1)
Yes, so \(q_\theta = \mathbf{d}(\theta): \mathbb{H}^* \to \mathbb{N}\). Forgive me, but I will write \(q_\theta\). Note that all of these variables depend on each other--this is a system of variables which depend on each other.
I apologize that the period is not obvious to you. Since \(q_\theta \in \mathbb{N}\). We have that:
\[
e^{i\theta q_\theta\left(s + \frac{2 \pi}{\theta q_\theta}\right)} = e^{i\theta q_\theta s} e^{i\theta q_\theta \frac{2\pi}{\theta q_\theta}} = e^{i\theta q_\theta s} e^{2 \pi i} = e^{i\theta q_\theta s}\\
\]
This is also the minimal period; every periodic element \(l\), as you wrote, is a multiple : \(l \in \frac{2 \pi}{q_\theta \theta} \mathbb{Z}\)... There's no smaller period. This is because \(p\) and \(q\) are coprime.
2)
Now it's natural to be confused at this point, maybe I should have gone slower.
You write:
\[
\left(e^{i\theta q_\theta}\right)^s\\
\]
And say this is an exponential satisfying my list of conditions--which you are absolutely correct!
BUT!
\[
e^{i\theta q_\theta} = \lambda_\theta^q \in (0,1)\\
\]
While \(0 < \lambda_\theta < 1\). This is precisely the problem, so your confusion is in the right direction at least
. We want to turn \(\lambda_\theta^{qs}\) into \(e^{i\theta s}\)...
I should also note, that when I said "period" I meant "minimal period". I apologize, that's standard terminology in complex analysis. So yes \(4\pi i\) is a period of \(e^{s}\) but it's not THE period, which is \(2 \pi i\). So when I say "the period" I actually mean the minimal period. It's just a poor affectation of language, lol. I apologize.
3) Okay, so the set \(\mathcal{N}\) is difficult to describe fully. I was just giving a sketch. So let's take a compact set \(\mathcal{K} \subset \mathbb{H}\). Let's let \(|e^{i\theta s}| < 1\). For \(\theta \in \mathcal{K} \cap \mathbb{H}^*\), and \(s \in \mathcal{S}\) such that \(|e^{i\theta s}| < 1\)--then \(\delta\) depends on \(\mathcal{K}\) and \(\mathcal{S}\). That alone.
Essentially what I meant, is that for \(\theta\) in a neighborhood \(\mathcal{K}\), and \(|e^{i\theta s}|<1\)--which restricts the variable \(s \in \mathcal{S}=\mathcal{S}(\mathcal{K})\)--there exists a value \(\delta > 0\), such that \(f_\theta^{\circ s}(z) : \mathcal{K} \times \mathcal{S} \times \mathcal{N} \to \mathcal{N}\), where \(\mathcal{N}\) is the domain \(|z| < \delta\)...
I apologize, I was being a tad too implicit here. It can get pretty symbol heavy for this, I was just giving a quick sketch...
4)
Okay, so this is the big reveal....
\[
\frac{d^s}{dx^s} \Big{|}_{x=0} \sum_{n=0}^\infty f_\theta^{\circ q_\theta n}(z) \frac{x^n}{n!} = \lambda^{q_\theta s}z + O (z^2)\\
\]
Where yes, a linear substitution will work fine if \(\theta \in \mathbb{H}^*\)--and \(\lambda^{s + as} = e^{i\theta s}\). But Mphlee....
How the fuck to I do this when \(\mathbb{H}^* \to \mathbb{H}\), where \(q_\theta \to \infty\) ??????
The sum will converge to \(0\)! I know this may seem trivial, but it's very much not. We have to be able to make an educated guess on the growth of \(q_\theta\)--this will relate to a lot of number theory...
We have to perform the action \(\theta \to q\theta\) before we perform the mellin transform. Then we perform the linear substitution pull back, to get \(\theta\) again. But when \(q \to \infty\)--we have to understand "how fast it goes to infinity" and "how fast the pullback pulls us back to reality".
As \(\mathbb{H}^* \to \mathbb{H}\), it is not entirely obvious that \(f_\theta^{\circ s}\) converges... We know that there are solution sets--but are they Mellin transformable!? We don't know that yet!
PLEASE ASK MORE QUESTIONS! I'm happy to answer and I think I can explain better when somebody asks me questions...
Also the way this relates to Euler's formula is a tad difficult to explain again. It only works for \(q = 2\)--for arbitrary \(q\) it relates to roots of unity. But:
\[
\sum_{n=0}^\infty f^{\circ n}(z) \frac{x^n}{n!} = \vartheta[f](x,z)\\
\]
Then:
\[
\vartheta[f](ix,z) = \sum_{n=0}^\infty f^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!} + i\sum_{n=0}^\infty f^{\circ 2n+1}(z) \frac{(-1)^n x^{2n+1}}{2n+1!}\\
\]
Both sums on the right are mellin transformable, insuring the object on the left is mellin transformable. We don't need this entirely, I was mistaken--I thought the Gamma functions would pull out, but they actually cancel out... I think I have a rough idea of how to find square roots though, which look like this...
