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Nowhere analytic superexponential convergence
#13
(01/30/2011, 06:48 PM)tommy1729 Wrote: in fact even log(exp(z)) is not entire.

Well thats an interpretation question.
If you consider log(exp(z)) in the strictest sense, the logarithm with imaginary part between -pi and pi, then log(exp(z)) is not a holomorphic function, because it is not continuous.
However you know that a holomorphic function is determined globally by just being defined in a small neighborhood of a point, thats the so called analytic continuation.
So on the neighborhood of some point on the positive real axis log(exp(z)) = id. This is the holomorphic function and you continue it just to the whole complex plane.

But I see your point in mentioning that there may be function sequences with singularities getting close to the real axis, whose limit has still non-zero convergence radius. Thats actually why I say that it is not a proof, but it very strongly hints toward non-analyticity. But it seems Sheldon is on a way to a proof, I am also in the phase of proof search.
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RE: Nowhere analytic superexponential convergence - by bo198214 - 01/31/2011, 04:00 AM

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