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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?

it's a hire car baby
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    When people speak of the zeta function, they refer to the analytic continuation of a function defined for Re$( s)>1$. – lulu Sep 13 '17 at 20:08
  • The Euler product $\prod_p \frac{1}{1-p^{-s}}$ as well as $\sum_{n=1}^\infty n^{-s}$ both converge only for $\Re(s) >1$. For $\Re(s) > 0$, $\zeta(s)$ is given by $\frac{1}{1-2^{1-s}} \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ or $\frac{s}{s-1} +s \int_1^\infty (\lfloor x \rfloor-x) x^{-s-1}dx$ or many other formulas – reuns Sep 13 '17 at 20:11
  • First they're not terms , they're factors. Second, there's not a finite number of factors, so the limit of the finite products may very well be $0$, albeit each finite product is different from $0$. – Bernard Sep 13 '17 at 20:13
  • How can a summation of infinite rational numbers sum up to an irrational number? Because we are taking a limit. Much like in your case, the factors never are zero, but keep getting closer and closer to zero, ergo is the limit. – Kaynex Sep 13 '17 at 20:17
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    (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. – Noah Schweber Sep 13 '17 at 21:00
  • Let me clarify why this is inappropriate as an answer: while it does address the OP's "base" confusion, it doesn't answer the question "for what reason(s) can we have $\zeta(z)=0$?," since $\zeta$ is not, in fact, defined as an infinite product in general. It does, however, answer the underlying question of "how can an infinite product of nonzero terms be zero?." – Noah Schweber Sep 13 '17 at 21:11
  • @RobertFrost The analytic continuation of the function is not convergent, but it is based on the convergent half of the function. Understanding why a product that is nowhere zero suddenly becomes zero is essentially what's missing. Then, taking that result and using it in the non-analytic half is the next, obvious step. – Kaynex Sep 13 '17 at 21:58
  • @Bernard: (1) The terms $1 - p^{-s}$ are not the factors of the OP's product. (2) It is standard to consider an infinite product to diverge if the limit of the finite products is $0$. – Rob Arthan Sep 13 '17 at 22:33
  • @RobArthan/ Yes, they're the denominators of the factors. Sorry for the ellipsis. – Bernard Sep 13 '17 at 22:52
  • Consider the product $1/10 \times1/10\times\cdots \times 1/10$. If you let the number of factors go to infinity, you get $0$ and none of the factors are zero. – YoTengoUnLCD Sep 14 '17 at 04:11
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    (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 –  Sep 14 '17 at 08:49
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    (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). –  Sep 14 '17 at 08:49
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    (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. –  Sep 14 '17 at 08:49

2 Answers2

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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).
davidlowryduda
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  • 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 $$\pi(x) -\underbrace{ \sum_{2 \le k \le x} \frac{1}{\log k}}_{ = \ \text{Li}(x)+\mathcal{O}(1)} = \mathcal{O}(x^{1/2+\epsilon})$$

reuns
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    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? – it's a hire car baby Sep 13 '17 at 20:58
  • @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 $\zeta(s)$. – reuns Sep 13 '17 at 21:01
  • 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, – it's a hire car baby Sep 13 '17 at 21:08