For c 09068997 6 taylor and maclaurin series 0 we

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online