# finding f(x) for f(f(x))=g(x) where f(x) and g(x) are continuous

• May 26th 2010, 09:06 AM
magus
finding f(x) for f(f(x))=g(x) where f(x) and g(x) are continuous
Hi,

I've been doodling around and I was pondering the solutions to the recursive function $\displaystyle f(f(x))=g(x)$ and I've got three things I'm working on. If you guys want I can put down my scratch but I think the only things I've come up with so far are pretty obvious.

1. Let $\displaystyle f''(x)=g(x)$ then an implicit formula of $\displaystyle f'(x)$ can be found with the Lambert W function.
2. If g(x) is a bijection, so is f(x)
3. For $\displaystyle g(x)=x^n, f(x)=x^{\sqrt(n)}$, for $\displaystyle g(x)=cx, f(x)=\sqrt(c)f(x)$, for $\displaystyle g(x)$ and that for the above formula on the nth recursion change the square root to the nth root.

I'm working on a formula for $\displaystyle g(x)=cx^n$and finding $\displaystyle f(x)=bx^a$ but my algebra is not letting it work the other way given $\displaystyle cx^n=b^{a+1}x^{a^2}$.

For n=1 of course it's trivial and for n<0 and n is an integer there is another class of solutions. in fact the given solution above is not unique and $\displaystyle g(x)=x^n$ has a solution for $\displaystyle f(x)=1/x^n$.

Postulates.

1. All of $\displaystyle f(x)$ must be surjections.
2. All of $\displaystyle f(x)$ of closed form are of the two forms stated above. (I've checked most other elementary functions I can think of and they won't fly)

If there anything wrong here that you can point out so i don't end upi spinning my wheels? Is there any study of recursive continuous functions I should be aware of?