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


Re: Iteration formula transformation
Posted:
Feb 11, 2009 3:06 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) = 2*x = integral dx / d/d0 f^0(x) ??? > > i dont get it. > > some examples plz > > regards > > tommy1729 ( corrected typo )

