[math-fun] strange approximations for F(x) = partition function. regarding A000041.
Hello, I stumbled on these 2 approximations regarding F(x), the partition function, where p(n) = the number of partitions of n, as usual. aka A000041. F(x) = sum(p(n)*x^n, n=0..infinity): Instead I use F*(x) = sum(p(n)*x^(n+1), n=0..infinity): Then here is the strange thing, for x = exp(-2*Pi/5) then the value is 1/sqrt(5), well almost ; the precision is 13 digits. For x = exp(-4*Pi/5) the value is 1/2+3/2/sqrt(5)-sqrt(1/2*(1+3/sqrt(5))) the precision is 28 decimal digits. I find this quite surprising. I was sure it was exact, it is NOT. I verified with large values. Also, apparently these are the only 2 examples I have found within F60 : the Farey fractions up to denominators = 60. Also when x = exp(-Pi/5) = apparently nothing algebraic of a low degree. caution : do not mistake these values for the standard F(x) which goes 0 for the exponent too, it is not the same. I added these 2 values in the formulas of A000041 of course. Does anybody have an idea why these values just pop out like that and apparently no other ???! Bonne journée. Simon Plouffe
Hi Simon, I would bet that your values are related to this: http://www.mathstat.carleton.ca/~williams/papers/pdf/220.pdf Victor On Wed, Feb 23, 2011 at 12:11 PM, Simon Plouffe <simon.plouffe@gmail.com>wrote:
Hello,
I stumbled on these 2 approximations regarding F(x), the partition function, where p(n) = the number of partitions of n, as usual. aka A000041.
F(x) = sum(p(n)*x^n, n=0..infinity):
Instead I use F*(x) = sum(p(n)*x^(n+1), n=0..infinity):
Then here is the strange thing, for x = exp(-2*Pi/5) then the value is 1/sqrt(5), well almost ; the precision is 13 digits.
For x = exp(-4*Pi/5) the value is 1/2+3/2/sqrt(5)-sqrt(1/2*(1+3/sqrt(5))) the precision is 28 decimal digits. I find this quite surprising. I was sure it was exact, it is NOT. I verified with large values.
Also, apparently these are the only 2 examples I have found within F60 : the Farey fractions up to denominators = 60. Also when x = exp(-Pi/5) = apparently nothing algebraic of a low degree.
caution : do not mistake these values for the standard F(x) which goes 0 for the exponent too, it is not the same.
I added these 2 values in the formulas of A000041 of course.
Does anybody have an idea why these values just pop out like that and apparently no other ???!
Bonne journée. Simon Plouffe
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Hello mr Miller, hum, interesting, we surely do not use the same methods at all, me : I look at the numbers directly, I do not see where it connects, the arguments I use are rationals, he uses algebraic arguments also the approx. where does it comes from ??, I do not see it. I have found these approximations when I did a series of mistakes, I first used the non-standard starting point (n+1) instead of n, starting at 1 and not 0. Then I used the wrong settings for digits, and PSLQ bugged, PSLQ has lots of problems with values that are approximations by the way, then I used pari-gp, more reliable, again with the wrong settings for the precision, and even there that false identity was still popping out !, I could not understand the error, this is where I started to investigate what was the problem, then I found the one for 2/5 and then later the one for 4/5 and no other, and still wondering why these values are so good and unique ??. In all this, if we use F(x) the partition function which is the 'base' function, usualy a property of that function will propagate to the others of the same type, eta-dedekind, ramanujan Tau, theta series, etc. it is all related to the same base function which is F(x) the partitions. (ref. G.H. Hardy, collected papers, Vol. V). This is why I am investigating that one. Yesterday, I have found a very simple trick to compute the partitions and those related functions with only a good knowledge of some math. constants like Pi, log(Pi), log(2) and log(GAMMA(1/4)), with these at hand, I can compute p(n) directly , I do not use Rademacher formula, the recurrence formula at all, it is very simple. I am not saying that my algorithm is faster, it is not. It is very simple, direct and not much theory behind actualy. I read Hardy's papers for years, inspired by the circle method and all that, followed by the Rademacher formula and or the recurrence which is efficient but not simple enough for me. best regards, and thank you for that reference of Van der Poorten et al. Simon Plouffe
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Simon Plouffe -
Victor Miller