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