July 2009 - mathfun@mailman.xmission.com

[math-fun] On to 67 and 163 !
by rwg＠sdf.lonestar.org 23 Jul '09

23 Jul '09

Gack! Apologies Fred, I meant The INsignificance of your irreverence cannot be underestimated! Bit of good fortune: The Jacobi symbol collapses those gruesome Gamma products in the eta identities. See Out[24] in http://gosper.org/nuetas.html . --rwg

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Re: [math-fun] pebbles via computer
by James Propp 22 Jul '09

22 Jul '09

I wrote: >Is there any kind of virtual environment in which one can play with >puzzles like this? Here's an example of the kind of puzzle I'd like to play with in a non-physical way: the states of the puzzle are the permutations of the numbers 1 through n=6 (though you can use other values of n if you like), and a legal move swaps two numbers if there is a number that is between them positionally and between them numerically (a "see-saw move"). E.g., in the permutation 1 4 2 5 3 6, you can't swap the 2 and the 3, because the only number that's between them positionally (5) isn't between them numerically, but you can swap the 2 and the 6 because the 5 is between them both positionally and numerically. (Note that the inverse of a see-saw move is again a see-saw move, so this puzzle is reversible.) Steve Linton wrote a computer program that checked that all the permutations of 1 through 6 can be obtained from one another via see-saw moves, but I'd like to come up with a non-brute force proof, and the most natural way to do that is to develop some intuition by playing around. If Conway were still posting to math-fun, he'd probably reply that I'd be better off just making numbered counters and moving them around. His early work on the Game of Life was done by hand, and he's of the opinion that some of the insights he gained from this couldn't have been gained if he'd just had a computer do the simulations. But I don't want to lose track of how I got to where I am. At the same time, I want the ability to backtrack. And I want to be able to mess around at the speed of thought, without having to write stuff down. Jim Propp

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[math-fun] Mathematica Package for Complex Differential Forms
by Ki Song 18 Jul '09

18 Jul '09

Hi Math-Fun, I was wondering if there was a mathematica package that can handle complex differential forms and Dolbeault operators. It seems like RICCI can do real differential forms... Best, Ki Song -- If I am not working, then I must be depressed.

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Re: [math-fun] There's one side to every story
by Dan Asimov 17 Jul '09

17 Jul '09

For any point x of the (say open) Moebius band M in R^3, there is an open 2-disk neighborhood D of x in M, and D lies in a 3-ball B in R^3 that is "cylindrical" in that it is homeomorphic to D x (-1,1), and such that D x {0} corresponds to B \int M. The two components -- or more accurately the *fact* of the two components -- D x (-1,0) and D x (0,1) of B - Dx{0} is why the Moebius band (or any surface embedded in R^3 ) is LOCALLY two-sided. But if you try to do this GLOBALLY, any embedded normal bundle of M in R^3 (aka a "tubular neighborhood"), *minus* the 0-section (the part corresponding to M itself), has only ONE component. This fact is why it's called one-sided (rigorously). --Dan P.S. Plus, if you make one, any 5-year-old can plainly see it has only one side. ARRIVISTE A STRIVER, I rwg wrote: << We like to say that a Moebius strip embedded in 3-space has one side, whereas a unit disk there has two. How can we say that? What are the "sides" of a 2-dimensional set of points? It is its own boundary. >> _____________________________________________________________________ "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] pebbles via computer
by James Propp 17 Jul '09

