Looking for references essentially corroborating (to authoritatively satisfy some editors) the sketch below of the relationship between even power (2,4,...) sums (traces) of the imaginary part of the complex zeros above the real axis of the Riemann zeta function $\zeta(s)$ and the derivatives evaluated at $t =0$ of $\Omega(t) = \xi(1/2+it)/\xi(1/2)$ where $\xi(s)$ is the Landau Riemann xi function. Please inform me if you feel there are unleapable gaps or flaws in the analysis below.

The Landau Riemann function $\xi(s)$ investigated in "Relations and positivity results for the derivatives of the Riemann ξ function" by Coffey can be used to define the real-valued, entire, even (recall $\xi(s)=\xi(1-s)$) function

$$\Omega(t) = \xi(1/2 + it)/\xi(1/2)$$

which can be expanded in the Taylor series

$$\Omega(t) = \sum_{n \geq 0} (-1)^n \frac{\xi^{(2n)}(1/2)}{\xi(1/2)} \frac{t^{2n}}{(2n)!}.$$

The numerical values for the first few derivatives are given in Coffey (as well as some ways to compute them and others).

The log of the Hadamard product (Weierstrass factorization) for $\xi$ allows the derivatives of $\Omega(t)$ to be expressed in terms of the even power sums (traces $Tr_{2n}$) of the reciprocals of the imaginary parts of the zeros of the Riemann zeta above the real axis, and conversely, the power sums can be calculated in terms of the derivatives, all through the Newton identities.

For example, for a polynomial

$$p(x) = \sum_m c_m x^m = \prod_m (1-x/x_m),$$

$$=\exp[\log(p(x))] = \exp[\sum_m\log(1-x/x_m)]$$

$$=\exp[\sum_{k \geq1} (-\sum_m \frac{1}{x_m^k})\frac{x^k}{k}] =\exp[\log(1-Tr.x)]$$

$$= \sum_n ST1_n(Tr_1,Tr_2,..,Tr_n) \frac{x^n}{n!},$$

and this is precisely the exponential generating function for the signed Stirling partition polynomials of the first kind, a.k.a., the cycle index partition polynomials for the symmetric groups, given in A036039, known more commonly as the Newton identity expressing the elementary symmetric polynomials in terms of the power sums. The Faber polynomials give the Newton identity expressing the power sums in terms of the elementary symmetric polynomials.

Now for some sanity checks:

$Tr_{2n} = \sum_m 1/z_m^{2n}$ for $n=1,2,3,4$ for the first several thousands of the zeros were calculated and presented by Gottfried Helms in an MSE question. $m$ indexes the zeros above the real axis and $z_m$ denotes the imaginary part.

Using the Newton identities with all odd-indexed indeterminates nulled, I have the numerical consistency checks between Coffey's and Helms' values:

A) From Coffey's derivative estimates,

$$Tr_2 = \xi^{(2)}(1/2) / (2\xi(1/2)) \simeq .022972/(2 \cdot .497) \simeq .02311,$$

B) and Helms' sum is

$$Tr_2 = .02307.$$

C) From Coffey's estimates,

$$\xi^{(4)}(1/2)/\xi(1/2) \simeq .0.002963/.497 \simeq .005962,$$

D) and from Helms,

$$3(-2Tr_2)^2 + 6(-2Tr_4) = 12(Tr_2^2-Tr_4) \simeq 12((.02311)^2-.0000372) \simeq.005962.$$

E) From Coffey's estimates,

$$2 Tr_4 = -4 \Omega^{(4)}(0)/4! + 2 (\Omega^{(2)}(0)/2)^2$$

$$\simeq -4 \cdot .005962/4! + 2 (-.04622/2)^2 \simeq .00007448,$$

F) and from Helms

$$2Tr_4 \simeq 2 \cdot .000037173 = .00007435.$$

If there is doubt about the order of $\xi$ and whether $Tr_2$ converges, the analysis here, since the sums are truncated in the computation by Helms, can be regarded as for a corresponding truncation of the Hadamard product for $\xi$. Also, the power sum of order two over the full complex zeros is convergent and can be checked with estimates of derivatives of $\xi(s)$ at $s=0$ by using precisely the Newton identities as above.

