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

 Messages: [ Previous | Next ]
 alainverghote@gmail.com Posts: 812 Registered: 5/31/08
Re: Iteration formula transformation
Posted: Feb 11, 2009 6:19 AM

On 11 fév, 01:24, Matt <matt271829-n...@yahoo.co.uk> wrote:
> 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  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)

>
> > Alain- Masquer le texte des messages précédents -
>
> - Afficher le texte des messages précédents -- Masquer le texte des messages précédents -
>
> - Afficher le texte des messages précédents -

Dear Matt,

OK, I come back to my thread.

Notice: for iterated functions f^[r](x) = g(r,x)
you have:f^[0](x) = g(0,x) = x ,
plainly, if you don't use the mapping f you've still got
the argument x.

It remains some operators uses:
Taylor expansion of f(x) , f(x+a) may be written
exp(a*d/dx) = (Id+a*d/dx + (a*d/dx)^2/2!+ (a*d/dx)^3/3! ... )
Id for identity . Try upon f(x)=x^2
Id o x^2 = x^2 , a*d/dx O x^2 = 2*a*x and
(a*d/dx)^2/2! o x^2 = a*d/dx O 2*a*x /2 = a^2
here, no need to go farther than power two!
so with f(x) = x^2 ,f(x+a) = (x+a)^2 .

f^[r](x) = exp(r/phi'(x)* d/dx ) O x
Example: f^[r](x) = x/(1+*rx) ,phi(x) = 1/x ,phi'(x) = -1/x^2
f^[0](x) = x and exp(0) o x = Id o x = x
f^[r](x) = exp(-rx^2* d/dx ) O x ,
{Id -rx^2*d/dx +(rx^2*d/dx)^2/2!- (rx^2*d/dx)^3/3! ...} o x
= x -r*x^2 + r^2*x^3 - ...

Just check it,

au revoir,
Alain