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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#43
bo198214 Wrote:
Quote:So the problem is now turned to the properties of continuous application of logarithms with base e^pi/2 on -1?
Which in turn is calculated in each step via normal logarithms by formula:
log b (z) = (ln (z) + 2pik)/ ln (b)?
Yes, yes Smile

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.
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Messages In This Thread
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/09/2007, 08:30 AM
RE: Tetration below 1 - by Gottfried - 09/09/2007, 07:04 AM
RE: The Complex Lambert-W - by Gottfried - 09/09/2007, 04:54 PM
RE: The Complex Lambert-W - by andydude - 09/10/2007, 06:58 AM

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