对(1)式两边求导有 f(x)=L(((on(3)+(+m(x) (n+1) 由于ξ与x有关[f()将很难确定 但是当x=x时,f(x2)可以求出 f(xk)=ln( k)+ [fm+( (n+1) On,+(xk)+ (5) on+(k) (n+1) (n+1) (n+1) (5) (n+1) n+1(~k k=0,1,…,n (n+1) Ln(k)+ (5)m (n+1)!1 k i≠k对(1)式两边求导,有 ( ) ( 1)! ( ) ( ) ( 1)! [ ( )] ( ) ( ) 1 ( 1) 1 ( 1) x n f x n f f x L x n n n n n + + + + ¢ + + + ¢ ¢ = ¢ + w x w x 由于x与x有关,[ f (n+1) (x )]¢将很难确定 但是当x = xk时, f ¢(xk )可以求出 ( ) ( 1)! ( ) ( ) ( 1)! [ ( )] ( ) ( ) 1 ( 1) 1 ( 1) n k n n k n k n k x n f x n f f x L x + + + + ¢ + + + ¢ ¢ = ¢ + w x w x ( ) ( 1)! ( ) ( ) 1 ( 1) n k n n k x n f L x + + ¢ + = ¢ + w x Õ ¹ = + - + = ¢ + n j k j k j n n k x x n f L x 0 ( 1) ( ) ( 1)! ( ) ( ) x --------(2) k = 0,1,L,n