So let's write the function we care about:
\[
Y(\theta^*,x,z) = \sum_{n=0}^\infty f_{\theta^*}^{\circ q(\theta^*)n}(z) \frac{x^n}{n!}\\
\]
Then:
\[
f_{\theta^*}^{\circ s}(z) = \frac{d^{\frac{s}{q}(1+2\pi i p)}}{dx^{\frac{s}{q}(1+2\pi i p)}}\Big{|}_{x=0} Y(\theta^*,x,z)\\
\]
The value \(\Re \theta^* = 2\pi \frac{p}{q}\) so that: \(p = \frac{\Re\theta^* q(\theta^*)}{2\pi}\). So we reduce everything to:
\[
f_{\theta^*}^{\circ s}(z) = \frac{d^{\frac{s}{q} + i s\Re\theta^*}}{dx^{\frac{s}{q} + i s\Re\theta^*}} \Big{|}_{x=0} Y(\theta^*,x,z)\\
\]
The value \(\frac{s}{q} \to 0\) as \(\theta^* \to \theta\) (\(q(\theta^*) \to \infty\)). The value \(i\Re \theta^* \to 2\pi i \xi = 2 \pi i\lim_{p,q\to\infty} \frac{p}{q}\). For some \(\xi \in (0,1)\) and \(\xi \in \mathbb{R}/\mathbb{Q}\).
So with these exact incantations and variable dependencies/structures--I'll write what I'm trying to say, and why it's so difficult:
\[
\begin{align}
\lim_{q\to\infty} f_{\theta^*}^{\circ s}(z) &= f_\theta^{\circ s}(z)\\
\frac{d^{\frac{s}{q} + is\Re \theta^*}}{dx^{\frac{s}{q} + is \Re\theta^*}}&= \frac{d^{\frac{s}{q}}}{dx^{\frac{s}{q}}}\frac{d^{is\Re \theta^*}}{dx^{is \Re\theta^*}}\\
&= \frac{d^{\frac{s}{q}}}{dx^{\frac{s}{q}}}\frac{d^{2\pi i s \frac{p}{q}}}{dx^{2\pi i s \frac{p}{q}}}
\end{align}
\]
The exponential \(e^{\lambda^qx}\) is the first order expansion of \(\vartheta[f_{\theta^*}^{\circ q(\theta^*)}](x,z)\)--for \(\lambda = e^{-\Im(\theta^*)} = e^{-t}\), which is fixed. By which let's look at:
\[
\frac{d^{\frac{s}{q}}}{dx^{\frac{s}{q}}} e^{\lambda^qx} = \lambda^{s} e^{\lambda^qx}\\
\]
And:
\[
\begin{align}
\frac{d^{\frac{s}{q}}}{dx^{\frac{s}{q}}}\frac{d^{2\pi i s \frac{p}{q}}}{dx^{2\pi i s \frac{p}{q}}} e^{\lambda^qx}&= e^{i\theta^* s} e^{\lambda^qx}\\
\lim_{q\to\infty} \frac{d^{\frac{s}{q}}}{dx^{\frac{s}{q}}}\frac{d^{2\pi i s \frac{p}{q}}}{dx^{2\pi i s \frac{p}{q}}} e^{\lambda^qx} &= e^{i\theta s}\\
\end{align}
\]
The equation on the left turns into a double integral as \(q \to \infty\)... As \(e^{\lambda^q x} \to 1\), the differintegrals to similar divergence...
We can write this in a plain manner:
\[
e^{2\pi i s \frac{p}{q} - st} e^{\lambda^qx} = e^{i\theta^* s} e^{\lambda^qx} = \frac{1}{\Gamma(-s\frac{1+2\pi i p}{q})} \int_0^\infty Y_q(-x,z)x^{-s\frac{1+2\pi i p}{q}-1}\,dx\\
\]
Letting \(q \to \infty\) produces an integral, we get something like \(\frac{1}{n} \sum_{j=0}^n a_{jn}\).... To do the nit and gritty, cold as concrete proofs, we will need \(q\)'s growth in relation to \(\theta^*\), hence Dirichlet's Approximation Theorem, is the simplest proof from analytic number theory we will need. Hopefully we don't need more; but we might need some fineesed results derived from Dirchlet
.
