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
∑
n
a
(
n
)
n
s
{\displaystyle \sum _{n}{\frac {a(n)}{n^{s}}}\,}
is equal to
∏
p
P
(
p
,
s
)
for
Re
(
s
)
>
1.
{\displaystyle \prod _{p}P(p,s)\quad {\text{for }}\operatorname {Re} (s)>1.}
where the product is taken over prime numbers p , and P (p , s ) is the sum
∑
k
=
0
∞
a
(
p
k
)
p
k
s
=
1
+
a
(
p
)
p
s
+
a
(
p
2
)
p
2
s
+
a
(
p
3
)
p
3
s
+
⋯
{\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 }
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
. Then
P
(
p
,
s
)
=
1
1
−
a
(
p
)
p
s
,
{\displaystyle P(p,s)={\frac {1}{1-{\frac {a(p)}{p^{s}}}}},}
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
Re
(
s
)
>
C
,
{\displaystyle \operatorname {Re} (s)>C,}
that is, in some right
. 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
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
P
{\displaystyle \mathbb {P} }
for the set of all primes, that is:
P
=
{
p
∈
N
|
p
is prime
}
.
{\displaystyle \mathbb {P} =\{p\in \mathbb {N} \,|\,p{\text{ is prime}}\}.}
The Euler product attached to the Riemann zeta function ζ (s ) , also using the sum of the geometric series, is
∏
p
∈
P
(
1
1
−
1
p
s
)
=
∏
p
∈
P
(
∑
k
=
0
∞
1
p
k
s
)
=
∑
n
=
1
∞
1
n
s
=
ζ
(
s
)
.
{\displaystyle {\begin{aligned}\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{aligned}}}
while for the Liouville function λ (n ) = (−1)ω (n ) , it is
∏
p
∈
P
(
1
1
+
1
p
s
)
=
∑
n
=
1
∞
λ
(
n
)
n
s
=
ζ
(
2
s
)
ζ
(
s
)
.
{\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)}}.}
Using their reciprocals, two Euler products for the Möbius function μ (n ) are
∏
p
∈
P
(
1
−
1
p
s
)
=
∑
n
=
1
∞
μ
(
n
)
n
s
=
1
ζ
(
s
)
{\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)}}}
and
∏
p
∈
P
(
1
+
1
p
s
)
=
∑
n
=
1
∞
|
μ
(
n
)
|
n
s
=
ζ
(
s
)
ζ
(
2
s
)
.
{\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)}}.}
Taking the ratio of these two gives
∏
p
∈
P
(
1
+
1
p
s
1
−
1
p
s
)
=
∏
p
∈
P
(
p
s
+
1
p
s
−
1
)
=
ζ
(
s
)
2
ζ
(
2
s
)
.
{\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)}}.}
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) = π 2 / 6 , ζ (4) = π 4 / 90 , and ζ (8) = π 8 / 9450 , then
∏
p
∈
P
(
p
2
+
1
p
2
−
1
)
=
5
3
⋅
10
8
⋅
26
24
⋅
50
48
⋅
122
120
⋯
=
ζ
(
2
)
2
ζ
(
4
)
=
5
2
,
∏
p
∈
P
(
p
4
+
1
p
4
−
1
)
=
17
15
⋅
82
80
⋅
626
624
⋅
2402
2400
⋯
=
ζ
(
4
)
2
ζ
(
8
)
=
7
6
,
{\displaystyle {\begin{aligned}\prod _{p\,\in \,\mathbb {P} }\left({\frac {p^{2}+1}{p^{2}-1}}\right)&={\frac {5}{3}}\cdot {\frac {10}{8}}\cdot {\frac {26}{24}}\cdot {\frac {50}{48}}\cdot {\frac {122}{120}}\cdots &={\frac {\zeta (2)^{2}}{\zeta (4)}}&={\frac {5}{2}},\\[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)}}&={\frac {7}{6}},\end{aligned}}}
and so on, with the first result known by
Ramanujan
. This family of infinite products is also equivalent to
∏
p
∈
P
(
1
+
2
p
s
+
2
p
2
s
+
⋯
)
=
∑
n
=
1
∞
2
ω
(
n
)
n
s
=
ζ
(
s
)
2
ζ
(
2
s
)
,
{\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)}},}
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
∏
p
∈
P
1
1
−
χ
(
p
)
p
s
=
∑
n
=
1
∞
χ
(
n
)
n
s
.
{\displaystyle \prod _{p\,\in \,\mathbb {P} }{\frac {1}{1-{\frac {\chi (p)}{p^{s}}}}}=\sum _{n=1}^{\infty }{\frac {\chi (n)}{n^{s}}}.}
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
∏
p
∈
P
(
x
−
1
p
s
)
≈
1
Li
s
(
x
)
{\displaystyle \prod _{p\,\in \,\mathbb {P} }\left(x-{\frac {1}{p^{s}}}\right)\approx {\frac {1}{\operatorname {Li} _{s}(x)}}}
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
π
4
=
∑
n
=
0
∞
(
−
1
)
n
2
n
+
1
=
1
−
1
3
+
1
5
−
1
7
+
⋯
{\displaystyle {\frac {\pi }{4}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+\cdots }
can be interpreted as a
superparticular ratios
(fractions where numerator and denominator differ by 1):
π
4
=
(
∏
p
≡
1
(
mod
4
)
p
p
−
1
)
(
∏
p
≡
3
(
mod
4
)
p
p
+
1
)
=
3
4
⋅
5
4
⋅
7
8
⋅
11
12
⋅
13
12
⋯
,
{\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)={\frac {3}{4}}\cdot {\frac {5}{4}}\cdot {\frac {7}{8}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdots ,}
where each numerator is a prime number and each denominator is the nearest multiple of 4.[1]
Other Euler products for known constants include:
∏
p
>
2
(
1
−
1
(
p
−
1
)
2
)
=
0.660161...
