Graphics Programs Reference
In-Depth Information
EXAMPLE 5.1
Given the evenly spaceddatapoints
x
0
0
.
1
0
.
2
0
.
3
0
.
4
f
(
x
)
0
.
0000
0
.
0819
0
.
1341
0
.
1646
0
.
1797
compute
f
(
x
) and
f
(
x
) at
x
=
0 and 0
.
2using finite difference approximationsof
O
(
h
2
).
Solution
From the forward difference formulas in Table 5.3a we get
−
3
f
(0)
+
4
f
(0
.
1)
−
f
(0
.
2)
−
3(0)
+
4(0
.
0819)
−
0
.
1341
f
(0)
=
=
=
.
0
967
2(0
.
1)
0
.
2
2
f
(0)
−
5
f
(0
.
1)
+
4
f
(0
.
2)
−
f
(0
.
3)
f
(0)
=
(0
.
1)
2
2(0)
−
5(0
.
0819)
+
4(0
.
1341)
−
0
.
1646
=
=−
3
.
77
(0
.
1)
2
The central difference approximations in Table 5.1 yield
−
f
(0
.
1)
+
f
(0
.
3)
−
0
.
0819
+
0
.
1646
f
(0
.
=
=
=
.
2)
0
4135
2(0
.
1)
0
.
2
f
(0
.
1)
−
2
f
(0
.
2)
+
f
(0
.
3)
0
.
0819
−
2(0
.
1341)
+
0
.
1646
f
(0
.
2)
=
=
=−
2
.
17
(0
.
1)
2
(0
.
1)
2
EXAMPLE 5.2
Use the data in Example 5.1 to compute
f
(0) as accuratelyas you can.
Solution
One solutionistoapplyRichardson extrapolation to finite difference ap-
proximations.Westart with twoforward difference approximationsfor
f
(0): one
using
h
1. Referring to the formulasof
O
(
h
2
) in
=
0
.
2 and the other one using
h
=
0
.
Table 5.3a, we get
−
3
f
(0)
+
4
f
(0
.
2)
−
f
(0
.
4)
3(0)
+
4(0
.
1341)
−
0
.
1797
g
(0
.
2)
=
=
=
0
.
8918
.
.
2(0
2)
0
4
−
3
f
(0)
+
4
f
(0
.
1)
−
f
(0
.
2)
−
3(0)
+
4(0
.
0819)
−
0
.
1341
g
(0
.
1)
=
=
=
0
.
9675
2(0
.
1)
0
.
2
where
g
denotes the finite difference approximation of
f
(0).Recalling that the error
in both approximations is of the form
E
(
h
)
=
c
1
h
2
+
c
2
h
4
+
c
3
h
6
+···
,
wecan use
Richardson extrapolation to eliminate the dominanterror term. With
p
=
2weobtain
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