ガンマ関数を含む極限
ガンマ関数を含む極限値
(1)
\[ \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)}=1 \](2)
\[ \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}=\sqrt{2} \](1)
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)} & =\lim_{n\rightarrow\infty}\sqrt{\sqrt{n}\frac{\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)}\sqrt{n+\frac{1}{2}}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(n+1\right)}}\\ & =\lim_{n\rightarrow\infty}\sqrt{\frac{\sqrt{n}\sqrt{n+\frac{1}{2}}}{n}}\\ & =\lim_{n\rightarrow\infty}\left(1+\frac{1}{2n}\right)^{\frac{1}{4}}\\ & =1 \end{align*}(1)-2
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)} & =\lim_{n\rightarrow\infty}\sqrt{\frac{n+1}{2}}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}\\ & =\frac{1}{\sqrt{2}}\lim_{n\rightarrow\infty}\frac{\sqrt{n+1}}{\sqrt{n}}\sqrt{n}\frac{2}{\Gamma\left(\frac{1}{2}\right)}\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta\\ & =\sqrt{\frac{2}{\pi}}\lim_{n\rightarrow\infty}\frac{\sqrt{n+1}}{\sqrt{n}}\sqrt{n}\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta\\ & =\sqrt{\frac{2}{\pi}}\sqrt{\frac{\pi}{2}}\\ & =1 \end{align*}(2)
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)} & =\lim_{n\rightarrow\infty}\sqrt{\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}\sqrt{n+1}\frac{\Gamma\left(\frac{n+2}{2}\right)}{\Gamma\left(\frac{n+3}{2}\right)}}\\ & =\lim_{n\rightarrow\infty}\sqrt{\frac{2\sqrt{n}\sqrt{n+1}}{n+1}}\\ & =\sqrt{2}\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{-\frac{1}{4}}\\ & =\sqrt{2} \end{align*}(2)-2
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)} & =\frac{2}{\Gamma\left(\frac{1}{2}\right)}\lim_{n\rightarrow\infty}\sqrt{n}\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta\\ & =\frac{2}{\sqrt{\pi}}\sqrt{\frac{\pi}{2}}\\ & =\sqrt{2} \end{align*}ページ情報
| タイトル | ガンマ関数を含む極限 |
| URL | https://www.nomuramath.com/hkc3vxq1/ |
| SNSボタン |
第1種・第2種不完全ガンマ関数の微分
\[
\frac{\partial\Gamma\left(a,x\right)}{\partial x}=-x^{a-1}e^{-x}
\]
ガンマ関数のルジャンドル倍数公式
\[
\Gamma(2z)=\frac{2^{2z-1}}{\sqrt{\pi}}\Gamma(z)\Gamma\left(z+\frac{1}{2}\right)
\]
ガンマ関数の微分
\[
\frac{d}{dz}\Gamma(z)=\Gamma(z)\psi(z)
\]
階乗と階乗の逆数の母関数
\[
\frac{x^{a}}{a!}=e^{x}\left(\frac{\Gamma\left(a+1,x\right)}{\Gamma\left(a+1\right)}-\frac{\Gamma\left(a,x\right)}{\Gamma\left(a\right)}\right)
\]