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 )215 156 1 11 33 1 2 33 3 13 4 23 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 ( xx j ) Q 4 j 6 = i ( x ix j ) Given data from the table f ( x ) =. 625( x +1) x ( x1)( x2)+0 . 75( x +2)( x +1)( x1)( x2). 125( x +2)( x +1) x ( x1) 1...
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 Spring '11
 LEE
 Numerical Analysis, Newton polynomial, 4 j, 3 Newton, 0 4 j

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