Gamma Function of Negative Half-Integer

Theorem
where:
 * $-m + \dfrac 1 2$ is a half-integer such that $m > 0$
 * $\Gamma$ denotes the Gamma function.

Proof
Proof by induction:

For all $m \in \Z_{> 0}$, let $\map P m$ be the proposition:
 * $\map \Gamma {-m + \dfrac 1 2} = \dfrac {\paren {-1}^m 2^{2 m} m!} {\paren {2 m}!} \sqrt \pi$

Basis for the Induction
$\map P 1$ is the case:

and so $P(1)$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\map \Gamma {-k + \dfrac 1 2} = \dfrac {\paren {-1}^k 2^{2 k} k!} {\paren {2 k}!} \sqrt \pi$

Then we need to show:
 * $\map \Gamma {-\paren {k + 1} + \dfrac 1 2} = \dfrac {\paren {-1}^{k + 1} 2^{2 \paren {k + 1} } \paren {k + 1}!} {\paren {2 \paren {k + 1} } !} \sqrt \pi$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Finally:

Therefore: