MA 3021: Numerical Analysis I Numerical Differentiation and Integration

Transcription

1 MA 3021: Numerical Analysis I Numerical Differentiation and Integration Suh-Yuh Yang ( 楊肅煜 ) Department of Mathematics, National Central University Jhongli District, Taoyuan City 32001, Taiwan syyang@math.ncu.edu.tw syyang/ Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 1/35

2 Introduction 1 If the values of function f are given at a few points x 0, x 1,, x n, can that information be used to estimate a derivative f (c) or an integral b a f (x)dx? 2 Taylor s Theorem: Let f C n+1 [a, b] and x 0 [a, b]. Then for every x [a, b], ξ(x) between x and x 0 such that f (x) = P n (x) + R n (x), where the n-th Taylor polynomial P n (x) is given by P n (x) = n 1 k! f (k) (x 0 )(x x 0 ) k k=0 and the remainder (error) term R n (x) is given by R n (x) = 1 (n + 1)! f (n+1) (ξ(x))(x x 0 ) n+1 (Lagrange s form). Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 2/35

3 Numerical differentiation 1 f f (x (x 0 ) = lim 0 +h) f (x 0 ) h 0 h, if the limit exists. Intuitively, we have f (x 0 ) f (x 0+h) f (x 0 ) h if h is small. 2 Assume that h > 0 and f C 2 [x 0, x 0 + h]. By Taylor s Theorem, f (x 0 + h) = f (x 0 ) + hf (x 0 ) + h2 2 f (ξ), for some ξ (x 0, x 0 + h). Rearranging the expansion, we obtain f (x 0 ) = 1 h ( f (x0 + h) f (x 0 ) ) h 2 f (ξ). If h 2 f (ξ) is small, then we have an approximation of f (x 0 ), f (x 0 ) 1 h ( f (x0 + h) f (x 0 ) ), called the forward-difference formula. The term h 2 f (ξ) is called the truncation error, O(h). (When h < 0, we only have to change the assumption to f C 2 [x 0 + h, x 0 ] = backward-difference formula) Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 3/35

4 Higher order methods 1 Assume that h > 0 and f C 3 [x 0 h, x 0 + h]. By Taylor s Theorem, we have f (x 0 + h) = f (x 0 ) + hf (x 0 ) + h2 2 f (x 0 ) + h3 6 f (ξ 1 ), f (x 0 h) = f (x 0 ) hf (x 0 ) + h2 2 f (x 0 ) h3 6 f (ξ 2 ), for some ξ 1 (x 0, x 0 + h) and ξ 2 (x 0 h, x 0 ). After subtracting and rearranging, we have f (x 0 ) = 1 2h ( f (x0 + h) f (x 0 h) ) h2 6 1 ( f (ξ 2 1 ) + f (ξ 2 ) ). 2 This is a more favorable result, because of the h 2 term in the error. Notice that, however, the presence of f in the error term. Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 4/35

5 The truncation error From the Intermediate Value Theorem, we have that there is a ξ (x 0 h, x 0 + h), such that Hence, Therefore f (x 0 ) = 1 2h f (ξ) = 1 2 ( f (ξ 1 ) + f (ξ 2 ) ). ( f (x0 + h) f (x 0 h) ) h2 6 f (ξ). f (x 0 ) 1 ( f (x0 + h) f (x 0 h) ), 2h which is a second-order formula, O(h 2 ). Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 5/35

6 Approximation of f (x 0 ) Assume that h > 0 and f C 4 [x 0 h, x 0 + h]. From Taylor s Theorem, f (x 0 + h) = f (x 0 ) + hf (x 0 ) + h2 2! f (x 0 ) + h3 3! f (3) (x 0 ) + h4 4! f (4) (ξ 1 ), f (x 0 h) = f (x 0 ) hf (x 0 ) + h2 2! f (x 0 ) h3 3! f (3) (x 0 ) + h4 4! f (4) (ξ 2 ), for some ξ 1 (x 0, x 0 + h) and ξ 2 (x 0 h, x 0 ). After sum and rearrangement, we obtain the following central difference formula for the 2nd derivative at x 0 : f (x 0 ) = 1 ( f (x0 h 2 + h) 2f (x 0 ) + f (x 0 h) ) h2 1 ( f (4) (ξ ) + f (4) (ξ 2 ) ) = 1 h 2 ( f (x0 + h) 2f (x 0 ) + f (x 0 h) ) h2 12 f (4) (ξ), where at the last equality we use the Intermediate Value Theorem again. Thus, we have a second-order approximation of f (x 0 ) f (x 0 ) 1 h 2 ( f (x0 + h) 2f (x 0 ) + f (x 0 h) ). Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 6/35

