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# HW14 - momin(rrm497 Homework 14 cheng(58520 This print-out...

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momin (rrm497) – Homework 14 – cheng – (58520) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Suppose T 4 ( x ) = 8 - 4( x - 2) + 8( x - 2) 2 - 6( x - 2) 3 + 2( x - 2) 4 is the degree 4 Taylor polynomial centered at x = 2 for some function f . What is the value of f (3) (2)? 1. f (3) (2) = 6 2. f (3) (2) = 36 3. f (3) (2) = - 6 4. f (3) (2) = 2 5. f (3) (2) = - 36 correct 6. f (3) (2) = - 2 Explanation: Since T 4 ( x ) = f (2) + f (2)( x - 2) + f ′′ (2) 2! ( x - 2) 2 + f (3) (2) 3! ( x - 2) 3 + f (4) (2) 4! ( x - 2) 4 , we see that f (3) (2) = - 3! × 6 = - 36 . 002 10.0 points Find the Taylor series centered at x = 0 for f ( x ) = cos 5 x. 1. summationdisplay n =0 ( - 1) n 5 2 n n ! x 2 n 2. summationdisplay n =0 ( - 1) n 5 2 n n ! x n 3. summationdisplay n =1 5 n (2 n )! x n 4. summationdisplay n =0 ( - 1) n 5 2 n (2 n )! x 2 n correct 5. summationdisplay n =0 ( - 1) n 5 2 n (2 n )! x n 6. summationdisplay n =1 5 n (2 n )! x 2 n Explanation: The Taylor series centered at x = 0 for any f is f (0) + f (0) x + 1 2! f ′′ (0) x 2 + ... = summationdisplay n =0 1 n ! f ( n ) (0) x n . But when f ( x ) = cos(5 x ), n f ( n ) ( x ) f ( n ) (0) 0 cos(5 x ) 1 1 - 5 sin(5 x ) 0 2 - 25 cos(5 x ) - 25 3 125 sin(5 x ) 0 4 625 cos(5 x ) 625 . . . . . . in other words, f (0) = f ′′′ (0) = f (5) (0) = ... = 0 , while f (0) = 1 , f ′′ (0) = - 5 2 , f (4) (0) = 5 4 . Thus in general, f (2 n +1) (0) = 0 , f (2 n ) (0) = ( - 1) n 5 2 n .

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momin (rrm497) – Homework 14 – cheng – (58520) 2 Consequently, the Taylor series of cos 5 x cen- tered at x = 0 is summationdisplay n =0 ( - 1) n 5 2 n (2 n )! x 2 n . keywords: 003 10.0 points Find the degree 2 Taylor polynomial of f centered at x = 2 when f ( x ) = 5 x ln x. 1. 10 + 5 ln 2( x - 2) + 5 2 ( x - 2) 2 2. 10 ln 2 + 5(ln 2 + 1)( x - 2) + 5 4 ( x - 2) 2 correct 3. 10 ln 2 + 5 ln 2( x - 2) + 5 4 ( x - 2) 2 4. 10 + 5(ln 2 + 1)( x - 2) + 5 4 ( x - 2) 2 5. 10 ln 2 + 5(ln 2 + 1)( x - 2) + 5 2 ( x - 2) 2 6. 10 + 2 ln 5( x - 2) + 5 4 ( x - 2) 2 Explanation: The degree 2 Taylor polynomial of f cen- tered at x = 2 is given by T 2 ( x ) = f (2) + f (2)( x - 2) + 1 2! f ′′ (2)( x - 2) 2 . When f ( x ) = 5 x ln x , therefore, f ( x ) = 5 ln x + 5 , f ′′ ( x ) = 5 x . But when f (2) = 10 ln 2, f (2) = 5(ln 2 + 1) , f ′′ (2) = 5 2 .
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HW14 - momin(rrm497 Homework 14 cheng(58520 This print-out...

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