Gamma Function of Positive Half-Integer
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Theorem
\(\ds \map \Gamma {m + \frac 1 2}\) | \(=\) | \(\ds \frac {\paren {2 m}!} {2^{2 m} m!} \sqrt \pi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 \times 3 \times 5 \times \cdots \times \paren {2 m - 1} } {2^m} \sqrt \pi\) |
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 $\map P m$ be the proposition:
- $\map \Gamma {m + \dfrac 1 2} = \dfrac {\paren {2 m}!} {2^{2 m} m!} \sqrt \pi$
Basis for the Induction
$\map P 1$ is the case:
\(\ds \map \Gamma {1 + \frac 1 2}\) | \(=\) | \(\ds \frac 1 2 \map \Gamma {\frac 1 2}\) | Gamma Difference Equation | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt \pi} 2\) | Gamma Function of One Half | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 4 \sqrt \pi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2!} {2^{2 \times 1} 1!} \sqrt \pi\) | Definition of Factorial | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 m}!} {2^{2 m} m!} \sqrt \pi\) | where $m = 1$ |
and so $\map P 1$ 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:
- $\map \Gamma {k + \dfrac 1 2} = \dfrac {\paren {2 k}!} {2^{2 k} k!} \sqrt \pi$
Then we need to show:
- $\map \Gamma {k + 1 + \dfrac 1 2} = \dfrac {\paren {2 \paren {k + 1} }!} {2^{2 \paren {k + 1} } \paren {k + 1}!} \sqrt \pi$
Induction Step
This is our induction step:
\(\ds \map \Gamma {k + 1 + \frac 1 2}\) | \(=\) | \(\ds \paren {k + \frac 1 2} \map \Gamma {k + \frac 1 2}\) | Gamma Difference Equation | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + \frac 1 2} \dfrac {\paren {2 k}!} {2^{2 k} k!} \sqrt \pi\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 k + 1} \paren {2 k}!} {2 \times 2^{2 k} k!} \sqrt \pi\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 k}! \paren {2 k + 1} \paren {2 k + 2} } {2 \paren {2 k + 2} 2^{2 k} k!} \sqrt \pi\) | multiplying top and bottom by $2 k + 2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 k + 2}!} {2^{2 k + 1} \paren {2 \paren {k + 1} } k!} \sqrt \pi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 \paren {k + 1} }!} {2^{2 \paren {k + 1} } \paren {k + 1}!} \sqrt \pi\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Finally:
\(\ds \frac {\paren {2 m}!} {2^{2 m} m!}\) | \(=\) | \(\ds \frac {1 \times 2 \times 3 \times \cdots \times 2 m} {2^{2 m} \ 1 \times 2 \times 3 \times \cdots \times m}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 \times 2 \times 3 \times \cdots \times \paren {2 m - 1} \times 2 m} {2^m \ 2^m \ \paren {1 \times 2 \times 3 \times \cdots \times m} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 \times 2 \times 3 \times \cdots \times \paren {2 m - 1} \times 2 m} {2^m \paren {\paren {2 \times 1} \times \paren {2 \times 2} \times \paren {2 \times 3} \times \cdots \times 2 m} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 \times 2 \times 3 \times \cdots \times \paren {2 m - 1} \times 2 m} {2^m \paren {2 \times 4 \times 6 \times \cdots \times 2 m} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 \times 3 \times 5 \times \cdots \times \paren {2 m - 1} } {2^m}\) |
Therefore:
\(\ds \forall m \in \Z_{>0}: \, \) | \(\ds \map \Gamma {m + \frac 1 2}\) | \(=\) | \(\ds \frac {\paren {2 m}!} {2^{2 m} m!} \sqrt \pi\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 \times 3 \times 5 \times \cdots \times \paren {2 m - 1} } {2^m} \sqrt \pi\) |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $16.6$: Special Values for the Gamma Function
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): gamma function
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): gamma function