# Approximation for the $n$th nontrivial zero of $\zeta(s)$

For all positive integers $$n$$, let $$t_n = \frac{1}{2\pi} \operatorname{Im} \rho_n,$$ where $$\rho_n$$ donates the $$n$$th nontrivial zero of the Riemann zeta function in the upper half plane (listed in increasing order of imaginary part, counted to multiplicity, and including potential zeros off of the critical line). Due to the Riemann-Von Mangoldt explicit formula for $$N(T)$$, one has the good approximation $$t_n \sim \frac{n-11/8}{W((n-11/8)/e)} \ (n \to \infty)$$ of $$t_n$$, where $$W$$ is the Lambert $$W$$ function. (The $$11/8$$ comes as the average of $$7/8$$ and $$7/8+1$$.) I discovered this approximation recently, but haven't found it in the literature anywhere. My question is, what is a good $$O$$ bound for the error $$t_n-\frac{n-11/8}{W((n-11/8)/e)}$$? I suspect that one ought to be able to prove that $$t_n-\frac{n-11/8}{W((n-11/8)/e)} = O_t((\log n)^t) \ (n \to \infty)$$ for all $$t> 1$$, and in fact it might even be true that $$t_n-\frac{n-11/8}{W((n-11/8)/e)} = O_t((\log n)^t) \ (n \to \infty)$$ for all $$t>-1/2$$. I am curious what you think is the infimum of all $$t \in\mathbb{R}$$ such that the $$O$$ estimate above holds. It should definitely be the case at least that $$t_n-\frac{n-11/8}{W((n-11/8)/e)} = O_t(n^t) \ (n \to \infty)$$ for all $$t>0$$, but I don't know how to prove this. Probably also the error is $$O(1)$$, and maybe even $$o(1)$$. I'm not great at series inversion, but my guess is that it should be verifiable from the Riemann-Von Mangoldt formula and some series inversion.

Note that, for any fixed $$a \in \mathbb{R}$$, the inverse of the function $$T\log T-T+a$$ is $$\frac{n-a}{W((n-a)/e)}$$, and I believe that the Riemann-Von Mangoldt explicit formula for $$N(T)$$ implies that $$a = 11/8$$ is optimal for approximating $$t_n$$. By contrast, $$a = 7/8$$ is optimal for approximating the average of $$t_{n+1}$$ and $$t_n$$. One can also use this method to well-approximate the $$n$$th gap $$t_{n+1}-t_n$$ as $$t_{n+1}-t_n \approx \frac{n-11/8+1}{W((n-11/8+1)/e)}-\frac{n-11/8}{W((n-11/8)/e)} \approx \frac{1}{1+{W((n-7/8)/e)}}.$$

In analogy with the prime counting function, I would say that $$\operatorname{Ri}(x)$$ and $$\frac{x}{\log x}$$ are to $$\pi(x)$$ as $$\frac{n-11/8}{W((n-11/8)/e)}$$ and $$\frac{n}{\log n}$$ are to $$t_n$$, where $$\operatorname{Ri}(x)$$ is Riemann's approximation to $$\pi(x)$$. Please correct me if you think I'm wrong, or point out where someone might have observed this before! I think the idea of Gram spacing is related, but he didn't quite say the same thing as I'm saying here.

Here is a related question I asked: Riemann-Von Mangoldt formula, revised question. The reason I asked that question is because I wanted to understand the error better in the approximation $$N(T) \approx 1+\frac{1}{\pi}\theta(T)$$, where $$\theta$$ is the Riemann-Siegel theta function, which is the analogue of Riemann's approximation $$\pi(x) \approx \operatorname{Ri}(x)$$.

An alternative approach is to utilize the inverse of the Riemann-Siegel theta function, but that I don't know how to compute. At least the Lambert $$W$$ function is easily computable using Mathematica. Also the error in the approximation $$1+\frac{1}{\pi}\theta(2\pi T) \approx T\log T-T+\frac{7}{8}$$ is $$O(T^{-1})$$, so the inverses are close enough that it doesn't really matter too much which you use.

This is current research of mine, and I am interested in getting some informed opinions about it. Many thanks!

• Please use a high-level tag like "nt.number-theory". I added this tag now. Nov 10, 2021 at 6:03
• Note that locating the position of the $n$th zero is equivalent to counting nontrivial zeros with imaginary part not excedding a given value; to this end, we can apply [zero-counting results](aimath.org/WWN/rh/articles/html/71a). It turns out that your $O((\log(n)^t)$ estimate is correct for $t > 1$. Unfortunately, nothing better (in the sense of reducing $t$) is known. Nov 10, 2021 at 10:16
• I took the liberty to edit your question so as to fix the parentheses issue and make it more readable. Nov 10, 2021 at 10:25
• @LeechLattice. I think I've proved that the difference is $O(1)$. Please check out my answer below and let me know if I did something wrong. Nov 11, 2021 at 11:47
• Sorry, it's my fault. The last sentence should be corrected into "It turns out that your $O((\log(n)^t)$ estimate is correct for $t \geq 0$. Unfortunately, nothing better (in the sense of reducing $t$) is known." Nov 11, 2021 at 15:44

I think I might now have an answer to my question, in that the approximation I gave should be within $$O(1)$$ of $$t_n$$, and it can be related to the function $$N(T)$$. However, I am not yet sure in what sense if any $$11/8$$ is optimal.

