November 2009 - mathfun@mailman.xmission.com

[math-fun] No. of words of length n in certain group
by N. J. A. Sloane 22 Nov '09

22 Nov '09

To Math Fun, Seq Fan: I ran into my distinguished (former) colleague Colin Mallows yesterday, and he said that it would be nice if someone would extend his sequence A154638. Take the infinite reflection group with 5 generators S_1, ..., S_5, satisfying (S_i)^2 = (S_i S_j)^3 = I, and let a(n) be number of distinct elements whose minimal representation as a product of generators has length n: 1, 5, 20, 70, 240, 780, 2730, .. (for n>= 0) Can anybody help extend this sequence? Is there a generating function? What about other groups? This might be a gaping hole in the OEIS! There must be a huge literature on this problem. The books that I know about that might be related, Lyndom & Schupp, Combinatorial Group Theory, Magnus, Karrass, Solitar, same title, Johnson, Presentations of Groups, aren't exactly full of sequences, as far as I can tell. Can some expert please help? Neil

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[math-fun] Mysterious filtered message
by rcs＠xmission.com 21 Nov '09

21 Nov '09

For reasons known only to itself, the mailing list software trapped the message below. --Rich ------------------ Date: Mon, 16 Nov 2009 07:57:46 -0800 Subject: Re: [math-fun] incredible images From: "Stephen B. Gray" <stevebg(a)roadrunner.com> To: math-fun <math-fun(a)mailman.xmission.com> Simon Plouffe wrote: > hello all, > > These are incredible images of the Mandelbrot set > in 3 dimensions. > > http://www.skytopia.com/project/fractal/mandelbulb.html > > similar to the Romanesco cauliflower > > http://images.google.fr/images?q=chou+romanesco&oe=utf-8&rls=com.ubuntu:fr:… > > simon plouffe > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > Absolutely astounding, a kind of art I've never seen before. Thrilling! Steve Gray

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[math-fun] Booby-trapped recursive codes
by Henry Baker 21 Nov '09