7 Richardson s extrapolation 1 Richardson extrapolation is a general procedure to improve accuracy. 2 Assume that f is sufficiently smooth and f (x 0 + h) = 1 k! hk f (k) 1 (x 0 ), f (x 0 h) = k! ( 1)k h k f (k) (x 0 ). k=0 k=0 After subtraction and rearrangement, we obtain f (x 0 ) = 1 ( f (x0 + h) f (x 0 h) ) 2h ( h 2 or in an abstract form 3! f (3) (x 0 ) + h4 5! f (5) (x 0 ) + h6 7! f (7) (x 0 ) + M = N(h) + ( k 2 h 2 + k 4 h 4 + k 6 h 6 + ), where M := f (x 0 ) and N(h) := ( f (x 0 + h) f (x 0 h) ) /(2h). ), Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 7/35

8 Richardson s extrapolation (cont d ) 1 In general, suppose that M = N(h) + ( k 1 h + k 2 h 2 + k 3 h 3 + ) (1) M N(h) = O(h) M = N( 2 h ) + k 1 h 2 + k h2 ) 2 ( 2( + h2 ) 3 k3 + (2) 2 (2) (1) M = 2N( h 2 ) N(h) + k ( h h2) ( + k h h3) + Define N 1 (h) := N(h) and N 2 (h) := N 1 ( h {N 2 ) + 1 ( h 2 ) N 1(h). M = N 2 (h) k 2 2 h 2 3k 3 4 h3 (3) M N 2 (h) = O(h 2 ). 2 This formula is the first step in Richardson extrapolation. It shows that a simple combination of N 1 (h) and N 1 ( h 2 ) furnishes an estimate of M with an accuracy of O(h 2 ). Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 8/35

9 Richardson s extrapolation (cont d ) From (3), we have M = N 2 ( h 2 ) k 2 8 h2 3k 3 32 h3 (4). 4 (4) (3) 3M = 4N 2 ( h 2 ) N 2(h) + 3k 3 8 h3 + M = N 2 ( h 2 ) + 1 { N 2 ( h 3 2 ) N 2(h)) + 3k 3 8 h3 + Define N 3 (h) = N 2 ( h 2 ) {N 2( h 2 ) N 2(h). M N 3 (h) = O(h 3 ). Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 9/35

10 Richardson s extrapolation (cont d ) Using the techniques, we have N 4 (h) = N 3 ( h 2 ) M N 4 (h) = O(h 4 ), N 5 (h) = N 4 ( h 2 ) M N 5 (h) = O(h 5 ), m 1 In general, if M = N 1 (h) +. j=1 { N 3 ( h 2 ) N 3(h), { N 4 ( h 2 ) N 4(h), k j h j + O(h m ) then for j = 2, 3,, m, N j (h) = N j 1 ( h 2 ) + 1 { 2 j 1 N j 1 ( h 1 2 ) N j 1(h), M = N j (h) + O(h j ). Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 10/35

11 Example f (x 0 ) = 1 ( f (x0 + h) f (x 0 h) ) 1 2h 6 h2 f (3) (x 0 ) h4 f (5) (x 0 ) f (x 0 ) = N 1 (h) + O(h 2 ) f (x 0 ) = N 1 (h) 1 6 h2 f (3) (x 0 ) h4 f (5) (x 0 ) f (x 0 ) = N 1 ( h 2 ) 1 24 h2 f (3) (x 0 ) h4 f (5) (x 0 ) 4f (x 0 ) = 4N 1 ( h 2 ) 1 6 h2 f (3) (x 0 ) h4 f (5) (x 0 ) 3f (x 0 ) = 4N 1 ( h 2 ) N 1(h) h4 f (5) (x 0 ) f (x 0 ) = N 1 ( h 2 ) + 1 { N1 ( h 3 2 ) N 1(h) h4 f (5) (x 0 ) f (x 0 ) := N 2 (h) + O(h 4 ) Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 11/35

