Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Riemann zeta function"

m
(No difference)

Revision as of 11:42, 6 July 2007

The zeta-function is a function very important to the Riemann Hypothesis. The function is $\zeta (s)=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^s}=1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots$ The series is convergent iff $\Re(s)>1$. Leonhard Euler showed that when $x=2$, the sum is equal to $\frac{\pi^2}{6}$. Euler also found that since every number is the product of a certain combination of prime numbers, the zeta-function can also be expressed as ${\zeta}(s) = \left(\frac{1}{(2^0)^s}+\frac{1}{(2^1)^s}+\frac{1}{(2^2)^s}+\cdots\right) \left(\frac{1}{(3^0)^s}+\frac{1}{(3^1)^s}+\frac{1}{(3^2)^s}+\cdots\right) \left(\frac{1}{(5^0)^s}+\frac{1}{(5^1)^s}+\frac{1}{(5^2)^s}+\cdots\right) \cdots$. By summing up each of these geometric series in parentheses, we have the following identity, the so-called Euler Product: $\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}=\prod_{p\ \mathrm{prime}} (1-p^{-s})^{-1}$.

However, 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 $\Re(s)>0$. 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 $\mathbb{C}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Riemann_zeta_function&oldid=15358"