November 2007 - mathfun@mailman.xmission.com

Re: [math-fun] Different values of the dot product
by Dan Asimov 04 Nov '07

04 Nov '07

Stefan asks: << Suppose we have the m different vectors x_1, x_2, ..., x_m in the R^n. No two vectors are the same and no vector is the zero vector. Furthermore we calculate for every 1<=i<j<=m the normal dot product x_i.x_j How many different numbers are we at least going to get, regardless of the choice of x_1, x_2, ..., x_m? . . . [W]hat if m>n? Is anything known about this problem? Any ideas, results, bounds? >> Also of course if m = n+1, the vectors can be the n+1 vertices of a regular n-simplex centered at the origin, having only 1 value of x_i*x_j for any i <>j. I find the problem more natural if the vectors are assumed to be of constant length. Then the question is essentially: "Given m distinct points on the surface of the unit sphere in R^n, what is the smallest size that the set of their mutual distances can be?" A natural generalization of this version would be to consider a set of m points that all lie in any other highly symmetric space S -- such as S = the n-torus (or S = any homogeneous space), and again ask how small the set of distinct distances can be. Very nice geometric questions here. It may be possible -- and fun -- to find good answers numerically by taking any starting configuration of the m points on the metric space S, and sum over the (m(m-1)/2)(m(m-1)/2 - 1)/2 pairs of distinct ordered pairs (i,j),(i',j') for 1 <= i < j <= m and 1 <= i' < j' <= m. Then minimize this sum via some robust minimization algorithm, like conjugate gradient, using many starting configurations.) The spheres S^2 and S^3, the projective spaces P^2 and P^3, and the tori T^2 and T^3 would be especially interesting to work on, as well as complex projective spaces like CP^2 and compact connected Lie groups like SO(4). --Dan

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[math-fun] Different values of the dot product
by Stefan Steinerberger 04 Nov '07

04 Nov '07

Hello, I already posted this problem at sci.math but didn't get a lot of response. Maybe someone on this list can offer some genuine insight. Suppose we have the m different vectors x_1, x_2, ..., x_m in the R^n. No two vectors are the same and no vector is the zero vector. Furthermore we calculate for every 1<=i<j<=m the normal dot product x_i.x_j How many different numbers are we at least going to get, regardless of the choice of x_1, x_2, ..., x_m? Certainly, if m<=n, we can set x_1 = (1,0,0...,0)^T, x_2 = (0,1,0...)^T, ... - each dot product will be zero, we therefore only get one number as a result. This, of course, also works for any orthogonal basis of R^n. But what if m>n? Is anything known about this problem? Any ideas, results, bounds? Stefan

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[math-fun] Magic squares generalized?
by Dan Asimov 04 Nov '07

04 Nov '07

One thing that doesn't appeal to me in the usual definition of a magic square (an nxn matrix of distinct integers such that the sum of each row, column, *and main diagonal* must be equal) is that no permutation of the matrix elements can carry a row or column onto a main diagonal. (Except in the 2x2 case, which admits no solutions even without the main diagonals rule.) SO: What I'd find much more interesting is the following: Let a "magic template" T = (K,S) "of shape (r,s)" denote a set K of r unknowns R_i, 1 <= i <= r, and a set S of subsets S_j of K, each of size s, such that: * for any two of the subsets, there is a permutation of K carrying one onto the other. * a "magic pattern" (nomenclature pending) modeled on T is defined as an assignment of distinct integers to the K(i)'s such that each subset S_j has the same sum. -------------------------------------------------------------- The more symmetric the configuration of subsets, the more appealing the the magic template. QUESTION: What are good ways to find highly-symmetric magic templates, especially ones that have at least one magic pattern? (I guess the cases where the number of unknowns is no smaller than the number of subsets -- i.e., equations -- have the best chance of being realizable.) Some candidates might include a) Use an n-dimensional cube of side L (so L^n unknowns) and its C(n,d)*L^(n-d) e-dimensional slices (e-cubes of size L^e) parallel to the coordinate axes. b) Use n unknowns and all subsets of size s -- the subsimplices of fixed dimension of an (n-1)-simplex. c) If these don't work, then perhaps highly-symmetrical subsets of the sets of subsets in a) or b). Are there standard ways of finding such things? --Danne thing that doesn't appeal to me in the usual definition of a magic square (an nxn matrix of distinct integers such that the sum of each row, column, *and main diagonal* must be equal) is the main diagonals rule. One way to think of a magic square template is as a set of n^2 integer unknowns K_(i,j), 1 <= i,j <=n required to all be unequal, and having 2n+2 specified subsets (of size n) required to all have the same sum. Clearly, for any two row or column subsets S_1, S_2, there's at least one permutation of the K(i,j)'s carrying S_1 onto S_2. But (except for the 2x2 case -- which is magic-squarefree even without the main diagonals rule) there is no permutation of the K(i,j)'s taking a row or column onto a main diagonal. This lack of symmetry is what I find less than interesting about traditional (ordinary) magic squares. SO: What I'd find much more interesting is the following: Let a "magic template" T = (K,S) "of shape (r,s)" denote a set K of r unknowns R_i, 1 <= i <= r, and a set S of subsets S_j of K, each of size s, such that: * for any two of the subsets, there is a permutation of K carrying one onto the other. * a "magic pattern" (nomenclature pending) modeled on T is defined as an assignment of distinct integers to the K(i)'s such that each subset S_j has the same sum. -------------------------------------------------------------- The more symmetric the configuration of subsets, the more appealing the the magic template. QUESTION: What are good ways to find highly-symmetric magic templates, especially ones that have at least one magic pattern? (I guess the cases where the number of unknowns is no smaller than the number of subsets -- i.e., equations -- have the best chance of being realizable.) Some candidates might include a) Use an n-dimensional cube of side L (so L^n unknowns) and its C(n,d)*L^(n-d) e-dimensional slices (e-cubes of size L^e) parallel to the coordinate axes. b) Use n unknowns and all subsets of size s -- the subsimplices of fixed dimension of an (n-1)-simplex. c) If these don't work, then perhaps highly-symmetrical subsets of the sets of subsets in a) or b). Are there standard ways of finding such things? --Dan P.S. I'm well aware that people have come up with such generalizations of magic squares before -- e.g., in Clifford Pickover's book "The Zen of Magic Squares, Circles, and Stars: An Exhibition of Surprising Structures Across Dimensions".

