```Date: Feb 10, 2009 7:24 PM
Author: matt271829-news@yahoo.co.uk
Subject: Re: Iteration formula transformation

On Feb 10, 10:56 am, "alainvergh...@gmail.com"<alainvergh...@gmail.com> wrote:> 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  xBonjour Alain, I don't understand your notation here. What does "d/dx"by itself mean? What does "O  x" mean?>          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
```