I want to describe the following identity--which we will use and abuse. Allez-y:
\[
\theta^* = 2 \pi \frac{p}{q} + it\\
\]
Where \(t \in \mathbb{R}^+\), and \((p,q) =1\) are coprime integers. This implies that \(\theta^* \in 2\pi \mathbb{Q} + i\mathbb{R}^+\). Let \(\theta^* \to \theta\), where we assume: \(\theta \in 2\pi \left(\mathbb{R}/\mathbb{Q}\right) + i\mathbb{R}^+\). We are going to additionally assume that \(\Re \theta^* \in (0,1)\). This assumption isn't necessary but simplifies our language. And additionally, avoids us talking about "different" branches.
As Mphlee pointed out, which I thought was apparent, is that \(q = q(\theta^*)\), is a function of \(\theta^*\). It takes \(2\pi \mathbb{Q} + i\mathbb{R}^+ = \mathbb{H}^* \to \mathbb{N}\)--and is therefore a type of functional/projection type function.
By Dirichlet's Approximation Theorem:
https://www.expii.com/t/dirichlets-appro...eorem-2468
This is pretty basic number theory/analytic number theory, so you guys should be able to get it (no zeta function nonsense, lmao). It's literally the Pigeon Hole Principle
--still, it's an indispensable theorem.
\[
|\theta - \theta^*| < \frac{1}{q^2}\\
\]
Thereby we can write:
\[
\theta = \theta^* + O\left(\frac{1}{q(\theta^*)^2}\right)\\
\]
This means that:
\[
|f_\theta(z) - f_{\theta^*}(z)| = O\left(\frac{1}{q(\theta^*)^2}\right)\\
\]
This O-term looks exactly like:
\[
O\left(\frac{1}{q(\theta^*)^2}\right) = \dfrac{\frac{d}{d\theta^*} f_{\theta^*}(z)}{q(\theta^*)^2} + O(\frac{1}{q(\theta^*)^4})\\
\]
Because:
\[
f_\theta(z) = f_{\theta^*}(z) + \frac{d}{d\theta^*} f_{\theta^*}(z) (\theta - \theta^*) + O(\theta - \theta^*)^2\\
\]
Which is just by taylor's theorem. Equally we have the formula:
\[
f_\theta^{\circ n}(z) = f_{\theta^*}^{\circ n}(z) + \dfrac{\frac{d}{d\theta^*} f_{\theta^*}^{\circ n}(z)}{q(\theta^*)^2} + O(\frac{1}{q(\theta^*)^4})
\]
As for the variable \(z\), we are restricting \(|z| < \delta\), where \(\delta\) can be chosen uniformly for all \(|\theta - \theta^*| < \epsilon\). This means, quite beautifully that:
\[
\vartheta[f_\theta](x,z) = \sum_{n=0}^\infty f_\theta^{\circ n}(z) \frac{x^n}{n!}\\
\]
That:
\[
\vartheta[f_\theta](x,z) = \vartheta[f_{\theta^*}](x,z) + \frac{d}{d\theta^*} \vartheta[f_{\theta^*}](x,z) \frac{1}{q(\theta^*)^2} + O(\frac{1}{q(\theta^*)^4})\\
\]
The bound on this O-term IS NOT UNIFORM IN \(x\). But it is uniform on integrable domains. So if \(\vartheta[f_{\theta^*}]\) is differintegrable in a sector, we can make this O term uniform for that sector.
But this doesn't give us our answer. Instead we want to write:
\[
\vartheta[f_{\theta^*}^{\circ q(\theta^*)}](x,z) = \sum_{n=0}^\infty f_{\theta^*}^{\circ q(\theta^*)n} \frac{x^n}{n!}\\
\]
This causes instantly one of the \(q(\theta^*)\)'s to cancel out...
\[
\frac{d}{d\theta^*} \vartheta[f_{\theta^*}^{\circ q(\theta^*)}](x,z) \frac{1}{q(\theta^*)^2} = \frac{d}{d\theta^*} \vartheta[f_{\theta^*}](x,z) O\left(\frac{1}{q(\theta^*)}\right)
\]
I need to fine tune this argument, but we should get something that expands as:
\[
f_{\theta}^{\circ s}(z) = f_{\theta^*}^{\circ s}(z) + O(\frac{1}{q(\theta^*)})\\
\]
This will cause an integral to arise as \(\theta^* \to \theta\), that I'm not sure of yet. It's evading me. I need more mulling time. But we should expect a result very close to this... So essentially, when we differintegrate, we'll probably lose a square on the error (not necessarily, but these are the best bounds I can think of).
