solution1_pdf

# solution1_pdf - dang(rtd356 – Wk 5 Check Knowledge –...

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dang (rtd356) – Wk 5 Check Knowledge – Cepparo – (57045) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points ±ind the coe²cient oF x 6 in the Taylor series expansion about x = 0 For f ( x ) = sin ( x 2 ) . 1. coe²cient = 1 120 2. coe²cient = 1 3. coe²cient = - 1 6 correct 4. coe²cient = 0 5. coe²cient = 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 coe²cient 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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solution1_pdf - dang(rtd356 – Wk 5 Check Knowledge –...

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