Sometimes, Mphlee, it helps to work with an example. So let's do exactly that.
Let \(\xi\) be an irrational number, \(\xi \in \mathbb{R}/\mathbb{Q}\)--let's additionally ask that \(0 < \xi < 1\).
Then, to approximate \(\xi\), we take a sequence of rational numbers: \(\xi_m = \frac{p_m}{q_m}\)-- where \((p_m,q_m) = 1\), and additionally \(p_m < q_m\). Now, let's move forward, and look at:
\[
\mathbb{H}^* \to \mathbb{H}\\
\]
But we will restrict: \(0 < \Re(\theta) < 2 \pi\). Let us write:
\[
\theta_m = 2 \pi \frac{p_m}{q_m} + it\\
\]
Where we're given the identity \(\theta_m \to 2 \pi \xi + it\) as \(m \to \infty\)...
It is not entirely obvious how to show that \(f_{\theta_m}^{\circ s}(z) \to f_\theta^{\circ s}\) under the integral... There are solutions, but it's not obvious: which solution the integral chooses, and what manner it converges. In order to do this, we need to understand the maximal growth of \(q_m\). Now, there are restricted forms of irrational numbers which we write as:
\[
|\xi - \frac{p_m}{q_m}| < \frac{M}{q_m^{2+\epsilon}}\\
\]
This is a result that we are constantly finding better and better bounds on. But, we know that a choice \(\epsilon \approx 1/4\) should work fine for all \(0 < \xi < 1\). This is a very deep result, I am not going to elaborate until I have all my jigsaw pieces in order.
So what we want to do, is use that \(\theta_m = \theta + O(1/q^{2+\epsilon})\). We need to use this growth. This is derived using a lot of number theory/ analytic number theory... And has been used in complex dynamics before... Any discussion of non parabolic neutral fixed points will involve something similar. (Any thing that talks about \(g(z) = e^{2 \pi i \xi} z + O(z^2)\), and its dynamics will have some kind of discussion like this... )
So, the question becomes:
\[
f_\theta^{\circ s}(z) = \frac{d^{\tau_\theta s}}{dx^{\tau_\theta s}}\Big{|}_{x=0} \sum_{n=0}^\infty f_\theta^{q_\theta n}(z) \frac{x^n}{n!}\\
\]
As \(q \to \infty\) (because \(m \to \infty\)\), and \(\tau \to \infty \in \widehat{\mathbb{C}}\) (because \(q \to \infty\)) does this object converge? Where we can bound \(\theta_m \to \theta\) as \(\frac{1}{q_m^{1+\epsilon}}\)... Everything should work out fine, but I need to get the exact details right. And getting the exact details right, involves discussing how we approximate \(\theta \in \mathbb{H}\) using elements \(\theta^* \in \mathbb{H}^*\). And doing such, involves looking at Gauss sums and the discussion of \( j \equiv p \pmod{q}\).
Now we can pretty much avoid talking about this, if we just prepackage our discussion to:
\[
|\theta^* - \theta| < \frac{M}{q^{2+\epsilon}}\\
\]
But just know, to properly bound \(\epsilon\)... we need analytic number theory. I'm happy to just reference satisfactory results, but any result will be done using Gauss sums/analytic number theory. We can comfortably set \(\epsilon =0\), but just know, we needed Dirichlet's analytic number theory to prove that \(\epsilon = 0\) always works.
This relates deeply to number theory, and algebraic number theory also. Liouville's transcendental number is written as:
\[
A = \sum_{n=1}^\infty 10^{-n!}\\
\]
Because this value converges so damn fast, with the partial sum values \(A_m = \frac{p_m}{q_m}\), we can guess that \(|A_m - A| < \frac{M}{q_m^m}\)... this means it is transcendental. Because we cannot bound an algebraic number in this manner. To get every element of \(\xi \in \mathbb{R}\), and create a bound for \(\xi^* \in \mathbb{Q}\), so that \(|\xi - \xi^*| < \frac{M}{|q|^{2+\epsilon}}\), we have to be super fucking careful. I believe, for all algebraic/transcendental numbers the correct \(\epsilon\) is about \(1/4\)--but I may be mistaken. It is definitely \(\epsilon > 0\), and that's all I need, lol. But taking \(\epsilon = 0\), produces the boundary level perfectly. And is an old timey result which will suffice, lol.
