3
$\begingroup$

Take the following Hurwitz zeta:

$$\zeta_{H}(s,z)$$

with $s=\sigma \pm ti$ and $\displaystyle z=1 \pm \frac{i}{a}$ and $t,a \in \mathbb{R}$.

In the critical strip $0 \lt \sigma \lt 1$, this Hurwitz zeta appears to have zeros $\rho_H$ for combinations of $(s,a)$ for all values of $\sigma$, with the exception of $\sigma = \frac12$. However, contrary to $\zeta(s)$, these zeros do not seem to obey the rule that when $s=\rho_H$ then the function must vanish also at $1-\rho_H$ .

Example:

$$\zeta_{H}(0.8-24.96910...i, 1+ \frac{i}{44.82238...} = 0 \ne \zeta_{H}(0.2+24.96910...i, 1+\frac{i}{44.82238...})$$

$$\zeta_{H}(0.2+25.05217...i, 1+ \frac{i}{44.84243...} = 0 \ne \zeta_{H}(0.8-25.05217...i, 1+\frac{i}{44.84243...})$$

Would a proof for all $\rho_H$ that:

$$\zeta_{H}(\rho_H,z) \ne \zeta_{H}(1- \rho_H,z)$$

for all $\sigma \ne \frac12$ in the critical strip, imply the RH when $\displaystyle \lim_{a \to +\infty}$?

My logic towards a 'yes' to this question, would be that the contradiction between 'all zeros off the critical line cannot be symmetrical for values of $a < \infty$' and 'all non trivial zeros of $\zeta(s)$ must be symmetrical', can only be solved when 'there cannot be zeros of $\zeta(s)$ lying off the critical line'. This would also imply that both functions quite naturally complement each other (i.e. all zeros lie off the critical line versus all zeros are on it). Not sure though about what happens to the properties of both functions at the tilting point when $\displaystyle \lim_{a \to +\infty} \zeta_{H}(s, 1 \pm \frac{i}{a})$ 'morphs' into $\zeta(s)$.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

The functional equation for the Hurwitz zeta function is far from being of the type $\zeta_{H}(s) = \Phi(s) \zeta_{H}(1-s)$ for some 'easy' $\Phi(s)$. I believe that the fact that $1 - \rho$ is not a zero $\zeta_{H}(s)$ when $\rho$ is, to be exclusively related to the shape of the functional equation of the Hurwitz zeta function, which is very different from that of the Riemann zeta function. It seems unlikely to me that the shape of the functional equation for the Hurwitz zeta function will have implications for the location of the zeros of $\zeta(s)$.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ You might be right. The functional equation for the Hurwitz zeta is different: $\zeta_{H}(1-s,\frac{m}{n})=\frac{2\,\Gamma \left( s \right)}{\left( 2\,\pi \,n \right) ^{s}} \sum _{k=1}^{n}\left[ \cos \left(\frac{\pi \,s}{2}+{\frac {2 \pi km}{n}} \right) \zeta_{H} \left(s,{\frac {k}{n}} \right) \right]$, but unfortunately it is only valid for real fractions $\frac{m}{n}$ as the second argument. When the latter becomes $1+ \frac{i}{a}$ I am keen to find an explanation for why there are only zeros in the strip for all $\sigma \ne \frac12$ (i.e. the exact complement of the RH). $\endgroup$
    – Agno
    May 28, 2012 at 13:23
  • $\begingroup$ I need to learn a bit more about hurwitz zeta when the shift is irrational. $\endgroup$
    – kolik
    May 28, 2012 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.