# exam1sol - Math 471 Midterm 1 17 October 2007 6-8 pm Name...

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Math 471 Midterm 1 17 October 2007, 6-8 pm Name: Instructor: Show all work and circle your final answers. If you need additional space, continue on the back of the page or on the extra sheet at the end of the exam. No calculators allowed. Problem Possible Points Score 1 20 2 15 3 15 4 10 5 20 6 20 Total 100 1

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1. (True/False) Justify your answer for full credit. (a) [5 pts] Suppose the value of a function f at certain points x is given by x f ( x ) 1.2 3.64 0.6 3.16 0.3 3.04 0.15 3.01 The data suggest that f is converging to 3 as x 0 with rate of convergence O( x ). False: lim x 0 | f ( x ) - 3 | ≈ 4 9 x 2 . So the rate of convergence should be O ( x 2 ). (b) [5 pts] Set f ( x ) = cos( x ) - 1 and note f (2 π ) = 0. Using g ( x ) = - x 2 2! + x 4 4! in place of f ( x ) is an effective way to reduce the error from cancellation for x close to 2 π . False: g ( x ) is the first two terms of the Taylor expansion of f ( x ) near 0, NOT 2 π . So if we want to reduce the error near 2 π , we should replace x in g ( x ) with ( x - 2 π ). 2
(c) [5 pts] Using three digit rounding decimal arithmetic, the value of 2 × 10 2 · (4 × 10 3 - 3 × 10 0 ) + 1 × 10 3 is 8 . 01 × 10 5 . True (Details skipped here but you should write them out) (d) [5 pts] Newton’s method will perform better finding the unique root of f ( x ) = x 3 - 2 than in finding the unique root of g ( x ) = ( x - 2) 3 .

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