The composition operator is a well know and quite often used method in integration and differentiation, think u-substitution. However, given a composition like $$f(f(f(...f(x)...)))$$ Where there are n compositions more compactly, $f^n(x)$ don't confuse this with the exponent operator.

Define the nth decomposition of a function $g(x)$ to be a function $f(x)$ such that $f^n(x)=g(x)$. Anyone know how to evaluate this "decomposition operator"? Also, if the above is to simple or restricting, how about evaluating if the composition series has an index variable k? This would allow an extra variable to range from 1 at the first composition, inner most part, to n at the outermost part. Any references or novel ideas are appreciated. It should also be noted that if there are an infinite number of solutions to the problem, I'd like ways to always find at least one of them. Although, I'm just saying the antiderivative isn't exactly hindered by its arbitrary number of solutions. Although, finding the antiderivative is also a bit of an abstract and most of the time impossible process.

For context purposes, my idea was to use the ideas of self similarity to decompose the function, much in the same way one breaks a fractal into its corresponding similarities. However I've only gotten the method to work in heuristic ways. In other words, the function has to be obviously self similar.

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    $\begingroup$ Without any restriction on the function $f$, this is extremely underdetermined. For example, take functions $\mathbb R\to\mathbb R$, $n=2$ and $g(x)=x$. Then $f(x)$ can be any permutation that exchanges pairs of real numbers. Examples are: $f(x)=x$, $f(x)=-x$, $f(x)=1-x$, $f(x)=x^{-1}$ for $x\ne 0$ and $f(0)=0$, … $\endgroup$ – celtschk Apr 11 '15 at 22:08
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    $\begingroup$ This is a well-studied problem. A good place to start might be the wikipedia page. $\endgroup$ – George V. Williams Apr 11 '15 at 22:11
  • $\begingroup$ @GeorgeV.Williams now I know that it's been studied. Any preferred references? $\endgroup$ – Zach466920 Apr 11 '15 at 22:18
  • $\begingroup$ @celtschk that should be a good thing. I only want a possible solution, not all solutions. Stated another way, an infinite number of solutions should vastly simplify the process of finding one solution. $\endgroup$ – Zach466920 Apr 11 '15 at 22:21

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