8 1 Lagrange Polynomial >插值余项/ Remainder 设节点a≤x<x<…<x≤b,且f满足条件∫eCn列l, 在a,b内存在考察截断误差R(x)=()-L) Rn(x)至少有+1个根→→R(x)=K(x)I(x-x) 注意这里是对t求导叫()=(0-()(=x) i=0 g(x)有+2个不同的根x…xnx今p"(5)=0,5 ∈(a (1(5x)-+(5)-K(x)(n+1)!=R(5)-k(x)(m+1! (n+1) ( K(x)= ( r,(x)= wIS ITe-x,) (n+1)! (n+1) i=0§1 Lagrange Polynomial 插值余项 /* Remainder */ 设节点 (n1) f 在[a , b]内存在, 考察截断误差 R (x) f (x) L (x) n = - n f C [a,b] n a x x x b 0 1 n ,且 f 满足条件 , Rolle’s Theorem: 若 充分光滑, ,则 存在 使得 。 (x) ( ) ( ) 0 x0 = x1 = ( , ) 0 1 x x ( ) = 0 推广:若(x0 ) =(x1 ) =(x2 ) = 0 ( , ), ( , ) 0 0 1 1 1 2 x x x x 使得 ( ) ( ) 0 0 = 1 = ( , ) 0 1 使得 ( ) = 0 ( x0 ) = = ( xn ) = 0 存在 (a, b) 使得 ( ) 0 ( ) = n Rn (x) 至少有 n+1个根 = = - n i Rn x K x x xi 0 ( ) ( ) ( ) 任意固定 x xi (i = 0, …, n), 考察 = = - - n i xi t Rn t K x t 0 ( ) ( ) ( ) ( ) (x)有 n+2 个不同的根 x0 … xn x ( ) 0, ( , ) ( 1) a b x x n = ( ) ( ) ( 1) ! ( 1) - R x K x n n n 注意这里是对 t 求导 - - = ( ) ( ) ( )( 1) ! ( 1) ( 1) f L x K x n n x n n ( 1)! ( ) ( ) ( 1) = n f K x x n = - = n i i x n n x x n f R x 0 ( 1) ( ) ( 1)! ( ) ( )