ベータ関数の関数等式
ベータ関数の関数等式
(1)
\[ xB(x,y+1)=yB(x+1,y) \](2)
\[ B(x,y)=\frac{y-1}{x}B(x+1,y-1) \](3)
\[ B(x,y)=\frac{x+y}{x}B(x+1,y) \](4)
\[ B(x,y)=\frac{x-1}{x+y-1}B(x-1,y) \](5)
\[ B(x,y)=B(x+1,y)+B(x,y+1) \](6)
\[ B(x,y)=\frac{\Gamma(x)}{Q(y,x)} \](1)
\begin{align*} xB(x,y+1) & =x\frac{\Gamma(x)\Gamma(y+1)}{\Gamma(x+y+1)}\\ & =y\frac{\Gamma(x+1)\Gamma(y)}{\Gamma(x+y+1)}\\ & =yB(x+1,y) \end{align*}(2)
\begin{align*} B(x,y) & =\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\\ & =\frac{y-1}{x}\frac{\Gamma(x+1)\Gamma(y-1)}{\Gamma(x+y)}\\ & =\frac{y-1}{x}B(x+1,y-1) \end{align*}(2)-2
\begin{align*} B(x,y) & =\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt\\ & =\left[\frac{1}{x}t^{x}(1-t)^{y-1}\right]_{t=0}^{t=1}+\frac{y-1}{x}\int_{0}^{1}t^{x}(1-t)^{y-2}dt\\ & =\frac{y-1}{x}B(x+1,y-1) \end{align*}(3)
\begin{align*} B(x,y) & =\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\\ & =\frac{x+y}{x}\frac{\Gamma(x+1)\Gamma(y)}{\Gamma(x+y+1)}\\ & =\frac{x+y}{x}B(x+1,y) \end{align*}(4)
\begin{align*} B(x,y) & =\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\\ & =\frac{x-1}{x+y-1}\frac{\Gamma(x-1)\Gamma(y)}{\Gamma(x+y-1)}\\ & =\frac{x-1}{x+y-1}B(x-1,y) \end{align*}(5)
\begin{align*} B(x,y) & =\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\left(\frac{x}{x+y}+\frac{y}{x+y}\right)\\ & =\frac{\Gamma(x+1)\Gamma(y)}{\Gamma(x+y+1)}+\frac{\Gamma(x)\Gamma(y+1)}{\Gamma(x+y+1)}\\ & =B(x+1,y)+B(x,y+1) \end{align*}(5)-2
\begin{align*} B(x,y) & =\int_{0}^{1}(t+(1-t))t^{x-1}(1-t)^{y}dt\\ & =\int_{0}^{1}t^{x}(1-t)^{y}dt+\int_{0}^{1}t^{x}(1-t)^{y+1}dt\\ & =B(x+1,y)+B(x,y+1) \end{align*}(6)
\begin{align*} B(x,y) & =\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\\ & =\frac{\Gamma(x)}{Q(y,x)} \end{align*}ページ情報
タイトル | ベータ関数の関数等式 |
URL | https://www.nomuramath.com/q55jf94u/ |
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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
\[
B(x,y)=\sum_{k=0}^{\infty}\frac{C(k-y,k)}{x+k}
\]
ベータ関数の微分
\[
\frac{\partial}{\partial x}B(x,y)=B(x,y)\left\{ \psi(x)-\psi(x+y)\right\}
\]
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