Fibonacci as iteration of fractional linear function

[JmsNxn]

OKAY! Absolutely beautiful! Yes this should be a LFT! I think this implicitly solves \(\theta(t)\), I'm just not sure how yet. Definitely agree this will get it though!

I think this is very important, because it shows how the iteration of an LFT is Fibonacci, and shows how a restriction to reals of an LFT creates a real Fibonacci.

It is not a Fibonacci! As Gottfried pointed out already it only agrees on the even Fibonacci indexes.
Hence it is not a theta transforming the non-realness (of the original Fibonacci interpolation) into a realness - which honestly I can not imagine that it exists, because of the pole in the middle between the fixed points - but one never knows. That's why Gottfried was excluding these poles in the middle (which only occurs for uneven iterates) and banning it to the left side of the left fixed point (where it occurs for even iterates, or to the right side of the right fixed point when we consider negative even iterates) by demanding that it only agrees with Fibonacci on even iterates.
Even you pointed out to me already once that the iterative root of the iterative square is probably not the original function anymore.
But the trick with changing the sign of the eigenvalue is quite naughty, Gottfried!
But yes, it boils down to taking the regular iteration of \(f^{\circ 2}\) and just taking 1/2 of it \((f^{\circ 2})^{\mathfrak{R} t/2}\).

OHHH OKAY!

I was thinking that primarily! That this would be something like that. Especially when he said he was squaring the eigenvalue. I thought it was going to be one of those \(f^{\circ 2 \circ 1/2} \neq f\) problems, but I guess I got blinded like a moth to the flame. Thinking there was the solution. lol.

Okay, so real valued fibonacci equations are still on the table! I wonder if the poles you are seeing in your iterate, translate directly to "no real valued fibonacci" though. Maybe the poles are just artifacts of the translation; not the actual fibonacci sequence.

I really believe there must be a "crescent iteration" fibonacci. It's probably not injective on \(\mathbb{R}^+\), but there's gotta be one. It just seems to simple. We're just looking for a \(\theta(z)\phi(z)\) that is real valued.

We are just solving an equation:

\[
\sum_{k=-\infty}^\infty a_ke^{2\pi i k z} \phi(z) : \mathbb{R}^+ \to \mathbb{R}^+\\
\]

Come on, that's gotta be doable right!? Where additionally \(\phi(z)\) is just a super position of exponentials. Come on...