Okay, so this post is a more refined study of the last. Let's look at:
\[
\vartheta[f_{\theta^*}^{\circ q}](x,z) = \sum_{k=0}^\infty f_{\theta^*}^{\circ qn}(z) \frac{x^n}{n!}\\
\]
Where we recall \(q = q(\theta^*)\), and is the minimal integer value such that \(f_{\theta^*}^{\circ q}(z) = \lambda z + O(z^2)\) for \(\lambda \in (0,1)\).
Now let's take the derivative in \(z\):
\[
\frac{d}{dz}\vartheta[f_{\theta^*}^{\circ q}](x,z) = \sum_{k=0}^\infty \left(\frac{d}{dz}f_{\theta^*}^{\circ qn}(z) \right)\frac{x^n}{n!}\\
\]
We can show that under the integral:
\[
f_{\theta^*}^{\circ s}(z) = f_{\theta^*}^{\circ s}(z_0) + (z-z_0) \frac{d^{\frac{s}{q} + 2\pi i \frac{p}{q}}}{dx^{\frac{s}{q} + 2\pi i \frac{p}{q}}}\Big{|}_{x=0} \frac{\partial}{\partial z} \vartheta[f_{\theta^*}^{\circ q}](x,z_0) + O(z-z_0)^2\\
\]
Where, this reduction is done as:
\[
\begin{align}
f_{\theta^*}^{\circ s}(z) &= \frac{d^{\frac{s}{q} + 2\pi i \frac{p}{q}}}{dx^{\frac{s}{q} + 2\pi i \frac{p}{q}}}\Big{|}_{x=0} \vartheta[f_{\theta^*}^{\circ q}](x,z)\\
&=\frac{d^{\frac{s}{q} + 2\pi i \frac{p}{q}}}{dx^{\frac{s}{q} + 2\pi i \frac{p}{q}}}\Big{|}_{x=0}\left(\vartheta[f_{\theta^*}^{\circ q}](x,z_0) + \frac{\partial}{\partial z}\vartheta[f_{\theta^*}^{\circ q}](x,z_0) + O(z-z_0)^2\right)\\
&= f_{\theta^*}^{\circ s}(z_0) + \frac{\partial}{\partial z}f_{\theta^*}^{\circ s}(s,z_0)(z-z_0) + O(z-z_0)^2\\
\end{align}
\]
Now, we can show that \(f_{\theta^*}^{\circ s}(z_0)\) converges uniformly as \(\theta \to \theta^*\), we can also show that \(\frac{\partial}{\partial z}f_{\theta^*}^{\circ s}(z_0)\) converges similarly. This is enough to say that \(\frac{d}{dz} f_{\theta^*}^{\circ s}(z)\) converges uniformly, and therefore \(f_{\theta}^{\circ s}(z)\) is a holomorphic function. Additionally it shows that the integral expression converges, with confirmation through the Newman method. If you can take a complex derivative of an object, you can state it's holomorphic. So the convergence of the complex derivative is enough to ensure holomorphy.
If:
\[
f_n(z) \to f(z)
\]
Where \(f_n\) and \(f\) are holomorphic--but this convergence is pointwise. And:
\[
f'_n(z) \to f'(z)\\
\]
But this convergence is pointwise, then:
Then \(f_n \to f\) uniformly; which is an odd consequence of cauchy's theorem. And since we have uniform convergence, we have that the convergence under the integral is uniform.... the \(\vartheta\)'s converge under the integral uniformly.
This allows us to write, for \(0 < \lambda < 1\):
\[
\lim_{m\to\infty} \frac{d^{\frac{s}{m}}}{dx^{\frac{s}{m}}} e^{\lambda^m x} = \lambda^s\\
\]
In a much much more general setting....