This goes in conjunction that \(\frac{d^s}{dx^s} \frac{d}{d\theta} = \frac{d}{d\theta}\frac{d^s}{dx^s}\)--because I am only considering the best kind of converging integral transforms. None of these weird Caputo style fractional derivatives. My derivative is just fancy fourier transforms
--differentiating/integrating under a Fourier transform is actually pretty easy if you are in a normal space!
\[
\theta^* = 2 \pi \frac{p}{q} + it\\
\]
Where \(t \in \mathbb{R}^+\), and \((p,q) =1\) are coprime integers. This implies that \(\theta^* \in 2\pi \mathbb{Q} + i\mathbb{R}^+\). Let \(\theta^* \to \theta\), where we assume: \(\theta \in 2\pi \left(\mathbb{R}/\mathbb{Q}\right) + i\mathbb{R}^+\). We are going to additionally assume that \(\Re \theta^* \in (0,1)\). This assumption isn't necessary but simplifies our language. And additionally, avoids us talking about "different" branches.
As Mphlee pointed out, which I thought was apparent, is that \(q = q(\theta^*)\), is a function of \(\theta^*\). It takes \(2\pi \mathbb{Q} + i\mathbb{R}^+ = \mathbb{H}^* \to \mathbb{N}\)--and is therefore a type of functional/projection type function.
By Dirichlet's Approximation Theorem:
https://www.expii.com/t/dirichlets-appro...eorem-2468
This is pretty basic number theory/analytic number theory, so you guys should be able to get it (no zeta function nonsense, lmao). It's literally the Pigeon Hole Principle
--still, it's an indispensable theorem.\[
|\theta - \theta^*| < \frac{1}{q^2}\\
\]
Thereby we can write:
\[
\theta = \theta^* + O\left(\frac{1}{q(\theta^*)^2}\right)\\
\]
This means that:
\[
|f_\theta(z) - f_{\theta^*}(z)| = O\left(\frac{1}{q(\theta^*)^2}\right)\\
\]
This O-term looks exactly like:
\[
O\left(\frac{1}{q(\theta^*)^2}\right) = \dfrac{\frac{d}{d\theta^*} f_{\theta^*}(z)}{q(\theta^*)^2} + O(\frac{1}{q(\theta^*)^4})\\
\]
Because:
\[
f_\theta(z) = f_{\theta^*}(z) + \frac{d}{d\theta^*} f_{\theta^*}(z) (\theta - \theta^*) + O(\theta - \theta^*)^2\\
\]
Which is just by taylor's theorem. Equally we have the formula:
\[
f_\theta^{\circ n}(z) = f_{\theta^*}^{\circ n}(z) + \dfrac{\frac{d}{d\theta^*} f_{\theta^*}^{\circ n}(z)}{q(\theta^*)^2} + O(\frac{1}{q(\theta^*)^4})
\]
As for the variable \(z\), we are restricting \(|z| < \delta\), where \(\delta\) can be chosen uniformly for all \(|\theta - \theta^*| < \epsilon\). This means, quite beautifully that:
\[
\vartheta[f_\theta](x,z) = \sum_{n=0}^\infty f_\theta^{\circ n}(z) \frac{x^n}{n!}\\
\]
That:
\[
\vartheta[f_\theta](x,z) = \vartheta[f_{\theta^*}](x,z) + \frac{d}{d\theta^*} \vartheta[f_{\theta^*}](x,z) \frac{1}{q(\theta^*)^2} + O(\frac{1}{q(\theta^*)^4})\\
\]
The bound on this O-term IS NOT UNIFORM IN \(x\). But it is uniform on integrable domains. So if \(\vartheta[f_{\theta^*}]\) is differintegrable in a sector, we can make this O term uniform for that sector.
But this doesn't give us our answer. Instead we want to write:
\[
\vartheta[f_{\theta^*}^{\circ q(\theta^*)}](x,z) = \sum_{n=0}^\infty f_{\theta^*}^{\circ q(\theta^*)n} \frac{x^n}{n!}\\
\]
This causes instantly one of the \(q(\theta^*)\)'s to cancel out...
\[
\frac{d}{d\theta^*} \vartheta[f_{\theta^*}^{\circ q(\theta^*)}](x,z) \frac{1}{q(\theta^*)^2} = \frac{d}{d\theta^*} \vartheta[f_{\theta^*}](x,z) O\left(\frac{1}{q(\theta^*)}\right)
\]
I need to fine tune this argument, but we should get something that expands as:
\[
f_{\theta}^{\circ s}(z) = f_{\theta^*}^{\circ s}(z) + O(\frac{1}{q(\theta^*)})\\
\]
This will cause an integral to arise as \(\theta^* \to \theta\), that I'm not sure of yet. It's evading me. I need more mulling time. But we should expect a result very close to this... So essentially, when we differintegrate, we'll probably lose a square on the error (not necessarily, but these are the best bounds I can think of).
This goes in conjunction that \(\frac{d^s}{dx^s} \frac{d}{d\theta} = \frac{d}{d\theta}\frac{d^s}{dx^s}\)--because I am only considering the best kind of converging integral transforms. None of these weird Caputo style fractional derivatives. My derivative is just fancy fourier transforms
--differentiating/integrating under a Fourier transform is actually pretty easy if you are in a normal space!