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


Re: Iteration formula transformation
Posted:
Feb 11, 2009 3:05 PM


> 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
regards
tommy1729
phi(x) = x + 2 =

