Riemann zeta function

Revision as of 08:33, 19 April 2008 by JBL (talk | contribs) (huh, but it auto-replaces them anyhow. oh, well)

The Riemann zeta function is a function very important in number theory. In particular, the Riemann Hypothesis is a conjecture about the roots of the zeta function.

The function is defined by

\[\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots\]

when the real part $\Re(s)$ is greater than 1. (When $\Re(s) \leq 1$ the series does not converge, but it can be extended to all complex numbers except $s = 1$ -- see below.)

Leonhard Euler showed that when $s=2$, the sum is equal to $\frac{\pi^2}{6}$. Euler also found that since every number is the product of a unique combination of prime numbers, the zeta function can be expressed as and infinite product,

\[\zeta(s) = \left(\frac{1}{(2^0)^s}+\frac{1}{(2^1)^s}+\frac{1}{(2^2)^s}+\cdots\right) \left(\frac{1}{(3^0)^s}+\frac{1}{(3^1)^s}+\frac{1}{(3^2)^s}+\cdots\right) \left(\frac{1}{(5^0)^s}+\frac{1}{(5^1)^s}+\frac{1}{(5^2)^s}+\cdots\right) \cdots.\]

By summing up each of these geometric series in parentheses, we arrive at the following identity (the Euler Product):

\[\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}=\prod_{p\ \mathrm{prime}} (1-p^{-s})^{-1}.\]

This gives a hint of why an analytic object like the zeta function could be related to number theoretic results.


Extending the zeta function

The most important properties of the zeta function are based on the fact that it extends to a meromorphic function on the full complex plane which is holomorphic except at $s=1$, where there is a simple pole of residue 1. Let us see how this is done.

First, we wish to extend $\zeta(s)$ to the strip $\Re(s)>0$. To do this, we introduce the alternating zeta function $\zeta_a(s)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}$, which is convergent on $\Re(s)>0$. (This follows from one of the standard convergence tests for alternating series.) We then have $\zeta(s)=\zeta_a(s)+\frac{2}{2^s}+\frac{2}{4^s}+\frac{2}{6^s}+\cdots=\zeta_a(s)+2^{1-s}{\zeta(s)}$. We therefore have $\zeta(s)=(1-2^{1-s})^{-1}\zeta_a(s)$ when $\Re(s)>0$.

The next step is the functional equation: Let $\xi(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s)$. Then $\xi(s)=\xi(1-s)$. This gives us a meromorphic continuation of $\zeta(s)$ to all of $\mathbb{C}$. Template:Wikify