多重階乗同士の関係
多重階乗同士の関係
\(q\in\mathbb{N}_{0}\)とする。
\(q\in\mathbb{N}_{0}\)とする。
(1)
\[ \left(qn+1\right)!_{n}=\left(qn+1\right)!^{n} \](2)
\[ \left(qn+r\right)!^{n}=r!^{n}\frac{\left(qn+r\right)!_{n}}{r!_{n}} \]*
\(x!_{n}\)は多重階乗、\(x!^{n}\)は拡張多重階乗。(1)
\begin{align*} \left(qn+1\right)!_{n} & =n^{q}\frac{\left(q+\frac{1}{n}\right)!}{\left(\frac{1}{n}\right)!}\\ & =n^{\frac{qn+1-1}{n}}\frac{\left(\frac{qn+1}{n}\right)!}{\left(\frac{1}{n}\right)!}\\ & =\left(qn+1\right)!^{n} \end{align*}(2)
\begin{align*} \left(qn+r\right)!^{n} & =r!^{n}\prod_{k=1}^{q}\frac{\left(kn+r\right)!^{n}}{\left(kn+r-n\right)!^{n}}\\ & =r!^{n}\prod_{k=1}^{q}\left(kn+r\right)\\ & =r!^{n}\prod_{k=1}^{q}\frac{\left(kn+r\right)!_{n}}{\left(kn+r-n\right)!_{n}}\\ & =r!^{n}\frac{\left(qn+r\right)!_{n}}{r!_{n}} \end{align*}ページ情報
| タイトル | 多重階乗同士の関係 |
| URL | https://www.nomuramath.com/hcnmgy5n/ |
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階乗の多重階乗表示
\[
n!=\prod_{k=0}^{j-1}\left(n-k\right)!_{j}
\]
ウォリス積分の拡張2重階乗表示
\[
\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta=\frac{\left(n-1\right)!^{2}}{\left(n\right)!^{2}}\sqrt{\frac{\pi}{2}}
\]
拡張多重階乗の漸化式
\[
x!^{n}=x\left(x-n\right)!^{n}
\]
(拡張)多重階乗の逆数和
\[
\sum_{k=0}^{n}\frac{1}{\left(ak+b\right)!_{a}}=\frac{e^{\frac{1}{a}}a^{\frac{b}{a}}\Gamma\left(\frac{b}{a}+1\right)}{b!_{a}}\left(\frac{\Gamma\left(n+\frac{b}{a}+1,\frac{1}{a}\right)}{\Gamma\left(n+\frac{b}{a}+1\right)}-\frac{\Gamma\left(\frac{b}{a},\frac{1}{a}\right)}{\Gamma\left(\frac{b}{a}\right)}\right)
\]