
Re: Iteration formula transformation
Posted:
Feb 11, 2009 7:18 PM


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 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. > > ?? > > 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)

