(03/14/2023, 07:51 AM)Caleb Wrote: Yes sorry-- I will shut up for now to let you work on the paper. I'm super excited to read it, so I'll bide my time doing something else for a few days. Good luck on the writing-- I can't wait to read it!
Thanks again for your work,
Caleb
Don't be sorry! This is some of the weirdest math I've ever seen! You have sparked Ramanujan's identity:
\[
-\sum_{k=1}^\infty f(-k)z^{-k} = \sum_{k=1}^\infty f(k) z^k\\
\]
Where:
\[
\begin{align*}
\sum_{k=1}^\infty f(k) z^k &= c(z) - c(0)\\
-\sum_{k=1}^\infty f(-k)z^{k} &= c(1/z) - c(\infty)\\
\end{align*}
\]
You typically prove this by showing that \(f(k)\) is subject to Carlson's Theorem for \(k \in \mathbb{C}\).
We are dodging all of these requirements! I don't quite get how, yet. But we are avoiding this entirely!
My suspicion is because we are writing this as a "zeta" function, because we are writing it as a divisor sum; rather than a normal sum! So, some "zeta-like" sums pop out. And the average zeta-function is subject to Carlson's theorem. But this still doesn't make sense! Either way though; these functions satisfy Ramanujan's identity.
Beautiful work, Caleb! Absolutely fucking beautiful!

