f( f(x) ) = exp(x) solved ! ! !

[bo198214]

well , i assume the existence of tetration ( knesers exp(F(x)) = F(x+1)) leads to existence of f(f(x)) = exp(x).

Yes, as we assume \( F \) to be strictly increasing we can take the inverse function \( F^{-1} \). Then you can verify that
\( f(x):=F\left(\frac{1}{2}+F^{-1}(x)\right) \) satisfies \( f(f(x))=\exp(x) \)

however , im not sure that f(f(x)) = exp(x) is not unique by analyticity alone ?

care to explain ?

Now if we have one analytic solution \( F \) then we have a lot of other analytic solutions given by
\( F_\theta(x):=F(x+\theta(x)) \) for any 1-periodic analytic function \( \theta \) (prove that \( F_\theta(z+1)=\exp(F_\theta(x)) \) too!)

If we make the amplitude of those \( \theta \) sufficiently small, then \( x+\theta(x) \) is strictly increasing and \( F_\theta \) is too.

Finally:
\( f_\theta(x):=F_\theta\left(\frac{1}{2}+F^{-1}_\theta(x)\right) \)
is another analytic solution of \( f(f(x))=F(x) \).