Difference between revisions of "Riemann zeta"

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Consider the function $\zeta$ defined by the series
 
Consider the function $\zeta$ defined by the series
$$\zeta(z) = \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{n^z}.$$
+
$$\zeta(z) = \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^z}.$$
  
 
[[File:Riemannzeta.png|500px]]
 
[[File:Riemannzeta.png|500px]]

Revision as of 06:17, 11 February 2015

Consider the function $\zeta$ defined by the series $$\zeta(z) = \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^z}.$$

Riemannzeta.png

500px

Properties

Proposition: If $\mathrm{Re} \hspace{2pt} z > 1$, then the series defining $\zeta(z)$ converges.

Proof:

Proposition (Euler Product): $\zeta(z)=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^z} = \displaystyle\prod_{p \mathrm{\hspace{2pt} prime}} \dfrac{1}{1-p^{-z}}$

Proof:

Proposition: Let $n$ be a positive integer. Then $$\zeta(2n)=(-1)^{n+1}\dfrac{B_{2n}(2\pi)^{2n}}{2(2n)!},$$ where $B_n$ denotes the Bernoulli numbers.

Proof:

External links