January 2011 - mathfun@mailman.xmission.com

[math-fun] Fwd: One more conjectural series for 1/pi
by rcs＠xmission.com 30 Jan '11

30 Jan '11

ZW Sun has posted three conjectured series for 1/pi to the NMBRTHRY list. I posted his first series to MathFun a couple of weeks ago. Since then, he's come up with two more. -- Rich ----- Forwarded message from zwsun(a)nju.edu.cn ----- Date: Sat, 29 Jan 2011 12:30:17 -0600 From: Zhi-Wei SUN <zwsun(a)nju.edu.cn> Reply-To: Number Theory List <NMBRTHRY(a)listserv.nodak.edu>, Zhi-Wei SUN <zwsun(a)nju.edu.cn> Subject: One more conjectural series for 1/pi To: NMBRTHRY(a)listserv.nodak.edu Dear number theorists, For integers b and c let T_n(b,c) denote the coefficient of x^n in the expansion of (x^2+bx+c)^n. I have announced two conjectural series for 1/pi: Sum_{k=0,1,2,...}(30k+7)binom(2k,k)^2*T_k(1,16)/(-256)^k = 24/pi and Sum_{k=0,1,2,..}(15k+2)binom(2k,k)binom(3k,k)T_k(18,6)/972^k = 45*sqrt(3)/(4*pi). Here I release a new conjectural series for 1/pi. CONJECTURE (Zhi-Wei Sun, Jan. 29, 2011). We have Sum_{k=0,1,2,...}(40k+3)binom(4k,2k)binom(2k,k)T_k(98,1)/12544^k = 70*sqrt(21)/(9*pi). Note that this new series also converges rapidly. Let s(n)=sum_{k=0}^n(40k+3)binom(4k,2k)binom(2k,k)T_k(98,1)/12544^k. By Mathematica, if n is at least 340 then |s(n)*9pi/(70*sqrt(21))-1|<10^{-100}. I have included this conjecture and the related congruences in the latest version of my preprint (arXiv:1101.0600) which will be available from the next Tuesday. In general, I suggest the investigation of series for 1/pi of the following forms (with a,b,c,d,m integers, d positive and m(b^2-4c) nonzero): Type I. Sum_{k=0,1,2,...}(a+d*k)binom(2k,k)^2*T_k(b,c)/m^k; Type II. Sum_{k=0,1,2,...}(a+d*k)binom(2k,k)binom(3k,k)*T_k(b,c)/m^k; Type III. Sum_{k=0,1,2,...}(a+d*k)binom(4k,2k)binom(2k,k)*T_k(b,c)/m^k; Type IV. Sum_{k=0,1,2,...}(a+d*k)binom(6k,3k)binom(3k,k)*T_k(b,c)/m^k. My three conjectural series belong to type I,II,III respectively. I have not been able to find a series for 1/pi of type IV. Since T_k(b,c) (with b^2-4c nonzero) is involved, series of the above types are not hyper-geometric and this is one of the main reasons why such conjectural series for 1/pi are more difficult than Ramanujan-type series. There should be VERY FEW series for 1/pi of the above 4 types. If there are some such series that I have missed, you are welcome to pick them up. In my opinion, the three conjectural series for 1/pi might remain open for many years. Zhi-Wei Sun http://math.nju.edu.cn/~zwsun ----- End forwarded message -----

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[math-fun] The people you know...
by James Propp 28 Jan '11

28 Jan '11

I'm thinking of submitting the following bit of silliness to Mathematics Magazine: http://jamespropp.org/galois.pdf Comments are welcome! In particular, I'm pretty sure that statements fairly close to "The people you know are the people who know all the people who know all the people you know" already occur in print, and I'd like to give credit where credit is due (even if that prevents my article from fitting onto one page). Jim Propp

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Re: [math-fun] The people you know...
by Dan Asimov 28 Jan '11

28 Jan '11

Hmm, interesting. If we added a point at infinity to R^n as a topological space, there is a natural way to view this as S^n, the n-sphere. But what is the rigorous mathematical object corresponding to the edges of the integer grid in R^n (n >= 2) after a "point at infinity" has been added. Just curious. --Dan Veit wrote: << Connect the boundary of a large hypercubic resistor grid to "ground". Ground acts as a current source at zero potential. Now apply a voltage V to the node at the origin such that 1 amp is extracted there. By symmetry, the current flowing in each of the resistors connected to the origin node is 1/(2D) amp, where D is the dimension. The potential across these resistors is 1/(2D) volt since their resistance is 1 ohm. Now consider what happens when instead, voltage -V is applied to a neighboring node, say [1,0]. We have the same symmetrical distribution of currents, only the center is shifted and the current directions are reversed. On the other hand, the voltage across the resistor R connecting [0,0] to [1,0] is the same (in magnitude and sign) as it was before. Now superimpose these two current/voltage distributions. The voltage across R will be 2/(2D) and exactly one amp will flow out of [0,0] and into [1,0]. The equivalent resistance is therefore 1/D. >> Those who sleep faster get more rest.

