Gamma and factorial

**phycdude** · March 1st 2010, 03:57 AM

Is this expression true?
$(n+s)! \Gamma (s+1)= s! \Gamma (n+s+1)$
i was looking at the proof of $J_{-n} (x)= (-1)^n J_n (x)$ and i deduced that expression must be true for the proof provided to be sound, the proof was not from a rigourous source so it may have cut corners & have botched up somewhere.

**CaptainBlack** · March 1st 2010, 05:23 AM

Originally Posted by phycdude

Is this expression true?
$(n+s)! \Gamma (s+1)= s! \Gamma (n+s+1)$ .

For this to make sense requires that $s$ and $n$ be natural numbers. In which case the gammas may be replaced by factorials and both sides are equal to $s!(n+s)!$

CB

**Laurent** · March 1st 2010, 05:32 AM

Originally Posted by phycdude

Is this expression true?
$(n+s)! \Gamma (s+1)= s! \Gamma (n+s+1)$

What makes sense if s is non-integer and n is positive integer is the following: $\Gamma(n+s+1)=(n+s)\Gamma(n+s)=(n+s)(n+s-1)\Gamma(n+s-1)$ $=\cdots=(n+s)\cdots (s+1)\Gamma(s+1)$ . On the other hand, it is not uncommon to define $s!=\Gamma(s+1)$ for non-integer $s$ , in which case your expression is trivially true.

**phycdude** · March 1st 2010, 06:03 AM

Originally Posted by Laurent

What makes sense if s is non-integer and n is positive integer is the following: $\Gamma(n+s+1)=(n+s)\Gamma(n+s)=(n+s)(n+s-1)\Gamma(n+s-1)$ $=\cdots=(n+s)\cdots (s+1)\Gamma(s+1)$ . On the other hand, it is not uncommon to define $s!=\Gamma(s+1)$ for non-integer $s$ , in which case your expression is trivially true.

thanks , makes lots of sense now

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