n为简便起见取节点为等分 b-a h x1=a+jhj=0,1, ■现在关键是求 cb f(xdx= Ln(x)dx+R(xb
◼ 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 ∑l(x)y)tx=∑叮J”1(x) b (b-a∑l l, ()dxf(xi) b (b-a∑C("f( 因此就归结为求权 X-x 4, ()dx b i=0
= = = = = − − − = − = = − − = − = = 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
b-a 由h a+hj=0,1,22…,n AHx,=a+ih, x=a+th, dx= hdt x-x1=(t-i)h,x1-x=(j-1)h x=a时t=0;x=b时t=n。 因此 bamx二x=m工二m ∏ I(t-i)dt
= = = = − − = − − = − − − = = = = = − = − − = − = + = + = = + = − = 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
7=DT(= l≠ 当n=时,仅有两个节点 1-0 (t-Ddt 1(t-1)21_1 l×0×(1-0 12 (-1) 1-1 (t-o)dt l×1×(1-1)
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：
in=2时 (2) (t-1)(t-2)at 2×O!×(2-0)! 4J【(-2)2+(t-2)dt [(t-2)+(t-2)2 同理可得<(2) v(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