为简便起见,取节点为等分 b-a h= x1=a+jhj=0,1,2,…,n 现在关键是求 C f()dx= L, (x)dx+R,(x)x
◼ 为简便起见,取节点为等分 ◼ 现在关键是求 x a j h j n n b a h j , = + = 0,1,2,..., − = f x dx L x dx R x dx b a n b a b a n ( ) = ( ) + ( )
b b T L,(x)dx=r(>I(x)y )dx=2I L, (xdxlv =(b-a)∑[ L, edxrf(x) (b-a∑Cmf( =0 因此就归结为求权 (n) l, (xdx= b a b =0x;-x
= = = = = − − − = − = = − − = − = = b a n i j i j i i b a j n j j n j n j j n j b a j j n j b a j b a n j j j b a n dx x x x x b a l x dx b a C b a C f x l x dx f x b a b a L x dx l x y dx l x dx y 0 ( ) 0 ( ) 0 0 0 1 ( ) 1 ( ) ( ) ( ) ] ( ) 1 ( ) [ ( ) ( ( ) ) [ ( ) ] 因此就归结为求权
由h x,=a+ h 0.1.2 AHx =a+ih, x=a+th, dx= hdt x-x1=(t-i)h,x1-x1=(j-1)h, =a时t=0;x=b时t=n 因此 b i=0 h fii ho ∏I 「"II(- i=0
= = = = − − = − − = − − − = = = = = − = − − = − = + = + = = + = − = n n i j i n i j i n n i j i b a n i j i j i n i j i j i i j t i dt n j i hdt j i t i nh dx x x x x b a C x a t x b t n x x t i h x x j i h x a ih x a t h dx hdt x a j h j n n b a h 0 0 0 0 0 0 ( ) ( ) 1 1 1 1 0; ( ) , ( ) , , , , 0,1,2,..., 因此 时 时 。 知 ， 由
n-J 八(n-)!J ∏(t-t 当n=1时,仅有两个节点 1(t-1) 1×0×(1-0) 12 2 (t-O)dt 1×1l(1-1) 12102
2 1 1 2 1 ( 0) 1 1! (1 1)! ( 1) 2 1 2 ( 1) 1 1 ( 1) 1 0! (1 0)! ( 1) 1 ( ) !( )! ( 1) 1 0 2 1 0 1 1 (1) 1 1 0 2 1 0 1 0 (1) 0 0 0 − = = − − = = − − − = − − = = − − − = − − = − t C t dt t C t dt n t i dt nj n j n n i j i n j 当 时，仅有两个节点：
当n=2时 (-1)2° C0=2×O×(2-0)(-1)(t-2)at (t-2)2+(t-2)]dla (-2)3+2(t-2)210 司理可得C(2) (2)
6 1 6 4 6 1 ( 2) ] 2 1 ( 2) 3 1 [ 4 1 [( 2) ( 2)] 4 1 ( 1)( 2) 2 0! (2 0)! ( 1) 2 (2) 2 (2) 1 2 0 3 2 2 0 2 2 0 2 0 (2) 0 = = = − + − = = − + − − − − − = = − C C t t t t dt C t t dt n 同理可得 ， 当 时