So, let's start now with:
\[
f_{\theta^*}^{\circ s}(z) = \frac{d^{\frac{s}{q} + 2\pi is\frac{p}{q}}}{ds^{\frac{s}{q} + 2\pi is\frac{p}{q}}}\Big{|}_{x=0} \sum_{n=0}^\infty f_{\theta^*}^{qn}(z)\frac{x^n}{n!}\\
\]
The value \(\theta^* = 2\pi \frac{p}{q} +it\), for \(p,q \in \mathbb{Z}\), coprime--and \(t>0\). The function:
\[
f_\theta(z) = e^{e^{i\theta}z}-1\\
\]
The value \(\theta^* \to \theta = 2\pi \xi + it\), where the error is given as \(O(1/q^2)\). We will refer to \(q = q(\theta^*)\). Where as \(\theta^* \to \theta\) we have \(q(\theta^*) \to \infty\). By which we have the convenient formula \(\theta = \theta^* + O(1/q(\theta^*)^2)\). To set the stage, we will observe the linear state first, for \(m \in \mathbb{N}\):
\[
f_{\theta^*}^{\circ m}(z) = e^{i\theta^*m} z + O(z^2)\\
\]
The value:
\[
f_\theta^{\circ m}(z) - f_{\theta^*}^{\circ m}(z) = \left(e^{i\theta^* m} - e^{i\theta m}\right) z + O(z^2)\\
\]
The first term is precisely:
\[
e^{i\theta^* m} - e^{i\theta m} = mi (\theta^* - \theta) + O(\theta^* - \theta)^2
\]
Setting \(m = q(\theta^*)\) we get:
\[
iq(\theta^*)(\theta^* - \theta) = O(1/q(\theta^*))\\
\]
Thereby:
\[
f_{\theta^*}^{\circ q(\theta^*)}(z) = ze^{i\theta q(\theta^*)} + zO(1/q(\theta^*)) + O(z^2)\\
\]
And by linear operations:
\[
f_{\theta^*}^{\circ q(\theta^*)n}(z) = e^{i\theta q(\theta^*)n}z + nzO(1/q(\theta^*)) + O(z^2)\\
\]
Thereby the function:
\[
\vartheta[f_{\theta^*}^{\circ q(\theta^*)}](x,z) = z \sum_{n=0}^\infty \left(e^{i\theta q(\theta^*)n} +nO(1/q(\theta^*))\right)\frac{x^n}{n!} + O(z^2)\\
\]
Bounding the error term in \(x\) on this O notation is very difficult, I'm just giving a rough idea at the moment. But it should be fine, if we choose \(x\) belonging to a differintegrable sector. This whole nonsense reduces to:
\[
\vartheta[f_{\theta^*}^{\circ q(\theta^*)}](x,z) = z\left(e^{e^{i\theta q(\theta^*)}x} + O(1/q(\theta^*)) \right)+ O(z^2)\\
\]
Now, let's take, the correct differintegral:
\[
f_{\theta^*}^{\circ s}(z) = \frac{d^{\frac{s}{q} + 2\pi i s \frac{p}{q}}}{ds^{\frac{s}{q} + 2\pi i s\frac{p}{q}}}\Big{|}_{x=0} \vartheta[f_{\theta^*}^{\circ q}](x,z)
\]
Inadvertently, we are performing an infinite sum of neglible terms \(\vartheta \to 0\), just as all of it's terms as \(q \to \infty\). But, we are approaching infinity on the left side like \(q \to \infty\). We are literally getting an equation that looks like:
\[
\lim_{q\to\infty} \sum_{j=0}^{q} O (1/q) = O(1)\\
\]
..... Man I thought compositional integration was hard. This shit has me scratching my head so much more, lmao. I'm getting closer. And I'm pissed off now, lmao. First order expansion should work fine though:
\[
\frac{d}{dz}\Big{|}_{z=0} f_{\theta^*}^{\circ s}(z) \to \frac{d}{dz}\Big{|}_{z=0} f_{\theta}^{\circ s}(z)
\]
And this will be a weird double integral where you mellin transform than do some weird ass Gauss limit, lmao. Which just equates to using exponential functions, instead of Varthetas, lol. Nothing more than the formula:
\[
\lim_{q\to\infty} \frac{1}{\Gamma(s/q)} \int_0^\infty e^{-e^{i\theta q}x}x^{\frac{s}{q}-1}\,dx = e^{-i\theta s}\\
\]
This limit converges in a very ugly way; it is a normal sequence, but letting \(q = m\) for arbitrary \(m\), causes this to converge to a different value.... We must choose the sequence \(q \to \infty\), where \(\theta = 2 \pi \frac{p}{q} + it + O(1/q^2)\), and is the sole sequence to satisfy these bounds. Another sequence might change our value here.
