Date: Feb 11, 2009 7:18 PM
Author: matt271829-news@yahoo.co.uk
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

arbitrary constant)

phi^-1(x) = 2*(x - C)

phi^-1(n + phi(x)) = x + 2*n = f^n(x)

Another example:

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)