Fibonacci as iteration of fractional linear function

[JmsNxn]

Would we have to solve:

\[
f_d(t) = \frac{F_{\text{Gottfried}}(t) - F_{\text{Gottfried}}(t-1)z}{F_{\text{Gottfried}}(t+1) - F_{\text{Gottfried}}(t)z}\\
\]

Or can you see a simpler way that I'm not seeing, where \(F_{\text{Gottfried}}(t) = \theta(t)\phi(t)\)?

Hmm, first remark: the "dual-" function \( f_d(x) \) is identical to \( f(x)\) for real \(x\) and even iteration height - this is simply because it is the other functional root of \( f(f(x)) \) where the (Schroeder-) multiplier \( \Psi \) is positive instead negative as it is with \(f(x)\) .  

second remark: didn't think of more properties of \( f_d(x) \) because I thought, it is simply a cheap exemplar of a real-to-real function, where Henryk could apply his derivations and which is still a near relative to the rational \( 1/(1+x) \) - function. (The Schroeder-function, for instance is of course the same - except the sign change on the \( \Psi \))

Cannot say more at the moment, especially nothing about a \( \theta()\)-component, sorry...

OHHHHH

Okay, so you've found the other functional square root of \(f(f(x))\) where \(f = \frac{1}{1+x}\). Yes, this probably wouldn't produce the Fibonacci sequence in the same natural way. \(f\) wouldn't even be a linear fractional transformation would it?

Yes, so these are two different questions. I think Daniel was much closer to what I was asking, but I'm still unsure of how he's pulling out a real valued Fibonacci iteration.