Gamma Function of Positive Half-Integer

From ProofWiki
Jump to navigation Jump to search

Theorem

\(\displaystyle \Gamma \left({m + \frac 1 2}\right)\) \(=\) \(\displaystyle \frac {\left({2 m}\right)!} {2^{2 m} m!} \sqrt \pi\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {1 \times 3 \times 5 \times \cdots \times \left({2 m - 1}\right)} {2^m} \sqrt \pi\)

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 $P \left({m}\right)$ be the proposition:

$\Gamma \left({m + \dfrac 1 2}\right) = \dfrac {\left({2 m}\right)!} {2^{2 m} m!} \sqrt \pi$


Basis for the Induction

$P \left({1}\right)$ is the case:

\(\displaystyle \Gamma \left({1 + \frac 1 2}\right)\) \(=\) \(\displaystyle \frac 1 2 \Gamma \left({\frac 1 2}\right)\) Gamma Difference Equation
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt \pi} 2\) Gamma Function of One Half
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 4 \sqrt \pi\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2!} {2^{2 \times 1} 1!} \sqrt \pi\) Definition of Factorial
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({2 m}\right)!} {2^{2 m} m!} \sqrt \pi\) where $m = 1$

and so $P(1)$ holds.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.


So this is our induction hypothesis:

$\Gamma \left({k + \dfrac 1 2}\right) = \dfrac {\left({2 k}\right)!} {2^{2 k} k!} \sqrt \pi$


Then we need to show:

$\Gamma \left({k + 1 + \dfrac 1 2}\right) = \dfrac {\left({2 \left({k + 1}\right)}\right)!} {2^{2 \left({k + 1}\right)} \left({k + 1}\right)!} \sqrt \pi$


Induction Step

This is our induction step:

\(\displaystyle \Gamma \left({k + 1 + \frac 1 2}\right)\) \(=\) \(\displaystyle \left({k + \frac 1 2}\right) \Gamma \left({k + \frac 1 2}\right)\) Gamma Difference Equation
\(\displaystyle \) \(=\) \(\displaystyle \left({k + \frac 1 2}\right) \dfrac {\left({2 k}\right)!} {2^{2 k} k!} \sqrt \pi\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({2 k + 1}\right) \left({2 k}\right)!} {2 \times 2^{2 k} k!} \sqrt \pi\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({2 k}\right)! \left({2 k + 1}\right) \left({2 k + 2}\right)} {2 \left({2 k + 2}\right) 2^{2 k} k!} \sqrt \pi\) multiplying top and bottom by $2 k + 2$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({2 k + 2}\right)!} {2^{2 k + 1} \left({2 \left({k + 1}\right)}\right) k!} \sqrt \pi\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({2 \left({k + 1}\right)}\right)!} {2^{2 \left({k + 1}\right)} \left({k + 1}\right)!} \sqrt \pi\)


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


Finally:

\(\displaystyle \frac {\left({2 m}\right)!} {2^{2 m} m!}\) \(=\) \(\displaystyle \frac {1 \times 2 \times 3 \times \cdots \times 2 m} {2^{2 m} \ 1 \times 2 \times 3 \times \cdots \times m}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {1 \times 2 \times 3 \times \cdots \times \left({2 m - 1}\right) \times 2 m} {2^m \ 2^m \ \left({1 \times 2 \times 3 \times \cdots \times m}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {1 \times 2 \times 3 \times \cdots \times \left({2 m - 1}\right) \times 2 m} {2^m \left({\left({2 \times 1}\right) \times \left({2 \times 2}\right) \times \left({2 \times 3}\right) \times \cdots \times 2 m}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {1 \times 2 \times 3 \times \cdots \times \left({2 m - 1}\right) \times 2 m} {2^m \left({2 \times 4 \times 6 \times \cdots \times 2 m}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {1 \times 3 \times 5 \times \cdots \times \left({2 m - 1}\right)} {2^m}\)


Therefore:

\(\displaystyle \forall m \in \Z_{>0}: \ \ \) \(\displaystyle \Gamma \left({m + \frac 1 2}\right)\) \(=\) \(\displaystyle \frac {\left({2 m}\right)!} {2^{2 m} m!} \sqrt \pi\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {1 \times 3 \times 5 \times \cdots \times \left({2 m - 1}\right)} {2^m} \sqrt \pi\)

$\blacksquare$


Sources