6 06 x x 06 x 09 dx 06 09 x x 06 x 09 dx 09 1 x x 06 x 09 dx \u00b2 1 4 x 06 4 01 x

6 06 x x 06 x 09 dx 06 09 x x 06 x 09 dx 09 1 x x 06

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6 ; 0 0.6 x x " 0.6   x " 0.9   dx " ; 0.6 0.9 x x " 0.6   x " 0.9   dx ± ; 0.9 1 x x " 0.6   x " 0.9   dx ² 1 4 x " 0.6   4 ± 0.1 x " 0.6   3 " 0.09 x " 0.6   2 | 0 0.6 " 1 4 x " 0.6   4 ± 0.1 x " 0.6   3 " 0.09 x " 0.6   2 | 0.6 0.9 ± 1 4 x " 0.9   4 ± 0.4 x " 0.9   3 ± 0.27 2 x " 0.9   2 | 0.9 1 ² 1 4 0.6   4 ± 0.1 0.6   3 " 0.09 0.6   2 " 1 4 0.3   4 " 0.1 0.3   3 ± 0.09 0.3   2 ± 1 4 0.1   4 ± 0.4 0.1   3 ± 0.27 2 0.1   2 ² 0.0268979 3. Neville’s Iterated Interpolation: Naturally if P n x   is not a good approximation to f x   we like to add more points x n ± 1 , ... to construct an interpolation polynomial with larger degree. Can we construct an interpolation polynomial iteratively ? The answer is yes. Neville’s Method does so. Let m 1 , m 2 , ..., m k be k distinct integers where 0 t m i t n for each i . Let P m 1 , m 2 ..., m k x   be the Lagrange polynomial that agrees with f at the k points: x m 1 , x m 2 , ..., x m k . For example, P 1,2,4 x   ² f x 1   x " x 2   x " x 4   x 1 " x 2   x 1 " x 4   ± f x 2   x " x 1   x " x 4   x 2 " x 1   x 2 " x 4   ± f x 4   x " x 1   x " x 2   x 4 " x 1   x 4 " x 2   is the Lagrange polynomial that agrees with f x   at 3 points: x 1 , x 2 , x 4 : P 1,2,4 x 1   ² f x 1   , P 1,2,4 x 2   ² f x 2   , P 1,2,4 x 3   ² f x 3   . Theorem 3 . 5 Let f be defined at x 0 , x 1 , ..., x k , and x j and x k be two distinct numbers. Then P 0,1, .... , k x   ² x " x j   P 0,1,..., j " 1, j ± 1,..., k x   " x " x i   P 0,1,..., i " 1, i ± 1,..., k x   x i " x j . Proof We know that both polynomials P 0,1,..., j " 1, j ± 1,..., k x   and P 0,1,..., i " 1, i ± 1,..., k x   are degree k " 1. So, the degree of P 0,1, .... , k x   is k . Check if P 0,1, .... , k x l   ² f x l   for l ² 0, 1,..., k : since P 0,1,..., j " 1, j ± 1,..., k x l   ² f x l   if l p j and P 0,1,..., i " 1, i ± 1,..., k x l   ² f x l   if l p i For l ² 0,1,..., k , but l p i and l p j , P 0,1, .... , k x l   ² x l " x j   P 0,1,..., j " 1, j ± 1,..., k x l   " x l " x i   P 0,1,..., i " 1, i ± 1,..., k x l   x i " x j ² x l " x j   f x l   " x l " x i   f x l   x i " x j   ² x i " x j   x i " x j   f x l   ² f x l   For l ² i : P 0,1, .... , k x i   ² x i " x j   P 0,1,..., j " 1, j ± 1,..., k x i   " x i " x i   P 0,1,..., i " 1, i ± 1,..., k x i   x i " x j ² P 0,1,..., j " 1, j ± 1,..., k x i   ² f x i   For l ² j : P 0,1, .... , k x i   ² x j " x j   P 0,1,..., j " 1, j ± 1,..., k x j   " x j " x i   P 0,1,..., i " 1, i ± 1,..., k x j   x i " x j ² P 0,1,..., i " 1, i ± 1,..., k x j   ² f x j   So, P 0,1, .... , k x   the k th degree interpolating polynomial. 5
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Example Consider x 0 , x 1 ,..., x n . Then P 1,2 x   ² f x 1   x " x 2   x 1 " x 2   ± f x 2   x " x 1   x 2 " x 1   P 2,4 x   ² f x 2   x " x 4   x 1 " x 4   ± f x 4   x " x 4   x 2 " x 4   P 1,2,4 x   ² x " x 4   P 1,2 x   " x " x 1   P 2,4 x   x 1 " x 4   is the Lagrange polynomial that agrees with f x   at 3 points: x 1 , x 2 , x 4 . Neville’s Iterated Interpolation: Steps of Computing P 0,1,2,..., k x ³   : Given x i , f x i   for i ² 0,1,..., k and / , let P i ² f x i   x i P i x ³   P i , i ± 1 x ³   P i , i ± 1, i ± 2 x ³   P i , i ± 1, i ± 2, i ± 3 x ³   P i , i ± 1, i ± 2, i ± 3, i ± 4 x ³   x 0 P 0 x 1 P 1 P 0,1 x ³   x 2 P 2 P 1,2 x ³   P 0,1,2 x ³   x 3 P 3 P 2,3 x ³   P 1,2,3 x ³   P 0,1,2,3 x ³   x 4 P 4 P 3,4 x ³   P 2,3,4 x ³   P 1,2,3,4 x ³   P 0,1,2,3,4 x ³   B B B B B B
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