December 2007 - mathfun@mailman.xmission.com

[math-fun] Smallest magic squares of powers
by Christian Boyer 01 Sep '08

01 Sep '08

Oct 7th, 2007: the smallest known magic square of 4th powers is now a 44x44 square constructed by Jaroslaw Wroblewski (Poland) and Hugo Pfoertner (Germany). Jaroslaw built the semi-magic square in June, then Hugo needed a very impressive computing job of 3 months, on 57 processors of a SGI computer http://www.sgi.com/products/remarketed/origin3000/, in order to find two magic diagonals in this square! A big progress, because the previous smallest known was a 243x243 square, coming from the 243x243 tetramagic square of Pan Fengchu (China). The best known results, and their first authors, are now: Magic square Smallest known ------------ --------------- Of squares 4x4 (Euler, 1770) Of cubes 9x9 (Boyer, 2006) Of 4th powers 44x44 (Wroblewski-Pfoertner, 2007) Of 5th powers 729x729 (from the pentamagic square of Li Wen, 2003) ON ANY POWER, WE DO NOT KNOW WHAT THE SMALLEST POSSIBLE MAGIC SQUARE IS! For example, as you know, we still do not know if a 3x3 magic square of squares is possible... 4x4, 5x5, 6x6, 7x7, and 8x8 magic squares of cubes are unknown... And the problem of the diagonals is extremely difficult, as proved by the impressive computing search of Hugo. It remains a big gap between the smallest known magic and semi-magic squares (=with or without magic diagonals) of 4th powers: 44x44 versus 4x4... Here is the 4x4 semi-magic square of 4th powers (Morgenstern-Boyer, 2006): 14101^4 938^4 931^4 37762^4 35866^4 20881^4 21038^4 13393^4 24806^4 30191^4 30418^4 9263^4 413^4 32026^4 31787^4 1106^4 Christian.

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Re: [math-fun] How we "do mathematics"
by Eugene Salamin 02 Feb '08

02 Feb '08

----- Original Message ---- From: David Gale <gale(a)math.berkeley.edu> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Thursday, August 9, 2007 3:17:48 AM Subject: RE: [math-fun] How we "do mathematics" ... >A (very) old one I like : > >Two Death Valley marathonians friends complete their run >together and seek shadow in their tent nearby the finish >line. > >The first one had prepared a fresh 5-liter tank of water >in a cooling box and his friend a similar 3-liter one. > >They were about to start drinking when a journalist enters >the tent desperately asking for water. > >OK, the water is divided into three equal parts -- and drunk. > >The thankful journalist insists to pay his share and leaves >8 dollars to be split between the two friends. > >How? > >---- Why is this problem so fascinating? The two friends are paid $8. The guy who contributed 5 l gets $5, and the guy who contributed 3 l gets $3. Or maybe the $8 goes towards paying for their campsite. I must be missing something. Gene ____________________________________________________________________________________ Pinpoint customers who are looking for what you sell. http://searchmarketing.yahoo.com/

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Re: [math-fun] Compute a Function From its Fourier Series?
by Bill Gosper 17 Jan '08

17 Jan '08

GFee>See the following paper. >Authors: Gottlieb and Shu Title: On the Gibb's phenomenon and its resolution Journal: SIAM Review Issue: Vol.39, No. 4, pages 644-668, December 1997 Also see http://www.tweedledum.com/rwg/gibbs.htm --rwg --------------------------------- Tonight's top picks. What will you watch tonight? Preview the hottest shows on Yahoo! TV.

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Re: [math-fun] xkcd points out dangers of math fun
by Fred lunnon 03 Jan '08

03 Jan '08

On 12/15/07, Steve Witham <sw(a)tiac.net> wrote: > Here's a comic with an interesting random-walk problem: > > http://www.xkcd.com/356/ > Apparently quoted from Google Labs aptitude test --- On an infinite square lattice with equal unit resistances along each edge, find the resistance between two nodes at a knight's move from one other. I got an extraordinary answer for this --- has anybody else had a go at it? WFL

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[math-fun] Proofs without words
by Dan Asimov 01 Jan '08

01 Jan '08

Can someone please point me to proofs without words for a) nth tetrahedral number = n(n+1)(n+2)/6 b) sum of the first n squares = n(n+1)(2n+1)/6 c) sum of the first n cubes = (sum of first n numbers)^2 = (T_n)^2 (T_n = nth triangular #) ? Thanks, Dan

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[math-fun] d x e = r and now concatenate e and r
by Eric Angelini 31 Dec '07

31 Dec '07

Hello Math-Fun and SeqFans, still a few hours to go in 2007, in Brussels, Belgium -- and thus time for our last (uninteresting) seq of the year! S=1,11,22,24,28,216,318,324,416,424,624,728,756,848,864, 936,954,1144,1248,1296,1378,13104,26104,27108,27216,... All integers of S are the concatenation of e and r, r being the "r"esult of (d x e) and d being the last "d"igit of the previous integer. Integer a(n+1) must be bigger than a(n) but the gap between a(n) and a(n+1) must be the smallest possible one. No integer of S can end with a zero. Example: Start with a(1)=1 then d=1 and, obviously, e=1 and r=1 --> S=1,11,... Now d=1 again and we must take e=2 ; thus r=2 and --> S=1,11,22,... Now d=2 and the closest integer to the last one (22) can be found with e=2 ; thus r=4 and --> S=1,11,22,24,... Etc. In other words, we multiply the last digit of a(n) by something, we concatenante the result immediately after this something, and we look for the smallest integer resulting in this concatenation but bigger than a(n). Voilà! Cheers, happy 2008, mind the headaches! É.