Where if \(\theta^* = 2 \pi \frac{p}{q} + it\), that:
\[
\frac{d^{\frac{s}{q} + 2 \pi i \frac{p}{q}}}{dx^{\frac{s}{q} + 2 \pi i \frac{p}{q}}} e^{e^{i\theta^*} x} = e^{i\theta^* s}e^{e^{i\theta^*}x}\\
\]
And that, as \(q\to \infty\) with \(\theta^* \to \theta\):
\[
\lim_{q\to\infty} \frac{d^{\frac{s}{q} + 2 \pi i \frac{p}{q}}}{dx^{\frac{s}{q} + 2 \pi i \frac{p}{q}}} e^{e^{i\theta^*} x} = e^{i\theta s}\\
\]
As we morph this, by moving \(z\), we get normal families, and we can be sure that the limit converges.. Since the limit converges, we can be sure the function underneath the integral converges. Thanks to Newman's proof about the Laplace transform. (Again, through a change of variables, Laplace is Mellin.)
I think playing a little game with phonics can help here. Lets write:
\[
e^{f_{\theta^*}^{\circ q}x} = \sum_{n=0}^\infty f_{\theta^*}^{\circ qn}(z) \frac{x^n}{n!}\\
\]
Where: \(f_{\theta^*}^{\circ q}(z) = \lambda z + O(z^2)\), where \(\lambda \in (0,1)\). The differintegral, exactly as Ramanujan, Riemann, Feynman, Schrodinger (etc etc) wrote it, is that:
\[
\frac{d^s}{dx^s} e^{f_{\theta^*}^{\circ q}x} = f_{\theta^*}^{\circ qs}(z) = F(s,z)\\
\]
For some solution of the equation:
\[
F(n) = f_{\theta^*}^{\circ qn}(z)\\
\]
Where:
\[
F(s,z) = \lambda^{qs}z + O(z^2)\\
\]
Where \(F(s_0,F(s_1,z)) = F(s_0+s_1,z)\). Our job is to modify the differintegral, and manage the variable changes, so that:
\[
F(\ell(s),z) = e^{i\theta^* s}z + O(z^2)\\
\]
Which is no more than turning \(\lambda^{qs}\) into \(e^{i\theta^*s}\), using nothing more than a linear transformation \(\ell\)... which means \(\ell(s) = cs\), for specific \(c \in \mathbb{C}\)... But as \(\theta^* \to \theta\) (as we approach irrational values using rational values). We can expect \(c\to 0\), and a weird limit on the integral..l
Regards, James.
\[
\vartheta[f_{\theta^*}^{\circ q}](x,z) = \sum_{k=0}^\infty f_{\theta^*}^{\circ qn}(z) \frac{x^n}{n!}\\
\]
Where we recall \(q = q(\theta^*)\), and is the minimal integer value such that \(f_{\theta^*}^{\circ q}(z) = \lambda z + O(z^2)\) for \(\lambda \in (0,1)\).
Now let's take the derivative in \(z\):
\[
\frac{d}{dz}\vartheta[f_{\theta^*}^{\circ q}](x,z) = \sum_{k=0}^\infty \left(\frac{d}{dz}f_{\theta^*}^{\circ qn}(z) \right)\frac{x^n}{n!}\\
\]
We can show that under the integral:
\[
f_{\theta^*}^{\circ s}(z) = f_{\theta^*}^{\circ s}(z_0) + (z-z_0) \frac{d^{\frac{s}{q} + 2\pi i \frac{p}{q}}}{dx^{\frac{s}{q} + 2\pi i \frac{p}{q}}}\Big{|}_{x=0} \frac{\partial}{\partial z} \vartheta[f_{\theta^*}^{\circ q}](x,z_0) + O(z-z_0)^2\\
\]
Where, this reduction is done as:
\[
\begin{align}
f_{\theta^*}^{\circ s}(z) &= \frac{d^{\frac{s}{q} + 2\pi i \frac{p}{q}}}{dx^{\frac{s}{q} + 2\pi i \frac{p}{q}}}\Big{|}_{x=0} \vartheta[f_{\theta^*}^{\circ q}](x,z)\\
&=\frac{d^{\frac{s}{q} + 2\pi i \frac{p}{q}}}{dx^{\frac{s}{q} + 2\pi i \frac{p}{q}}}\Big{|}_{x=0}\left(\vartheta[f_{\theta^*}^{\circ q}](x,z_0) + \frac{\partial}{\partial z}\vartheta[f_{\theta^*}^{\circ q}](x,z_0) + O(z-z_0)^2\right)\\
&= f_{\theta^*}^{\circ s}(z_0) + \frac{\partial}{\partial z}f_{\theta^*}^{\circ s}(s,z_0)(z-z_0) + O(z-z_0)^2\\
\end{align}
\]
Now, we can show that \(f_{\theta^*}^{\circ s}(z_0)\) converges uniformly as \(\theta \to \theta^*\), we can also show that \(\frac{\partial}{\partial z}f_{\theta^*}^{\circ s}(z_0)\) converges similarly. This is enough to say that \(\frac{d}{dz} f_{\theta^*}^{\circ s}(z)\) converges uniformly, and therefore \(f_{\theta}^{\circ s}(z)\) is a holomorphic function. Additionally it shows that the integral expression converges, with confirmation through the Newman method. If you can take a complex derivative of an object, you can state it's holomorphic. So the convergence of the complex derivative is enough to ensure holomorphy.
