Difference between revisions of "Riemann zeta"

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{{:Euler product for Riemann zeta}}
<strong>Proposition (Euler Product):</strong> $\zeta(z)=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^z} = \displaystyle\prod_{p \mathrm{\hspace{2pt} prime}} \dfrac{1}{1-p^{-z}}$
 
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<strong>Proof:</strong> █
 
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Revision as of 05:11, 4 September 2015

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

Properties

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

Proof:

Theorem

The following formula holds for $\mathrm{Re}(z)>1$: $$\zeta(z)=\displaystyle\prod_{p \mathrm{\hspace{2pt} prime}} \dfrac{1}{1-p^{-z}},$$ where $\zeta$ denotes Riemann zeta.

Proof

References

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:

Theorem

The following formula holds: $$P(z)=\displaystyle\sum_{k=1}^{\infty} \dfrac{\mu(k)}{k} \log \zeta(kz),$$ where $P$ denotes the Prime zeta function, $\mu$ denotes the Möbius function, $\log$ denotes the logarithm, and $\zeta$ denotes the Riemann zeta function.

Proof

References

Videos

Riemann Zeta function playlist

External links