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[math-fun] Vi Hart's doodles and my own contribution
by Mike Stay 28 Jan '11

28 Jan '11

I'm sure many of you have seen these by now, but just in case: http://www.youtube.com/user/Vihart I was so taken by them that I made one of my own: http://www.youtube.com/watch?v=RTVeMWJ8lbI -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

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Re: [math-fun] The people you know...
by James Propp 27 Jan '11

27 Jan '11

Lots of helpful responses; thanks, everyone! I was indeed assuming that "A knows B" is a symmetric relation. And it's not clear to me that this agrees with most people's informal use of the word. I'll have to think about this. Jim

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Re: [math-fun] The people you know...
by James Propp 27 Jan '11

27 Jan '11

I should mention that even if knowing is asymmetric, one can show that K(S) = K(L(K(S))), where L(S) is the set of all people who ARE KNOWN BY everyone in S. (But that doesn't go to the tune of "The Irish Washerwoman". :-) ) Jim

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Re: [math-fun] The people you know...
by Dan Asimov 27 Jan '11

27 Jan '11

This is interesting. I think the article would be at least slightly enhanced if the theorem is preceded by a rigorous definition of "know" (given a graph) -- even if the obvious guess is the right one -- and if the allusion to a Galois connection were accompanied by a definition thereof. (For example, I'd like to know whether a person automatically "knows" themself, even if this doesn't affect the theorem and its proof.) --Dan Jim wrote: << I'm thinking of submitting the following bit of silliness to Mathematics Magazine: http://jamespropp.org/galois.pdf Comments are welcome! In particular, I'm pretty sure that statements fairly close to "The people you know are the people who know all the people who know all the people you know" already occur in print, and I'd like to give credit where credit is due (even if that prevents my article from fitting onto one page). >> Those who sleep faster get more rest.

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[math-fun] "Implicitization" of non-rational parameterizations
by Henry Baker 27 Jan '11

27 Jan '11

I'm still hacking away at the baseball baserunner's problem, where paths/curves in 2D are produced by constant absolute value accelerations. The paths/curves in _velocity_ space are portions of catenaries (y=cosh(x), etc.), but the paths/curves in _position_ space aren't so easy to characterize. If s=asinh(t) (t time), then t=sinh(s) [sinh() is bijective]. acceleration a=[sech(s),tanh(s)] velocity v=[s-vx,cosh(s)-vy] (vx, vy are constants.) position z=[(s-vx)*sinh(s)-cosh(s),((cosh(s)-vy)*sinh(s)+s-vy*sinh(s))/2] Thus, z=[x,y] traces out a curve in 2D space parameterized by s (or alternatively t). Normally, we prefer a parameterized curve to a curve implicitly defined by a function f(x,y)=0, but sometimes it is nice to be able to see the curve w/o the extraneous parameter. If the parameterization utilized only rational functions, then it would be relatively easy to compute the implicit function using resultants, but the parameterization above goes beyond rational functions. I don't expect that the implicit function f(x,y) that I'm looking for is a polynomial, but I'd be interested if there exists an f(x,y) constructed from elementary functions that defines the same curve. Is there any straight-forward way to find such a function or prove that it doesn't exist?

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[math-fun] Re^2: FW: A conjectural series for 1/pi of a new type
by Bill Gosper 27 Jan '11

