• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Nowhere analytic superexponential convergence bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 01/29/2011, 11:53 PM (This post was last modified: 01/29/2011, 11:55 PM by bo198214.) (01/29/2011, 08:59 PM)tommy1729 Wrote: well according to Bo it is not real-analytic and it doesnt have a radius , and according to mike it does have a radius. I was not talking about $\log^{[n]}(\exp^{[n]}(x))$ which is for each $n$ the identity function. Of course it is real-analytic, even entire. I was talking about tommysexp construction and the basechange. In each step it has a radius of convergence, which though diminishes to 0 for $n\to\infty$. Quote:so opinions and/or terminology differs They dont. Quote: , i hope at least everyone agrees on : on the real line f(x) = lim n -> oo log^[n] ( exp^[n] (x) ) is real-analytic because it simply reduces to f(x) = x for real x , which is clearly real-analytic. Of course. Quote:and for real-analytic i prefer to say : analytic on an interval , rather than a radius , for imho a nonzero-radius is for analytic taylor series - analytic on on a disk on the complex plane with nonzero-radius. "analytic" is defined for a point (in a neighborhood, so one can take derivatives). If one says "analytic on an interval", it means "for each point on the interval". If one says "analytic on a disk", it means "for each point of the disk". However "analytic at a point" always implies being analytic on the open disk of convergence, i.e. being analytic at each point of the disk of convergence. I hope this brings some light into your confusion tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 01/30/2011, 06:48 PM (01/29/2011, 11:53 PM)bo198214 Wrote: (01/29/2011, 08:59 PM)tommy1729 Wrote: well according to Bo it is not real-analytic and it doesnt have a radius , and according to mike it does have a radius. I was not talking about $\log^{[n]}(\exp^{[n]}(x))$ which is for each $n$ the identity function. Of course it is real-analytic, even entire. I was talking about tommysexp construction and the basechange. In each step it has a radius of convergence, which though diminishes to 0 for $n\to\infty$. and here you are wrong. you say $\log^{[n]}(\exp^{[n]}(x))$ is entire. it is not. i suggest you plot $\log^{[2]}(\exp^{[2]}(z))$ MINUS z on the complex plane ; and see ' how entire it actually is '. in fact even log(exp(z)) is not entire. hint : log(exp(z)) is periodic and id(z) is not. regards tommy1729 bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 01/31/2011, 04:00 AM (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. sheldonison Long Time Fellow Posts: 626 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 sheldonison Long Time Fellow Posts: 626 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 »

 Possibly Related Threads... Thread Author Replies Views Last Post [repost] A nowhere analytic infinite sum for tetration. tommy1729 0 930 03/20/2018, 12:16 AM Last Post: tommy1729 Analytic matrices and the base units Xorter 2 2,046 07/19/2017, 10:34 AM Last Post: Xorter Non-analytic Xorter 0 1,218 04/04/2017, 10:38 PM Last Post: Xorter A conjectured uniqueness criteria for analytic tetration Vladimir Reshetnikov 13 9,779 02/17/2017, 05:21 AM Last Post: JmsNxn Is bounded tetration is analytic in the base argument? JmsNxn 0 1,138 01/02/2017, 06:38 AM Last Post: JmsNxn Are tetrations fixed points analytic? JmsNxn 2 2,511 12/14/2016, 08:50 PM Last Post: JmsNxn The bounded analytic semiHyper-operators JmsNxn 2 3,220 05/27/2016, 04:03 AM Last Post: JmsNxn Periodic analytic iterations by Riemann mapping tommy1729 1 1,933 03/05/2016, 10:07 PM Last Post: tommy1729 Bounded Analytic Hyper operators JmsNxn 25 17,779 04/01/2015, 06:09 PM Last Post: MphLee Real-analytic tetration uniqueness criterion? mike3 25 19,205 06/15/2014, 10:17 PM Last Post: tommy1729

Users browsing this thread: 1 Guest(s)