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Homework 15

# Homework 15 - Rehman(aar638 HW15 sachse(56620 This...

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Rehman (aar638) – HW15 – sachse – (56620) 1 This print-out should have 15 questions. Multiple-choice 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 0 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 0 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 0 ( - 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 0 . 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)?

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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 . Consequently, f (0 . 1) 4 . 7484 .
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Homework 15 - Rehman(aar638 HW15 sachse(56620 This...

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