負の整数の階乗の商
負の整数の階乗の商
\(m,n\in\mathbb{Z}\)とする。
\begin{align*} \frac{\left(-m\right)!}{\left(-n\right)!} & =\left(-1\right)^{n-m}\frac{\Gamma\left(n\right)}{\Gamma\left(m\right)}\\ & =\left(-1\right)^{n-m}\frac{\left(n-1\right)!}{\left(m-1\right)!} \end{align*}
\(m,n\in\mathbb{Z}\)とする。
\begin{align*} \frac{\left(-m\right)!}{\left(-n\right)!} & =\left(-1\right)^{n-m}\frac{\Gamma\left(n\right)}{\Gamma\left(m\right)}\\ & =\left(-1\right)^{n-m}\frac{\left(n-1\right)!}{\left(m-1\right)!} \end{align*}
(0)
\begin{align*} \frac{\left(-m\right)!}{\left(-n\right)!} & =P\left(-m,-m+n\right)\\ & =\left(-1\right)^{n-m}Q\left(m,n-m\right)\\ & =\left(-1\right)^{n-m}P^{-1}\left(m-1,m-n\right)\\ & =\left(-1\right)^{n-m}\frac{\left(m-1-\left(m-n\right)\right)!}{\left(m-1\right)!}\\ & =\left(-1\right)^{n-m}\frac{\left(n-1\right)!}{\left(m-1\right)!}\\ & =\left(-1\right)^{n-m}\frac{\Gamma\left(n\right)}{\Gamma\left(m\right)} \end{align*}(0)-2
\begin{align*} \frac{\left(-m\right)!}{\left(-n\right)!} & =\lim_{\epsilon\rightarrow0}\frac{\left(-\left(m+\epsilon\right)\right)!}{\left(-\left(n+\epsilon\right)\right)!}\\ & =\lim_{\epsilon\rightarrow0}\frac{\Gamma\left(1-\left(m+\epsilon\right)\right)}{\Gamma\left(1-\left(n+\epsilon\right)\right)}\\ & =\lim_{\epsilon\rightarrow0}\frac{\Gamma\left(\left(m+\epsilon\right)\right)\pi\sin\left(\pi\left(n+\epsilon\right)\right)}{\Gamma\left(\left(n+\epsilon\right)\right)\pi\sin\left(\pi\left(m+\epsilon\right)\right)}\\ & =\lim_{\epsilon\rightarrow0}\frac{\Gamma\left(\left(m+\epsilon\right)\right)\pi\cos\left(\pi\left(n+\epsilon\right)\right)}{\Gamma\left(\left(n+\epsilon\right)\right)\pi\cos\left(\pi\left(m+\epsilon\right)\right)}\\ & =\frac{\Gamma\left(m\right)\cos\left(\pi n\right)}{\Gamma\left(n\right)\cos\left(\pi m\right)}\\ & =\left(-1\right)^{n-m}\frac{\Gamma\left(n\right)}{\Gamma\left(m\right)}\\ & =\left(-1\right)^{n-m}\frac{\left(n-1\right)!}{\left(m-1\right)!} \end{align*}ページ情報
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ガンマ関数の相反公式
\[
\Gamma(z)\Gamma(1-z)=\pi\sin^{-1}(\pi z)
\]
ガンマ関数の1/2値
\[
\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}
\]
ガンマ関数の絶対収束条件
ガンマ関数$\Gamma\left(z\right)$は$\Re\left(z\right)>0$で絶対収束
ガンマ関数を含む極限
\[
\lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)}=1
\]