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 levy ecalle koenigs ? tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 07/08/2010, 11:31 PM for parabolic iteration i use the carleman matrix method. what other methods exist ? bo mentioned " levy and ecalle " but i dont know about their parabolic iteration formula's ... one can get many series expansions from the carleman matrix method , but i would like to see " totally different from carleman methods ". in fact i would like to see a limit like superfunction type of solution to parabolic iteration , something similar to koenigs non-parabolic iteration solution. does there exist a method for all fixpoints ( parabolic or non-parabolic ) in terms of a limit , not using carleman ? im sorry , i tried looking on the internet , but found nothing apart from stuff that is equivalent to carleman ... if i recall correct , koenigs is not necc analytic. are there similar methods that only work when the solution is analytic ? sorry , im a number theorist getting old regards tommy1729 bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 07/09/2010, 05:46 AM (07/08/2010, 11:31 PM)tommy1729 Wrote: for parabolic iteration i use the carleman matrix method. What you mean by carleman method? We have Andrews intuitive Abel function which uses the Carleman matrix as well as Gottfrieds matrix power method. Also I dont know how you do that if you develop at a fixed point, or dont you? Matrix power method at a fixed point is equal to regular/parabolic iteration and yields a series expansion that has convergence radius 0. Quote:what other methods exist ? bo mentioned " levy and ecalle " but i dont know about their parabolic iteration formula's ... ... in fact i would like to see a limit like superfunction type of solution to parabolic iteration , something similar to koenigs non-parabolic iteration solution. Its explained in my overview paper (see the FAQ thread). Quote:does there exist a method for all fixpoints ( parabolic or non-parabolic ) in terms of a limit , not using carleman ? something like that. Quote:im sorry , i tried looking on the internet , but found nothing apart from stuff that is equivalent to carleman ... look at my overview paper, everything is explained there though its quite raw yet. If you have specific question you can always ask me. Quote:if i recall correct , koenigs is not necc analytic. are there similar methods that only work when the solution is analytic ? koenigs is analytic (in vicinity of the hyperbolic fixed point, though some authors also call the Levy formula for parabolic fixed point "koenigs"). sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 07/09/2010, 11:48 AM (This post was last modified: 07/09/2010, 11:49 AM by sheldonison.) (07/08/2010, 11:31 PM)tommy1729 Wrote: .....in fact i would like to see a limit like superfunction type of solution to parabolic iteration , something similar to koenigs non-parabolic iteration solution. does there exist a method for all fixpoints ( parabolic or non-parabolic ) in terms of a limit , not using carleman ?$\eta=e^{1/e}$ is parabolic. Khoustenov gives the equation for the assymptotic behavior, although convergence is slow. http://en.citizendium.org/wiki/Tetration > At b=e^(1/e) , the limiting value L=e, and, asymptotically, > F(z)=e - 2e/z + error term tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 07/09/2010, 12:27 PM why is the radius zero again ? cant find your paper about parabolic iteration ... bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 07/10/2010, 05:17 AM (This post was last modified: 07/10/2010, 09:31 AM by bo198214.) (07/09/2010, 12:27 PM)tommy1729 Wrote: why is the radius zero again ? cant find your paper about parabolic iteration ... Actually I dont know exactly the proof, but in most cases the convergence radius is 0 for parabolic iteration, particularly for the iteration of e^x-1 which is in turn equivalent to iteration of e^(x/e). For parabolic Abel function there is a very old formula by Lévy (2.20 in the overview paper), which is: $\alpha_u(v)=\lim_{n\to\infty}\frac{f^{[n]}(v) - f^{[n]}(u)}{f^{[n+1]}(u)-f^{[n]}(u)}$ which however is not usable for numeric calculation as it is too slow. A formula given by Ecalle (2.22 in the overview paper) is much more usable and works for both cases hyperbolic and parabolic. It is $\alpha(z)=\lim_{n\to\infty} \tilde{\alpha}(f^{[n]}(z))-n$ where $\tilde{\alpha}$ is a sum of some negative powers (none in the hyperbolic case) and a logarithm for example for e^x-1 we get: $\tilde{\alpha}(x)={-2x^{-1}+\frac{1}{3}\ln(x)$. Another formula (2.29 in the overview paper) that kinda combines hyperbolic and parabolic is: $\lim_{n\to\infty} \frac{f^{[n]}(v)-f^{[n]}(z)}{f^{[n+1]}(z)-f^{[n]}(z)}=w\frac{1-\lambda^{w}}{1-\lambda}$, $f^{[w]}(z)=v$, $\alpha_z(v)=w$ where $\lambda$ is the derivative at the fixed point 0, which is 1 in the parabolic case and you take the limit of lambda->1. I am in a hurry a bit. So perhaps more detailed later. tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 07/21/2010, 10:41 PM im waiting and hoping for those " more details " dear bo. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 07/24/2010, 02:59 AM (This post was last modified: 07/24/2010, 03:11 AM by bo198214.) (07/10/2010, 05:17 AM)bo198214 Wrote: Another formula (2.29 in the overview paper) that kinda combines hyperbolic and parabolic is: $\lim_{n\to\infty} \frac{f^{[n]}(v)-f^{[n]}(z)}{f^{[n+1]}(z)-f^{[n]}(z)}=w\frac{1-\lambda^{w}}{1-\lambda}$, $f^{[w]}(z)=v$, $\alpha_z(v)=w$ where $\lambda$ is the derivative at the fixed point 0, which is 1 in the parabolic case and you take the limit of lambda->1. I am in a hurry a bit. So perhaps more detailed later. (07/21/2010, 10:41 PM)tommy1729 Wrote: im waiting and hoping for those " more details " dear bo. Well, there is no more much to add, if you take the limit of the above right side: $\lim_{\lambda\to 1} w\frac{1-\lambda^{w}}{1-\lambda}$ you get $w$ if I not err, which is then the parabolic Levý formula. If you could invert $h(w)=w\frac{1-\lambda^{w}}{1-\lambda}$ for $\lambda\neq 1$ then $h^{-1}\left(\frac{f^{[n]}(v)-f^{[n]}(z)}{f^{[n+1]}(z)-f^{[n]}(z)}\right)$ would be another formula for the hyperbolic Abel function. If you can't (numerically/symbolically whatever) invert then you still have a different formula for the hyperbolic (and Lévy's formla for $\lambda\to 1$) superfunction/iteration: $f^{[w]}(z)=f^{[-n]}(h(w)*(f^{[n+1]}(z)-f^{[n]}(z)) + f^{[n]}(z))$ « Next Oldest | Next Newest »

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