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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#45
bo198214 Wrote:
Ivars Wrote: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.

Actually it does not matter with which value (instead of -1) you start the iterations of as long as you dont take some hyperpowers of for example would lead to 0 after 4 times application of the and taking again would give , so the formula would not work with this starting value.

So we have just limit log b log b log b log b ... log b (+-i*2pik) to be concerned with for branches in numerator. Right? But how do we get real part as well from this? because there is real value in denominator if we use ln to compute log b?

So we have lim k, n-> infintylog b log b log b ...... log b ( +-i*2pik/(pi/2+- i*2pi/k))

So the simplest starting value is then +-i*2pik/(pi/2+-i*2pik)?
Are these k in numerator and denominator the same or running independently ?

Ivars
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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, 11:51 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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