g(x,y) variables separation and iteration...

[alainv...@gmail.com]

I think there are some links between separability and iterates forms.

Be g(x,y) a smooth function and exists f such as :
f(g(x,y)) = f(x) + h(y) , f bijective .
Example: g(x,y) = x/(1 +yx)
gives 1/g = 1/x + y
If f(u) =1/u , g(x,y) = f^ [-1] (f(x)+y) = (x/(1+x))^[y]
[y] y th iterate ,

In case f(g(x,y)) = m(x) + h(y) , m and h non constant functions,
we may write g(m^-[1](f(x) ,y) = f^[-1](f(x) + h(y))
= { f^[-1](f(x) +
1)) } ^ [h(y)] ,

Any comments ?

Alain

[amy666]

alain wrote :

> I think there are some links between separability and
> iterates forms.
>
> Be g(x,y) a smooth function and exists f such as :
> f(g(x,y)) = f(x) + h(y) , f bijective .
> Example: g(x,y) = x/(1 +yx)
> gives 1/g = 1/x + y
> If f(u) =1/u , g(x,y) = f^ [-1] (f(x)+y) =
> (x/(1+x))^[y]
> [y] y th iterate ,
>
> In case f(g(x,y)) = m(x) + h(y) , m and h non
> constant functions,
> we may write g(m^-[1](f(x) ,y) = f^[-1](f(x) + h(y))
> = {

> = {
> = {
> = { f^[-1](f(x)

> = { f^[-1](f(x) +
> 1)) } ^ [h(y)] ,
>
> Any comments ?
>
> Alain

intresting.

[alainv...@gmail.com]

> intresting.- Masquer le texte des messages précédents -
>
> - Afficher le texte des messages précédents -

Bon Dimanche,

It is a very old hobby of mines...
I try ways to legitimate continuous iterations...

Amicalement,
Alain