# A series of the form 17 matches d 8 1 s2 an x n n c 0

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Unformatted text preview: 2 x 2 x 2 r n x 2 x 4 1 2 x)1 ... 1 1 x 1 2 0 1 This series converges on (b) 2 x2 4 x2 4 x2 4 a 12 x2 1 n 16 5. Writing f x in the form a 1 2 x 1 2, 2 . x3 8 x 2 x 2 x 2 1 2 1 2 xn x2 x 4 3. (a) r n 1x 02 2 n a 1 This series converges on (b) 2 483 Representation of Functions by Power Series 1 2 Representation of Functions by Power Series 3 02 1 n xn 2n 1 2 < x < 2. 1 2 x. Therefore, 1 x 2 3 2n r 0 n xn , 2 484 13. Chapter 9 3x x x2 Infinite Series 2 2 1 x 2 2 x 1 2 1 1x x 1 1 2x 1 1 1 x Writing f x as a sum of two geometric series, we have 3x x x2 2 n n 1 x 2 0 n un 1 un n→ 15. 1 lim 0 n 2 2 2 n→ n 1 n 1 xn 1 n 2 1 xn x2 n x 0 n n 1 0 n n 2 1 1 1 1 lim xn 2 2 n→ n 1 n 2 x x. 1 xn 2x 2n. n 0 1 < x < 1 since lim n→ n lim n→ 2 x2n 2 2 x2 n x2 . 0 x n 0 1 1 n n x 1 n 2 hx 1 x2 1 2 0x 1 2n x n xn 0 n 1 x 2x2 1 1 n xn x 0x3 0 n 2x4 xn 0 n 0x5 1 0 n 19. By taking the first derivative, we have 1 1 x d dx 2 d 1 dx x 1 1 n xn n 2x2n, n 1 ln x x 1 n 1 1 25. Since, n 0 n 0 1 x2 , n 0 1 < x < 1. 0 1 , 1 < x ≤ 1. 0. Therefore, 1 < x ≤ 1. 1<x<1 0 1 n x n, we have n 1 n xn n1 C 1 n x 2n, n 1 x 1x n1 1 1 x n, n 1 . Therefore, 0 and conclude that C 1 1 1 n nn n n ln x 0 To solve for C, let x ln x dx 1 n x n dx 1 1 0 1 21. By integrating, we have 1 < x < 1 (See Exercise 15.) 0 1 n 1 xn 1 . Therefore, 12 x 1 n nx n 0 n 0 ... 2x6 n x2 un 1 un 1 n xn x 1 n 0 The interval of convergence is x 2 < 1 or 23. 1 x n. n 1 1 2 1 x2 1 x 1 x Writing f x as a sum of two geometric series, we have 1 17. 0 1 < x < 1 since The interval of convergence is lim 1 2 n 1x 1 4x 2 1 1 n 0 n 4x 2 n 1 n 4n x 2n n 0 1 n 0 n 2x 2n, 1 1 <x< . 2 2 S ection 9.9 x2 ≤ ln x 2 27. x x2 2 1 ≤x x3 3 x x 5 S3 f 8 x S2 −3 1 n n 1 x 1 n x n 1 (a) 1 x 1 1 2 x 2 1 3 2 3 (b) From Example 4, x n 1 n 1 n n x 1 1 n 0 n 0.182 0.336 0.470 0.588 0.693 0.183 0.341 0.492 0.651 0.833 n 1 1 33. g x In Exercises 35-37, arctan x 1 n 1 n n 0 arctan x 2 x 14 2n 1 xn 2n 1 1 n 1n 1 42n 2n 0 1 0 n x2 2n 2n 1 1 n 0 arctan x 2 dx x n 0 1 1 4n arctan x dx x Since 1 12 n x5 5 x2n 1 . 2n 1 n 0 2 0 n x3 3 x 1 4 1 192 1 5120 ... 1 1 Since 5120 < 0.001, we can approximate the series by its first two terms: arctan 4 12 n Matches (a) Matches (c) 37. 12 n 1. x line 1 4 0.500 1 0 < x ≤ 2, R ln x, 35. arctan 0.480 n 0.693147 (d) This is an approximation of ln 1 . The error is approxi2 mately 0. The error is less than the first omitted term, 1 51 251 8.7 10 18. −3 31. g x 0.420 0.5: n n=2 1 0.320 1 n=6 n 0.180 ... 