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 Nowhere analytic superexponential convergence sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 01/31/2011, 04:47 AM (This post was last modified: 01/31/2011, 06:22 AM by sheldonison.) (01/31/2011, 04:00 AM)bo198214 Wrote: (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.Henryk, I've run into some gaps (maybe the same gap you mention), with the ideas I was pursuing in this post, so I don't feel like I have a proof, just lots of interesting observations. At some point, I'll post some more, when the ideas are more clear to me. In the meantime, I need to visit the local university library, and read some of these papers (by Walker, and Levy, and Kneser, and a book by Kuczma). I look forward to seeing your proof! Tommy is correct, that one needs to be very careful in taking the logarithm of these functions in the complex plane. In the computations I've posted, there wasn't any problem figuring out which logarithm to take of 2sinh^[z] as long as all of the points between the real axis and z-1, have $\Re(\text{2sinh}^{[z-1]})>0$. Of course the contour where $\Re(\text{2sinh}^{[z-1]})=0$ is a problem when $\Im(\text{2sinh}^{[z-1]})=n\pi$, then $\text{2sinh}^{[z]}=\exp(n\pi)-\exp(-n\pi)=0$ and $\log(\text{2sinh}^{[z]})$ is a singularity. Otherwise, if you stay below the contour where the singularities are, then the unique logarithm seems to always be determined by the requirement that $-\pi<\Im(\log(\text{2sinh}^{[z]})-\text{2sinh}^{[z-1]})<\pi$. This also works for the base change function. - Sheldon « Next Oldest | Next Newest »

 Messages In This Thread Nowhere analytic superexponential convergence - by sheldonison - 01/20/2011, 05:23 PM RE: Nowhere analytic superexponential convergence - by tommy1729 - 01/21/2011, 12:21 AM RE: Nowhere analytic superexponential convergence - by sheldonison - 01/26/2011, 12:49 AM RE: Nowhere analytic superexponential convergence - by tommy1729 - 01/26/2011, 11:17 PM RE: Nowhere analytic superexponential convergence - by bo198214 - 01/28/2011, 11:29 PM RE: Nowhere analytic superexponential convergence - by tommy1729 - 01/29/2011, 12:03 AM RE: Nowhere analytic superexponential convergence - by bo198214 - 01/29/2011, 10:21 AM RE: Nowhere analytic superexponential convergence - by tommy1729 - 01/29/2011, 03:26 PM RE: Nowhere analytic superexponential convergence - by mike3 - 01/29/2011, 08:41 PM RE: Nowhere analytic superexponential convergence - by tommy1729 - 01/29/2011, 08:59 PM RE: Nowhere analytic superexponential convergence - by bo198214 - 01/29/2011, 11:53 PM RE: Nowhere analytic superexponential convergence - by tommy1729 - 01/30/2011, 06:48 PM RE: Nowhere analytic superexponential convergence - by bo198214 - 01/31/2011, 04:00 AM RE: Nowhere analytic superexponential convergence - by sheldonison - 01/31/2011, 04:47 AM RE: Nowhere analytic superexponential convergence - by sheldonison - 02/10/2011, 07:22 AM

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