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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:

  1. What is this kind of problem named? Can I have some papers about it?
  2. How we can get the solution by letting $n=\frac{1}{m}$ while $n$ is supposed to be an integer?
  3. Can you demonstrate the method above rigorously?
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  • $\begingroup$ fair amount of information at mathoverflow.net/questions/45608/… Let's see, there are often solutions on intervals, bounded by critical points or fixpoints of $g(x).$ I liked A History of Complex Dynamics by D. S. Alexander. Also Dynamics of One Complex Variable, by Milnor. Then Iterative Functional Equations by Kuczma, Choczewski, Ger. The famous paper is H. Kneser (1950), pages 56-67 gdz.sub.uni-goettingen.de/id/… $\endgroup$
    – Will Jagy
    Commented Jan 8, 2023 at 1:16
  • $\begingroup$ What is $m?{}{}{}{}$ $\endgroup$ Commented Jan 8, 2023 at 3:30
  • $\begingroup$ It's not clear how you can define $h(1/m)$ if you have $h(n)$ defined for $n\in\mathbb Z^{+}.$ Fractional derivatives don't let you define fractional composition. $\endgroup$ Commented Jan 8, 2023 at 3:32
  • $\begingroup$ @ThomasAndrews You pointed out the part where I stuck. The equality $f(x)=h(\frac{1}{m})$ suggests that $h(n)$ might have been defined for all real-valued $n$? Weird. I didn't know where I messed up. $\endgroup$ Commented Jan 8, 2023 at 10:40

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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.

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