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

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Posts: 1,859
Registered: 6/27/07
Re: Iteration formula transformation
Posted: Feb 11, 2009 3:05 PM
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> 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.


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



phi(x) = x + 2 =

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