December 2011 - mathfun@mailman.xmission.com

Re: [math-fun] Gamma function (formulas collection)
by Bill Gosper 10 Jan '12

10 Jan '12

wds>many formulas about gamma function are in my paper:http://rangevoting.org/WarrenSmithPages/homepage/gammprox.pdf _______________________________________________ Interesting stuff. Random remarks: P4 middle is garbled, at least in my Preview. Did you mean asymptosy vs asymptopia? P6, In[16]:= F[0, c] == Assuming[c > 0,Integrate[ x^(2*n + 1)/(1 + x^2)^(n + 1)*E^(-c*x) /. n -> 0, {x, 0, \[Infinity]}]] Out[16]= F[0,c] == (-Cos[c])*CosIntegral[c] + (1/2)*Sin[c]*(Pi - 2*SinIntegral[c]) In[12]:= G[0, c] == Assuming[c > 0, Integrate[ x^(2*n)/(1 + x^2)^(n + 1)*E^(-c*x) /. n -> 0, {x, 0, \[Infinity]}]] Out[12]= G[0, c] == CosIntegral[c]*Sin[c] + (1/2)*Cos[c]*(Pi - 2*SinIntegral[c]) In[14]:= H[0, c] == Assuming[c > 0, Integrate[ x^(2*n)/(1 + x^2)^(n)*E^(-c*x) /. n -> 0, {x, 0, \[Infinity]}]] Out[14]= H[0, c] == 1/c Eliminating G and H from the F recurrence, 4*(n-1)*n*F(n)=2*(n-1)*(6*n-5)*F(n-1)-(12*n^2-32*n+c^2+22)*F(n-2)+2*(n-2)*(2*n-3)*F(n-3) Mma gets only MeijerG for n>0 in all of these. P7 (40), see http://www.tweedledum.com/rwg/idents.htm near bottom of 1st screen. (42) and (43) raise the interesting question of evaluating Product[((1/(12*n)) + 1)*((n - 1/2)*(E^(1/n) - 1))^n,{n, 1, Infinity}] (i.e., prod(((n-1/2)*(%e^(1/n)-1))^n*(1+1/12/n),n,1,inf) --"cwg"

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Re: [math-fun] Life's Garden of Eden
by Bill Gosper 09 Jan '12

09 Jan '12

acw> I read this, and it is, unfortunately, not the discussion of the proof that I remembered. Could it have been that Conway presented the proof at the talk I went to? I know it was discussed. His "parlor trick" around then was to challenge people to draw Life patterns, and he would construct a predecessor. He was uncannily good at it. Record small GoE: http://homepage.tudelft.nl/37b0y/ Just confirmed by Achim Flammenkamp. --rwg

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Re: [math-fun] Puzzler
by N. J. A. Sloane 07 Jan '12

07 Jan '12

> > It is 72, I think, for x = 125 > Yes, you are right. It seems likely that 72 should suffice for all integers x. What about the sequence: a(n) = smallest k such that ... ? With a(125) = 72. Could you please submit it? (With keyword "base", of course) Neil

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[math-fun] Mental factoring
by Dan Asimov 07 Jan '12

07 Jan '12

Take a number like 2257 (say < 10000), easily checked to have no prime factors < 13. To find the least prime factor of such a number N, is there on average a faster way (after having excluded primes < 13) than the naïve method of dividing it by each prime in order <= sqrt(N) until one divides N evenly? --Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Countdown to 2012
by Hans Havermann 06 Jan '12

06 Jan '12

10*9*8+7+6-5+4*321 Happy New Year!

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[math-fun] hyperelliptic curves and AGM?
by Warren Smith 02 Jan '12

02 Jan '12

Victor Miller's cite of PhD thesis by Robert Carls was pretty useless to me -- seems to be a hellish marriage of number theory and topology (??!) without what I wanted, which is complex or real analysis. It seems amazing that they've somehow translated Gauss's AGM into the p-adic number theory world and now use it to do things like count points on finite field versions of "curves"... but at the same time, for mundane analysis purposes this seems useless. However, google then revealed these papers: Takayuki Kato and Keiji Matsumoto: http://www.math.sci.hokudai.ac.jp/~matsu/pdf/FD.pdf we define a (generalized) arithmetic-geometric mean among four terms and express it by Lauricella’s hypergeometric function FD of three variables. Kenji Koike & Hironori Shiga: http://mitizane.ll.chiba-u.jp/metadb/up/irtoroku2/06001.pdf A 3-variable AGM-like iteration yields a certain hyperelliptic integral, also an Appell 2-variable hypergeometric fn. There also are other papers by these same authors. They seem to in some of them be claiming a more general theory. This all seems an impressive breakthrough in analysis which probably will (or should) lead to various new "superfast" algorithms. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Gamma function (formulas collection)
by Warren Smith 01 Jan '12

