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Part5_ANS - Answers to Selected Exercises Chapter 30 2a 121...

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Unformatted text preview: Answers to Selected Exercises Chapter 30 2a. 121 81 2b. 9,841 6,561 2c. 3 2 2d. 1 162 2e. 3 5 2f. 665 32 2g. Diverges 2h. 14 2i. − 5 2j. 15 4 4a. 1 + ∞ summationdisplay k = 1 − 1 k ( k + 1 ) x k 4b. x 2 + ∞ summationdisplay k = 4 ( − 1 ) k k x k 4c. ∞ summationdisplay n = 4 3 ( n − 3 ) 2 ( x − 5 ) n 4d. ∞ summationdisplay n = 1 ( n + 1 ) nx n 4e. − 6 + ∞ summationdisplay n = 1 [− 2 ( n + 1 )( n + 3 ) ] x n 4f. 6 + ∞ summationdisplay n = 1 [ 2 ( n + 1 )( 2 n + 3 ) ] x n 4g. ∞ summationdisplay n = 4 ( n − 3 ) 2 a n − 3 x n 4h. a x + 2 a 1 x 2 + ∞ summationdisplay n = 3 n ( a n − 1 − a n + 1 ) x n 4i. ∞ summationdisplay n = 2 a n − 2 ( x − 1 ) n 4j. ∞ summationdisplay n = bracketleftbig ( n + 1 ) a n + 1 + 5 a n bracketrightbig x n 4k. − 4 a − 4 a 1 x + ∞ summationdisplay k = 2 bracketleftbig k 2 − k − 4 bracketrightbig a k x k 4l. 2 a 2 + 6 a 3 x + ∞ summationdisplay n = 2 bracketleftbig ( n + 2 )( n + 1 ) a n + 2 − 3 a n − 2 bracketrightbig x n 5a. ∞ summationdisplay k = ( k + 1 ) x k for | x | < 1 5b. ∞ summationdisplay k = 1 2 ( k + 2 )( k + 1 ) x k for | x | < 1 6a. ∞ summationdisplay k = 1 k ! x k 6b. ∞ summationdisplay k = ( − 1 ) k ( 2 k ) ! x 2 k 6c. ∞ summationdisplay k = ( − 1 ) k ( 2 k + 1 ) ! x 2 k + 1 6d. ∞ summationdisplay k = 1 ( − 1 ) k − 1 k ( x − 1 ) k 7a. 1 + 1 2 x − 1 8 x 2 + 1 16 x 3 − 5 128 x 4 7b i. 1 − 1 2 x + 3 8 x 2 − 5 16 x 3 7b ii. 1 − 1 2 x 2 + 3 8 x 4 − 5 16 x 6 8b i. 1 + 2 x + 2 x 2 + 4 3 x 4 + 4 15 x 5 + 4 45 x 6 + 8 315 x 7 + 2 315 x 8 + 4 2835 x 9 8b ii. 1 + 1 2 x 2 + 5 24 x 4 + 61 720 x 6 + 277 8064 x 8 + 50521 3628800 x 10 8b iii. 3 − 4 3 ( x − 2 ) + 1 27 ( x − 2 ) 2 − 4 243 ( x − 2 ) 3 + 31 4374 ( x − 2 ) 4 − 58 19683 ( x − 2 ) 5 + 139 118098 ( x − 2 ) 6 − 238 531441 ( x − 2 ) 7 9a. ∞ summationdisplay k = 2 k x k with R = 1 / 2 9b. ∞ summationdisplay k = ( − 1 ) k x 2 k with R = 1 9c. ∞ summationdisplay k = 1 2 k x k with R = 2 9d. ∞ summationdisplay k = k + 1 2 k + 1 x k with R = 2 9e. ∞ summationdisplay k = ( − 1 ) k k ! x 2 k with R = ∞ 9f. ∞ summationdisplay k = ( − 1 ) k ( 2 k + 1 ) ! x 4 k + 2 with R = ∞ 10d. Taylor series = 8/14/2011 797 798 Answers to Selected Exercises 10e. Because f ( x ) does not equal its Taylor series about 0 except right at x = 0 . Chapter 31 2. S k is the statement that a k = F ( k ) for a given function F and a given sequence A , A 1 , A 2 , . . . . 3a. a k = 2 k a k − 1 for k ≥ 1 , y ( x ) = a ∞ summationdisplay k = 2 k k ! x k = a ∞ summationdisplay k = 1 k ! ( 2 x ) k = a e 2 x 3b. a k = 2 k a k − 2 for k ≥ 2 (with a 1 = 0) , y ( x ) = a ∞ summationdisplay m = 1 m !...
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Part5_ANS - Answers to Selected Exercises Chapter 30 2a 121...

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