To begin, I'll present the first non obvious theorem, for this; I will start with \(f^{\circ 2}_\pi(z)\). So if \(f_\pi(z) = e^{-z}-1\), \(f^{\circ 2}_\pi = f_\pi(f_\pi(z))\) And by such a manner, I will deconstruct the function \(\vartheta_\pi\). As we begin we have:
\[
H_\pi(x,z) = \sum_{n=0}^\infty f^{\circ 2(n+1)}_\pi(z) \frac{(-x)^n}{n!}
\]
Now, as noted above, there's an abel function for \(f^{\circ 2}_\pi\), simply by nature of it having a neutral fixed point at \(z=0\), and has a multiplier \(\lambda = 1\). And therefore it has an iteration \((f^{\circ 2})^{\circ s}_\pi\). Which we can write as:
\[
\Gamma(1-s) (f^{\circ 2})^{\circ s}_\pi(z) = \int_0^\infty H_\pi(x,z)x^{s-1}\,dx
\]
We write \((f^{\circ 2})^{\circ s}_\pi\), because this powering operation is non-associative. The value \((f^{\circ 2})^{\circ 1/2}_\pi \neq f\). Instead, we solely have:
\[
\begin{align}
F(s,z) &=(f^{\circ 2})^{\circ s}_\pi(z)\\
F(1,z) &= f^{\circ 2}_\pi(z)\\
F(s+y,z) &= F(s,F(y,z))
\end{align}
\]
We now write, our first alteration, by which we wish to use this expansion to get the true \(f^{\circ s}_\pi\). We begin this by writing an integral transform:
\[
Q(x,z) = \frac{1}{2\pi i} \int_{-\frac{1}{2}-i\infty}^{-\frac{1}{2} + i \infty} \frac{\Gamma(1-s)\Gamma(s)}{\Gamma(1-2s)} (f^{\circ 2})^{\circ s}_\pi(z)x^{-2s}\,ds
\]
IF (Now this is a big if, which I plan on justifying, but momentarily, follow me--it is a slow convergence but converges upon massaging) this object converges it converges exactly to:
\[
Q(x,z) = \sum_{n=1}^\infty f^{\circ 2(n+1)}_\pi(z) \frac{(-1)^nx^{2n}}{(2n!)}
\]
But this is precisely:
\[
\frac{\vartheta_\pi(ix,z) + \vartheta_\pi(-ix,z)}{2} - f_\pi(z)= Q(x,z)
\]
Whereby this is mellin transformable--and puts us exactly at the point where the mellin transform of \(\vartheta_\pi\) can be found, up to a multiplicative factor. Explaining this is pretty advanced, so I'm going to start with an example every single one of you have seen--though perhaps not in this late staged format.
TRIGGER WARNING: This puts us right next door to Hard Analytic Number Theory, and additionally the work used to make the Riemann Hypothesis. I apologize, I was always one of those hubris young kids/nerds who studied everything in and around the Riemann hypothesis--by such, I've developed much of the same tools. I'm a nerd for this shit. We're like 1/2 half way before relating Hard Analytic Number Theory to complex dynamics.
----------------------------------------------------------------
Let's begin by writing:
\[
e^{-x} = \sum_{n=0}^\infty \frac{(-x)^n}{n!}
\]
This when applied under the mellin transform:
\[
\Gamma(1-s) = \int_0^\infty e^{-x}x^{-s}\,dx
\]
Now first of all, we know that:
\[
h(s) = 1
\]
Is a bounded function on \(\mathbb{C}_{\Re(s) > 0}\). (Like duh--but note in this proof we only use that it is bounded in this half plane, that's all we care about).
So we can instead write this as:
\[
\Gamma(1-s)h(s) = \int_0^\infty e^{-x}x^{-s}\,dx
\]
Now let's write:
\[
q(x) = \frac{1}{2\pi i} \int_{-\frac{1}{2}-i\infty}^{-\frac{1}{2}+i\infty} \frac{\Gamma(1-s)\Gamma(s)}{\Gamma(1-2s)}x^{-s}\,ds
\]
We can start, by proving this converges. It's not as hard as you might ask:
\[
\Gamma(x+iy) = O(e^{ -\frac{\pi}{2} |y|} |y|^{x-1/2})
\]
And this result holds asymptotically if the constant of proportion is chosen correctly. Whereby, we can also say:
\[
\frac{1}{\Gamma(x+iy)} = O(e^{ \frac{\pi}{2} |y|} |y|^{-x+1/2})
\]
Therefore:
\[
\frac{\Gamma(1-x-iy)\Gamma(x+iy)}{\Gamma(1-2x-2iy)} = O(\frac{|y|^{x-1/2}|y|^{1-x-1/2}}{|y|^{1-2x-1/2}}) = O(|y|^{2x - 1/2})
\]
We have set \(\Re(s) = x = -1/2\) in this integral, thereby.
