Method for solving "Iterated" functional equation While learning fractional derivative, I have found a way to solve the functional equation:
$f$$m$($x$)=$g(x)$ (where $m$ is a given positive integer, $f$$m$($x$) is the $m$-th iterate of the function $f(x)$-the function we are finding; $g(x)$ is a given function)
Here is a sketch of the method (not rigorous):
Define ($a$$n$): $a$$n+1$$=g$($a$$n$) and $a$$0$$=x$, assume we can find the closed-form of ($a$$n$): $a$$n$=$h(n)$, then $f(x)=h(\frac{1}{m})$.
I know we can use the table in https://en.wikipedia.org/wiki/Iterated_function to look up for $f(x)$ and most of the time $f(x)$ doesn't exist but what I want to ask is:

*

*What is this kind of problem named? Can I have some papers about it?

*How we can get the solution by letting $n=\frac{1}{m}$ while $n$ is supposed to be an integer?

*Can you demonstrate the method above rigorously?

 A: I'll take your questions in a different order. To put it simply, you want to solve the functional equation $f^{m}(x)=g(x)$ for a given function $g$, where the $m$ power denotes the $m^\mathrm{th}$ composition with itself.
(1)
This kind of problem comes down to study the iteration of the function $f$, that is why one speaks of "iterated function", especially when considering non-integer values of $m$, as you intend to do. Nevertheless, when $m$ is an integer, it is often named "functional root", particularly "functional square root" when $m=2$. By the way, the functional square root of the exponential is a well-known topic.
(3)
The sequence you suggest above is not the good one; you should have $a_{n+1}=f(a_n)$, with the initial condition $a_0=x$ and the additional constraint $a_m=g(x)$; then, the solution will be $a_1=f(x)$.
(2)
Obviously, the method in (3) doesn't work for non-integer $m$. In that case, the basic relation is Abel's equation $f(x) = h^{-1}(h(x)+1)$. Then, the $n^\mathrm{th}$ iterate of $f$ is simply given by $f^n(x) = h^{-1}(h(x)+n)$; this relation permits to generalize/define non-integer iterates through $f^t(x) = h^{-1}(h(x)+t)$ with $t\in\mathbb{R}$ (or even $\mathbb{C}$) $-$ note that $t=-1$ corresponds to the reciprocal. In your case, you would have to solve the equation $g(x) = f^m(x) = h^{-1}(h(x)+m)$ for $h$, whence the solution $f(x) = h^{-1}(h(x)+1)$.
Addendum
The function $h$ is usually quite hard to determine (functional equations are almost always a nightmare to solve). Sometimes, it is found more easily thanks to Schöder's or Böttcher's equations, which are obtained from Abel's equation after a change of variable. Finally, the function $h$ is related to the translation/composition operator in Lie theory, whose tools may help you in your developments.
