Date: Feb 11, 2009 3:06 PM
Author: amy666
Subject: Re: Iteration formula transformation
> > 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.

>

> ??

>

> 2*x is the nth iterate of x+2.

>

> phi(x) = 2*x = integral dx / d/d0 f^0(x) ???

>

> i dont get it.

>

> some examples plz

>

> regards

>

> tommy1729

( corrected typo )