Factorial of Half/Proof 2
Jump to navigation
Jump to search
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
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.5$: Permutations and Factorials: Exercise $18$