solhw1

solhw1 - Solutions for Homework 1 Foundations of...

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Unformatted text preview: Solutions for Homework 1 Foundations of Computational Math 2 Spring 2011 Problem 1.1 Consider the data points ( x, y ) = { (0 , 2) , (0 . 5 , 5) , (1 , 8) } Write the interpolating polynomial in both Lagrange and Newton form for the given data. Solution: For the Lagrange form we have p 2 ( x ) = 2 ( x . 5)( x 1) ( . 5)( 1) + 5 ( x )( x 1) ( . 5)(0 . 5) + 8 ( x )( x . 5) (1)(0 . 5) = 6 x + 2 For Newtons form we have f [0 , . 5] = 6 f [0 . 5 , 1] = 6 f [0 , . 5 , 1] = 0 p 2 ( x ) = 2 + ( x 0) f [0 , . 5] + ( x 0)( x . 5) f [0 , . 5 , 1] = 6 x + 2 Problem 1.2 Use this divided difference table for this problem. Justify all of your answers. i 1 2 3 4 5 x i 1 2 4 5 6 f i 13 2 14 18 67 91 f [ , ] 11 8 16 49 24 f [ , , ] 1 6 11 25 / 2 f [ , , , ] 1 1 47 / 8 f [ , , , , ] 55 / 48 f [ , , , , , ] 55 / 336 1 1.2.a Use the divided difference information about the unknown function f ( x ) and consider the unique polynomial, denoted p 1 , 5 ( x ), that interpolates the data given by pairs ( x 1 , f 1 ), ( x 2 , f 2 ), ( x 3 , f 3 ), ( x 4 , f 4 ) , and ( x 5 , f 5 ). Use two different sets of divided differences to express p 1 , 5 ( x ) in two distinct forms. Solution: Two of the standard paths are the left and right edges of the triangle of divided differences defined by the pairs ( x 1 , f 1 ), ( x 2 , f 2 ), ( x 3 , f 3 ), ( x 4 , f 4 ) , and ( x 5 , f 5 ). The left side uses the points in standard order ( x 1 , f 1 ), ( x 2 , f 2 ), ( x 3 , f 3 ), ( x 4 , f 4 ) , and ( x 5 , f 5 ): p 1 , 5 ( x ) = f 1 + ( x x 1 ) f [ x 1 , x 2 ] + ( x x 1 )( x x 2 ) f [ x 1 , x 2 , x 3 ] + ( x x 1 )( x x 2 )( x x 3 ) f [ x 1 , x 2 , x 3 , x 4 ] + ( x x 1 )( x x 2 )( x x 3 )( x x 4 ) f [ x 1 , x 2 , x 3 , x 4 , x 5 ] = 2 8 x + 6 x ( x 2) + x ( x 2)( x 4) 55 48 x ( x 2)( x 4)( x 5) Note that p 1 , 5 (6) = 2 48 + 144 + 48 55 48 48 = 91 and the other points can also be verified. The right side uses the points in reverse order ( x 5 , f 5 ), ( x 4 , f 4 ), ( x 3 , f 3 ), ( x 2 , f 2 ) , and ( x 1 , f 1 ): p 1 , 5 ( x ) = f 5 + ( x x 5 ) f [ x 5 , x 4 ] + ( x x 5 )( x x 4 ) f [ x 5 , x 4 , x 3 ] + ( x x 5 )( x x 4 )( x x 3 ) f [ x 5 , x 4 , x 3 , x 2 ] + ( x x 5 )( x x 4 )( x x 3 )( x x 2 ) f [ x 5 , x 4 , x 3 , x 2 , x 1 ] p 1 , 5 ( x ) = f 5 + ( x x 5 ) f [ x 4 , x 5 ] + ( x x 5 )( x x 4 ) f [ x 3 , x 4 , x 5 ] + ( x x 5 )( x x 4 )( x x 3 ) f [ x 2 , x 3 , x 4 , x 5 ] + ( x x 5 )( x x 4 )( x x 3 )( x x 2 ) f [ x 1 , x 2 , x 3 , x 4 , x 5 ] due to the equivalence of divided differences independent of ordering. We therefore have p 1 , 5 ( x ) = f 5 + ( x x 5 ) f [ x 4 , x 5 ] + (...
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solhw1 - Solutions for Homework 1 Foundations of...

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