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Math128_hw3

# Math128_hw3 - Math 128A Spring 2007 Homework 3 Solution...

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Math 128A, Spring 2007 Homework 3 Solution 4.1.12 (a) L 0 ( x ) + L 1 ( x ) + L 2 ( x ) + L 3 ( x ) is the unique polynomial of degree 3 interpolating the data ( x 0 , 1) , ( x 1 , 1) , ( x 2 , 1) , ( x 3 , 1). Since the polynomial 1 also has degree 3 and interpolates the data, we conclude L 0 ( x ) + L 1 ( x ) + L 2 ( x ) + L 3 ( x ) = 1 . (b) In general, L 0 ( x ) + · · · L n ( x ) and 1 are both degree n interpolants of ( x 0 , 1) , . . . , ( x n , 1), and therefore must be equal. 4.1.13 For j 3, note that x j 0 L 0 ( x )+ x j 1 L 1 ( x )+ x j 2 L 2 ( x )+ x j 3 L 3 ( x ) and x j are both polynomials of degree 3 interpolating ( x 0 , x j 0 ) , ( x 1 , x j 1 ) , ( x 2 , x j 2 ) , ( x 3 , x j 3 ). Since such interpolants are unique, the two polynomials must be equal. For j > 3, note that each L i ( x ) has degree 3, so a linear combination of them can’t have the same degree as x j . 4.1.15 Note: this is the complicated solution. In analogy to the Lagrange basis { L i ( x ) } , we want to find a basis { H i ( x ) } for the (vector space of) polynomials of degree 3 so that the coordinates of P ( x ) with respect to { H i ( x ) } are ( y 0 , y 0 , y 1 , y 1 ). The analogous versions of the requirements L i (

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