If \(e^{i\theta} = e^{-t}= \lambda \in (0,1)\), then this is easier, and any sequence \(m\to\infty\) satisfies:
\[
\lim_{m\to\infty} \frac{1}{\Gamma(s/m)} \int_0^\infty e^{-\lambda^{ m}x}x^{\frac{s}{m}-1}\,dx = \lambda^s\\
\]
The trouble is in the complex plane, we are rotating domains, and hitting boundaries, so it's not as straightforward... The sequence \(q \to \infty\) is unique in that it satisfies \(e^{-e^{i\theta q}x} = e^{-\lambda^q(1+O(\frac{1}{q}))x}\), which makes the limit feasible...... Additionally, there are linear substitutions which "turn \(\lambda^s\) into \(e^{i\theta s}\)"..... So we can actually expect \(e^{2 \pi i \frac{p}{q}} = 1 + O(1/q)\) to be a reasonable complex approximation.
At this point we should move the integral from \(\int_0^\infty\), to the differintegral which is much more general, and is impartial to the direction of the integral.... But we don't have to, I'm fixing everything on \(\int_0^\infty\) here. And every step of the limit converges, and the mellin transformed limit converges, and the \(\vartheta\) converges the same. So we're fine....at least for the first terms in \(z\), lmfaooo!!!!!
\[
f_{\theta^*}^{\circ s}(z) = \frac{d^{\frac{s}{q} + 2\pi is\frac{p}{q}}}{ds^{\frac{s}{q} + 2\pi is\frac{p}{q}}}\Big{|}_{x=0} \sum_{n=0}^\infty f_{\theta^*}^{qn}(z)\frac{x^n}{n!}\\
\]
The value \(\theta^* = 2\pi \frac{p}{q} +it\), for \(p,q \in \mathbb{Z}\), coprime--and \(t>0\). The function:
\[
f_\theta(z) = e^{e^{i\theta}z}-1\\
\]
The value \(\theta^* \to \theta = 2\pi \xi + it\), where the error is given as \(O(1/q^2)\). We will refer to \(q = q(\theta^*)\). Where as \(\theta^* \to \theta\) we have \(q(\theta^*) \to \infty\). By which we have the convenient formula \(\theta = \theta^* + O(1/q(\theta^*)^2)\). To set the stage, we will observe the linear state first, for \(m \in \mathbb{N}\):
\[
f_{\theta^*}^{\circ m}(z) = e^{i\theta^*m} z + O(z^2)\\
\]
The value:
\[
f_\theta^{\circ m}(z) - f_{\theta^*}^{\circ m}(z) = \left(e^{i\theta^* m} - e^{i\theta m}\right) z + O(z^2)\\
\]
The first term is precisely:
\[
e^{i\theta^* m} - e^{i\theta m} = mi (\theta^* - \theta) + O(\theta^* - \theta)^2
\]
Setting \(m = q(\theta^*)\) we get:
\[
iq(\theta^*)(\theta^* - \theta) = O(1/q(\theta^*))\\
\]
Thereby:
\[
f_{\theta^*}^{\circ q(\theta^*)}(z) = ze^{i\theta q(\theta^*)} + zO(1/q(\theta^*)) + O(z^2)\\
\]
And by linear operations:
\[
f_{\theta^*}^{\circ q(\theta^*)n}(z) = e^{i\theta q(\theta^*)n}z + nzO(1/q(\theta^*)) + O(z^2)\\
\]
Thereby the function:
\[
\vartheta[f_{\theta^*}^{\circ q(\theta^*)}](x,z) = z \sum_{n=0}^\infty \left(e^{i\theta q(\theta^*)n} +nO(1/q(\theta^*))\right)\frac{x^n}{n!} + O(z^2)\\
\]
Bounding the error term in \(x\) on this O notation is very difficult, I'm just giving a rough idea at the moment. But it should be fine, if we choose \(x\) belonging to a differintegrable sector. This whole nonsense reduces to:
