Continuing from the thread Bo posted about mapping the fibonacci iteration to the iteration of an LFT: https://math.eretrandre.org/tetrationfor...p?tid=1607

Which, ultimately, is an isomorphism. Represents a solution to:

$$

\phi(t) = \phi(t-1) + \phi(t-2)\\

$$

And its relations to iterates of

$$

f(z) = \frac{1}{1+z}\\

$$

We are going to try and relate these objects in a more general sense. This means, we are going to consider:

$$

f^{\circ t}(z) = \frac{\phi(t) +\phi(t-1)z}{\phi(t+1) + \phi(t)z}\\

$$

And we are going to ask, when we can produce an iteration \(z \in \mathbb{R}\) and \(t \in \mathbb{R}\) that is real valued (disregarding if there are poles).

We are first of all, going to assume that \(\phi(t)\) is entire in \(t\). We don't care about not entire solutions in \(t\). And additionally, we will always assume that:

$$

\phi(0) = 0\,\,\,\phi(1) = 1\,\,\,\phi(t) = \phi(t-1) + \phi(t-2)\\

$$

-------------------------------------------------------------------------

To begin, we can write every entire solution to this equation. (This was asked by myself on MathOverflow a long time ago, about the uniqueness of entire linear recurrence relations. I can find the linked article and the reasoning--but until asked I'm too lazy to find it.) Ultimately this isn't that hard to prove for Fibonacci, but it is an advanced result for general linear recurrences.

Firstly we take \(\Phi\) and \(\Psi\) as Bo described, which are the golden ratio, and its conjugate pair for its minimal polynomial. Then we write:

$$

\mu^+_k = \log(\Phi) + 2\pi i k\,\,\,\mu^-_j = \log\Psi + 2 \pi i j\,\,\,\,\text{for}\,\,j,k \in \mathbb{Z}\\

$$

Then there exists the fundamental solutions of the fibonacci equation, for entire functions, which is written as:

$$

\phi_{kj}(t) = \frac{e^{\mu^+_k t} + e^{\mu^-_j t}}{\Phi - \Psi}\\

$$

This is clearly not real valued, which you can always check. But, each version is a more chaotic version of Bo's iteration of \(\frac{1}{1+z}\), where:

$$

f^{\circ t}(z) = \frac{\phi_{kj}(t) + \phi_{kj}(t-1)z}{\phi_{kj}(t+1) - \phi_{kj}(t)z}\\

$$

So, this creates a selection of \(\mathbb{Z}^2\) iterates, which are never real valued. Now we superimpose these solutions on each other. So let's let \(a_{jk} \in \mathbb{C}\) and additionally, is summable across both indices. Then:

$$

\phi(t) = \sum_{j=-\infty}^\infty \sum_{k=-\infty}^\infty a_{jk} \phi_{kj}(t)\\

$$

While \(\phi(0) = 0\) and \(\phi(1) = \sum_{j=-\infty}^\infty \sum_{k=-\infty}^\infty a_{jk} = 1\).

This describes every possible entire solution to the fibonacci equation. And this translates to not every iteration of \(f = \frac{1}{1+z}\), but for sake of brevity, the nicest ones; where the singularities, are in some sense, local.

$$

f^{\circ t}(z) = \frac{\phi(t) + \phi(t-1)z}{\phi(t+1) - \phi(t)z}\\

$$

DISCLAIMER: What I had previously written as \(\theta(z)\phi(z)\), I mean by the above solution. What I had written before is a valid method, but it proves to be extraneous and non necessary; and additionally would produce Bo's iteration by fact:

$$

\frac{\theta(t)\phi(t) + \theta(t-1)\phi(t-1)z}{\theta(t+1)\phi(t+1) - \theta(t)\phi(t)z} = \frac{\theta(t)\phi(t) + \theta(t)\phi(t-1)z}{\theta(t)\phi(t+1) - \theta(t)\phi(t)z} = f^{\circ t}(z)\\

$$

This means for every fibonacci iteration, there is an iteration of \(\frac{1}{1+z}\), but for every iteration of \(\frac{1}{1+z}\) there are many fibonacci iterations. So this isn't a 1-1 correspondence that Bo's described, it's many to one.

