Functional Square Root

[Catullus]

The equation is of the form \(f=sfsf\). Here \(s=\sqrt{}\)

If all the functions are granted to be bijective, Tommy's answer is evident since \({\rm id}=sfs\) thus \(s^{-2}=f\). Let \(s={\rm sqrt}\) you get Tommy's solution. Asking less means we transform the functional equation in easier f.eq. For example.

• if we look for just \(f\) surjective, then we are solving for \(\) \(x^2=f(\sqrt{x})\);
  
• if we ask \(fs=sf\) then we are solving \(f(x)^4=f(f(x))\);
  

Is x^4 the only answer to it? A more interesting question might be that f(x) is analytic and  is analytic (Or at least piece-wise analytic.), and square roots each term of the Taylor series about any point. If , then  would have to equal . But is not the answer.