Date: Feb 11, 2009 3:06 PM
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
> > 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
> > 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
( corrected typo )