假设P(x0)=f(x)k=12,…, n=f(x),1.a1=f(x,2a2=f"(x) 得ak=,f(x0)(k=0,1,2,…,n) ! 代入P(x)中得 P(x)=f(x)+f(x)(x-x)+(x0)(x-b)2+假设 P x f x k n k k n ( ) ( ) 1,2, , 0 ( ) 0 ( ) = =  ( ), 0 x0 a = f 代入P (x) n 中得 n n n x x n f x x x f x P x f x f x x x ( ) ! ( ) ( ) 2! ( ) ( ) ( ) ( )( ) 0 0 ( ) 2 0 0 0 0 0 + − − +  = +  − +  得 ( ) ( 0,1,2, , ) ! 1 0 ( ) f x k n k a k k = =  1 ( ), 1 0  a = f  x 2! ( ) 2 x0 a = f    , ! ( ) 0 ( ) n a f x n  n =