Engineering Mathematics: Oct 28, 2011

Friday, October 28, 2011

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)

Multiple Sine Integration

MSI.ID.1: Recurrence Relation

$I_{n}= \int \sin ^{n}x\, dx$

(1)

$I_{n}= \int \sin ^{n}x\, dx = -\frac{1}{n}\sin ^{n-1}x\cos x + \frac{n-1}{n}I_{n-2}$

(2)

Proof

$I_{n}= \int \sin ^{n}x\, dx = \int \sin ^{n-2}x (1-\cos^{2}x)\, dx$

(3)

$I_{n}= \int \sin ^{n}x \, dx = \int \sin ^{n-2}x \, dx - \int \sin ^{n-2}x \cos^{2}x\, dx$

(4)

$I_{n}= \int \sin ^{n}x \, dx = \int \sin ^{n-2}x \, dx - \int \sin ^{n-2}x \cos x\cos x\, dx$

(5)

$u= \cos x \, , \, du= -\sin x \, dx$

(6)

$dv= \sin^{n-2}x \, dx \, \cos x \, ,\, v= \frac{\sin^{n-1}x}{n-1}$

(7)

$I_{n}= \int \sin ^{n}x \, dx = \int \sin ^{n-2}x \, dx - \frac{1}{n-1}\sin ^{n-1}\cos x- \int \frac{\sin^{n-1}x}{n-1} \, dx$

(8)

$\int \sin ^{n}x \, dx + \frac{1}{n-1}\int \sin^{n-1}x \, dx = \int \sin ^{n-2}x \, dx - \frac{1}{n-1}\sin ^{n-1}\cos x$

(9)

$\frac{n}{n-1}\int \sin^{n-1}x \, dx = \int \sin ^{n-2}x \, dx - \frac{1}{n-1}\sin ^{n-1}\cos x$

(10)

$\int \sin^{n-1}x \, dx = \frac{n}{n-1}\int \sin ^{n-2}x \, dx - \frac{1}{n}\sin ^{n-1}\cos x$

(11)

From (11) and (1)

$I_{n}= \int \sin ^{n}x\, dx = -\frac{1}{n}\sin ^{n-1}x\cos x + \frac{n-1}{n}I_{n-2}$

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