Okay, I have a proof of two statements. I am not entirely there yet; but I'm pretty certain at this point. I like to take these moments to refresh everyone on what every function means.
We are going to start with \(\nu\) as I am more confident; and then I will move on to \(\zeta_G\). I would like to use these two symbols to denote our main functions. To begin we have that:
\[
\zeta_G(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{n^{-s}}\\
\]
This is to mean, the leading term is \(1\). This is Gottfried's zeta function (or Gottfried's function for short). The conjecture at hand, by Caleb, is that this converges for \(\Re(s) > -1\). Where there is a divergence along the line:
\[
\sum_{n=1}^\infty (-1)^{n+1} n^{n^{1-it}}\\
\]
Where this series does not converge for \(t \in \mathcal{D} \subset \mathbb{R}\) where \(\overline{\mathcal{D}} = \mathbb{R}\)--so that \(\mathcal{D}\) is dense. Which means; for a dense amount of points we may begin to see singular behaviour. This is still open. We are going to always refer to \(s\) as the complex variable of a zeta function; and we'll adopt the Riemann convention of \(s = \sigma + it\) for \(\sigma \in \mathbb{R}\) and \(t \in \mathbb{R}\).
I refer to the earlier parts of this thread for clarifications on notation; but I'll try to remind everybody. The goal of this post is to show that:
\[
\zeta_G(s) \,\,\text{is holomorphic for}\,\,\Re(s) > 0\\
\]
And is represented in this half plane, by the zeta series:
\[
\zeta_G(s) = \sum_{m=1}^\infty g(m) m^{-s}\\
\]
While simultaneously showing that:
\[
A(x) = \sum_{1 \le m \le \lfloor x \rfloor} g(m) = O(x^{\epsilon})\\
\]
For all \(\epsilon > 0\). Which I conjecture is some kind of log growth; but could be even slower... who knows.
The function:
\[
q(m) = (-1)^m |\sqrt{\mathcal{I}_m}|\\
\]
Where
\[
\sqrt{\mathcal{I}_m} = \{n \in \mathbb{N}\,|\,n^k = m,\,\,k\in\mathbb{N}\}\\
\]
Where the indicator function \(\chi_m(n)\) expresses:
\[
q(m) = (-1)^m \sum_{1 \le n\le m} \chi_m(n)\\
\]
such that \(\chi_m(n) =1\) if and only if \(n^k = m\), and is zero otherwise. Upon which we can write our function:
\[
\nu(s) = \sum_{m=1}^\infty q(m)m^{-s}\\
\]
The function \(\nu\) doesn't look too related to \(\zeta_G\)--but they are very similar behaved. Looking at:
\[
Q(x) = \sum_{1\le m \le \lfloor x \rfloor} q(m)\\
\]
We equally look like \(O(x^{\epsilon})\)--just like \(A(x)\). This is supported by numerical evidence strongly; if anything I am being modest with these estimations.
The most obvious comparison between \(\nu\) and \(\zeta_G\) is:
\[
g(m) = (-1)^m \sum_{1 \le n \le m} \frac{\log(n)^{\frac{\log(m)}{\log(n)}}}{\frac{\log(m)}{\log(n)}!} \chi_{m}(n)\\
\]
The value \(\frac{\log(n)^{\frac{\log(m)}{\log(n)}}}{\frac{\log(m)}{\log(n)}!} = \frac{\log(n)^k}{k!}\); where it clearly tends to zero as \(m\) and \(n\) go to infinity. So Gottfried's function has more decay in these coefficients than \(\nu\). The \(\nu\) zeta is a sort of natural boundary; where we replace \(\frac{\log(n)^k}{k!}\le 1\). This is a generous bound; but can tell us the "flavour" of Gottfried's function. This may look stupid; but we do replacements like this in zeta function talk all time. We are "guessing" that the asymptotics should be similar, and then we prove it.
