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Unformatted text preview: Rehman (aar638) HW15 sachse (56620) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Use the Taylor series representation cen tered at the origin for e x 2 to evaluate the definite integral I = integraldisplay 2 5 e x 2 dx. 1. I = n summationdisplay k = 0 1 k !(2 k + 1) 5 2 2 k +1 2. I = summationdisplay k = 0 ( 1) k k ! 5 2 2 k 3. I = summationdisplay k = 0 1 k ! 5 2 2 k 4. I = summationdisplay k = 0 ( 1) k k !(2 k + 1) 5 2 2 k +1 correct 5. I = summationdisplay k = 0 ( 1) k 2 k + 1 5 2 2 k +1 Explanation: The Taylor series for e x is given by e x = 1 + x + 1 2! x 2 + . . . + 1 n ! x n + . . . and its interval of convergence is ( , ). Thus we can substitute x  x 2 for all val ues of x , showing that e x 2 = summationdisplay k =0 ( 1) k k ! x 2 k everywhere on ( , ). Thus I = integraldisplay 2 5 parenleftBig summationdisplay k =0 ( 1) k k ! x 2 k parenrightBig dx. But we can change the order of summation and integration on the interval of convergence, so I = 5 summationdisplay k =0 parenleftBig integraldisplay 2 ( 1) k k ! x 2 k parenrightBig dx = 5 summationdisplay k =0 bracketleftBig ( 1) k k !(2 k + 1) x 2 k +1 bracketrightBig 2 . Consequently, I = summationdisplay k =0 ( 1) k k !(2 k + 1) 5 2 2 k +1 . 002 (part 1 of 4) 10.0 points Suppose p 4 ( x ) = 5 3 x + 5 x 2 2 x 3 + 4 x 4 is the degree 4 Taylor polynomial centered at x = 0 for some function f . (i) What is the value of f (0)? 1. f (0) = 5 2. f (0) = 6 3. f (0) = 7 4. f (0) = 6 5. f (0) = 5 correct Explanation: Since p 4 ( x ) = f (0) + f (0) x + f (0) 2! x 2 + f (3) (0) 3! x 3 + f (4) (0) 4! x 4 , we see that f (0) = 5 . 003 (part 2 of 4) 10.0 points (ii) What is the value of f (3) (0)? Rehman (aar638) HW15 sachse (56620) 2 1. f (3) (0) = 2 3 2. f (3) (0) = 2 3 3. f (3) (0) = 12 4. f (3) (0) = 12 correct 5. f (3) (0) = 2 Explanation: Since p 4 ( x ) = f (0) + f (0) x + f (0) 2! x 2 + f (3) (0) 3! x 3 + f (4) (0) 4! x 4 , we see that f (3) (0) = 3! 2 = 12 . 004 (part 3 of 4) 10.0 points (iii) Use p 4 to estimate the value of f (0 . 1). 1. f (0 . 1) 5 . 0484 2. f (0 . 1) 4 . 6484 3. f (0 . 1) 4 . 8484 4. f (0 . 1) 4 . 9484 5. f (0 . 1) 4 . 7484 correct Explanation: Since p 4 ( x ) is an approximation for f ( x ) we see that f (0 . 1) 5 3 10 + 5 10 2 2 10 3 + 4 10 4 ....
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This note was uploaded on 02/01/2011 for the course MATH 408L taught by Professor Gogolev during the Fall '09 term at University of Texas at Austin.
 Fall '09
 GOGOLEV
 Calculus

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