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 [MSE] Shape of orbit of iterations with base b on Shell-Thron-region sheldonison Long Time Fellow Posts: 663 Threads: 23 Joined: Oct 2008 11/08/2019, 09:09 PM (This post was last modified: 11/08/2019, 09:33 PM by sheldonison.) (11/08/2019, 05:55 PM)Ember Edison Wrote: (11/08/2019, 03:20 PM)sheldonison Wrote: 1. no sexpeta; and cheta are both generated from the Ecalle assymptotic series solution for $f(z)=\exp(z)-1$, which has two sectors.  The Ecalle assymptotic generates two different solutions, depending on whether you approach the fixed point of zero from z>0 by iterating $f^{-1}(z)=\log(z+1)$ before evaluating the series, which generates the cheta upper superfunction, or whether you approach the fixed point from z<0 by iterating f(z) before evaluating the series which generates the sexpeta lower superfunction.  That two different analytic functions can be generated from the same assymptotic series makes sense when you realize Ecalle's solution is a divergent assymptotic series, and you need to iterate f(z) or f^-1(z) enough times so that |z| of z is small enough to generate an accurate result. The conjecture is that the limit of Kneser, as the base approaches eta from above would be sexpeta; I don't know if the conjecture has been proven. 1.What base can't merged? Is the all Shell-Thron-region rational base and Singularity base can't merged, or just Singularity base, or just eta? 2.What does "limit" mean? Is the ${\lim_{\delta \to 0^+}}fatou.gp.sexp_{\eta+I*\delta}(z)$ is merge superfunction? upper? lower? The first half of my answer was only discussing sexp generated via Ecalle for base eta=exp(1/e).  Base exp(1/e) has only one neutral fixed point.  Kneser isn't defined for base exp(1/e).  Kneser is defined for base=e, and if you gradually modify the real valued Kneser base, then you get to a singularity at base exp(1/e) where there is only one fixed point.  Even though Kneser isn't defined at base exp(1/e), the conjecture is that the limit as you approach exp(1/e) from bases a little bit bigger than exp(1/e) is defined, and in the limit Kneser's slog would equal Ecalle's Abel function for base exp(1/e) plus a constant. $\lim_{b\to\eta^{+}}\text{slog}_b(z)=\alpha(z)+k\;\;$ where $\alpha(z)$ is Ecalle's Abel function for base exp(1/e) from the attracting sector, and k is a constant You can't compare Ecalle's solution for other rational bases on the Shell Thron region to Kneser since Ecalle is the Abel function for $f^{\circ n}(z)$ where n is the denominator of the rational root of unity that is the derivative of f(z) at the fixed point whereas Kneser's slog could be compared to the Abel function for f(z).  Does that answer your question? - Sheldon Ember Edison Junior Fellow Posts: 38 Threads: 5 Joined: May 2019 11/10/2019, 06:13 PM (11/08/2019, 09:09 PM)sheldonison Wrote:  Does that answer your question? Is very helpful. My last question is fatou.gp how to merge upper and lower superfunction. sheldonison Long Time Fellow Posts: 663 Threads: 23 Joined: Oct 2008 11/11/2019, 05:05 PM (This post was last modified: 11/11/2019, 05:29 PM by sheldonison.) (11/10/2019, 06:13 PM)Ember Edison Wrote: (11/08/2019, 09:09 PM)sheldonison Wrote:  Does that answer your question? Is very helpful. My last question is fatou.gp how to merge upper and lower superfunction. There is post#15 in the fatou.gp thread which includes some equations and a picture of how I setup a 60x60 system of equations to solve for the base_e slog.  https://math.eretrandre.org/tetrationfor...98#pid8998 You need the two Schroder functions for each fixed point, and then you can get Koenig's Abel function by taking the logarithm of the Schroder function, from each fixed point.  That gives you an upper and a lower Abel complex valued function; each Abel function has a 1-cyclic mapping that goes from the Abel function to Kneser's slog.  The 1-cyclic mapping for the upper Abel function decays to a constant as imag(z) to infinity, which is the uniqueness criteria.  Ditto for the lower Abel function.  So I have three mathematically identical representations for Kneser's slog which work best in different parts of the complex plane. slog(z)=$\sum_{n = 0}^{\infty} a_n (z-\Re(L))^n\;\;$ The Taylor series representation slog(z)=$\alpha_u(z)+\theta_u(\alpha_u(z))\;\;$ the upper Abel function plus theta mapping slog(z)=$\alpha_l(z)+\theta_l(\alpha_l(z))\;\;$ the lower Abel function plus theta mapping So, the system of equations requires these three functions to be identical over carefully chosen subsets of points in the picture, and this leads to a well behaved linear set of equations that converges very nicely. - Sheldon Daniel Fellow Posts: 79 Threads: 29 Joined: Aug 2007 12/13/2019, 12:00 AM (This post was last modified: 12/13/2019, 01:00 AM by Daniel.) I was working on Mathematica software to display the dynamics of the exponential map close to the Shell Thron boundary and I discovered the Galanakis Tetrational Code viewer which does much of what I wanted to achieve. - Daniel Ember Edison Junior Fellow Posts: 38 Threads: 5 Joined: May 2019 12/13/2019, 02:33 PM (12/13/2019, 12:00 AM)Daniel Wrote: I was working on Mathematica software to display the dynamics of the exponential map close to the Shell Thron boundary and I discovered the Galanakis Tetrational Code viewer which does much of what I wanted to achieve. - Daniel This program is more useful than fatou.gp.  fatou.gp is very very slowly when some output is overflow and can't change period. « Next Oldest | Next Newest »

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