3
$\begingroup$

The simple zero conjecture says that all zeros of the Riemann zeta function are simple.

Suppose the conjecture is not true. Namely there is an $s$ in the complex plane such that $\zeta(s)=0$ and the first derivative of $\zeta$ at $s$ is zero. If $s$ is a double zero of $\zeta$ (i.e. the second derivative isn't zero), is it true that the third derivative at $s$ also isn't zero?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ There is no underlying symmetry to draw such a conclusion. $\endgroup$
    – KConrad
    Commented Jan 15, 2014 at 19:58
  • 3
    $\begingroup$ For a zero with imaginary part around $T$, we don't know anything beyond the fact that the multiplicity is bounded by a constant times $\log T$. Even RH would only allow the slightly better result that the multiplicity is bounded by a constant $\log T/\log \log T$. The analogous problem for the multiplicity of a zero at $1/2$ for the $L$-function of an elliptic curve has been extensively studied (e.g. by Brumer). $\endgroup$
    – Lucia
    Commented Jan 15, 2014 at 20:45
  • $\begingroup$ I deleted my answer. I was confusing something. So you ask :$\zeta(s)=0, \zeta'(s)=0, \zeta''(s) \neq 0 \Rightarrow \zeta'''(s) =0$ where I was reading $\zeta(s)=0, \zeta'(s)=0, \Rightarrow \zeta''(s)\neq 0$ $\endgroup$
    – Marc Palm
    Commented Feb 4, 2014 at 12:56
  • $\begingroup$ I ask the following: $\zeta(s)=0, \zeta^{'}(s)=0, \zeta^{''}(s) \neq 0 \rightarrow \zeta^{'''}(s) \neq 0.$ $\endgroup$
    – Mahdi Teymuri Garakani
    Commented Feb 5, 2014 at 14:03

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .