How do we know we're finding all of the zeros to Riemann zeta function?

Question

I've seen it reported that we've found all of the nontrivial zeros to the Riemann zeta function up to some large height, and that all of them have real part $1/2$; i.e. a counterexample would need imaginary part larger than that. But how do we know that we are not missing any zeros?

I found a formula for counting the number of zeros $N(T)$ up to some height $T$:

$$N(T) = \frac{T}{2\pi}\left(\log\frac{T}{2\pi} - 1\right) + O(\log T)$$

Wikipedia says (without citation) and explicit form of the error is:

$$N(T) < \frac{T}{2\pi}\left(\log\frac{T}{2\pi} - 1\right) - \frac{7}{8} + 0.137\log T + 0.443\log\log T + 4.350$$

For illustration, the 3rd zero is reported as having $T=25.01085...$ If we plug that in to the inequality, we get $N(T)<5.95$. So since we only found three zeros, who's to say there aren't 2 zeros hiding out somewhere with real part $\neq 1/2$, imaginary part $<25$? Obviously this gets much more loose when we're talking about zeros up to $10^{20}$ or whatever.

Answer 1

The zeta function is so well-understood in many ways that we can actually accomplish this task. Here is the general strategy (of course, actual computational implementations are much more complicated in an effort to gain as much speed as possible). First, we:

• Make a function that can evaluate the completed zeta function $\xi(s)$ to however much precision you need. Not trivial (especially for fast calculation), but definitely doable.
• Note that the combination of the functional equation and Schwarz reflection gives us the identity $\xi(1-\bar s) = \overline{\xi(s)}$. In particular, $\xi(s)$ is real-valued on the critical line $\sigma=\frac12$.
• That means we can sample $\xi(s)$ at many values on the critical line and look for sign changes; every time we have a sign change between two sampled values, it is guaranteed that $\xi(s)$ (and hence $\zeta(s)$) has at least one zero on the critical line between the two sample points, by the usual intermediate value theorem.
• In this way, for example, we can show that $\zeta(s)$ has at least $29$ zeros up to height $H=100$. (There might be multiple zeros in those sampled intervals or zeros off of the critical line; but we can be sure that there are at least $29$ zeros.)

Then, independently we:

• Make a function that can evaluate the argument of the zeta function, and hence $S(T) = \frac1\pi \arg\zeta\big( \frac12+iT \big)$, to a reasonable precision. Again, not trivial but definitely doable.
• Derive a version of the formula $N(T) = \frac T{2\pi} \log\frac T{2\pi e} + \frac78 + S(T) + O\big( \frac1T \big)$ with an explicit constant---that is, with an error term that is at most $\frac CT$ in absolute value for some explicit $C>0$.
• Then, simply calculate the right-hand side of that explicit version of the formula. Note the answer must be an integer, so we don't even need our calculations to be all that precise!
• In this way, for example, we calculate $N(100)$ and the answer “just happens” to be $29$. Therefore we know that there are no zeros other than the ones we found on the critical line (and, for that matter, that all those zeros are simple; it is indeed also conjectured that all zeros of $\zeta(s)$ are simple zeros). We just verified the Riemann hypothesis up to height $100$!