just to add to what Marty said, it seems to me that any scheme glued from schemes of the form $\mathbb{Z}[M]$ will be geometrically rational when you base change to $\overline{\mathbb{Q}}$. Since $M_g$ is definitely irrational when $g$ is large, it seems to me it cannot be of this form. – Jason Starr Jan 2 2012 at 22:08 3 Others have mentioned that it seems unlikely that you can define the whole moduli space over $\mathbb{F}_1$. However, it looks like you can define a formal neighborhood of any maximally degenerate point of the boundary, since such loci describe gluings of three-pointed genus zero curves. This suggests that objects like the Tate curve, $n$-gons, and Néron models of maximally degenerating Jacobians are definable over $\mathbb{F}_1$. – S. Carnahan♦ Jan 3 2012 at 5:51 show 5 more comments ## 2 Answers This "answer" will basically restate the comments of Marty and Jason Starr. Any variety covered by schemes of the form $\mathrm{Spec}(\mathbf Z[M_i])$, or any torified variety, is rational. And indeed Severi conjectured at one point that $M_g$ is rational for any $g$! But we know a lot about the Kodaira dimension of $M_g$ by work of Harris--Mumford, Farkas, Eisenbud, Verra, ... in particular we know that Severi's conjecture is maximally false. We have $\kappa(M_g) = -\infty$ for $g \leq 16$, we don't know anything for $17 \leq g \leq 21$, and $\kappa(M_g) \geq 0$ for $g \geq 22$. In fact we know that $M_{g}$ is of general type for $g = 22$ and $g \geq 24$. But if one only wants an example of a $g$ for which $M_g$ is not rational, then I think one can find examples much earlier (maybe $g \approx 6,7$?). See http://arxiv.org/abs/0810.0702 for a survey by Farkas. Moreover, the cohomology of a torified scheme can only contain mixed Tate motives. In particular, this implies properties like: the number of $\mathbf F_q$-points is a polynomial function of $q$. As Marty says there are no problems in genus zero; these might be honest-to-God $\mathbf F_1$ schemes. The cohomology of $M_{1,n}$ contains only mixed Tate motives when $n \leq 10$, but for $n \geq 11$ one finds motives associated to cusp forms for $\mathrm{SL}(2,\mathbf Z)$. (The number 11 arises as one less than the smallest weight of a nonzero cusp form; the discriminant form $\Delta$.) This implies that the polynomiality behaviour changes drastically -- the number of $\mathbf F_q$-points is now given by Fourier coefficients of modular forms, which are far more complicated and contain lots of arithmetic information. The connection with birational geometry is also very visible here. The cohomology classes on $M_{1,n}$ associated to the cusp forms of weight $n+1$ are of type $(n,0)$ and $(0,n)$ in the Hodge realization, since given such a cusp form one can explicitly write down a corresponding differential form in coordinates. So by the very definition of Kodaira dimension we can not have $\kappa = -\infty$ anymore. Bergström computed the Euler characteristic of $M_{2,n}$ in the Grothendieck group of $\ell$-adic Galois representations by point counting techniques. That is, he found formulas for the number of $\mathbf F_q$-points of $M_{2,n}$ by working really explicitly with normal forms for hyperelliptic curves. These formulas turned out to be polynomial in $q$ for $n \leq 7$ (and conjecturally for $n \leq 9$), just as they would be if we knew that the cohomology of $M_{2,n}$ contained only mixed Tate motives. By thinking a bit about the stratification of topological type, one concludes that this holds also for $\overline M_{2,n}$ when $n \leq 7$, which is smooth and proper over the integers. Then a theorem of van den Bogaart and Edixhoven implies that the cohomology of $\overline M_{2,n}$ is all of Tate type and with Betti numbers given by the coefficients of the polynomials. There are similar results by Bergström for $M_{3,n}$ when $n \leq 5$ and Bergström--Tommasi for $M_4$. But the general phenomenon is that increasing either $g$ or $n$ will rapidly take you out of the world of $\mathbf F_1$-schemes, at least if this is taken to mean "commutative monoids". However, I don't know enough $\mathbf F_1$--geometry to say what the answer is if one takes $\mathbf F_1$-schemes in the sense of Borger. The first nontrivial case to answer would be: are the motives associated to cusp forms for $\mathrm{SL}(2,\mathbf Z)$ defined over $\mathbf F_1$ in his set-up? To be clear, I don't believe that this is true, but I know almost nothing about $\lambda$-schemes. Let me also make two small remarks: (i) Everything I have written above is necessary conditions for being defined over $\mathbf F_1$. (ii) The fact that $M_{g,n}$ is smooth over the integers says that the cohomology of $M_{g,n}$ is at least quite "special", even though it is not defined over $\mathbf F_1$. Smoothness is a strong restriction on the Galois representations that can occur in the cohomology. - Very nice - thanks for the survey of examples of $M_{g,n}$ for various small $g$ and $n$ too! – Marty Jan 3 2012 at 6:57 Thanks for the very clear explanation. – Jeffrey Giansiracusa Jan 3 2012 at 8:59 ### 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. Here a few remarks from the `$\Lambda$` point of view on `$\mathbf{F}_1$`. (There we say a scheme is defined over `$\mathbf{F}_1$` if it admits a `$\Lambda$`-structure. If the scheme is flat over `$\mathbf{Z}$`, a `$\Lambda$`-structure is equivalent to a commuting family of endomorphisms `$\psi_p$`, one for each prime number `$p$`, such that each `$\psi_p$` agrees with the Frobenius map modulo `$p$`.) I prove in http://arxiv.org/abs/0906.3146 that if a scheme of finite type over `$\mathbf{Z}$` is defined over `$\mathbf{F}_1$`, then its motive is pure Tate (or rather becomes so after base change to some cyclotomic field - it's actually false in general without this qualification). So as in Dan Petersen's answer, `$M_{g,n}$` won't admit a `$\Lambda$`-structure unless the pair `$(g,n)$` is sufficiently small. I haven't really thought about the small cases, except just a bit when `$(g,n)=(1,1)$`. Then `$M_{g,n}$` is just the affine line, so it has many `$\Lambda$`-structures. As far as I know, you can't really say that they have any meaning, but I suspect there is still something interesting to say. 1) There is a `$\Lambda$`-structure on the completion of `$\bar{M}_{1,1}$` at the point at infinity which does have a meaning. The completion is `$\mathbf{Z}[[q-1]]$`, and the `$\Lambda$`-structure is defined by `$\psi_p(q)=q^p$`, for all `$p$`. The interest in this is that this `$\Lambda$`-structure prolongs to a `$\Lambda$`-structure on the universal generalized elliptic curve over `$\mathbf{Z}[[q-1]]$` (i.e. the Tate curve). From what I remember, the idea is that a point `$z\in\mathbf{C}^/q^{\mathbf{Z}}$` should be mapped to its image `$z^p\in\mathbf{C}^/q^{p\mathbf{Z}}$`. Making this rigorous is not hard, though of course you have to have an actual definition of the Tate curve over `$\mathbf{Z}[[q-1]]$`, a point that is often glided over. What I think could be interesting is to study the extent to which these `$\Lambda$`-structures extend to the whole moduli space. Here is one possible interpretation. A `$\Lambda$`-structure on a scheme `$X$` is equivalent to a section of the canonical projection `$W_(X)\to X$` satisfying a certain associativity property. (`$W_$` is the right adjoint to the Witt vector functor `$W^$`. It is an arithmetic analogue of a jet space.) Now consider the map `$T\to X$`, where `$T$` is the Tate curve above and `$X$` is the total space of the universal elliptic curve. (This is a stack. One would have to check that all these `$\Lambda$` and `$W$` concepts make sense for stacks.) Then the projection `$W_(X)\to X$` has a natural section over `$T$`. Then we can ask: what is the Zariski closure `$Z$` of this section in `$W_(X)$`? The extent to which `$Z$` fails to be a section of `$W_(X)\to X$` should be some measure of the extent to which the `$\Lambda$`-structure on `$T$` fails to extend to one on `$X$`. 2) One could try to do something similar at CM points, instead of at the boundary of the moduli space. CM elliptic curves admit a certain generalized version of a `$\Lambda$`-structure. (See the end of my paper cited above.) Do these generalized `$\Lambda$`-structures extend to the formal neighborhood of the elliptic curve in the universal elliptic curve? If so, one could look at closures of formal sections (as above) to see the extent to which they fail to extend to generalized `$\Lambda$`-structures on the entire universal elliptic curve. For all this, you'd have to fix an imaginary quadratic field, so all this is probably less fundamental than in 1). 3) Thomas Scanlon tried to convince me a few years ago that the Hecke correspondences have some kind of `$\Lambda$` nature. From what I remember, there were certain `$\Lambda$`-schemes closely related to modular varieties which are not of finite type but which are still finite-dimensional in some model-theoretic sense. I never got around to understanding what he meant (though I always intended to). Perhaps it is related to what I wrote above. So there are a few pieces of evidence that something is going on with `$\Lambda$`-structures and modular curves, though it remains to be seen if it's really an identifiable phenomenon and, if so, how interesting it is. - I see in the comments to the original post that Scott Carnahan made predictions about the Tate curve that are similar to what I said above. – James Borger Jan 4 2012 at 3:43	{"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": 90, "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": 0.9474338293075562, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/66183/finding-the-center-of-a-group-with-magma?answertab=active	# Finding the center of a group with MAGMA Given a finitely presented group $G = \langle S \| R \rangle$, is there a command in MAGMA that computes the center of G ? - If Center() won't work on G, you can use one of the functions available to covert G into a group of a type on which Center() will work. – mt_ Sep 20 '11 at 21:52 This is unlikely to be possible unless $G$ is finite, in which case you should use a command like CosetImage to convert it to a group of permutations. – Derek Holt Sep 21 '11 at 8:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.3, "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": 0.899063229560852, "perplexity_flag": "middle"}

http://math.stackexchange.com/questions/44139/how-many-solutions-are-there-to-fn-m-n2nmm2-q/

# How many solutions are there to $F(n,m)=n^2+nm+m^2 = Q$? Let $n,m$ be two positive integers, we consider: $$F(n,m)=n^2+nm+m^2$$ Let $Q$ be one value reach by $F(n,m)$. How many different pairs $(n,m)$ verify $F(n,m)=Q$? - 6 The trick in this instance is that $n^2+nm+m^2$ is the norm in the ring $\mathbb{Z}[\omega]$ where $\omega=\frac{1+\sqrt{-3}}2$. Then $N(n+m\omega)=n^2+nm+m^2$. Since $\mathbb{Z}[\omega]$ is a unique factorization domain, you get results like the result for the sum of two squares. – Thomas Andrews Jun 8 '11 at 17:46 @Eric and Ray: Sorry about my erroneous answer; I was too quick to "jump the gun" and as I pursued the problem after posting, I realized I overlooked the restriction to positive m, n...and further, I realized there are far more solutions than my erroneous post revealed. – amWhy Jun 8 '11 at 17:47 @Thomas Andrews: That is an excellent point! I wish had seen that comment before typing my answer. – Eric♦ Jun 8 '11 at 18:35 @Amwhy: It happens to all of us, no worries. (I have jumped the gun like this many times!) – Eric♦ Jun 8 '11 at 18:42 ## 1 Answer As it turns out, we can give a complete answer to this question. The exact number of solutions depends on the prime factorization of $Q$ Specifically, it is a function of the exponents of the prime factors which are congruent to $1$ mod $3$, with the condition that all the factors congruent to $2$ modulo $3$ have their prime factors appear with even multiplicity. Note: I have not yet provided a proof, the result that primes of the form $1+3k$ can be represented is a theorem of Jacobi. Let $$Q=\prod_i q_i^{\alpha_i} \prod_i p_i^{\beta_i}$$ where the $q_i$ are $1$ mod $3$ and the $p_i$ are $2$ mod $3$. Our equation $n^2+nm+m^2$ has solutions if and only if each $\beta_i$ is even. Proof: Take the equation modulo $3$. By case analysis for $n,m$ the right hand side cannot be congruent to $2$, and hence the statement follows. Remark: Notice the following similarity to the sum of squares problem (points on a circle). The Answer: Suppose that as before $$Q=\prod_i q_i^{\alpha_i} \prod_i p_i^{\beta_i}$$ where the $q_i$ are $1$ mod $3$ and the $p_i$ are $2$ mod $3$. Suppose as well that all of the $\beta_i$ are even. (Otherwise we can have no solutions) Let $$B=(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_n+1).$$ Then the number of non-negative integer solutions to $$m^2+mn+n^2=Q$$ is exactly $$\left\lceil\frac{B}{2}\right\rceil.$$ (Again notice the similarity to the sum of squares function) In particular, if $l(n)$ is the number of representations where $n,m$ are any integers, (that is positive or negative) then $$l(n)=2B.$$ Hope that helps. - 3 This is a very nice answer. One comment: I'm not sure why you're diagonalizing -- you don't seem to use it anywhere. More fundamentally, the given quadratic form is the norm form of the maximal order of $\mathbb{Q}(\sqrt{-3})$, whereas $x^2 + 3y^2$ is the norm form of the nonmaximal order $\mathbb{Z}[\sqrt{-3}]$, which is not a UFD. So in fact if you wanted to study the form $x^2 + 3y^2$ you would do it by comparing to $x^2 + xy + y^2$, not the other way around. Note also that even without the change of variables you are looking for lattice points on an ellipse. – Pete L. Clark Jun 8 '11 at 18:45 2 – Pete L. Clark Jun 8 '11 at 19:11 @Pete: Ok, thanks for the note! I don't know why I thought it wasn't. – Eric♦ Jun 8 '11 at 19:27

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http://nrich.maths.org/6484/index?nomenu=1

Imagine that I plot the pdfs for two normal distributions on the same axes. Could I choose the parameters so that the curves intersect $0$, $1$ or $2$ times? Imagine that I plot the cdfs for two normal distributions on the same axes. Could I choose the parameters so that the curves intersect $0$, $1$ or $2$ times?

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http://crypto.stackexchange.com/questions/tagged/elgamal-signature+forgery

# Tagged Questions 0answers 77 views ### Why do we need to hash both the message and the $h$ value in ElGamal signature? The professor left us a question on ElGamal signatures: Given the hash function $H$ and message $M$, choose a random $r$ and compute $h=g^r$ and $H(M||h)$. Show that, if $H(M)$ is used instead of ... 0answers 37 views ### ElGamal signature: Forging a signature of a specific form I have a question I can't solve from one of the courses I'm currently taking: Show that given a legitimate ElGamal signature $(S,R)$ on a given message $m$, an attacker can compute a signature ...

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http://www.nag.com/numeric/CL/nagdoc_cl23/html/F12/f12fdc.html

NAG Library Function Documentnag_real_symm_sparse_eigensystem_option (f12fdc) Note: this function uses optional arguments to define choices in the problem specification. If you wish to use default . 1 Purpose nag_real_symm_sparse_eigensystem_option (f12fdc) is an option setting function in a suite of functions consisting of nag_real_symm_sparse_eigensystem_init (f12fac), nag_real_symm_sparse_eigensystem_iter (f12fbc), nag_real_symm_sparse_eigensystem_sol (f12fcc), nag_real_symm_sparse_eigensystem_option (f12fdc) and nag_real_symm_sparse_eigensystem_monit (f12fec), and may be used to supply individual optional arguments to nag_real_symm_sparse_eigensystem_iter (f12fbc) and nag_real_symm_sparse_eigensystem_sol (f12fcc). The initialization function nag_real_symm_sparse_eigensystem_init (f12fac) must have been called prior to calling nag_real_symm_sparse_eigensystem_option (f12fdc). 2 Specification #include <nag.h> #include <nagf12.h> void nag_real_symm_sparse_eigensystem_option (const char *str, Integer icomm[], double comm[], NagError *fail) 3 Description nag_real_symm_sparse_eigensystem_option (f12fdc) may be used to supply values for optional arguments to nag_real_symm_sparse_eigensystem_iter (f12fbc) and nag_real_symm_sparse_eigensystem_sol (f12fcc). It is only necessary to call nag_real_symm_sparse_eigensystem_option (f12fdc) for those arguments whose values are to be different from their default values. One call to nag_real_symm_sparse_eigensystem_option (f12fdc) sets one argument value. Each optional argument is defined by a single character string consisting of one or more items. The items associated with a given option must be separated by spaces, or equals signs $\left[=\right]$. Alphabetic characters may be upper or lower case. The string ```'Iteration Limit = 500' ``` is an example of a string used to set an optional argument. For each option the string contains one or more of the following items: – a mandatory keyword; – a phrase that qualifies the keyword; – a number that specifies an Integer or double value. Such numbers may be up to $16$ contiguous characters in C's d or g format. nag_real_symm_sparse_eigensystem_option (f12fdc) does not have an equivalent function from the ARPACK package which passes options by directly setting values to scalar arguments or to specific elements of array arguments. nag_real_symm_sparse_eigensystem_option (f12fdc) is intended to make the passing of options more transparent and follows the same principle as the single option setting functions in Chapter e04. The setup function nag_real_symm_sparse_eigensystem_init (f12fac) must be called prior to the first call to nag_real_symm_sparse_eigensystem_option (f12fdc) and all calls to nag_real_symm_sparse_eigensystem_option (f12fdc) must precede the first call to nag_real_symm_sparse_eigensystem_iter (f12fbc), the reverse communication iterative solver. A complete list of optional arguments, their abbreviations, synonyms and default values is given in Section 10. 4 References Lehoucq R B (2001) Implicitly restarted Arnoldi methods and subspace iteration SIAM Journal on Matrix Analysis and Applications 23 551–562 Lehoucq R B and Scott J A (1996) An evaluation of software for computing eigenvalues of sparse nonsymmetric matrices Preprint MCS-P547-1195 Argonne National Laboratory Lehoucq R B and Sorensen D C (1996) Deflation techniques for an implicitly restarted Arnoldi iteration SIAM Journal on Matrix Analysis and Applications 17 789–821 Lehoucq R B, Sorensen D C and Yang C (1998) ARPACK Users' Guide: Solution of Large-scale Eigenvalue Problems with Implicitly Restarted Arnoldi Methods SIAM, Philidelphia 5 Arguments 1: str – const char *Input On entry: a single valid option string (as described in Section 3 and Section 10). 2: icomm[$\mathit{dim}$] – IntegerCommunication Array Note: the dimension, dim, of the array icomm must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{licomm}}\right)$ (see nag_real_symm_sparse_eigensystem_init (f12fac)). On initial entry: must remain unchanged following a call to the setup function nag_real_symm_sparse_eigensystem_init (f12fac). On exit: contains data on the current options set. 3: comm[$\mathit{dim}$] – doubleCommunication Array Note: the dimension, dim, of the array comm must be at least $60$. On initial entry: must remain unchanged following a call to the setup function nag_real_symm_sparse_eigensystem_init (f12fac). On exit: contains data on the current options set. 4: fail – NagError *Input/Output The NAG error argument (see Section 3.6 in the Essential Introduction). 6 Error Indicators and Warnings NE_BAD_PARAM On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value. NE_INTERNAL_ERROR An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance. NE_INVALID_OPTION Ambiguous keyword: $〈\mathit{\text{value}}〉$ Keyword not recognized: $〈\mathit{\text{value}}〉$ Second keyword not recognized: $〈\mathit{\text{value}}〉$ Not applicable. None. 9 Example This example solves $Ax=\lambda Bx$ in ${\mathbf{Shifted Inverse}}$ mode, where $A$ and $B$ are obtained from the standard central difference discretization of the one-dimensional Laplacian operator $\frac{{\partial }^{2}u}{\partial {x}^{2}}$ on $\left[0,1\right]$, with zero Dirichlet boundary conditions. Data is passed to and from the reverse communication function nag_real_symm_sparse_eigensystem_iter (f12fbc) using pointers to the communication array. 9.1 Program Text Program Text (f12fdce.c) 9.2 Program Data Program Data (f12fdce.d) 9.3 Program Results Program Results (f12fdce.r) 10 Optional Arguments Several optional arguments for the computational functions nag_real_symm_sparse_eigensystem_iter (f12fbc) and nag_real_symm_sparse_eigensystem_sol (f12fcc) define choices in the problem specification or the algorithm logic. In order to reduce the number of formal arguments of nag_real_symm_sparse_eigensystem_iter (f12fbc) and nag_real_symm_sparse_eigensystem_sol (f12fcc) these optional arguments have associated default values that are appropriate for most problems. Therefore, you need only specify those optional arguments whose values are to be different from their default values. The remainder of this section can be skipped if you wish to use the default values for all optional arguments. The following is a list of the optional arguments available and a full description of each optional argument is provided in Section 10.1. Optional arguments may be specified by calling nag_real_symm_sparse_eigensystem_option (f12fdc) before a call to nag_real_symm_sparse_eigensystem_iter (f12fbc), but after a call to nag_real_symm_sparse_eigensystem_init (f12fac). One call is necessary for each optional argument. All optional arguments you do not specify are set to their default values. Optional arguments you do specify are unaltered by nag_real_symm_sparse_eigensystem_iter (f12fbc) and nag_real_symm_sparse_eigensystem_sol (f12fcc) (unless they define invalid values) and so remain in effect for subsequent calls unless you alter them. 10.1 Description of the Optional Arguments For each option, we give a summary line, a description of the optional argument and details of constraints. The summary line contains: • the keywords, where the minimum abbreviation of each keyword is underlined; • a parameter value, where the letters $a$, $i\text{ and }r$ denote options that take character, integer and real values respectively; • the default value, where the symbol $\epsilon $ is a generic notation for machine precision (see nag_machine_precision (X02AJC)). Keywords and character values are case and white space insensitive. Optional arguments used to specify files (e.g., ${\mathbf{Advisory}}$ and ${\mathbf{Monitoring}}$) have type Nag_FileID. This ID value must either be set to $0$ (the default value) in which case there will be no output, or will be as returned by a call of nag_open_file (x04acc). Advisory Default $\text{}=0$ (See Section 3.2.1.1 in the Essential Introduction for further information on NAG data types.) Advisory messages are output to Nag_FileID ${\mathbf{Advisory}}$ during the solution of the problem. Defaults This special keyword may be used to reset all optional arguments to their default values. Exact Shifts Default Supplied Shifts During the Lanczos iterative process, shifts are applied internally as part of the implicit restarting scheme. The shift strategy used by default and selected by the ${\mathbf{Exact Shifts}}$ is strongly recommended over the alternative ${\mathbf{Supplied Shifts}}$ (see Lehoucq et al. (1998) for details of shift strategies). If ${\mathbf{Exact Shifts}}$ are used then these are computed internally by the algorithm in the implicit restarting scheme. If ${\mathbf{Supplied Shifts}}$ are used then, during the Lanczos iterative process, you must supply shifts through array arguments of nag_real_symm_sparse_eigensystem_iter (f12fbc) when nag_real_symm_sparse_eigensystem_iter (f12fbc) returns with ${\mathbf{irevcm}}=3$; the real and imaginary parts of the shifts are supplied in y and mx respectively. This option should only be used if you are an experienced user since this requires some algorithmic knowledge and because more operations are usually required than for the implicit shift scheme. Details on the use of explicit shifts and further references on shift strategies are available in Lehoucq et al. (1998). Iteration Limit $i$ Default $\text{}=300$ The limit on the number of Lanczos iterations that can be performed before nag_real_symm_sparse_eigensystem_iter (f12fbc) exits. If not all requested eigenvalues have converged to within ${\mathbf{Tolerance}}$ and the number of Lanczos iterations has reached this limit then nag_real_symm_sparse_eigensystem_iter (f12fbc) exits with an error; nag_real_symm_sparse_eigensystem_sol (f12fcc) can still be called subsequently to return the number of converged eigenvalues, the converged eigenvalues and, if requested, the corresponding eigenvectors. Largest Magnitude Default Both Ends Largest Algebraic Smallest Algebraic Smallest Magnitude The Lanczos iterative method converges on a number of eigenvalues with given properties. The default is for nag_real_symm_sparse_eigensystem_iter (f12fbc) to compute the eigenvalues of largest magnitude using ${\mathbf{Largest Magnitude}}$. Alternatively, eigenvalues may be chosen which have ${\mathbf{Largest Algebraic}}$ part, ${\mathbf{Smallest Magnitude}}$, or ${\mathbf{Smallest Algebraic}}$ part; or eigenvalues which are from ${\mathbf{Both Ends}}$ of the algebraic spectrum. Note that these options select the eigenvalue properties for eigenvalues of $\mathrm{OP}$ (and $B$ for ${\mathbf{Generalized}}$ problems), the linear operator determined by the computational mode and problem type. Nolist Default List Normally each optional argument specification is not printed to ${\mathbf{Advisory}}$ as it is supplied. Optional argument ${\mathbf{List}}$ may be used to enable printing and optional argument ${\mathbf{Nolist}}$ may be used to suppress the printing. Monitoring Default $\text{}=-1$ (See Section 3.2.1.1 in the Essential Introduction for further information on NAG data types.) Unless ${\mathbf{Monitoring}}$ is set to $-1$ (the default), monitoring information is output to Nag_FileID ${\mathbf{Monitoring}}$ during the solution of each problem; this may be the same as ${\mathbf{Advisory}}$. The type of information produced is dependent on the value of ${\mathbf{Print Level}}$, see the description of the optional argument ${\mathbf{Print Level}}$ in this section for details of the information produced. Please see nag_open_file (x04acc) to associate a file with a given Nag_FileID. Print Level $i$ Default $\text{}=0$ This controls the amount of printing produced by nag_real_symm_sparse_eigensystem_option (f12fdc) as follows. $=0$ No output except error messages. If you want to suppress all output, set ${\mathbf{Print Level}}=0$. $>0$ The set of selected options. $=2$ Problem and timing statistics on final exit from nag_real_symm_sparse_eigensystem_iter (f12fbc). $\ge 5$ A single line of summary output at each Lanczos iteration. $\ge 10$ If ${\mathbf{Monitoring}}$ is set, then at each iteration, the length and additional steps of the current Lanczos factorization and the number of converged Ritz values; during re-orthogonalisation, the norm of initial/restarted starting vector; on a final Lanczos iteration, the number of update iterations taken, the number of converged eigenvalues, the converged eigenvalues and their Ritz estimates. $\ge 20$ Problem and timing statistics on final exit from nag_real_symm_sparse_eigensystem_iter (f12fbc). If ${\mathbf{Monitoring}}$ is set, then at each iteration, the number of shifts being applied, the eigenvalues and estimates of the symmetric tridiagonal matrix $H$, the size of the Lanczos basis, the wanted Ritz values and associated Ritz estimates and the shifts applied; vector norms prior to and following re-orthogonalisation. $\ge 30$ If ${\mathbf{Monitoring}}$ is set, then on final iteration, the norm of the residual; when computing the Schur form, the eigenvalues and Ritz estimates both before and after sorting; for each iteration, the norm of residual for compressed factorization and the symmetric tridiagonal matrix $H$; during re-orthogonalisation, the initial/restarted starting vector; during the Lanczos iteration loop, a restart is flagged and the number of the residual requiring iterative refinement; while applying shifts, some indices. $\ge 40$ If ${\mathbf{Monitoring}}$ is set, then during the Lanczos iteration loop, the Lanczos vector number and norm of the current residual; while applying shifts, key measures of progress and the order of $H$; while computing eigenvalues of $H$, the last rows of the Schur and eigenvector matrices; when computing implicit shifts, the eigenvalues and Ritz estimates of $H$. $\ge 50$ If ${\mathbf{Monitoring}}$ is set, then during Lanczos iteration loop: norms of key components and the active column of $H$, norms of residuals during iterative refinement, the final symmetric tridiagonal matrix $H$; while applying shifts: number of shifts, shift values, block indices, updated tridiagonal matrix $H$; while computing eigenvalues of $H$: the diagonals of $H$, the computed eigenvalues and Ritz estimates. Note that setting ${\mathbf{Print Level}}\ge 30$ can result in very lengthy ${\mathbf{Monitoring}}$ output. Random Residual Default Initial Residual To begin the Lanczos iterative process, nag_real_symm_sparse_eigensystem_iter (f12fbc) requires an initial residual vector. By default nag_real_symm_sparse_eigensystem_iter (f12fbc) provides its own random initial residual vector; this option can also be set using optional argument ${\mathbf{Random Residual}}$. Alternatively, you can supply an initial residual vector (perhaps from a previous computation) to nag_real_symm_sparse_eigensystem_iter (f12fbc) through the array argument resid; this option can be set using optional argument ${\mathbf{Initial Residual}}$. Regular Default Regular Inverse Shifted Inverse Buckling Cayley These options define the computational mode which in turn defines the form of operation $\mathrm{OP}\left(x\right)$ to be performed when nag_real_symm_sparse_eigensystem_iter (f12fbc) returns with ${\mathbf{irevcm}}=-1$ or $1$ and the matrix-vector product $Bx$ when nag_real_symm_sparse_eigensystem_iter (f12fbc) returns with ${\mathbf{irevcm}}=2$. Given a ${\mathbf{Standard}}$ eigenvalue problem in the form $Ax=\lambda x$ then the following modes are available with the appropriate operator $\mathrm{OP}\left(x\right)$. ${\mathbf{Regular}}$ $\mathrm{OP}=A$ ${\mathbf{Shifted Inverse}}$ $\mathrm{OP}={\left(A-\sigma I\right)}^{-1}$ where $\sigma $ is real Given a ${\mathbf{Generalized}}$ eigenvalue problem in the form $Ax=\lambda Bx$ then the following modes are available with the appropriate operator $\mathrm{OP}\left(x\right)$. ${\mathbf{Regular Inverse}}$ $\mathrm{OP}={B}^{-1}A$ ${\mathbf{Shifted Inverse}}$ $\mathrm{OP}={\left(A-\sigma B\right)}^{-1}B$, where $\sigma $ is real ${\mathbf{Buckling}}$ $\mathrm{OP}={\left(B-\sigma A\right)}^{-1}A$, where $\sigma $ is real ${\mathbf{Cayley}}$ $\mathrm{OP}={\left(A-\sigma B\right)}^{-1}\left(A+\sigma B\right)$, where $\sigma $ is real Standard Default Generalized The problem to be solved is either a standard eigenvalue problem, $Ax=\lambda x$, or a generalized eigenvalue problem, $Ax=\lambda Bx$. The optional argument ${\mathbf{Standard}}$ should be used when a standard eigenvalue problem is being solved and the optional argument ${\mathbf{Generalized}}$ should be used when a generalized eigenvalue problem is being solved. Tolerance $r$ Default $\text{}=\epsilon $ An approximate eigenvalue has deemed to have converged when the corresponding Ritz estimate is within ${\mathbf{Tolerance}}$ relative to the magnitude of the eigenvalue. Vectors Default $\text{}=\text{RITZ}$ The function nag_real_symm_sparse_eigensystem_sol (f12fcc) can optionally compute the Schur vectors and/or the eigenvectors corresponding to the converged eigenvalues. To turn off computation of any vectors the option ${\mathbf{Vectors}}=\mathrm{NONE}$ should be set. To compute only the Schur vectors (at very little extra cost), the option ${\mathbf{Vectors}}=\mathrm{SCHUR}$ should be set and these will be returned in the array argument v of nag_real_symm_sparse_eigensystem_sol (f12fcc). To compute the eigenvectors (Ritz vectors) corresponding to the eigenvalue estimates, the option ${\mathbf{Vectors}}=\mathrm{RITZ}$ should be set and these will be returned in the array argument z of nag_real_symm_sparse_eigensystem_sol (f12fcc), if z is set equal to v (as in Section 9) then the Schur vectors in v are overwritten by the eigenvectors computed by nag_real_symm_sparse_eigensystem_sol (f12fcc).

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http://mathoverflow.net/questions/107030/relationships-between-properties-of-model-categories

## relationships between properties of model categories ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I've recently found myself running up against all sorts of adjectives that can describe a model category: cofibrantly generated, combinatorial, tractable, stable, locally (finitely) presentable, (left and/or right) proper, simplicial, admits Bousfield localizations, et al. I'm hoping for a reference whose explicit goal is to give an intuitive explanation of these concepts; presumably, such a reference would also tell me what each of these adjectives can buy me. (As a bonus, this reference might even have a diagram analogous to the one on pp. 576-7 of Gortz & Wedhorn's book Algebraic Geometry, which summarizes the relationships between different properties that a morphism of schemes may satisfy. But that'd just be for fun.) - In the case of algebraic geometry there's something to prove for many of the indicated relations. For model categories, and for the properties you mention, one only has to look at the definition and see whether one property is required in the definition of another property. This is why I think this question is somewhat empty as currently stated. – Fernando Muro Sep 13 at 0:52 I've never come across such a list, and I agree with Fernando that it wouldn't be very useful for the properties that you've listed. However, a few things are known. If you add "admits left Bousfield Localizations" to your list then left proper and combinatorial implies this. So does left proper and cellular. If you replace simplicial by "quillen equivalent to a simplicial model category" then Dan Dugger has a paper saying which of the other properties imply this: pages.uoregon.edu/ddugger/smod.html. There's also "permits small object argument" $\Rightarrow$ cofibrantly generated. – David White Sep 13 at 1:17 Thanks for the feedback. The reason I asked is that I've seen these sorts of adjectives running around before, but only recently have I begun to actually wrestle with them. I wasn't sure a diagram would be helpful, but the one in Gortz-Wedhorn popped into my head when I was thinking about it. I suppose the more important part is the last sentence. Certainly nLab has been a good resource, but I think I'd benefit from something a little more narrative. I'll edit. – Aaron Mazel-Gee Sep 13 at 11:33 Well, the classical reference is Hovey, but he doesn't cover a lot of the adjectives on your list. Hirschhorn covers more and is very thorough, so that's where I would (and did) start. To my knowledge those are the only books on model categories, though there are large sections of Lurie's DAG which cover the subject and you could do a CTRL+F search on those for the adjectives of interest (esp tractable, finitely presentable, etc). See also Chorny's recent work on class-combinatorial & class accessible categories. Otherwise I recommend Dugger. His writing is very clear and down to earth. – David White Sep 13 at 13:52 Ah, so there's no magic pill that I can take and suddenly understand all this stuff? ;o) I'll give this another day or so, and if nobody else responds then I'll accept your comment as an answer. Already Hirschhorn's introduction (as well as his introductions to the various Parts and Chapters) has been quite helpful. – Aaron Mazel-Gee Sep 14 at 22:08 show 1 more comment

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http://unapologetic.wordpress.com/2009/11/02/parallelepipeds-and-volumes-i/?like=1&source=post_flair&_wpnonce=8149a33672

# The Unapologetic Mathematician ## Parallelepipeds and Volumes I And we’re back with more of what Mr. Martinez of Harvard’s Medical School assures me is onanism of the highest caliber. I’m sure he, too, blames me for not curing cancer. Coming up in our study of calculus in higher dimensions we’ll need to understand parallelepipeds, and in particular their volumes. First of all, what is a parallelepiped? Or, more specifically, what is a $k$-dimensional parallelepiped in $n$-dimensional space? It’s a collection of points in space that we can describe as follows. Take a point $p$ and $k$ vectors $\left\{v_i\right\}_{i=1}^k$ in $\mathbb{R}^n$. The parallelepiped is the collection of points reachable by moving from $p$ by some fraction of each of the vectors $v_i$. That is, we pick $k$ values $t^i$, each in the interval $\left[0,1\right]$, and use them to specify the point $p+t^iv_i$. The collection of all such points is the parallelepiped with corner $p$ and sides $v_i$. One possible objection is that these sides may not be linearly independent. If the sides are linearly independent, then they span a $k$-dimensional subspace of the ambient space, justifying our calling it $k$-dimensional. But if they’re not, then the subspace they span has a lower dimension. We’ll deal with this by calling such a parallelepiped “degenerate”, and the nice ones with linearly independent sides “nondegenerate”. Trust me, things will be more elegant in the long run if we just deal with them both on the same footing. Now we want to consider the volume of a parallelepiped. The first observation is that the volume doesn’t depend on the corner point $p$. Indeed, we should be able to slide the corner around to any point in space as long as we bring the same displacement vectors along with us. So the volume should be a function only of the sides. The second observation is that as a function of the sides, the volume function should commute with scalar multiplication in each variable separately. That is, if we multiply $v_i$ by a non-negative factor of $\lambda$, then we multiply the whole volume of the parallelepiped by $\lambda$ as well. But what about negative scaling factors? What if we reflect the side (and thus the whole parallelepiped) to point the other way? One answer might be that we get the same volume, but it’s going to be easier (and again more elegant) if we say that the new parallelepiped has the negative of the original one’s volume. Negative volume? What could that mean? Well, we’re going to move away from the usual notion of volume just a little. Instead, we’re going to think of “signed” volume, which includes the possibility of being positive or negative. By itself, this sign will be less than clear at first, but we’ll get a better understanding as we go. As a first step we’ll say that two parallelepipeds related by a reflection have opposite signs. This won’t only cover the above behavior under scaling sides, but also what happens when we exchange the order of two sides. For example, the parallelogram with sides $v_1=a$ and $v_2=b$ and the parallelogram with sides $v_1=b$ and $v_2=a$ have the same areas with opposite signs. Similarly, swapping the order of two sides in a given parallelepiped will flip its sign. The third observation is that the volume function should be additive in each variable. One way to see this is that the $k$-dimensional volume of the parallelepiped with sides $v_1$ through $v_k$ should be the product of the $k-1$-dimensional volume of the parallelepiped with sides $v_1$ through $v_{k-1}$ and the length of the component of $v_k$ perpendicular to all the other sides, and this length is a linear function of $v_k$. Since there’s nothing special here about the last side, we could repeat the argument with the other sides. The other way to see this fact is to consider the following diagram, helpfully supplied by Kate from over at f(t): The side of one parallelogram is the (vector) sum of the sides of the other two, and we can see that the area of the one parallelogram is the sum of the areas of the other two. This justifies the assertion that for parallelograms in the plane, the area is additive as a function of one side (and, similarly, of the other). Similar diagrams should be apparent to justify the assertion for higher-dimensional parallelepipeds in higher-dimensional spaces. Putting all these together, we find that the $k$-dimensional volume of a parallelepiped with $k$ sides is an alternating multilinear functional, with the $k$ sides as variables, and so it lives somewhere in the exterior algebra $\Lambda(V^*)$. We’ll have to work out which particular functional gives us a good notion of volume as we continue. ### Like this: Posted by John Armstrong | Analytic Geometry, Geometry ## 5 Comments » 1. [...] and Volumes II Yesterday we established that the -dimensional volume of a parallelepiped with sides should be an [...] Pingback by | November 3, 2009 | Reply 2. [...] An Example of a Parallelogram Today I want to run through an example of how we use our new tools to read geometric information out of a parallelogram. [...] Pingback by | November 5, 2009 | Reply 3. [...] Now that we’ve used exterior algebras to come to terms with parallelepipeds and their transformations, let’s come back to apply these ideas to the [...] Pingback by | November 11, 2009 | Reply 4. [...] to come back down again, covering the same region twice with opposite signs. This is related to the signed volumes we talked about, where (in one dimension) an interval can be traversed (integrated over) from left [...] Pingback by | January 5, 2010 | Reply 5. [...] because of this correspondence. In fact, it’s not hard to see that they’re related to signed volumes. This is the starting point from which all integration on manifolds emerges, and everything will [...] Pingback by | August 2, 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.

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http://mathoverflow.net/questions/82058/is-sln-mathbbz-a-cat0-group/82065

## Is $SL(n,\mathbb{Z})$ a CAT(0) group? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is it possible to find a CAT(0) space on which the matrix group $SL(n,\mathbb{Z})$ acts properly discontinuously and cocompactly? Note: when the cocompactness is dropped , it is possible. - ## 1 Answer If n=2, yes: it acts on its Bass-Serre tree. If n>2, no: your group contains distorted elements, i.e. elements conjugated to a proper power of themselves (look at unipotent matrices). Such an element will have zero displacement length, which is impossible for an infinite order element in a group acting discretely cocompactly. For this and much more, see the monograph of Bridson and Haefliger. For even more restrictions on SL_n actions, see Theorem 1.14 in: Caprace-Monod, Isometry groups of non-positively curved spaces: structure theory Journal of Topology 2 No. 4 (2009), 661–700 -

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http://mathoverflow.net/revisions/64064/list

Return to Question 2 added 231 characters in body A $G$-structure $\pi : B_G \rightarrow M$ is said to be of $finite$ $type$ if $\mathfrak{g}^{(k)} = 0$ for some $k \in \mathbb{N}$, where $\mathfrak{g}^{(k)}$ denotes the $k$th prolongation of the Lie algebra $\mathfrak{g} = T_{e}G$. For finite type $G$-structures, let us call the first $k$ for which $\mathfrak{g}^{(k)} = 0$ the $order$ of the $G$-structure. For example, : • $O(n)$-structures (Riemannian metrics) are of finite type and order $1$, because $\mathfrak{o} (n)^{(1)} = 0$, but . • But $Sp(n)$-structures (symplectic structures) are not of finite type because the group of symplectomorphisms is infinite dimensional. • It can also be shown that $CO(n)$-structures (conformal structures) are of finite type and order $2$ (except if the dimension is $n > 2$)2\$, in which case it is not of finite type). Are there any finite type $G$-structures of order greater than $2$? More generally, are there $G$-structures of any order? Thanks. 1 $G$-structures of finite type. A $G$-structure $\pi : B_G \rightarrow M$ is said to be of $finite$ $type$ if $\mathfrak{g}^{(k)} = 0$ for some $k \in \mathbb{N}$, where $\mathfrak{g}^{(k)}$ denotes the $k$th prolongation of the Lie algebra $\mathfrak{g} = T_{e}G$. For finite type $G$-structures, let us call the first $k$ for which $\mathfrak{g}^{(k)} = 0$ the $order$ of the $G$-structure. For example, $O(n)$-structures (Riemannian metrics) are of finite type and order $1$, because $\mathfrak{o} (n)^{(1)} = 0$, but $Sp(n)$-structures are not of finite type. It can also be shown that $CO(n)$-structures are of finite type and order $2$ (if $n > 2$). Are there any $G$-structures of order greater than $2$? More generally, are there $G$-structures of any order? Thanks.

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http://www.physicsforums.com/showthread.php?t=246195

Physics Forums Thread Closed Page 1 of 2 1 2 > ## Axiom of Choice Can someone give me a list of problems which at first sight require the axiom of choice, but do not? 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 Quote by Dragonfall Can someone give me a list of problems which at first sight require the axiom of choice, but do not? Here are two problems. One will fit on your list, the other won't. Preliminaries: R denotes set of real numbers N denotes set of natural numbers 1) f:R->N arbitrary surjection. Show there exists an injection g:N->R s.t. f(g(b)) = b for every b in N. 2) f:R->R arbitrary surjection, continuous and non-decreasing. Show there exists an injection g:R->R s.t. f(g(b)) = b for every b in R. Well for 2, since continuous functions map compact sets to compact sets, and the intervals where f is constant are compact, g(x) can be defined as the minimal element of f inverse of x. 1 is not possible without AC. ## Axiom of Choice Here's one where 'experience' may tell you to use the axiom of choice. Suppose that S and T are infinite sets. Let f:S-->T be a surjection, and let g:T-->S be another surjection. Write down a bijection between S and T. That requires the axiom of choice. Now change the word surjection to injection. Does the proof require AC, now? I think this is quite interesting, from wikipedia, about a result of Banach & Tarski which they believed required the axiom of choice. Vitali's and Hausdorff's constructions depend on Zermelo's axiom of choice ("AC"), which is also crucial to the Banach–Tarski paper, both for proving their paradox and for the proof of another result: Two Euclidean polygons, one of which strictly contains the other, are not equidecomposable. They remark: Le rôle que joue cet axiome dans nos raisonnements nous semble mériter l'attention (The role this axiom plays in our reasoning seems, to us, to deserve attention) and point out that while the second result fully agrees with our geometric intuition, its proof uses AC in even more substantial way than the proof of the paradox. Thus Banach and Tarski imply that AC should not be rejected simply because it produces a paradoxical decomposition. Indeed, such an argument would also reject some geometrically intuitive statements! Ironically, in 1949 A.P.Morse showed that the statement about Euclidean polygons can be proved in ZF set theory and thus does not require the axiom of choice. Quote by n_bourbaki Now change the word surjection to injection. Does the proof require AC, now? Nope, since you can define h:T->S by $h(t)=f^{-1}(t)$ if $t\in f(S)$ and [/itex]h(t)=g(t)[/itex] otherwise. 1) f is invertible is it? 2) How do you know that h is an injection? In fact your h *cannot* be an injection, unless f is already a bijection (but you assumed that by writing down it's inverse). If we assume that you're taking a one sided inverse, then the map from h from f(S) to S is surjective, thus if there is any element in t in T\f(S), then necessarily g(t)=h(t') for some t' in f(S). You certainly don't need the axiom of choice, though not for the reasons you wrote. I thought you might like an example of two superficially similar statements one using and one not using the axiom of choice. I seem to recall Conway having a similar thing for 'dividing a set into 3'. Quote by Dragonfall Well for 2, since continuous functions map compact sets to compact sets, and the intervals where f is constant are compact, g(x) can be defined as the minimal element of f inverse of x. 1 is not possible without AC. If by "the minimal element of f inverse of x" you mean "the min of the pre-image of the unit set {x} under f", then OK. Of course, I don't what else you could have possibly meant. "1 is not possible without AC." Fine. Can't say I'd disagree with you. With a slight generalization it can be shown to be equivalent to AC. Quote by n_bourbaki Here's one where 'experience' may tell you to use the axiom of choice. Suppose that S and T are infinite sets. Let f:S-->T be a surjection, and let g:T-->S be another surjection. Write down a bijection between S and T. That requires the axiom of choice. Now change the word surjection to injection. Does the proof require AC, now? If I understand you correctly, we now have injections, and you ask for a bijection? Schroder-Bernstein. The classical proof of this theorem is a constructive-existence proof (no AC). But contrary to what was suggested in another post, the construction of the bijection is non-trivial (my opinion). Here's one you might think about: f:A->B arbitrary injection (A,B arbitrary sets). Show there exists a surjection g:B->A s.t. g(f(a)) = a for every a in A. Can we get by without AC? Recognitions: Homework Help Science Advisor There exists a subset of $\mathbb{R}$ which is not Lebesgue measurable. Quote by fopc If I understand you correctly, we now have injections, and you ask for a bijection? Yes. Schroder-Bernstein. The classical proof of this theorem is a constructive-existence proof (no AC). Yes, and the result for surjections does (I believe) need the axiom of choice. This is why I included it. Quote by morphism There exists a subset of $\mathbb{R}$ which is not Lebesgue measurable. I always thought that did require the axiom of choice, something to do with viewing R as a vector space over Q? What's the non AC method? Quote by fopc Schroder-Bernstein Ah yes, now I remember. Back when I took set theory I wrote that "if there are injections from S to T and from T to S, then S and T are in bijection" is "obvious" in an exercise. I lost many, many points. Quote by n_bourbaki I always thought that did require the axiom of choice, something to do with viewing R as a vector space over Q? What's the non AC method? Actually, that does require the Axiom of Choice (or something similar). Maybe he meant that there exists a subset of R which is not Borel measurable. That doesn't require AC. Here is the Conway paper I was thinking of. http://citeseer.ist.psu.edu/cache/pa...n-by-three.pdf it also gives a discussion of what is entailed in avoiding AC, and why one might wish to do it without just going 'ugh, it's false'. Recognitions: Homework Help Science Advisor Quote by gel Actually, that does require the Axiom of Choice (or something similar). Maybe he meant that there exists a subset of R which is not Borel measurable. That doesn't require AC. No, I meant Lebesgue measurable. Apparently all you need to construct such a set is the Hahn-Banach theorem (which is strictly weaker than choice); see this paper by Foreman and Wehrung. Quote by morphism No, I meant Lebesgue measurable. Apparently all you need to construct such a set is the Hahn-Banach theorem (which is strictly weaker than choice); see this paper by Foreman and Wehrung. ok, well that shows that it isn't equivalent to AC, but I assumed that the OP wanted things that can be proved using standard ZF axioms. Otherwise you could just say the Hahn-Banach theorem doesn't require AC as another example. In the wikipedia link I posted above, it mentions that the "ultrafilter lemma" is enough to prove the Banach-Tarski paradoxical decomposition, which would also give non Lebesgue measurable sets (which I why I added the disclaimer ...or something similar... to my prev post). Thread Closed Page 1 of 2 1 2 > Thread Tools | | | | |--------------------------------------|--------------------------------------------|---------| | Similar Threads for: Axiom of Choice | | | | Thread | Forum | Replies | | | General Math | 9 | | | Set Theory, Logic, Probability, Statistics | 2 | | | Set Theory, Logic, Probability, Statistics | 3 | | | General Math | 9 | | | Set Theory, Logic, Probability, Statistics | 9 |

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http://math.stackexchange.com/questions/184301/is-there-a-non-contradictory-non-trivial-axiomatic-system-in-which-godels-theor/184307

# Is there a non-contradictory non-trivial axiomatic system in which Gödel's theorem is undecidable? Gödel's Incompleteness Theorem states that in any non-contradictory non-trivial axiomatic system there are certain statements or theorems whom cannot be proved in that system, For example in ZFC, assuming it is non-contradictory, there is the famed Continuum Hypothesis and Whitehead's problem. My question is, is there any such system (but different from ZFC) in which Gödel's Incompleteness Theorem itself is an unprovable statement? - 2 Are you sure that the Riemann Hypothesis is unprovable in ZFC (if ZFC is consistent)? You should write that up for publication -- you'll be famous! – Henning Makholm Aug 19 '12 at 14:51 This is not really about set theory as much as it is about logic and incompleteness. I retagged it appropriately. (And it's not very soft either) – Asaf Karagila Aug 19 '12 at 14:52 @HenningMakholm Ah yes you're right. Then it means that I probably read about another famous theorem that was proved unprovable within ZFC. And Asaf, yes, you're right, thanks, and I originally tagged it as soft as I did not expect a proof or anything very concrete, given the question asked. – andreas.vitikan Aug 19 '12 at 14:59 @andreas.vitikan: The most famous statements that are proved independent of ZFC are the Continuum Hypothesis and the Generalized Continuum Hypothesis. Since those have "hypothesis" in their names, they might be what you're thinking about. – Henning Makholm Aug 19 '12 at 15:01 @Henning Ah yes, that was it, thanks for spotting it! – andreas.vitikan Aug 19 '12 at 15:03 ## 2 Answers Robinson arithmetic is non-trivial enough that the incompleteness theorem applies to it, but as far as I know not strong enough to prove the incompleteness theorem itself. - Just a footnote to Henning Makholm's answer. You indeed need to look at very weak systems for examples of the phenomenon you want. For a start, as soon as you add a little bit of induction to Robinson Arithmetic (officially, induction for $\Sigma_1$ wffs) you get a theory that, for any recursively axiomatized $T$ which includes a bit of arithmetic and any sensible system of Gödel-coding, can show that $Con_T \to \neg Prov_T(\overline{\ulcorner G_T\urcorner)}$, i.e. it can prove the formalized version of (half of) Gödel's incompleteness theorem for $T$. (It was rather important to Gödel that the incompleteness theorem is elementary in this sort of way.) -

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http://mathhelpforum.com/algebra/15895-algebra-polynomial.html

Thread: 1. Algebra polynomial The degree three polynomial f(x) with real coefficients and leading coefficient 1, has 4 and 3 + i among its roots. Express f(x) as a product of linear and quadratic polynomials with real coefficients. choices f(x) = (x+4)(x^2+6x+10) f(x) = (x-4)(x^2-6x-9 f(x) = (x-4)x^2-6x+10) f(x) = (x-4)(x^2-6x+9) 2. Originally Posted by wvmcanelly@cableone.net The degree three polynomial f(x) with real coefficients and leading coefficient 1, has 4 and 3 + i among its roots. Express f(x) as a product of linear and quadratic polynomials with real coefficients. All such non-zero polynomials difference by a constant multiple. So we will find one which has 1 as its leading coefficient (called monic). Now if f(x) is a polynomial in R[x] (meaning the real numbers) and if a+bi is a zero then a-bi is also a zero. So if 3+i is a zero that means 3-i is a zero. Thus, $f(x) = (x-4)(x-(3+i))(x-(3-i))$ Multiply the last two factors together, $f(x)=(x-4)(x^2 - 6x + 10)$ 3. thanks, good not have figured that one out without you.

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http://mathoverflow.net/questions/23770/why-are-normal-crossing-divisors-nice/23776

## Why are normal crossing divisors nice? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This question is going to be extremely vague. It seems that wherever I go (especially about Grothendieck's circle of ideas) the higher-dimensional analogue of a curve minus a finite number of points is a scheme minus a normal crossing divisor. Why is that? What's so special about a normal crossing divisor that it simulates a curve minus a finite number of points better? - 1 A normal crossing divisor looks like a bunch of coordinate planes, and has monomial equations. What could be better?! :) – Mariano Suárez-Alvarez Oct 8 2010 at 0:44 ## 2 Answers It mostly has to do with finding nice compactifications. Compactifications of varieties are a good thing as they allow us to control what happens at "infinity". If the variety itself is smooth it seems a good idea (and it is!) to demand that the compactification also be smooth. However, you need the situation to be nice at infinity in order to make the study of asymptotic behaviour at infinity to be as easy as possible. The best behaviour at infinity would be if the complement were smooth but that is in general not possible. What is always possible is to demand that the complement be a divisor with normal crossings. In practice it works essentially as well as having a smooth complement: You have a bunch of smooth varieties intersecting in as nice a manner as possible. - "What is always possible is to demand that the complement be a divisor with normal crossings." - can you elaborate on that? – Makhalan Duff May 6 2010 at 20:03 13 Hironaka proved that a smooth variety can always be realized as an open subvariety of a smooth projective variety in such a way that the boundary is a divisor with normal crossings. – JS Milne May 6 2010 at 20:43 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. A related point is that if say $X\subset Y$ and you want an embedded resolution of $X$, then you can of course ask that you want a birational morphism $\pi:Z\to Y$ such that the strict transform of $X$ is smooth, but a better thing to ask is the the entire pre-image of $X$ is as nice as possible. Unfortunately (in general) you cannot make the preimage of $X$ smooth as the exceptional set will add additional components and where they meet is going to be a singular point. So, you can ask for the next best thing: normal crossings. You could even say that normal crossings is the reducible analogue of smooth. Anyway, this is the result of Hironaka, JS Milne referred to above: for any $X\subset Y$ (plus some reasonable assumptions) there exists a projective birational $\pi$ such that $Z$ is smooth and $\pi^{-1}X$ is a normal crossing divisor. If $Y\setminus X$ is smooth, then you may even require that $\pi$ is an isomorphism outside $X$. The compactification result is a simple consequence of this: if $U$ is open (say quasi-projective), pick a projective compactification $Y$ and let $X=Y\setminus U$. Perform Hironaka's embedded resolution of singularities and you get $U\subset Z$ with the complement being a normal crossing divisor. One, slightly independent word on normal crossings. There seems to be some confusion in the literature about what normal crossings mean. Or rather what the difference is between normal crossings and simple normal crossings. Well, the point is that (nowadays) the latter is understood in the Zariski topology while the former in the analytic or formal topology. In other words, simple normal crossings mean that each irreducible component is smooth and they meet transversally, while normal crossings allows for a component to meet itself transversally. In particular, a nodal curve has normal crossings but not simple normal crossings. In the above discussion and in the other answers before this one, you can always put simple normal crossings in place of normal crossings and the statements remain true. It is possible that back when Hironaka proved his famous theorem, this distinction had not been made so in older texts the meaning might be different. At the same time, according to Miles Reid, it was the Japanese who invented the term simple normal crossings. -

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http://mathhelpforum.com/calculus/179105-trig-substitution-print.html

# Trig substitution Printable View • April 30th 2011, 09:05 PM Hardwork Trig substitution How can we find the integral $\int_{0}^{\pi}\cos{x}\sin{x}\,{dx}$ with the substitution $u = sinx$? I know how to find the integral, but the limits become the same sin(0) = sin(pi) = 0. What should I do in general when a given substitution makes upper bound = lower bound? • April 30th 2011, 09:21 PM TheChaz Quote: Originally Posted by Hardwork How can we find the integral $\int_{0}^{\pi}\cos{x}\sin{x}\,{dx}$ with the substitution $u = sinx$? I know how to find the integral, but the limits become the same sin(0) = sin(pi) = 0. What should I do in general when a given substitution makes upper bound = lower bound? Using a different property, you can analyse the integrand as a function with period "pi", the (signed) area of which equals zero after each period. • April 30th 2011, 09:24 PM chisigma Quote: Originally Posted by Hardwork How can we find the integral $\int_{0}^{\pi}\cos{x}\sin{x}\,{dx}$ with the substitution $u = sinx$? I know how to find the integral, but the limits become the same sin(0) = sin(pi) = 0. What should I do in general when a given substitution makes upper bound = lower bound? Is... http://quicklatex.com/cache3/ql_0a43...4bcb647_l3.png http://quicklatex.com/cache3/ql_5801...1eb2dab_l3.png ... so that, no matter which 'substitution' You do, the integral is 0... Kind regards $\chi$ $\sigma$ • May 2nd 2011, 01:32 AM Hardwork Okay. I do see how the integral is zero. I just want to know what to do in general when a substitution makes the limits the same. All times are GMT -8. The time now is 05:54 PM.

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http://mathoverflow.net/questions/90522?sort=newest

## Is there a really big ring of differential operators in characteristic p? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) $k$ is a field of characteristic $p$. $k[t]$ has canonical first-order differential operator $\partial$ As an endomorphism of $k[t]$, $\partial^p=0$. First way to fix it: Use the divided power differential operator $\partial^{[p]}$. Shortfall: As an endomorphism of $k[t]$, ${\partial^{[p]}}^p=0$ Second way to fix it: Use crystalline differential operators. Shortfall: No higher order operators on $k[t]$. Question: Is there a really big ring of differential operators which contains the divided powers $\partial^[n]$ for all $n$ and which also has a natural evaluation map to $End_k(k[t])$? - 2 There is the ring of diferential operators à la Grothendieck, which is infinite dimensional and has elements of all orders. It is an increasing union of matrix rings in the case of the line, though, so it is a bit strange. – Mariano Suárez-Alvarez Mar 8 2012 at 0:28 The comment of Mariano and resulting answer by Lars are good. However, you may be interested to look here: mathoverflow.net/questions/56860/… – B. Bischof Mar 8 2012 at 18:10 ## 1 Answer Let me give some more details on Mariano's comment: The ring of differential operators a la EGA4 in this particular case will be a free $k[t]$-algebra generated by the following operators: We write $$\partial_t^{(n)}$$ for the operator which is defined by $$\partial_t^{(n)}(t^m)={m\choose n}t^{m-n}.$$ Because of this, sometimes the notation $$\partial_t^{(n)}=\frac{1}{n!}\frac{\partial^n}{\partial t^n}$$ is used. Actually, to generate the ring, the operators $\partial_t^{(p^n)}$ suffice. Now this ring is not noetherian, but it is an increasing union of noetherian subalgebras, lets denote them by $D^{(m)}$, which are the subalgebras generated by operators of degree $\leq p^m$. Using partially divided powers, Berthelot abstractly defines rings $\mathcal{D}^{(m)}$ such that the full ring of differential operators $\mathcal{D}$ is the direct limit of the $\mathcal{D}^{(m)}$. The image of $\mathcal{D}^{(m)}$ in $\mathcal{D}$ is then precisely the $D^{(m)}$ that I defined ad-hoc above. The crystalline operators that you defined in the question correspond to Berthelot's $\mathcal{D}^{(0)}$. - YoungMathematician asks (in a deleted answer) whether ${\partial^{[p]}}^p=0$ holds in Berthelot's ring. – S. Carnahan♦ Mar 9 2012 at 7:44 Ah, I wonder which of Bertholot's rings he means, and what precisely is $\delta^{[p]}$. In general, if $\delta$ is an operator of order 1, then $\delta^p=0$ in `$\mathcal{D}^{(n)}$` for $n>1$, but not necessarily `$\mathcal{D}^{(0)}$`. Modules over`$\mathcal{D}^{(0)}$` are connections. Now if by $\delta^{[p]}$` you mean what I wrote as `$\delta^{(p}_t$,then that′s an operator of order $p$, and cannot be considered as an elementof `$\mathcal{D}^{(0)}$`, but if I remember correctly, the same reasoning applies. – Lars Mar 9 2012 at 14:13

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http://mathforum.org/mathimages/index.php?title=Fibonacci_Numbers&oldid=13375

Fibonacci Numbers From Math Images Revision as of 10:21, 28 June 2010 by Iris (Talk | contribs) Fibonacci Spiral Field: Number Theory Created By: [[Author:| ]] Website: [ ] Fibonacci Spiral The spiral curve of the Nautilus sea shell follows the pattern of the spiral drawn in a Fibonacci rectangle, a collection of squares with sides that have the length of Fibonacci numbers. Basic Description The Fibonacci sequence is the sequence $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots,$ where the first two numbers are 1s and every later number is the sum of the two previous numbers. So, given two $1$'s as the first two terms, the next terms of the sequence follows as : $1+1=2, 1+2=3, 2+3=5, 3+5=8, \dots$ Image 1 The Fibonacci numbers can be discovered in nature, such as the spiral of the Nautilus sea shell, the petals of the flowers, the seed head of a sunflower, and many other parts of the nature. The seeds at the head of the sunflower, for instance, are arranged so that one can find a collection of spirals in both clockwise and counterclockwise ways. The number of spirals differs depending on whether one counts in a clockwise or a counterclockwise way because different patterns of spirals are formed depending on the counting direction, as shown by Image 1. The two numbers of spirals are always consecutive numbers in the Fibonacci sequence. Nature prefers this way of arranging the seeds because it seems to allow the seeds to be uniformly distributed. For more information about Fibonacci patterns in nature, see Fibonacci Numbers in Nature Origin The Fibonacci sequence was studied by Leonardo of Pisa, or Fibonacci (1770-1240). In his work Liber Abacci, he introduced a problem involving the growth of the rabbit population. The assumptions were • there is one pair of baby rabbits placed in an enclosed place on the first day of January • this pair will grow for one month before reproducing and produce a new pair of baby rabbits on the first day of March • each new pair will mature for one month and produce a new pair of rabbits on the first day of their third month • the rabbits never die, so after they mature, the rabbits produce a new pair of baby rabbits every month. The problem was to find out how many pairs of rabbits there will be after one year. Image 2 On January 1st, there is only 1 pair. On February 1st, the baby rabbits matured to be grown up rabbits, but they have not reproduced, so there will only be the original pair present. Now look at any later month. June is a good example. As you can see in Image 2, all 5 pairs of rabbits that were alive in May continue to be alive in June. Furthermore, all 3 pairs of rabbits that were also alive on April 1st, which all became or were adult rabbit pairs on May 1st, reproduce, creating 3 new pairs of rabbits born in June. This means that on June 1st, there are 8 pairs of rabbits. This is equal to the 5 pairs from May 1st plus the 3 new pairs, which is the number of pairs from April 1st. This same reasoning can be applied to any month, March or later, so the number of rabbits pairs at a certain point is the same as the sum of the number of rabbit pairs in the two previous months. This is exactly the rule that defines the Fibonacci sequence. As you can see in the image, the population by month begins: 1, 1, 2, 3, 5, 8, ..., which is the same as the beginning of the Fibonacci sequence. The population continues to match the Fibonacci sequence no matter how many months out you go. An interesting fact is that this problem of rabbit population was not intended to explain the Fibonacci numbers. This problem was originally intended to introduce the Hindu-Arabic numerals to Western Europe, where people were still using Roman numerals, and to help people practice addition. It was coincidence that the number of rabbits followed a certain pattern which people later named as the Fibonacci sequence. Fibonacci Numbers in Nature Template:Hideshow A More Mathematical Explanation [Click to view A More Mathematical Explanation] Symbolic Definition of Fibonacci Sequence The Fibonacci sequence is the sequence UNIQ76e5473510 [...] [Click to hide A More Mathematical Explanation] Symbolic Definition of Fibonacci Sequence The Fibonacci sequence is the sequence $F_1, F_2, F_3, \ldots, F_n, \ldots$ where $F_n = F_{n-1} + F_{n-2} \quad \hbox{ for } n>2$, and $F_1 = 1,\ F_2 = 1$. The Fibonacci sequence is recursively definedA recursively defined sequence is one in which each term is defined by preceding terms in the sequence. For instance, $x_n=(x_{n-1})^2-3x_{n-2}$ is recursively defined. because each term is defined in terms of its two immediately preceding terms. Identities and Properties Idenitities There are some interesting identities, including formula for the sum of first $n$ Fibonacci numbers, the sum of Fibonacci numbers with odd indices and sum of Fibonacci number with even indices. Note that all the identities and properties in this section can be proven in a more rigorous way through mathematical induction. Sum of first $n$ Fibonacci numbers The sum of first$n$ Fibonacci numbers is one less than the value of the ${(n+2)}^{\rm th}$ Fibonacci number: $F_1+F_2+\dots+F_n=F_{n+2}-F_2=F_{n+2}-1$ For example, the sum of first $5$ Fibonacci number is : $F_1+F_2+F_3+F_4+F_5= 1 + 1 + 2 + 3 +5=F_7-1=12$ The example is demonstrated below. The total length of red bars that each correspond to $F_1, F_2, F_3, F_4, F_5$ is one unit less than the length of $F_7$. Image 11 To see the proof, click below [Click to see more.] $F_1=F_3-F_2$ $F_2=F_4-F_3$ $F_3=F_5-F_4$ UNIQ76e547351052 [...] $F_1=F_3-F_2$ $F_2=F_4-F_3$ $F_3=F_5-F_4$ $\dots$ $F_{n-1}=F_{n+1}-F_n$ $F_n=F_{n+2}-F_{n+1}$ Adding up all the equations, we get : $F_1+F_2+\dots+F_n=F_{n+2}-F_2$. Sum of Fibonacci numbers with odd indices The sum of first $n$ Fibonacci numbers with odd indices is equal to the ${(2n)}^{\rm th}$ Fibonacci number: $F_1+F_3+F_5+\dots+F_{2n-1}=F_{2n}$ For instance, the sum of first $4$ Fibonacci numbers with odd indices is: $F_1+F_3+F_5+F_7=1+2+5+13=21=F_8$ This example is shown below. Image 12 To see the proof, click below [Click to see more.] $F_1=F_2$ $F_3=F_4-F_2$ $F_5=F_6-F_4$ UNIQ76e547351052 [...] $F_1=F_2$ $F_3=F_4-F_2$ $F_5=F_6-F_4$ $\dots$ $F_{2n-1}=F_{2n}-F_{2n-2}$ Adding all the equations, we get : $F_1+F_3+F_5+\dots+F_{2n-1}=(F_2-F_2)+(F_4-F_4)+(F_6-F_6)+\dots+(F_{2n-2}-F_{2n-2})+F_{2n}$ $=F_{2n}$ Except for $F_{2n}$, all the terms on the right side of the equation disappear because each term is canceled out by another term that has the opposite sign and the same magnitude. Sum of Fibonacci numbers with even indices The sum of first $n$ Fibonacci numbers with even indices is one less than the ${(2n+1)}^{\rm th}$ Fibonacci number: $F_2+F_4+\dots+F_{2n}=F_{2n+1}-1$ For example, the sum of first $4$ Fibonacci numbers with even indices is : $F_2+F_4+F_6= 1+3+8=F_7-1=13-1=12$ This example is shown below. Image 13 To see the proof, click below [Click to see more.] Subtracting , the sum of Fibonacci numbers with odd indices [...] Subtracting Eq. (2), the sum of Fibonacci numbers with odd indices, from Eq. (1), the sum of the first $n$ Fibonacci numbers, we get the identity of the sum of Fibonacci numbers with even indices. Sum of the squares of Fibonacci numbers The sum of the squares of the first $n$ Fibonacci numbers is the product of the $n^{\rm th}$ and the ${(n+1)}^{\rm th}$ Fibonacci numbers. Image 14 $\sum_{i=1}^n {F_i}^2=F_n F_{n+1}$ This identity can be proved by studying the area of the rectangles in Image 14. To see the proof, click below [Click to see more.] The rectangle is called a Fibonacci rectangle, which is further described in [[Fibonacci_Numbers#Fibonacci_Numbers_in_Nature| Fibonacci Numbers in Na [...] The rectangle is called a Fibonacci rectangle, which is further described in Fibonacci Numbers in Nature. The numbers inside each square indicate the length of one side of the square. Notice that the lengths of the squares are all Fibonacci numbers. Any rectangle in the picture is composed of squares with lengths that are Fibonacci numbers. In fact, any rectangle is composed of every square with side lengths $F_1$ through $F_n$. Moreover, the dimension of this rectangle is $F_n$ by $F_{n+1}$. We can prove this identity by computing the area of the rectangle in two different ways. The first way of finding the area is to add the area of each squares. That is, the area of the rectangle will be : ${F_1}^2+{F_2}^2+{F_3}^2+\dots+{F_n}^2$. Another way of computing the area is by multiplying the width by the height. Using this method, the area will be : $F_n F_{n+1}$. Because we are computing the area of the same rectangle, the two methods should give the same results. Thus, ${F_1}^2+{F_2}^2+{F_3}^2+\dots+{F_n}^2=F_n F_{n+1}$. For example, for the red rectangle, the width is $5$ and the height is $8$. Since $5$ is the $5^{\rm th}$ Fibonacci number and $8$ is the $6^{\rm th}$ Fibonacci number, let $n=5$. The area of the rectangle is : $1^2+1^2+2^2+3^2+5^2={F_1}^2+{F_2}^2+{F_3}^2+{F_4}^2+{F_5}^2=\sum_{i=1}^5 {F_i}^2=40$, or $5 * 8 = F_5 F_{5+1} = F_5 F_6 = 40$. Thus, $\sum_{i=1}^5 {F_i}^2=F_5 F_{5+1}$. Properties Greatest Common Divisor The greatest common divisor of two Fibonacci numbers is the Fibonacci number whose index is the greatest common divisor of the indices of the original two Fibonacci numbers. In other words, $\gcd(F_n,F_m) = F_{\gcd(n,m)}$. For instance, $\gcd(F_9,F_6)=\gcd(34,8)=2=F_3=F_{\gcd(9,6)}$. In a special case where $F_n$ and $F_m$ are consecutive Fibonacci numbers, this property says that $\gcd(F_n, F_{n+1})=F_{\gcd(n,n+1)}=F_1=1$. That is, $F_n$ and $F_{n+1}$are always relatively prime. To see the proof for this special case, click below [Click to see more.] Assume that $F_n$ and $F_{n+1}$ have some integer UNIQ76e547351052273a-math-000 [...] Assume that $F_n$ and $F_{n+1}$ have some integer $k$ as their common divisor. Then, both $F_{n+1}$ and $F_n$ are each multiples of $k$: $F_{n+1}=ka$ $F_n=kb$ Subtracting Eq. (4) from Eq. (3), we get : $F_{n-1}=k(a-b)$, which means that if two consecutive Fibonacci numbers, $F_n$ and $F_{n+1}$, have $k$ as their common divisor, then the previous Fibonacci number, $F_{n-1}$ must also be a multiple of $k$. In that case, $F_{n-1}$ and $F_n$, which are also two consecutive Fibonacci numbers, will have $k$ as a common divisor. Then, it follows that $F_{n-2}$ must also be a multiple of $k$. Repeating the subtraction of consecutive Fibonacci numbers, we can conclude that the very first Fibonacci number, $F_1 = 1$ must also be a multiple of $k$. So $k=1$, and the only common divisor between two consecutive Fibonacci numbers is 1. Thus, two consecutive Fibonacci numbers are relatively prime. Finite Difference of Fibonacci Numbers One of the interesting properties of Fibonacci numbers is that the sequence of differences between consecutive Fibonacci numbers also forms a Fibonacci sequence, as shown in the table below. For more information about the difference table, click Difference Tables. Because the first sequence of differences of the Fibonacci sequence also includes a Fibonacci sequence, the second difference also includes a Fibonacci sequence. The Fibonacci sequence is thus reproduced in every sequence of differences. We can see that the sequence of differences is composed of Fibonacci numbers by looking at the definition of Fibonacci numbers : $F_n = F_{n-1} + F_{n-2}$. The difference between two consecutive Fibonacci numbers is : $F_n - F_{n-1} = F_{n-2}$. Thus, the difference between two consecutive Fibonacci numbers, $F_n$ and $F_{n-1}$, is equal to the value of the previous Fibonacci number, $F_{n-2}$. Golden Ratio Image 9 The golden ratio appears in paintings, architecture, and in various forms of nature. Two numbers are said to be in the golden ratio if the ratio of the smaller number to the larger number is equal to the ratio of the larger number to the sum of the two numbers. In Image 9, the width of A and B are in the golden ratio if$a : b = (a+b) : a$. The golden ratio is represented by the Greek lowercase phi ,$\varphi$, and the exact value is $\varphi=\frac{1 + \sqrt{5}}{2} \approx 1.61803\,39887\dots\,$ This value can be found from the definition of the golden ratio. To see an algebraic derivation of the exact value of the golden ratio, go to Golden Ratio : An Algebraic Representation. An interesting fact about golden ratio is that the ratio of two consecutive Fibonacci numbers approaches the golden ratio as the numbers get larger, as shown by the table below. $\frac{F_{n+1}}{F_n}$ $\frac{1}{1}$=1 $\frac{2}{1}$=2 $\frac{3}{2}$=1.5 $\frac{5}{3}$=1.66667 $\frac{8}{5}$=1.6 $\frac{13}{8}$=1.625 $\frac{21}{13}$=1.61538 $\frac{34}{21}$=1.61904 $\frac{55}{34}$=1.61765 $\frac{89}{55}$=1.61818 Lets assume that the ratio of two consecutive Fibonacci numbers have a limit and verify that this limit is, in fact, the golden ratio. Let $r_n$ denote the ratio of two consecutive Fibonacci numbers, that is, $r_n=\frac{F_{n+1}}{F_n}$. Then, $r_{n-1}=\frac{F_n}{F_{n-1}}$. $r_n$ and $r_{n-1}$ are related by : $r_n=\frac{F_{n+1}}{F_n}=\frac{F_n+F_{n-1}}{F_n}=1+\frac{F_{n-1}}{F_n}=1+\cfrac{1}{{F_n}/{F_{n-1}}}=1+\frac{1}{r_{n-1}}$. Assuming that the ratio $r_n$ has a limit, let $r$ be that limit: $\lim_{n \to \infty} r_n=\lim_{n \to \infty}\frac{F_{n+1}}{F_n}=r$. Then, $\lim_{n \to \infty} r_n = \lim_{n \to \infty} r_{n-1} = r$. Taking the limit of $r_n=1+\frac{1}{r_{n-1}}$ we get : $r=1+\frac{1}{r}$ Multiplying both sides by $r$, we get ${r}^2=r+1$ which can be written as: $r^2 - r - 1 = 0$. Applying the quadratic formula An equation, $\frac{-b \pm \sqrt {b^2-4ac}}{2a}$, which produces the solutions for equations of form $ax^2+bx+c=0$ , we get $r = \frac{1 \pm \sqrt{5}} {2}$. Because the ratio has to be a positive value, $r=\frac{1 + \sqrt{5}}{2}$ which is the golden ratio. Thus, if $r_n$ has a limit, then this limit is the golden ration. That is, as we go farther out in the sequence, the ratio of two consecutive Fibonacci numbers approaches the golden ratio. In fact, it can be proved that $r_n$ does have a limit; one way is to use Binet's formula in the next section. For a different proof using infinite continued fraction A continued fraction is a fraction in which the denominator is composed of a whole number and a fraction. An infinite continued fraction of the golden ratio has the form : $\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1} {1 + \ddots\,}}}}$go to Continued Fraction Representation and Fibonacci Sequences Image 10: Vitruvian man Many people find the golden ratio in various parts of nature, art, architecture, and even music. However, there are some people who criticize this viewpoint. They claim that many mathematicians are wishfully trying to make the connection between the golden ratio and other parts of the world even though there is no real connection. One example of the golden ratio that mathematicians found in nature is the human body. According to many, an ideal human body have proportions that show the golden ratio, such as: • distance between the foot and navel : distance between the navel and the head • distance between the finger tip and the elbow : distance between the wrist and the elbow • distance between the shoulder line and top of the head : length of the head. Leonardo da Vinci's drawing Vitruvian man shown in Image 10 emphasizes the proportion of human body. This drawing shows the proportions of an ideal human body that was studied by a Roman architect Vitruvius in his book De Architectura. In the drawing, a man is simultaneously inscribed in a circle and a square. The ratio of the square side to the radius of the circle in the drawing reflects the golden ratio, although the drawing deviates from the real value of the golden ratio by 1.7 percent. The proportions of the body of the man is also known to show the golden ratio. Although people later found the golden ratio in the painting, there is no evidence whether Leonardo da Vinci was trying to use the golden ratio in his painting or not. For more information about the golden ratio, go to Golden Ratio Binet's Formula for Fibonacci Numbers Binet's Formula gives a formula for the $n^{\rm th}$ Fibonacci number as : $F_n=\frac{{\varphi}^n-{\bar{\varphi}}^n}{\sqrt5}$, where $\varphi$ and $\bar{\varphi}$ are the two roots of Eq. (5), that is, $\varphi=\frac{1 + \sqrt{5}}{2},\quad \bar{\varphi}=\frac{1-\sqrt{5}}{2}$. Here is one way of verifying Binet's formula through mathematical induction, but it gives no clue about how to discover the formula. Let $F_n=\frac{{\varphi}^n-{\bar{\varphi}}^n}{\sqrt5}$ as defined above. We want to verify Binet's formula by showing that the definition of Fibonacci numbers holds true even when we use Binet's formula. First, we will show through inductive stepAn inductive step is one of the two parts of mathematical induction (base case and inductive step) where one shows that if a statement holds true for some $n$, then the statement also holds true for $n+1$ that: $F_n=F_{n-1}+F_{n-2}\quad\hbox{ for } n>2$ and then we will show the base caseA base case is one of the two parts of mathematical induction (base case and inductive step) where one shows that a statement holds true for the lowest value of $n$, usually $n=0$ or $n=1$, depending on situation. that: $F_1=1,\quad F_2=1$. First, according to Binet's fromula, $F_{n-1}+F_{n-2} = \frac{{\varphi}^{n-1}-{\bar{\varphi}}^{n-1}}{\sqrt5}+ \frac{{\varphi}^{n-2}-{\bar{\varphi}}^{n-2}}{\sqrt5}$ $=\frac{({\varphi}^{n-1}+{\varphi}^{n-2})-({\bar{\varphi}}^{n-1}+{\bar{\varphi}}^{n-2})}{\sqrt5}$ $=\frac{({\varphi}+1){\varphi}^{n-2}-(\bar{\varphi}+1){\bar{\varphi}}^{n-2}}{\sqrt5}$. Because $\varphi$ and $\bar{\varphi}$ are the two roots of Eq. (1), the above equation becomes : $F_{n-1}+F_{n-2}=\frac{{{\varphi}^2}{\varphi}^{n-2}-{{\bar{\varphi}}^2}{\bar{\varphi}}^{n-2}}{\sqrt5}$ $=\frac{{\varphi}^n-{\bar{\varphi}}^n}{\sqrt5}$ $=F_n$, as desired. Now, because $\varphi=\frac{1 + \sqrt{5}}{2},\quad \bar{\varphi}=\frac{1-\sqrt{5}}{2}$, $F_1=\frac{\varphi-\bar{\varphi}}{\sqrt5}=\frac{1}{\sqrt5}\left (\frac{1 + \sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}\right)=\frac{1}{\sqrt5} {\sqrt5} = 1$ $F_2=\frac{{\varphi}^2-{\bar{\varphi}}^2}{\sqrt5}=\frac{(\varphi+\bar{\varphi})(\varphi-\bar{\varphi})}{\sqrt5}=\frac{1*{\sqrt5}}{\sqrt5}=1$. Binet's formula thus is a correct formula of Fibonacci numbers. Fibonacci Numbers and Fractals Fibonacci Numbers and the Mandelbrot Set Image 15 The Mandelbrot set is a set of points in which the boundary forms a fractal. It is a set of all complex numbers $c$ for which the sequence :$z_{n+1}=(z_n)^2+c \quad \hbox{ for } n=0,1,2,\dots$ does not go to infinity, starting with $z_0=0$. For instance, $c=0$ is included in the Mandelbrot set because $z_1=(z_0)^2+c=0^2+0 = 0$ $z_2=(z_0)^2+c=0^2+0=0$ $\dots$ $z_n=0^2+0=0$ for any $n$. Thus, the sequence defined by $c=0$ is bounded and $0$ is included in the Mandelbrot set. On the other hand, when we test$c=1$, Image 16 $z_1=(z_0)^2+c=1$ $z_2=(z_1)^2+c=2$ $z_3=(z_2)^2+c=5$ $\dots$ The terms of this sequence will increase to infinity. Thus, $c=1$ is not included in the Mandelbrot set. Image 17 People have been drawn to study the Mandelbrot set because of its aesthetic beauty. It is surprising to many people how a simple formula like Eq. (6) can generate a complex structure of the Mandelbrot set. The Fibonacci sequence is related to the Mandelbrot set through the period of the main cardioid and some large primary bulbs. For each bulb, there are many antennas, and the largest antenna is called the main antenna. The number of spokes in the main antenna is the period of the bulb. The period of the main cardioid is considered to be 1. In Image 17, the main antenna has five spokes, including the one connecting the primary bulb and the junction point of the antenna. The period of this bulb is five. Now, we will consider the period of the largest primary bulbs that are attached to the main cardioid and are in between two larger bulbs. In Image 18, the largest bulb between the bulb of period 1 and the bulb of period 2 is the bulb of period 3, and this bulb was found by looking for the largest bulb on the periphery of the main cardioid. The largest bulb between the bulb of period 2 and period 3 is the bulb of period 5, and the one between bulb of period 3 and period 5 is the bulb of period 8. The sequence generated in this way proceeds as 1, 2, 3, 5, 8, 13, ..., following the pattern of Fibonacci sequence. Image 18 Teaching Materials There are currently no teaching materials for this page. Add teaching materials. References http://www.worldproutassembly.org/archives/2007/08/the_mathematica.html http://www.jimloy.com/algebra/fibo.htm http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibnat.html http://milan.milanovic.org/math/english/division/division.html http://dougerino.blogspot.com/2010/03/fibonacci-phi-and-kepler.html http://www.mathematicianspictures.com/FIBONACCI/Fibonacci.htm http://www.chabad.org/library/article_cdo/aid/463900/jewish/Deciphering-Natures-Code.htm http://hiddenlighthouse.wordpress.com/2010/04/07/nature-fibonacci/ http://en.wikipedia.org/wiki/Mandelbrot_set http://plus.maths.org/issue40/features/devaney/ } Future Directions for this Page Things to add(possible ideas for future) • Fibonacci numbers and Pascal's triangle • A helper page for recursively defined sequence • A section describing the Fibonacci numbers with negative subscripts. Things to 'not' add • A derivation of the exact value of the golden ratio. The derivation is redundant with the information in the golden ratio page. Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page. [[Category:]]

