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Related to this question, if we try and narrow it down a bit. One possible constraint to reduce the number of candidate functions that I could think of is to try and minimize some kind of distance to unity.

Example: the real identity function $$f(x)=x$$ has the non-trivial compositional square root $$f^{\circ \frac 1 2}(x)=k-x$$ This is easy to show: $$( f^{\circ \frac 1 2}\circ f^{\circ \frac 1 2})(x) = f^{\circ \frac 1 2}(f^{\circ \frac 1 2}(x)) = k-(k-x) = k-k+x = x = f(x)$$ But a simpler square root is of course $f$ itself, since $f(f(x)) = x$ if $f(x) = x$!

Can we create some distance measure between functions that will make us find some kind of a "simplest" compositional $n$-th root?

A naive example that would work here for our very simple example above is $$d(x,f^{\circ \frac 1 n}) = \sqrt[p]{\int_{-\infty}^\infty \left|f^{\circ \frac 1 n}(x)-x \right|^pdx}$$

But would that make sense in general? If not, what could we use instead?

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mathreadler
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    Your metric is very unlikely to converge in the general case. Sadly, when finding compositional roots, it usually only works in very specific instances, and secondly is rarely defined on all of $\mathbb{R}$ (without being constructed for the sole purpose of being defined on $\mathbb{R}$). Local inner products are a much better option, and I have used this often. Typically I just let the space be the space of all fractional iterates of $f$. Then define an inner-product $(h,g) = \int_L h(x)g(x),dx$, where $L$ is the maximal domain the iterates of $f$ are defined. –  Apr 16 '17 at 15:36
  • Cont'd... This inner product will define a metric, and distance from the identity is trivial from here. But sometimes, the $n$'th root of $f$ doesn't tend to the identity (be wary of this), this is true for $\sin$ for example. All of $\sin$'s roots are periodic, so the limit tends to the triangle wave, not the identity function... Plus, mathematicians are usually more worried about somehow constructing fractional iterates, than what to do with them when they're constructed. So a lot of machinery is lacking, in terms of inner product spaces, orthogonality relationships, and the such and such. –  Apr 16 '17 at 15:40
  • Last comment, if your $n$'th roots are defined on all of $\mathbb{R}$ you'll definitely need a weight function. The gaussian can sometimes work. But sometimes $f^{\circ n}(x)$ grows ridiculously fast though and kills the Gaussian, so finding a very fast decaying function can be a hassle. –  Apr 16 '17 at 15:46
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    That's lot of useful knowledge. Feel free to write an answer if you want to. I would gladly upvote. – mathreadler Apr 16 '17 at 15:48

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