Question

help pls... gamma functions..

how $$\frac{3}{2}$$ obtained in
$$\Gamma(n+\frac{1}{2})=[n+\frac{1}{2}-1] [n+\frac{1}{2}-2]...........\frac{3}{2}\frac{1}{2}\Gamma(\frac{1}{2})$$

Formula used : $$\Gamma(n+a)=[n+a-1] [n+a-2]......a\Gamma(a)$$

And,how to eliminate $$\frac{3}{2}$$ to get the Ans : $$\frac{1.3.5......(2n-1)\sqrt\pi}{2^n}$$

Answer 1

Is \(n\) a positive integer ?

Answer 2

yes, n is a positive integer

Answer 3

then, as you keep subtracting 1, last few terms look like below
\[\ldots (2+\frac{1}{2}) (1+\frac{1}{2})(0+\frac{1}{2})\Gamma(\frac{1}{2})\]

yes ?

Answer 4

basically we're using below recurrence relation
\[\Gamma(t+1)=t\Gamma(t)\]

Answer 5

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

Answer 6

n + 1/2 -1 = (2n-1) / 2

Answer 7

yes , i got it . n= 0, 1,2,3 from last term ..... thank you!!

Answer 8

Gamma(1/2) = sqrt(pi)

Answer 9

yes , gamma functions is not defined for n=0

Answer 10

okay looks neat :)

Answer 11

actually you can keep going, for instance
Gamma( n + 1/2) = (n + 1/2 -1)...1/2 * -1/2 *-3/2 Gamma(-3/2)

But you might as well stop at 1/2, since that is well known value.