Bout to go on my 2-3 month free time vacation. I will get this out clearly.
\[
Y(\theta^*,x,z) = \sum_{n=0}^\infty f_{\theta^*}^{\circ q(\theta^*)n}(z) \frac{x^n}{n!}\\
\]
Then:
\[
f_{\theta^*}^{\circ s}(z) = \frac{d^{\frac{s}{q}(1+2\pi i p)}}{dx^{\frac{s}{q}(1+2\pi i p)}}\Big{|}_{x=0} Y(\theta^*,x,z)\\
\]
The value \(\Re \theta^* = 2\pi \frac{p}{q}\) so that: \(p = \frac{\Re\theta^* q(\theta^*)}{2\pi}\). So we reduce everything to:
\[
f_{\theta^*}^{\circ s}(z) = \frac{d^{\frac{s}{q} + i s\Re\theta^*}}{dx^{\frac{s}{q} + i s\Re\theta^*}} \Big{|}_{x=0} Y(\theta^*,x,z)\\
\]
The value \(\frac{s}{q} \to 0\) as \(\theta^* \to \theta\) (\(q(\theta^*) \to \infty\)). The value \(i\Re \theta^* \to 2\pi i \xi = 2 \pi i\lim_{p,q\to\infty} \frac{p}{q}\). For some \(\xi \in (0,1)\) and \(\xi \in \mathbb{R}/\mathbb{Q}\).
So with these exact incantations and variable dependencies/structures--I'll write what I'm trying to say, and why it's so difficult:
\[
\begin{align}
\lim_{q\to\infty} f_{\theta^*}^{\circ s}(z) &= f_\theta^{\circ s}(z)\\
\frac{d^{\frac{s}{q} + is\Re \theta^*}}{dx^{\frac{s}{q} + is \Re\theta^*}}&= \frac{d^{\frac{s}{q}}}{dx^{\frac{s}{q}}}\frac{d^{is\Re \theta^*}}{dx^{is \Re\theta^*}}\\
&= \frac{d^{\frac{s}{q}}}{dx^{\frac{s}{q}}}\frac{d^{2\pi i s \frac{p}{q}}}{dx^{2\pi i s \frac{p}{q}}}
\end{align}
\]
The exponential \(e^{\lambda^qx}\) is the first order expansion of \(\vartheta[f_{\theta^*}^{\circ q(\theta^*)}](x,z)\)--for \(\lambda = e^{-\Im(\theta^*)} = e^{-t}\), which is fixed. By which let's look at:
\[
\frac{d^{\frac{s}{q}}}{dx^{\frac{s}{q}}} e^{\lambda^qx} = \lambda^{s} e^{\lambda^qx}\\
\]
And:
\[
\begin{align}
\frac{d^{\frac{s}{q}}}{dx^{\frac{s}{q}}}\frac{d^{2\pi i s \frac{p}{q}}}{dx^{2\pi i s \frac{p}{q}}} e^{\lambda^qx}&= e^{i\theta^* s} e^{\lambda^qx}\\
\lim_{q\to\infty} \frac{d^{\frac{s}{q}}}{dx^{\frac{s}{q}}}\frac{d^{2\pi i s \frac{p}{q}}}{dx^{2\pi i s \frac{p}{q}}} e^{\lambda^qx} &= e^{i\theta s}\\
\end{align}
\]
The equation on the left turns into a double integral as \(q \to \infty\)... As \(e^{\lambda^q x} \to 1\), the differintegrals to similar divergence...
We can write this in a plain manner:
\[
e^{2\pi i s \frac{p}{q} - st} e^{\lambda^qx} = e^{i\theta^* s} e^{\lambda^qx} = \frac{1}{\Gamma(-s\frac{1+2\pi i p}{q})} \int_0^\infty Y_q(-x,z)x^{-s\frac{1+2\pi i p}{q}-1}\,dx\\
\]
Letting \(q \to \infty\) produces an integral, we get something like \(\frac{1}{n} \sum_{j=0}^n a_{jn}\).... To do the nit and gritty, cold as concrete proofs, we will need \(q\)'s growth in relation to \(\theta^*\), hence Dirichlet's Approximation Theorem, is the simplest proof from analytic number theory we will need. Hopefully we don't need more; but we might need some fineesed results derived from Dirchlet
.Bout to go on my 2-3 month free time vacation. I will get this out clearly.