Homework 11 - = x 2-x 2 , l 1 ( x ) = ( x-(-1))( x-1)...

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Homework 11: Answer Key MATH 3315 / CSE 3365 – Fall Semester 2007 November 18, 2007 The homework assignment consists of answering the following problems from the lecture notes: Problem 4.4.4 : (5 marks) The least sqaures Normal Equations for a straight line Ft p 1 ( x ) = a 0 + a 1 x to the data ( x i , f i ) N i =0 are " N i =0 1 N i =0 x i N i =0 x i N i =0 x 2 i # ± a 0 a 1 ² = " N i =0 f i N i =0 f i x i # . ±or the data in this problem we get ± 5 20 20 100 ² ± a 0 a 1 ² = ± 5 20 ² . Solving the above linear system we get a 0 = 1 and a 0 = 0 giving p 1 ( x ) = 1. This is expected because all the data have constant value 1 so we can Ft them exactly with a constant. Problem 5.2.12 : (5 marks) In Lagrange form, the quadratic interpolating polynomial to the data (-1,f(-1)), (0,f(0)), and (1,f(1)) is p 2 ( x ) = f ( - 1) l 0 ( x ) + f (0) l 1 ( x ) + f (1) l 2 ( x ), where the Lagrange basis polynomials are l 0 ( x ) = ( x - 0)( x - 1) ( - 1 - 0)( - 1 - 1)
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Unformatted text preview: = x 2-x 2 , l 1 ( x ) = ( x-(-1))( x-1) (0-(-1))(0-1) = x 2-1-1 , l ( x ) = ( x-(-1))( x-0) (1-(-1))(1-0) = x 2 + x 2 . The Simpson’s rule is derived as R S ( f ) = R 1-1 p 2 ( x )d x = f (-1) R 1-1 l ( x )d x + f (0) R 1-1 l 1 ( x )d x + f (1) R 1-1 l 2 ( x )d x . 1 We have Z 1-1 l ( x )d x = Z 1-1 x 2-x 2 d x = 1 2 ± x 3 3-x 2 2 ² 1-1 = 1 3 , Z 1-1 l 1 ( x )d x = Z 1-1 x 2-1-1 d x =-± x 3 3-x ² 1-1 = 4 3 , Z 1-1 l 2 ( x )d x = Z 1-1 x 2 + x 2 d x = 1 2 ± x 3 3 + x 2 2 ² 1-1 = 1 3 . So Simpson’s rule for I ( f ) = R 1-1 f ( x )d x is R S ( f ) = 1 3 f (-1) + 4 3 f (0) + 1 3 f (1). 2...
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This homework help was uploaded on 04/07/2008 for the course MATH 3337 taught by Professor Xu during the Fall '07 term at Southern Methodist.

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Homework 11 - = x 2-x 2 , l 1 ( x ) = ( x-(-1))( x-1)...

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