By Carlo Viola (auth.)

ISBN-10: 3319413449

ISBN-13: 9783319413440

ISBN-10: 3319413457

ISBN-13: 9783319413457

The matters handled during this ebook were particularly selected to symbolize a bridge connecting the content material of a primary direction at the trouble-free thought of analytic capabilities with a rigorous remedy of a few of an important distinct capabilities: the Euler gamma functionality, the Gauss hypergeometric functionality, and the Kummer confluent hypergeometric functionality. Such unique capabilities are integral instruments in "higher calculus" and are often encountered in just about all branches of natural and utilized arithmetic. the single wisdom assumed at the a part of the reader is an knowing of easy recommendations to the extent of an common direction masking the residue theorem, Cauchy's crucial formulation, the Taylor and Laurent sequence expansions, poles and crucial singularities, department issues, and so on. The ebook addresses the wishes of complicated undergraduate and graduate scholars in arithmetic or physics.

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**Sample text**

12). 6) by proving Stirling’s formula for the Euler gamma-function, with an asymptotic expansion of the remainder term. 1 Stirling’s Formula for n! 7) we obtain an asymptotic formula for the Bernoulli numbers B2k . Since ∞ ∞ n −2k < n=2 x −2k dx = 1 1 1 = O 2k − 1 k we get ∞ 1 < ζ(2k) = 1 + 1 , k n −2k < 1 + O n=2 whence ζ(2k) = 1 + O 1 . 7), |B2k | = 2 √ 2k 4kπ (2π)2k e 2k 1+O 1 k 1 k . 2 Partial Summation and Euler’s Constant Our next task is to prove a partial summation formula, a variant of which was introduced by Abel in connection with the proof of Abel’s theorem on uniform convergence of power series.

3) if and only if ord F(z) ≤ A. 1 If ord F1 (z) = α1 and ord F2 (z) = α2 , then ord (F1 (z) + F2 (z)) ≤ max{α1 , α2 } and ord (F1 (z)F2 (z)) ≤ max{α1 , α2 }. Proof Let α = max{α1 , α2 }. Then F1 (z) ε exp(|z|α+ε ) and F2 (z) ε exp(|z|α+ε ), whence |F1 (z) + F2 (z)| ≤ |F1 (z)| + |F2 (z)| and |F1 (z)F2 (z)| = |F1 (z)| |F2 (z)| ε ε exp(|z|α+ε ) exp(2|z|α+ε ) ε exp(|z|α+2ε ). 1 Since |z|n nomial P(z), ord P(z) = 0. If deg P(z) = n, then ord e P(z) = n. To see this, note exp(|z|n+ε ), so that that |P(z)| |z|n , whence e P(z) = eRe P(z) ≤ e|P(z)| P(z) n n−1 ≤ n.

Because 1 < ζ(2k) ≤ ζ(2) = π2 . 18) 44 4 Bernoulli Numbers and Polynomials Since π 2 (2k)! (2k + 2)! 18) we obtain −B8 < B10 < −B12 < . . 14) we see that |B2k | decreases for k ≤ 3 and increases for k ≥ 3. 13). 16) shows that, for each k = 1, 2, 3, . . , ζ(2k) is a rational multiple of π 2k and hence is transcendental by Lindemann’s theorem (1882) on the transcendence of π. For each k = 1, 2, 3, . . , ζ(2k + 1) is conjectured to be transcendental, but very little has been proved so far about the arithmetical nature of ζ(2k + 1).