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http://www.haskell.org/haskellwiki/index.php?title=User:Michiexile/MATH198/Lecture_2&diff=30193&oldid=30183

# User:Michiexile/MATH198/Lecture 2 ### From HaskellWiki (Difference between revisions) | | | | | |----------|------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|----------|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------| | | | | | | Line 40: | | Line 40: | | | | We call a left cancellable arrow in a category a ''monomorphism''. | | We call a left cancellable arrow in a category a ''monomorphism''. | | | | | | | - | ====In concrete categories==== | + | ====In Set==== | | | | | | | | Left cancellability means that if, when we do first <math>g_1</math> and then <math>f</math> we get the same as when we do first <math>g_2</math> and then <math>f</math>, then we had equality already before we followed with <math>f</math>. | | Left cancellability means that if, when we do first <math>g_1</math> and then <math>f</math> we get the same as when we do first <math>g_2</math> and then <math>f</math>, then we had equality already before we followed with <math>f</math>. | | Line 49: | | Line 49: | | | | | | | | | ====Subobjects==== | | ====Subobjects==== | | | | + | | | | | + | Consider the subset <math>\{1,2\}\subset\{1,2,3\}</math>. This is the image of an accordingly chosen injective map from any 2-element set into <math>\{1,2,3\}</math>. Thus, if we want to translate the idea of a subset into categorical language, it is not enough talking about monomorphisms, though the fact that inclusion is an injection indicates that we are on the right track. | | | | + | | | | | + | The trouble that remains is that we do not want to view <math>\{1,2\}</math> when it occurs as the image of <math>\{1,2\}</math> as a different subset from <math>\{5,6\}</math> mapping to <math>\{1,2\}</math>. So we need some way of figuring out how to catch these situations and parry for them. | | | | + | | | | | + | We'll say that a morphism <math>f</math> ''factors through'' a morphism <math>g</math> if there is some morphism <math>h</math> such that <math>f=gh</math>. | | | | + | | | | | + | We can also talk about a morphism <math>f:A\to C</math> factoring through an ''object'' <math>B</math> by requiring the existence of morphisms <math>g:A\to B, h:B\to C</math> that compose to <math>f</math>. | | | | + | | | | | + | Now, we can form an equivalence relation on monomorphisms into an object <math>A</math>, by saying <math>f\sim g</math> if <math>f</math> factors through <math>g</math> and <math>g</math> factors through <math>f</math>. The arrows implied by the factoring are inverse to each other, and the source objects of equivalent arrows are isomorphic. | | | | + | | | | | + | Equipped with this equivalence relation, we define a ''subobject'' of an object <math>A</math> to be an equivalence class of monomorphisms. | | | | | | | | ===Epimorphisms=== | | ===Epimorphisms=== | | Line 58: | | Line 70: | | | | A right cancellable arrow in a category is an ''epimorphism''. | | A right cancellable arrow in a category is an ''epimorphism''. | | | | | | | - | ====In concrete categories==== | + | ====In Set==== | | | | | | | | For epimorphims the interpretation in set functions is that whatever <math>f</math> does, it doesn't hide any part of the things <math>g_1</math> and <math>g_2</math> do. So applying <math>f</math> first doesn't influence the total available scope <mamth>g_1</math> and <math>g_2</math> have. | | For epimorphims the interpretation in set functions is that whatever <math>f</math> does, it doesn't hide any part of the things <math>g_1</math> and <math>g_2</math> do. So applying <math>f</math> first doesn't influence the total available scope <mamth>g_1</math> and <math>g_2</math> have. | | | | | | | - | ===Terminal objects=== | + | ===More on factoring=== | | | | | | | - | ====Pointless sets and global constants==== | + | In Set, and in many other categories, any morphism can be expressed by a factorization of the form <math>f=ip</math> where <math>i</math> is a monomorphism and <math>p</math> is an epimorphism. For instance, in Set, we know that a function is surjective onto its image, which in turn is a subset of the domain, giving a factorization into an epimorphism - the projection onto the image - followed by a monomorphism - the inclusion of the image into the domain. | | | | | | | - | ===Initial objects=== | + | ---- | | | | + | | | | | + | Note that in Set, every morphisms that is both a mono and an epi is immediately an isomorphism. We shall see in the homework that the converse does not necessarily hold. | | | | + | | | | | + | ===Initial and Terminal objects=== | | | | + | | | | | + | An object <math>0</math> is ''initial'' if for every other object <math>C</math>, there is a unique morphism <math>0\to C</math>. Dually, an object <math>1</math> is ''terminal'' if there is a unique morphism <math>C\to 1</math>. | | | | + | | | | | + | First off, we note that the uniqueness above makes initial and terminal objects unique up to isomorphism whenever they exist: we shall perform the proof for one of the cases, the other is almost identical. | | | | + | | | | | + | '''Proposition''' Initial (terminal) objects are unique up to isomorphism. | | | | + | | | | | + | '''Proof''': Suppose <math>C</math> and <math>C'</math> are both initial (terminal). Then there is a unique arrow <math>C\to C'</math> and a unique arrow <math>C'\to C</math>. The compositions of these arrows are all endoarrows of one or the other. Since ''all'' arrows from (to) an initial (terminal) objects are unique, these compositions have to be the identity arrows. Hence the arrows we found between the two objects are isomorphisms. QED. | | | | + | | | | | + | * In Sets, the empty set is initial, and any singleton set is terminal. | | | | + | * In the category of Vector spaces, the single element vector space 0 is both initial and terminal. | | | | | | | | ===Zero objects=== | | ===Zero objects=== | | | | | | | - | ===Internal and external hom=== | + | This last example is worth taking up in higher detail. We call an object in a category a ''zero object'' if it is simultaneously initial and terminal. | | | | | | | - | * Isomorphisms and existence of inverses. | | | | - | * Epi- and mono-morphisms and cancellability. | | | | - | ** Examples in concrete categories. | | | | - | ** Monomorphisms and subobjects: | | | | - | *** Factoring through. Equivalence relation by mutual factoring. | | | | - | *** Subobjects as equivalence classes of monomorphisms. | | | | - | ** Splitting and the existence of inverses. | | | | - | * Terminal and initial objects. | | | | - | ** Constants. Pointless sets. | | | | | | | | | | | | | | | | + | ====Pointless sets and generalized elements==== | | | | + | | | | | + | Arrows to initial objects and from terminal objects are interesting too - and as opposed to the arrows from initial and to the terminals, there is no guarantee for these arrows to be uniquely determined. Let us start with arrows <math>A\to 0</math> into initial objects. | | | | + | | | | | + | In the category of sets, such an arrow only exists if <math>A</math> is already the empty set. | | | | + | | | | | + | In the category of all monoids, with monoid homomorphisms, we have a zero object, so such an arrow is uniquely determined. | | | | + | | | | | + | For arrows <math>1\to A</math>, however, the situation is significantly more interesting. Let us start with the situation in Set. <math>1</math> is some singleton set, hence a function from <math>1</math> picks out one element as its image. Thus, at least in Set, we get an isomorphism of sets <math>A = Hom(1,A)</math>. | | | | + | | | | | + | As with so much else here, we build up a general definition by analogy to what we see happening in the category of sets. Thus, we shall say that a ''global element'', or a ''point'', or a ''constant'' of an object <math>A</math> in a category with terminal objects is a morphism <math>x:1\to A</math>. | | | | + | | | | | + | This allows us to talk about elements without requiring our objects to even be sets to begin with, and thus reduces everything to a matter of just morphisms. This approach is fruitful both in topology and in Haskell, and is sometimes called ''pointless''. | | | | + | | | | | + | The important point here is that we can replace ''function application'' <math>f(x)</math> by the already existing and studied ''function composition''. If a constant <math>x</math> is just a morphism <math>x:1\to A</math>, then the value <math>f(x)</math> is just the composition <math>f\circ x:1\to A\to B</math>. Note, also, that since <math>1</math> is terminal, it has exactly one point. | | | | + | | | | | + | In the idealized Haskell category, we have the same phenomenon for constants, but slightly disguised: a global constant is 0-ary function. Thus the type declaration | | | | + | <haskell> | | | | + | x :: a | | | | + | </haskell> | | | | + | can be understood as syntactic sugar for the type declaration | | | | + | <haskell> | | | | + | x :: () -> a | | | | + | </haskell> | | | | + | thus reducing everything to function types. | | | | + | | | | | + | ---- | | | | + | | | | | + | Similarly to the ''global'' elements, it may be useful to talk about ''variable elements'', by which we mean non-specified arrows <math>f:T\to A</math>. Allowing <math>T</math> to range over all objects, and <math>f</math> to range over all morphisms into <math>A</math>, we are able to recover some of the element-centered styles of arguments we are used to. We say that <math>f</math> is ''parametried over <math>T</math>''. | | | | + | | | | | + | Using this, it turns out that <math>f</math> is a monomorphism if for any variable elements <math>x,y:T\to A</math>, if <math>x\neq y</math> then <math>f\circ x\neq f\circ y</math>. | | | | + | | | | | + | ===Internal and external hom=== | | | | + | | | | | + | If <math>f:B\to C</math>, then <math>f</math> induces a set function <math>Hom(A,f):Hom(A,B)\to Hom(A,C)</math> through <math>Hom(A,f)(g) = f\circ g</math>. Similarly, it induces a set function <math>Hom(f,A):Hom(C,A)\to Hom(C,B)</math> through <math>Hom(f,A)(g) = g\circ f</math>. | | | | | | | | | + | Using this, we have an occasionally enlightening | | | | | | | | | + | '''Proposition''' An arrow <math>f:B\to C</math> is | | | | + | # a monomorphism if and only if <math>Hom(A,f)</math> is injective for every object <math>A</math>. | | | | + | # an epimorphism if and only if <math>Hom(f,A)</math> is injective for every object <math>A</math>. | | | | + | # a split monomorphism if and only if <math>Hom(f,A)</math> is surjective for every object <math>A</math>. | | | | + | # a split epimorphism if and only if <math>Hom(A,f)</math> is surjective for every object <math>A</math>. | | | | + | # an isomorphism if and only if any one of the following equivalent conditions hold: | | | | + | ## it is both a split epi and a mono. | | | | + | ## it is both an epi and a split mono. | | | | + | ## <math>Hom(A,f)</math> is bijective for every <math>A</math>. | | | | + | ## <math>Hom(f,A)</math> is bijective for every <math>A</math>. | | | | | | | | | + | ---- | | | | | | | - | ==Morphisms== | + | For any <math>A,B</math> in a category, the homset is a ''set'' of morphisms between the objects. For many categories, though, homsets may end up being objects of that category as well. | | - | The arrows of a category are called ''morphisms''. This is derived from ''homomorphisms''. | + | | | | | | | | - | Some arrows have special properties that make them extra helpful; and we'll name them: | + | As an example, the set of all linear maps between two fixed vector spaces is itself a vector space. | | | | | | | - | ;Endomorphism:A morphism with the same object as source and target. | + | Alternatively, the function type <hask>a -> b</hask> is an actual Haskell type, and captures the morphisms of the idealized Haskell category. | | - | ;Monomorphism:A morphism that is left-cancellable. Corresponds to injective functions. We say that ''f'' is a monomorphism if for any <math>g_1,g_2</math>, the equation <math>fg_1 = fg_2</math> implies <math>g_1=g_2</math>. In other words, with a concrete perspective, ''f'' doesn't introduce additional relations when applied. | + | | | - | ;Epimorphism:A morphism that is right-cancellable. Corresponds to surjective functions. We say that ''f'' is an epimorphism if for any <math>g_1,g_2</math>, the equation <math>g_1f = g_2f</math> implies <math>g_1=g_2</math>. | + | | | - | Note, by the way, that cancellability does not imply the existence of an inverse. Epi's and mono's that have inverses realizing their cancellability are called ''split''. | + | | | - | ;Isomorphism:A morphism is an isomorphism if it has an inverse. Split epi and split mono imply isomorphism. Specifically, <math>f:v\to w</math> is an isomorphism if there is a <math>g:w\to v</math> such that <math>fg=1_w</math> and <math>g=1_v</math>. | + | | | - | ;Automorphism:An automorphism is an endomorphism that is an isomorphism. | + | | | | | | | | - | ==Objects== | + | We shall return to this situation later, when we are better equipped to give a formal scaffolding to the idea of having elements in objects in a category act as morphisms. For now, we shall introduce the notations <math>[A\to B]</math> or <math>B^A</math> to denote the ''internal'' hom - where the morphisms between two objects live as an object of the category. This distinguishes <math>B^A</math> from <math>Hom(A,B)</math>. | | - | In a category, we use a different name for the vertices: ''objects''. This comes from the roots in describing concrete categories - thus while objects may be actual mathematical objects, but they may just as well be completely different. | + | | | | | | | | - | Just as with the morphisms, there are objects special enough to be named. An object ''v'' is | + | To gain a better understanding of the choice of notation, it is worth noting that <math>|Hom_{\textsf{Set}}(A,B)|=|B|^|A|</math>. | | - | ;''Initial'':if ''[v,w]'' has exactly one element for all other objects ''w''. | + | | | - | ;''Terminal'': if ''[w,v]'' has exactly one element for all other objects ''w''. | + | | | - | ;''A Zero object'': if it is both initial and terminal. | + | | | | | | | | - | All initial objects are isomorphic. If <math>i_1,i_2</math> are both initial, then there is exactly one map <math>i_1\to i_2</math> and exactly one map <math>i_2\to i_1</math>. The two possible compositions are maps <math>i_1\to i_1</math> and <math>i_2\to i_2</math>. However, the initiality condition holds even for the morphism set ''[v,v]'', so in these, the only existing morphism is <math>1_{i_1}</math> and <math>1_{i_2}</math> respectively. Hence, the compositions have to be this morphism, which proves the statement. | + | ===Homework=== | | | | | | | - | ==Dual category== | + | Passing mark requires at least 6 of 11. | | | | | | | - | The same proof carries over, word by word, to the terminal case. This is an illustration of a very commonly occurring phenomenon - dualization. | + | # Suppose <math>g,h</math> are two-sided inverses to <math>f</math>. Prove that <math>g=h</math>. | | | | + | # (requires some familiarity with analysis) There is a category with object <math>\mathbb R</math> (or even all smooth manifolds) and with morphisms smooth (infinitely differentiable) functions. Prove that being a bijection does not imply being an isomorphisms. Hint: What about <math>x\mapsto x^3?</math>. | | | | + | # (do this if you don't do 2) In the category of posets, with order-preserving maps as morphisms, show that not all bijective homomorphisms are isomorphisms. | | | | + | # Consider the partially ordered set <math>P</math> as a category. Prove: every arrow is both monic and epic. Is every arrow thus an isomorphism? | | | | + | # What are the terminal and initial objects in a poset? Give an example of a poset that has both, either and none. Give an example of a poset that has a zero object. | | | | + | # What are the terminal and initial objects in the category with objects graphs and morphisms graph homomorphisms? | | | | + | # Prove that if a category has one zero object, then all initial and all terminal objects are all isomorphic and all zero objects. | | | | + | # Prove that the composition of two monomorphisms is a monomorphism and that the composition of two epimorphisms is an epimorphism. If <math>g\circ f</math> is monic, do any of <math>g,f</math> have to be monic? If the composition is epic, do any of the factors have to be epic? | | | | + | # Verify that the equivalence relation used in defining subobjects really is an equivalence relation. Further verify that this fixes the motivating problem. | | | | + | # Describe a representative subcategory each of: | | | | + | #* The category of vectorspaces over the reals. | | | | + | #* The category of finite sets. | | | | + | #* The category formed by the poset of the integers <math>\mathbb Z</math> and the order relation <math>a\leq b</math> if <math>a|b</math>. | | | | + | # * An arrow <math>f:A\to A</math> in a category <math>C</math> is an ''idempotent'' if <math>f\circ f = f</math>. We say that <math>f</math> is a ''split idempotent'' if there is some <math>g:A\to B, h:B\to A</math> such that <math>h\circ g=f</math> and <math>g\circ h=1_B</math>. Show that in Set, <math>f</math> is idempotent if and only if its image equals its set of fixed points. Show that every idempotent in Set is split. Give an example of a category with a non-split idempotent. | ## Revision as of 20:01, 23 September 2009 IMPORTANT NOTE: THESE NOTES ARE STILL UNDER DEVELOPMENT. PLEASE WAIT UNTIL AFTER THE LECTURE WITH HANDING ANYTHING IN, OR TREATING THE NOTES AS READY TO READ. ## Contents ### 1 Morphisms and objects Some morphisms and some objects are special enough to garner special names that we will use regularly. In morphisms, the important properties are • cancellability - the categorical notion corresponding to properties we use when solving, e.g., equations over $\mathbb N$: $3x = 3y \Rightarrow x = y$ • existence of inverses - which is stronger than cancellability. If there are inverses around, this implies cancellability, by applying the inverse to remove the common factor. Cancellability, however, does not imply that inverses exist: we can cancel the 3 above, but this does not imply the existence of $1/3\in\mathbb N$. Thus, we'll talk about isomorphisms - which have two-sided inverses, monomorphisms and epimorphisms - which have cancellability properties, and split morphisms - which are mono's and epi's with correspodning one-sided inverses. We'll talk about what these concepts - defined in terms of equationsolving with arrows - apply to more familiar situations. And we'll talk about how the semantics of some of the more wellknown ideas in mathematics are captured by these notions. For objects, the properties are interesting in what happens to homsets with the special object as source or target. An empty homset is pretty boring, and a large homset is pretty boring. The real power, we find, is when all homsets with the specific source or target are singleton sets. This allows us to formulate the idea of a 0 in categorical terms, as well as capturing the roles of the empty set and of elements of sets - all using only arrows. ### 2 Isomorphisms An arrow $f:A\to B$ in a category C is an isomorphism if it has a twosided inverse g. In other words, we require the existence of a $g:B\to A$ such that fg = 1B and gf = 1A. #### 2.1 In concrete categories In a category of sets with structure with morphisms given by functions that respect the set structure, isomorphism are bijections respecting the structure. In the category of sets, the isomorphisms are bijections. #### 2.2 Representative subcategories Very many mathematical properties and invariants are interesting because they hold for objects regardless of how, exactly, the object is built. As an example, most set theoretical properties are concerned with how large the set is, but not what the elements really are. If all we care about are our objects up to isomorphisms, and how they relate to each other - we might as well restrict ourselves to one object for each isomorphism class of objects. Doing this, we get a representative subcategory: a subcategory such that every object of the supercategory is isomorphic to some object in the subcategory. #### 2.3 Groupoids A groupoid is a category where all morphisms are isomorphisms. The name originates in that a groupoid with one object is a bona fide group; so that groupoids are the closest equivalent, in one sense, of groups as categories. ### 3 Monomorphisms We say that an arrow f is left cancellable if for any arrows g1,g2 we can show $fg_1 = fg_2 \Rightarrow g_1=g_2$. In other words, it is left cancellable, if we can remove it from the far left of any equation involving arrows. We call a left cancellable arrow in a category a monomorphism. #### 3.1 In Set Left cancellability means that if, when we do first g1 and then f we get the same as when we do first g2 and then f, then we had equality already before we followed with f. In other words, when we work with functions on sets, f doesn't introduce relations that weren't already there. Anything non-equal before we apply f remains non-equal in the image. This, translated to formulae gives us the well-known form for injectivity: $x\neq y\Rightarrow f(x)\neq f(y)$ or moving out the negations, $f(x)=f(y) \Rightarrow x=y$. #### 3.2 Subobjects Consider the subset $\{1,2\}\subset\{1,2,3\}$. This is the image of an accordingly chosen injective map from any 2-element set into {1,2,3}. Thus, if we want to translate the idea of a subset into categorical language, it is not enough talking about monomorphisms, though the fact that inclusion is an injection indicates that we are on the right track. The trouble that remains is that we do not want to view {1,2} when it occurs as the image of {1,2} as a different subset from {5,6} mapping to {1,2}. So we need some way of figuring out how to catch these situations and parry for them. We'll say that a morphism f factors through a morphism g if there is some morphism h such that f = gh. We can also talk about a morphism $f:A\to C$ factoring through an object B by requiring the existence of morphisms $g:A\to B, h:B\to C$ that compose to f. Now, we can form an equivalence relation on monomorphisms into an object A, by saying f˜g if f factors through g and g factors through f. The arrows implied by the factoring are inverse to each other, and the source objects of equivalent arrows are isomorphic. Equipped with this equivalence relation, we define a subobject of an object A to be an equivalence class of monomorphisms. ### 4 Epimorphisms Right cancellability, by analogy, is the implication $g_1f = g_2f \Rightarrow g_1 = g_2$ The name, here comes from that we can remove the right cancellable f from the right of any equation it is involved in. A right cancellable arrow in a category is an epimorphism. #### 4.1 In Set For epimorphims the interpretation in set functions is that whatever f does, it doesn't hide any part of the things g1 and g2 do. So applying f first doesn't influence the total available scope <mamth>g_1</math> and g2 have. ### 5 More on factoring In Set, and in many other categories, any morphism can be expressed by a factorization of the form f = ip where i is a monomorphism and p is an epimorphism. For instance, in Set, we know that a function is surjective onto its image, which in turn is a subset of the domain, giving a factorization into an epimorphism - the projection onto the image - followed by a monomorphism - the inclusion of the image into the domain. Note that in Set, every morphisms that is both a mono and an epi is immediately an isomorphism. We shall see in the homework that the converse does not necessarily hold. ### 6 Initial and Terminal objects An object 0 is initial if for every other object C, there is a unique morphism $0\to C$. Dually, an object 1 is terminal if there is a unique morphism $C\to 1$. First off, we note that the uniqueness above makes initial and terminal objects unique up to isomorphism whenever they exist: we shall perform the proof for one of the cases, the other is almost identical. Proposition Initial (terminal) objects are unique up to isomorphism. Proof: Suppose C and C' are both initial (terminal). Then there is a unique arrow $C\to C'$ and a unique arrow $C'\to C$. The compositions of these arrows are all endoarrows of one or the other. Since all arrows from (to) an initial (terminal) objects are unique, these compositions have to be the identity arrows. Hence the arrows we found between the two objects are isomorphisms. QED. • In Sets, the empty set is initial, and any singleton set is terminal. • In the category of Vector spaces, the single element vector space 0 is both initial and terminal. ### 7 Zero objects This last example is worth taking up in higher detail. We call an object in a category a zero object if it is simultaneously initial and terminal. #### 7.1 Pointless sets and generalized elements Arrows to initial objects and from terminal objects are interesting too - and as opposed to the arrows from initial and to the terminals, there is no guarantee for these arrows to be uniquely determined. Let us start with arrows $A\to 0$ into initial objects. In the category of sets, such an arrow only exists if A is already the empty set. In the category of all monoids, with monoid homomorphisms, we have a zero object, so such an arrow is uniquely determined. For arrows $1\to A$, however, the situation is significantly more interesting. Let us start with the situation in Set. 1 is some singleton set, hence a function from 1 picks out one element as its image. Thus, at least in Set, we get an isomorphism of sets A = Hom(1,A). As with so much else here, we build up a general definition by analogy to what we see happening in the category of sets. Thus, we shall say that a global element, or a point, or a constant of an object A in a category with terminal objects is a morphism $x:1\to A$. This allows us to talk about elements without requiring our objects to even be sets to begin with, and thus reduces everything to a matter of just morphisms. This approach is fruitful both in topology and in Haskell, and is sometimes called pointless. The important point here is that we can replace function application f(x) by the already existing and studied function composition. If a constant x is just a morphism $x:1\to A$, then the value f(x) is just the composition $f\circ x:1\to A\to B$. Note, also, that since 1 is terminal, it has exactly one point. In the idealized Haskell category, we have the same phenomenon for constants, but slightly disguised: a global constant is 0-ary function. Thus the type declaration `x :: a` can be understood as syntactic sugar for the type declaration `x :: () -> a` thus reducing everything to function types. Similarly to the global elements, it may be useful to talk about variable elements, by which we mean non-specified arrows $f:T\to A$. Allowing T to range over all objects, and f to range over all morphisms into A, we are able to recover some of the element-centered styles of arguments we are used to. We say that f is parametried over T. Using this, it turns out that f is a monomorphism if for any variable elements $x,y:T\to A$, if $x\neq y$ then $f\circ x\neq f\circ y$. ### 8 Internal and external hom If $f:B\to C$, then f induces a set function $Hom(A,f):Hom(A,B)\to Hom(A,C)$ through $Hom(A,f)(g) = f\circ g$. Similarly, it induces a set function $Hom(f,A):Hom(C,A)\to Hom(C,B)$ through $Hom(f,A)(g) = g\circ f$. Using this, we have an occasionally enlightening Proposition An arrow $f:B\to C$ is 1. a monomorphism if and only if Hom(A,f) is injective for every object A. 2. an epimorphism if and only if Hom(f,A) is injective for every object A. 3. a split monomorphism if and only if Hom(f,A) is surjective for every object A. 4. a split epimorphism if and only if Hom(A,f) is surjective for every object A. 5. an isomorphism if and only if any one of the following equivalent conditions hold: 1. it is both a split epi and a mono. 2. it is both an epi and a split mono. 3. Hom(A,f) is bijective for every A. 4. Hom(f,A) is bijective for every A. For any A,B in a category, the homset is a set of morphisms between the objects. For many categories, though, homsets may end up being objects of that category as well. As an example, the set of all linear maps between two fixed vector spaces is itself a vector space. Alternatively, the function type a -> b is an actual Haskell type, and captures the morphisms of the idealized Haskell category. We shall return to this situation later, when we are better equipped to give a formal scaffolding to the idea of having elements in objects in a category act as morphisms. For now, we shall introduce the notations $[A\to B]$ or BA to denote the internal hom - where the morphisms between two objects live as an object of the category. This distinguishes BA from Hom(A,B). To gain a better understanding of the choice of notation, it is worth noting that Failed to parse (unknown function\textsf): |Hom_{\textsf{Set}}(A,B)|=|B|^|A| . ### 9 Homework Passing mark requires at least 6 of 11. 1. Suppose g,h are two-sided inverses to f. Prove that g = h. 2. (requires some familiarity with analysis) There is a category with object $\mathbb R$ (or even all smooth manifolds) and with morphisms smooth (infinitely differentiable) functions. Prove that being a bijection does not imply being an isomorphisms. Hint: What about $x\mapsto x^3?$. 3. (do this if you don't do 2) In the category of posets, with order-preserving maps as morphisms, show that not all bijective homomorphisms are isomorphisms. 4. Consider the partially ordered set P as a category. Prove: every arrow is both monic and epic. Is every arrow thus an isomorphism? 5. What are the terminal and initial objects in a poset? Give an example of a poset that has both, either and none. Give an example of a poset that has a zero object. 6. What are the terminal and initial objects in the category with objects graphs and morphisms graph homomorphisms? 7. Prove that if a category has one zero object, then all initial and all terminal objects are all isomorphic and all zero objects. 8. Prove that the composition of two monomorphisms is a monomorphism and that the composition of two epimorphisms is an epimorphism. If $g\circ f$ is monic, do any of g,f have to be monic? If the composition is epic, do any of the factors have to be epic? 9. Verify that the equivalence relation used in defining subobjects really is an equivalence relation. Further verify that this fixes the motivating problem. 10. Describe a representative subcategory each of: • The category of vectorspaces over the reals. • The category of finite sets. • The category formed by the poset of the integers $\mathbb Z$ and the order relation $a\leq b$ if a | b. 11. * An arrow $f:A\to A$ in a category C is an idempotent if $f\circ f = f$. We say that f is a split idempotent if there is some $g:A\to B, h:B\to A$ such that $h\circ g=f$ and $g\circ h=1_B$. Show that in Set, f is idempotent if and only if its image equals its set of fixed points. Show that every idempotent in Set is split. Give an example of a category with a non-split idempotent.

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http://mathoverflow.net/questions/2390/why-does-non-abelian-group-cohomology-exist/15970

## Why does non-abelian group cohomology exist? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If K is a non-abelian group on which a group G acts via automorphisms, we can define 1-cocycles and 1-coboundaries by mimicking the explicit formulas coming from the bar resolution in ordinary group cohomology, and thus we have a reasonable notion of H^1(G, K). It turns out we have a part of the expected long exact sequence, until this construction breaks down for building H^i when i > 1, where the long exact sequence stops. There are other analogues to ordinary group cohomology as well. The only proof I've ever seen of any of this is by hand. Is there some deeper explanation of why non-abelian group cohomology exists (and then ceases to exist)? - ## 6 Answers Topologically, you could say that this is true because K(A,1) exists for nonabelian groups A. When the action of G on A is trivial, at least, H^1(G,A) should be homotopy classes of maps from K(G,1) to K(A,1) (the same way H^n(G,A)=H^n(K(G,1);A)=homotopy classes of maps K(G,1) \to K(A,n) for A abelian). In a similar way, H^0 is defined with coefficients in any pointed set. - 7 And the reason, in turn, that K(A,1) exists for nonabelian groups but not K(A,n) for higher n is the Eckmann-Hilton argument showing that pi_n is abelian for n>1. Likewise, K(A,0) exists for any pointed set, but not K(A,n) for n>0 since pi_n is a group for n>0. (-: – Mike Shulman Oct 25 2009 at 2:45 ### 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. To elaborate on Eric's answer, I believe that H1-n(G, A) is πn of the homotopy fixed point space K(A, 1)hG. That exact sequence which ends at H1--which is only a set, while H0 is a group--is actually the long exact sequence of the fibration K(A, 1)hG -> K(B, 1)hG -> K(C, 1)hG in disguise. That reindexing 1-n is related to the 1 in K(A, 1); if A is abelian, then we can replace all the occurrences of 1 by r for any r ≥ 0, giving arbitrarily long segments of the long exact sequence. (Or you can use the language of spectra: H-n(G, A) = πn((HA)hG). (HA)hG has nonzero homotopy groups only in non*positive* dimension.) K(A, 1)hG is a groupoid (it only has homotopy in dimensions 0 and 1) and I expect it should be the groupoid of some kind of extensions of G by A (where morphisms are isomorphisms of extensions which are the identity on G and A). H1(G, A) would then classify those extensions, and H0(G, A) = AG would be the automorphism group of the basepoint, which I guess is the semidirect product (this is easy to check). - I don't know if this is a "deep" or "shallow" explanation, but if anyone is still reading this thread, here is a different explanation. I'll start with the preliminary comment that the cohomology of a group $K$ is a special case of cohomology of topological spaces. In topology in general, you get the same phenomenon that $H^k(X,G)$ is well-defined either when $G$ is abelian or $k=1$. Consider the definition of simplicial cohomology for locally finite simplicial complexes. Or, more generally, CW cohomology for locally finite, regular complexes — regular means mainly that each attaching map is embedded. You can define a $k$-cochain with coefficients in a group $G$ (or even in any set) as a function from the $k$-cells to $G$. In attempting to define the coboundary of a cochain $c$ on a $k+1$-cell $e$, you should multiply together the values of $c$ on the facets of $e$. The obvious problem is that if $G$ is non-abelian, the product is order-dependent. However, if $k=1$, geometry gives you a gift: The facets are cyclically ordered, and what you mainly wanted to know is whether the product is trivial. The criterion of whether a cyclic word is trivial is well-defined in any group, not just abelian groups. A similar but simpler phenomenon occurs for the notion of a coboundary: If $e$ is an oriented edge and $c$ is a 0-cochain, there is a non-abelian version of modifying a 1-cochain by $c$ because the vertices of $e$ are an ordered pair. So far this is just a more geometric version of Eric Wofsey's answer. It is very close to the fact that $\pi_1$ is non-abelian while higher homotopy groups are abelian — and therefore non-commutative classifying spaces exist only for $K(G,1)$. However, in this version of the explanation, something extra appears when $X=M$ is a 3-manifold. If $M$ is a 3-manifold, then not only are the edges of a face cyclically ordered, the faces incident to an edge are also cyclically ordered. It turns out that, at least at the level of computing the cardinality of $H^1(M,G)$, you can let $G$ be both non-commutative and non-cocommutative. In other words, $G$ can be replaced by a finite-dimensional Hopf algebra $H$ which does not need to be commutative or cocommutative. Finiteness is necessary because it is a counting invariant. The resulting invariant $\#(M,H)$ was a topic of my PhD thesis and is explained here and here. Although the motivation is original, the invariant is a special case of more standard quantum invariants defined by other people. (The same construction was also later found by three physicists, but I can't remember their names at all right now.) Many 3-manifolds are also classifying spaces of groups, so for these groups there is the same notion of noncommutative, non-cocommutative group cohomology. - The topological viewpoint of Eric's answer applies to cohomology with nontrivial actions, too. If the action of G on A is not trivial, then the group cohomology H(G;A) can be identified with the topological cohomology of K(G,1) with local coefficients in the coefficient system (= locally constant sheaf) which is A with its action of π1(K(G,1)) = G. So the real question is then, why is the sheaf cohomology H1(X;A) defined with coefficients in a sheaf of nonabelian groups, but not Hn for n>1? This then essentially follows from the same argument (K(A,1) exists for any group A, but other K(A,n)s only for A abelian) but applied in the category (or (∞,1)-category, or model category, or whatever) of sheaves of spaces over X. There is also a sort of "higher nonabelian cohomology." For a nonabelian group A, you can't make K(A,2) but you can make B(hAut(K(A,1))) where hAut denotes the topological monoid of self-homotopy-equivalences, and you can think about homotopy classes of maps from a space X (such as K(G,1)) into B(hAut(K(A,1))) as a sort of "nonabelian H2." If A happens to be abelian, this construction contains the usual abelian H2 via the map K(A,2) = B(K(A,1)) --> B(hAut(K(A,1))) given by letting K(A,1) act on itself via left translation (since it is a topological group whenever A is abelian). But even in this case, the "nonabelian H2" contains much more than the usual abelian H2, so it's a little misleading to call it "nonabelian H2." - 2 Yes, I was viewing group cohomology as the derived version -^hG of the fixed point functor -^G, but the "global sections over K(G,1)" functor is probably more fundamental in a sense. – Reid Barton Oct 25 2009 at 3:33 A concrete and arithmetically useful way to interpret it without appeal to explicit cocycle formulas is to express everything in the language of torsors. More specifically, for arithmetic purposes if the group G is Gal(F'/F) for a Galois extension F'/F and if the group K is H(F') for an F-group scheme H of finite type and K is equipped with the evident left G-action then ${\rm{H}}^1(G,K)$ is the set of isomorphism classes of H-torsors over F which split over F' (i.e.,, admit an F'-rational point). The low-degree exact sequence can then be expressed entirely in such terms, using pushouts and pullbacks with torsors. (Implicit in the argument is effectivity of Galois descent for H-torsors, which uses that H is quasi-projective over F.) This is useful in settings as varied as H an abelian variety and H a linear algebraic group, and even the non-smooth case. In fact, when using non-smooth H it is rather restrictive to use Galois cohomology (but not unnatural if studying Tate-Shafarevich sets with coefficients in an Aut-functor, such as for a projective variety), and in such cases the "right" variant that is often more useful is to work with torsors for the fppf topology over F. The torsor viewpoint also gives a useful perspective when working over richer base rings than fields, such as rings of S-integers in a global field, even in the case of a smooth coefficient group (over the ring of S-integers), for which the etale topology is "enough". See section 5.3 in Chapter I of Serre's book on Galois cohomology for the Galois case, Milne's "Etale cohomology" book for generalization with flat and \'etale topologies, and Appendix B in my paper on "Finiteness theorems for algebraic groups over function fields" for a concrete fleshing out of the dictionary between the torsor and Galois languages (where I work with affine group schemes, due to the context of that paper). Some papers of Mazur and Grothendieck (not as co-authors...) on abelian varieties make creative use of the torsor viewpoint when working with Tate-Shafarevich groups. The exact sequence for Brauer groups in global class field theory also has a useful interpretation via torsors; see Grothendieck's papers on Brauer groups for more in that direction (and somewhere in there he also discusses Tate-Shafarevich). Beware that when the base is not a field (or even when it is a field but we relax "quasi-projective" to "locally finite type" on $H$, such as for Aut-schemes of projective or proper varieties) then effectivity of descent for torsors is not at all clear, even with quasi-projective hypotheses, and so the torsors often need to be understood to be taken in the category of algebraic spaces (for which fppf descent is always effective). In the N\'eron Models book they have a discussion (somewhere in Chapter 6, I think) on effectivity of descent for torsors if one wishes to avoid algebraic spaces (under suitable hypotheses on the "coefficient group"), but working with algebraic spaces isn't so bad once one gets used to them and it is a more natural setting due to their better general behavior with respect to descent. - Extensions exist for non-abelian groups too. -

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http://mathoverflow.net/questions/100469?sort=newest

Opinions on the Multiplication of Measures Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) A few questions, hopefully to spark some discussion. How can one define a product of measures? 1. We could use Colombeau products by embedding the measures into the distributions? I'm not sure why this approach is frowned on? 2. Since measures are map from sets into numbers, can we not define the product as the pointwise product on sets? I would like to see the counter example here. 3. Another product I can think of is to take the product measure and then use the fact that there exists a bijection from R to R^2. I guess that there exist no measurable bijection, which is why that might not work. Thanks, Daniel - 6 Before tackling this general problem, take a very special case: both measures are the Dirac delta at zero in the real line. Now "define" the product. I think you can find that question asked here more than once already... – Gerald Edgar Jun 23 at 19:44 2 Answers Counterexample for suggestion 2: Take a measure on a 2-point set, assigning to each singleton measure 1 and therefore assigning to the whole set measure 2. The pointwise product of this measure with itself still gives the singletons measure 1 but gives the whole space measure 4, so it's not a measure. For suggestion 3: There are measurable bijections from $R^2$ to $R$ when $R$ is the real line, but under the "natural" such bijections (e.g., interleaving binary expansions) the image of the product measure (Lebesgue measure times itself, on the plane) would just be Lebesgue measure on the line. By using other, specially constructed bijections, you could get the image to be any of lots of other measures on the line. - Thank you for your answer. Any opinion on whether there is any sort of product? – dcs24 Jun 23 at 16:22 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. Not a definitive answer, just some thoughts that I hope you might find useful. Firstly, there is a celebrated example of Schwartz that shows that you can't do this globally in the sense of getting a ring structure on the space of measures which extends that on the space of continuous functions. It can, however, often be fruitful to try a local approach, i.e., find pairs of measures which {\it can} be multiplied in a sensible way. For example, the product of two dirac measures is only a problem if their singularities coincide. One can, trivially, always multiply a continuous function and a measure (and with a little bit of integration theory this can be extended to more general functions). One way to deal with less trivial situations is to say that if we can cover $I$ (for the sake of simplicity, I will assume that the measures are defined on the unit interval $I$) with a finite family of relatively open subintervals so that on each interval at least one of the the measures is a continuous function (which one depending on the interval, of course), then we can define the product. This covers the above case of the product of dirac measures with distinct singularities. One would have to know more about the kind of measures that you want to multiply in order to decide if this is of much help. -

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http://mathhelpforum.com/differential-geometry/197549-proof-parametrized-curve-orthogonal-certain-vector-v.html

1Thanks • 1 Post By HallsofIvy # Thread: 1. ## Proof that a parametrized curve orthogonal to a certain vector v Let $\alpha\colon I\rightarrow R^3$ be a parametrized curve and let $\mathbf{v} \in R^3$ be a fixed vector. Assume that $\alpha'(t)$ is orthogonal to $\mathbf{v}$ for all $t \in I$ and that $\alpha(0)$ is also orthogonal to $\mathbf{v}$ Prove that $\alpha(t)$ is orthogonal to $\mathbf{v}$ for all $t \in I$ We are given that $\alpha'(t)$ and $\alpha(0)$ is orthogonal to $\mathbf{v}$ Let $\mathbf{v} = (v_1, v_2, v_3)$ and $\alpha(t) = (x(t), y(t), z(t))$ $\therefore \alpha'(t) = (x'(t), y'(t), z'(t))$ So we have: $\alpha'(t) \cdot\, \mathbf{v} = (x'(t) * \, v_1 +\,\, y'(t) * v_2 + \,\, z'(t) * v_3) = 0 \text{ [Given]}$ $\alpha(0) \cdot\, \mathbf{v} = (x(0) *\, v_1 +\,\, y(0) * v_2 +\,\, z(0) * v_3) = 0 \text{ [Given]}$ Now how do I use this info and prove that $\alpha(t)$ is orthogonal to $\mathbf{v}$ for all $t \in I$? Is it possible for anyone to kindly give some hints on how to prove this? 2. ## Re: Proof that a parametrized curve orthogonal to a certain vector v I think I solved it. For those who are interested: Since $\alpha'(t) \cdot\, \mathbf{v} = (x'(t) *\, v_1)+\,\,( y'(t) * v_2) +\,\, (z'(t) * v_3) = 0 \text{ [Given]}$ Particularly: $(x'(t) *\, v_1)+\,\, (y'(t) * v_2) +\,\, (z'(t) * v_3) = 0$ Taking integration of both sides: $(v_1 * x(t)) + C_1 +\,\,( v_2 * (y(t)) + C_2 +\,\, (v_3 * z(t)) + C_3 = C$ By taking all constants to the right side we get: $(v_1 * x(t)) + (v_2 * y(t)) + (v_3 * z(t)) = C - C_1 - C_2 - C_3 \text{ ................(1)}$ We also know that: $\alpha(0) \cdot\, \mathbf{v} = (x(0) *\, v_1) +\,\, (y(0) * v_2) +\,\, (z(0) * v_3) = 0 \text{ [Given]}$ So pluging in $t = 0$ into eq(1) we get: $C - C_1 - C_2 - C_3 = 0$ therefore: $(v_1 * x(t)) + (v_2 * y(t)) + (v_3 * z(t)) = 0$ But this means: $\alpha(t) \cdot\,\, \mathbf{v} = 0$ So $\alpha(t)$ is orthogonal to $\mathbf{v}$.[As was to be shown] 3. ## Re: Proof that a parametrized curve orthogonal to a certain vector v Looks good but I think I would go the "other way"- that is, differentiate $\alpha$ rather than integrate $\alpha'$. Let $u(t)= \alpha(t)\cdot v$. Then, since v is a constant vector, $u'= \alpha\cdot v= 0$. Since the derivative of u is 0, u is a constant. And since $u(0)= \alpha(0)\cdot v= 0$ it follows that $u(t)= \alpha(t)\cdot v= 0$ for all t. 4. ## Re: Proof that a parametrized curve orthogonal to a certain vector v Originally Posted by HallsofIvy Looks good but I think I would go the "other way"- that is, differentiate $\alpha$ rather than integrate $\alpha'$. Let $u(t)= \alpha(t)\cdot v$. Then, since v is a constant vector, $u'= \alpha\cdot v= 0$. Since the derivative of u is 0, u is a constant. And since $u(0)= \alpha(0)\cdot v= 0$ it follows that $u(t)= \alpha(t)\cdot v= 0$ for all t. Thank you for showing another way of solving this problem. I believe your way is the correct and better method. Again thanks.

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http://mathoverflow.net/questions/105293/proofs-of-ergodicity-of-sl2-z-action-on-r2-without-using-duality/108744

## proofs of ergodicity of Sl(2, Z) action on R^2 without using duality ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The group $G=SL(2, R)$ acts linearly on $\mathbb R^2$. The Lebesgue measure of $\mathbb R^2$ is invariant and ergodic for $G$. There is a proof using duality theorem: Let $U$ be the upper triangular unipotent subgroup of $G$ and let $\Gamma=SL(2, \mathbb Z)$. Then $U$ acts ergodically on $G/\Gamma$. Therefore duality tells us that $\Gamma$ acts ergodically on $G/U\cong \mathbb R^2\backslash 0$. This method is good but it can not be extended to the group action on non homogeneous manifolds. Are there any other proofs? It is better that I can see different proofs to learn methods. - The most basic proof of the ergodicity result mentionned is by Fourier series but of course it does not generalize so it is not a good answer ... – unknown (google) Aug 26 at 11:50 ## 2 Answers What you call "duality" (that for two commuting actions under reasonable conditions ergodic properties of either action on the space of orbits of the other one are the same) goes back to the work of Furstenberg in the 60s. This idea is especially useful when dealing with the so-called geometric flows: the horocycle and the geodesic ones. However, usually one goes in the direction opposite to the one you mention, namely first studies the action on the space of $\Gamma$-orbits and then deduces the corresponding properties of geometric flows. You are referring to the situation with the horocycle flow: its ergodicity on the quotient of the hyperbolic plane $\mathbf H^2$ by a discrete subgroup $\Gamma$ is equivalent to ergodicity of the action of $\Gamma$ on the space of horocycles in $\mathbf H^2$ (the latter one being isomorphic to the linear action of $\Gamma$ on $\mathbb R^2\setminus{0}$). In the case of the geodesic flow the space of its orbits, i.e., the space of geodesics in the hyperbolic plane, is isomorphic to the square of the boundary circle with removed diagonal $\partial\mathbf H^2\times\partial\mathbf H^2\setminus \text{diag}$. In particular, invariant measures of the geodesic flow are in natural one-to-one correspondence with the so-called geodesic currents, i.e., invariant Radon measures on $\partial\mathbf H^2\times\partial\mathbf H^2\setminus \text{diag}$. For geodesic flows the case of "big" surfaces and manifolds (with infinite volume) was being considered from the very beginning of the theory in the 30s (and in the constant curvature case finalized in the so-called Hopf-Tsuji-Sullivan theorem in the late 70s), whereas for the horocycle flow it wasn't really considered (and settled) until the late 90s. Generally speaking, proving ergodicity of the horocycle flow on a general surface consists of two components: (1) establishing ergodicity of the boundary action, (2) proving ergodicty of the so-called Busemann cocycle (which determines the extension from the boundary circle to the space of horocycles). For more details see the works of Babillot - Ledrappier MR1699356, Kaimanovich MR1738739, MR1919405, MR2731695, Pollicott MR1739598, Coudene MR1836430 , Solomyak MR1860491, Ledrappier MR1871151, Hamenstadt MR1926280, Ledrappier - Pollicott MR1953296, Babillot MR2087786, Ledrappier - Sarig MR2226490, Sarig MR2827866. - ### 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 recommand the work of Thomas Roblin "Ergodicité et équidistribution en courbure négative" Mémoires de la SMF 95 (2003), in addition to the references mentioned in the first answer. In particular, he does not need any compactness or finite volume assumption, he works with variable negative curvature, and even on CAT(-1) spaces $X$, not only on manifolds. In this framework ergodic properties of the horospherical foliation are related (by duality) to ergodic properties of the action of $\Gamma$ on the space of horospheres $\partial\tilde{X}\times \mathbb{R}$ (which identifies with $\mathbb{R}^2\setminus {0 }/\pm$ in the case of hyperbolic surfaces). These ergodic properties are strongly related to the mixing property of the geodesic flow. As mentioned in the beginning of the first answer, in the much earlier approach of Furstenberg, harmonic analysis is used on $\mathbb{R}^2$ to get unique ergodicity of the action of a cocompact fuchsian group $\Gamma$ on $\mathbb{R}^2\setminus{0}$, and the unique ergodicity of the horocyclic flow on $SL(2,\mathbb{R})/\Gamma$ is deduced by duality. The modern approach (as in Roblin, or Coudène, or others) through mixing is more dynamical. -

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http://alanrendall.wordpress.com/2010/11/

Hydrobates A mathematician thinks aloud Archive for November, 2010 Me on TV November 26, 2010 Recently I was interviewed by TV journalists for a documentary of the channel 3Sat called “Rätsel Dunkle Materie” [The riddle of dark matter]. It was broadcast yesterday. Before I say more about my experience with this let me do a flashback to the only other time in my life I appeared on TV. On that occasion the BBC visited our school. I guess I was perhaps twelve at the time although I do not know for sure. I was filmed reading a poem which I had written myself. I was seen sitting in a window of the Bishops’ Palace in Kirkwall, looking out. I suppose only my silhouette was visible. I no longer have the text of the poem. All I know is that the first line was ‘Björn, adventuring at last’ and that later on there was some stuff about ravens. At that time I was keen on Vikings. The poem was no doubt very heroic, so that the pose looking out the window was appropriate. Coming back to yesterday, the documentary consisted of three main elements. There was a studio discussion with three guests – the only one I know personally is Simon White. There were some clips illustrating certain ideas. Thirdly there were short sequences from interviews with some other people. I was one of these people. They showed a few short extracts of the interview with me and I was quite happy with the selection they made. This means conversely that they nicely cut out things which I might not have liked so much. I was answering questions posed by one of the journalists and which were not heard on TV. They told me in advance that this would be the case. They told me that for this reason I should not refer to the question during my answers. I found this difficult to do and I think I would need some practice to do it effectively. Fortunately it seems that they efficiently cut out these imperfections. I did not know the questions in advance of the filming and this led to some hesitant starts in my answers. This also did not come through too much in what was shown. Summing up, it was an interesting experience and I would do it again if I had the chance. Of course being a studio guest would be even more interesting … I found the documentary itself not so bad. I could have done without the part about religion at the end. Perhaps the inclusion of this is connected with the fact that the presenter of the series, Gert Scobel, studied theology and also has a doctorate in hermeneutics. (I had to look up that word to have an idea what it meant.) An aspect of the presentation which was a bit off track was that it gave the impression that the idea of a theory unifying general relativity and quantum theory was solely due to Stephen Hawking. Before ending this post I should perhaps say something about my own point of view on dark matter and dark energy. Of course they are symptoms of serious blemishes in our understanding of reality. I believe that dark matter and dark energy are better approaches to explaining the existing observational anomalies than any other alternative which is presently available. In the past I have done some work related to dark energy myself. The one thing that I do not like about a lot of the research in this area is that while people are very keen on proposing new ‘theories’ (which are often just more or less vague ideas for models) there is much less enthusiasm for working out these ideas to obtain a logically sound proposal. Of course that would be more difficult. A case study in this direction was carried out in the diploma thesis of Nikolaus Berndt which was done under my supervision. The theme was to what extent the so-called Cardassian models (do not) deserve to be called a theory. We later produced a joint publication on this. It has not received much attention in the research community and as far as I know has only been cited once. Posted in Uncategorized | 3 Comments » Twisting Gowdy November 11, 2010 I have previously written about the Gowdy solutions as the simplest model case of spatially inhomogeneous solutions of the Einstein vacuum equations and about the generalization of this to the Einstein-Maxwell equations. I have also written about spatially homogeneous solutions (Bianchi models). I have just posted a paper where I explore connections between these two classes of solutions (Gowdy and Bianchi) in the hope of learning more about both. The basic idea is simple. A Bianchi model has Killing vectors which form a three-dimensional Lie algebra. Most of these (all except types VIII and IX) have a two-dimensional Abelian subalgebra. When this is the case we can simply forget about one Killing vector and end up with a spacetime with two commuting Killing vectors. This looks very like a Gowdy spacetime. There are, however, two potential problems. The first is that Gowdy spacetimes have a certain discrete symmetry in addition to their two Killing vectors. Almost all Bianchi types do satisfy this condition, the exception being type ${\rm VI}{}_{\frac19}$. In what follows I ignore this exceptional type. The other problem is that a Gowdy spacetime is supposed to have a compact Cauchy surface and the Killing vectors are assumed to exist globally. If these conditions are weakened then more alternatives are allowed. Let me call this situation a locally Gowdy spacetime. (This is not standard terminology.) Assume that the Killing vector fields have no zeroes. Then in the Gowdy case the Cauchy surface has the topology of a three-torus $T^3$. In the locally Gowdy case the existence of a compact Cauchy surface is closely linked to those Bianchi types belonging to Bianchi Class A. I now restrict to that class. Then the additional Bianchi types which are possible are II, ${\rm VI}{}_0$ and ${\rm VII}{}_0$. In Gowdy spacetimes there are two of the Einstein equations for functions $P(t,\theta)$ and $Q(t,\theta)$ which are of central importance. I call them the Gowdy equations. In order to describe Gowdy spacetimes the functions $P$ and $Q$ should be periodic in $\theta$. To accomodate Bianchi models of types II and ${\rm VI}{}_0$ more complicated boundary conditions are required. The twisted Gowdy solutions are solutions of the Gowdy equations which satisfy these boundary conditions but need not be homogeneous. They represent finite inhomogeneous perturbations of the corresponding Bianchi models. Periodic boundary conditions can accommodate type ${\rm VII}{}_0$, which then corresponds to what are called circular loop spacetimes. In the twisted cases the topology of a compact Cauchy surface is more complicated than that of $T^3$. For type II and type ${\rm VI}{}_0$ it corresponds to a manifold admitting geometric structures of type Nil and Sol in the sense of Thurston, respectively. These are twisted topologies which are torus bundles over a circle. A lot is understood about the global dynamics of Gowdy solutions, mainly due to the work of Hans Ringström. By uncovering certain connections between different classes of solutions I have been able to transfer some of these results to the twisted Gowdy case. Unfortunately I was not able to obtain a general analysis of the late-time behaviour of twisted Gowdy solutions. The analytical techniques which were successful in the untwisted case do not seem to adapt well. This leaves a challenge for the future.

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http://medlibrary.org/medwiki/Clifford_algebras

