# Get An Introduction to Special Functions PDF

By Carlo Viola (auth.)

ISBN-10: 3319413449

ISBN-13: 9783319413440

ISBN-10: 3319413457

ISBN-13: 9783319413457

The matters taken care of during this publication were specifically selected to symbolize a bridge connecting the content material of a primary direction at the undemanding concept of analytic features with a rigorous remedy of a few of crucial targeted features: the Euler gamma functionality, the Gauss hypergeometric functionality, and the Kummer confluent hypergeometric functionality. Such particular services are critical instruments in "higher calculus" and are often encountered in just about all branches of natural and utilized arithmetic. the one wisdom assumed at the a part of the reader is an figuring out of uncomplicated options to the extent of an straightforward direction masking the residue theorem, Cauchy's indispensable formulation, the Taylor and Laurent sequence expansions, poles and crucial singularities, department issues, and so forth. The ebook addresses the wishes of complicated undergraduate and graduate scholars in arithmetic or physics.

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**Extra resources for An Introduction to Special Functions**

**Example text**

8) Since we have identically ez z z −z z + = −z − , −1 2 e −1 2 the function z/(e z − 1) + z/2 is even and takes the value 1 at z = 0. 8) yields B0 = 1, 1 B1 = − , 2 B2k+1 = 0 (k = 1, 2, 3, . . ). 8) we get, for |z| < 2π, ∞ 1 = Bn n=0 ez − 1 = z zn n! ∞ Bn n=0 zn n! ∞ n=1 ∞ n−1 n=1 k=0 = z n−1 n! Bk z n−1 . k! (n − k)! Thus, for n ≥ 2, n−1 k=0 Bk = 0, k! (n − k)! , n−1 k=0 n Bk = 0 k (n = 2, 3, 4, . . ). 10) (n = 0, 2, 3, 4, . . 10) yields n k=0 n Bk = Bn k which can symbolically be written as Bn = (1 + B)n (n = 1), where in the binomial expansion of (1 + B)n the symbolic powers B k are replaced by Bk .

2 Bernoulli Numbers 43 1 1 1 1 , B4 = − , B6 = , B8 = − , 6 30 42 30 5 691 7 3617 , B12 = − , B14 = , B16 = − ,.... 9), in the disc |z| < π we get the Laurent expansion cot z = i ei z + e−i z 1 2i z 1 e2i z + 1 = i+ = i+ = i i z −i z 2i z 2i z e −e e −1 z e −1 z = i+ 1 1 − iz + z ∞ (−1)k B2k k=1 (2z) (2k)! 2k = 1 + z ∞ Bn n=0 (2i z)n n! ∞ (−1)k 22k B2k k=1 z 2k−1 , (2k)! whence π 1 − cot(πz) = 2z 2 ∞ z 2k−1 2 (2k)! (−1)k−1 (2π)2k B2k k=1 (|z| < 1). 15) we obtain Euler’s formulae ζ(2k) = (−1)k−1 (2π)2k B2k 2 (2k)!

19), for |z| < 2π we have ∞ n=0 zn n! m−1 k=0 = k Bn x + m xz ze ez − 1 m−1 ∞ = k=0 n=0 m−1 e z/m k=0 ∞ = m k Bn (mx) n=0 = k zn = Bn x + m n! m−1 k=0 z e(x+k/m)z ez − 1 e −1 (z/m) e x z ze = m z/m z z/m e −1 e −1 e −1 xz (z/m)n = n! ∞ n=0 z z n Bn (mx) . n! m n−1 Hence we get the multiplication formula for the Bernoulli polynomials: m−1 Bn x + Bn (mx) = m n−1 k=0 k m (n = 0, 1, 2, . . ; m = 1, 2, . . ). 1 Stirling’s Formula for n! For n ∈ N let π/2 Sn : = sinn x dx. 0 Integrating by parts we get, for n ≥ 2, Sn = − sin n−1 x cos x π/2 0 π/2 + (n − 1) cos2 x sinn−2 x dx 0 π/2 = (n − 1) 1 − sin2 x sinn−2 x dx = (n − 1)Sn−2 − (n − 1)Sn , 0 whence the recurrence formula Sn = n−1 Sn−2 n (n ≥ 2).

### An Introduction to Special Functions by Carlo Viola (auth.)

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