# Translations are possible at x 2 8n 71 let f be an

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Unformatted text preview: k n nx 1 x 3 n 480 Chapter 9 Infinite Series xn , 2 45. (a) f x n 0 x 2 n 1, S2 S1 53. g 1 8 55. g S1 x n 1 1 1 ,0 &lt; x ≤ 2 1 2&lt;x&lt;2 1 1 n, 0 &lt; x &lt; 2 1 n 1n x 1 n n x 1 (c) f x 2≤x&lt;2 , (d) f x dx n 1 9 1 3 1 0 2 , n n (b) f x ... 3.1 3 51. g 3.1 n 0 1 1 1 n x 1, n 1 1n 0&lt;x&lt;2 2 2 ,0 ≤ x ≤ 2 n diverges. Matches (b). 1.33. Matches (c). n n 1 8 0 1, S2 9 16 0 n n x 2 x 12 n 1 2 S1 2&lt;x&lt;2 1 n 1 3 0 0 1 n n 2 n n , 2 f x dx n 1 n 2 n 2 (c) f x 49. g 1 n 1 n 1 47. (a) f x n n 2 (b) f x (d) 2 &lt; x &lt; 2 (Geometric) n n 0 1 4 1 1 16 . . ., converges 1.3125 Matches (b). 1.25, S3 2 0 n 1 4 n 9 16 n 9n , diverges 8 0 57. A series of the form 17 Matches (d). 8 1, S2 an x n n c 0 is called a power series centered at c. 59. A single point, an interval, or the entire real line. 61. Answers will vary. xn converges for n 1n 1 ≤ x &lt; 1. At x xn converges for n2 1 1 ≤ x ≤ 1. At x n 1 n x 2n 1 , 2n 1 ! 63. (a) f x n 0 ± 1, the convergence is absolute. (c) g x &lt;x&lt; n 1 1 n x 2n 1 2n 1 ! n (See Exercise 29.) n n 0 n (b) f x n 65. y n 0 0 &lt;x&lt; x 2n 1 2n ! 1 n x 2n 1 2n 1 ! n n 2n 1 x2n 2n 1 ! n 0 y 1 y n 1n 2n 1 1 n 1 1x 2n 2n x 2n ! n 0 1 1 n x 2n 1 2n 1 ! 1x 2n 1 ! 0 sin x and g x n 1 1 67. y n 1 nx 2n 2n ! 2n n 1 1! n 2n (d) f x n 2n 1n 2n 0 1x 2n 1! 1 fx cos x gx 1 y y n 2n 1x , 2n ! gx 1 diverges. n 1, the convergence is conditional because n 0 1 2n x 2n 2n ! 1 1x 2n 1! y n 1 0 y y 0 1 x 2n 1 2n 1 ! n n 1 x2n 1! 2n 2n y 1x 2n 1 ! 1n 2n 0 x 2n 1 2n 1 ! x 2n 2n ! 1 n 1 0 x 2n 1 2n 1 ! y 1 S ection 9.8 69. y n x 2n n 0 2 n! 2nx 2n n 1 2 n! y n 1 n xy y n n 1 x 2n n n! 2 2 2n 2n y 2n 2n 1 x 2n 2n n! 2 1 1 2n n 0 71. J0 x k (a) lim k→ 0 n 0 0 2n 1 x2n 2n n! 1 lim 2k 2 k→ 2 k 1 x 2k 2 k 1!2 2n 2n 1 1 1 0 22k k! 2 1 k x 2k 1 k 1 J0 1 k 1 k k k x2 4 1 x4 64 k 1 x 2J0 xJ0 k 0 J0 k k 1 J0 (b) 0 k x 2k 4k k! 2kx 2k 4k k! k→ 1 k k 2 2 0 1 k x 2k 4k k! 2 1 2k 2 x 2k 1 4k 1 k 1 ! 2 1 0 2 1 k 2 2k 1 x 2k 2 4k 1 k 1 !k! 0 0 2 . 1 2 2k 2k 1 4k k! 2 1 k x2k 4k k! 2 2 2 x 2k 1 1 x2 2k 1 lim &lt;x&lt; Therefore, the interval of convergence is (c) P6 x n 1 k x 2k 22k k! 2 uk 1 uk x 2J0 x 2n n 0 2 n! 1 x 2n 2n n! 1 x 2n 2n 1 2n 2n 1 n 1 ! 2n n 1 2n x2n 2 2n n! 2nx 2n n 1 2 n! n 2 2n 1 1n 1! 2n 1 x 2n 2n 2n y Power Series 2 2k 4k 1 k k 2 4k 4k 2x 2k 2 k 1 !k! 2 4 4 1 0 0k 0 3 −6 0 x 2k 2 4k k! 2 0 J0dx (d) k k 1 1 1 x6 2304 1 1 4k 4k 4k 4 1 1 0 1 2 2k 1 x 2k k 1!2 2k 4k 1 0 1 1 4k 2 4k 4 k k 6 k −5 0 0 4k 1 k x 2k dx k! 2 4k 1 k x 2k 1 k! 2 2k 1 4k k! 1k 2k 2 1 12 1 1 0 1 1 320 0.92 (integral is approximately 0.9197304101) 73. f x 1 n 0 n x 2n 2n ! cos x 1 nx n 75. f x 2 n −2 x 0 n 1 2 1 1 x 1 x 3 −2 −1 1 0 n Geometric 0 for 1&lt;x&lt;1 481 482 Chapter 9 n n x 2 77. 