若将x≠x,(=0,1…,m)视为一个节点,则 f[x,x0,x1,…xk f[x,x0,x1,…,xk1]一f[x,x12…,xk] f[x,x0,x1…,xk-1]=f[x,x1…,x]+f[x,x2,x12…,xk](x-xk) 因此可得f(x)=f0+f[x,xl(x-x0) fo +(fLox,+ fx, ro, X,(x-xi)(x-xo fo+foxi(x-xo)+f[x, xo,,()(x-x1) =+∑/1,x,…x小( x-x)+1x,x1…,Xn[ , , , , ] 0 1 k f x x x L x [ , , , , ] 0 1 k-1 f x x x L x 若将x ¹ xi ,(i = 0,1,L,n)视为一个节点,则 因此可得 ( ) [ , ]( ) 0 0 0 f x = f + f x x x - x ( [ , ] [ , , ]( ))( ) 0 0 1 0 1 1 0 = f + f x x + f x x x x - x x - x [ , ]( ) [ , , ]( )( ) 0 0 1 0 0 1 0 1 = f + f x x x - x + f x x x x - x x - x LL å Õ Õ = = - = = + - + - n j n j n k k j k j f f x x x x x f x x x x x x 0 0 1 1 1 0 0 0 1 [ , ,L, ] ( ) [ , , ,L, ] ( ) k k k x x f x x x x f x x x - - = - [ , , , , ] [ , , , ] 0 1 L 1 0 1 L [ , , , ] [ , , , , ]( ) 0 1 k 0 1 k k = f x x L x + f x x x L x x - x