4
$\begingroup$

Here is a super naive question from a physicist:

Given the zeros of the Riemann zeta function, $\zeta(s) = \sum_{n=1}^\infty n^{-s}$, how do I actually evaluate them?

On this web page I found a list of zeros. Well I guess if the values, call one ${a_\text{zero}}$, are just given in decimal expansion, then I'd have to run a program and see how it approaches zero $$\zeta({a_\text{zero}}) = \sum_{n=1}^\infty n^{-{a_\text{zero}}} = \frac{1}{1^{a_\text{zero}}} + \frac{1}{2^{a_\text{zero}}} + \frac{1}{3^{a_\text{zero}}} + \cdots\approx 0. \qquad ;\Re(a)>1$$

But are there also analytical solutions, let's call one ${b_\text{zero}}$, which I can plug in and see $\zeta({b_\text{zero}}) = 0$ exactly? Specifically the non-trivial ones with imaginary part are of interest. What I find curious is that these series, with each term being a real number multiplied by some $n^{-i\cdot \text{Im}(b_\text{zero})}$, then also kind of resemble Fourer series.

$\endgroup$
4
  • $\begingroup$ Try the trivial ones first. $\endgroup$ Jul 24, 2013 at 10:55
  • $\begingroup$ That is not the definition of the Riemann zeta function. $\endgroup$ Jul 24, 2013 at 10:58
  • 1
    $\begingroup$ It should be noted that, while the zeroes are found by approximation, the fact that their real part is exactly $\frac12$ is rigorously known for each (so far). $\endgroup$ Jul 24, 2013 at 11:15
  • $\begingroup$ The earlier threads here and here could help. $\endgroup$ Jul 24, 2013 at 11:33

2 Answers 2

5
$\begingroup$

I've never seen anyone vocalize any expectation for any of the nontrivial zeros to have an explicit symbolic expression, presumably from which one could in principle then mathematically prove said expression is a zero of $\zeta(s)$ ("plug in and see"). That sounds too good to be true. Perhaps there are cheap ways to express them analytically in terms of $\zeta(s)$ that basically amount to defining them to be zeros of the function in the first place in some roundabout fashion.

The trivial zeros though one can show are in fact zeros, using the functional equation and the fact that the gamma function has no zeros in the complex plane but it has poles at negative integers.

Beyond this, and the values of $\zeta$ at positive even integers (which derive from harmonic analysis essentially), you'll have to settle for numerical approximation instead of symbolic computation.

Chris notes in the comments that the series $\sum n^{-s}$ only defines $\zeta(s)$ for ${\rm Re}(s)>1$; obviously it doesn't converge to the left of this by comparison with the harmonic series. The Riemann zeta function is defined as the analytic continuation of the function defined by this series. It is the magical functional equation that relates $\zeta$'s values on the left half-plane to its values on the right.

For reference, the functional equation is given by $\xi(s)=\xi(1-s)$, where $\xi$ is the "completed" zeta function defined by $\xi(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s)$ (sometimes a factor like $s(1-s)$ is introduced to make the proofwork easier). The explanation for the mysterious "gamma factor" lies in heavy algebraic number theory machinery, which is the content of Tate's thesis.

The theory of zeta functions definitely has connections with Fourier theory. Coefficients can be twisted by characters, zeta functions can be attached to modular forms which themselves can be defined by $q$-series, the explicit formula for the growth of primes in terms of the nontrivial zeros can be interpreted as "the music of the primes," and so on. I am not sure how high the observation that the $n^{-s}$-series looks like a Fourier series in $\tau={\rm Im}(s)$ ranks on the ladder of significance.

As a physicist, you may also be interested to know that the zeros have been empirically observed to obey certain statistical spectral laws, keywords: "GUE ensemble," "random matrix theory." There's a nice collection of information related to these ideas here. Another direction of interest is that on the Riemann hypothesis the nontrivial zeros form a one-dimensional quasi-crystal, see here.

Hagen notes in the comments that even though we are forced (in the critical strip in particular) to settle with numerical approximation, we can in fact test whether discovered nontrivial zeros have real part exactly half. See this MSE question for the relevant math behind this.

(Sorry I can't be of much help.)

$\endgroup$
1
$\begingroup$

The Riemann Zeta function has zeros as follows:

(1) From the Euler Product follows $\zeta(s)\not=0$ for $\mathrm{Re}\,s>1$. Taking the functional equation into account results that the only zeros outside the critical strip $\{ s\in\mathbb C\mid 0\leq\mathrm{Re}\,s\leq1\}$ are the trivial zeros $-2,-4,-6,\ldots$

(2) Beyond the trivial zeros the Zeta-Function has also zeros within the critical strip $S = \{ s \in \mathbb{C} \mid 1 > \mathrm{Re} \, s > 0 \} $. These are the non-trivial zeros. Little is yet known about these and your link to the numerical calculations of Andrew Odlyzko refer exactly to those zeros. There is not an analytical solution yet available.

To your point on the Fourier series analogy. It is possible to write out the Zeta function for $\mathrm{Re}\,s>1$ as a so called partition function that (resembles the form of a Fourier):

$$Z(T):= \sum_{n=1}^\infty \exp \left(\frac{-E(n)}{k_B T}\right) = \sum_{n=1}^\infty \exp \left(\frac{-E_0 \log n}{k_B T}\right) \equiv \sum_{n=1}^\infty \exp \left(-s \log n\right) = \sum_{n=1}^\infty \frac{1}{n^s} = \zeta (s)$$

and so forth see here>>>. But this was not the question I guess.

$\endgroup$
1
  • 1
    $\begingroup$ Why was this answer downvoted? $\endgroup$
    – anon
    Jul 24, 2013 at 22:43

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.