--------------------------------------------------------------------------------

And largely why I think Bo's solution is unique by construction. Every entire function \(\phi(t)\) which satisfies the fibonacci equation; must satisfy a strict representation. Let's call Bo's solution, which is the standard from Binet's formula. That:

$$

\phi_{00}(t) = \frac{\Phi^t - \Psi^t}{\Phi - \Psi}\\

$$

Then our solution \(\phi(t)\) looks like:

$$

\phi(t) = \sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^\infty a_{jk} \frac{e^{2 \pi i k t} \Phi^t + e^{2 \pi i j t}\Psi^t}{\Phi - \Psi}\\

$$

Which equates to:

$$

\phi(t) = \frac{\theta^+(t) \Phi^t + \theta^-(t) \Psi^t}{\Phi - \Psi}\\

$$

And now we ask, where this thing is real valued. In order for that to happen, we can reduce to:

$$

\overline{\phi(t)} = \phi(\overline{t})\\

$$

Which devolves into solving n'th order polynomial solutions for \(t \approx 0\); and solving the Matrix equation for \(n\times n\).

By which we are just solving a second order equation in \(\theta^{\pm}\). Anyone more proefficient in matrix coefficients is welcome to jump in here.

This is a strong statement in more scenarios, but for us it relates a standard linear recurrence relationship, to the complex iteration of a function.

-------------------------------------------------------------

We should always be able to solve for \(\theta^+\) and \(\theta^-\), and by consequence solve for \(a_{jk}\). Uniqueness is another problem. But this would produce the real valued version of Bo's fibonacci iteration of \(1/1+z\). By which, we now have:

$$

\begin{align}

f^{\circ t}(z) &: \mathbb{C} \times \widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\\

f^{\circ t}(z) &: \mathbb{R} \times \mathbb{R} \to \mathbb{R}\cup\{\infty\}\\

\end{align}

$$

I'm hoping someone else can run the numbers. But I'd call this the "crescent iteration of \(f= 1/1+z\)". This is related, in more ways than one, to the crescent iteration as we think of it. When you take n'th order solutions to Kneser's Taylor series at \(s=0\), we end up doing something very similar to finding an n'th order expansion of \(\phi(t)\) at \(t=0\). These are inextricably linked.

Which, ultimately, is an isomorphism. Represents a solution to:

$$

\phi(t) = \phi(t-1) + \phi(t-2)\\

$$

And its relations to iterates of

$$

f(z) = \frac{1}{1+z}\\

$$

We are going to try and relate these objects in a more general sense. This means, we are going to consider:

$$

f^{\circ t}(z) = \frac{\phi(t) +\phi(t-1)z}{\phi(t+1) + \phi(t)z}\\

$$

And we are going to ask, when we can produce an iteration \(z \in \mathbb{R}\) and \(t \in \mathbb{R}\) that is real valued (disregarding if there are poles).

We are first of all, going to assume that \(\phi(t)\) is entire in \(t\). We don't care about not entire solutions in \(t\). And additionally, we will always assume that:

$$

\phi(0) = 0\,\,\,\phi(1) = 1\,\,\,\phi(t) = \phi(t-1) + \phi(t-2)\\

$$

-------------------------------------------------------------------------

To begin, we can write every entire solution to this equation. (This was asked by myself on MathOverflow a long time ago, about the uniqueness of entire linear recurrence relations. I can find the linked article and the reasoning--but until asked I'm too lazy to find it.) Ultimately this isn't that hard to prove for Fibonacci, but it is an advanced result for general linear recurrences.

Firstly we take \(\Phi\) and \(\Psi\) as Bo described, which are the golden ratio, and its conjugate pair for its minimal polynomial. Then we write:

$$

\mu^+_k = \log(\Phi) + 2\pi i k\,\,\,\mu^-_j = \log\Psi + 2 \pi i j\,\,\,\,\text{for}\,\,j,k \in \mathbb{Z}\\

$$

Then there exists the fundamental solutions of the fibonacci equation, for entire functions, which is written as:

$$

\phi_{kj}(t) = \frac{e^{\mu^+_k t} + e^{\mu^-_j t}}{\Phi - \Psi}\\

$$

This is clearly not real valued, which you can always check. But, each version is a more chaotic version of Bo's iteration of \(\frac{1}{1+z}\), where:

