Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...

104 CHAPTER 8. INTEGRALS AND QUADRATURE
Quite often Romberg’s Algorithm is used to compute columns of this array. Subtractive cancelling
or unbounded higher derivatives of f(x) can make successive approximations less accurate.
For this reason, entries in ever rightward columns are usually not calculated, rather lower entries
in a single column are calculated instead. That is, the user calculates the array:
R(k, 0)
R(k + 1, 0) R(k + 1, 1)
R(k + 2, 0) R(k + 2, 1) R(k + 2, 2)
.
.
.
. ..
R(k + n, 0) R(k + n, 1) R(k + n, 2) · · · R(k + n, n)
R(k + n + 1, 0) R(k + n + 1, 1) R(k + n + 1, 2) · · · R(k + n + 1, n)
R(k + n + 2, 0) R(k + n + 2, 1) R(k + n + 2, 2) · · · R(k + n + 2, n)
R(k + n + 3, 0) R(k + n + 3, 1) R(k + n + 3, 2) · · · R(k + n + 3, n)
.
.
Then R(k + n + l, n) makes a fine approximation to the integral as l → ∞. Usually n is small,
like 2 or 3.
8.3.1 Recursive Trapezoidal Rule
It turns out there is an efficient way of calculating R(n + 1, 0) given R(n, 0); first notice from the
above example that
R(0, 0) = b − a 1
1 2
1
R(1, 0) = b − a
2
2
.
[f(a) + f(b)] ,
[
f(a) + f( a + b
2 ) + f(a + b ]
2 ) + f(b) .
It would be best to calculate R(1, 0) without recalculating f(a) and f(b). It turns out this is possible.
Let h n = b−a
2
, and recall that
n [
]
2
f(a) + f(b) ∑
n −1
R(n, 0) = φ(n) = h n + f (a + ih n ) .
2
Thus
⎡
R(n + 1, 0) = φ(n + 1) = h n+1
⎣
f(a) + f(b)
2
= 1 2 h n
= 1 2 h n
[
f(a) + f(b)
2
∑
2 n+1 −1
+
i=1
2 n −1
+
i=1
⎤
f (a + ih n+1 ) ⎦ ,
∑
f (a + (2i − 1)h n+1 ) + f
i=1
[
2
f(a) + f(b) ∑
n −1
+
2
= 1 2 R(n, 0) + h n+1
2∑
n −1
i=1
i=1
2∑
n −1
f (a + ih n ) +
i=1
f (a + (2i − 1)h n+1 ) .
.
(
a + (2i) 1 2 h n) ] ,
f (a + (2i − 1)h n+1 )
]
,
Then calculating R(n + 1, 0) requires only 2 n − 1 additional evaluations of f(x), instead of the
2 n+1 + 1 usually required.