chapter10 - Chapter 10 Solutions to 10.2 1. lim n a n+1 a n...

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Unformatted text preview: Chapter 10 Solutions to 10.2 1. lim n a n+1 a n = lim n 1 2 2n+2 2 2n 1 = 1 4 R = 4. 3. lim n a n+1 a n = lim n 2 n+1 n + 1 n 2n = lim n 2n n + 1 = 2 R = 1 2 . 5. lim n a n+1 a n = lim n 5 n+1 (n + 1)! n! 5 n = lim n 5 n + 1 = 0 R = . 7. The zeroes of q(x) = x 2 + 1 are x = i. The distance from x = 0 to the nearest zero of q(x) is 1; thus, R = 1. 9. The zeroes of q(x) = x 2 2x + 5 = 0 are x = 1 2i. the distance from x = 0 to to x = 1 2i is 1 2 + 2 2 = 5; consequently, R = 5. 11. a) By Theorem 10.2.1, f(x) = 1 x 2 1 is analytic for all x R such that x 2 1 0. Therefore, f(x) is analytic for all x such that x 1. b) The zeroes of q(x) = x 2 1 are x = 1. For 0 x < 1, R = 1 x = 1 |x |, and for 1 < x < 0, R = |1 x | = x + 1 = 1 |x |. For |x | > 1, R = |x | 1. Thus, R = 1 |x |, if |x | < 1, |x | 1, if |x | > 1. 13. n=0 n(n + 2)a n x n + n=1 (n 3)a n1 x n = 0 n=1 [n(n + 2)a n + (n 3)a n1 ]x n = n(n + 2)a n + (n 3)a n1 = 0 a n = 3 n n(n + 2) a n1 for n = 1, 2, 3, n = 1: a 1 = 2 3 a ; n = 2: a 2 = 1 4 2 a 1 = 2 4! a ; and a n = 0 for n = 3, 4, 5, Thus, f(x) = a + 2 3 a x + a 12 x 2 = a (1 + 2 3 x + 1 12 x 2 ). 293 294 Ch. 10 Series Solutions to Differential Equations 15. n=1 (n + 1)(n + 2)a n+1 x n n=1 na n1 x n = 0 n=1 [(n + 1)(n + 2)a n+1 na n1 ]x n = (n + 1)(n + 2)a n+1 na n1 a n+1 = n (n + 1)(n + 2) a n1 for n = 1, 2, 3, . For odd n : n = 1: a 2 = 1 3! a ; n = 3: a 4 = 3 5 4 a 2 = 3 5! a ; n = 5: a 6 = 5 7 6 a 4 = 5 3 7! a , and in general a 2k = 1 3 5 (2k 1) (2k + 1)! a for k = 1, 2, 3, . For even n : n = 2: a 3 = 2 4 3 a 1 = 2 2 4! a 1 ; n = 4: a 5 = 4 6 5 a 3 = 2 3 2! 6! a 1 ; n = 6: a 7 = 6 8 7 a 3 = 2 4 3! 8! a 1 , and in general a 2k+1 = 2 k+1 k! (2k + 2)! a 1 for k = 1, 2, 3, . Solutions to 10.3 1. The point x = 0 is an ordinary point. p(x) = 0 and q(x) = 1 are both polynomials and their power series expansions about 0 are valid for all x so by Theorem 10.3.1 try for a solution of the form y(x) = n=0 a n x n y = n=1 na n x n1 and y = n=2 n(n 1)a n x n2 . So y y = n=2 n(n1)a n x n2 n=0 a n x n = 0 k=0 (k + 2)(k + 1)a k+2 x k k=0 a k x k = k=0 [(k + 2)(k + 1)a k+2 a k ]x k = 0 (k + 2)(k + 1)a k+2 a k = 0 for k = 0, 1, a k+2 = a k (k + 2)(k + 1) for k = 0, 1, 2, ....
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chapter10 - Chapter 10 Solutions to 10.2 1. lim n a n+1 a n...

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