1次式の総乗と階乗
1次式の総乗と階乗
\(a,b\in\mathbb{Z}\)とする。
\[ \prod_{k=a}^{b}\left(kn+r\right)=n^{b-a+1}\frac{\left(b+\frac{r}{n}\right)!}{\Gamma\left(a+\frac{r}{n}\right)} \]
\(a,b\in\mathbb{Z}\)とする。
\[ \prod_{k=a}^{b}\left(kn+r\right)=n^{b-a+1}\frac{\left(b+\frac{r}{n}\right)!}{\Gamma\left(a+\frac{r}{n}\right)} \]
(0)
\begin{align*} \prod_{k=a}^{b}\left(kn+r\right) & =n^{b-a+1}\prod_{k=a}^{b}\left(k+\frac{r}{n}\right)\\ & =n^{b-a+1}\prod_{k=a}^{b}\frac{\left(k+\frac{r}{n}\right)!}{\left(k+\frac{r}{n}-1\right)!}\\ & =n^{b-a+1}\frac{\left(b+\frac{r}{n}\right)!}{\Gamma\left(a+\frac{r}{n}\right)} \end{align*}(0)-2
\begin{align*} \prod_{k=a}^{b}\left(kn+r\right) & =\prod_{k=a}^{-1}\left(kn+r\right)\prod_{k=0}^{b}\left(kn+r\right)\\ & =\prod_{k=0}^{a-1}\left(kn+r\right)^{-1}\prod_{k=0}^{b}\left(kn+r\right)\\ & =\left\{ n^{a-1}r\frac{\left(a-1+\frac{r}{n}\right)!}{\frac{r}{n}!}\right\} ^{-1}n^{b}r\frac{\left(b+\frac{r}{n}\right)!}{\frac{r}{n}!}\\ & =n^{b-a+1}\frac{\left(b+\frac{r}{n}\right)!}{\left(a+\frac{r}{n}-1\right)!}\\ & =n^{b-a+1}\frac{\left(b+\frac{r}{n}\right)!}{\Gamma\left(a+\frac{r}{n}\right)} \end{align*}ページ情報
| タイトル | 1次式の総乗と階乗 |
| URL | https://www.nomuramath.com/f039zr6h/ |
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第2種不完全ガンマ関数とガンマ関数の比の極限
\[
\lim_{k\rightarrow0}\frac{\Gamma\left(k,x\right)}{\Gamma\left(k\right)}=\delta_{0x}
\]
負の整数の階乗の商
\[
\frac{\left(-m\right)!}{\left(-n\right)!}=\left(-1\right)^{n-m}\frac{\Gamma\left(n\right)}{\Gamma\left(m\right)}
\]
ガンマ関数の微分
\[
\frac{d}{dz}\Gamma(z)=\Gamma(z)\psi(z)
\]
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