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 Fractional iteration of x^2+1 at infinity and fractional iteration of exp tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 06/08/2011, 01:18 PM (06/08/2011, 09:55 AM)bo198214 Wrote: It is the iteration at infinity. To see whats happening there one moves the fixpoint at infinity to 0, by conjugating with 1/x: $g(x)=\frac{1}{(\left(\frac{1}{x}\right)^2+1}=\frac{x^2}{x^2+1}$ in thread tid 403 I , Bo , Mike and Ben already discussed moving fixpoints in particular - as the title says - f(f(x)) = exp(x) + x. this thread seems similar. however , i consider this slightly different. exp(x) + x has a " true " fixpoint at oo. with " true " i mean that it touches the id(x) line. in this case of x^2 + 1 , i would not call oo a fixpoint , but rather say that Bo uses so-called " linearization " : g^[-1](f^[n](g(x))) = [g^[-1](f(g(x))]^[n] Quote:$g^t(x)=\lim_{n\to\infty} \beta^{-1}(\beta(x)^{2^t})$. you take lim n -> oo ... and then i see no n. so this needs a correction. Quote:Again similar to the Schröder case, we have an alternative expression: $g^t(x)=\lim_{n\to\infty} g^{-n}\left(g^n(x)^{2^t}\right)$ If we even roll back our conjugation with $1/x$ we get: $f^t(x)=\lim_{n\to\infty} f^{-n}\left(f^n(x)^{2^t}\right)$. i dont get how you arrive at this ... g and f on both sides ? Quote:This may lead into a new way of computing fractional iterates of exp, because we just approximate exp(x) with polynomials $\sum_{n=0}^N \frac{x^n}{n!}$ and approximate the half-iterate of exp with the half-iterate of these polynomials. hmm ... i wonder if (x^2 + 1)^[h] is analytic at x = 0 for small h. the reason is that lim h-> 0 (x^2 + 1)^[h] = abs(x) and abs(x) is not analytic at 0. im not sure about being analytic elsewhere either , though maybe levy böttcher schröder imply so ( i still do not know enough about them ). also approximating exp with polynomials might be troublesome ; does the n'th approximation of the half-iterate converge when n -> oo ? do we really get an analytic function at n = oo ? as a bad example that does not make sense in the case of exp(x) but gives an idea of what i mean ( i dont have a good one for the moment ) for instance if an n'th polynomial satisfies f(x) = f(-x) we are in trouble. polynomials also have zero's and those zero's will need to drift towards oo fast if the sequence of polynomials wants to approximate exp(x) well. so i suggest working with the (n^2)'th polynomial approximations , rather then compare the n'th with the (n+1)'th. despite much comment , dont get me wrong , i like this idea. regards tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by bo198214 - 06/08/2011, 09:55 AM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by Gottfried - 06/08/2011, 01:09 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by tommy1729 - 06/08/2011, 01:18 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by bo198214 - 06/08/2011, 07:59 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by JmsNxn - 06/08/2011, 07:26 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by mike3 - 06/08/2011, 09:31 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by bo198214 - 06/08/2011, 09:48 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by mike3 - 06/08/2011, 10:08 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by bo198214 - 06/08/2011, 10:12 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by mike3 - 06/08/2011, 10:58 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by bo198214 - 06/09/2011, 05:56 AM

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