12 Differentiation via polynomial interpolation 1 Suppose that f C 2 [a, b], x 0 (a, b) and x 1 := x 0 + h [a, b]. Then ξ(x) [a, b] such that f (x) = x x 1 f (x x 0 0 ) + x x 0 f (x x 1 x 1 x 1 ) + f (ξ(x)) 0 2! = x x 0 h h If D xf (ξ(x)) 2! + D xf (ξ(x)) 2! f (x 0 ) + x x 0 h (x x 0 )(x x 1 ) f (x 0 + h) + f (ξ(x)) (x x 0 )(x x 0 h) 2! exists = f (x) = 1 h f (x 0) + 1 h f (x 0 + h) 2 We have f (x 0 ) = f (x 0 + h) f (x 0 ) h (x x 0 )(x x 0 h) + 2(x x 0) h f (ξ(x)). 2! h 2! f (ξ(x 0 )) f (x 0 ) f (x 0 + h) f (x 0 ), with error bound h h 2 max f (x). x [a,b] h > 0 : the forward-difference formula h < 0 : the backward-difference formula Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 12/35

13 General case Suppose that x 0, x 1,, x n I distinct & f C n+1 (I). Then f (x) = n k=0 f (x k )L k (x) + f (n+1) (ξ(x)) (x x (n + 0 )(x x 1)! 1 ) (x x n ), where ξ(x) I. { f (n+1) (ξ(x)) If D x exists = f (x) = (n + 1)! n f (x k )L k (x) k=0 { f (n+1) (ξ(x)) +D x (x x (n + 0 )(x x 1)! 1 ) (x x n ) + f (n+1) (ξ(x)) { D x (x x0 )(x x (n + 1)! 1 ) (x x n ). = f (x j ) = n f (x k )L k (x j) + f (n+1) (ξ(x j )) n (n + 1)! (x j x k ). k=0 k=0,k =j We obtain an (n + 1)-point formula. Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 13/35

14 Three point formula: x 0, x 1, x 2 L 0 (x) = (x x 1)(x x 2 ) (x 0 x 1 )(x 0 x 2 ) L 0 (x) = (x x 2) + (x x 1 ) (x 0 x 1 )(x 0 x 2 ) 2x x = 1 x 2 (x 0 x 1 )(x 0 x 2 ), L 1 (x) = (x x 0)(x x 2 ) (x 1 x 0 )(x 1 x 2 ) L 1 (x) = 2x x 0 x 2 (x 1 x 0 )(x 1 x 2 ), L 2 (x) = (x x 0)(x x 1 ) (x 2 x 0 )(x 2 x 1 ) L 2 (x) = 2x x 0 x 1 (x 2 x 0 )(x 2 x 1 ). = f (x j ) = 2x j x 1 x 2 (x 0 x 1 )(x 0 x 2 ) f (x 0) + 2x j x 0 x 2 (x 1 x 0 )(x 1 x 2 ) f (x 1) + 2x j x 0 x 1 (x 2 x 0 )(x 2 x 1 ) f (x 2) + f (3) (ξ j ) 2 3! (x j x k ), k=0,k =j where ξ j := ξ(x j ). Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 14/35

15 Equal spaced: x 0, x 1 = x 0 + h, x 2 = x 0 + 2h f (x 0 ) = 1 { 3 h 2 f (x 0) + 2f (x 0 + h) 1 2 f (x 0 + 2h) h2 f (3) (ξ 0 ), f (x 0 + h) = 1 { 1 h 2 f (x 0) f (x 0 + 2h) 1 6 h2 f (3) (ξ 1 ), f (x 0 + 2h) = 1 { 1 h 2 f (x 0) 2f (x 0 + h) f (x 0 + 2h) h2 f (3) (ξ 2 ). = f (x 0 ) = 1 { 3f (x 0 ) + 4f (x 0 + h) f (x 0 + 2h) + 1 2h 3 h2 f (3) (ξ 0 ), ( ) f (x 0 ) = 1 { f (x 0 h) + f (x 0 + h) 1 2h 6 h2 f (3) (ξ 1 ), f (x 0 ) = 1 { f (x 0 2h) 4f (x 0 h) + 3f (x 0 ) + 1 2h 3 h2 f (3) (ξ 2 ). ( ) ( ) and ( ) are essentially the same! (h > 0 or h < 0, respectively) Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 15/35

