Aritalab:Lecture/Math/Function

Revision as of 21:23, 20 July 2011

$\Gamma(z) = \int^\infty_0 e^{-t} t^{z-1} dt$

は階乗の一般化で $\Gamma(z+1) = z \Gamma(z)\,$ を満たす。z が正の整数の場合は $\Gamma(z + 1) = z!\,$

例: $\Gamma(1) = 1\,$; $\Gamma(1/2) = \sqrt{\pi}\,$; $\Gamma(3/2) = \sqrt{\pi}/2\,$; $\Gamma(z+1/2) = \frac{(2n)!}{2^{2n}n!}\sqrt{\pi}$

$\Beta(x,y) = \int^1_0 t^{x-1} (1-t)^{y-1} dt$

例: $\Beta(x,y) = \Beta(y,x)\,$; $\Beta(x,y) = \frac{x+y}{y} \Beta(x,y+1)\,$; $\Beta(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x + y)}$; $\Beta(\frac{1}{2},\frac{1}{2}) = \pi\,$

@@ Line 14: / Line 14: @@
 :<math>\Gamma(z+1/2) = \frac{(2n)!}{2^{2n}n!}\sqrt{\pi}</math>
-;近似
+;Stirling の公式
-:<math>\Gamma(z+1) = z! \sim \sqrt{2\pi z} e^{-z}z^z </math> Stirling の公式
+:<math>\Gamma(z+1) = z! \sim \sqrt{2\pi z} e^{-z}z^z </math>
 ==ベータ関数==