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

Oct 2009

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 \(\displaystyle f(x)=1/x^n\).


  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?