Zeros of successive higher order derivatives of the Riemann zeta function seem to cluster along roughly horizontal lines.

Is there a heuristic explanation of why this happens (especially inside the critical strip)?

R. Spira, in Zero free regions of $\zeta^{(k)}(s)$ wrote "The zeros of $\zeta^{\prime\prime}$ have imaginary part almost exactly equal to those of $\zeta^\prime$, and lie to the right of them." Figure 1 from his paper shows zeros of $\zeta^\prime$ marked with triangles, and zeros of $\zeta^{\prime\prime}$ marked with squares.

Plot of zeros of zeta, zeta prime and zeta double prime

Below is a Mathematica plot of the argument of $\zeta^{\prime\prime}/\zeta^\prime(s)$ for $1/2\le \sigma\le 1$ and $10^6\le t\le 10^6+20$ (Five strips of height $4$.). The poles at the zeros of $\zeta^\prime(s)$ have to opposite orientation of colors that the zeros of $\zeta^{\prime\prime}(s)$ have.

Logarithmic derivative of zeta prime(s)

S.L Skorokhodov, in "Pade Approximants and Numerical Analysis of the Riemann zeta function", Computational Mathematics and Mathematical Physics vol 43 (2003) pp. 1277-1299, formula (7.7) differentiates the Dirichlet series expansion for $\zeta(s)$ term by term (for $\sigma>1$), and deduces "Therefore, it is again natural to expect a zero of the derivative $\zeta^\prime(s)$ to be located on the right [sic] of each finite zero of $\zeta^{\prime\prime}(s)$."

I don't understand this, particularly in regards to zeros in the critical strip.

Binder, Pauli, and Saidak, in "Zeros of High Derivatives of the Riemann Zeta Function", Rocky Mountain J. Math, vol 45 (2015), pp. 903-926 have similar results (Theorem 2.3), again using the Dirichlet series. They see chains of zeros up to order 90:

Figure 5 from Binder et al

From Theorem 3 in the seminal paper of Levinson and Montgomery, we can see this happens on average over an interval of size, say, $U=\log T$: $$ \frac{2\pi}{\log T}\left(\sum_{T<\gamma^{(k+1)}<T+\log T}(\beta^{(k+1)}-1/2)-\sum_{T<\gamma^{(k)}<T+\log T}(\beta^{(k)}-1/2)\right) =\log\log T/2\pi+O(1). $$ The right side is independent of $k$.

Edit/Correction: The corollary of the Levinson-Montgomery result speaks to the horizontal spacing of consecutive higher derivatives, on average independent of $k$. But it does not address Spira's observation about the vertical spacing: consecutive higher derivatives at very nearly the same height.

  • $\begingroup$ (Tangentially related) There is a theorem which states that zeroes of derivatives of a polynomial must lie in the convex of hull of the original polynomial’s zeroes I believe. There might be a generalization of this to analytic functions somehow which could explain your finding. $\endgroup$ Commented Aug 25, 2022 at 1:44
  • 1
    $\begingroup$ @SidharthGhoshal This is the Gauss-Lucas Theorem. There are generalizations to entire functions by Morris Marden, see the accepted answer to this MO question mathoverflow.net/questions/242381/… $\endgroup$
    – Stopple
    Commented Aug 25, 2022 at 15:52

2 Answers 2


Consider the absolute value of the $n$th derivative $\zeta^{(n)}$. It is large (in magnitude) to the left and it slopes down to be asymototic to 0 as you move to the right. A zero of that function is a sharp valley interrupting the overall trend. A zero of its derivative is a saddle point of that surface. So one should expect the saddle point to be directly to the right of the valley (because from the left the land slopes down, and as you emerge moving to the right from the valley it slopes up before resuming its overall downward trend). That might not happen if there were other nearby valleys, causing the topography to be weird. But that happens rarely.

The above picture might not be completely accurate in the critical strip, but all those pictures of zeros of successive derivatives are in the region of absolute convergence.

  • $\begingroup$ Maybe what is different in the critical strip is that the zeros (valleys) can be close together. $\endgroup$ Commented Aug 29, 2022 at 13:24
  • $\begingroup$ It is worth noting that the gradient of $|f|^2$ being zero, together with the Cauchy Riemann equations, is enough to imply that $f^\prime$ is zero. $\endgroup$
    – Stopple
    Commented Aug 29, 2022 at 16:36
  • $\begingroup$ This argument seems to apply to any Dirichlet series of the form $\sum_{n>1} a_nn^{-s}$. $\endgroup$
    – Stopple
    Commented Aug 29, 2022 at 16:37

Here's another approach. Assume the Riemann Hypothesis for simplicity. As $\sigma\to +\infty$, $\zeta^{\prime\prime}/\zeta^\prime(s)\to -\log(2)$. Meanwhile, on the critical line, $\zeta^{\prime\prime}/\zeta^\prime(s)$ will be dominated by terms in the sum $\sum_{\rho^\prime} 1/(s-\rho^\prime)$ with $\rho^\prime$ near $s$, and this implies the real part of $\zeta^{\prime\prime}/\zeta^\prime(s)$ will again be negative.

Now fix a zero $\rho^\prime$ of $\zeta^\prime(s)$, and consider the level curves $\arg(\zeta^{\prime\prime}/\zeta^\prime(s))=C$ exiting the pole at $\rho^\prime$, in particular the curve $C=0$ (i.e. Im($\zeta^{\prime\prime}/\zeta^\prime(s)$)=0 and Re($\zeta^{\prime\prime}/\zeta^\prime(s)$)$>0$.). By the above observations, this contour can't cross the critical line, nor extend too far into the right half plane. The only possibility is that it terminates in a zero of $\zeta^{\prime\prime}(s)$. Since $$ \frac{\zeta^{\prime\prime}}{\zeta^\prime}(s)=\frac{1}{s-\rho^\prime}+O(1), $$ the contour has to exit the pole to the right, and so the zeros of $\zeta^{\prime\prime}$ will typically be to the right of the zeros of $\zeta^\prime$.

The graphics below shows, for $1/2\le\sigma\le 5/2$ and $10^6\le t\le 10^6+10$, the argument of $\zeta^{\prime\prime}/\zeta^\prime(s)$ on the left, and only those curves corresponding to the four coordinate axes on the right. We are focusing on the red curves. (NB the graphic on the left is the same as in the original question, but I've fixed a bug in the Mathematica function ComplexPlot[] which shifts the phase by $\pi$)

phase of zeta''/zeta'

This same argument works for higher derivatives as well.

  • $\begingroup$ Your question made me wonder about the zeros of the fractional derivatives of zeta(s). Would these just ‘fill the gaps’ on the horizontal lines between two integer derivatives? Found this paper that seems related to the subject: link.springer.com/chapter/10.1007/978-3-030-52200-1_9 $\endgroup$
    – Agno
    Commented Aug 30, 2022 at 23:34
  • $\begingroup$ If you make a contour plot instead of a colored plot showing the phases, you can see the saddle points and how they emerge from the overall trend and the valleys from the zeros. I prefer plotting Log[Abs[Zeta[sig + I t]]] because then you can use equally spaced contours. $\endgroup$ Commented Sep 2, 2022 at 17:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.