advertisement

LECTURE 21: NUMERICAL INTEGRATION MINGFENG ZHAO March 02, 2015 Numerical integration Definition 1. Let f be integrable on [a, b] and n be a positive integer, then the regular partition of [a, b] with n subintervals is: ∆x = b−a , n xk = a + k∆x, ∀k = 0, 1, · · · , n. Then b Z I. The Midpoint Rule approximation to f (x) dx using n equally spaced subintervals on [a, b] is: a n X xk−1 + xk f ∆x, M (n) := [f (m1 ) + f (m2 ) + · · · + f (mn )] ∆x = 2 k=1 where mk = xk−1 + xk for all k = 1, · · · , n. 2 Z II. The Trapezoid Rule approximation to b f (x) dx using n equally spaced subintervals on [a, b] is: a " # n−1 X 1 1 T (n) := f (x0 ) + f (xk ) + f (xn ) ∆x. 2 2 k=1 1 2 MINGFENG ZHAO Z III. If n is an even integer, the Simpson’s Rule approximation to b f (x) dx using n equally spaced subintervals a on [a, b] is: S(n) := [f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + · · · + 4f (xn−1 ) + f (xn )] ∆x . 3 Definition 2. Suppose c is a computed numerical solution to a problem having an exact solution x, then absolute error relative error = |c − x| |c − x| |x| = if x 6= 0. 1 Example 1. Let c = 0.111 be the approximate value of x = , then 9 1 absolute error = 0.111 − ≈ 0.000111 9 0.111 − 1 9 relative error = ≈ 0.000999 ≈ 0.1% 1 9 LECTURE 21: NUMERICAL INTEGRATION 3 4 Z x2 dx using the Midpoint Rule with n = 4 subintervals, and find the absolute error and Example 2. Approximate 2 relative error. Since a = 2, b = 4 and n = 4, then ∆x = 4−2 = 0.5, 4 x0 = 2, x1 = 2.5, x2 = 3, x3 = 3.5, and x4 = 4. Then the midpoints are: m1 = x0 + x1 = 2.25, 2 m2 = x1 + x2 = 2.75, 2 m3 = x2 + x2 = 3.25, 2 and m4 = x3 + x4 = 3.75. 2 So M (4) = [f (m1 ) + f (m2 ) + f (m3 ) + f (m4 )] ∆x = [2.252 + 2.752 + 3.252 + 3.752 ] · 0.5 = 18.625. 4 56 1 3 x = , so the absolute error is: 3 2 3 2 18.625 − 56 absolute error = 18.625 − ≈ 0.0417, and relative error = 56 3 3 Z Notice that 4 x2 dx = Z Example 3. Approximate 56 3 ≈ 0.00223 = 0.223%. 4 x2 dx using the Trapezoid Rule with n = 4 subintervals, and find the absolute error and 2 relative error. Since a = 2, b = 4 and n = 4, then ∆x = 4−2 = 0.5, 4 x0 = 2, x1 = 2.5, x2 = 3, x3 = 3.5, and x4 = 4. So T (4) = = f (x0 ) f (x4 ) + f (x1 ) + f (x2 ) + f (x3 ) + ∆x 2 2 1 2 1 2 2 2 2 · 2 + 2.5 + 3 + 3.5 + · 4 · 0.5 2 2 = 18.75 4 4 1 56 Notice that x2 dx = x3 = , so the absolute error is: 3 3 2 2 18.75 − 56 absolute error = 18.75 − ≈ 0.083, and relative error = 56 3 3 Z 56 3 ≈ 0.004 = 0.4%. 4 MINGFENG ZHAO 4 Z x2 dx using the Simpson’s Rule with n = 4 subintervals, and find the absolute error and Example 4. Approximate 2 relative error. Since a = 2, b = 4 and n = 4, then ∆x = 4−2 = 0.5, 4 x0 = 2, x1 = 2.5, x2 = 3, x3 = 3.5, and x4 = 4. So S(4) = = = Z Notice that 2 4 [f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + f (x4 )] ∆x 3 2 0.5 2 + 4 · 2.52 + 2 · 32 + 4 · 3.52 + 42 · 3 56 3 4 56 1 3 x dx = x = , so the absolute error is: 3 2 3 56 − 56 56 and relative error = 3 56 absolute error = − = 0, 3 3 3 2 56 3 = 0%. Theorem 1. For any positive integer n, then T (2n) = T (n) + M (n) , 2 and S(2n) = 4T (2n) − T (n) T (n) + 2M (n) = 3 3 Theorem 2. Let f be a forth order differentiable function on [a, b], EM , ET and ES be the absolute errors corresponding Z b to the Midpoint, Trapezoid and Simpson’s Rule approximation of f (x) dx using n subintervals. Assume that |f 00 (x)| ≤ a k and |f (4) (x)| ≤ K for all a ≤ x ≤ b, then |EM | ≤ k(b − a)3 , 24n2 |ET | ≤ k(b − a)3 , 12n2 and |ES | ≤ K(b − a)5 if n is even. 180n4 Remark 1. Let f be an integrable function on [a, b], then a. If f (x) is a linear function, e.g., f (x) = 2x + 1, then both absolute and relative errors of Midpoint Rule and Z b Trapezoid Rule approximation of f (x) dx are 0. a b. If f (x) is a cubic or quadratic or linear function, e.g., f (x) = x3 − 2x + 1, then both absolute and relative errors Z b of Simpson’s Rule approximation of f (x) dx are 0. a Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C. Canada V6T 1Z2 E-mail address: [email protected]