
Re: Iteration formula transformation
Posted:
Feb 11, 2009 9:06 AM


Am 09.02.2009 06:32 schrieb Matt: > 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. Hmmm,
but how do you find the derivative of f^n(x) wrt to n without having the fractional iterate before? As I understand this would be
df^n(x) f^(n+h)(x)  f^n(x)  =  for h>0 dn h
or df^n(x) f^(n+h/2)(x)  f^(nh/2)(x)  =  for h>0 dn h
Gottfried