\[
\frac{\Gamma(1-s)\Gamma(s)}{\Gamma(1-2s)} = O(|y|^{-\frac{3}{2}})
\]
.......................................
Now what does this object converge to?
\[
q(x) = \cos(\sqrt{x}) -1 = \sum_{n=1}^\infty \frac{(-1)^nx^n}{2n!}
\]
Where, as we noted above:
\[
q(x^2) = \frac{e^{ix} + e^{-ix}}{2} - 1 = \sum_{n=1}^\infty \frac{(-1)^n x^{2n}}{2n!}
\]
Now, when we take, for \(-1 <\Re(s) < 0\):
\[
\int_0^\infty q(x)x^{s-1}\,dx = \frac{\Gamma(1-s)\Gamma(s)}{\Gamma(1-2s)}
\]
Which through plain substitution:
\[
\int_0^\infty \left( \cos(u) - 1\right)u^{s-1}\,du = \cos(s\frac{\pi}{2})\Gamma(s)
\]
Where this is the duplication formula of the Gamma function in action--though with no obvious mention of anything of Gauss. We attribute this to Legendre, but to see the Cosine function, it's helpful to look at the general case by Gauss.
I am simply trying to do the same thing, but instead of the transformation with \(1 \to -1\), I am trying to write \(e^z -1 \to e^{-z}-1\). Surprisingly, under these mellin transforms, this is always possible. Though, the real trick becomes "How we take the mellin transform".
Anyway, that's all you'll have from me tonight. I'm going to use this thread as a note dump primarily; and I plan to expand on my previous rough paper. I don't have the time at the moment to develop these arguments fully, but numbers don't lie![[Image: biggrin.gif]](https://math.eretrandre.org/tetrationforum/images/smilies/biggrin.gif)
'
I SCREWED UP SOME OF THE INDICES IN THIS POST
\[
H_\pi(x,z) = \sum_{n=0}^\infty f^{\circ 2(n+1)}_\pi(z) \frac{(-x)^n}{n!}
\]
Now, as noted above, there's an abel function for \(f^{\circ 2}_\pi\), simply by nature of it having a neutral fixed point at \(z=0\), and has a multiplier \(\lambda = 1\). And therefore it has an iteration \((f^{\circ 2})^{\circ s}_\pi\). Which we can write as:
\[
\Gamma(1-s) (f^{\circ 2})^{\circ s}_\pi(z) = \int_0^\infty H_\pi(x,z)x^{s-1}\,dx
\]
We write \((f^{\circ 2})^{\circ s}_\pi\), because this powering operation is non-associative. The value \((f^{\circ 2})^{\circ 1/2}_\pi \neq f\). Instead, we solely have:
\[
\begin{align}
F(s,z) &=(f^{\circ 2})^{\circ s}_\pi(z)\\
F(1,z) &= f^{\circ 2}_\pi(z)\\
F(s+y,z) &= F(s,F(y,z))
\end{align}
\]
We now write, our first alteration, by which we wish to use this expansion to get the true \(f^{\circ s}_\pi\). We begin this by writing an integral transform:
\[
Q(x,z) = \frac{1}{2\pi i} \int_{-\frac{1}{2}-i\infty}^{-\frac{1}{2} + i \infty} \frac{\Gamma(1-s)\Gamma(s)}{\Gamma(1-2s)} (f^{\circ 2})^{\circ s}_\pi(z)x^{-2s}\,ds
\]
IF (Now this is a big if, which I plan on justifying, but momentarily, follow me--it is a slow convergence but converges upon massaging) this object converges it converges exactly to:
\[
Q(x,z) = \sum_{n=1}^\infty f^{\circ 2(n+1)}_\pi(z) \frac{(-1)^nx^{2n}}{(2n!)}
\]
But this is precisely:
\[
\frac{\vartheta_\pi(ix,z) + \vartheta_\pi(-ix,z)}{2} - f_\pi(z)= Q(x,z)
\]
Whereby this is mellin transformable--and puts us exactly at the point where the mellin transform of \(\vartheta_\pi\) can be found, up to a multiplicative factor. Explaining this is pretty advanced, so I'm going to start with an example every single one of you have seen--though perhaps not in this late staged format.
