5
$\begingroup$

How can the Zeta function be zero?

If the zeta function is the Euler product:

$$\zeta(s)=\prod_p \frac{1}{1-p^{-s}}$$

Then being a product my first thought was that it could only be zero if one or more of its terms were zero.

This would require $\frac{1}{1-p^{-s}}$ to be zero for some prime $p$

So there would have to be some prime $p$ for which $p^{-s}$ is infinite.

Clearly I'm misunderstanding something. Are the zeroes where the terms $(1-p^{-s})$ diverge?

|cite|improve this question
$\endgroup$
  • 3
    $\begingroup$ When people speak of the zeta function, they refer to the analytic continuation of a function defined for Re$( s)>1$. $\endgroup$ – lulu Sep 13 '17 at 20:08
  • 4
    $\begingroup$ (This doesn't address the specific case of the Riemann zeta function, but is aimed at the OP's more general confusion around infinite products:) An infinite product can be zero without any term being zero - remember that its value is the limit of the finite partial products. So think about $\prod_{k\in\mathbb{N}}{1\over 2^k}$: each finite partial product is nonzero (since each term is nonzero), but those finite partial products get arbitrarily small. $\endgroup$ – Noah Schweber Sep 13 '17 at 21:00
  • 1
    $\begingroup$ (1/3) From an informal viewpoint: when you want to ask abut zeros of the Riemann Zeta function you need to know the theorems that say to you where has zeros the Riemann Zeta function $\zeta(s)$ and how is defined the Riemann Zeta function in the whole complex plane $\mathbb{C}$, or in some specific region. You need to check both conditions, see these informal examples: A) $\prod_p \frac{1}{1-p^{-s}}=0$ for some complex number with $\Re s>1$ is a contradiction because $\zeta(s)$ has no zeros in such region and $\sum_{n=1}^\infty\frac{1}{n^s}=0$ for $0<\Re s<1$ is a contradiction because this $\endgroup$ – user243301 Sep 14 '17 at 8:49
  • 1
    $\begingroup$ (2/3) Euler product doesn't converge in such region. B) Equate to $0$ the Mellin transform in $(11)$ of this MathWorld for $0<\Re s<1$ is right because we met both conditions, there are zeros in this region (non-trivial zeros) and both sides of $(11)$ are well defined (are convergent, since it is a theorem). $\endgroup$ – user243301 Sep 14 '17 at 8:49
  • 1
    $\begingroup$ (3/3) C) Finally in this section of the Digital Library of Mathematical Functions you've more representations for different regions of the complex plane, and if you equate to zero some of such representations you can to dilucidate when it has mathematical meaning. Good luck. $\endgroup$ – user243301 Sep 14 '17 at 8:49
3
$\begingroup$

  • The Euler product only converges for $\Re(s) > 1$.

  • The Riemann hypothesis is that $$\phi(s) = \prod_{k=1}^\infty \frac{1-(k \log k)^{-s}}{1-p_k^{-s}}$$ converges for $\Re(s) > 1/2$ (it is easy to show it converges for $\Re(s) > 1$ and the prime number theorem is that it converges for $\Re(s) \ge 1$)

  • That it converges means to an analytic function, ie. the series $\sum_{k=1}^\infty \log \frac{1-(k \log k)^{-s}}{1-p_k^{-s}}$ converges to an analytic function so that $\log \phi(s)$ is finite and $\phi(s)$ has no zeros.

    Since the analytic continuation of $\psi(s) = \sum_{k=1}^\infty \log (1-(k \log k)^{-s})$ is known to have no zeros for $\Re(s) > 0$, it implies $\log \zeta(s)-\psi(s)$ and hence $\log \zeta(s)$ have no zeros for $\Re(s) > 1/2$

  • That $\phi(s)$ converges for $\Re(s) > 1/2$ is equivalent to the number theoretic statement $$\pi(x) -\underbrace{ \sum_{2 \le k \le x} \frac{1}{\log k}}_{ = \ \text{Li}(x)+\mathcal{O}(1)} = \mathcal{O}(x^{1/2+\epsilon})$$

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ Thanks, this is all good stuff but the question asks how can it be zero. I'm not sure you answer that. Is it because the parts $(1-p_k^{-s})^{-1}$ become arbitrarily small like Noah says below? $\endgroup$ – samerivertwice Sep 13 '17 at 20:58
  • $\begingroup$ @RobertFrost No. See a course on complex analysis and the links of mixedmath answer. $2+\sum_{n=1}^\infty z^n$ converges only for $|z| < 1$, but writing it as $2+\frac{z}{1-z}$ shows its analytic continuation has a zero at $z=2$. This is exactly the same for $\zeta(s)$. $\endgroup$ – reuns Sep 13 '17 at 21:01
  • $\begingroup$ ok thanks. There's a lot of talk about convergence but I'm not seeing much about zeroes. It sounds like I need to better understand what it means for something's analytic continuation to have a zero, $\endgroup$ – samerivertwice Sep 13 '17 at 21:08
7
$\begingroup$

The Euler product only converges for $\mathrm{Re} (s) > 1$. For $\mathrm{Re} (s) \leq 1$, you need to consider a different representation of the zeta function.

The different representation comes from the "analytic continuation" of the zeta function. This has been written about extensively on this site. See for instance

  1. Riemann zeta function's analytic continuation
  2. What exactly is the Riemann zeta function?
  3. How are zeta function values computed in the critical strip (which is where all the interesting zeroes are).
|cite|improve this answer
$\endgroup$
  • $\begingroup$ And the only way to "see" the non-trivial zeros is with the functional equation allowing to define the real function $Z(t)$ corresponding to $\zeta(1/2+it)$ $\endgroup$ – reuns Sep 13 '17 at 21:15

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.