HW14 - momin (rrm497) Homework 14 cheng (58520) 1 This...

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Unformatted text preview: 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) cos(5 x ) 1 1- 5 sin(5 x ) 2- 25 cos(5 x )- 25 3 125sin(5 x ) 4 625cos(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 . 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 ln2( x- 2) + 5 2 ( x- 2) 2 2. 10ln 2 + 5(ln 2 + 1)( x- 2) + 5 4 ( x- 2) 2 correct 3. 10ln 2 + 5 ln2( x- 2) + 5 4 ( x- 2) 2 4. 10 + 5(ln 2 + 1)( x- 2) + 5 4 ( x- 2) 2 5. 10ln 2 + 5(ln2 + 1)( x- 2) + 5 2 ( x- 2) 2 6. 10 + 2 ln5( 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!...
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This note was uploaded on 12/11/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.

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HW14 - momin (rrm497) Homework 14 cheng (58520) 1 This...

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