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By differentiating $\xi$ and solving for $\zeta(1-x)$:

$$ \zeta(1-x) = \frac{2(\zeta'(x)\Gamma(x/2)+\Gamma((1-x)/2) \zeta'(1-x)\pi^{x-1/2}) )}{\Gamma((1-x)/2) \pi^{-1/2+x}(2\log\pi -\psi((1-x)/2)-\psi(x/2))} \qquad (1)$$

and unless the denominator vanishes (like near $\frac12 \pm 6 i$) the non-trivial zeros of zeta are the zeros of

$$\zeta'(x)\Gamma(x/2)+\Gamma((1-x)/2) \zeta'(1-x)\pi^{x-1/2} = 0$$

Is something similar possible for Riemann $\xi$ or symmetrized zeta $\zeta^*$?

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1  
@Joro. Did you notice that your formula for the non-trivial zeros does also induce a zero at $\frac12 \pm 6.289835989i...$? I believe this is the value where: $|\chi(s)|=|2^s \pi^{s-1}\sin(\pi s/2) \Gamma(1-s)|=1$ when $\Re(s) \rightarrow \frac12$. See here: mathoverflow.net/questions/85351/… –  Agno Jun 24 '13 at 13:11
    
@Agno Thank you. At 6.28... the denominator is zero too, so it cancels. I suppose this is not possible for other values, not sure. –  joro Jun 24 '13 at 14:01
    
@Agno edited because of your comment. (1) doesn't induce zero there because of $0/0$. –  joro Jun 24 '13 at 14:11
    
@Agno the plots always coincide, here is a plot on the critical line: s23.postimg.org/ktoyver0b/zeta_via_deriv.png –  joro Jun 24 '13 at 14:46
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@joro. Nice formula indeed! You now have $\zeta(1-x)$ fully expressed in terms of its derivatives. Did you try converting it to $\zeta(s)$ using $\zeta(1-x)=\dfrac{\zeta(x)}{\chi(x)}$? What is immediately clear from the formula is that $\zeta'(x)$ and $\zeta'(1-x)$ cannot both be zero at the same time, unless this coincides with a (non)-trivial zero of $\zeta(x)$ or they are annihilated by the denominator. –  Agno Jun 24 '13 at 15:57
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