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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 12:45 PM
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On 12 fév, 13:22, amy666 <tommy1...@hotmail.com> wrote:
> > On Feb 11, 8:05 pm, amy666 <tommy1...@hotmail.com>
> > wrote:

> > > > 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
>
> > 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)

>
> hmm
>
> how do you work out f(x) = e*x ?- Masquer le texte des messages précédents -
>
> - Afficher le texte des messages précédents -


Bonsoir,

f(x) = exp(1)*x ,f^[r](x)= (exp(1))^r*x = g(r,x)
I prefer rewriting the formula : phi(x) = int(c to x ,du/g(0,u) )
Since f^[r](x)= (exp(1))^r*x = e^r*x = g(r,x)
d/dr g(r,x) = ln(e)*e^r*x (logarithmic derivation)
g'[r](0,x) = x ,
int(c to x du/u) = ln(x)-ln(c) = phi(x)

Alain



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