16 Three-point and five-point formulas 1 Three-point formula: f (x 0 ) = 1 { 3f (x 0 ) + 4f (x 0 + h) f (x 0 + 2h) + 1 2h 3 h2 f (3) (ξ 0 ), for some ξ 0 between x 0 and x 0 + 2h, f (x 0 ) = 1 { f (x 0 + h) f (x 0 h) 1 2h 6 h2 f (3) (ξ 1 ), for some ξ 1 between x 0 h and x 0 + h. 2 Five-point formula: f 1 { (x 0 ) = f (x 0 2h) 8f (x 0 h) + 8f (x 0 + h) f (x 0 + 2h) 12h f (x 0 ) = + h4 30 f (5) (ξ), for some ξ between x 0 2h and x 0 + 2h, ( ) 1 { 25f (x 0 ) + 48f (x 0 + h) 36f (x 0 + 2h) 12h +16f (x 0 + 3h) 3f (x 0 + 4h) + (h 4 /5)f (5) (ξ), for some ξ between x 0 and x 0 + 4h. Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 16/35

17 Use Taylor s Theorem + extrapolation to derive ( ) f (x 0 + h) = f (x 0 ) + hf (x 0 ) + h2 2! f (x 0 ) + h3 3! f (3) (x 0 ) + h4 4! f (4) (x 0 ) + h5 120 f (5) (ξ 1 f (x 0 h) = f (x 0 ) hf (x 0 ) + h2 2! f (x 0 ) h3 3! f (3) (x 0 ) + h4 4! f (4) (x 0 ) h5 120 f (5) (ξ 2 where ξ 1 between x 0 and x 0 + h, ξ 2 between x 0 and x 0 h. f (x 0 + h) f (x 0 h) = 2hf (x 0 ) + h3 3 f (x 0 ) + h5 120 { f (5) (ξ 1 ) + f (5) (ξ 2 ) { f (x 0 ) = 1 2h f (x 0 + h) f (x 0 h) h2 6 f (x 0 ) 120 h4 f (5) ( ξ), (Eqn1) Replacing h by 2h, we have { f (x 0 ) = 1 4h f (x 0 + 2h) f (x 0 2h) 4h2 6 f (x 0 ) 16h4 120 f (5) ( ξ), (Eqn2) where ξ between x 0 h and x 0 + h, ξ between x 0 2h and x 0 + 2h., Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 17/35

18 Use Taylor s Theorem + extrapolation to derive ( ) (cont d ) 4 (Eqn1) (Eqn2) = 3f (x 0 ) = 2 h { f (x 0 + h) f (x 0 h) 1 { f (x 0 + 2h) f (x 0 2h) h4 4h 30 f (5) ( ξ) + 2h4 15 f (5) ( ξ). If f C 5 [x 0 2h, x + 2h], h > 0, then we have 4h 4 30 f (5) ( ξ) h4 30 f (5) ( ξ) = h4 30 {4f (5) ( ξ) f (5) ( ξ) = h4 30 3f (5) (ξ). (why?) Therefore, we have f (x 0 ) = 1 { f (x 0 2h) 8f (x 0 h) + 8f (x 0 + h) f (x 0 + 2h) + h4 12h 30 f (5) (ξ), for some ξ [x 0 2h, x 0 + 2h]. Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 18/35

19 Homework The forward-difference formula can be expressed as f (x 0 ) = 1 h { f (x 0 + h) f (x 0 ) h 2 f (x 0 ) h2 6 f (x 0 ) + O(h 3 ). Use extrapolation to derive an O(h 3 ) formula for f (x 0 ). Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 19/35

20 Numerical integration Numerical quadrature: b a f (x)dx n i=0 a i f (x i ). Let x 0, x 1,, x n [a, b] be n + 1 distinct nodes. Let P n (x) = n i=0 f (x i )L i (x) be the nth Lagrange polynomial. Then b a f (x)dx = := b a n i=0 n i=0 b f f (x i )L i (x)dx + (n+1) (ξ(x)) a (n + 1)! a i f (x i ) + E(f ), n i=0 (x x i )dx where a i = b a L i (x)dx. Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 20/35

