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*}
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| タイトル | ２項係数とベータ関数の関係 |
| URL | https://www.nomuramath.com/l2u2rt4a/ |
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ベータ関数になる積分
\[
\int_{0}^{\frac{\pi}{2}}\sin^{x}t\cos^{y}tdt=\frac{1}{2}B\left(\frac{x+1}{2},\frac{y+1}{2}\right)
\]
ベータ関数と2項係数の逆数の級数表示
\[
B(x,y)=\sum_{k=0}^{\infty}\frac{C(k-y,k)}{x+k}
\]
ベータ関数とガンマ関数の関係
\[
B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}
\]
ベータ関数の絶対収束条件
ベータ関数$B\left(p,q\right)$は$\Re\left(p\right)>0\;\land\;\Re\left(q\right)>0$で絶対収束