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 levy ecalle koenigs ? bo198214 Administrator Posts: 1,389 Threads: 90 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. « Next Oldest | Next Newest »

 Messages In This Thread levy ecalle koenigs ? - by tommy1729 - 07/08/2010, 11:31 PM RE: levy ecalle koenigs ? - by bo198214 - 07/09/2010, 05:46 AM RE: levy ecalle koenigs ? - by tommy1729 - 07/09/2010, 12:27 PM RE: levy ecalle koenigs ? - by bo198214 - 07/10/2010, 05:17 AM RE: levy ecalle koenigs ? - by sheldonison - 07/09/2010, 11:48 AM RE: levy ecalle koenigs ? - by tommy1729 - 07/21/2010, 10:41 PM RE: levy ecalle koenigs ? - by bo198214 - 07/24/2010, 02:59 AM

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