The whole idea.............generally speaking of what Euler did with his product...............was equate a series of sums with one of products............................he equated addition to multiplication........

Euler product

From Wikipedia, the free encyclopedia

Jump to: navigation, search

In number theory, an Euler product is an expansion of a Dirichlet series into an infinite product indexed by prime numbers. The original such product was given for the sum of all positive integers raised to a certain power as proven by Leonhard Euler. This series and its continuation to the entire complex plane would later become known as the Riemann zeta function.

Contents

 [hide] 

• 1 Definition
• 2 Convergence
• 3 Examples
• 4 Notable constants
• 5 Notes
• 6 References
• 7 External links

Definition[edit]

In general, if ${\displaystyle a}$ is a multiplicative function, then the Dirichlet series

${\displaystyle \sum _{n}a(n)n^{-s}\,}$

is equal to

${\displaystyle \prod _{p}P(p,s)\,}$

where the product is taken over prime numbers ${\displaystyle p}$, and ${\displaystyle P(p,s)}$ is the sum

${\displaystyle 1+a(p)p^{-s}+a(p^{2})p^{-2s}+\cdots .}$

In fact, if we consider these as formal generating functions, the existence of such a formal Euler product expansion is a necessary and sufficient condition that ${\displaystyle a(n)}$ be multiplicative: this says exactly that ${\displaystyle a(n)}$ is the product of the ${\displaystyle a(p^{k})}$ whenever ${\displaystyle n}$ factors as the product of the powers ${\displaystyle p^{k}}$ of distinct primes ${\displaystyle p}$.
An important special case is that in which ${\displaystyle a(n)}$ is totally multiplicative, so that ${\displaystyle P(p,s)}$ is a geometric series. Then

${\displaystyle P(p,s)={\frac {1}{1-a(p)p^{-s}}},}$

as is the case for the Riemann zeta-function, where ${\displaystyle a(n)=1}$, and more generally for Dirichlet characters.

Convergence[edit]

In practice all the important cases are such that the infinite series and infinite product expansions are absolutely convergent in some region

${\displaystyle \operatorname {Re} (s)>C,}$

that is, in some right half-plane in the complex numbers. This already gives some information, since the infinite product, to converge, must give a non-zero value; hence the function given by the infinite series is not zero in such a half-plane.
In the theory of modular forms it is typical to have Euler products with quadratic polynomials in the denominator here. The general Langlands philosophy includes a comparable explanation of the connection of polynomials of degree m, and the representation theory for GL_m.

Examples[edit]

The Euler product attached to the Riemann zeta function ${\displaystyle \zeta (s),}$ using also the sum of the geometric series, is

${\displaystyle \prod _{p}(1-p^{-s})^{-1}=\prod _{p}{\Big (}\sum _{n=0}^{\infty }p^{-ns}{\Big )}=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=\zeta (s).}$

while for the Liouville function ${\displaystyle \lambda (n)=(-1)^{\Omega (n)},}$ it is

${\displaystyle \prod _{p}(1+p^{-s})^{-1}=\sum _{n=1}^{\infty }{\frac {\lambda (n)}{n^{s}}}={\frac {\zeta (2s)}{\zeta (s)}}.}$

Using their reciprocals, two Euler products for the Möbius function ${\displaystyle \mu (n)}$ are

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

and

${\displaystyle \prod _{p}(1+p^{-s})=\sum _{n=1}^{\infty }{\frac {|\mu (n)|}{n^{s}}}={\frac {\zeta (s)}{\zeta (2s)}}.}$

Taking the ratio of these two gives

${\displaystyle \prod _{p}{\Big (}{\frac {1+p^{-s}}{1-p^{-s}}}{\Big )}=\prod _{p}{\Big (}{\frac {p^{s}+1}{p^{s}-1}}{\Big )}={\frac {\zeta (s)^{2}}{\zeta (2s)}}.}$