5. Integration 2. Quadrature as Box Counting 3. Algorithm: Trapezoid Rule 4. Algorithm: Simpson’s Rule 5. Integration Error 6. Algorithm: Gaussian Quadrature 7. Empirical Error Estimate 8. Experimentation 9. Higher Order Rules 5.2. Quadrature as Box Counting b Riemann: a ba / h f x d x lim h f xi h0 i 1 b Numerical: N f x d x w f x a i 1 i i w = weight • Keeping N finite can still give exact results, e.g., polynomials. • Aim: accurate result for small N. • No universal “best” algorithm. Tips • Remove singularities first. • By putting them at endpoints of sub-intervals • By change of variable. 1 0 1 1 1 0 x f x d x x f x d x x f x d x 1 x 1 1/3 d x 3y3 d y 0 1 0 y x1/3 0 f x 1 x2 1 dx 2 0 f 1 y 2 2 y2 dy y2 1 x • Speed up or slow down (by change of variable or step size) in slowly or rapidly varying region. Algorithms for Evenly Spaced Points b N f x d x w a i 1 i fi f xi fi xi a i 1 h Evenly spaced points : i 1, Name Degree wi Trapezoid 1 1 h 1 , 1 2 Simpson’s 2 1 h 1 , 4 , 1 3 3/8 3 3 h 1 , 3 , 3 , 1 8 Milne 4 h ,N 1 h 14 , 64 , 24 , 64 , 14 45 ba N 8.3. f x ax b h 0 b a Algorithm: Trapezoid Rule f0 b f1 a h b 1 f x d x h f 0 f1 f1 f 2 2 a N f x d x wi f xi i 1 f x xi 1 1 1 f x d x f1 f 0 h f 0 h h f 0 f1 2 2 1 h f 0 f1 f 2 2 b f1 f 0 h b f0 a xi 1 f x d x h f i f i 1 2 f N 2 f N 1 f N 1 f N f N 2 f N 1 1 fN 2 1 wi h , 1 , 2 1 ,1 , 2 f1 f 0 x f0 h Algorithm: Simpson’s Rule 5.4. f 1 ah bh c f x ax 2 bx c f0 c f1 ah 2 bh c h h 2 1 1h f 4f f h f 1 f 1 f 0 2 f 0 1 0 1 3 3 3 2 f x d x a h3 2 c h 3 1 f 1 f1 f 0 2 1 bh f 1 f1 2 c f0 ah 2 2 xi 2 xi 1 f x d x h f i 4 f i 1 f i 2 3 xi 2 xi b a 1 f x d x h f i 4 f i 1 f i 2 3 1 f x d x h f 0 4 f1 f 2 f 2 4 f 3 f 4 3 b a h f 0 4 f1 2 f 2 4 f 3 2 f 4 3 N f x d x wi f xi i 1 N int f N 4 4 f N 3 f N 2 f N 2 4 f N 1 f N 4 f N 3 2 f N 2 4 f N 1 f N 1 wi h 1 , 4 , 2 , 4 , 2 , 3 , 4 , 2 , 4 , 1 1 N 1 N = odd 2 N 1 w h N int 1 4 1 N 1 h i 3 i 1 ba 5.5. Integration Error Expand f at middle of interval x [ h, h ] : 1 1 1 4 4 f x f 0 f 0 x f 0 x 2 f 0 x 3 f0 x 2 3! 4! h h f x d x 2 f0 h h 2 3 2 f0 h f 0 4 h 5 23 4! 5 Error, Trapezoid : Et O N int f 0 h 3 O 1 f b a 3 N2 0 int Error, Simpson : Es O N int f 0 4 h5 Relative error : 1 5 O 4 f 0 4 b a N int t, s Et , s f f n n 1 n b a N f 1 ba 2 N int n data points in each interval n=2 n=4 t, s f n n 1 n b a N f n f n 1 b a N m Nn f Set scale to Trapezoid : Simpson : ba 1 N 1015 N 1015 2/5 2/9 ~ 107 ( single prec) ~ 1015 ( double prec) d tot 1 1 0 n t , s m dN N 2 Min tot : f n 1 f m machine precision ro N m Round-off error is random : tot t , s ro n = 2,4 for t , s n f n 1 N b a f m N m 106 1010/3 2154 2 2 n 1 t,s m 2 2 n 1 tot N m m tot 1015 4/5 tot 1015 8/9 1 1 2 n 1 1012 1040/3 4.6 1014 Conclusions • Simpson’s rule is better than trapezoid. • It’s possible to get tot m . • Best result is obtained for N ~ 1000 instead of . 5.6. Algorithm: Gaussian Quadrature b b n a k 1 Ak dx I d x f ( x ) w( x ) S Ak f ( xk ) a ( x) ( x xk ) '( xk ) where is the nth degree member of a complete set of orthogonal polynomials, and { xk } are its roots. I = S if f is an 2n-1 degree polynomial. Integral Polynomial weight Limits d x f ( x) Legendre 1 ( 1, 1 ) d x e x f ( x) Hermite exp( x2 ) ( , ) Laguerre exp( x ) ( , ) ( 1x2 )1/2 ( 1, 1 ) 1 1 2 dx e x f ( x) 1 1 dx 1 1 x 2 f ( x ) Chebyshev I Proof : 5.6.1. Mapping Integration Points b An integral F y d y 1 can be transformed into 1 a by the linear transform y x a Thus b 1 b a 2 1 b a 2 f x f x d x 1 b a F y 2 y 1 y b a x b a 2 1 b a x b a 2 dy 1 b a d x 2 An integral F y d y 1 can be transformed into 1 0 y by the linear transform 0 Thus 2 0 1 x f x f x d x 2 0 1 x 2 1 1 x 1 y 2 1 x 1 x 2 dy F y y 1 x 2 1 x 1 x 2 dx Similarly, one get the following transforms 1 F y d y f x d x 1 interval Interval [a,b] [0,] W W W [ , ] W [b,] W [0,b] W y 1 b a 2 1 x 2 1 x 2 1 x 2 1 x 2 2 1 x 1 b a x b a 2 1 x 2 2x 1 x2 1 x b 2 1 x 2 b 2 1 2 x 2 f x W F y x 1 1 x b 1 2 x 0 5.7. Empirical Error Estimate (Ex.5.1) Answer Relative error, numerical exact exact , as a function of N for the trapezoid, Simpson, & Gaussian methods, in the calculation of 1 I e t d t 1 e1 0 5.8. Experimentation Evaluate 2 2 sin 1000 x d x 0 and sin 100 x d x x 0 What’s wrong ? 5.9. Higher Order Rules Let A(h) be the numerical evaluation of an integral with leading error h2, i.e., b A h f x d x h2 h4 a b 1 1 h A f x d x h2 h4 4 16 2 a 4 h 1 1 A Ah f x d x h4 3 2 3 4 a b Romberg’s extrapolation see Burden, § 4.5