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# set2 - \$ Set 2 Polynomial Interpolation Part 2 Kyle A...

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a39 a38 a36 a37 Set 2: Polynomial Interpolation – Part 2 Kyle A. Gallivan Department of Mathematics Florida State University Foundations of Computational Math 2 Spring 2011 1

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a39 a38 a36 a37 Overview Complexity measured in terms of number of computations, i.e., sequential computation Evaluation of a polynomial in mononmial form Interpolation polynomials For each form consider: Complexity of constructing the polynomial, i.e., the parameters Complexity of evaluating the polynomial at a point x negationslash = x i Complexity of updating the polynomial to include a new point 2
a39 a38 a36 a37 Horner’s Rule Assume the polynomial, p n ( x ) , is given in terms of monomials, x i p n ( x ) = α 0 + α 1 x + α 2 x 2 + · · · + α n x n For example, let n = 4 p 4 ( x ) = α 0 + α 1 x + α 2 x 2 + α 3 x 3 + α 4 x 4 p 4 ( x ) = α 0 + x ( α 1 + x ( α 2 + x ( α 3 + ( 4 ))) Repeated application leads to evaluation of derivatives at x . Adaptable to other similar forms, e.g., Newton form. 3

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a39 a38 a36 a37 Lagrange Form Storing only x i , y i , i.e., no preprocessing, yields an O ( n 2 ) to evaluate p n ( x ) via repeated linear interpolation, e.g., Aitken’s method. Using the basic Lagrange form defined earlier yields O ( n 2 ) to compute the n + 1 coefficients that define p n ( x ) O ( n 2 ) to evaluate p n ( x ) Rewriting yields improvements Barycentric form 1 Barycentric form 2 4
a39 a38 a36 a37 Lagrange Form Recall, the standard form is sum of n -degree polynomials m ( n ) i ( x ) = n productdisplay j =0 ,j negationslash = i ( x x j ) ( n ) i ( x ) = m ( n ) i ( x ) m ( n ) i ( x i ) p n ( x ) = n summationdisplay i =0 y i ( n ) i ( x ) 5

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a39 a38 a36 a37 Lagrange Form Lemma. Given ( x i , y i ) for 0 i n and defining m ( n ) i ( x ) = n productdisplay j =0 ,j negationslash = i ( x x j ) and ω j ( x ) = j 1 productdisplay i =0 ( x x i ) we have m ( n ) i ( x ) = ω n +1 ( x ) ( x x i ) and m ( n ) i ( x i ) = ω n +1 ( x i ) and the Lagrange characteristic functions can be written i ( x ) = ω n +1 ( x ) ( x x i ) ω n +1 ( x i ) . 6
a39 a38 a36 a37 Lagrange Form Proof. Let i = n . We have ω n +1 ( x ) = ω n ( x )( x x n ) ω n +1 ( x ) = ω n ( x )( x x n ) + ω n ( x ) ω n +1 ( x n ) = ω n ( x n ) n ( x ) = ω n ( x ) ω n ( x n ) = ω n +1 ( x ) ( x x n ) ω n +1 ( x n ) . This adapts trivially to i ( x ) for i negationslash = n . 7

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a39 a38 a36 a37 Barycentric Form 1 Definition 2.1. (Berrut and Trefethen, Siam Review Vol. 46 No. 3) Given ( x i , y i ) for 0 i n , the Barycentric interpolation formula form 1 is p n ( x ) = ω n +1 ( x ) n summationdisplay i =0 y i ( x x i ) ω n +1 ( x i ) = ω n +1 ( x ) n summationdisplay i =0 y i γ i ( x x i ) where γ i = 1 n +1 ( x i ) and ω n +1 ( x ) = n productdisplay i =0 ( x x i ) 8
a39 a38 a36 a37 Barycentric Form 1 Construction of p n ( x ) requires the computation of γ i , for 0 i n where γ 1 i = ω n +1 ( x i ) = m ( n ) i ( x i ) = n productdisplay j =0 ,j negationslash = i ( x i x j ) Construction of p n ( x ) does not depend on y i .

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set2 - \$ Set 2 Polynomial Interpolation Part 2 Kyle A...

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