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Unformatted text preview: pokharel (yp624) HW15 Radin (57410) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine an expression for b 9 if f ( x ) = summationdisplay n = 0 b n ( x 6) n . 1. b 9 = f (6) (6) 9! 2. b 9 = f (6) (9) 6! 3. b 9 = f (9) (6) 9! correct 4. b 9 = f (9) (6) 9 5. b 9 = f (9) (9) 6! Explanation: We know that if f ( x ) = summationdisplay n = 0 b n ( x 6) n , then the series representation is the Taylor series representation of f centered at x = 6, in which case b n = f ( n ) (6) n ! . Consequently, b 9 = f (9) (6) 9! . 002 10.0 points Find the Taylor series centered at x = 0 for f ( x ) = cos 3 x . 1. summationdisplay n = 0 ( 1) n 3 2 n (2 n )! x n 2. summationdisplay n = 0 ( 1) n 3 2 n n ! x 2 n 3. summationdisplay n = 0 ( 1) n 3 2 n (2 n )! x 2 n correct 4. summationdisplay n = 0 ( 1) n 3 2 n n ! x n 5. summationdisplay n = 1 3 n (2 n )! x n 6. summationdisplay n = 1 3 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(3 x ), n f ( n ) ( x ) f ( n ) (0) cos(3 x ) 1 1 3 sin(3 x ) 2 9 cos(3 x ) 9 3 27 sin(3 x ) 4 81cos(3 x ) 81 . . . . . . in other words, f (0) = f (0) = f (5) (0) = . . . = 0 , while f (0) = 1 , f (0) = 3 2 , f (4) (0) = 3 4 . Thus in general, f (2 n +1) (0) = 0 , f (2 n ) (0) = ( 1) n 3 2 n . pokharel (yp624) HW15 Radin (57410) 2 Consequently, the Taylor series of cos 3 x cen tered at x = 0 is summationdisplay n = 0 ( 1) n 3 2 n (2 n )! x 2 n . 003 10.0 points Find the Taylor series representation for f centered at x = 1 when f ( x ) = 4 + 2 x 3 x 2 . 1. f ( x ) = 4 + 2( x 1) 3( x 1) 2 2. f ( x ) = 3 4( x 1) 6( x 1) 2 3. f ( x ) = 4 4( x 1) + 6( x 1) 2 4. f ( x ) = 3 4( x 1) 3( x 1) 2 correct 5. f ( x ) = 3 + 2( x 1) + 3( x 1) 2 6. f ( x ) = 4 + 2( x 1) 6( x 1) 2 Explanation: For a function f the Taylor series represen tation centered at x = 1 is given by f ( x ) = summationdisplay n = 0 1 n ! f ( n ) (1)( x 1) n . Since f is a polynomial of degree 2, however, f ( n ) = 0 for all n 3, so we have only to calculate derivatives of f up to order 2: f ( x ) = 2 6 x, f ( x ) = 6 . Thus f (1) = 3 , f (1) = 4 , f (1) = 6 ....
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This note was uploaded on 12/23/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.
 Fall '08
 RAdin

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