# Clifford algebras Welcome to MedLibrary.org. For best results, we recommend beginning with the navigation links at the top of the page, which can guide you through our collection of over 14,000 medication labels and package inserts. For additional information on other topics which are not covered by our database of medications, just enter your topic in the search box below: In mathematics, Clifford algebras are a type of associative algebra. As K-algebras, they generalize the real numbers, complex numbers, quaternions and several other hypercomplex number systems.[1][2] The theory of Clifford algebras is intimately connected with the theory of quadratic forms and orthogonal transformations. Clifford algebras have important applications in a variety of fields including geometry, theoretical physics and digital image processing. They are named after the English geometer William Kingdon Clifford. The most familiar Clifford algebra, or orthogonal Clifford algebra, is also referred to as Riemannian Clifford algebra.[3] ## Introduction and basic properties Specifically, a Clifford algebra is a unital associative algebra which contains and is generated by a vector space V equipped with a quadratic form Q. The Clifford algebra Cℓ(V, Q) is the "freest" algebra generated by V subject to the condition[4] $v^2 = Q(v)1\ \text{ for all } v\in V,$ where the product on the left is that of the algebra, and the 1 is its multiplicative identity. The definition of a Clifford algebra endows it with more structure than a "bare" K-algebra: specifically it has a designated or privileged subspace that is isomorphic to V. Such a subspace cannot in general be uniquely determined given only a K-algebra isomorphic to the Clifford algebra. If the characteristic of the ground field K is not 2, then one can rewrite this fundamental identity in the form $uv + vu = 2\lang u, v\rang \text{ for all }u,v \in V,$ where ⟨u, v⟩ = (Q(u + v) − Q(u) − Q(v))/2 is the symmetric bilinear form associated with Q, via the polarization identity. The idea of being the "freest" or "most general" algebra subject to this identity can be formally expressed through the notion of a universal property, as done below. Quadratic forms and Clifford algebras in characteristic 2 form an exceptional case. In particular, if char(K) = 2 it is not true that a quadratic form determines a symmetric bilinear form, or that every quadratic form admits an orthogonal basis. Many of the statements in this article include the condition that the characteristic is not 2, and are false if this condition is removed. ### As a quantization of the exterior algebra Clifford algebras are closely related to exterior algebras. In fact, if Q = 0 then the Clifford algebra Cℓ(V, Q) is just the exterior algebra Λ(V). For nonzero Q there exists a canonical linear isomorphism between Λ(V) and Cℓ(V, Q) whenever the ground field K does not have characteristic two. That is, they are naturally isomorphic as vector spaces, but with different multiplications (in the case of characteristic two, they are still isomorphic as vector spaces, just not naturally). Clifford multiplication together with the privileged subspace is strictly richer than the exterior product since it makes use of the extra information provided by Q. More precisely, Clifford algebras may be thought of as quantizations (cf. quantization (physics), Quantum group) of the exterior algebra, in the same way that the Weyl algebra is a quantization of the symmetric algebra. Weyl algebras and Clifford algebras admit a further structure of a *-algebra, and can be unified as even and odd terms of a superalgebra, as discussed in CCR and CAR algebras. ## Universal property and construction Let V be a vector space over a field K, and let Q: V → K be a quadratic form on V. In most cases of interest the field K is either R, C or a finite field. A Clifford algebra Cℓ(V, Q) is a unital associative algebra over K together with a linear map i : V → Cℓ(V, Q) satisfying i(v)2 = Q(v)1 for all v ∈ V, defined by the following universal property: given any associative algebra A over K and any linear map j : V → A such that j(v)2 = Q(v)1A for all v ∈ V (where 1A denotes the multiplicative identity of A), there is a unique algebra homomorphism f : Cℓ(V, Q) → A such that the following diagram commutes (i.e. such that f ∘ i = j): Working with a symmetric bilinear form ⟨·,·⟩ instead of Q (in characteristic not 2), the requirement on j is j(v)j(w) + j(w)j(v) = 2⟨v, w⟩ for all v, w ∈ V. A Clifford algebra as described above always exists and can be constructed as follows: start with the most general algebra that contains V, namely the tensor algebra T(V), and then enforce the fundamental identity by taking a suitable quotient. In our case we want to take the two-sided ideal IQ in T(V) generated by all elements of the form $v\otimes v - Q(v)1$ for all $v\in V$ and define Cℓ(V, Q) as the quotient algebra Cℓ(V, Q) = T(V)/IQ. The ring product inherited by this quotient is sometimes referred to as the Clifford product[5] to differentiate it from the inner and outer products. It is then straightforward to show that Cℓ(V, Q) contains V and satisfies the above universal property, so that Cℓ is unique up to a unique isomorphism; thus one speaks of "the" Clifford algebra Cℓ(V, Q). It also follows from this construction that i is injective. One usually drops the i and considers V as a linear subspace of Cℓ(V, Q). The universal characterization of the Clifford algebra shows that the construction of Cℓ(V, Q) is functorial in nature. Namely, Cℓ can be considered as a functor from the category of vector spaces with quadratic forms (whose morphisms are linear maps preserving the quadratic form) to the category of associative algebras. The universal property guarantees that linear maps between vector spaces (preserving the quadratic form) extend uniquely to algebra homomorphisms between the associated Clifford algebras. ## Basis and dimension If the dimension of V is n and {e1, …, en} is a basis of V, then the set $\{e_{i_1}e_{i_2}\cdots e_{i_k} \mid 1\le i_1 < i_2 < \cdots < i_k \le n\mbox{ and } 0\le k\le n\}$ is a basis for Cℓ(V, Q). The empty product (k = 0) is defined as the multiplicative identity element. For each value of k there are n choose k basis elements, so the total dimension of the Clifford algebra is $\dim C\ell(V,Q) = \sum_{k=0}^n\begin{pmatrix}n\\ k\end{pmatrix} = 2^n.$ Since V comes equipped with a quadratic form, there is a set of privileged bases for V: the orthogonal ones. An orthogonal basis is one such that $\langle e_i, e_j \rangle = 0 \qquad i\neq j. \,$ where ⟨·,·⟩ is the symmetric bilinear form associated to Q. The fundamental Clifford identity implies that for an orthogonal basis $e_ie_j = -e_je_i \qquad i\neq j. \,$ This makes manipulation of orthogonal basis vectors quite simple. Given a product $e_{i_1}e_{i_2}\cdots e_{i_k}$ of distinct orthogonal basis vectors of V, one can put them into standard order while including an overall sign determined by the number of pairwise swaps needed to do so (i.e. the signature of the ordering permutation). ## Examples: real and complex Clifford algebras The most important Clifford algebras are those over real and complex vector spaces equipped with nondegenerate quadratic forms. It turns out that every one of the algebras Cℓp,q(R) and Cℓn(C) are isomorphic to A or A⊕A, where A is a full matrix ring with entries from R, C, or H. For a complete classification of these algebras see classification of Clifford algebras. ### Real numbers Main article: Geometric algebra The geometric interpretation of real Clifford algebras is known as geometric algebra. Every nondegenerate quadratic form on a finite-dimensional real vector space is equivalent to the standard diagonal form: $Q(v) = v_1^2 + \cdots + v_p^2 - v_{p+1}^2 - \cdots - v_{p+q}^2$ where n = p + q is the dimension of the vector space. The pair of integers (p, q) is called the signature of the quadratic form. The real vector space with this quadratic form is often denoted Rp, q. The Clifford algebra on Rp, q is denoted Cℓp, q(R). The symbol Cℓn(R) means either Cℓn,0(R) or Cℓ0,n(R) depending on whether the author prefers positive definite or negative definite spaces. A standard orthonormal basis {ei} for Rp,q consists of n = p + q mutually orthogonal vectors, p of which have norm +1 and q of which have norm −1. The algebra Cℓp,q(R) will therefore have p vectors that square to +1 and q vectors that square to −1. Note that Cℓ0,0(R) is naturally isomorphic to R since there are no nonzero vectors. Cℓ0,1(R) is a two-dimensional algebra generated by a single vector e1 that squares to −1, and therefore is isomorphic to C, the field of complex numbers. The algebra Cℓ0,2(R) is a four-dimensional algebra spanned by {1, e1, e2, e1e2}. The latter three elements square to −1 and all anticommute, and so the algebra is isomorphic to the quaternions H. Cℓ0,3(R) is an 8-dimensional algebra isomorphic to the direct sum H ⊕ H called split-biquaternions. ### Complex numbers One can also study Clifford algebras on complex vector spaces. Every nondegenerate quadratic form on a complex vector space is equivalent to the standard diagonal form $Q(z) = z_1^2 + z_2^2 + \cdots + z_n^2$ where n = dim V, up to isomorphism so there is only one nondegenerate Clifford algebra for each dimension n. We will denote the Clifford algebra on Cn with the standard quadratic form by Cℓn(C). The first few cases are not hard to compute. One finds that Cℓ0(C) ≅ C, the complex numbers Cℓ1(C) ≅ C ⊕ C, the bicomplex numbers Cℓ2(C) ≅ M(2, C), the biquaternions where M(n, C) denotes the algebra of n×n matrices over C. ## Examples: constructing quaternions and dual quaternions ### Quaternions In this section, Hamilton's quaternions are constructed as the even sub algebra of the Clifford algebra Cℓ0,3(R). Let the vector space V be real three dimensional space R3, and the quadratic form Q be derived from the usual Euclidean metric. Then, for v, w in R3 we have the quadratic form, or dot product, $\mathbf{v}\cdot\mathbf{w}= v_1w_1 + v_2w_2 + v_3w_3.$ Now introduce the Clifford product of vectors v and w given by $\mathbf{v}\mathbf{w} + \mathbf{w}\mathbf{v} = -2 (\mathbf{v}\cdot \mathbf{w}).\!$ This formulation uses the negative sign so the correspondence with quaternions is easily shown. Denote a set of orthogonal unit vectors of R3 as e1, e2, and e3, then the Clifford product yields the relations $\mathbf{e}_2 \mathbf{e}_3 = -\mathbf{e}_3 \mathbf{e}_2, \,\,\, \mathbf{e}_3 \mathbf{e}_1 = -\mathbf{e}_1 \mathbf{e}_3,\,\,\, \mathbf{e}_1 \mathbf{e}_2 = -\mathbf{e}_2 \mathbf{e}_1,\!$ and $\mathbf{e}_1 ^2 = \mathbf{e}_2^2 =\mathbf{e}_3^2 = -1. \!$ The general element of the Clifford algebra Cℓ0,3(R) is given by $A = a_0 + a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 + a_3 \mathbf{e}_3 + a_4 \mathbf{e}_2 \mathbf{e}_3 + a_5 \mathbf{e}_3 \mathbf{e}_1 + a_6 \mathbf{e}_1 \mathbf{e}_2 + a_7 \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3.\!$ The linear combination of the even rank elements of Cℓ0,3(R) defines the even sub algebra Cℓ00,3(R) with the general element $Q = q_0 + q_1 \mathbf{e}_2 \mathbf{e}_3 + q_2 \mathbf{e}_3 \mathbf{e}_1 + q_3 \mathbf{e}_1 \mathbf{e}_2. \!$ The basis elements can be identified with the quaternion basis elements i, j, k as $i= \mathbf{e}_2 \mathbf{e}_3, j= \mathbf{e}_3 \mathbf{e}_1, k = \mathbf{e}_1 \mathbf{e}_2,$ which shows that the even sub algebra Cℓ00,3(R) is Hamilton's real quaternion algebra. To see this, compute $i^2 = (\mathbf{e}_2 \mathbf{e}_3)^2 = \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_2 \mathbf{e}_3 = - \mathbf{e}_2 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_3 = -1,\!$ and $ij = \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_3 \mathbf{e}_1 = -\mathbf{e}_2 \mathbf{e}_1 = \mathbf{e}_1 \mathbf{e}_2 = k.\!$ Finally, $ijk = \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_3 \mathbf{e}_1 \mathbf{e}_1 \mathbf{e}_2 = -1.\!$ ### Dual quaternions In this section, dual quaternions are constructed as the even Clifford algebra of real four dimensional space with a degenerate quadratic form.[6][7] Let the vector space V be real four dimensional space R4, and let the quadratic form Q be a degenerate form derived from the Euclidean metric on R3. For v, w in R4 introduce the degenerate bilinear form $d(\mathbf{v}, \mathbf{w})= v_1w_1 + v_2w_2 + v_3w_3.$ This degenerate scalar product projects distance measurements in R4 onto the R3 hyperplane. The Clifford product of vectors v and w is given by $\mathbf{v}\mathbf{w} + \mathbf{w}\mathbf{v} = -2 \,d(\mathbf{v}, \mathbf{w}).\!$ Note the negative sign is introduced to simplify the correspondence with quaternions. Denote a set of orthogonal unit vectors of R4 as e1, e2, e3 and e4, then the Clifford product yields the relations $\mathbf{e}_m \mathbf{e}_n = -\mathbf{e}_n \mathbf{e}_m, \,\,\, m \ne n,\!$ and $\mathbf{e}_1 ^2 = \mathbf{e}_2^2 =\mathbf{e}_3^2 = -1, \,\, \mathbf{e}_4^2 =0.\!$ The general element of the Clifford algebra Cℓ(R4,d) has 16 components. The linear combination of the even ranked elements defines the even sub algebra Cℓ0(R4,d) with the general element $H = h_0 + h_1 \mathbf{e}_2 \mathbf{e}_3 + h_2 \mathbf{e}_3 \mathbf{e}_1 + h_3 \mathbf{e}_1 \mathbf{e}_2 + h_4 \mathbf{e}_4 \mathbf{e}_1 + h_5 \mathbf{e}_4 \mathbf{e}_2 + h_6 \mathbf{e}_4 \mathbf{e}_3 + h_7 \mathbf{e}_1 \mathbf{e}_2\mathbf{e}_3 \mathbf{e}_4. \!$ The basis elements can be identified with the quaternion basis elements i, j, k and the dual unit ε as $i=\mathbf{e}_2 \mathbf{e}_3, j=\mathbf{e}_3 \mathbf{e}_1, k = \mathbf{e}_1 \mathbf{e}_2, \,\, \varepsilon = \mathbf{e}_1 \mathbf{e}_2\mathbf{e}_3 \mathbf{e}_4. \!$ This provides the correspondence of Cℓ00,3,1(R) with dual quaternion algebra. To see this, compute $\varepsilon ^2 = (\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4)^2 = \mathbf{e}_1 \mathbf{e}_2\mathbf{e}_3 \mathbf{e}_4 \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4 = -\mathbf{e}_1 \mathbf{e}_2\mathbf{e}_3 (\mathbf{e}_4 \mathbf{e}_4 ) \mathbf{e}_1 \mathbf{e}_2\mathbf{e}_3 = 0,\!$ and $\varepsilon i = (\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4) \mathbf{e}_2 \mathbf{e}_3 = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4 \mathbf{e}_2 \mathbf{e}_3 = \mathbf{e}_2\mathbf{e}_3 (\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4) = i\varepsilon.\!$ The exchanges of e1 and e4 alternate signs an even number of times, and show the dual unit ε commutes with the quaternion basis elements i, j, and k. ## Properties ### Relation to the exterior algebra Given a vector space V one can construct the exterior algebra Λ(V), whose definition is independent of any quadratic form on V. It turns out that if K does not have characteristic 2 then there is a natural isomorphism between Λ(V) and Cℓ(V, Q) considered as vector spaces (and there exists an isomorphism in characteristic two, which may not be natural). This is an algebra isomorphism if and only if Q = 0. One can thus consider the Clifford algebra Cℓ(V, Q) as an enrichment (or more precisely, a quantization, cf. the Introduction) of the exterior algebra on V with a multiplication that depends on Q (one can still define the exterior product independent of Q). The easiest way to establish the isomorphism is to choose an orthogonal basis {ei} for V and extend it to a basis for Cℓ(V, Q) as described above. The map Cℓ(V, Q) → Λ(V) is determined by $e_{i_1}e_{i_2}\cdots e_{i_k} \mapsto e_{i_1}\wedge e_{i_2}\wedge \cdots \wedge e_{i_k}.$ Note that this only works if the basis {ei} is orthogonal. One can show that this map is independent of the choice of orthogonal basis and so gives a natural isomorphism. If the characteristic of K is 0, one can also establish the isomorphism by antisymmetrizing. Define functions fk: V × … × V → Cℓ(V, Q) by $f_k(v_1, \cdots, v_k) = \frac{1}{k!}\sum_{\sigma\in S_k}{\rm sgn}(\sigma)\, v_{\sigma(1)}\cdots v_{\sigma(k)}$ where the sum is taken over the symmetric group on k elements. Since fk is alternating it induces a unique linear map Λk(V) → Cℓ(V, Q). The direct sum of these maps gives a linear map between Λ(V) and Cℓ(V, Q). This map can be shown to be a linear isomorphism, and it is natural. A more sophisticated way to view the relationship is to construct a filtration on Cℓ(V, Q). Recall that the tensor algebra T(V) has a natural filtration: F0 ⊂ F1 ⊂ F2 ⊂ … where Fk contains sums of tensors with rank ≤ k. Projecting this down to the Clifford algebra gives a filtration on Cℓ(V, Q). The associated graded algebra $Gr_F C\ell(V,Q) = \bigoplus_k F^k/F^{k-1}$ is naturally isomorphic to the exterior algebra Λ(V). Since the associated graded algebra of a filtered algebra is always isomorphic to the filtered algebra as filtered vector spaces (by choosing complements of Fk in Fk+1 for all k), this provides an isomorphism (although not a natural one) in any characteristic, even two. ### Grading In the following, assume that the characteristic is not 2.[8] Clifford algebras are Z2-graded algebras (also known as superalgebras). Indeed, the linear map on V defined by v ↦ −v (reflection through the origin) preserves the quadratic form Q and so by the universal property of Clifford algebras extends to an algebra automorphism α: Cℓ(V, Q) → Cℓ(V, Q). Since α is an involution (i.e. it squares to the identity) one can decompose Cℓ(V, Q) into positive and negative eigenspaces of α $C\ell(V,Q) = C\ell^0(V,Q) \oplus C\ell^1(V,Q)$ where Cℓi(V, Q) = {x ∈ Cℓ(V, Q) | α(x) = (−1)ix}. Since α is an automorphism it follows that $C\ell^{\,i}(V,Q)C\ell^{\,j}(V,Q) = C\ell^{\,i+j}(V,Q)$ where the superscripts are read modulo 2. This gives Cℓ(V, Q) the structure of a Z2-graded algebra. The subspace Cℓ0(V, Q) forms a subalgebra of Cℓ(V, Q), called the even subalgebra. The subspace Cℓ1(V, Q) is called the odd part of Cℓ(V, Q) (it is not a subalgebra). This Z2-grading plays an important role in the analysis and application of Clifford algebras. The automorphism α is called the main involution or grade involution. Elements that are pure in this Z2-grading are simply said to be even or odd. Remark. In characteristic not 2 the underlying vector space of Cℓ(V, Q) inherits an N-grading and a Z-grading from the canonical isomorphism with the underlying vector space of the exterior algebra Λ(V).[9] It is important to note, however, that this is a vector space grading only. That is, Clifford multiplication does not respect the N-grading or Z-grading, only the Z2-grading: for instance if Q(v) ≠ 0, then v ∈ Cℓ1(V, Q), but v2 ∈ Cℓ0(V, Q), not in Cℓ2(V, Q). Happily, the gradings are related in the natural way: Z2 ≅N/2N≅ Z/2Z. Further, the Clifford algebra is Z-filtered: Cℓ≤i(V, Q) ⋅ Cℓ≤j(V, Q) ⊂ Cℓ≤i+j(V, Q). The degree of a Clifford number usually refers to the degree in the N-grading. The even subalgebra Cℓ0(V, Q) of a Clifford algebra is itself isomorphic to a Clifford algebra.[10][11] If V is the orthogonal direct sum of a vector a of norm Q(a) and a subspace U, then Cℓ0(V, Q) is isomorphic to Cℓ(U, −Q(a)Q), where −Q(a)Q is the form Q restricted to U and multiplied by −Q(a). In particular over the reals this implies that $C\ell_{p,q}^0(\mathbf{R}) \cong C\ell_{p,q-1}(\mathbf{R})$ for q > 0, and $C\ell_{p,q}^0(\mathbf{R}) \cong C\ell_{q,p-1}(\mathbf{R})$ for p > 0. In the negative-definite case this gives an inclusion Cℓ0,n−1(R) ⊂ Cℓ0,n(R) which extends the sequence R ⊂ C ⊂ H ⊂ H⊕H ⊂ …; Likewise, in the complex case, one can show that the even subalgebra of Cℓn(C) is isomorphic to Cℓn−1(C). ### Antiautomorphisms In addition to the automorphism α, there are two antiautomorphisms which play an important role in the analysis of Clifford algebras. Recall that the tensor algebra T(V) comes with an antiautomorphism that reverses the order in all products: $v_1\otimes v_2\otimes \cdots \otimes v_k \mapsto v_k\otimes \cdots \otimes v_2\otimes v_1.$ Since the ideal IQ is invariant under this reversal, this operation descends to an antiautomorphism of Cℓ(V, Q) called the transpose or reversal operation, denoted by xt. The transpose is an antiautomorphism: (xy)t = yt xt. The transpose operation makes no use of the Z2-grading so we define a second antiautomorphism by composing α and the transpose. We call this operation Clifford conjugation denoted $\bar x$ $\bar x = \alpha(x^t) = \alpha(x)^t.$ Of the two antiautomorphisms, the transpose is the more fundamental.[12] Note that all of these operations are involutions. One can show that they act as ±1 on elements which are pure in the Z-grading. In fact, all three operations depend only on the degree modulo 4. That is, if x is pure with degree k then $\alpha(x) = \pm x \qquad x^t = \pm x \qquad \bar x = \pm x$ where the signs are given by the following table: k mod 4 0 1 2 3 $\alpha(x)\,$ + − + − (−1)k $x^t\,$ + + − − (−1)k(k−1)/2 $\bar x$ + − − + (−1)k(k+1)/2 ### Clifford scalar product When the characteristic is not 2, the quadratic form Q on V can be extended to a quadratic form on all of Cℓ(V, Q) (which we also denoted by Q). A basis independent definition of one such extension is $Q(x) = \lang x^t x\rang$ where ⟨a⟩ denotes the scalar part of a (the grade 0 part in the Z-grading). One can show that $Q(v_1v_2\cdots v_k) = Q(v_1)Q(v_2)\cdots Q(v_k)$ where the vi are elements of V – this identity is not true for arbitrary elements of Cℓ(V, Q). The associated symmetric bilinear form on Cℓ(V, Q) is given by $\lang x, y\rang = \lang x^t y\rang.$ One can check that this reduces to the original bilinear form when restricted to V. The bilinear form on all of Cℓ(V, Q) is nondegenerate if and only if it is nondegenerate on V. It is not hard to verify that the transpose is the adjoint of left/right Clifford multiplication with respect to this inner product. That is, $\lang ax, y\rang = \lang x, a^t y\rang,$ and $\lang xa, y\rang = \lang x, y a^t\rang.$ ## Structure of Clifford algebras In this section we assume that the vector space V is finite dimensional and that the bilinear form of Q is non-singular. A central simple algebra over K is a matrix algebra over a (finite dimensional) division algebra with center K. For example, the central simple algebras over the reals are matrix algebras over either the reals or the quaternions. • If V has even dimension then Cℓ(V, Q) is a central simple algebra over K. • If V has even dimension then Cℓ0(V, Q) is a central simple algebra over a quadratic extension of K or a sum of two isomorphic central simple algebras over K. • If V has odd dimension then Cℓ(V, Q) is a central simple algebra over a quadratic extension of K or a sum of two isomorphic central simple algebras over K. • If V has odd dimension then Cℓ0(V, Q) is a central simple algebra over K. The structure of Clifford algebras can be worked out explicitly using the following result. Suppose that U has even dimension and a non-singular bilinear form with discriminant d, and suppose that V is another vector space with a quadratic form. The Clifford algebra of U+V is isomorphic to the tensor product of the Clifford algebras of U and (−1)dim(U)/2dV, which is the space V with its quadratic form multiplied by (−1)dim(U)/2d. Over the reals, this implies in particular that $C\ell_{p+2,q}(\mathbf{R}) = M_2(\mathbf{R})\otimes C\ell_{q,p}(\mathbf{R})$ $C\ell_{p+1,q+1}(\mathbf{R}) = M_2(\mathbf{R})\otimes C\ell_{p,q}(\mathbf{R})$ $C\ell_{p,q+2}(\mathbf{R}) = \mathbf{H}\otimes C\ell_{q,p}(\mathbf{R}).$ These formulas can be used to find the structure of all real Clifford algebras and all complex Clifford algebras; see the classification of Clifford algebras. Notably, the Morita equivalence class of a Clifford algebra (its representation theory: the equivalence class of the category of modules over it) depends only on the signature (p − q) mod 8. This is an algebraic form of Bott periodicity. ## Clifford group In this section we assume that V is finite dimensional and the quadratic form Q is nondegenerate. The invertible elements of the Clifford algebra act on it by twisted conjugation: conjugation by x maps y ↦ xy α(x)−1. The Clifford group Γ is defined to be the set of invertible elements x that stabilize vectors, meaning that for all v in V we have: $x v \alpha(x)^{-1}\in V$ This formula also defines an action of the Clifford group on the vector space V that preserves the norm Q, and so gives a homomorphism from the Clifford group to the orthogonal group. The Clifford group contains all elements r of V of nonzero norm, and these act on V by the corresponding reflections that take v to v−2⟨v,r⟩r/Q(r) (In characteristic 2 these are called orthogonal transvections rather than reflections.) The Clifford group Γ is the disjoint union of two subsets Γ0 and Γ1, where Γi is the subset of elements of degree i. The subset Γ0 is a subgroup of index 2 in Γ. If V is a finite dimensional real vector space with positive definite (or negative definite) quadratic form then the Clifford group maps onto the orthogonal group of V with respect to the form (by the Cartan-Dieudonné theorem) and the kernel consists of the nonzero elements of the field K. This leads to exact sequences $1 \rightarrow K^* \rightarrow \Gamma \rightarrow \mbox{O}_V(K) \rightarrow 1,\,$ $1 \rightarrow K^* \rightarrow \Gamma^0 \rightarrow \mbox{SO}_V(K) \rightarrow 1.\,$ Over other fields or with indefinite forms, the map is not in general onto, and the failure is captured by the spinor norm. ### Spinor norm For more details on this topic, see Spinor_norm#Galois_cohomology_and_orthogonal_groups. In arbitrary characteristic, the spinor norm Q is defined on the Clifford group by $Q(x) = x^tx.\,$ It is a homomorphism from the Clifford group to the group K* of non-zero elements of K. It coincides with the quadratic form Q of V when V is identified with a subspace of the Clifford algebra. Several authors define the spinor norm slightly differently, so that it differs from the one here by a factor of −1, 2, or −2 on Γ1. The difference is not very important in characteristic other than 2. The nonzero elements of K have spinor norm in the group K*2 of squares of nonzero elements of the field K. So when V is finite dimensional and non-singular we get an induced map from the orthogonal group of V to the group K*/K*2, also called the spinor norm. The spinor norm of the reflection of a vector r has image Q(r) in K*/K*2, and this property uniquely defines it on the orthogonal group. This gives exact sequences: $1 \to \{\pm 1\} \to \mbox{Pin}_V(K) \to \mbox{O}_V(K) \to K^*/K^{*2},\,$ $1 \to \{\pm 1\} \to \mbox{Spin}_V(K) \to \mbox{SO}_V(K) \to K^*/K^{*2}.\,$ Note that in characteristic 2 the group {±1} has just one element. From the point of view of Galois cohomology of algebraic groups, the spinor norm is a connecting homomorphism on cohomology. Writing μ2 for the algebraic group of square roots of 1 (over a field of characteristic not 2 it is roughly the same as a two-element group with trivial Galois action), the short exact sequence $1 \to \mu_2 \rightarrow \mbox{Pin}_V \rightarrow \mbox{O}_V \rightarrow 1\,$ yields a long exact sequence on cohomology, which begins $1 \to H^0(\mu_2;K) \to H^0(\mbox{Pin}_V;K) \to H^0(\mbox{O}_V;K) \to H^1(\mu_2;K).\,$ The 0th Galois cohomology group of an algebraic group with coefficients in K is just the group of K-valued points: H0(G; K) = G(K), and H1(μ2; K) ≅ K*/K*2, which recovers the previous sequence $1 \to \{\pm 1\} \to \mbox{Pin}_V(K) \to \mbox{O}_V(K) \to K^*/K^{*2},\,$ where the spinor norm is the connecting homomorphism H0(OV); K) → H1(μ2; K). ## Spin and Pin groups In this section we assume that V is finite dimensional and its bilinear form is non-singular. (If K has characteristic 2 this implies that the dimension of V is even.) The Pin group PinV(K) is the subgroup of the Clifford group Γ of elements of spinor norm 1, and similarly the Spin group SpinV(K) is the subgroup of elements of Dickson invariant 0 in PinV(K). When the characteristic is not 2, these are the elements of determinant 1. The Spin group usually has index 2 in the Pin group. Recall from the previous section that there is a homomorphism from the Clifford group onto the orthogonal group. We define the special orthogonal group to be the image of Γ0. If K does not have characteristic 2 this is just the group of elements of the orthogonal group of determinant 1. If K does have characteristic 2, then all elements of the orthogonal group have determinant 1, and the special orthogonal group is the set of elements of Dickson invariant 0. There is a homomorphism from the Pin group to the orthogonal group. The image consists of the elements of spinor norm 1 ∈ K*/K*2. The kernel consists of the elements +1 and −1, and has order 2 unless K has characteristic 2. Similarly there is a homomorphism from the Spin group to the special orthogonal group of V. In the common case when V is a positive or negative definite space over the reals, the spin group maps onto the special orthogonal group, and is simply connected when V has dimension at least 3. Further the kernel of this homomorphism consists of 1 and −1. So in this case the spin group, Spin(n), is a double cover of SO(n). Please note, however, that the simple connectedness of the spin group is not true in general: if V is Rp,q for p and q both at least 2 then the spin group is not simply connected. In this case the algebraic group Spinp,q is simply connected as an algebraic group, even though its group of real valued points Spinp,q(R) is not simply connected. This is a rather subtle point, which completely confused the authors of at least one standard book about spin groups. ## Spinors Clifford algebras Cℓp,q(C), with p+q=2n even, are matrix algebras which have a complex representation of dimension 2n. By restricting to the group Pinp,q(R) we get a complex representation of the Pin group of the same dimension, called the spin representation. If we restrict this to the spin group Spinp,q(R) then it splits as the sum of two half spin representations (or Weyl representations) of dimension 2n−1. If p+q=2n+1 is odd then the Clifford algebra Cℓp,q(C) is a sum of two matrix algebras, each of which has a representation of dimension 2n, and these are also both representations of the Pin group Pinp,q(R). On restriction to the spin group Spinp,q(R) these become isomorphic, so the spin group has a complex spinor representation of dimension 2n. More generally, spinor groups and pin groups over any field have similar representations whose exact structure depends on the structure of the corresponding Clifford algebras: whenever a Clifford algebra has a factor that is a matrix algebra over some division algebra, we get a corresponding representation of the pin and spin groups over that division algebra. For examples over the reals see the article on spinors. ### Real spinors For more details on this topic, see spinor. To describe the real spin representations, one must know how the spin group sits inside its Clifford algebra. The Pin group, Pinp,q is the set of invertible elements in Cℓp, q which can be written as a product of unit vectors: ${\mbox{Pin}}_{p,q}=\{v_1v_2\dots v_r |\,\, \forall i\, \|v_i\|=\pm 1\}.$ Comparing with the above concrete realizations of the Clifford algebras, the Pin group corresponds to the products of arbitrarily many reflections: it is a cover of the full orthogonal group O(p, q). The Spin group consists of those elements of Pinp, q which are products of an even number of unit vectors. Thus by the Cartan-Dieudonné theorem Spin is a cover of the group of proper rotations SO(p,q). Let α : Cℓ → Cℓ be the automorphism which is given by the mapping v ↦ −v acting on pure vectors. Then in particular, Spinp,q is the subgroup of Pinp, q whose elements are fixed by α. Let $C\ell_{p,q}^0 = \{ x\in C\ell_{p,q} |\, \alpha(x)=x\}.$ (These are precisely the elements of even degree in Cℓp, q.) Then the spin group lies within Cℓ0p, q. The irreducible representations of Cℓp, q restrict to give representations of the pin group. Conversely, since the pin group is generated by unit vectors, all of its irreducible representation are induced in this manner. Thus the two representations coincide. For the same reasons, the irreducible representations of the spin coincide with the irreducible representations of Cℓ0p, q To classify the pin representations, one need only appeal to the classification of Clifford algebras. To find the spin representations (which are representations of the even subalgebra), one can first make use of either of the isomorphisms (see above) Cℓ0p,q ≈ Cℓp,q−1, for q > 0 Cℓ0p,q ≈ Cℓq,p−1, for p > 0 and realize a spin representation in signature (p,q) as a pin representation in either signature (p,q−1) or (q,p−1). ## Applications ### Differential geometry One of the principal applications of the exterior algebra is in differential geometry where it is used to define the bundle of differential forms on a smooth manifold. In the case of a (pseudo-)Riemannian manifold, the tangent spaces come equipped with a natural quadratic form induced by the metric. Thus, one can define a Clifford bundle in analogy with the exterior bundle. This has a number of important applications in Riemannian geometry. Perhaps more importantly is the link to a spin manifold, its associated spinor bundle and spinc manifolds. ### Physics Clifford algebras have numerous important applications in physics. Physicists usually consider a Clifford algebra to be an algebra spanned by matrices γ0,…,γ3 called Dirac matrices which have the property that $\gamma_i\gamma_j + \gamma_j\gamma_i = 2\eta_{ij}\,$ where η is the matrix of a quadratic form of signature (1,3). These are exactly the defining relations for the Clifford algebra Cℓ1,3(C) (up to an unimportant factor of 2), which by the classification of Clifford algebras is isomorphic to the algebra of 4 by 4 complex matrices. The Dirac matrices were first written down by Paul Dirac when he was trying to write a relativistic first-order wave equation for the electron, and give an explicit isomorphism from the Clifford algebra to the algebra of complex matrices. The result was used to define the Dirac equation and introduce the Dirac operator. The entire Clifford algebra shows up in quantum field theory in the form of Dirac field bilinears. The use of Clifford algebras to describe quantum theory has been advanced among others by Mario Schönberg,[13] by David Hestenes in terms of geometric calculus, by David Bohm and Basil Hiley and co-workers in form of a hierarchy of Clifford algebras, and by Elio Conte et al.[14][15] ### Computer Vision Recently, Clifford algebras have been applied in the problem of action recognition and classification in computer vision. Rodriguez et al.[16] propose a Clifford embedding to generalize traditional MACH ﬁlters to video (3D spatiotemporal volume), and vector-valued data such as optical flow. Vector-valued data is analyzed using the Clifford Fourier transform. Based on these vectors action filters are synthesized in the Clifford Fourier domain and recognition of actions is performed using Clifford Correlation. The authors demonstrate the effectiveness of the Clifford embedding by recognizing actions typically performed in classic feature ﬁlms and sports broadcast television. ## Notes 1. W. K. Clifford, "Preliminary sketch of bi-quaternions, Proc. London Math. Soc. Vol. 4 (1873) pp. 381-395 2. W. K. Clifford, Mathematical Papers, (ed. R. Tucker), London: Macmillan, 1882. 3. see for ex. Z. Oziewicz, Sz. Sitarczyk: Parallel treatment of Riemannian and symplectic Clifford algebras. In: Artibano Micali, Roger Boudet, Jacques Helmstetter (eds.): Clifford Algebras and their Applications in Mathematical Physics, Kluwer Academic Publishers, ISBN 0-7923-1623-1 [Amazon-US | Amazon-UK], 1992, p. 83 4. Mathematicians who work with real Clifford algebras and prefer positive definite quadratic forms (especially those working in index theory) sometimes use a different choice of sign in the fundamental Clifford identity. That is, they take v2 = −Q(v). One must replace Q with −Q in going from one convention to the other. 5. Thus the group algebra K[Z/2] is semisimple and the Clifford algebra splits into eigenspaces of the main involution. 6. The Z-grading is obtained from the N grading by appending copies of the zero subspace indexed with the negative integers. 7. Technically, it does not have the full structure of a Clifford algebra without a designated vector subspace. 8. We are still assuming that the characteristic is not 2. 9. The opposite is true when using the alternate (−) sign convention for Clifford algebras: it is the conjugate which is more important. In general, the meanings of conjugation and transpose are interchanged when passing from one sign convention to the other. For example, in the convention used here the inverse of a vector is given by v−1 = vt / Q(v) while in the (−) convention it is given by v−1 = v / Q(v). 10. See the references to Schönberg's papers of 1956 and 1957 as described in section "The Grassmann–Schönberg algebra $G_n$" of:A. O. Bolivar, Classical limit of fermions in phase space, J. Math. Phys. 42, 4020 (2001) doi:10.1063/1.1386411 11. E. Conte: The solution of EPR paradox in quantum mechanics, in: Fundamental Problems of Natural Sciences and Engineering, pp. 271–204, Saint Petersburg, 2002 12. Elio Conte: On some considerations of mathematical physics: May we identify Clifford algebra as a common algebraic structure for classical diffusion and Schrödinger equations? Adv. Studies Theor. Phys., vol. 6, no. 26 (2012), pp. 1289–1307 13. Rodriguez, Mikel; Shah, M (2008). "Action MACH: A Spatio-Temporal Maximum Average Correlation Height Filter for Action Classification". Computer Vision and Pattern Recognition (CVPR). ## References • Bourbaki, Nicolas (1988), Algebra, Berlin, New York: Springer-Verlag, ISBN 978-3-540-19373-9 [Amazon-US | Amazon-UK], section IX.9. • Carnahan, S. Borcherds Seminar Notes, Uncut. Week 5, "Spinors and Clifford Algebras". • Garling, D. J. H. (2011), Clifford algebras. An introduction, London Mathematical Society Student Texts 78, Cambridge: Cambridge University Press, ISBN 978-1-107-09638-7 [Amazon-US | Amazon-UK], Zbl 1235.15025 • Lam, Tsit-Yuen (2005), Introduction to Quadratic Forms over Fields, Graduate Studies in Mathematics 67, American Mathematical Society, ISBN 0-8218-1095-2 [Amazon-US | Amazon-UK], MR 2104929, Zbl 1068.11023 • Lawson, H. Blaine; Michelsohn, Marie-Louise (1989), Spin Geometry, Princeton, NJ: Princeton University Press, ISBN 978-0-691-08542-5 [Amazon-US | Amazon-UK]. An advanced textbook on Clifford algebras and their applications to differential geometry. • Lounesto, Pertti (2001), Clifford algebras and spinors, Cambridge: Cambridge University Press, ISBN 978-0-521-00551-7 [Amazon-US | Amazon-UK] • Porteous, Ian R. (1995), Clifford algebras and the classical groups, Cambridge: Cambridge University Press, ISBN 978-0-521-55177-9 [Amazon-US | Amazon-UK] • Jagannathan, R., On generalized Clifford algebras and their physical applications, arXiv:1005.4300 • Sylvester, J. J., (1882), Johns Hopkins University Circulars I: 241-242; ibid II (1883) 46; ibid III (1884) 7-9. Summarized in The Collected Mathematics Papers of James Joseph Sylvester (Cambridge University Press, 1909) v III .online and further. Content in this section is authored by an open community of volunteers and is not produced by, reviewed by, or in any way affiliated with MedLibrary.org. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Clifford algebras", available in its original form here: http://en.wikipedia.org/w/index.php?title=Clifford_algebras • ## Finding More You are currently browsing the the MedLibrary.org general encyclopedia supplement. To return to our medication library, please select from the menu above or use our search box at the top of the page. In addition to our search facility, alphabetical listings and a date list can help you find every medication in our library. • ## Questions or Comments? If you have a question or comment about material specifically within the site’s encyclopedia supplement, we encourage you to read and follow the original source URL given near the bottom of each article. You may also get in touch directly with the original material provider. • ## About This site is provided for educational and informational purposes only and is not intended as a substitute for the advice of a medical doctor, nurse, nurse practitioner or other qualified health professional.

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http://en.m.wikipedia.org/wiki/Specific_detectivity

# Specific detectivity Specific detectivity, or D*, for a photodetector is a figure of merit used to characterize performance, equal to the reciprocal of noise-equivalent power (NEP), normalized per unit area. Specific detectivity is given by $D^*=\frac{\sqrt{A}}{NEP}$, where $A$ is the area of the photosensitive region of the detector. Its common units are $cm \cdot \sqrt{Hz}/ W$, also called the Jones in honor of R. Clark Jones who defined this magnitude.[1][2] Given that noise-equivalent power can be expressed as a function of the responsivity $\mathfrak{R}$ (in units of $A/W$ or $V/W$) and the noise spectral density $S_n$ (in units of $A/Hz^{1/2}$ or $V/Hz^{1/2}$) as $NEP=\frac{S_n}{\mathfrak{R}}$, it's common to see the specific detectivity expressed as $D^*=\frac{\mathfrak{R}\cdot\sqrt{A}}{S_n}$. The unit Jones is now commonly used with the D* figure of merit. It is often useful to express the specific detectivity in terms of relative noise levels present in the device. A common expression is given below. $D^* = \frac{q\lambda \eta}{hc} \left[\frac{4kT}{R_0 A}+2q^2 \eta \Phi_b\right]^{-1/2}$ With q as the electronic charge, $\lambda$ is the wavelength of interest, h is Planck's constant, c is the speed of light, k is Boltzmann's constant, T is the temperature of the detector, $R_0A$ is the zero-bias dynamic resistance area product (often measured experimentally, but also expressible in noise level assumptions), $\eta$ is the quantum efficiency of the device, and $\Phi_b$ is the total flux of the source (often a blackbody) in photons/sec/cm². ## References 1. R. C. Jones, "Quantum efficiency of photoconductors," Proc. IRIS 2, 9 (1957) 2. R. C. Jones, "Proposal of the detectivity D** for detectors limited by radiation noise," J. Opt. Soc. Am. 50, 1058 (1960), doi:10.1364/JOSA.50.001058) This article incorporates public domain material from the General Services Administration document "Federal Standard 1037C". This physics-related article is a stub. You can help Wikipedia by expanding it. ↑Jump back a section

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http://mathhelpforum.com/geometry/167244-finding-area-shaded-region.html

# Thread: 1. ## Finding the area of the shaded region. Hi I'm a college student and I'm having a bit of difficulty with my Solid Mensuration homework. Can someone help me out? I'm trying to find the area of the shaded region in the problems which I've posted pictures of in this thread. I'm having a hard time because I think the data given is insufficient, but maybe I'm wrong so I'm asking for help Attached Thumbnails 2. Originally Posted by purplesky828 Hi I'm a college student and I'm having a bit of difficulty with my Solid Mensuration homework. Can someone help me out? I've attached a picture of the problems I'm confused by. What exactly are you trying to do? Please post the whole question, not just the pictures. Edit: The OP has since edited his/her post. 3. Originally Posted by purplesky828 I'm trying to find the area of the shaded region in the problems which I've posted pictures of in this thread. I'm having a hard time because I think the data given is insufficient, but maybe I'm wrong so I'm asking for help I agree with you that technically the data is insufficient. However, a good guess on #1 would be $(85)(42.5)-\dfrac{(42.5)^2}{2}$. Do you see why? 4. Originally Posted by Plato I agree with you that technically the data is insufficient. However, a good guess on #1 would be $(85)(42.5)-\dfrac{(42.5)^2}{2}$. Do you see why? Yes I see how it would be a good guess. I was thinking of the same thing but I was unsure because the data wasn't given. Thank you very much for your help! 5. Originally Posted by purplesky828 Hi I'm a college student and I'm having a bit of difficulty with my Solid Mensuration homework. Can someone help me out? I'm trying to find the area of the shaded region in the problems which I've posted pictures of in this thread. I'm having a hard time because I think the data given is insufficient, but maybe I'm wrong so I'm asking for help

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http://math.stackexchange.com/questions/38867/finite-neighborhood-base?answertab=active

# finite neighborhood base My question might seem obvious but still I want to sure if the result I shall state below is true: if the neighborhood base in each point of a space say X is finite. Then we have a discrete topology. Many thanks and apologies in case I just stated some insanity above :) - ## 2 Answers No, this is not true. It would imply, for instance, that every finite topological space is discrete, but this is far from the case. For a trivial example, take the trivial (or indiscrete) topology on any finite set $X$ of cardinality greater than one. Or take the Sierpinski space, i.e., the unique two-point space (up to homeomorphism) with one open point and one closed point.

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http://mathoverflow.net/questions/33420/norms-and-operator-algebras

## Norms and Operator Algebras ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) We know that two unital matrix algebras are isomorphic only if completely isometric. But when is a unital subspace of $N \times N$ matrices completely isometric to a matrix algebra? Can we describe this using matrix norms? - What (if anything) does Paulsen's book say? – Yemon Choi Jul 26 2010 at 18:06 3 It may also be worth giving a brief account of your motivation for this question, which sources you're familiar with, or what level you are at. This would make it easier for people to focus their answers to your question. – Yemon Choi Jul 26 2010 at 18:09

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http://mathoverflow.net/questions/70993?sort=votes

## Proof systems and their hierarchy ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Why ZFC is placed in top of the proof system hierarchy? How it can p-simulate other systems? - 2 I really would appreciate some definition links to "proof system hierarchy" and "p-simulation". – Hans Stricker Jul 23 2011 at 15:18 2 @Hans Stricker, check the Wikipedia pages for "proof complexity" (en.wikipedia.org/wiki/Proof_complexity) and "propositional proof system" (en.wikipedia.org/wiki/Propositional_proof_system). – Kaveh Aug 2 2011 at 6:19 ## 2 Answers It is an open problem if there is an optimal propositional proof system. Therefore we don't know if ZFC as a propositional proof system is optimal either. ZFC as propositional proof system can p-simulate any propositional proof system whose soundness (if there is proof for a formula then the formula is true) as a propositional proof system is provable in ZFC. The trick (which I think is due to Steve Cook) is based on the fact that proofs are concrete finite objects, if there is a proof for a formula in a proof system, then ZFC can prove its existence (i.e. ZFC is $\Sigma_1$-complete), combining this with the provability of the soundness we derive the truth of the encoded formula in ZFC. The rest of the argument is translating this first order proof in ZFC to a propositional proof in ZFC as propositional proof system and proving the equivalence of a propositional formula with the formula itself in the proof system. This can be done in $\mathsf{TC^0}$-Frege and any system that contains it. See Logical Foundations of Proof Complexity, 2010 by Cook and Nguyen for the details. - 1 Strictly speaking, it's the function $F$ that maps ZFC-proofs of $T$ to $T$ (where $T$ is a tautology) that is the propositional proof system, not ZFC itself. Note also that the assertion that $F$ is a propositional proof system isn't provable in ZFC itself; you need to assume the soundness of ZFC. This is why people put ZFC at the top; conceivably, there could be "stronger" propositional proof systems $F$ that ZFC can't prove are actually propositional proof systems. But if you think that ZFC represents the outer limit of mathematical knowledge, you might have qualms about such $F$. – Timothy Chow Jul 22 2011 at 22:24 @Timothy, there are two equivalent definitions for a pps, one is what you described, the other one is a proof checker program (which the one I found more natural). Yes, strictly speaking, but I didn't want to go into too much details. One can add axioms to obtains a possibly stronger pps, e.g. large cardinal axioms. Note that we don't need the full soundness, we only need a very restricted form of it. In fact, as you know :), some proof complexity theorist conjecture that EF is a plausible candidate for the optimal proof system, in which case ZFC will be just equal to it. – Kaveh Aug 2 2011 at 6:13 ### 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. Iam really sorry for this delayed reply...I just digged up a bit on this topic. Here are 'my views' , Iam novice in this area, if I am wrong correct me. Frege systems ( also hilbert systems) is a better proof system than Gentzen systems as it captures "Human thinking" also. Now, if we add some rules which , say act as lemmas, they readily help us in proving a problem in the proof system. This lemma need not be proved in this problem again , we can just quote it.The Frege + rules = EXTENDED FREGE system. There is strong parallelism between Proof system and Circuit complexity proof system wherein we can simulate one problem into other.The circuit complexity class heirarchy is like Ac , NC , P/poy , TC . We concentrate on P\Poly what it means. P\poly solves the problem in polynomial time given "some polynomial time advice function". If we closely observe these advice functions, they are similar to 'lemmas' or 'rules' in extended frege system. Thus P\poly is equivalent to Extended Frege system.Is this the best proof system ? I think no because if we want to compare two outputs produced by applying different lemmas in this system we don;t have any parameter to compare. So we have Zorn's Lemma in our hand where it says that if we have upper bound in every chain, then we have maximal element in that set. This for us nothing but the "Axiom Of Choice". The proof system with the axoim of choice is ZFC. IS THIS THE ULTIMATUM ? No one knows...... - 2 As a non-expert in this particular area, I much prefer to read answers of experts than to write my own attempts. – Andrej Bauer Aug 26 2011 at 9:45 I am unable to see how this addresses the question (which, according to Kaveh's answer, seems to be based on false premises). – S. Carnahan♦ Aug 29 2011 at 2:10

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http://mathhelpforum.com/differential-geometry/179726-null-sequences.html

# Thread: 1. ## null sequences. Prove or disprove: If $(x_ny_n)$ is a null sequence then one or both of $(x_n), \, (y_n)$ is(are) null. My attempt: if $(x_n)\, and \, (y_n)$ both are null then certainly $(x_ny_n)$ is null, no problem with that. I assumed that $(x_n)$ is not null, $(y_n)$ is not null and as given, $(x_ny_n)$ is null. I tried to bring a contradiction since i have a feeling that the statement in the question is true. since its assumed that $(x_n)$ is not null so: there exists $\epsilon_1>0$ such that for all $N$ there exists $n \geq N$ such that $|x_n|>\epsilon_1$ similarly: there exists $\epsilon_2>0$ such that for all $N$ there exists $n \geq N$ such that $|y_n|>\epsilon_2$ and since $(x_ny_n)$ is null: there exists $N$ such that $n \geq N \Rightarrow |x_ny_n|<\epsilon_1 \epsilon_2$ now since i can't say that $|x_n|>\epsilon_1 \, and \, |y_n|>\epsilon_2$ occur at the same $n$ i can't find a contradiction. someone help. 2. Originally Posted by abhishekkgp Prove or disprove: If $(x_ny_n)$ is a null sequence then one or both of $(x_n), \, (y_n)$ is(are) null. Assuming both converge, suppose that neither is a null sequence. Then $(x_n)\to X\ne 0$ and $(y_n)\to Y\ne 0$. What can you say about $(x_ny_n)\to~ ?$ What does that tell you. 3. Originally Posted by Plato Assuming both converge, suppose that neither is a null sequence. Then $(x_n)\to X\ne 0$ and $(y_n)\to Y\ne 0$. What can you say about $(x_ny_n)\to~ ?$ What does that tell you. probably $(x_ny_n) \rightarrow XY \neq 0$ and probably null sequences converge to 0 and hence a contradiction. But convergence has not yet been discussed in the book i am reading(first course in real analysis- stirling K berberian) so i can only guess. 4. Originally Posted by abhishekkgp But convergence has not yet been discussed in the book i am reading(first course in real analysis- stirling K berberian) so i can only guess. How very odd. I know Berberian's book on measure theory and find it odd. Pray tell then what is a null sequence? 5. Originally Posted by Plato How very odd. I know Berberian's book on measure theory and find it odd. Pray tell then what is a null sequence? a sequence $(x_n)$ in $\mathbb{R}$ is said to be null if: for all $\epsilon>0$ there exists $N$ such that $n \geq N \Rightarrow |x_n|<\epsilon$ 6. Originally Posted by abhishekkgp a sequence $(x_n)$ in $\mathbb{R}$ is said to be null if: for all $\epsilon>0$ there exists $N$ such that $n \geq N \Rightarrow |x_n|<\epsilon$ OK go back to the OP. Let $\varepsilon = \frac{{\varepsilon _1 \varepsilon _2 }}{2}$. Make $|x_ny_n|<\varepsilon$ then $\varepsilon _1 \varepsilon _2\le |x_n||y_n|=|x_ny_n|<\varepsilon$. That is a contradiction. 7. Originally Posted by Plato OK go back to the OP. Let $\varepsilon = \frac{{\varepsilon _1 \varepsilon _2 }}{2}$. Make $|x_ny_n|<\varepsilon$ then $\varepsilon _1 \varepsilon _2\le |x_n||y_n|=|x_ny_n|<\varepsilon$. That is a contradiction. the problem is that i can't( or yet not able to) say that there exists an $n$ such that $|x_n||y_n|> \epsilon_1 \epsilon_2$. I can only say that $|x_{n_1}||y_{n_2}|>\epsilon_1 \epsilon_2$ and that's why i can't contradict anything. 8. Originally Posted by abhishekkgp the problem is that i can't( or yet not able to) say that there exists an $n$ such that $|x_n||y_n|> \epsilon_1 \epsilon_2$. I can only say that $|x_{n_1}||y_{n_2}|>\epsilon_1 \epsilon_2$ and that's why i can't contradict anything. I think that I misread the OP. Consider this example. $x_n = \left\{ {\begin{array}{rl} {1,} & {\text{n is even}} \\ {\frac{1}{n},} & {\text{ n is odd}} \\ \end{array} } \right.$ and $y_n = \left\{ {\begin{array}{rl} {1,} & {\text{n is odd}} \\ {\frac{1}{n},} & {\text{ n is even}} \\ \end{array} } \right.$ What is $x_ny_n~?$ 9. Originally Posted by Plato I think that I misread the OP. Consider this example. $x_n = \left\{ {\begin{array}{rl} {1,} & {\text{n is even}} \\ {\frac{1}{n},} & {\text{ n is odd}} \\ \end{array} } \right.$ and $y_n = \left\{ {\begin{array}{rl} {1,} & {\text{n is odd}} \\ {\frac{1}{n},} & {\text{ n is even}} \\ \end{array} } \right.$ What is $x_ny_n~?$ so it can happen that if we have $(x_n)$ not null, $(y_n)$ not null, we still have $(x_ny_n)$ as null!! thank you for this.

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http://skullsinthestars.com/2011/10/09/faster-than-a-speeding-photon-precursors-test-whether-light-can-be-faster-than-light/

About these ads The intersection of physics, optics, history and pulp fiction ## Faster than a speeding photon? Precursors test whether light can be faster than light Posted on October 9, 2011 by Over the past two weeks, the biggest physics news has been the apparent observation of neutrinos (nearly undetectable subatomic particles) moving faster than the vacuum speed of light. At first glance, this would seem to violate Einstein’s special theory of relativity, which fixes the vacuum speed of light at $c = 3\times 10^8$ meters per second, and as a consequence makes it in principle impossible to travel faster than that speed. The theoretical implications are in fact a bit more subtle, but before we worry too much about those implications the experimental results will need to be checked carefully and independently verified. While we wait, it is worth noting that in June 0f 2011 a group of researchers performed an experiment to see if light itself could move faster than light! In particular, the scientists used a little known optical phenomenon known as an optical precursor to see if individual photons might travel faster than $c$ while propagating in a material. In the end, the experiment suggests that these single photons did not in fact violate Einstein’s speed limit, though the results still got a significant amount of press. The response of many physicists to the news was a collective, “Well, duh!” The prevailing attitude seems to have been: “What’s so interesting about proving something we already knew?” In this post I’d like to explore that question a little bit, and explain how some uncertainty remains about the behavior of light in materials. Along the way, we’ll introduce the fascinating phenomenon of precursors, and see how they can be used both to probe the nature of matter as well as the nature of light. To start, we need to understand a bit about how light and matter interact with one another, and how this relates to the speed of light in matter. This is a surprisingly tricky subject, which I’ve covered several times previously on this blog; we’ll review the important aspects, with a few additional details. Let’s talk about how we measure the speed of an object first. If we’re looking at the motion of a rigid object, like a speeding car or a thrown baseball, the speed $v$ can be determined simply by measuring how much time $\Delta t$ it takes for an object to travel a distance $\Delta x$. The speed is simply the distance divided by the time: $\mbox{speed} = \mbox{distance}/\mbox{time} = \Delta x/\Delta t$. There’s a small subtlety to this definition: cars and baseballs are extended objects! To accurately measure an object’s speed, we have to be consistent in how we define its position. For a car driving down the road, for instance, we should do all measurements of its position from a fixed position, such as the front bumper, to measure the speed. But what do we do when the object doesn’t have a fixed position on it? For example, what is the best way to measure the speed of a hurled bucketful of water? Water is, of course, not a rigid body, and the volume of water changes shape and spreads out as it travels. It is strange to realize, after some thought, that a blob of water like this doesn’t really have a well-defined speed. We arrive at the same problem with a pulse of light traveling in matter: in general, a light pulse has no “fixed point” upon it, and it can change shape as it travels. There isn’t a single definition for the speed of light in matter that is useful in all cases. There are three different descriptions of the speed* of light used in physics, each with its own significance and usefulness. We can introduce each of these by looking at similar definitions for our bucketful of water. We imagine that our volume of water may be broken up into a collection of well-defined droplets. The most straightforward and thorough thing to do is to measure the speed of each individual droplet: the velocity of each droplet will be referred to as its phase velocity. If there is only one or a handful or drops, this set of numbers will be a good description, but for a large number of droplets this collection of numbers will not necessarily tell us anything about the motion of the entire bulk of the water. A better option is to define the speed of the body of water by its center of mass. This definition, referred to as the group velocity, works quite well in many cases, but can also be misleading. If we’re interested, for instance, in the time it takes for water to first reach its target (say a fire), there may be droplets well ahead of the center of mass and the use of the group velocity can overestimate the time. We may also choose to define the speed of the body of water as the speed of the fastest drop in the collection; this definition is called the signal velocity. One problem with the signal velocity is the opposite of that of the group velocity: if we need to know when the bulk of a body of water reaches a fire, the signal velocity will underestimate the time. Physicists use roughly analogous definitions to characterize the speed of light in matter. Instead of “droplets”, a pulse of light can be decomposed into a collection of waves of different frequencies (colors). It can be shown that a narrow pulse of light necessarily consists of a very broad spectrum (range of colors), while a very broad, slowly varying pulse has a narrow spectrum. Also, and significant for our later discussion, a pulse with a very sharp edge necessarily has a very broad spectrum, as well. This idea is illustrated crudely below: Atoms respond differently to light of different frequencies, in a phenomenon known as dispersion. The result of this is that some frequencies of light are absorbed more than others, and in general every frequency of light travels at a different speed. The term phase velocity is used to refer to the different speeds of each frequency of light. For a pulse narrow in frequency (like the upper one above), this can sometimes serve as an approximation to the overall speed of light. However, in many circumstances this phase velocity can be faster than the vacuum speed of light $c$, and therefore does not accurately represent the speed of the pulse. Because of difficulties with the phase velocity, the group velocity is the standard method of defining the velocity of a pulse in a medium. It may be considered, in essence, the speed of the “center of mass” of the pulse. However, if the shape of the pulse is highly distorted as it propagates, the group velocity can lose all useful meaning. Also, it has been shown that the group velocity can be greater than the vacuum speed of light, a result that generated quite a bit of controversy at first. The absolute speed of light in a medium is the signal velocity, which is loosely defined as the speed at which information can be conveyed in a medium, or the fastest possible speed that the front of a light signal can travel. This is generally thought to be no faster than the vacuum speed of light $c$, in agreement with Einstein’s special theory of relativity. But is $c$ the fastest speed of light in matter? It seems likely, but a few gaps in our knowledge leave it a partially open question. The quantum-mechanical nature of light-matter interactions leaves open the possibility that some “quantum weirdness” allows Einstein’s speed limit to be broken in a subtle way. A more concrete concern involves our understanding of how light propagates in matter. It is well-known that the “causality” of a light signal — and the absolute speed of light — is built into the frequency-by-frequency response of light to matter. If one knows the speed of light, and the absorption properties of light, for every frequency, one can determine whether or not $c$ is the top speed. But we don’t know these properties for every frequency! In particular, we don’t know exactly what happens to light in matter for arbitrary high frequencies (frequency approaches infinity) or for arbitrary low frequencies (frequency approaches zero). There is always the possibility that some small violation of relativity is “hiding” in these extreme frequency ranges. This is where the idea of optical precursors becomes useful. Let’s consider the temporal and spectral properties of another pulse, with a square envelope: This pulse has a large central peak in frequency, but also has long frequency “tails” that stretch out, to some degree, to arbitrarily high frequencies (to infinity) and arbitrarily low frequencies (to zero). These tails arise due to the instantaneous start and end of the pulse in time: the rapid rise and fall that give the “square pulse” its name. At the beginning of the 20th century, physicists Arnold Sommerfeld and Léon Brillouin independently performed theoretical work to investigate what happens to such sharp-edged pulses as they propagate a long distance through matter. In 1914, they each published results** showing that the main body of the pulse (traveling at the group velocity) is preceded by faster-moving waves, now known as “precursors”. The precursors are general broken into two types: the Sommerfeld precursor, which actually travels at the vacuum speed of light $c$, and the Brillouin precursor, which travels at the speed of light $c/n(0)$, where $n(0)$ is the refractive index of the medium at zero frequency. A crude illustration of the arrangement of precursors is shown below: The most remarkable thing about the precursor fields is that they are very weakly absorbed by matter, much less so than the main body of the pulse. If we send our original square pulse through a very thick piece of absorbing material, the main body will be almost completely absorbed while the precursors will be absorbed only weakly. This fact has led to some researchers suggesting that precursors could be used to “see” through normally opaque objects, like clouds. It took quite a few years for Sommerfeld and Brillouin’s theoretical work to be confirmed. The first observation of precursors was performed in 1969, using microwaves***. Since then, they have been observed for a variety of wavelengths, and have been observed even in “mundane” materials such as water****. What is the origin of these precursors? I wasn’t able to find a simple explanation for them in the literature, perhaps not surprising in terms of their mathematical complexity. I would like to introduce a (hopefully accurate) description of them, in what I will refer to as the “pendulum slapping” model! We have all been taught, at some point in our education, the planetary model of the atom. In this model, an atom consists of a positively-charged nucleus surrounded by one or more orbiting negatively-charged electrons, much like the planets in the solar system orbit the Sun: With our modern understanding of quantum mechanics, we know that this model isn’t a terribly good one: electrons act much more like “clouds” of negative charge centered on the nucleus, and don’t “orbit” in a well-defined sense. However, the planetary model does get one thing right: electrons have a characteristic frequency associated with their motion, much like the planets each orbit the Sun with a characteristic frequency (the Earth orbits the Sun with a frequency of 1/365 days). In a very loose sense, we can picture atoms (and molecules) as collections of oscillating electron pendulums, each vibrating with a characteristic frequency: What happens when we apply an oscillating electric field (i.e. a light wave) to our pendulum? The electric field applies a force to the electron, driving its oscillation. The effectiveness of the electric field in moving the pendulum depends on how close its frequency is to the pendulum frequency. The pendulum will have the largest swing when the electric field matches its own frequency; this is like a child pumping his legs on a swing to increase his motion. Any other frequency of the electric field will be less effective in moving the pendulum, but pretty much every frequency will force some oscillation in it. The energy that is imparted to the pendulum is lost by the electric field, and consequently the light wave; this is the origin of the absorption of light by matter. There are two extreme limits of frequency in which no energy is transferred to the pendulum: the limit of infinite frequency and the limit of zero frequency. We can visualize what happens in these cases by imaging that we are “slapping” a pendulum with our hands on either side (or you can try it yourself, if you have a pendulum). When we slap a pendulum at a frequency much higher than its characteristic frequency, a push on the left is almost immediately canceled by a push on the right — the pendulum doesn’t move, and we’ve transferred no energy to it! In the limit of very low frequency, the pendulum moves, but doesn’t oscillate: we “lift” it to the right with our left hand, then slowly lower it back to its rest position and then “lift” it to the left with our right hand. At no point does the pendulum oscillate freely, and therefore we have transferred no energy to it. These limits, in which no energy is transferred to the pendulum, are the origins of the precursors. We have seen that a square wave has components of arbitrarily high frequency and arbitrarily low frequency, and these components are only very weakly absorbed by the medium. The high frequency components of the wave result in the Sommerfeld precursor, and the low frequency components of the wave result in the Brillouin precursor. For tests of the absolute speed of light, the Sommerfeld precursor is of particular interest, because (a) it tests the whether a violation of relativity is “hiding” at high frequencies, and (b) the speed of a precursor is theoretically supposed to be equal to $c$, making “faster than light” violations relatively easy to spot. All of this brings us back at last to the June paper in Physical Review Letters on the “Optical precursor of a single photon”, by a research group in Hong Kong. A photon is a single quantum particle of light; all of the discussion of precursors up to this point have concerned pulses consisting of many, many photons. Two questions arise when considering precursors and single photons: (1) Do precursors even exist for a single photon? One would naturally be inclined to say “yes”, but it may be that, on a quantum level, precursors inherently involve the interaction of many photons at once. (2) Can single photons travel faster than $c$? One would be inclined to say “no” in this case, but again the behavior of single photon precursors (if they exist) might be subtly different than a group of many photons. The Hong Kong researchers investigated these possibilities by producing coupled pairs of photons using the following experimental configuration (adapted and simplified from the article): The fundamental components of the system are a pair of magneto-optical traps (MOT) containing Rubidium atoms. Magneto-optical traps use light and magnetic fields to localize a group of atoms and cool them to a low temperature, forming a low-density medium. The first MOT is excited by a pair of lasers — a pump beam and a coupling beam — and through a complicated light-matter interaction pairs of photons are produced. The physics of this production is too complicated to discuss here, but the result is a higher frequency “Stokes photon” and a lower-frequency “anti-Stokes photon”. These photons are produced at the same time and are therefore correlated in time. The Stokes photon is picked up by detector 1 and sends a signal to a function generator, which triggers an electro-optic modulator that the anti-Stokes photon passes through. This modulator “chops off” the front end of the anti-Stokes photon signal, producing a sharp-edged pulse necessary for precursor generation*****. This sharp-edged pulse propagates into the second MOT, in which its waveform is expected to take a precursor shape. It is essential for this experiment that a single anti-Stokes photon be measured at a time; after the second MOT, the light signal is split by a beam-splitter and sent to a pair of detectors, detectors 2 and 3. Because a single photon can only arrive at a single detector, the simultaneous tripping of both detectors implies the presence of more than one photon and the event is thrown out. The time of arrival of the single anti-Stokes photons can be tallied, producing a profile of the average behavior of the photons. The coupling laser illuminating the second MOT could modify the optical properties of the Rubidium atoms into one of two configurations. In the first configuration, the absorption properties of the medium can be completely suppressed, in a technique called electromagnetic induced transparency (EIT). With EIT, both the main body of the pulse and the precursor signal make it through the MOT and can be measured by the detectors. Furthermore, the main signal is slowed considerably in speed and is separated in time from the optical precursor. In the second configuration, the group velocity of the main signal is faster than $c$, i.e. the group velocity is “superluminal”. However, this main signal is highly absorbed. What were the results of the experiments? The researchers measured the speed of the main body of the pulse and the precursor for both MOT configurations. The precursor was clearly visible in both cases, and its speed was found to be exactly $c$, regardless of the MOT behavior. Therefore no true “faster than light” behavior was observed, even when the group velocity was greater than the vacuum speed of light. An important aspect of this result is a partial answer to a debate in quantum information theory — how fast does a single photon transmit information? As we’ve noted, there are multiple definitions of the speed of light in matter, and it has remained a bit of a mini-mystery (at least to some) which of these described the speed at which information is transmitted. The observation that a photon does have a precursor signal suggests that single photons can travel at the vacuum speed of light in matter, at least under the right conditions. The result isn’t as earth-shattering as the possible discovery of superluminal neutrinos, but it does highlight a number of unusual aspects of optics, and further strengthens our understanding of both relativity and quantum mechanics! ************************ * In physics, “velocity” is used to refer to the vectorial motion of an object: not only how fast it is going, but in what direction. ”Speed” is used to refer to the magnitude of velocity, or simply how fast the object is going. We use the terms interchangeably here in the sense of “speed”. ** A. Sommerfeld, Ann. Phys. (Leipzig) 349 (1914), 177. L. Brillouin, Ann. Phys. (Leipzig) 349 (1914), 203. *** P. Pleshko and I. Palócz, “Experimental observation of Sommerfeld and Brillouin precursors in the microwave domain,” Phys. Rev. Lett. 22 (1969), 1201. **** S-H. Choi and U. Österberg, “Observation of optical precursors in water,” Phys. Rev. Lett. 92 (2004), 193903. ***** Actually, the anti-Stokes photon will already have a precursor signal due to its propagation in MOT 1! This precursor is also chopped off by the EOM. ************************* Zhang, S., Chen, J., Liu, C., Loy, M., Wong, G., & Du, S. (2011). Optical Precursor of a Single Photon Physical Review Letters, 106 (24) DOI: 10.1103/PhysRevLett.106.243602 About these ads ### Like this: Like Loading... This entry was posted in Optics, Physics, Relativity. Bookmark the permalink. ### 13 Responses to Faster than a speeding photon? Precursors test whether light can be faster than light 1. Pingback: Faster than a speeding photon? Precursors test whether light can be faster than light | benjamin junior 2. Pingback: RB Editor’s Selections: The Expanding Universe, Triggered Earthquakes, Statistical Badness, and More Neutrinos 3. CHRIS says: Typo in last paragraph. “are” should be “our”. • skullsinthestars says: Ah, thanks! Fixed. 4. Ray Kilburn says: Previous measurements of light velocity traveling through matter is akin to the photoelectric effect for which Einstein received his Noble prize. The current description is incomplete as electrons are incorrectly described, as in this blog, as a :”pendulum” or a spring is no more accurate than the orbital model. When we correlate the wave description of the electron’s orbit with its particle description, we realize a different view of the model of the atom. Rutherford’s model of the atom, and all subsequent variations, propose an orbital model of sorts with material electrons flying around the proton nucleus. Subatomic physics and quantum theory reduced this model to electron clouds that supported the quantum uncertainty of finding the electron in any particular position around the proton. Quantum Field Theory proposes that perhaps, the electron is not orbiting the proton at all, in this sense, but that the proton is actually inside of the electron! Each subsequent electron “shell” encompasses the previous shell. When light encounters an atom, corresponding wavelengths of the photonic wave are absorbed by the electron bubble of the corresponding frequency. While the time delay between absorption and emission explains normal ionic activity, apparent increase light velocity is directly proportional to the uncertainty of the speed of light to begin with. Keep in mind that Einstein’s velocity of light is ONLY accurate in a vacuum but not accurate within a gravitation field which effect light’s true path thru space as confirmed by relativity. Therefore light could be said to NEVER travel in a straight line by any terrestrial measurement than can be made and therefore its accurate velocity can never be determinate. In absence of a quantum description of gravity, it could be said that no gravity exists within an atom as we see no gravitational effect on particles with the atom and the gravitational effect is clearly only present outside of an atomic system. Therefore, one could postulate that our measure of light velocity within matter cannot be accurately determinate. Unlike a classical particle having mass, the Photon does not emit a rotating spherically symmetrical wave system with a continuous degree of freedom as do material particles. Nor is it normally polarized in a specific directional time orientation as observed in Electrons and Protons. Instead, it appears to interact with particles and antiparticles equally in both time domains. For this reason the Photon is said to be its own antiparticle. Following this reasoning further, one could postulate that one or more of the following is true: 1) the photon exists simultaneously in both time domains, 2) it rotates back and forth between time domains, or 3) is unaffected by time domains. When a Photon is compared to a particle with mass in motion, the Photon cannot be said to have a momentum in a single vector direction, however it appears to be propagated as a directional wave packet. However the Photon appears to propagate its momentum along a three-directional vector corresponding to physical construction of the space-time lattice structure with a definitive wave structure. Along each directional vector there is an angular momentum in the direction of motion called a heliocity. But just like particles having mass, this heliocity is restricted normally to integer values. The exception to this observation is in the case of a polarized photon (neutrino/antineutrino) which has a heliocity value of half-integer values. Because virtual particles imply both the inversion of space and time, both positive and negative energy states could be said to exist. It is only a mathematical trick that requires massless particles to be anti-unitary and anti-linear because the Standard Model of classical physics does not allow negative energy states, that is, energy less than that of the so-called vacuum. Because of this inexplicable paradox, classical physics does not allow the reversal of its time components either. Both cases to be incorrect but too lengthy to describe here. 5. Сергей Морозов says: May I translate it in my blog (moogoo.ru)? 6. Pingback: Links for Tuesday « Galileo's Pendulum 7. Pingback: The Value of Experimenting With What We “Know” | Citizen Scientists League 8. Primoz Gabrijelcic says: I don’t understand how the electro-optic modulator can react quick enough to clip the photon, which is traveling at c. That would imply that the signal that triggers the modulator travels with velocity > c. Or is the anti-Stokes photon traveling through some materil with low “speed of light”? • skullsinthestars says: One of the things I left out of my simple picture of the experimental setup is the delay line the anti-Stokes photon passes through! This photon is coupled into a long optical fiber, and with a long enough fiber the photon can be delayed as long as one likes compared to the modulator trigger — within reason! 9. Ray Kilburn says: The result of any observation of a “superluminal neutrinos” may result in the realization that neutrinos may not really be an independent particle at all as currently considered by the Standard Model and may explain the difficulty in detecting them. Consider how the neutrino was “invented” to explain differences in mass between a neutron and its resulting beta decay model which supposedly results in an electron, proton and a neutrino that apparently flies away at light speed in violation of relativity and doesn’t interact with any matter. What is generally ignored is that the additional mass measured in the neutron may be a result of the retrograde spin of the neutron (which could be considered a micro-model of a hydrogen atom (proton+electron)) and therefore may be relativistic mass and not a real mass at which explains why the neutrino is always considered massless. During beta decay, when the proton and electron are released from the neutron, the remaining relativistic mass spins off as a photon that is moving as a helicoid instead of a plane-wave photon due to the angular momentum imparted upon it from the spin of the originating neutron. Since a helicoid looks like a spring, the photon’s angular momentum tunnels the so-called “neutrino” through all mass without stopping. The is the same principal used by electrons to tunnel thru matter as well. I think that upon further examination we will find that the neutrino is really photon traveling as a helicoid and therefore unpolarized and incapable of interaction with matter. Plane-wave photons which are polarized along two planes run into everything. Real super-luminal communication of information is probably the result that the photon as a discreet entity really doesn’t travel anywhere. Consider that the photon is really at both its origin and destination at the same time, more akin to a string connecting point A to B. The phontonic wave we observe is the jiggling of the string, nothing is really moving anywhere except the wave along the string, the string doesn’t move anywhere. From an outside observer we see the wave moving along the string and measure its velocity at c. From the perspective of the photon, it is already there since zero time dilation occurs and the photon appears (from its perspective) to arrive at the same moment it departed it source. Another point to consider in the measurement of light speed in a vacuum is that true vacuums exist nowhere in reality. Even when we peer into the deepest darkest part of space, we still measure background microwave radiation in 3micron wavelength. Therefore a photon at some energy level exists at all points in space and no point in space could be said not to contain either traditional matter or a photon at some energy level. So no true vacuum could be said to exist nor can be created and therefore light true velocity will always be indeterminate in all true circumstances. Our current measurement of light’s velocity is an approximation. Theories surrounding the big bang suggests that the speed of light varies as the universes expands….the string is getting longer? 10. zephirawt says: My stance is, OPERA results are real and they should be published ASAP. IMO we can model the space-time (brane) with density gradient at the phase interface of two elastic fluids. After then a two kinds of solitons will appear: A) the one, which corresponds to photons and it spreads with slightly lower speed, than the transverse surface waves (which are serving as an analogy of light waves) B) the faster one, which corresponds the neutrinos and it will spread with slightly higher speed, than the surface ripples. The first kind of solitons results from coupling of surface ripples with longitudinal bulk waves of more dense phase, the second one from coupling of surface ripples with longitudinal waves of less dense phase. From this perspective the neutrinos would behave like the superpartners of gamma ray photons, i.e. like the lightweight photinos. 11. Anirudh Kumar Satsangi says: Dear Mr. Jug Suraiya Congratulations for your very thought provoking article ‘Einstein Won’t Mind’ (The Speaking Tree, Oct. 9, 2011, pg. 7). Einstein is great, but Newton is all time great. I would like to quote your few excellent lines from this article before I come to main theme of this article: “Faith and religious beliefs are destinations reached; science and skepticism are journeys without end”. But it’s to be modified. In Veda it is written, ‘Neti, Neti’, ‘Not this, Not this’. So in Vedic religion journeys never end. I consider Vedas the most honest scriptures which do not limit the scope of further exploration of truth. Now I come to the speed of light. This has long been proved since the time of discovery of black holes that the speed of light is not the fastest. Black holes do not allow even light to escape. It means the escape velocity at the black holes is much higher than the speed of light. Black holes are the infinitely dense ball of gravitation force. All creational forces of the universe have originated from the gravitational force field and will end up in it. The speed of light is no doubt fastest in our solar system. The source of light is Sun in our solar system. But how this light is originated? We should study the various stages involved in the formation of a star. Our Sun is also a star. The starting material for the formation of a star is mainly hydrogen gas and helium gas. If the hydrogen cloud contains a very large number of atoms, each atom feels the gravitational pull of all the atoms in the hydrogen cloud. (Here is NO LIGHT) The gas cloud becomes a permanent entity, held together by the mutual attraction of all the atoms present in it. The cloud then begins to contract under its own gravity setting off the process which will convert this huge condensed gas cloud into a star. Such a tight contracting cluster of atoms held in the grip of its own gravity, is called a protostar. The protostar is not yet a star and does NOT emit LIGHT. The temperature of this star is as low as -173 degree C. The force of gravity acting on different atoms in the protostar draws every atom towards centre. As a result, the protostar shrinks in size and its density increases. As the atoms in the protostar fall towards the centre, they pick up speed. Because of the high speed and greater density of atoms, the atoms in the gas cloud collide with one another more frequently, thereby raising its temperature from -173 degree C to about 10 ^7 degree C. At these extremely high temperatures the proton (hydrogen nuclei) at the centre of the protostar collide together and undergo a nuclear fusion to form helium nuclei. In this reaction a tremendous amount of energy is released. This further raises the temperature and pressure. The release of nuclear energy marks the birth of the star. The protostar now beings to GLOW and becomes a STAR. Here at this stage LIGHT is ORIGINATED. Thus light is NOT ETERNAL. It has a beginning and an end. So LIGHT cannot be claimed as Cosmic Constant. However, Gravitation Force is eternal. It is evident from the above description that light is latent before the birth of star. Light originates and become kinetic only after the action of gravitation force. So the speed of light can never exceed the speed of gravitation force. It cannot be ruled out that the speed of gravitation force is infinitely greater than the speed of light at black holes. 1. Gravitation Force is the Ultimate Creator 2. In Scientific Terminology Source of Gravitational Wave is God About these ads • ### Search Skulls in the Stars: • The author of Skulls in the Stars is an associate professor of physics, specializing in optical science, at UNC Charlotte. The blog covers topics in physics and optics, the history of science, classic pulp fantasy and horror fiction, and the surprising intersections between these areas. • ### Twitter Updates • RT @hitRECordJoe: RT @rupertmurdoch: "IRS scandal makes perfect case for flat tax. Nothing could be fairer...” If by "fairer" you mean bet… 39 minutes ago • RT @FreakyFrogger: It's raining diamonds on Neptune and Uranus?!?! spacedaily.com/news/carbon-99… 8 hours ago • RT @sfpelosi: Congress couldve spent the week creatig jobs or ending #sequester that cut embassy security funding. #DoYourJobGOP #WhoDoct… 19 hours ago • I think I finally understand differential geometry enough to do some productive research. Beautiful subject, really. 19 hours ago • ### Blogroll %d bloggers like this:

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http://mathhelpforum.com/advanced-statistics/62051-uniform-distribution-random-independent-variables.html

# Thread: 1. ## uniform distribution of random independent variables okay. my exam is approaching in a few days and I have one last question. Let X1, X2, ... X7 be a set of seven independent random variables, each having the uniform distribution of values from the [3, 6] interval (this is called random independent sample of size 7 from this distribution). (a) Find the expected value and standard deviation of their sum T ≡ X1 + X2 + ... + X7. (b) What is the probability that more than three of these seven random variables will have a value smaller than 1 3? (c) Find the skewness and kurtosis of X1. answers: (a) E(x) = 7(6+3)/2 = 31.5 Var(x) = 7*[(6-3)^2]/2 standard deviation = 2.2913 (b) not sure (c) would skewness be 0 since its uniform? what about kurtosis 2. Originally Posted by chrisc okay. my exam is approaching in a few days and I have one last question. Let X1, X2, ... X7 be a set of seven independent random variables, each having the uniform distribution of values from the [3, 6] interval (this is called random independent sample of size 7 from this distribution). (a) Find the expected value and standard deviation of their sum T ≡ X1 + X2 + ... + X7. (b) What is the probability that more than three of these seven random variables will have a value smaller than 1 3? Mr F says: The question is not clear to me. Is it meant to be 13, 1 or something else? (c) Find the skewness and kurtosis of X1. answers: (a) E(x) = 7(6+3)/2 = 31.5 Var(x) = 7*[(6-3)^2]/2 standard deviation = 2.2913 (b) not sure (c) would skewness be 0 since its uniform? what about kurtosis (a) You're expected to know or derive that $E(X_i) = \frac{6+3}{2} = \frac{9}{2}$ and $Var(X_i) = \frac{(6-3)^2}{12} = \frac{3}{4}$. Then: $E\left(\sum_{i=1}^{7} X_i\right) = \sum_{i=1}^{7} E(X_i) = 7 \cdot \frac{9}{2} = \frac{63}{2}$. $Var\left(\sum_{i=1}^{7} X_i\right) = \sum_{i=1}^{7} Var(X_i) = 7 \cdot \frac{3}{4} = \frac{21}{4}$ (summing the variances follows because the X's are independent). (b) Is there a typo? (c) What definition/type of kurtosis are you using? 3. a) okay, confirmed b) was a typo, its 1/3, not 1 c) nothing else was specified outside of this question 4. Originally Posted by chrisc a) okay, confirmed b) was a typo, its 1/3, not 1 Mr F says: Well if that's the case, what's the probability that any of the X's can be smaller than 1/3 .....?! Look at the interval over which the pdf is not equal to zero! c) nothing else was specified outside of this question c) That's why you go to your class notes or textbook, find the definition/type that's been used and then report the answer in this thread. I had thought that's what you'd do. 5. this is the one thing we were given about for kurtosis, part of our exponential stuff (not uniform like the question asks) kurtosis: [E(X4)−4μE(X3)+6μ2E(X2)−3μ4]/σ4 = ... [24β4−4×6β4+6×2β4−3β4]/β4 = 9 does this help at all? if not, ill just hope this specific type of question doesnt make its way to the exam 6. Originally Posted by chrisc this is the one thing we were given about for kurtosis, part of our exponential stuff (not uniform like the question asks) kurtosis: [E(X4)−4μE(X3)+6μ2E(X2)−3μ4]/σ4 = ... [24β4−4×6β4+6×2β4−3β4]/β4 = 9 does this help at all? if not, ill just hope this specific type of question doesnt make its way to the exam Then calculate each of the bits in the formula: $E(X^n) = \int_3^6 x^n \cdot 1 \, dx$. $\mu = E(X_1) = \frac{9}{2}$. $\sigma = s.d.(X_1) = \frac{\sqrt{3}}{2}$. Now substitute into the formula.

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http://mathhelpforum.com/differential-geometry/90859-separation-axioms.html

# Thread: 1. ## separation axioms prove that : Every $\displaystyle \tau \scriptstyle3$ space is $\displaystyle \tau \scriptstyle 2 \frac{1}{2}$ 2. A good way to start: write out the definitions of $T_3$ and $T_{2 1/2}$ 3. Originally Posted by flower3 prove that : Every $\displaystyle \tau \scriptstyle3$ space is $\displaystyle \tau \scriptstyle 2 \frac{1}{2}$ Definition. A space X is a $T_{2\frac{1}{2}}$-space provided that for each pair x,y of distinct points of X, there exists open sets U and V with disjoint closures such that $x \in U$ and $y \in V$. Lemma 1. X is regular Hausdorff ( $T_3$) if and only if given a point x of X and a neighborhoood U of x, there is a neighborhood W of x such that $\bar{W} \subset U$. Assume X is regular Hausdorff ( $T_3$). Since X is also Hausdorff, for each pair x,y of distinct points of X, there exists disjoint open sets U and V containing x and y, respectively. By hypothesis, it follows that there is a neighborhood W of x such that $\bar{W} \subset U$ and a neighborhood Y of y such that $\bar{Y} \subset V$ by lemma 1. Since U and V are disjoint, we see that $\bar{W}$ and $\bar{Y}$ are disjoint. Now, for each pair x,y of distinct points of X, there exists open sets W and Y with disjoint closures such that $x \in W$ and $y \in Y$.Thus, X is $T_{2\frac{1}{2}}$-space. The converse ("Every $T_{2\frac{1}{2}}$-space is $T_3$") is not necessarily true. The example where a space X is $T_{2\frac{1}{2}}$-space but not $T_3$ can be found in the K-topology on $\mathbb{Re}$, where a closed set $K=\{1/n | n \in \mathbb{Z}^+\}$ (Note that K is closed in the K-topology on $\mathbb{Re}$ ) and {0} cannot be separated by disjoint open sets.