The results I want to present are actually pretty simple. I want to show that:
\[
\begin{align}
A(x) &= O(x^{\epsilon})\,\,\text{for all}\,\,\epsilon > 0\\
Q(x) &= O(x^{\epsilon})\,\,\text{for all}\,\,\epsilon > 0\\
\end{align}
\]
So.... When we make the equivalence:
\[
\log(n)^k/k! := 1\\
\]
We perform the following:
\[
\zeta_G(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{n^{-s}} = \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=0}^\infty \frac{\log(n)^k}{k!}n^{-ks}\\
\]
becomes:
\[
\nu(s) = \sum_{n=1}^\infty \sum_{k=1}^\infty n^{-ks} = \sum_{n=1}^\infty (-1)^{n+1}\frac{n^{-s}}{1-n^{-s}}\\
\]
This series is anomalous as FUCK! but it converges in a similar manner that Gottfried's original series does. Again, we're looking at this conditional convergence nonsense kind of series. But Gottfried's used \(n^{n^{-s}}\); and \(\nu\) just kind of flat lines it to \(\frac{1}{1-n^{-s}}\). The cancellation in these sums is unfounded. I'm honestly blown away by how well these things cancel.
All numerical evidence points to two things right off the bat:
\[
\begin{align}
\zeta_G(s) &= \sum_{m=1}^\infty g(m)m^{-s}\,\,\text{converges absolutely as a series for}\,\,\sigma > 1\\
\nu(s) &= \sum_{m=1}^\infty q(m)m^{-s}\,\,\text{converges absolutely as a series for}\,\,\sigma > 1\\
\end{align}
\]
This isn't hard to prove because we have the modest bounds:
\[
\begin{align}
|g(m)| &\le \log(m) \Pi(m)\\
|q(m)| &\le \Pi(m)\\
\end{align}
\]
Where \(\Pi(m)\) was talked of before; and being of \(O(m^{\epsilon})\) growth for all \(\epsilon > 0\). (It's almost actually \(O(\log(m)^\epsilon)\) but I couldn't prove it; but it's closer to this bound). The hard thing to prove, which we are proving; is that:
\[
\begin{align}
\zeta_G(s) &= \sum_{m=1}^\infty g(m)m^{-s}\,\,\text{converges conditionally as a series for}\,\,0 < \sigma \le 1\\
\nu(s) &= \sum_{m=1}^\infty q(m)m^{-s}\,\,\text{converges conditionally as a series for}\,\,0 < \sigma \le 1\\
\end{align}
\]
I was racking my brain trying to figure out how to prove this; and what actually happens is very simple. The values:
\[
\begin{align}
\text{sign}(g(m)) = (-1)^{m+1}\\
\text{sign}(q(m)) = (-1)^{m+1}\\
\end{align}
\]
Where \(\text{sign}(a)\) gives whether \(a>0\) or \(a < 0\). The sequence \(g(m)m^{-\epsilon}\to 0\) and \(q(m)m^{-\epsilon} \to 0\). Therefore, since they oscillate between positive and negative perfectly; the alternating series test is enough to prove the zeta functions converge for \(\Re(s) > 0\).
And finally. We present our two results! We start with Perron's formula:
\[
A(x) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{\zeta_G(s) x^s}{s}\,ds\\
\]
But we can take any value \(c > 0\) here. Write: \(c = \epsilon > 0\). And:
\[
|A(x)| \le \frac{x^\epsilon}{2\pi}\left|\int_{-\infty}^{\infty} \frac{\zeta_G(\epsilon+it) x^{it}}{\epsilon+it}\,dt\right|\\
\]
The integral in this bound converges; but I'm not going to prove it for you. You guys gotta do some work!!! (Hint: Mellin Transform).
Same result follows for \(Q(x)\)...
In conclusion we have two plain as day results!
\[
\begin{align}
\zeta_G(s)\,\,&\text{is holomorphic for at least}\,\,\Re(s) > 0\\
A(x) &= O(x^{\epsilon})\,\,\text{for all} \,\,\epsilon > 0\\
\end{align}
\]
\[
\begin{align}
\nu(s)\,\,&\text{is holomorphic for at least}\,\,\Re(s) > 0\\
Q(x) &= O(x^{\epsilon})\,\,\text{for all}\,\, \epsilon > 0\\
\end{align}
\]
I've got more graphs incoming!! Maybe 1 or 2 days! It looks like \(\zeta_G\)'s zeroes ARE ON THE CRITICAL LINE!!!!!! Also when I say alternating series test; I mean Dirichlet's series test for the complex plane, which is just fancy alternating series test. Where when you study zeta functions "alternating series test" is meant to mean "alternating series test for zeta functions"--which there's an easy way to prove that since it converges for real numbers, it must converge for imaginary numbers too...
