Date: Feb 10, 2009 7:24 PM
Subject: Re: Iteration formula transformation
On Feb 10, 10:56 am, "alainvergh...@gmail.com"
> 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
Bonjour 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)