May 2014 - mathfun@mailman.xmission.com

Re: [math-fun] sum 1/Fib[n]
by Bill Gosper 25 May '14

25 May '14

George Andrews>This is the negative of the generating function for the divisor numbers and is not reducible to simpler functions. It is possible to transform it. For example:, it equals inf 2 ==== n n \ q (1 + q ) > -------------. / n ==== q - 1 n = 1 However, this only speeds up convergence [symbolically, even--rwg] but does not provide a simpler form. George Well, it does if we admit dGamma_q[x]/dx as a "simpler" form. To repeat, Sum[1/Fibonacci[n], {n,∞}] == Sqrt[5]/(4*Log[ϕ])* (2*Im[QPolyGamma[0, 1/2 + I*Pi/(4*Log[ϕ]), 1/ϕ^2]] - QPolyGamma[0, 1, ϕ^2] + Re[QPolyGamma[0, 1 - (I*Pi)/(2*Log[ϕ]), ϕ^2]]) Mathworld seems to accept qGamma', and in light of the 𝜗 connection at the bottom of this mail, perhaps we all should. Most of us accept Gamma_q[x] as closed form. When we accept a special function, do we accept its derivatives? In some cases, e.g. dZeta_s(a)/ds, it seems more like just making up a name than evaluating the sum. But it's good to have a name. We can use it until such happy time as someone finds the answer in less exotic terms. I wish I could remember whether I used to think otherwise --rwg The qGamma reflection and tuplication formulæ haven't yet appeared at functions.wolfram.com, but I strongly suspect they will. Likewise for QPolyGamma. ⥛ * ∞ = 8 ----- Original Message ----- From: "Bill Gosper" <billgosper(a)gmail.com> To: math-fun(a)mailman.xmission.com Sent: Friday, May 23, 2014 3:00:22 AM Subject: sum 1/Fib[n] Argh! From: rwg(a)sdf.lonestar.org < rwg(a)sdf.lonestar.org > [Jul 12. 2008 16:01]: [...] but still nothing about plain old inf ==== n \ q > ------, / n ==== q - 1 n = 1 with which we could sum the reciprocal Fibonacci numbers. Jörg>Have you checked Borwein/Borwein "Pi and the AGM"? pp.91-101 might make you happy. rwg>Sadly not)-: They denote this Lambert series -L(q), but never give it in terms of Thetas or ThetaPrimes. Despite the section heading, they never deliver Sum 1/Fib(n) except in terms of L. It may well be inexpressible without some new special function. (They do give Sum 1/Fib(2*n+1) in Thetas, which is easy.)<rwg But did they give it in terms of q-digamma and I somehow missed or rejected it? http://mathworld.wolfram.com/ReciprocalFibonacciConstant.html claims they do! Could it have been added in a later edition? My (autographed) first edition is unhandy. Here I thought I had capped my 40 yr search, and the answer is sitting in Mathworld. The only consolation is that my answer doesn't use 𝜗s, so we get this peculiarity: EllipticTheta[2, 0, ϕ^-2]^2 = 2 Im[QPolyGamma[0, 1/2 + I π/(4 Log[ϕ]), ϕ^-2]]/Log[ϕ] ~ 3.26379 Can this generalize, or is it only about the Golden Ratio? --rwg

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[math-fun] sum 1/Fib[n]
by Bill Gosper 25 May '14

25 May '14

Argh! From: rwg(a)sdf.lonestar.org <rwg(a)sdf.lonestar.org> [Jul 12. 2008 16:01]: [...] but still nothing about plain old inf ==== n \ q > ------, / n ==== q - 1 n = 1 with which we could sum the reciprocal Fibonacci numbers. Jörg>Have you checked Borwein/Borwein "Pi and the AGM"? pp.91-101 might make you happy. rwg>Sadly not)-: They denote this Lambert series -L(q), but never give it in terms of Thetas or ThetaPrimes. Despite the section heading, they never deliver Sum 1/Fib(n) except in terms of L. It may well be inexpressible without some new special function. (They do give Sum 1/Fib(2*n+1) in Thetas, which is easy.)<rwg But did they give it in terms of q-digamma and I somehow missed or rejected it? http://mathworld.wolfram.com/ReciprocalFibonacciConstant.html claims they do! Could it have been added in a later edition? My (autographed) first edition is unhandy. Here I thought I had capped my 40 yr search, and the answer is sitting in Mathworld. The only consolation is that my answer doesn't use 𝜗s, so we get this peculiarity: EllipticTheta[2, 0, ϕ^-2]^2 = 2 Im[QPolyGamma[0, 1/2 + I π/(4 Log[ϕ]), ϕ^-2]]/Log[ϕ] ~ 3.26379 Can this generalize, or is it only about the Golden Ratio? --rwg

