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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#81
One application of h(1/(n^n)) = 1/n as n-> infinity

I will try to apply my infinite tetration formula h(1/(n^n)) = 1/n to see what happens:

So as n-> infinity

(e^n)(n!)/((n^n)(sqrt n))=sqrt 2pi (Clawson)

1/(n^n) = (sgrt(2pi*n)) / ((e^n)*n!)

h(1/(n^n)) = h( sgrt(2pi*n)/(e^n)*n!) = 1/n as n-> infinity which is? 0?

If I knew how h works on composite arguments, may be it would be possible to study this deaper.
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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 - 12/30/2007, 01:09 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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