$f^n(x)$ where $n \in \mathbb{R}^+$ Given a function $f(x)$ whose image is a subset of its domain, we can define
$$
f^n(x) = \underbrace{f(f(f(\dots f(x) \dots )))}_{n \text{ times}}
$$

This makes sense when $n$ is a nonnegative integer.

Can we extend this definition to continuous values of $n$? Such as $f^{\frac{1}{2}}(x)$?
Let $f(x)=x$. You would want $f^{1/2}$ defined so that $$\left(f^{1/2}\circ f^{1/2}\right)(x)=f(x)=x$$

Here are five options, for starters:


$f^{1/2}(x)=x$
$f^{1/2}(x)=-x$
$f^{1/2}(x)=\sqrt[3]{1-x^3}$
The implicit solution to
$x+f^{1/2}(x)=e^{-\left(x-f^{1/2}(x)\right)^2}$
The implicit solution to
$x+f^{1/2}(x)=\frac12\cos\mathopen{}\left(x-f^{1/2}(x)\right)\mathclose{}$


What would be the basis to choose one of these? I could understand a preference for the first option here. But with general $f$, if this is evidence that there are many possibilities for $f^{1/2}$, you might want to have a definition that somehow picks one canonically.

Or maybe you just let $f^{1/2}$ stand for the class of all functions that meet the condition $f^{1/2}\circ f^{1/2}=f$.
You see here $n$ is a whole number. If you take it as 0.5 then problems will arise. For example is $n=2$ then that means we input function into itself two times but what $n=0.5$ would mean. Would it mean that we only input half of f or the first term of f or the last term.This becomes ridiculous when n goes out of range