21 Nov '09

One of the coolest things I worked with in high school was the IBM IBSYS "MAP" (Macro Assembly Program), which ran on the 709/7090/704/7040. Fletcher utilized this macro facility to solve a combinatorial problem; as I understand, the generated assembly code itself was thrown away, because all of the computation was done within the macro expansion phase! Fletcher, John G. "A program to solve the pentomino problem by the recursive use of macros". CACM 8, 10 (Oct. 1965), 621-3. http://portal.acm.org/citation.cfm?id=365654&dl=GUIDE&coll=GUIDE&CFID=62648… Similar hacks have been used with the preprocessor for the C language. These hacks are far more difficult with the C preprocessor due to the constraints installed to prevent looping & stack overflows. Nevertheless, some pretty cool things can still be done. See the "Obfuscated C" home page for more info: http://www.ioccc.org/ I noticed while Googling that you can still run 7090 code on your Windoze machine, should you wish to do so: http://www.piercefuller.com/oldibm-shadow/709x.html http://www.cozx.com/~dpitts/ibm7090.html At 11:53 AM 11/12/2009, Fred lunnon wrote: >My PDF viewer did cope with the PostScript original --- after several minutes of >frantic computation and memory-swapping, and the production of a 3 Mbyte file! > >Many years ago, I wrote similarly booby-trapped recursive PostScript for Penrose >kites-and-darts tilings. [This improved upon a more primitive earlier program, >which had itself been regarded with such awe by the mainframe maintenance >team that they used it to test the flatbed plotter.] > >The program achieved some notoriety when a firm of toilet-paper manufacturers >employed it to decorate their product. Sir Roger Penrose was not amused, and >threatened to take them to court. I have no recollection of how the case was >finally resolved, but it did lead at the time to a certain amount of unkind ribaldry >at his expense, in connection with some rather controversial cosmological speculation. > >Fred Lunnon > >On 11/12/09, Mike Stay <metaweta(a)gmail.com> wrote: >> That's much better; my pdf viewer didn't render anything but the three circles. >> >> On Thu, Nov 12, 2009 at 10:16 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> > It is indeed a fractal. Look more closely -- there are individual dots. >> > >> > Try printing this Postscript version, which actually computes the fractal inside your printer using embedded Postscript code: >> > >> > http://home.pipeline.com/~hbaker1/sigplannotices/sigcol07.ps.gz >> > >> > (".gz" means "gzip"; I believe that "7-zip" can ungzip this file for you. >> > >> > There is a small program called "PrintFile" which can send Postscript files on Windows to your Postscript printer. >> > >> > If worst comes to worst, install Ghostscript on your computer & look at the output that way. >> > >> > At 10:06 AM 11/12/2009, Mike Stay wrote: >> >>Figure three seems wrong in this rendering--shouldn't it be a fractal? >> >> >> >>On Thu, Nov 12, 2009 at 9:47 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> >>> That was my paper! >> >>> >> >>> http://home.pipeline.com/~hbaker1/sigplannotices/sigcol07.pdf >> >>> >> >>> At 09:19 AM 11/12/2009, mcintosh(a)servidor.unam.mx wrote: >> >>>>Re: [math-fun] Cube root of a complex number >> >>>> >> >>>>somewhat tangential to the original question, MÃ¶bius >> >>>>transformations map three points into three points. So, >> >>>>why not map the three roots of the cubic into the three >> >>>>complex roots of unity? You only get to use the coefficients >> >>>>of the polynomial. >> >>>> >> >>>>I recall a paper in an ACM journal humorously dated >> >>>>March 32 some years ago where someond did that; I don't >> >>>>remember if the solution was relevant to the present >> >>>>inquiry. >> >> >> >>-- >> >>Mike Stay - metaweta(a)gmail.com >> >>http://math.ucr.edu/~mike >> >>http://reperiendi.wordpress.com >> >> -- >> Mike Stay - metaweta(a)gmail.com >> http://math.ucr.edu/~mike >> http://reperiendi.wordpress.com

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Re: [math-fun] AP news quote
by Dan Asimov 20 Nov '09

20 Nov '09

Marc wrote: << This AP clipping maybe isn't as innumerate as it first seems, but still... "British clothing sizes are normally two sizes above clothing sold in the U.S., so a 'size zero' in the United States would be a size 4 in Britain." >> Right. See < http://www.size-0.com/ >. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] AP news quote
by Marc LeBrun 20 Nov '09

20 Nov '09

This AP clipping maybe isn't as innumerate as it first seems, but still... "British clothing sizes are normally two sizes above clothing sold in the U.S., so a 'size zero' in the United States would be a size 4 in Britain."

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Re: [math-fun] incredible images
by Dan Asimov 17 Nov '09

17 Nov '09

A computer graphics student of mine, around 1985, made some pictures of the quaternionic Mandelbrot set as his term project. They certainly *appeared* more complicated than the ordinary M. set. But it's true that any non-real quaternion, together with 1, spans a complex subspace of the quaternions, whose arithmetic is identical to the complex numbers, and so would give the usual Mandelbrot set. (Using the standard definition, non-commutativity plays no role.) So the different appearance of the quaternionic M. set might be due to its being a 4D object that was projected to 3D in order to view it. (Can *that* be what the cool computer graphics are doing?) --Dan ------------- Gareth wrote: << It's alleged (on a page linked to from the one we're talking about) that the obvious quaternionic generalization of the Mandelbrot set is basically the ordinary Mandelbrot set plus some rotational symmetry and has no "extra" fractal structure to it. This seems plausible enough on the face of it, but I've not made any attempt to check. I share your feeling that it's all a bit ad hoc and humdrum, and that it should be possible to get as much or more aesthetic beauty with more mathematical elegance. >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] incredible images
by Simon Plouffe 16 Nov '09

