Gamma Function of 4
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Theorem
Let $\Gamma$ denote the Gamma function.
Then:
- $\map \Gamma 4 = 6$
Proof
\(\ds \map \Gamma 4\) | \(=\) | \(\ds \map \Gamma {3 + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3 \map \Gamma 3\) | Gamma Difference Equation | |||||||||||
\(\ds \) | \(=\) | \(\ds 3 \times 2\) | Gamma Function of 3 |
$\blacksquare$
Sources
- 1980: Angela Dunn: Mathematical Bafflers (revised ed.) ... (previous) ... (next): $1$. Say it with Letters: Algebraic Amusements: Four Fours