570 Chapter 20Numerical methods
EXAMPLE 20.9Newton’s method of divided differences fore
xFive points on the graph ofe
xare given in the first two columns of Table 20.6. The
divided differences have been computed on a standard 10-digit calculator, and are
quoted to 8 decimal places.
Table 20.5 Divided difference interpolation table
xy D
1
D
2
D
3
D
4
x
0
y
0
f[x
0
, x
1
]
x
1
y
1
f[x
0
, x
1
, x
2
]
f[x
1
, x
2
] f[x
0
, x
1
, x
2
, x
3
]
x
2
y
2
f[x
1
, x
2
, x
3
] f[x
0
, x
1
, x
2
, x
3
, x
4
]
f[x
2
, x
3
] f[x
1
, x
2
, x
3
, x
4
]
x
3
y
3
f[x
2
, x
3
, x
4
]
f[x
3
, x
4
]
x
4
y
4
Table 20.6 Divided difference table for e
xxy D
1
D
2
D
3
D
4
0.80 2.22554093
2.27065125
0.84 2.31636698 1.15833750
2.36331825 0.39393233
0.88 2.41089971 1.20560938 0.10058544
2.45976700 0.41002600
0.92 2.50929039 1.25481250
2.56015200
0.96 2.61169647
Using the numbers in boldfacein the table,
p
1(x) 1 = 1 2.22554093 1 + 1 2.27065125(x 1 − 1 0.8) for 0.80 1 < 1 x 1 < 1 0.84
p
2(x) 1 = 1 p
1(x) 1 + 1 1.15833750(x 1 − 1 0.8)(x 1 − 1 0.84) 0.80 1 < 1 x 1 < 1 0.88
p
3(x) 1 = 1 p
2(x) 1 + 1 0.39393233(x 1 − 1 0.8)(x 1 − 1 0.84)(x 1 − 1 0.88) 0.80 1 < 1 x 1 < 1 0.92
p
4(x) 1 = 1 p
3(x) 1 + 1 0.10058544(x 1 − 1 0.8)(x 1 − 1 0.84)(x 1 − 1 0.88)(x 1 − 1 0.92) 0.80 1 < 1 x 1 < 1 0.96