8 1 Newton-Cotes Formulae n=1:c=7 2 9r. Excuses for not zoidal Rule doing homework HW I could only get arbitrarily RLA Fb-a,用中 p170 close to my textbook. #3 I couldnt actually reach it 2:C (2) 0 Simpson’ s Rule n为偶数阶的 Newton-Cotes 公式至少有n+1次代数精度。 代数精度=3 b-a RIfT ∈(a,b),h= n=3: Simpson338Ru、成数精度=3,刚/1=80/(5) n=4: Cotes rule,代数精度=5,fl 8 g4sh'f(6(5)§1 Newton-Cotes Formulae 2 1 , 2 1 (1) 1 (1) n = 1: C0 = C = [ ( ) ( )] 2 ( ) f a f b b a f x dx b a + −   Trapezoidal Rule x a x b dx f R f b a x ( )( ) 2! ( ) [ ] − −  =   /* 令 x = a+th, h = b−a, 用中 值定理 */ 1 ( ) , [ , ], 12 1 3 b a h f a b h − = −     = 代数精度 = 1 n = 2: 6 1 , 3 2 , 6 1 (2) 2 (2) 1 (2) C0 = C = C = [ ( ) 4 ( ) ( )] 6 ( ) 2 f a f f b b a f x dx a b b a + + −  +  Simpson’s Rule 代数精度 = 3 2 ( ) , ( , ) , 90 1 [ ] 5 (4) b a R f h f a b h − = −    = n = 3: Simpson’s 3/8-Rule, 代数精度 = 3, ( ) 80 3 [ ] 5 (5) R f = − h f  n = 4: Cotes Rule, 代数精度 = 5, ( ) 945 8 [ ] 7 (6) R f = − h f  Excuses for not doing homework I could only get arbitrarily close to my textbook. I couldn't actually reach it. HW: p.170 #3 n 为偶数阶的Newton-Cotes 公式至少有 n+1 次代数精度