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Re: [math-fun] Golyhedra
by Adam P. Goucher 25 May '14

25 May '14

Greetings. It would appear that even 15 is suboptimal, since Alexey Nigin has subsequently found a 12-faced golyhedron! https://yadi.sk/d/dmqoxL8kR6fSC A remarkable property of this golyhedron -- which must necessarily be possessed by an 11-faced golyhedron, if one exists -- is that it is `monolithic'. In particular, it is possible to build it from unit cubes with no adhesive such that the structure does not collapse under gravity. Equivalently, it can be easily built in layers by a 3D printer. 1 1 8 1 5 5 Monolithic polyhedra can be described by an configuration of numbers* giving the height (in cubes) of each `tower'. The configuration above is a description of Alexey's 12-faced golyhedron. (* formally, a finitely-supported function from Z^2 to the non-negative integers) This makes exhaustively searching for 11-faced golyhedra suddenly tractable, since monolithic polyhedra are easy to represent. I expect that it will not be long before we either prove Alexey's golyhedron optimal or find an 11-faced example. Here is the current timeline of golyhedra discovered so far: 2014-04-29: 32 (Goucher) 2014-05-09: 15 (Nigin) 2014-05-16: 19* (Friedman) 2014-05-24: 12 (Nigin) * majorised by an earlier but independent result. Sincerely, Adam P. Goucher Sent: Friday, May 16, 2014 at 2:48 PM From: "Erich Friedman" <efriedma(a)stetson.edu> To: "Adam P. Goucher" <apgoucher(a)gmx.com> Subject: Re: [math-fun] Golyhedra adam thanks for the update. i'm going to guess 15 is optimal. erich On May 16, 2014, at 9:36 AM, Adam P. Goucher wrote: > Erich, > > Very nice. You seem to have also had the excellent idea of using only one club foot; Alexey Nigin found a 15-face golyhedron using a 4-by-2 club foot and a similar method: > > http://cp4space.wordpress.com/2014/05/11/golyhedron-update/ > > This means that the minimum number of faces is either 11, 12 or 15. > > Both your golyhedron and Alexey's are considerably better than my 32-face behemoth. > > > Sincerely, > > > Adam P. Goucher > >> ----- Original Message ----- >> From: Erich Friedman >> Sent: 05/16/14 01:42 AM >> To: Adam P. Goucher, Ed Pegg, alexanderbellos(a)googlemail.com >> Subject: Re: [math-fun] Golyhedra >> >> adam >> >> i found a golyhedron with 19 faces. george sicherman deserves credit for giving me the good idea of a 2x3 club foot. and you deserve credit for your snake idea. there is no golyhedron of the same form with 15 faces, but i haven't checked all different club feet. or there may be some other construction possible for smaller n. >> >> a picture is enclosed. i would post this to math-fun, but i can't post for some reason. feel free to forward it if you think people care. >> >> erich >

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[math-fun] Atropa Belladonna
by Warren D Smith 24 May '14

