I fixed a small error in the post above; I'm correcting it here....
We are going to do another change of variables now. I wasn't sure to show this off yet; but I like it:
\[
\nu(s) = \sum_{m=1}^\infty (-1)^{m+1}\sum_{n\le m} \chi_m(n)m^{-s}\\
\]
Now let's interchange this sum so that \(n\) is first...
\[
\sum_{m=1}^\infty \sum_{n\le m} (-1)^{m+1}\chi_m(n)m^{-s} = \sum_{n=1}^\infty \sum_{m=n}^\infty (-1)^{m+1}\chi_m(n)m^{-s}\\
\]
The sum:
\[
\sum_{m=n}^\infty \chi_m(n)m^{-s} = \sum_{k=1}^\infty n^{-ks}\\
\]
And the ACTUAL EXPANSION OF \(\nu\) is:
\[
\nu(s) = \sum_{n=1}^\infty (-1)^{n+1} \frac{n^{-s}}{1-n^{-s}}\\
\]
I forgot a \(n^{-s}\) before, I apologize. It doesn't change anything about the results provided; just for future work; I had a typo
This is the stupid result that I remembered. And why I even started a lot of this conversation. I knew how it worked. I've seen this formula before; and I was getting deja vu. I still don't remember which mathematician proved this. But I know I've seen this before. Someone pull out the ChatGPT or Google BARD to find out. lmao
I know these stupid types of "zeta" functions have a name. Where Gottfried's function would be of this type. They are all written in a similar manner.
My solution is still a considerable speed up for Gottfried and \(\nu\). Gottfried's expansion for \(\zeta_G\) and \(\nu\) are good when we're away from problem values. The zeta sums are more uniform with their convergence speed. More regular so to speak. So problem values aren't really problem values; but fast convergence areas, are slower. My code is much slower; but this is because I'm able to grab super high precision using different code. I can write 3-digit precision if you'd like; and my code would technically be faster. Just based off of how simple the zeta functions are.
Anyway; any questions, Caleb; keep asking them. Happy to answer ten times over!!!!!
The function \(\nu\) compiles much faster. So I'm posting these graphs first.
This is \(0 < \Re(s) < 1\) and \(0 < \Im(s) < 8\) (the critical strip). This is \(\nu\) WHICH HAS ZEROES EVERYWHERE!!!!! You can see them popping up near the boundary of the critical strip. They look like blackholes. If you see a "whitehole" it's a singularity--no white holes here though.
The second graph I am showing is of \(0 < \Re(s) < 5\) and \(-2.5 < \Im(s) < 2.5\).
Which shows just how beautiful, tame, well behaved, calm, this function is:
....... Therefore \(Q(x) = O(\log(x))\)
EDIT: Also just to be transparent; the graphs of this function are done through the pari calculator as:
\[
\nu(s) = \sum_{m=1}^{300} q(m)m^{-s} + O(300^{-s})\\
\]
Where the \(O\) time is about \(O(N^{\epsilon})\) but it can blow up, lol. But these are still ridiculously accurate results. You won't see it in the graphs. But the first five digits are found for every value. Where at worst we border 4 digits. So, my code is super slow; but super accurate...Which has always been my achilles heel, lmao.
EDIT:
So here is a graph of \(\zeta_G(s)\) on the domain \(0 \le \Re(s) \le 1\) and \(0 \le \Im(s) \le 8\):
It looks like the zeroes are near \(\Re(s) = 1/2\), but they actually move towards \(\Re(s) =0\) as \(t \to \infty\). So I spoke too soon. The values \(\zeta_G(s_0) = 0\) slowly move to \(\sigma_0 \to 0\). They aren't fixed like the Riemann hypothesis. I apologize, I got too excited.
I think the correct answer is \(\zeta_G(s) \neq 0\) for \(\Re(s) > 1/2\)--and this is the best we're going to get....
