Euler product

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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.

Definition

In general, if a is a bounded multiplicative function, then the Dirichlet series

[math]\displaystyle{ \sum_{n} \frac{a(n)}{n^s}\, }[/math]

is equal to

[math]\displaystyle{ \prod_{p} P(p, s) \quad \text{for } \operatorname{Re}(s) \gt 1 . }[/math]

where the product is taken over prime numbers p, and P(p, s) is the sum

[math]\displaystyle{ \sum_{k=0}^\infty \frac{a(p^k)}{p^{ks}} = 1 + \frac{a(p)}{p^s} + \frac{a(p^2)}{p^{2s}} + \frac{a(p^3)}{p^{3s}} + \cdots }[/math]

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 a(n) be multiplicative: this says exactly that a(n) is the product of the a(pk) whenever n factors as the product of the powers pk of distinct primes p.

An important special case is that in which a(n) is totally multiplicative, so that P(p, s) is a geometric series. Then

[math]\displaystyle{ P(p, s)=\frac{1}{1-\frac{a(p)}{p^s}}, }[/math]

as is the case for the Riemann zeta function, where a(n) = 1, and more generally for Dirichlet characters.

Convergence

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

[math]\displaystyle{ \operatorname{Re}(s) \gt C, }[/math]

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 GLm.

Examples

The following examples will use the notation [math]\displaystyle{ \mathbb{P} }[/math] for the set of all primes, that is:

[math]\displaystyle{ \mathbb{P}=\{p \in \mathbb{N}\,|\,p\text{ is prime}\}. }[/math]

The Euler product attached to the Riemann zeta function ζ(s), also using the sum of the geometric series, is

[math]\displaystyle{ \begin{align} \prod_{p\, \in\, \mathbb{P}} \left(\frac{1}{1-\frac{1}{p^s}}\right) &= \prod_{p\ \in\ \mathbb{P}} \left(\sum_{k=0}^{\infty}\frac{1}{p^{ks}}\right) \\ &= \sum_{n=1}^{\infty} \frac{1}{n^s} = \zeta(s). \end{align} }[/math]

while for the Liouville function λ(n) = (−1)ω(n), it is

[math]\displaystyle{ \prod_{p\, \in\, \mathbb{P}} \left(\frac{1}{1+\frac{1}{p^s}}\right) = \sum_{n=1}^{\infty} \frac{\lambda(n)}{n^{s}} = \frac{\zeta(2s)}{\zeta(s)}. }[/math]

Using their reciprocals, two Euler products for the Möbius function μ(n) are

[math]\displaystyle{ \prod_{p\, \in\, \mathbb{P}} \left(1-\frac{1}{p^s}\right) = \sum_{n=1}^{\infty} \frac{\mu (n)}{n^{s}} = \frac{1}{\zeta(s)} }[/math]

and

[math]\displaystyle{ \prod_{p\, \in\, \mathbb{P}} \left(1+\frac{1}{p^s}\right) = \sum_{n=1}^{\infty} \frac{|\mu(n)|}{n^{s}} = \frac{\zeta(s)}{\zeta(2s)}. }[/math]

Taking the ratio of these two gives

[math]\displaystyle{ \prod_{p\, \in\, \mathbb{P}} \left(\frac{1+\frac{1}{p^s}}{1-\frac{1}{p^s}}\right) = \prod_{p\, \in\, \mathbb{P}} \left(\frac{p^s+1}{p^s-1}\right) = \frac{\zeta(s)^2}{\zeta(2s)}. }[/math]

Since for even values of s the Riemann zeta function ζ(s) has an analytic expression in terms of a rational multiple of πs, then for even exponents, this infinite product evaluates to a rational number. For example, since ζ(2) = π2/6, ζ(4) = π4/90, and ζ(8) = π8/9450, then

[math]\displaystyle{ \begin{align} \prod_{p\, \in\, \mathbb{P}} \left(\frac{p^2+1}{p^2-1}\right) &= \frac53 \cdot \frac{10}{8} \cdot \frac{26}{24} \cdot \frac{50}{48} \cdot \frac{122}{120} \cdots &= \frac{\zeta(2)^2}{\zeta(4)} &= \frac52, \\[6pt] \prod_{p\, \in\, \mathbb{P}} \left(\frac{p^4+1}{p^4-1}\right) &= \frac{17}{15} \cdot \frac{82}{80} \cdot \frac{626}{624} \cdot \frac{2402}{2400} \cdots &= \frac{\zeta(4)^2}{\zeta(8)} &= \frac76, \end{align} }[/math]

and so on, with the first result known by Ramanujan. This family of infinite products is also equivalent to

[math]\displaystyle{ \prod_{p\, \in\, \mathbb{P}} \left(1+\frac{2}{p^s}+\frac{2}{p^{2s}}+\cdots\right) = \sum_{n=1}^\infty \frac{2^{\omega(n)}}{n^s} = \frac{\zeta(s)^2}{\zeta(2s)}, }[/math]

where ω(n) counts the number of distinct prime factors of n, and 2ω(n) is the number of square-free divisors.

