負の多重階乗
負の多重階乗
\(q\in\mathbb{N}_{0}\)とする。
\[ \left(-\left(qn+r\right)\right)!_{n}=\frac{\left(-1\right)^{q}}{\left(qn-\left(n-r\right)\right)!_{n}} \]
\(q\in\mathbb{N}_{0}\)とする。
(1)
\[ \left(-\left(qn+r\right)\right)!_{n}=\frac{(n-r)!_{n}\left(-1\right)^{q}}{\left(n-r\right)\left(qn-\left(n-r\right)\right)!_{n}}\cnd{n\ne r} \](2)
\(0<r<n\)のとき\[ \left(-\left(qn+r\right)\right)!_{n}=\frac{\left(-1\right)^{q}}{\left(qn-\left(n-r\right)\right)!_{n}} \]
(1)
\begin{align*} \left(-\left(qn+r\right)\right)!_{n} & =(n-r)!_{n}\prod_{k=0}^{q}\frac{\left(-\left(nk+r\right)\right)!_{n}}{\left(-\left(nk+r\right)+n\right)!_{n}}\\ & =(n-r)!_{n}\prod_{k=0}^{q}\frac{\left(-\left(nk+r\right)\right)!_{n}}{\left(-\left(nk+r\right)+n\right)\left(-\left(nk+r\right)\right)!_{n}}\\ & =(n-r)!_{n}\prod_{k=0}^{q}\frac{-1}{n\left(k-1\right)+r}\\ & =(n-r)!_{n}\left(-1\right)^{q}\frac{1}{n-r}\prod_{k=1}^{q}\frac{1}{n\left(k-1\right)+r}\\ & =(n-r)!_{n}\left(-1\right)^{q}\frac{1}{n-r}\prod_{k=0}^{q-1}\frac{1}{nk+r}\\ & =\frac{(n-r)!_{n}\left(-1\right)^{q}}{\left(n-r\right)\left(qn-\left(n-r\right)\right)!_{n}} \end{align*}(2)
\(0<r<n\)のとき\(0<n-r<n\)となるので\(\left(n-r\right)!_{n}=n-r\)となる。これより、
\begin{align*} \left(-\left(qn+r\right)\right)!_{n} & =(n-r)!_{n}\left(-1\right)^{q}\frac{1}{\left(n-r\right)\left(qn-\left(n-r\right)\right)!_{n}}\cmt{(1)\text{より}}\\ & =\frac{\left(-1\right)^{q}}{\left(qn-\left(n-r\right)\right)!_{n}} \end{align*}
ページ情報
タイトル | 負の多重階乗 |
URL | https://www.nomuramath.com/s2hb44fp/ |
SNSボタン |
拡張多重階乗の漸化式
\[
x!^{n}=x\left(x-n\right)!^{n}
\]
多重階乗と拡張多重階乗の定義
\[
\left(x\right)!^{n}=n^{\frac{x-1}{n}}\frac{\left(\frac{x}{n}\right)!}{\left(\frac{1}{n}\right)!}
\]
2重階乗の逆数和
\[
\sum_{k=0}^{n}\frac{1}{\left(2k\right)!!}=\sqrt{e}\frac{\Gamma\left(n+1,\frac{1}{2}\right)}{\Gamma\left(n+1\right)}
\]
多重階乗同士の関係
\[
\left(qn+r\right)!^{n}=r!^{n}\frac{\left(qn+r\right)!_{n}}{r!_{n}}
\]