As Bo has started to talk a lot about linear fractional iterations, I thought I'd throw in my hat on some results I know about Linear fractional relations, and how they compare to what Bo is talking about.

I'd like to put a little disclaimer though, that these iterations are wildly different beasts than Tetration--and wildly different than Bo's regular iterations. But with that out of the way, we can get started.

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

INFINITE COMPOSITIONS

Infinite compositions behave a lot differently for automorphisms of \(\widehat{\mathbb{C}}\). We can remember that an automorphism of \(\widehat{\mathbb{C}}\) looks like:

$$

\mu(z) = \frac{az + b}{cz + d}\\

$$

Which, as Bo has pointed out, looks like a matrix applied to the \((x,y)\) coordinate--\(z=x+iy\):

$$

\begin{pmatrix}

a & c\\

b & d

\end{pmatrix}

$$

So long as the determinant \(\det(\mu) \neq 0\)--i.e \(ad - bc \neq 0\)--this is an automorphism of \(\mathbb{R}^2\). Now, this tends to be diagonalizable/ then iterated through its eigenvalues from the diagonalization; I'm not going to do that. That's a horse that's been beaten one too many times. Let's instead, write:

$$

\mu_n(z) = \frac{a_nz + b_n}{c_nz + d_n}\\

$$

Such that each matrix satisfies \(\det(\mu_n) \neq 0\). We are going to look at the function:

$$

U_N(z) = \Omega_{n=1}^N \mu_n(z)\bullet z = \mu_1(\mu_2(...\mu_N(z)))\\

$$

And ask where it converges as \(N\to\infty\). By which, if I proclaim that:

$$

\sum_{n=1}^\infty \det(\mu_n - I) < \infty\\

$$

For the identity matrix \(I\). Then the above function converges compactly on \(\widehat{\mathbb{C}}\). This follows from a proof that's 4 years gone by now on how infinite compositions converge. I had originally proved this for entire functions; and it becomes much more placated for linear fractional transformations.

The more natural way I would write this, is if for \(K \subset \widehat{\mathbb{C}}\) is compact:

$$

\sum_{n=1}^\infty \sup_{z \in K}|\mu_n(z) - z| < \infty\\

$$

Then \(\lim_{N\to\infty} U_N(z) = U(z)\) everywhere on \(\widehat{\mathbb{C}}\).

When we talk about the Beta method, we are not talking about something that looks like this. We are talking about a very different beast. We are asking instead, that \(det(\mu_n) \to 0 \) as \(n \to \infty\)

This introduces the formula:

$$

\sum_{n=1}^\infty \sup_{z\in K} |\mu_n(z)| < \infty

$$

The function \(U\) in this scenario always converges to a CONSTANT in z. So it is holomorphic in \(z\), but it's constant. The beta method then asks, that instead we write:

$$

\mu_n(z) = \mu(-n,z)\\

$$

Where \(\mu(s,z)\) is holomorphic in \(s\) but maps \(\widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\). Then, if, on compact sets in \(s\), along with compact sets in \(z\) we have:

$$

\sum_{n=1}^\infty \sup_{s\in S,z\in K} |\mu(s-n,z)| < \infty

$$

Then we've created a beta function which looks like:

$$

U(s) = \Omega_{n=1}^\infty \mu(s-n,z)\bullet z\\

$$

Which satisfies:

$$

U(s+1) = \mu(s,U(s))\\

$$

Since we've allowed for poles, and poles are natural, this function \(U(s)\) maps \(\widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\), so long as the summation in \(s\) converges for the extended plane. Which isn't as scary as it seems.

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

Reinventing the beta function

Let's take Bo's iteration of:

$$

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

$$

Which he described using Fibonacci's sequence. Let's play a bit looser though. Let's write, for \(\Re(\lambda) > 0\):

$$

\mu(s,z) = \frac{1}{(1+e^{-\lambda s})(1+z)}\\

$$

Then this function satisfies:

$$

\sum_{n=1}^\infty \sup_{s\in S,z\in K} \left|\frac{1}{(1+e^{\lambda(j- s)})(1+z)}\right| < \infty\\

$$

This is obvious away from \(z \approx -1\), we have to do a change of variables near \(-1\). So, additionally, where there are poles, we can change a different coordinate space and the sum converges. The benefit of working with a perfect Riemann surface \(\widehat{\mathbb{C}}\) .

