March 7, 2017

Table of Integrals, Series, and Products, Seventh Edition by Alan Jeffrey, Visit Amazon's Daniel Zwillinger Page, search

By Alan Jeffrey, Visit Amazon's Daniel Zwillinger Page, search results, Learn about Author Central, Daniel Zwillinger,

A piece demanding to decipher at the start, yet as soon as I bought used to it this can be the easiest math reference ebook i have ever had. i'm going to by no means want one other one except I put on this one out (not likely...SERIOUS binding) and if I do i will need to get an analogous publication back. attention-grabbing historical past published within the introductions to earlier models.

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Table of Integrals, Series, and Products, Seventh Edition

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Example text

1) = 2n n = (−1)n 2n + 1 k (−1)k 3. 3) (2n − 1)! 15810 n 1. 2k 2n − k n−k − 2k+1 2n − k − 1 n−k−1 k = 4n − 2k 2n − k n−k − 2k+1 2n − k − 1 n−k−1 k 2 = 4n − 2k 2n − k n−k − 2k+1 2n − k − 1 n−k−1 k 3 = (6n + 13)4n − 18n 2k 2n − k n−k − 2k+1 2n − k − 1 n−k−1 k 4 = (32n2 + 104n) k=1 n 2. k=1 n 3. k=1 n 4. 15910 n 1. k=0 2n n−k − 2n n−k−1 k= 1 n 4 − 2 2n n 2n n 2n n 3 · 4n 2n n 2n n − (60n + 75)4n 6 Numerical Series and Infinite Products n 2. k=0 n 3. 16010 2n 1. k=n+1 n (−1)r 2. 211 uk = u1 + u2 + u3 + .

157 n n k 1. k=0 2 2n 2n k (−1)k 2. k=0 2n+1 k=0 n 4. 1) = 2n n = (−1)n 2n + 1 k (−1)k 3. 3) (2n − 1)! 15810 n 1. 2k 2n − k n−k − 2k+1 2n − k − 1 n−k−1 k = 4n − 2k 2n − k n−k − 2k+1 2n − k − 1 n−k−1 k 2 = 4n − 2k 2n − k n−k − 2k+1 2n − k − 1 n−k−1 k 3 = (6n + 13)4n − 18n 2k 2n − k n−k − 2k+1 2n − k − 1 n−k−1 k 4 = (32n2 + 104n) k=1 n 2. k=1 n 3. k=1 n 4. 15910 n 1. k=0 2n n−k − 2n n−k−1 k= 1 n 4 − 2 2n n 2n n 2n n 3 · 4n 2n n 2n n − (60n + 75)4n 6 Numerical Series and Infinite Products n 2. k=0 n 3.

P) kp 2 3 ∞ (−1)k+1 2. 10 k=1 1 22n−1 π 2n |B2n |, = k 2n (2n)! ∞ (−1)k+1 4. k=1 ∞ 5. k=1 (−1)k+1 k=1 ∞ k=1 1 π2 = k2 6 1 (22n−1 − 1)π 2n |B2n | = k 2n (2n)! 1 (22n − 1)π 2n |B2n | = 2n (2k − 1) 2 · (2n)! ∞ 6. 1 = (1 − 21−p ) ζ(p) kp 1 π 2n+1 |E2n | = 2n+2 2n+1 (2k − 1) 2 (2n)! 234 Examples of numerical series ∞ (−1)k+1 1. k=1 ∞ 2. k=1 ∞ 3. k=0 ∞ 4. k=1 ∞ 5. k=1 ∞ 6. ∗ k=1 ∞ EU 1 π2 = 2 (2k − 1) 8 EU (−1)k =G (2k + 1)2 FI II 482 (−1)k+1 π3 = (2k − 1)3 32 EU 1 π4 = 4 (2k − 1) 96 EU (−1)k+1 5π 5 = (2k − 1)5 1536 EU (−1)k+1 7.

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