27 Jan '11

Simon>hello everybody, about that formula for 1/pi, SP>it is nice , the coefficient a(k) is a binomial term SP>The sequence a(k) is : 1, 1, 33, 97, 1729, 8001, 105441, 627873, 6989697, 48363649, 488206753, 3701949153, 35289342529, 283146701761, 2610495177057, 21695983405857, 196218339243777, 1667338615773441, 14917038493453089, 128562758660255073, 1143482133220664769, 9946278255903268929, 88205310329762729697, 771946983805271894433, 6837125121111415598721,... SP>a(k] has this g.f. 1 -------------------- = 2 1/2 (1 - 2 x - 63 x ) How did you get this? SP>Now, this has a closed binomial expression, of course Julian used the trinomial theorem to get a(k)==Hypergeometric2F1[1/2 - k/2, -(k/2), 1, 64] Did you mean something different? Nicer?? Julian's yields numerous triangular double sums for 24/pi, e.g. Sum[((Sum[((Binomial[2 * (2 * i - k), 2 * i - k])^2 * (2 * i - k)! * ( - 30 * k + 60 * i + 7) * ( - 1)^k * 2^(4 * k)/(((i - k)!)^2 * k!)), List[k, 0, i]])/(2^(12 * i))), List[i, 0, Infinity]], Sum[(((30 * k + 7) * k! * (Binomial[2 * k, k])^2 * Sum[((16^i)/((i!)^2 * (k - 2 * i)!)), List[i, 0, Floor[k/2]]])/(( - 256)^k)), List[k, 0, Infinity]], Sum[Sum[((16^i * (30 * (k + i) + 7) * (k + i)! * (Binomial[2 * (k + i), k + i])^2)/((i!)^2 * ( - 256)^(k + i) * (k - i)!)), List[i, 0, k]], List[k, 0, Infinity]], and this wild thing: 160/(7*Pi) + (28*EllipticK[(1/2)*(1 - (3*Sqrt[7])/8)]^2)/Pi^2 - (25*Gamma[1/7]^2*Gamma[2/7]^2*Gamma[4/7]^2)/(14*Sqrt[7]*Pi^4) - 30*Sum[(k*Binomial[2*k, k]^2* HypergeometricPFQ[{1 - k, 1/2 + k, 1/2 + k}, {1, 1 + k}, 1])/(-256)^k, {k, 1, Infinity}] (We're trying to get 24/pi and it gives us something plus 160/(7 pi).) But unless one of these can be written with the inner summand free of the outer index, these have an asymptotic convergence rate of *zero* bits/term. The >0 estimates cheat by assuming a(k) is free, or rather constant(k). It remains, however an intriguing conjecture. --rwg

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[math-fun] Re^2: Plouffe's 2 surprising approximations and 1 exact formula for pi.
by Bill Gosper 26 Jan '11

26 Jan '11

Holy bleep! It seems that SUM(1/(N*(%E^(%PI*K*N)-1)),N,1,INF) = -LOG(ETA(%E^-(%PI*K)))-%PI*K/24 inf ==== \ 1 - %pi k %pi k > ----------------- = - log(eta(%e )) - ----- / %pi k n 24 ==== n (%e - 1) n = 1 E.g., SUM(1/(N*(%E^(%PI*N/SQRT(3))-1)),N,1,INF) = -3*LOG(GAMMA(1/3))/2+LOG(%PI)+LOG(SQRT(3)+1)/4-3*LOG(3)/8+9*LOG(2)/8-%PI/(24*SQRT(3)) inf 1 ==== 3 log(gamma(-)) \ 1 3 log(sqrt(3) + 1) 3 log(3) 9 log(2) %pi > ----------------- = - --------------- + log(%pi) + ---------------- - -------- + -------- - ---------- / %pi n 2 4 8 8 24 sqrt(3) ==== ------- n = 1 sqrt(3) n (%e - 1) (so I misspoke about the 3/4 log(pi)-log(gamma(1/4)), which only appears for rational k). This can't be hard. S. Plouffe>Hello mr. Gosper, now that is interesting, you have the explicit formula for sum(1/n/(exp(Pi*n*k)-1),n=1..infinity) when k=1,2,3,4,5,6. rwg>In principle, k = sqrt(any rational). SP> Here with PSLQ or LLL I could get 1,2 and 4 only separatly in terms of log(Pi), Pi, log(2) and log(GAMMA(1/4)), I did not know that one could get those explicit algebrico-log-gamma expressions . What is surprising is the <cancel out> of the algebraic expression when k = 1/5, 2/5 and 4/5 and also the approximations when k = 2/13, 2/7 or 2/163, I tried to find other fractions for k and found only that explicit 1/5, 2/5 and 4/5. Nevertheles, I have found a simpler formulation of the formula for pi ; it appears on my home page at http://www.plouffe.fr/ <http://www.plouffe.fr/>rwg>Ah, but you still show only the approximate value of Sum[n^3/(E^(2*Pi*n/7)-1),{n,Infinity}== 1/4 3/4 1 8 (301 + 210 Sqrt[2] 7 + 120 Sqrt[7] + 90 Sqrt[2] 7 ) Gamma[-] 1 4 -(---) + ------------------------------------------------------------------ 240 6 5120 Pi (that quadrinomial probably factors some.) sp> I should switch to macsyma! best regards, Simon Plouffe I'm afraid the best hope for Macsyma at this point is for Noftsker to get it out of probate and the Maxima guys swallow their pride and cannibalize as much as possible. --rwg

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