17 Jul '09

Following up on my post with the subject line "(x^6-1)/(x^2-x+1) via pebbles", Joshua Zucker gave the URL of the video I was asking about: >http://www.mathlesstraveled.com/?p=81 Thanks, Joshua! So now that I have that link, I can ask my real question: What's the thoroughly modern way to work on puzzles like this? One can make physical models, but that's just so twentieth century. (Or maybe I should say "so nineteenth century": think of all those white plaster models of arcane surfaces that are sitting around in seldom-open display-cases in dozens of math departments around the country.) Is there any kind of virtual environment in which one can play with puzzles like this? I'm thinking of something like Harold and his purple crayon, but Harold could only draw static objects. I think I've heard about a "purple crayon physics" computer game in which one draws objects that come to life; that's more the sort of thing one would need. One advantage of a virtual environment is that it could keep track of what I've done, so I wouldn't be in the frustrating situation of having solved a puzzle but not remembering how I did it! (Which reminds me: I once managed to tie two mirror-image trefoil knots in a rope and pass one of them through the other in such a way as to make them cancel, leaving a completely unknotted rope. Now of course this is topologically impossible, as Conway showed in a very elegant fashion; so what I really came up with is some way of *appearing* to disprove Conway's theorem. This would make a fun little party trick for confusing drunk topologists. Trouble is, I can't remember how I did the trick!) Jim Propp

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[math-fun] clean 16-th order AGM iteration
by Joerg Arndt 17 Jul '09

17 Jul '09

AGM16(a,b)= \\ return AGM(a,b) { local(P,M,R); a = sqrt(a); b = sqrt(b); while ( abs(a-b) > eps, \\ eps == 10^(-precision) P=(a+b)/2; M=(a-b)/2; R=(P^4-M^4)^(1/4); a=(P+R)/2; b=(P*R*(P^2+R^2)/2)^(1/4); ); return( a^2 ); } The rate of convergence is _very_ impressive: ? t = AGM16(1.0, 0.5) [1.00000000000000, 0.292893218813452] \\ == [a, a-b] [0.853460904507906, 2.94235252084151 E-17] [0.853460904507906, 3.16563558120117 E-273] [0.853460904507906, 1.02028230345421 E-4368] [0.853460904507906, 1.38315649004407 E-69896] [0.853460904507906, 0.E-100009] \\ using a precision of 100,000 digits 0.728395515523453 \\ returned value ? t-agm(a,b) -4.08056715461103 E-100010 \\ check

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[math-fun] (x^6-1)/(x^2-x+1) via pebbles
by James Propp 16 Jul '09

16 Jul '09

Can anyone steer me toward an online (You-tube?) video of someone using pebble-moves to prove that x^6 - 1 is a multiple of x^2 - x + 1, by starting with a single pebble and managing to move it six spaces to the left (or maybe it was to the right?) with moves that replace a pebble at n with pebble at each of n-1 and n+1, and other moves that do the reverse? Thanks, Jim Propp

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Re: [math-fun] On to 67 and 163 !
by Dan Asimov 16 Jul '09

16 Jul '09

Let me be the first to predict that, in the future, Ramanujan will be known as the Bill Gosper of the 20th century. --Dan rwg wrote: << Bit of good fortune: The Jacobi symbol collapses those gruesome Gamma products in the eta identities. See Out[24] in http://gosper.org/nuetas.html . >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] clean 16-th order AGM iteration
by Dan Asimov 16 Jul '09

16 Jul '09

(Just playing around with the standard definition of AGM, I see that it tends to converge to 15 decimal places almost always within 5 or 6 iterations -- so it already has impressive convergence.) Define an equivalence relation via (x,y) ~ (x',y') whenever AGM(x,y) = AGM(x',y'). Then what do the equivalence classes look like as subsets of (0,oo)^2 ? They're most likely all curves; what are their equations? --Dan Also -- and this is probably well-known -- it seems that the "weighted AGM": a(n+1) = p*a(n) + q*g(n) g(n+1) = a(n)^p * g(n)^q where 0 < p < 1, p+q= 1, always converges for positive a(0),g(0) as well. _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: Springer Verlag Publishes âProofâ of Goldbachâs Conjecture (& FLT)
by James Propp 15 Jul '09

15 Jul '09

I wrote >They mean Flt (Fermat's little theorem), not FLT (Fermat's Last Theorem). but I retract this. (I was skimming through the page and jumped to conclusions.) Sorry! Jim

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