## 一個關於階乘的恆等式

$\displaystyle \sum_{i=0}^n(-1)^i\binom{n}{i}(x-i)^n=n!$

$\displaystyle \sum_{i=0}^n(-1)^i\binom{n}{i}p(i)=0$

$\displaystyle f_n(x)=\sum_{i=0}^n(-1)^i\binom{n}{i}(x-i)^n$

\displaystyle\begin{aligned} f'_{k+1}(x)&=\frac{d}{dx}\left(\sum_{i=0}^{k+1}(-1)^i\binom{k+1}{i}(x-i)^{k+1}\right)\\ &=(k+1)\sum_{i=0}^{k+1}(-1)^i\binom{k+1}{i}(x-i)^{k}. \end{aligned}

$\displaystyle \binom{k+1}{i}=\binom{k}{i}+\binom{k}{i-1}$

\displaystyle \begin{aligned} f'_{k+1}(x)&=(k+1)\left(\sum_{i=0}^k(-1)^i\binom{k}{i}(x-i)^k+\sum_{i=1}^k(-1)^i\binom{k}{i-1}(x-i)^k+(-1)^{k+1}(x-k-1)^k\right)\\ &=(k+1)\left(\sum_{i=0}^k(-1)^i\binom{k}{i}(x-i)^k-\sum_{j=0}^k(-1)^{j}\binom{k}{j}(x-j-1)^k\right)\\ &=(k+1)\left(f_k(x)-f_k(x-1)\right)\\ &=(k+1)(k!-k!)=0, \end{aligned}

\displaystyle\begin{aligned} f_{k+1}(x)&=\sum_{i=0}^{k+1}(-1)^i\binom{k+1}{i}(k+1-i)^{k+1}\\ &=\sum_{i=0}^{k}(-1)^i\frac{(k+1)!(k+1-i)}{i!(k+1-i)!}(k+1-i)^{k}\\ &=(k+1)\sum_{i=0}^{k}(-1)^i\binom{k}{i}(k+1-i)^{k}\\ &=(k+1)f_k(k+1)\\ &=(k+1)k!=(k+1)!, \end{aligned}

$\displaystyle f_n(x)=\sum_{i=0}^n(-1)^i\binom{n}{i}(x-i)^n=n!$

$f_n(x)$$j$ 次導數，

$\displaystyle f_n^{(j)}(x)=n(n-1)\cdots(n-(j-1))\sum_{i=0}^n(-1)^i\binom{n}{i}(x-i)^{n-j}=0$

$\displaystyle \sum_{i=0}^n(-1)^i\binom{n}{i}(x-i)^{n-j}=0,~~1\le j\le n$

$\displaystyle \sum_{i=0}^n(-1)^i\binom{n}{i}i^{n-j}=0,~~1\le j\le n$

\displaystyle\begin{aligned} \sum_{i=0}^n(-1)^i\binom{n}{i}p(i)&=\sum_{i=0}^n(-1)^i\binom{n}{i}(c_ri^r+\cdots+c_1i+c_0)\\ &=\sum_{j=n-r}^nc_{n-j}\sum_{i=0}^n(-1)^i\binom{n}{i}i^{n-j}=0. \end{aligned}

This entry was posted in 特別主題 and tagged , . Bookmark the permalink.