Euler product

The Euler product is a term from the mathematical branch of analysis and in particular number theory . It is a representation of a Dirichlet series by means of an infinite product indexed over the set of prime numbers . The Euler product is named after Leonhard Euler , who investigated the infinite product with regard to the Dirichlet series of the Riemann zeta function .

definition

Let be a multiplicative number theoretic function and the corresponding Dirichlet series of the number theoretic function. If this series converges absolutely for a complex number , then we have ${\displaystyle f}$ ${\displaystyle \textstyle F(s):=\sum _{n=1}^{\infty }{\frac {f(n)}{n^{s}}}}$ ${\displaystyle s}$

${\displaystyle F(s)=\prod _{p\ {\rm {prim}}}\sum _{k=0}^{\infty }{\frac {f(p^{k})}{p^{ks}}}}$.

In the case of a fully multiplicative function , this product is simplified to ${\displaystyle f}$

${\displaystyle F(s)=\prod _{p\ \operatorname {prim} }{\frac {1}{1-f(p)p^{-s}}}}$.

These infinite products over all prime numbers are called Euler products. The value of these products is defined as the limit of the sequence of finite products , which is created by this product on prime numbers below a barrier N extends. ${\displaystyle \textstyle \lim _{N\to \infty }P_{N}}$ ${\displaystyle P_{N}}$

proof

There are several pieces of evidence for the validity of the Euler product.

First of all, it is clear that with absolute convergence of the series , every factor also converges absolutely. It follows that for each the partial product ${\displaystyle F(s)}$ ${\displaystyle \textstyle \sum _{k=0}^{\infty }{\frac {f(p^{k})}{p^{ks}}}}$ ${\displaystyle N}$

${\displaystyle F_{N}(s)=\prod _{p\leq N \atop p\ {\text{Primzahl}}}\sum _{k=0}^{\infty }{\frac {f(p^{k})}{p^{ks}}}}$

exists. With the Cauchy product formula and the ascending sequence of the prime numbers , one can immediately see : ${\displaystyle 2=p_{1}<p_{2}<\cdots <p_{j}\leq N<p_{j+1}}$

${\displaystyle F_{N}(s)=\sum _{k_{1}=0}^{\infty }\cdots \sum _{k_{j}=0}^{\infty }{\frac {f(p_{1}^{k_{1}})\cdots f(p_{j}^{k_{j}})}{(p_{1}^{k_{1}}\cdots p_{j}^{k_{j}})^{s}}}=\sum _{k_{1}=0}^{\infty }\cdots \sum _{k_{j}=0}^{\infty }{\frac {f(p_{1}^{k_{1}}\cdots p_{j}^{k_{j}})}{(p_{1}^{k_{1}}\cdots p_{j}^{k_{j}})^{s}}}.}$

In the second step the multiplicativity of was used. So follows ${\displaystyle f}$

${\displaystyle F_{N}(s)=\sum _{n\leq N}{\frac {f(n)}{n^{s}}}+\sum _{n>N}{'}{\frac {f(n)}{n^{s}}},}$

The line at the second sum indicates that the sum is only made over all whose prime divisors are all . It follows that for each there is a with ${\displaystyle n>N}$ ${\displaystyle \leq N}$ ${\displaystyle \varepsilon >0}$ ${\displaystyle N>0}$

${\displaystyle |F(s)-F_{N}(s)|\leq \sum _{n=N+1}^{\infty }\left|{\frac {f(n)}{n^{s}}}\right|<\varepsilon .}$

Thus the sequence of partial products for each converges in the region of absolute convergence to (even uniformly on compact subsets ) and the theorem is shown. ${\displaystyle F_{N}(s)}$ ${\displaystyle s}$ ${\displaystyle F(s)}$

The Euler product of the Riemann zeta function

formulation

In the case for all it is apparently completely multiplicative. It therefore applies to everyone ${\displaystyle f(n)=1}$ ${\displaystyle n\in \mathbb {N} }$ ${\displaystyle f}$ ${\displaystyle \operatorname {Re} (s)>1}$

${\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=\prod _{p}{\frac {1}{1-p^{-s}}}.}$

The function is also known as the Riemann zeta function . ${\displaystyle \zeta (s)}$

Derived from Euler

The idea of ​​this derivation path was already used by Euler. Take a subset and a prime such that and . So if it is , it also follows . Then applies quite generally to ${\displaystyle M\subset \mathbb {N} }$ ${\displaystyle p}$ ${\displaystyle 1\in M}$ ${\displaystyle pM\subset M}$ ${\displaystyle m\in M}$ ${\displaystyle pm\in M}$ ${\displaystyle \operatorname {Re} (s)>1}$

${\displaystyle \left(1-{\frac {1}{p^{s}}}\right)\sum _{m\in M}{\frac {1}{m^{s}}}=\sum _{m\in M}{\frac {1}{m^{s}}}-\sum _{m\in M}{\frac {1}{(pm)^{s}}}=\sum _{m\in M\setminus pM}{\frac {1}{m^{s}}}.}$

Let us now denote the sequence of prime numbers in ascending order, and the set of numbers that are not divisible by (e.g. ). Also put . Then each has the property above with the next prime number and it holds . So: ${\displaystyle p_{n}}$ ${\displaystyle M_{k}}$ ${\displaystyle p_{1},\dots ,p_{k}}$ ${\displaystyle M_{1}=\{1,3,5,7,\dots \}}$ ${\displaystyle M_{0}:=\mathbb {N} }$ ${\displaystyle M_{k}}$ ${\displaystyle p_{k+1}}$ ${\displaystyle M_{k+1}=M_{k}\setminus p_{k+1}M_{k}}$

${\displaystyle \left(1-{\frac {1}{p_{k+1}^{s}}}\right)\sum _{m\in M_{k}}{\frac {1}{m^{s}}}=\sum _{m\in M_{k+1}}{\frac {1}{m^{s}}}}$

and thus inductive

${\displaystyle \left(1-{\frac {1}{2^{s}}}\right)\left(1-{\frac {1}{3^{s}}}\right)\cdots \left(1-{\frac {1}{p_{k}^{s}}}\right)\zeta (s)=\sum _{m\in M_{k+1}}{\frac {1}{m^{s}}}.}$

If you form the Limes on both sides, this results

${\displaystyle \lim _{k\to \infty }\left(1-{\frac {1}{2^{s}}}\right)\left(1-{\frac {1}{3^{s}}}\right)\cdots \left(1-{\frac {1}{p_{k}^{s}}}\right)\zeta (s)=\lim _{k\to \infty }\sum _{m\in M_{k+1}}{\frac {1}{m^{s}}}=1,}$

since 1 is the only natural number that is not divisible by any prime number.

Web links

• Eric W. Weisstein : Euler Product . In: MathWorld (English).
• SA Stepanov: Euler product . In: Michiel Hazewinkel (Ed.): Encyclopaedia of Mathematics . Springer-Verlag , Berlin 2002, ISBN 978-1-55608-010-4 (English, online ).

Individual evidence

1. ^ Euler product . In: Guido Walz (Ed.): Lexicon of Mathematics . 1st edition. Spectrum Academic Publishing House, Mannheim / Heidelberg 2000, ISBN 3-8274-0439-8 .
2. ↑ Rainer Schulze-Pillot: Introduction to Algebra and Number Theory . 2nd corrected and enlarged edition. Springer-Verlag, Berlin, Heidelberg 2008, ISBN 978-3-540-79569-8 , pp. 53 .