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[math-fun] "Greek calculator" puzzle
by Marc LeBrun 30 Dec '07

30 Dec '07

Given a straightedge, compass, unit segment and another of length X, please finitely construct sqrt(X), or show why not.

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[math-fun] Surfer Theory of Everything ?
by Henry Baker 20 Dec '07

20 Dec '07

Any thoughts? An Exceptionally Simple Theory of Everything http://arxiv.org/abs/0711.0770

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[math-fun] atan(tan(x)/2)
by Bill Gosper 18 Dec '07

18 Dec '07

For real x, tan(x) 3 x x 1 x 1 x atan(------) = atan(tan (-)) - pi floor(-- + -) + pi floor(---- + -) + -, 2 2 pi 2 2 pi 2 2 implying that 3 2 atan(x) 1 x atan(x ) = pi floor(--------- + -) + atan(------) - atan(x), pi 2 2 1 - x and 2 x atan(x) sqrt(x + 1) - 1 3 atan(-) = ------- + atan((----------------) ) 2 2 x I was fiddling with GFee's Fourier query. A while ago I mentioned gene_salamin's observation that you could get rational function nth harmonics to shrink like n^-k, for any positive integer k, by *composing* the approximated function with one that k-smoothly (actually (k-2)-smoothly) stops and resumes at each discontinuity. The catch was large numerical coefficients delaying the onset of the faster convergence with increasing k. E.g., suppose you draw a square with epicyclic rotors. If you jump from corner to corner, you get k=1 convergence (and Gibbs ringing) and need thousands of rotors (harmonics) for screen resolution. If instead you sweep the square at constant speed, you get k=2 and need only dozens. If you come to a smooth stop at each corner, you get k=3, etc., needing fewer and fewer rotors, leading to the absurd limit that eventually you would need only one or zero! But what actually happens when you get too greedy (k too big) is the postponement of the amplitude dropoff as a function of n. So each resolution target (error bound) has its optimal k, which presumably varies with the function being approximated. But GFee didn't mention a speed-smoother option, so we're left with matching k derivatives at 0 and -1 == 1, which, I think, exhibits the same behavior. So I was looking at patch functions like 2/pi*atan(tan(t*pi/2)^(2*k+1)) for sneaking from 0 to 1 with 2k vanishing derivatives at the endpoints. Or even tanh(tan(t*pi/2)), with *all* derivatives vanishing. I don't really have time to investigate this, but the temptation may be overwhelming. Oh, hell, I gave in and tried f(t) := t 2 2 %e ((%e - 1) tanh(cot(%pi t)) + %e + 1) ------------------------------------------ 2 on [-1,0], which matches all derivatives at both ends, and got g(t) = 1.97473445780695d0 - 0.8981270448759d0 cos(pi t) - 0.41753308527721d0 sin(pi t) - 0.1247188401558d0 cos(2 pi t) + 0.23434279646825d0 sin(2 pi t) + 0.00330568635158d0 cos(3 pi t) + 0.03901681282034d0 sin(3 pi t) + 0.00107335808438d0 cos(4 pi t) + 0.04714152469382d0 sin(4 pi t) + 0.02622378383461d0 cos(5 pi t) + 0.00949607729559d0 sin(5 pi t) + 0.00589694508514d0 cos(6 pi t) + 7.34937084126405d-4 sin(6 pi t) + 0.00925330770208d0 cos(7 pi t) + 7.6772973803263d-6 sin(7 pi t) + 0.00221574684403d0 cos(8 pi t) - 0.00414748855076d0 sin(8 pi t) + 0.00175756423966d0 cos(9 pi t) - 0.00109985016356d0 sin(9 pi t) + 4.19777476400997d-4 cos(10 pi t) - 0.00251568862979d0 sin(10 pi t) - 2.97177678883817d-4 cos(11 pi t) - 6.82066527883218d-4 sin(11 pi t) which is rather disappointing: g(0) = 1.0017, g(1) = 2.7175 . Possibly the numerical integration is poor, but Maple 5.4 gave 13 place agreement. Strangely, I couldn't coax more than six places out of Mma 5.2. Maybe an atan(tan^k) hack would be better. --rwg Input forms: atan(tan(x)/2) = atan(tan(x/2)^3)-%pi*floor(x/%pi+1/2)+%pi*floor(x/(2*%pi)+1/2)+x/2 atan(x^3) = %pi*floor(2*atan(x)/%pi+1/2)+atan(x/(1-x^2))-atan(x) atan(x/2) = atan(x)/2+atan(((sqrt(x^2+1)-1)/x)^3) PS: Mathematica knows I from (-I: In[1]:= N[DedekindEta[I]] Out[1]= 0.7682254223260567 - 1.7568472086707517*^-19*I In[2]:= N[DedekindEta[-I]] Out[2]= DedekindEta[0. - 1.*I] (* Undefined. Diverges. :*) Clear thinking is much too hard. Ask any reporter. --------------------------------- Get easy, one-click access to your favorites. Make Yahoo! your homepage.

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[math-fun] xkcd points out dangers of math fun
by Steve Witham 15 Dec '07

15 Dec '07

Here's a comic with an interesting random-walk problem: http://www.xkcd.com/356/ --Steve

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