May 2007 - mathfun@mailman.xmission.com

[math-fun] Wolfram 2,3 Turing Machine Research Prize
by Eric Angelini 16 May '07

16 May '07

Universal Turing Machine ? http://www.wolframscience.com/prizes/tm23/ É.

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Re: [math-fun] Torus tiling question
by James Propp 15 May '07

15 May '07

>Can someone prove the converse: > >"If T can be tiled by N congruent squares, then N is the sum >of 2 square numbers." . . . > >. . . or come up with a counterexample ??? > >--Dan This is a really nice question; has anyone made progress with it? For instance, can someone show that you can't tile a torus with 3 congruent squares? If nobody reports any progress with Dan's question, then I'll mention it on the domino forum (see jamespropp.org/ about-domino); I'm sure someone there will have some ideas. Jim

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[math-fun] Mathematical ringtones for your cellphone
by Henry Baker 14 May '07

14 May '07

Mathematical ringtones for your phone: http://www.cs4fn.org/rtones/

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[math-fun] Mayan Calendar and the Chinese Remainder Theorem
by wouter meeussen 13 May '07

13 May '07

playing with the Mayan calendar, cfr http://hermetic.nofadz.com/cal_stud/maya/chap1.htm I noticed that some web pages allow conversion of the Gregorian date to the Mayan Tzolkin-Haab date, but never give the inverse. Calculation of this inverse needs the Chinese Remainder Theorem: given integers u, v and w, find n so that mod(n,20)=u, mod(n,13)=v, mod(n,365)=w (with the added complication that 20 and 365 share a common factor of 5). Would the Mayan priests have had a trick to solve this (in Roman times)? for Mathematica fans: http://users.pandora.be/Wouter.Meeussen/mayan_calendar.nb enjoy, Wouter.

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Re: [math-fun] Torus tiling question
by Michael Reid 13 May '07

13 May '07

> T = R^2/Z^2 > "If T can be tiled by N congruent squares, then N is the sum > of 2 square numbers." . . . i think this is relatively easy to prove: lift such a tiling to a tiling of R^2 . it is easy to show that any tiling of R^2 by congruent squares is invariant under some translation, with length equal to the side length of the squares. let (x, y) be this translation. since the tiling is lifted from T , it is also invariant under (1, 0) and (0, 1) translations. the group of translational invariants must be a discrete subgroup of R^2 , so this means that (x, y) is commensurate with the lattice generated by (1, 0) and (0, 1) . in other words, some (non-zero) multiple of (x, y) is in the lattice ; say M(x, y) is that multiple. now we may write x = a/M , y = b/M for some integers a, b, M . the side length of the square is 1 / sqrt(N) , because N of them tile T . thus we have (a^2 + b^2) N = M^2 . from this equation, we see that any prime = 3 mod 4 must divide N with even multiplicity. therefore N is a sum of two squares. qed mike

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[math-fun] Torus tiling question
by D. Asimov 13 May '07

13 May '07

It's probably well known in some quarters that for any integer N > 0 that's the sum of 2 square numbers, the torus T = R^2/Z^2 with the induced metric -- aka the "square torus" -- can be tiled by N congruent square tiles. (Note that such N include all square numbers.) It's a blast to work out how this goes, so I won't spoil your fun. I'm not sure how to prove the converse -- or certain that it's true (though this seems likely). Can someone prove the converse: "If T can be tiled by N congruent squares, then N is the sum of 2 square numbers." . . . . . . or come up with a counterexample ??? --Dan

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[math-fun] Canonical sequence of rational approximants to an irrational
by Daniel Asimov 08 May '07

08 May '07

Given any irrational c > 0, there's a canonical sequence of rational approximants to c, defined as the maximal sequence of rationals with increasing denominators that get increasingly close to c. I calculated the first few terms for pi, e, and sqrt(2). I wasn't surprised to find the denominator sequence for pi was already in OEIS, but not those for e or sqrt(2) . . . and no written references were mentioned. Is there a name for such sequences? Is anything much known about them (e.g., speeds of approximation, growth rates of denominators, relationship to continued fractions) ? (Seems that these sequences could be used to characterize properties of the irrational, though it's not clear whether they could be any more useful than the continued fraction.) Here are the first 19 terms of each sequence: approximants to pi: 3/1, 13/4, 16/5, 19/6, 22/7, 179/57, 201/64, 223/71, 245/78, 267/85, 289/92, 311/99, 333/106, 355/113, 52163/16604, 52518/16717, 52873/16830, 53228/16943, 53583/17056, ... approximants to e: 3/1, 5/2, 8/3, 11/4, 19/7, 49/18, 68/25, 87/32, 106/39, 193/71, 685/252, 878/323, 1071/394, 1264/465, 1457/536, 2721/1001, 12341/4540, 15062/5541, 17783/6542, ... approximants to sqrt2: 1/1, 3/2, 4/3, 7/5, 17/12, 24/17, 41/29, 99/70, 140/99, 239/169, 577/408, 816/577, 1393/985, 3363/2378, 4756/3363, 8119/5741, 11482/8119, 19601/13860, 47321/33461, ... --Dan

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[math-fun] Re: Winding line puzzle
by James Propp 04 May '07

04 May '07

>As is well know, for any irrational number c, there exists >an injective immersion of the reals into the square torus, >I_c: R -> T = R^2/Z^2 given by > > I_c(x) = q(x, cx) > >where q: R^2 -> R^2/Z^2 is the quotient mapping. (Such an >image I_c(R) is called a winding line in the torus.) > >The image I_c(R) inherits a the subspace topology from T. > >PUZZLE: Given any irrational numbers c <> d, are the spaces >I_c(R) and I_d(R) (with the subspact topologies) necessarily >homeomorphic? Another great question! Jim

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[math-fun] Winding line puzzle
by Daniel Asimov 04 May '07

04 May '07

As is well know, for any irrational number c, there exists an injective immersion of the reals into the square torus, I_c: R -> T = R^2/Z^2 given by I_c(x) = q(x, cx) where q: R^2 -> R^2/Z^2 is the quotient mapping. (Such an image I_c(R) is called a winding line in the torus.) The image I_c(R) inherits a the subspace topology from T. PUZZLE: Given any irrational numbers c <> d, are the spaces I_c(R) and I_d(R) (with the subspact topologies) necessarily homeomorphic? --Dan

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[math-fun] Correction re simple group of order 168
by Daniel Asimov 04 May '07

04 May '07

In a recent post on the Klein quartic I stated that the simple group of order 168 (call it G) is a subgroup of PSL(2,C) (PSL(2,C) is one way to express the group of all fractional linear transformations over C, aka the group of conformal automorphisms of S^2 =C{^1.) This is incorrect; this group is *not* a subgroup of PSL(2,C). (The finite subgroups of PSL(2,C) are the same as the finite subgroups of SO(3): the cyclic groups C_n, dihedral groups D_n, and rotation groups of the Platonic solids: A_4, S_4,and A_5). G is, however, a subgroup of PSL(3,C) (the group of projective transformations of CP^2, where the Klein quartic naturally lives). (Oddly enough, as I learned from an algebraic geometer friend, the conformal automorphisms of projective varieties in CP^2 of degree <= 4 always extend to CP^2 itself, but this is not necessarily true for degree >= 5.) --Dan

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