I have come across a strange approximation for the Riemann-Siegel theta function involving the Bernoulli numbers - namely that

$$\frac{1}{2} \log \left| B_{2 n}\right|\approx \vartheta (2n)\ ,\quad n \in \mathbb{N}$$

where $B_n$ is the $n$th Bernoulli number, and $\vartheta,$ the Riemann-Siegel theta function. More accurately, it seems that

$$\frac{1}{2} \left(\log \left(\frac{\left| B_{2 n}\right| }{\sqrt{2 n}}\right)-\left(\frac{2}{3}\right)^{2/3} \pi \right)$$

is closer to the actual value of $\vartheta (2n),$ and it may be true to say that

$$\lim_{n\rightarrow\infty}\frac{1}{2} \left(\log \left(\frac{\left| \sqrt{2}n \zeta(1-2n)\right| }{\sqrt{ n}}\right)-\left(\frac{2}{3}\right)^{2/3} \pi \right)=\vartheta (2n),$$

and it seems likely that large values of $|B_{2n}|$ can be estimated with

$$\sqrt{2 n} \exp \left(2 \vartheta (2 n)+\pi \left(\frac{2}{3}\right)^{2/3}\right).$$

I haven't seen any references anywhere alluding to any connection between the absolute values of the Bernoulli numbers at even $n,$ and the Riemann-Siegel theta function previously. Is the above statement correct, and if it is, what is the connection?


1 Answer 1


Summary: Your approximation is very good, but not quite perfect. The absolute error does not go to $0$ as $n\to\infty$. But it is astonishingly close: the error is approximately $1.3\times 10^{-8}+o(1)$.

Detailed answer: we can easily check your proposed approximation for $\vartheta(2n)$, since both $|B_{2n}|$ and $\vartheta(t)$ have well-known asymptotic expansions. In the case of $|B_{2n}|$, we have $$ B_{2n} = (-1)^{n-1} \zeta(2n) \frac{2(2n)!}{(2\pi)^{2n}} = (1+O(2^{-2n})) (-1)^{n-1} \frac{2(2n)!}{(2\pi)^{2n}}, $$ so, using this together with Stirling's approximation in the form $$ \log(m!) = m\log m-m+\frac12 \log(2\pi m)+O(m^{-1}) $$ we have that \begin{align} \frac12\log \frac{|B_{2n}|}{\sqrt{2n}} &= \frac12 \Big(\log 2+ 2n\log(2n)-2n+\frac12\log(4\pi n)+O(n^{-1}) - (2n)\log(2\pi) \\ & \qquad - \frac12\log(2n) +O(2^{-2n}) \Big) \\ &= n \log\frac{n}{\pi}-n + \frac14 \log(8\pi) + O(n^{-1}). \end{align} On the other hand, for $\vartheta(t)$ we have the asymptotic formula $$ \vartheta(t) = \frac{t}{2}\log\frac{t}{2\pi} - \frac{t}{2} - \frac{\pi}{8} + O(t^{-1}). $$ which for $t=2n$ gives $$ \vartheta(2n) = n\log\frac{n}{\pi} - n - \frac{\pi}{8} + O(n^{-1}). $$ It follows that the difference $\frac12\log \frac{|B_{2n}|}{\sqrt{2n}}-\vartheta(2n)$ satisfies $$ \lim_{n\to\infty}\left( \frac12\log \frac{|B_{2n}|}{\sqrt{2n}}-\vartheta(2n)\right) = \frac{\pi}{8} + \frac14 \log(8\pi) \neq \frac12\left(\frac23\right)^{2/3}\pi. $$

Now note that, amazingly, $\frac{\pi}{8} + \frac14 \log(8\pi)\approx 1.19874193858$ and $\frac12\left(\frac23\right)^{2/3}\pi \approx 1.19874195162$. The difference between these two constants is $\approx 1.304\times 10^{-8}$.

To summarize, there is certainly something interesting going on here, and I am very curious to know how you obtained your approximation. Was it just a lucky guess? If not, there seems to be some mysterious mechanism at play causing the constant $$ \frac12\left(\frac23\right)^{2/3}\pi - \frac{\pi}{8} - \frac14 \log(8\pi) $$ to assume a surprisingly small numerical value.

Note however that my approach of using the standard asymptotic expansions for $|B_{2n}|$ and $\vartheta(t)$ says nothing about a possible connection between the Bernoulli numbers and the Riemann-Siegel theta function, other than maybe the trivial fact that the asymptotics of the Bernoulli numbers is governed by factorials (aka, the gamma function), and the Riemann-Siegel $\vartheta(t)$ is also defined in terms of the gamma function.

  • $\begingroup$ great answer - thanks :) Yes, I though the gamma funcion might play a central role, and yes, it was just a lucky guess! $\endgroup$
    – martin
    Sep 24, 2015 at 8:25
  • 3
    $\begingroup$ @martin I am still curious how you found the approximation. Was it based on numerical evidence? What made you think of trying to find a connection between the Bernoulli numbers and the Riemann-Siegel theta function? How did you come up with the "lucky guess" of $\frac{\pi}{2} (2/3)^{2/3}$? There seems much that is mysterious about your cute little discovery. Of course, if this is part of an ongoing research project and you don't care to reveal the details, I understand. In that case it would be nice if you come back and update the question after you've finished the paper. $\endgroup$
    – Dan Romik
    Sep 24, 2015 at 22:43

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