Euler product
If f is a multiplicative function, then
∞∑n=1f(n)=∏p is prime(1+f(p)+f(p2)+⋯) | (1) |
provided the sum on the left converges absolutely. The product
on the right is called the Euler product for the sum on the
left.
Proof of (1).
Expand partial products on right of (1) to
obtain by fundamental theorem of arithmetic
∏p<y(1+f(p)+f(p2)+⋯) | =∑k1f(pk11)∑k2f(pk22)⋯∑ktf(pktt) | ||
=∑k1,k2,…,ktf(pk11)f(pk22)⋯f(pktt) | |||
=∑k1,k2,…,ktf(pk11pk22⋯pktt) | |||
=∑P+(n)<yf(n) |
where p1,p2,…,pt are all the primes between 1 and y, and P+(n) denotes the largest prime factor of n. Since every natural number less than y has no factors exceeding y we have that
|∞∑n=1f(n)-∑P+(n)<yf(n)|≤∞∑n=y|f(n)| |
which tends to zero as y→∞. ∎
Examples
-
•
If the function
f is defined on prime powers by f(pk)=1/pk for all p<x and f(pk)=0 for all p≥x, then allows one to estimate ∏p<x(1+1/(p-1))
∏p<x(1+1p-1)=∏p<x(1+1p+1p2+⋯)=∑P+(n)<x1n>∑n<x1n>lnx. One of the consequences of this formula is that there are infinitely many primes.
-
•
The Riemann zeta function
is defined by the means of the series
ζ(s)=∞∑n=1n-s for ℜs>1. Since the series converges absolutely, the Euler product for the zeta function
is
ζ(s)=∏p11-p-s for ℜs>1. If we set s=2, then on the one hand ζ(s)=∑n1/n2 is π2/6 (proof is here (http://planetmath.org/ValueOfTheRiemannZetaFunctionAtS2)), an irrational number, and on the other hand ζ(2) is a product of rational functions of primes. This yields yet another proof of infinitude of primes.
Title | Euler product |
---|---|
Canonical name | EulerProduct |
Date of creation | 2013-03-22 14:10:58 |
Last modified on | 2013-03-22 14:10:58 |
Owner | bbukh (348) |
Last modified by | bbukh (348) |
Numerical id | 8 |
Author | bbukh (348) |
Entry type | Definition |
Classification | msc 11A05 |
Classification | msc 11A51 |
Related topic | MultiplicativeFunction |
Related topic | RiemannZetaFunction |