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http://math.stackexchange.com/questions/155409/how-do-i-solve-a-certain-characteristic-system/155494

# How do I solve a certain characteristic system? I am studying PDEs and have the following (seemingly simple) problem: Find a surface that passes through the curve $$x^2+y^2=z=1$$ and is orthogonal to the family of surfaces $$z(x+y)=c(3z+1)\qquad(c\in\Bbb R)$$ After writing down the orthogonality condition (assuming my calculations are correct $(*)$), this yields the following equation: $$u(3u+1)(u_x+u_y)-x-y=0$$ We usually solve such equations by using the method of characteristics, which tells us (using assumption $(*)$ again) to solve the following characteristic system: $$\begin{align}\dot x=&u(3u+1)\\\dot y=&u(3u+1)\\\dot u =&x+y\end{align}$$ Differentiating the last equation of this system with respect to $t$ gives us $\ddot u=\dot x+\dot y$, which using the first two equations gives us $$\ddot u = 2u(3u+1)$$ After staring at this equation for some time, I decided to ask Wolfram|Alpha. The result seems pretty ugly, so the following questions arise: Did I make a mistake/am I missing something? Is my approach correct? How do I proceed? Thanks. - – Weltschmerz Jun 8 '12 at 7:00 ## 1 Answer Here's what I do (I'm probably wrong, but we can compare our procedures. I'm not using the comment box because there's not enough space in it). The intersection of the surfaces would be the set of points $(x,y,z)$ such that \begin{cases}u(x,y)-z=0\\z(x+y)-3cz-c=0,\end{cases} that is, those that lie both on the solution surface and the given surface (determined by $c$, and this must hold for every $c$). The surfaces are orthogonal means (I'm guessing) their respective gradients at the intersection points must be orthogonal. The gradient of the solution surface is $(u_x,u_y,-1)$, and the gradient of the given surface(s) is $(z,z,x+y-3c)$. So the condition is $zu_x+zu_y-x-y+3c=0$. Since this occurs when $z=u$ we can write it as $$uu_x+uu_y=x+y-3c.$$ We can clear $c$ from the intersection equations: $c=\frac{(x+y)u}{1+3u}$, and plugging it in the last equation gives us $$uu_x+uu_y=(x+y)(\frac{1}{1+3u}).$$ So we agree there. As for what follows, the method of characteristics, I think you proceeded correctly. - Thanks for your thoughts. – Dejan Govc Jun 8 '12 at 10:00

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http://physics.stackexchange.com/questions/4334/could-life-survive-a-pole-shift-caused-by-an-asteroid-collision/4424

# Could life survive a pole shift caused by an asteroid collision? Could life on earth survive a large pole shift caused by an asteroid collision? I became aware that there are people who believe that the earth's pole suddenly shifts. That is, its rotational axis changes rather than its magnetic axis. This is listed in wikipedia as follows: Cataclysmic pole shift hypothesis The cataclysmic pole shift hypothesis is the conjecture that there have been rapid shifts in the relative positions of the modern-day geographic locations of the poles and the axis of rotation of a planet. For the Earth, such a dynamic change could create calamities such as floods and tectonic events. ... The field has attracted pseudoscientific authors offering a variety of evidence, including psychic readings. ... http://en.wikipedia.org/wiki/Cataclysmic_pole_shift_hypothesis While watching a chess game at the local club a fellow member told me that he'd found the solution for where oil came from. It was long and involved. One of its features required that the earth's axis be changed by 90 degrees due to the collision of an asteroid a few million years ago. The idea was that the earth's axis used to be in the plane of the earth's orbit, but then it was knocked into its current rotation axis tilt of around 23 degrees. This changed the climate and buried a lot of stuff. The axis shift seemed to be an obvious weak link with the theory in that it involves the simplest physics. I told him that any asteroid large enough to change the earth's axis an appreciable amount (i.e. 70 degrees!) would cause so much damage as to destroy all life on the surface of the planet including its oceans. I'd have forgotten about it, but the other night, while listening to the radio, a guest appeared who had an idea that was similar. So this seems like a common target for pseudoscience. And so my question is this: Would an asteroid collision sufficient to immediately change the earth's rotation axis wipe out all life on the planet? - 4 Some people have evidently never played with a gyroscope or tried to turn a spinning bicycle wheel out of plane with their hands. What mechanism is proposed to damp the ferocious precession that would be the inevitable aftermath of twisting the axis around like that? Bah! Humbug! – dmckee♦ Feb 1 '11 at 1:24 1 What sort of asteroid could knock Earth's axis off by 70 degrees without ripping the planet to shreds, for that matter!? – Noldorin Feb 1 '11 at 2:35 @Noldorin: Yeah. I started trying to estimate the energies involved, but floundered when I realized there was no impulse-like way to get from there to here. – dmckee♦ Feb 1 '11 at 2:56 1 Some very simple calculation would answer this question. – Georg Feb 1 '11 at 10:34 1 Two votes to close because it (a) has nothing to do with physics, and (b) does not have a solution. However, I've provided a physics calculation as a solution so these objections are misplaced. – Carl Brannen Feb 3 '11 at 3:08 show 4 more comments ## 3 Answers Quite clearly this man’s theory is balderdash. A massive collision could change the tilt of the Earth’s axis, but that would be one hell of a torque. The collision would not be with an asteroid either, but with another planet that might be as big as Mars. Clearly nothing like this happened in recent geological history, though the Earth did suffer a collision of this magnitude a few 100 million years after gravitational accretion. A chunk from the collision spalled off to form the moon. This other planet probably was in some resonance condition around a Lagrange point. This would be a devastating collision, which would doubtless change this planet into something else. The solar system has not experienced such collsions since its early evolution. The solar system is now staid and respectable. Velikovskian things just do not happen. - I fully agree. If you could simply change the rotational axis via magic (with actual physics you gotta find a source of a huge amount of torque), there would be major climate change, and seas would shift around by thousands of feet, so it would be pretty tramatic, but the new earth should still support life. The real issue is that any concievable hit of that magnitude would be catastropic for nearly all life. [It is possible that simple bacteria living a couple of kilometers undergroud would outlive the heat spike (the surface would be a lava lake for years).] – Omega Centauri Feb 1 '11 at 5:09 Let's begin by choosing coordinates. Let the orbit of the earth define the x-y plane. Assume that just before the collision the earth's orbital axis is pointing in the x-direction, that is $(1,0,0)$. I'll assume that the earth absorbs the asteroid, and that after the collision the axis has been changed to $(\sin(27),0,\cos(27))$ where 27 is the degree measure of the earth's axis currently. So the relative change in angular momentum for the earth is $(1,0,0) - (\sin(27),0,\cos(27)) = (0.55,0,-.89)$ which is a vector of length 1.05 and so the angular momentum of the asteroid has to be 1.05 times the angular momentum of the earth. Make the approximation that the earth is a sphere of constant density. Then its angular momentum is given by $0.4 M_e\omega\;R_e^2$ where $M_e,R_e$ are the mass and radius of the earth and $\omega$ is its angular rate of rotation. We have $M_e = 6\times 10^{24}\;kg\;\;\;R_e = 6.4\times 10^6\;m,\;\;\;\omega = 2\pi/(24\;hours) = 2\pi/ (86400\;sec)$. I'm using sloppy approximations here; the earth is not constant density, I'm not using the sidereal day, etc. This gives $7\times 10^{33}\;kg\;m^2/s$ as the approximate angular momentum contributed by the asteroid. The formula for angular momentum is mass x velocity x radius. This is maximum when the radius is maximum; for an asteroid hitting the earth this happens when the radius is equal to the earth's radius, i.e. $6.4\times 10^6\;m$. Dividing the asteroid's angular momentum by this give its linear momentum: $P_a = 7\times 10^{33} / 6.4\times 10^6 = 10^{27}\;kg\;m/s$ The reason a relatively small asteroid can wipe out all life on earth is due to the kinetic energy it carries. Upon collision, the kinetic energy is converted to heat. If the asteroid is large and fast enough, the heat will increase the temperature of the atmosphere enough to boil off the oceans and even vaporize the salt deposits left over. Since kinetic energy is proportional to the square of velocity while momentum is only proportional to velocity, we will assume that our asteroid is moving as slowly as possible. The escape velocity of the earth is 11 km per second; to achieve a slower velocity an asteroid would have to be very lucky. So in calculating the kinetic energy of the asteroid, we'll assume a very conservative speed of 5 km/sec. Putting $10^{27} = 5000 m$ we find the mass of the asteroid as $2\times 10^{23}$ kilograms. Assuming a specific density of 5, or 5000 kg per cubic meter, this gives a radius for the asteroid of 2000 kilometers or a diameter of 5000 km. This is far more than enough to destroy all life on the planet. The asteroid's kinetic energy is $0.5\times 2\times 10^{23}\times (5000)^2 = 2.5\times 10^{30}$ Joules $= 6\times 10^{14}$ megatons. In other words, the collision would result in the release of kinetic energy equal to 600 million million hydrogen bombs going off, each with an equivalent energy of one million tons of high explosives. The earth's surface area is 500 million square kilometers or 500 million million square meters. So the energy release is equivalent to having a 1 megaton bomb going off on each square meter of the earth's surface. And since the vaporized rock from the blast is lighter than rock, this vapor will condense on the surface and apply its heat to the surface. For a video of a 500 km asteroid hitting the earth (1/1000 of the volume necessary to change the earth's axis) see:

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http://mathhelpforum.com/differential-geometry/159054-isometric-print.html

# Isometric? Printable View • October 10th 2010, 10:23 AM Showcase_22 Isometric? Quote: Let $X=\{A,\ B,\ C, \ D\}$ with $d(A,D)=2$, but all other distances equal to 1. d is a metric. Prove that the metric space X is not isometric to any subset of $\mathbb{E}^n$ for any n. Hint: can you realise $X$ as a subset of a sphere $S^2$ of appropriate radius, with the spherical "great circle" metric? I think I have to take two points in $X$ and show that any mapping from $X$ to $\mathbb{E}^n$ would result in a different distance between them. Here's my attempt: Take a sphere of radius 1 and the points A and D. Since $d(A,D)=2$, A and D are on opposite sides of the sphere. Suppose there exists an isometry $\phi:X \rightarrow \sigma$ where $\sigma$ is is the plane used in stereographic projection (also a subset of $\mathbb{E}^n$). Then $\phi(A)=\phi(D)$. Hence $d(\phi(A),\phi(D))=0$ which is not equal to $d(A,D)=2$. Hence $\phi$ cannot be an isometry. • October 10th 2010, 11:26 AM Opalg Quote: Originally Posted by Showcase_22 I think I have to take two points in $X$ and show that any mapping from $X$ to $\mathbb{E}^n$ would result in a different distance between them. Here's my attempt: Take a sphere of radius 1 and the points A and D. Since $d(A,D)=2$, A and D are on opposite sides of the sphere. Suppose there exists an isometry $\phi:X \rightarrow \sigma$ where $\sigma$ is is the plane used in stereographic projection (also a subset of $\mathbb{E}^n$). Then $\phi(A)=\phi(D)$. Hence $d(\phi(A),\phi(D))=0$ which is not equal to $d(A,D)=2$. Hence $\phi$ cannot be an isometry. I'm not sure that the hint is very helpful, and I don't think that stereographic projection is needed. If such an isometry $\phi$ exists then the points $\phi (A),\ \phi (B),\ \phi (C)$ form an equilateral triangle with side 1, in some 2-dimensional subspace of $\mathbb{E}^n$. So do the points $\phi (B),\ \phi (C),\ \phi (D)$. Thus the images of the four points lie in some 3-dimensional subspace of $\mathbb{E}^n$, so we may as well assume that n=3. The points $\phi (A)$ and $\phi (D)$ must both lie at a distance $\sqrt3/2$ from the midpoint of the line segment joining $\phi (B)$ and $\phi (C)$. So the distance from $\phi (A)$ to $\phi (D)$ is at most $\sqrt3<2$, contradicting the condition that $\phi$ is an isometry. All times are GMT -8. The time now is 11:22 PM.

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http://unapologetic.wordpress.com/2011/08/04/integrals-are-independent-of-parameterization/

# The Unapologetic Mathematician ## Integrals are Independent of Parameterization If $c:[0,1]^k\to M$ is a singular $k$-cube and $\omega$ is a $k$-form on the image of $c$, then we know how to define the integral of $\omega$ over $c$: $\displaystyle\int\limits_{c([0,1]^k)}\omega=\int\limits_{[0,1]^k}c^*\omega$ On its face, this formula depends on the function $c$ used to parameterize the region of integration. But does it really? What if we have a different function $d:[0,1]^k\to M$ with the same image? For convenience we’ll only consider singular $k$-cubes that are diffeomorphisms onto their images — any singular $k$-cube can be broken into pieces for which this is true, and we’ll soon deal with how to put these together. Anyway, if $c([0,1]^k)=d([0,1]^k)$, then given our assumptions there is some diffeomorphism $f=c^{-1}\circ d:[0,1]^k\to[0,1]^k$ such that $d=c\circ f$. If $J_f$ is everywhere positive, then we say that $d$ is an “orientation-preserving reparameterization” of $c$. And I say that the integrals of $\omega$ over $c$ and $d$ are the same. Indeed, we calculate: $\displaystyle\begin{aligned}\int\limits_{c\circ f}\omega&=\int\limits_{[0,1]^k}(c\circ f)^*\omega\\&=\int\limits_{[0,1]^k}f^*c^*\omega\\&=\int\limits_fc^*\omega\\&=\int\limits_{f([0,1]^k)}\left[c^*\omega\right]\left(\frac{\partial}{\partial u^1},\dots,\frac{\partial}{\partial u^k}\right)\,d(u^1,\dots,u^k)\\&=\int\limits_{[0,1]^k}\left[c^*\omega\right]\left(\frac{\partial}{\partial u^1},\dots,\frac{\partial}{\partial u^k}\right)\,d(u^1,\dots,u^k)\\&=\int\limits_c\omega\end{aligned}$ where we use our expression for the integral of $\omega$ over the image $f([0,1]^k)$ in passing from the third to the fourth line. Thus the integral is a geometric quantity, depending only on the image $c([0,1]^k)$ and the $k$-form $\omega$ rather than on any detail of the parameterization itself. About these ads ### Like this: Like Loading... Posted by John Armstrong | Differential Topology, Topology ## 2 Comments » 1. [...] it all works out for the same reason parameterization invariance and the change of variables formula do. Passing from the boundary of the singular cube back to the [...] Pingback by | August 20, 2011 | Reply 2. [...] we use to reparameterize our integral. Of course, this function may not be defined on all of , but it’s defined on , [...] Pingback by | September 5, 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. • ## RSS Feeds RSS - Posts RSS - Comments • ## Feedback Got something to say? Anonymous questions, comments, and suggestions at Formspring.me! %d bloggers like this:

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http://math.stackexchange.com/questions/4894/why-does-symplectic-geometry-have-many-applications-in-mathematics?answertab=active

# Why does symplectic geometry have many applications in mathematics It is not quite intuitive , at least from its origin. Could any one can give me an intuitive explanation?Thank you! - ## 4 Answers Any two of (a) symplectic structure, (b) almost-complex structure, and (c) Riemannian structure determine the third. This reflects the fact that $$U(n) = O(2n) \cap GL(n,\mathbb{C}) \cap Sp(2n),$$ but the intersection of any two is already $U(n)$. This gives rise to the theory of Kähler manifolds (cf. wikipedia), which are central to mirror symmetry (and the mirror symmetry conjecture). - I suppose a question to ask here is : what kinds of applications? One of the areas where symplectic geometry is used is Hamiltonian dynamics, where the geometric framework allows one to generalize away from cotangent bundles, but this direction is directly related to the origins of the subject. Another direction comes from algebraic geometry, where the fact that complex and Riemannian structure determine a symplectic one (see Aaron's answer) makes all complex projective (and affine) algebraic varieties symplectic. Viewing them as such allows one to use differential-geometric methods, while the symplectic form provides some control (in the theory of holomorphic curves, symplectic form gives the crucial energy bounds without which things fall apart; this is why there is no Gromow-Witten theory of almost-complex manifolds). Add a bit of mirror symmetry to the mix and you have interaction with many other areas of math. There is also the fact that the wedge product is anti-symmetric, which leads the space of connections on a principal bundle over 2-d Riemannian manifold to be an (infinite dimensional) symplectic manifold (the gauge group is Hamiltonian, and curvature is the moment map, leading to flat connections being the reduction!). This leads, for G=SO(3) and the surface coming from a Heegard splitting, to Atiyah-Floer conjecture and some related stuff leads to Heegard-Floer theory. Thus connections to low-dimensional topology. The "naive" connection here is the anti-symmetry underlying symplectic form and wedge product, but probably there is a deeper reason known to wise people... - There is a simple "philosophical" answer: Symplectic forms allow you to only measure two-dimensional quantities, not one-dimensional ones (you can measure area infinitesimally). That's the basic difference with Riemannian geometry, where you can measure length infinitesimally. Furthermore symplectic geometry pops up in connection with the cotangent bundle. For instance if you have a (scalar) first order differential operator on a manifold, then the principal symbol transforms like a covector (when changing coordinates). Note that this does not require a metric. - Symplectic geometry is applicable in lot of areas such as Classical mechanics, Hamiltonian mechanics. While browsing through the web i have found a reason as to why SG is good for classical mechanics. You may read it here: http://research.microsoft.com/en-us/um/people/cohn/thoughts/symplectic.html - 3 I think the question is more about why symplectic geometry has found applications beyond classical mechanics. – Qiaochu Yuan Sep 18 '10 at 5:34 @Qiaochu Yuan: I answered the question, w.r.t my thinking. – anonymous Sep 18 '10 at 7:21

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http://mathhelpforum.com/trigonometry/39976-radians.html

# Thread: 1. ## Radians Hi, need help understanding 2 problems. Give all the solutions to the following problems, answer in radians. $\sin(3x)=\sqrt{3}/2$ and $\cos(x-pi/4)=1/\sqrt{2}$ Thanks. 2. Originally Posted by Smifsret Hi, need help understanding 2 problems. Give all the solutions to the following problems, answer in radians. $\sin(3x)=\sqrt{3}/2$ and $\cos(x-pi/4)=1/\sqrt{2}$ Thanks. You need to know what angle for sine gives the value $\frac{\sqrt{3}]{2}$. This happens to be $\frac{\pi}{3}$ Solve for $3x = \frac{\pi}{3}$ 3. Thanks, is the second one pi / 4 then? Does pi/4 count as an answer (if its right) or do they want me to write a number? 4. sin(3x)= pi/3 x1. sin(x) = pi/9 + 2 * pi * n? x2. sin(x) = pi-pi/9 + 2 * pi * n? Is that correct? 5. Ahh the question is to answer EXACTLY in radians and I was told it isn't exactly unless you write it like that. Also the formula + n * 2 * pi I was told to add because n is the number of times it has gone around the circle. Hope I explained that decently. But I understand that I shouldn't have addad sin to the solution, thank you, as always I wasn't aware that it mattered. 6. For the second part x1 = pi/2 x2 = -pi/2 Correct? 7. Originally Posted by Smifsret Ahh the question is to answer EXACTLY in radians and I was told it isn't exactly unless you write it like that. Also the formula + n * 2 * pi I was told to add because n is the number of times it has gone around the circle. Hope I explained that decently. But I understand that I shouldn't have addad sin to the solution, thank you, as always I wasn't aware that it mattered. I'm sorry, I should have realized that your x1 and x2 were solutions to the first problem, not solutions to problems 1 and 2. In that case, yes, those solutions are correct (with x replacing your sin(x).) -Dan 8. You wrote that $x - \frac{\pi}{4} = \frac{\pi}{6}$ i think that its $1/\sqrt{2}=0.70711 = 45 degrees = \frac{\pi}{4}$ Or am I completely wrong? 9. Originally Posted by Smifsret You wrote that $x - \frac{\pi}{4} = \frac{\pi}{6}$ i think that its $1/\sqrt{2}=0.70711 = 45 degrees = \frac{\pi}{4}$ Or am I completely wrong? My apologies again. You are correct. I'm going back to bed. -Dan 10. Thank you very much for the help, does that mean that my previous answers were correct?

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http://mathhelpforum.com/calculus/136215-use-cylindrical-shells.html

Thread: 1. use cylindrical shells... use cylindrical shells to find the volume of the solid generated when the region bounded by xy = 4, x + y = 5 is revolved about the x -axis. i did the work for this problem i was wondering if someone could check my work and let me know if my setup for this problem is right? thanks in advance...IMG.pdf i cant add the second page, but my answer, the final answer for the problem was volume = (58/6)pi 2. I came up with $9\pi$ but I worked it in a hurry! 3. Is this the integral? $2\pi \int_{1}^{4} y(5y - y^2 - 4) dy$ 4. um not thats not quite what i got..one of of the functions xy = 4, so one of the functions should be divided by x or y, right? 5. I'm not sure what your asking, but using the shell method and rotating the solid about the x-axis would require the equation: $V = \int_{c}^{d} p(y)h(y) dy$ right? And if so, p(y) = y, the distance from the axis of revolution to the representaive rectangle (incremental slice) and h(y) would represent the length of the representative rectangle which would be the difference of the two given functions x = 5 - y minus x = 4/y. 6. I'm sorry, the integral that I gave still had "y" in it. That was an error, it should have been: $\int_{1}^{4}(5y - y^2 -4) dy$ 7. The work you submitted is correct: However, there may be a mistake in your algebra. Check out my solution and compare. $2\pi[\frac{5y^2}{2} - \frac{y^3}{3} - 4y]_{1}^{4}$ $2\pi[(40 - \frac{64}{3} - 16) - (\frac{5}{2} - \frac{1}{3} - 4)]$ $2\pi[(\frac{120 - 64 - 48}{3} - (\frac{15 - 2 - 24}{6})]$ $2\pi(\frac{8}{3} + \frac{11}{6})<br />$ = $2\pi(\frac{27}{6})$ = $9\pi$ 8. thanks im going to go back and check the algebra on that one, thanks again for the catch..

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http://math.stackexchange.com/questions/208888/every-automorphism-of-a-tree-with-an-odd-number-of-vertices-has-a-fixed-point?answertab=votes

# Every automorphism of a tree with an odd number of vertices has a fixed point If $T$ is a tree, and $T$ has an odd number of vertices, then $\forall f$, where $f$ is automorphism $\Rightarrow \exists$ fixed point (vertex). What it means: Formally, an automorphism of a tree $T$ is a permutation $\sigma$ of the vertex set $V$ of tree, such that the pair of vertices $(u, v)$ form an edge if and only if the pair $(\sigma(u),\sigma(v))$ also form an edge. If $T$ is a tree with vertices ($v1,..vn$), then after applying $f$, some vertex will be always stay in place. if somebody want to illustrate this, I will do. - I removed the automorphic-forms tag since that's something totally different. – Ted Oct 7 '12 at 20:28 I don't understand the last sentence. – joriki Oct 7 '12 at 20:36 @joriki what sentence? – jofisher Oct 7 '12 at 20:41 As I said, the last one. – joriki Oct 7 '12 at 21:01 "if somebody want to illustrate this, I will do." if you don't understand the problem above, I can illustrate this for more understanding. – jofisher Oct 7 '12 at 21:14 ## 1 Answer Call the original tree $T_0$. Remove all the leaves of $T_0$ to get a tree $T_1$. Then remove all the leaves of $T_1$ to get $T_2$. After finitely many iterations, you get a tree $T_n$ which is all leaves, and therefore consists of either a single vertex, or two vertices connected by an edge. Any automorphism $\sigma$ of the original tree $T_0$ must take every $T_i$ to itself. If $T_n$ consists of a single vertex, we are done. If $T_n$ consists of 2 vertices, and $\sigma$ fixes them, we are done. Suppose $\sigma$ reverses the two vertices of $T_n$. Then $\sigma$ fixes the edge $e$ of $T_n$. Remove the edge $e$ from the original tree $T_0$. We get a forest with 2 components, and $\sigma$ must be an automorphism of this forest which reverses the 2 components. But this is impossible, because $T_0$ has an odd number of vertices and therefore the number of vertices in the two components must be different (one must be odd and the other must be even). - 1 Very handy. Thank you! – jofisher Oct 7 '12 at 21:47 I have a small question. The automorphism transfers vertex to the vertex with the same degree (in common case). But in your case transferred the vertices only at the same level from the bottom. Why, if a vertex from ${T}_{i}$ that can not be transferred to a vertex from ${T}_{j}$..? This is the only question. – jofisher Oct 9 '12 at 12:57 1 Are you asking why $\sigma$ must map $T_i$ to itself? Because we know an automorphism must map leaves of $T_0$ to other leaves of $T_0$. Therefore, it must also map the rest of $T_0$ to the rest of $T_0$. But the rest of $T_0$ is $T_1$. Therefore $\sigma$ must map $T_1$ to $T_1$, so $\sigma$ is also an automorphism of $T_1$. Now apply this argument repeatedly to see that $\sigma$ maps every $T_i$ to itself. – Ted Oct 9 '12 at 16:29 Yep, I thought the same. But scared a little bit. Thank you. – jofisher Oct 9 '12 at 17:31

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http://en.wikipedia.org/wiki/Exterior_covariant_derivative

# Exterior covariant derivative In mathematics, the exterior covariant derivative, sometimes also covariant exterior derivative, is a very useful notion for calculus on manifolds, which makes it possible to simplify formulas which use a principal connection. ## Definition Let P → M be a principal G-bundle on a smooth manifold M. If ϕ is a tensorial k-form on P, then its exterior covariant derivative is defined by $D\phi(X_0,X_1,\dots,X_k)=\mathrm{d}\phi(h(X_0),h(X_1),\dots,h(X_k))$ where h denotes the projection to the horizontal subspace, Hx defined by the connection, with kernel Vx (the vertical subspace) of the tangent bundle of the total space of the fiber bundle. Here Xi are any vector fields on P. Dϕ is a tensorial (k + 1)-form on P. ## Properties Unlike the usual exterior derivative, which squares to 0 (that is d2 = 0), we have $D^2\phi=\Omega\wedge\phi$ where Ω denotes the curvature form. In particular D2 vanishes for a flat connection. ## References • Kobayashi, Shoshichi and Nomizu, Katsumi (1996 (New edition)). . Wiley-Interscience. ISBN 0-471-15733-3. This Differential geometry related article is a stub. You can help Wikipedia by expanding it.

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http://mathhelpforum.com/advanced-algebra/32894-sylow-p-subgroups-print.html

# Sylow p-subgroups Printable View • April 1st 2008, 09:49 PM math_science_dude Sylow p-subgroups Hello, I'm trying to show that if H is a subgroup of a finite group G, and p is a prime that the number of sylow p-subgroups of H $n_H$ is less thanor equal to the number of sylow p-subgroups of $n_G$. This is all I have: Clearly if p does not divide |G| then p does not divide |H| and so $n_H = n_G = 0$. Let $p^j$ and $p^k$ be the greatest powers of p that divides |H| and |G| respectively. Case 1: j = k. In this case any sylow p-subgroup of H is a sylow p-subgroup of G and so clearly $n_H \le n_G$. Case 2: j < k. This is where i got stuck. I am at a point where I need to show that each sylow p-subgroup of G can contain only one sylow p-subgroup of H. I do not know if this is right path, maybe I overlooked a simple theorem. Any assistance is greatly appreciated. Thank you. m_s_d All times are GMT -8. The time now is 06:34 AM.

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http://math.stackexchange.com/questions/71750/can-we-say-for-sure-that-the-function-is-increasing-to-mean-that-the-rst-de/71754

# Can we say (for sure) that "the function is increasing” to mean that the ﬁrst derivative is positive? Can we say (for sure) that "the function is increasing” to mean that the ﬁrst derivative is positive? Whenever $f'$ (the first derivative) is positive the function is increasing,but does that imply if a function is increasing the first derivative must be positive? - 2 No. First, increasing functions need not to be everywhere differentiable (for example $f\colon \mathbb R\to \mathbb R$ given by $f(x) = x$ for $x \ge 0$ and $f(x) = 2x$ for $x \le 0$), second: If $f$ is differentiable, you only need to have $f' \ge 0$, not $f' > 0$, for example: $g(x) = x^3$ has $g'(0) =0$. – martini Oct 11 '11 at 16:29 3 Increasing functions don't even have to be continuous! – barrycarter Oct 11 '11 at 16:41 ## 3 Answers No. Consider a dense countable subset $Q$ of $\mathbb R$ and a family $(a_q)_{q\in Q}$ of positive real numbers such that $\sum\limits_{q\in Q}a_q(1+|q|)$ converges. For every $x$, let $(x)^+=\max\{x,0\}$ denote the positive part of $x$. Then, the function $f$ defined by $$f(x)=\sum\limits_{q\in Q}a_q(x-q)^+$$ is well defined for every real number $x$. The left and right derivatives of $f$ exist everywhere, with $$f'_\ell(x)=\sum\limits_{q<x}a_q\qquad\text{and}\qquad f'_r(x)=\sum\limits_{q\leqslant x}a_q.$$ Thus, $f$ is strictly increasing and strictly convex, differentiable at every point not in $Q$, and not differentiable at every point in $Q$. To prove the existence of $f'_\ell$ and $f'_r$ at every point, one can come back to the definitions of the left and right derivatives as the limits, if these exist, of $\pm(f(x\pm h)-f(x))/h$ when $h\to0^+$. Or one can use directly the fact that each function $g_q$ defined by $g_q(x)=(x-q)^+$ has left and right derivatives $(g_q)'_\ell(x)=[x>q]$ and $(g_q)'_r(x)=[x\geqslant q]$. - No. Consider function $f(x)=x^3$. It is increasing on $\mathbb R$ but $f'(0)=0$. - Someone mentioned $f(x)=x^3$, for which the derivative at one point is $0$ but the function is everywhere increasing. It can also happen that the derivative at one point is undefined and the function is everywhere increasing. For example, let $$f(x) = \begin{cases} x & \text{if }x<0, \\ 2x & \text{if }x\ge 0. \end{cases}$$ The derivative is undefined at $x=0$. The function is everywhere increasing. -

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http://en.wikipedia.org/wiki/Solution_point

# Equation (Redirected from Solution point) This article may be expanded with text translated from the corresponding article in the French Wikipedia. (February 2013) Click [show] on the right to read important instructions before translating. Google's machine translation is a useful starting point for translations, but translators must revise errors as necessary and confirm that the translation is accurate, rather than simply copy-pasting machine-translated text into the English Wikipedia. Do not translate text that appears unreliable or low-quality. If possible, verify the text with references provided in the foreign-language article. After translating, `{{Translated|fr|Équation}}` must be added to the talk page to ensure copyright compliance. The first use of an equals sign, equivalent to 14x+15=71 in modern notation. From The Whetstone of Witte by Robert Recorde (1557). In mathematics an equation is an expression of the shape A = B, where A and B are expressions containing one or several variables called unknowns. An equation looks like an equality, but has a very different meaning: An equality is a mathematical statement that asserts that the left-hand side and the right-hand side of the equals sign (=) are the same or represent the same mathematical object; for example 2 + 2 = 4; is an equality. On the other hand, an equation is not a statement, but a problem consisting in finding the values, called solutions, that, when substituted to the unknowns, transform the equation into an equality. For example, 2 is the unique solution of the equation x + 2 = 4, in which the unknown is x.[1] Equation may also refer to a relation between some variables that is expressed by the equality of some expressions of their values. For example the equation of the unit circle is x2 + y2 = 1, which means that a point belongs to the circle if and only if its coordinates are related by this equation. Most physical laws are expressed by equations. One of the most popular ones is Einstein's equation E=mc2. The = symbol was invented by Robert Recorde (1510–1558), who considered that nothing could be more equal than parallel straight lines with the same length. Centuries ago, the word "equation" frequently meant what we now usually call "correction" or "adjustment". This meaning is still occasionally found, especially in names which were originally given long ago. The "equation of time", for example, is a correction that must be applied to the reading of a sundial in order to obtain mean time, as would be shown by a clock. ## Parameters and unknowns See also: Expression (mathematics) Equations often contain other variables than the unknowns. These other variables that are supposed to be known are usually called constants, coefficients or parameters. Usually, the unknowns are denoted by letters at the end of the alphabet, x, y, z, w, …, while coefficients are denoted by letters at the beginning, a, b, c, d, … . For example, the general quadratic equation is usually written ax2 + bx + c = 0. The process of finding the solutions, or in case of parameters, expressing the unknowns in terms of the parameters is called solving the equation. Such expressions of the solutions in terms of the parameters are also called solutions. A system of equations is a set of simultaneous equations, usually in several unknowns, for which the common solutions are sought. Thus a solution to the system is a set of one value for each unknown, which is a solution to each equation in the system. For example, the system $\begin{align} 3x+5y&=2\\ 5x+8y&=3 \end{align}$ has the unique solution x = -1, y = 1. ## Analogous illustration Illustration of a simple equation; x, y, z are real numbers, analogous to weights. The analogy often presented is a weighing scale, balance, seesaw, or the like. Each side of the balance corresponds to each side of the equation. Different quantities can be placed on each side, if they are equal the balance corresponds to an equality (equation), if not then an inequality. In the illustration, x, y and z are all different quantities (in this case real numbers) represented as circular weights, each of x, y, z has a different weight. Addition corresponds to adding weight, subtraction corresponds to removing weight from what is already placed on. The total weight on each side is the same. ## Types of equations Equations can be classified according to the types of operations and quantities involved. Important types include: • An algebraic equation or polynomial equation is an equation in which both sides are polynomials (see also system of polynomial equations). These are further classified by degree: • linear equation for degree one • quadratic equation for degree two • cubic equation for degree three • quartic equation for degree four • quintic equation for degree five • A Diophantine equation is an equation where the unknowns are required to be integers • A transcendental equation is an equation involving a transcendental function of its unknowns • A parametric equation is an equation for which the solutions are sought as functions of some other variables, called parameters appearing in the equations • A functional equation is an equation in which the unknowns are functions rather than simple quantities • A differential equation is a functional equation involving derivatives of the unknown functions • An integral equation is a functional equation involving the antiderivatives of the unknown functions • An integro-differential equation is a functional equation involving both the derivatives and the antiderivatives of the unknown functions • A difference equation is an equation where the unknown is a function f which occurs in the equation through f(x), f(x-1), ..., f(x-k), for some whole integer called the order of the equation. If x is restricted to be an integer, a difference equation is the same as a recurrence relation ## Identities Main articles: Identity (mathematics) and List of trigonometric identities An identity is a statement resembling an equation which is true for all possible values of the variable(s) it contains. Many identities are known, especially in trigonometry. Probably the best known example is: $\sin^2(\theta)+\cos^2(\theta)=1,$, which is true for all values of θ. In the process of solving an equation, it is often useful to combine it with an identity to produce an equation which is more easily soluble. For example, to solve the equation: $3\sin(\theta) \cos(\theta)= 1,$ where θ is known to be between zero and 45 degrees, use the identity: $\sin(2 \theta)=2\sin(\theta) \cos(\theta),$ so the above equation becomes: $\frac{3}{2}\sin(2 \theta) = 1$ Whence: $\theta = \frac{1}{2} \arcsin\left(\frac{2}{3}\right),$ which comes to about 20.9 degrees. ## Properties Two equations or two systems of equations are equivalent if they have the same set of solutions. The following operations transform an equation or a system into an equivalent one: • Adding or subtracting the same quantity to both sides of an equation. This shows that every equation is equivalent to an equation in which the right-hand side is zero. • Multiplying or dividing both sides of an equation by a non-zero constant. • Applying an identity to transform one side of the equation. For example, expanding a product or factoring a sum. • For a systems: adding to both sides of an equation the corresponding side of another, equation multiplied by the same quantity. If some functions is be applied to both sides of an equation, the resulting equation has the solutions of the initial equation among its solutions, but may have further solutions called extraneous solutions. If the function is not defined everywhere, (like 1/x that is not defined for x=0) some solutions may be lost. Thus caution must be exercised when applying such a transformation to an equation. For example, the equation $x=1$ has the solution $x=1.$ Raising both sides to the exponent of 2 (which means, applying the function $f(s)=s^2$ to both sides of the equation) changes our equation into $x^2=1$, which not only has the previous solution but also introduces the extraneous solution, $x=-1.$ Above transformations are the basis of most elementary methods for equation solving and some less elementary ones, like Gaussian elimination For more details on this topic, see Equation solving. ## References 1. "Equation". Dictionary.com. Dictionary.com, LLC. Retrieved 2009-11-24.

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http://stats.stackexchange.com/questions/31321/testing-periodogram-peaks-sine-like-wave-or-ar-ma-arma-noise

# Testing periodogram “peaks”: sine-like wave or AR/MA/ARMA noise? I'm performing an harmonic fit to data I know (from physical constraints) come from a periodic source of the form $$\sum_j^M \sum_i^N a_{i,j}\sin(2\pi f_it)+b_{i,j}\cos(2\pi f_it)$$ using the Lomb-Scargle periodogram (I know this is not the optimal thing to do, but it's fast and reliable so far), where I'm also trying to estimate $N$ and $M$, the number of frequencies and the number of harmonics per frequency to fit. The number of harmonics is really not a very hard problem to solve once the number of frequencies is fixed, because I can perform methods like Ridge Regression, LASSO, etc. in order to find the best subset of harmonics, so my problem is really on estimating the optimal number of frequencies. The thing is that I know red noise is almost always present, and certain harmonic fits give AR-like spectra on the residuals (and, at least to me, it seems like I could even have ARMA-like spectra). The "classical" Lomb-Scargle periodogram, on the other hand, assumes white noise in order to test the significance of the peaks, so the significance test on the peaks doesn't seem like a good idea in order to test if there are any residual frequencies on the periodogram. I've been trying to derive significance tests assuming ARMA-like noise, but the problem of defining the order $p,q$ of the process arises: do you know of a way of differencing between correlated noise and sine wave-like spectra? This paper shows a way of doing this assuming AR(1) noise (in fact something more like a CAR-type noise?), but the problem of it is that it actually knows there is red noise. I think I can "see" how to do it: sine wave-like spectra has almost-symmetric decreasing peaks around the main peak because of the (uneven) sampling, which ARMA-like noise doesn't have (the amplitudes are random with an apparent upper limit dictated by the form of the spectra). I've also been thinking in performing fits with ARMA and sine-like waves, and perform model selection via AIC...any suggestions? - ## 1 Answer A general test for periodicity is Fisher's periodogram test. I assume you are familar with it. But in case you aren't here are some links to recent papers related to it. http://users.ics.aalto.fi/harrila/publications/RobFisherPeriodicity.pdf

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http://mathoverflow.net/revisions/74311/list

## Return to Answer 2 added 257 characters in body Especially if your interested in dynamical systems, I highly recommend Abraham--Marsden--Ratiu, Manifolds, tensor analysis, and applications. For a more Riemannian-geometric/global-analytic focus, you might want to try Klingenberg, Riemannian Geometry, or Lang, Differentiable Manifolds. There is a standard way to construct a canonical topology on $C^r(M,N)$ for $M$ compact, one that turns $C^r(M, N)$ into a Banach manifold. But I don't think there is a canonical metric on $C^r(M, N)$ unless you put some additional structure on $N$. 1 Especially if your interested in dynamical systems, I highly recommend Abraham--Marsden--Ratiu, Manifolds, tensor analysis, and applications. For a more Riemannian-geometric/global-analytic focus, you might want to try Klingenberg, Riemannian Geometry, or Lang, Differentiable Manifolds.

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http://computationalblackbody.wordpress.com/

# Model of Atmosphere with CO2 shows Small Emissivity Posted on February 10, 2013 by Here is a another argument indicating that the effect of the atmospheric trace gas CO2 on the radiation balance of the Earth is small. Our model of blackbody radiation consists of collection of oscillators with small damping with equal oscillator internal energy T representing temperature, with oscillator resonance frequencies n varying from 1 to a cut-off set at T and each oscillator radiating • $E_n = \gamma T n^2$ where $\gamma$ is a universal constant, which is Planck’s Law. Summing over n from 1 to T, we obtain the total radiance • $E =\sum_n \gamma T n^2 = \sigma T^4$ which is Stefan-Boltzmann’s law with $\sigma = \gamma /3$. In the case of only one resonance frequency $n = T$, the radiance would be reduced to • $e = \gamma T T^2 =\gamma T^3 = 3 E/T$ with the reduction factor $1/T$. The radiance of an atmosphere which is fully opaque over the entire spectrum would radiate $E$, while an atmosphere opaque only for a specific frequency near cut-off $T$, would radiate $e\approx 3E/T$ with a reduction factor $1/T$. We conclude that the emissivity of transparent atmosphere with a trace gas like CO2 with only a few isolated resonances, would scale like 1/T and thus be small as soon as T is bigger than say 100 K. We thus find theoretical evidence from a basic model that the emissivity of the Earth’s atmosphere with the trace gas CO2 would be small, and thus that CO2 would have little effect on the Earth’s radiation balance. We thus find theoretical evidence from a basic model that the emissivity of the Earth’s atmosphere with the trace gas CO2 would be small, and thus that CO2 would have little effect on the Earth’s radiation balance. Posted in Uncategorized | # Blackbody as Transformer of Radiation Posted on September 9, 2012 by Our blackbody model, which has a high frequency cut-off scaling with temperature, absorbs all incoming radiation (forcing), re-emits frequencies below cut-off and stores frequencies above cut-off as internal energy, which is assumed to be equally distributed over frequencies below cut-off. The result is that a blackbody subject to forcing with frequencies above cut-off, will heat up with a corresponding increase of cut-off until the highest frequency of the forcing is reached, assuming the blackbody is not somehow cooled. The effect is that the blackbody acts as a transformer of radiation from high to low frequencies by absorbing frequencies above current cut-off while emitting frequencies below cut-off. The Earth subject to forcing from the Sun acts as such a transformer, see Blackbody: Transformer of Radiation. Posted in Uncategorized | # Kirchhoff’s Law 2 Posted on September 8, 2012 by Kirchhoff’s Law states that for a blackbody emissivity = absorptivity. In our wave equation with small damping as a model of blackbody in radiative equilibrium with an exterior forcing, this is built into the model for frequencies below high-frequency cut-off. This is because for these frequencies the outgoing radiation simply reproduces the part of the incoming radiation which is not reflected. Or put differently: In radiative equilibrium there is below cut-off no distinction between incoming radiation and outgoing radiation as explained in the previous post. However, if the incoming waves (the forcing) contain frequencies above the present cut-off of the absorbing body, then the game changes, as these frequencies contribute to the internal energy of the body increasing its temperature and are not (re-)emitted. This is a case with positive absorptivity and zero emissivity of the frequencies above cut-off. This leads to coefficients of absorptivity depending on temperature, which complicates the picture. The effect is that different bodies with different cut-off at the same distance to the Sun, can take on different equilibrium temperatures, as evidenced in e.g. Satellite Thermal Control Engineering and What Is the Temperature of Space? Posted in absorptivity, emissivity, high frequency cut-off | # String and Soundboard Posted on September 8, 2012 by Our model of blackbody radiation has an acoustic analog in the form of a string interacting with a soundboard, with the following key components (as total or for each frequency in a spectral decomposition): • internal string energy • string amplitude • soundboard amplitude • force balance between string and soundboard = wave equation • small damping in the interaction expressed by the wave equation. The small damping is an important feature of the model since it creates an “optimal” interaction between the string and the soundboard in the sense that the tone generated by the soundboard will be both loud and have long sustain. With zero damping the interaction will small and the tone weak, and with too big damping the sustain will be short. In radiation the damping is small. The effect of the small damping is that the string velocity is out-of-phase with the forcing, which is the key to optimal interaction between string and soundboard. Radiative equilibrium corresponds to acoustic equilibrium, which can be interpreted in two different ways: 1. The vibration of the soundboard is sustained by the vibration of the string = outgoing acoustic waves. 2. The vibration of the string is sustained by the vibration of the soundboard = incoming acoustic waves. The same equilibrium can thus be interpreted as incoming or outgoing, which can be expressed as absorptivity = emissivity, which thus is built into the model. This argument applies to frequencies below cut-off with only radiative damping being active. For frequencies above cut-off the interaction between string and soundboard is different with internal heating of the string, and emissivity in general different from absorptivity. There is also non-equilibrium dynamics below cut-off with e.g. the string energy being transferred to the soundboard as the string amplitude decreases. Posted in acoustics, resonance | # Equidistribution 1 Posted on September 7, 2012 by Equidistribution in our wave model of blackbody radiation means that the internal energy of different frequencies is the same, that is, all frequencies have the same temperature. More specifically, the internal energy $IE_\nu$ of the wave component of frequency $\nu$ with amplitude $U_\nu$ is measured by (with subindices indicating differentiation with respect to time $t$ and space $x$ and integration in space and time over a period) • $IE_\nu = \frac{1}{2}\int (U_{\nu ,t}^2+U_{\nu ,x}^2)dxdt$. Equidistribution means that $IE_\nu$ is the same for all frequencies and then defines a common temperature $T =IE_\nu$. In the piano model equidistribution corresponds to pressing all the keys with the same force to give all strings of the piano the same string energy. Planck’s Law $E(\nu ,T) =\gamma T\nu^2$ assumes equidistribution with all frequencies having the same internal energy measured by temperature $T$. With equidistribution as a characteristic of blackbody radiation the radiation intensity spectrum shows the characteristic quadratic increase with frequency. Without equidistribution the spectrum can show different dependence on frequency. Posted in equidistribution, temperature | # Kirchhoff’s Law 1 Posted on September 7, 2012 by Kirchhoff’s Law of Thermal Radiation: 150 Years starts off by: • Kirchhoff’s law is one of the simplest and most misunderstood in thermodynamics. Let us see what we can say about Kirchhoff’s Radiation Law stating that the emissivity and absorptivity of a radiating body are equal, in the setting of the wave model with damping presented in Computational Blackbody Radiation and Mathematical Physics of Blackbody Radiation: • $U_{tt} - U_{xx} - \gamma U_{ttt} - \delta^2U_{xxt} = f,$ where the subindices indicate differentiation with respect to space $x$ and time $t$, and 1. $U_{tt} - U_{xx}$ models a vibrating material string with \$U\$ displacement 2. $- \gamma U_{ttt}$ is a dissipative term modeling outgoing radiation 3. $- \delta^2U_{xxt}$ is a dissipative term modeling internal heating by friction 4. $f$ is the amplitude of the incoming forcing, 5. $T$ is temperature with $T^2=\int\frac{1}{2}(\vert U_t^2+U_x^2)dxdt$, 6. the wave equation expresses a balance of forces, where $\gamma$ and $\delta^2$ are certain small damping coefficients defined by spectral decomposition as follows in a model case: • $\gamma = 0$ if the frequency $\nu > \frac{1}{\delta}$ • $\delta = 0$ if the frequency $\nu < \frac{1}{\delta}$, where $\delta = \frac{h}{T}$ represents a “smallest coordination length” depending on temperature $T$ and $h$ is a fixed smallest mesh size (representing some atomic dimension). This represents a switch from outgoing radiation to internal heating as the frequency $\nu$ passes the threshold $\frac{T}{h}$, with the threshold increasing linearly with $T$. The idea is that a hotter vibrating string is capable of radiating higher frequencies as coherent outgoing radiation. The switch acts as a band filter with frequencies outside the band being stored as internal heat instead of being radiated: The radiator is then muted and heats up internally instead of delivering outgoing radiating. A spectral analysis, assuming that all frequencies share a common temperature, shows an energy balance between incoming forcing $f$ measured as • $F = \int f^2(x,t)\, dxdt$ assuming periodicity in space and time and integrating over periods, and (rate of) outgoing radiation \$R\$ measured by • $R = \int \gamma U_{tt}^2\, dxdt$, and (rate of) internal energy measured by • $IE = \int \delta^2U_{xt}^2\, dxdt$, together with the oscillator energy • $OE =T^2 = \frac{1}{2}\int (U_t^2 + U_x^2)\, dxdt$ with the energy balance in stationary state with $OE$ constant taking the form • $F = \kappa (R + IE)$ with $\kappa\lessapprox 1$ is a constant independent of $T$, $\gamma$, $\delta$ and $\nu$. In other words, • incoming energy = $\kappa\times$ outgoing radiation energy for $\nu <\frac{1}{\delta}$ • incoming energy = $\kappa\times$ stored internal energy for $\nu >\frac{1}{\delta}$, which can be viewed as an expression of Kirchhoffs’ law that emissivity equals absorptivity. The equality results from the independence of the coefficient $\kappa$ of the damping coefficients $\gamma$ and $\delta^2$, and frequency. Summary: The energy of damping from outgoing radiation or internal heating is the same even if the damping terms represent different physics (emission and absorption) and have different coefficients ($\gamma$ and $\delta^2$). PS: Note that internal heat energy accumulating under (high-frequency) forcing above cut-off eventually will be transformed into low-frequency outgoing radiation, but this transformation is not part of the above model. | # Helmholtz Reciprocity 1 Posted on September 7, 2012 by Helmholtz Reciprocity states that certain phenomena of propagation of light and interaction between light and matter, are reversible. Let us study reversibility in our model of blackbody radiation: • $U_{tt} - U_{xx} - \gamma U_{ttt} - \delta^2U_{xxt} = f$ where the subindices indicate differentiation with respect to space $x$ and time $t$, and 1. $U_{tt} - U_{xx}$ represents a vibrating string with U displacement 2. $- \gamma U_{ttt}$ is a dissipative term modeling outgoing radiation = emission 3. $- \delta^2U_{xxt}$ is a dissipative modeling internal heating = absorption 4. $f$ is incoming forcing/microwaves, where $\gamma$ and $\delta^2$ are positive constants connected to dissipative losses as outgoing radiation = emission and internal heating = absorption. We see emission represented by $-\gamma U_{ttt}$ and absorption by $-\delta^2U_{xxt}$. We now ask: 1. How is the distinction between emission and absorption expressed in this model? 2. Is Helmholtz Reciprocity valid (emission and absorption are reverse processes)? 3. Is Kirchhoff’s Radiation Law (emissivity = absorptivity) valid? Before seeking answers let us recall the basic energy balance between incoming forcing $f$ measured as • $F = \int f^2(x,t)\, dxdt$ assuming periodicity in space and time and integrating over periods, and (rate of) outgoing radiation = emission \$R\$ measured by • $R = \int \gamma U_{tt}^2\, dxdt$, the oscillator energy $OE$ measured by • $OE =\frac{1}{2}\int (U_t^2 + U_x^2)\, dxdt$ and (rate of) internal energy = absorption measured by • $IE = \int \delta^2U_{xt}^2\, dxdt$ • $F = R + IE$ • incoming energy = emission + absorption. The model has a frequency switch switching from emission to absorption as the frequency increases beyond a certain threshold proportional to temperature in accordance with Wien’s displacement law. We now return to questions 1 – 3: Both terms generate dissipative effects when multiplied with $U_t$ as $R = \int \gamma U_{tt}^2\, dxdt$ and $IE = \int \delta^2U_{xt}^2\, dxdt$, but the terms involve different derivatives with $U_{tt}$ acting only in time and $U_{xt}$ acting also in space. The absorption $U_{xt}^2$ represents a smoothing effect in space, which is irreversible and thus cannot be reversed into emission as reversed absorption. The emission $U_{tt}^2$ represents a smoothing effect in time, which is irreversible and thus cannot be reversed into absorption as reversed emission. In other words, in the model both absorption and emission are time irreversible and thus cannot be reversed into each other. We conclude that the model does not satisfy Helmholtz reciprocity. Nevertheless, the model satisfies Kirchhoff’s law as shown in a previous post. Conclusion: The space derivative in $U_{xt}$ models absorption as process of smoothing in space with irreversible transformation of high frequencies in space into low frequencies with a corresponding increase of internal energy as heat energy. Absorbed high frequencies can with increasing temperature be rebuilt through (resonance in) the wave equation into high frequency emission. Absorption and emission are not reverse processes, but my be transformed into each other through (resonance in) the wave equation and the switch. We may compare absorption with a catabolic process of destroying (space-time) structure and emission with an anabolic process of building structure, with the wave equation as a transformer. For more details see Mathematical Physics of Blackbody Radiation and The Clock and the Arrow. Posted in Uncategorized | • 1,335 hits

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http://math.stackexchange.com/questions/90340/proving-p-to-q-to-r-to-p-to-q-to-p-to-r

# Proving $(p \to (q \to r)) \to ((p \to q) \to (p \to r))$ I'm looking for a way to prove $$(p \to (q \to r)) \to ((p \to q) \to (p \to r))$$ from the axioms $$\begin{align} & p \to (q \to p) \\ & (p \to (p \to q)) \to (p \to q) \\ & (p \to q) \to ((q \to r) \to (p \to r)) \\ & (\sim p \to \, \sim q) \to (q \to p) \\ \end{align}$$ using universal substitution and modus ponens I suspect the fourth axiom is not necessary for the proof. I have been working in Tarski's Introduction to Logic and am trying to establish the equivalence of his axiom system with the axioms used at us.metamath.org $$\begin{align} & p \to (q \to p) \\ & (p \to (q \to r)) \to ((p \to q) \to (p \to r)) \\ & ( \sim p \to \, \sim q) \to (q \to p) \\ \end{align}$$ which will allow me to connect Tarski's 4 axioms with all sorts of interesting proofs on that site. - 1 Do you have the deduction theorem available for your system? – Henning Makholm Dec 11 '11 at 2:32 @toph: I agree that the fourth axiom is not relevant here. The proposition you are trying to prove is essentially an internal form of modus ponens, which is valid intuitionitically, but the fourth axiom is essentially double negation elimination. – Zhen Lin Dec 11 '11 at 2:55 This appears to be hard -- are you sure the four axioms you've given are all there is? The first three ones are the K, W, and (almost) B axioms of the BCKW system of combinatory logic, but the C axiom is missing. That makes me doubt that the system you present is complete. – Henning Makholm Dec 11 '11 at 4:45 2 The axioms are taken from page 147 of Tarski's Introduction to Logic. I must (guiltily) add that there are three other axioms: (p <-> q) -> (p -> q), (p <-> q) -> (q -> p), (p -> q) -> ((q -> p) -> (p <-> q). I couldn't see how these axioms, which basically define equivalence, would be necessary. But following your remarks I suspect the last axiom above in particular might be crucial. Thanks for your help so far! – toph Dec 11 '11 at 13:11 3 Aha! Wikipedia's List of logic systems lists your three (original) axioms as Hilbert's second system for "Postive implicational calculus" (i.e., the implicational fragment of intuitionistic logic). Since your target sentence is intuitionistically valid, it indeed ought to be derivable. Still beats me how, though. – Henning Makholm Dec 11 '11 at 15:23 show 7 more comments ## 4 Answers As you suspected, this can be proved from the first three axioms only. I couldn't find a short proof, though – I tried brute force enumeration of the theorems deducible from the three axioms (by taking all pairs of theorems already proved and unifying one with the premise of the other), but didn't find your target in the first $80000$ theorems proved. I then found some guidance in the article on relevance logic in the Handbook of Philosophical Logic. Relevance logic focuses on the fragment of logic in which, roughly speaking, the premises are relevant to the conclusions. It doesn't include the axiom $p\to(q\to p)$, which allows us to add an irrelevant premise to a theorem already proved without that premise, and is thus strictly weaker than the system you're using, but we can nevertheless make use of the results cited in that article. I'll first describe the structure of the proof and how I found it, and then give the proof in detail. Here are the names I'll use for the axioms; the first column names the corresponding axioms of combinator logic, for comparison with the discussion in the comments under the question: $$\begin{array}{c|l|l} \mathbf I&\text{self-implication}&p\to p\\ \mathbf K&\text{weakening}& p \to (q \to p) \\ \hline \mathbf B&\text{prefixing}& (p \to q) \to ((r \to p) \to (r \to q)) \\ \mathbf A&\text{suffixing}& (p \to q) \to ((q \to r) \to (p \to r)) \\ \hline \mathbf W&\text{contraction}& (p \to (p \to q)) \to (p \to q) \\ \mathbf S&\text{self-distribution}&(p \to (q \to r)) \to ((p \to q) \to (p \to r))\\ \hline \mathbf C&\text{permutation}&(p\to(q\to r))\to(q\to(p\to r))\\ &\text{assertion}&p\to((p\to q)\to q) \end{array}$$ (The names are the ones used in the article, except I use "weakening" instead of "positive paradox", since it's shorter and makes more sense to me.) Theorem $1$ of the article states that, with modus ponens (and implicitly universal substitution), the axiom sets formed by self-implication and one each from the three pairs prefixing/suffixing, contraction/self-distribution and permutation/assertion lead to the same theory. What you have is weakening, suffixing and contraction. Self-implication can be deduced from weakening and contraction in a single step (by substituting $p$ for $q$ everywhere). Thus, if we can deduce assertion in your system, the theorem will tell us that we can deduce everything else, including your target, self-distribution. I did find a proof for assertion by brute force search. The article doesn't give a proof of its Theorem $1$ and only says that it can be proved by consulting a book that isn't available online and doing some "fiddling", so we still have to show how to get from self-implication, suffixing, contraction and assertion to self-distribution. I found a deduction of self-distribution online that uses prefixing and permutation. It turns out that prefixing is deducible in a single step from suffixing and permutation, so the problem remained only to deduce permutation. Again, I found a proof for this by brute force search. So here's the entire proof put together, starting with your three axioms and ending with your target. First, a high-level description similar to the actual calls in my Java code: ````assertion = t (t (weakening,suffixing),contraction); permutation = t (suffixing,m (assertion,suffixing)); prefixing = m (suffixing,permutation); target = t (m (prefixing,prefixing),t (permutation,m (contraction,prefixing))); ```` Each call to `m` is an application of modus ponens, in which the first argument is $A$, the second argument is $A\to B$ and the most general unifier that makes the $A$s coincide is used. Each call to `t` is an invocation of transitivity (i.e. deducing $A\to C$ from $A\to B$ and $B\to C$), which can be implemented as ````t (A->B,B->C) = m (B->C,m (A->B,suffixing)) ```` using suffixing, or as ````t (A->B,B->C) = m (A->B,m (B->C,prefixing)) ```` once prefixing is available. Here's the proof spelled out in $14$ steps. The first table shows the theorems used to generate the antecedents $A$ and the implications $A\to B$ for modus ponens, as well as the resulting theorems $B$: $$\begin{array}{c|c|c|c|c} &&A&A\to B&B\\\hline \text{(a)}&\text{weakening}&&&p \to (q \to p)\\ \text{(b)}&\text{suffixing}&&&(p \to q) \to ((q \to r) \to (p \to r))\\ \text{(c)}&\text{contraction}&&&(p \to (p \to q)) \to (p \to q)\\ \hline \text{(d)}&\text{*}&\text{(a)}&\text{(b)}&((p \to q) \to r) \to (q \to r)\\ \text{(e)}&&\text{(b)}&\text{(d)}&p \to ((p \to q) \to (r \to q))\\ \text{(f)}&\text{*}&\text{(e)}&\text{(b)}&(((p \to q) \to (r \to q)) \to s) \to (p \to s)\\ \text{(g)}&\text{assertion}&\text{(c)}&\text{(f)}&p \to ((p \to q) \to q)\\ \text{(h)}&&\text{(g)}&\text{(b)}&(((p \to q) \to q) \to r) \to (p \to r)\\ \text{(i)}&\text{*}&\text{(b)}&\text{(b)}&(((p \to q) \to (r \to q)) \to s) \to ((r \to p) \to s)\\ \text{(j)}&\text{permutation}&\text{(h)}&\text{(i)}&(p \to (q \to r)) \to (q \to (p \to r))\\ \text{(k)}&\text{prefixing}&\text{(b)}&\text{(j)}&(p \to q) \to ((r \to p) \to (r \to q))\\ \text{(l)}&&\text{(k)}&\text{(k)}&(p \to (q \to r)) \to (p \to ((s \to q) \to (s \to r)))\\ \text{(m)}&&\text{(c)}&\text{(k)}&(p \to (q \to (q \to r))) \to (p \to (q \to r))\\ \text{(n)}&\text{*}&\text{(j)}&\text{(b)}&((p \to (q \to r)) \to s) \to ((q \to (p \to r)) \to s)\\ \text{(o)}&&\text{(m)}&\text{(n)}&(p \to (q \to (p \to r))) \to (q \to (p \to r))\\ \text{(p)}&\text{*}&\text{(l)}&\text{(b)}&((p \to ((q \to r) \to (q \to s))) \to t) \to ((p \to (r \to s)) \to t)\\ \text{(q)}&\text{self-distribution}&\text{(o)}&\text{(p)}&(p \to (q \to r)) \to ((p \to q) \to (p \to r))\\ \end{array}$$ The asterisks mark intermediate steps in invocations of transitivity. Note that most theorems with more than three variables occur only in such intermediate steps. Substitutions are being performed as late as possible; by performing them as early as possible, the proof could be written using only theorems with at most three variables. The second table shows the substitutions used; you can also find these automatically by unification. The variables are named such that they appear in alphabetical order in the resulting theorems. $$\begin{array}{c|l|l} &A&A\to B\\\hline \text{(d)}& p\mapsto q,q\mapsto p& p\mapsto q,q\mapsto (p \to q),r\mapsto r\\ \text{(e)}& p\mapsto r,q\mapsto p,r\mapsto q& p\mapsto r,q\mapsto p,r\mapsto ((p \to q) \to (r \to q))\\ \text{(f)}& p\mapsto p,q\mapsto q,r\mapsto r& p\mapsto p,q\mapsto ((p \to q) \to (r \to q)),r\mapsto s\\ \text{(g)}& p\mapsto (p \to q),q\mapsto q& p\mapsto p,q\mapsto q,r\mapsto (p \to q),s\mapsto ((p \to q) \to q)\\ \text{(h)}& p\mapsto p,q\mapsto q& p\mapsto p,q\mapsto ((p \to q) \to q),r\mapsto r\\ \text{(i)}& p\mapsto r,q\mapsto p,r\mapsto q& p\mapsto (r \to p),q\mapsto ((p \to q) \to (r \to q)),r\mapsto s\\ \text{(j)}& p\mapsto q,q\mapsto r,r\mapsto (p \to r)& p\mapsto (q \to r),q\mapsto r,r\mapsto p,s\mapsto (q \to (p \to r))\\ \text{(k)}& p\mapsto r,q\mapsto p,r\mapsto q& p\mapsto (r \to p),q\mapsto (p \to q),r\mapsto (r \to q)\\ \text{(l)}& p\mapsto q,q\mapsto r,r\mapsto s& p\mapsto (q \to r),q\mapsto ((s \to q) \to (s \to r)),r\mapsto p\\ \text{(m)}& p\mapsto q,q\mapsto r& p\mapsto (q \to (q \to r)),q\mapsto (q \to r),r\mapsto p\\ \text{(n)}& p\mapsto q,q\mapsto p,r\mapsto r& p\mapsto (q \to (p \to r)),q\mapsto (p \to (q \to r)),r\mapsto s\\ \text{(o)}& p\mapsto q,q\mapsto p,r\mapsto r& p\mapsto q,q\mapsto p,r\mapsto (p \to r),s\mapsto (q \to (p \to r))\\ \text{(p)}& p\mapsto p,q\mapsto r,r\mapsto s,s\mapsto q& p\mapsto (p \to (r \to s)),q\mapsto (p \to ((q \to r) \to (q \to s))),r\mapsto t\\ \text{(q)}& p\mapsto p,q\mapsto (p \to q),r\mapsto r& p\mapsto p,q\mapsto p,r\mapsto q,s\mapsto r,t\mapsto ((p \to q) \to (p \to r))\\ \end{array}$$ - 1 Well done! ${}$ – Henning Makholm Dec 21 '11 at 13:48 Would you mind if I add a column to your first table showing the corresponding combinator names for comparison with the discussion between me and Zhen Lin in the comments? – Henning Makholm Dec 21 '11 at 13:53 @Henning: That would be great, thanks! – joriki Dec 21 '11 at 13:55 @Henning: Interesting -- so the answer actually fits well with your discussion, in that the hard part that I did by computer search was to deduce $\mathbf C$/permutation (up to step (j)), and the rest was known. – joriki Dec 21 '11 at 14:29 It seems to me that a much simpler and human readable proof is possible, unless I’m misunderstanding something. Using the Deduction Theorem, the result is relatively straightforward to prove. This motivated me to prove the Deduction Theorem for this logical system, which I found to be less straightforward, but still not particularly difficult. To make sure we’re all on the same page, the logical system in question consists of the inference rule modus ponens (MP) and the following three axiom schema: axiom 1 $\;\;\;\;\; p \; \rightarrow \; (q \rightarrow p)$ axiom 2 $\;\;\;\;\;(p \rightarrow q) \;\; \rightarrow \;\; [\; (q \rightarrow r) \rightarrow (p \rightarrow r) \; ]$ axiom 3 $\;\;\;\;\;[\; p \rightarrow (p \rightarrow q) \; ] \;\; \rightarrow \;\; (p \rightarrow q)$ We want to show that the following wff (well formed formula) is provable in this logical system: $$[p \rightarrow (q \rightarrow r)] \;\; \rightarrow \;\; [(p \rightarrow q) \rightarrow (p \rightarrow r)]$$ By 3 applications of the Deduction Theorem (proved further below), it suffices to prove $r$ under the following assumptions: $p \rightarrow (q \rightarrow r),$ $p \rightarrow q,$ and $p.$ That is, it suffices to prove $$p \rightarrow (q \rightarrow r), \; p \rightarrow q, \; p \;\vdash \; r$$ (1) $\;\;\;p \rightarrow q$ (2) $\;\;\;$(line 1) $\rightarrow [\;(q \rightarrow r) \rightarrow (p \rightarrow r)\;]$ (3) $\;\;\;(q \rightarrow r) \rightarrow (p \rightarrow r)$ (4) $\;\;\;p \rightarrow (q \rightarrow r)$ (5) $\;\;\;p$ (6) $\;\;\;q \rightarrow r$ (7) $\;\;\;p \rightarrow r$ (8) $\;\;\;r$ Reasons for the above steps (1) $\;\;\;$assumption (2) $\;\;\;$axiom 2 (3) $\;\;\;$MP (lines 1, 2) (4) $\;\;\;$assumption (5) $\;\;\;$assumption (6) $\;\;\;$MP (lines 5, 4) (7) $\;\;\;$MP (lines 6, 3) (8) $\;\;\;$MP (lines 5, 7) In trying to prove the Deduction Theorem for this logical system (i.e. $\Gamma, \;p \vdash q$ implies $\Gamma \vdash p \rightarrow q$), I simply followed the standard proof (which makes use of axiom 1 and the wff we originally wanted to prove), and noted that the standard proof only requires us to make use of the following 3 results: 1. If $q$ is an axiom or a member of $\Gamma$, then for any wff $p$ we can prove $p \rightarrow q$ in our logical system. 2. We can prove $p \rightarrow p$ in our logical system. 3. Given $p \rightarrow r$ and $p \rightarrow (r \rightarrow q)$, we can prove $p \rightarrow q$ in our logical system. proof of 1: $\;\;\;$Apply MP to $q$ and $q \rightarrow (p \rightarrow q)$ (axiom 1). proof of 2: $\;\;\;$Apply MP to $p \rightarrow (p \rightarrow p)$ (axiom 1) and axiom 3. proof of 3: $\;\;\;$This is the difficult part. Below is a proof of what’s needed, namely $$p \rightarrow r, \; p \rightarrow (r \rightarrow q) \;\vdash \; p \rightarrow q$$ (1) $\;\;\;p \rightarrow r$ (2) $\;\;\;p \rightarrow (r \rightarrow q)$ (3) $\;\;\;$(line 1) $\rightarrow \; [(r \rightarrow q) \rightarrow (p \rightarrow q)]$ (4) $\;\;\;(r \rightarrow q) \rightarrow (p \rightarrow q)$ (5) $\;\;\;$(line 2) $\;\;\rightarrow \;\; \{\;$(line 4)$\rightarrow [p \rightarrow (p \rightarrow q)] \; \}$ (6) $\;\;\;$(line 4) $\;\rightarrow \; [p \rightarrow (p \rightarrow q)]$ (7) $\;\;\;p \rightarrow (p \rightarrow q)$ (8) $\;\;\;[p \rightarrow (p \rightarrow q)] \; \rightarrow \; (p \rightarrow q)$ (9) $\;\;\;p \rightarrow q$ Reasons for the above steps (1) $\;\;\;$assumption (2) $\;\;\;$assumption (3) $\;\;\;$axiom 2 ($r$ is $q$) (4) $\;\;\;$MP (lines 1, 3) (5) $\;\;\;$axiom 2 ($r$ is $p \rightarrow q$) (6) $\;\;\;$MP (lines 2, 5) (7) $\;\;\;$MP (lines 4, 6) (8) $\;\;\;$axiom 3 (9) $\;\;\;$MP (lines 7, 8) Here is how I discovered the above proof. Working backwards, I noticed that the conclusion of axiom 3 was what I wanted, so I made note of the fact that it would be enough to obtain $p \rightarrow (p \rightarrow q).$ Then I tried working forward. First, I applied axiom 2 followed by MP to the assumption $p \rightarrow r,$ using $q$ as the introduced suffix. (Since I already had $p$ and $r$ appearing, this seemed to be a natural way to get $q$ to appear.) Then I tried applying axiom 2 followed by MP to the assumption $p \rightarrow (r \rightarrow q).$ At some point (perhaps my 3rd attempt), I used $p \rightarrow q$ as the introduced suffix, motivated by the fact that this got line 4 to show up. After this, the proof immediately fell into place, since in line 5 the conclusion of the conclusion is $p \rightarrow (p \rightarrow q),$ which I had previouly noted was sufficient. Incidentally, the logical system above is the same (in the sense of having the same set of provable wffs) as the logical system with the inference rule MP and the following two axioms: axiom 1 and the wff we originally wanted to prove. Each of these logical systems is also equal to the logical system with MP and Deductive Theorem as inference rules and no axioms (thus, one might call this system “DT Logic”). I think logicians call this the positive implicational fragment of intuitionistic propositional logic, but I like “DT Logic” better. Other axiomatizations of DT Logic can be found at the Wikipedia page “List of logic systems” under the category “Positive implicational calculus”. For completeness, here’s a proof that DT Logic can be characterized by no axioms along with the inference rules MP and DT (and also the Rule of Assumptions, I suppose). It suffices to prove, in this no-axiom logical system, axiom 1 and the wff we were proving in this thread. 1. $\;\;\;p,\; q \vdash p\;$ implies $\;p \vdash q \rightarrow p\;$ implies $\;\vdash p \rightarrow (q \rightarrow p)$ 2. $\;\;\;p \rightarrow (q \rightarrow r), \; p \rightarrow q, \; p \; \vdash \; r\;\;\;$ (MP, 3 times) implies $\;p \rightarrow (q \rightarrow r), \; p \rightarrow q \; \vdash \; p \rightarrow r\;\;\;$ (DT) implies $\;p \rightarrow (q \rightarrow r) \; \vdash \; (p \rightarrow q) \rightarrow (p \rightarrow r)\;\;\;$ (DT) implies $\;\vdash \; [p \rightarrow (q \rightarrow r)] \;\; \rightarrow \;\; [(p \rightarrow q) \rightarrow (p \rightarrow r)] \;\;\;$ (DT) - 1 Well done too. In combinator-logical notation, this amounts to using $\mathbf W(\mathbf A\;M\;(\mathbf A\;N))$ instead of $\mathbf S\;M\;N$ in the translation of an application -- which is indeed so simple that we ought to have been able to find it by hand. – Henning Makholm Dec 27 '11 at 18:10 1 Incidentally, your "MP+DT" formulation is known (at least in computer-sciency contexts) as "natural deduction" for $\to$, and corresponds exactly, via the Curry-Howard isomorphism, to the simply typed lambda calculus. – Henning Makholm Dec 27 '11 at 18:14 I only saw this now; great! – joriki May 14 '12 at 7:55 For comparison, here is Joriki's solution in the combinator language we used in the comment thread between me and Zhen Lin: $$\begin{align} \mathbf X &= \mathbf A (\mathbf A \; \mathbf K \; \mathbf A) \mathbf W \\ \mathbf C &= \mathbf A \; \mathbf A (\mathbf A \; \mathbf X) \\ \mathbf B &= \mathbf C \; \mathbf A \\ \mathbf S &= \mathbf A (\mathbf B \; \mathbf B) \; (\mathbf A \; \mathbf C (\mathbf B \; \mathbf W)) \end{align}$$ where $\mathbf X$ is an ad-hoc name for Joriki's "assertion" formula. Zhen Lin's constuction for the final line $$\mathbf{S} = \mathbf{A A} ( \mathbf{A} ( \mathbf{B W} ) ( \mathbf{A A} ) )$$ is slightly more efficient than Joriki's because it contains only one $\mathbf B$ that needs to be unfolded. This yields the final term $$\mathbf S = \mathbf A\; \mathbf A\;(\mathbf A ( \mathbf A \; \mathbf A (\mathbf A \; (\mathbf A (\mathbf A \; \mathbf K \; \mathbf A) \mathbf W) ) \mathbf A \; \mathbf W) \; (\mathbf A\; \mathbf A))$$ which encodes a Hilbert-style proof with 15 fully substituted axiom instances and 14 modus ponens steps. - Thanks very much for this! An impressive demonstration of the conciseness of combinator logic :-) – joriki Dec 21 '11 at 15:21 This isn't much at all, but I think too long for a comment. Managing parenthesis for long formulas becomes difficult for me, so I'll use Polish notation. Somewhere in the proof (which most fittingly I would think would come as the penultimate step of the proof) we have to have one of the following: 1. C CCpqCpr C CpCqr CCpqCpr (substitution instance of axiom 1). 2. CC CpCqr C CpCqr CCpqCpr C CpCqr CCpqCpr (substitution instance of axiom 2). 3. C CpCqr x C x CCpqCpr C CpCqr CCpqCpr (substitution instance of axiom 3). 4. CC NCCpqCpr NCpCqr C CpCqr CCpqCpr (substitution instance of axiom 4). Now, CCpqCpr is not a theorem in a complete classical propositional calculus. So, cross out 1. The antecedent of 2; CCpCqrCCpCqrCCpqCpr can't get proven using the rules here, since we'd have to have CpCqr as a theorem. Edit: so, we can't have 2 as the penultimate step of the proof. We might still have CCpCqrCCpCqrCCpqCpr as a theorem (if we don't have it as a theorem we know right then and there the system is incomplete with respect to the full, classical propositional calculus), but we'd have to have CCpCqrCCpCqrCCpqCpr as the consequent of some other theorem, along with the antecedent of this other theorem as a theorem also. So, either a step of one of the forms of 3, or 4 will appear in the proof somewhere. The antecedent of 3 I'd guess contingent, while the antecedent of 4 does hold as a tautology... though I'm not sure it qualifies as a theorem in this system. - 1 Please don't leave downvotes without stating the reason. From the FAQ: "Add comments indicating what, specifically, is wrong." – joriki Dec 13 '11 at 4:58 – Carl Mummert Dec 13 '11 at 19:14 2 I added a second downvote because writing formulas in Polish notation only serves to obfuscate an otherwise very nice post. – Carl Mummert Dec 13 '11 at 19:17 @Carl: Many thanks for the link. If I remember correctly this is the second time I've learned something important from you about this site. Also thanks for explaining your downvote ;-) – joriki Dec 13 '11 at 19:22 3 @CarlMummert So goes your opinion. You can find hints in the logical literature that some have disagreed at points at least, such as Rose and Rosser (unless they wanted to obfuscate things), as described by Chang jstor.org/pss/20015986. After all, "C" comes as an abbreviation for "conditional" while "->" isn't like that. So, I don't see how they can "only serve to obfuscate", though they may obfuscate to you for sure. – Doug Spoonwood Dec 14 '11 at 4:33

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http://math.stackexchange.com/questions/2158/division-of-factorials/2165

# Division of Factorials I have a partition of a positive integer (p). How can I prove that the factorial of p can always be divided by the product of the factorials of the parts? As a quick example 9!/(2!3!4!) = 1260 (no remainder), where 9=2+3+4. I can nearly see it by looking at factors, but I can't see a way to guarantee it. - 12 @Andrew: Then why mention it? :-) – ShreevatsaR Aug 11 '10 at 17:43 @ShreevatsaR I apologise if it was spam. It was just to show that there is yet another way of proving it, and I may return to this page after publication. – Andrew Aug 11 '10 at 17:58 2 So @Andrew, were you really intending to post your proof here? – J. M. Oct 29 '11 at 11:05 ## 8 Answers The key observation is that the product of $n$ consecutive integers is divisible by $n!$. This can be proved by induction. - A quicker proof than induction would be to note that (by definition) $\binom{k}{n}$ is an integer, hence $n!|k\cdot (k-1)\ldots (k-n+2)\cdot (k-n+1)$. – Andrew Aug 11 '10 at 17:24 Is that because given n (or more) consecutive integers one of them will always be divisible by n? – Guillermo Phillips Aug 13 '10 at 13:05 The "high-level" way to see this is to recall that whenever a finite group $G$ has a subgroup $H$, we know that $|H|$ divides $|G|$. Then note that $S_n$ clearly contains $S_{n_1} \times ... \times S_{n_k}$ as a subgroup for any partition $n_1 + ... + n_k = n$. (This is actually the same as the combinatorial interpretation in Robin Chapman's answer, since what we are counting is the number of cosets $G/H$, and these cosets are precisely what the multinomial theorem is counting.) This basic lemma is surprisingly useful. For example, it is not hard to use it to show that $m! (n!)^m$ divides $(mn)!$. - 3 Among many nice answers, I like this one the best. – Pete L. Clark Aug 11 '10 at 20:17 Below is a sketch of a little-known purely arithmetical proof that binomial coefficients are integral. I purposely constructed the proof so that it would be comprehensible to an educated layperson. The proof gives an algorithm to rewrite a binomial coefficient as a product of fractions whose denominators are coprime to any given prime. The method of proof is best comprehended by applying the algorithm to a specific example. [Note: It may prove helpful to first read this simpler example before proceeding to the exposition below]. E.g. consider $\ \ \binom{39}{17}\: =\: \frac{39!}{22! \; 17!}\: =\: \frac{23 \cdot 24 \cdots\; 39}{1 \cdot 2 \cdots\; 17}\:.\$ When this fraction is reduced to lowest terms $\rm\:a/b\:,\$ no prime $\rm\ p > 17\$ can divide its denominator $\rm\: b\:,\:$ since $\rm\ b\:|\:17\:!\ \:$ Hence, to show that $\rm\ a/b\$ is an integer, it suffices to show that no prime $\rm\ p \le 17\$ divides its denominator $\rm\: b\:$. E.g. we show that $2$ doesn't divide $\rm b$. The highest power of $2$ in the denominator terms is $16 < 17$. Now align the numerator and denominator terms $\rm (mod\ 16)$ by shifting the 1st numerator term so it lies above its value $\rm (mod\ 16)$, viz. $23 \equiv 7 \pmod{16}$ so right-shift the numerator terms until $23\:$ lies above $7$ $$\frac{}{1}\frac{}{2}\frac{}{3}\frac{}{4}\frac{}{5}\frac{}{6}\frac{23}{7}\frac{24}{8}\frac{25}{9}\frac{26}{10}\frac{27}{11}\frac{28}{12}\frac{29}{13}\frac{30}{14}\frac{31}{15}\frac{32}{16}\frac{33}{17}\frac{34}{}\frac{35}{}\frac{36}{}\frac{37}{}\frac{38}{}\frac{39}{}$$ Now I claim that $2$ doesn't divide the reduced denominator of each aligned fraction. Indeed $\ 24/8 = 3$, $\: 26/10 = 13/5$, $\:\ 28/12 = 7/3$, $\:\ 30/14 = 15/7$, $\:\ 32/16 = 2$. This holds because these fractions $\rm\; c/d \;$ satisfy $\rm c \equiv d\ (mod\ 16)\:$ i.e. $\rm c = d + 16\: n \;$ so $\rm 2|d \Rightarrow 2|c$, $\rm\; 4|d \Rightarrow 4|c$, $\:\cdots$, $\rm\; 16|d \Rightarrow 16|c$, i.e. any power of $2\:$ below $16$ that divides $\rm d$ must also divide $\rm c$, so it cancels out upon reduction. Therefore to prove that $2$ doesn't divide the reduced denominator of $\binom{39}{17}$ it suffices to prove the same for the "leftover" fraction composed of the above non-aligned terms $(34 \cdots 39)/(1 \cdots 6) = \binom{39}{6}$. Being an $\rm\binom{n}{k}$ with smaller $\rm k = 6 < 17\:,\:$ this follows by induction. Since the same proof works for any prime $\rm p$, we conclude that no prime divides the reduced denominator of $\binom{39}{17}$, therefore it is an integer.$\quad$ QED Informally, the reason that this works is because the denominator sequence starts at $1$, which is coprime to every prime $\rm p$. This ensures that it is the "greediest" possible contiguous sequence of integers, in the sense that its product contains the least power of $\rm\:p\:$ compared to other contiguous sequences of equal length. The algorithm extends to multinomials by using the simple reduction of multinomials to products of binomials mentioned in my prior post here. - Very elegant idea! – Andres Caicedo Nov 9 '10 at 5:24 These quotients are integers since they solve counting problems. For instance, how many nine-letter words are there with 2 As, 3 Bs and 4 Cs? For the full story see the multinomial theorem. - 2 While true, it's not a good answer - usually one needs to be convinced that the quotients are integers before they can understand why the results are the solution to a counting problem... – BlueRaja - Danny Pflughoeft Aug 11 '10 at 16:19 11 @BlueRaja: Not at all! Showing that some ratio is the solution to a counting problem is my favourite way of proving divisibility. You don't need to be convinced that the quotient is an integer here; once you show it's the number of [something], it has to be an integer. (You don't usually check divisibility and compare prime factors when you show that the number of ways of choosing k out of n is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.) – ShreevatsaR Aug 11 '10 at 17:37 If you believe (:-) in the two-part Newton case, then the rest is easily obtained by induction. For instance (to motivate you to write a full proof): $$\frac{9!}{2! \, 3! \,4!} = \frac{9!}{5!\, 4!} \frac{5!}{2!\, 3!}$$ - I wanted to indent the equality, but the simplest standard TeX spacing "\ \ \ \ " didn't work. – wlod Aug 11 '10 at 15:48 use two dollars for centered LaTeX, (single dollar for inline). [; \cdots ;] doesn't work on this site.... – anon Aug 11 '10 at 15:58 We use dollar signs to denote TeX equations like `$this$`. – KennyTM Aug 11 '10 at 15:58 1 In terms of multinomial coefficients: choosing (2, 3, and 4) out of 9 is the same as first choosing 4 from the original 9, then choosing 2 and 3 out of the remaining 5. Etc. – ShreevatsaR Aug 11 '10 at 17:42 Here is another proof: We use Legendre's formula for the exact power of a prime $p$ which divides $n!$ which is given by $$\sum_{k=1}^{\infty} [\frac{n}{p^k}]$$ where $[x]$ is the integer part of $x$. Coupled with $$\sum_{i=1}^{n} [x_i] \le [\sum_{i=1}^{n} x_i]$$ we get that if $\sum_{i=1}^{n} a_i = N$ then $$\sum_{i=1}^{n} [\frac{a_i}{p^k}] \le [\frac{N}{p^k}]$$ And so any prime power which divides $(a_1)! \dots (a_n)!$ divides $N!$ and so $\frac{N!}{(a_1)! \dots (a_n)!}$ is an integer. - Besides the obvious combinatorial interpretations, one also has the has the following reduction from multinomial coefficient to products of binomial coefficients. Namely for $n = i+j+k+\cdots + m$ $$\frac{n!}{i!j!k!\cdots m!} = \binom{n}{i} \frac{(n-i)!}{j!k!\cdots m!} = \binom{n}{i}\binom{n-i}{j} \frac{(n-i-j)!}{k!\cdots m!} = \;\cdots$$ - 4 This is implicitly what wlod's answer does, with an example. – ShreevatsaR Aug 13 '10 at 16:57

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http://mathhelpforum.com/algebra/155259-relations-set.html

# Thread: 1. ## Relations in a set Hello, Given is a positive integer $k$. Prove that out of every set of integers which has more than $3^k$ elements one can pick out a sub-set $S$ with $k+1$ elements and the following quality: for any two subsets $A,B\subseteq S$ the sum of all elements in A is different from the sum of all elements in B. I've been trying to play with base-system representation for a while but it leads me nowhere. Your help will be appreciated. 2. I'm not sure about this, but I do have an idea about how you could approach this. Maybe you can prove that there is a subset you can make such that all numbers are "out of reach". For example, for k = 3 and the set is all numbers up until 27, the numbers 1, 4, 9 and 27 satisfy S. Maybe if you order them and pick the $3^0$th, $3^1$th, $3^2$th, $3^3$th, etc. ? 3. Ok, but these numbers don't have to start with 1 and increment by 1, they can be any numbers, including negatives. 4. Yeah, I know. Hence the "maybe if you order them". I have no idea how to actually prove this, but this was something that sprang to mind.