24 May '14

Q1. Prove that (a) Every point has some knight-move neighbour at smaller knight-move distance from the origin if and only if (b) Every point has some knight-move neighbour at greater knight-move distance from the origin. --first of all (a) is false, a counterexample is (0,0). However, (a') is true, meaning (a) for every point other than (0,0). This is simply because for ANY connected graph, and ANY two unequal vertices A and B in it, there exists a point 1-nearer to A than B. (Here we use the knight-adjacency graph.) And you can find it by breadth-first search from A to find all points at distance 1, then all at 2, then 3, etc. Now I don't think you really were trying to get me to prove (b) is false. You wanted me to prove (b) true. It seems to me that (a') is obvious as above but (b) is not so obvious. However, in your paper draft there is a result that not only is (a') true, but indeed a 1-nearer neighbor exists in a certain angular wedge. Given that lemma, and perhaps some manual examination of cases at small distance (say distance<9) too, it seems to me that (b) then follows because the set of points "1-nearer", after taking union with all angular rotations and also union with all small-distance cases verified manually, is the entire universe. >Q2. Prove(!) that there is just a single shortest knight path from (0,0) to (2r,r) taking distance r steps. --this path is a "geodesic." That is, the furthest EUCLIDEAN distance you can hop in r steps, is sqrt(5)*r, obviously, which happens iff all steps are exactly parallel, which proves uniqueness. >Q3. Denote e(x, y) the number of shortest knight paths from (0,0) to (x,y) . Prove that e(1999999999, 3) = e(1500000000, 1499999998) . --Let me chop out 7 digits because otherwise I have trouble counting them: "Prove that e(199, 3) = e(150, 148)." Well... think I'll skip that one for now. If you edit the draft to incorporate my last round of abuse, I'll ponder the new draft. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Dick Hess email address?
by James Propp 24 May '14

24 May '14

Does anyone have a current address for Dick Hess? Please reply privately, since the fact that Dick hasn't made his email address readily available on the web might mean he's trying to make it available only to friends, friends-of-friends, and so on. (Of course, you might rightly reply "Maybe Dick doesn't want YOU to have his email address, Jim!") Jim P.S. I was going to give this email the subject line "Dick Hess", but I realized that when I get email addresses whose subject line is just someone's name, a significant portion of the time it means that the person has died, and I didn't want to give anyone any momentary worries. It's kind of like the feeling I get when I pass a radio and overhear Terry Gross interviewing someone who hasn't done anything newsworthy lately; my first thought is always "Oh no, did he/she just die?"

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[math-fun] A Not-So-Random Walk
by Bill Gosper 22 May '14

22 May '14

Good grief: In[239]:= Position[Sign[Convergents[2*Zeta[3], 777]/Convergents[Zeta[3], 777] - 2], 0] Out[239]= {{1}, {2}, {3}, {4}, {678}, {679}, {720}, {722}, {723}, {724}, {725}, {727}} vs In[240]:= Position[Sign[Convergents[2*Zeta[5], 77]/Convergents[Zeta[5], 77] - 2], 0] Out[240]= {{1}, {18}, {19}, {21}, {22}, {23}, {25}, {31}, {39}} --rwg FromHans Havermann: In preparation for next month's tau-day I've been playing with the continued fraction thereof. Consider the sign of nth convergent of 2*pi divided by the nth convergent of pi, minus 2. The first sixty are: 0, 1, -1, 1, -1, 1, -1, 0, 1, 0, 0, 0, 0, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 0, 0, 1, 0, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, 0, 0, 0, -1, 1, -1, 1, -1. I wasn't particularly surprised that stretches of non-zeros alternated sign. What impressed me was that the alternation continued *in step* after arbitrary-length stretches of zeros. _______________________________________________

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[math-fun] A Not-So-Random Walk
by Hans Havermann 22 May '14

22 May '14

In preparation for next month's tau-day I've been playing with the continued fraction thereof. Consider the sign of nth convergent of 2*pi divided by the nth convergent of pi, minus 2. The first sixty are: 0, 1, -1, 1, -1, 1, -1, 0, 1, 0, 0, 0, 0, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 0, 0, 1, 0, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, 0, 0, 0, -1, 1, -1, 1, -1. I wasn't particularly surprised that stretches of non-zeros alternated sign. What impressed me was that the alternation continued *in step* after arbitrary-length stretches of zeros.

