128a_su09_hw5_sol

128a_su09_hw5_sol - MATH 128A, SUMMER 2009: HOMEWORK 5...

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Unformatted text preview: MATH 128A, SUMMER 2009: HOMEWORK 5 SOLUTIONS (1) Given a quadrature rule of the form R b a f ( x ) dx n i =0 c i f ( x i ), let f ( x ) be [( x- x ) ... ( x- x n )] 2 , a polynomial of degree 2( n + 1). Since f ( x ) is a continuous function, positive except at its n +1 roots, the exact value of R b a f ( x ) dx must be positive. Since f ( x ) is zero when x = x i , the estimate for R b a f ( x ) dx given by the quadrature rule is zero. Since the quadrature cant integrate f ( x ) (a polynomial of degree 2( n + 1)) exactly, its degree of precision must be less than 2( n + 1). (2) Use the inner product h f,g i = R e- x f ( x ) g ( x ) dx for functions on [0 , ). Let f ( x ) be a polynomial of degree 2 n- 1 (or less). use polynomial division to express f ( x ) in the form q ( x ) L n ( x )+ r ( x ), where q ( x ) (the quotient) is a polynomial of degree at most (2 n- 1)- n = n- 1, and r ( x ) (the remainder) is a polynomial of degree less than n . The quadrature rule estimates R e- x q ( x ) L n ( x ) dx exactly due to the choice of nodes: Since deg q ( x ) n- 1, q ( x ) is orthogonal to L n ( x ): the value of R e- x q ( x ) L n ( x ) dx is zero. Since q ( x ) L n ( x ) = 0 when x = x i , n i =1 c n,i f ( x i ) is also zero. The quadrature rule estimates R r ( x ) dx exactly due to the choice of coefficients: Let P n,i ( x ) = Q n j =1 j 6 = i x- x j x i- x j be the Legendre interpolating polynomial (with P n,i ( x i ) = 1 and P n,i ( x j ) = 0 for...
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128a_su09_hw5_sol - MATH 128A, SUMMER 2009: HOMEWORK 5...

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