Date: Feb 11, 2009 9:50 AM
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.
> 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
phi(x) = log(log(alpha))/log(2)