What happens when infinite values of $\zeta_{H}(s,z)$ approach $\zeta(s)$ ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:03:35Z http://mathoverflow.net/feeds/question/98144 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98144/what-happens-when-infinite-values-of-zeta-hs-z-approach-zetas What happens when infinite values of $\zeta_{H}(s,z)$ approach $\zeta(s)$ ? Agno 2012-05-27T19:16:51Z 2013-05-14T10:22:00Z <p>Take the following Hurwitz zeta:</p> <p>$$\zeta_{H}(s,z)$$</p> <p>with $s=\sigma \pm ti$ and $\displaystyle z=1 \pm \frac{i}{a}$ and $t,a \in \mathbb{R}$.</p> <p>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 $1-\rho_H$ . </p> <p>Example:</p> <p>$$\zeta_{H}(0.8-24.96910...i, 1+ \frac{i}{44.82238...} = 0 \ne \zeta_{H}(0.2+24.96910...i, 1+\frac{i}{44.82238...})$$</p> <p>$$\zeta_{H}(0.2+25.05217...i, 1+ \frac{i}{44.84243...} = 0 \ne \zeta_{H}(0.8-25.05217...i, 1+\frac{i}{44.84243...})$$</p> <p>Would a proof for all $\rho_H$ that:</p> <p>$$\zeta_{H}(\rho_H,z) \ne \zeta_{H}(1- \rho_H,z)$$ </p> <p>for all $\sigma \ne \frac12$ in the critical strip, imply the RH when $\displaystyle \lim_{a \to +\infty}$?</p> <p>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 $a &lt; \infty$' 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 $\zeta(s)$.</p> http://mathoverflow.net/questions/98144/what-happens-when-infinite-values-of-zeta-hs-z-approach-zetas/98172#98172 Answer by kolik for What happens when infinite values of $\zeta_{H}(s,z)$ approach $\zeta(s)$ ? kolik 2012-05-28T06:12:42Z 2012-05-28T06:12:42Z <p>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 $\zeta(s)$. </p>