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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#22
Gotfried,

Thanks for the article,I tried to avoid Latin texts of Euler, as I do know 0 Latin. However:

I looked at article; Euler never specifies the sign of sgrt(-1). We do, when we use i.
When he sets angle to pi/2, he gets r=e^pi/2. Power tower e^pi/2 = sgrt(-1) = e^sgrt(-1) pi/2

May be I miss something, but as long as he shows no signs , it is OK,if we always take the right sign = the same in both cases. Though he did put a constraint on angle pi, but I could not understand what that means.

However, when we calculate h(e^pi/2) = -W(-ln(e^pi/2)/ln(e^pi/2) we specify that we will use +i via W(-pi/2) = i*pi/2; and we get h(e^pi/2)=-i; Euler would have written W(-pi/2)= sgrt(-1) *pi/2, and got h(e^pi/2)=sgrt(-1).No problems.

If we put W(-pi/2) = -i*pi/2, we would get h(e^pi/2) = +i.

If we look at power series for W(-pi/2):

i*pi/2= - pi/2-p^2/4-pi^3/2*2^3 - 8/3*pi^4/2^4 - 125/24*pi^5/2^5 .....

i= -1-pi/2 - pi^2/2*2^2 - 8/3*pi^3/2^3 - 125/24*pi^4/2^4......= e^(-pi/2)*sum ( -1^(n)*(n^(n-2))


-i = 1+pi/2+pi^2/2*2^2+ 8/3*pi^3/2^3 +125/24*pi^4/2^4.... = (e^pi/2)*sum(n^(n-2))?

These both should be divergent, but definitely not to the same value,

Ivars Fabriciuss
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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 - 11/04/2007, 12:10 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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