21 Trapezoidal rule Let x 0 = a, x 1 = b, h = b a. Then b a f (x)dx = x 1 { x x1 x 0 x 0 x f (x 1 0 ) + x x 0 x 1 x 0 f (x 1 ) dx + 1 x1 2 x f (ξ(x))(x 0 x 0 )(x x 1 )dx ( ) (x x1 ) 2 2(x 0 x 1 ) f (x 0) + (x x 0) 2 x1 2(x 1 x 0 ) f (x 1) + f (ξ) x 2 0 for some ξ (x 0, x 1 ) = = 1 2 (x 1 x 0 )f (x 1 ) 1 2 (x 0 x 1 )f (x 0 ) f (ξ) = 1 2 (x 1 x 0 )(f (x 0 ) + f (x 1 )) f (ξ)( 1 6 )(x 1 x 0 ) 3 = h 2 { f (x 0 ) + f (x 1 ) 1 12 h3 f (ξ). x1 x 0 (x x 0 )(x x 1 )dx, ( ) x 3 3 (x 0+x 1 )x 2 x1 2 + x 0 x 1 x x 0 If f (x) is a polynomial with degree(f ) 1, then the Trapezoidal Rule gives exact result! Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 21/35

22 Simpson s rule Let x 0 = a, x 1 = a + h, x 2 = b, h = (b a)/2. Then b a f (x)dx = x2 { (x x1 )(x x 2 ) x 0 (x 0 x 1 )(x 0 x 2 ) f (x 0) + (x x 0)(x x 2 ) (x 1 x 0 )(x 1 x 2 ) f (x 1) + (x x 0)(x x 1 ) (x 2 x 0 )(x 2 x 1 ) f (x 2) dx + x2 x 0 = O(h 4 ) error term. f (3) (ξ(x)) (x x 0 )(x x 3! 1 )(x x 2 )dx Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 22/35

23 Alternative approach: Taylor s Theorem Let x [x 0, x 2 ]. Then ξ(x) (x 0, x 2 ) such that f (x) = f (x 1 ) + f (x 1 )(x x 1 ) + f (x 1 ) 2 (x x 1 ) 2 + f (x 1 ) 6 (x x 1 ) 3 + f (4) (ξ(x)) 24 (x x 1 ) 4. = ( x 2 x f (x)dx 0 = f (x 1 )(x x 1 ) + f (x 1 ) 2 (x x 1 ) 2 + f (x 1 ) x2 x 0 6 (x x 1 ) 3 + f (x 1 ) 24 (x x 1 ) 4) x 2 f (4) (ξ(x))(x x 1 ) 4 dx x2 x f (4) (ξ(x))(x 0 x 1 ) 4 dx = f (4) (ξ 1 ) 24 x2 x 0 = f (4) (ξ 1 ) 24 5 (x x 1) 5 x 2 x 0 x2 x 0 (x x 1 ) 4 dx x 0, for some ξ 1 (x 0, x 2 ). f (x)dx = 2hf (x 1 ) + h3 3 f (x 1 ) + f (4) (ξ 1 ) 2h Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 23/35

24 Simpson s rule/degree of precision x2 x 0 f (x)dx = 2hf (x 1 ) + h3 3 = h 3 { 1 ( ) h 2 f (x 0 ) 2f (x 1 ) + f (x 2 ) h2 12 f (4) (ξ 2 ) + f (4) (ξ 1 ) h 5 60 ( ) f (x 0 ) + 4f (x 1 ) + f (x 2 ) h5 = h ( f (x 0 ) + 4f (x 3 1 ) + f (x 2 ) for some ξ (x 0, x 2 ). { 1 3 f (4) (ξ 2 ) 1 5 f (4) (ξ 1 ) 12 ) h5 90 f (4) (ξ), Definition: The degree of accuracy (precision) of a quadrature formula is the largest positive integer n such that the formula is exact for x k, k = 0, 1,, n. Note: Trapezoidal rule: degree of precision = 1; Simpson s rule: degree of precision = 3. Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 24/35

