Graphics Programs Reference
In-Depth Information
EXAMPLE 6.5
Show that
R
k
,
2
in Romberg integrationis identical to the composite Simpson's 1/3rule
in Eq. (6.10) with 2
k
−
1
panels.
Solution
Recall that in Romberg integration
R
k
,
1
=
I
k
denoted the approximate in-
tegralobtainedbythe composite trapezoidal rule with 2
k
−
1
panels.Denoting the
abscissas of the nodes by
x
1
,
x
2
,...,
x
n
, wehavefrom the composite trapezoidal rule
in Eq. (6.5)
f
(
x
1
)
2
f
(
x
n
)
h
n
−
1
1
R
k
,
1
=
I
k
=
+
2
f
(
x
i
)
+
2
i
=
2
Whenwehalve the number of panels (panel width 2
h
), only the odd-numberedab-
scissasenter the composite trapezoidal rule, yielding
f
(
x
1
)
f
(
x
n
)
h
n
−
2
R
k
−
1
,
1
=
I
k
−
1
=
+
2
f
(
x
i
)
+
i
=
3
,
5
,...
Applying Richardson extrapolationyields
4
3
R
k
,
1
−
1
3
R
k
−
1
,
1
R
k
,
2
=
1
3
f
(
x
1
)
3
f
(
x
n
)
h
n
−
1
n
−
2
4
3
2
3
1
=
+
+
+
f
(
x
i
)
f
(
x
i
)
i
=
2
,
4
,...
i
=
3
,
5
,...
which agrees with Simpson's rule in Eq. (6.10).
EXAMPLE 6.6
Use Romberg integration to evaluate
0
=
f
(
x
)
dx
, where
f
(
x
)
sin
x
.Work with four
decimal places.
Solution
From the recursivetrapezoidal rule in Eq. (6.9b) we get
2
[
f
(0)
R
1
,
1
=
π
=
+
π
=
I
(
)
f
(
)]
0
1
2
I
(
+
2
f
(
R
2
,
1
=
π/
=
π
π/
=
.
I
(
2)
)
2)
1
5708
1
2
I
(
+
4
R
3
,
1
=
π/
=
π/
π/
+
π/
=
.
I
(
4)
2)
[
f
(
4)
f
(3
4)]
1
8961
1
2
I
(
+
8
[
f
(
R
4
,
1
=
π/
=
π/
π/
+
π/
+
π/
+
π/
I
(
8)
4)
8)
f
(3
8)
f
(5
8)
f
(7
8)]
=
1
.
9742
Search WWH ::
Custom Search