Yes, so \(q_\theta = \mathbf{d}(\theta): \mathbb{H}^* \to \mathbb{N}\). Forgive me, but I will write \(q_\theta\). Note that all of these variables depend on each other--this is a system of variables which depend on each other.
I apologize that the period is not obvious to you. Since \(q_\theta \in \mathbb{N}\). We have that:
\[
e^{i\theta q_\theta\left(s + \frac{2 \pi}{\theta q_\theta}\right)} = e^{i\theta q_\theta s} e^{i\theta q_\theta \frac{2\pi}{\theta q_\theta}} = e^{i\theta q_\theta s} e^{2 \pi i} = e^{i\theta q_\theta s}\\
\]
This is also the minimal period; every periodic element \(l\), as you wrote, is a multiple : \(l \in \frac{2 \pi}{q_\theta \theta} \mathbb{Z}\)... There's no smaller period. This is because \(p\) and \(q\) are coprime.
2)
Now it's natural to be confused at this point, maybe I should have gone slower.
You write:\[
\left(e^{i\theta q_\theta}\right)^s\\
\]
And say this is an exponential satisfying my list of conditions--which you are absolutely correct!
BUT!
\[
e^{i\theta q_\theta} = \lambda_\theta^q \in (0,1)\\
\]
While \(0 < \lambda_\theta < 1\). This is precisely the problem, so your confusion is in the right direction at least
. We want to turn \(\lambda_\theta^{qs}\) into \(e^{i\theta s}\)... I should also note, that when I said "period" I meant "minimal period". I apologize, that's standard terminology in complex analysis. So yes \(4\pi i\) is a period of \(e^{s}\) but it's not THE period, which is \(2 \pi i\). So when I say "the period" I actually mean the minimal period. It's just a poor affectation of language, lol. I apologize.
3) Okay, so the set \(\mathcal{N}\) is difficult to describe fully. I was just giving a sketch. So let's take a compact set \(\mathcal{K} \subset \mathbb{H}\). Let's let \(|e^{i\theta s}| < 1\). For \(\theta \in \mathcal{K} \cap \mathbb{H}^*\), and \(s \in \mathcal{S}\) such that \(|e^{i\theta s}| < 1\)--then \(\delta\) depends on \(\mathcal{K}\) and \(\mathcal{S}\). That alone.
Essentially what I meant, is that for \(\theta\) in a neighborhood \(\mathcal{K}\), and \(|e^{i\theta s}|<1\)--which restricts the variable \(s \in \mathcal{S}=\mathcal{S}(\mathcal{K})\)--there exists a value \(\delta > 0\), such that \(f_\theta^{\circ s}(z) : \mathcal{K} \times \mathcal{S} \times \mathcal{N} \to \mathcal{N}\), where \(\mathcal{N}\) is the domain \(|z| < \delta\)...
I apologize, I was being a tad too implicit here. It can get pretty symbol heavy for this, I was just giving a quick sketch...
4)
Okay, so this is the big reveal....
\[
\frac{d^s}{dx^s} \Big{|}_{x=0} \sum_{n=0}^\infty f_\theta^{\circ q_\theta n}(z) \frac{x^n}{n!} = \lambda^{q_\theta s}z + O (z^2)\\
\]
Where yes, a linear substitution will work fine if \(\theta \in \mathbb{H}^*\)--and \(\lambda^{s + as} = e^{i\theta s}\). But Mphlee....
How the fuck to I do this when \(\mathbb{H}^* \to \mathbb{H}\), where \(q_\theta \to \infty\) ??????
The sum will converge to \(0\)! I know this may seem trivial, but it's very much not. We have to be able to make an educated guess on the growth of \(q_\theta\)--this will relate to a lot of number theory...
We have to perform the action \(\theta \to q\theta\) before we perform the mellin transform. Then we perform the linear substitution pull back, to get \(\theta\) again. But when \(q \to \infty\)--we have to understand "how fast it goes to infinity" and "how fast the pullback pulls us back to reality".
As \(\mathbb{H}^* \to \mathbb{H}\), it is not entirely obvious that \(f_\theta^{\circ s}\) converges... We know that there are solution sets--but are they Mellin transformable!? We don't know that yet!
PLEASE ASK MORE QUESTIONS! I'm happy to answer and I think I can explain better when somebody asks me questions...
Also the way this relates to Euler's formula is a tad difficult to explain again. It only works for \(q = 2\)--for arbitrary \(q\) it relates to roots of unity. But:
\[
\sum_{n=0}^\infty f^{\circ n}(z) \frac{x^n}{n!} = \vartheta[f](x,z)\\
\]
Then:
\[
\vartheta[f](ix,z) = \sum_{n=0}^\infty f^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!} + i\sum_{n=0}^\infty f^{\circ 2n+1}(z) \frac{(-1)^n x^{2n+1}}{2n+1!}\\
\]
Both sums on the right are mellin transformable, insuring the object on the left is mellin transformable. We don't need this entirely, I was mistaken--I thought the Gamma functions would pull out, but they actually cancel out... I think I have a rough idea of how to find square roots though, which look like this...