# For c 09068997 6 taylor and maclaurin series 0 we

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Unformatted text preview: 11 264 2 2 15 2 10,800 2 19 2 766,080 0.7040. 61. From Exercise 21, we have 1 2 1 e x2 2 1 2 dx 0 1 0n 1 nx 2n dx 2nn! 0 1 2 1 2 63. fx x cos 2x n P5 x 2x3 x 0 1 n4nx2n 2n ! n 1 0 1 nx 2n 1 2n 1 2nn! 2 1 1 1 1 2 0 1 2! 22 3 n 0 2n n! 1 2n 1 3! 7 23 5 n 1 0.3414. 1 65. fx x ln x, c P5 x 2x5 3 x 1 x 1 1 24 3 x 1 4 24 71 x 1 1920 3 2 g P5 f −3 −2 3 P5 4 −2 −2 The polynomial is a reasonable approximation on the interval 3 , 3 . 44 The polynomial is a reasonable approximation on the interval 1 , 2 . 4 67. See Guidelines, page 680. 69. (a) Replace x with (b) Replace x with 3x. x. (c) Multiply series by x. (d) Replace x with 2x, then replace x with 2x, and add the two together. 5 S ection 9.10 71. y g kv0 cos tan g ln 1 k2 x tan x gx kv0 cos g k2 tan x gx kv0 cos x gx2 2v0 cos2 e 0, 73. f x 1 x2, (a) x x 2 kx 1 2 v0 cos kx v0 cos gx 2 2v0 cos2 kx 1 4 v0 cos gk2x 4 4v04 cos4 3 k2gx 4 4v04 cos4 3 3 kx 1 3 v0 cos gkx3 3v0 cos3 2 kgx3 3v0 cos3 2 493 kx v0 cos gx kv0 cos tan Taylor and Maclaurin Series 4 ... ... ... 0 0 (b) f 0 y fx x lim x→0 f0 0 lim e 1 x2 x→0 0 x 2 1 Let y x −3 −2 −1 1 2 n f0 75. By the Ratio Test: lim f 0 x2 2! f 0x 1! xn 1 n 1! n 0 0n x n! n→ 5 3 77. 5 43 3! 60 6 n! xn x→0 e n→ 0 x n e 1 x2 lim x x→0 1 x2 0 and we have f 0 ... lim . Then lim ln Thus, y f 1 x2 x x→0 3 ln y (c) e lim 1 ln x 0 which shows that n 79. 0.5 4 0.5 1.5 x k n Example: 1 0 kn x n x 2 n 83. g x x x 1 x 1 x a0 x a1 0 2n x n a0 x2 a1x x2 a0 a1x a0 x a2 1 a2 x 2 a2 x 2 a1 x2 2x ... ... a0 x2 a3 a2 a1 x3 ... Equating coefficients, a0 0 a1 a0 1 ⇒ a1 a2 a1 a0 0 ⇒ a2 1 a3 a2 a1 0 ⇒ a3 2 a4 a3 a2 3, etc. In general, an an 1 1 an 2. The coefficients are the Fibonacci numbers. 2.5 4! 0.0390625 81. 1 0 only. xn converges for all x. 0 n! 0.5 x 2 ln x x2 0. f x This series converges to f at x 10 1 lim x→0 5 128 . 494 Chapter 9 Infinite Series Review Exercises for Chapter 9 1 n! 1. an 2 : 6, 5, 4.67, . . . n Matches (a) 3. an 5n 7. an 2 10 0.3 5. an 4 n 1 : 10, 3, . . . Matches (d) 8 n The sequence seems to converge to 5. lim an 5n lim n→ n→ 2 n 0 2 n lim 5 n→ n 9. lim 1 5 11. lim 0 n2 n→ 12 0 n→ n3 2 n 1 Converges 13. lim n n→ sin 15. lim 1 n n lim n n→ 1 n n n 1 1 n n 17. An 0 n→ n Converges lim n→ 1 1 n 5000 1 n Converges 0 n n 0.05 4 5000 1.0125 1, 2, 3 (a) A1 5062.50 A5 5320.41 A2 5125.78 A6 5386.92 A3 5189.85 A7 5454.25 A4 5254.73 A8 5522.43 (b) A40 19. (a) 5 10 15 20 25 Sk 13.2 113.3 873.8 6448.5 3 2 0 > 1. k 5 10 15 20 0.4597 0.4597 0.4597 0.4597 (b) 25 0.82, r < 1. 