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http://mathoverflow.net/questions/100652/efficient-algorithm-for-projection-onto-a-convex-set/100703

## Efficient Algorithm For Projection Onto A Convex Set ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given $\mathbf{x} \in \mathbb{R}^n$ and $\tau$ a scalar, I would like to solve the following Euclidean projection problem: ```$\underset{\mathbf{p}}{\mathrm{argmin}} \; \|\mathbf{p}-\mathbf{x}\|_2 \;\; \mathrm{s.t.} \;\; \displaystyle \sum_{i}{ \left \| \left [ \begin{array}{c} \mathbf{f}_i^\mathrm{T} \\ \mathbf{g}_i^\mathrm{T} \end{array} \right] \cdot \mathbf{p} \right \|_2 } \leq \tau $```, where $\mathbf{f}_i,\mathbf{g}_i \in \mathbb{R}^n$. The above is a convex function over a convex set and as such should have a unique solution. Moreover, we can find the upper bound on the summation as follows: ```$\displaystyle \sum_{i}{ \left \| \left [ \begin{array}{c} \mathbf{f}_i^\mathrm{T} \\ \mathbf{g}_i^\mathrm{T} \end{array} \right] \cdot \mathbf{p} \right \|_2 } \leq \|\mathbf{p}\|_2 \cdot \sum_i \sigma_i $```, where $\sigma_i$ is the operator norm of the $2 \times n$ matrix $[\mathbf{f}_i \;\; \mathbf{g}_i]^{\mathrm{T}}$. I have been using CVX to solve the above, but it's just too slow in its current form. I have not figured out how to make use of them, but the operator norms are easily found before-hand. Can anyone suggest a re-formulation of the above or an algorithm that is tailored to these types of problems? - Have you tried minConf? di.ens.fr/~mschmidt/Software/minConf.html – Vidit Nanda Jun 26 at 1:22 @Vel Nias: I looked at minConf but unless I have misread the page, it requires one to provide a function to compute the projection onto the convex set, which is the problem I am trying to find an efficient solution for. – AnonSubmitter85 Jun 26 at 1:28 I'm confused: aren't you trying to minimize the distance to a fixed $x \in \mathbb{R}^n$ constrained by a single convex inequality? It sounds like "distance to $x$" is the function that you should use. Have I misunderstood? – Vidit Nanda Jun 26 at 1:51 1 @Vel Nias: I think you understand things correctly. However, in order to use either minConf_SPG() or minConf_SPG() I have to supply a projection function of the form: funProj(x) = argmin_y ||x - y||_2, subject to y is in X. Is this not the very problem I trying to solve? I am left looking at minConf thinking it is for problems for which my problem above is only a sub-problem. – AnonSubmitter85 Jun 26 at 2:04 1 Either you can solve it as an SOCP (presumably that's something you've already tried), or you can use an ADMM style method to solve it. The details are somewhat messy, but it seems that one can do a fairly good job of efficiently solving this. Some of the key ideas that you can exploit, are discussed in the paper: Barbero and Sra (2011), people.kyb.tuebingen.mpg.de/suvrit/work/papers/… – S. Sra Jun 26 at 14:29 show 3 more comments ## 3 Answers It's easier to just write $A_i = [f_i \ g_i]^T$ so then the problem is: $$\min_p \frac{1}{2}\|x-p\|^2 \quad \mbox{s.t. } \sum_i \|A_ip\| \le \tau$$ Forming the Lagrangian and minimizing shows that the optimal $p^*$ satisfies: $$p^* = x - \lambda^* \sum_i A_i^T \frac{ A_i p^* }{ \|A_i p^* \|}$$ So one obvious thing to try is just iterating this equation. We can solve for $\lambda^*$ with a 1-dimensional root-finding procedure since we know that $\sum_i \|A_i p^*\| = \tau$ (assuming that $x$ is not already in the interior of the feasible region). In summary, we iterate: $$d_n = \sum_i A_i^T \frac{ A_i p_{n-1} }{ \|A_i p_{n-1} \|}$$ $$p_n = x - \lambda_n d_n, \mbox{ where } \lambda_n \mbox{ is given by } \sum_i \|A_i(x - \lambda_n d_n)\| = \tau$$ Of course, you need to be careful not to divide zero by zero in case $A_i p_n=0$. Unfortunately, I don't know if this procedure converges, but it might be sufficient for your purposes. - This iteration is a simple version of what is called "the method of multipliers" in optimization. Its convergence isn't robust, but can be improved by using an augmented Lagrangian to stabilize the algorithm. – Brian Borchers Jun 27 at 14:54 I don't see how we know that $sum_{i}{\| A_i (x-\lambda_n d_n )\|} = \tau$ has a solution. – AnonSubmitter85 Jul 21 at 20:36 ### 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. You haven't told us anything about the size of your problem instances. How many terms are there in the sum of norms? What is $n$? Your problem is an example of a "sum of norms" optimization problem. Searching with Google Scholar will lead you to published research on this class of problems. CVX is using a standard approach for solving this problem by reformulating it as a second order cone programming (SOCP) problem and then using a primal-dual interior point method (SeDuMi's or SDPT3) to solve the resulting SOCP. For small problem instances, this should be a very robust and reasonably fast approach to solving the problem, but there are faster primal-dual codes for SOCP available (both CPLEX and MOSEK can be used to solve SOCP's.) You could ask CVX to extract the SOCP problem and export it in SeDuMi format and then try to use another solver on the SOCP. You might also look at first order methods to solve the SOCP- For example, I believe that TFOCS could be used to solve the problem. - I am working with images, so each $n$ is the total number of pixels and there will be $n+p-1$ terms in the sum of norms, where $p$ will probably be less than 20 . I am currently trying to get a rough cut working, so I am using smaller images, say, $100 \times 100 \Rightarrow n = 10000$. However, I'd like to use more realistic image sizes, so $n$ could be over a million. – AnonSubmitter85 Jun 27 at 6:07 For problems of that size, you'll definitely want to look at first order methods such as the alternating direction method of multipliers. – Brian Borchers Jun 27 at 14:51 Check out Dattorro's convex optimization book, page 748. -

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http://cms.math.ca/cmb/v44/

Canadian Mathematical Society www.cms.math.ca | | Site map | CMS store location: Publications → journals → CMB « 2000 (v43) 2002 (v45) » Volume 44 (2001) Page Contents 3 Alexandru, Victor; Popescu, Nicolae; Zaharescu, Alexandru The generating degree $\gdeg (A)$ of a topological commutative ring $A$ with $\Char A = 0$ is the cardinality of the smallest subset $M$ of $A$ for which the subring $\Z[M]$ is dense in $A$. For a prime number $p$, $\C_p$ denotes the topological completion of an algebraic closure of the field $\Q_p$ of $p$-adic numbers. We prove that $\gdeg (\C_p) = 1$, \ie, there exists $t$ in $\C_p$ such that $\Z[t]$ is dense in $\C_p$. We also compute $\gdeg \bigl( A(U) \bigr)$ where $A(U)$ is the ring of rigid analytic functions defined on a ball $U$ in $\C_p$. If $U$ is a closed ball then $\gdeg \bigl( A(U) \bigr) = 2$ while if $U$ is an open ball then $\gdeg \bigl( A(U) \bigr)$ is infinite. We show more generally that $\gdeg \bigl( A(U) \bigr)$ is finite for any {\it affinoid} $U$ in $\PP^1 (\C_p)$ and $\gdeg \bigl( A(U) \bigr)$ is infinite for any {\it wide open} subset $U$ of $\PP^1 (\C_p)$. 12 Anisca, Razvan; Ilie, Monica This paper is concerned with the structure of the arithmetic sum of a finite number of central Cantor sets. The technique used to study this consists of a duality between central Cantor sets and sets of subsums of certain infinite series. One consequence is that the sum of a finite number of central Cantor sets is one of the following: a finite union of closed intervals, homeomorphic to the Cantor ternary set or an $M$-Cantorval. 19 Brindza, B.; Pintér, Á.; Schmidt, W. M. In this note the multiplicities of binary recurrences over algebraic number fields are investigated under some natural assumptions. 22 Evans, Ronald Precise, elegant evaluations are given for Gauss sums of orders six and twelve. 27 Goodaire, Edgar G.; Milies, César Polcino We show that an $\RA$ loop has a torsion-free normal complement in the loop of normalized units of its integral loop ring. We also investigate whether an $\RA$ loop can be normal in its unit loop. Over fields, this can never happen. 36 Kapovich, Michael; Millson, John J. The Hamiltonian potentials of the bending deformations of $n$-gons in $\E^3$ studied in \cite{KM} and \cite{Kl} give rise to a Hamiltonian action of the Malcev Lie algebra $\p_n$ of the pure braid group $P_n$ on the moduli space $M_r$ of $n$-gon linkages with the side-lengths $r= (r_1,\dots, r_n)$ in $\E^3$. If $e\in M_r$ is a singular point we may linearize the vector fields in $\p_n$ at $e$. This linearization yields a flat connection $\nabla$ on the space $\C^n_*$ of $n$ distinct points on $\C$. We show that the monodromy of $\nabla$ is the dual of a quotient of a specialized reduced Gassner representation. 61 Kats, B. A. The paper is dealing with determination of the integral $\int_{\gamma} f \,dz$ along the fractal arc $\gamma$ on the complex plane by terms of polynomial approximations of the function~$f$. We obtain inequalities for polynomials and conditions of integrability for functions from the H\"older, Besov and Slobodetskii spaces. 70 Lempert, László; Szőke, Róbert Motivated by deformation theory of holomorphic maps between almost complex manifolds we endow, in a natural way, the tangent bundle of an almost complex manifold with an almost complex structure. We describe various properties of this structure. 80 Levin, Michael Applying the Sullivan conjecture we construct compacta of certain cohomological and extensional dimensions. 87 Lieman, Daniel; Shparlinski, Igor Let $p$ be prime and let $\vartheta\in\Z^*_p$ be of multiplicative order $t$ modulo $p$. We consider exponential sums of the form $$S(a) = \sum_{x =1}^{t} \exp(2\pi i a \vartheta^{x^2}/p)$$ and prove that for any $\varepsilon > 0$ $$\max_{\gcd(a,p) = 1} |S(a)| = O( t^{5/6 + \varepsilon}p^{1/8}) .$$ 93 Neumann, B. H. A challenge by R.~Padmanabhan to prove by group theory the commutativity of cancellative semigroups satisfying a particular law has led to the proof of more general semigroup laws being equivalent to quite simple ones. 97 Ou, Zhiming M.; Williams, Kenneth S. An asymptotic formula is obtained for the number of cyclic quartic fields over $Q$ with discriminant $\leq x$. 105 Pilipović, Stevan The singular spectrum of $u$ in a convolution equation $\mu * u = f$, where $\mu$ and $f$ are tempered ultradistributions of Beurling or Roumieau type is estimated by $$SS u \subset (\mathbf{R}^n \times \Char \mu) \cup SS f.$$ The same is done for $SS_{*}u$. 115 Roy, Damien The purpose of this paper is to show the limitations of the conjectures of algebraic approximation. For this, we construct points of $\bC^m$ which do not admit good algebraic approximations of bounded degree and height, when the bounds on the degree and the height are taken from specific sequences. The coordinates of these points are Liouville numbers. 121 Wojciechowski, Michał Simple necessary conditions for weak type $(1,1)$ of invariant operators on $L(\rr^d)$ and their applications to rational Fourier multiplier are given. 126 Zeron, E. Santillan Around 1995, Professors Lupacciolu, Chirka and Stout showed that a closed subset of $\C^N$ ($N\geq 2$) is removable for holomorphic functions, if its topological dimension is less than or equal to $N-2$. Besides, they asked whether closed subsets of $\C^2$ homeomorphic to the real line (the simplest 1-dimensional sets) are removable for holomorphic functions. In this paper we propose a positive answer to that question. 129 Currás-Bosch, Carlos In this paper the germ of neighborhood of a compact leaf in a Lagrangian foliation is symplectically classified when the compact leaf is $\bT^2$, the affine structure induced by the Lagrangian foliation on the leaf is complete, and the holonomy of $\bT^2$ in the foliation linearizes. The germ of neighborhood is classified by a function, depending on one transverse coordinate, this function is related to the affine structure of the nearly compact leaves. 140 Gotay, Mark J.; Grabowski, Janusz We prove an algebraic no-go theorem'' to the effect that a nontrivial \pa\ cannot be realized as an associative algebra with the commu\-ta\-tor bracket. Using it, we show that there is an obstruction to quantizing the \pa\ of polynomials generated by a nilpotent \ba\ on a \sm. This result generalizes \gr 's famous theorem on the impossibility of quantizing the Poisson algebra of polynomials on $\r^{2n}$. Finally, we explicitly construct a polynomial quantization of a \sm\ with a solvable \ba, thereby showing that the obstruction in the nilpotent case does not extend to the solvable case. 150 Jakóbczak, Piotr Let $B_N$ be the unit ball in $\mathbb{C}^N$ and let $f$ be a function holomorphic and $L^2$-integrable in $B_N$. Denote by $E(B_N,f)$ the set of all slices of the form $\Pi =L\cap B_N$, where $L$ is a complex one-dimensional subspace of $\mathbb{C}^N$, for which $f|_{\Pi}$ is not $L^2$-integrable (with respect to the Lebesgue measure on $L$). Call this set the exceptional set for $f$. We give a characterization of exceptional sets which are closed in the natural topology of slices. 160 Langlands, Robert P. James Arthur was awarded the Canada Gold Medal of the National Science and Engineering Research Council in 1999. This introduction to his work is an attempt to explain his methods and his goals to the mathematical community at large. 210 Leung, Man Chun We construct unbounded positive $C^2$-solutions of the equation $\Delta u + K u^{(n + 2)/(n - 2)} = 0$ in $\R^n$ (equipped with Euclidean metric $g_o$) such that $K$ is bounded between two positive numbers in $\R^n$, the conformal metric $g=u^{4/(n-2)}g_o$ is complete, and the volume growth of $g$ can be arbitrarily fast or reasonably slow according to the constructions. By imposing natural conditions on $u$, we obtain growth estimate on the $L^{2n/(n-2)}$-norm of the solution and show that it has slow decay. 223 Marshall, M. A generalization of Schm\"udgen's Positivstellensatz is given which holds for any basic closed semialgebraic set in $\mathbb{R}^n$ (compact or not). The proof is an extension of W\"ormann's proof. 231 Rosenblatt, Joseph M.; Willis, George A. Let $G$ be an infinite discrete amenable group or a non-discrete amenable group. It is shown how to construct a net $(f_\alpha)$ of positive, normalized functions in $L_1(G)$ such that the net converges weak* to invariance but does not converge strongly to invariance. The solution of certain linear equations determined by colorings of the Cayley graphs of the group are central to this construction. 242 Schueller, Laura Mann The zeta function of a nonsingular pair of quadratic forms defined over a finite field, $k$, of arbitrary characteristic is calculated. A.~Weil made this computation when $\rmchar k \neq 2$. When the pair has even order, a relationship between the number of zeros of the pair and the number of places of degree one in an appropriate hyperelliptic function field is 257 Abánades, Miguel A. Let $X$ be a reduced nonsingular quasiprojective scheme over ${\mathbb R}$ such that the set of real rational points $X({\mathbb R})$ is dense in $X$ and compact. Then $X({\mathbb R})$ is a real algebraic variety. Denote by $H_k^{\alg}(X({\mathbb R}), {\mathbb Z}/2)$ the group of homology classes represented by Zariski closed $k$-dimensional subvarieties of $X({\mathbb R})$. In this note we show that $H_1^{\alg} (X({\mathbb R}), {\mathbb Z}/2)$ is a proper subgroup of $H_1(X({\mathbb R}), {\mathbb Z}/2)$ for a nonorientable hyperelliptic surface $X$. We also determine all possible groups $H_1^{\alg}(X({\mathbb R}), {\mathbb Z}/2)$ for a real ruled surface $X$ in connection with the previously known description of all possible topological configurations of $X$. 266 Cencelj, M.; Dranishnikov, A. N. We show that every compactum has cohomological dimension $1$ with respect to a finitely generated nilpotent group $G$ whenever it has cohomological dimension $1$ with respect to the abelianization of $G$. This is applied to the extension theory to obtain a cohomological dimension theory condition for a finite-dimensional compactum $X$ for extendability of every map from a closed subset of $X$ into a nilpotent $\CW$-complex $M$ with finitely generated homotopy groups over all of $X$. 270 Cheung, Wai-Shun; Li, Chi-Kwong Let $c = (c_1, \dots, c_n)$ be such that $c_1 \ge \cdots \ge c_n$. The $c$-numerical range of an $n \times n$ matrix $A$ is defined by $$W_c(A) = \Bigl\{ \sum_{j=1}^n c_j (Ax_j,x_j) : \{x_1, \dots, x_n\} \text{ an orthonormal basis for } \IC^n \Bigr\},$$ and the $c$-numerical radius of $A$ is defined by $r_c (A) = \max \{|z| : z \in W_c (A)\}$. We determine the structure of those linear operators $\phi$ on algebras of block triangular matrices, satisfying $$W_c \bigl( \phi(A) \bigr) = W_c (A) \text{ for all } A \quad \text{or} \quad r_c \bigl( \phi(A) \bigr) = r_c (A) \text{ for all } A.$$ 282 Lee, Min Ho; Myung, Hyo Chul Jacobi-like forms for a discrete subgroup $\G \subset \SL(2,\mbb R)$ are formal power series with coefficients in the space of functions on the Poincar\'e upper half plane satisfying a certain functional equation, and they correspond to sequences of certain modular forms. We introduce Hecke operators acting on the space of Jacobi-like forms and obtain an explicit formula for such an action in terms of modular forms. We also prove that those Hecke operator actions on Jacobi-like forms are compatible with the usual Hecke operator actions on modular forms. 292 McKay, Angela There is a theorem, usually attributed to Napoleon, which states that if one takes any triangle in the Euclidean Plane, constructs equilateral triangles on each of its sides, and connects the midpoints of the three equilateral triangles, one will obtain an equilateral triangle. We consider an analogue of this problem in the hyperbolic plane. 298 Muić, Goran In this paper the author prove that standard modules of classical groups whose Langlands quotients are generic are irreducible. This establishes a conjecture of Casselman and Shahidi for this important class of groups. 313 Reverter, Amadeu; Vila, Núria We give an explicit recipe for the determination of the images associated to the Galois action on $p$-torsion points of elliptic curves. We present a table listing the image for all the elliptic curves defined over $\QQ$ without complex multiplication with conductor less than 200 and for each prime number~$p$. 323 Schuman, Bertrand Soit $H_0 = \frac{x^2+y^2}{2}$ un hamiltonien isochrone du plan $\Rset^2$. On met en \'evidence une classe d'hamiltoniens isochrones qui sont des perturbations polynomiales de $H_0$. On obtient alors une condition n\'ecessaire d'isochronisme, et un crit\ere de choix pour les hamiltoniens isochrones. On voit ce r\'esultat comme \'etant une g\'en\'eralisation du caract\ere isochrone des perturbations hamiltoniennes homog\`enes consid\'er\'ees dans [L], [P], [S]. Let $H_0 = \frac{x^2+y^2}{2}$ be an isochronous Hamiltonian of the plane $\Rset^2$. We obtain a necessary condition for a system to be isochronous. We can think of this result as a generalization of the isochronous behaviour of the homogeneous polynomial perturbation of the Hamiltonian $H_0$ considered in [L], [P], [S]. 335 Stacey, P. J. Irrational rotation $C^*$-algebras have an inductive limit decomposition in terms of matrix algebras over the space of continuous functions on the circle and this decomposition can be chosen to be invariant under the flip automorphism. It is shown that the flip is essentially the only toral automorphism with this property. 337 Vinet, Luc; Zhedanov, Alexei We continue to study the simplest closure conditions for chains of spectral transformations of the Laurent biorthogonal polynomials ($\LBP$). It is shown that the 1-1-periodic $q$-closure condition leads to the $\LBP$ introduced by Pastro. We introduce classes of semi-classical and Laguerre-Hahn $\LBP$ associated to generic closure conditions of the chain of spectral transformations. 346 Wang, Wei In this paper, we establish the existence of positive solution of a nonlinear subelliptic equation involving the critical Sobolev exponent on the Heisenberg group, which generalizes a result of Brezis and Nirenberg in the Euclidean case. 355 Weaver, Nik We examine Hilbert bimodules which possess a (generally unbounded) involution. Topics considered include a linking algebra representation, duality, locality, and the role of these bimodules in noncommutative differential geometry 370 Weston, Anthony Motivated by a question of Per Enflo, we develop a hypercube criterion for locating linear isometric copies of $\lone$ in an arbitrary real normed space $X$. The said criterion involves finding $2^{n}$ points in $X$ that satisfy one metric equality. This contrasts nicely to the standard classical criterion wherein one seeks $n$ points that satisfy $2^{n-1}$ metric equalities. 376 Zhang, Xi 385 Ballantine, Cristina M. Let $F$ be a totally real number field and let $\GL_{n}$ be the general linear group of rank $n$ over $F$. Let $\mathfrak{p}$ be a prime ideal of $F$ and $F_{\mathfrak{p}}$ the completion of $F$ with respect to the valuation induced by $\mathfrak{p}$. We will consider a finite quotient of the affine building of the group $\GL_{n}$ over the field $F_{\mathfrak{p}}$. We will view this object as a hypergraph and find a set of commuting operators whose sum will be the usual adjacency operator of the graph underlying the hypergraph. 398 Cardon, David A.; Ram Murty, M. We find a lower bound on the number of imaginary quadratic extensions of the function field $\F_q(T)$ whose class groups have an element of a fixed order. More precisely, let $q \geq 5$ be a power of an odd prime and let $g$ be a fixed positive integer $\geq 3$. There are $\gg q^{\ell (\frac{1}{2}+\frac{1}{g})}$ polynomials $D \in \F_q[T]$ with $\deg(D) \leq \ell$ such that the class groups of the quadratic extensions $\F_q(T,\sqrt{D})$ have an element of order~$g$. 408 Falbel, E. We classify finite subgroups of $\SO(4)$ generated by anti-unitary involutions. They correspond to involutions fixing pointwise a Lagrangian plane. Explicit descriptions of the finite groups and the configurations of Lagrangian planes are obtained. 420 Gauthier, P. M.; Pouryayevali, M. R. Functions defined on closed sets are simultaneously approximated and interpolated by meromorphic functions with prescribed poles and zeros outside the set of approximation. 429 Henniger, J. P. In answer to a question posed in \cite{G}, we give sufficient conditions on a Lie nilmanifold so that any ergodic rotation of the nilmanifold is metrically conjugate to its inverse. The condition is that the Lie algebra be what we call quasi-graded, and is weaker than the property of being graded. Furthermore, the conjugating map can be chosen to be an involution. It is shown that for a special class of groups, the condition of quasi-graded is also necessary. In certain examples there is a continuum of conjugacies. 440 Hironaka, Eriko In this paper we find a formula for the Alexander polynomial $\Delta_{p_1,\dots,p_k} (x)$ of pretzel knots and links with $(p_1,\dots,p_k, \nega 1)$ twists, where $k$ is odd and $p_1,\dots,p_k$ are positive integers. The polynomial $\Delta_{2,3,7} (x)$ is the well-known Lehmer polynomial, which is conjectured to have the smallest Mahler measure among all monic integer polynomials. We confirm that $\Delta_{2,3,7} (x)$ has the smallest Mahler measure among the polynomials arising as $\Delta_{p_1,\dots,p_k} (x)$. 452 Ishihara, Hironobu Let $\ce$ be an ample vector bundle of rank $r$ on a projective variety $X$ with only log-terminal singularities. We consider the nefness of adjoint divisors $K_X + (t-r) \det \ce$ when $t \ge \dim X$ and $t>r$. As an application, we classify pairs $(X,\ce)$ with $c_r$-sectional genus zero. 459 Kahl, Thomas Nous montrons que la A-cat\'egorie d'un espace simplement connexe de type fini est inf\'erieure ou \'egale \a $n$ si et seulement si son mod\ele d'Adams-Hilton est un r\'etracte homotopique d'une alg\ebre diff\'erentielle \a $n$ \'etages. Nous en d\'eduisons que l'invariant $\Acat$ augmente au plus de 1 lors de l'attachement d'une cellule \`a un espace. We show that the A-category of a simply connected space of finite type is less than or equal to $n$ if and only if its Adams-Hilton model is a homotopy retract of an $n$-stage differential algebra. We deduce from this that the invariant $\Acat$ increases by at most 1 when a cell is attached to a space. 469 Marcoux, Laurent W. In this article it is shown that every bounded linear operator on a complex, infinite dimensional, separable Hilbert space is a sum of at most eighteen unilateral (alternatively, bilateral) weighted shifts. As well, we classify products of weighted shifts, as well as sums and limits of the resulting operators. 482 Mezo, Paul We prove an identity between weighted orbital integrals of the unit elements in the Hecke algebras of $\GL(r)$ and its $n$-fold metaplectic covering, under the assumption that $n$ is relatively prime to any proper divisor of every $1 \leq j \leq r$. 491 Wang, Weiqiang We give canonical resolutions of singularities of several cone varieties arising from invariant theory. We establish a connection between our resolutions and resolutions of singularities of closure of conjugacy classes in classical Lie algebras. 504 Zhang, Yong We show that, if a Banach algebra $\A$ is a left ideal in its second dual algebra and has a left bounded approximate identity, then the weak amenability of $\A$ implies the ($2m+1$)-weak amenability of $\A$ for all $m\geq 1$. 509 2001, for 2001 - pour No abstract.

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http://www.physicsforums.com/showthread.php?p=4092761

Physics Forums Recognitions: Gold Member ## Dirac delta in curved space The Dirac delta function is defined as: $$\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = 1$$ Or more generally the integral is, $$\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {dx'} )dx}$$ But if the metric varies with x, then the integral becomes, $$\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx}$$ But from the wikipedia.com article we have, $$\delta (f(x)) = \frac{{\delta (x - {x_z})}}{{\left| {f'({x_z})} \right|}}$$ where xz is such that $f({x_z}) = 0$. And since, $$\frac{d}{{dx}}(\int_{{x_0}}^x {\sqrt {g(x')} dx'} ) = \sqrt {g(x)}$$ and $$f(x) = \int_{{x_0}}^x {\sqrt {g(x')} dx'} = 0$$ when x= x0 so that xz= x0. Then the above becomes, $$\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} = \int_{ - \infty }^{ + \infty } {\frac{{\delta (x - {x_0})}}{{\left| {\sqrt {g({x_0})} } \right|}}\sqrt {g(x)} dx}$$ Now obviously if g(x) is constant, we recover the original Dirac delta function. But what if g(x) is not constant, can I still recover the original Dirac delta function? Then the Dirac delta would behave the same in any space. Have I evaluated ${\left| {\sqrt {g({x_0})} } \right|}$ wrong? Could there be some sense is saying this should simply by ${\sqrt {g(x)} }$ so that it cancels out? Maybe x never really equals x0, but only approaches it, so that we use some other property of the Dirac delta. 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 OK, I think I got it. Starting with $$\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} = \int_{ - \infty }^{ + \infty } {\frac{{\delta (x - {x_0})}}{{\left| {\sqrt {g({x_0})} } \right|}}\sqrt {g(x)} dx}$$ I suppose it is safe to assume that the metric g(x) is real and always positive so the squareroot is positive, and we can drop the absolute sign. And since we can bring the denominator outside the integral, we have, $$\frac{1}{{\sqrt {g({x_0})} }}\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})\sqrt {g(x)} dx}$$ And then the sifting property lets us bring the function multiplied by the Dirac delta outside the integral with an x value of x0, to get, $$\frac{1}{{\sqrt {g({x_0})} }}\sqrt {g({x_0})} \int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx }$$ But since the squareroots of g cancel, this is just $\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx}$. So, $$\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} = \int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = 1$$ Or the Dirac delta function is invariant with changes of coordinates, metric, curvature or spaces. But I suppose if we think in higher dimensional terms, the integral inside the Dirac delta is a line integral, and the integral outside the delta is a volume integral, right? So would we still get the cancellation of squareroots of g in higher dimensions? Or would the g(x) term in the line integral not be squarerooted like the volume integral outside the delta? Well, the Dirac distribution satisifes: $$\int \delta(x-x_0) f(x) dx = f(x_0)$$ That pretty much answers your question. ## Dirac delta in curved space BTW, if I am not mistaken we should take minus the squareroot of the det of the metric in the change of variables in integration. Recognitions: Gold Member I'm not satisfied because I'm not sure how this generalizes in multidimensional space. What does $$\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx}$$ look like in 3D? The $${\int_{{x_0}}^x {\sqrt {g(x')} dx'} }$$ part would be a line integral. How does that generalize in 3D with a changing metric? The $$\int_{ - \infty }^{ + \infty } {\delta (......)\sqrt {g(x)} dx}$$ part would be a volume integral where the squareroot of the metic is commonly used. So I guess my biggest question is how to use the metric in a line integral. And since you can have more than one line from x0 to x in 3D, how do you determine which line to take, if it even matters? Would you integrate the line integral as it travels over the geodesic? Well it should be similar to the one dimensional case. I assume you know that the n-dimensional Dirac Delta distribution is made up of a product of 1-dimensional Dirac Delta distribution. So you do the same trick you did before and get something like: $$\int_{\mathbb{R}^n} \delta ^n (\bar{x}-\bar{x_0}) \frac{\sqrt{g(\bar{x})}}{\sqrt{g(\bar{x_0})}} d\bar{x}$$ Ok, in your case I think (though not certain), that th line integral from one dimensional becomes if $$\bar{x} =(x^1,x^2,...,x^n)$$ then the line integral becomes: $$\int_{x^1_0}^{x^1}...\int_{x^n_0}^{x^n} \sqrt{g(x')}dx'^1...dx'^n$$ Recognitions: Gold Member The differential line segment squared is: $$d{s^2} = {g_{ij}}(\vec x)d{x^i}d{x^j}$$ So the line integral in a general curved space is: $$\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} }$$ which is usually parameterized as: $$\int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t))\frac{{d{x^i}}}{{dt}}\frac{{d{x^j}}}{{dt}}} } \,\,\,dt$$ And when we put this inside the Dirac delta to get $$\delta (\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } )$$ I don't know how this can be put in the form $$\delta (f(\vec x)) = \frac{{\delta (\vec x - {{\vec x}_z})}}{{\left| {f'({{\vec x}_z})} \right|}}$$ where $f({{\vec x}_z}) = 0$. Can ${g_{ij}}(\vec x)d{x^i}d{x^j}$ be put in the form of a determinant of the metric time differentials so that we can evaluate $g(\vec x)$ at ${{\vec x}_0}$ and bring it outside the delta in the denominator where it cancels as before? Recognitions: Homework Help Quote by friend The differential line segment squared is: $$d{s^2} = {g_{ij}}(\vec x)d{x^i}d{x^j}$$ So the line integral in a general curved space is: $$\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} }$$ which is usually parameterized as: $$\int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t))\frac{{d{x^i}}}{{dt}}\frac{{d{x^j}}}{{dt}}} } \,\,\,dt$$ And when we put this inside the Dirac delta to get $$\delta (\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } )$$ I don't know how this can be put in the form $$\delta (f(\vec x)) = \frac{{\delta (\vec x - {{\vec x}_z})}}{{\left| {f'({{\vec x}_z})} \right|}}$$ where $f({{\vec x}_z}) = 0$. Can ${g_{ij}}(\vec x)d{x^i}d{x^j}$ be put in the form of a determinant of the metric time differentials so that we can evaluate $g(\vec x)$ at ${{\vec x}_0}$ and bring it outside the delta in the denominator where it cancels as before? Recognitions: Gold Member In one dimensional space, $\vec x = (x)$, which means ${x^i} = x$ and $d{x^i} = dx$, so that $$\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} = \sqrt {{g_{ii}}(\vec x)d{x^i}d{x^i}} = \sqrt {g(x)} dx$$ and then $$\delta (\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } ) = \delta (\int_{{x_0}}^x {\sqrt {g(x')} dx} ') = \frac{{\delta (x - {x_0})}}{{\sqrt {g({x_0})} }}$$ And this ${\sqrt {g({x_0})} }$ cancels with the one in the volume integral. And the Dirac delta is invariant in 1D. In 2D, we have $\vec x = ({x^1},{x^2})$, and $${g_{ij}}(\vec x)d{x^i}d{x^j} = {g_{11}}({x^1},{x^2})d{x^1}d{x^1} + {g_{12}}({x^1},{x^2})d{x^1}d{x^2} + {g_{22}}({x^1},{x^2})d{x^2}d{x^2} + {g_{21}}({x^1},{x^2})d{x^2}d{x^1}$$ And in flat space, ${g_{12}} = {g_{21}} = 0$, so that the above becomes, $${g_{ij}}(\vec x)d{x^i}d{x^j} = {g_{11}}d{x^1}d{x^1} + {g_{22}}d{x^2}d{x^2}$$ Then if ${g_{11}} = {g_{22}} = g$, we get $${g_{ij}}(\vec x)d{x^i}d{x^j} = g{(d{x^1})^2} + g{(d{x^2})^2}$$ And, $$\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} = \sqrt g \sqrt {{{(d{x^1})}^2} + {{(d{x^2})}^2}} = \sqrt g ds$$ So that the $\sqrt g$ cancels out to make the integral of the Dirac delta invariant with changes in rescaling of flat space. I wonder what this means. Is the Dirac delta invariant with Lorentz transformation since those are transformation from one flat spacetime to another flat spacetime? If we parameterize the line integral with $\vec x = \vec x(t)$ to get, $$\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } = \int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t'))\frac{{d{x^i}}}{{dt'}}\frac{{d{x^j}}}{{dt'}}} } \,\,\,dt'$$ Can we show that the time variable must be orthogonal to the flat space dimensions? Can we prove the necessity of Minkowski metric for an invariant integral of a parameterized Dirac delta function? Recognitions: Gold Member Quote by friend I wonder what this means. Is the Dirac delta invariant with Lorentz transformation since those are transformation from one flat spacetime to another flat spacetime? If we parameterize the line integral with $\vec x = \vec x(t)$ to get, $$\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } = \int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t'))\frac{{d{x^i}}}{{dt'}}\frac{{d{x^j}}}{{dt'}}} } \,\,\,dt'$$ Can we show that the time variable must be orthogonal to the flat space dimensions? Can we prove the necessity of Minkowski metric for an invariant integral of a parameterized Dirac delta function? Notice, for example, that in $$\int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t))\frac{{d{x^i}}}{{dt}}\frac{{d{x^j}}}{{dt}}} } \,\,\,dt$$ if we make ${g_{ij}}(\vec x) = 0$ whenever i≠j, then we get $$\int_{{t_0}}^t {\sqrt {{g_i}(\vec x(t)){{\left( {\frac{{d{x^i}}}{{dt}}} \right)}^2}} } \,\,\,dt$$ Now, in order to make $\sqrt {g(x)}$ come outside the Dirac delta and cancel as before, we need $${\left( {\frac{{d{x^i}}}{{dt}}} \right)^2} = 1$$ Or, we can put this in a different way by saying $${(ds)^2} = {(dt)^2} - {(dx)^2} = 0$$ This looks like the Minkowski metric. But I don't know what it would mean to have ds≠0. Recognitions: Gold Member Now that I think about it, when I write: $$\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = \int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {dx'} )dx} = \int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} = 1$$ I don't actually know whether the ${g(x)}$ in the volume form of $\int_{ - \infty }^{ + \infty } {\delta (...)\sqrt {g(x)} dx}$ outside the dirac delta is really the same as the ${g(x')}$ in $\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )$ in the line integral inside the dirac delta. The volume form ${g(x)}$ outside the delta is $$\sqrt {g(x)} = \det ({J^T}J)$$ where $$J = \left( {\begin{array}{*{20}{c}} {\frac{{\partial {u^1}}}{{\partial {x^1}}}}&{\frac{{\partial {u^1}}}{{\partial {x^2}}}}\\ {\frac{{\partial {u^2}}}{{\partial {x^1}}}}&{\frac{{\partial {u^2}}}{{\partial {x^2}}}} \end{array}} \right)$$ for a change of coordinates from u to x. But the ${g(x')}$ in the line integral inside the delta is $${g_{ij}}(\vec x)d{x^i}d{x^j} = {g_{11}}d{x^1}d{x^1} + {g_{12}}d{x^1}d{x^2} + {g_{22}}d{x^2}d{x^2} + {g_{21}}d{x^2}d{x^1}$$ And then when ${g_{12}} = {g_{21}} = 0$ and ${g_{11}} = {g_{22}} = g$, the ${g(x')}$ is just a constant ${g}$. Are these two g's actually the same thing... even in curved space? How does that work? Recognitions: Gold Member Previously, I considered: Quote by friend $$\delta (x - {x_0})=\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx} ') = \frac{{\delta (x - {x_0})}}{{\sqrt {g({x_0})} }}$$ But on second thought, this is probably not useful since I could do the same thing with the dirac delta at the right as I did on the left. I could iterate the process to get another squareroot of g in the denominator making it equal the dirac delta divided by the squarroot of g squared. In fact, I could iterate this an infinite number of times to get a squarroot of g to the power of infinity. This only works if g=1 for flat space, and I have not converted the flat dirac delta to a dirac delta in curved space. I will have to look for something else. Recognitions: Homework Help Perhaps this will help? http://www.fen.bilkent.edu.tr/~ercelebi/mp03.pdf It gives the form of the dirac delta in curvilinear coordinates (in 3d). I would guess that to achieve your purposes you would need to generalize to arbitrary dimension and modify the scale factors ##h_i## to account for different metrics. Recognitions: Gold Member Quote by Mute Perhaps this will help? http://www.fen.bilkent.edu.tr/~ercelebi/mp03.pdf Thank you. That's a start. My main goal is to investigate how the properties of the dirac delta function might be used to determine the properties of the background spacetime metric. For example, if the dirac delta describes the density of a mass point, is that enough to determine the properties of the metric? Or, if the dirac delta represents causality, is that enough to determine the Minkowski metric? So I consider equation such as: $$\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx}=\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {dx'} )dx}=\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx}=\int_{ - \infty }^{ + \infty } {\delta (\int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t))\frac{{d{x^i}}}{{dt}}\frac{{d{x^j}}}{{dt}}} } \,\,\,dt){\sqrt {g(x)} }dx}=1$$ Or, if I also apply the first variation to all this and set it equal to zero, is there enough information to specify the metric of Special Relativity or General Relativity? Wouldn't it be interesting if it did? Quote by friend Thank you. That's a start. My main goal is to investigate how the properties of the dirac delta function might be used to determine the properties of the background spacetime metric. For example, if the dirac delta describes the density of a mass point, is that enough to determine the properties of the metric? My gut reaction would be yes. On ℝ at least, metrics define Borel measures and every Borel measure is a Schwartz distribution. IIRC, Schwartz distributions can be broken up into sums of other singular distributions related to Dirac. Honestly, this sounds like the domain of current theory, so ask mathematician who specialise in that? Recognitions: Gold Member Quote by pwsnafu My gut reaction would be yes. On ℝ at least, metrics define Borel measures and every Borel measure is a Schwartz distribution. IIRC, Schwartz distributions can be broken up into sums of other singular distributions related to Dirac. Honestly, this sounds like the domain of current theory, so ask mathematician who specialise in that? God! I hope it's not that complicated. Recognitions: Gold Member Quote by pwsnafu My gut reaction would be yes. On ℝ at least, metrics define Borel measures and every Borel measure is a Schwartz distribution. IIRC, Schwartz distributions can be broken up into sums of other singular distributions related to Dirac. Honestly, this sounds like the domain of current theory, so ask mathematician who specialise in that? Hopefully, it is easier than this. For example, the dirac delta function is a continuous version of the Kronecker delta which is used as the metric in flat space. But the spacetime metric of relativity is also flat spacetime metric. Is it also discrete version of a delta function? Thread Tools | | | | |--------------------------------------------------|------------------------------|---------| | Similar Threads for: Dirac delta in curved space | | | | Thread | Forum | Replies | | | Quantum Physics | 6 | | | Calculus | 3 | | | Special & General Relativity | 7 | | | Differential Geometry | 6 | | | Advanced Physics Homework | 1 |

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http://physics.stackexchange.com/questions/14276/why-cant-quantum-teleportation-be-used-to-transport-information

# Why can't quantum teleportation be used to transport information? Kaku Michio says in an interview that we've teleported photons, cesium atoms and beryllium atoms. Having watched a lot of Kaku as well as way too many astrophysics documentaries in general, I know he's exaggerating. From what I understand, some state of those particles were teleported. Here's my question: what state exactly is being teleported, and why can't it be used to transfer classical information? How can we know experimentally that a teleportation occured, if the transported information can't be read classically? Most of the news articles for the general public just throw around the word "state" without explaining the idea (or flat out tell us like Kaku that atoms were teleported), and Wikipedia uses math and terminology that's way over my head. Is there a layman's explanation by any chance? - 2 +1, damn good question – Arjang Sep 2 '11 at 10:26 2 @Arjang Thanks, I was a bit nervous about posting because this is really not my field at all :) – Rei Miyasaka Sep 2 '11 at 10:53 ## 2 Answers To elaborate on Misha's answer: here is a spiel on teleportation from quantum information theory. I'm going to do some actual computations, but they will be made as simple as possible for novice readers. ### Beginning of the sketch Let us simplify the discussion by limiting ourselves to the teleportation of a single bit's worth of information, e.g. a spin ½ particle. Let us denote spin-down by the bit-value 0, and spin-up by 1. A general (pure, i.e. non-randomized) state of the spin ½ particle is given by $|\sigma\rangle_s = \alpha_0 |0\rangle_s + \alpha_1 |1\rangle_s$: a superposition of the two spin-values, where $|\alpha_0|^2$ and $|\alpha_1|^2$ must sum to 1 and are otherwise unconstrained complex numbers. If you wanted to store a single bit, you could just use $\alpha_0 = 1$ or $\alpha_1 = 1$ to represent "0" or "1" respectively. (The subscript s just means that this indicates the state of a spin "s".) For teleportation between two distant locations A and B (where A is in the same general area as the spin s), you need an entangled state as a resource, such as $$|\beta_{00}\rangle_{A,B} = \frac{1}{\sqrt 2}\Bigl[\; |0\rangle_A|0\rangle_B + |1\rangle_A|1\rangle_B \Bigr]$$ which is a superposition of two "classical" states: one where both A and B are in the state 0, and one where both are in the state 1. This is a notation which represents a vector, but that's not important for this post. • You should think of the symbols $|u\rangle_{\!\!X}$ as being like variables that you can multiply together in whichever order you like; each one describes the state of one part of the system; and combinations of these products can be used to describe superpositions. (This "multiplication" is actually a tensor product; but that's also not important for this post.) • The factor of $1/\!\sqrt{2}\,$ is just for the sake of book-keeping, so that we can compute probabilities. We won't be doing a lot of probability calculations, but there's an interesting comparison coming below which makes it worthwhile to mention normalization. —— If you're happy with the notion of a vector, what this is doing is ensuring that the vector has norm 1; corresponding roughly to being in this entangled two-particle state with certainty. (Shorter vectors represent states which occur only with some probability: we will see an example later.) The terms $|0\rangle_A|0\rangle_B$ and $|1\rangle_A|1\rangle_B$ are also unit vectors, and are orthogonal to one another because they represent perfectly distinguishable classical states. Their sum has length $\sqrt 2$, so we divide by this to get another unit vector. States in quantum info are typically represented by unit vectors.) Another way of reading this is that if we measured this state to get bit values, we would have a probability of $(1/\!\sqrt 2)^2 = \frac12$ change of obtaining 00 for A and B, and the same chance of obtaining 11 for A and B. Now, I'm going to make a sharp right turn, and talk for a little about cryptography, because the protocoal for teleportation bears a striking resemblance to the protocol for transmitting a message secretly with a one-time pad. ### A digression into classical cryptography A one-time pad is nothing more than two copies of the same sequence of random numbers, e.g. strings of random digits. When you want to encrypt a message which is n bits long, to send from Alice (A) to Bob (B), the two sequences of random bits must have length at least n. For example, if you want to encode a single bit of information, it is enough for A and B to share a single random bit. In a manner of speaking, the random bits in their position must be in a joint state of the form $$\mathbf{Pads} = \frac{1}{2} \Bigl[ \Bigl(\langle 0\rangle_A \text{ and } \langle 0 \rangle_B\Bigr) + \Bigl(\langle 1 \rangle_A \text{ and } \langle 1 \rangle_B\Bigr) \Bigr]$$ where $\langle x \rangle_{\!A/B}\,$ just describes the scenario of Alice's/Bob's bit being x, the addition represents a sort of disjunction between the two (exclusive) events, and where the scaling of 1/2 represents the probabilities with which these two events occur. (I've written it this way deliberately; the mathematics for writing this as a classical probability distribution is almost identical to what one would use for writing quantum states such as $|\beta_{00}\rangle$.) If Alice has a single bit s ∈ {0,1} that she wants to send to Bob, she measures — yes, measures; you could also say compute if you like — the parity of s with her bit A, $$s' = s \oplus A = \begin{cases} 0,& \text{if $s = A$} \\ 1,& \text{if $s \ne A$} \end{cases}$$ where $\oplus$ is addition modulo two [$0 \oplus 0 = 1 \oplus 1 = 0$, and $0 \oplus 1 = 1 \oplus 0 = 1$] and then sends the result s' to Bob, who either flips his own bit B according to whether s' is zero or one: $$B' = B \oplus s'\;.$$ Exercise for the reader: show that B' = s, Alice's bit prior to encryption. Of course, before obtaining s', Bob has no way of guessing what the value of B' will be: no amount of staring at his copy of the original shared bit will reveal any operations Alice has done. Back to teleportation! ### How teleportation actually works (mathematically speaking) In order to teleport $|\sigma\rangle_s$, Alice performs a joint measurement on s and A. There are four possible outcomes, so the most information Alice can obtain out of her measurement is two bits. The measurement she performs is in the basis $$\begin{align*} |\beta_{00}\rangle_{s,A} \;&=\; \tfrac{1}{\sqrt 2}\Bigl[\; |0\rangle_s|0\rangle_A + |1\rangle_s|1\rangle_A \Bigr], \\ |\beta_{01}\rangle_{s,A} \;&=\; \tfrac{1}{\sqrt 2}\Bigl[\; |0\rangle_s|1\rangle_A + |1\rangle_s|0\rangle_A \Bigr], \\ |\beta_{10}\rangle_{s,A} \;&=\; \tfrac{1}{\sqrt 2}\Bigl[\; |0\rangle_s|0\rangle_A - |1\rangle_s|1\rangle_A \Bigr], \\ |\beta_{11}\rangle_{s,A} \;&=\; \tfrac{1}{\sqrt 2}\Bigl[\; |0\rangle_s|1\rangle_A - |1\rangle_s|0\rangle_A \Bigr]. \end{align*}$$ There is a reason why we choose these states in particular; the high-level summary for why we choose these states is given in the final paragraph where I complete the analogy to the one-time pad. (This is not the only possible choice of basis, but it's the simplest.) She will obtain one of the states $$|\beta_{zx}\rangle_{s,A} = \tfrac{1}{\sqrt 2}\Bigl[\; |0\rangle_s|x\rangle_A + (-1)^z |1\rangle_s|1 \oplus x\rangle_A \Bigr]$$ as a result, which can be represented succinctly by the two bits x, z ∈ {0,1}. How do we compute the effect of this measurement? By applying the operator $$\langle \beta_{zx} |_{s,A} = \tfrac{1}{\sqrt 2}\Bigl[\; \langle 0|_s\langle x|_A + (-1)^z \langle 1|_s \langle 1 \oplus x|_A \Bigr]$$ to the input state $|\sigma\rangle_s |\beta_{00}\rangle_{A,B}$, distributing the $\langle \text{whatever}|_X$ symbols over addition like linear operators (which they are), expanding the "multiplication" of the state $|\sigma\rangle_s$ with $|\beta_{00}\rangle_{A,B}$, and using the combination rule $\langle u |_{\!\!X}\, | v \rangle_{\!\!X} = \langle u | v\rangle_{\!\!X} = \delta_{u,v}$ for symbols with a common subscript (this is mechanical calculation, skip to the end if you prefer): $$\begin{align*} \langle \beta_{xz} |_{s,A} &\Bigl[ |\sigma\rangle_s | \beta_{00}\rangle_{A,B} \Bigr] \\\\\\=&\; \tfrac{1}{\sqrt 2}\langle \beta_{xz} |_{s,A}\Bigl[\; \alpha_0 |0\rangle_s |0\rangle_s|0\rangle_A + \alpha_0 |0\rangle_s |1\rangle_s|1\rangle_A + \alpha_1 |1\rangle_s |0\rangle_s|0\rangle_A + \alpha_1 |1\rangle_s |1\rangle_s|1\rangle_A \Bigr] \\\\\\=&\; \tfrac{1}{2} \Bigl[\; \langle 0|_s\langle x|_A + (-1)^z \langle 1|_s \langle 1 \oplus x|_A \Bigr] \times \\&\qquad\qquad \Bigl[\; \alpha_0 |0\rangle_s |0\rangle_A|0\rangle_B + \alpha_0 |0\rangle_s |1\rangle_A|1\rangle_B + \alpha_1 |1\rangle_s |0\rangle_A|0\rangle_B + \alpha_1 |1\rangle_s |1\rangle_A|1\rangle_B \Bigr] \\\\\\=&\; \tfrac{1}{2} \Bigl[\; \alpha_0 \langle 0|0\rangle_s \langle x|0\rangle_A|0\rangle_B + \alpha_0 \langle 0|0\rangle_s \langle x|1\rangle_A|1\rangle_B \\&\qquad + \alpha_1 \langle 0|1\rangle_s \langle x|0\rangle_A|0\rangle_B + \alpha_1 \langle 0|1\rangle_s \langle x|1\rangle_A|1\rangle_B \\&\qquad + (-1)^z \alpha_0 \langle 1|0\rangle_s \langle 1 \oplus x |0\rangle_A|0\rangle_B + (-1)^z \alpha_0 \langle 1|0\rangle_s \langle 1 \oplus x |1\rangle_A|1\rangle_B \\&\qquad + (-1)^z \alpha_1 \langle 1|1\rangle_s \langle 1 \oplus x |0\rangle_A|0\rangle_B + (-1)^z \alpha_1 \langle 1|1\rangle_s \langle 1 \oplus x |1\rangle_A|1\rangle_B\Bigr] \\\\\\=&\; \tfrac{1}{2} \Bigl[\; \alpha_0 \langle x|0\rangle_A|0\rangle_B + \alpha_0 \langle x|1\rangle_A|1\rangle_B \\&\qquad + (-1)^z \alpha_1 \langle 1 \oplus x |0\rangle_A|0\rangle_B + (-1)^z \alpha_1 \langle 1 \oplus x |1\rangle_A|1\rangle_B\Bigr] \\\\\\=&\; \tfrac{1}{2} \delta_{x,0} \Bigl[\; \alpha_0 |0\rangle_B + (-1)^z \alpha_1 |1\rangle_B \Bigr] + \tfrac{1}{2} \delta_{x,1} \Bigl[\; \alpha_0 |1\rangle_B + (-1)^z \alpha_1 |0\rangle_B \Bigr] \\\\\\=&\; \tfrac{1}{2} \Bigl[\; \alpha_0 |x\rangle_B + (-1)^z \alpha_1 |1 \oplus x\rangle_B \Bigr]. \end{align*}$$ This calculation is doing nothing more than computing the marginal state on B, conditioned on obtaining the state $|\beta_{zx}\rangle_{\!s,A}$ on s and A. People still argue about the interpretation of this concept, but the mathematics itself is identical to what you would do for computing classical probability distributions on B conditioned on some property of s and A holding (such as the parity of s and A being some bit x ∈ {0,1}). The scalar factor of $\frac12$ just represents the fact that the outcome $|\beta_{zx}\rangle$ of the measurement arises, independently of the values of x and z, with probability $(\frac12)^2 = \frac14$; the final state is some state on B which depends on the values of x and z. Without knowing what x and z are, Bob doesn't know the result of the measurement collapse — any more than he could guess Alice's message in the one-time pad protocol by staring at his own random bit. In fact, we can show that without knowledge of x and z, Bob's state looks perfectly random to him, just as it would have if he had measured B before Alice did anything, and just as the random bit in a one-time pad looks. But if Alice were to send x and z to him, he could perform operations on his state to obtain the state $|\sigma\rangle_B = \alpha_0 |0\rangle_B + \alpha_1 |1\rangle_B$. Which operations he performs depends on the values of x and z, just as whether Bob flips his bit in the one-time pad protocol depends on the message Alice sends him. ### Completing the analogy to the one-time pad The measurement above, in the basis of $|\beta_{zx}\rangle$ states, corresponds exactly to a parity measurement in the classical one-time pad scenario: the two bits x and z represent measured correlations between the spins s and A, in the standard basis and in the basis $|+\rangle \propto |0\rangle + |1\rangle$ and $|-\rangle \propto |0\rangle - |1\rangle$. In particular, you'll notice that whenever we obtain an outcome of $|\beta_{zx}\rangle$ where x = 0, Bob's final state is incorrect only by the sign of the $|1\rangle$ component; and when x = 1, then mapping each component of Bob's state from $|0\rangle$ to $|1\rangle$ (and vice-vera) gives him a state which is again correct up to a sign. The x bit corresponds exactly to the bit-parity measurement in a one-time pad classically; and without that bit, you could never hope to communicate classical information by teleportation. But on the plus side, although you have to send a bit of communication, whichever state s was in to begin with has been transmitted to Bob with perfect security! - Thanks for all that. I'm going to spend a while to try to learn as much of the notation and terminology as I can to see if I can fit even a small piece of this inside my head! – Rei Miyasaka Sep 3 '11 at 8:05 Of course you can "teleport" information. The problem is that to make this you still need classical channel which is limited by the speed of light. If you have no classical channel, teleportation process itself can not help to give any new information at recepient side due to its quantum nature. If you have no information of what happens on the other end of teleportation process from the other sources, results are just random for you. It is complicated, and I see no way to explain it without quantum mechanics. But the whole idea has a bit different formulation than stated in your question. Probably, you should read something about Quantum key distribution. It is more technical area and may be formulated in more technical terms. -

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http://mathhelpforum.com/advanced-algebra/167744-matrix-equation.html

# Thread: 1. ## Matrix equation Let M = { [a -b] , a,b E R } ..............[b a] Solve X^4+X^2+I=0, where X E M I attempted this question a while ago. I remember finding the eigenvalues and eigenvectors (X^4+X^2=-I and treat X^4+X^2 as a matrix itself) but it led to nowhere. I then tried to use a linear map but no luck either. I would love to hear ideas on how to solve this one. Thanks. 2. You would solve for the eigenvalues by solving the numeric equation $x^4+ x^2- 1= 0$. If we let $y= x^2$, then we have $y^2+ y- 1= 0$ which, by the quadratic formula, has roots $\frac{-1\pm i\sqrt{3}}{2}$. The eigenvalues of the original equation are the roots of those and so will be very messy! 3. $M = \begin{bmatrix}a & -b \\ b & a\end{bmatrix} = \gamma\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$ Where $\gamma = \sqrt{a^2 + b^2}$ and $\theta = \tan^{-1}(b/a)$. So M is a rotation by $\theta$ with a scaling by $\gamma$. Immediately using a geometric interpretation, we can see that a rotation by 60 degrees is a solution for X. Also a rotation by 120 degrees is another solution. There are more solutions, but hopefully this helps you get started finding roots. 4. An alternative: we can use the natural field isomorphism: $\varphi: \mathcal{M}\rightarrow{\mathbb{C}},\quad \varphi\begin{bmatrix}{a}&{-b}\\{b}&{\;\;a}\end{bmatrix}=a+bi$ so, solving $X^4+X^2+I=0$ in $\mathcal{M}$ it is equivalent to solve $z^4+z^2+1=0$ in $\mathbb{C}$. But, $z^4+z^2+1=(z^2+1)^2-z^2=(z^2+z+1)(z^2-z+1)=0$ $z=\dfrac{-1\pm\sqrt{3}i}{2},\;z=\dfrac{1\pm\sqrt{3}i}{2}$ As a consequence the roots of the given equation are the elements of the set: $\mathcal{S}=\left\{{\dfrac{1}{2}\begin{bmatrix}{-1}&{-\sqrt{3}}\\{\sqrt{3}}&{-1}\end{bmatrix},\;\dfrac{1}{2}\begin{bmatrix}{-1}&{\sqrt{3}}\\{-\sqrt{3}}&{-1}\end{bmatrix}, \;\dfrac{1}{2}\begin{bmatrix}{1}&{-\sqrt{3}}\\{\sqrt{3}}&{1}\end{bmatrix},\;\dfrac{1} {2}\begin{bmatrix}{1}&{-\sqrt{3}}\\{\sqrt{3}}&{1}\end{bmatrix}}\right\}$ Fernando Revilla 5. Thanks for your help guys. FernandoRevilla, I haven't learn anything about natural field isomorphism so I don't understand what you did there...Is that some kind of a linear map? Is there any link about it so I can have a read? I tried google but nothing really came up. EDIT: you can actually prove that its a linear map. =) Thanks that's a very pretty solution! snowtea, could you please explain how you got 60 as a solution using geometric interpretation? I didn't quite see that =S I get everything up to "So M is a rotation by with a scaling by gamma." X^4 and X^2 mean rotations by 4*theta and 2*theta respectively(with scalings too). But what about X^4+X^2+I? How did you see that 60 is a solution so quickly? 6. Let X be a rotation by 60 degrees (no scaling). Draw a vector v. Multiplying by I keeps it the same. Multiplying it by X^2 rotates it by 120 degrees. Multiplying it by X^4 rotates it by 240 degrees. (X^4+X^2+I) v is the sum of all the above vectors. You can show this is always 0 using some simple trigonometry. 7. Originally Posted by snowtea Let X be a rotation by 60 degrees (no scaling). Draw a vector v. Multiplying by I keeps it the same. Multiplying it by X^2 rotates it by 120 degrees. Multiplying it by X^4 rotates it by 240 degrees. (X^4+X^2+I) v is the sum of all the above vectors. You can show this is always 0 using some simple trigonometry. Thanks that makes sense now.

