Difference between revisions of "Euler product for Riemann zeta"
From specialfunctionswiki
| Line 1: | Line 1: | ||
| − | + | ==Theorem== | |
| − | + | The following formula holds for $\mathrm{Re}(z)>1$: | |
$$\zeta(z)=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^z} = \displaystyle\prod_{p \mathrm{\hspace{2pt} prime}} \dfrac{1}{1-p^{-z}},$$ | $$\zeta(z)=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^z} = \displaystyle\prod_{p \mathrm{\hspace{2pt} prime}} \dfrac{1}{1-p^{-z}},$$ | ||
where $\zeta$ is the [[Riemann zeta function]]. | where $\zeta$ is the [[Riemann zeta function]]. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | ||
| − | + | ==References== | |
| + | * {{BookReference|The Zeta-Function of Riemann|1930|Edward Charles Titchmarsh|prev=Riemann zeta|next=Series for log(riemann zeta)}}: § Introduction (2) | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Unproven]] | ||
Revision as of 19:46, 9 June 2016
Theorem
The following formula holds for $\mathrm{Re}(z)>1$: $$\zeta(z)=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^z} = \displaystyle\prod_{p \mathrm{\hspace{2pt} prime}} \dfrac{1}{1-p^{-z}},$$ where $\zeta$ is the Riemann zeta function.
Proof
References
- 1930: Edward Charles Titchmarsh: The Zeta-Function of Riemann ... (previous) ... (next): § Introduction (2)
