Factorial of Half/Proof 2

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Theorem

$\left({\dfrac 1 2}\right)! = \dfrac {\sqrt \pi} 2$


Proof

From Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta:

$\displaystyle \prod_{n \mathop \ge 1} \dfrac {\left({n + \alpha_1}\right) \cdots \left({n + \alpha_k}\right)} {\left({n + \beta_1}\right) \cdots \left({n + \beta_k}\right)} = \dfrac {\Gamma \left({1 + \beta_1}\right) \cdots\Gamma \left({1 + \beta_1}\right)} {\Gamma \left({1 + \alpha_1}\right) \cdots\Gamma \left({1 + \alpha_k}\right)}$

where:

$\alpha_1 + \cdots + \alpha_k = \beta_1 + \cdots + \beta_k$
none of the $\beta$s is a negative integer.


Setting:

$k = 2$
$\alpha_1 = \alpha_2 = 0$
$\beta_1 = -\dfrac 1 2, \beta_2 = \dfrac 1 2$

we see that:

$\alpha_1 + \alpha_2 = \beta_1 + \beta_2$


So this reduces to:

\(\ds \prod_{n \mathop \ge 1} \dfrac n {n - \frac 1 2} \dfrac n {n + \frac 1 2}\) \(=\) \(\ds \dfrac {\Gamma \left({1 - \frac 1 2}\right) \Gamma \left({1 + \frac 1 2}\right)} {\Gamma \left({1}\right) \Gamma \left({1}\right)}\)
\(\ds \) \(=\) \(\ds \Gamma \left({\frac 1 2}\right) \Gamma \left({1 + \frac 1 2}\right)\) as $\Gamma \left({1}\right) = 0! = 1$
\(\ds \) \(=\) \(\ds 2 \Gamma \left({1 + \frac 1 2}\right)^2\) Gamma Difference Equation

We then note that by Wallis's Product:

$\displaystyle \prod_{n \mathop \ge 1} \dfrac n {n - \frac 1 2} \dfrac n {n + \frac 1 2} = \frac \pi 2$

Thus:

\(\ds \frac \pi 2\) \(=\) \(\ds 2 \Gamma \left({1 + \frac 1 2}\right)^2\)
\(\ds \leadsto \ \ \) \(\ds \frac \pi 4\) \(=\) \(\ds \left({\left({\frac 1 2}\right)!}\right)^2\) Gamma Function Extends Factorial
\(\ds \leadsto \ \ \) \(\ds \left({\frac 1 2}\right)!\) \(=\) \(\ds \frac {\sqrt \pi} 2\)

$\blacksquare$


Sources