Understanding \(f(z,\theta) = e^{e^{i\theta}z} - 1\)

[JmsNxn]

OKAY!

We got the big boy guns proof right now!

I knew it was staring me right in the face!

Let:

\[
f_{\theta}^{\circ s}(z) = z\lim_{q\to\infty} \frac{d^{\frac{s}{q} + 2 \pi i s \frac{p}{q}}}{dx^{\frac{s}{q} + 2 \pi i s \frac{p}{q}}}\Big{|}_{x=0} e^{e^{i\theta q}x} + O(z^2)\\
\]

This produces a solution to cauchy's equation:

\[
\frac{\partial f_\theta^{\circ s}}{\partial \overline{z}} = 0\\
\]

Or rather that:

\[
\frac{d}{dz} f^{\circ s}_{\theta}(z)
\]

Always exists for \(|z| < \delta\) for small enough \(\delta>0\).

This works for \(\Im(\theta) > 0\)!!!!!! And proves this limit always converges!

This is a fairly deep referenced result, especially as I've set things up. But if a mellin/laplace/fourier transform, is holomorphic in a neighborhood of a point, then the integral converges. This is one of the biggest selling points of these transforms. If \(f_{\theta^*}(z)\) is holomorphic at \(z=0\) (it has a holomorphic derivative in a neighborhood), and additionally the holomorphic derivative converges to \(f_\theta(z)\), then the actual function converges under the transform. Point and Case is a paper on the Prime Number Theorem in recent years (https://people.mpim-bonn.mpg.de/zagier/f...lltext.pdf  LOOK SPECIFICALLY AT THE ANALYTIC THEOREM AND PROOF OF ANALYTIC THEOREM; we're in the same boat but with the mellin transform instead of Laplace, and we have to track another variable (which is easy as it looks like)). They use a result stating that if a function \(F(s)\) is holomorphic for \(\Re(s) > 0\), and is Laplace representable, then if it can be analytically continued to \(\Re(s) \ge 0\), then the transform converges on the line \(\Re(s) = 0\).

So if:

\[
H(y) = \int_{0}^\infty h(x)e^{-yx}\,dx\\
\]

And the integral converges for \(\Re(y) > 0\).... Assume that \(H(y)\) is analytically continuable to \(\Re(y) \ge 0\), so that at each point \(y_0\) where \(\Re(y_0) =0\), the function \(H(y)\) is holomorphic at \(y_0\). Then the following integral converges:

\[
H(y_0) = \int_0^\infty h(x) e^{-y_0x}\,dx\\
\]

Which is to say, if the extended Laplace transform is holomorphic on the whole boundary, then the integral converges at each point on the boundary.... Mellin is just fancy Laplace/ fancy Fourier.....

This is a statement that on the boundary, if there is a holomorphic function on the boundary, that at the boundary, not only does the function converge, it converges to the holomorphic boundary function.

So since \(f_{\theta^*}\) can be mellin transformed, and everything is holomorphic and perfect. And we can construct perfect holomorphic function extensions for \(f_\theta\); we must have that \(f_{\theta^*} \to f_\theta\), not only as holomorphic functions, but also underneath the integral. So that the integral expression converges.

Now, all we need to do this in this case, is to ensure that the derivative \(\frac{d}{dz}\) is a viable action; which solves all of our problems.

I'm being kind of vague right now, but I think this does it. And solves entirely the \(\theta^* \to \theta\) problem, at least for \(\Im(\theta) > 0\)--we might have problems for \(\Im(\theta) = 0\). But a similar work around should work.

[JmsNxn]

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.

[tommy1729]

Im reading your stuff James.

Not sure what to say , but I just want to let you know im reading so you dont feel lonely. lol.

regards

tommy1729

[JmsNxn]

Im reading your stuff James.

Not sure what to say , but I just want to let you know im reading so you dont feel lonely. lol.

regards

tommy1729

Thanks, TOMMY!

That means more than you think   I'm kind of note dumping right now. But I think I'm on to something, lmao!

Sincere regards, James