If:
\[
f_n(z) \to f(z)
\]
Where \(f_n\) and \(f\) are holomorphic--but this convergence is pointwise. And:
\[
f'_n(z) \to f'(z)\\
\]
But this convergence is pointwise, then:
Then \(f_n \to f\) uniformly; which is an odd consequence of cauchy's theorem. And since we have uniform convergence, we have that the convergence under the integral is uniform.... the \(\vartheta\)'s converge under the integral uniformly.
This allows us to write, for \(0 < \lambda < 1\):
\[
\lim_{m\to\infty} \frac{d^{\frac{s}{m}}}{dx^{\frac{s}{m}}} e^{\lambda^m x} = \lambda^s\\
\]
In a much much more general setting....
Where if \(\theta^* = 2 \pi \frac{p}{q} + it\), that:
\[
\frac{d^{\frac{s}{q} + 2 \pi i \frac{p}{q}}}{dx^{\frac{s}{q} + 2 \pi i \frac{p}{q}}} e^{e^{i\theta^*} x} = e^{i\theta^* s}e^{e^{i\theta^*}x}\\
\]
And that, as \(q\to \infty\) with \(\theta^* \to \theta\):
\[
\lim_{q\to\infty} \frac{d^{\frac{s}{q} + 2 \pi i \frac{p}{q}}}{dx^{\frac{s}{q} + 2 \pi i \frac{p}{q}}} e^{e^{i\theta^*} x} = e^{i\theta s}\\
\]
As we morph this, by moving \(z\), we get normal families, and we can be sure that the limit converges.. Since the limit converges, we can be sure the function underneath the integral converges. Thanks to Newman's proof about the Laplace transform. (Again, through a change of variables, Laplace is Mellin.)
I think playing a little game with phonics can help here. Lets write:
\[
e^{f_{\theta^*}^{\circ q}x} = \sum_{n=0}^\infty f_{\theta^*}^{\circ qn}(z) \frac{x^n}{n!}\\
\]
Where: \(f_{\theta^*}^{\circ q}(z) = \lambda z + O(z^2)\), where \(\lambda \in (0,1)\). The differintegral, exactly as Ramanujan, Riemann, Feynman, Schrodinger (etc etc) wrote it, is that:
\[
\frac{d^s}{dx^s} e^{f_{\theta^*}^{\circ q}x} = f_{\theta^*}^{\circ qs}(z) = F(s,z)\\
\]
For some solution of the equation:
\[
F(n) = f_{\theta^*}^{\circ qn}(z)\\
\]
Where:
\[
F(s,z) = \lambda^{qs}z + O(z^2)\\
\]
Where \(F(s_0,F(s_1,z)) = F(s_0+s_1,z)\). Our job is to modify the differintegral, and manage the variable changes, so that:
\[
F(\ell(s),z) = e^{i\theta^* s}z + O(z^2)\\
\]
Which is no more than turning \(\lambda^{qs}\) into \(e^{i\theta^*s}\), using nothing more than a linear transformation \(\ell\)... which means \(\ell(s) = cs\), for specific \(c \in \mathbb{C}\)... But as \(\theta^* \to \theta\) (as we approach irrational values using rational values). We can expect \(c\to 0\), and a weird limit on the integral..l
Regards, James.