amy666
Posts:
1,859
Registered:
6/27/07


Re: Iteration formula transformation
Posted:
Feb 12, 2009 7:22 AM


> 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)
hmm
how do you work out f(x) = e*x ?

