The algorithm stops and f x \u00b3 X P 012 k x \u00b3 whenever P 012 k x \u00b3 P 012 k 1 x \u00b3

The algorithm stops and f x ³ x p 012 k x ³

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The algorithm stops and f x ³   X P 0,1,2,..., k x ³   whenever P 0,1,2,..., k x ³   " P 0,1,2,..., k " 1 x ³   ´ / . An alternative way: 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 0,2 x ³   P 0,1,2 x ³   x 3 P 3 P 0,3 x ³   P 0,2,3 x ³   P 0,1,2,3 x ³   x 4 P 4 P 0,4 x ³   P 0,3,4 x ³   P 0,2,3,4 x ³   P 0,1,2,3,4 x ³   B B B B B B Example Suppose that x j ² j for j ² 0,1,2,3 and it is known that P 0,1 x   ² x ± 1, P 1,2 x   ² 3 x " 1, and P 1,2,3 1.5   ² 4. Find P 0,1,2,3, 1.5   . 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   0 P 0 1 P 1 P 0,1 1.5   ² 2.5 2 P 2 P 1,2 1.5   ² 3.5 P 0,1,2 1.5   ² 3.25 3 P 3 P 0,3 1.5   P 1,2,3 1.5   ² 4 P 0,1,2,3 1.5   ² 5.4375 6
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P 0,1,2 1.5   ² 1.5 " 2   P 0,1 1.5   " 1.5 " 0   P 1,2 1.5   0 " 2   ² " 0.5   2.5   " 1.5   3.5   " 2   ² 3.25 P 0,1,2,3 1.5   ² 1.5 " 3   P 0,1,2 1.5   " 1.5 " 0   P 1,2,3 1.5   0 " 3   ² " 1.5   3.25   " 1.5   4   " 2   ² 5. 4375 Example Neville’s method is used to approximate f 0.4   as follows. Complete the table. x i P i x ±   P i , i ± 1 x ±   P i , i ± 1, i ± 2 x ±   P i , i ± 1, i ± 2, i ± 3 x ±   0 1 0.25 2 P 0,1 0.4   ² 2.6 0.5 P 2 P 1,2 0.4   P 0,1,2 0.4   0.75 8 P 2,3 0.4   ² 2.4 P 1,2,3 0.4   ² 2.96 P 0,1,2,3 0.4   ² 3.016 P 2,3 0.4   ² 0.4 " 0.75   P 2 " 0.4 " 0.5   P 3 0.5 " 0.75 ² 2.4, P 2 ² 2.4   " 0.25   " 0.1 8   " 0.35 ² 4 P 1,2 0.4   ² 0.4 " 0.5   P 1 " 0.4 " 0.25   P 2 0.25 " 0.5 ² 0.4 " 0.5   2   " 0.4 " 0.25   4   0.25 " 0.5 ² 3.2 P 0,1,2 0.4   ² 0.4 " 0.5   P 0,1 0.4   " 0.4 " 0   P 1,2 0.4   0 " 0.5 ² 0.4 " 0.5   2.6   " 0.4 " 0   3.2   0 " 0.5 ² 3.08 Check: P 0,1,2,3 0.4   ² 0.4 " 0.75   3.08   " 0.4 " 0   2.96   0 " 0.75 ² 3. 016 Example Let f x   ² sin lnx   , x 0 ² 2.0, x 1 ² 2.4, x 2 ² 2.6. Find a bound for the absolute error on 2.0, 2.6 . Approximate f 2.5   . R 2 x   ² f UUU c   3! x " 2   x " 2.4   x " 2.6   , where c is in 2, 2.6 f U x   ² cos ln x   x , f UU x   ² " sin ln x   1 x   x   " cos ln x   x 2 ² " sin ln x   " cos ln x   x 2 f UUU x   ² ¡ " cos ln x   1 x   ± sin ln x   1 x  ¢ x 2 ± 2 x ¡ sin ln x   ± cos ln x  ¢ x 4 ² ¡ " cos ln x   ± sin ln x  ¢ ± 2 ¡ sin ln x   ± cos ln x  ¢ x 3 ² 3sin ln x   ± cos ln x   x 3 7
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0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 2 2.1 2.2 2.3 2.4 2.5 2.6 f UUU x   0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 2 2.1 2.2 2.3 2.4 2.5 2.6 x y ² | x " 2   x " 2.4   x " 2.6   | f UUU x   t f UUU 2   ² 3sin ln2   ± cos ln2   2 3 ² 0.336 R 2 x   ² f UUU c   3! x " 2   x " 2.4   x " 2.6   t 0.336 6 0.02   ² 0.00112 ±± xv ² [2;2.4;2.6]; ±± yv ² sin(log(xv)); ±± [yout,yall] ² neville(xv,yv,2.5,3) yout ² 0.7935 yall ² 0.6390 0.8001 0.7935 0.7678 0.7922 0 0 0.8166 0 0 8
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