Fibonacci as iteration of fractional linear function

[bo198214]

So basically an old idea you are reviving.

Not really. "reinventing" would be a better word, because it was completely new to me at the time of the writing.
But of course Gottfried - the old veteran - was dealing with it already 10 years before me

[JmsNxn]

OKAY! Absolutely beautiful! Yes this should be a LFT! I think this implicitly solves \(\theta(t)\), I'm just not sure how yet. Definitely agree this will get it though!

I think this is very important, because it shows how the iteration of an LFT is Fibonacci, and shows how a restriction to reals of an LFT creates a real Fibonacci.

It is not a Fibonacci! As Gottfried pointed out already it only agrees on the even Fibonacci indexes.
Hence it is not a theta transforming the non-realness (of the original Fibonacci interpolation) into a realness - which honestly I can not imagine that it exists, because of the pole in the middle between the fixed points - but one never knows. That's why Gottfried was excluding these poles in the middle (which only occurs for uneven iterates) and banning it to the left side of the left fixed point (where it occurs for even iterates, or to the right side of the right fixed point when we consider negative even iterates) by demanding that it only agrees with Fibonacci on even iterates.
Even you pointed out to me already once that the iterative root of the iterative square is probably not the original function anymore.
But the trick with changing the sign of the eigenvalue is quite naughty, Gottfried!
But yes, it boils down to taking the regular iteration of \(f^{\circ 2}\) and just taking 1/2 of it \((f^{\circ 2})^{\mathfrak{R} t/2}\).

OHHH OKAY!

I was thinking that primarily! That this would be something like that. Especially when he said he was squaring the eigenvalue. I thought it was going to be one of those \(f^{\circ 2 \circ 1/2} \neq f\) problems, but I guess I got blinded like a moth to the flame. Thinking there was the solution. lol.

Okay, so real valued fibonacci equations are still on the table! I wonder if the poles you are seeing in your iterate, translate directly to "no real valued fibonacci" though. Maybe the poles are just artifacts of the translation; not the actual fibonacci sequence.

I really believe there must be a "crescent iteration" fibonacci. It's probably not injective on \(\mathbb{R}^+\), but there's gotta be one. It just seems to simple. We're just looking for a \(\theta(z)\phi(z)\) that is real valued.

We are just solving an equation:

\[
\sum_{k=-\infty}^\infty a_ke^{2\pi i k z} \phi(z) : \mathbb{R}^+ \to \mathbb{R}^+\\
\]

Come on, that's gotta be doable right!? Where additionally \(\phi(z)\) is just a super position of exponentials. Come on...

[bo198214]

I thought it was going to be one of those \(f^{\circ 2 \circ 1/2} \neq f\) problems, but I guess I got blinded like a moth to the flame. Thinking there was the solution. lol.

You thought absolutely right. And this \(f^{\circ 2\circ 1/2}\) is the function that only coincides with every second fibonacci number.

I really believe there must be a "crescent iteration" fibonacci. It's probably not injective on \(\mathbb{R}^+\), but there's gotta be one. It just seems to simple. We're just looking for a \(\theta(z)\phi(z)\) that is real valued.

You (and Sheldon) are the ones with the numerical equipment of tinkering with the theta function, isn't it?

Pole or not but at the right fixed point the function looks very similar to the eta minor case - I mean in the sense of negative multiplier. And there you could derive a real super function, so why not here ...

[JmsNxn]

You (and Sheldon) are the ones with the numerical equipment of tinkering with the theta function, isn't it?

Don't put me on Sheldon's level, I am nowhere near Sheldon's level, lol. Especially when it comes to the crescent iteration. I am just indebted to his language when describing the beta method. I think it should be possible though, perhaps by beta iterating and letting the period be infinite.

Any function:

\[
\theta(z)\phi(z)\\
\]

Wouldn't be periodic, so it would be similar to "stretching the period" of the beta iteration I described for this LFT problem. This might be a good place to check if we can do "a crescent kind of iteration" with fibonacci, but using the "beta method". I will have to look into this. This seems really interesting. I mean, I always figured it possible, never thought I'd be able to hammer it down. And additionally was never that interested. This iterated function comparison has got me far more interested though.

