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 the fractional iterate in the first place. It assumes that you already know f^n(x) and that this is, as I say, a "valid continuous function iteration", into which you can plug non-integer n to get non-integer iterates. All the formula does is provide a mechanical way of transforming 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))/4
then it might not be immediately obvious what form the function phi should take. The formula makes it mechanical to calculate that