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:

$\ds \prod_{n \mathop \ge 1} \dfrac {\paren {n + \alpha_1} \cdots \paren {n + \alpha_k} } {\paren {n + \beta_1} \cdots \paren {n + \beta_k} } = \dfrac {\map \Gamma {1 + \beta_1} \cdots \map \Gamma {1 + \beta_k} } {\map \Gamma {1 + \alpha_1} \cdots \map \Gamma {1 + \alpha_k} }$

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:

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

We then note that by Wallis's Product:

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

Thus:

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

$\blacksquare$


Sources

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