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Difference between revisions of "Riemann zeta function"
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The '''zeta-function''' is a function very important to the [[Riemann Hypothesis]].  The function is <math>\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</math>
 
The '''zeta-function''' is a function very important to the [[Riemann Hypothesis]].  The function is <math>\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</math>
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The series is convergent [[iff] <math>\Re(s)>1</math>.  [[Euler]] showed that when <math>x=2</math>, the sum is equal to <math>\frac{\pi^2}{6}</math>.  Euler also found that since every number is the product of a certain combination of [[prime number]]s, the zeta-function can also be expressed as <math>{\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 \left(\frac{1}{(p^0)^s}+\frac{1}{(p^1)^s}+\frac{1}{(p^2)^s}+\cdots\right)\cdots</math>. By summing up each of these [[geometric series]] in parentheses, we have the following identity, the so called [[Euler Product]]: <math>\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}=\prod_{p\ \mathrm{prime}} (1-p^{-s})^{-1}</math>.
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The series is convergent [[iff] <math>\Re(s)>1</math>.  [[Euler]] showed that when <math>x=2</math>, the sum is equal to <math>\frac{\pi^2}{6}</math>.  Euler also found that since every number is the product of a certain combination of [[prime number]]s, the zeta-function can also be expressed as <math>{\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 </math>. By summing up each of these [[geometric series]] in parentheses, we have the following identity, the so called [[Euler Product]]: <math>\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}=\prod_{p\ \mathrm{prime}} (1-p^{-s})^{-1}</math>.
  
 
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 <math>s=1</math>, where there is a [[simple pole]] of [[residue]] 1. Let us see how this is done: First, we wish to extend <math>\zeta(s)</math> to <math>\Re(s)>0</math>. To do this, we introduce the alternating zeta function <math>\zeta_a(s)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}</math>, which is convergent on <math>\Re(s)>0</math>. (This follows from one of the standard [[convergence test]]s for alternating series.) We then have <math>\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)}</math>. We therefore have <math>\zeta(s)=(1-2^{1-s})^{-1}\zeta_a(s)</math> when <math>\Re(s)>0</math>.
 
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 <math>s=1</math>, where there is a [[simple pole]] of [[residue]] 1. Let us see how this is done: First, we wish to extend <math>\zeta(s)</math> to <math>\Re(s)>0</math>. To do this, we introduce the alternating zeta function <math>\zeta_a(s)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}</math>, which is convergent on <math>\Re(s)>0</math>. (This follows from one of the standard [[convergence test]]s for alternating series.) We then have <math>\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)}</math>. We therefore have <math>\zeta(s)=(1-2^{1-s})^{-1}\zeta_a(s)</math> when <math>\Re(s)>0</math>.
  
 
The next step is the [[functional equation for the zeta function|functional equation]]: Let <math>\xi(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s)</math>. Then <math>\xi(s)=\xi(1-s)</math>. This gives us a meromorphic continuation of <math>\zeta(s)</math> to all of <math>\mathbb{C}</math>.
 
The next step is the [[functional equation for the zeta function|functional equation]]: Let <math>\xi(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s)</math>. Then <math>\xi(s)=\xi(1-s)</math>. This gives us a meromorphic continuation of <math>\zeta(s)</math> to all of <math>\mathbb{C}</math>.

Revision as of 13:38, 28 June 2006

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] $\Re(s)>1$. 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 $\mathbb{C}$.

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