这是一份manchester曼切斯特大学 MATH20701作业代写的成功案例
$$
\Gamma(\alpha)=(\alpha-1) \Gamma(\alpha-1) .
$$
In particular, if $\alpha$ is an integer, repeated applications of recursive produce
$$
\Gamma(\alpha)=(\alpha-1)(\alpha-2) \ldots \Gamma(1)
$$
But $\Gamma(1)=\int_{0}^{\infty} e^{-y} d y=1$, so that
$$
\Gamma(\alpha)=(\alpha-1)(\alpha-2) \ldots 1=(\alpha-1) !
$$
For later reference, we mention here that, by integration, we obtain:
$$
\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}
$$
MATH20701 COURSE NOTES ：
The median of the distribution of a r.v. $X$ is usually defined as a point, denoted by $x_{0.50}$, for which
$$
P\left(X \leq x_{0.50}\right) \geq 0.50 \quad \text { and } \quad P\left(X \geq x_{0.50}\right) \geq 0.50
$$
or, equivalently,
$$
P\left(X<x_{0.50}\right) \leq 0.50 \quad \text { and } \quad P\left(X \leq x_{0.50}\right) \geq 0.50
$$
If the underlying distribution is continuous, the median is (essentially) unique and may be simply defined by:
$$
P\left(X \leq x_{0.50}\right)=P\left(X \geq x_{0.50}\right)=0.50
$$