Engineering Mathematics: Gamma-Catalan Relation

GC.ID.1: Catalan Number Identity

$C_{n}=\frac{4^{n }\Gamma (n+\frac{1}{2})}{\sqrt{\pi }\Gamma(n+2)}$

Proof

based on recurrence relation of gamma

$\Gamma (n+\frac{1}{2}) = (n-\frac{1}{2})\Gamma (n-\frac{1}{2})$

(1)

$\Gamma (n+\frac{1}{2}) = (n-\frac{1}{2})(n-\frac{3}{2})\Gamma (n-\frac{3}{2})$

(2)

$\Gamma (n+\frac{1}{2}) = (n-\frac{1}{2})\cdot \cdot \cdot (n-\frac{2n-1}{2})\Gamma (n-\frac{2n-1}{2})$

(3)

$\Gamma (n+\frac{1}{2}) = (\frac{2n-1}{2})\cdot \cdot \cdot (\frac{2n-2n+1}{2})\Gamma (\frac{1}{2})$

(4)

$\Gamma (n+\frac{1}{2}) = (\frac{2n-1}{2})\cdot \cdot \cdot (\frac{1}{2})\Gamma (\frac{1}{2})$

(5)

$\Gamma (n+\frac{1}{2}) = \frac{1.3.5....(2n-1)}{2^{n}}\Gamma (\frac{1}{2})$

(6)

$\Gamma (n+\frac{1}{2}) = \frac{1.3.5....(2n-1)}{2^{n}}\sqrt{\pi }$

(7)

$\Gamma (n+\frac{1}{2}) = \frac{(2n)!}{2^{n (2.4.6....(2n))}}\sqrt{\pi }$

(8)

$\Gamma (n+\frac{1}{2}) = \frac{(2n)!}{2^{n }(2.4.6....2n)}\sqrt{\pi }$

(9)

$\Gamma (n+\frac{1}{2}) = \frac{(2n)!}{4^{n }(1.2.3....n)}\sqrt{\pi }$

(10)

$\Gamma (n+\frac{1}{2}) = \frac{(2n)!}{4^{n }n!}\sqrt{\pi }$

(11)

$\frac{\Gamma (n+\frac{1}{2})}{(n+1)!} = \frac{(2n)!}{4^{n }n!(n+1)!}\sqrt{\pi }$

(12)

$\frac{\Gamma (n+\frac{1}{2})}{\Gamma(n+2)} = \frac{(2n)!}{4^{n }n!(n+1)!}\sqrt{\pi }$

(13)

based on the Catalan number identity , therefore ,

$C_{n}=\frac{(2n)!}{n!(n+1)!}$

(14)

$\frac{\Gamma (n+\frac{1}{2})}{\Gamma(n+2)} = \frac{C_{n}}{4^{n }}\sqrt{\pi }$

(15)

$C_{n}=\frac{4^{n }\Gamma (n+\frac{1}{2})}{\sqrt{\pi }\Gamma(n+2)}$

(16)

Engineering Mathematics