01 Jan '12

many formulas about gamma function are in my paper: http://rangevoting.org/WarrenSmithPages/homepage/gammprox.pdf

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[math-fun] Year 2012 per the OEIS
by Marc LeBrun 31 Dec '11

31 Dec '11

Searching on just [2012] in the OEIS returns 13 pages of results--too many for me to sift for pearls right now. But [year 2012] returns just a single page, which includes the following sequences: A141039 Years in which there are five Wednesdays in the month of February. [next is 2040] A190653 Years with exactly three "Friday the 13ths", starting from 1901. [next is 2015] A171467 Years in which a transit of Venus (as seen from Earth) took place or is expected to occur, according to the catalog by Fred Espenak. [next is 2117] A125680 Blue moon years > 2000. [next is 2015] A162244 Years in which January first falls on a Sunday (since 2001) [next is 2017] Happy 2012!

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[math-fun] RIES and tan(1) (was re: Gamma(p/q) values)
by Robert Munafo 31 Dec '11

31 Dec '11

I'm just getting back to this now: 2011/12/29 Bill Gosper <billgosper(a)gmail.com> > [...] > But things like sin(1) and cos(2) are exceptionally rare, so sin and cos > should maybe be replaced with sin pi # and cos pi #. Irony: tan(1) is a > bit more > interesting due its continued fraction, but we'd need (sinpi > 1/pi)/(cospi 1/pi) ! > [...] tan(1) = 1.5574077246549022305069748..., with continued fraction convergents [1, 1, 1, 3, 1, 5, 1, 7, 1, 9, ...] RWG's sinpi, cospi and tanpi suggestion made so much sense that I went ahead and made the change (as RWG and a few others already know) but in the mean-time, using the version of RIES that you have now, it of course finds cos(1) x = sin(1) when given as few as 5 digits (1.5574). There is no simpler equation with a root within 10^-4 of that value. With the new version it finds x = tanpi(1/pi) equally quickly. To reproduce old formulas one can use either the -NT (no tangent) or --trig-argument-scale 1 (use radians) options, or both. Even if I hadn't added tanpi(), it still would have worked, but what it finds is: sqrt(x^2+1) = 1/cospi(1/pi) RIES finds an identity that converts tan(1) to sec(1), allowing it to match 1/cos(1). Why not RWG's hypothetical "(sinpi 1/pi)/(cospi 1/pi)"? In an equation RIES would find this first as "cospi(1/pi) x = sinpi(1/pi)", but it happens to consider "sqrt(x^2+1) to be a less complicated expression than "x cospi(1/pi)". Note the complexity scores {47}, {41}, etc: ries 1.5574077246549 --eval-expression xs1+q prCr 'xprC*' prS Evaluating postfix expression 'xs1+q' with x=1.5574077246549 x = 1.5574077246549001 (d/dx = 1) x^2 = 2.4255188208147529 (d/dx = 3.1148154493098001) 1 = 1 x^2+1 = 3.4255188208147529 (d/dx = 3.1148154493098001) sqrt(x^2+1) = 1.8508157176809237 (d/dx = 0.84147098480789617) [xs1+q] = 1.85081571768092; d/dx = 0.841470984807896, complexity = {47} Evaluating postfix expression 'prCr' pi = 3.1415926535897931 1/pi = 0.31830988618379069 cospi(1/pi) = 0.54030230586813977 1/cospi(1/pi) = 1.8508157176809255 [prCr] = 1.85081571768093; d/dx = 0, complexity = {41} Evaluating postfix expression 'xprC*' with x=1.5574077246549 x = 1.5574077246549001 (d/dx = 1) pi = 3.1415926535897931 1/pi = 0.31830988618379069 cospi(1/pi) = 0.54030230586813977 cospi(1/pi) x = 0.84147098480789539 (d/dx = 0.54030230586813977) [xprC*] = 0.841470984807895; d/dx = 0.54030230586814, complexity = {53} Evaluating postfix expression 'prS' pi = 3.1415926535897931 1/pi = 0.31830988618379069 sinpi(1/pi) = 0.8414709848078965 [prS] = 0.841470984807897; d/dx = 0, complexity = {34} - Robert -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com

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Re: [math-fun] Gamma function (formulas collection)
by Dan Asimov 31 Dec '11

31 Dec '11

Some other values might be checked. E.g., since Gamma(z) Gamma(1-z) == pi/sin(pi z), setting z = 1/2 + i gives |Gamma(1/2 + i)|^2 = 2pi /(e^pi + e^-pi). --Dan David Makin wrote: << Thanks Warren, will take a look to verify I got the UF version at least roughly correct - of course I had no simple way of checking it beyond the obvious positive integer values - which I seem to remember it got correct to a minimum of around 10 or 11 sf, not that great but good enough to play with. >> ________________________________________________________________________________________ It goes without saying that .

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