February 2011 - mathfun@mailman.xmission.com

[math-fun] Continued fractions and the hyperbolic plane
by Mike Stay 03 Feb '11

03 Feb '11

It's well known that continued fractions and the (3,infty) tiling of the hyperbolic plane are deeply connected. What happens if we move to the (4,infty) tiling of the plane? It's the Cayley graph of the free group on two generators; as a gasket, you place two "equal" (in the hyperbolic sense) circles in the triangular gap instead of one. Do we get a notion of continued fraction with binary strings instead of natural numbers? What's analogous to the modular group? -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

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Re: [math-fun] exact and not exact formulas
by Bill Gosper 03 Feb '11

03 Feb '11

Hello mr. Gosper, yes, very interesting, explicit formulas : this is what I need. I included your formula for the case k=2/7 on the 3rd version of my preprint here : http://arxiv.org/abs/1101.6066 (tell me if there is an error ?). thanks for all this, très respectueusement, Simon Plouffe ----------- Yes, that's a better way to write it. You could mention that all sums Sum[n^3/(E^(2*Pi*n/k) - 1), {n, Infinity}] for rational k seem to be 3 2 n 1 Pi A[k] Sum[----------------, {n, Infinity}] == -(---) + --------- (2 n Pi)/k 240 3 8 -1 + E Gamma[-] 4 for some algebraic number A[k]. E.g., for k=6, 3 n Sum[--------------, {n, Infinity}] == (n Pi)/3 -1 + E 1/4 3/4 2 1 (231 + 120 Sqrt[2] 3 + 140 Sqrt[3] + 60 Sqrt[2] 3 ) Pi -(---) + ------------------------------------------------------------ 240 3 8 320 Gamma[-] 4 so they are all readily found with PSLQ. For increasing k, they become very close to rationals, e.g., In[474]:= ContinuedFraction[%[[2]], 7] Out[477]= {5, 2, 1, 1, 9, 7896393272, 8} For k=13 (your 2.2), you could write Sum[n^3/(-1 + E^((2*n*Pi)/13)), {n, Infinity}] == -(1/240) + (1/Gamma[3/4]^8)*(Pi^2*Root[-2543222153227391999 + 1780150088904190080*#1 - 5828105673767424000*#1^2 - 1264509990338560000*#1^3 - 1050320673177600000*#1^4 - 48338514739200000*#1^5 + 1073741824000000*#1^6 & , 2]) I see by your remark following 2.7 that you already knew Sum[(1/(n * (E^(k * n * Pi) - 1))),{n, 1, Infinity}==Log[F(k)] ! Maybe you should state that formula explicitly. And you can redo 2.3 thru 2.7, since we know the values of all those F[k]. (We can even do a fair number with k=sqrt(rational). This should all go into FunctionExpand.) --rwg

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[math-fun] exact and not exact formulas
by Bill Gosper 03 Feb '11

03 Feb '11

Simon, all you need is Jacobi's imaginary transformation F(y) = F(4/y)*e^(pi*(1/y-y/4)/6)*sqrt(y)/sqrt(2) or his aequatio satis abstrusa 1/F(2*y)^24 = 1/(F(y)^16*F(4*y)^8)+16*e^-(pi*y)/(F(y)^8*F(4*y)^16) to get F(1/2) from the results you quote. These, with 1/(F(y)^18*F(2*y)^8) = 256*e^-(pi*y)/(F(2*y)^24*F(5*y)^2)+1/(F(y)^24*F(5*y)^2)-250*e^-(pi*y)/(F(y)^12*F(2*y)^8*F(5*y)^6)-3125*e^-(2*pi*y)/(F(y)^6*F(2*y)^8*F(5*y)^12) and maybe 1/F(y)^6 = 1/(F(y/5)^5*F(5*y))+5*e^-(pi*y/5)/(F(y/5)^4*F(5*y)^2)+15*e^-(2*pi*y/5)/(F(y/5)^3*F(5*y)^3)+25*e^-(3*pi*y/5)/(F(y/5)^2*F(5*y)^4)+25*e^-(4*pi*y/5)/(F(y/5)*F(5*y)^5) should give you exact values for all the rest. --rwg Hello, I am collecting these approximations, I have some which are of interest , let's say that F is the partition function of Euler, infinity --------' ' | | 1 F(x) = | | ------ | | n | | 1 - x n = 1 now I put x-> exp(-Pi*x), then I have this EXACT formula (well tested to 1000 digits). 5 20 28 F(1/5) F(4/5) F(2) ---------------- = ------------ 5 31 7 F(2/5) F(1) F(4) Now here is the funny thing, 4 2 F(1/12) F(1/36) ----------------- = exp(Pi) 2 4 F(1/9) F(1/24) This formula is valid up to 48 decimal digits. From the theory of partitions and other theories about these kind of functions, the values of F(1), F(2) and F(4) are well known (sort of). Here are the EXACT known values : 3/8 2 GAMMA(3/4) --------------- = F(1) Pi 1/4 exp(----) Pi 24 1/2 2 GAMMA(3/4) --------------- = F(2) Pi 1/4 exp(----) Pi 12 7/8 2 GAMMA(3/4) --------------- = F(4) Pi 1/4 exp(----) Pi 6 But on the other hand for 1/2 : 16 F(1/2) = 1/65536 exp(5 Pi) is an approximation to 6 decimals. Interesting isn't ? I made a search for simpler formulas, I could not, these are the simplest terms I know. Bonne journée, Simon Plouffe

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[math-fun] just rubber bands
by James Propp 02 Feb '11

02 Feb '11

What sorts of structures can be made out of rubber bands (and nothing else)? Making balls is easy, and it's possible to make an approximate cube, as http://boingboing.net/2005/11/26/cube-made-from-rubbe.html demonstrates, but what about other shapes? Jim Propp

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[math-fun] exact and not exact formulas
by Simon Plouffe 02 Feb '11

02 Feb '11

Hello, I am collecting these approximations, I have some which are of interest , let's say that F is the partition function of Euler, infinity --------' ' | | 1 F(x) = | | ------ | | n | | 1 - x n = 1 now I put x-> exp(-Pi*x), then I have this EXACT formula (well tested to 1000 digits). 5 20 28 F(1/5) F(4/5) F(2) ---------------- = ------------ 5 31 7 F(2/5) F(1) F(4) Now here is the funny thing, 4 2 F(1/12) F(1/36) ----------------- = exp(Pi) 2 4 F(1/9) F(1/24) This formula is valid up to 48 decimal digits. From the theory of partitions and other theories about these kind of functions, the values of F(1), F(2) and F(4) are well known (sort of). Here are the EXACT known values : 3/8 2 GAMMA(3/4) --------------- = F(1) Pi 1/4 exp(----) Pi 24 1/2 2 GAMMA(3/4) --------------- = F(2) Pi 1/4 exp(----) Pi 12 7/8 2 GAMMA(3/4) --------------- = F(4) Pi 1/4 exp(----) Pi 6 But on the other hand for 1/2 : 16 F(1/2) = 1/65536 exp(5 Pi) is an approximation to 6 decimals. Interesting isn't ? I made a search for simpler formulas, I could not, these are the simplest terms I know. Bonne journÃ©e, Simon Plouffe

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