4 1 1.0 n=1 0 n 0.8 3 (c) x 1 0.6 0.000 x3 0.4 0.000 1 x2 0.2 0.000 x2 2 3 n=3 485 0.0 ln x −4 29. Representation of Functions by Power Series n 0 1 x 2 2n 1 4n 1 2 2n 1 24n n n 1 192 0.245. x 4n 1 2n 1 2 4n 1 2 1 4 C Note: C 2 1 8 0 1 1152 1 < 0.001, we can approximate the series by its first term: 1152 ... 12 0 arctan x 2 dx x 0.125. 486 Chapter 9 Infinite Series In Exercises 39– 43, use 39. 41. 1 1 n d dx 1 2 x x n, x < 1. x 1 d dx 1 x 1 1 1 1 0 x x x2 1 2 x n xn n 1 1 nx n 0 n 1, x <1 1 x xn , x <1 n 1 2 43. P n 2 En nP n 1 n 0 1 21 1 x n, x < 1 2n n xn n 1 n 1 12 n 1 2 n 1 1 2n n 1 n 1 2 1 2 2 Since the probability of obtaining a head on a single toss 1 is 2 , it is expected that, on average, a head will be obtained in two tosses. 45. Replace x with 49. Let arctan x x. . Then, arctan y tan arctan x x and multiply the series by 5. 47. Replace x with arctan y tan tan arctan x tan arctan y 1 tan arctan x tan arctan y tan x 1 arctan y xy x 1 tan y xy . Therefore, arctan x 51. (a) 2 arctan 1 2 arctan 1 2 arctan 1 2 2 arctan 1 2 arctan 1 7 arctan 4 3 arctan 1 7 8 1 2 arctan (b) 1 2 8 arctan 4 arctan 1 2 1 1 2 12 1 7 0.5 3 arctan y 3 2 arctan 0.5 5 5 arctan 1 0.5 7 y for xy xy 1. 4 3 arctan 43 x 1 17 43 17 7 4 arctan 1 7 17 3 3 25 25 arctan 1 17 5 5 4 7 17 7 3.14 53. From Exercise 21, we have ln x 1 n 0 1 n xn n1 n 1 n 1 n 1xn n 1 1 n 1xn n 1 n 1 n 1 5n 1 n ln 1 1 1 2n n 1 n ln n 1 n 12 n 1 1 2 ln 1 3 2 n 12 2n 0.4055. 57. From Exercise 56, we have 2n n n n . 55. From Exercise 53, we have 1 1 Thus, n 25 n 1 n 1 2 5 1 1 n ln 7 5 0.3365. 0 n 2 2n 1 1 2n 1 1 n 0 arctan 1 2 2n 1 1 0.4636. S ection 9.10 59. f x arctan x is an odd function (symmetric to the origin). Taylor and Maclaurin Series 61. The series in Exercise 56 converges to its sum at a slower rate because its terms approach 0 at a much slower rate. 63. Because the first series is the derivative of the second series, the second series converges for x 1 < 4 and perhaps at the endpoints, x 3 and x 5. 1 65. n 0 3n n 2n 1 From Example 5 we have arctan x n 1 n n n 0 3 2n 1 n 1n 2n 2n 3 0 n 1 2n 0 0 3 1 2n 3 arctan 1. For c 0.9068997 6 Taylor and Maclaurin Series 0, we have: fx e2x n f 2n e2x ⇒ f x e2x 3. For c 1 2x n 2n 0 4x 2 2! 8x3 3! 16x 4 4! 4 2 2 ... n 0 2x n . n! 4, we have: fx cos x f sin x f fx cos x 4 f x f 4 sin x x cos x 2 2 4 2 2 4 f f 2 2 4 fx f 1 1 3 3 Section 9.10 x2n 1 . 2n 1 3 3 1 n 1 3 n 1 2 2 4 and so on. Therefore, we have: f c...
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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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