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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 n–1 x n = 0 ⇒ ∑ n=1 ∞ [n(n + 2)a n + (n – 3)a n–1 ]x n = ⇒ n(n + 2)a n + (n – 3)a n–1 = 0 ⇒ a n = 3 – n n(n + 2) a n–1 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 n–1 x n = 0 ⇒ ∑ n=1 ∞ [(n + 1)(n + 2)a n+1 – na n–1 ]x n = ⇒ (n + 1)(n + 2)a n+1 – na n–1 ⇒ a n+1 = n (n + 1)(n + 2) a n–1 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 n–1 and y´´ = ∑ n=2 ∞ n(n – 1)a n x n–2 . So y´´ – y = ⇒ ∑ n=2 ∞ n(n–1)a n x n–2 – ∑ 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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This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue.
 Spring '08
 Harris

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