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).

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Page 1: 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).