Riemann zeta function - one identity

Click For Summary
SUMMARY

The discussion centers on the Riemann zeta function, specifically the identity relating the number of Non-Isomorphic Abelian Groups by order n, denoted as p_n, to the zeta function. It establishes that for R(s) > 1, the equation ∑_{n=1}^{∞} p_n/n^s = ∏_{j=1}^{∞} ζ(js) holds true, where ζ(s) = ∑_{n=1}^{∞} 1/n^s. The conversation also references the fundamental theorem of arithmetic and the Euler product representation of the zeta function, emphasizing the relationship between partitions and generating functions.

PREREQUISITES
  • Understanding of the Riemann zeta function, specifically ζ(s).
  • Familiarity with Non-Isomorphic Abelian Groups and their properties.
  • Knowledge of generating functions and partition functions.
  • Basic concepts of infinite series and products in mathematics.
NEXT STEPS
  • Study the proof of the identity ∑_{n=1}^{∞} p_n/n^s = ∏_{j=1}^{∞} ζ(js).
  • Explore the properties of the partition function p(n) and its generating function.
  • Investigate the Euler product formula for the Riemann zeta function.
  • Learn about the implications of the fundamental theorem of arithmetic in number theory.
USEFUL FOR

Mathematicians, number theorists, and students studying advanced topics in algebra and analytic number theory, particularly those interested in the Riemann zeta function and its applications in group theory.

Karamata
Messages
59
Reaction score
0
Let p_n be number of Non-Isomorphic Abelian Groups by order n. For R(s)>1 with \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s} we define Riemann zeta function. Fundamental theorem of arithmetic is biconditional with fact that \zeta(s)=\prod_{p} (1-p^{-s})^{-1} for R(s)>1. Proove that for R(s)>1 is: \sum_{n=1}^{\infty}\frac{p_n}{n^s}=\prod_{j=1}^{∞}\zeta(js).

Do you know where can I found this proof (or maybe you know it :smile:)

Sorry for bad English :biggrin:
 
Physics news on Phys.org
The number of abelian groups of order n = ∏piai is ∏p(ai) where p is the partition function.

The generating function for p is Ʃp(n)xn=(1+x+x2+..)(1+x2+x4+..)(1+x3+..).. and we only need to know for later how to see that (the exponent taken from the first parantheses corresponds to the number of 1s in the partition etc).

Now ∏ ζ(js) = ζ(s)ζ(2s)ζ(3s) = ∏(1+1/ps+1/p2s+..)(1+1/p2s+..).. where the last product is over all primes, and we have written the Euler product of each factor ζ(js).

Write out some more terms, then study the expression, and you will see that the coefficient of 1/ns, after multiplying, is precisely what you want.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K