Hi,

I've been doodling around and I was pondering the solutions to the recursive function 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 f''(x)=g(x) then an implicit formula of f'(x) can be found with the Lambert W function.
  2. If g(x) is a bijection, so is f(x)
  3. For g(x)=x^n, f(x)=x^{\sqrt(n)}, for g(x)=cx, f(x)=\sqrt(c)f(x), for 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 g(x)=cx^n and finding f(x)=bx^a but my algebra is not letting it work the other way given 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 g(x)=x^n has a solution for f(x)=1/x^n.

Postulates.


  1. All of f(x) must be surjections.
  2. All of 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?