Edit (June 5, 2020):

Affirming the convergence, see the comments in the MSE question, extracted from the Titchmarsh references.

  • $\begingroup$ From the example, $$-\log(g(x)) = -\log[1+a_1x+a_3x^2+ \cdots] = \sum_{n > 0} F_n(a_1,..,a_n) \frac{x^n}{n}= \sum_{n>0} Tr_n \frac{x^n}{n}$$ where $F_n$ are the Faber polynomials. Using the Appell raising op, a simple recursion relation can be developed for the $a_n$ in terms of a convolution of lower order $a_k$ and $Tr_k$, another form of the Newton identities. $\endgroup$ Commented Jun 8, 2020 at 1:27
  • $\begingroup$ mathoverflow.net/questions/111165/… is an example of this type of analysis, for the entire reciprocal gamma function. $\endgroup$ Commented Jun 8, 2020 at 1:33
  • $\begingroup$ See also e.g.f. versions--the logarithmic polynomials of oeis.org/A263634 and the cumulant expansion polynomials of oeis.org/A127671 --of the Faber polynomials oeis.org/A263916. $\endgroup$ Commented Jun 8, 2020 at 17:43
  • $\begingroup$ A note by Terry Tao on the sector through which the RH has been numerically checked: google.com/amp/s/terrytao.wordpress.com/2018/09/06/… $\endgroup$ Commented Jun 15, 2020 at 11:44

1 Answer 1


It is not a good idea to compute $$\sum_{n=1}^\infty \gamma_n^{-k}$$ by computing a partial sum of several thousands of terms. The series converges but too slowly for this. (see "Computation of the secondary zeta function" is on arXiv now: https://arxiv.org/abs/2006.04869).

For simplicity assume that the Riemann Hypothesis is true, then $\gamma_n$ are the ordinates of non-trivial zeros $\rho$ of $\zeta(s)$ with $\textrm{Im}(\rho)>0$.

Here I can not give a complete account of how to compute these sums efficiently. But I implemented this in mpmath some years ago. You can use this in Sage. For example writing

from mpmath import *


for n in range(2,6):

$\quad$ print secondzeta(n)

You get the approximate values with 50 digits (it is true the last four of five will not be good, as can be seen computing to more precision)

$$\begin{array}{c} 0.023104993115418970788933810430339014003381760397422\\ 0.00072954827270970421587551856909397050335150570355395\\ 0.000037172599285269686164866262471740578453650889730014\\ 0.0000022311886995021033286406286918371933760764310879243 \end{array} $$

I will try to revise my old paper about this, and put it on arXiv in a few days.

In mpmath the function secondzeta(s)$=\sum_n \gamma_n^{-s}$. This extends to a meromorphic function on the complex plane. So secondzeta(1) is the value of this extended function. The series do not converge. For comparison we should give the values for second zeta(2n) these are:

$$\begin{array}{c} 0.000037172599285269686164866262471740578453650889730014\\ 0.00000014417393140097327969538155609482090703688300853254\\ 0.00000000066303168025299086987327208196135724847369284211165\\ 0.0000000000032136641506166012161021165998346551415628219091519 \end{array}$$

  • $\begingroup$ @Tom Copeland Sage is free. from mpmath import *, load mpmath to Sage. mp.dps=50, put your computation to 50 digits. secondzeta(s) compute the value of this function at the point s. Even in cases the series is divergent, because it computes the only meromorphic extension. $\endgroup$
    – juan
    Commented Jun 4, 2020 at 11:41
  • $\begingroup$ @ Tom Copeland No, my numbers is those in which you are interested. If $\rho = \frac12+i\gamma$ is a zero of zeta then I compute $\sum_{n=1}^\infty \gamma_n^{-s}$. $\endgroup$
    – juan
    Commented Jun 4, 2020 at 17:11
  • $\begingroup$ @ Tom Copeland In fact my numbers coincide with those of Helm, as you can check from my values. $\endgroup$
    – juan
    Commented Jun 4, 2020 at 17:13
  • $\begingroup$ Thanks for revising and clarifying that for me. I see you use "ordinate" in the revised post rather than "coordinate" and your results are consistent with Helms. Thanks for corroborating that. $\endgroup$ Commented Jun 4, 2020 at 17:53
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – juan
    Commented Jun 4, 2020 at 19:44

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