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So I was playing around with function composition, and started wondering if there was a way to split up a function in to a repeated application of another function.

Notation: Let $f^1 = f$ and $f^n = f \circ f^{n-1}$. (Basically, apply the function $n$ times)

Given a function $f: X \rightarrow X$, I am trying to find a function $g: X \rightarrow X$, such that $f = g^2$. Let's say $f$ is like differentiable and all this pretty stuff.

As a trivial example, $f(x) = x+1$ would mean $g(x) = x+\frac12$, or a solution for $f(x) = 2x$ would be $g(x) = \sqrt2 x$.

How would you go about proving that, for a given function $f$, such a $g$ does or does not exist? What about a more general case, $f = g^n$? (I suppose we could extend the notation to $f^{\frac1n}$, such that $(f^{\frac1n})^n = f$)

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    These go under the name of 'functional square root' or 'half-iterate' of a given function. Wikipedia has an article about them, and they've shown up on Stack sites a number of times before. For a list of such questions on MathOverflow, see here: http://mathoverflow.net/questions/tagged/fractional-iteration – Semiclassical Jan 06 '17 at 15:49
  • The inverse you propose is trivial as its just the desired function applied $n$ times. – Ian Miller Jan 06 '17 at 15:53

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This is a difficult question.

Let just have a look at the very simple f(f(x))=x

It has already many solutions and they look very different.

It is a case by case kind of problem, for affine functions $ax+b$ or monomials $x^n$ you can find peaceful solutions, but it becomes already harder for general polynomials.

For instance, have a look at : f(f(x))=x²+c

This is a trivial polynomial, yet some solutions are already crazy.

In general, there is no unique solution, so you have to define what you are expecting, does $f$ has to have the same limits in $\pm\infty$ than $f^2$, or does it has to be continuous, and so on, in order to restrict the search.

There are also techniques to find analytical solutions (on the form of power series), let have a look at this one for instance : f(f(z))=e^z

This subject has a lot of entries on stackexchange, search with keyword half-iterate to convince yourself that this is not an easy task in general.

zwim
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