$$

f^{\circ t}(z) = \frac{\phi_{kj}(t) + \phi_{kj}(t-1)z}{\phi_{kj}(t+1) - \phi_{kj}(t)z}\\

$$

So, this creates a selection of \(\mathbb{Z}^2\) iterates, which are never real valued. Now we superimpose these solutions on each other. So let's let \(a_{jk} \in \mathbb{C}\) and additionally, is summable across both indices. Then:

$$

\phi(t) = \sum_{j=-\infty}^\infty \sum_{k=-\infty}^\infty a_{jk} \phi_{kj}(t)\\

$$

While \(\phi(0) = 0\) and \(\phi(1) = \sum_{j=-\infty}^\infty \sum_{k=-\infty}^\infty a_{jk} = 1\).

This describes every possible entire solution to the fibonacci equation. And this translates to not every iteration of \(f = \frac{1}{1+z}\), but for sake of brevity, the nicest ones; where the singularities, are in some sense, local.

$$

f^{\circ t}(z) = \frac{\phi(t) + \phi(t-1)z}{\phi(t+1) - \phi(t)z}\\

$$

DISCLAIMER: What I had previously written as \(\theta(z)\phi(z)\), I mean by the above solution. What I had written before is a valid method, but it proves to be extraneous and non necessary; and additionally would produce Bo's iteration by fact:

$$

\frac{\theta(t)\phi(t) + \theta(t-1)\phi(t-1)z}{\theta(t+1)\phi(t+1) - \theta(t)\phi(t)z} = \frac{\theta(t)\phi(t) + \theta(t)\phi(t-1)z}{\theta(t)\phi(t+1) - \theta(t)\phi(t)z} = f^{\circ t}(z)\\

$$

This means for every fibonacci iteration, there is an iteration of \(\frac{1}{1+z}\), but for every iteration of \(\frac{1}{1+z}\) there are many fibonacci iterations. So this isn't a 1-1 correspondence that Bo's described, it's many to one.

--------------------------------------------------------------------------------

And largely why I think Bo's solution is unique by construction. Every entire function \(\phi(t)\) which satisfies the fibonacci equation; must satisfy a strict representation. Let's call Bo's solution, which is the standard from Binet's formula. That:

$$

\phi_{00}(t) = \frac{\Phi^t - \Psi^t}{\Phi - \Psi}\\

$$

Then our solution \(\phi(t)\) looks like:

$$

\phi(t) = \sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^\infty a_{jk} \frac{e^{2 \pi i k t} \Phi^t + e^{2 \pi i j t}\Psi^t}{\Phi - \Psi}\\

$$

Which equates to:

$$

\phi(t) = \frac{\theta^+(t) \Phi^t + \theta^-(t) \Psi^t}{\Phi - \Psi}\\

$$

And now we ask, where this thing is real valued. In order for that to happen, we can reduce to:

$$

\overline{\phi(t)} = \phi(\overline{t})\\

$$

Which devolves into solving n'th order polynomial solutions for \(t \approx 0\); and solving the Matrix equation for \(n\times n\).

By which we are just solving a second order equation in \(\theta^{\pm}\). Anyone more proefficient in matrix coefficients is welcome to jump in here.

This is a strong statement in more scenarios, but for us it relates a standard linear recurrence relationship, to the complex iteration of a function.

-------------------------------------------------------------

We should always be able to solve for \(\theta^+\) and \(\theta^-\), and by consequence solve for \(a_{jk}\). Uniqueness is another problem. But this would produce the real valued version of Bo's fibonacci iteration of \(1/1+z\). By which, we now have:

$$

\begin{align}

f^{\circ t}(z) &: \mathbb{C} \times \widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\\

f^{\circ t}(z) &: \mathbb{R} \times \mathbb{R} \to \mathbb{R}\cup\{\infty\}\\

\end{align}

$$

I'm hoping someone else can run the numbers. But I'd call this the "crescent iteration of \(f= 1/1+z\)". This is related, in more ways than one, to the crescent iteration as we think of it. When you take n'th order solutions to Kneser's Taylor series at \(s=0\), we end up doing something very similar to finding an n'th order expansion of \(\phi(t)\) at \(t=0\). These are inextricably linked.