{\displaystyle \prod _{p>2}\left(1-{\frac {1}{\left(p-1\right)^{2}}}\right)=0.660161...}
π
4
∏
p
≡
1
(
mod
4
)
(
1
−
1
p
2
)
1
2
=
0.764223...
1
2
∏
p
≡
3
(
mod
4
)
(
1
−
1
p
2
)
−
1
2
=
0.764223...
{\displaystyle {\begin{aligned}{\frac {\pi }{4}}\prod _{p\equiv 1{\pmod {4}}}\left(1-{\frac {1}{p^{2}}}\right)^{\frac {1}{2}}&=0.764223...\\[6pt]{\frac {1}{\sqrt {2}}}\prod _{p\equiv 3{\pmod {4}}}\left(1-{\frac {1}{p^{2}}}\right)^{-{\frac {1}{2}}}&=0.764223...\end{aligned}}}
∏
p
(
1
+
1
(
p
−
1
)
2
)
=
2.826419...
{\displaystyle \prod _{p}\left(1+{\frac {1}{\left(p-1\right)^{2}}}\right)=2.826419...}
∏
p
(
1
−
1
(
p
+
1
)
2
)
=
0.775883...
{\displaystyle \prod _{p}\left(1-{\frac {1}{\left(p+1\right)^{2}}}\right)=0.775883...}
∏
p
(
1
−
1
p
(
p
−
1
)
)
=
0.373955...
{\displaystyle \prod _{p}\left(1-{\frac {1}{p(p-1)}}\right)=0.373955...}
∏
p
(
1
+
1
p
(
p
−
1
)
)
=
315
2
π
4
ζ
(
3
)
=
1.943596...
{\displaystyle \prod _{p}\left(1+{\frac {1}{p(p-1)}}\right)={\frac {315}{2\pi ^{4}}}\zeta (3)=1.943596...}
∏
p
(
1
−
1
p
(
p
+
1
)
)
=
0.704442...
{\displaystyle \prod _{p}\left(1-{\frac {1}{p(p+1)}}\right)=0.704442...}
and its reciprocal OEIS : A065489 :
∏
p
(
1
+
1
p
2
+
p
−
1
)
=
1.419562...
{\displaystyle \prod _{p}\left(1+{\frac {1}{p^{2}+p-1}}\right)=1.419562...}
1
2
+
1
2
∏
p
(
1
−
2
p
2
)
=
0.661317...
{\displaystyle {\frac {1}{2}}+{\frac {1}{2}}\prod _{p}\left(1-{\frac {2}{p^{2}}}\right)=0.661317...}
∏
p
(
1
−
1
p
2
(
p
+
1
)
)
=
0.881513...
{\displaystyle \prod _{p}\left(1-{\frac {1}{p^{2}(p+1)}}\right)=0.881513...}
∏
p
(
1
+
1
p
2
(
p
−
1
)
)
=
1.339784...
{\displaystyle \prod _{p}\left(1+{\frac {1}{p^{2}(p-1)}}\right)=1.339784...}
∏
p
>
2
(
1
−
p
+
2
p
3
)
=
0.723648...
{\displaystyle \prod _{p>2}\left(1-{\frac {p+2}{p^{3}}}\right)=0.723648...}
∏
p
(
1
−
2
p
−
1
p
3
)
=
0.428249...
{\displaystyle \prod _{p}\left(1-{\frac {2p-1}{p^{3}}}\right)=0.428249...}
∏
p
(
1
−
3
p
−
2
p
3
)
=
0.286747...
{\displaystyle \prod _{p}\left(1-{\frac {3p-2}{p^{3}}}\right)=0.286747...}
∏
p
(
1
−
p
p
3
−
1
)
=
0.575959...
{\displaystyle \prod _{p}\left(1-{\frac {p}{p^{3}-1}}\right)=0.575959...}
∏
p
(
1
+
3
p
2
−
1
p
(
p
+
1
)
(
p
2
−
1
)
)
=
2.596536...
{\displaystyle \prod _{p}\left(1+{\frac {3p^{2}-1}{p(p+1)\left(p^{2}-1\right)}}\right)=2.596536...}
∏
p
(
1
−
3
p
3
+
2
p
4
+
1
p
5
−
1
p
6
)
=
0.678234...
{\displaystyle \prod _{p}\left(1-{\frac {3}{p^{3}}}+{\frac {2}{p^{4}}}+{\frac {1}{p^{5}}}-{\frac {1}{p^{6}}}\right)=0.678234...}
∏
p
(
1
−
1
p
)
7
(
1
+
7
p
+
1
p
2
)
=
0.0013176...
{\displaystyle \prod _{p}\left(1-{\frac {1}{p}}\right)^{7}\left(1+{\frac {7p+1}{p^{2}}}\right)=0.0013176...}
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)
(Provides an introductory discussion of the Euler product in the context of classical number theory.)
(Chapter 17 gives further examples.)
George E. Andrews, Bruce C. Berndt, Ramanujan's Lost Notebook: Part I , Springer (2005),
G. Niklasch, Some number theoretical constants: 1000-digit values"
External links
International National Other