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http://math.stackexchange.com/questions/186607/does-an-orientable-subbundle-of-an-orientable-vector-bundle-always-have-a-orient

# Does an orientable subbundle of an orientable vector bundle always have a orientable complement? If I have an orientable vector bundle $E$ and a subbundle $F$ on a manifold $M$, where both the bundles are orientable, does $F$ have a complement in $E$ which is also orientable? Does it have a complement bundle at all? That is, a subbundle of $E$ that is pointwise a complement of $F$. What if $F$ is of codimension $1$? Is the complement always trivial in this case? Does anything change if $E$ is specifically the tangent bundle $TM$? - ## 2 Answers If $E=F\oplus G$, then $\Lambda^\det E=\Lambda^\det F\otimes\Lambda^\det G$. If both $\Lambda^\det E$ and $\Lambda^\det G$ are trivial, then so is $\Lambda^\det F$, because the three being line bundles, you can «divide by $\Lambda^\det G$» in that equality. - By $\Lambda^\det$ I mean the maximal exterior power here. – Mariano Suárez-Alvarez♦ Aug 25 '12 at 3:36

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http://mathoverflow.net/questions/78278/is-su-infty-amenable

## Is $SU(\infty)$ amenable? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) We can write the finitary special unitary group $SU(\infty)$ as the direct limit $\varinjlim SU(n)$ of ordinary special unitary groups. These groups $SU(n)$ are compact, thus amenable. In other words each of them has an invariant mean (in this case, from Haar measure). Is $SU(\infty)$ amenable? Of course one then asks the same question for the full unitary group $U(\infty)$, the special orthogonal group $SO(\infty)$ and the unitary symplectic (quaternion unitary) group $Sp(\infty)$. - I think that it contains free groups, isn't it? – Valerio Capraro Oct 16 2011 at 19:44 @Valerio Capraro Certainly it does, but that only makes it non-amenable as a discrete group. – John Wiltshire-Gordon Oct 16 2011 at 21:12 5 Sorry if this is silly but what exactly do you mean by amenability for non locally compact groups such as $SU(\infty)$? I'm not at all an expert on this but one definition I guess could be that if $G$ acts on a compact space then there is an invariant probability measure. With that definition I think any direct limit of amenable (in that sense) groups is amenable by weak compactness of probability measures on a compact space. – Vitali Kapovitch Oct 16 2011 at 21:53 ## 1 Answer The answer is that $G=SU(\infty)$ (with the direct limit topology of the usual Hilbert-Schmidt topologies) is extremely amenable. This means (by definition) that every continuous action of $G$ on a compact set has a fixed point. This was proved as an application of the isoperimetric inequality by Gromov and Milman M. Gromov and V.D. Milman, A topological application of the isoperimetric inequality, Amer. J. Math. 105 (1983), 843–854. Since $SU(\infty)$ is not locally compact, various characterizations of amenability have to be adapted. One way to define amenability for such groups is to say that there exists a $G$-invariant mean on the algebra of bounded uniformly continuous real-valued functions on $G$. Extreme amenability then gives even the existence of a $G$-invariant character on this algebra. This is somewhat unintuitive, since obviously no compact group can admit such a character on the algebra of continuous functions on it. Extreme amenability is a concept which is related to phenomena of measure concentration. Gromov and Milman proved (as an application of lower bounds on the Ricci curvature) that for every sequence of measurable subsets $A_n \subset SU(n)$ with $\liminf_{n \to \infty} \mu_n(A_n) \neq 0$, one has $$\lim_{n \to \infty} \mu_n(A_{n,\epsilon}) = 1.$$ Here, $\mu_n$ denotes the normalized Haar measure and $A_{n,\varepsilon}$ the $\varepsilon$-neighborhood of $A_n$ in the unnormalized Hilbert-Schmidt metric. This concentration phenomenon can be used to prove extreme amenability of $SU(\infty)$. - I found another affirmative answer in Example 3.4 of a paper "Functional Analytic Background for a Theory of Infinite-Dimensional Reductive Lie Groups" by Daniel Beltita. He shows that if $G$ is a topological group and there is a directed system $\{G_\alpha\}_{\alpha \in A}$ of amenable topological subgroups whose union is dense in $G$ then $G$ is amenable. Here "amenable" means that there is a left-invariant mean. – Joseph Wolf Oct 25 2011 at 18:20

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http://mathoverflow.net/questions/4994?sort=oldest

## Fundamental Examples ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is not unusual that a single example or a very few shape an entire mathematical discipline. Can you give examples for such examples? (One example, or few, per post, please) I'd love to learn about further basic or central examples and I think such examples serve as good invitations to various areas. (Which is why a bounty was offered.) Related MO questions: What-are-your-favorite-instructional-counterexamples, Cannonical examples of algebraic structures, Counterexamples-in-algebra, individual-mathematical-objects-whose-study-amounts-to-a-subdiscipline, most-intricate-and-most-beautiful-structures-in-mathematics, counterexamples-in-algebraic-topology, algebraic-geometry-examples, what-could-be-some-potentially-useful-mathematical-databases, what-is-your-favorite-strange-function To make this question and the various examples a more useful source there is a designated answer to point out connections between the various examples we collected. In order to make it a more useful source, I list all the answers in categories, and added (for most) a date and (for 2/5) a link to the answer which often offers more details. (~year means approximate year, *year means a year when an older example becomes central in view of some discovery, year? means that I am not sure if this is the correct year and ? means that I do not know the date. Please edit and correct.) Of course, if you see some important example missing, add it! Logic and foundations: $\aleph_\omega$ (~1890), Russell's paradox (1901), Halting problem (1936), Goedel constructible universe L (1938), McKinsey formula in modal logic (~1941), 3SAT (*1970), The theory of Algebraically closed fields (ACF) (?), Physics: Brachistochrone problem (1696), Ising model (1925), The harmonic oscillator,(?) Dirac's delta function (1927), Heisenberg model of 1-D chain of spin 1/2 atoms, (~1928), Feynman path integral (1948), Real and Complex Analysis: Harmonic series (14th Cen.) {and Riemann zeta function (1859)}, the Gamma function (1720), li(x), The elliptic integral that launched Riemann surfaces (*1854?), Chebyshev polynomials (?1854) punctured open set in C^n (Hartog's theorem *1906 ?) Partial differential equations: Laplace equation (1773), the heat equation, wave equation, Navier-Stokes equation (1822),KdV equations (1877), Functional analysis: Unilateral shift, The spaces $\ell_p$, $L_p$ and $C(k)$, Tsirelson spaces (1974), Cuntz algebra, Algebra: Polynomials (ancient?), Z (ancient?) and Z/6Z (Middle Ages?), symmetric and alternating groups (*1832), Gaussian integers ($Z[\sqrt -1]$) (1832), $Z[\sqrt(-5)]$,$su_3$ ($su_2)$, full matrix ring over a ring, $\operatorname{SL}_2(\mathbb{Z})$ and SU(2), quaternions (1843), p-adic numbers (1897), Young tableaux (1900) and Schur polynomials, cyclotomic fields, Hopf algebras (1941) Fischer-Griess monster (1973), Heisenberg group, ADE-classification (and Dynkin diagrams), Prufer p-groups, Number Theory: conics and pythagorean triples (ancient), Fermat equation (1637), Riemann zeta function (1859) eliptic curves, transendental numbers, Fermat hypersurfaces, Probability: Normal distribution (1733), Brownian motion (1827), The percolation model (1957), The Gaussian Orthogonal Ensemble, the Gaussian Unitary Ensemble, and the Gaussian Symplectic Ensemble, SLE (1999), Dynamics: Logistic map (1845?), Smale's horseshoe map(1960). Mandelbrot set (1978/80) (Julia set), cat map, (Anosov diffeomorphism) Geometry: Platonic solids (ancient), the Euclidean ball (ancient), The configuration of 27 lines on a cubic surface, The configurations of Desrague and Pappus, construction of regular heptadecagon (*1796), Hyperbolic geometry (1830), Reuleaux triangle (19th century), Fano plane (early 20th century ??), cyclic polytopes (1902), Delaunay triangulation (1934) Leech lattice (1965), Penrose tiling (1974), noncommutative torus, cone of positive semidefinite matrices, the associahedron (1961) Topology: Spheres, Figure-eight knot (ancient), trefoil knot (ancient?) (Borromean rings (ancient?)), the torus (ancient?), Mobius strip (1858), Cantor set (1883), Projective spaces (complex, real, quanterionic..), Poincare dodecahedral sphere (1904), Homotopy group of spheres, Alexander polynomial (1923), Hopf fibration (1931), The standard embedding of the torus in R^3 (*1934 in Morse theory), pseudo-arcs (1948), Discrete metric spaces, Sorgenfrey line, Complex projective space, the cotangent bundle (?), The Grassmannian variety,homotopy group of spheres (*1951), Milnor exotic spheres (1965) Graph theory: The seven bridges of Koenigsberg (1735), Petersen Graph (1886), two edge-colorings of K_6 (Ramsey's theorem 1930), K_33 and K_5 (Kuratowski's theorem 1930), Tutte graph (1946), Margulis's expanders (1973) and Ramanujan graphs (1986), Combinatorics: tic-tac-toe (ancient Egypt(?)) (The game of nim (ancient China(?))), Pascal's triangle (China and Europe 17th), Catalan numbers (mid 19th century), (Fibonacci sequence (12th century; probably ancient), Kirkman's schoolgirl problem (1850), surreal numbers (1969), alternating sign matrices (1982) Algorithms and Computer Science: Newton Raphson method (17th century), Turing machine (1937), RSA (1977), universal quantum computer (1985) Social Science: Prisoner's dilemma (1950) (and also the chicken game, chain store game, and centipede game), the model of exchange economy, second price auction (1961) - 15 I think that this should be community wiki. – Andrew Stacey Nov 11 2009 at 7:55 12 @Jose: Hard to say exactly. My instinct is that the kind of answers that this question will garner are those that didn't involve much actual thought, and the votes up or down will be more an assessment of whether the voter liked the example rather than whether the voter liked the answer (which, ideally, should contain an explanation of why that example shaped the discipline); both of these indicate that the answerers should not gain reputation for their answers, hence community wiki. – Andrew Stacey Nov 11 2009 at 9:50 4 I've hit this with the wiki hammer. – Scott Morrison♦ Nov 11 2009 at 19:34 7 I can't imagine a counterexample to the following rule: Any question whose purpose is to produce a sorted list of resources (i.e. the question includes, or should include, "one per post please") should be community wiki. – Anton Geraschenko♦ Nov 12 2009 at 8:03 8 Why does this question have a bounty anyway? – Kevin Lin Nov 21 2009 at 17:33 show 17 more comments ## 129 Answers The harmonic oscillator is a fundamental example in both classical and quantum mechanics. - 1 Of course, for this question and many others there is no meaning to "correct answer". In fact I liked all the answers to the question and I hope more answers will come along. Jose was the most valuable partner to this endeavor and he contributed several great answers both before and after the boundy was announced. – Gil Kalai Nov 26 2009 at 12:39 1 @Gil, now that the bounty has been delivered you might "unaccept" this answer. – Scott Morrison♦ Nov 30 2009 at 16:20 1 Is it possible? Is it moral? – Gil Kalai Dec 1 2009 at 14:53 ### 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. There's always the venerable normal distribution (for probability theory). - The complex projective space is the fundamental example in toric geometry, symplectic and GIT quotients,... - The cotangent bundle is the fundamental example of symplectic manifold/phase space. - It has often been said that if you understand $su_3$ you understand all simple Lie algebras, so that should make it the fundamental example. (Personally I think that it suffices to understand $su_2$!) - 3 On the same line, SU(2) (compared to U(1), which is abelian) already shows a lot of features of the compact non abelian groups (Peter-Weil theorem, irreducible representations of (every) dimension greater than 1, ...). – Gian Maria Dall'Ara Nov 11 2009 at 9:11 1 I guess you could say that the heat equation "shaped" functional analysis since so many of the early tools were developed to study just that equation (more so if you say fourier analysis). Similarly (as has been noted) the spheres are currently shaping algebraic topology since so many of the tools are developed to get at the stable homotopy of spheres. I didn't put my comment on your "harmonic oscillator" example because that has played a role in shaping quantum mechanics. Part of it is purely timing: good examples have a chance to shape a subject if they are encountered in its infancy. – Andrew Stacey Nov 11 2009 at 10:33 show 2 more comments I do think Milnor's exotic sphere distinguish differential topology from general topology, but i don't know if this is an example of the kind you want. Answered by: Yuhao Huang - In the theory of holomorphic functions of several variables, Hartogs's theorem that any holomorphic function on a punctured open set of ( ) can holomorphically be continued through the deleted point really started the subject and is still the most spectacular divide between and [ where it is completely false : on look at 1/z or, worse, exp(1/z): these functions clearly can't be continued holomorphically through zero] - 1 Dear Gil and Jonas:thanks for the comments and, yes, you are both right. Hartogs's example-theorem shows that there exist in C^n domains which are not regions of holomorphy,a phenomenon impossible in dimension one.This launched the notion of pseudoconvex domains (which exclude Hartogs-type extensions of holomorphic functions) and ultimately led to the concept of Stein manifolds, central in complex geometry (the analogues of affine varieties in algebraic geometry) – Georges Elencwajg Nov 11 2009 at 10:28 show 3 more comments The full matrix rings of any order over another ring (and their direct union of row-finite, column-finite matrices) are a fundamental example for Noncommutative Ring Theory: they are simple enough to be easily understood "in a glimpse" but complex enough to highlight many interesting concepts of the theory. - Laplace's equation is the fundamental example of a PDE. If I could broaden the question to allow a triumvirate of examples, I'd say Laplace's equation, the heat equation, and the wave equation are the canonical examples of PDEs, representing elliptic, parabolic, and hyperbolic equations respectively. - The torus is THE example in many branches of math. Algebraic Topology: it is the example where you can compute explicitly its fundamental group as well as its covering spaces, universal cover and everything else. Algebraic geometry: provided it is smooth, is a cubic which is not rational. It is a compact Lie group (and then a non so trivial example of a trivial tangent bundle). It is compact in and has zero curvature (Riemannian geometry). It is a Riemann Surface. Actually one can compute very explicitly the moduli space of such Riemann surfaces as well as (very explicitly) the mapping class group (which is a beautiful example). It is an elliptic curve, an abelian variety... and it has all those properties I don't know about...meaning, an endless amount of properties. - 3 But in what way has the torus shaped any of these subjects? It's certainly a good example demonstrating many of the features, but I don't see (from your answer) that it has played a significant role in shaping them. – Andrew Stacey Nov 11 2009 at 9:51 5 "Algebraic Topology: it is the example where you can compute explicitly its fundamental group as well as its covering spaces, universal cover and everything else." I don't see how this distinguishes the torus from any other closed surface, from the algebro-topological point of view. Indeed, if you're interested in the fundamental group, the fact that it's abelian in this case makes it highly *un*representative. – HW Nov 11 2009 at 17:36 The Fano plane in finite geometry http://en.wikipedia.org/wiki/Fano_plane - The Prisoner's Dilemma in Game Theory. - 2 Prisoner's Dilemma is certainly a fundamental example but there are other games which I think have been as rich in encouraging interesting research: Chicken en.wikipedia.org/wiki/Chicken_game Chain store en.wikipedia.org/wiki/Chainstore_paradox centipede en.wikipedia.org/wiki/Centipede_game One reason these are interesting games is that help one try to understand what is "rational" behavior and help distinguish between what people do in practice as compared with some "abstract model" of rationality. – Joseph Malkevitch Dec 25 2009 at 15:10 show 2 more comments A lot of algebraic topology was developed with computing the higher homotopy groups of spheres in mind. - - Ravi Vakil gives interesting examples in algebraic geometry: "The existence of some of these pathologies is common knowledge'', but I had never known what they were.". - The Noncommutative Torus in non-commutative geometry. (Maybe it has just shaped the subject because it is about the only thing one can handle explicitly) - Answered by Agol: The figure eight knot (complement) is the starting point for much of hyperbolic geometry. Although other hyperbolic manifolds were discovered before it, the figure eight knot complement has one of the simplest hyperbolic structures to analyze. Thurston first proved his hyperbolic Dehn surgery theorem for the figure eight knot complement - after understanding the proof in this case, the general case is not much harder to understand. It is the simplest knot for which every 3-manifold is a branched cover over it. It was one of the first (non-torus) knots for which the knot-complement problem was proven. It has the most number of non-hyperbolic Dehn-fillings over any one-cusped hyperbolic 3-manifold. It is the smallest volume orientable hyperbolic manifold with one cusp. It was the first knot proven that all non-trivial Dehn fillings have a finite-sheeted cover with positive first betti number. It was the first knot for which the volume conjecture has been verified. (see also this Wikipedia article.) - The KdV equation in integrable systems. It was through a numerical study of KdV that the word soliton was coined. This numerical study lead to much analytical work, including the development of Lax Pairs. (Answer by Aaron Hoffman) - The ring ${Z}[\sqrt{-5}]$ is a fundamental example of non-unique factorization in rings in algebraic integers. Perhaps of more historical relevance is the example of p=37 that shows that Lamé's "proof" of Fermat's Last Theorem fails. I'm not sure Kummer used p=37 though. In any case, examples of non-unique factorization in rings in algebraic integers lead to the whole theory of ideal numbers and later Dedekind domains ans set the tone for algebraic number theory. - 1 Unique prime factorization fails for the ring of integers underlying the p = 23 case of FLT. Kummer's method of proof works for p = 23, but not p = 37 (as Kummer was well aware of). – Jonah Sinick Nov 16 2009 at 19:35 show 1 more comment For a picture that launched a thousand papers, I'd nominate the bifurcation diagram of the logistic map. (image via Wikipedia). Answered by Martin M. W. - 1 If pictures are allowed then the Mandelbrot set and its Julia sets are also quite fundamental as prime examples of what lies hidden in dynamical systems. Of course, as the Wikipedia page on the Mandelbrot set says, there is a close relation between the Mandelbrot set and the bifurcation diagram of the logistic map. – lhf Nov 11 2009 at 18:37 3 Roughly speaking, the bifurcation diagram is what you get by going through the Mandelbrot set along the real axis. – Lasse Rempe-Gillen Mar 3 2010 at 10:54 show 1 more comment The Petersen Graph in graph theory - The Fermat Equation xn + yn - zn = 0. This has truly been much more than an example in both algebra and number theory: it was one of the main motivations to develop the theory of unique factorization domains, Dedekind domains, class numbers, regular primes, etc. in the 19th century. In the late 20th century it provided a motivation for Wiles to work on modularity of elliptic curves. In the 21st century, the equation c1 xa + c2 yb - c3 zc = 0 is similarly motivational for things like Q-curves, Galois representations, hypergeometric abelian varieties... Answer by Peter L. Clark: - The integral $\int \frac{dx}{\sqrt{x(x-1)(x-\lambda)}}$ for $\lambda\neq 0,1$ essentially launched both complex analysis and algebraic geometry, via Riemann's discovery of the Riemann surface that is the natural domain of a function, leading to both analytic theory of Riemann surfaces and to the study of algebraic curves, leading to...complex analysis including in several variables and complex algebraic geometry, as we now know it. - 9 This great answer have earned Charles the first gold badge in mathoverflow history. Congratulations, Charles! – Gil Kalai Nov 27 2009 at 12:38 In convex geometry, the Euclidean ball. In fact (as I think Gil knows but many other readers here probably don't) a huge portion of (high-dimensional) convex geometry consists of results that show that arbitrary high-dimensional convex bodies behave like the Euclidean ball in various ways. And if I may be permitted to add another complementary example or two, the simplex and cube are for many purposes the least "Euclidean ball-like" convex bodies, so they are useful for understanding the limitations of the Euclidean ball as a prototype for arbitrary bodies. - The cyclic polytope in the study of convex polytopes in high dimensions. It is the convex hull of n points on the moment curve (t,t^2,t^3,...,t^d). It is simplicial and has the property that every [d/2] points form a face. (So, for example, in 4 dimension every two vertices form an edge.) - Recursion theory has the halting problem. From it, you obtain the first example of an interesting Turing degree, and generalizing it, you get the Turing jump operator. The halting problem also plays an important role in (the boundary of) computational complexity theory, since it's one of the main tools for demonstrating the insolubility of computational problems. - In geometry, group theory and other areas: The Leech Lattice: - In complex dynamics: The Mandelbrot set - In modal logic there is a particularly simple formula, called McKinsey formula: ◻⋄p→⋄◻p. It is so simple, yet it defines a frame property which cannot be expressed in first-order logic. Also, with the right selection of other formulas, it gives rise to frame incompleteness examples (logics that are consistent, but are not logics of any class of frames whatsoever). - Relevant to many areas (but mostly topology) is the Cantor set. It is an example of a set with properties too numerous to list here. To name a few: it is uncountable, compact, nowhere dense and it has Lebesgue measure 0. - 5 Another good reason: Every compact Hausdorff space is a quotient of the Cantor set!! See Tom Leinster's answer here mathoverflow.net/questions/5357/… It's just incredible - I still can't get over it. – Peter Arndt Nov 20 2009 at 15:30 5 @Peter Arndt: Every compact Hausdorff metrizable (equivalently, second-countable) space is a quotient of the Cantor set. There are compact Hausdorff spaces of greater than continuum cardinality, and these evidently are not quotients of the Cantor set. – Pete L. Clark Jan 15 2010 at 10:30

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http://math.stackexchange.com/questions/190146/what-is-the-inverse-function-of-fx-192x-16x2

# What is the Inverse function of $f(x)=192x-16x^2$? What is the inverse function of: $$f(x)=192x-16x^{2}$$ I have been finding myself going in a circle in trying to complete this problem which otherwise looks simple, but for some reason I am at a block. Could someone help explaining how to go about this problem? - This function is not monotone since $f'(12)=0$ hence it is not $1-1$ thus not invertible – Belgi Sep 2 '12 at 19:14 1 @Belgi Even though your four pieces of fact are correct, the "hence"-part is wrong. There are loads of non-monotone functions which are 1-1 and perfectly invertible, even assuming everywhere-differentiability. $x^3$ is the easiest example. – Arthur Sep 2 '12 at 19:26 @Arthur $x^3$ is monotone. What it fails is having a nonvanishing derivative. If you want a non-monotonic injective function, by IVT it will have to be discontinuous. – Logan Maingi Sep 2 '12 at 19:31 @Arthur: $f(x)=x^3$ is monotone. It’s even strictly monotone. – Brian M. Scott Sep 2 '12 at 19:32 I'm sorry, it's the "since"-part that is wrong, then. – Arthur Sep 2 '12 at 19:36 show 2 more comments ## 4 Answers Instead of completing the square, as in lab bhattacharjee’s answer, you can use the quadratic formula. You have $y=192x-16x^{2}$; rewrite it as $16x^2-192x+y=0$, and treat $y$ as the constant term to get $$x=\frac{192\pm\sqrt{192^2-64y}}{32}=6\pm\frac{\sqrt{64(576-y)}}{32}=6\pm\frac{\sqrt{576-y}}4\;.$$ This gives you two functions, both defined for $y\le 576$: $$x=6+\frac{\sqrt{576-y}}4\;,$$ and $$x=6-\frac{\sqrt{576-y}}4\;.$$ The first corresponds to the righthand side of the parabola $y=192x-16x^2$, and the second to its lefthand side. - The answers of @nayrb and @lab bhatacharjee are correct as far as they go, but as a (retired) teacher, I feel that I must address a fault of the math education system, here in the States as well as, perhaps, elsewhere. To describe a function, you need to mention the domain and the target space, and that is particularly important here. Sketch the graph! You see that there’s a maximum at $(6,576)$, and that on either side of the vertical line $x=6$, there are points on the graph at equal height. So the function fails the “horizontal line test” unless you restrict the domain. Let’s restrict to the interval $\langle-\infty,6]$, i.e. the closed half-line to the left of $6$. But what about our inverse function? It’s not defined for $y>576$, and that means that its domain has to be no bigger than $\langle-\infty,576]$. It’s only now, once we’ve restricted both the domain and the target space of our original function, that we really have an inverse function, and its formula is indeed the one given by @lab, with the minus sign of course. - Hi Jonathan! This originated in my calculus class here at UConn. Thanks for the consultation. – Jeremy Teitelbaum Sep 2 '12 at 19:43 A function is an object with three parts: a domain, a range (allowable set out outputs) and a rule tying each element of the domain to some element of the range. Failure to specify these is failure to completely specify a function. If you are asking if a function is 1-1 or onto, this is especially important. – ncmathsadist Sep 2 '12 at 20:18 @ncmathsadist, thanks for the confirmation. Only thing is, in high-schools nowadays, it seems that the word “range” is used for what I learned to call the image, and no word at all is used for what I have just called the target space. So I think that the word “range” can be misunderstood by many. – Lubin Sep 3 '12 at 20:49 People fail to make this important distinction. The range is the set of allowable outputs. The image is the set of realized outputs. When the image and the range coincide, the function is onto. I have taught HS math and computer science. I always have a discussion about the importance of specifying domain and range in functions in math and CS classes. We call them "pure functions" in a CS class. – ncmathsadist Sep 3 '12 at 20:51 Sorry to rejoin this discussion late. @ncmathsadist, you and I are on the same side, but I don’t think we have seen the same phenomena. In the high-school where I volunteer, the “range” is the set of values taken on. And I have to say that more than 50 years ago, George Mackey, a far greater mathematician then either of us, routinely said in class, “Let $f$ be a function with domain $S$, and with range in $\mathbb{R}$.” To me, this says clearly that for him, the range was the set of values taken on. For this reason, I sedulously avoid the use of the word. – Lubin Sep 5 '12 at 0:59 show 1 more comment Let $f(x)=y=192x-16x^2\implies -y=16(x^2-12x)=16(x-6)^2-16\cdot 36$ $\implies 576-y=16(x-6)^2$ $\implies x=f^{-1}(y)=6±\frac{\sqrt{576-y}}{4}$ Clearly $y ≤576$ to make $f(y)$ invertible in real numbers. - you lost me at 16(x-6)^2. where does that come from? I appreciate the help – Carl Strum Sep 2 '12 at 19:20 @CarlStrum $-y=16(x^2-12x)= 16(x^2 -12x +36 -36) = 16((x-6)^2 -36) =16(x-6)^2-16\cdot 36$ – Arthur Sep 2 '12 at 19:22 Do you think you could use words to explain what you just did? How did you just know to choose 36 to add/subtract to that equation. (i mean i know it doesn't change the problem, but how did you know to use the specific number 36?) – Carl Strum Sep 2 '12 at 19:27 $x^2-12x=(x)^2-2(x)(6)+(6)^2-(6)^2=(x-6)^2-36$, is it ok? – lab bhattacharjee Sep 2 '12 at 19:29 I appreciate all the help but I feel like I'm being misunderstood. I am interested in how you knew to use whatever technique it was to get those answers. I.E. I am a little behind on the concept of "completing the square" if that is what's being used to solve the problem. – Carl Strum Sep 2 '12 at 19:33 show 1 more comment The function will only have an inverse if you restrict it to a region where it is one-to-one. You can look at this graph: and see that you need to restrict your function to one of the regions where $6\le x\le \infty$ and $-\infty<y\le 576$ or $-\infty<x\le 6$ and the same $y$-region. The quadratic formula (as indicated above) can give you the proper formula. Wolfram Alpha works it out nicely here: Wolfram Alpha Knows the Quadratic Formula You can "show steps" to see how Wolfram Alpha does this; it's essentially completing the square. The two solutions correspond to the two branches of the inverse function. - Looks like you’re relatively new here. Welcome! - jdL – Lubin Sep 2 '12 at 19:47 I'm trying to see if I can get my Calc students to use this site as a resource. – Jeremy Teitelbaum Sep 2 '12 at 19:51 It’s more than a resource, ’cause they’ll see mathematics at many different levels. Should be an eye-opener! – Lubin Sep 2 '12 at 19:52

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http://mathhelpforum.com/trigonometry/127744-proof-required.html

# Thread: 1. ## proof required... if cosA+cosB+cosC=0=sinA+sinB+sinC prove that cos^2 A + cos^2 B + cos^2 C = sin^2 A + sin^2 B + sin^2 C = 3/2 I have no idea how to equate the two. This was part of the questions in the chapter on complex numbers, so perhaps I should have included it in number theory. But any proof is welcome. 2. Originally Posted by nahduma if cosA+cosB+cosC=0=sinA+sinB+sinC prove that cos^2 A + cos^2 B + cos^2 C = sin^2 A + sin^2 B + sin^2 C = 3/2 I have no idea how to equate the two. This was part of the questions in the chapter on complex numbers, so perhaps I should have included it in number theory. But any proof is welcome. Let $u = e^{iA}\ (=\cos A + i\sin A)$, and similarly $v = e^{iB}$ and $w = e^{iC}$. Then $u+v+w = 0$. Also, u, v and w all have absolute value 1, so their inverses are equal to their complex conjugates: $u^{-1} = \overline{u}$, and similarly for v and w. Then $uv+vw+wu = uvw\bigl(u^{-1} + v^{-1} + w ^{-1}\bigr) = uvw\bigl(\overline{u+v+w}\bigr) = 0$. Thus $0 = (u+v+w)^2 = u^2+v^2+w^2 + 2(uv+vw+wu)$, and it follows that $u^2+v^2+w^2 = 0$. Take the real part of that to see that $\cos(2A) + \cos(2B) + \cos(2C) = 0$. The results then follow from the facts that $\cos^2A = \tfrac12\bigl(1+\cos(2A)\bigr)$ and $\sin^2A = \tfrac12\bigl(1-\cos(2A)\bigr)$.

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http://physics.stackexchange.com/questions/19218/certain-material-heating-water-in-a-recipient/19234

# Certain material heating water in a recipient I don't know how to resolve a problem, but I don't want the answer since I'm almost going to have it resolved. What the problem says is we have 85 liters of water at 7ºC in an iron pot of 29kg. We want the water to be at 86ºC. The temperature of the iron pot is 12ºC. The water is heated by fire wood (65% of the thermal energy is wasted in combustion), and has a heating value of 12 MJ/kg. We need to determine the amount of wood needed to heat the water. I would apply the normal formula of energy transfer: m1*c1*(t2-t1)=m2*c2*(t2-t1). But since we are using fire wood there, I don't know how to resolve this problem, probably because I'm missing something there... and I don't know what should I do. Which formula or principle would I need to use when having the material and the heating value of the material used? - 1 Hi Anne - it sounds like you're talking about a homework problem, right? I added the tag for it. As for your question, it's not bad, but I think you're omitting some information that we need in order to help you out. In fact, you come across as if you're trying to avoid revealing the true nature of the problem. Unless there really is some reason to keep your work super-secret, it's better if you don't do that. Give us everything, including the full text of the problem. The more information we have to work with, the better we can help you. – David Zaslavsky♦ Jan 7 '12 at 6:06 1 (cont.) For example, what values do you have that you can plug into the equation you mentioned? Once you've figured that out, what values do you still need, and why aren't you able to get them? – David Zaslavsky♦ Jan 7 '12 at 6:08 Hi David! Well, it's not actual homework but let's take it like that. I didn't posted the full problem since I didn't wanted an answer. What the problem says is we have 85 liters of water at 7ºC in an iron pot of 29kg. We want the water to be at 86ºC. The temperature of the iron pot is 12ºC. The water is heated by fire wood (65% of the thermal energy is wasted in combustion), and has a heating value of 12 MJ/kg. We need to determine the amount of wood needed to heat the water. And I don't know how I should understand this with the formula I posted... – Anne Jan 7 '12 at 6:12 1 OK, well, the homework tag is for homework-like questions so it probably applies even if this isn't actually homework. Also, we're not going to give you a full answer even if you do post the whole problem - it's right there in our homework FAQ that giving complete answers to homework-like questions is against the site policy. So don't worry about that! I'd suggest just editing the text of the problem (basically what you just put in your comment) into your question. – David Zaslavsky♦ Jan 7 '12 at 6:14 1 ... and it's done! Thanks David. I felt guilty last time because of that and didn't want to make the same mistake again. I'm trying to learn, that's why I don't want the final answer, but how to continue resolving my problem, since I'm unable to plug all the data in the formula - so probably I'm missing something as I said. – Anne Jan 7 '12 at 6:18 show 5 more comments ## 3 Answers If you know the specific heat of water and iron you can work out how much energy is needed to bring the water and the pot to 86°C. You need to work out how many kg of wood you need to burn to produce that amount of energy. The question even tells you how much of the energy produced by burning the wood is wasted and doesn't go towards heating the pot and water. Whoever set the question is probably looking for the simple answer, but there are lots of extra bits you could have fun with. For example you're told the volume of the water not it's mass, but specific heat is normally quoted per unit mass. That means you need to know the density of water at 7°C to work out it's mass, and the density isn't exactly 1g/cm$^3$. Also the specific heat of both water and iron are functions of temperature. - Which formula or principle would I need to use when having the material and the heating value of the material used? I would suggest starting with the conservation of energy. Have you tried drawing a block diagram showing where the energy (heat) is going from/to within the limits of the setup? Lastly keep an eye on the first law of thermodynamics... - I would apply the normal formula of energy transfer: m1*c1*(t2-t1)=m2*c2*(t2-t1). That's not the "normal formula of energy transfer". That's just the formula used when you mix liquids. A more generalized formula for energy transfer would be "heat lost=heat gained", or the first law of thermodynamics $$\Delta Q=\Delta U+\Delta W$$ Where $\Delta Q$ is heat supplied, $\Delta U$ is change in internal energy, and $\Delta W$ is work done by system. In the case of your problem, as there is no work being done, just balance the heats. See how much heat goes into the system, and then compare this with the total heat required (sum of all $mc(T_2-T_1)$) to raise the temperature of the system. -

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http://terrytao.wordpress.com/tag/convex-geometry/

What’s new Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao # Tag Archive You are currently browsing the tag archive for the ‘convex geometry’ tag. ## (Gil Kalai) The weak epsilon-net problem 22 April, 2007 in guest blog, math.MG, question | Tags: convex geometry, entropy, epsilon-net, Gil Kalai | by Gil Kalai | 13 comments [This post is authored by Gil Kalai, who has kindly “guest blogged” this week’s “open problem of the week”. - T.] This is a problem in discrete and convex geometry. It seeks to quantify the intuitively obvious fact that large convex bodies are so “fat” that they cannot avoid “detection” by a small number of observation points. More precisely, we fix a dimension d and make the following definition (introduced by Haussler and Welzl): • Definition: Let $X \subset {\Bbb R}^d$ be a finite set of points, and let $0 < \epsilon < 1$. We say that a finite set $Y \subset {\Bbb R}^d$ is a weak $\epsilon$-net for X (with respect to convex bodies) if, whenever B is a convex body which is large in the sense that $|B \cap X| > \epsilon |X|$, then B contains at least one point of Y. (If Y is contained in X, we say that Y is a strong $\epsilon$-net for X with respect to convex bodies.) For example, in one dimension, if $X = \{1,\ldots,N\}$, and $Y = \{ \epsilon N, 2 \epsilon N, \ldots, k \epsilon N \}$ where k is the integer part of $1/\epsilon$, then Y is a weak $\epsilon$-net for X with respect to convex bodies. Thus we see that even when the original set X is very large, one can create a $\epsilon$-net of size as small as $O(1/\epsilon)$. Strong $\epsilon$-nets are of importance in computational learning theory, and are fairly well understood via Vapnik-Chervonenkis (or VC) theory; however, the theory of weak $\epsilon$-nets is still not completely satisfactory. One can ask what happens in higher dimensions, for instance when X is a discrete cube $X = \{1,\ldots,N\}^d$. It is not too hard to cook up $\epsilon$-nets of size $O_d(1/\epsilon^d)$ (by using tools such as Minkowski’s theorem), but in fact one can create $\epsilon$-nets of size as small as $O( \frac{1}{\epsilon} \log \frac{1}{\epsilon} )$ simply by taking a random subset of X of this cardinality and observing that “up to errors of $\epsilon$“, the total number of essentially different ways a convex body can meet X grows at most polynomially in $1/\epsilon$. (This is a very typical application of the probabilistic method.) On the other hand, since X can contain roughly $1/\epsilon$ disjoint convex bodies, each of which contains at least $\epsilon$ of the points in X, we see that no $\epsilon$-net can have size much smaller than $1/\epsilon$. Now consider the situation in which X is now an arbitrary finite set, rather than a discrete cube. More precisely, let $f(\epsilon,d)$ be the least number such that every finite set X possesses at least one weak $\epsilon$-net for X with respect to convex bodies of cardinality at most $f(\epsilon,d)$. (One can also replace the finite set X with an arbitrary probability measure; the two formulations are equivalent.) Informally, f is the least number of “guards” one needs to place to prevent a convex body from covering more than $\epsilon$ of any given territory. • Problem 1: For fixed d, what is the correct rate of growth of f as $\epsilon \to 0$? Read the rest of this entry »

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http://unapologetic.wordpress.com/2007/06/18/cones-and-cocones/?like=1&source=post_flair&_wpnonce=97257dee4e

# The Unapologetic Mathematician ## Cones and cocones There are a few auxiliary concepts we’ll need before the next major topic. Let’s start with two categories $\mathcal{J}$ and $\mathcal{C}$, and the category of functors $\mathcal{C}^\mathcal{J}$. If the following seems very complicated, consider $\mathcal{J}$ to be any particular toy category you’d like, so this category of functors is a diagram category. Now, for every object $C\in\mathcal{C}$ there is a constant functor that sends every object of $\mathcal{J}$ to $C$, and every morphism to the identity on $C$. Actually, this assignment of the constant functor to an object of $C$ is a functor from $\mathcal{C}$ to $\mathcal{C}^\mathcal{J}$. Indeed, given a morphism $f:C\rightarrow D$ we get a natural transformation from the one constant functor to the other, whose component at each object of $\mathcal{J}$ is $f$. We call this the “diagonal” functor $\Delta:\mathcal{C}\rightarrow\mathcal{C}^\mathcal{J}$. That is, $\Delta(C)$ is the constant functor with value $C$. Let $F:\mathcal{J}\rightarrow\mathcal{C}$ be some particular functor in $\mathcal{C}^\mathcal{J}$. A cone is an object in the comma category $(\Delta\downarrow F)$. Let’s unpack this definition. Since $\Delta$ has $\mathcal{C}$ as its source, and $F$ is a fixed object — the same thing as a functor from the category $\mathbf{1}$ to $\mathcal{C}^\mathcal{J}$, we know what objects of this category look like. An object of the category $(\Delta\downarrow F)$ consists of an object $C$ of $\mathcal{C}$ and an arrow from $\Delta(C)$ to $F$. But an arrow in $\mathcal{C}^\mathcal{J}$ is a natural transformation of functors. That is, for each object $J$ of $\mathcal{J}$ we need an arrow in $\mathcal{C}$ from $\left[\Delta(C)\right](J)$ to $F(J)$. But $\left[\Delta(C)\right](J)=C$ for all objects $J$. So we just need an object $C\in\mathcal{C}$ and an arrow $C\rightarrow F(J)$ for each $J\in\mathcal{J}$. Of course, there’s also naturality conditions to be concerned with. If $j:J_1\rightarrow J_2$ is an arrow in $\mathcal{J}$, and $c_1:C\rightarrow F(J_1)$ and $c_2:C\rightarrow F(J_2)$ are the required arrows from $C$, then naturality requires that $c_2=F(j)\circ c_1$. So we need a collection of arrows from $C$ to the objects in the image of $F$ that are compatible with the arrows from $\mathcal{J}$. Such a collection defines an object in the comma category $(\Delta\downarrow F)$ — a cone on $F$. Cocones are defined similarly. A cocone on $F$ is an object in the comma category $(F\downarrow\Delta)$. That is, it’s an object $C\in\mathcal{C}$ and a collection of arrows from the objects in the image of $F$ that are compatible with the arrows from $\mathcal{J}$. The description may seem a little odd, but try writing it out for some very simple categories $\mathcal{J}$. For example, let $\mathcal{J}$ be a set. Then try letting it be an ordinal, or another preorder. After you write down the definition of a cone and a cocone on some simple categories the general idea should seem to make more sense. ### Like this: Posted by John Armstrong | Category theory ## 4 Comments » 1. [...] a functor the limit of (if it exists) is a couniversal cone on — a terminal object in the comma category . That is, it consists of an object and arrows [...] Pingback by | June 19, 2007 | Reply 2. [...] are Adjoints When considering limits, we started by talking about the diagonal functor . This assigns to an object the “constant” functor that sends each object of to and [...] Pingback by | July 20, 2007 | Reply 3. 2 ‘formula does not parse’ latex errors here Comment by Avery | January 5, 2008 | Reply 4. Yeah, those have been creeping in. WordPress changed something in their parser and now brackets give trouble unless they’re given as \left[ \right] pairs. I don’t know which pages have the problems, but when I stumble across them I fix them. Thanks for pointing this example out to me. Comment by | January 6, 2008 | Reply « Previous | Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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http://en.wikipedia.org/wiki/Multiset

# Multiset In mathematics, the notion of multiset (or bag) is a generalization of the notion of set in which members are allowed to appear more than once. For example, there is a unique set that contains the elements a and b and no others, but there are many multisets with this property, such as the multiset that contains two copies of a and one of b or the multiset that contains three copies of both a and b. The term "multiset" was coined by Nicolaas Govert de Bruijn in the 1970s.[1] The use of multisets in mathematics and beyond predates the name "multiset" by many centuries: Knuth (1998) attributes the first study of multisets to the Indian mathematician Bhascara Acharya (circa 1150), who described permutations of multisets. ## Overview The number of times an element belongs to the multiset is the multiplicity of that member. The total number of elements in a multiset, including repeated memberships, is the cardinality of the multiset. For example, in the multiset {a, a, b, b, b, c} the multiplicities of the members a, b, and c are respectively 2, 3, and 1, and the cardinality of the multiset is 6. To distinguish between sets and multisets, a notation that incorporates brackets is sometimes used: the multiset {2,2,3} can be represented as [2,2,3].[2] In multisets, as in sets and in contrast to tuples, the order of elements is irrelevant: The multisets {a, a, b} and {a, b, a} are equal. ## History Blizard (1989) traced multisets back to the very origin of numbers, arguing that “in ancient times, the number n was often represented by a collection of n strokes, tally marks, or units.” These and similar collections of objects are multisets as strokes, tally marks, or units are considered indistinguishable. This shows that people implicitly used multisets even before mathematics emerged. This shows that necessity in this structure has been always so urgent that multisets have been several times rediscovered and appeared in literature under different names (Blizard, 1991). For instance, they are called bags by Peterson (1981), who attributed this term to Cerf, Gostelow, Estrin, and Volanski (1971). A multiset has been also called an aggregate, heap, bunch, sample, weighted set, occurrence set, and fireset (finitely repeated element set) (cf. (Blizard, 1991; Singh, et al, 2007)). Although multisets were implicitly utilized from ancient times, their explicit exploration happened much later. The first known study of multisets is attributed to the Indian mathematician Bhascara Acharya (circa 1150), who described permutations of multisets (Knuth, 1998). Angelelli (1965) discovered another early reference to the concept of multiset in the work of Marius Nizolius (1498-1576). Much later Kircher (1650) found the number of multiset permutations when one element can be repeated. Prestet (1675) published a general rule for multiset permutations. Wallis (1685) explains this rule in more detail. In the explicit form, multisets appeared in the work of Richard Dedekind "Was sind und was sollen die Zahlen?" (1888). However, mathematicians formalized multisets and began to study them as a precise mathematical object (structure) only in the 20th century. ## Formal definition Within set theory, a multiset may be formally defined as a 2-tuple $\left(A, m\right)$ where $A$ is some set and $m \colon A \to \mathbb{N}_{\geq 1}$ is a function from $A$ to the set $\mathbb{N}_{\geq 1} = \left\{1, 2, 3, \dots\right\}$ of positive natural numbers. The set $A$ is called the underlying set of elements. For each $a$ in $A$ the multiplicity (that is, number of occurrences) of $a$ is the number $m\!\left(a\right)$. If a universe $U$ in which the elements of $A$ must live is specified, the definition can be simplified to just a multiplicity function $m_U\colon U \to \mathbb{N}$ from $U$ to the set $\mathbb{N} = \left\{0, 1, 2, 3, \dots\right\}$ of natural numbers, obtained by extending $m$ to $U$ with values 0 for each element not in $A$. This extended multiplicity function is the multiplicity function called $1_A$ below. Like any function, the function $m$ may be defined as its graph: the set of ordered pairs $\left\{\left(a, m\left(a\right)\right) : a \in A\right\}$. With these definitions the multiset written as $\left\{a, a, b\right\}$ is defined as $\left(\left\{a, b\right\}, \left\{\left(a, 2\right), \left(b, 1\right)\right\}\right)$, and the multiset $\left\{a, b\right\}$ is defined as $\left(\left\{a, b\right\}, \left\{\left(a, 1\right), \left(b, 1\right)\right\}\right)$. The concept of a multiset is a generalization of the concept of a set. A multiset corresponds to an ordinary set if the multiplicity of every element is one (as opposed to some larger natural number). An indexed family, $\left(a_i\right)$, where $i$ is in some index-set, may define a multiset, sometimes written $\left\{a_i\right\}$, in which the multiplicity of any element $x$ is the number of indices $i$ such that $a_i = x$. The condition for this to be possible is that no element occurs infinitely many times in the family: even in an infinite multiset, the multiplicities must be finite numbers. It is possible to extend the definition of a multiset by allowing multiplicities of individual elements to be infinite cardinals instead of natural numbers. Not all properties carry over to this generalization however, and this article uses the definition above, with finite multiplicities. ## Multiplicity function The set indicator function of a normal set is generalized to the multiplicity function for multisets. The set indicator function of a subset A of a set X is the function $\mathbf{1}_A : X \to \lbrace 0,1 \rbrace \,$ defined by $\mathbf{1}_A(x) = \begin{cases} 1 &\text{if }x \in A, \\ 0 &\text{if }x \notin A. \end{cases}$ The set indicator function of the intersection of sets is the minimum function of the indicator functions $\mathbf{1}_{A\cap B}(x) = \min\{\mathbf{1}_A(x),\mathbf{1}_B(x)\}.$ The set indicator function of the union of sets is the maximum function of the indicator functions $\mathbf{1}_{A\cup B}(x) = \max\{{\mathbf{1}_A(x),\mathbf{1}_B(x)}\}.$ The set indicator function of a subset is smaller than or equal to that of the superset $A\subseteq B \Leftrightarrow \forall x \mathbf{1}_{A}(x) \le \mathbf{1}_{B}(x).$ The set indicator function of a cartesian product is the product of the indicator functions of the cartesian factors $\mathbf{1}_{A \times B}(x,y) = \mathbf{1}_A(x) \cdot\mathbf{1}_B(y).$ The cardinality of a (finite) set is the sum of the indicator function values $|A|=\sum_{x\in X} \mathbf{1}_{A}(x).$ Now generalize the concept of set indicator function by releasing the constraint that the values are 0 and 1 only and allow the values 2, 3, 4 and so on. The resulting function is called a multiplicity function and such a function defines a multiset. The concepts of intersection, union, subset, cartesian product and cardinality of multisets are defined by the above formulas. The multiplicity function of a multiset sum, is the sum of the multiplicity functions $\mathbf{1}_{A \uplus B}(x) = \mathbf{1}_A(x) + \mathbf{1}_B(x).$ The multiplicity function of a multiset difference is the zero-truncated subtraction of the multiplicity functions $\mathbf{1}_{A \setminus B}(x) = \max(0, \mathbf{1}_A(x) - \mathbf{1}_B(x)).$ The scalar multiplication of a multiset by a natural number n may be defined as: $\mathbf{1}_{n \otimes A}(x) = n \times \mathbf{1}_A(x). \,$ A small finite multiset, A, is represented by a list where each element, x, occurs as many times as the multiplicity, 1A(x), indicates. $\{1,1,1,3\} \cap \{1,1,2\} = \{1,1\} \,$ $\{1,1\} \cup \{1,2\} = \{1,1,2\}\,$ $\{1,1\} \subseteq \{1,1,1,2\}\,$ $\left|\{1,1\}\right|=2 \,$ $\{1,1\} \times \{1,2\} = \{(1,1), (1,1), (1,2), (1,2)\} \,$ $\{1,1\} \uplus \{1,2\} = \{1,1,1,2\} \,$ ## Examples One of the simplest and most natural examples is the multiset of prime factors of a number n. Here the underlying set of elements is the set of prime divisors of n. For example the number 120 has the prime factorization $120 = 2^3 3^1 5^1\,$ which gives the multiset {2, 2, 2, 3, 5}. A related example is the multiset of solutions of an algebraic equation. A quadratic equation, for example, has two solutions. However, in some cases they are both the same number. Thus the multiset of solutions of the equation could be { 3, 5 }, or it could be { 4, 4 }. In the latter case it has a solution of multiplicity 2. A special case of the above are the eigenvalues of a matrix, if these are defined as the multiset of roots of its characteristic polynomial. However a choice is made here: the (usual) definition of eigenvalues does not refer to the characteristic polynomial, and other possibilities give rise to different notions of multiplicity, and therefore to different multisets. The geometric multiplicity of λ as eigenvalue of a matrix A is the dimension of the kernel of A - λI, which is often smaller than its multiplicity as root of the characteristic polynomial (the algebraic multiplicity) when the latter is larger than 1. The set of eigenvalues of A is also the set of roots of its minimal polynomial, but the multiset of those roots need not be the same either as the one defined using algebraic multiplicity, or using the geometric multiplicity. ## Free commutative monoids The free commutative monoid on a set X (see free object) can be taken to be the set of finite multisets with elements drawn from X, with the monoid operation being multiset sum and the empty multiset as identity element. Such monoids are also known as (finite) formal sums of elements of X with natural coefficents. The free commutative semigroup is the subset of the free commutative monoid which contains all multisets with elements drawn from X except the empty multiset. Free abelian groups are formal sums (i.e. linear combinations) of elements of X with integer coefficients. Equivalently, they may be seen as signed finite multisets with elements drawn from X. ## Counting multisets Bijection between 3-subsets of a 7-set (left) and 3-multisets with elements from a 5-set (right) So this illustrates that $\textstyle {7 \choose 3} = \left(\!\!{5 \choose 3}\!\!\right)$. See also: Stars and bars (combinatorics) The number of multisets of cardinality k, with elements taken from a finite set of cardinality n, is called the multiset coefficient or multiset number. This number is written by some authors as $\textstyle\left(\!\!{n\choose k}\!\!\right)$, a notation that is meant to resemble that of binomial coefficients; it is used for instance in (Stanley, 1997), and could be pronounced "n multichoose k" to resemble "n choose k" for $\tbinom nk$. Unlike for binomial coefficients, there is no "multiset theorem" in which multiset coefficients would occur, and they should not be confused with the unrelated multinomial coefficients that occur in the multinomial theorem. The value of multiset coefficients can be given explicitly as $\left(\!\!{n\choose k}\!\!\right) = {n+k-1 \choose k} = \frac{(n+k-1)!}{k!\,(n-1)!} = {n(n+1)(n+2)\cdots(n+k-1)\over k!},$ where the second expression is as a binomial coefficient; many authors in fact avoid separate notation and just write binomial coefficients. So, the number of such multisets is the same as the number of subsets of cardinality k in a set of cardinality n + k − 1. The analogy with binomial coefficients can be stressed by writing the numerator in the above expression as a rising factorial power $\left(\!\!{n\choose k}\!\!\right) = {n^{\overline{k}}\over k!},$ to match the expression of binomial coefficients using a falling factorial power: ${n\choose k} = {n^{\underline{k}}\over k!}.$ There are for example 4 multisets of cardinality 3 with elements taken from the set {1,2} of cardinality 2 (n=2, k=3), namely : {1,1,1}, {1,1,2}, {1,2,2}, {2,2,2}. And there are also 4 subsets of cardinality 3 in the set {1,2,3,4} of cardinality 4 (n+k-1 = 4), namely : {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}. One simple way to prove the equality of multiset coefficients and binomial coefficients given above, involves representing multisets in the following way. First, consider the notation for multisets that would represent { a, a, a, a, a, a, b, b, c, c, c, d, d, d, d, d, d, d } (6 as, 2 bs, 3 cs, 7 ds) in this form: $\bullet \bullet \bullet \bullet \bullet \bullet \mid \bullet \bullet \mid \bullet \bullet \bullet \mid \bullet \bullet \bullet \bullet \bullet \bullet \bullet$ This is a multiset of cardinality k=18 made of elements of a set of cardinality n=4. The number of characters including both dots and vertical lines used in this notation is 18 + 4 − 1. The number of vertical lines is 4 − 1. The number of multisets of cardinality 18 is then the number of ways to arrange the 4 − 1 vertical lines among the 18 + 4 − 1 characters, and is thus the number of subsets of cardinality 4 − 1 in a set of cardinality 18 + 4 − 1. Equivalently, it is the number of ways to arrange the 18 dots among the 18 + 4 − 1 characters, which is the number of subsets of cardinality 18 of a set of cardinality 18 + 4 − 1. This is ${4+18-1 \choose 4-1}={4+18-1 \choose 18} = 1330,$ thus is the value of the multiset coefficient and its equivalencies: $\left(\!\!{4\choose18}\!\!\right)={21\choose18}=\frac{21!}{18!\,3!}={21\choose3}=\left(\!\!{19\choose3}\!\!\right),$ $=\frac{{\color{Periwinkle}\mathfrak{ 4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15\cdot16\cdot17\cdot18}} \mathbf{\cdot19\cdot20\cdot21}}{\mathbf{1\cdot2\cdot3}{\color{Periwinkle}\mathfrak{\cdot 4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15\cdot16\cdot17\cdot18}}},$ $=\frac{ 1\cdot2\cdot3\cdot4\cdot5\cdots16\cdot17\cdot18\;\mathbf{\cdot\;19\cdot20\cdot21}} {\,1\cdot2\cdot3\cdot4\cdot5\cdots 16\cdot17\cdot18\,\times\,\mathbf{1\cdot2\cdot3\quad}},$ $=\frac{19\cdot20\cdot21}{1\cdot2\cdot3}.$ One may define a generalized binomial coefficient ${n \choose k}={n(n-1)(n-2)\cdots(n-k+1) \over k!}$ in which n is not required to be a nonnegative integer, but may be negative or a non-integer, or a non-real complex number. (If k = 0, then the value of this coefficient is 1 because it is the empty product.) Then the number of multisets of cardinality k in a set of cardinality n is $\left(\!\!{n\choose k}\!\!\right)=(-1)^k{-n \choose k}.$ This fact led Gian-Carlo Rota to ask "Why are negative sets multisets?".[citation needed] He considered that question worthy of the attention of philosophers of mathematics. ### Recurrence relation A recurrence relation for multiset coefficients may be given as $\left(\!\!{n\choose k}\!\!\right) = \left(\!\!{n\choose k - 1}\!\!\right) + \left(\!\!{n-1\choose k}\!\!\right) \quad \mbox{for } n,k>0$ with $\left(\!\!{n \choose 0}\!\!\right) = 1,\quad n\in\N, \quad\mbox{and}\quad \left(\!\!{0 \choose k}\!\!\right) = 0,\quad k>0.$ The above recurrence may be interpreted as follows. Let [n] := {1, ..., n} be the source set. There is always exactly one (empty) multiset of size 0, and if n = 0 there are no larger multisets, which gives the initial conditions. Now, consider the case in which n,k > 0. A multiset of cardinality k with elements from [n] might or might not contain any instance of the final element n. If it does appear, then by removing n once, one is left with a multiset of cardinality k − 1 of elements from [n], and every such multiset can arise, which gives a total of $\left(\!\!{n\choose k - 1}\!\!\right)$ possibilities. If n does not appear, then our original multiset is equal to a multiset of cardinality k with elements from [n − 1], of which there are $\left(\!\!{n-1\choose k}\!\!\right).$ Thus, $\left(\!\!{n\choose k}\!\!\right) = \left(\!\!{n\choose k - 1}\!\!\right) + \left(\!\!{n-1\choose k}\!\!\right).$ ## Polynomial notation The set {x} may be represented by the monomial x. The set of subsets, { {}, {x} }, is represented by the binomial 1 + x. The set {x,y} may be represented by the monomial x·y. The set of subsets, { {}, {x}, {y}, {x,y} }, is represented by the polynomial (1 + x)·(1 + y) = 1 + x + y + x·y. The multiset {x,x} may be represented by the monomial x·x = x2. The multiset of submultisets, { {}, {x}, {x}, {x,x} }, is represented by the polynomial (1 + x)2 = 1 + x + x + x·x = 1 + 2·x + x2. The multiset of submultisets of the multiset xn is $(1+x)^n=\sum_{k=0}^n{n \choose k}\cdot x^k.$ That is why the binomial coefficient counts the number of k-combinations of an n-set. The multiset xK·yN−K, containing N elements, K of which are hits, is called a statistical population, and a submultiset is called a statistical sample. The set of samples is (1 + x)K·(1 + y)N−K which by the binomial theorem equals $\sum_{n=0}^N\sum_{k=0}^K{K \choose k}\cdot{N-K \choose n-k}\cdot x^k\cdot y^{n-k}.$ So the number of n-samples with k hits is ${K \choose k}\cdot{N-K \choose n-k}.$ See hypergeometric distribution and inferential statistics for further on the distribution of hits. The infinite set of finite multisets of elements taken from the set {x}: { {}, {x}, {x,x}, {x,x,x}, ... } may be represented by the formal power series S = 1 + x + x2 + x3 + ... = 1 + xS . The formal solution, S = (1 − x)−1, makes sense as a set of multisets, but the intermediate result, 1−x, does not make sense as a set of multisets. The infinite set of finite multisets of elements taken from the set x·y is (1 − x)−1·(1 − y)−1 = 1 + (x + y) + (x2 + x·y + y2) + ... The special case y=x : The infinite multiset of finite multisets of elements taken from the multiset x2 is (1 − x)−2 = 1 + 2·x + 3·x2 + ... The general case: The infinite multiset of finite multisets of elements taken from the multiset xn is $(1-x)^{-n}=\sum_{k=0}^\infty{-n \choose k} \cdot(-x)^k$ . This explains why "multisets are negative sets". The negative binomial coefficients count the number of k-multisets of elements taken from an n-set. ## Cumulant generating function A non-negative integer, n, can be represented by the monomial xn . A finite multiset of non-negative integers, say {2, 2, 2, 3, 5}, can likewise be represented by a polynomial f(x), say f(x) = 3·x2 + x3 + x5 . It is convenient to consider the cumulant generating function g(t) = log(f(et)), say g(t) = log(3·e2·t + e3·t + e5·t) . • The cardinality of the multiset is eg(0) = f(1), say 3 + 1 + 1 = 5. • The derivative g is g '(t) = f(et)−1·f '(et)·et, say g '(t) = (3·e2·t + e3·t + e5·t)−1·(6·e2·t + 3·e3·t + 5·e5·t) . • The mean value of the multiset is μ = g '(0) = f(1)−1·f '(1), say μ = (3+1+1)−1·(6+3+5) = 2.8 . • The variance of the multiset is σ2 = g ' '(0) . The numbers ( μ, σ2, ··· ) = ( g '(0), g ' '(0), ··· ) are called cumulants. The infinite set of non-negative integers {0, 1, 2, ···} is represented by the formal power series 1 + x + x2 + ··· = (1 − x)−1. The mean value and standard deviation are undefined. Nevertheless it has a cumulant-generating function, g(t) = −log(1−et). The derivative of this cumulant-generating function is g '(t) = (e−t−1)−1. A finite multiset of real numbers, A = { Ai }, is represented by the cumulant generating function $g_A(t) = \log \left(\sum_i e^{t \cdot A_i}\right) .$ This representation is unique: different multisets have different cumulant generating functions. See partition function (statistical mechanics) for the case where the numbers in question are the energy levels of a physical system. The cumulant-generating function of a multiset of n real numbers having mean μ and standard deviation σ is: g(t) = log(n) + μ·t + 2−1·(σ·t)2 + ··· , and the derivative is simply: g '(t) = μ + σ2·t + ··· The cumulant-generating function of set, {k}, consisting of a single real number, k, is g(t) = k·t , and the derivative is the number itself: g '(t) = k . So the concept of the derivative of the cumulant generating function of a multiset of real numbers is a generalization of the concept of a real number. The cumulant-generating function of a constant multiset, {k, k, k, k, ··· , k} of n elements all equal to the same real number k, is g(t) = log(n)+k·t , and the derivative is the number itself: g '(t) = k , irrespective of n. The cumulant-generating function of the multiset of sums of elements of two multisets of numbers is the sum of the two cumulant-generating functions: $\begin{align} g_{A+B}(t) & = \log \left(\sum_i \sum_j e^{t\cdot(A_i+B_j)}\right) = \log \left(\sum_i\sum_j e^{t\cdot A_i}\cdot e^{t\cdot B_j}\right) \\ & = \log \left(\sum_i e^{t\cdot A_i}\cdot \sum_j e^{t\cdot B_j}\right) = \log \left(\sum_i e^{t\cdot A_i}\right)+ \log\left(\sum_j e^{t\cdot B_j}\right) \\ & = g_A(t) + g_B(t). \end{align}$ There is unfortunately no general formula for computing the cumulant generating function of a product $g_{A\cdot B}(t) = \log \left(\sum_i\sum_j e^{t\cdot A_i\cdot B_j}\right) \,$ but the special case of a constant times a multiset of numbers is: $g_{k\cdot A}(t) = \log \left(\sum_i e^{t\cdot (k\cdot A_i)}\right) = \log \left(\sum_i e^{(t\cdot k)\cdot A_i}\right) = g_A\left(k\cdot t\right).$ The multiset 2·A = {2·Ai} is not the same multiset as 2×A =A+A = {Ai+Aj}. For example, 2·{+1,−1} = {+2,−2} while 2×{+1,−1} = {+1,−1} + {+1,−1} = {+1+1,+1−1,−1+1,−1−1} = {+2,0,0,−2}. $g_{k\times A}(t) = k\cdot g_{A}(t).$ The standard normal distribution is like a limit of big multisets of numbers. $\lim_{k \rarr \infty}k^{-1}\cdot (k^2\times \{+1,-1\}).$ This limit does not make sense as a multiset of numbers, but the derivative of the cumulant generating functions of the multisets in question makes sense, and the limit is well defined. $\begin{align} \lim_{k \rarr \infty} g'_{k^{-1}\cdot (k^2\times \{+1,-1\})}(t) & = \lim_{k \rarr \infty} \frac{d(k^2\cdot \log(e^{+t\cdot k^{-1}}+e^{-t\cdot k^{-1}}))}{dt} \\ & = \lim_{k \rarr \infty} \frac{d(k^2\cdot \log(2)+2^{-1}\cdot t^2+\cdots)}{dt}=t. \end{align}$ The constant term k2·log(2) vanishes by differentiation. The terms ··· vanish in the limit. So for the standard normal distribution, having mean 0 and standard deviation 1, the derivative of the cumulant generating function is simply g '(t) = t . For the normal distribution having mean μ and standard deviation σ, the derivative of the cumulant generating function is g '(t) = μ + σ2·t . See also random variable. ## Applications Multisets have various applications (Singh, et al, 2007). They are becoming the main structure of combinatorics because in its search for higher rigorousness, modern combinatorics has been developed not for sets but for multisets (Aigner, 1979; Anderson, 1987; Stanley, 1999). Now multisets has become an important tool in databases (Libkin and Wong, 1994; 1995; Grumbach and Milo, 1996). For instance, multisets are often used to implement relations in database systems. Multisets also play an imperative role in computer science (Knuth, 1997; 1998). There are also other applications. For instance, Rado (1975) uses multisets as a device to investigate the properties of families of sets. He writes, "The notion of a set takes no account of multiple occurrence of any one of its members, and yet it is just this kind of information which is frequently of importance. We need only think of the set of roots of a polynomial f(x) or the spectrum of a linear operator." ## Generalizations Different generalizations of multisets have been introduced, studied and applied to solving problems. Yager (1986) introduced fuzzy multisets under the name fuzzy bags. Grzymala-Busse (1987) introduced the concept of a rough multiset, using multirelations. Blizard (1989) generalized multisets to real valued multisets, i.e., multisets in which multiplicity of an element can be any non-negative real number. Blizard (1990) also introduced multisets with negative multiplicity. Loeb (1992) introduced hybrid sets as a generalization of multisets in which multiplicity of an element is any integer number. Miyamoto (2001) further generalized multisets introducing multisets in which multiplicity of an element is a any real-valued step function. Alkhazaleh, Salleh and Hassan (2011) introduced soft multisets, while Alkhazaleh and Salleh (2012) introduced fuzzy soft multisets. Burgin (1990; 1992; 2004; 2011) unified multisets and all their generalizations by the concept of a named set, which also comprises all other generalizations of sets. ## Notes 1. Knuth, Donald E. (1998). (3rd edition ed.). Addison Wesley. p. 694. ISBN 0-201-89684-2. Knuth also lists other names that were proposed for multisets, such as list, bunch, bag, heap, sample, weighted set, collection, and suite. 2. Hein, James L. (2003). Discrete mathematics. Jones & Bartlett Publishers. pp. 29–30. ISBN 0-7637-2210-3. ## References 1. Aigner, M. Combinatorial Theory, Springer Verlag, New York/Berlin, 1979 2. Alkhazaleh, S. and Salleh, A.R. Fuzzy Soft Multiset Theory, Abstract and Applied Analysis, 2012, article ID 350600, 20 p. 3. Alkhazaleh, S., Salleh, A.R. and Hassan, N. Soft Multisets Theory, Applied Mathematical Sciences, v. 5, No. 72, 2011, pp. 3561–3573 4. Anderson, I. Combinatorics of Finite Sets, Clarendon Press, Oxford, 1987 5. Angelelli, I. (1965) Leibniz's misunderstanding of Nizolius' notion of 'multitudo', Notre Dame Journal of Formal Logic, v. 6, pp. 319–322 6. Blizard, Wayne D. (1989) "Multiset theory," Notre Dame Journal of Formal Logic, Volume 30, Number 1, Winter: pp. 36–66. doi:10.1305/ndjfl/1093634995 http://projecteuclid.org/euclid.ndjfl/1093634995 MR990203 0668.03027 7. Blizard, W.D. (1989) Real-valued Multisets and Fuzzy Sets, Fuzzy Sets and Systems, v. 33, pp. 77–97 8. Blizard, W. (1990) Negative Membership, Notre Dame Journal of Formal Logic, v. 31, No. 1, pp. 346–368 9. Blizard, W.D. (1991) The Development of Multiset Theory, Modern Logic, v. 1, No.4, pp. 319–352 10. Bogart, Kenneth P. (2000). Introductory combinatorics, 3rd. ed. San Diego CA: Harcourt. 11. Burgin M. Theory of Named Sets as a Foundational Basis for Mathematics, In: Structures in Mathematical Theories, San Sebastian, 1990, pp. 417–420 (http://www.blogg.org/blog-30140-date-2005-10-26.html) 12. Burgin, M. (1992) On the concept of a multiset in cybernetics, Cybernetics and System Analysis, No. 3, pp. 165–167 13. Burgin, M. Unified Foundations of Mathematics, Preprint Mathematics LO/0403186, 2004, 39 p. (electronic edition: http://arXiv.org) 14. Burgin, M. (2011), Theory of Named Sets, Mathematics Research Developments, Nova Science Pub Inc, ISBN 978-1-61122-788-8, books.google.com 15. Dedekind R. Was sind und was sollen die Zahlen?, Vieweg, Braunschweig, 1888 16. Gessel, Ira M., and Stanley, Richard P. (1995) "Algebraic enumeration" in Graham, R. L., Grötschel, M., & Lovász, L., eds., Handbook of combinatorics, Vol. 2. Elsevier: 1021–1061. ISBN 0-444-82351-4, 0-444-88002-X, 0-262-07171-1, 0-262-07169-X. 17. Grzymala-Busse, J. Learning from examples based on rough multisets, in Proceedings of the 2nd International Symposium on Methodologies for Intelligent Systems, Charlotte, NC, USA, 1987, pp. 325–332 18. Grumbach, S. and Milo, T. (1996) Towards tractable algebras for bags, Journal of Computer and System Sciences, v. 52 , No. 3, pp. 570–588 19. Hickman, J. L. (1980) "A note on the concept of multiset," Bulletin of the Australian Mathematical Society 22: 211–17. 20. Kircher, A. Musurgia Universalis, Corbelletti, Rome, 1650 21. Knuth, Donald E. (1998). The Art of Computer Programming – Vol. 2: Seminumerical Algorithms Addison Wesley. p. 694. ISBN 0-201-89684-2. 22. Knuth, D. The Art of Computer Programming, v.3: Sorting and Searching, Addison-Wesley, Reading, Mass., 1998 23. Libkin, L. and Wong, L. (1994) Some properties of query languages for bags, in Proceedings of the Workshop on Database Programming Languages, Springer Verlag, pp. 97–114 24. Libkin, L. and Wong, L. (1995) On representation and querying incomplete information in databases with bags, Information Processing Letters, v. 56, No. 4, pp. 209–214 25. Loeb, D. (1992) Sets with a negative numbers of elements, Advances in Mathematics, v. 91, pp. 64–74 26. Miyamoto, S. Fuzzy Multisets and their Generalizations, in ‘Multiset Processing’, LNCS 2235, pp. 225–235, 2001 27. Prestet, J. Elemens des Mathematiques, André Pralard, Paris, 1675 28. Singh, D. (1994) A Note on the Development of Multiset Theory, Modern Logic, v. 4, pp. 405–406 29. Singh, D., Ibrahim, A. M., Yohanna, T. and Singh, J. N. (2007) An overview of the applications of multisets, Novi Sad Journ. Math., v. 37, No. 2, pp. 73–92 30. Stanley, Richard P. (1997, 1999) Enumerative Combinatorics, Vols. 1 and 2. Cambridge University Press. ISBN 0-521-55309-1, 0-521-56069-1. 31. Syropoulos, Apostolos (2001) "Mathematics of Multisets" in C. S. Calude et al., eds., Multiset processing: Mathematical, computer science, and molecular computing points of view, LNCS 2235. Springer-Verlag: 347–358 32. Yager, R. R. (1986) On the Theory of Bags, International Journal of General Systems, v. 13, pp. 23–37 33. Wallis, A treatise of algebra, John Playford, London, 1685

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http://www.physicsforums.com/showthread.php?p=4209731

Physics Forums Is Energy proptional to frequency(standing wave) Here is an extract from my Physics textbook. According to classical physics, each of these standing waves carries energy, and as their frequency increases, so does their energy . I don't understand how this is true . I think that the classical wave theory says that the energy is proportional to the intensity while on the other hand it is the Quantum theory which predicts : E = $h$ f So, is there an error in the book regarding this issue ? 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 According to classical physics, each of these standing waves carries energy, and as their frequency increases, so does their energy . See http://www.egglescliffe.org.uk/physi...ody/bbody.html on how the standing wave analysis of standing waves in a cavity led to an incorrect result of the radiation emmitted by a blackbody. Yes, in classical physics way back then it was known that the energy of an electromagnetic wave increases with frequency. http://galileo.phys.virginia.edu/cla...radiation.html gives some more history on the subject. Quote by 256bits See http://www.egglescliffe.org.uk/physi...ody/bbody.html on how the standing wave analysis of standing waves in a cavity led to an incorrect result of the radiation emmitted by a blackbody. Yes, in classical physics way back then it was known that the energy of an electromagnetic wave increases with frequency. http://galileo.phys.virginia.edu/cla...radiation.html gives some more history on the subject. I don't think you get my point. I want to know how is it possible that classical physics which was formulated 400 years ago predicted what is now known to be true . We now know that the energy of a photon of E.M wave is equal to E = h f In other words it is dependent upon frequency. From classical physics, i think we all know that for a wave, its energy is propotional to its intensity. What exactly did classical physics say ? (frequency or intensity) Recognitions: Homework Help Science Advisor Is Energy proptional to frequency(standing wave) It is not the energy of radiation which increases with frequency. It is the photon energy - the step size of possible energies. This quantization is not classical - you do not get it in classical physics. I read this link http://www.egglescliffe.org.uk/physi...ody/bbody.html I don't really understand how the introduction of quantized (special and discrete) frequency of oscillation of electrons solved the Ultra Violet Catastrophe According to classical physics, the energy emitted by the black body would increase to infinity as the wavelength decreases. (how is this problem solved by quantized energy levels) Also, I am facing another problem : If the wavelength decreases ie frequency increases(to infinity); according to quantum theory, wouldn't the energy of each photon tend to infinity as well ? and hence give the exact same result as the classical theory ? Mentor Quote by hms.tech What exactly did classical physics say ? (frequency or intensity) It's not either-or. A classical wave has energy proportional to its frequency and also proportional to the square of its amplitude (sometimes called intensity). Recognitions: Homework Help I agree with Vanadium. In particular, for the example of standing waves on a string, the frequency is quantised, so it is highly analogous to quantum physics. But what they didn't know back then was that light is only emitted in quanta. That is a more recent development (along with the rest of quantum mechanics). Recognitions: Homework Help Science Advisor @hms.tech: In classical physics, radiation is not quantized. A black body could emit x-rays (or any other frequency) without the requirement of a minimal amount of energy. As every frequency should receive the same intensity and the frequency range is not limited, the total radiation should be infinite. Quantum mechanics solves this issue with the photon energy: The minimal energy required to emit radiation of a specific frequency. For x-rays, that energy threshold is so high that materials at room temperature do not reach it (or so extremely rare that it does not matter), and the used frequency range is finite. If the wavelength decreases ie frequency increases(to infinity); according to quantum theory, wouldn't the energy of each photon tend to infinity as well ? It does, but the number of photons drops quicker. Quote by hms.tech I read this link http://www.egglescliffe.org.uk/physi...ody/bbody.html I don't really understand how the introduction of quantized (special and discrete) frequency of oscillation of electrons solved the Ultra Violet Catastrophe According to classical physics, the energy emitted by the black body would increase to infinity as the wavelength decreases. (how is this problem solved by quantized energy levels) Also, I am facing another problem : If the wavelength decreases ie frequency increases(to infinity); according to quantum theory, wouldn't the energy of each photon tend to infinity as well ? and hence give the exact same result as the classical theory ? the problem you are facing was first solved exactly by Max plank. Recognitions: Homework Help Quote by mfb Quantum mechanics solves this issue with the photon energy: The minimal energy required to emit radiation of a specific frequency. For x-rays, that energy threshold is so high that materials at room temperature do not reach it (or so extremely rare that it does not matter), and the used frequency range is finite. Yeah, the way I think of it, there is a bunch of molecules flying around, also with rotational degrees of freedom. And for a molecule to go from one state to the other, it can absorb or emit a photon. So to get a high frequency photon, you would need a molecule to go from a very high energy state to a much lower energy state. So this requires a molecule in a high energy state, but this doesn't happen very often, therefore high energy photons do not happen very often. Of course, this is a very qualitative, hand-wavey description, but I think it is roughly a good idea of what is going on. According to Wikipedia : (on Ultraviolet Catastrophe) Max Planck solved the problem by postulating that electromagnetic energy did not follow the classical description, but could only be emitted in discrete packets of energy proportional to the frequency, as given by Planck's law. This has the effect of reducing the number of possible excited modes with a given frequency in the cavity described above, and thus the average energy at those frequencies Can someone explain the bolded part of the extract from this article The probability of a very high frequency mode with very high energy (according to quantum theory) getting a big fraction of the available energy, at the expense of the huge number of less greedy modes, is very low. Hence the non-occurrence of the u-v catastrophe. [How hand-waving can you get?] The word you want is 'emboldened', by the way! Thread Tools Similar Threads for: Is Energy proptional to frequency(standing wave) Thread Forum Replies Introductory Physics Homework 15 Advanced Physics Homework 1 Classical Physics 7 Introductory Physics Homework 4 Introductory Physics Homework 4

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http://mathoverflow.net/questions/54720?sort=newest

## Verifying coefficients of modular forms ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi, I was wondering about good techniques that one can use to show that given certain coefficients, they are the Fourier coefficients of a cusp form, assuming we know the desired weight and level. I am aware of Weil's "converse theorem", but am not aware of any examples of it being used to prove something is a modular form. So, even a pointer to an example of that would be useful. I'm curious if there are other more direct methods as well. Thanks! - ## 5 Answers Maybe I can add something: 1. Wiles famous proof of Fermat's Last Theorem (and the later proof of modularity of all elliptic curves over $\mathbf{Q}$ by Breuil, Conrad, Diamond, Taylor) is a sort of ultimate example in which "given certain coefficients [associated to an elliptic curve by point count], they are [proved to be] the Fourier coefficients of a cusp form". This theorem also produces a general strategy for showing that a list of numbers are the coefficients of a cusp form in some cases -- relate those numbers to the point counts of an elliptic curve. Similar remarks apply for abelian varieties these days, due to Ribet's theorem that Serre's conjecture implies modularity of all abelian varieties over $\mathbf{Q}$ of ${\rm GL}_2$-type. 2. There is something called the Faltings-Serre method, which is an alternative approach to proving modularity statements in particular cases. A google search for "Faltings Serre" will yield many hits. Here's a particular example: http://www.math.wisc.edu/~boston/ell.pdf . There is also a paper by Edray Goins that explains how to use the Faltings-Serre method to get new results in particular cases that go beyond (1), and it has a good list of references: http://homepage.mac.com/ehgoins/papers/serre-faltings.pdf 3. Another example is http://wstein.org/papers/artin/, which is a paper Kevin Buzzard and I wrote, in which we show that lists of numbers (traces of Frobenius) associated to certain A_5 Galois representations are the coefficients of weight 1 modular forms. One can get the same result by "pure thought" these days as an application of Khare-Wintenberger's proof of Serre's conjecture. But still, there is some value in the computational techniques introduced in that paper. There are similar ideas in Joe Buhler's Ph.D. thesis and also in Springer Lecture Notes in Math Volume 1585. 4. There is a Theorem of Sturm (see, e.g., Theorem 9.18 of my modular forms book http://wstein.org/books/modform/stein-modform.pdf and the references) that tends to be very relevant to your question in practice. Sturm's result basically tells you that if the first "few" coefficients of two $q$-expansions of modular forms agree modulo a prime, then all terms agree. This list could go on and on. Also, there are similar ideas and techniques for Hilbert modular forms. - How sharp is Sturm's result? – Dror Speiser Feb 8 2011 at 20:12 1 @Dror: Sturm's result in general says one has to check about $2d$ coefficients, where $d$ is the dimension. For two arbitrary cusp forms, that's pretty good (massively better than "effective Chebotarev", say), but for eigenforms it is far from optimal. Under some hypothesis, Sturm reduces the number of coefficients to roughly $2d/2^v$, for some $v$ (this $v$ is $\leq$ the number of primes dividing the level). There is a result in the Buzzard-Stein paper I mentioned that improves on Sturm's result in the case of eigenforms with nontrivial character. – William Stein Feb 9 2011 at 7:56 ### 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. ( tried posting this earlier today, apologies if it appears twice) For an example of a direct application of Weil's converse theorem, see Shimura's original formulation of the Shimura lift (Annals 1973). He constructs a map from modular forms of half-integral weight to modular forms of integral weight by defining an L-series built out of the Fourier coefficients of the input form, and spends most of the paper proving the analytic properties needed for the converse theorem. - If you assume that the modular form is a newform with prescribed weight and level (EDIT : you also need to know the eigenvalue of the Atkin-Lehner involution), then its Fourier coefficients can sometimes be guessed using the fact that the associated L-series satisfies a functional equation. The point is that using the functional equation, the L-series can be computed by a rapidly convergent series, so it is only necessary to have few Fourier coefficients at hand to compute an approximation of it. This even gives a very pratical and powerful method for finding modular forms, at least when the level is small. You should look at the following example written by Tim Dokchitser : http://magma.maths.usyd.edu.au/magma/handbook/text/1392#15270 Dokchitser's algorithm for computing L-functions has been implemented in Sage and Magma. Note that the method outlined here is (to my knowledge) purely experimental, in the sense that it doesn't prove that the resulting q-expansion is modular. - 1 This "approximate functional equation" technique is surprisingly useful in practice. For instance: sometimes you have the L-function of a variety at all primes except, say, 2, because the special fiber at 2 is horrible. But the method described here allows you to guess the Euler factor at 2 with moral certainty if you know (a long list of) the other Euler factors. – JSE Feb 8 2011 at 14:36 @JSE : in the situation you describe, isn't the method even perfectly rigourous (because Fourier coefficients are integers, it suffices to find approximate them within $1/2$). Ah, maybe the problem is that in general the functional equation of the L-function is only conjectural ? – François Brunault Feb 8 2011 at 15:20 Right, that was the case in the applications I seem to remember (though of course so many more Galois representations are provably modular these days...) – JSE Feb 9 2011 at 5:57 I think Ramsey's answer hits the important points, but I wanted to comment on the converse theorem. The converse theorem takes as an input a Dirichlet series (more generally, an $L$-function) whose analytic continuation (plus quite a bit more) is already known and tells you that it actually came from an automorphic form. Unfortunately, the only methods that I know for proving analytic continuation require knowing that it already comes from an automorphic form. So it certainly sounds circular, and I've only seen it applied to proving global functoriality (for quasi-split classical groups, small symmetric powers, etc). Let me quickly(!) explain the argument: Basically (eliding the hard details to the point where I may say things that are literally, but hopefully not morally, incorrect) say you start out with an automorphic representation $\pi$ on some classical group $G$, living inside $GL(N)$. The Langlands-Shahidi method proves that the twisted $L$-functions $L(s,\pi\times \tau)$ are nice (entire, functional equation, etc), with $\tau$ an automorphic representation of $GL(m)$ with $m < N$. Local functoriality gives lifts from $\pi_\nu$ on $G_\nu$ to $\Pi_\nu$ on $GL(N)$ such that the local $L$-factors coincide, in that $L_\nu(s,\pi_\nu\times\tau_\nu)=L_\nu(s,\Pi_\nu\times\tau_\nu)$. Since $L(s,\Pi\times\tau)=\prod_\nu L_\nu(s,\Pi_\nu\times\tau_\nu)=L(s,\pi\times\tau)$, it is nice for all $\tau$, so the converse theorem applies and $\Pi$ is an automorphic representation of $GL(N)$. Modularity / potential automorphy methods (e.g. Serre's Conjecture, Modularity Theorem) exist, too. In the weight 2 case, given enough information about the coefficients and a sufficient amount of pluck, it is not inconceivable that you could show that the coefficients actually come from an elliptic curve (or several elliptic curves if your modular form is not an eigenform), then use modularity to deduce that they come from a modular form. - It depends on what you mean by "certain coefficients" and how they're given. This sort of sounds like you have a (finite) handful of coefficients in a $q$-expansion and want to try to "fill them out" to a modular form of prescribed weight and level. If this is the case, it is pretty easy. This is because the space of such forms is finite-dimensional, and a basis of the space is readily computed (using, for example, the SAGE software). This reduces the problem to basic linear algebra. If, on the other hand, you have an entire $q$-expansion, it's not so clear. You can truncate and employ the above procedure. I'll just mention three things: 1) If your expansion isn't a modular form, then this will eventually (by taking larger chunks) tell you so. 2) If it is a modular form (but you have no a priori knowledge of this), then it just gives you a lot of evidence that your expansion is a modular form, but doesn't prove it. Of course, if you know more about your coefficients you might be able to go back and prove that it is the form evidenced above. 3) If you do happen to know that your expansion is a modular form of given weight and level and just want to know "what it is," then this method will eventually provably determine what it is (in terms of the basis) by taking large enough chunks. As for the converse theorem... you'd certainly need all the coefficients, but I have a hard time imagining that method to be very practical (but I'm not very analytically minded). - Regarding your point 2, you can run into undecidability problems if you don't restrict how the coefficients are presented to you (but I haven't heard of this being a problem very often in practice). – S. Carnahan♦ Feb 8 2011 at 5:17 @Ramsey: Point 1) seems vague and only practical if there is something that limits the order of the expansion that needs to be considered. Otherwise one eventually will run into computer resource limitations - which are in fact quite severe in SAGE even at modest weights and levels. – Laie Feb 8 2011 at 17:33 2 @Laie: It is vague, and relies on the fact that if a collection of $q$-expansions are linearly independent, then so are their truncations to the first $N$ terms for large $N$. Perhaps the $N$ required can be huge, I'm not sure. I'd be curious (for several reasons related to these answers) to see what the OP actually has. – Ramsey Feb 8 2011 at 19:28

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http://mathhelpforum.com/math-topics/128508-calculating-speed.html

# Thread: 1. ## Calculating speed. I currently have a machine that can calculate the time it takes for an object to travel one meter. (two lasers first one starts a timer second one stops it, The lasers are positioned exactly 1 meter apart.) So if an object was going 10 M/s (Meters per second) it would take 0.1 of a second (10 nanoseconds) to travel one meter. If it was going 1 meter per second it would take 1 second, obviously. What would the equation needed to turn the time taken into the speed (In Meters per second preferably) For those interested the program used is called Algodoo 2. Originally Posted by Mystery I currently have a machine that can calculate the time it takes for an object to travel one meter. (two lasers first one starts a timer second one stops it, The lasers are positioned exactly 1 meter apart.) So if an object was going 10 M/s (Meters per second) it would take 0.1 of a second (10 nanoseconds) to travel one meter. If it was going 1 meter per second it would take 1 second, obviously. What would the equation needed to turn the time taken into the speed (In Meters per second preferably) For those interested the program used is called Algodoo $v = \frac{\Delta x}{\Delta t}$ also ... 0.1 sec does not equal 10 nanoseconds. 10 nanoseconds = $10^{-8}$ sec 0.1 sec = 100 milliseconds 3. So basically Velocity is the Distance Divided by the time taken. thanks. 4. However, it doesn't matter in this particular problem, because the laser travels one straight path. But generally in these problems you want the average speed. The formula $v = \frac{\Delta x}{\Delta t}$ gives you the "average velocity" of the object, $\Delta x$ is the object's displacement (change in position given by $\Delta x = x_f - x_i$). So the formula is not always the same thing as $\frac{d}{\Delta t}$ which gives you the "average speed"! Because in the latter formula, "d" means the total distance traveled which isn't always the same as displacement. Here's an example: If a runner runs a distance d of 10 km and yet ends up at his starting point, his displacement is zero, so his average velocity is zero! But his speed is clearly NOT zero!

