Date: Feb 9, 2009 12:32 AM
Author: matt271829-news@yahoo.co.uk
Subject: 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.