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Unformatted text preview: Version 101 Exam01 gilbert (55485) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points To apply the ratio test to the infinite series summationdisplay n a n , the value = lim n a n +1 a n has to be determined. Compute for the series summationdisplay n =1 4 n 3 n 4 + 8 . 1. = 4 correct 2. = 0 3. = 4 3 4. = 1 2 5. = 4 11 Explanation: By algebra, a n +1 a n = 4 n +1 4 n bracketleftbigg 3 n 4 + 8 3( n + 1) 4 + 8 bracketrightbigg . But 3 n 4 + 8 3( n + 1) 4 + 8 = 3 + 8 n 4 3 parenleftbigg n + 1 n parenrightbigg 4 + 8 n 4 . Since parenleftbigg n + 1 n parenrightbigg 4 = parenleftbigg 1 + 1 n parenrightbigg 4 1 as n , we see that lim n a n +1 a n = lim n 4 parenleftBig 3 + 8 n 4 parenrightBig 3 parenleftBig n + 1 n parenrightBig 4 + 8 n 4 = 4 . Consequently, = 4 . 002 10.0 points Compute the degree 2 Taylor polynomial for f centered at x = 4 when f ( x ) = x . 1. T 2 ( x ) = 4 + 1 8 ( x 4) + 1 32 ( x 4) 2 2. T 2 ( x ) = 2 + 1 4 ( x 4) 1 64 ( x 4) 2 correct 3. T 2 ( x ) = 2 1 8 ( x 4) + 1 64 ( x 4) 2 4. T 2 ( x ) = 2 + 1 4 ( x 4) 1 32 ( x 4) 2 5. T 2 ( x ) = 2 1 4 ( x 4) + 1 64 ( x 4) 2 6. T 2 ( x ) = 4 1 8 ( x 4) 1 32 ( x 4) 2 Explanation: The degree 2 Taylor polynomial centered at x = 4 for a general f is given by T 2 ( x ) = f (4)+ f (4) ( x 4)+ f (4) 2! ( x 4) 2 . Now when f ( x ) = x , f ( x ) = 1 2 x , f ( x ) = 1 4 x x , in which case, f (4) = 2 , f (4) = 1 4 , Version 101 Exam01 gilbert (55485) 2 while f (4) 2! = 1 64 . Consequently, T 2 ( x ) = 2 + 1 4 ( x 4) 1 64 ( x 4) 2 . 003 10.0 points Which one of the following properties does the series summationdisplay k =1 ( 1) k 1 3 1 + k 2 have? 1. divergent 2. conditionally convergent correct 3. absolutely convergent Explanation: The given series has the form of an alter nating series summationdisplay k =1 ( 1) k 1 a k , a k = 3 1 + k 2 . To check for absolute convergence we apply the Limit Comparison Test with a k = 3 1 + k 2 , b k = 1 k . For then lim k a k b k = lim k 3 k 1 + k 2 = 3 > . Thus the series a k converges if and only if the series summationdisplay k = 1 1 k converges. But, by the pseries test with p = 1, this last series diverges; in particular, the given series does not converge absolutely....
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This note was uploaded on 05/02/2011 for the course M 408 D taught by Professor Textbookanswers during the Spring '07 term at University of Texas at Austin.
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