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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 9:37 PM

Am 12.02.2009 01:18 schrieb Matt:
> Very simple example:
>
> f(x) = x + 2
> f^n(x) = x + 2*n (= nth iterate of f(x))
> g(n,x) = d/dn f^n(x) = 2
> phi(x) = Integral dx/g(0,x) = Integral dx/2 = x/2 + C (C is an
> arbitrary constant)
> phi^-1(x) = 2*(x - C)
> phi^-1(n + phi(x)) = x + 2*n = f^n(x)
>
> Another example:
>
> f(x) = a*x/(x + a) (a is any constant)
> f^n(x) = a*x/(n*x + a)
> g(n,x) = d/dn f^n(x) = -a*x^2/(n*x + a)^2
> phi(x) = Integral dx/g(0,x) = Integral -a*dx/x^2 = a/x + C
> phi^-1(x) = a/(x - C)
> phi^-1(n + phi(x)) = a*x/(n*x + a) = f^n(x)

Very nice!

Gottfried