Date: Feb 11, 2009 7:18 PM
Subject: Re: Iteration formula transformation
On Feb 11, 8:05 pm, amy666 <tommy1...@hotmail.com> wrote:
> > 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) = x + 2 = integral dx / d/d0 f^0(x) ???
> i dont get it.
> some examples plz
Very simple example:
f(x) = x + 2
f^n(x) = x + 2*n (= nth iterate of f(x))
g(n,x) = d/dn f^n(x) = 2
phi(x) = Integral dx/g(0,x) = Integral dx/2 = x/2 + C (C is an
phi^-1(x) = 2*(x - C)
phi^-1(n + phi(x)) = x + 2*n = f^n(x)
f(x) = a*x/(x + a) (a is any constant)
f^n(x) = a*x/(n*x + a)
g(n,x) = d/dn f^n(x) = -a*x^2/(n*x + a)^2
phi(x) = Integral dx/g(0,x) = Integral -a*dx/x^2 = a/x + C
phi^-1(x) = a/(x - C)
phi^-1(n + phi(x)) = a*x/(n*x + a) = f^n(x)