Fibonacci as iteration of fractional linear function

[tommy1729]

1) This identity is like hundreds of years old and mentioned a thousand times and even on wiki.

Well, I mean you reading this thread, seeing the people asking for a real extension, obviously knowing the formula and didn't say a thing (before)?

2) the fibonacci sequence clearly is not an iteration since we have 0,1,1,2,... the occurence of 1 twice makes it not an iteration.

So what, we were looking in the context of LFT here where it is an iteration. Anyways you can also consider it an iteration in 2d vectors.
I also wanted to follow up how the corresponding LFTs look.

3) that identity does not satisfy the recursion f(x+1) = f(x) + f(x-1). It is just a lame cos used for a dubious unmotivated interpolation.

The equation \(f(x+2)=f(x+1)+f(x)\) boils down to \(\Phi^2=\Phi+1\) and
\begin{align}
\cos(\pi t + 2\pi)|\Psi|^2 &= \cos(\pi t + \pi)|\Psi| + \cos(\pi t)\\
\cos(\pi t)\Psi^2 &= - \cos(\pi t) |\Psi| + \cos(\pi t)\\
\Psi^2 &= \Psi + 1
\end{align}
And \(\Phi\) and \(\Psi\) are exactly the solutions of \(x^2=x+1\). It's even written in "the wiki" as you call it that it satisfies the Fibonacci identity. So your reasoning rather seems lame, dubious and unmotivated ...

ok , i take part 3 back although i still have some issues with it.

regards

tommy1729