F(f(x))=exp(exp(x)).

[JmsNxn]

A quick way to remedy this problem is stick to a fixed point.

Call \(G(x) = \exp^{\circ n}(x)\), and we want to know how many \(n\) roots there are about the fixed point. The general rule is that there are \(n\) \(n\)-roots about a fixed point.

So if:

\[
\Psi(G(z)) = \lambda \Psi(z)\\
\]

There are precisely \(n\) functions \(g_i\) such that:

\[
\Psi(g_i(z)) = \sqrt[n]{\lambda} \Psi(z)\\
\]

These are given as:

\[
g_i(z) = \Psi^{-1}\left(\sqrt[n]{|\lambda|}\zeta_i\Psi(z)\right)\\
\]

Where \(\zeta_i\) is an \(n\)'th root of unity.

Ok but hold on a minute...

What about the 1 periodic function ??


regards

tommy1729

Hey tommy, this is about the fixed point only!

I apologize, thought I made that clear. So these are the only local solutions about a fixed point.

If we use something like Kneser, then you can make a whole bunch of solutions, but they won't be holomorphic about the fixed point--the Kneser fractional iteration isn't holomorphic at any fixed point/periodic point.


A proof of this is pretty standard. If you take the functional:

\[
\frac{d}{dz}\Big{|}_{z=L} g(z)\\
\]

It maps to the multiplicative group under composition \(g_1(g_2(z)) \mapsto g_1'(L)\cdot g_2'(L)\)

oh yes i see.

But that would not help in solving the equation then I think ...

I might be wrong though ...

IIRC it was asked informally on mathoverflow as a thread and in chat but without a definite answer.
So it might be harder than it seems.

Well whats new , a hard problem lol.



regards

tommy1729

Oh yes, there are many complicated ways to answer the question. I was just giving my two cents when we're focused on fixed points. There's probably infinite solutions in actuality--say on the real line. But who's got time for that. One could potentially be, just do everything Kneser does with \(\exp\) but do it with the squareroot of \(\exp\exp\) that is not \(\exp\). Then you get a brand new kneser like iteration of the not \(\exp\) square root, which is real valued, and satisfies the equation. My head spins just thinking about it, lmao.