Use Induction to prove

**maxgunn555** · January 10th 2012, 03:22 PM

By Induction prove
$\Gamma (n+\frac{1}{2})\ =\ \frac{1.3.5\cdots (2n-1)}{2^n}\sqrt{\pi}$
Recall $\Gamma (\frac{1}{2})\ =\ \sqrt{\pi}$

I know proof by induction is something to do with putting a +1 by the n's however i am fairly clueless as to this example. Thanks.

ps: the dots between 1.3.5 represent multiplying 1 3 and 5. and the dots after are meant to symbolise a series continuing. i'm also a bit confused as to what that series is... 1, 3, 5 perhaps it's just a difference of two each time?

**FernandoRevilla** · January 10th 2012, 11:21 PM

Using a well known property of the gamma function, write $\Gamma\left(n+1+\frac{1}{2}\right)=\left(n+\frac{1 }{2}\right)\Gamma\left(n+\frac{1}{2}\right)$ .

**chisigma** · January 11th 2012, 02:12 AM

Originally Posted by maxgunn555

By Induction prove
$\Gamma (n+\frac{1}{2})\ =\ \frac{1.3.5\cdots (2n-1)}{2^n}\sqrt{\pi}$
Recall $\Gamma (\frac{1}{2})\ =\ \sqrt{\pi}$

I know proof by induction is something to do with putting a +1 by the n's however i am fairly clueless as to this example. Thanks.

ps: the dots between 1.3.5 represent multiplying 1 3 and 5. and the dots after are meant to symbolise a series continuing. i'm also a bit confused as to what that series is... 1, 3, 5 perhaps it's just a difference of two each time?

Using the basic property of the Gamma Function...

$\Gamma(x+1)=x\ \Gamma(x)$ (1)

... and setting $a_{n}= \Gamma(n+\frac{1}{2})$ You arrive to write the difference equation...

$a_{n+1}= (n+\frac{1}{2})\ a_{n}\ ,\ a_{0}=\sqrt{\pi}$ (2)

As explained in...

http://www.mathhelpforum.com/math-he...-i-188482.html

... the solution of (2) is...

$a_{n}= a_{0}\ \prod_{k=0}^{n-1} \frac{2k+1}{2}= \sqrt{\pi}\ \frac {1 \cdot 3 \cdot ... (2n-1)}{2^{n}}$ (3)

Kind regards

$\chi$ $\sigma$

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