Theorem (Composite Trapezoidal Rule) 2 C
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82 4. NUMERICAL INTEGRATION AND DIFFERENTIATION Theorem (Composite Trapezoidal Rule). Suppose f 2 C 2 [a, b]. Then for some μ 2 (a, b) we have Z b a f (x) dx = h 2 h f (a)+ f (b)+2 n-1 X j =1 f (x j ) i - (b - a)h 2 12 f 00 (μ). Composite Midpoint Rule If f 2 C 2 [a, b], then a number ⇠ in (x -1 ,x 1 ) exists with Z x 1 x -1 f (x) dx =2hf (x 0 )+ h 3 3 f 00 (⇠ ) where h = x 1 - x -1 2 . Divide [a, b] into n + 2 intervals, n even, h = b - a n +2 , x j = a +(j + 1)h. Then apply the Midpoint Rule to successive pairs of intervals. 0 1 0 0 0 1 1 0 Thus the composite weight pattern is 0 - 1 - 0 - 1 -···- 0 - 1 - 0. Theorem (Composite Midpoint Rule). Suppose f 2 C 2 [a, b]. Then for some μ 2 (a, b) we have Z b a f (x) dx =2h n/2 X j =0 f (x 2j ) - (b - a)h 2 6 f 00 (μ).
Transcript of Theorem (Composite Trapezoidal Rule) 2 C
82 4. NUMERICAL INTEGRATION AND DIFFERENTIATION
Theorem (Composite Trapezoidal Rule). Suppose f 2 C2[a, b]. Then forsome µ 2 (a, b) we haveZ b
af(x) dx =
h
2
hf(a) + f(b) + 2
n�1Xj=1
f(xj)i� (b� a)h2
12f 00(µ).
Composite Midpoint Rule
If f 2 C2[a, b], then a number ⇠ in (x�1, x1) exists withZ x1
x�1
f(x) dx = 2hf(x0) +h3
3f 00(⇠)
where h =x1 � x�1
2.
Divide [a, b] into n + 2 intervals, n even, h =b� a
n + 2, xj = a + (j + 1)h. Then
apply the Midpoint Rule to successive pairs of intervals.
0 1 0 00 1 1 0
Thus the composite weight pattern is
0� 1� 0� 1� · · ·� 0� 1� 0.
Theorem (Composite Midpoint Rule). Suppose f 2 C2[a, b]. Then forsome µ 2 (a, b) we have
Z b
af(x) dx = 2h
n/2Xj=0
f(x2j)�(b� a)h2
6f 00(µ).