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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2項係数とベータ関数の関係
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