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Calculus Math107 II Fall 2007 University of Nebraska-Lincoln Name: Section #: (print legibly) Exam 2 Instructions Turn off all communication devices. If you do not do so, then you will not receive any credit for your exam. There are 7 pages in this exam with 6 problems. Before you begin, make sure that your exam has all 7 pages. The examination period is from 12:30pm to 1:20pm. If you wish to receive credit for your exam, then make sure that your exam is submitted for grading by 1:20pm. You may NOT use a calculator during the exam. You may not use a text, notes, nor any other reference. To receive full credit for a problem, you must provide a correct answer and a sufficient amount of work so that it can be determined how you arrived at your answer. Clearly indicate what your solutions are and any work that you do not want to be included in the grading process. Write your solutions in an explicit form whenever possible. If you wish to speak with a proctor during the exam, then raise your hand and a proctor will come to you. Each problem will be graded out of 20 points. If it is determined that you have given or received any unauthorized aid during this exam, then you will receive no credit for your exam. Problem Score 1 2 3 4 5 6 Total 1 1. For the following problems, let an = (-1) n-1 1 2 n for n = 1, 2, 3, . . . . (a) Compute the first five terms in the sequence {an }. Solution: a1 = 1 1 1 1 1 , a2 = - , a3 = , a4 = - , a5 = 2 4 8 16 32 (b) If the sequence {an } converges, compute lim an . Otherwise, explain why the n sequence diverges. Solution: For each n = 1, 2, 3, . . . , we have that - Since n 1 2 n an 1 2 n . lim 1 1 = lim - n = 0. n n 2 2 n By the squeeze theorem, or sandwich theorem, we conclude that lim an = 0. 2 7 2. (a) Rewrite SWER. n=3 Solution: We have that 1 - 3 n without Sigma notation. DO NOT SIMPLIFY YOUR AN- 7 - n=3 1 3 n = - 1 3 3 + 1 3 4 - 1 3 5 + 1 3 6 - 1 3 7 . (b) Is n=3 Solution: 1 - 3 n a p-series, geometric series, alternating series, or none of these. This is a geometric series and an alternating series. (c) Does the series n=3 Solution: We can use the theorem for geometric series. From part (a), we see that 1 - 3 n converge? JUSTIFY YOUR ANSWER. - n=3 1 3 n = n=1 - 1 3 3 - 1 3 n-1 . Since - 1 = 3 1 3 < 1, the series converges ; indeed - n=3 1 3 n =- 1 33 1+ 1 3 = 1 . 36 We could also use the alternating series test. For each n 3, we clearly have that 1 n+1 3 1 n 3 = 1 1, 3 so the sequence 1 n 3 is non-increasing (for n 3). Also lim 1 3 n n = lim n 1 = 0. 3n Thus, the alternating series test implies that the series converges. 3 3. These problems refer to the power series n=0 1 (x - 3)n . n 3 (a) What is the center of this power series? Solution: The center is at x = 3. (b) What is the radius of convergence for this power series? Solution: We us the ratio test: 1 n+1 3n+1 |x - 3| 1 n n 3n |x - 3| lim = lim n 1 |x - 3|; 3 By the ratio test, the series converges if 1 |x - 3| < 1 and it diverges if 1 |x - 3| > 1. 3 3 It follows that the series converges if |x - 3| < 3 and it diverges if |x - 3| > 3. The radius of convergence is 3. (c) What is the interval of convergence for power this series? Solution: From (b), we know that the series converges for x in (0, 6) and diverges if x < 0 or x > 6. It only remains to see what happens at x = 0 and at x = 6. At x = 0, the power series equals 1 (0 - 3)n = (-1)n . 3n n=0 n=0 This series fails the n-th term test, so it diverges. At x = 6, the power series equals 1 (6 - 3)n = 1. 3n n=0 n=0 This series also fails the n-th term test, so it diverges. The interval of convergence is (0, 6). 4 4. Find the 3rd order Taylor polynomial P3 (x) generated by f (x) = x5 at x = -1. Solution: We need to compute the derivatives up to order 3 for f and evaluate them at -1: f (x) = x5 f (x) = 5x 4 f (x) = 20x3 f (x) = 60x2 polynomial yields f (-1) = -1 f (-1) = 5 f (-1) = -20 f (-1) = 60. Assembling the Taylor P3 (x) = -1 + 5(x + 1) - 20(x + 1)2 + 60(x + 1)3 . 5 5. For this problem, recall that sin x = n=0 (-1)n 2n+1 x , (2n + 1)! for all x in (-, ). Recall also that the radius of convergence for this power series is . (a) Provide a power series representation, with a center at x = 3, for sin(x - 3)2 . What is its radius of convergence? Solution: Substituting (x - 3)2 for x in the power series representation for sin x, we find that sin(x - 3)2 = (-1)n (x - 3)4n+2 . (2n + 1)! n=0 Since the radius of convergence for the power series for sin x is infinity, the radius of convergence for the power series for sin(x - 3)2 is also . (b) Provide a power series representation, with a center at x = 3, for What is its radius of convergence? Solution: The power series for sin(x-3)2 (x-3)2 sin(x-3)2 (x-3)2 . is (-1)n (x - 3)4n . (2n + 1)! n=0 We can use the ratio test to find the radius of convergence for this series: 1 4n+4 (2n+3)! |x - 3| lim 1 4n n (2n+1)! |x - 3| = lim n 1 |x - 3|4 = 0. (2n + 2)(2n + 3) This the series converges for all x and the radius of convergence is . Integrating term-by-term, we have sin(x - 3)2 dx = (x - 3)2 (-1)n (-1)n (x - 3)4n dx = (2n + 1)! (2n + 1)! n=0 n=0 (-1)n (x - 3)4n+1 . (4n + 1)(2n + 1)! n=0 (x - 3)4n dx = C+ From a theorem in the test, we know that the radius of convergence for this new series is also . One could also use the ratio test to determine this. 6 6. These problems refer to the series n=1 (-1)n . JUSTIFY YOUR ANSWERS. n + 10 (a) Does this series converge absolutely? Solution: We need to analyze 1 n=1 n+10 . We see that = lim n 1 n+10 lim 1 n n n = 1. n + 10 the limit comparison test tells us 1 Hence the series n=1 n diverges. Since this is a finite and positive number, (-1)n that the series diverges, since n=1 n+10 does not converge absolutely. (b) Does this series converge conditionally? Solution: We try the alternating series test. We see that 1 n+11 1 n+10 n + 10 = 1 n + 11 1 n+10 for all n 1. In particular, the sequence 1 limn n+10 is non-increasing. We also see that (-1)n n=1 n+10 = 0. Thus the alternating series test implies that the series converges. Thus the series does converge conditionally. (c) Provide, if possible, an estimate (upperbound) for n=1 Solution: We can use the Alternating Series Estimation Theorem. According to this theorem, (-1)n - n + 10 n=1 (-1)n - n + 10 89 n=1 (-1)n . n + 10 (-1)n = |R89 | n + 10 n=1 89 (-1)90 90 + 10 = 1 . 10 7

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