What if this follows for more advanced recursions; what if iterations of functions \(p : \widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\), produces a correspondence to more general difference equations than Fibonacci. (Where I come from we make a change of variables to Fibonacci so that all linear difference equations are based on the operator \(\Delta = a_n - a_{n-1}\)). I know you explained a similar phenomenon for first order difference equations. But what if we increase the degree of \(p\) as a rational function--would it have a similar kind of correspondence? Wow, sounds super interesting. If anything, bo. You've given me A LOT to think about.

But technically, I have created a theta mapping which changes the period for this fibonacci iteration. It should be real valued. But, it'll probably have singularities on the real line, or about the real line. The trouble would be moving those singularities away from the real line, excepting a glaringly obvious pole somewhere on the real line (to account for the \(z=-1\) problem in the initial function). I think these beta solutions are more at home to mappings of \(\widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\) ). So it may be real valued but less than desired. Again, I'm asking for an ENTIRE \(\theta(z)\), which makes this much more difficult. Only a crescent kind of iteration, which could or could not be constructed by "stretching the period" of the iteration.

This sounds like a good toy model though. I really like your use of LFTs, bo. This has given me something more "hold in your hands" to experiment with. I don't mean toy model offensively either, I hope you know that

[bo198214]

Just to see what we are talking about with this fib2 variant:
\[\text{fib}_2(t) = \frac{\Phi^t - \left|\Psi\right|^t}{\Phi-\Psi}\]
   
The black dots are the real Fibonacci numbers.

Maybe it is also an option to somehow theta-distort the fib2 to fit fib. But then again, if we have one solution then there are too many more.

[bo198214]

OMG, its as easy as this:
\[\text{fib}_\text{alt}(t) = \frac{\Phi^t - \cos(\pi t)\left|\Psi\right|^t}{\Phi-\Psi}\]

   

[tommy1729]

OMG, its as easy as this:
\[\text{fib}_\text{alt}(t) = \frac{\Phi^t - \cos(\pi t)\left|\Psi\right|^t}{\Phi-\Psi}\]

To be honest im getting a bit irritated.

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

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.

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.

4) the real issue is that one of the eigenvalues has a negative value.
that is analogue to solving for real functions that are half-iterates of exp(- z).

5) without realizing 1 till 4 and adressing it or looking at or defining differently the fibonacci sequence we will not get anywhere.

6) if we take the matrix representation for other similar recursions that have only positive eigenvalues the solution is simple. Then again there was no issue , problem or question solved in that case.

7) there are some nice definitions or recursions for -say- the even fibonacci numbers. 
things like f(x+1) = f(x) + f(x) + f(x-1) + f(x-2)  + f(x-3) + f(x-4) + ...
( continuum sum recursion ideas flashbacks * intensifies * )
nice ... and not nice ; we get positive eigenvalues then and the case returns to trivial.

8. if you want something a bit new :
you might want to consider iterations of a^z + b^z, that seems closest to fibonnacci and analytic continu iterations.
or you might wanna find recursions for such exp sums.  Well my suggestion ...

basically i felt the whole debate was running or circles or stating the obvious and well known.



regards

tommy1729

[bo198214]

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 ...

[bo198214]

As announced here the motion picture of the LFT corresponding to the alternative Fibonacci extension:
   
The values at 0 are \(\frac{\phi_t}{\phi_{t+1}}\) and going \(\to \frac{1}{\Phi}\approx\) 0.618 for \(t\to \infty\).
One can very well see that it is not a regular iteration at the left fixed point \(z_1\), because there is no interval of length \(>1\) such that \(f^{\circ t}\) is analytic at \(z_1\) for all t in this interval.
However I wonder why it seems to be regular at the right fixed point ...

[bo198214]

However I wonder why it seems to be regular at the right fixed point ...

It's because it is not an iteration at all. It satisfies \(f^{\circ t+1} = \frac{1}{1+f^{\circ t}} \) (which corresponds to the Fibonacci identity) but it does not satisfy \(f^{\circ s+t}=f^{\circ s}\circ f^{\circ t}\), while the original Fibonacci extension is a true (and regular) iteration.
So the question still stands to find a real valued super function \(\frac{1}{1+z}\).