[NT] more zeta stuff for the fans

[tommy1729]

On mathstack the following conjecture about a dirichlet series was written.

https://math.stackexchange.com/questions...dot3-s-5-c

Since there are " zeta fans " here ...


Regards

tommy1729

I don't think this is extendable beyond \(\Re(s) > 2\)--and if it were it would be very non-obvious. To ask for zeroes before analytically continuing is a recipe for disaster. The sum DOES NOT converge for \(1 < \Re(s) < 2\)--so to say there are zeroes here; when the sum diverges; without an analytic continuation is... nonsense.

The standard way to analytically continue is useless here too. If I write:

\[
P(x) = \sum_{n\le x} p_n\\
\]

Then the tight bounds are:

\[
P(x) = \frac{x^2}{2} \left(\log(x) + \log \log(x) - 3/2 + o(1)\right)\\
\]

So from here we have:

\[
\zeta_P(s) = -s\int_1^\infty P(x)x^{-s-1}\,ds\\
\]

This integral converges for \(\Re(s) > 2\); but once you account for \(P(x)\)'s asymptotics--there's no obvious way to extend it. The error term \(o(1)\) decays too slowly-- about \(1/log^2 x\) to meaningfully allow for a simple continuation. Unless mick has a magical reflection formula up his sleeve, this is pretty pointless...

But that integral transform is designed not to have analytic continuation.

If P(x) = x^3 this would also not converge , yet its related zeta function does as it can be expressed by riemann zeta functions.



In fact if we write p_n as a taylor series 

p(n) = a_0 + a_1 n + a_2 n^2 + ...

Then we get expressions like

A(s) = zeta(s) + a_1 zeta(s-1) + a_2 zeta(s-2) + a_3 zeta(s-3) + ...

which might have analytic continuation then, since zeta does.


Notice this is not like an euler product or the prime zeta function.



And even if we have an natural boundary , that does not explain why the zero's approach the boundaries.


regards

tommy1729

I mean, I kind of see where your coming from.

But if \(|A_M(s) - A(s)| < \epsilon/2\) and \(|A(s)| > \epsilon\)--then:

\[
\begin{align}
|A(s)| &\le |A_M(s)| + |A_M(s) - A(s)|\\
|A_M|(s) &\ge \epsilon/2\\
\end{align}
\]

Where \(\epsilon\) only depends on \(M\) and \(\Re(s)>2\).

---

This is exactly what they teach you NOT TO DO! to solve the Riemann Hypothesis. If I take the sum:

\[
\zeta^M(s) = \sum_{n=1}^M n^{-s}\\
\]

The zeroes of this function coallesce near \(\Re(s) = 1\); because the sum goes to infinity and back near here. But it has no marker on the actual zeroes of \(\zeta(s)\). Because \(\zeta\) NEEDS TO BE analytically continued first... Without an analytic continuation; we can't speak on the zeroes. The partial sums tell us nothing about the zeroes; because the partial sums converge to infinity as \(M\to\infty\) in the critical strip--there is no domain of normality.

Yes yes very good point.

And it might destroy that way of thinking for the riemann zeta.

But in general just because the partial sums converge to infinity locally does not mean we cannot detect zero's beyond it.
We may get some fake zero's and we might get only approximations of where the actual zero's are.

But with some luck it might still work.

I must say that the method i * actually * used was a bit more complex than the story.
Which probably explains why it worked.

Numerically it is very unstable because the strip converges to infinity almost everywhere , but maybe not at all the zero's.

So if we find a zero is it surrounded by large numbers going to infinity.
Which means derivatives going to infinity as well, resembling discontinu , chaos and divergeance in root finding algorithms etc.

and phenomena like double limits and numerical errors and such.

I remember that my algorithm gave 1/2 + 14.135 i as a zero but an overflow error at 0.49999999999 + 14.135 i.

It was crazy.

I remember the algorithm did not give zero's beyond a natural boundary.

I tested it on functions where the natural boundary was the unit circle like f(x)  = x^2 + x^2^2 + x^2^3 + x^2^4 + ...

Maybe I should look into those old methods again.

I have to give it more thought that is for sure, I guess we agree on that lol.
Im very busy at the moment though.

regards

tommy1729