Pre-Calc Homework Solutions 449

Pre-Calc Homework Solutions 449 - Cumulative Review 2 449...

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Unformatted text preview: Cumulative Review 2 449 96. 0 dx 3 1 dx 3 2 dx 3 101. The first six terms of the Maclaurin series are 1 1 x6 . By the Alternating 6! x7 1 Series Estimation Theorem, error 0.001. 7! 7! x 1 b 0 0 x 1 1 2 x dx lim b1 dx 3 1 x lim 1 a1 2/3 b a x2 2! x3 3! x4 4! x5 5! x 3 x lim b1 3 (x 2 1) lim 0 a1 3 (x 2 1)2/3 2 0. a The whole integral converges. 97. We know that 1 1 x 102. f (0) f (0) 1, f (0) 2 9(0 1 3(0 1)2/3 1)2/3 1 , 3 1 1 x 1 2x x x x2 x3 ... ( 1)nx n ... ... f (n)(0) ... 2 , 9 2 5 ... (3n ( 1)n 1 3n 4) , so the for 1 1 1. Substituting 2x for x yields 2x 4x 2 8x 3 ... ( 1)n2nx n Taylor series is 1 1 x 3 2 5 8 4 2 5 3 x x 4! 34 3! 33 2 5 ... (3n 4) n ... ( 1)n 1 x . n! 3n 2 x2 2! 32 2x for 1 2 1 1, so the interval of convergence is ... 1 . 2 98. (a) cos t 2 Since by the Ratio Test 1 1 (t 2)2 (t 2)4 (t 2)6 ... (t 2)2n ( 1)n 2! 4! 6! (2n)! 4n t 12 t4 t8 n t ... ( 1) 6! 2! 4! (2n)! ... lim n an 1 an 2 5 ... (3n 4)(3n (n 1)!3n 1 (3n 1)x (n 1)(3) 1)x n 1 Integrating each term with respect to t from 0 to x yields x (b) x 5(2!) 5 lim n n!3n 2 5 ... (3n 4)x n lim x 9(4!) 9 x , the radius of convergence is 1. an 1 an x 13(6!) 13 ... ( 1)n x (4n 4n 1 1)(2n)! .... n 103. Using the Ratio Test, lim n lim n 2 3 n 1 3n 2 1 , so 3 x ; Since the cosine series converges for all the series converges. 104. Note that an Test, since n 1 real numbers, so does the integrated series, by, the term-by-term integration theorem (Section 9.1, Theorem 2). 99. ln (2 ln 2 2x) x ln [2(x x 2 2 1 for every n. By the Direct Comparison n 1 2 diverges, so does . n n 1 n 105. Use the alternating series test. 1)] 3 ln 2 ... ln (x ( 1)n lim n 1) 1x n x 3 x 4 4 n n 1 ... x x , the Note that n 0 ( 1)n n 1 n 0 ( 1)nun, where un un 1 1 n 1 . an 1 Since by the Ratio Test lim an n Since each un is positive, un n for all n, and n lim un 0, the original series converges. series converges for 1 x 1. 1 does not. n 106. Using the Ratio Test, lim n n 1 ( 1)n converges, but n n 1 an 1 an lim n 3n 1 (n 1)! n! 3n lim n 3 n 1 0, and the 100. Let f (x) f (x) sin x. Then f (x) cos x, f (4)(x) cos x, f (x) sin x, 2 the series converges. sin x, and so on. At x sine terms are zero and the cosine terms alternate between 1 and (x 1, so the Taylor series is 2 )3 (x 2 )5 3! 5! (x 2 )2n 1 .... ( 1)n (2n 1)! (x 2 ) ... ...
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