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[math-fun] Closed form question
by Bill Gosper 22 May '14

22 May '14

David Wilson>If you considered >f(n, x) = SUM(n = 0..inf, 1 / (x^n + 1)) I think you mean f(1,x). >to be a closed form, could we find a closed form for >f(a, b) = SUM(n = 0..inf, 1 / (a^n + b^n))? Mathematica doesn't think so, even with its all-purpose In[86]:= Sum[1/(1 - a*x^n), n] Out[86]= -((-n Log[x] + QPolyGamma[0, n - Log[1/a]/Log[x], x])/Log[x]) (Note indefinite sum.) Which can even, with some work, express the ever-nagging Out[139]= (1/(4 ArcCsch[2])) Sqrt[5] (-QPolyGamma[0, 1, GoldenRatio^2] + I (QPolyGamma[0, 1/2 - (I π)/(4 ArcCsch[2]), 1/GoldenRatio^2] - QPolyGamma[0, 1/2 + (I π)/(4 ArcCsch[2]), 1/GoldenRatio^2]) + Re[QPolyGamma[0, 1 - (I π)/(2 ArcCsch[2]), GoldenRatio^2]]) In[140]:= N[{%, Sum[1/Fibonacci[n], {n, ∞}]}] Out[140]= {3.35989 + 0. I, 3.35989} --rwg

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[math-fun] quasiperiodic analytic functions, quasicrystals, and abelian functions
by Warren D Smith 21 May '14

21 May '14

OK, given any elliptic function, i.e. doubly-periodic meromorphic function W(z) of complex z, e.g. a Weierstrass function if you want concreteness, the function of a real variable x W((a+b*i)*x) will in general be quasiperiodic. Next, we can construct quadruply-periodic analytic functions of two complex variables. More interestingly, there is a 100+ year-old theory of quadruply periodic functions of 2 complex variables, whose output also is a 2-tuple of complex numbers, and each output is meromorphic in each input (or any fixed complex linear combination of outputs is meromorphic in any fixed complex linear combination of inputs). There also are generalizations to any dimension p -- the input is a complex p-tuple and so it the output, and the function has 2p period vectors -- these are called "Abelian functions." I will call them "Abelian functions of order p." The case p=1 is elliptic functions. The case p=2 is the quadruply-periodic things in 2 complex dimensions. It was used by Jacobi 1834, Georg Rosenhain 1851 and Adolph G"opel 1847 to solve the inversion problem for "hyperelliptic integrals of the first kind" (integrands have denominator which is square root of polynomial of degree 5 or 6). B.Riemann and K.Weierstrass and Felix Klein then redid and superseded this work. Henry Frederick Baker: An Introduction to the Theory of Multiply Periodic Functions, Cambridge Univ Press 1907 now available as google e-book http://books.google.com/books?id=0EQLAAAAYAAJ Multiply periodic functions, Cambridge Univ Press 2007 [I assume is a re-issue of the same book, perhaps with updates?] Henry Frederick Baker: Abelian Functions: Abel's Theorem and the Allied Theory, Including the Theory of the Theta Functions. Cambridge, England: Cambridge University Press, 1995 e-book: https://archive.org/details/abeltheoralltheor00bakerich A more recent book is David Mumford: Curves and their Jacobians, Uni Michigan Press 1975. A promising-sounding modern re-do of all this is here (online book): http://arxiv.org/abs/1208.0990 But the only source I actually read is this encyc. of maths article: http://www.encyclopediaofmath.org/index.php/Abelian_function So anyway, if you take any abelian function of order 2 i.e. F1(z1,z2) and F2(z1,z2) and take a generic 1-complex dimensional "cross section" of it, i.e. F1(a*z+b, c*z+d), F2(a*z+b, c*z+d) with a,b,c,d, generic complex constants, then you are going to get a pair of meromorphic functions which are both "quasicrystalline" in nature in the complex z-plane, like the Penrose tiling. For example the locations of the zeros, or poles, both should be sets of points forming a "quasicrystal." If you plotted contourish plots of them, pretty pictures should result. If the zeros were the locations of positive ions and the poles of negative ions presumably this would be a fake quasicrystalline chemical substance. (CRAZY QUESTION: do these functions have anything to do with the "Painleve transcendents"?) -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] More spurious correlations
by Henry Baker 21 May '14

21 May '14

FYI -- "Big Data" = a myth is a good as a mile. http://www.tylervigen.com/

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