把a、a1代入(*)式,移项后得 ∫(x)-f(x0)-f(x)(x-x0) ∑n(x-xn)2+0(x-x)"2 C-d 2 f(x)-∫(x lir n)-f(xn(x=x)=lm∑1(x-x)y2+0(x-xn)"2 x-x (x-x0) L Hospital lim f(x)-f'(ro)1 X→x02(x-x0) f"(x0)=a2 依此类推可得 ∫(x)k=0, !5 ( ) ( ( ) ) ( ) ( ) ( ) ( ) ( ) 2 0 2 0 2 2 0 0 0 0 i n n i i a x x o x x x x f x f x f x x x lim [ ( ) ( ( ) ) ] ( ) ( ) ( ) ( ) ( ) lim 2 0 2 0 2 2 0 0 0 0 0 0 i n n i i x x x x a x x o x x x x f x f x f x x x LHospital 2( ) ( ) ( ) lim 0 0 0 x x f x f x x x ( ) 2 1 x0 f a2 f x k n k a k k ( ) 0,1, , ! 1 0 ( ) 0 1 把 a a 、 代入 ( ) 式,移项后得 依此类推可得