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Topic: Iteration formula transformation
Replies: 14   Last Post: Feb 12, 2009 12:45 PM

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Gottfried Helms

Posts: 1,926
Registered: 12/6/04
Re: Iteration formula transformation
Posted: Feb 11, 2009 9:06 AM
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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 trial-and-error, 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.


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

df^n(x) f^(n+h/2)(x) - f^(n-h/2)(x)
------- = ---------------------------- for h->0
dn h


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