This means a reflection formula is very unlikely. And means that the Gottfried sum, is the maximal analytic continuation. So that Caleb's comments on Polya are correct. There is probably a natural boundary. Good job
We are going to do another change of variables now. I wasn't sure to show this off yet; but I like it:
\[
\nu(s) = \sum_{m=1}^\infty (-1)^{m+1}\sum_{n\le m} \chi_m(n)m^{-s}\\
\]
Now let's interchange this sum so that \(n\) is first...
\[
\sum_{m=1}^\infty \sum_{n\le m} (-1)^{m+1}\chi_m(n)m^{-s} = \sum_{n=1}^\infty \sum_{m=n}^\infty (-1)^{m+1}\chi_m(n)m^{-s}\\
\]
The sum:
\[
\sum_{m=n}^\infty \chi_m(n)m^{-s} = \sum_{k=1}^\infty n^{-ks}\\
\]
And the ACTUAL EXPANSION OF \(\nu\) is:
\[
\nu(s) = \sum_{n=1}^\infty (-1)^{n+1} \frac{n^{-s}}{1-n^{-s}}\\
\]
I forgot a \(n^{-s}\) before, I apologize. It doesn't change anything about the results provided; just for future work; I had a typo
This is the stupid result that I remembered. And why I even started a lot of this conversation. I knew how it worked. I've seen this formula before; and I was getting deja vu. I still don't remember which mathematician proved this. But I know I've seen this before. Someone pull out the ChatGPT or Google BARD to find out. lmao
I know these stupid types of "zeta" functions have a name. Where Gottfried's function would be of this type. They are all written in a similar manner.
My solution is still a considerable speed up for Gottfried and \(\nu\). Gottfried's expansion for \(\zeta_G\) and \(\nu\) are good when we're away from problem values. The zeta sums are more uniform with their convergence speed. More regular so to speak. So problem values aren't really problem values; but fast convergence areas, are slower. My code is much slower; but this is because I'm able to grab super high precision using different code. I can write 3-digit precision if you'd like; and my code would technically be faster. Just based off of how simple the zeta functions are.
Anyway; any questions, Caleb; keep asking them. Happy to answer ten times over!!!!!
The function \(\nu\) compiles much faster. So I'm posting these graphs first.
This is \(0 < \Re(s) < 1\) and \(0 < \Im(s) < 8\) (the critical strip). This is \(\nu\) WHICH HAS ZEROES EVERYWHERE!!!!! You can see them popping up near the boundary of the critical strip. They look like blackholes. If you see a "whitehole" it's a singularity--no white holes here though.
The second graph I am showing is of \(0 < \Re(s) < 5\) and \(-2.5 < \Im(s) < 2.5\).
Which shows just how beautiful, tame, well behaved, calm, this function is:
....... Therefore \(Q(x) = O(\log(x))\)
EDIT: Also just to be transparent; the graphs of this function are done through the pari calculator as:
\[
\nu(s) = \sum_{m=1}^{300} q(m)m^{-s} + O(300^{-s})\\
\]
Where the \(O\) time is about \(O(N^{\epsilon})\) but it can blow up, lol. But these are still ridiculously accurate results. You won't see it in the graphs. But the first five digits are found for every value. Where at worst we border 4 digits. So, my code is super slow; but super accurate...Which has always been my achilles heel, lmao.
EDIT:
So here is a graph of \(\zeta_G(s)\) on the domain \(0 \le \Re(s) \le 1\) and \(0 \le \Im(s) \le 8\):
It looks like the zeroes are near \(\Re(s) = 1/2\), but they actually move towards \(\Re(s) =0\) as \(t \to \infty\). So I spoke too soon. The values \(\zeta_G(s_0) = 0\) slowly move to \(\sigma_0 \to 0\). They aren't fixed like the Riemann hypothesis. I apologize, I got too excited.
I think the correct answer is \(\zeta_G(s) \neq 0\) for \(\Re(s) > 1/2\)--and this is the best we're going to get....
This means a reflection formula is very unlikely. And means that the Gottfried sum, is the maximal analytic continuation. So that Caleb's comments on Polya are correct. There is probably a natural boundary. Good job
