[MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n))
#18
I AM GOING TO DO SOMETHING I DONT KNOW THE IMPORTANCE OF, JUST FOR CALEB! LISTEN!


We are going to define the Dirichlet \(\eta\) function, as the function:

\[
\eta(s) = \sum_{n=1}^\infty (-1)^{n-1} n^{-s}\\
\]

This function is holomorphic for \(\Re(s) > 0\). Additionally, we know that it is \(O(s)\) bounded for \(\Re(s) > 0\). Which means, there exists a constant \(C_\delta >0\) such that \(|\eta(s)| < C_\delta |s|\) for \(\Re(s) > \delta\) when \(\delta>0\).

Now your first instinct might tell you to "fractionally differentiate" \(\eta\). But that's wrong. We instead take:

\[
\vartheta(w) = \sum_{n=0}^\infty \eta(1+n) \frac{(-w)^n}{n!} = \frac{1}{2\pi i} \int_{1/2-i\infty}^{1/2+i\infty} \Gamma(s) \eta(1-s) w^{-s}\,ds\\
\]

Where:

\[
\frac{d^{s-1}}{dw^{s-1}}\Big{|}_{w=0} \vartheta(w) = \eta(s)\\
\]



Here is where advanced number theory pokes its head. I am just trying to describe the problem, I am not trying to solve it. For you to describe singularities within the function \(\eta(\infty)\); within these fucking crazy beautiful residues your looking at.... You're solving very deep problems that I don't think you are aware of; and that's why I'm doubtful of some of the shenanigans you're doing.

BUT BY GOD IF YOU'RE NOT AT LEAST ON TO SOMETHING!

Dirichlet's function \(\eta(s)\) is meromorphic on \(\mathbb{C}\)......... Let's just move some integrals around and pull out some residues  Tongue  Tongue  Tongue  Tongue  Tongue



This is true and unique and Hilbert Special and fucking JON VON NEUMANNN UNIQUE so long as we assume that these functions are "Ramanujan bounded". Or we subject ourselves to advanced fourier theory. CARLSON WON A FIELDS MEDAL FOR FINDING THE FOURIER RIGOR VERSION OF RAMANUJAN'S RAW.

So I Fucking love this Caleb! But just know, the fucking devil's in the details. And you're going to have to go case by case; because this relates to highly advanced analytic number theory.

Fucking awesome, regards, James

EDIT:

Forgot to mention; all it takes to turn your integrals into my integrals is a variable change. I mean,

\[
\begin{align}
\frac{d^s}{dw^s}\big{|}_{w=0}f(w^2) &= \frac{1}{\Gamma(-s)} \int_{\gamma} f(w^2) w^{-s-1}\,dw\\
&=\frac{1}{2\Gamma(-s)} \int_{\gamma} f(u) u^{-(s+1)/2-1/2}\,du\\
&= \frac{\Gamma(-s/2)}{2\Gamma(-s)} \frac{d^{s/2}}{dw^{s/2}}\big{|}_{w=0}f(w)
\end{align}
\]

I'm still just rapidly posting these things but fuck this is beautiful.
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RE: [MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n)) - by JmsNxn - 04/06/2023, 05:30 PM



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