# Final 'Quiz' Solution - 472.pdf - MATH 472 002/992 Final...

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MATH 472 002/992 Final Exam December 16, 2020 Read the following instructions carefully: There are seven problems and you have 120 minutes . Each submission must be a single pdf file including your solutions in the order they were assigned. Do not use red ink or color in your submission. Write your name at the top right of each page. Each problem should start on a fresh page in your submission pdf. You may use one double-sided sheet ( 8 . 5 × 11 inches) of notes in your own handwriting . No other notes or books are allowed. You may use a calculator, but all intermediate steps in your calculations must be compre- hensible from your notes. Simplify your results as far as possible. All results have to be justified by appropriate intermediate steps and/or arguments . You can, however, use standard formulas and results known from the lecture without deriva- tion (unless explicitly stated otherwise). You will have 15 minutes to upload your submission to Canvas after the end of the exam . Good luck! Sheet 1 of 15
Problem 1 (12 Points) Name: Assume that you are to use Chebyshev interpolation to find a degree 2 interpolating polynomial Q 2 ( x ) that approximates the function f ( x ) = x 1 / 3 (= 3 x ) on the interval [3 , 4] . (a) (5 Points) Write down the ( x, y ) points that will serve as interpolation nodes for Q 2 . (b) (4 Points) Find a worst-case estimate for the error | f ( x ) - Q 2 ( x ) | that is valid for all x in the interval [3 , 4] . (c) (3 Points) How many digits after the decimal point will be correct when Q 2 is used to approximate f ? Solution: (a) For a 2 - degree interpolating polynomial, we need n = 3 nodes. (1P) The Chebyshev inter- polation nodes on [ - 1 , 1] are given by ¯ x 1 := cos ( π 6 ) = 3 2 , ¯ x 2 := cos ( 3 π 6 ) = cos ( π 2 ) = 0 and ¯ x 3 := cos ( 5 π 6 ) = - 3 2 . (1P) Therefore, for the interval [ a, b ] = [3 , 4] the interpolating nodes are x i := a + b 2 + b - a 2 ¯ x i = 7 2 + 1 2 ¯ x i , for i = 1 , 2 , 3 . (2P) Their approximations are x 1 3 . 933 , x 2 = 3 . 5 and x 3 3 . 067 . In total, the required nodes are ( x 1 , f ( x 1 ) ) = (3 . 933 , 3 3 . 933) = (3 . 933 , 1 . 578528) , ( x 2 , f ( x 2 ) ) = (3 . 5 , 3 3 . 5) = (3 . 5 , 1 . 5183) and (1P) ( x 3 , f ( x 3 ) ) = (3 . 067 , 3 3 . 067) = (3 . 067 , 1 . 4529) . (b) The interpolation error formula gives sup x [3 , 4] | f ( x ) - Q 2 ( x ) | = sup x [3 , 4] n | ( x - x 1 )( x - x 2 )( x - x 3 ) | 3! | f (3) ( x ) | o (1P) 1 2 5 × 3! × 0 . 019784 0 . 000103 . (1P) where we have used that | ( x - x 1 )( x - x 2 )( x - x 3 ) | ≤ b - a 2 3 2 2 = 1 2 5 (1P) and f (3) ( x ) = 1 3 x - 2 3 00 = - 2 9 x - 5 3 0 = 10 27 x - 8 3 with 0 f (3) ( x ) f (3) (3) = 0 . 019784 for x [3 , 4] (1P) .
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