23. Converges. Geometric series, r 0 2 3 S 1 0.4597 0 The series converges by the Alternating Series Test. n 12 − 10 Sk 27. 120 50,500.3 The series diverges geometric r 21. (a) 8218.10 (b) k n n Geometric series with a a 1 r 1 1 23 1 13 1 and r 3 2 . 3 12 − 0.1 25. Diverges. nth-Term Test. lim an 0. n→ 29. n 0 1 2n 1 3n n 1 0 1 2 n 1 12 n 0 1 1 3 n 1 13 2 3 2 1 2 R eview Exercises for Chapter 9 31. (a) 0.09 0.09 0.0009 ... 0.000009 0.09 1 0.01 ... 0.0001 0.09 0.01 n (b) 0.09 33. D1 1 0.09 0.01 35. See Exercise 110 in Section 9.2. 0.7 8 D 8 0.7 8 4 16 0.7 16 0.7 n x 16 0.7 16 0.7 8 37. ... 2 n 8 ln x 3x3 lim b→ 1 16 0.7 16 0.7 1 0 ln x dx 1 9x3 ... n 451 meters 3 b \$5087.14 1 9 1 9 0 1 n 1 39. n lim 1 1 n→ n 2n n3 2 n3 2 lim n→ n3 2n 1 n n1 2 en 1 en2 n 1 en2 2n 1n lim 1 e2n n→ n→ n→ 01 2 1 3 2 1 1 n 5 . . . 2n 1 4 6 . . . 2n 2n 3 1n lim n→ n an 1 an lim n→ 2n 1 n1 3 n3 2n 1 n 0<1 By the Ratio Test, the series converges. 53. (a) Ratio Test: lim n→ (b) an 1 an lim n→ 2n3 n1 Therefore, by the Ratio Test, the series diverges. n 1 n 47. Diverges by the nth-Term Test. en2 lim 1 n2 5 . . . 2n 1 1 1 > 4 2n 2 2n 2n 1 1 1 Since diverges (harmonic series), 2n 1n n 1 2n so does the original series. 51. lim n→ an 1 an 1 3 2 n n2 1e lim n 5 . . . 2n 1 4 6 . . . 2n 3 1 an 45. Converges by the Alternating Series Test. (Conditional convergence) n 1 n 1 n2 1 43. By a limit comparison test with the convergent p-series 1 the series converges. 3 2, n 1n 49. 1 Since the second series is a divergent p-series while the first series is a convergent p-series, the difference diverges. 2n n3 1 1 200 e 0.06 2 1 e0.06 12 1 1 n3 P e rt er 12 A By the Integral Test, the series converges. 41. n 0 1 11 8 D2 495 lim n n→ 1 35 n3 5n n 1 lim n n→ x 5 10 15 20 25 Sn 2.8752 3.6366 3.7377 3.7488 1 n 3 5 3 < 1, Converges 5 (c) 4 3.7499 0 (d) The sum is approximately 3.75. −1 12 3 2 496 Chapter 9 Infinite Series 1 x 1 dx x2 55. (a) N 1 N N N 5 n N 1 4x 4 1 dx x5 (b) N 1 4N4 N 20 30 40 1 n2 1.4636 1.5498 1.5962 1.6122 1.6202 0.2000 0.1000 0.0500 0.0333 0.0250 5 1 10 1 dx x2 N 10 20 30 40 1 n5 1.0367 1.0369 1.0369 1.0369 1.0369 1 dx x5 0.0004 0.0000 0.0000 0.0000 0.0000 N N n 1 N The series in part (b) converges more rapidly. The integral values represent the remainders of the partial sums. 57. f x x2 e 1 e 2 fx x2 1x e 4 1 e 8 fx fx 95 1809 59. Since x2 63. f x n 65. n 0 n n 1 z x 10 1 x 2 1 x 2 1 n lim n→ 95 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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