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

#### magus

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?