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 Nowhere analytic superexponential convergence sheldonison Long Time Fellow Posts: 633 Threads: 22 Joined: Oct 2008 02/10/2011, 07:22 AM (This post was last modified: 02/10/2011, 03:02 PM by sheldonison.) (01/31/2011, 04:47 AM)sheldonison Wrote: ... 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).This is all getting much clearer to me, and I'm getting much much closer, but it will take one or two additional posts to present all of the equations and ideas that I think are required for a proof. This post is mostly overview, along with a counter example showing the problems. Consider the base change superfunction of exp(z)-1, or the superfunction for 2sinh(z). I'm going to call this function s(z). S(z) is any real valued superexponential function. Then define theta(z) as follows: $\theta(z)=\text{slog_e}(s(z))$ For s(z) as the superfunction of 2sinh, then $\text{tommysexp(z)}=\lim_{n \to \infty}\text{sexp_e(z+\theta(z+n))$ For s(z) as the superfunction of exp(z)-1, $\theta(z)=\text{slog_e}(s(z))$, $\text{basechangesexp(z)}=\lim_{n \to \infty}\text{sexp_e(z+\theta(z+n))$ This is equivalent to ideas discussed earlier, where the idea is to show that there are singularities in theta(z), corresponding approximately to the contour where $\Re(s(z))=0$ and $\Im(s(z))$ is growing, because then s(z+1) will have zeroes, and there will be corresponding singularities for log(s(z+1)). The locations of some of those singularities were listed earlier, and it was observed that they get arbitrarily close to the real axis, and arbitrarily close together. For theta(z+3), the radius of convergence is approximately 0.458. For theta(z+4), the radius of convergence is 0.0347. Theta(z) is interesting because it very quickly converges to a 1-cyclic function. Theta(z) is also inifinitely differentiable. And as z increases, theta(z+1)-theta(z) superexponentially approaches zero, and theta(z) superexponentially converges to a 1-cyclic function, whose limit will be shown to be nowhere analytic. But first though, here is an example f(z) of a "gap" or counterexample function, which is much like theta(z), but is analytic. $f(z)=sin(2\pi z)+1/\text{sexp_e}(z)$ Like theta(z), f(z) misbehaves as z increases. Like theta(z), this misbehavior get superexponentially closer together, and approach arbitrarily close to the real axis as z increases. And, like theta(z), the limiting behavior of f(z) is a 1-cyclic function. But, the limiting behavior of f(z) is a completely well defined analytic function, sin(2pi*z)! So more is required to show that limiting behavior of theta(z) is truly nowhere analytic! To prove that theta(z) is nowhere analytic a criteria is required to get around the counterexample. It is suggested that as z increases, there is a value of n, such that the terms of the taylor series of theta(z) <=n for $a_n z^n$, such that the truncated taylor series still misbehaves in the same way as it did for theta(z-1), where the function misbehaves at the same radius as in the previous iteration. In addition, there is also a larger number p, such that the series for theta(z) with p terms misbehaves in a new way, for a smaller radius. And for z+1, there is a new q, corresponding to the new set of singularities. But the Taylor series for z+1, truncated to n terms, will misbehave at the original radius, and truncated to p terms will misbehave at the smaller radius, and truncated to q terms, will misbehave at an even smaller radius. In this way, it will be shown that taylor series for the limiting behavior of theta(z+n) as n goes to infinity, will have misbehavior at radii corresponding to the nearest singularity of theta(z+3), and corresponding to the nearest singularity of theta(z+4), and corresponding to the nearest singulariy of theta(z+5), and that thus theta(z) will truly be nowhere analytic. The definition of misbehavior used will be that the taylor series terms, scaled to a radius just a bit larger than the singularity radius, will increase to 1, which will lead to misbehavior in the truncated taylor series of theta(z). To prove theta(z) is nowhere analytic, we need to find values for n, p, and q etc, and the corresponding radius of misbehavior, which requires understanding theta(z), and understanding how theta(z) changes from one iteration to the next. The delta for theta(z) from one iteration to the next will be shown to have a lot of similarity to 1/sexp(z), or rather a power series in 1/sexp(z), and thus theta(z) will be shown to have similarities to the superexponential reciprical sum presented earlier. - 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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