practicesolution

practicesolution - i +2 ,x i +1 ,x i ) f ( x i +3 ,x i +2...

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Practice Solution Given a set of data provided in Table 2 below: (a) Write the general form of the Newton’s interpolating polynomial that passes through five data points. (b) Make a table with columns of i , x i , f ( x i ), f [ x i +1 ,x i ], f [ x i +2 ,x i +1 ,x i ], etc., and complete the table. List all coefficients clearly and write down the Newton’s interpolating polynomial. (c) Derive the Lagrange form of the equation that passes through the same data points. i x i f ( x i ) 1 -2 -15 2 -1 0 3 0 3 4 1 0 5 2 -3 Solution: (a) The Newton form of the equation y = f ( x ) = b 0 + 4 X i =1 b i i - 1 Y j =0 ( x - x j ) (b) The result is shown as following table i x i f ( x i ) f ( x i +1 ,x i ) f ( x
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Unformatted text preview: i +2 ,x i +1 ,x i ) f ( x i +3 ,x i +2 ,x i +1 ,x i ) f ( x i +4 ,x i +3 ,x i +2 ,x i +1 ,x i )-2-15 15-6 1 1-1 3-3 1 2 3-3 3 1-3 4 2-3 Newton’s interpolating polynomial: f ( x ) =-15 + 15( x + 2)-6( x + 2)( x + 1) + x ( x + 2)( x + 1) (c) Lagrange polynomial form f ( x ) = 4 X i =0 f ( x i ) Q 4 j 6 = i ( x-x j ) Q 4 j 6 = i ( x i-x j ) Given data from the table f ( x ) =-. 625( x +1) x ( x-1)( x-2)+0 . 75( x +2)( x +1)( x-1)( x-2)-. 125( x +2)( x +1) x ( x-1) 1...
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This note was uploaded on 11/14/2011 for the course EAME 250 taught by Professor Lee during the Spring '11 term at Case Western.

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