Fibonacci as iteration of fractional linear function

[tommy1729]

I realize that my investigation of the relationship between Fibonacci Extensions and corresponding LFTs are barely thorough. Just for recapitulation
We took the LFT \(f(z)=\frac{1}{1+z}\) realized that \(f^{\circ n}(z)=\frac{\phi_{n-1}z+\phi_n}{\phi_n z+\phi_{t+1}}\), where \(\phi_n\) is the \(n\)-th Fibonacci number. So there could be a relation between non-integer iteration of \(f\) and non-integer extension of \(\phi\).

However I just took the iteration of the LFT and plugged in the Fibonacci extension and claimed that it was the regular iteration of the LFTs. But I din't prove anything and just wondered that the alternative (real valued) Fibonacci extension didn't work (not an iteration group).

Being an iteration group (though I say "group" I always also include semi-group in the meaning, it is just too long to write it out each time) means the following:
\[ f^{\circ s}\circ f^{\circ t} = f^{\circ s+t}, \quad f^{\circ 1} = f \]
So lets have a look at corresponding LFTs
\begin{align}
f^{\circ s}(f^{\circ t}(z)) &= \frac{\phi_s+\phi_{s-1}\frac{\phi_t + \phi_{t-1}z}{\phi_{t+1} + \phi_t z}}{\phi_{s+1}+\phi_s\frac{\phi_t + \phi_{t-1}z}{\phi_{t+1} + \phi_t z}}\\
&=\frac{\phi_s(\phi_{t+1} + \phi_t z)+\phi_{s-1}(\phi_t + \phi_{t-1}z)}{\phi_{s+1}(\phi_{t+1} + \phi_t z)+\phi_s(\phi_t + \phi_{t-1}z)}\\
&=\frac{\phi_s\phi_{t+1} + \phi_{s-1}\phi_t + (\phi_s\phi_t+\phi_{s-1}\phi_{t-1})z}
{\phi_{s+1}\phi_{t+1} + \phi_s\phi_t + (\phi_{s+1}\phi_t+\phi_s\phi_{t-1})z}
\end{align}
So being an iteration group in terms of Fibonacci extension means the following peculiar identity:
\[ \phi_s\phi_{t+1} + \phi_{s-1}\phi_t = \phi_{s+t} \]
which reduces to the Fibonacci identity for t=1! So lets call this the extended Fibonacci identity.

So lets check this at least for the standard Fibonacci extension \(\phi_t = \frac{\Phi^t - \Psi^t}{\Phi-\Psi}\)
\begin{align}
\phi_s\phi_{t+1} + \phi_{s-1}\phi_t &= \frac{\Phi^s-\Psi^s}{\Phi-\Psi}\frac{\Phi^{t+1}-\Psi^{t+1}}{\Phi-\Psi}
+\frac{\Phi^{s-1}-\Psi^{s-1}}{\Phi-\Psi}\frac{\Phi^t-\Psi^t}{\Phi-\Psi}\\
(\phi_s\phi_{t+1} + \phi_{s-1}\phi_t)(\Phi-\Psi)^2 &=
\Phi^s\Phi^{t+1}-\Phi^s\Psi^{t+1}-\Psi^s\Phi^{t+1} + \Psi^s\Psi^{t+1} + \Phi^{s-1}\Phi^t - \Phi^{s-1}\Psi^t - \Psi^{s-1}\Phi^t + \Psi^{s-1}\Psi^s\\
&=\Phi^{s+t+1}-\Phi^s\Psi^t\Psi-\Psi^s\Phi^t\Phi + \Psi^{s+t+1}+ \Phi^{s+t-1} - \frac{1}{\Phi}\Phi^s\Psi^t - \frac{1}{\Psi}\Psi^s\Phi^t + \Psi^{s+t-1}\\
\frac{1}{\Phi}&=-\Psi\\
&=\Phi^{s+t+1} + \Psi^{s+t+1}+ \Phi^{s+t-1}  + \Psi^{s+t-1}\\
&=\Phi\Phi^{s+t} + \Psi\Psi^{s+t} - \Psi\Phi^{s+t} - \Phi\Psi^{s+t}\\
&=(\Phi-\Psi)\Phi^{s+t} - (\Phi-\Psi)\Psi^{s+t}\\
&=(\Phi-\Psi)^2\phi_{s+t}
\end{align}

So it is indeed true! The standard extension corresponds to an iteration group, and it is the regular one, because it is analytic at the fixed points.

But - and now, Tommy, your aversion against the real-valued Fibonacci extensions becomes justified - this identity is not true for the real-valued Fibonacci extension \(\phi'_t=\frac{\Phi^t+\cos(\pi t)(-\Psi)^t}{\Phi-\Psi}\). Though it even satisfies what Leo.W. calls the Abelian property!: \(f^{\circ s}\circ f^{\circ t} = f^{\circ t}\circ f^{\circ s}\) or in terms of Fibonacci
\[ \phi'_s\phi'_{t+1} + \phi'_{s-1}\phi'_t = \phi'_t\phi'_{s+1} + \phi'_{t-1}\phi'_s \]
But for example s=0.5 and t=0.5 do not sum up to 1 for the real-valued extension.

So who accepts the challenge and finds a real valued solution to the extended Fibonacci identity?!
I think one can not find a single valued solution because this would mean to have a single valued real iteration close to a fixed point of negative multiplier.
See also this thread.

I will continue here with finding a real (and multi) valued iteration group.
But time is already too limited, so one later time.
Sorry for not answering your replies, has also to be postponed.

yes !

that is what i was aiming at in part.

this combines many ideas from the regular posters here.

the fibonacci addition formula was explictly recently mentioned by me in one of the recent related threads.

it is way more general than the fibonacci standard recursion.

in fact it is almost analogue to

solving

f(2x) = f(x)^2

or

f(x+1) = e f(x)

where both have f(x) = exp(x) as a solution.

notice the fibo addition formula has less freedom in its solutions.

especially no " if f is a solution so is constant times f ".

if we add ( to the addition formula ) the condition of boundedness on a region we have uniqueness of the fibo.
this relates to the analogue of the uniqueness of the gamma function , and was mentioned and proved ( but still not accepted or ignored ) by the TPID I PROVED  about exp functions and their uniqueness.
this relates to james ideas of fractional integrals , integral transforms and ramanujan. or kouznetsov.

basically all no surprise : fibo is just in essense iterating 2 exponentials ; the eigenvalues.

as for the real part with sine or cosine , you see that it does not work out nice ... but my other argument was that it fails on the complex plane.

and that is probably true for all those "sine replacements of (-c)^t".
You started a topic about those sine replacements working for iterating - x.
Nice pictures and it works for the reals ... but probably fails for the complex.

and the reason is they are just taking the real part of the actual complex solution.

taking real part is not analytic in a sense ( even if the sine or cosine is )

So im still very skeptical

and the idea is that the fibo will just have the same uniqueness conditions as the exponentials do ...

because they ARE two exponentials with base the eigenvalue.

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more interesting is how tribonacci might be related to analogues of the linear fraction iterations ??

if fibo is essentially iterations of LF then what is tribonacci essentially an iteration of ????????


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fractional derivates remains a road of possible interest though.

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regards

tommy1729