\[
\vartheta[f_{\theta^*}^{\circ q(\theta^*)}](x,z) = z\left(e^{e^{i\theta q(\theta^*)}x} + O(1/q(\theta^*)) \right)+ O(z^2)\\
\]
Now, let's take, the correct differintegral:
\[
f_{\theta^*}^{\circ s}(z) = \frac{d^{\frac{s}{q} + 2\pi i s \frac{p}{q}}}{ds^{\frac{s}{q} + 2\pi i s\frac{p}{q}}}\Big{|}_{x=0} \vartheta[f_{\theta^*}^{\circ q}](x,z)
\]
Inadvertently, we are performing an infinite sum of neglible terms \(\vartheta \to 0\), just as all of it's terms as \(q \to \infty\). But, we are approaching infinity on the left side like \(q \to \infty\). We are literally getting an equation that looks like:
\[
\lim_{q\to\infty} \sum_{j=0}^{q} O (1/q) = O(1)\\
\]
..... Man I thought compositional integration was hard. This shit has me scratching my head so much more, lmao. I'm getting closer. And I'm pissed off now, lmao. First order expansion should work fine though:
\[
\frac{d}{dz}\Big{|}_{z=0} f_{\theta^*}^{\circ s}(z) \to \frac{d}{dz}\Big{|}_{z=0} f_{\theta}^{\circ s}(z)
\]
And this will be a weird double integral where you mellin transform than do some weird ass Gauss limit, lmao. Which just equates to using exponential functions, instead of Varthetas, lol. Nothing more than the formula:
\[
\lim_{q\to\infty} \frac{1}{\Gamma(s/q)} \int_0^\infty e^{-e^{i\theta q}x}x^{\frac{s}{q}-1}\,dx = e^{-i\theta s}\\
\]
This limit converges in a very ugly way; it is a normal sequence, but letting \(q = m\) for arbitrary \(m\), causes this to converge to a different value.... We must choose the sequence \(q \to \infty\), where \(\theta = 2 \pi \frac{p}{q} + it + O(1/q^2)\), and is the sole sequence to satisfy these bounds. Another sequence might change our value here.
If \(e^{i\theta} = e^{-t}= \lambda \in (0,1)\), then this is easier, and any sequence \(m\to\infty\) satisfies:
\[
\lim_{m\to\infty} \frac{1}{\Gamma(s/m)} \int_0^\infty e^{-\lambda^{ m}x}x^{\frac{s}{m}-1}\,dx = \lambda^s\\
\]
The trouble is in the complex plane, we are rotating domains, and hitting boundaries, so it's not as straightforward... The sequence \(q \to \infty\) is unique in that it satisfies \(e^{-e^{i\theta q}x} = e^{-\lambda^q(1+O(\frac{1}{q}))x}\), which makes the limit feasible...... Additionally, there are linear substitutions which "turn \(\lambda^s\) into \(e^{i\theta s}\)"..... So we can actually expect \(e^{2 \pi i \frac{p}{q}} = 1 + O(1/q)\) to be a reasonable complex approximation.
At this point we should move the integral from \(\int_0^\infty\), to the differintegral which is much more general, and is impartial to the direction of the integral.... But we don't have to, I'm fixing everything on \(\int_0^\infty\) here. And every step of the limit converges, and the mellin transformed limit converges, and the \(\vartheta\) converges the same. So we're fine....at least for the first terms in \(z\), lmfaooo!!!!!