So essentially for all \(\theta^* \in \mathbb{H}^*\) and \(0<\Re\theta^* < 2\pi\), where \(\Re \theta^* = 2 \pi\frac{p}{q}\). Then as \(\theta^* \to \theta\), we converge in the exact manner:
\[
|\theta^* - \theta| < \frac{M}{q^{2+\epsilon}}\\
\]
For some \(\epsilon > 0\)..........
(\(M\) is just a constant depending solely on the \(\theta^* \to \theta\)...)
Regards, James
EDIT:
For fuck's sakes I mixed up some stuff. The absolute bound is that:
\[
|\theta^* - \theta| < \frac{M}{q^2}\\
\]
This doesn't affect any of my results. If anything it makes my results better. I was just playing it safe by thinking \(\epsilon = 1/4\); when Dirichlet proved that \(\epsilon =0\) is an absolute bound. I need to refresh myself. I realize that what I was worked up about is the case \(2 + \epsilon\), where we try to find \(\epsilon\) here. If we just fix the exponent at \(2\) we are fine.
So we know that:
\[
|\frac{p_m}{q_m} - \xi| < \frac{1}{q_m^2}\\
\]
This is enough to do everything we need to do... but again you need Dirichlet's analytic number theory to prove this in the first place... lol
But this tells us the "growth of \(q_\theta\)" as a function of \(\theta\).... Because we can write it as:
\[
|\Re\theta^* - \Re\theta| < \frac{2\pi}{q^2} = \frac{2\pi}{\mathbf{d}(\theta^*)^2}\\
\]
For \(\Re \theta^* = 2\pi\frac{p}{q}\) and \(\Re\theta = 2\pi \xi\). The values \(0 < p < q\) and \((p,q)=1\) are co-prime. This means we write it as:
\[
\Re\theta = \Re\theta^* + O(\mathbf{d}(\theta^*)^{-2}) = \Re\theta^* + O(q^{-2})\\
\]
EDIT: For a quick read I suggest https://en.wikipedia.org/wiki/Liouville_..._constant)
Liouville really set the stage and we are using basic bounds developed by Dirichlet...
Let \(\xi\) be an irrational number, \(\xi \in \mathbb{R}/\mathbb{Q}\)--let's additionally ask that \(0 < \xi < 1\).
Then, to approximate \(\xi\), we take a sequence of rational numbers: \(\xi_m = \frac{p_m}{q_m}\)-- where \((p_m,q_m) = 1\), and additionally \(p_m < q_m\). Now, let's move forward, and look at:
\[
\mathbb{H}^* \to \mathbb{H}\\
\]
But we will restrict: \(0 < \Re(\theta) < 2 \pi\). Let us write:
\[
\theta_m = 2 \pi \frac{p_m}{q_m} + it\\
\]
Where we're given the identity \(\theta_m \to 2 \pi \xi + it\) as \(m \to \infty\)...
It is not entirely obvious how to show that \(f_{\theta_m}^{\circ s}(z) \to f_\theta^{\circ s}\) under the integral... There are solutions, but it's not obvious: which solution the integral chooses, and what manner it converges. In order to do this, we need to understand the maximal growth of \(q_m\). Now, there are restricted forms of irrational numbers which we write as:
\[
|\xi - \frac{p_m}{q_m}| < \frac{M}{q_m^{2+\epsilon}}\\
\]
This is a result that we are constantly finding better and better bounds on. But, we know that a choice \(\epsilon \approx 1/4\) should work fine for all \(0 < \xi < 1\). This is a very deep result, I am not going to elaborate until I have all my jigsaw pieces in order.
