# Pair correlation for the Riemann zeros and $(\zeta^\prime(s)/\zeta(s))^\prime$

Added Background: The pair correlation of the zeros of the Riemann zeta function is influenced by the the derivative of the logarithmic derivative $$(\zeta^\prime(s)/\zeta(s))^\prime$$; see for example the answers to this question

I'm looking for references for computation of $$\zeta^\prime(s)/\zeta(s)$$, as well as the its derivative, in the critical strip $$0<\text{Re}(s)<1$$. (I'm sure they're out there but google scholar/MathSciNet searches return way too many irrelevant hits.)

Of course, both $$\zeta(s)$$ and $$\zeta^\prime(s)$$ are implemented in packages like Sage, one can just take the quotient and then use this to numerically estimate the derivative via the difference quotient, but this seems computationally wasteful. We have that $$\frac{\zeta^\prime(s)}{\zeta(s)}=\log(2\pi)-1-\gamma/2-\frac{1}{s-1}-\frac12\frac{\Gamma^\prime(s/2+1)}{\Gamma(s/2+1)}+\sum_\rho\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right).$$

Similarly, one gets $$(\zeta^\prime(s)/\zeta(s))^\prime$$ upon differentiating term by term.

The digamma function $$\Gamma^\prime/\Gamma$$ as well as the Riemann zeros $$\rho$$ are implemented in Mathematica. So I think what I'm asking is a reference to answer the following:

Given $$\epsilon$$, how many zeros do I need to take as a as a function of $$t=\text{Im}(s)$$ so the error is bounded by $$\epsilon$$?

Added: In Theorem 9.6(A) in Titchmarsh's "Theory of the Riemann Zeta Function", one can compute the relevant constants to show that $$\left|\frac{\zeta^\prime(s)}{\zeta(s)}-\sum_{|\rho-s|\le 6}\frac{1}{s-\rho}\right|\le 4\log t.$$ So in answer to Joro's question below, yes the sum is dominated by the zeros close to $$s$$.

• For large $t$ wouldn't the sum be dominated by the zeros closest to $t$ because of $s - \rho$? If so finding large zeros doesn't seem computationally effective to me. Might be wrong.
– joro
Jan 30, 2013 at 7:13
• btw, How long do you think it will take to compute for $s=1/2 + i 10^{12}$? mpmath in sage does it for less than 13 seconds on an old pc.
– joro
Jan 30, 2013 at 12:26
• @Joro, see edit above. I'm interested in computing for a large number of values, and the question in general. Jan 30, 2013 at 16:18
• If your approach is efficient will it be unconditional? You need to compute the zeros. IIRC there are optimizations for zeros on the critical line. In what complexity do you disprove (or find) hypothetical zeros off the critical line sufficiently close to large $t$?
– joro
Jan 31, 2013 at 8:38
• I'm sure you've come across arXiv/1105.4312 --- this straightforward approach, numerical differentiation, is not necessarily less efficient than the route via the zeros. Feb 9, 2013 at 15:48

I believe the comments of joro and Carlo Benakker right under your question is to the point. Since the zeroes close to $s$ will be the ones that contributes in the sum, the zeroes close to s must be computed first and in order to do that several values of the zeta-function must certainly be calculated and the time for each such computation (and zero) will certainly not be less than computing any other particular value of $\zeta(s)$ or $\zeta'(s)$.

As for calculating a particular value $\zeta(s)$ the recent method of Ghaith Hiary http://arxiv.org/abs/0711.5005 "Fast methods to compute the Riemann zeta function" (published in Annals of Mathematics 2011) contains the fastest method known. He shows a method that calculates the value of $\zeta(1/2+it)$ with an error term less than $|t|^{-N}$ in time $O_{\varepsilon,N}(t^{4/13+\varepsilon})$. The method of Odlyzko and Schönhage "Fast algorithms for multiple evaluations of the Riemann zeta function", doi:10.2307/2000939 is faster if sufficiently many values of $t$ needs to be calculated, i.e. it can calculate $\sqrt T$ values in the range $[T,T+\sqrt T]$ in time $O_\varepsilon(T^{1/2+\varepsilon})$.

Now, In Hiary's paper it might look like he only considers the critical line. However in fact his method works for any value in the critical strip. Indeed in part of his argument he considers just the line Re$(s)=0$, but then he writes "It is clear that the restriction $\sigma=0$ is not important and a similar conclusions can be drawn for other values of $\sigma$" (page 6, second to last paragraph in Hiary's paper). It is true that his main interest is the critical line, but the algorithm holds for any $s$ in any vertical strip such as the critical strip.

Now to calculate $\zeta'(s)$ up to error of order $t^{-N}$ it is sufficient to calculate $\zeta(s+ih)$ and $\zeta(s-ih)$ with error less than $t^{-3N/2-1/4}$ for $h=t^{-N/2-1/4}$ and consider the difference quotient $(\zeta(s+ih)-\zeta(s-ih))/h$ by also using that $\zeta'''(s) =O(\sqrt t)$ in the critical strip and some version of the mean value theorem inequality. It is simple to use finite difference quotients to calculate $\zeta^{(k)}(s)$ up to any desired degree of accuracy. Thus we can certainly use this method to calculate for example $$\frac d {ds} \frac{\zeta'(s)}{\zeta(s)}= \frac{\zeta''(s)\zeta(s)-\zeta'(s)^2}{\zeta(s)^2}$$ by this method. This method should not be computionally wasteful.

I'm sure you've come across arXiv/1105.4312 --- this straightforward approach, numerical differentiation, is not necessarily less efficient than the route via the zeros. Alternatively, for that route, I think you'll find much of what you need in the work of Mark Coffey, in particular:

The $j$-th logarithmic derivative of the function $\xi(s)=(s/2)(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)$ is given by

$$\left[\frac{\xi'(s)}{\xi(s)}\right]^{(j)}=-\sum_{k=j+1}^{\infty}(k-1)(k-2)\cdots(k-j)\sigma_{k}s^{k-j-1}$$

where $\sigma_{j}=\sum_\rho \rho^{-j}$ is the sum of reciprocal powers of the complex zeros of the Riemann zeta function $\zeta(s)$. These coefficients are rigorously bounded by $O[(1+1/\pi)^j/j$, but actually decay much more rapidly. An efficient way to compute them numerically is given in the Appendix to the 2004 paper, in terms of a related coefficient $\eta_k$, defined by $\sigma_k=(-1)^k\eta_{k-1}-(1-2^{-k})\zeta(k)+1$, $k\geq 2$.