I missed where Warren's paper probably says that, for "unrestricted" (arbitrary precision) numerics, the divergence of the log Gamma and polygamma series is rather irrelevant. A plausible algorithm is simply to add some positive integer n to z so that the Stirling series z+n swoops within the desired accuracy. Then divide by (z+1)(z+2)...(z+n). Windschitl's and Warren's improvements make this even more attractive. Looking to provide Gamma numerics for Robert's RIES, I accidentally overkilled the problem by Remezing up f[z]:=(0.011987912746518014561506488044010 + z^2 (-0.028678142684262583619372798888834 + z^2 (-0.10376325088888750555218325558841 + z^2 (-0.034612435282031905677869242849747 - 0.0024691358024691358024691358024691 z^2))))/(z^7 * (45.984169004371332413538940188341 + z^2 (54.768197084214553421696753890699 + z^2 (14.960893432079977923321408051895 + z^2)))) Then plotting z! /(Sqrt[2 Pi] (E^(-f[z] - 4 z) z^(2 + 6 z) Sinh[1/z]^(2 z))^(1/4) -1 9 < z < oo shows maximum error 6*10^-26 (at only one ripple because I neglected the weight function). So up to nine divides could be required for this method. But RIES only needs 10^-17, for which f[z] could be a lot simpler and n smaller than 9. To squeeze some actual fun out of this: Why is f an odd function? --rwg PS, I wasn't confused in that 2001 Remez posting--just confusing. In the next sentences I explained why truncating the Chebys doesn't minimax. I was just using "Chebychef optimal" to mean the best you could do by subtracting off Chebys. On Fri, Dec 30, 2011 at 1:57 PM, Bill Gosper <billgosper@gmail.com> wrote:
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"