莱布尼兹（Leibniz）公式

一、内容

$$(uv)^{(n)}=\sum_{i=0}^n C_n^i u^{(n-i)}v^{(i)}$$

二、证明

用数学归纳法证明

1. $n=1$时，即$$(uv)'=u'v+uv'$$命题成立

2. 假设$n=k(k\ge 1)$时成立，即$$(uv)^{(k)}=\sum_{i=0}^k C_k^i u^{(k-i)}v^{(i)}$$成立
   当$n=k+1$时，

   $$ \begin{aligned} (uv)^{(k+1)} &= \left[\sum_{i=0}^k C_k^i u^{(k-i)}v^{(i)}\right]' \\ &= \sum_{i=0}^k C_k^i \left[u^{(k-i)}v^{(i)}\right]' \\ &= \sum_{i=0}^k C_k^i \left[u^{(k-i+1)}v^{(i)}+u^{(k-i)}v^{(i+1)}\right] \\ &= \sum_{i=0}^k C_k^i u^{(k-i+1)}v^{(i)} + \sum_{i=0}^k C_k^i u^{(k-i)}v^{(i+1)} \\ &= C_k^0 u^{(k+1)}v + \sum_{i=1}^k C_k^i u^{(k-i+1)}v^{(i)} + \sum_{i=0}^{k-1} C_k^i u^{(k-i)}v^{(i+1)} + C_k^k uv^{(k+1)} \\ &= C_k^0 u^{(k+1)}v + \sum_{i=1}^k C_k^i u^{(k-i+1)}v^{(i)} + \sum_{i=1}^{k} C_k^{i-1} u^{(k-i+1)}v^{(i)} + C_k^k uv^{(k+1)} \\ &= C_{k+1}^0 u^{(k+1)}v + \sum_{i=1}^k (C_k^i + C_k^{i-1}) u^{(k-i+1)}v^{(i)} + C_{k+1}^{k+1} uv^{(k+1)} \\ &= C_{k+1}^0 u^{(k+1)}v + \sum_{i=1}^k C_{k+1}^i u^{(k-i+1)}v^{(i)} + C_{k+1}^{k+1} uv^{(k+1)} \\ &= \sum_{i=0}^{k+1} C_{k+1}^i u^{(k-i+1)}v^{(i)} \end{aligned} $$

   由1.2.知$n\ge 1$时命题成立