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Re: [math-fun] Different values of the dot product
by N. J. A. Sloane 03 Nov '07

03 Nov '07

I think it makes a nicer problem to include the case i=j For m <= n then the answer is 2 For m from n+1 to 2n you can get the answer 3 by taking a coordinate frame +- e_i, i = 1..n For larger m there are some bounds in the work of Delsarte, Goethals and Seidel that are relevant. See P. Delsarte, J. M. Goethals and J. J. Seidel, Bounds for systems of lines, and Jacobi polynomials, {\em Philips Research Reports}, {\bf 30} (1975), 91--105. Neil

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[math-fun] Esthetic expressions
by wouter meeussen 03 Nov '07

03 Nov '07

Sum[1/Product[1+1/k^2 ,{k, n}]/n^2,{n,Infinity}] == 1 -1/Product[1+1/k^2,{k,Infinity}] ==1-Pi/Sinh[Pi] and, playing in the same garden, I suppose the following is generally known (among those that know such things) Sum[HarmonicNumber[n]/n^w,{n,\[Infinity]}] == Sum[-Zeta[k]Zeta[1+w-k],{k,2,w-1}]/2+(w+2)/2 Zeta[w+1] but I can't find it in http://mathworld.wolfram.com/RiemannZetaFunction.html nor in http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/ With thanks to Olivier Gerard and Bill Gosper for their patient tutoring. Wouter.

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Re: [math-fun] PDF searchable text
by Eugene Salamin 03 Nov '07

03 Nov '07

I use OpenOffice. I prefer it's equation editor to the one in MS Word, and documents can be converted to PDF for the convenience of others who do't use OpenOffice. I haven't yet tried searching the generated text. OpenOffice is downloadable for free from http://wwwopenoffice.org , and runs under Windows, Linux, and Mac. ----- Original Message ---- From: Steve Gray <stevebg(a)adelphia.net> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Friday, November 2, 2007 5:52:19 PM Subject: Re: [math-fun] PDF searchable text When I need a PDF for math I write the paper in Word (and MS Equation) and use BullZip Printer to make a PDF file. I use Tex only as a very last resort because I can't remember how to use it for equations, graphs, etc. from one use to the next, and I dislike the fact that user-visible embedded codes should have disappeared 25 years ago. BullZip seems to work well for what I've used it for so far, and it's about $20. I am ready to be called an iconoclastic crank for this opinion. Steve Gray Fred lunnon wrote: > I compile TeX directly to PDF using free software called TeXShop --- > I've found this system robust and reliable, if lacking fancy facilities > such as inline WYSIWIG availalble on some other (PC) systems. > It runs on pretty much any available harware/software, and the > output produced is searchable by other PDF viewers (I tried). > > WFL > > _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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[math-fun] Martin Gardner's Letter to Santa at YouTube
by Thane Plambeck 03 Nov '07

