Non-trivial zeros play an important (main) role in the distribution of prime numbers. Are there theorems in which trivial zeros play an important (main) role?

  • 7
    $\begingroup$ Sure. Look at the explicit formula that expresses Chebyshev's $\psi(x)$ (well, the modified version $\psi_0(x)$) in terms of a sum over the nontrivial zeros, e.g., at the start of the question mathoverflow.net/questions/262868/…. The proof of the formula also involves accounting for the trivial zeros, which is where the last term $\log(1 - 1/x^2)$ comes from. $\endgroup$ – KConrad Jul 22 '17 at 18:34
  • 4
    $\begingroup$ Basically, it is necessary to take them into account for things to work. They are understood well-enough that they make nearly no difference overall. $\endgroup$ – Wojowu Jul 22 '17 at 19:28
  • 3
    $\begingroup$ For exact formulas (such as Explicit Formulas) they must be included, but the size of their contribution is much less than the critical-strip zeros (since they are far to the left). $\endgroup$ – paul garrett Jul 22 '17 at 19:45

Since you tagged this 'reference request', I'm going to answer with a theorem in my own paper "Euler, the symmetric group, and the Riemann zeta function."

Let $\pi$ be a permutation in the symmetric group $S_n$. An ascent is an occurrence of $\pi(j)<\pi(j+1)$ for $1\le j\le n-1$. For example, the permutation $(24513)$ has $3$ ascents. The Eulerian number $\genfrac{\langle}{\rangle}{0pt}{1}{n}{k}$ is defined to be the number of permutations in $S_n$ with exactly $k$ ascents. (The Eulerian numbers are not to be confused with the Euler numbers $E_n$.) There is a surprising identity for alternating sums of Eulerian numbers: For integer $n\ge 1$ we have $$ \zeta(-n)=\frac{ \sum_{k=1}^n (-1)^{k} \genfrac{\langle}{\rangle}{0pt}{0}{n}{k}}{2^{n+1}(1-2^{n+1})}. $$

Of course, $\zeta(-n)$ can be expressed in closed form in terms of the Bernoulli numbers by $$ \zeta(-n)=-\frac{B_{n+1}}{n+1}, $$ so the theorem is also an identity relating Eulerian numbers to Bernoulli numbers. However, the proof is direct. It's also not not deep; it consists of identities in Concrete Mathematics by Graham, Knuth and Patashnik, along with Abel summation.


A conjecture of Quillen-Lichtenbaum

$$ \lim_{s \to n} (n-s)^{-\mu_n} \zeta_F (-s) = \pm \frac{\mid{K_{2n} (O_F)_{tor} }\mid}{\mid{K_{2n+1} (O_F)_{tor}}\mid} R_{F,n} * 2^{?}\\ $$

where $F$ is the number field like $\mathbb{Q}$ and $O_F$ is the integers therein generalizing $\mathbb{Z}$ in the case of Riemann zeta. $\mu_n$ is the multiplicity of the zero there. That makes sure you don't just get $0$. $K_\bullet (O_F)_{tor} $ means the torsion subgroup of algebraic K theory of that ring. $R_{F,n}$ is a so-called regulator and $2^?$ is for an unknown power of $2$. So you see the left hand side is leading information about when you have zeroes at negative integers.

This is proven in a bunch of cases via Voevodsky. In the $O_F=\mathbb{Z}$ case you just get the numerators and denominators of the Bernoulli numbers. That recovers the $-\frac{B_{n+1}}{n+1}$ above up to signs, powers of 2 and regulators.

So if you are interested in computing anything in the RHS, zooming in on the zeroes of the associated $\zeta_F$ are the most important.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy