
Iteration formula transformation
Posted:
Feb 9, 2009 12:32 AM


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 trialanderror, 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.

