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# solexam1s11 - Foundations of Computational Math II Exam 1...

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Foundations of Computational Math II Exam 1 In-class Exam Open Notes, Textbook, Homework Solutions Only Calculators Allowed Wednesday February 23, 2011 Question Points Points Possible Awarded 1. Interpolation 40 2. Piecewise Interpolation 30 3. Splines and Basics 30 Total 100 Points Name: SOLUTIONS 1

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Problem 1 (40 points) Assume you are given the data points ( x 0 , f ( x 0 )) = (1 , 7) , ( x 1 , f ( x 1 )) = (3 , 59) ( x 2 , f ( x 2 )) = (4 , 124) , ( x 3 , f ( x 3 )) = (6 , 402) 1.a (10 points) Find the unique cubic interpolating polynomial, p 3 ( x ), in Newton form and evaluate it at x = 0. Solution: The divided difference table is i x i f ( x i ) f 1 [ * , * ] f 2 [ * , * , * ] f 3 [ * , * , * , * ] 0 1 7 26 1 3 59 13 65 7/3 2 4 124 74/3 139 3 6 402 Various paths can be chosen. p 3 ( x ) = 7 + 26( x - 1) + 13( x - 1)( x - 3) + 7 3 ( x - 1)( x - 3)( x - 4) p 3 ( x ) = 402 + 139( x - 6) + 74 3 ( x - 6)( x - 4) + 7 3 ( x - 6)( x - 4)( x - 3) p 3 ( x ) = - 8 + 55 3 x - 17 3 x 2 + 7 3 x 3 = 1 3 ( - 24 + 55 x - 17 x 2 + 7 x 3 ) p 3 (0) = - 8 2
1.b (15 points) Given that the divided difference table used to construct the Netwon form above uses the points in the increasing order given in the data, i.e., x 0 < x 1 < x 2 < x 3 , determine the Newton form of the polynomial that evaluates p 3 (2 . 75) using an heuristic designed to keep the numerical error in the evaluation small. Solution: Since we are to use the given the divided difference table computed with the ordering x 0 < x 1 < x 2 < x 3 the Leja ordering cannot be used. It would require recomputing the divided difference table. Hence, we must use the ordering given but are free to choose the path through the table based on the value of x . The heuristic is to add points to the polynomial in increasing order of distance from the argument x = 2 . 75. We have x 1 - 2 . 75 = 0 . 25 x 2 - 2 . 75 = 1 . 25 x 0 - 2 . 75 = - 1 . 75 x 3 - 2 . 75 = 3 . 25 ordering is: x 1 , x 2 , x 0 , x 3 p 3 ( x ) = f 1 + ( x - x 1 ) f [ x 1 , x 2 ] + ( x - x 1 )( x - x 2 ) f [ x 0 , x 1 , x 2 ] + ( x - x 1 )( x - x 2 )( x - x 0 ) f [ x 0 , x 1 , x 2 , x 3 ] = 59 + 65( x - 3) + 13( x - 3)( x - 4) + 7 3 ( x - 3)( x - 4)( x - 1) 3

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1.c (15 points) Suppose the interpolating cubic p 3 ( x ) is stored by keeping f 3 , f [ x 2 , x 3 ] , f [ x 1 , x 2 , x 3 ] , f [ x 0 , x 1 , x 2 , x 3 ] and x 0 , x 1 , x 2 , x 3 . Now suppose you are given values for x 4 and f [ x 0 , x 1 , x 2 , x 3 , x 4 ]. Describe an efficient algorithm that updates the information stored for p 3 ( x ) to specify similar infomation that determines the cubic interpolating polynomial q 3 ( x
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solexam1s11 - Foundations of Computational Math II Exam 1...

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