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http://math.stackexchange.com/questions/109181/what-can-i-do-with-what-i-know

# What can I do with what I know? [closed] This might be difficult for me to put into words, but bear with me because I think it's an important question. Among the many people who study math, I am one of them. I'm not particularly advanced but I enjoy it. I'm working through calculus 2 in my free time. I do it for fun. It occurred to me: I have a certain amount of ideas in my head: calculus, trig, geometry. These are all tools that required hundreds of years of development. I've developed my ability to reason as best as I can so far. However, I find that I'm not capable of very much, other than I can work through a textbook, I can solve the problems, maybe even sometimes in clever ways that surprise me. But what can I do? If I am capable of more, I don't know it. The question, more specifically, is this: if someone has studied up to, say, calc 2, what should they be capable of? Mathematicians and physicists over the years have achieved much more with much less. So what I'm saying is, with what I know, I feel like I should be capable of more and I'm sure there must be students out there that feel the same. I'd like to think that with knowledge comes more ability than just regurgitating their studies or passing contrived tests. Insight, intuition should lead them to discover new-found abilities. Well I hope I've made my question clear and, if I have not, feel free to delete it. Hopefully if I've left any gaps, then the reader can read between the lines and understand what I'm getting at. Okay think of it this way. Let's say someone has a powerful computer and they're using it as a paperweight. How could they use this resource to the fullest? I'm expecting question of a format like, "well, if you have trig under your belt, I'd expect you to be able to give me a good estimate of the size of the Earth. You have the tools for it." Thanks for reading, it's a bit long-winded. - 7 I feel this is a bit too soft a question, but: what you can do is really up to you. You encounter some problem/application that interests you in the course of reading different things, and you see if you have the tools necessary for tackling. If you don't, you prepare; if you do, then you can start experimenting. With luck, your experiments might turn out to be something special. – J. M. Feb 14 '12 at 5:42 9 One unexpected and very important side effect of having some background in more advanced mathematics is that you get true mastery of the elementary stuff. Many real world problems require only complete control of very basic mathematics. But not all that many people, even among those who "did well" in high school mathematics but did not go on, have enough control over that material to be able to solve real problems with confidence. – André Nicolas Feb 14 '12 at 5:54 3 What can you do with your math knowledge? You mean like doing even more math, or real-world jobs, or taking over the world, or what? This is way too broad in my opinion. – anon Feb 14 '12 at 6:00 2 I don’t think that it’s a bad question to ask, but it’s almost impossible to answer. Ideally you should be able to spot problems that can be solved by the tools that you’ve studied, but in practice @André’s observation is absolutely correct. Typically you acquire real mastery of a set of tools only when you start using them routinely in some more advanced context. – Brian M. Scott Feb 14 '12 at 6:50 3 One thing that's fun to do is write your own physics simulations. – Rahul Narain Feb 14 '12 at 8:14 show 4 more comments ## closed as not constructive by Austin Mohr, anon, lhf, Asaf Karagila, Zev Chonoles♦Feb 15 '12 at 1:37 As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or specific expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, see the FAQ for guidance. ## 1 Answer In my opinion, Calculus (SV, MV, Diff. Eqns) and Linear Algebra forms the basis of most things in Engineering (I mean basic engineering. Not high end stuff, you can get there later though) However, mathematics acts as hygiene and not a motivator. Let me explain what I mean: Pick up any field of engineering. Let us assume you chose Mechanical Engineering. The major components of ME are Heat Transfer/Fluid Mechanics and maybe newer topics like Robotics, CAD CAM or whatever. Among all of these, the aforementioned mathematical topics suffice to understand all that is being said. But effort is required to link an equation with the "real world". In you multivariable/differential equation course, you probably learnt how to solve $x\frac{dy}{dx} = 0$. You may be answers but they won't have any physical intuition. This same equation might show up in heat transfer where $y$ would signify the temperature at a given point. If you know math, you are 1 good book away from knowing the basics of an engineering field. The absence of math knowledge (no matter how good your engineering intuition is) is going to cause problems in the long run. To cite another example: Say you pick up "rocket science", the prerequisites for basic astrodynamics is Calculus with a little physics (Thermodynamics). There will be people who point out that Calculus and LA are not enough and that one needs other subjects to succeed in Engineering. This is not entirely true. You might need a few basic concepts from Probability/Statistics/Topology etc. but they can be learnt on an ad-hoc basic. Note: While the theme of SE is not to spurt out opinions but facts, this question invited one so I provided. -

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http://mathoverflow.net/questions/29202?sort=newest

## Does a “Chern character” exist for any generalized cohomology theory? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The Chern character is a ring homomorphism from the complex K-theory to the usual cohomology. 1) I wonder if there are "Chern character"-like ring homomorphisms from other generalized cohomology theory to the usual cohomlogy. Are they related with Atiyah-Hirzebruch? 2) And if there are such nice homomorphisms, what is the "Todd genus" in these cases, making the generalization of that famous diagram in Grothendieck–Hirzebruch–Riemann–Roch commute? When I think about it, I cannot even recall seeing anything like this in real K-theory, but that is probably because I dont really know real K-theory at all. Thank you very much. - 1 I think, your question is related to the orientability of cohomology theories. There are many multiplicative transformations between bordism theories and other (co)homology theories such as oriented bordism to singular homology, spin bordism to K-theory, string bordism to TMF. As you see, the natural home for such a morphism is not always singular cohomology, but eg in the case of spin bordism you can, of course, compose the morphism to K-theory with the chern character. – Lennart Meier Jun 23 2010 at 8:39 ## 2 Answers In Oscar Randall-Williams' answer "connective" is unnecessary. Also, there is no need to choose that isomorphism (from a rational theory to the ordinary theory with the same coefficient groups); it is canonical. The generalization of the Todd genus or Todd class arises when the multiplicative theory $E$ has a "complex orientation": a multiplicatively well-behaved way of producing Thom isomorphisms in $E^*$-theory for all complex vector bundles. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. For any (connective) spectrum $E$ one may rationalise it to get a rational spectrum $E_\mathbb{Q}$, and a map $E \to E_\mathbb{Q}$. Now rational spectra split as wedges of Eilenberg-Mac Lane spectra, so one may choose an isomorphism `$E_\mathbb{Q}^*(X) \simeq H^*(X;\pi_{-*}(E)\otimes \mathbb{Q})$`, and the rationalisation gives a map `$$ch_E : E^*(X) \longrightarrow H^*(X;\pi_{-*}(E)\otimes \mathbb{Q}).$$` For complex K-theory this gives the Chern character, and for real K-theory it gives the Pontrjagin character. Of course, if $E$ is a ring spectrum so is $E_\mathbb{Q}$, and one must identify the induced ring structure on `$H^*(X;\pi_{-*}(E)\otimes \mathbb{Q})$`. -

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http://physics.stackexchange.com/questions/11800/can-energy-be-taken-out-of-the-qft-vacuum/31173

# Can energy be taken out of the QFT vacuum? There have been recent questions about the vacuum. In my simplified knowledge the vacuum is like a ground state energy level, and also that there might even exist other lower energy levels than the vacuum we find ourselves in. The sea is a soup of created and annihilated pairs of virtual particles with virtual energies and momenta. In a normal sea on earth, which also represents a ground state of water, energy can be taken out of random waves by the clever construction of valves that allow only one way motion of water. Is it conceivable that a gadget of similar function could be found for the vacuum sea, or is it forbidden by conservation laws? My intuition tells me that it might be possible if GR is taken into account, but my physics knowledge does not stretch to support this. Edit : An explanation of why I am asking this question: Let me expand on the example of the sea. The energy from the waves comes from either tides, i.e. gravitational forces, or wind (temperature differentials). If these were missing the oceans would be like glass representing a unique ground state of the gravitational well of the earth. In an analogy, a gravitational wave going through the vacuum would be supplying energy to the vacuum sea. I have been thinking of this analogy ever since cold fusion surfaced and refuses to die out, the most recent one being discussed here too. Approached from nuclear physics orders of magnitude the claims seem preposterous. There are people though who believe they have results of extra energy over input energy, much more than chemical reactions could supply. This set me thinking on vacuum energy and the analogy with getting energy from the sea. A crystal is a prime candidate for any exploration of such concepts and in all cold fusion "successful" results crystals have been used. Now if the effect depended on the vacuum and how much distorted it was by a passage of a gravitational wave at the time of the experiment, or the exact orientation of the crystal, or the type of impurities in the crystal ( F centers etc) one would expect to get haphazard results, and non repeatable by other experimenters. Of course this would be the first experimental evidence of gravitational waves :). Edit 20/7/12 Maybe I should clarify that an acceptable answer in the negative would be one based on conservation laws. I believe that data trumps theory, and next in line are conservation laws,because they are the distillation of an enormous amount of data. Some people seem to think that theoretical definitions can substitute for proof in physics, but physical theories change, solid data do not, and this is physics, not axiomatic mathematics. - 1 Wouldn't that be very similar to Maxwell's demon? – Willie Wong Jul 2 '11 at 14:44 If the vacuum is the ground state, what does it mean to have a lower state. Analogy would be a value smaller than the minimum. Also in the particle interpretation, what would it correspond to, negative number of particles? – MBN Jul 2 '11 at 16:32 – anna v Jul 2 '11 at 19:01 I looked at it but it doesn't help my confusion. What is the vacuum state? I thought it is the state killed by all the anihilation operators. What would a lower than that mean? – MBN Jul 3 '11 at 0:08 1 @Marek The question is not about pumping energy from different vacua, although it might be related, if, for example, vacua had an "uncertainty principle" type of existence. The question relates to the marrying of QFT and gravity. Whether a gravitational wave would change the vacuum energy level as it is propagating in time so that energy might be captured of its passage, as in the sea wave analogy. – anna v Jul 4 '11 at 3:58 show 7 more comments ## 7 Answers Well, let us be honest here. This question was not supposed to be answered from the very beggining. First of all we don't know how to mix quantum mechanics and gravity. There is no good consistent theory for that. Another thing is that today "the Dirac sea" analogy is not considered to be a very good thing. It is a pre-QFT naiive picture. Finally we are supposed to talk about "waves" in this "sea"... While the "sea" itself is an obsolete analogy... And all that is in a context of non-existing theory... Come on... Now. There is actually a formal way to answer the question, because the question is about the "QFT vaccuum". And the QFT vacuum have a precise definition. Which basically says that it is "something you cannot take energy from". Actually, we start from that definition to build QFT. So the answer is: "you cannot take energy out of the QFT vacuum by definition". Maybe we are wrong to start from this definition. Maybe for quantum gravity we need another starting point. But then it wouldn't be a QFT -- it would be a new theory which will have QFT as a limiting case. And in the range of validity of QFT that "formal" answer will hold. - 1 Thanks for regurgitating my answer.............. – Chris Gerig Jul 19 '12 at 22:12 See my comment to @ChrisGerig . There are theories mooted which do have several vacua, QFT would be working on one of those, at the moment the one we find ourselves in. I believe my simplistic analogy of the tides should give an intuition. – anna v Jul 20 '12 at 3:09 "it would be a new theory which will have QFT as a limiting case." This is what I am exploring with this question. "And in the range of validity of QFT that "formal" answer will hold." Do you have a strong argument that you know how that limiting case is approached? For example, superconductivity , quantum mechanics on kilometer scales, shows that it is not dimension that defines absolutely the limit between QM and classical.There is the added ingredient of complexity of bulk matter and its different responses. – anna v Jul 21 '12 at 4:04 @Kostya , if you expand a bit your last paragraph I would accept your answer. I am in disagreement with the answer chosen by the bounty setter. It is not the answer of a physicist, but of a mathematician. A theory can be beautiful and fully consistent in its definitions. The results may agree with data to the available limit of measurements, but that does not mean that the definitions define nature, as progress in physics has shown us again and again. It is at the limits of the validity of a theory that it may be tested against nature with surprising results, for the theory. – anna v Jul 25 '12 at 3:10 @annav I've tried to expand my last paragraph. But I can't. I just really have nothing else to say on the point: whatever I try, I either already said that or don't know what I'm talking about. I should note that I'm a bit formal here as well -- I'm answering the question. I'm not participating in any discussion or "exploration" or whatever. Let me finally stress that I'm against this distinction you draw between "mathematicians" and "physicists". Intuition is great. But if rationality tells that intuition is wrong -- have courage to leave it. – Kostya Jul 25 '12 at 14:29 show 1 more comment No you cannot take energy out of the vacuum, BY DEFINITION. Using this analogy with a "sea" is nonsense, because the statement "vacuum is a sea of virtual particles" is also an ambiguous statement not to be taken literally. The definition of vacuum is ground-state energy, and so if you could take energy away from it, then the energy of the vacuum would be lowered, contradicting the fact that it's already in the lowest possible state. This is also the reason why spontaneous absorption does not occur with electrons/atoms (whereas spontaneous emission can occur). I want to point out why this question is not closed yet: Anna's train of thought is "The answer does not satisfy me because it is dependent on definitions which change as theories change." BUT, the question itself is dependent on definitions... you can't ask about apples and then say 'well maybe they can be oranges'. If you can pose a question about 'energy' and 'vacuum' then they have to be defined. I now vote to close! - The ground state energy is the energy of the quantum mechanical system whose potentials one knows and has solved. It is different for different initial set ups. I am talking of something like the Theory of Everything, which in some speculative theories does have several ground states. – anna v Jul 20 '12 at 3:05 p.s. BTW, nature does not follow definitions, it is observers who make them, and new definitions can supersede or embed old ones. – anna v Jul 20 '12 at 3:20 3 Well, usually the physics of tomorrow hides in the metaphysics of today, though I do not think of this as metaphysics. Rather at the frontier of physics. Your answer does not satisfy me, because it says: these are the definitions, do not bother me with facts. We would have never gotten quantum mechanics if questioning were stop by current definitions. – anna v Jul 20 '12 at 5:57 1 The answer does not satisfy me because it is dependent on definitions which change as theories change. Are you saying that gravitational waves do not carry energy by "definition" and it is futile to try and discover them? – anna v Jul 20 '12 at 6:30 2 For you to even begin talking to anyone, you have to specify what a vacuum is and what energy is, otherwise your question doesn't even make sense, and no logical statements can be made! Therefore, once you bring in the words "energy" and "vacuum", then a priori they have a definition which your question must adhere to. And this question has an answer... the end. – Chris Gerig Jul 20 '12 at 6:38 show 5 more comments Peter Milonni has devoted a lot of work to this area, i will recommend this book: The Quantum Vacuum, it basically gives a overview on lot of physical systems where vacuum field effects are dominant, and how certain boundary conditions can effect them to produce unexpected effects the expansion of fields around the vacuum in harmonic modes with quantized particle population of the modes is only valid when the physics can be fairly approximated in terms of the vacuum eigenmodes (which are the regular field wave functions). Of course, when there are special boundary conditions, this doesn't change too much since we still have eigenmodes (which correspond to the boundary geometry) but still can talk about particle populations when the system (meaning, the boundary conditions) are dynamic (moving mirrors, or superconductor walls moving over the superconductivity phase) i don't think there is an authoritative answer about the validity of these approximations. So in short, there could be interesting things to say about the vacuum dynamics when a more friendly framework exists to make computations in highly dynamical regimes I made a somewhat relevant question a while ago about casimir walls that melted and froze back with an oscillatory transversal magnetic field. i'm sorry, i'm not addressing your larger question, as to what relevant things change when taking GR in consideration. Hopefully someone more knowledgeable on the matter will. - -1: This is nonsense. You can't take energy out of the vacuum, and this is conservation laws. – Ron Maimon Sep 4 '11 at 23:22 1 @Ron, there is no such thing as an energy conservation law in GR. – lurscher Sep 5 '11 at 13:53 you aren't doing GR. – Ron Maimon Sep 5 '11 at 13:56 GR effects should manifest locally as out-of-equilibrium boundary conditions, which is why my answer tried to make those points. – lurscher Sep 5 '11 at 14:17 GR effects do not manifest locally--- the lack of conservation in GR can be attributed to new material entering a region on cosmological scales, or leaving it. The conservation of energy in QFT vacuum is a fact, and it is impossible to pump energy out unless the vacuum is unstable and you then wreck the whole universe by tunneling to a lower energy state. – Ron Maimon Sep 5 '11 at 14:25 show 3 more comments The ground state (=vacuum) is an eigenstate of the system Hamiltonian. The wave function is unique, no energy uncertainty exist. In this sense there is no fluctuation of energy to harvest. "Fluctuations" exist for non commuting with the Hamiltonian variables but it cannot be used to get some energy for the reason given above. In GR the energy is not conserved and this is not connected to the vacuum. Edit: in heat machines one tries to increase the temperature difference $T_{hot}-T_{cold}$ to increase the machine efficiency. Vacuum is the coldest body ever known. So the vacuum "cleaner" efficiency cannot exceed zero. - 2 Your last sentence is the crux of the matter. Why wouldn't general relativity create fluctuations in the vacuum, in a TOE? – anna v Jul 2 '11 at 18:40 Because GR is not a unique theory of gravity. There are theories compatible with a flat space-time with the energy/momentum conservation laws. – Vladimir Kalitvianski Jul 2 '11 at 18:59 It certainly is not a heat machine. @Vladimir , the system of comments does not accept your name in the beginning of the comment!! – anna v Jul 3 '11 at 19:18 In theory, QFT vacuum state is isotropic and invariant for all observers, as the base of all ladder operator algebras. It does not have any features other than the fluctuations. Hence, you cannot extract any more energy from it. In practice, extracting energy out of the vacuum is not only possible, but it has been already achieved in laboratory using optical parametric amplifiers (NOPA); they are used for squeezing input light, but when there is nothing in the input, the output is a field that has lower average standard deviation of the energy than the normal vacuum in the range of frequencies where the amplifier is active. If we assign to the normal vacuum zero energy, then this "squeezed vacuum" field must have negative energy. Of course, let's make perfectly clear that this mechanism cannot be used to extract useful energy, since the energy spent pumping the amplifiers greatly exceed anything that you could extract. “In theory, theory and practice are the same. In practice, they are not.” ― Albert Einstein - The concept of squeezing does NOT truly beat the "standard quantum limit", because although it squeezes one variable, the other canonical-variable increases. – Chris Gerig Jul 19 '12 at 23:02 @ChrisGerig, "it squeezes one variable, the other canonical-variable increases" that is correct. But do not forget you can choose what canonical variables you use. In most experimental setups, the variables chosen are amplitude versus phase. In this case, amplitude-squeezed vacuum has smaller amplitude fluctuations, which implies smaller average energy – diffeomorphism Jul 20 '12 at 16:15 @ChrisGerig, there is however, a valid critisism, if we accept that some form of quantum inequalities must hold, then the negative energy regions need to be overcompensed by regions with extra positive energy, so even the time-averaged energy density is nonnegative or the space-averaged energy density is nonnegative, but not both – diffeomorphism Jul 20 '12 at 16:17 There has been a proposal to use Casimir force and meta materials to build a vacuum energy extractor. Casimir force applied on parallel plates made of "normal" material pushes the plates outwardly and inwardly for metamaterials. I do not know if the idea has yet been tested. http://physicsworld.com/cws/article/news/2007/may/02/casimir-force-could-drive-tiny-ratchets - The vacuum state has a property named passivity, which means that any local operation onto the vacuum state does not extract but inject energy to the system. This implies that energy cannot be taken out of the vacuum only by local operations. However, if we adopt local operations and classical communication (LOCC), a part of zero-point energy of the vacuum state can be extracted. The scheme is called quantum energy teleportation. More information is available in wikipedia and a review article by Hotta, who first proposed the concept: http://www.tuhep.phys.tohoku.ac.jp/~hotta/extended-version-qet-review.pdf . - ## protected by Qmechanic♦Jan 28 at 23:03 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.

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http://nrich.maths.org/7302/solution

### Chocolate There are three tables in a room with blocks of chocolate on each. Where would be the best place for each child in the class to sit if they came in one at a time? ### Plants Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this? ### Gran, How Old Are You? When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is? # Count the Digits ##### Stage: 1 and 2 Challenge Level: We had quite a few responses from pupils who had found out things by exploring this situation. Calum, Christopher and Matthew from St. Andrews in Scotland wrote to say: We found out that using the digits $1, 2$ and $3$ making a four-digit number you would start doubling eventually when you reach line $10$ every time no matter what order you put the digits in. We found out that using the digits $1, 2$ and $4$ making a four-digit number you would start doubling eventually when you reach line $7$ every time no matter what order you put the digits in. We found out that using the digits $1, 3$ and $4$ making a four-digit number you would also start doubling eventually when you reach line $7$ every time no matter what order you put the digits in. We found out that using the digits $2, 3$ and $4$ making a four-digit number you would start doubling eventually when you reach line $6$ every time no matter what order you put the digits in. Sayeed from St. Michael's London also sent in a well thought out reponse:- This is my answer when starting with $4$ digits: $3243, 122314, 21221314, 31321314, 31123314, 31123314$ It will continue as the same number forever. I notice: Each row has two more digits then the previous row until the rows have the maximum amount of digits ($8$) possible. In each row with the maximum digits, the second digit always has a $1$, the fourth digit always has a $2$, and the sixth digit always has a $3$ and so on. Every row ends in $14$ excluding the starting row. This is my answer when starting with $5$ digits: $22411, 212214, 213214, 21221314, 31321314, 31123314, 31123314$ It will continue as the same number forever. I notice the same thing that happens with the starting $4$ digit row except the rows getting two more digits each time until there are maximum digits. I also notice that with the starting four digit row it takes four counts till the number continues as the same number forever. But for the starting five digit row it takes five counts till the number continues as the same number forever! Thomas from Colet Court School said the following and attached his numbers. When you start with $4$ digits the series converges to $21322314$ whatever the first $4$ digits. When you start with $5$ digits it gives the same result. Finally, Miss Stanley's Numeracy group from Greystoke Leicester wrote:- We liked this challenge and worked very hard. Trying other numbers using the same rules we found that we could continue until the numbers were the same, because that number would keep repeating. We discovered some have shorter sequences and some have longer sequences until the same number repeats, but we're not sure why yet. Caitlin and Millie found that some numbers ($4122$) didn't seem to have an end because we spotted the pattern that it kept repeating itself, so we decided to stop. Some of us even moved onto extending this challenge to $5, 6$,and even $7$ digits. Many of us spotted that the larger the number of digits in the starting number, the shorter the sequence was to get to the end. Thank you, we enjoyed this challenge. Well done to all the contributors, it sounds as if you really enjoyed this. (A late arrival came from Adam at Cypress School who noted something special about $1$ & $4$. I am wondering if that was because it was not a usual thing you'd find going on in many Mathematics lessons? On the last day of the month we recieved this excellent presentation of Oscar from Spain. You have to take out one digit of the $1,2,3,4$ which are the possible digits to make the starting number. If you take out a number and want to get a $4$ digit number, you have to repeat one of the other $3$ numbers. If you take out $1$, you have possible starting numbers $2234, 2334, 2344$ and other possible numbers that you get changing the order of the digits in each of those $3$ numbers. As the order does not affect digit counting, those give the same counting sequence. The counting is: $A 2 B 3 C 4$ and A-B-C have to be $2-1-1$ (for $2234$) or $1-2-1$ (for $2334$) or $1-1-2$ (for $2344$). The next counting in all cases is $2 1 2 2 1 3 1 4$ and sequence is: $3 1 3 2 1 3 1 4$ $3 1 1 2 3 3 1 4$ and this last number stays the same if you count the digits. If you take out $2$ you get the number:$2 1 3 2 2 3 1 4$ If you take out $3$ you get the same as if you take out $1$, and if you take out $4$ you get the same as if you take out $2$. 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.

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http://mathoverflow.net/revisions/33373/list

Return to Answer 2 deleted 2 characters in body When property P is universal ($\forall ...$) it is likely to correspond to closed sets, and thus be preserved under intersection. Examples: axioms of a group, ring, field, directed graph; having symmetry under a given group. However, if P is existential ($\exists ...$) it corresponds to open sets and is more likely to be preserved by unions (or coproducts)products), not intersections. Examples: being algebraically closed, having at least 53 elements. (Well, algebraic closure is $\forall \exists$ so of course it is even more complicated. But falling out of the pure $\forall$ class it fails the intersection property.) The first situation is possibly more common because we want structures to satisfy some, well, structural properties. Properties expressed by equations usually correspond to closed sets. To some extent this is formalized in Birkhoff's theorem On equational presentations. Any book on Universal Algebra will discuss it. Also, the sample of concepts is biased, because definitions that become standard are often selected for their useful formal properties. Concepts not having stability under intersection (or union, or inheritance by sub- or super-structures) are less likely to be used. 1 When property P is universal ($\forall ...$) it is likely to correspond to closed sets, and thus be preserved under intersection. Examples: axioms of a group, ring, field, directed graph; having symmetry under a given group. However, if P is existential ($\exists ...$) it corresponds to open sets and is more likely to be preserved by unions (or coproducts), not intersections. Examples: being algebraically closed, having at least 53 elements. (Well, algebraic closure is $\forall \exists$ so of course it is even more complicated. But falling out of the pure $\forall$ class it fails the intersection property.) The first situation is possibly more common because we want structures to satisfy some, well, structural properties. Properties expressed by equations usually correspond to closed sets. To some extent this is formalized in Birkhoff's theorem On equational presentations. Any book on Universal Algebra will discuss it. Also, the sample of concepts is biased, because definitions that become standard are often selected for their useful formal properties. Concepts not having stability under intersection (or union, or inheritance by sub- or super-structures) are less likely to be used.

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http://en.wikipedia.org/wiki/Third_derivative

# Third derivative Calculus Definitions Concepts Rules and identities Integral calculus Definitions Integration by Formalisms Definitions Specialized calculi In calculus, a branch of mathematics, the third derivative is the rate at which the second derivative, or the rate of change of the rate of change is changing. The third derivative of a function y=f(x) can be denoted by $\frac{d^3y}{dx^3},\quad f'''(x),\quad\text{or }\frac{d^3}{dx^3}[f(x)].$ Other notations can be used, but the above are the most common. ## Mathematical definitions Let $f(x)=x^4$. Then $f'(x)=4x^3$, and $f''(x)=12x^2$. Therefore, the third derivative of f(x) is, in this case, $f'''(x)=24x$ or, using Leibniz notation, $\frac{d^3}{dx^3}[x^4]=24x$ Now for a more general definition. Let $f(x)$ be any function of x. Then the third derivative of $f(x)$ is given by the following: $\frac{d^3}{dx^3}[f(x)]=\frac{d}{dx}[f''(x)]$ The third derivative is the rate at which the second derivative (f''(x)) is changing. ## Applications in physics Main article: Jerk (physics) In physics, particularly kinematics, jerk is defined as the third derivative of the position function of an object. It is basically the rate at which acceleration is changing. In mathematical terms: $\bold{j}(t)=\frac{d^3\bold{r}}{dt^3}$ where j(t) is the jerk function with respect to time, and r(t) is the position function of the object with respect to time. ## Economic example President Richard Nixon, when campaigning for a second term in office announced that the rate of increase of inflation was decreasing, which has been noted as "the first time a sitting president used the third derivative to advance his case for reelection."[1] Since inflation is itself the negation of a derivative—the rate at which the purchasing power of money decreases—then the rate of increase of inflation is the derivative of inflation, or the negation of the second derivative of the function of purchasing power of money with respect to time. Stating that a function is decreasing is equivalent to stating that its derivative is negative, so Nixon's statement is that the second derivative of inflation—or the negation of the third derivative of purchasing power—is negative. Thus, Nixon's statement is that the third derivative of the purchasing power is positive. Nixon's statement allowed for the rate of inflation to increase, however, so his statement was not as indicative of economic growth as it sounds. ## References 1. Rossi, Hugo (October 1996). "Mathematics Is an Edifice, Not a Toolbox". Notices of the American Mathematical Society 43 (10): 1108. Retrieved 13 November 2012.

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http://mathoverflow.net/questions/24658/construct-scheme-from-quivers/25023

## construct scheme from quivers? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I heard from some guys working in noncommutative geometry talking about the idea that one can construct the noncommutative space from quivers. I feel it is rather interesting. However, I can not image how one can construct a non-affine scheme or space. For affine case, one might consider the path algebra of a quiver $Q$, say, $KQ$. It is a hereditary algebra(or quasi-free algebra, formally smooth algebra). How can one construct a non-affine space from quiver? Is it possible to construct a noncommutative space using different interesting quivers as "local data"? If it is possible, the algebraic geometry machine working on this space will be helpful to study the representations of quiver? Thanks - ## 3 Answers It is possible to construct more general schemes (even 'projective' schemes) starting from quiver-representations. The idea is to consider a stability structure and consider the corresponding moduli spaces of semi-stable quiver representations as introduced by Alastair King (Moduli of representations of finite dimensional algebras, Quat. J. Math. Oxford 45 (1994) 515-530.) If one starts with a quiver without oriented cycles, these moduli spaces are projective and so it makes sense to take the collection of all these moduli spaces (or multiples of a fixed dimension vector compatible with the stability structure) as a noncommutative projective space. This point of view was advocated in 'Noncommutative compact manifolds constructed from quivers'. More generally, allowing cycles, the collection of moduli spaces (or multiples of fixed dim vector) for all dim vectors compatible with the stabilitu structure can be viewed as a noncommutative scheme. Here the main idea is that polynomial semi-invariants of quivers are known to be generated by so called determinental semi-invarinats (a result by Aidan Schofield and Michel Van den Bergh 'Semi-invariants of quivers for arbitrary dimension vectors'. What this says is that these collections of moduli spaces have an affine cover determined by universal localizations of the path algebra. This point of view is expressed here. EDIT : Here's an example to illustrate all of this : noncommutative projective n-space. Consider the quiver with 2 vertices and n+1 arrows, all from the first to the second vertex, Take stability structure (-1,1), then we consider only representations of dimension vector (m,m). A semi-stable representation is one given by the n+1 mxm matrices X0,...,Xn such that there are numbers ai with det(a0X0+a1X1+...+anXn) is non-zero. For example, consider the open piece where Xi is invertible, then one can use Xi to identify the two vertex-spaces and the other arrows now become loops in that common vertex. That is, this open piece gives all representations of the free algebra in n variables. Hence, we can cover the moduli space by affine pieces, all iso to the reps of a one-vertex quiver with n loops. Clearly one can generalize this procedure to glue certain representations of two very different quivers. Formally invert one arrow in your quiver, this comes down to taking a stability structure with 0's at all other vertices and (-1,1) at start and end-point of your arrow. Then the 'Zariski open piece' of repQ consists the representations of a new quiver Q' in which one identifies the two vertices, and turns all other arrows between these two vertices into loops at the new vertex. In this way to quivers can be 'glued' to form more general schemes provided they are isomorphic after doing this 'invert one arrow' trick on each of them a finite number of times. Universal localizations corresponding to 'determinental semi-invariants' are a generalization of this idea, allowing paths (or even collections of linear combinations of paths) to be 'inverted'. (End EDIT) A recent incarnation of this strain of ideas (restricting to stable rather than semi-stable representations) is taken by Markus Reineke in his study of 'noncommutative Hilbert schemes' which have been used by Markus recently in 'Cohomology of quiver moduli, functional equations, and integrality of Donaldson-Thomas type invariants'. As to the local structure of these noncommutative schemes. Their 'local rings' are indeed again path algebras of 'local quivers' which can be determined from the stable components of the semi-stable representation. The details are here. If you want to extend all of this to general formally smooth algebras (rather than just path algebras) a place to start is with 'One quiver to rule them all'. Finally, Ive tried to collect all of this in a book as Daniel mentioned in his answer (one might replace 'blablabla' by 'noncommutative geometry'...). The latest version of it can be downloaded here. - Excellent! – Victor Protsak May 17 2010 at 19:12 ### 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. As far as I know a "non-commutative space" is for many people just an $A_\infty$-category. Every scheme determines a non-commutative space: its derived category of coherent sheaves (with an $A_\infty$ enrichment). Also to an (non-commuative) algebra $A$ you can associate a non-commutative space, the derived category of modules over $A$. In some cases these two types of non-commutative space are isomorphic. The classical example (due to Belinson) is projective space whose derived category is equivalent to the derived category of modules over $End(O \oplus O(1) \oplus \dots O \oplus O(n) )$. This algebra has a description as path algebra over a quiver with relations. E.g. for $P^1$ you get the Kronecker quiver (two vertices two arrows in the same direction). - I think you should start looking at the webpage of Lieven Le Bruyn. He most definitely has some notes that will quell your thirst in this respect. For one he has written a book that is freely available on his webpage ("Noncommutative Geometry and Cayley-smooth Orders"). Another good person whose papers and webpage it's reasonable to pay a visit to if you're interested in this area is Markus Reineke. Victor Ginzburg's notes on Noncommutative geometry arXiv:math/0506603 can maybe also be of interest even though I don't know if there are any quivers lurking there. -

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http://mathhelpforum.com/math-topics/201411-velocity-time-curve-problem.html

# Thread: 1. ## velocity-time curve problem A train start to move from rest and then travel its first part with constant acceleration of $f_1$then travel the second part of the constant velocity of V then travel the third part of the constant deceleration of $f_2$ then it comes to rest.So draw a velocity - time curve and it travel constant velocity with the $\frac{4}{3}t$ of the total time and the average speed of the total travelling is KV.so find out the value of k? so please give me the ideas to do that. 2. ## Re: velocity-time curve problem Originally Posted by srirahulan A train start to move from rest and then travel its first part with constant acceleration of $f_1$then travel the second part of the constant velocity of V then travel the third part of the constant deceleration of $f_2$ then it comes to rest.So draw a velocity - time curve and it travel constant velocity with the $\frac{4}{3}t$ of the total time and the average speed of the total travelling is KV.so find out the value of k? so please give me the ideas to do that. At constant acceleration $f_1$, after time t, the speed will be $f_1t$. You say "the first part" but you don't say what the time is or if "the first part" is the same length of the time as the other "parts". The graph of that is a straight line with slope $f_1$. Calling the length of time for the first part " $T_1$", the velocity at the end of the first part will be $f_1T_1$ which, presumably, is the constant speed of the "second part". Taking the length of time for the "second part" " $T_2$, the second part of the graph is a horizontal straight line with height V, from $T_1$ to $T_2$. The third part of the graph will be a straight line with slope $-f_1$. Of course, the train will stop after time interval $T_2$ where $V- f_3T_3- T_2)= 0$ I don't uderstand what the "it travel constant velocity with the $\frac{4}{3}t$ of the total time" means. You did not mention "t" before and 4/3 is larger than 1. 3. ## Re: velocity-time curve problem I am really sorry i assume the total time for travelling is t.when it travel in the constant velocity of V with 3/4t of the total time (sorry for mention 4/3)

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http://mathhelpforum.com/advanced-applied-math/198820-partitions-proof.html

# Thread: 1. ## Partitions proof Let the number of partitions of n into parts each less than or equal to m be denoted P(n,m). Prove P(n,m) = P(n, m-1) + P(n-m,m) 2. ## Re: Partitions proof Let the number of partitions of n into parts each less than or equal to m be denoted P(n,m). Prove P(n,m) = P(n, m-1) + P(n-m,m) I have written on this question. So I tell you whoever wrote your question has left out some cases. What happens if $m=1~?$ What happens if $m=n~?$ Here is the MathCad way of doing it. [ATTACH=CONFIG]23869[/ATTACH] Attached Thumbnails 3. ## Re: Partitions proof what is mathcad? >.> ... i'm not really following that equation I'm a little confused because there is a formula in the book about partitioning n into exactly r parts where order counts = C(n-1,r-1) but then there's another one that says "the number of ways of writing n as the sum of r or fewer integers where the order of the summands matters is C(r+n-1,n)" but i didn't think they applied because order shouldn't count, should it? I was also trying to play with the concept that the number of partitions of n into r or less parts = the number of partitions of n into parts that are r or less, but have gotten nowhere so far :\ Originally Posted by Plato $m=1~?$ What happens if $m=n~?$ if m = 1 there should only be one partition, n itself, right? and if m = n, then all the partitions = 1 4. ## Re: Partitions proof could you be more specific please?

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http://mathoverflow.net/questions/118515?sort=oldest

## Derivative indicator function ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am wondering what is the derivative of the following function with respect to $x(t)$ in sense of distributions. $$I\left(\int_0^t x(\tau)d\tau \leq c\right)$$ where $I$ is the indicator function and $c$ is a constant. - What is the indicator function? – Liviu Nicolaescu Jan 10 at 11:01 Distributions on a space of functions? Even measures there are notoriously difficult (cylindrical measures of L. Schwartz). – Peter Michor Jan 10 at 11:36 Could you add a bit more details? For example, it is important what you mean by indicator function -- the standard definition I know is extended-real-valued, and hence has no derivative in the sense of classical analysis. There are other derivative concepts that are applicable here, but it would help to know the context of this question. – Christian Clason Jan 10 at 13:01 ## 2 Answers Let me try. Your mapping is: $1 - H(x-c)$ , where $H$ is the Heaviside function, derivative is $-\delta(x-c)$. composed with $\int_0^t : C^\infty(\mathbb R) \to \mathbb R$, which is bounded and linear, surjective. No way with the chain rule, in fact, there is no chain rule in this situation. Simpler approach, but it might not answer your question: Suppose $s\mapsto x(\tau,s)$ is a smooth variation of $x(\tau)$, i.e., $x\in C^\infty(\mathbb R^2)$. Then $$s\mapsto 1-H\Big(\int_0^t x(\tau,s)d\tau -c\Big)$$ can be viewed as a distribution in $s$. Take a test function $f(s)$ and consider $$\int \Big(1-H\Big(\int_0^t x(\tau,s)d\tau -c\Big)\Big)(-f'(s))ds$$ Now try formal partial integration with respect to $s$. Since the left hand integrand is 1 or 0, by the fundamental theorem it will be the difference of the values of $f$ at those $s$ where this jumps. If you fix $x(\tau,s)$ you can work this out. - ### 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. You are asking for the derivative of a non-linear function on an infinite dimensional space. (You do not specify the latter but the space locally integrable functions seems a natural candidate). The derivative can only exist in a very weak sense so it is natural to go for the directional derivative which has two inputs---the point $x$ where the derivative is computed and the direction $y$ in which the rate of change takes place. A back of the envelope calculation suggests that the derivative should then be $$-\sum \frac {Y(a)}{x(a)}\delta_a$$ where we use capitals to indicate the primitives which occur in the formulation and the sum is taken over the $a$ in the pre-image of $c$ under $X$. Some general remarks: the question was posed in such a vague manner that it is not really possible to give a precise, rigorous answer. Presumably you have some concrete application in view and I suggest that you check the above formula there to see if it leads to the expected result (ones I looked at were the functions $t$---with primitive $t^2$---and $\sin t$ for the function $x$). As is evident from the above formula, the derivative is in a much weaker sense than the usual concepts of functional analysis---one requires special conditions on $x$ (beyond just smoothness) for the above expression to make sense and the limit of the difference quotients used to define the derivative does not take place in the underlying function space but in a larger space (of distributions---hence the Dirac functions in the formula. I presume that this is the reason for the reference to distributions in your question). - 1 am having TeX problems in editing this answer. the denominator in the formula should be encased in absolute value symbols. – jbc Feb 10 at 6:30

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http://cms.math.ca/Competitions/MOCP/2002/prob_jun.mml

Canadian Mathematical Society www.cms.math.ca | | | | | | |----------|----|-----------|----|----------| | | | | | | | Français | | Site map | | | CMS store | | | location: PROBLEMS FOR JUNE Valeria Pandelieva 641 Kirkwood Avenue Ottawa, ON K1Z 5X5 no later than July 21, 2002. 151. Prove that, for any natural number $n$, the equation $x\left(x+1\right)\left(x+2\right)\dots \left(x+2n-1\right)+\left(x+2n+1\right)\left(x+2n+2\right)\dots \left(x+4n\right)=0$ does not have real solutions. 152. Andrew and Brenda are playing the following game. Taking turns, they write in a sequence, from left to right, the numbers 0 or 1 until each of them has written 2002 numbers (to produce a 4004-digit number). Brenda is the winner if the sequence of zeros and ones, considered as a binary number (i.e., written to base 2), can be written as the sum of two integer squares. Otherwise, the winner is Andrew. Prove that the second player, Brenda, can always win the game, and explain her winning strategy (i.e., how she must play to ensure winning every game). 153. (a) Prove that, among any 39 consecutive natural numbers, there is one the sum of whose digits (in base 10) is divisible by 11. (b) Present some generalizations of this problem. 154. (a) Give as neat a proof as you can that, for any natural number $n$, the sum of the squares of the numbers $1,2,\dots ,n$ is equal to $n\left(n+1\right)\left(2n+1\right)/6$. (b) Find the least natural number $n$ exceeding 1 for which $\left({1}^{2}+{2}^{2}+\dots +{n}^{2}\right)/n$ is the square of a natural number. 155. Find all real numbers $x$ that satisfy the equation ${3}^{\left[\left(1/2\right)+\mathrm{log}{}_{3}\left(\mathrm{cos}x+\mathrm{sin}x\right)\right]}-{2}^{\mathrm{log}{}_{2}\left(\mathrm{cos}x-\mathrm{sin}x\right)}=\sqrt{2} .$ [The logarithms are taken to bases 3 and 2 respectively.] 156. In the triangle $\mathrm{ABC}$, the point $M$ is from the inside of the angle $\mathrm{BAC}$ such that $\angle \mathrm{MAB}=\angle \mathrm{MCA}$ and $\angle \mathrm{MAC}=\angle \mathrm{MBA}$. Similarly, the point $N$ is from the inside of the angle $\mathrm{ABC}$ such that $\angle \mathrm{NBA}=\angle \mathrm{NCB}$ and $\angle \mathrm{NBC}=\angle \mathrm{NAB}$. Also, the point $P$ is from the inside of the angle $\mathrm{ACB}$ such that $\angle \mathrm{PCA}=\angle \mathrm{PBC}$ and $\angle \mathrm{PCB}=\angle \mathrm{PAC}$. (The points $M$, $N$ and $P$ each could be inside or outside of the triangle.) Prove that the lines $\mathrm{AM}$, $\mathrm{BN}$ and $\mathrm{CP}$ are concurrent and that their intersection point belongs to the circumcircle of the triangle $\mathrm{MNP}$.

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http://mathoverflow.net/questions/108903/imaginary-quadratic-field-contained-in-hecke-orbit-field/108950

## Imaginary quadratic field contained in Hecke orbit field? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $\tau$ in the upper half plane lie in an imaginary quadratic field $K$. Then is $K \subset \mathbb{Q}({j(g \tau) \ | \ g \in GL_2^+(\mathbb{Q}) })$? (here $j$ is the modular $j$-function, and $GL_2^+(\mathbb{Q})$ means positive determinant - i.e. we adjoined all $j$ values of elliptic curves with CM by an order in $K$). If the above isn't true, then is $K$ contained in $\mathbb{Q}(S)$ where $S$ is the set of $j$ values of all CM elliptic curves? I'm also interested in whether the appropriate statement is true for a general Shimura variety. - ## 1 Answer "No" for both questions about CM elliptic curves and "I'm not even sure I know what the question would be" about general Shimura varieties. Basic idea: If $j$ is the $j$-invariant of a CM elliptic curve, then there is some imaginary quadratic discriminant $\Delta \equiv 0$ or $1 \bmod 4$ such that $\mathbf{Q}(j) \cong \mathbf{Q}[X]/H_\Delta(X)$ where $H_\Delta(X) \in \mathbf{Z}[X]$ is the Hilbert Class Polynomial of discriminant $\Delta$, whose roots are the $j$-invariants of elliptic curves over the complex numbers (actually $\overline{\mathbf{Q}}$ is enough) with CM by $\mathbf{Z}\left[\dfrac{ \Delta + \sqrt \Delta}{2}\right]$. The point is that now we can see that there is an embedding $\mathbf{Q}(j) \hookrightarrow \mathbf{R}$, for all possible $j$. Therefore there is an embedding $\mathbf{Q}(S) \hookrightarrow \mathbf{R}$. To see this, it's enough to note that for any two CM $j$-invariants $j_1$ and $j_2$ that there exists an embedding $\mathbf{Q}(j_1,j_2)\hookrightarrow \mathbf{R}$. Let $J_1$ and $J_2$ be the canonical image of $j_1$ and $j_2$ in the real numbers. Then $\mathbf{Q}(j_1)$ embeds into the real numbers as $\mathbf{Q} + \mathbf{Q}J_1 + \dots + \mathbf{Q}J_1^{h_1 -1}$ and $\mathbf{Q}(j_2)$ embeds into the real numbers as $\mathbf{Q} + \mathbf{Q}J_2 + \dots + \mathbf{Q}J_2^{h_2 -1}$. Therefore $\mathbf{Q} + \mathbf{Q}J_1 + \mathbf{Q}J_2 + \dots + \mathbf{Q}J_1^{h_1 -1}J_2^{h_2 -1}$ is a copy of $\mathbf{Q}(j_1,j_2)$ inside of $\mathbf{R}$. Notice that I didn't use direct sums because $\mathbf{Q}(j_1)$ and $\mathbf{Q}(j_2)$ might not be linearly disjoint over $\mathbf{Q}$! This is also the reason I didn't use a tensor product argument. In any case, this inductive step allows us to work with direct limits and embed $\mathbf{Q}(S)$ into $\mathbf{R}$. Therefore if we assume that there is an embedding $K\hookrightarrow \mathbf{Q}(S)$ then there must be an embedding $K\hookrightarrow\mathbf{Q}(S) \hookrightarrow \mathbf{R}$, which is absurd. Therefore, there is no embedding $K\hookrightarrow \mathbf{Q}(S)$. To show that we have an embedding $\mathbf{Q}(j)\hookrightarrow\mathbf{R}$, consider that there is some $\tau \in \mathcal{H}$ of the form $\dfrac{1}{2}\sqrt\Delta$ or $\dfrac{ 1 + \sqrt \Delta}{2}$ such that $j(\tau)$ is a root of $H_\Delta(X)$. But then $j(\tau)$ is real, because the inverse image of the reals under the $j$-function contains the lines $\lbrace iy : y \ge 1\rbrace$ and $\lbrace 1/2 + iy : y \ge (1/2)\sqrt 3\rbrace$. Therefore we have our embedding $\mathbf{Q}(j) \hookrightarrow \mathbf{R}$. - @stankewicz:Fair enough, you have an embedding $\mathbb{Q}(j)$ into $\mathbb{R}$. But could you please give more details on how you embed $\mathbb{Q}(S)$ into $\mathbb{R}$ (surely $H_{\Delta}(X)$ could have some complex roots)?. – Adam Harris Oct 8 at 13:25 @stankewicz: Isn't it true that if the class of $\mathfrak{a}$ in the ideal class group has order greater than 2 then $j(\mathfrak{a})$ isn't real? – Adam Harris Oct 8 at 13:33 I edited the answer to be more explicit. It is very possible for Hilbert Class Polynomials to have complex roots and it's easy to find examples by quickly playing around with MAGMA or sage. It just doesn't matter, because as soon as there is ONE real embedding, any subfield has to also have a real embedding. – stankewicz Oct 8 at 15:13 @stankewicz: Thanks, but I'm still confused: What is this `canonical image' of $j_i$ in the reals? If it is $j(\mathcal{O})$ where $\mathcal{O}$ is your order then what if the canonical images are equal i.e. $J_1=J_2$? i.e. how do you embed the splitting field of $H_D(X)$ in the reals if it has a complex root? – Adam Harris Oct 8 at 20:08 I don't embed the splitting field of $H_D(X)$ into the reals because you can't. Is your Hecke orbit field the union of $\mathbf{Q}(j(E))$ over the set of elliptic curves with CM by an order inside $K$, or is it the union of the Galois closures? If you mean Galois closures, the answer becomes yes and in fact the Galois closure for $E$ with CM by ANY order in $K$ will work. – stankewicz Oct 8 at 21:25 show 9 more comments

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http://unapologetic.wordpress.com/2008/09/27/analytic-functions/?like=1&_wpnonce=9d779cb1c0

# The Unapologetic Mathematician ## Analytic Functions Okay, we know that power series define functions, and that the functions so defined have derivatives, which have power series expansions. And thus these derivatives have derivatives themselves, and so on. Thus a function defined by a power series in a given disk is actually infinitely differentiable within that same disk. What about the other way? Say if we have a function $f$ with arbitrarily high derivatives at the point $z_0$. We know that if this function has a power series about $z_0$, then the only possible sequence of coefficients is given by the formula $\displaystyle a_k=\frac{f^{(k)}(z_0)}{k!}$ But does this sequence actually give a power series expansion of $f$? That is, does the (in)famous “Taylor series” $\displaystyle\sum\limits_{k=0}^\infty\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k$ converge to the function $f$ in any neighborhood of $z_0$? If so, we’ll call the function “analytic” at $z_0$. So, are all infinitely-differentiable functions analytic? Are all functions for which the Taylor series at $z_0$ actually the limit of said Taylor series near $z_0$? Well, the fact that we have a special name should give a hint that the answer isn’t always “yes”. We’ve been working with complex power series, but let’s specialize now to real power series. That is, all the coefficients are real, we center them around real points, and they converge within a real disk — an interval — of a given radius. Now in this context we can consider the function defined by $f(x)=e^{-x^{-2}}$ away from $x=0$. It’s straightforward to calculate $\displaystyle\lim\limits_{x\rightarrow0}e^{-x^{-2}}=0$ And if we define $f(0)=0$ it turns out to even be differentiable there. The derivative turns out to be $\left(2x^{-3}\right)e^{-x^{-2}}$. And we can also calculate $\displaystyle\lim\limits_{x\rightarrow0}2x^{-3}e^{-x^{-2}}=0$ And so on. The $n$th-derivative will be $P_n(x^{-1})e^{-x^{-2}}$, where $P_n$ is a polynomial. We can calculate $\frac{d}{dx}P_n(x^{-1})e^{-x^{-2}}=\left(P'_n(x^{-1})(-x^{-2})+P_n(x^{-1})(2x^{-3})\right)e^{-x^{-2}}$ That is, we can set $P_0=1$ and $P_{n+1}=2X^3P_n-X^2P_n'$, and thus recursively define a sequence of polynomials. Thus for each degree we have a polynomial in $x^{-1}$ multiplied by $e^{-x^{-2}}$, and in the limit the latter clearly wins the race. For each derivative we can fill in the “gap” at $x=0$ by defining $f^{(n)}(0)=0$. But now when we set up the Taylor series around $x_0=0$ what happens? The series is $\displaystyle\sum\limits_{k=0}^\infty\frac{f^{(n)}(0)}{k!}x^k=\sum\limits_{k=0}^\infty\frac{0}{k!}x^k$ Which clearly converges to the constant function ${0}$. That is, the Taylor series of this function at $x_0=0$ converges to nothing like the function itself. This function is infinitely differentiable at ${0}$, but it is not analytic there. There are a lot of theorems about what conditions on an infinitely-differentiable function make it analytic, but I’m going to leave them alone for now. ### Like this: Posted by John Armstrong | Analysis, Calculus ## 5 Comments » 1. [...] I’ve decided I really do need one convergence result for Taylor series. In the form we’ll consider today, it’s an extension of the ideas in the Fundamental [...] Pingback by | September 30, 2008 | Reply 2. [...] we can construct the Taylor series at . The coefficient formula tells [...] Pingback by | October 7, 2008 | Reply 3. [...] algebra. Instead, I set off on power series and how power series expansions can be used to express analytic functions. Then I showed how power series can be used to solve certain differential equations, which led us [...] Pingback by | October 16, 2008 | Reply 4. [...] it might seem like a really weird digression, but let’s look at the Taylor series for the function . Yes, that was purely a product of our work on analysis over , but let’s [...] Pingback by | March 10, 2009 | Reply 5. [...] this the end? Not quite. Just like in one variable we have analytic functions. Once we’re in and we have all higher derivatives we can use [...] Pingback by | October 21, 2009 | 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.

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http://mathhelpforum.com/discrete-math/137064-graph-theory-proof.html

# Thread: 1. ## graph theory proof Prove that if G is a graph with no cycles of even length, then every cycle in G is an induced subgraph of G. any help would be great thanks. 2. Suppose G has no cycles of even length. Let $H=\{ h_1,h_2,\ldots,h_n\}$ be a cycle of G (of odd length) and suppose it is not an induced subgraph of G (assume wlog that $h_i E_H h_{i+1}$ and $h_n E_H h_1$. Then there is some i, j (wlog i<j) such that $h_i E_G h_j$ but $\neg(h_i E_H h_j)$. Consider the two cycles $h_1\rightarrow \cdots\rightarrow h_i\rightarrow h_j\rightarrow h_{j+1}\rightarrow \cdots \rightarrow h_n\rightarrow h_1$ and $h_i\rightarrow h_{i+1}\rightarrow \cdots\rightarrow h_{j-1}\rightarrow h_j\rightarrow h_i$ A moment shows (and drawing a picture will help) that one of these has even length. Contradiction. 3. thanks for the reply wgunther, im kind of new to the world of graph theory, what do you mean by wlog?? and i can see quite clearly that this statement is obviously true and can show it with a diagram but how to show proof of it in mathematical terms is where i get abit lost. 4. Wlog means without loss of generality. When I say E sub H I mean the edge relation on the graph H. It's pretty simple to prove from here. Count the number of vertices in each. One has n+j-i and the other is i-j. They sum to n. n is odd, thus one of the two is even

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http://www.physicsforums.com/showthread.php?p=3810906

Physics Forums ## How to normalize wave functions in QFT? such as \lambda \phi 4 theory? In quantum mechanics, most wave functions are normalized with \int |\phi|^2 dx^3 =1. But I did not see any field in the quantum field theory is normalized. I understand they maybe just plain waves and does not need to be normalized. But in some cases, if we do not expand the field as plain wave, how to normalize them? For instance, in the \lambda \phi 4 theory, the field \phi has the dimension of GeV. Should we use \int |\phi|^4 dx^4 =1 instead of \int |\phi|^2 dx^3 =1 or \int |\phi|^4 dx^3 =1? 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 The wave function in quantum mechanics represents the state of the system. The field in quantum field theory is not the state, so why do you expect that it be normalized? On the other hand the states in QFT are assumed normalized. If we want to solve the \lambda \phi 4 equation in the same way as solve the Dirac equation, I think we should normalize the result. So we can say that the trivial expresion of <initial | final> = <\phi|\phi>=1 means, when no perturbation, the amplitude from \phi to its own is 1. For instance, the free particle solution of the Dirac equation is normalized to 1 by a factor \sqrt{m/E}. Blog Entries: 19 Recognitions: Science Advisor ## How to normalize wave functions in QFT? such as \lambda \phi 4 theory? Normalization of a field is needed only when that field ITSELF is interpreted as a probability density amplitude. The field \phi in \phi^4 is usually not interpreted in that way. Recognitions: Science Advisor In QFT the normalization of the field operators are defined via the equal-time commutation relations, $[\phi(t,\vec{x}),\Pi(t,\vec{y})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y}).$ Here $\Pi$ is the canonical field momentum for $\phi$. In $\phi^4$ theory, it's $\Pi(x)=\frac{\partial \mathcal{L}}{\partial \dot{\phi}(x)}=\dot{\phi}(x).$ For the asymptotically free states, symbolized by external legs in Feynman diagrams, the states are to be normalized in the usual way to $\delta$ distributions (supposed you have taken account of wave-function renormalization in the external legs, i.e., left out all self-energy insertions in them, see Weinberg, QT of Fields, vol. 1). Thread Tools | | | | |---------------------------------------------------------------------------------------------|---------------------------|---------| | Similar Threads for: How to normalize wave functions in QFT? such as \lambda \phi 4 theory? | | | | Thread | Forum | Replies | | | Advanced Physics Homework | 17 | | | Advanced Physics Homework | 5 | | | Advanced Physics Homework | 4 | | | Advanced Physics Homework | 8 |

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http://math.stackexchange.com/questions/tagged/closed-form+exponentiation

# Tagged Questions 0answers 37 views ### can´t solve this kind of exponential equations I have no clue how can i solve this kind of exponential equation in closed form: $a^n - b^n \le c$ where $a > 1$ and $-1 < b < 0$ thank you very much for your help 0answers 84 views ### How to solve these exponential equations for D? I'm curious if this is even possible to solve for D. D is the only variable, x, y, z, and w are all constants, and e is the mathematical constant e. (\frac{x+yD^{2}}{zD})^{\sqrt{2D}} = ... 3answers 324 views ### How do I find if $\frac{e^x}{x^3} = 2x + 1$ has an algebraic solution? Is there some way of solving $$\frac{e^x}{x^3} = 2x + 1$$ non-numerically? How would I go about proving if there exists a closed form solution? Similarly how would I go about proving if there exists ...

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http://mathhelpforum.com/trigonometry/78556-sine-rule.html

# Thread: 1. ## sine rule (Give non-exact answers to 3 significant figures.) In a triangle, the largest side has length 2 cm and one of the other sides has length √2 cm. Given that the area of the triangle is 1cm2, show that the triangle is right-angled and isosceles. Am not sure how to state off, can anyone just start me off . thanks. 2. Hi Using cross product properties we know that the area of the triangle is half the product of 2 lengths of the triangle multiplied by the absolute value of sine of the angle formed by the two sides $A = \frac12\:||\overrightarrow{AB}\wedge\overrightarro w{AC}|| = \frac12\:AB\:AC\:|sin(BAC)|$ With this formula you can compute the angle BAC, the length of the other side and conclude 3. $1 = \frac{1}{2} \times 2 \times \sqrt{2} \times sin\theta$ $\theta = 45$ where do I go from here? how do I show it is a right angled triangle and isosceles.? 4. Hello, Tweety In a triangle, the largest side has length 2 cm and one of the other sides has length √2 cm. Given that the area of the triangle is 1 cm², show that the triangle is right-angled and isosceles. Code: ``` * * * _ * * √2 * * * * * θ * * * * * * * * 2``` Formula: . $A \;=\;\frac{1}{2}\,bc\sin A$ The area is one-half the product of two sides and the sine of the included angle. We have: . $\frac{1}{2}\left(\sqrt{2}\right)(2)\sin\theta \:=\:1 \quad\Rightarrow\quad \sin\theta \:=\:\frac{1}{\sqrt{2}} \quad\Rightarrow\quad \theta \:=\:45^o$ Hence, the triangle looks like this: The right half is identical to the left half. Code: ``` * * | * _ * | * _ √2 * |1 * √2 * | * * 45° | 45° * * * * * * * * * * 1 1``` Therefore, the triangle is an isosceles right triangle.

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http://math.stackexchange.com/questions/67814/inverse-of-the-curl

# Inverse of the curl Given $\nabla \cdot u = 0$ and $w = \nabla \times u$ equation 9 of http://projecteuclid.org/euclid.cmp/1103941230 has the identity $u = - \nabla \times (\nabla^{-1} w)$. What is $\nabla^{-1}$ exactly and how was this identity derived? Thanks a lot. - 3 @QED: it would be helpful if you told us where you got this from... – Mariano Suárez-Alvarez♦ Sep 27 '11 at 2:39 QED is working through a 1984 article in Comm. Math. Phys. by Kato, Majda, and Beale, link under my answer. – Will Jagy Sep 27 '11 at 4:14 @Will: either your knowledge of the literature is simpy outstanding or you have great changes of getting the site's Mind-Reader of the Year Award. :) – Mariano Suárez-Alvarez♦ Sep 27 '11 at 4:21 Mariano, I am a pretty good mind-reader, but my first indication was the OP putting a comment with a link to the reference under my answer. – Will Jagy Sep 27 '11 at 4:29 Oh. Well, you'll have to wait a bit more for the award, then! – Mariano Suárez-Alvarez♦ Sep 27 '11 at 4:41 show 3 more comments ## 2 Answers The fact that $u$ is divergence free does mean that $u$ is the curl of something, locally at least. The fact that we have, for some $v,$ that $u = \nabla \times v$ amounts to little more than the fact that mixed partial derivatives commute, and in general is called Poincare's Lemma. Furthermore, one can replace any such $v$ by $v + \nabla \cdot f$ for some function $f.$ As in the comments, there is therefore little reason to talk about an operator $\nabla^{-1},$ such a thing is not going to be well defined. If you want to give an exact reference and convince us that a responsible person wrote the material you are quoting, things might be different. - 2 – user16697 Sep 27 '11 at 3:11 There is no such thing as $\nabla^{-1}$. What is true is that $\nabla \times (\nabla \times u) = \nabla(\nabla \cdot u) - \nabla^2 u$. - 2 QED is working his way through a 1984 article in Comm. Math. Phys. by Kato, Majda, and Beale, link under my answer. Yesterday's question was about equation (7), today equation (9) – Will Jagy Sep 27 '11 at 3:37 2 So presumably this was a misprint and they meant $(\nabla^2)^{-1} w$, i.e. some vector field $f$ such that $\nabla^2 f = w$, instead of $\nabla^{-1} w$. This is nonunique because you could add any vector field whose components are harmonic functions. – Robert Israel Sep 28 '11 at 22:21 @Robert, That could be it but if so how do I prove $u = - \nabla \times f$? I tried to show the divergence and curl of $- \nabla \times f$ match that of $u$ but could only do divergence.. and I'm not sure if both those are enough to prove it. – user16697 Sep 28 '11 at 22:47 – user16697 Sep 28 '11 at 22:59 1 – Hans Lundmark Sep 29 '11 at 8:57 show 1 more comment

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http://math.stackexchange.com/questions/tagged/partitions?page=5&sort=newest&pagesize=15

# Tagged Questions Questions related to the different ways of expressing an integer as a sum of integers; or, questions related to the subdivision of a set into smaller disjoint sets; questions related to the subdivision of an interval into smaller intervals that intersect only at the endpoints. 1answer 70 views ### HAKMEM 18(B): Cubic Partitions Taken from HAKMEM 18. Quoting... A partition of $N$ is a finite string of non-increasing integers that add up to $N$. Thus 7 3 3 2 1 1 1 is a partition of 18. Sometimes an infinite string of ... 2answers 248 views ### A question on partitions of n Let $P$ be the set of partitions of n. Let $\lambda$ denote the shape of a particular partition. Let $f_\lambda(i)$ be the frequency of $i$ in $\lambda$ and let \$a_\lambda(i) := \# \lbrace j : ... 3answers 396 views ### An Inequality Involving Bell Numbers: $B_n^2 \leq B_{n-1}B_{n+1}$ The following inequality came up while trying to resolve a conjecture about a certain class of partitions (the context is not particularly enlightening): $$B_n^2 \leq B_{n-1}B_{n+1}$$ for \$n \geq ... 2answers 292 views ### How can I reduce a number? I'm trying to work on a program and I think I've hit a math problem (if it's not, please let me know, sorry). Basically what I'm doing is taking a number and using a universe of numbers and I'm ... 1answer 296 views ### Number of solutions of Frobenius equation I have one problem which needs to count the number of solution of the equation $$2x+7y+11z=42$$ where $x,y,z \in \{0,1,2,3,4,5,\dots\}$. My attempt: I noticed that that maximum value of $z$ could ... 1answer 261 views ### Recurrence for the partition numbers I'm reading Analytic Combinatorics [PDF] book by Flajolet and Sedgewick, and I can't figure out one of the steps in the derivation of the $P_n$ — number of partitions of size $n$ (or coefficients in ... 2answers 200 views ### Two combinatorics problems. I'm not 100% confident in my answers These are two problems from my combinatorics assignment that I'm not quite confident in my answer. Am I thinking of these the right way? Problem 1: On rolling 16 dice. How many of the $6^{16}$ ... 2answers 298 views ### Learning about partitions and modular forms I'm interested in learning about partitions and modular forms. I already know algebra and analysis (complex and real). Can any one suggest me books or other materials from where I can learn these ... 0answers 121 views ### Jordan Measure, Semi-closed sets and Partitions From earlier question here. Consider $$\mu \left( (0,1) \cap \mathbb Q \right) = 0$$ where $$(0,1) = (0,\frac{1}{3})\cup [\frac{1}{3}, \frac{2}{3}] \cup (\frac{2}{3}, 1)$$ so \mu ((0,1) \cap ... 2answers 248 views ### Jordan Measures, Open sets, Closed sets and Semi-closed sets I cannot understand: $$\bar{\mu} ( \{Q \cap (0,1) \} ) = 1$$ and (cannot understand this one particularly) $$\underline{\mu} ( \{Q \cap (0,1) \} ) = 0$$ where $Q$ is rational numbers, why? I ... 1answer 744 views ### Ellipse 3-partition: same area and perimeter Inspired by the question, "How to partition area of an ellipse into odd number of regions?," I ask for a partition an ellipse into three convex pieces, each of which has the same area and the same ... 1answer 971 views ### The number of partitions of $n$ into distinct parts equals the number of partitions of $n$ into odd parts This seems to be a common result. I've been trying to follow the bijective proof of it, which can be found easily online, but the explanations go over my head. It would be wonderful if you could give ... 1answer 369 views ### Partition number problem Denote by $I_m=\{0,1,2,…m\}$, by $N_s=\{1,2,…,s\}$ , by $\overline s$ least common multiple of elements of set $N_s$ and by $p(k,N_s)$ the number of partitions of natural number $k$ in parts used ... 1answer 152 views ### Number of ways to sum square numbers to yield a given number I would like to know how many choices of $x_i$ there are such that $$\sum_{i=1}^{n}x_i^2=m$$ where $n$, $m$ are given. The $x_i$ can be any nonnegative integer and need not be unique and the order is ... 2answers 438 views ### Partition Problem, verifying solution in polynomial time I add a look at the partition problem, this problem is know as the Easiest hard problem since it is NP-complete and seems pretty easy. From wikipedia on NP-complete: In computational complexity ... 1answer 712 views ### Hardy Ramanujan Asymptotic Formula for the Partition Number I am needing to use the asymptotic formula for the partition number, $p(n)$ (see here for details about partitions). The asymptotic formula always seems to be written as, \$ p(n) \sim ... 1answer 93 views ### Something basic in “l-adic properties of the partition function” paper I am trying to understand the basic result in this paper: http://www.aimath.org/news/partition/folsom-kent-ono.pdf My problem is with the example at the end of page 2. I understand it's supposed to ... 1answer 198 views ### Closed-form Expression of the Partition Function $p(n)$ I feel like I have seen news that a paper was recently published, at most a few months ago, that solved the well-known problem of finding a closed-form expression for the partition function $p(n)$ ... 2answers 81 views ### List all 3 part compositions of 5 I am looking at a past exam written by a student. There was a question I believed he got correct but received only 1/4. The marker wrote down "4 more compositions, order matters". This is the ... 2answers 150 views ### Trouble understanding proof that the unit interval cannot be partitioned in a certain way From the book "Putnam and Beyond." The problem: Show that the interval [0, 1] cannot be partitioned into two disjoint sets A and B such that B = A + a for some real number a. Proof: Assume ... 0answers 45 views ### Prove that single machine can reduce from 2-partition problem I'm given the following problem: Prove that single machine \sum U_i (number of tardy job) with release date constraints problem can reduce from 2-partition problem ... 2answers 97 views ### Generating sequences of numeric partitions Definition: A tuple $\lambda = (\lambda_1, \ldots, \lambda_k)$ of natural numbers is called a numeric partition of $n$ if $1 \leq \lambda_1 \leq \cdots \leq \lambda_k$ and \$\lambda_1 + \cdots + ... 1answer 293 views ### identity proof for partitions of natural numbers Definition: A tuple $\lambda = (\lambda_1, \cdots, \lambda_k)$ of Natural Numbers is called a numeric partition of n if $1 \leq \lambda_1 \leq \cdots \leq \lambda_k$ and \$\lambda_1 + \cdots + ... 1answer 295 views ### How to find the coefficient of a term in this expression How to determine the coefficient of z3q100 in I stumbled upon this problem while trying to solve this type of partition problem: Find the number of integer solutions to x + y + z = 100 such that 3 ... 1answer 326 views ### Number of permutations with a given partition of cycle sizes Part of my overly complicated attempt at the Google CodeJam GoroSort problem involved computing the number of permutations with a given partition of cycle sizes. Or equivalently, the probability of a ... 1answer 153 views ### Why is there a derivative in this formula? This is a very simple question. Why is Rademacher's formula presented with d/dx in it? Why not just "do" the derivative? Then replace x with n? Is it so there is only one transcendental ... 1answer 427 views ### Feeding real or even complex numbers to the integer partition function $p(n)$? Like most people, when I first encountered $n!$ in grade school, I graphed it, then connected the dots with a smooth curve and reasoned that there must be some meaning to $\left(\frac43\right)!$ — ... 1answer 131 views ### What is the standard way of writing this partition function? [closed] What is the standard way of writing the number of partitions of n where each number is less than or equal to k. Is it $p(n,k)$ or $p(k,n)$? So far in my reading, have seen both. 3answers 261 views ### Partition an integer $n$ by limitation on size of the partition According to my previous question, is there any idea about how I can count those decompositions with exactly $i$ members? for example there are $\lfloor \frac{n}{2} \rfloor$ for decompositions of $n$ ... 3answers 166 views ### Decomposition by subtraction In how many ways one can decompose an integer $n$ to smaller integers at least 3? for example 13 has the following decompositions: \begin{gather*} 13\\ 3,10\\ 4,9\\ 5,8\\ 6,7\\ 3,3,7\\ 3,4,6\\ ... 1answer 143 views ### Seeking some details about what is denoted by the partition function $P(n,k)$ Quoting from Wolfram MathWorld, "$P(n,k)$ denotes the number of ways of writing $n$ as a sum of exactly $k$ terms or, equivalently, the number of partitions into parts of which the largest is exactly ... 2answers 180 views ### Asymptotics for partitions of $n$ with largest part at most $k$ (or into at most $k$ parts) Let $\bar p_k(n)$ be the number of partitions of $n$ with largest part at most $k$ (equivalently, into at most $k$ parts). Is there an elementary formula for the asymptotic behavior of $\bar p_k(n)$ ... 1answer 175 views ### Seeking a textbook proof of a formula for the number of set partitions whose parts induce a given integer partition Let $t \geq 1$ and $\pi$ be an integer partition of $t$. Then the number of set partitions $Q$ of $\{1,2,\ldots,t\}$ for which the multiset $\{|q|:q \in Q\}=\pi$ is given by \[\frac{t!}{\prod_{i \geq ... 1answer 88 views ### Validity of a q-series theorem Define the $q$-analog $(a;q)_n = \prod_{k=0}^n \left(1 - aq^k\right)$. I want to prove the identity $\frac{(q^2;q^2)_\infty}{(q;q)_\infty}=\frac{1}{(q;q^2)_\infty}$. I viewed the LHS this way: ... 2answers 263 views ### Natural set to express any natural number as sum of two in the set Any natural number can be expressed as the sum of three triangular numbers, or as four square numbers. The natural analog for expressing numbers as the sum of two others would apparently be the sum ... 1answer 179 views ### What is the number of ways to choose x groups from y items? (partitions with x cells of a multiset) Where a group can consist of 1 or more items, groups don't have to be equally sized and items can be duplicates. Example - Choose 3 groups: Items: 1 2 2 3 Groups: (1) (2 2) (3) (1 2) (2) (3) (3 ... 2answers 233 views ### Graph coloring problem (possibly related to partitions) Given an undirected graph I'd like to color each node either black or red such that at most half of every node's neighbors have the same color as the node itself. As a first step, I'd like to show ... 1answer 436 views ### Upper bound on integer partitions of n into k parts Recent news piqued my interest in integer partitions again. I'm working my way back through an old text and I'm completely hung up on this problem: Recall that $p_k(n)$ is the number of partitions ... 1answer 46 views ### Notation for “duplicating” partitions I'm using Macdonald's "Symmetric Functions and Hall Polynomials" as a reference and did not find what I was looking for -- apologies if I only missed it. As an example, let us consider the partition ... 1answer 772 views ### On problems of coins totaling to a given amount I don't know the proper terms to type into Google, so please pardon me for asking here first. While jingling around a few coins, I realized that one nice puzzle might be to figure out which $n$ or so ... 0answers 173 views ### Visualizing the Partition numbers (suggestions for visualization techniques) So Ken Ono says that the partition numbers behave like fractals, in which case I'd like to try to find an appropriately illuminating way of visualizing them. But I'm sort of stuck at the moment, so ... 4answers 2k views ### Algorithm for generating integer partitions I'm looking for a fast algorithm for generating all the partitions of an integer up to a certain maximum length; ideally, I don't want to have to generate all of them and then discard the ones that ... 6answers 2k views ### Making Change for a Dollar (and other number partitioning problems) I was trying to solve a problem similar to the "how many ways are there to make change for a dollar" problem. I ran across a site that said I could use a generating function similar to the one quoted ... 1answer 108 views ### Number of distributions leaving none of $n$ cells empty The solution for the number of distributions leaving none of the $n$ cells empty (with unlike cells and $r$ unlike objects) is given by ... 3answers 284 views ### “Converting” equivalence relations to partitions There is a direct relationship between equivalence relations and partitions. Is there a way to simply use an equivalence relation's definition to get the matching partition? And what about the other ... 1answer 323 views ### Matrix representation of a partition Is there a natural way to represent all the partitions of an integer set {1,2,3,...,n} as a matrix in the similar way permutations can be mapped to group of matrices?

