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The Hurwitz zeta function: $$\zeta_{H}(s,a)$$ reduces to $\zeta(s)$ when $a=1$ and to $(2^s-1)\zeta(s)$ when $a=\frac12$. However, I stumbled upon a peculiar third connection: $$\zeta_{H}(s,a) + \zeta_{H}(s,1-a)$$ that seems to exactly produce the non-trivial zeros of $\zeta(s)$, when $a=\frac12$ (obviously), but also (and apparently only) when $a=\frac13, \frac14$ or $\frac16.$ Why does it only work for these values? Is there any reference to this in the literature? Thanks. |
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Found the answer, the question can be closed. answer It boils down to: $$\zeta_{H}(s,a) + \zeta_{H}(s,1-a) = \frac{4}{(2\pi)^{1-s}}\Gamma(1-s)C(1-s,a)$$ $$C(s,a)=\sum_{n=1}^\infty \frac{\cos(2n\pi a)}{n^s}$$ And $C(s,a)$ reducing to: $a=\frac12 \rightarrow$ $(2^{1-s}-1)\zeta(s)$ $a=\frac13 \rightarrow$$\dfrac12(3^{1-s}-1)\zeta(s)$ $a=\frac14 \rightarrow$$2^{-s}(2^{1-s}-1)\zeta(s)$ $a=\frac16 \rightarrow$$\dfrac12(1-2^{1-s})(1-3^{1-s})\zeta(s)$ hence the non-trivial zeros. |
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