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

[tommy1729]

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 \( f(f(x))=F(x) \).

i had expected such a reply.

but i disagree.

let F( real ) map to reals.

and let f( real ) map to reals.

assuming those are satisfied ,

i feel that F(x+1) = exp(F(x))

should also satisfy F(x+1/2) = f(F(x))

where f(f(x)) = exp(x) , if it wants to be tetration.

in general F(x+a) = f_a(F(x))

where f_a satisfies f_a(((... a times ...(x)))) = exp(x)

should be satisfied.

furthermore the inverse of F might be multivalued !!

but that doesnt mean f(x) is all the possible results of F(1/2 + invF(x))

taking that into account , its probably clear that i will only accept examples of 2 distinct analytic solutions f(x) that map all reals to a subset of reals and satisfy f(f(x)) = exp(x).


im not trying to be difficult.

but this is important.

working with F(x) seems like overkill , instead i focus on f : f(f(x)) = exp(x).

( as an analogue : there are multiple functions that satisfy f(x+1) = e*f(x) but that doesnt mean there are multiple functions that are a solution to exp( log(x) + 1 ) ( being e*x ) )


regards

tommy1729