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.
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.