But either way, this is no different than:

$$

\sum_{n=1}^\infty |\det(\mu(s-n))| < \infty\\

$$

So when we take the infinite composition we get:

$$

U(s) = \Omega_{n=1}^\infty \mu(s-n,z)\,\bullet z = \mu(s-1,\mu(s-2,...\mu(s-n,...)))\\

$$

This function satisfies:

$$

\frac{1}{1+U(s)} \cdot \frac{1}{1+e^{-\lambda s}} = U(s+1)\\

$$

And you wouldn't be that hard pressed to prove this is a meromorphic function in \(\widehat{\mathbb{C}}\).

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

So what does this have to do with bo?

Since \(\mu(+\infty,z) = \mu(z) = \frac{1}{1+z}\), then this is a bijection of \(\widehat{\mathbb{C}}\). Then \(\mu^{-1}(z) = h(z) = \frac{1-z}{z}\). So the beta method for all of bo's fancy iterations is precisely:

$$

\lim_{n\to\infty} h^{\circ n}(U(s+n)) = F_\lambda(s)\\

$$

But in order for this to converge, we write:

$$

F_\lambda(s) = U(s) + \tau(s)\\

$$

Where we're taking advantage of the asymptotic formula:

$$

h(U(s+1)) = U(s) + O(e^{-\lambda s})\\

$$

That these objects are asymptotically equivalent (in a fairly strong manner) to the iterations you are writing.

Note, that this iteration is periodic with period \(2\pi i/\lambda\). It will by its nature have a measure zero "no-go" zone, but it will definitely converge everywhere else.

I am mostly just making this post to describe the \(\beta\) method, but using linear fractional transformations. I'm doing this, I hope for everyone's benefit. And especially, because for linear fractional transformations, the beta method is much more straight forward.

So, as Bo is drawing a parallel using LFTs, I am drawing a parallel using LFTs. And I am doing so specific to what makes the beta method interesting. Additionally, I'm hoping to explain this construction more simply to Bo, who I don't think has grasped the beta method and how it works.

How am I doing, bo

EDIT! Also, much of my work on stuff like this is not published in a peer reviewed manner. But I'm fairly well known at U of T for a lot of good work, and much of my work has been peer reviewed by professors at U of T. So it's not all coming out of my ass like your regular poster. A bit is coming outta my ass; but I've been "peer reviewed" on a lot of my work at u of t.

I'd like to put a little disclaimer though, that these iterations are wildly different beasts than Tetration--and wildly different than Bo's regular iterations. But with that out of the way, we can get started.

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

INFINITE COMPOSITIONS

Infinite compositions behave a lot differently for automorphisms of \(\widehat{\mathbb{C}}\). We can remember that an automorphism of \(\widehat{\mathbb{C}}\) looks like:

$$

\mu(z) = \frac{az + b}{cz + d}\\

$$

Which, as Bo has pointed out, looks like a matrix applied to the \((x,y)\) coordinate--\(z=x+iy\):

$$

\begin{pmatrix}

a & c\\

b & d

\end{pmatrix}

$$

So long as the determinant \(\det(\mu) \neq 0\)--i.e \(ad - bc \neq 0\)--this is an automorphism of \(\mathbb{R}^2\). Now, this tends to be diagonalizable/ then iterated through its eigenvalues from the diagonalization; I'm not going to do that. That's a horse that's been beaten one too many times. Let's instead, write:

$$

\mu_n(z) = \frac{a_nz + b_n}{c_nz + d_n}\\

$$

Such that each matrix satisfies \(\det(\mu_n) \neq 0\). We are going to look at the function:

$$

U_N(z) = \Omega_{n=1}^N \mu_n(z)\bullet z = \mu_1(\mu_2(...\mu_N(z)))\\

$$

And ask where it converges as \(N\to\infty\). By which, if I proclaim that:

$$

\sum_{n=1}^\infty \det(\mu_n - I) < \infty\\

$$

For the identity matrix \(I\). Then the above function converges compactly on \(\widehat{\mathbb{C}}\). This follows from a proof that's 4 years gone by now on how infinite compositions converge. I had originally proved this for entire functions; and it becomes much more placated for linear fractional transformations.

The more natural way I would write this, is if for \(K \subset \widehat{\mathbb{C}}\) is compact:

$$

\sum_{n=1}^\infty \sup_{z \in K}|\mu_n(z) - z| < \infty\\

$$

Then \(\lim_{N\to\infty} U_N(z) = U(z)\) everywhere on \(\widehat{\mathbb{C}}\).

