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Just asking...
#18
martin Wrote:
Ivars Wrote:If in series there is 1/n!, then formula becomes:
n= ln(2) * ((1-exp(-x))/x)

if just 1/n, then:
n= ln(2) * (ln(1+x)/x)

Nope, neither of them worked. But hey, it was worth trying.

I may have been mistaken with formulas which I derived from series, the series of type ln(2)*(1- a*x +b*x^2 - c*x^3+..) itself looked more promisingSad

May be it is just ln(2)-ln(2)/2*x.

Ivars
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Messages In This Thread
Just asking... - by martin - 07/14/2008, 09:49 PM
RE: Just asking... - by bo198214 - 07/14/2008, 10:53 PM
RE: Just asking... - by martin - 07/15/2008, 11:40 AM
RE: Just asking... - by bo198214 - 07/16/2008, 08:00 PM
RE: Just asking... - by martin - 07/16/2008, 09:52 PM
RE: Just asking... - by bo198214 - 07/16/2008, 10:36 PM
RE: Just asking... - by bo198214 - 07/16/2008, 10:52 PM
RE: Just asking... - by martin - 07/17/2008, 11:28 AM
RE: Just asking... - by bo198214 - 07/18/2008, 01:48 PM
RE: Just asking... - by martin - 07/18/2008, 06:37 PM
RE: Just asking... - by bo198214 - 07/18/2008, 10:02 PM
RE: Just asking... - by Gottfried - 07/19/2008, 03:48 AM
RE: Just asking... - by Ivars - 07/19/2008, 07:04 AM
RE: Just asking... - by martin - 07/19/2008, 09:03 AM
RE: Just asking... - by Ivars - 07/19/2008, 07:45 PM
RE: Just asking... - by andydude - 07/21/2008, 04:05 AM
RE: Just asking... - by martin - 07/21/2008, 06:31 PM
RE: Just asking... - by Ivars - 07/22/2008, 06:07 AM



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