[math-fun] A000041, the partitions : the classical example of p(200), macmahon, hardy and ramanujan
Simon Plouffe>Hello, I have been musing around the expansion of certain numbers in the exp(Pi) base. Some results are I think interesting. If I use F(x) = product(1/(1-x^n),n=1..infinity), then when x -> exp(-Pi*x) the value for F(10) is easily found to be 1/2 1/2 GAMMA(3/4) (5 + 5 5 ) -------------------------- = F(10). 1/4 5 Pi Pi exp(----) 12 and lprinted : F(10) = GAMMA(3/4)*(5+5*5^(1/2))^(1/2)/(Pi^(1/4)*exp(5/12*Pi)) If I expand this number in base exp(Pi*10) then the first 205 coefficients can be obtained. In other words, I can compute p(n) for up to n=205 using 1 mathematical constant and enough precision. (rwg)>Clever! SP> note : all the coefficients from n=1 to n=205 are computed at once. The <enough precision> is 2800 digits, unfortunately. I can only think at the Major Percy MacMahon that took 6 months to compute p(200) by hand and later Hardy and Ramanujan, using a very clever method to get that number! The method I use is not very clever, I only use a numerical trick, but it does work. With the base exp(Pi*16) I could obtain the algebraic value also and get : F(16) = 2*2^(3/4)*(560+396*2^(1/2)+3*(69708+49291*2^(1/2))^(1/2))^(1/8)*GAMMA(3/4)/Pi^(1/4)/exp(2/3*Pi) (rwg)>F(16) = GAMMA(3/4)*2^(57/32)*%E^-(Pi*2/3)/(Pi^(1/4)*(-3*2^(1/2)-4*2^(1/4)+9)^(1/8)) Also, F(24) = 2^(25/48)*3^(3/8)*((3^(1/4)+SQRT(2))*SQRT(3)+3^(1/4)+2)*Pi^(3/4)/ (exp(Pi)*(SQRT(2)-1)^(3/4)*(SQRT(3)-1)^(1/6)*GAMMA(1/4)) isn't too bad. SP>well, by using 11000 digits, I can get the first 500 values, not bad. NOTE 2) the computation of F(x) when x is small is more suitable when I use the infinite product, it converges quite fast, As fast as F(x) = 1/SUM((-1)^n*(1-exp(-(2*n+1)*Pi*x))*exp(-n*(3*n+1)*Pi*x/2),n,0,INF) ? it is nice to have the exact value of F(x) but faster to use the classical formula. Best regards, Simon Plouffe (rwg)>If you use a series or product formula, you can use F(arbitrarily large).
(With much help from Julian, who says this is good thru p(2791).) F(40) = sqrt(sqrt(2) + 1) * (5^(1/4) + 1)^(3/2) * sqrt(5^(1/4) + sqrt(2)) * sqrt(5) * (sqrt(5) + 1)^(11/4) * (sqrt(10) + 3)^(1/4) * pi^(3/4) * exp( - 5 * pi/3)/(2^(27/16) * GAMMA(1/4)) --rwg By-product: 3/4 (d83) 6 sqrt(2) 5 + (69 sqrt(2) + 12) sqrt(5) = 1/4 102 sqrt(2) 5 + 21 sqrt(2) + 28 (c84) DFLOAT(%); (d84) 273.402222551711d0 = 273.40222255171d0 (c85) BFLOAT(D83); (d85) 2.7340222255171092466b2 = 2.7340222255171007641b2 On Mon, Mar 14, 2011 at 3:32 PM, Bill Gosper <billgosper@gmail.com> wrote:
Simon Plouffe>Hello,
I have been musing around the expansion of certain numbers in the exp(Pi) base.
Some results are I think interesting.
If I use F(x) = product(1/(1-x^n),n=1..infinity), then
when x -> exp(-Pi*x) the value for F(10) is easily found to be
1/2 1/2 GAMMA(3/4) (5 + 5 5 ) -------------------------- = F(10). 1/4 5 Pi Pi exp(----)
12
and lprinted : F(10) =
GAMMA(3/4)*(5+5*5^(1/2))^(1/2)/(Pi^(1/4)*exp(5/12*Pi))
If I expand this number in base exp(Pi*10) then the first 205 coefficients can be obtained. In other words, I can
compute p(n) for up to n=205 using 1 mathematical constant and enough precision.
(rwg)>Clever!
SP> note : all the coefficients from n=1 to n=205 are computed at once.
The <enough precision> is 2800 digits, unfortunately.
I can only think at the Major Percy MacMahon that took 6 months to compute p(200) by hand and later Hardy and Ramanujan, using a very clever method to get that number!
The method I use is not very clever, I only use
a numerical trick, but it does work.
With the base exp(Pi*16) I could obtain the algebraic value also and get :
F(16) = 2*2^(3/4)*(560+396*2^(1/2)+3*(69708+49291*2^(1/2))^(1/2))^(1/8)*GAMMA(3/4)/Pi^(1/4)/exp(2/3*Pi)
(rwg)>F(16) = GAMMA(3/4)*2^(57/32)*%E^-(Pi*2/3)/(Pi^(1/4)*(-3*2^(1/2)-4*2^(1/4)+9)^(1/8)) Also, F(24) = 2^(25/48)*3^(3/8)*((3^(1/4)+SQRT(2))*SQRT(3)+3^(1/4)+2)*Pi^(3/4)/ (exp(Pi)*(SQRT(2)-1)^(3/4)*(SQRT(3)-1)^(1/6)*GAMMA(1/4))
isn't too bad.
SP>well, by using 11000 digits, I can get the first 500 values, not bad.
NOTE 2) the computation of F(x) when x is small is more suitable when I use the infinite product, it converges
quite fast,
As fast as
F(x) = 1/SUM((-1)^n*(1-exp(-(2*n+1)*Pi*x))*exp(-n*(3*n+1)*Pi*x/2),n,0,INF) ?
it is nice to have the exact value of F(x) but faster to use the classical formula.
Best regards,
Simon Plouffe
(rwg)>If you use a series or product formula, you can use F(arbitrarily large).
Yes, with F(40) (wow), with such a value you can get further, but a little detail, you need over 152319 digits to get there.! exp(Pi*40)^2791 = 0.946548568816 x 10^152319. I took the simplest value enough to compute the classical example of MacMahon. still, I am surprised by the relative simplicity of that algebraic number, Simon Plouffe Le 2011-03-16 10:55, Bill Gosper a écrit :
(With much help from Julian, who says this is good thru p(2791).) F(40) = sqrt(sqrt(2) + 1) * (5^(1/4) + 1)^(3/2) * sqrt(5^(1/4) + sqrt(2)) * sqrt(5) * (sqrt(5) + 1)^(11/4) * (sqrt(10) + 3)^(1/4) * pi^(3/4) * exp( - 5 * pi/3)/(2^(27/16) * GAMMA(1/4)) --rwg By-product: 3/4 (d83) 6 sqrt(2) 5 + (69 sqrt(2) + 12) sqrt(5) =
1/4 102 sqrt(2) 5 + 21 sqrt(2) + 28
(c84) DFLOAT(%);
(d84) 273.402222551711d0 = 273.40222255171d0
(c85) BFLOAT(D83);
(d85) 2.7340222255171092466b2 = 2.7340222255171007641b2
Hello mr Gosper, Why you do not publish all of your results ? I was wondering that the other day, you must have tons of formulas ? You could prepare a template article in word- latex-whatever suitable formula preparation program and submit all of it to the arXiv ? let me guess, you have too many formulas ? just a naïve question, simon plouffe
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Simon Plouffe