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Unformatted text preview: dang (rtd356) Wk 5 Check Knowledge Cepparo (57045) 1 This printout should have 7 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the coefficient of x 6 in the Taylor series expansion about x = 0 for f ( x ) = sin ( x 2 ) . 1. coefficient = 1 120 2. coefficient = 1 3. coefficient = 1 6 correct 4. coefficient = 0 5. coefficient = 1 6 Explanation: f ( x ) = 2 x sin x 2 f ( x ) = 2 sin x 2 + 4 x 2 cos x 2 f (3) ( x ) = 4 x cos x 2 + 8 x cos x 2 8 x 3 sin x 2 = 12 x cos x 2 8 x 3 sin x 2 f ( iv ) ( x ) = 12 cos x 2 24 x 2 sin x 2 24 x 2 sin x 2 16 x 4 cos x 2 = (12 16 x 4 ) cos x 2 48 x 2 sin x 2 f ( v ) ( x ) = 64 x 3 cos x 2 (24 x 32 x 5 ) sin x 2 96 x sin x 2 96 x 3 cos x 3 = 160 x 3 cos x 2 + (32 x 5 120 x ) sin x 2 f ( vi ) ( x ) = 480 x 2 cos x 2 + 320 x 4 sin x 2 + (160 x 4 120) sin x 2 + (64 x 6 240 x 2 ) cos x 2 = (64 x 6 720 x 2 ) cos x 2 + (480 x 4 120) sin x 2 The coefficient is a 6 = f ( vi ) (0) 6! = 120 6! = 1 6 . 002 (part 1 of 2) 10.0 points (i) Compute the degree 2 Taylor polynomial for f centered at x = 1 when f ( x ) = x . 1. T 2 ( x ) = 1 1 2 ( x 1) + 1 8 ( x 1) 2 2. T 2 ( x ) = 1 + 1 2 ( x 1) 1 8 ( x 1) 2 correct 3. T 2 ( x ) = 1 + 1 2 ( x 1) 1 4 ( x 1) 2 4. T 2 ( x ) = 1 + 1 4 ( x 1) + 1 4 ( x 1) 2 5. T 2 ( x ) = 1 1 4 ( x 1) 1 4 ( x 1) 2 6. T 2 ( x ) = 1 1 4 ( x 1) + 1 8 ( x 1) 2 Explanation: The degree 2 Taylor polynomial centered at x = 1 for a general f is given by T 2 ( x ) = f (1)+ f (1) ( x 1)+ f (1) 2!...
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 Fall '08
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