02/10/2023, 07:25 AM
(02/10/2023, 07:09 AM)Caleb Wrote: Damn this is cool-- zeroes on the critical line would be super exciting! Also, are you familiar with the argument prinicipal in complex analysis? We could use something like that to fairly precisely locate some zeroes of the function, which gives us a computational way to check \( \zeta_G \) is actually zero at the locations we think its zero. Anyways, please share the graphs you come up with in the coming days, I'm excited to see what you come up with!
I am familiar with the argument principle, but I'm not proficient with it. I mean, I can prove it, and understand it; but I can't use it in my own proofs; because I don't reaaaaaally get it. As far as I see--just going off the graphs; the zeroes seem to be exactly at \(\Re(s) = 1/2\). So I'm just eyeballing it. A good way to test--as I see it. Is to take:
\[
\frac{1}{\zeta_G(s)} = \sum_{m=1}^\infty g^{\mu}(m) m^{-s}\\
\]
And check if:
\[
A^{\mu}(x) = \sum_{1\le m \le \lfloor x \rfloor} g^{\mu}(m) = O(\sqrt{x})\\
\]
And that this is a tight bound. This proves all the zeroes are less than \(\Re(s) = 1/2\).
........
I think I can find a reflection formula too, I'm not sure yet. But I believe; at least on the critical strip; there should be something like \(\zeta_G(1-s) = G(s) \zeta_G(s)\)--but this is still just a hunch. If you have that, and the growth conditions on \(A^\mu\)--we have that all the zeroes are on \(\Re(s) = 1/2\). And this is just a mock up of the Riemann hypothesis; lmfao.
