2項係数とベータ関数の関係
2項係数とベータ関数の関係
(1)
\[ B(x,y)=\frac{x+y}{xyC(x+y,x)} \](2)
\[ C(x,y)=\frac{1}{(x+1)B(x-y+1,y+1)} \](3)
\[ B(x,y)=\frac{-\pi}{x\sin(\pi x)B(x+y,-x)} \](4)
\[ C(x,y)=-\frac{y\sin(\pi y)}{C(x-y,x)\pi} \](5)
\[ B(x,y)=\frac{C(y-1,-x)\pi}{\sin(\pi x)} \](6)
\[ C(x.y)=\frac{-B(x+1,-y)\sin(\pi y)}{\pi} \](1)
\begin{align*} B(x,y) & =\frac{(x-1)!(y-1)!}{(x+y-1)!}\\ & =\frac{x+y}{xy}\frac{x!y!}{(x+y)!}\\ & =\frac{x+y}{xyC(x+y,x)} \end{align*}(2)
\begin{align*} C(x,y) & =\frac{x!}{y!(x-y)!}\\ & =\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}\\ & =\frac{1}{x+1}\frac{\Gamma(x+2)}{\Gamma(y+1)\Gamma(x-y+1)}\\ & =\frac{1}{(x+1)B(x-y+1,y+1)} \end{align*}(3)
\begin{align*} B(x,y) & =\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\\ & =\Gamma(x)\Gamma(1-x)\frac{\Gamma(y)}{-x\Gamma(-x)\Gamma(x+y)}\\ & =\frac{-\pi}{x\sin(\pi x)B(x+y,-x)} \end{align*}(4)
\begin{align*} C(x,y) & =\frac{\Gamma(x+1)}{\Gamma(x-y+1)\Gamma(y+1)}\\ & =\frac{\Gamma(x+1)\Gamma(1-y)}{\Gamma(x-y+1)\Gamma(y+1)\Gamma(1-y)}\\ & =\frac{1}{C(x-y,x)y\Gamma(y)\Gamma(1-y)}\\ & =\frac{\sin(\pi y)}{C(x-y,x)\pi y} \end{align*}(5)
\begin{align*} B(x,y) & =\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\\ & =\frac{(y-1)!}{(x+y-1)!(-x)!}\Gamma(x)\Gamma(1-x)\\ & =\frac{C(y-1,-x)\pi}{\sin(\pi x)} \end{align*}(6)
\begin{align*} C(x.y) & =\frac{x!}{y!(x-y)!}\\ & =\frac{\Gamma(x+1)\Gamma(1-y)}{y\Gamma(y)\Gamma(1-y)\Gamma(x-y+1)}\\ & =\frac{-y\Gamma(x+1)\Gamma(-y)}{y\Gamma(y)\Gamma(1-y)\Gamma(x-y+1)}\\ & =\frac{-B(x+1,-y)\sin(\pi y)}{\pi} \end{align*}ページ情報
タイトル | 2項係数とベータ関数の関係 |
URL | https://www.nomuramath.com/l2u2rt4a/ |
SNSボタン |
ウォリスの公式
\[
\prod_{k=1}^{\infty}\left(\frac{(2k)^{2}}{(2k-1)(2k+1)}\right)=\frac{\pi}{2}
\]
ゼータ関数の交代級数
\[
\sum_{k=1}^{\infty}\left(\zeta\left(2k\right)-\zeta\left(2k+1\right)\right)=\frac{1}{2}
\]
偏角・対数と符号関数の関係
\[
\Arg\left(z\right)=-i\Log\left(\sgn\left(z\right)\right)
\]
運用による資産推移
\[
x=\begin{cases}
\left(x_{0}+\frac{b}{\log a}\right)a^{t}-\frac{b}{\log a} & a\ne1\\
x_{0}+bt & a=1
\end{cases}
\]