# Why does the Riemann zeta function have zeros in the complex plane? How is it possible to find them?

I ask this because, according to Euler's product formula, Riemann's zeta function =(1/something), so how could that be zero? Also, how could one find zeros that are on the negative side and find a zero in the "critical strip", that is, when the real part of the input is between 0 and 1?

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Unfortunately, I don't have enough points to post comments, but just to tell you that you need to consider the analytic continuation of the function beyond the strip, to find the zeros. –  MAK Sep 30 '11 at 3:23
Euler's product formula is only valid when $\mathrm{Re}(s)>1$; indeed, the original $p$-series doesn't even converge unless $\mathrm{Re}(s)>1$, so one needs to perform analytic continuation. Otherwise Weierstrass factorization gives the Hadamard product $$2\pi^{-s/2}(s-1)\Gamma(1+s/2)\zeta(s)=\prod\left(1-\frac{s}{\rho}\right)$$ taken over the nontrivial zeroes $\rho$. I believe this product is globally convergent. –  anon Sep 30 '11 at 3:36
Quite related... –  Guess who it is. Sep 30 '11 at 8:02

Wherever it converges, the Riemann zeta function $\zeta(s)$ is equal to the $p$-series - and it converges only when the real part of $s$ is greater than one. The Euler product analogously only converges when $\mathrm{Re}(s)>1$, so the reasoning that $\zeta=1/\text{something}\ne0$ does not apply where the product is an invalid representation of zeta.

Moreover, the idea is flawed all by itself: the "$\text{something}$" is an infinite product, so if that product diverges to infinity in the limit, we have $\text{something}^{-1}\to0$. For example, formally we have

$$0\le\frac{1}{(1+1)(1+1/2)(1+1/3)\cdots}\le\frac{1}{1+1/2+1/3+\cdots}=\frac{1}{\infty}=0.$$

In general, properties (e.g. inequalities) satisfied by expressions with a finite number of terms do not necessarily carry over to an infinite number of terms (but we can still apply logic to the partial sums if need be).

So if neither the original series nor the product converges outside the right side of $1$, how can it be then that $\zeta(s)$ can be computed on or to the left of this? The answer is analytic continuation, which I gave a brief introductory explanation to just recently. But then precisely how can it be explicitly extended left of $1$? The simplest, but limited, way is through the Dirichlet eta function:

$$\eta(s)=1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}-\cdots=\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\cdots\right)-2\left(\frac{1}{2^s}+\frac{1}{4^s}+\cdots\right)$$ $$=\zeta(s)-\frac{2}{2^s}\zeta(s)=(1-2^{1-s})\zeta(s).$$ The eta function converges for $\mathrm{Re}(s)>0$, so if we write $\zeta(s)=(1-2^{1-s})^{-1}\eta(s)$ we have a way to evaluate zeta to the left of where we originally could (by $1$ at most). Analytic continuation beyond this requires more sophisticated machinery - a functional equation was first established by Bernhard Riemann in the paper "On the Number of Primes Less Than a Given Magnitude,"

$$\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s).$$ This means that what $\zeta$ evaluates to on the left of $s=1/2$ is determined by its evaluation on the same point reflected across $s=1/2$. With this and the $\eta$ continuation we are given the ability to compute $\zeta$ anywhere we like. Note that with the presence of the $\sin$ above, it is trivially true that $s=-2n$ are zeros of $\zeta$ for $n=1,2,3,\dots$ Moreover, since $\zeta(s)$ has no zeros right of $\mathrm{Re}(s)=1$, the functional equation predicts it has no other nontrivial zeros to the left of $\mathrm{Re}(s)=0$: this means all of the nontrivial zeros lie in the critical strip.

The Riemann Hypothesis is that all nontrivial zeros have real part $1/2$, but it has not been proven - we can, however, calculate the zeros to arbitrary accuracy and prove when particular zeros have real part exactly half, see this question linked by J.M.

As a sidenote, while the Euler product doesn't converge globally, I believe the Hadamard product (via Weierstrass factorization) does globally converge - see my comment above.

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+1! I'd have taken this example of an infinite product equal to $0$: $$\frac{1}{2}\cdot\frac{1}{2}\cdots$$ I'm lazier than you... –  Pierre-Yves Gaillard Oct 1 '11 at 1:23
@Pierre: Heh, for some reason I was unconsciously shooting for a product where the factors approach $1$ in the limit, to keep it analogous to the Euler product. –  anon Oct 1 '11 at 1:28

Inferences about location of zeros or poles of a function from a product formula are valid only in a region where the product converges to the function.

The constant function $f(x)=1$ has no zeros or poles but can be written as a product

$$f(x) = (1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})\dots$$

which superficially predicts zeros of $f(x)$, and poles of $1/f(x)$, at all $2^n$-th roots of unity. The formula holds for $|x|<1$ so there is no contradiction.

For $\zeta(s)$ the non-trivial zeros are known to exist only through Riemann's explicit formula and calculations on the line $s=1/2$. I don't know if there is any other argument for why zeros should exist with real part in $(0,1)$, except for analogy with other types of zeta functions (that did not exist in Riemann's time).

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