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http://mathoverflow.net/questions/104134?sort=votes

## Is a colimit of fibration still a fibration? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $\mathcal{C}$ a category. Let $Fib(\mathcal{C})$ the category of fibrations (on $\mathcal{C}$) with morphisms the cartesian functors $T: (\mathcal{A}, P)\to (\mathcal{B}, Q)$ i.e. $T: \mathcal{A}\to \mathcal{B}$ with $Q\circ T=P$ (where $P: \mathcal{A}\to \mathcal{C}$, $P: \mathcal{B}\to \mathcal{C}$ are fibrations on $\mathcal{C}$). From literature (es. "Categorical logic and Type THeory" B.Jacobs) the pullbak of two fibration (i.e. the product in $Fib(\mathcal{C})$) is still a fibration, and this has a easy generalization to a multi-pullback i.e. a the inclusion $Fib(\mathcal{C})\subset CAT\downarrow \mathcal{C}$ create (small) product, the some is true for the kernel of a couple, then the inclusion above create all (small) limits. THe some is easly true for coproducts. I ask: do the inclusion $Fib(\mathcal{C})\subset CAT\downarrow \mathcal{C}$ create also cokernels? Counterexamples?. Observation $Fib(\mathcal{C})$ is equivalent to the category of pseudo-functors $P: \mathcal{C}^{op}\to CAT$ with pseudo-transformations as morphisms, and the punctual limits or colimits of pseudofunctors is still a pseudofuntors (I seems yes, anyway this is true for funtors and natural tranformations considering the category $sFib(\mathcal{C})$ of fibrations with split clevages and clevage preserving functors ), then $Fib(\mathcal{C})$ (or at least $sFib(\mathcal{C})$) is complete and cocomplete. - ## 1 Answer $\mathrm{Fib}(\mathcal{C})$ is both monadic and comonadic over $\mathrm{Cat}/\mathcal{C}$, in the pseudo sense. That is, there are a monad $T$ and a comonad $G$ on $\mathrm{Cat}/\mathcal{C}$ such that $\mathrm{Fib}(\mathcal{C})$ is equivalent to the 2-category of strict $T$-algebras and pseudo $T$-morphisms, and also to the 2-category of strict $G$-coalgebras and pseudo $G$-morphisms. By definition $T$ is the span-composite with $\mathcal{C} \leftarrow \mathcal{C}^{\mathbf{2}} \to \mathcal{C}$, while $G$ is its right adjoint. Therefore, by a theorem of Blackwell-Kelly-Power and its dual, the forgetful functor creates strict PIE-limits and strict PIE-colimits, hence also (weak) 2-limits and 2-colimits. - Thank you very much. I guess you first answere is from the R. Street article on Sydney category seminar (LNm 420). WHere I can find references about Blackwell-Kelly-Power theorem? – Buschi Sergio Aug 6 at 21:13 The first observation is made in multiple places, another one is Street's "Fibrations in bicategories". The Blackwell-Kelly-Power paper is "Two-dimensional monad theory". – Mike Shulman Aug 7 at 6:35

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http://math.stackexchange.com/questions/76594/throwing-dice-a-randomly-determined-number-of-times

# Throwing dice a randomly determined number of times Let $X$ be the number that shows up when rolling a die. Now throw another dice $X$ times $(Y_1, ..., Y_X)$ and calculate the sum $Z = \sum_{k=1}^X Y_k$. What kind of Stochastic Process is this? How do you calculate mean value and variance of $Z$? - ## 3 Answers Look up Wald's Lemma. This lemma is applicable here as $X$ and $Y_k$ are independent. Variance also can be found using this lemma and using the fact that $Z^2=\sum_{k=1}^{X}Y_k^2+\sum_{i<j}Y_i Y_j$ where we have $X(X-1)/2$ terms in the second sum. - Does this help find the variance? – opt Oct 28 '11 at 6:42 Yes, so long as $X$ and $Y_k$ are independent which is actually the case here. – Ashok Oct 28 '11 at 7:49 @Ashok Thank you for the answer. However, I wonder why the Wikipedia article does not mention that independence of X and Y_k is necessary. – artistoex Oct 28 '11 at 15:26 Condition (3) is essentially saying about the amount of dependence is allowed. See the discussion of assumptions below the statement in the Wikipedia page. – Ashok Oct 29 '11 at 3:03 It is a compound probability distribution. You can marginalize by taking the average of the six pmfs, giving a distribution that looks like $p(1) = 1/6^2,\dots, p(36) = 1/6^7$ and then you can find the variance of this finite marginal distribution directly. For the mean you can just take the mean of the six distribution means. - Your random variable is also a branching process $(Z_n)$ observed at time $n=2$, where the offspring distribution is given by a single roll of the die. You want to substitute $n=2$ below, where $\mu$ and $\sigma^2$ are the mean and variance for a single die roll $$\mathbb{E}(Z_n)=\mu^n\quad\mbox{ and }\quad\mathbb{V}(Z_n)={\sigma^2\mu^n(\mu^n-1)\over\mu^2-\mu}.$$ - Glad to know this interpretation! – Ashok Oct 28 '11 at 13:46

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http://www.mpi-inf.mpg.de/~fietzke/benford.html

# An illustration of Benford's Law Benford's law, also known as the Newcomb-Benford law or the first digit law, states that in sets of numbers from many real-life sources of data, the first digits follow a logarithmic distribution. More specifically, the probability that the first digit is $$n$$ is given by $$\log_{10}\left( 1 + \frac{1}{n} \right).$$ Such distributions can arise for different reasons, including exponential growth processes, or as properties of finite sets of rational numbers representing results of physical measurements. Here we focus on exponential growth. In the application below, you can define a sequence of numbers by giving initial value(s) and a rule. The diagram on the left shows the distribution of first digits in the sequence. The Benford distribution is shown in the background. The diagram on the right shows a plot of the sequence. The dark areas correspond to numbers starting with 1. Click Run to see the distribution of first digits evolve. Observe how, as the exponential curve gets steeper and steeper, it spends more time in lower first-digit areas than in higher first-digit areas, at any order of magnitude. Some sequences to try: • Simple exponential growth: $$f(0)=1, f(n)=f(n-1) * 1.1$$ • Fibonacci numbers: $$f(0)=0, f(1)=1, f(n)=f(n-1)+f(n-2)$$ • Linear growth: $$f(0)=0, f(n)=f(n-1) + 1$$ Linearly growing sequences don't follow Benford's law, but lower first digits are still more probable than higher first digits "most of the time". This explains why lower first digits can also be more frequent in data like street numbers. ## Try it out! Initial value(s): $$f(0), f(1), \dots$$ Rule: $$f(n) =$$ $$f($$$$)=$$ First-digit distribution 50% 100% 1 2 3 4 5 6 7 8 9 Plot y scale: 1.0 Author: Arnaud Fietzke

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http://www.mashpedia.com/Partition_of_a_set

# Partition of a set Language A partition of a set into 6 parts: an Euler diagram representation The 52 partitions of a set with 5 elements The traditional Japanese symbols for the chapters of the Tale of Genji are based on the 52 ways of partitioning five elements. In mathematics, a partition of a set X is a division of X into non-overlapping and non-empty "parts" or "blocks" or "cells" that cover all of X. More formally, these "cells" are both collectively exhaustive and mutually exclusive with respect to the set being partitioned. ## Definition A partition of a set X is a set of nonempty subsets of X such that every element x in X is in exactly one of these subsets. Equivalently, a set P is a partition of X if and only if all of the following conditions hold: 1. It does not contain the empty set. 2. The union of the elements of P is equal to X. (The elements of P are said to cover X.) 3. The intersection of any two distinct elements of P is empty. (We say the elements of P are pairwise disjoint.) In mathematical notation, these conditions can be represented as 1. $\varnothing \notin P$ 2. $\bigcup P = X$ 3. $(A \in P \and B\in P \and A \neq B) \Rightarrow A \cap B = \varnothing$ where $\varnothing$ is the empty set. The elements of P are called the blocks, parts or cells of the partition.[1] The rank of P is |X| − |P|, if X is finite. ## Examples • Every singleton set {x} has exactly one partition, namely { {x} }. • For any nonempty set X, P = {X} is a partition of X, called the trivial partition. • For any non-empty proper subset A of a set U, the set A together with its complement form a partition of U, namely, {A, U−A}. • The set { 1, 2, 3 } has these five partitions: • { {1}, {2}, {3} }, sometimes written 1|2|3. • { {1, 2}, {3} }, or 12|3. • { {1, 3}, {2} }, or 13|2. • { {1}, {2, 3} }, or 1|23. • { {1, 2, 3} }, or 123 (in contexts where there will be no confusion with the number). • The following are not partitions of { 1, 2, 3 }: • { {}, {1, 3}, {2} } is not a partition (of any set) because one of its elements is the empty set. • { {1, 2}, {2, 3} } is not a partition (of any set) because the element 2 is contained in more than one block. • { {1}, {2} } is not a partition of {1, 2, 3} because none of its blocks contains 3; however, it is a partition of {1, 2}. ## Partitions and equivalence relations For any equivalence relation on a set X, the set of its equivalence classes is a partition of X. Conversely, from any partition P of X, we can define an equivalence relation on X by setting x ~ y precisely when x and y are in the same part in P. Thus the notions of equivalence relation and partition are essentially equivalent.[2] The axiom of choice guarantees for any partition of a set X the existence of a subset of X containing exactly one element from each part of the partition. This implies that given an equivalence relation on a set one can select a canonical representative element from every equivalence class. ## Refinement of partitions Partitions of a 4-set ordered by refinement A partition α of a set X is a refinement of a partition ρ of X—and we say that α is finer than ρ and that ρ is coarser than α—if every element of α is a subset of some element of ρ. Informally, this means that α is a further fragmentation of ρ. In that case, it is written that α ≤ ρ. This finer-than relation on the set of partitions of X is a partial order (so the notation "≤" is appropriate). Each set of elements has a least upper bound and a greatest lower bound, so that it forms a lattice, and more specifically (for partitions of a finite set) it is a geometric lattice.[3] The partition lattice of a 4-element set has 15 elements and is depicted in the Hasse diagram on the left. Based on the cryptomorphism between geometric lattices and matroids, this lattice of partitions of a finite set corresponds to a matroid in which the base set of the matroid consists of the atoms of the lattice, the partitions with $n-2$ singleton sets and one two-element set. These atomic partitions correspond one-for-one with the edges of a complete graph. The matroid closure of a set of atomic partitions is the finest common coarsening of them all; in graph-theoretic terms, it is the partition of the vertices of the complete graph into the connected components of the subgraph formed by the given set of edges. In this way, the lattice of partitions corresponds to the graphic matroid of the complete graph. Another example illustrates the refining of partitions from the perspective of equivalence relations. If D is the set of cards in a standard 52-card deck, the same-color-as relation on D – which can be denoted ~C – has two equivalence classes: the sets {red cards} and {black cards}. The 2-part partition corresponding to ~C has a refinement that yields the same-suit-as relation ~S, which has the four equivalence classes {spades}, {diamonds}, {hearts}, and {clubs}. ## Noncrossing partitions A partition of the set N = {1, 2, ..., n} with corresponding equivalence relation ~ is noncrossing provided that for any two 'cells' C1 and C2, either all the elements in C1 are < than all the elements in C2 or they are all > than all the elements in C2. In other words: given distinct numbers a, b, c in N, with a < b < c, if a ~ c (they both are in a cell called C), it follows that also a ~ b and b ~ c, that is b is also in C. The lattice of noncrossing partitions of a finite set has recently taken on importance because of its role in free probability theory. These form a subset of the lattice of all partitions, but not a sublattice, since the join operations of the two lattices do not agree. ## Counting partitions The total number of partitions of an n-element set is the Bell number Bn. The first several Bell numbers are B0 = 1, B1 = 1, B2 = 2, B3 = 5, B4 = 15, B5 = 52, and B6 = 203. Bell numbers satisfy the recursion $B_{n+1}=\sum_{k=0}^n {n\choose k}B_k$ and have the exponential generating function $\sum_{n=0}^\infty\frac{B_n}{n!}z^n=e^{e^z-1}.$ The number of partitions of an n-element set into exactly k nonempty parts is the Stirling number of the second kind S(n, k). The number of noncrossing partitions of an n-element set is the Catalan number Cn, given by $C_n={1 \over n+1}{2n \choose n}.$ ## Notes 1. Brualdi, pp. 44–45 2. Schechter, p. 54 3. Birkhoff, Garrett (1995), Lattice Theory, Colloquium Publications 25 (3rd ed.), American Mathematical Society, p. 95, ISBN 9780821810255 . ## References • Brualdi, Richard A. (2004). Introductory Combinatorics (4th edition ed.). Pearson Prentice Hall. ISBN 0-13-100119-1. • Schechter, Eric (1997). Handbook of Analysis and Its Foundations. Academic Press. 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http://math.stackexchange.com/questions/194236/problem-with-a-circumference/194243

# Problem with a circumference I have the following equation for a circumference: $$9 X^2 + 25 Y^2 - 36 X - 50 Y = 154.$$ So far I only used this general equation: $X^2 + Y^2 + A X + B Y + C = 0$, but now $X^2$ and $Y^2$ are not alone and are being multiplied by an integer. I would like to know how to find out the center and the radius of this equation - What is your question? – Clive Newstead Sep 11 '12 at 16:59 Get the center and the radius of this equation – Pacha Sep 11 '12 at 16:59 This is an ellipse, not a circle, so does not have a 'radius'; but it does have major and minor axes. I'll elaborate in an answer. – Clive Newstead Sep 11 '12 at 17:01 you might be right! – Pacha Sep 11 '12 at 17:16 ## 2 Answers First of all, you ask for the center and radius, but you have an ellipse and not a circle. You can talk about a center but not really a radius. If what you want is to know the general shape of what you have, here is what you do. Start by completing the square for both $X$ and $Y$. This will give you something of the form $$9(X - H)^2 + 25(Y - K)^2 = S$$ Now, divide both sides by $S$ to get something of the form $$\frac{(X - H)^2}{A^2} + \frac{(Y - K)^2}{B^2} = 1$$ This is a general form of the equation of an ellipse with center $(H, K)$ (not the most general form, by the way). Then $A$ represents the distance from the center to the edge of the ellipse, if traveling only left or right. And $B$ represents the distance from the center to the edge if traveling up or down. So, the $A$ and $B$ represent something similar to a radius, but not a real radius. - Your equation is that of an ellipse, not a circle. However, if you substitute $x=3X$ and $y=5Y$, then your equation describes a circle in the $(x,y)$-plane. You can then find the properties (centre, major and minor axes) of the ellipse in the $(X,Y)$-plane by considering the properties of the circle, and applying graph transformations $(x,y) \mapsto (X,Y)$. For instance, if the centre of the circle in the $(x,y)$-plane is $(a,b)$, then the centre of the ellipse in the $(X,Y)$-plane is $(\frac{a}{3}, \frac{b}{5})$, since $X=\dfrac{x}{3}$ and $Y=\dfrac{y}{5}$. You can do something similar to find the major and minor axes if the ellipse. -

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http://scicomp.stackexchange.com/questions/1909/fem-singularity-of-the-stiffness-matrix

# FEM: singularity of the stiffness matrix I'm solving the differential equation $$\left( \sigma^{2}(x) u ''(x) \right)'' = f(x), \;\;\; 0 \leqslant x \leqslant 1$$ with initial conditions $u(0) = u(1) = 0$, $u''(0) = u''(1) = 0$. Here $\sigma(x) \geqslant \sigma_{0} > 0$ is parameter. In operator form we can rewrite the differential equation as $Au = f$, where operator $A$ is positive definite. Following FEM scheme, I reduce my problem to an optimisation problem $$J(u) = (Au,u) - 2(f,u) \to \min_{u}$$ I introduce finite elements $h_{k}(x)$ as $$v_{k}(x) = \left\{ \begin{array}{rl} 1 - \left( \frac{x-x_{k}}{h} \right)^2, & x \in [x_{k-1},x_{k+1}] \\ 0, & \text{otherwise} \end{array} \right.$$ for any $k = 1,\ldots,n-1$, where $x_{k} = hk$, $h = \frac{1}{n}$. Finite elements $v_{0}(x)$ and $v_{n}(x)$ are introduced similarly. I try to find numericaly the vector $\alpha$ such that $u(x) = \sum_{k=0}^{n} \alpha_{k} v_{k}(x)$ solves the optimisation problem. We have $$J(u) = \sum\limits_{i=0}^{n} \sum\limits_{j=0}^{n} \alpha_{i} \alpha_{j} (Av_{i},v_{j}) - \sum\limits_{i=0}^{n} 2\alpha_{i} (v_{i},f) = \alpha^{T} V \alpha - 2\alpha^{T} b \to \min\limits_{\alpha},$$ where $b_{i} = (f,v_{i})$ and $V_{i,j} = (Av_{i},v_{j})$. After differentiation with respect to $\alpha$ I receive $$V\alpha = b,$$ but here the stiffness matrix $V$ is singular. So what I have to do? Maybe I have to choose other finite elements? - Hi, Nimza, do you have a test problem that you know the exact solution? If yes, try solving $V^T V \alpha = V^T b$ first to test if your basis is correct inside the domain, if everything looks correct, then maybe it is the incorrectly posed BC makes the matrix singular. But the BC seems OK to me though. – Shuhao Cao Apr 12 '12 at 14:49 ## 2 Answers In decreasing order of likelihood 1. Incorrect basis. From you description, it appears that you have exactly two quadratic functions with support on each element. That space is not a partition of unity and is not $C^1$ (continuous first derivatives). To discretize your fourth order problem directly (instead of reducing it to a system of second order equations, for example), you will need a $C^1$ basis. Note that the $C^1$ basis should be able to exactly reproduce all linear functions. 2. Insufficient boundary conditions. This will be blatantly obvious if you compute and plot the null space. 3. Incorrect assembly. Check the map from elements to assembled ordering to confirm that it is what you expected, for example that it isn't reversing the orientation of elements. 4. Incorrect local assembly. In 1D, you can analytically compute what the element stiffness matrix looks like (perhaps for a simplified case) and check that the code reproduces it. - Thank you. 1. I think that I will need a $C^2$ basis because $(Au,v) = \int_{0}^{1} \sigma^{2}(x) u''(x) v''(x) dx$. Then, if I consider only functions that satisfy boundary conditions then $\ker A = \{ 0 \}$. – Nimza Apr 11 '12 at 21:11 1 – Jed Brown Apr 11 '12 at 23:25 Ok. I corrected my basis: now $v_{k}(x) = \cos^2 \left(\frac{\pi}{2h}(x-x_{i}) \right)$ on $[x_{i-1},x_{i+1}]$ and $i = 1,\ldots,n-1$. Now it is $C^1$. But method still doesn't work. – Nimza Apr 12 '12 at 8:58 The $C^1$ basis should be able to reproduce linear functions, but this cannot. Once you fix that, check that integrals are being performed correctly, then check boundary conditions. – Jed Brown Apr 12 '12 at 13:30 Clearly the problem has an ODD order derivative. More specifically for larger Péclet numbers, the stiffness matrix might not maintain 'fine' shape, which creates zeros during assembly and hence gets singular or sometimes very small determinant that are noticeable by the oscillations in solution plot. The solution to these kind of problem is the use of penalty, among other methods. More specifically this is called Petrov-Galerkin method. Sorry for my bad English comprehension. -

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http://mathhelpforum.com/number-theory/9183-ph-others-so-inclined.html

# Thread: 1. ## for PH and others so inclined "Show that $\frac{x^{5}}{5}+\frac{x^{3}}{3}+\frac{7x}{15}$ is always an integer for integral values of x". 2. It is relatively easy (but messy) to show that for each positive integer N $3N^5 + 5N^3 + 7N$ is a multiple of 15. an other 3. That was a fast response, Plato . I noticed, since the semesters have come to a close, there hasn't been much posting. Anyway, maybe I am off base, but $\frac{x^{5}}{5}+\frac{x^{3}}{3}+\frac{7x}{15}=\fra c{x^{5}-5}{5}+\frac{x^{3}-3}{3}+x$ Now, use Fermat's theorem: $x^{5}-x\equiv{0} \;\ mod(5)$ and $x^{3}-x\equiv{0} \;\ mod (3)$ Therefore, it's an integer. Seems rather logical. But, I am not that much of a number theorist. Have been learning it, though. 4. Originally Posted by Plato It is relatively easy (but messy) to show that for each positive integer N $3N^5 + 5N^3 + 7N$ is a multiple of 15. an other Hello Plato, use mathematical induction to prove this property: Prove that $3n^5+5n^3+7n=15 \cdot k\ , \ k \in \mathbb{Z}$ $step\ 1:\ n=1\ :\ 3+5+7=15=15 \cdot 1 \text{ is true}$ $step \ 2:\ assumption: 3n^5+5n^3+7n=15 \cdot k \text{ is true}$ $step\ 3:\ conclusion: 3(n+1)^5+5(n+1)^3+7(n+1)=$ $3n^5+15n^4+30n^3+30n^2+15n+3+5n^3+15n^2+15n+5+7n+7 =$ $\underbrace{3n^5+5n^3+7n}_{k\cdot 15}+15(n^4+2n^3+3n^2+2n+1)+3+5+7=$ The 1rst summand is a multiple of 15 according assumption, the 2nd summand is a multiple of 15 because it contains the factor 15 and the last 3 summands add up to 15. Thus $3(n+1)^5+5(n+1)^3+7(n+1)=$ is multiple of 15 and therefore the asumption is true for all $n \in \mathbb{N}$ EB Merry Christmas and a happy New Year.

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http://mathoverflow.net/questions/25512/determining-the-field-from-the-topology-of-the-affine-space-over-it

## Determining the field from the topology of the affine space over it. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The inspiration is from another question in which it is remarked that the passage to Zariski topology is a very interesting functor from Commutative Rings to Topological spaces. I am trying to understand how "faithful" this functor is. Any two fields have the same topology. But, with polynomial rings, we have more hope. So. Let $K$, $L$ be two fields. Then if $Spec\ K[X]$ and $Spec\ L[X]$ are homeomorphic, does it follow that $K$ and $L$ are isomorphic? Similarly: Let $K$, $L$ be two fields. Then if $Spec\ K[X_1, \ldots , X_n]$ and $Spec\ L[X_1, \ldots , X_n]$ are homeomorphic, does it follow that $K$ and $L$ are isomorphic? - 1 The affine line over a field carries the cofinite topology, so its homeomorphism type depends only on its cardinality. – Qiaochu Yuan May 21 2010 at 17:52 Ahh, posted as an answer as you posted your comment! – Charles Siegel May 21 2010 at 17:53 Surely the answer is no (by which I mean I do not have a proof). At the very least, if $K$ is the algebraic closure of $\mathbb{Q}$ and $L$ is the algebraic closure of $\mathbb{Q}(t)$, then $\operatorname{Spec} K[t_1,\ldots,t_n]$ and $\operatorname{Spec} L[t_1,\ldots,t_n]$ should be homeomorphic for all $n$. It seems reasonable to conjecture that the only invariant of $K$ that these spectra see is the cardinality of $K$. – Pete L. Clark May 21 2010 at 17:59 ## 2 Answers The following paper of Hrushovski-Zilber shows that if we restrict our attention to algebraically closed fields $F$, then $F$ is uniquely determined up to isomorphism by its "Zariski geometry". Presumably an examination of the proof will show that an algebraically closed field $F$ is determined by $Spec(F[x_{1}, \cdots F[x_{n}])$ for sufficiently large $n$. Hrushovski, Ehud; Zilber, Boris (1996). "Zariski Geometries". Journal of the American Mathematical Society 9: 1–56. Another "reference": http://en.wikipedia.org/wiki/Zariski_geometry - Interesting. The relevant result seems to be Proposition 1.1. If you take $C$ to be the affine line over an algebraically closed field $F$ and $C'$ to be the affine line over an algebraically closed field $F'$, then the result says that an isomorphism of Zariski geometries from $C$ to $C'$ induces an isomorphism from $F$ to $F'$. It also says that a morphism of Zariski geometries is an isomorphism if it induces a homeomorphism on $C^n$ for all $n$.... – Pete L. Clark May 21 2010 at 20:17 ...But what I wasn't immediately able to see was whether a morphism of Z.G.'s from $C$ to $C'$ involves more data than just a collection of such homeomorphisms -- e.g. whether various geometric compatibility conditions need to be satisfied. I would be very interested to see these details worked out. – Pete L. Clark May 21 2010 at 20:19 The experts in this area (such as Dave Marker) would know all about this. Hopefully one of them will stumble across this question and save us the trouble of working out the details ourselves! – Simon Thomas May 21 2010 at 20:32 In this type of argument you usually reconstruct the field from a 2-dimensional family of curves on $C\times C$. So probably you can get by knowing that the homeomorphism between $C$ and $C_1$ lifts to homeomorphisms of the Zariski topologies of $C^n$ and $C_1^n$ for $n=2,3,4$. – Dave Marker May 24 2010 at 6:56 1 A simple version of this idea would be if $K$ and $L$ are algebraically closed fields and we had a homeomorphism between $P_K^2\timesP_K^2$' and $P_L^2\times P_L^2$. Let $V\subset P^2_K \times P^2_K$ be the incidence variety for lines in $P_2$. Desargues ideas allow us to reconstruct $K$ from $V$. A homeomorphism would give us an isomorphic variety in $P_L^2\times P_L^2$. This would allow us to interpret $K$ in $L$. But by a result of Poizat then $K$ and $L$ would be isomorphic. – Dave Marker May 24 2010 at 12:53 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This is false. Let $K=\bar{\mathbb{Q}}$ and $L=\bar{\mathbb{F}}_2$. These are clearly both algebraically closed, of different characteristics, so $K\not\cong L$. However, if we ONLY look at the topology, $\mathrm{Spec}(K[x])$ and $\mathrm{Spec}(L[x])$ will be be countable sets with the finite complement topology on the closed points, with a single generic point, so they're homeomorphic. For algebraically closed fields, the TOPOLOGY on the affine line over the field is determined by the cardinality. For higher dimensions, it's less clear to me, because you might be able to recover characteristic (I'm a char 0 kind of person, so I don't know) from how the various curves/hypersurfaces sit inside it. - @CS: I don't know for sure, but I am skeptical that you can recover the characteristic of an infinite field from the topologies on affine spaces over it. – Pete L. Clark May 21 2010 at 18:01 I'm also skeptical, but my positive characteristic intuition is dismal enough that I'm not willing to come out and say it. – Charles Siegel May 21 2010 at 18:11 @Charles: um, well, in that case, maybe you should...never mind, I'm sure we'll turn out to be right. :) – Pete L. Clark May 21 2010 at 18:56

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http://crypto.stackexchange.com/questions/1881/realize-a-mac-using-a-pseudo-random-function/1883

Realize a MAC using a Pseudo-random function? Given a pseudo-random function and assuming that we do not have any other tools, How can we construct a MAC? I believe this can be done. Would like to know if there is more than one way of doing this. - 3 Answers If you have a PRF (with larger input than output), you can use it as compression function in a Merkle-Damgård structure, yielding a hash function which you can subsequently turn into a MAC with HMAC. Indeed, the security proof of HMAC relies on indistinguishability of the compression function from a PRF. There are still an awful lot of details, though. And the PRF assumption is quite strong. - Just to quibble over minor points, I think NMAC makes more sense than HMAC here (since you're starting with a compression function, rather than a hash function), and I believe the requirement is that the output has to be at least as long as the PRF key (since you'd be chaining through the key input). – Seth Feb 19 '12 at 10:03 If you have a keyed PRF, where the key is sufficient to turn the function into an independent PRF, you can turn it into a keyed PRP using a Luby Rackoff construct. If you got a keyed PRP, you can use it in CBC-MAC mode. - I am not sure if $\mathsf{PRF}$ alone can be used to construct; however there is a natural way to construct $\mathsf{MAC}$ using $\mathsf{PRF}$ and $\mathsf{UHF}$. You can see the following papers for more detail: 1. LFSR-based hashing and authentication. In CRYPTO 1994. 2. Bucket hashing and its application to fast message authentication. In CRYPTO 1995. 3. On fast and provably secure message authentication based on universal hashing. In CRYPTO 1996. I just found a very easy way to construct $\mathsf{MAC}$ from $\mathsf{PRF}$ in the lecture notes of Barak. More concisely, it says that if $\{ f_k \}_{k \in \mathcal{K}}$ is a family of $\mathsf{PRF}$, then the following is a $\mathsf{MAC}$: $$\mathsf{SIGN}_k(x) = f_k(x) \qquad \mathsf{VERIFY}_k(x,s)=1 \text{ iff } f_k(x)=s.$$ I think this is pretty neat. - I think the exact details depend from the used definition of pseudo-random function. If you have one that allows input of any size (like a hash function and the $f_k$ used here), it is a lot easier than if you have one with fixed input size. – Paŭlo Ebermann♦ Feb 20 '12 at 20:17

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http://mathoverflow.net/questions/26607/what-does-the-semiring-of-ideals-of-a-ring-r-tell-us-about-r/26668

## What does the semiring of ideals of a ring R tell us about R? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Here is something I've wondered about since I was an undergraduate. Let $R$ be a ring (commutative, let's say, although the generalization to noncommutative rings is obvious). Ideals of $R$ can be multiplied and can be added (the ideal $I+J$ is the ideal generated by $I$ and $J$), and multiplication distributes over addition. Therefore we can consider the semiring $S$ of ideals of $R$. The question is, does the structure of $S$ tell us anything interesting about the structure of $R$? Or vice versa? I asked this question on sci.math.research last year and got a few replies but nothing very substantive. http://mathforum.org/kb/thread.jspa?messageID=6599151 For a more concrete question: Give an interesting sufficient condition for $S$ to be finitely generated. Conversely, if $S$ is finitely generated, does that imply anything interesting about $R$? - 1 Just one example which comes into my mind: If R is a direct product of n fields, then the semiring of ideals is the power set of $1,...,n$ with union as sum and intersection as product (zero element is the empty subset, unit element is the whole set). You cannot recover the fields. – Martin Brandenburg Jun 1 2010 at 0:05 So basically you want to know which properties can be covered? – Martin Brandenburg Jun 1 2010 at 0:08 Sorry you didn't like my answers... – Pete L. Clark Jun 1 2010 at 2:31 @Pete: I appreciated your answers, but am still hoping for something more substantive. – Timothy Chow Jun 1 2010 at 4:35 1 To multiply ideals shouldn't they be two-sided ideals? There are interesting (noncommutative) rings with no nontrivial two-sided ideals (for example, the first Weyl algebra, which is differential operators in 1 variable with polynomial coefficients), so in this generality I believe not very much can be said. – Peter Samuelson Jul 29 2010 at 17:57 ## 4 Answers Here are a few observations. None of them require our ring to be commutative. First, notice that one can recover the natural partial ordering of the ideals via addition, because for any two ideals $I$ and $J$ of $R$, $I\subseteq J$ if and only if $I+J=J$. (More generally, $I+J$ is the smallest ideal containing both $I$ and $J$.) Second, this allows us to recover the prime ideals of $R$. This is because an ideal $P$ of $R$ is prime if and only if, for any ideals $I$ and $J$ of $R$, $IJ\subseteq P$ implies $I\subseteq P$ or $J\subseteq P$. (The same can be said for the semiprime ideals of $R$, which are the radical ideals of $R$ in case $R$ is commutative.) Third, we can recover the Zariski topology on the prime spectrum of $R$ because it is defined using the natural partial ordering on the ideals of $R$. - The Zariski topology doesn't require the ring to be commutative? – Pete L. Clark Jun 1 2010 at 3:48 No, the Zariski topology makes sense for any ring! This is because, for ideals $I$ and $J$, $V(I)\cap V(J) = V(IJ)$ and $\bigcup_j V(I_j) = V(\sum I_j)$. What you lose, however, is the fact that ring homomorhpisms determine continuous functions between prime spectra. – Manny Reyes Jun 1 2010 at 4:13 For a noncommutative ring and in the case of $\textit{primitive}$ ideals, this is known as "Jacobson topology". (Primitive ideals are proper analogues of maximal ideals in the noncommutative setting.) It's a terminological issue, I simply cannot remember whether the same term applies to the whole prime spectrum, but certainly, "Zariski topology" implies that the ring is commutative. – Victor Protsak Jun 1 2010 at 5:15 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This is sort of a sideways answer, but: in many ways the monoid $\operatorname{Prin}(R)$ of principal ideals carries more information. If $R$ is a domain $\operatorname{Prin}(R)$ is a cancellative monoid so injects into its group completion, the group of divisibility $K^{\times}/R^{\times}$ of $R$. Many of the factorization properties of $R$ can be gracefully rephrased in terms of $\operatorname{Prin}(R)$ and/or $K^{\times}/R^{\times}$. See for instance Section 4.1 of http://www.math.uga.edu/~pete/factorization2010.pdf for more on this point of view. - I've been interested in this lately. Hopefully you have seen this? Golan, Jonathan S.(IL-HAIF) Semirings for the ring theorist. Rev. Roumaine Math. Pures Appl. 35 (1990), no. 6, 531–540. Golan cautions that treating semirings as 'poor man's rings' is not always good. They can really be different animals altogether. I think someone noted above that the semiring of ideals is additively idempotent. In a sense, this is as far as you can get from an additive group. The compensation for the loss of the additive group is the complete lattice structure compatible with the multiplication. That is, if A\leq B, then AC\leq BC and CA\leq CB. Ring theorists have been saying things about rings via the lattice of one sided ideals for years! The lattices of onesided ideals are almost as nice, except you lose compatibility of multiplication with the order, and there is no longer a twosided identity for the semiring. These are called quantales in some places. - This is pretty simple too, but here it goes. I take R to be commutative and have 1. If S is generated by just one ideal P, then all the ideals of R are of the form P^k. Thus R is local. If P^2 is not all of P then any p in P\P^2 generates P, and each P^k is generated by p^k. Hence the only prime ideal is P, and it is exactly the ideal of nilpotents (since these are the intersection of all prime ideals). It follows that some P^k = 0. If P^2 is all of P then again there is just the one prime ideal P, but now P = P^k = 0, so R is a field. So either R is a field or there is a nilpotent p in R s.t. all x in R are of the form u*p^k for some unit u and non-negative integer k. (Just consider the biggest k for which x is a multiple of p^k.) Sometimes, but not always (see below), we can identify R with a quotient of the polynomial algebra (R/P)[X] (note that R/P is a field), namely (R/P)[X]/(X^k) where k is the smallest s.t. P^k = 0. Conversely, any quotient F[X]/(X^k), F a field, has its semiring of ideals generated by (X). - The identification of $R$ with a quotient of $(R/P)[X]$ is a little shaky. How does $\mathbb{Z}/9\mathbb{Z}$ fit into your classification scheme? About the best you can say, I think, is that if $S$ is one-generated, then $R \cong V/t^k$ for some discrete valuation ring $(V,t)$. – Graham Leuschke Jun 1 2010 at 19:00 You're right, of course. I've edited accordingly. I wasn't sure myself but I wanted it to be true so badly! – Pietro KC Jun 1 2010 at 20:31 From an ideal-theoretic point of view, DVRs are just as good as F[X]/(X^k) -- in fact better, since you don't have to worry about silly arithmetic issues with the field. Might as well just say that $S$ is one-generated iff $R$ is a non-trivial image of a DVR. Cleaner that way. – Graham Leuschke Jun 1 2010 at 22:44

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http://stats.stackexchange.com/questions/38471/strategies-for-introducing-advanced-statistics-to-various-audiences

# Strategies for introducing advanced statistics to various audiences I work mainly with non-statisticians in fields such as medicine, social sciences and education. Whether I am consulting with graduate students, helping researchers with articles or reviewing articles for journals, I often have the problem that someone (client, author, dissertation committee, journal editor) wants to use some relatively well-known technique when it is either entirely inappropriate or when better but lesser-known methods exist. Often, I will explain the alternative technique but then be told "everybody does it the other way". I'd be interested in how others deal with this sort of difficulty. ADDITIONS @MichaelChernick suggested I could share some stories, so I will Currently I am working with one person who is duplicating a previous paper and adding one independent variable to see if it helps. The previous paper is, frankly, terrible. It treats dependent data as if they were independent; it is tremendously overfit and there are other problems too. Yet he (my client) submitted an earlier version as a dissertation and not only got his degree but was widely praised for the research. Many times I have tried to convince people not to dichotomize variables. This comes up very often in medicine. I patiently point out that dicohotomizing (say) birthweight into low and normal (usually at 2,500 g) means treating a 2,499 g baby as just like a 1,400 g one; but treating the 2,501 gram baby quite differently. The clinician agrees with me that this is silly. Then says to do it that way. I had a graduate student client long ago whose committee insisted on a cluster analysis. The student did not understand the method, the method did not answer useful questions, but that's what the committee wanted, so that's what they got. The entire field of statistical graphics is one where, for many, "this is how grandpa did it" is enough. Then there are people who seem to just push buttons. I remember one presentation (not by someone I helped!) who had taken an entire questionnaire and factor analyzed it. One of the variables she included was ID number! Oy. - 6 Peter, Meta is for questions about this site. I suspect you may have been thinking about "Community Wiki" status, which is for useful, interesting questions unlikely to have an objectively best answer (or which are likely to need collaborative efforts to answer at all). Accordingly, I have interpreted your suggestion as a CW request and implemented that. – whuber♦ Oct 2 '12 at 16:36 Thanks @whuber that is what I meant – Peter Flom Oct 2 '12 at 20:32 1 A quick note on dichotomisation: I actually think this is a misplaced intuition from people who know "a little bit" of stats. In the fields you speak of I would imagine a lot of analysis would be geared around decision making (eg, should I start treatment for disease X or Y?). This is dichotomus - and often a useful dichotomy. If you only have a small number of variables, then a proper decision analysis or hypothesis test may well mimic this - we may have "if x>10 take option 1, otherwise take option 2". – probabilityislogic Oct 3 '12 at 7:44 1 Also an analysis based on dichotomised variables is very easy to remember if you don't have access to a computer. – probabilityislogic Oct 3 '12 at 7:46 1 – Chris Beeley Oct 4 '12 at 19:29 ## 5 Answers This is a tricky question! First, some thoughts on why this happens. I work in an area which does (or at least should) make extensive use of statistics, but where most practitioners are not statistical experts. Consequently one sees a lot of "I put a vector into excel's t-test function and this number fell out. Therefore my paper is supported by statistics." The main reason I see for this happening is that lack of statistics knowledge starts at the top. If your reviewers and thesis committee don't keep up to date on statistical techniques, then you need to justify use of anything that is "unconventional". For example, in a thesis, I opted to use violin plots instead of box plots to show the shape of a distribution. The use of this technique required extensive documentation in the thesis, as well as a prolonged discussion in my defense where all of the committee members wanted to know what this strange plot meant, despite both the descriptions in the text and the references to the source material. Had I just used a box plot (which shows strictly less information in this case, and can easily deceive the viewer about the shape of a distribution if it is multi-modal) no one would have said anything, and my defense would have been easier. The point is, in non-stats fields practitioners face a difficult choice: We can read about and then use the correct methods, which entails a bunch of work that none of our higher ups are interested in; or we can just go with the flow, get the rubber stamp on our papers and theses, and keep using incorrect but conventional methods. I think a good approach is to emphasize the consequences of failing to use correct techniques. This might entail: • Giving a real world example of how someone in their field experienced the consequences of poor inference. This is easier in some fields than others. Examples where careers were damaged are especially good. • Explaining that doing incorrect analysis can leave you in a situation where your results are very unlikely to transfer to the real world, which could cause harms (e.g. In my field, if your A.I. system prototype appears statistically better than the competition, but in fact is the same, then spending the next 6 months building a full implementation is a really bad idea. • Pick techniques which will save the users lots of time. Enough time so that they can spend what they save explaining the techniques to the higher ups. - 1 Good discussion and good answer +1. – Michael Chernick Oct 2 '12 at 18:39 Good points @John – Peter Flom Oct 2 '12 at 20:33 1 +1 for pointing out the consequences. It can do wonders for getting people to switch to better methods. – Leo Oct 3 '12 at 6:35 Speaking from the perspective of a psychologist with only slight statistical sophistication: When you introduce the method, also introduce the tools. If you tell most researchers in my field a long story about a great new method, they're going to spend the whole time worried that the punchline is "and all you have to do is brush up on your differential calculus and then take a two week training course!" (or "and buy a \$2000 stats package!" or "and adapt 5000 lines of Python and R code!"). Whereas if there's an implementation of the method available in the stats package they already use, or in a piece of free software with a comprehensible GUI, and they can get up to speed on it in a day or two, they might be willing to give it a try. I'm aware that this approach can seem venal and unscientific, but it's easy for people to fall into when they're worried about grants and publications, and don't see learning huge amounts of math as likely to help them keep their jobs. - 2 @octem Couldn't the investigator trust the statistician to do that part of the collaboration. Why does the investigator need the tool. I liken this with the physician to reversing the tables. How would he feel if I said give me a quick tutorial on how to do this surgery and I will go ahead and do it for you. I think shock and dsimay and it is illegal for me to practice mewdicine without a license. That is probably a good thing. But doesn't the statistician deserve equal respect. Why expect that I can just give him the tool and let him hack away with his lack of training. – Michael Chernick Oct 2 '12 at 18:37 1 In social psychology, it's not usual to have a statistician (because it's not usual to have enough funding to pay a statistician). Now I'm in public health / health psychology. Big grants usually include salary for a statistician, but a lot of our work is done on stingy little pilot grants where we can't even afford salary for the PI. That's the perspective I'm coming from... if you're in a field where most projects include a trained statistician, then I agree, this kind of resistance would not be reasonable. – octern Oct 2 '12 at 18:45 1 @octem Thanks for the response. Just to follow-up with the analogy. There was a time when abortion was illegal in the US and some women went to other countries or had it done in a back room illegally without sterile conditions and great health risks. It may not seem like a great analogy but not being able to afford a statistician justifies doing an inferior job? I know medicine is more of a life or death thing but bad science has bad consequences too? Misuse of data could lead to practicing bad medicine because an unsafe drug gets used when it shouldn't be. – Michael Chernick Oct 2 '12 at 19:11 @MichaelChernick Look, I'm not happy about the state of statistical analysis in the social sciences either. But the question was how to get researchers to adopt a new statistical approach, and I'm giving an answer that's appropriate to a large subpopulation of researchers -- whether we like it or not. – octern Oct 2 '12 at 19:33 1 @octem Fair enough and I think I understood that. I gave you an upvote before I asked the question. The attitude of the investigators to think that the cheap way is the way to go is what I am questioning and not the fact that you recognize the problem exists. It does and I agree with you there. But somehow in the long run I think we just need to get more respect and recognition that our job is not trivial. – Michael Chernick Oct 2 '12 at 19:39 Thanks for this nice question Peter. I work at a medical research institution and deal with physicians who do research and publish in the medical journals. Often they are more interested in getting their paper published than "doing the statistics completely right". So when I propose an unfamilar technique they will point to a similar paper and say "look they did it this way and got their results published." There is a problem I think when the published paper is really bad and has mistakes. It is difficult to argue even though I have a great reputation. Some docs have big egos and think they can learn almost anything. So they think they understand the statistics when they don't and can be insistent. It can get frustrating. When it is a t test and Wilcoxon is more appropriate I get them to do a Wilk Shapiro test and if normality is rejected we include both methods and explain why Wilcoxon is better. I sometimes can convince them and often they depend on me for statistics, so I have a little more clout then a general consultant might have. I also ran into a situation where I did Kaplan-Meier curves for them and we used the log rank test but Wilcoxon gave a different result. It was hard for me to decide and in such situations I think it is best to present both methods and explain why they differ. The same goes for using Peto vs Greenwood confidence intervals for the survival curve. Explaining the Cox proportion hazard assumption can be difficult and they often misinterpret odds ratios and relative risk. There is no simple answer. I had a boss here who was a top medical researcher in cardiology and he sometimes referees for journals. He was looking at a paper that dealt with diagnosis and used AUC as a measure. He had never seen an AUC curve before and came to me to see if I thought it was valid. He had doubts. It turned out to be appropriate and I explained it to him as best I could. I have tried to lecture on biostatistics to physicians and have taught biostatistics in public health schools. i try to do it better than others have and produced a book for health science majors introductory course in 2002 with an epidemiologist as coauthor. Wiley wants me to do a second edition now. In 2011 I published a more concise book that I tried to cover just the essentials so that busy MDs might take the time to reasd it and reference it. That is how I deal with it. Maybe you can share your stories with us. - These are good points @Michael. I will add some stories – Peter Flom Oct 2 '12 at 20:35 1 @PeterFlom I think we might have some very similar experiences. I alos think you are getting some other very good responses in other people's answers. – Michael Chernick Oct 2 '12 at 20:38 There are some nice comments already made here, but I'll throw in my 2 cents. I'll preface this all by saying that I'm assuming we're talking about a situation where using the traditional "canned" techniques will damage the substantive conclusions reached by the analysis. If that's not the case, then I think that sometimes doing an overly simplistic analysis is excusable both for brevity and for ease of understanding when the target audience are laymen. Is it really such a crime to assume independence when the intraclass correlation is .02 or to assume linearity when the truth is $\log(x); \ x \in (1,2)? \$ I'd say no. In my career I do a lot of interdisciplinary research and has lead me to work with closely with substance abuse researchers, epidemiologists, biologists, criminologists and physicians at various times. This typically involved analysis of data where the usual "canned" approaches would fail for various reason (e.g. some combination of biased sampling and clustered, longitudinally and/or spatially indexed data). I also spent a couple years consulting part time in graduate school, where I worked with people from a large variety of fields. So, I've had to think about this a lot. My experience is that the most important thing is to explain why the usual canned approaches are inappropriate and appeal to the person's desire to do "good science". No respectable researcher wants to publish something that is blatantly misleading in its conclusions because of inappropriate statistical analysis. I've never encountered someone who says something along the lines of "I don't care whether the analysis is correct or not, I just want to get this published" although I'm sure such people exist - my response there would be to end the professional relationship if at all possible. As the statistician, it's my reputation that could be damaged if someone who actually knows what they're talking about happens to read the paper. I admit that it can be challenging to convince someone that a particular analysis is inappropriate, but I think that as statisticians we should (a) have the knowledge necessary to know exactly what can go wrong with the "canned" approach and (b) have the ability to explain it is a reasonably comprehensible way. Unless you're working as a statistics or math professor, a part of your job is going to be to work with non-statisticians (and even sometimes if you are a stat/math prof). Regarding (a), if the statistician doesn't have this knowledge, why would they be discouraging the canned approach? If the statistician is saying "use a random effects models" but can't explain why assuming independence is a problem, then aren't they guilty of giving in to dogma in the same way the client is? Any reviewer, statistician or not, can make pedantic critiques of a statistical modeling approach because, let's face it - all models are wrong. But, it requires expertise to know exactly what could go wrong. Regarding (b), I've found that graphical depictions of what could go wrong typically "hit home" the most. Examples: • In the example given by Peter about categorizing continuous data, the best way to show why this is a bad idea is to graph the data in its continuous form and compare it with its categorical form. For example, if you're making your response variable binary then plot the continuous variable vs. $x$, and, if it doesn't look an awful lot like a step function, then you know the discretization lost valuable information. If this difference isn't drastic or resulting in any changes in the substantive conclusions, you could also see this from the plot. • When the proposed "form" of the model (e.g. linear) is inappropriate. For example, if the regression function "plateaus" like $y = x$ for $x \in (0,1)$ but $y = 1$ for $x > 1$ then a linear fit's slope will be too shallow and, depending on the data, this could push the $p$-value below significance despite there being an obvious relationship between $x$ and $y$. • Another common situation (also mentioned by Peter) is explaining why assuming independence is a bad idea. For example, you can show with a plot that positive autocorrelation will typically produce data that is more "clustered" and the variance will be underestimate for that reason, giving some intuition of why the naive standard errors tend to be too small. Or, you could also plot the data with the fitted curve that assumes independence and one can visually see how the clusters influence the fit (effectively lowering the sample size) in a way that is not present in independent data. There are a million other examples but I'm working with space/time constraints here :) When pictures simply won't do for whatever reason (e.g. showing why one approach is underpowered) then simulation examples are also an option that I've employed from time to time. - Some random thoughts because this is a complex issue... I feel that a big problem is the lack of math education in a variety of professional disciplines and graduated programs. Without a mathematical understanding of statistics, it becomes a bunch of formulas to be applied according the case. Also, for getting a real understanding of the matter, professors should talk about the original problems that the original authors were facing at the time they published their approaches. One can learn more from that than from reading thousands books on the subject. Statistics is a toolbox for solving problems, but it is also an art and faces the same issues than any other art. One can learn how to make sounds with an instrument. But by being able of "playing" an instrument one does not become a musician. However, is not uncommon to find people that see themselves as musicians without having studied a single concept of rhythm, melody and harmony. In the same line, for getting papers published, most people don't need to know nor understand the concepts behind a formula... nowadays scientists just need to know what key they have to press and when it has to be pressed, period. So this has nothing to do with the "ego" of MDs. This is a subcultural problem, a problem more related with education, customs and values of the scientific community. What one can expect in an era in which there are thousands and thousands and thousands of useless papers and books being published for fulfilling some academic requisites/policies? In an era in which the amount of papers one publishes is more important than the quality of them? Mainstream scientists are not worried about the good science anymore. They are slaves of numbers. They are affected (or infected) by the administrative bug of our era... So, from my perspective, a good course in statistics should include the mathematical, historical and philosophical basis of the approach being studied, always highlighting the several paths one can take for solving a single problem. Finally, if I were a professor in statistics/probability my first lecture(s) would be dedicated to problems like shuffling cards or tossing a coin. That will put the audience in the right position for listening... probably. -

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http://math.stackexchange.com/questions/106070/if-a2-i-identity-matrix-then-a-pm-i?answertab=active

# If $A^2 = I$ (Identity Matrix) then $A = \pm I$ So I'm studying linear algebra and one of the self-study exercises has a set of true or false questions. One of the question is this: If $A^2 = I$ (Identity Matrix) Then $A = \pm I$ ? I'm pretty sure it is true but the answer say it's false. How can this be false (maybe its a typography error on the book)?. Thanks. - 15 Try $$A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$ – Dylan Moreland Feb 5 '12 at 20:13 1 I'd point out that it is true if you're working with $1$-by-$1$ matrices (over $\mathbb C$, or any other integral domain). But for $n \geq 2$ the ring of $n$-by-$n$ matrices over any non-trivial ring is not an integral domain: this means that $(A+I)(A-I) = 0$ doesn't necessarily imply that $A + I = 0$ or $A - I = 0$. – Matt Feb 5 '12 at 20:30 – Jonas Meyer Feb 5 '12 at 20:56 There's an entire family of so-called involutory matrices. Look up Householder reflectors, for instance. – J. M. Feb 6 '12 at 5:11 ## 5 Answers A simple counterexample is $$A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$$ We have $A \neq \pm I$, but $A^{2} = I$. - I know $2·\mathbb C^2$ many counterexamples, namely $$A=c_1\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}+c_2\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}\pm\sqrt{c_1^2+c_2^2\pm1}\begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix},$$ see Pauli Matrices $\sigma_i$. These are all such matrices and can be written as $A=\vec e· \vec \sigma$, where $\vec e^2=\pm1$. - "Most" (read: diagonalizable) matrices can be viewed simply as a list of numbers -- its eigenvalues -- in the right basis. When doing arithmetic with just this matrix (or with other matrices that diagonalize in the same basis), you just do arithmetic on the eigenvalues. So, to find diagonalizable solutions to $A^2 = I$, we just need to write down a matrix whose eigenvalues satisfy $\lambda^2 = 1$ -- and any such matrix will do. When thinking about matrices in this way -- as a list of independent numbers -- it makes it easy to think your way through problems like this. - Every matrix satisfying $A^2=I$ is diagonalizable, because either it is $\pm I$ or its minimal polynomial is $(x-1)(x+1)$. The general solution is obtained by taking all diagonal matrices with entries $\pm 1$ on the diagonal and conjugating by invertible matrices. – Jonas Meyer Feb 6 '12 at 5:03 1 Jonas Meyer, this is only true if $char F \ne 2$. Otherwise, there are such matrices which are not diagonalizable, – anonymous Feb 6 '12 at 8:18 1 @Jonas: That's a good point to mention as an appendix, but dealing properly with non-diagonalizable matrices in this fashion is somewhat more sophisticated. The only reason I mentioned the word was so that I didn't mislead Randolf into thinking this method works (unmodified) for all matrices; e.g. that the argument I gave isn't sufficient to tell us that this (or any!) equation has only diagonalizable solutions. – Hurkyl Feb 6 '12 at 9:53 @anonymous: Good point, e.g. $\begin{bmatrix}1&1\\ 0&1\end{bmatrix}$. @@Hurkyl: I agree, it is best as an appendix. I appreciate your caution, but wanted to point out that your method does lead to the general solution (in the characteristic $0$ case that the OP is probably working in). – Jonas Meyer Feb 6 '12 at 15:49 The following matrix is a conterexample $A = \left( {\begin{array}{cc} -1 & 0 \\ 0 & 1 \\ \end{array} } \right)$ - In dimension $\geq 2$ take the matrix that exchanges two basis vectors ("a transposition") - If you want to exchange the (standard) basis vectors $e_{i}$ and $e_{j}$ ($1 \leq i,j \leq n$), then use the matrix $A = [m_{ij}]$ with $m_{kk} = 1, k\neq i,j$, $m_{ij} = m_{ji} = 1$ and $m_{kl} = 0$ for all other values of $k$ and $l$. For example, if you want $e_2$ and $e_3$ exhanged in $\mathbb{R}^{3}$, take $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$ It is clear that such a matrix always satisfies $A^2 = I$, since applying it twice always gets you back to where you started. – Martin Wanvik Feb 5 '12 at 21:01 Thank you @Martin Wanvik, pretty clear explanation. – Randolf R-F Feb 5 '12 at 21:52

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http://mathoverflow.net/questions/5659/abstract-nonsense-versions-of-combinatorial-group-theory-questions/14229

## Abstract nonsense versions of “combinatorial” group theory questions ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In particular, I'm just curious whether there's a version of the Sylow theorems (which are very combinatorially-flavored) which allows horizontal and/or vertical categorification? Or at least can be stated in more category-theoretic terms? - I was wondering this just last week. I have prelims to start worrying about, and categorical statements always stick best. – alekzander Nov 16 2009 at 0:49 alexzander, that's my justification too. I never did like abstract algebra 'til I learned to start thinking of it in categorical terms. – Harrison Brown Nov 16 2009 at 3:06 Although in my case it's not prelims so much as undergraduate-level abstract algebra test, so I'm probably overthinking it. Still curious about the question on its own merits, though. – Harrison Brown Nov 16 2009 at 3:15 ## 5 Answers Sylow subgroups are an example of a type of object satisfying a sort of universal property. Exploring other objects with similar properties gave birth to the modern theory of finite soluble groups in the 1960s. If X is a class of finite groups and G is a finite group, then P is an X-covering subgroup of G if P is in X, and whenever P ≤ H ≤ G, N ⊴ H, and H/N in X, then PN=H. In other words, P covers every X-factor of G. If X is the class of finite p-groups, then X-covering subgroups of G and Sylow p-subgroups of G are the same thing. Indeed, if P is an X-covering group, and if H is a Sylow p-subgroup containing P and N=1, then we must have H=P. If P is a Sylow p-subgroup and H contains P with N ⊴ H and [H:N] a power of p, then [H:NP] is a divisor of [H:N] and [H:P], so must be 1. Notice how the "containment" part of the Sylow theorems is replaced with a "covering" condition that behaves better with the normal structure of the group. If G is a finite solvable group and X is the class of nilpotent groups, then there is a sort of "Sylow nilpotent subgroup", the X-covering groups or Carter subgroups. They were studied by R.W. Carter who described them as self-normalizing nilpotent subgroups. Like Sylow p-subgroups, there is exactly one conjugacy class of Carter subgroups, and they have some reasonable arithmetic properties. People tried to determine which classes X of groups are such that X-covering groups exist and are unique up to conjugacy. Roughly speaking, this was the dawn of the modern theory of finite soluble groups, with Gaschütz's (et al.) classification of such X as "saturated formations". This shifts focus away from the subgroup P to the class X. If X is sufficiently nice, then there will be a nicely embedded X-subgroup for any finite group. Sylow p-subgroups also satisfy a dual condition, they are also X-injectors for the class X of finite p-groups. If X is a class of finite groups, and G is a finite group, then P is an X-injector of G if for every subnormal subgroup N of G, P∩N is a maximal X-subgroup of N. The dual definition of covering group (for X a saturated formation) is that P is an X-covering group iff PN/N is a maximal X-subgroup of G/N for every N ⊴ G. If X is the class of finite nilpotent groups, then X-injectors are called Fischer subgroups and again form a single, well-behaved conjugacy class of subgroups. A Fischer subgroup of a finite soluble group is a nilpotent subgroup that contains every nilpotent subgroup that it normalizes. This is similar to the idea that a Sylow p-subgroup contains every p-group that it normalizes. X such that X-injectors form a unique conjugacy class are called Fitting classes, due to their similarity to Fitting's lemma on subnormal nilpotent subgroups. A very approachable introduction to these ideas is B.F. Wehrfritz's tiny textbook for a Second Course on Group Theory. Some of these ideas are described in Robinson's textbook for a Course in the Theory of Groups, but I believe it spends very little time on general formations. The standard textbook source for formations, especially in the soluble universe, is K. Doerk and T. Hawkes's book Finite Soluble Groups. Doerk&Hawkes explains several of Gaschütz's arithmetically defined Xs, which you might find a good contrast. - ### 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 may be splitting hairs, but I think the Sylow theorems are more arithmetic than combinatorial. Depending on your standards of good categorification, I think this makes it difficult to encode precise congruence behavior such as the index of the normalizer being congruent to 1 mod p, and so on. There is a homotopy-theoretic way to look at the Sylow theorems involving fusion systems. These are categories that encode the conjugation data between the p-Sylow subgroups of a finite group, and let you study the homotopy type of the p-completed classifying space of a finite or compact group. Bob Oliver has some papers on this subject. I don't know if this level of sophistication will help you learn the basic theorems, though. When I was studying Sylow, I personally preferred exercises, like showing that there are no simple groups of order 56. - I learned the Sylow theorems precisely like that. Homework problem in my abstract algebra class "Show that there are no nonabelian simple groups of order less than 60." – Charles Siegel Nov 16 2009 at 6:16 Yeah, okay, good point; it's probably too much to hope for there to be something easy enough to apply in day-to-day situations. But there's no obvious reason (to me, anyway) that there shouldn't be an extension even to finite groupoids. – Harrison Brown Nov 16 2009 at 16:09 1 J.L. Alperin's "Sylow Intersections and Fusion" (MR0215913; DOI: 10.1016/0021-8693(67)90005-1) is a nice group theoretic introduction to some of the key ideas behind fusion. It is a more precise version of Sylow's conjugacy theorem. Linckelmann's introduction (MR2336638) is one of the easiest ways to read the category theoretic version of fusion. Alperin's Local Representation Theory textbook introduces the Brauer subpairs theory nicely; that is the other main source of fusion systems. – Jack Schmidt Feb 5 2010 at 4:21 Philosophically I think of the Sylow theorems as a statement about "localization of groups" at a prime. For abelian groups one can make this precise, since abelian groups, as Z-modules, can be realized as sheaves over Spec Z and the stalk over p is precisely the Sylow p-subgroup (although for finitely generated abelian groups these are trivial to describe). The hard part would be trying to carry this picture over to general groups. - You might find the theory of fusion categories interesting (this is, unfortunately for the terminology, disjoint from Scott's mention of "fusion systems"). A fusion category is a type of tensor category which generalizes both Rep(G) and Vec_G, the latter being the category of vector spaces graded by a group G (so it's Groethendieck ring is the group algebra kG). These are tensor categories which are semi-simple and have finitely many isomorphism classes of simple objects. Each simple object has a notion of dimension, called Frobenius Perron dimension, and the categories themselves have a dimension associated to them, such that in the two cases Rep(G) and Vec_G, the dimension of the category is |G|. While I don't know of Sylow theorems for fusion categories per se, there are many familiar theorems, saying e.g. that every fusion category of order p^k is nilpotent, and classifying groups of small orders p, pq, pqr, p^k, etc. There is also a version of Burnside's theorem stating that every category of dimension p^aq^b is solvable. It's quite possible that there are some analogs of Sylow's theorems in this direction. I'd recommend "On Fusion Categories" http://arxiv.org/abs/math/0203060, by Etingof Nikshych and Ostrik, which introduces many of these ideas. More extensively, there are course notes from a class Etingof taught on this, which can be found on his webpage, http://math.mit.edu/~etingof. Papers by various subsets of (these three, union Gelaki) introduce the concepts and claims I mentioned above. As Scott mentioned, this kind of abstract nonsense won't help much with prelims questions, most likely, but might make for a fun distraction =]. - One of the simplest consequences of Sylow's first theorem is the Cauchy theorem, saying that a group of order divisible by a prime $p$ contains an element of order $p$. I'd like to point out that there is an analog of Cauchy's theorem for semisimple Hopf algebras, arXiv:math/0311199, which has been generalized to quasi-Hopf algebras (i.e., fusion categories with integer dimensions) in arXiv:math/0601012. -

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http://mathoverflow.net/questions/84705/is-the-moduli-space-of-curves-defined-over-the-field-with-one-element/84785

## Is the moduli space of curves defined over the field with one element? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) There are various frameworks around which enlarge the category of rings to include more exotic objects such as the 'field with one element,' `$\mathbb{F}_1$`. While these frameworks differ in their details, there are certain things this should be true of any object that deserves to be called `$\mathbb{F}_1$`. For example, The algebraic K-theory of `$\mathbb{F}_1$` should be sphere spectrum, and the theory of toric varieties should be defined over `$\mathbb{F}_1$`. Question 1: Is there a moral reason why the moduli space of curves should (or should not) be defined over Spec `$\mathbb{F}_1$`? EDIT: For anyone who would like to be more concrete, I'm happy to take the Toen-Vaquie definition of schemes over `$\mathbb{F}_1$`. (see arXiv:math/0509684). In this setup (and most of the other frameworks I know) an affine scheme over `$\mathbb{F}_1$` is just a commutative monoid $M$. After base change to $\mathbb{Z}$ this becomes the monoid ring $\mathbb{Z}[M]$. So here is a more precise question: Question 2: Does the moduli space of curves $\mathcal{M}_{g,n}$ (over $\mathbb{Z}$, say) admit a covering by affine charts of the form spec $\mathbb{Z}[M_i]$ for commutative monoids $M_i$? If so, can this covering be chosen so that (as in the case of toric varieties) the gluing is entirely determined by maps of monoids? - 3 I think what you mean is that spec $\mathbb{F}_1$ should be a final object, so everything maps to it. This is different from being defined over it. E.g. $\mathbb{Z}$ is initial among rings and so any scheme maps to spec $\mathbb{Z}$, but there are certainly things that are not defined over $\mathbb{Z}$ - i.e., there exists schemes over spec $k$ that are not the pullback (along spec $k$ $\to$ spec $\mathbb{Z}$) of a scheme over $\mathbb{Z}$. – Jeffrey Giansiracusa Jan 2 2012 at 12:28 2 My opinion is that this is not a valid question unless you specify at least one definition of the field with one element. Could you please give a reference to the definition you are using? – Jason Starr Jan 2 2012 at 13:33 8 Did you look at mathoverflow.net/questions/8190/…? Outside of $M_{0,n}$ (moduli of curves of genus zero with $n$ marked points), it seems doubtful to me that any of the other $M_{g,n}$ (except a few small g,n cases?) would be defined over $F_1$, whatever that means. For example, don't the L-functions of schemes over $F_1$ (after base change to schemes over $Z$) coincide with L-functions of mixed Tate motives? Maybe $M_{1,1}$ would be interesting, since the orbifold points could be definable over $F_1$ (or at least something "below" $Z$). – Marty Jan 2 2012 at 18:18 4 Just to add to what Marty said, it seems to me that any scheme glued from schemes of the form $\mathbb{Z}[M]$ will be geometrically rational when you base change to $\overline{\mathbb{Q}}$. Since $M_g$ is definitely irrational when $g$ is large, it seems to me it cannot be of this form. – Jason Starr Jan 2 2012 at 22:08 3 Others have mentioned that it seems unlikely that you can define the whole moduli space over $\mathbb{F}_1$. However, it looks like you can define a formal neighborhood of any maximally degenerate point of the boundary, since such loci describe gluings of three-pointed genus zero curves. This suggests that objects like the Tate curve, $n$-gons, and Néron models of maximally degenerating Jacobians are definable over $\mathbb{F}_1$. – S. Carnahan♦ Jan 3 2012 at 5:51 show 5 more comments ## 2 Answers This "answer" will basically restate the comments of Marty and Jason Starr. Any variety covered by schemes of the form $\mathrm{Spec}(\mathbf Z[M_i])$, or any torified variety, is rational. And indeed Severi conjectured at one point that $M_g$ is rational for any $g$! But we know a lot about the Kodaira dimension of $M_g$ by work of Harris--Mumford, Farkas, Eisenbud, Verra, ... in particular we know that Severi's conjecture is maximally false. We have $\kappa(M_g) = -\infty$ for $g \leq 16$, we don't know anything for $17 \leq g \leq 21$, and $\kappa(M_g) \geq 0$ for $g \geq 22$. In fact we know that $M_{g}$ is of general type for $g = 22$ and $g \geq 24$. But if one only wants an example of a $g$ for which $M_g$ is not rational, then I think one can find examples much earlier (maybe $g \approx 6,7$?). See http://arxiv.org/abs/0810.0702 for a survey by Farkas. Moreover, the cohomology of a torified scheme can only contain mixed Tate motives. In particular, this implies properties like: the number of $\mathbf F_q$-points is a polynomial function of $q$. As Marty says there are no problems in genus zero; these might be honest-to-God $\mathbf F_1$ schemes. The cohomology of $M_{1,n}$ contains only mixed Tate motives when $n \leq 10$, but for $n \geq 11$ one finds motives associated to cusp forms for $\mathrm{SL}(2,\mathbf Z)$. (The number 11 arises as one less than the smallest weight of a nonzero cusp form; the discriminant form $\Delta$.) This implies that the polynomiality behaviour changes drastically -- the number of $\mathbf F_q$-points is now given by Fourier coefficients of modular forms, which are far more complicated and contain lots of arithmetic information. The connection with birational geometry is also very visible here. The cohomology classes on $M_{1,n}$ associated to the cusp forms of weight $n+1$ are of type $(n,0)$ and $(0,n)$ in the Hodge realization, since given such a cusp form one can explicitly write down a corresponding differential form in coordinates. So by the very definition of Kodaira dimension we can not have $\kappa = -\infty$ anymore. Bergström computed the Euler characteristic of $M_{2,n}$ in the Grothendieck group of $\ell$-adic Galois representations by point counting techniques. That is, he found formulas for the number of $\mathbf F_q$-points of $M_{2,n}$ by working really explicitly with normal forms for hyperelliptic curves. These formulas turned out to be polynomial in $q$ for $n \leq 7$ (and conjecturally for $n \leq 9$), just as they would be if we knew that the cohomology of $M_{2,n}$ contained only mixed Tate motives. By thinking a bit about the stratification of topological type, one concludes that this holds also for $\overline M_{2,n}$ when $n \leq 7$, which is smooth and proper over the integers. Then a theorem of van den Bogaart and Edixhoven implies that the cohomology of $\overline M_{2,n}$ is all of Tate type and with Betti numbers given by the coefficients of the polynomials. There are similar results by Bergström for $M_{3,n}$ when $n \leq 5$ and Bergström--Tommasi for $M_4$. But the general phenomenon is that increasing either $g$ or $n$ will rapidly take you out of the world of $\mathbf F_1$-schemes, at least if this is taken to mean "commutative monoids". However, I don't know enough $\mathbf F_1$--geometry to say what the answer is if one takes $\mathbf F_1$-schemes in the sense of Borger. The first nontrivial case to answer would be: are the motives associated to cusp forms for $\mathrm{SL}(2,\mathbf Z)$ defined over $\mathbf F_1$ in his set-up? To be clear, I don't believe that this is true, but I know almost nothing about $\lambda$-schemes. Let me also make two small remarks: (i) Everything I have written above is necessary conditions for being defined over $\mathbf F_1$. (ii) The fact that $M_{g,n}$ is smooth over the integers says that the cohomology of $M_{g,n}$ is at least quite "special", even though it is not defined over $\mathbf F_1$. Smoothness is a strong restriction on the Galois representations that can occur in the cohomology. - Very nice - thanks for the survey of examples of $M_{g,n}$ for various small $g$ and $n$ too! – Marty Jan 3 2012 at 6:57 Thanks for the very clear explanation. – Jeffrey Giansiracusa Jan 3 2012 at 8:59 ### 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. Here a few remarks from the `$\Lambda$` point of view on `$\mathbf{F}_1$`. (There we say a scheme is defined over `$\mathbf{F}_1$` if it admits a `$\Lambda$`-structure. If the scheme is flat over `$\mathbf{Z}$`, a `$\Lambda$`-structure is equivalent to a commuting family of endomorphisms `$\psi_p$`, one for each prime number `$p$`, such that each `$\psi_p$` agrees with the Frobenius map modulo `$p$`.) I prove in http://arxiv.org/abs/0906.3146 that if a scheme of finite type over `$\mathbf{Z}$` is defined over `$\mathbf{F}_1$`, then its motive is pure Tate (or rather becomes so after base change to some cyclotomic field - it's actually false in general without this qualification). So as in Dan Petersen's answer, `$M_{g,n}$` won't admit a `$\Lambda$`-structure unless the pair `$(g,n)$` is sufficiently small. I haven't really thought about the small cases, except just a bit when `$(g,n)=(1,1)$`. Then `$M_{g,n}$` is just the affine line, so it has many `$\Lambda$`-structures. As far as I know, you can't really say that they have any meaning, but I suspect there is still something interesting to say. 1) There is a `$\Lambda$`-structure on the completion of `$\bar{M}_{1,1}$` at the point at infinity which does have a meaning. The completion is `$\mathbf{Z}[[q-1]]$`, and the `$\Lambda$`-structure is defined by `$\psi_p(q)=q^p$`, for all `$p$`. The interest in this is that this `$\Lambda$`-structure prolongs to a `$\Lambda$`-structure on the universal generalized elliptic curve over `$\mathbf{Z}[[q-1]]$` (i.e. the Tate curve). From what I remember, the idea is that a point `$z\in\mathbf{C}^*/q^{\mathbf{Z}}$` should be mapped to its image `$z^p\in\mathbf{C}^*/q^{p\mathbf{Z}}$`. Making this rigorous is not hard, though of course you have to have an actual definition of the Tate curve over `$\mathbf{Z}[[q-1]]$`, a point that is often glided over. What I think could be interesting is to study the extent to which these `$\Lambda$`-structures extend to the whole moduli space. Here is one possible interpretation. A `$\Lambda$`-structure on a scheme `$X$` is equivalent to a section of the canonical projection `$W_*(X)\to X$` satisfying a certain associativity property. (`$W_*$` is the right adjoint to the Witt vector functor `$W^*$`. It is an arithmetic analogue of a jet space.) Now consider the map `$T\to X$`, where `$T$` is the Tate curve above and `$X$` is the total space of the universal elliptic curve. (This is a stack. One would have to check that all these `$\Lambda$` and `$W$` concepts make sense for stacks.) Then the projection `$W_*(X)\to X$` has a natural section over `$T$`. Then we can ask: what is the Zariski closure `$Z$` of this section in `$W_*(X)$`? The extent to which `$Z$` fails to be a section of `$W_*(X)\to X$` should be some measure of the extent to which the `$\Lambda$`-structure on `$T$` fails to extend to one on `$X$`. 2) One could try to do something similar at CM points, instead of at the boundary of the moduli space. CM elliptic curves admit a certain generalized version of a `$\Lambda$`-structure. (See the end of my paper cited above.) Do these generalized `$\Lambda$`-structures extend to the formal neighborhood of the elliptic curve in the universal elliptic curve? If so, one could look at closures of formal sections (as above) to see the extent to which they fail to extend to generalized `$\Lambda$`-structures on the entire universal elliptic curve. For all this, you'd have to fix an imaginary quadratic field, so all this is probably less fundamental than in 1). 3) Thomas Scanlon tried to convince me a few years ago that the Hecke correspondences have some kind of `$\Lambda$` nature. From what I remember, there were certain `$\Lambda$`-schemes closely related to modular varieties which are not of finite type but which are still finite-dimensional in some model-theoretic sense. I never got around to understanding what he meant (though I always intended to). Perhaps it is related to what I wrote above. So there are a few pieces of evidence that something is going on with `$\Lambda$`-structures and modular curves, though it remains to be seen if it's really an identifiable phenomenon and, if so, how interesting it is. - I see in the comments to the original post that Scott Carnahan made predictions about the Tate curve that are similar to what I said above. – James Borger Jan 4 2012 at 3:43

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http://math.stackexchange.com/questions/66183/finding-the-center-of-a-group-with-magma?answertab=active

# Finding the center of a group with MAGMA Given a finitely presented group $G = \langle S | R \rangle$, is there a command in MAGMA that computes the center of G ? - If Center() won't work on G, you can use one of the functions available to covert G into a group of a type on which Center() will work. – mt_ Sep 20 '11 at 21:52 This is unlikely to be possible unless $G$ is finite, in which case you should use a command like CosetImage to convert it to a group of permutations. – Derek Holt Sep 21 '11 at 8:50

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