Deriving the Riemann non-trivial zeros from $\zeta_{H}(s,a) + \zeta_{H}(s,1-a)$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T07:18:19Z http://mathoverflow.net/feeds/question/98900 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98900/deriving-the-riemann-non-trivial-zeros-from-zeta-hs-a-zeta-hs-1-a Deriving the Riemann non-trivial zeros from $\zeta_{H}(s,a) + \zeta_{H}(s,1-a)$ Agno 2012-06-05T18:55:47Z 2012-06-05T22:23:02Z <p>The Hurwitz zeta function:</p> <p>$$\zeta_{H}(s,a)$$</p> <p>reduces to $\zeta(s)$ when $a=1$ and to $(2^s-1)\zeta(s)$ when $a=\frac12$.</p> <p>However, I stumbled upon a peculiar third connection:</p> <p>$$\zeta_{H}(s,a) + \zeta_{H}(s,1-a)$$</p> <p>that seems to exactly produce the non-trivial zeros of $\zeta(s)$,</p> <p>when $a=\frac12$ (obviously), but also (and apparently only) when $a=\frac13, \frac14$ or $\frac16.$</p> <p>Why does it only work for these values? Is there any reference to this in the literature?</p> <p>Thanks.</p> http://mathoverflow.net/questions/98900/deriving-the-riemann-non-trivial-zeros-from-zeta-hs-a-zeta-hs-1-a/98916#98916 Answer by Agno for Deriving the Riemann non-trivial zeros from $\zeta_{H}(s,a) + \zeta_{H}(s,1-a)$ Agno 2012-06-05T22:23:02Z 2012-06-05T22:23:02Z <p>Found the answer, the question can be closed. <a href="http://mathstat.carleton.ca/~williams/papers/pdf/179.pdf/" rel="nofollow">answer</a></p> <p>It boils down to:</p> <p>$$\zeta_{H}(s,a) + \zeta_{H}(s,1-a) = \frac{4}{(2\pi)^{1-s}}\Gamma(1-s)C(1-s,a)$$</p> <p>$$C(s,a)=\sum_{n=1}^\infty \frac{\cos(2n\pi a)}{n^s}$$</p> <p>And $C(s,a)$ reducing to:</p> <p>$a=\frac12 \rightarrow$ $(2^{1-s}-1)\zeta(s)$</p> <p>$a=\frac13 \rightarrow$$\dfrac12(3^{1-s}-1)\zeta(s) </p> <p>a=\frac14 \rightarrow$$2^{-s}(2^{1-s}-1)\zeta(s)$</p> <p>$a=\frac16 \rightarrow$$\dfrac12(1-2^{1-s})(1-3^{1-s})\zeta(s)$</p> <p>hence the non-trivial zeros.</p>