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 \(g_1(g_2(z)) \mapsto g_1'(L)\cdot g_2'(L)\)