We are going to start with \(\nu\) as I am more confident; and then I will move on to \(\zeta_G\). I would like to use these two symbols to denote our main functions. To begin we have that:
\[
\zeta_G(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{n^{-s}}\\
\]
This is to mean, the leading term is \(1\). This is Gottfried's zeta function (or Gottfried's function for short). The conjecture at hand, by Caleb, is that this converges for \(\Re(s) > -1\). Where there is a divergence along the line:
\[
\sum_{n=1}^\infty (-1)^{n+1} n^{n^{1-it}}\\
\]
Where this series does not converge for \(t \in \mathcal{D} \subset \mathbb{R}\) where \(\overline{\mathcal{D}} = \mathbb{R}\)--so that \(\mathcal{D}\) is dense. Which means; for a dense amount of points we may begin to see singular behaviour. This is still open. We are going to always refer to \(s\) as the complex variable of a zeta function; and we'll adopt the Riemann convention of \(s = \sigma + it\) for \(\sigma \in \mathbb{R}\) and \(t \in \mathbb{R}\).
I refer to the earlier parts of this thread for clarifications on notation; but I'll try to remind everybody. The goal of this post is to show that:
\[
\zeta_G(s) \,\,\text{is holomorphic for}\,\,\Re(s) > 0\\
\]
And is represented in this half plane, by the zeta series:
\[
\zeta_G(s) = \sum_{m=1}^\infty g(m) m^{-s}\\
\]
While simultaneously showing that:
\[
A(x) = \sum_{1 \le m \le \lfloor x \rfloor} g(m) = O(x^{\epsilon})\\
\]
For all \(\epsilon > 0\). Which I conjecture is some kind of log growth; but could be even slower... who knows.
The function:
\[
q(m) = (-1)^m |\sqrt{\mathcal{I}_m}|\\
\]
Where
\[
\sqrt{\mathcal{I}_m} = \{n \in \mathbb{N}\,|\,n^k = m,\,\,k\in\mathbb{N}\}\\
\]
Where the indicator function \(\chi_m(n)\) expresses:
\[
q(m) = (-1)^m \sum_{1 \le n\le m} \chi_m(n)\\
\]
such that \(\chi_m(n) =1\) if and only if \(n^k = m\), and is zero otherwise. Upon which we can write our function:
\[
\nu(s) = \sum_{m=1}^\infty q(m)m^{-s}\\
\]
The function \(\nu\) doesn't look too related to \(\zeta_G\)--but they are very similar behaved. Looking at:
\[
Q(x) = \sum_{1\le m \le \lfloor x \rfloor} q(m)\\
\]
We equally look like \(O(x^{\epsilon})\)--just like \(A(x)\). This is supported by numerical evidence strongly; if anything I am being modest with these estimations.
The most obvious comparison between \(\nu\) and \(\zeta_G\) is:
\[
g(m) = (-1)^m \sum_{1 \le n \le m} \frac{\log(n)^{\frac{\log(m)}{\log(n)}}}{\frac{\log(m)}{\log(n)}!} \chi_{m}(n)\\
\]
The value \(\frac{\log(n)^{\frac{\log(m)}{\log(n)}}}{\frac{\log(m)}{\log(n)}!} = \frac{\log(n)^k}{k!}\); where it clearly tends to zero as \(m\) and \(n\) go to infinity. So Gottfried's function has more decay in these coefficients than \(\nu\). The \(\nu\) zeta is a sort of natural boundary; where we replace \(\frac{\log(n)^k}{k!}\le 1\). This is a generous bound; but can tell us the "flavour" of Gottfried's function. This may look stupid; but we do replacements like this in zeta function talk all time. We are "guessing" that the asymptotics should be similar, and then we prove it.