03 Nov '07

I'm experimenting with the possibility of uploading videos of talks given at the Gathering for Gardner to YouTube. I'd be interested in what people think of the video quality of this talk fragment from a 2006 presentation by Dana Richards at G4G7 titled "Martin Gardner: The Tulsa Years." The video starts with a letter to Santa that Martin wrote when he was "eight or nine" years old. The whole video clip is about 5 minutes long. http://www.youtube.com/watch?v=UP4aT9sRUgs -- Thane Plambeck tplambeck(a)gmail.com http://www.plambeck.org/ehome.htm

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Re: [math-fun] Prime question
by Henry Baker 03 Nov '07

03 Nov '07

At 03:39 AM 11/2/2007, Joshua Zucker wrote: >http://mathworld.wolfram.com/CunninghamChain.html and >http://hjem.get2net.dk/jka/math/Cunningham_Chain_records.htm > >might be a good enough starting point for learning about these things. > >Enjoy, >--Joshua Zucker On the upper right hand corner of this page, there is a link called "Download PDF 9.5M". I clicked on it, and downloaded what appears to be the entire book of 308 pages. You can also click on "View plain text", which will give you some idea of how the character recognition program is working (used for indexing). Since algebraic equation recognition isn't doing so well, yet, the equations get trashed. It might be nice to be able to download both the pdf and the ascii text, so that you can search the book yourself, as most books have completely useless indices (the major exception being Knuth's). http://books.google.com/books?id=aC0PAAAAIAAJ&printsec=frontcover&dq=intitl… Good luck!

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[math-fun] nice little sin(a+b+c+...) formula
by Bill Gosper 03 Nov '07

03 Nov '07

n n ==== /===\ \ k | | %pi k > (- 1) | | 2 sin(a - -----) n / | | j n + 1 ==== ==== j = 0 \ k = 0 sin( > a ) = ------------------------------------, / k 2 (n + 1) ==== k = 1 where %pi n a := -----. 0 2 There are n^2 sins on the right, vs n2^(n-1) for the usual addition formula. But all n^2 sins are different, vs only 2n species for the usual formula. E.g., for n=3, %pi %pi sin(a + a + a ) = - sqrt(2) sin(a + ---) sin(a + ---) 3 2 1 1 4 2 4 %pi %pi %pi %pi sin(a + ---) - sqrt(2) sin(a - ---) sin(a - ---) sin(a - ---) 3 4 1 4 2 4 3 4 + 2 sin(a ) sin(a ) sin(a ) 1 2 3 = - sin(a ) sin(a ) sin(a ) + cos(a ) cos(a ) sin(a ) 1 2 3 1 2 3 + cos(a ) sin(a ) cos(a ) + sin(a ) cos(a ) cos(a ) 1 2 3 1 2 3 We can trade # of terms vs # of species by applying the new(?) formula recursively, e.g., a_1 = b_11+b_12+b_13, a_2= b_21+b_22+... . --rwg __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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[math-fun] Re: submarine betting
by Steve Witham 02 Nov '07

02 Nov '07

My previous guess about the minimum number of shots needed to shoot n x n subs, call that m, was: m = n ln n Which is way too high. Now that we have a better idea how many shots are needed, I have decided to revise my guess. In order to be "realistic," I need to know the number of distinct "strategies" of m shots, since out of n ^ m possible sequences of moves, many are symmetrical or even identical. So given a function s( n, m ) = the number of distinct strategies for m shots at n x n subs ask, what's the probability for a given m, that none of the possible shooting strategies will shoot all the subs? I get this: ( 1 - ( 1 - ( 1 - 1/n )^m ) ^ (n^2) ) ^ s( n, m ) (If s(n,m) were just n^m, then this expression in C would be pow(1-pow(1-pow(1-1/n,m),pow(n,2)),pow(n,m)) which surely should sink a lot of subs.) The m where this comes out to .5 is the median of where I should expect the minimal m. I calculated for n = 10, using my guess at the s() function: n m s( n, m ) P(success of any strategy) P(failure of all) 10 16 2.1e+12 1.3e-09 0 10 15 4.6e+11 9.7e-11 4.0e-20 10 14 1.0e+11 5.2e-12 0.59 10 13 2.3e+10 1.8e-13 0.996 So, it should probably take more than 14 shots, but not more than 15 (it takes 17). Michael Kleber <michael.kleber(a)gmail.com> responded to my high guess with something that's stuck with me: >But surely the structure of the problem should give us some >leverage and let us do better than that, right? If my new calculations are plausible, they seem to say the structure of the problem is working against us. http://www.tiac.net/~sw/math-fun/pfailall.c --Steve

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