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# soln3 - MATH 245 Linear Algebra 2 Solutions to Assignment 3...

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MATH 245 Linear Algebra 2, Solutions to Assignment 3 1: (a) Find the least-squares best fit quadratic f P 2 for the following data points. x i - 1 0 1 2 3 y i 0 2 3 2 - 2 Solution: Let A = 1 x 1 x 1 2 1 x 2 x 2 2 1 x 3 x 3 2 1 x 4 x 4 2 1 x 5 x 5 2 = 1 - 1 1 1 0 0 1 1 1 1 2 4 1 3 9 and y = y 1 y 2 y 3 y 4 y 5 = 0 2 3 2 - 2 . Then we have A t A = 1 1 1 1 1 - 1 0 1 2 3 1 0 1 4 9 1 - 1 1 1 0 0 1 1 1 1 2 4 1 3 9 = 5 5 15 5 15 35 15 35 99 . A t y = 1 1 1 1 1 - 1 0 1 2 3 1 0 1 4 9 0 2 3 2 - 2 = 5 1 - 7 , and ( A t A A t y ) = 5 5 15 5 15 35 15 35 99 5 1 - 7 1 1 3 0 10 20 0 20 54 1 - 4 - 22 1 1 3 0 1 2 0 0 14 1 - 2 5 - 14 1 0 1 0 1 2 0 0 1 7 5 - 2 5 - 1 1 0 0 0 1 0 0 0 1 12 5 8 5 - 1 . Thus the best fit quadratic is f ( x ) = 12 5 + 8 5 x - x 2 . (b) Let { p 1 , p 2 , · · · , p l } be a linearly independent set of polynomials in P m and let ( x 1 , y 1 ) , ( x 2 , y 2 ) , · · · , ( x n , y n ) be points in R 2 such that at least m +1 of the x -coordinates x i are distinct. Show that there exists a unique polynomial f Span { p 1 , p 2 , · · · , p l } which minimizes the sum n i =1 ( y i - f ( x i ) ) 2 . Solution: For p ( t ) = a 0 + a 1 t + · · · + a m t m P m , we write p ( x ) = ( p ( x 1 ) , p ( x 2 ) , · · · , p ( x n ) ) t R n . Note that p ( x ) = p ( x 1 ) p ( x 2 ) . . . p ( x n ) = a 0 + a 1 x 1 + · · · + a m x 1 m a 0 + a 1 x 2 + · · · + a m x 2 m . . . a 0 + a 1 x n + · · · + a m x n m = A [ p ] , where A = 1 x 1 x 1 2 · · · x 1 m 1 x 2 x 2 2 · · · x 2 m . . . 1 x n x n 2 · · · x n m M n × ( m +1) and [ p ]= a 0 a 1 . . . a m R m +1 is the coordinate vector of p . We need to find c R l such that the polynomial f = c 1 p 1 + c 2 p 2 + · · · + c l p l minimizes y - f ( x ) . Note that f ( x ) = c 1 p 1 ( x ) + · · · + c l p l ( x ) = c 1 A [ p 1 ] + · · · + c l A [ p l ] = ( A [ p 1 ] , · · · , A [ p l ] ) c = APc , where P = ( [ p 1 ] , [ p 2 ] , · · · , [ p l ] ) M ( m +1) × l . Since some m + 1 of the x i are distinct, it follows from question 5 on assignment 2 that some set of m + 1 rows of A is linearly independent, so A has rank m + 1 and hence A is 1:1. Since { p 1 , p 2 , · · · , p l } is linearly independent, it follows that [ p 1 ] , [ p 2 ] , · · · , [ p l ] is also linearly independent, so P has rank l and hence P is 1:1. Since P and A are both 1:1, so is the composite AP , so AP has rank l , and the l × l matrix ( AP ) t ( AP ) is invertible. To minimize the length y - f ( x ) = y - APc we must have APc = Proj Col( AP ) y , that is y = APc + z for some z

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soln3 - MATH 245 Linear Algebra 2 Solutions to Assignment 3...

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