(02/16/2023, 10:28 AM)Caleb Wrote: I like the graphs, and I think this generalization into arbitrary heights is exciting! I'm really interested in the behaviour near the natural boundary, and in the region \(-1<\Re(s) < 0\), is it possible you could create some graphs in that region? I haven't really thought of a reasonable computational way to analytically continue the function there, do you have any thoughts?
*cue ptsd vietnam noises*
*ALSO, I may have fucked up some negative signs or some shit. The story is the same, though.*
We need a good asymptotic on \(A(x)\). So if we write:
\[
g(m) = (-1)^{m+1} \sum_{n\le m} \frac{\log(n)^{\frac{\log(m)}{\log(n)}}}{\frac{\log(m)}{\log(n)}!} \chi_m(n)\\
\]
And:
\[
A(x) = \sum_{m \le \lfloor x \rfloor} g(m)\\
\]
I can show that \(A(x) = o(x^{\epsilon})\). Which is a fancy way of saying it is some kind of logarithmic. The crazy thing about \(A\), is that this is a fairly fast bound. I cannot show this; and I think it's fields' medal shit to show this, but:
\[
A(x) = C\log(x) + o(1)\\
\]
Now, from this, we can extend \(\zeta_G(s)\) to \(\Re(s) > -1\). But it's really fucking tricky; and we've already hoped for something, that may not be true. But, I'm going to continue like everything works out perfect
We write:
\[
\zeta_G(s) = -s \int_1^\infty \left(A(x) - C\log(x)\right)x^{-s-1}\,dx - C s \int_1^\infty \log(x) x^{-s-1}\,dx \\
\]
The second function converges like:
\[
\begin{align}
C s \int_1^\infty \log(x) x^{-s-1}\,dx &= -Cs \frac{d}{ds} \int_1^\infty x^{-s-1}\,dx\\
&= Cs \frac{d}{ds} \frac{1}{s}\\
&= -\frac{C}{s}\\
\end{align}
\]
The first integral, is a bit trickier. We know that it's \(o(1)\) ( I can prove this). But I don't know HOW \(o(1)\) it is. I believe it's something like \(o(1) = O(\frac{1}{x})\) just from my own experiments. But I can't prove it. And "even though you see numerical evidence, doesn't make it true". Something every string theorist dark matter nerd should have shouted in their face!
But, for the moment we have:
\[
\zeta_G(s) = -s\int_1^\infty o(1)x^{-s-1}\,dx - \frac{C}{s}\\
\]
I can find the value \(C\) too. It's not that hard to find. But, it may not be a constant. It may just be:
\[
C = -0.235005249546161 + o(1)\\
\]
The general shape should definitely look like this; and I believe I can prove it. But for the moment this is pretty fucking accurate. I believe I am exact, but I may be off by some bullshit logarithmic decay garbage. Either way, this is still a very strong heuristic that these functions behave this way.
The function:
\[
B(s) = -s \int_1^\infty \left(A(x) - C\log(x)\right) x^{-s-1}\,dx\\
\]
Converges for \(\Re(s) > -1\). And it appears to happen pretty fucking efficiently. I can't prove it, but the numbers seem to support this hypothesis. Whereupon we write:
\[
\zeta_G(s) = B(s) - \frac{C}{s}\\
\]
Which is holomorphic for \(\Re(s) > -1\).
To cement my idea; if you take:
\[
A(x) - C\log(x) = h(x)\\
\]
Then \(h(1000) = 1E-16\). It already has 16 zeroes. So this is at least \(1/1000\) in the error. The function:
\[
A(x) - C\log(x) = O(1/x)\\
\]
Is perfectly fucking reasonable, and the numbers support it. If anything, this is the bare minimum. They support something more like \(O(1/x^2)\)--but I don't want to go there yet. So when I write:
\[
\zeta_G(s) = -s\int_1^\infty h(x)x^{-s-1}\,dx - \frac{C}{s}\\
\]
We can expect that \(h(x) = O(1/x)\), and therefore this entire expression converges for \(\Re(s) > -1\)--at least. It may be better
Please note that this is just a rough guess of what looks like should happen. It may be more difficult. And may be more of a headache to pull out a result of meromorphic for \(\Re(s) > -1\). But all of my research is saying: \(\zeta_G(s)\) is meromorphic for at least \(\Re(s) > -1\) with a simple pole at \(s=0\)....
To explain the value \(C\) isn't that hard either; I have written:
\[
\begin{align}
\frac{A(x)}{\log(x)} = C + o(1)\\
\lim_{x\to\infty} \frac{A(x)}{\log(x)} = C\\
\end{align}
\]
If I could prove this limit everything in this post is valid! Once this happens pretty much everything happens as consequence. Where we can expect the error term to be \(O(1/x)\)....
Here's a graph of \(\zeta_G(s)\) for \(0 \le s \le 3\):
I'd bet my left nut the residue is \(C\); and subtracting this from the function, produces holomorphy for \(\Re(s) > -1\)-- at least.
I CANNOT STRESS ENOUGH; THIS IS JUST WHAT MY CALCULATOR IS POINTING TOWARDS. I HAVE NOT PROVEN THIS YET!
If \(C\) is not a constant, then it looks like \(C(x) = C_0 + \frac{C_1}{\log(x)} + \frac{C_2}{\log(x)^2}\). The majority of my discussion is still fine. But we have to deep dive more. We can't treat \(C\) as a constant, we have to treat it as an "almost constant". Which, speaking honestly, would produce singularities at the boundary in a dense manner. Which would explain the boundary of singularities Caleb is seeing at \(\Re(s) = -1\).
I need to sleep now. But, I think we're on to something. I'll produce better code soon. I just want to debug and make sure everything works, with any input.