So what we want to do, is use that \(\theta_m = \theta + O(1/q^{2+\epsilon})\). We need to use this growth. This is derived using a lot of number theory/ analytic number theory... And has been used in complex dynamics before... Any discussion of non parabolic neutral fixed points will involve something similar. (Any thing that talks about \(g(z) = e^{2 \pi i \xi} z + O(z^2)\), and its dynamics will have some kind of discussion like this... )
So, the question becomes:
\[
f_\theta^{\circ s}(z) = \frac{d^{\tau_\theta s}}{dx^{\tau_\theta s}}\Big{|}_{x=0} \sum_{n=0}^\infty f_\theta^{q_\theta n}(z) \frac{x^n}{n!}\\
\]
As \(q \to \infty\) (because \(m \to \infty\)\), and \(\tau \to \infty \in \widehat{\mathbb{C}}\) (because \(q \to \infty\)) does this object converge? Where we can bound \(\theta_m \to \theta\) as \(\frac{1}{q_m^{1+\epsilon}}\)... Everything should work out fine, but I need to get the exact details right. And getting the exact details right, involves discussing how we approximate \(\theta \in \mathbb{H}\) using elements \(\theta^* \in \mathbb{H}^*\). And doing such, involves looking at Gauss sums and the discussion of \( j \equiv p \pmod{q}\).
Now we can pretty much avoid talking about this, if we just prepackage our discussion to:
\[
|\theta^* - \theta| < \frac{M}{q^{2+\epsilon}}\\
\]
But just know, to properly bound \(\epsilon\)... we need analytic number theory. I'm happy to just reference satisfactory results, but any result will be done using Gauss sums/analytic number theory. We can comfortably set \(\epsilon =0\), but just know, we needed Dirichlet's analytic number theory to prove that \(\epsilon = 0\) always works.
This relates deeply to number theory, and algebraic number theory also. Liouville's transcendental number is written as:
\[
A = \sum_{n=1}^\infty 10^{-n!}\\
\]
Because this value converges so damn fast, with the partial sum values \(A_m = \frac{p_m}{q_m}\), we can guess that \(|A_m - A| < \frac{M}{q_m^m}\)... this means it is transcendental. Because we cannot bound an algebraic number in this manner. To get every element of \(\xi \in \mathbb{R}\), and create a bound for \(\xi^* \in \mathbb{Q}\), so that \(|\xi - \xi^*| < \frac{M}{|q|^{2+\epsilon}}\), we have to be super fucking careful. I believe, for all algebraic/transcendental numbers the correct \(\epsilon\) is about \(1/4\)--but I may be mistaken. It is definitely \(\epsilon > 0\), and that's all I need, lol. But taking \(\epsilon = 0\), produces the boundary level perfectly. And is an old timey result which will suffice, lol.
So essentially for all \(\theta^* \in \mathbb{H}^*\) and \(0<\Re\theta^* < 2\pi\), where \(\Re \theta^* = 2 \pi\frac{p}{q}\). Then as \(\theta^* \to \theta\), we converge in the exact manner:
\[
|\theta^* - \theta| < \frac{M}{q^{2+\epsilon}}\\
\]
For some \(\epsilon > 0\)..........
(\(M\) is just a constant depending solely on the \(\theta^* \to \theta\)...)
Regards, James
EDIT:
For fuck's sakes I mixed up some stuff. The absolute bound is that:
\[
|\theta^* - \theta| < \frac{M}{q^2}\\
\]
This doesn't affect any of my results. If anything it makes my results better. I was just playing it safe by thinking \(\epsilon = 1/4\); when Dirichlet proved that \(\epsilon =0\) is an absolute bound. I need to refresh myself. I realize that what I was worked up about is the case \(2 + \epsilon\), where we try to find \(\epsilon\) here. If we just fix the exponent at \(2\) we are fine.
So we know that:
\[
|\frac{p_m}{q_m} - \xi| < \frac{1}{q_m^2}\\
\]
This is enough to do everything we need to do... but again you need Dirichlet's analytic number theory to prove this in the first place... lol
But this tells us the "growth of \(q_\theta\)" as a function of \(\theta\).... Because we can write it as:
\[
|\Re\theta^* - \Re\theta| < \frac{2\pi}{q^2} = \frac{2\pi}{\mathbf{d}(\theta^*)^2}\\
\]
For \(\Re \theta^* = 2\pi\frac{p}{q}\) and \(\Re\theta = 2\pi \xi\). The values \(0 < p < q\) and \((p,q)=1\) are co-prime. This means we write it as:
\[
\Re\theta = \Re\theta^* + O(\mathbf{d}(\theta^*)^{-2}) = \Re\theta^* + O(q^{-2})\\
\]
EDIT: For a quick read I suggest https://en.wikipedia.org/wiki/Liouville_..._constant)
Liouville really set the stage and we are using basic bounds developed by Dirichlet...