11/09/2007, 08:30 AM
bo198214 Wrote:Quote:So the problem is now turned to the properties of continuous application of logarithms with base e^pi/2 on -1?Yes, yes
Which in turn is calculated in each step via normal logarithms by formula:
log b (z) = (ln (z) + 2pik)/ ln (b)?
So now we need analytic expressions for that infinite taking of logarithms with base e^pi/2 of (-1) as a function of k.
Any ideas? I am searching as well.
if we put e^(pi/2) = e^((sqrt(pi)/sqrt(2))^(sgrt(pi)/sqrt(2) ) then
we can use Gamma(1/2) = sqrt(pi), so
e^(pi/2) = e^(Gamma^2 (1/2)/2)
and ln (e^pi/2) = Gamma^2 (1/2) /2 +- 2pik
next ln would be 2 lnGamma(1/2) - ln 2
next ln2+ ln*lnGamma(1/2) - lnln2 etc. ln2 and lnln2 will cancel out, so
in the end we will have lnln............ln (Gamma(1/2) when n-> infinity.