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

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alainverghote@gmail.com

Posts: 812
Registered: 5/31/08
Re: Iteration formula transformation
Posted: Feb 12, 2009 2:22 AM
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On 12 fév, 03:37, Gottfried Helms <he...@uni-kassel.de> wrote:
> 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- Masquer le texte des messages précédents -
>
> - Afficher le texte des messages précédents -


Bonjour,

YES,it's a very nice formula,
easy to be proved too...

Alain



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