16 Nov '09

hello all, These are incredible images of the Mandelbrot set in 3 dimensions. http://www.skytopia.com/project/fractal/mandelbulb.html similar to the Romanesco cauliflower http://images.google.fr/images?q=chou+romanesco&oe=utf-8&rls=com.ubuntu:fr:… simon plouffe

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Re: [math-fun] incredible images
by Dan Asimov 16 Nov '09

16 Nov '09

<< On 11/16/09, Simon Plouffe <simon.plouffe(a)gmail.com> wrote: > hello all, > > These are incredible images of the Mandelbrot set > in 3 dimensions. > > http://www.skytopia.com/project/fractal/mandelbulb.html > > similar to the Romanesco cauliflower > > http://images.google.fr/images?q=chou+romanesco&oe=utf-8&rls=com.ubuntu:fr:… > > simon plouffe >> Amazing pictures. But it's a bit frustrating that there are so many links in the text on that page, that it seems impossible to figure out just exactly what the pictures are showing. Maybe if I had an hour or two to spare it would become clear. But could someone please just provide a link to text that explains just what the pictures (toward the bottom part of the page) are showing? Thanks, Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] characteristic polys of real symmetric matrices
by Henry Baker 16 Nov '09

16 Nov '09

We would like to represent the polynomial t^2+2*b*t-c as the characteristic polynomial of a 2x2 matrix. By translating the origin, we can get rid of b, which makes the following analysis more perspicuous. So we have p(t)=t^2-c as the polynomial we want to represent by the symmetric real matrix [x y] [y z] = M. This matrix M satisfies its own characteristic polynomial, so we have the two equations: y*(x+z)=0 x^2+y^2-c=0 The first equation gives z=-x, while the second gives x=+- sqrt(c-y^2). So, M is [-sqrt(c-y^2) y] [y sqrt(c-y^2)] Thus, so long as |y|<sqrt(c), M will represent p(t). In general, for 2x2 matrices, y^2 enters as an addition to the discriminant, so we have some freedom to choose y if the discriminant is positive. Unfortunately, we have effectively solved the polynomial p(t) already, prior to constructing M, so it would appear that constructing M is not easier than solving p(t).

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Re: [math-fun] Polynomials with real roots
by Henry Baker 15 Nov '09

15 Nov '09

Yes, you are right, I wasn't very clear. Let's assume that the polynomial has no repeated roots, since those can be detected with rational operations only. I'd like to construct an Hermitian matrix with the same real roots, but _without solving for the roots_; i.e., the elements of the matrix are in the field of the polynomial coefficients, prior to being extended all the way to the root field. So your answer gets close to the answer to my question -- thanks for the pointer. In the case where the answer is no, can we make modest extensions -- e.g., square roots -- that would get us such a matrix w/o extending to the full root field. I assume that the generality of Hermitian v. real symmetric doesn't help, does it? At 10:56 AM 11/14/2009, victor miller wrote: >Perhaps there's some condition you're not telling me. Since every real >diagonal matrix is Hermitian the answer is obviously yes (just put the roots >on the diagonal). There is a related arithmetic question -- Suppose that a >monic polynomial with integer coefficients has all real roots, is it the >characteristic polynomial of a symmetric integer matrix? The answer is no >-- there's a series of papers about this by Ed Bender and Norman Hertzberg >around 1970. > >Victor > >On Sat, Nov 14, 2009 at 10:21 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: > >> Stupid matrix question: >> >> I know that the characteristic polynomial of a Hermitian matrix must have >> all real roots (it is diagonalizable into a real diagonal matrix). >> >> But is every polynomial with only real roots the characteristic polynomial >> of a Hermitian matrix? >> >> If so, is there a construction that takes one from the polynomial (in the >> form of a vector of coefficients) to the Hermitian matrix? >> >> (The construction from the roots themselves is trivial: they form a >> diagonal matrix that can be rotated by any unitary matrix.) >> >> Presumably, such a construction would fail if/when not all of the roots are >> real.

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