25 Midpoint rule Consider the smooth function f on [a, b]. Let x 0 = a+b 2. 1 Intuition: b b f (x)dx f (x 0 )dx = f (x 0 )(b a) = f ( a + b )(b a). a a 2 2 Based on Taylor s Theorem: b a f (x)dx b a f (x 0) + f (x 0 )(x x 0 )dx = f ( a+b 2 )(b a) + f (x 0 ) 2 (x a+b b 2 )2 = f ( a+b 2 )(b a) + 0. a f (x) ( f (x 0 ) + f (x 0 )(x x 0 ) ) = f (ξ(x)) 2! (x x 0 ) 2. = b a+b a f (x)dx f ( 2 )(b a) = b f (ξ(x)) a 2! (x x 0 ) 2 dx = f (ξ) 2! b a (x x 0) 2 dx = f (ξ) (b a) = f (ξ) 24 (b a)3, ξ (a, b). Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 25/35

26 Composite numerical integration 1 Large integration interval = large h = inaccurate; small h = high-degree polynomial = inaccurate. 2 Example: Using Simpson s rule with h = 2, we have 4 0 e x dx 2 3 (e0 + 4e 2 + e 4 ) = (exact value = ) Composite Simpson s rule: (h = 1) e x dx = e x dx + e x dx (e0 + 4e 1 + e 2 ) (e2 + 4e 3 + e 4 ) = (h = 1/2) 4 0 e x dx = 1 0 e x dx e x dx 1 6 (e0 + 4e 1/2 + e 1 ) (e3 + 4e e 4 ) = Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 26/35

27 Composite Simpson s rule Let n be an even integer. Divide [a, b] into n subintervals. Let h = b a n b a n/2 h = j=1{ 3 = h 3 = h 3 and x j = a + jh for j = 0, 1,, n. Then n/2 x2j f (x)dx = f (x)dx j=1 x 2j 2 ( ) f (x 2j 2 ) + 4f (x 2j 1 ) + f (x 2j ) h5 90 f (4) (ξ j ) { f (x 0 ) + 4f (x 1 ) + f (x 2 ) + f (x 2 ) + 4f (x 3 ) + f (x 4 ) + + f (x n ) h5 n/2 90 { n/2 1 f (x 0 ) + 2 j=1 j=1 f (4) (ξ j ) n/2 f (x 2j ) + 4 j=1 f (x 2j 1 ) + f (x n ) h5 n/2 90 j=1 f (4) (ξ j ). Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 27/35

28 Composite Simpson s rule (cont d) If f C 4 [a, b] then min x [a,b] f (4) (x) f (4) (ξ j ) max x [a,b] f (4) (x). = n 2 min f (4) n/2 (x) x [a,b] j=1 = min f (4) (x) 2 n/2 x [a,b] n j=1 f (4) (ξ j ) n 2 max x [a,b] f (4) (x). f (4) (ξ j ) max x [a,b] f (4) (x). By the Intermediate Value Theorem, µ (a, b) such that f (4) (µ) = 2 n/2 n f (4) (ξ j ) = h5 n/2 90 f (4) (ξ j ) = h5 n 180 f (4) (µ). j=1 j=1 h = b a n n = b a h h5 90 n/2 j=1 f (4) (ξ j ) = h5 (b a) f (4) (µ) = 180h (b a) 180 h4 f (4) (µ). Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 28/35

29 Composite rules 1 Composite Simpson s rule: Let n be an even integer, h = b a n, x 0 = a < x 1 < < x n = b and x j = a + jh. If f C 4 [a, b] then µ (a, b) such that b f (x)dx = h a 3 { n/2 1 f (x 0 ) + 2 j=1 n/2 f (x 2j ) + 4 j=1 (b a) f (x 2j 1 ) + f (x n ) 180 h4 f (4) (µ). 2 Composite trapezoidal rule: Let h = b a n, x 0 = a < x 1 < < x n = b and x j = a + jh. If f C 2 [a, b] then µ (a, b) such that b f (x)dx = h { a 2 f (x 0 ) + 2 n 1 j=1 f (x j ) + f (x n ) (b a) h 2 f (µ) Composite midpoint rule: Let n be an even integer, h = b a n+2, x 1 = a < x 0 < x 1 < < x n < x n+1 = b and x j = a + (j + 1)h. If f C 2 [a, b] then µ (a, b) such that b a n/2 f (x)dx = 2h f (x 2j ) + j=0 (b a) h 2 f (µ). 6 Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 29/35