The results I want to present are actually pretty simple. I want to show that:
\[
\begin{align}
A(x) &= O(x^{\epsilon})\,\,\text{for all}\,\,\epsilon > 0\\
Q(x) &= O(x^{\epsilon})\,\,\text{for all}\,\,\epsilon > 0\\
\end{align}
\]
So.... When we make the equivalence:
\[
\log(n)^k/k! := 1\\
\]
We perform the following:
\[
\zeta_G(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{n^{-s}} = \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=0}^\infty \frac{\log(n)^k}{k!}n^{-ks}\\
\]
becomes:
\[
\nu(s) = \sum_{n=1}^\infty \sum_{k=1}^\infty n^{-ks} = \sum_{n=1}^\infty (-1)^{n+1}\frac{n^{-s}}{1-n^{-s}}\\
\]
This series is anomalous as FUCK! but it converges in a similar manner that Gottfried's original series does. Again, we're looking at this conditional convergence nonsense kind of series. But Gottfried's used \(n^{n^{-s}}\); and \(\nu\) just kind of flat lines it to \(\frac{1}{1-n^{-s}}\). The cancellation in these sums is unfounded. I'm honestly blown away by how well these things cancel.
All numerical evidence points to two things right off the bat:
\[
\begin{align}
\zeta_G(s) &= \sum_{m=1}^\infty g(m)m^{-s}\,\,\text{converges absolutely as a series for}\,\,\sigma > 1\\
\nu(s) &= \sum_{m=1}^\infty q(m)m^{-s}\,\,\text{converges absolutely as a series for}\,\,\sigma > 1\\
\end{align}
\]
This isn't hard to prove because we have the modest bounds:
\[
\begin{align}
|g(m)| &\le \log(m) \Pi(m)\\
|q(m)| &\le \Pi(m)\\
\end{align}
\]
Where \(\Pi(m)\) was talked of before; and being of \(O(m^{\epsilon})\) growth for all \(\epsilon > 0\). (It's almost actually \(O(\log(m)^\epsilon)\) but I couldn't prove it; but it's closer to this bound). The hard thing to prove, which we are proving; is that:
\[
\begin{align}
\zeta_G(s) &= \sum_{m=1}^\infty g(m)m^{-s}\,\,\text{converges conditionally as a series for}\,\,0 < \sigma \le 1\\
\nu(s) &= \sum_{m=1}^\infty q(m)m^{-s}\,\,\text{converges conditionally as a series for}\,\,0 < \sigma \le 1\\
\end{align}
\]
I was racking my brain trying to figure out how to prove this; and what actually happens is very simple. The values:
\[
\begin{align}
\text{sign}(g(m)) = (-1)^{m+1}\\
\text{sign}(q(m)) = (-1)^{m+1}\\
\end{align}
\]
Where \(\text{sign}(a)\) gives whether \(a>0\) or \(a < 0\). The sequence \(g(m)m^{-\epsilon}\to 0\) and \(q(m)m^{-\epsilon} \to 0\). Therefore, since they oscillate between positive and negative perfectly; the alternating series test is enough to prove the zeta functions converge for \(\Re(s) > 0\).
And finally. We present our two results! We start with Perron's formula:
\[
A(x) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{\zeta_G(s) x^s}{s}\,ds\\
\]
But we can take any value \(c > 0\) here. Write: \(c = \epsilon > 0\). And:
\[
|A(x)| \le \frac{x^\epsilon}{2\pi}\left|\int_{-\infty}^{\infty} \frac{\zeta_G(\epsilon+it) x^{it}}{\epsilon+it}\,dt\right|\\
\]
The integral in this bound converges; but I'm not going to prove it for you. You guys gotta do some work!!! (Hint: Mellin Transform).
Same result follows for \(Q(x)\)...
In conclusion we have two plain as day results!
\[
\begin{align}
\zeta_G(s)\,\,&\text{is holomorphic for at least}\,\,\Re(s) > 0\\
A(x) &= O(x^{\epsilon})\,\,\text{for all} \,\,\epsilon > 0\\
\end{align}
\]
\[
\begin{align}
\nu(s)\,\,&\text{is holomorphic for at least}\,\,\Re(s) > 0\\
Q(x) &= O(x^{\epsilon})\,\,\text{for all}\,\, \epsilon > 0\\
\end{align}
\]
I've got more graphs incoming!! Maybe 1 or 2 days! It looks like \(\zeta_G\)'s zeroes ARE ON THE CRITICAL LINE!!!!!! Also when I say alternating series test; I mean Dirichlet's series test for the complex plane, which is just fancy alternating series test. Where when you study zeta functions "alternating series test" is meant to mean "alternating series test for zeta functions"--which there's an easy way to prove that since it converges for real numbers, it must converge for imaginary numbers too...
