Functional Square Root

[tommy1729]

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 (x) is analytic (Or at least piece-wise analytic.), and square roots each term of the Taylor series about any point. If f(x) = x^0+x+x^/2, (x) would have to equal sqrt(x^2)+sqrt(x)+sqrt(x^2/). But f(x)=x^0+x+x^2/2 is not the answer.

Well x^4 is analytic everywhere.

And cosh(x) also not a solution.

Let g(x) = c * x^a.



f(x) = g(g(x)) = c * ( c^a x^(a^2) ) = c^(a+1) x^(a^2)

f(sqrt(x)) = c^(a+1) x^(a^2 /2) = g(x)

so c^(a+1) = c , a^2 = 2 * a.


So a = 0 or a = 2.

So we get 

g(x) = C for any C.

or 

g(x) = c x^2 for c^3 = c.

c^3 = c implies c = -1 or 0 or 1.

regards

tommy1729