82 Newton's Interpolation (k+1)阶差商: ∫[x n…,xk1=n,,…, ∫Lx1 195k5~k+1 xn-x 0 05··5~k-15k l-∫ 0,···9k-19~k+1 x;- k k+1 事实上几xn,,x1=∑/(x) k+1 其中a k+1 (x)=I(x-x),0+(x ∏ j=0 差商的值与x的顺序无关!§2 Newton’s Interpolation 1 0 1 0 1 1 0 1 0 1 1 1 0 1 [ , ... , , ] [ , ... , , ] [ , , ... , ] [ , ... , , ] [ , ... , ]              k k k k k k k k k k k x x f x x x f x x x x x f x x x f x x x f x x (k+1)阶差商：     k i k i i k x f x f x x 0 1 0 ( ) ( ) [ , ... , ]  事实上 其中 ( ) ( ) , 0 1     k i k x x xi        k j i j k i i j x x x 0 1  ( ) ( ) Warning: my head is exploding… Wh差at商is的the值po与int of this formula? xi 的顺序无关！