```Date: Feb 11, 2009 9:50 AM
Author: matt271829-news@yahoo.co.uk
Subject: Re: Iteration formula transformation

On Feb 11, 2:06 pm, Gottfried Helms <he...@uni-kassel.de> wrote:> Am 09.02.2009 06:32 schrieb Matt:> It's well known and straightforward that if f(x) = phi^-1(1 + phi(x))> > for some function phi, then f^n(x) = phi^-1(n + phi(x)), where "f^n"> > denotes iteration of f. But say we have an expression for f^n that we> > found some other way and we know is a valid continuous function> > iteration, then how to recover phi? As an alternative to guessing the> > correct form followed by some trial-and-error, I found>> >     phi(x) = Integral dx/g(0,x)>> > where g(n,x) = d/dn f^n(x).>> > Probably nothing new, but kind of cute I thought.>> Hmmm,>>  but how do you find the derivative of f^n(x) wrt to n>  without having the fractional iterate before?You don't. I don't claim that this formula helps to find thefractional iterate in the first place. It assumes that you alreadyknow f^n(x) and that this is, as I say, a "valid continuous functioniteration", into which you can plug non-integer n to get non-integeriterates. All the formula does is provide a mechanical way oftransforming a known f^n(x) into the form f^n(x) = phi^-1(n + phi(x)).For example, if you know   f^n(x) = (2*alpha^(2^n) + 2*alpha^(-2^n) - b)/(2*a)      with alpha = (2*a*x + b + sqr((2*a*x + b)^2 - 16))/4then it might not be immediately obvious what form the function phishould take. The formula makes it mechanical to calculate that   phi(x) = log(log(alpha))/log(2)
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