b 1 f(x)x≈Sn=b∑A[f(x)+4f(x-1)+f(x+1) k k=0 复合 Simpson公式 复合抛物线公式 =b2(a)+2/(x1)+2/()+(b k+ k=0 k=1 l=4时,可得复合 Cotes求积公式 复合 Cotes公式 b 八(xkCn=b∑cf(x) k=0i=0 如∑[7f(xk)+32(x1)+12(x2)+32f(x3)+7(xk+1 k=0 4 4 I7f(a)+∑[32f(x1)+12f(x,2)+32f(x 901 ∑f(x)+7/(b) k=0 k k 3力+14
复合Simpson公式 复合抛物线公式 n b a f x dx S ò ( ) » å - = + + = + + 1 0 1 2 1 [ ( ) 4 ( ) ( )] 6 1 n k k k k h f x f x f x [ ( ) 4 ( ) 2 ( ) ( )] 6 1 1 1 0 2 f a f x 1 f x f b n b a n k k n k k + + + - = å å - = - = + l = 4时,可得复合Cotes求积公式 åå - = = + = 1 0 4 0 4 (4) ( ) n k i i k i n h C f x b a f x dx » C ò ( ) [7 ( ) 32 ( ) 12 ( ) 32 ( ) 7 ( )] 90 1 1 0 4 3 4 2 4 1 + - = + + + = å + + + + k n k k k k k f x f x f x f x f x h [7 ( ) [32 ( ) 12 ( ) 32 ( )] 14 ( ) 7 ( )] 90 1 1 1 0 4 3 4 2 4 1 f a f x f x f x f x f b n b a n k k n k k k k + + + + + - = å å - = - = + + + 复合Cotes公式
(b T [f(a)+f(b)] 复合梯形公式分解 2 [f(a)+f(x1) [f(x1)+f(x2) X,,X3 T h 2 [f(x2)+f(x3) h [f(x)+f(x4) 2n-1 +f(x [f(xn21)+f(b) 2 =2U(a)+2/(x)+2/x)+…+2f(x1)+f(b)
[ ( ) ( )] 2 ( ) f a f b b a T + - = [ , ] 0 1 x x [ ( ) ( )] 2 1 (0) f a f x h T = + [ , ] 1 2 x x 2 (1) h T = [ ( ) ( )] 1 2 f x + f x [ , ] 2 3 x x 2 (2) h T = [ ( ) ( )] 2 3 f x + f x [ , ] 3 4 x x 2 (3) h T = [ ( ) ( )] 3 4 f x + f x [ , ] n-2 n-1 x x 2 ( 2) h T n = - [ ( ) ( )] n-2 + n-1 f x f x [ , ] n 1 n x x - 2 ( 1) h T n = - [ ( ) ( )] f xn-1 + f b 2 h Tn = [ f (a) + 2 f (x1 ) + 2 f (x2 ) + L+ 2 f (xn-1 ) + f (b)] 复合梯形公式分解
+ S==[f(a)+4∫ +∫(b 复合 Simpson公式分解 x0nS0)方 [f(a)+4f(x 1)+f(x1) 0 h 1 [f(x1)-4f(x1)+f(x2) [f(xn1)"4f(x1)+f(b)] 1+ Sn=Da)+4∑f(x1)+2∑f(x)+f( k=0 k=1
) ( )] 2 [ ( ) 4 ( 6 f b a b f a f b a S + + + - = [ , ] 0 1 x x 6 (0 ) h S = [ ( ) 4 ( ) ( )] 1 2 1 0 f a + f x + f x + [ , ] 1 2 x x 6 (1) h S = [ ( ) 4 ( ) ( )] 2 2 1 1 1 f x + f x + f x + [ , ] n 1 n x x - 6 ( 1) h S n = - [ ( ) 4 ( ) ( )] 2 1 1 1 f x f x f b n n + + - + - 6 h Sn = [ f (a) + å + - = + 1 0 2 1 4 ( ) n k k f x å + - = 1 1 2 ( ) n k k f x f (b)] 复合Simpson公式分解
例1.使用各种复合求积公式计算定积分I Sinx 解:为简单起见依次使用8阶复合梯形公式、4阶 复合 Simpson公式和2阶复合 Cotes公式 可得各节点的值如右表 Trapz Simp. Cotes f(,) 0 0.1250.99739787 复合求积 公式的程序 0+20.25 0.98961584 x3x2x0÷0.375 newtoncotes. m 0.97672674 0.5 0.95885108 函数程序 x22x+|06250.93615564 unc. m 11075 0.90885168 x2:1087508779257 0.84147098
例1. ò = 1 0 sin dx x x 使用各种复合求积公式计算定积分 I 解: 为简单起见,依次使用8阶复合梯形公式、4阶 复合Simpson公式和2阶复合Cotes公式 可得各节点的值如右表 0 1 0.125 0.99739787 0.25 0.98961584 0.375 0.97672674 0.5 0.95885108 0.625 0.93615564 0.75 0.90885168 0.875 0.87719257 1 0.84147098 ( ) i i x f x 8 7 6 5 4 3 2 1 0 x x x x x x x x x Trapz 4 2 1 3 3 2 1 2 2 2 1 1 1 2 1 0 0 . x x x x x x x x x Simp + + + + 2 4 3 1 2 1 1 4 1 1 1 4 3 0 2 1 0 4 1 0 0 x x x x x x x x x Cotes + + + + + + 复合求积 公式的程序 newtoncotes.m 函数程序 func.m
分别由复合 Trapz、 Simpson、 Cotes公式有 7=160)+2(x)+/(1)=0959868 24 [f(0)+4∑f(x1)+2∑f(x)+f(1) k=0 =0.94608331 18070+>32/(x12)+12/(x2)+32(x2)+142(x)+7 =0.94608307
T8 [ (0) 2 ( ) (1)] 16 1 7 1 å= = + + k k f f x f 分别由复合Trapz、Simpson、Cotes公式有 = 0.94569086 S4 [ (0) 4 ( ) 2 ( ) (1)] 24 1 3 1 3 0 2 1 f f x f x f k k k k = + å + å + = = + = 0.94608331 C2 [7 (0) [32 ( ) 12 ( ) 32 ( )] 14 ( ) 7 (1)] 180 1 1 1 1 0 4 3 4 2 4 1 f f x f x f x f x f k k k k k k = +å + + + å + = = + + + = 0.94608307