By Claude Brezinski

ISBN-10: 0824786165

ISBN-13: 9780824786168

Publication by way of Brezinski, Claude

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F (s) = n=0 This series converges absolutely in the set {s ∈ C : |s| < 1} and uniformly in the set {s ∈ C : |s| < 1 − δ} for any δ > 0 to the sum 1/(1 − s). Now let g(s) = 1 1−s in C. Then g is analytic in the set C \ {1}, g(s) = f (s) in the set {s ∈ C : |s| < 1}, and g has a pole at s = 1. So g can be viewed as an analytic continuation of f to C with a pole at s = 1. Returning to the functions ζ(s) and L(s, χ), we shall establish the following results on analytic continuation. THEOREM 4H. The function ζ(s) admits an analytic continuation to the half plane σ > 0, and is analytic in this half plane except for a simple pole at s = 1 with residue 1.

Suppose that f (s) is an entire function satisfying f (0) = 0, and that (14) holds for every α > 1. Suppose further that s1 , s2 , s3 , . . are the zeros of f (s), counted with multiplicities and where |s1 | ≤ |s2 | ≤ |s3 | ≤ . . Then for every α > 1, the series ∞ |sn |−α n=1 is convergent. Proof. Note that the right hand side of (16) is equal to R r−1 n(r) dr, 0 where, for every non-negative r ≤ R, n(r) denotes the number of zeros of f (s) in |s| ≤ r. To see this, note that R n−1 r−1 n(r) dr = 0 j=1 |sj+1 | r−1 j dr + |sj | R r−1 n dr |sn | n−1 j(log |sj+1 | − log |sj |) + n(log R − log |sn |) = j=1 = n log R − log |s1 | − .

M χ(m)Λ(m)χ m|n It follows from Theorems 3B and 3E that for σ > 1, we have ∞ L (σ) = − χ(n) log n =− nσ n=1 ∞ χ(n)Λ(n) nσ n=1 ∞ χ(n) nσ n=1 . Hence ∞ (−1) n=1 n odd n−1 2 Λ(n) nσ ∞ = χ(n)Λ(n) L (σ) =− . σ n L(σ) n=1 Now as σ → 1+, we expect L(σ) → L(1) = 1 − 1 1 1 + − + ... > 0 3 5 7 and L (σ) → log 3 log 5 log 7 − + − ... 3 5 7 which converges by the Alternating series test. We therefore expect the series (1) to converge to a ﬁnite limit. 3. Dirichlet Characters Dirichlet’s most crucial discovery is that for every q ∈ N, there is a family of φ(q) functions χ : N → C, known nowadays as the Dirichlet characters modulo q, which generalize the function χ in the special case and satisfy 1 φ(q) χ mod q χ(n) = χ(a) 1 if n ≡ a (mod q), 0 if n ≡ a (mod q), where the summation is over the φ(q) distinct Dirichlet characters modulo q.

### Biorthogonality and its applications to numerical analysis by Claude Brezinski

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