04/22/2009, 03:45 PM
(This post was last modified: 04/22/2009, 03:50 PM by nuninho1980.)
andydude Wrote:I used regular iteration from the fixed-point p =
1.000000000000000000010000000000000000000100000000000000000001500000000000000000023333333333333333333733333333333333333340408333333333333333463361111111111111113553591269841269841316667063492063492064403294973544973544991509125661375661376019282963213484046824576164483124899791712269571579076787413095423662162995000357635728649265105119539204243709252241024379290515068892945934184599484695347437210742175409191006184949392420206632843923492376014272124055526327911424878272
in the series
\( {}^{x}a = p + \ln(p)^x(1 - p) + \frac{p\ln(a)^2(1 - \ln(p)^x)\ln(p)^{x-1}}{2(1 - \ln(p))}(1 - p)^2 + \cdots \)
where a = 1.00000000000000000001
Andrew Robbins
this method is much faster. thanks!
p+ ... (1-p)^2+ ...
what do you add new last terms, if you want 15 or more terms?