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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#72
Ivars Wrote:h(1/n^n)=1/n
h(1) +h(1/4) + h(1/27) + h(1/256) ...+h(1/n^n)... = harmonic series
h(1)^2+h((1/4))^2+h((1/27))^2 +...+ h(1/n^n))^2...= pi^2/6 and so on for higher powers in series.
Multplying h(1/x^x) * h(x^1/x) = 1 at each x if -1/e<x<e.

To make a more general case, we can consider:

x= z^y so that 1/x = z^-y; then from h(1/x^x)=1/x ->
h((z^-y)^(z^y)) = z^-y

but we know that h(e^pi/2)^h(e^pi/2) = i^i = -i^-i = e^-pi/2
on other hand, i^-i =i^(1/i) = e^( ipi/2*-i)= e^pi/2

so:

h(e^pi/2)^1/h(e^pi/2) = e^pi/2

then we can generalize to z^y:

h(z^y)^h(z^y) = z^-y
h(z^y)^1/h(z^y) = z^y
so that:

h(z^y)^h(z^y)*h(z^y)^1/h(z^y) =1 or

h(z^y)^ ((h(z^y))^2+1/h(z^y)) =1 which is possible always if

(h(z^y))^2+1=0 -> h(z^y) = +- i; We know that the first solution is:

z^y = e^pi/2 or z=e, y=pi/2 and h(e^pi/2) = +- i .

But we also know that for z^y = e^pi/2 other branches of Lambert function leads to values h(e^pi/2) different from +-2pik.

Question:

Do there exist any other roots z and y for equations:

h(z^y) = e^(i*pi/2+-I*2pi*k) ?
for all k, or just for some single special k?
Reply


Messages In This Thread
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/29/2007, 10:37 PM
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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