```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 plzVery simple example:f(x) = x + 2f^n(x) = x + 2*n (= nth iterate of f(x))g(n,x) = d/dn f^n(x) = 2phi(x) = Integral dx/g(0,x) = Integral dx/2 = x/2 + C (C is anarbitrary 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)^2phi(x) = Integral dx/g(0,x) = Integral -a*dx/x^2 = a/x + Cphi^-1(x) = a/(x - C)phi^-1(n + phi(x)) = a*x/(n*x + a) = f^n(x)
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