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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 10, 2009 5:56 AM
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On 9 fév, 06:32, Matt <matt271829-n...@yahoo.co.uk> 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.


Bonjour Matt,

With smooth functions from phi(f^[r](x)=phi(x)+r , r real
we've got d/dr f^[r](x) = phi'(x)* d/dx f^[r](x).
There are other relations:
f^[r](x) = exp(1/phi'(x)* d/dx ) O x
phi^-1(L)= f^[L-phi(x)](x) = c constant,
Example f(x) = 2*x +1
phi^-1(L)= (2x+1)^[L-phi(x)] = c ,constant
2^(L-phi(x))*(x+1)-1 = c
all done phi(x) = ln(x+1)/ln(x) + k ,any ct.

ISOLATING POWER r in g(r,x) :
Ex. g(r,x) = 5^r*x/(1+(5^r-1)*x)
5^r*x/(x-1) = gr/(gr-1) and a little later :
ln(gr/(gr-1))/ln(5) = ln(x/(x-1))/ln(5) + r
So phi(x) = ln(x/(x-1))/ln(5) counts iterations of
f(x) = g(1,x)=5x/(1+4*x)

Alain



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