TRIGGER WARNING: This puts us right next door to Hard Analytic Number Theory, and additionally the work used to make the Riemann Hypothesis. I apologize, I was always one of those hubris young kids/nerds who studied everything in and around the Riemann hypothesis--by such, I've developed much of the same tools. I'm a nerd for this shit. We're like 1/2 half way before relating Hard Analytic Number Theory to complex dynamics.
----------------------------------------------------------------
Let's begin by writing:
\[
e^{-x} = \sum_{n=0}^\infty \frac{(-x)^n}{n!}
\]
This when applied under the mellin transform:
\[
\Gamma(1-s) = \int_0^\infty e^{-x}x^{-s}\,dx
\]
Now first of all, we know that:
\[
h(s) = 1
\]
Is a bounded function on \(\mathbb{C}_{\Re(s) > 0}\). (Like duh--but note in this proof we only use that it is bounded in this half plane, that's all we care about).
So we can instead write this as:
\[
\Gamma(1-s)h(s) = \int_0^\infty e^{-x}x^{-s}\,dx
\]
Now let's write:
\[
q(x) = \frac{1}{2\pi i} \int_{-\frac{1}{2}-i\infty}^{-\frac{1}{2}+i\infty} \frac{\Gamma(1-s)\Gamma(s)}{\Gamma(1-2s)}x^{-s}\,ds
\]
We can start, by proving this converges. It's not as hard as you might ask:
\[
\Gamma(x+iy) = O(e^{ -\frac{\pi}{2} |y|} |y|^{x-1/2})
\]
And this result holds asymptotically if the constant of proportion is chosen correctly. Whereby, we can also say:
\[
\frac{1}{\Gamma(x+iy)} = O(e^{ \frac{\pi}{2} |y|} |y|^{-x+1/2})
\]
Therefore:
\[
\frac{\Gamma(1-x-iy)\Gamma(x+iy)}{\Gamma(1-2x-2iy)} = O(\frac{|y|^{x-1/2}|y|^{1-x-1/2}}{|y|^{1-2x-1/2}}) = O(|y|^{2x - 1/2})
\]
We have set \(\Re(s) = x = -1/2\) in this integral, thereby.
\[
\frac{\Gamma(1-s)\Gamma(s)}{\Gamma(1-2s)} = O(|y|^{-\frac{3}{2}})
\]
.......................................
Now what does this object converge to?
\[
q(x) = \cos(\sqrt{x}) -1 = \sum_{n=1}^\infty \frac{(-1)^nx^n}{2n!}
\]
Where, as we noted above:
\[
q(x^2) = \frac{e^{ix} + e^{-ix}}{2} - 1 = \sum_{n=1}^\infty \frac{(-1)^n x^{2n}}{2n!}
\]
Now, when we take, for \(-1 <\Re(s) < 0\):
\[
\int_0^\infty q(x)x^{s-1}\,dx = \frac{\Gamma(1-s)\Gamma(s)}{\Gamma(1-2s)}
\]
Which through plain substitution:
\[
\int_0^\infty \left( \cos(u) - 1\right)u^{s-1}\,du = \cos(s\frac{\pi}{2})\Gamma(s)
\]
Where this is the duplication formula of the Gamma function in action--though with no obvious mention of anything of Gauss. We attribute this to Legendre, but to see the Cosine function, it's helpful to look at the general case by Gauss.
I am simply trying to do the same thing, but instead of the transformation with \(1 \to -1\), I am trying to write \(e^z -1 \to e^{-z}-1\). Surprisingly, under these mellin transforms, this is always possible. Though, the real trick becomes "How we take the mellin transform".
Anyway, that's all you'll have from me tonight. I'm going to use this thread as a note dump primarily; and I plan to expand on my previous rough paper. I don't have the time at the moment to develop these arguments fully, but numbers don't lie
'I SCREWED UP SOME OF THE INDICES IN THIS POST