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For polynomials with powers 1 and under, it's easy to find the functional square root.

i.e. if: $$f(f(x)) = f^{[2]}(x) = ax+b$$

then: $$f(x) = \sqrt{a}x+\frac{b}{\sqrt{a}+1}$$

Is it possible to find a general form for a quadratic function? I.e what is $f$, given that $$f^{[2]}(x)=ax^2+bx+c$$

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    $\begingroup$ You could try assuming $g$ has a Taylor Series expansion, and then see what conditions you get on the coefficients in order for $g \circ g$ to equal f. $\endgroup$ – Jonathan Feb 7 '18 at 23:21
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    $\begingroup$ See Find $f(x)$ such that $f(f(x)) = x^2 - 2$. Note in particular that the answer depends on what the domain requirements are for $f$. $\endgroup$ – dxiv Feb 7 '18 at 23:34
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There is no solution to such equations in general. For example, there is no solution to $f(f(x)) = x^2 - 2$ - see problem 7 here.

However, some of those equations do have a nontrivial solution, for example $f(f(x)) = x^2 +2$. Can you find one?

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