If χ(n) is a Dirichlet character of conductor N, so that χ is totally multiplicative and χ(n) only depends on n mod N, and χ(n) = 0 if n is not coprime to N, then

[math]\displaystyle{ \prod_{p\, \in\, \mathbb{P}} \frac{1}{1- \frac{\chi(p)}{p^s}} = \sum_{n=1}^\infty \frac{\chi(n)}{n^s}. }[/math]

Here it is convenient to omit the primes p dividing the conductor N from the product. In his notebooks, Ramanujan generalized the Euler product for the zeta function as

[math]\displaystyle{ \prod_{p\, \in\, \mathbb{P}} \left(x-\frac{1}{p^s}\right)\approx \frac{1}{\operatorname{Li}_s (x)} }[/math]

for s > 1 where Lis(x) is the polylogarithm. For x = 1 the product above is just 1/ζ(s).

Notable constants

Many well known constants have Euler product expansions.

The Leibniz formula for π

[math]\displaystyle{ \frac{\pi}{4} = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = 1 - \frac13 + \frac15 - \frac17 + \cdots }[/math]

can be interpreted as a Dirichlet series using the (unique) Dirichlet character modulo 4, and converted to an Euler product of superparticular ratios (fractions where numerator and denominator differ by 1):

[math]\displaystyle{ \frac{\pi}{4} = \left(\prod_{p\equiv 1\pmod 4}\frac{p}{p-1}\right)\left( \prod_{p\equiv 3\pmod 4}\frac{p}{p+1}\right)=\frac34 \cdot \frac54 \cdot \frac78 \cdot \frac{11}{12} \cdot \frac{13}{12} \cdots, }[/math]

where each numerator is a prime number and each denominator is the nearest multiple of 4.[1]

Other Euler products for known constants include:

[math]\displaystyle{ \prod_{p\gt 2} \left(1 - \frac{1}{\left(p-1\right)^2}\right) = 0.660161... }[/math]
[math]\displaystyle{ \begin{align} \frac{\pi}{4} \prod_{p \equiv 1\pmod 4} \left(1 - \frac{1}{p^2}\right)^\frac12 &= 0.764223... \\[6pt] \frac{1}{\sqrt{2}} \prod_{p \equiv 3\pmod 4} \left(1 - \frac{1}{p^2}\right)^{-\frac12} &= 0.764223... \end{align} }[/math]
[math]\displaystyle{ \prod_{p} \left(1 + \frac{1}{\left(p-1\right)^2}\right) = 2.826419... }[/math]
  • The strongly carefree constant ×ζ(2)2 OEISA065472:
[math]\displaystyle{ \prod_{p} \left(1 - \frac{1}{\left(p+1\right)^2}\right) = 0.775883... }[/math]
[math]\displaystyle{ \prod_{p} \left(1 - \frac{1}{p(p-1)}\right) = 0.373955... }[/math]
[math]\displaystyle{ \prod_{p} \left(1 + \frac{1}{p(p-1)}\right) = \frac{315}{2\pi^4}\zeta(3) = 1.943596... }[/math]
[math]\displaystyle{ \prod_{p} \left(1 - \frac{1}{p(p+1)}\right) = 0.704442... }[/math]
and its reciprocal OEISA065489:
[math]\displaystyle{ \prod_{p} \left(1 + \frac{1}{p^2+p-1}\right) = 1.419562... }[/math]
[math]\displaystyle{ \frac{1}{2}+\frac{1}{2} \prod_{p} \left(1 - \frac{2}{p^2}\right) = 0.661317... }[/math]
[math]\displaystyle{ \prod_{p} \left(1 - \frac{1}{p^2(p+1)}\right) = 0.881513... }[/math]
[math]\displaystyle{ \prod_{p} \left(1 + \frac{1}{p^2(p-1)}\right) = 1.339784... }[/math]
[math]\displaystyle{ \prod_{p\gt 2} \left(1 - \frac{p+2}{p^3}\right) = 0.723648... }[/math]
[math]\displaystyle{ \prod_{p} \left(1 - \frac{2p-1}{p^3}\right) = 0.428249... }[/math]
[math]\displaystyle{ \prod_{p} \left(1 - \frac{3p-2}{p^3}\right) = 0.286747... }[/math]
[math]\displaystyle{ \prod_{p} \left(1 - \frac{p}{p^3-1}\right) = 0.575959... }[/math]
[math]\displaystyle{ \prod_{p} \left(1 + \frac{3p^2-1}{p(p+1)\left(p^2-1\right)}\right) = 2.596536... }[/math]
[math]\displaystyle{ \prod_{p} \left(1 - \frac{3}{p^3}+\frac{2}{p^4}+\frac{1}{p^5}-\frac{1}{p^6}\right) = 0.678234... }[/math]
[math]\displaystyle{ \prod_{p} \left(1 - \frac{1}{p}\right)^7 \left(1 + \frac{7p+1}{p^2}\right) = 0.0013176... }[/math]

Notes

References

  • G. Polya, Induction and Analogy in Mathematics Volume 1 Princeton University Press (1954) L.C. Card 53-6388 (A very accessible English translation of Euler's memoir regarding this "Most Extraordinary Law of the Numbers" appears starting on page 91)
  • Apostol, Tom M. (1976), Introduction to analytic number theory, Undergraduate Texts in Mathematics, New York-Heidelberg: Springer-Verlag, ISBN 978-0-387-90163-3  (Provides an introductory discussion of the Euler product in the context of classical number theory.)
  • G.H. Hardy and E.M. Wright, An introduction to the theory of numbers, 5th ed., Oxford (1979) ISBN 0-19-853171-0 (Chapter 17 gives further examples.)
  • George E. Andrews, Bruce C. Berndt, Ramanujan's Lost Notebook: Part I, Springer (2005), ISBN 0-387-25529-X
  • G. Niklasch, Some number theoretical constants: 1000-digit values"

External links