S ection 97 fx 13 x e fx x e fx f f0 x p3 x x f f0

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Unformatted text preview: 1 for all x and all n. ≤ 1 x 6! 6 < 0.001 < 0.72 x < 0.9467 0.3936 < 0.3936, z < 0 ez 4 0.9467 < x < 0.9467 0.3936 < x < 0 Note: Use a graphing utility to graph y cos x 1 x2 2 x 4 24 in the viewing window 0.9467, 0.9467 0.001, 0.001 to verify the answer. 63. See definition on page 650. 61. The graph of the approximating polynomial P and the elementary function f both pass through the point c, f c and the slopes of P and f agree at c, f c . Depending on the degree of P, the nth derivatives of P and f agree at c, f c . 65. The accuracy increases as the degree increases (for values within the interval of convergence). 67. (a) f x ex P4 x 1 gx Q5 x x Q5 x 12 x 2 13 x 6 14 x 24 xe x x P4 x (b) f x P5 x gx Q6 x x 13 x 2 x2 15 x 24 sin x (c) g x x3 3! x sin x x 1 Px x5 1 x2 3! x4 5! x5 5! x sin x x2 x P5 x 2 69. (a) Q2 x 14 x 6 1 x4 3! x2 32 x6 5! 2 2 (b) R2 x 1 x6 32 2 (c) No. The polynomial will be linear. Translations are possible at x 2 8n. 71. Let f be an even function and Pn be the nth Maclaurin polynomial for f. Since f is even, f is odd, f is even, f is odd, etc. All of the odd derivatives of f are odd and thus, all of the odd powers of x will have coefficients of zero. Pn will only have terms with even powers of x. 73. As you move away from x c, the Taylor Polynomial becomes less and less accurate. S ection 9.8 Section 9.8 1 3. Centered at 2 n xn n n 0 n n lim n→ n 9. 0 L 1 n 1xn n2 lim n→ 1 x 2 x < 1⇒R n 7. 1 u lim n 1 n→ un L 1 n n→ n 2x lim 2n 2 2x n→ 2n 2x 2 2 2n 2n 1 n lim n→ 1xn 1 n 1 n xn 1 nx n When x 1, the p-series n 1 n 1 1 n 0 lim n→ x 2 un 1 un lim n→ 2n 2 !x n 2n 1 1 n 1 n lim n→ 1 converges. 1 < x ≤ 1. 2n 2n !x n lim 2n n→ 2 2n 2 1x 0. 1n x 4n Since the series is geometric, it converges only if x 4 < 1 or xn 1 n 1! x n 1 n! xn 0 The series converges for all x. Therefore, the interval of convergence is <x< . n n Therefore, the series converges only for x 19. n→ un 1 un 1 diverges. n Therefore, the interval of convergence is 2n ! lim lim n 17. xn 0 n! n→ 1, the alternating series When x 1 2 n x 2 x 1 2x 2 . n n n2 2x n 0 1 1<x<1 Interval: 0 15. lim 2n2x n1 2x n 1 n 12 Since the series is geometric, it converges only if x 2 < 1 or 2 < x < 2. 2n 2 ! 2n ! 1 n xn n n→ lim n→ 2 x < 1⇒R 11. un 1 un un 1 un lim Thus, the series converges for all x. R lim lim n→ lim 1 2x n n2 n→ x n→ n 1 L 1 un 1 un lim 13. 1 1 n xn 2x 2n 2n ! n→ 477 Power Series 1. Centered at 0 5. Power Series 4 < x < 4. 478 Chapter 9 n 1 21. n 1 1 x n5n n 5 un 1 un lim n→ R Infinite Series n 1 lim 2 n n→ x 5n 1 5n 1 1 1 n n5n 1x 5 nx 5n lim n n→ 1 x 5 5 1 5 5 Center: x 5 Interval: 5<x 5 < 5 or 0 < x < 10 When x 0, the p-series n 1 1 diverges. When x n 10, the alternating series n 1 1n n 1 converges. Therefore, the interval of convergence is 0 < x ≤ 10. n 1 23. n 1 x n 0 n→ R 1 1 un 1 un lim n 1 1 lim n 2 x n n→ 1 n 2 n 2 1 n 1 1 x 1 n 1 n lim n→ 1x n2 1 x 1 1 Center: x 1 Interval: 1<x 1 < 1 or 0 < x < 2 When x 0, the series 1 n When x 0n 1 diverges by the integral test. 1 2, the alternating series n n n 0 1 1 converges. Therefore, the interval of convergence is 0 < x ≤ 2. x 25. n 3 3 1 x 3 3 n 1 is geometric. It converges if <1⇒ x 3 < 3 ⇒ 0 < x < 6. Interval convergence: 0 < x < 6 n 27. n 1 n 1 u lim n 1 n→ un 2x n 1 29. n lim n n→ lim n→ R 1 2x n n 2 n n 2x n 1 nn 2 lim 1 2x n n→ 1 2x 1 , the series 2 n by the nth Term Test. n When x 1 n lim n→ n→ x 2n 3 2n 3 ! 2n x2 2 2n 2n 1 ! x 2n 1 3 Therefore, the interval of convergence is 1 1 <x< 2 2 When x un 1 un lim 2 1 2 Interval: 0 x 2n 1 2n 1 ! 1 1 , the alternating series 2 n diverges 1 Therefore, the interval of convergence is 1 n 1n diverges. n1 1 1 <x< . 2 2 0 <x< . S ection 9.8 2 31. n 4. . . n n! 3 1 lim n→ an 1 an 1 xn n lim n→ 1 33. n 13 un 1 un 1 lim n 2 1x 3 4n lim 3x 4 3 n 1 4n 3x 3 n 1 1 n 1 3 7 4n 11 . . . 4n 1x 3 0 Center: x 3 Therefore, the series converges only for x x 35. n x 11 . . . 4n 4n 1 7 n→ n→ R 2 x 1 1<x<1 11 . . . 4n 4n 7 1 n→ n n 1 < x < 1. Interval of convergence: lim lim n→ ± 1, diverges. At x n 1 2 xn 1 n 1 xn Converges if x < 1 ⇒ lim R n c cn 1 n→ 3. 1 37. 1 n un 1 un x lim n c cn cn n→ x 1 c n 1 1 x c 0 x k n Since the series is geometric, it converges only if x k < 1 or k < x < k.. c c Center: x c Interval: c<x When x 0, the series c < c or 0 < x < 2c 1 n When x n 1 diverges. 1 2c, the series 1 diverges. n 1 Therefore, the interval of convergence is 0 < x < 2c. 1...k n! kk 39. n 1 lim n→ R un 1 un 1...k n kk n→ n 1k 1! n xn 1 kk n! 1...k ± 1, the series diverges and the interval of convergence is 1...k n 1 2. . .n kk n lim 1 xn n n 1 xn lim n→ 1 When x 41. 479 1 xn n n Power Series xn 0 n! 1 x 1 x2 2 1 1 < x < 1. ≥1 ... n 1 xn 1 n 1! 43. n 0 x 2n 1 2n 1 ! Replace n with n n 1 1. x 2n 1 2n 1 !...
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This note was uploaded on 01/11/2013 for the course MATH 111 taught by Professor Man during the Spring '13 term at University of Washington-Tacoma Campus.

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