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Unformatted text preview: Ma 450: Mathematics for Multimedia Solution: to Homework Assignment 4 Prof. Wickerhauser Due Friday, April 2nd, 2010 1. Find the Lagrange polynomial through the points (1 , 3), (4 , 3). Solution: The Lagrange polynomial through the points (1 , 3), (4 , 3) is Λ 1 ( x ) = 3. This is a degenerate case of two points on a line of zero slope; it yields a constant interpolating polynomial, of degree 0. 2 2. Fix x and find a formula for the value y = f ( x ) of the Lagrange polynomial f through the points ( 1 , p ), (0 , q ), and (1 , r ), in terms of x , p , q , and r . Then find d 2 y/dx 2 . Solution: The Lagrange polynomial is Λ 2 ( x ) = q p x 2 + r x 2 + p x 2 2 q x 2 + r x 2 2 Its second derivative with respect to x is d 2 dx 2 Λ 2 ( x ) = p 2 q + r . 2 3. Fix an integer k > 0 and let { x j = j : j = 0 , 1 , ..., k } be an equispaced grid. For f ( x ) = x k , prove that Δ k f ( k ) = k !. (Hint: consider c k in Equation 4.7.) Solution: This may be done using Equation 4.7 after an application of Theorem 4.2. Since the Lagrange polynomial of degree k or less interpolating the k +1 points { ( j, j k ) : j = 0 , 1 , . . . , k } is unique, and the polynomial f ( x ) = x k evidently interpolates this set, it must be that Λ k ( x ) = f ( x ) = x k . But also, Λ k ( x ) = Λ k 1 ( x ) + c k k 1 Y i =0 ( x x i ) , where Λ k 1 ( x ) is the unique Lagrange polynomial of degree k 1 or less interpolating { ( j, j k ) : j = , 1 , . . . , k 1 } . But x i = i , so the abscissas are equispaced with h = 1, so the Newtonform coefficient is c k = Δ k f ( k ) /k !....
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This note was uploaded on 05/16/2010 for the course MATH 450 taught by Professor Wickerhauser during the Spring '10 term at Washington University in St. Louis.
 Spring '10
 Wickerhauser
 Math

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