0 (a) n Infinite Series 0 n 34 2 n n 3 8 0 1 38 1 8 5 1.6 34 2 (b) n 0 n n 3 8 0 n 1 1.80 1 8 11 38 0.7272 1.10 −1 6 0 (c) The alternating series converges more rapidly. The partial sums of the series of positive terms approach the sum from below. The partial sums of the alternating series alternate sides of the horizontal line representing the sum. −1 6 0.60 N (d) n 0 n 3 2 &gt;M M 10 N 100 1000 10,000 9 15 21 4 81. True; the radius of convergence is R 79. False; nx n n 0 1 n2n converges for x an 1 an 83. lim n→ 2 but diverges for x lim n1 1!n n n→ 2. p! xn 1 q! Thus, the series converges for all x : R cn x n, cn 85. (a) f x n 1 n n! n c0x3 c1x 4 c2x5 c0 1 x3 ... x3n c1x 1 x3 x3n c0 n x3n c1x 0 n (b) For x &lt; 1, f x R, the series is cn x0 n ... x3n c2 x 2 1 c1x x0 R ... 0 1 c2 x 2 x3 1 x3 x3n n 0 n cn n 1 1 c0 x3 1, 1 . c1x c2 x 2 . 1 x3 R n which converges. 0 0 cn R n, which diverges. R, the series is Hence, at x0 x3 1 n At x0 px 1q ... c0x6 1, and the interval of convergence is 1 c0 c2 x 2 0 Each series is geometric, R 87. At x0 1 1n cn 3 c2x 2 lim S3n n n 0 c1x n→ lim n→ . c0 S3n p! n x q! R, we have cn n R n converges, 0 ⇒ The series converges conditionally. cn n 0 R n cn R n diverges. n 0 x3n 0 1 for both series. Section 9.9 Section 9.9 1. (a) x 12 x2 1 1 2 )1 x n n 02 x3 8 x3 8 x3 8 1 x 3 r , we have 5 1 3 n 0 n 5 x n 3 0 0 1 3 5 2x 1 2x 1 5 5 0 11 or 2 r ar n 1 n 2x. Therefore, the n 3 2x 0 n 0 x a 3 1 1 11 n 0 0 n 2x 3 , 11n 1 17 5 &lt;x&lt; . 2 2 2 x r , we have 32 1 2x 1 a 1 which implies that a 3 2 and r the power series for f x is given by r 2 x 11 0 3 2 3 3. x 2 ar n n 0 n n 3 3 n x &lt; 2 or 0 1 1 &lt;x&lt; . 2 2 2x n, 2x &lt; 1 or 3 3 n 3&lt; r , we have a 1 11. Writing f x in the form a 1 3 n x x4 16 3 and r which implies that a power series for f x is given by 2x r , we have ar n n ... x3 8 x3 8 x3 8 3 2x 1 3 5 which implies that a 1 11 and r 2 11 x Therefore, the power series for f x is given by 2x 1 5 &lt; 3 or 2 &lt; x &lt; 8. 1 11 2 11 x 1 x2 4 x2 4 x2 4 n 11 1 5. n 1 x 3 ,x 9. Writing f x in the form a 1 2x 16 n n 2, 2 . 7. Writing f x in the form a 1 13 13 x 1 ar n x 8 1 n xn 2n 1 0 x3 x4 16 1 3 and r 13 x which implies that a Therefore, the power series for f x is given by 1 n x2 x...
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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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