(02/06/2023, 09:44 PM)Caleb Wrote: I have not looked in great detail at all the work you have done on this problem, but it seems really quite interesting. Excuse me for the simple question, but I am not well versed in number theory-- is the general idea for analytical continuation to obtain sufficently simple and tight bounds on \( A(x) \) so that \( A(x) \) minus the bound is small enough to be integrated, and then the integral over the subtracted bound can be analytically continued as well?
Also, I should add that from my own look into this series, it appears to have a pretty nasty natural boundary at \( \mathfrak{R}(s) =1 \). In my own eyes, it looks almost modular-form-like, except with the poles on the boundary have an extra 'tail' at some fixed angle that contain more poles. Anyway, if this type of argument is correct, it might be more natural to look at the function after applying the map \( z \to -i(z-1) \). Then you can study the function on the upper half plane which may make it more likely you run into some sort of familiar number theory object. Anyway, I look forward to what you come up with-- I'd love to see the number-theoretic significance of this function since the methods I tend to use don't usual to reveal those structures.
Well, that is a difficult question to answer Caleb! But I will say that, tight bounds on your function \(A(x)\) will translate to tight bounds on the zeta function; that's a given. The manner I employed in that post, is just one of many ways to analytically continue zeta functions. It dates to Riemann, and Dirichlet, most notably. There are definitely more sophisticated ways that people use now--but similar ideas are used. Usually more clever ideas. I'm kind of just throwing shit at the wall and seeing what sticks.
And yes, I absolutely believe that this function is meromorphic in a neighborhood of \(\Re(s) = 1\), and going to be a very ugly meromorphic. If it's a boundary at \(\Re(s) = 1\), then that's cool too! A boundary to me, means that there are a dense number of singularities on \(1 + i\mathbb{R}\). Then that means something super super coool!
It would be mean the tight bound is something like \(x\)-- which would be absolutely bananas! By, this, I'm going to run you through it
\[
\nu(s) = -s \int_1^\infty A(x)x^{-s-1}\,dx\\
\]
Let's assume there is a WALL of singularities at \(\Re(s) = 1\) (as opposed to me just thinking there are a bunch of singularities). Now let's pull out the oldest trick in the book in analytic number theory! This is known as Perron's formula https://en.wikipedia.org/wiki/Perron%27s_formula, which helps us analytically express \(A(x)\) from \(\nu\).
Let \(c>1\), then
\[
A(x) = \frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} \frac{\nu(s)x^{s}}{s}\,ds\\
\]
So if we can let \(c \to 1\) with no poles, then for all \(\epsilon > 0\):
\[
|A(x)| \le O(x^{1+\epsilon})
\]
So far everything is good here--and I can prove this. The thing is, if there is a WALL of singularities at \(1 + i\mathbb{R}\), then this is what's known as a TIGHT bound, because you cannot do any better. Which would mean the tight bound on \(A(x)\) is \(O(x)\). This could definitely help us a lot!!! (Remember \(\nu(s)\) isn't the function we care about, it's the lame version of Gottfried's function!). But this would mean that \(\nu(s)\) is landlocked to \(\Re(s) > 1\).
Normally, to get a better bound, say by moving \(\sigma < c < 1\), we get:
\[
A(x) = \frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} \frac{\nu(s)x^{s}}{s}\,ds - \sum \text{Res}\\
\]
And we pray the residues have nice asymptotics in \(x\)...
Fun fact! This is how you prove the equivalence of Mertens conjecture to the Riemann hypothesis https://en.wikipedia.org/wiki/Mertens_function. If the function:
\[
M(x) = \sum_{n \le x} \mu(n) = O(\sqrt{n})\\
\]
Where:
\[
\begin{align}
\frac{1}{\zeta(s)} &= \sum_{n=1}^\infty \mu(n) n^{-s}\\
&= -s\int_1^\infty M(x)x^{-s-1}\,dx\\
\end{align}
\]
Then Riemann's hypothesis is true. Because:
\[
M(x) = \frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} \frac{x^{s}}{\zeta(s)s}\,ds
\]
If we can move \(c \to 1/2\) without hitting any poles in \(1/\zeta(s)\)--then \(M(x)\) is bounded by \(\sqrt{x}\). But this just means, \(\zeta\) has no zeroes for \(1/2 < \Re(s) < 1\)--which is the Riemann Hypothesis, lmao.
A lot of this stuff can be very counterintuitive, and confusing as fuck. The complex analysis is usually what trips people up, because it's very non-conventional. And tends to look like magic. But a lot of it just follows somewhat basic building blocks, but they add together in very very weird (kind of like quantum physics, basic building blocks, but once you add it all up it adds up in a really weird way). It can get difficult to remember the rules.
Also, I'm out of practice. I've been doing so much complex dynamics, that I'm having to refresh my memories on so many things (I know it looks something like this, but I don't know the fine details, lol).
Thanks for your comments! Excited that you are excited! I should probably run some numerical trials myself, just haven't bothered, lmao!
Regards, James
EDIT:
Just realized you may be referring to Gottfried's function, when you talked about the wall of singularities, and not \(\nu\). Do you mind sharing more details? Are you using my change of variables \(x \mapsto -x\)? By this, when you say wall of singularities at \(\Re(s) = 1\) are you talking about \(\sum n^{n^s}\) or \(\sum n^{n^{-s}}\)? If there's a wall of singularities at \(\Re(s) =1\) for the latter case, the best result I can pull out is:
\[
\sum_{m\le x} \sum_{n\le m} \frac{\log^{\frac{\log(m)}{\log(n)}}(n)}{\frac{\log(m)}{\log(n)}!}\chi_m(n) = O(x)\\
\]
And we may be shit out of luck for a further analytic continuation.... But, that's still a fairly non-trivial result... To be honest, I don't think I've ever heard of zeta functions having walls of singularities; or natural boundaries in this manner. More typically than not, the sums and numerical procedures start to spaz out, and what looks like chaos is just chaos in your calculator, not the actual platonic form of the function. So I doubt it's a wall; probably just mad singularities (I.e. dense versus just a lot of singularities).
EDIT:
I'm pretty sure that zeta functions cannot have "walls of singularities" in the same way we see them in dynamics, where a natural boundary forms. At least, not if we ask that the coefficients of the zeta function \(f(n)\) satisfy \(\sum_{n\le x} f(n) = O(x)\)--what's far more likely is that it's just A LOT of singularities, and that this causes the calculator to spaz out and make it look more chaotic than it is. I will write my own calculator for Gottfried's function and report back
