Constructing a real valued Fibonacci iteration--its relation to \(1/1+z\)

[JmsNxn]

\[
\phi(t) = \frac{\theta^+(t) \Phi^t + \theta^-(t) \Psi^t}{\Phi - \Psi}\\
\]

And now we ask, where this thing is real valued.

No, we looking for a real-valued superfunction of f (as you did with eta minor).
Because the derivatives \(\frac{-1}{(1+x)^2}\) at the fixed points \(-\Phi,-\Psi\) are negative,
(also remember from here that the fixed point derivatives are the quotient of the eigenvalues, i.e. \(\frac{\Phi}{\Psi}\) and reciprocal, and \(\Psi\) is negative.)
but at least asymptotically (if not regular) you need to have \(\left(f^{\circ t}\right)'(z_0) = f'(z_0)^t \),
which means in a small vicinity it can not be real valued for non-integer t.

That's why I was so surprised about your real valued superfunction at eta minor!

Hmmm, I'm confused then.

If we have a fibonacci function, say \(\phi(t)\); which satisfies \(\phi(0) = 0\) and \(\phi(1) = 1\); and if it is entire; isn't it necessary that:

\[
f^{\circ t}(z) = \frac{\phi(t) + \phi(t-1)z}{\phi(t+1) + \phi(t)z}\\
\]

Is a fractional iteration?

And additionally, the solution I'm suggesting will behave like:

\[
f'(z_0)^t \theta(t)\\
\]

I'm pretty sure that's still okay... but only as a super function; not as an iterate.


AHHHHH I see now!

This only produces a super function, this doesn't produce a fractional iteration!!

So if you went about my route, you'd have:

\[
F(t,z) = \frac{\phi(t) + \phi(t-1)z}{\phi(t+1) + \phi(t)z}\\
\]

Where: \(f(F(t,z)) = F(t+1,z)\).

But you wouldn't have:

\[
F(s,F(t,z)) = F(t+s,z)\\
\]

Yes, okay now I see where my confusion is.

---

Absolutely we can use the regular iteration argument as that forcing our solution to be unique (to both fibonacci, and the iterate of \(1/1+z\)). But when we weaken, leaving the semi-group property, and only consider super functions; then we can make a real valued function super function. It will probably look a lot like \(\eta^-\) to be honest. I think the only way for me to convince you is to run some code though; so let's get at it!