The zeta Function, Multiplying zeta Functions

Multiplying zeta Functions

Let two zeta functions, ζ(f,s) and ζ(g,s), converge absolutely on some half plane to the right of x. In other words, x is the larger of the two values of s_a. The product of the two zeta functions can be described, at least to the right of x.

For a given s to the right of x, ζ(f,s) and ζ(g,s) are represented as absolutely convergent series. Their product can be converted into a series of products. Group these together based on n^s. For instance, f₃/3^s times g₅/5^s contributes to h₁₅/15^s. In general, let h_n be the sum, as d divides n, of f_d×g_n/d. Thus ζ(h,s) = ζ(f,s)×ζ(g,s).

f Multiplicative

Let f be multiplicative if f_a×f_b = f_ab, for a and b coprime. Thus f_n is the product over f_j, where j is a prime power dividing n. Since 1 is coprime to everything, f₁ = 1.

Remove the coprime constraint, and f becomes completely multiplicative. In this case f_n is the product of f_p, for each instance of p dividing n.

Bring in the denominators n^s, and the terms f_n/n^s remain multiplicative, or completely multiplicative.

Consider ζ(f,s), where f is a multiplicative function. Let x = s_a, the line of absolute convergence. Every subsequence converges absolutely, so if we set some of the terms to 0, zeta remains absolutely convergent to the right of x. For instance, set f_n = 0, except where n is a power of p, and the resulting zeta function is absolutely convergent to the right of x.

Since f is multiplicative, ζ(f,s) is the infinite product of zeta functions based on prime powers. Each term f_n/n^s appears exactly once in the product, since n is uniquely a product of prime powers.

If f is completely multiplicative, look at the subsequence based on powers of p. Replace each f_pⁱ with f_pⁱ. Let u = f_p, and the zeta function looks like 1 + u/p^s + u²/(p²)^s + u³/(p³)^s, and so on. Remember that this geometric series converges, hence it is equal to 1 over 1-u/p^s.

Put this all together and, for f completely multiplicative, ζ(f,s) is the infinite product, for each prime p, of 1 over 1-f_p/p^s.

Inverse of Riemann zeta

Recall that the riemann zeta function is based on f_n = 1 for all n. Let g_n be the mobius function μ(n). The sum of μ(d), where d divides n, is 0, unless n = 1. Therefore the product ζ(f,s)×ζ(g,s) = 1, and g builds the inverse of zeta.

Before we put this to bed, make sure g converges to the right of 1. Remember that zeta is absolutely convergent, and each g_n is 0, 1, or -1. Either the term goes away, or it has the same norm as its counterpart in f. Therefore g is absolutely convergent wherever f is, i.e. to the right of 1. We have indeed found the inverse of zeta.

Since zeta is invertible to the right of 1, it is never 0 to the right of 1.

Since f_n = 1 is completely multipplicative, zeta can be written as the infinite product, over all primes p, of 1 over 1-1/p^s.

Since μ is multiplicative, the inverse of zeta can be written as the infinite product, over all primes p, of 1-1/p^s. Set s = 2 and find the product of 1-1/p², which is the inverse of ζ(2). This was used in an earlier theorem.