拉格朗日有唔有型？

tommywong

$a_{k+1}=a$

$\displaystyle \sum_{i=1}^k \frac{1}{a-a_i}\prod_{j=1\atop j\neq i}^k \frac{1}{a_i-a_j}
=-\sum_{i=1}^k \prod_{j=1\atop j\neq i}^{k+1} \frac{1}{a_i-a_j}$

$\displaystyle =\prod_{j=1}^k \frac{1}{a-a_j}-\sum_{i=1}^{k+1} \prod_{j=1\atop j\neq i}^{k+1} \frac{1}{a_i-a_j}=\prod_{j=1}^k \frac{1}{a-a_j}$

$\displaystyle \sum_{i=1}^k \frac{1}{(a-a_i)^2}\prod_{j=1\atop j\neq i}^k \frac{1}{a_i-a_j}
=\left(\sum_{j=1}^k \frac{1}{a-a_j}\right)
\left(\prod_{j=1}^k \frac{1}{a-a_j}\right) ?$

$\displaystyle \sum_{i=1}^k \frac{1}{(a-a_i)^3}\prod_{j=1\atop j\neq i}^k \frac{1}{a_i-a_j}
= ?$

kuing

第一个我之前就是用拉格朗证的：forum.php?mod=viewthread&tid=6913

第二、三个就是第一个对 `a` 求导？