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lab2 - MA 2213 Numerical Analysis 1 Year 20122013 Semester...

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MA 2213 Numerical Analysis 1 Year 2012–2013 Semester II Laboratory 2 Objectives 1. Task 1: to reduce the interpolation error by adjusting the locations of the nodes 2. Task 2: to build a function that computes the coefficients of natural cubic splines Background 1. Let P n be the interpolating polynomial for f : [ a, b ] R using n + 1 distinct nodes x 0 , x 1 , . . . , x n . The interpolation error at x is given by | f ( x ) - P n ( x ) | . If we are allowed to choose the n + 1 nodes freely, then which choice is the “best”? A natural way to define “the best” is the one that makes the maximum absolute error in [ a, b ] (worst-case error throughout the interval [ a, b ]) max x [ a,b ] | f ( x ) - P n ( x ) | minimal. To simplify the problem, we let y j = a + j ( b a ) m for j = 0 , 1 , 2 , . . . , m with a large integer m (e.g. 2 × 10 5 ) and consider the minimization of E ( x 0 , x 1 , . . . , x n ) := max j =0 , 1 ,...,m | f ( y j ) - P n ( y j ) | . with respect to x 0 , x 1 , . . . , x n . In this task, you are asked to empirically determine a good distribution of the nodes (one that makes the error E ( x 0 , . . . , x n ) small). When you finish this task, upload the best error and the corresponding nodes that you found to IVLE. The best three solutions will be posted in the Forum for sharing. 2. We shall use notations that are consistent with Matlab. Let ( x i , f i ) for i = 1 , 2 , . . . , n be given data (note that the index starts from 1 instead of 0 because indices of Matlab arrays start from 1). Let h i = x i +1 - x i for i = 1 , 2 , . . . , n - 1 be the spacing of the i th subinterval. Then, the n × n linear system that determines the coefficients { c i } i =1 , 2 ,...,n of a natural spline is (the blanks depict zeros) 1 0 h 1 2( h 1 + h 2 ) h 2 h 2 2( h 2 + h 3 ) h 3 . . . . . . . . . . . . . . . . . . h n - 2 2( h n - 2 + h n - 1 ) h n - 1 0 1 c 1 c 2 c 3 . . . . . . c n - 1 c n = 0 3 h 2 ( f 3 f 2 ) 3 h 1 ( f 2 f 1 ) 3 h 3 ( f 4 f 3 ) 3 h 2 ( f 3 f 2 ) .
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