Let $$\tau_n = 2\pi t_n$$ denote the imaginary part of the $$n$$th zero in the upper half plane, and let $$r_n = 2\pi \frac{n-11/8}{W((n-11/8)/e)}$$ for all $$n$$. One then has \begin{align*} n-1-\frac{1}{\pi}\theta(r_n) & = n- \frac{r_n}{2\pi} \log \frac{r_n}{2\pi}-\frac{r_n}{2\pi}-\frac{7}{8} + O\left( \frac{1}{r_n} \right)\\ & = \frac{1}{2}+ O\left(\frac{\log n}{n} \right) \ (n \to \infty), \end{align*} while also $$S(\tau_n) = -\frac{1}{2}+n-1-\frac{1}{\pi}\theta(\tau_n) = O(\log \tau_n) = O(\log n) \ (n \to \infty),$$ and therefore $$\theta(\tau_n)-\theta(r_n) = O(\log n) \ (n \to \infty).$$ Since $$2\theta(t)$$ is approximated by $$t\log t-t+\log 2\pi -\frac{\pi}{4}$$ to within $$O(T^{-1})$$, where the latter function has derivative $$\log t$$, by the mean value theorem it follows that \begin{align*} (\tau_n-r_n)\log u_n & = (\tau_n \log \tau_n - \tau_n)-(r_n \log r_n -r_n) \\ & = O(\log n) \ (n \to \infty) \end{align*} for some $$u_n$$ lying between $$r_n$$ and $$\tau_n$$ and therefore $$\tau_n-r_n = O(1) \ (n \to\infty).$$ More generally, the argument above shows that $$\tau_n-r_n = O\left(\frac{S(\tau_n)}{\log n}\right) \ (n \to\infty),$$ so that various conjectures bounding $$S(T)$$ yield bounds on the approximation of $$\tau_n$$. I suspect the converse is true as well.

I still think that $$a = 11/8$$ is optimal, but the argument above does not prove that. Note that if $$s_n = 2\pi \frac{n-11/8}{W((n-11/8)/e)},$$ then $$n-1-\frac{1}{\pi}\theta(s_n) = \frac{1}{2}+O\left(\frac{\log n}{n} \right) \ (n \to \infty),$$ so the same argument above implies that $$\tau_n-s_n = O(1)$$. I still think that $$\tau_n-r_n = O((\log n)^t)$$ for some $$t < 0$$, but I still don't know how to prove that. A weaker conjecture would be that $$\tau_n-s_n = o(1)$$.

ADDITIONAL EDIT: I now know at least one sense in which $$a = \frac{11}{8}$$ is optimal. I proved the following.

Proposition. Let $$a \in \mathbb{R}$$, and let $$r_n = r_n(a) = 2\pi \frac{n-a}{W((n-a)/e)}.$$ Then one has \begin{align*} r_n-\tau_n = 2\pi\frac{\frac{11}{8}-a +S(\tau_n)}{\log r_n} + O \left( \frac{1}{r_n \log r_n}\right) \ (n \to \infty). \end{align*} Consequently, one has $$\frac{n-a}{W((n-a)/e)} = \frac{\tau_n}{2\pi} + O\left(\frac{S(\tau_n)}{\log \tau_n} \right) \ (n \to \infty)$$ and $$\frac{n-a}{W((n-a)/e)} = \frac{\tau_n}{2\pi} + O(1) \ (n \to \infty).$$ Moreover, for $$a = \frac{11}{8}$$, one has \begin{align*} r_n-\tau_n = 2\pi\frac{S(\tau_n)}{\log r_n} + O \left( \frac{1}{r_n \log r_n}\right) \ (n \to \infty). \end{align*}

It follows from the proposition above and known $$\Omega$$ results on $$S(T)$$ that $$2\pi\frac{S(\tau_n)}{\log r_n}$$ is the true order of growth of $$r_n-\tau_n$$ when $$a = \frac{11}{8}$$. Another consequence of the proposition is that $$r_n-\tau_n = o(1)$$ if and only if $$S(T) = o(\log T)$$.

• Have you considered engaging with the literature on this? For example there is a classic book by Titchmarsh you may wish to consult. Or the answer that Anurag Sahay gave to your previous question, which already contains information on what is known for $N(T)$. Inverting this function (which all that is being done here) is a simple exercise. Nov 11, 2021 at 14:50
• @Lucia. I have read the literature you mentioned. And I digested Anurag's answer. So, yes, I have been engaging with the literature. Why do you always sound condescending in your comments? If it's a "simple exercise", then why has it not been written down before? And can you answer my question as to whether or not $\tau_n-s_n = o(1)$? If it is a "simple exercise", then maybe you know the answer. Nov 11, 2021 at 15:05
• For the same reason that two ten digit numbers are not multiplied and written down anywhere. It is not known how to get an asymptotic with $o(1)$. This would follow from the Riemann hypothesis (in some ways the problem is close to the Lindelof hypothesis). RH would give $O(1/\log \log n)$ (a result of Littlewood); the conjecture of Farmer Gonek and Hughes would predict something like $O(1/\sqrt{\log n})$. Again all this is easily deduced from Sahay's answer; or from Titchmarsh. Nov 11, 2021 at 15:10
• @Lucia. I think my question and answer provides more useful information than multiplying two ten digit numbers. Just because something is obvious to you doesn't mean it is to others. I think it's a nice observation that $\frac{n-11/8}{W((n-11/8)/e)}$ is a good approximation of $\frac{\tau_n}{2\pi}$, whether or not it is "easily deduced" from what is known about $N(T)$. In fact, I predicted the infimum I'm asking about is $-1/2$, which according to you would give another way to test the Farmer, Gonek, and Hughes conjecture. I'm not sure how to verify your claims, but I will think on it. Nov 11, 2021 at 15:21
• I also think it's an interesting question as to whether $11/8$ is optimal. Nov 11, 2021 at 16:08