242hw9solns - Math 242 Homework 9 Solutions (may be...

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Unformatted text preview: Math 242 Homework 9 Solutions (may be somewhat terse) In questions 1 to 5, find the interval of convergence of the power series. ∞ 1. n=1 xn n3 an+1 xn+1 n3 n3 = · n = |x| · → |x|. So |x| < 1. Checking an (n + 1)3 x (n + 1)3 endpoints, we find that both are included. ANSWER: [−1, 1]. ∞ 2. √ nxn n=1 √ n + 1 xn+1 n+1 an+1 √n = = |x| → |x|. So |x| < 1. Checking endan n nx points, we find that neither is included. ANSWER: (−1, 1). ∞ 3. n=2 xn 4n ln n an+1 xn+1 4n ln n |x| ln(n) |x| = n+1 · = · → . So |x| < 4. an 4 ln(n + 1) xn 4 ln(n + 1) 4 Checking endpoints, we find that −4 is included and 4 is excluded. ANSWER: [−4, 4). ∞ (−1)n 4. n=0 x2n (2n)! an+1 x2n+2 (2n)! x2 = · 2n = → 0. The series ALWAYS an (2n + 2)! x (2n + 2)(2n + 1) converges. ANSWER: (−∞, ∞). ∞ (−1)n 5. n=0 (x − 3)n 2n + 1 1 an+1 (x − 3)n+1 2n + 1 2n + 1 · → |x − 3|. So |x − 3| < 1, = = |x − 3|· an 2n + 3 (x − 3)n 2n + 3 so −1 < x − 3 < 1, so 2 < x < 4. Checking endpoints, we find that 2 is excluded and 4 is included. ANSWER: (2, 4]. In questions 6 to 10, find a power series representation of the given function. 6. f (x) = x 1−x 1 = 1 + x + x 2 + x3 + · · · 1−x x = x + x2 + x3 + x4 + · · · 1−x 1 9+x 1 1 1 1 1 =· =· . Note 9+x 9 1 + x/9 9 1 − (−x/9) 7. f (x) = 1 x x2 x3 = 1 − + 2 − 3 + ··· 1 − (−x/9) 99 9 1 1 x x2 x3 = − 2 + 3 − 4 + ··· 9+x 99 9 9 x 9 + x2 1 1 1 Note =· . 2 9+x 9 1 − (−x2 /9) 8. f (x) = x2 x4 x6 1 =1− + 2 − 3 + ··· 1 − (−x2 /9) 9 9 9 1 1 x2 x4 x6 = − 2 + 3 − 4 + ··· 9 + x2 99 9 9 x x x3 x5 x7 = − 2 + 3 − 4 + ··· 9 + x2 99 9 9 2 9. f (x) = cos 3x ∞ (−1)n n=0 (3x)2n (2n)! 10. f (x) = xex ∞ n=0 xn+1 n! 3 ...
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242hw9solns - Math 242 Homework 9 Solutions (may be...

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