When we talk about the Beta method, we are not talking about something that looks like this. We are talking about a very different beast. We are asking instead, that \(det(\mu_n) \to 0 \) as \(n \to \infty\)

This introduces the formula:

$$

\sum_{n=1}^\infty \sup_{z\in K} |\mu_n(z)| < \infty

$$

The function \(U\) in this scenario always converges to a CONSTANT in z. So it is holomorphic in \(z\), but it's constant. The beta method then asks, that instead we write:

$$

\mu_n(z) = \mu(-n,z)\\

$$

Where \(\mu(s,z)\) is holomorphic in \(s\) but maps \(\widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\). Then, if, on compact sets in \(s\), along with compact sets in \(z\) we have:

$$

\sum_{n=1}^\infty \sup_{s\in S,z\in K} |\mu(s-n,z)| < \infty

$$

Then we've created a beta function which looks like:

$$

U(s) = \Omega_{n=1}^\infty \mu(s-n,z)\bullet z\\

$$

Which satisfies:

$$

U(s+1) = \mu(s,U(s))\\

$$

Since we've allowed for poles, and poles are natural, this function \(U(s)\) maps \(\widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\), so long as the summation in \(s\) converges for the extended plane. Which isn't as scary as it seems.

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

Reinventing the beta function

Let's take Bo's iteration of:

$$

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

$$

Which he described using Fibonacci's sequence. Let's play a bit looser though. Let's write, for \(\Re(\lambda) > 0\):

$$

\mu(s,z) = \frac{1}{(1+e^{-\lambda s})(1+z)}\\

$$

Then this function satisfies:

$$

\sum_{n=1}^\infty \sup_{s\in S,z\in K} \left|\frac{1}{(1+e^{\lambda(j- s)})(1+z)}\right| < \infty\\

$$

This is obvious away from \(z \approx -1\), we have to do a change of variables near \(-1\). So, additionally, where there are poles, we can change a different coordinate space and the sum converges. The benefit of working with a perfect Riemann surface \(\widehat{\mathbb{C}}\) .

But either way, this is no different than:

$$

\sum_{n=1}^\infty |\det(\mu(s-n))| < \infty\\

$$

So when we take the infinite composition we get:

$$

U(s) = \Omega_{n=1}^\infty \mu(s-n,z)\,\bullet z = \mu(s-1,\mu(s-2,...\mu(s-n,...)))\\

$$

This function satisfies:

$$

\frac{1}{1+U(s)} \cdot \frac{1}{1+e^{-\lambda s}} = U(s+1)\\

$$

And you wouldn't be that hard pressed to prove this is a meromorphic function in \(\widehat{\mathbb{C}}\).

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

So what does this have to do with bo?

Since \(\mu(+\infty,z) = \mu(z) = \frac{1}{1+z}\), then this is a bijection of \(\widehat{\mathbb{C}}\). Then \(\mu^{-1}(z) = h(z) = \frac{1-z}{z}\). So the beta method for all of bo's fancy iterations is precisely:

$$

\lim_{n\to\infty} h^{\circ n}(U(s+n)) = F_\lambda(s)\\

$$

But in order for this to converge, we write:

$$

F_\lambda(s) = U(s) + \tau(s)\\

$$

Where we're taking advantage of the asymptotic formula:

$$

h(U(s+1)) = U(s) + O(e^{-\lambda s})\\

$$

That these objects are asymptotically equivalent (in a fairly strong manner) to the iterations you are writing.

Note, that this iteration is periodic with period \(2\pi i/\lambda\). It will by its nature have a measure zero "no-go" zone, but it will definitely converge everywhere else.

I am mostly just making this post to describe the \(\beta\) method, but using linear fractional transformations. I'm doing this, I hope for everyone's benefit. And especially, because for linear fractional transformations, the beta method is much more straight forward.

So, as Bo is drawing a parallel using LFTs, I am drawing a parallel using LFTs. And I am doing so specific to what makes the beta method interesting. Additionally, I'm hoping to explain this construction more simply to Bo, who I don't think has grasped the beta method and how it works.

How am I doing, bo

EDIT! Also, much of my work on stuff like this is not published in a peer reviewed manner. But I'm fairly well known at U of T for a lot of good work, and much of my work has been peer reviewed by professors at U of T. So it's not all coming out of my ass like your regular poster. A bit is coming outta my ass; but I've been "peer reviewed" on a lot of my work at u of t.