30 Gaussian quadrature 1 Degree of precision + use values of function at equally spaced points, e.g. the trapezoidal rule. 2 Gaussian quadrature: b a f (x)dx n i=1 c i f (x i ), where c i R and x i [a, b] for i = 1, 2,, n = 2n parameters to choose. The greatest degree of precision 2n 1. 3 Example: Let [a, b] = [ 1, 1] and n = 2. We want to determine c 1, c 2 R, x 1, x 2 [ 1, 1] such that 1 1 f (x)dx c 1 f (x 1 ) + c 2 f (x 2 ) and gives exact value whenever f (x) is a polynomial with degree(f ) 3(= 2n 1). gives exact value when f (x) = 1, x, x 2, x 3. Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 30/35

31 Example (cont d) 2 = 1 1 1dx = c 1f (x 1 ) + c 2 f (x 2 ) = c 1 + c 2 (f (x) = 1), 0 = 1 1 xdx = c 1f (x 1 ) + c 2 f (x 2 ) = c 1 x 1 + c 2 x 2 (f (x) = x), 2 3 = 1 1 x2 dx = c 1 f (x 1 ) + c 2 f (x 2 ) = c 1 x c 2x 2 2 (f (x) = x 2 ), 0 = 1 1 x3 dx = c 1 f (x 1 ) + c 2 f (x 2 ) = c 1 x c 2x 3 2 (f (x) = x 3 ). = c 1 + c 2 = 2, c 1 x 1 + c 2 x 2 = 0, c 1 x c 2x 2 2 = 2 3, c 1x c 2x 3 2 = 0. = c 1 = 1, c 2 = 1, x 1 = 3 3, x 2 = = f (x)dx 1 f ( ) + 1 f ( 3 ). 1 This formula has degree of precision 3. Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 31/35

32 Legendre polynomials (Chapter 8) Some Legendre polynomials are given by p 0 (x) = 1, p 1 (x) = x, p 2 (x) = x 2 1 3, p 3 (x) = x x, p 4(x) = x x , For each n, p n (x) is a polynomial of degree n. 1 p(x)p n (x)dx = 0 whenever p(x) is a polynomial of degree 1 n 1. The roots of p n (x) are distinct, lie in ( 1, 1), have a symmetry with respect to 0. e.g., p 2 (x) = x has roots 3 3 and 3 3. Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 32/35

33 Theorem Suppose that x 1, x 2,, x n are the roots of the nth Legendre polynomial p n (x). For i = 1, 2,, n, c i := ( 1 x xj 1 n j=1,j =i x i x )dx. j If p(x) is a polynomial and degree(p(x)) < 2n. Then 1 1 p(x)dx = n i=1 c ip(x i ). Proof: Let R(x) be a polynomial and degree(r(x)) n 1. Let x 1, x 2,, x n ( be the roots of the nth ) Legendre polynomial p n (x). Then R(x) = n i=1 n x x j j=1,j =i x i x R(x j i ) + nth derivative of R(x) ( ) = n i=1 n x x j j=1,j =i x i x R(x j i ) R(x)dx = ( ) 1 1 n i=1 n x x j ( j=1,j =i x i x R(x j i ) dx = n ) 1 i=1 1 n j=1,j =i R(x i ) = n i=1 c ir(x i ). x x j x i x j dx Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 33/35

34 Proof (con d) Let p(x) be a polynomial with degree(p(x)) 2n 1. Then p(x) = Q(x)p n (x) + R(x) for some Q(x) and R(x) with degree(q(x)) n 1 and degree(r(x)) n 1. degree(q(x)) n Q(x)p n (x)dx = 0. x i is a root of p n (x) for i = 1, 2,, n p(x i ) = Q(x i )p n (x i ) + R(x i ) = R(x i ). 1 1 ( ) = p(x)dx = Q(x)p n (x) + R(x) dx = 1 n i=1 = c i R(x i ) = 1 n i=1 This completes the proof. c i p(x i ). 1 1 R(x)dx Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 34/35

35 Concluding remarks Change of variables: b a f (x)dx 1 1 f (t)dt. t = 2x a b x = 1 ( ) (b a)t + a + b. b a 2 b 1 ( 1 { ) b a f (x)dx = f (b a)t + a + b 2 2 dt. a Example: e x2 dx = 1 1 ( ) 2 e 12 (0.5t+2.5) dt = 1 1 e (t+5)2 16 dt. 4 1 Suh-Yuh Yang ( 楊肅煜 ), Math. Dept., NCU, Taiwan Differentiation and Integration 35/35