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How large can the derivative of the Riemann zeta function be at its zeros?

More specifically, let $\rho$ be a zero of the zeta function with $\Im(\rho)\in (0,T]$. What can we say about $|\zeta'(\rho)|$? Surely this is a known result? I can't find anything in the literature about it, though.

Moreover, even if there are extreme values of $\zeta'(\rho)$, is there a nice 'average' for $\sum_{0<\Im{\rho}<T}\zeta'(\rho)^k$? for integer $k$? I ask this as it would help with my understanding of the growth of the derivative.

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    $\begingroup$ Also: who's buried in Grant's father's tomb? $\endgroup$
    – PseudoNeo
    Commented Oct 1 at 15:44
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    $\begingroup$ @PseudoNeo what do you mean? $\endgroup$
    – user536681
    Commented Oct 1 at 15:58
  • $\begingroup$ Did you really intend for the averaged problem to not have absolute values? $\endgroup$ Commented Oct 1 at 20:12
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    $\begingroup$ I meant that a (stupid) interpretation of your question is that you are asking how big is $\zeta'$ at its zeros, and the answer is obviously $0$. Not a very funny remark, but the gods have imposed upon my stupid jokes the yoke of a foreign tongue that was not sung at my cradle. $\endgroup$
    – PseudoNeo
    Commented Oct 1 at 20:30

2 Answers 2

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Bounds for the value of zeta take the form $$ |\zeta(\sigma +it) | \ll t^{ f(\sigma)+\epsilon} $$ where $|t|>2$ and $f$ is some function of $\sigma$. We can take $f(1)=0$ and $f(0)=1/2$ and we can always take $f$ to be convex.

Any bound of this form for $\zeta$ implies an identical form for the derivative of $\zeta$: We can express the derivative of $\zeta$ as the integral of $\zeta$ over a circle and taking the radius of this circle to go to zero slowly the loss between the bound for $\zeta$ and the bound for $\zeta'$ goes to $0$.

There can't be a better bound for $\zeta'$ at an arbitrary point than $\zeta$, since we can also express $\zeta$ as an integral of $\zeta'$. There could conceivably be a better bound for $\zeta'(\rho)$ specifically at zeroes $\rho$ of $\zeta$ but this seems unlikely - we don't typically expect the derivative of a holomorphic function to be especially small at its zeroes.

In particular, if you have no information on the real part of the zero, then the worst case is when $\sigma$ is close to $0$ so $f$ is close to $1/2$ and the best bound will be of the form $t^{1/2+\epsilon}$. Zero-free regions are not currently sufficient to rule out this probably.

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As the other answer indicates, at a single zero you can't say much, but there are a lot of results on discrete moments. With $$ J_k(T)=\frac{1}{N(T)}\sum_{0<\gamma<T}|\zeta^\prime(\rho)|^{2k}, $$ Gonek showed that $$ J_1(T)\sim\frac{1}{12}\log(T)^3. $$ Milinovich and Ng showed $$ J_k(T)\gg_k \log(T)^{k(k+2)} $$ and Kirila showed that on RH, $$ J_k(T)\ll_k\log(T)^{k(k+2)}. $$ A search on "moments derivative Riemann zeta" turns up lots of papers.

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  • $\begingroup$ With your "As the above answer indicates, ...", a minor issue is it's dependent on the the order used to show the answers. Thus, if your answer has a higher score & the sort order is by highest score, then that answer would be below yours. FWIW, I usually provide a link to that answer, along with the user's name, e.g., "As Will Savin's answer indicates, ...". However, I've also seen quite a few cases where just a link is used, or only the answer's username (or possibly the first start), but that may be a problem, e.g., they change their username or have more than one answer. $\endgroup$ Commented Oct 1 at 18:09

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