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Thread: Gamma and factorial

  1. #1
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    Gamma and factorial

    Is this expression true?
    $\displaystyle (n+s)! \Gamma (s+1)= s! \Gamma (n+s+1) $
    i was looking at the proof of $\displaystyle 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.
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  2. #2
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    Quote Originally Posted by phycdude View Post
    Is this expression true?
    $\displaystyle (n+s)! \Gamma (s+1)= s! \Gamma (n+s+1) $.
    For this to make sense requires that $\displaystyle s$ and $\displaystyle n$ be natural numbers. In which case the gammas may be replaced by factorials and both sides are equal to $\displaystyle s!(n+s)!$

    CB
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  3. #3
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    Quote Originally Posted by phycdude View Post
    Is this expression true?
    $\displaystyle (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: $\displaystyle \Gamma(n+s+1)=(n+s)\Gamma(n+s)=(n+s)(n+s-1)\Gamma(n+s-1)$ $\displaystyle =\cdots=(n+s)\cdots (s+1)\Gamma(s+1)$. On the other hand, it is not uncommon to define $\displaystyle s!=\Gamma(s+1)$ for non-integer $\displaystyle s$, in which case your expression is trivially true.
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  4. #4
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    Quote Originally Posted by Laurent View Post
    What makes sense if s is non-integer and n is positive integer is the following: $\displaystyle \Gamma(n+s+1)=(n+s)\Gamma(n+s)=(n+s)(n+s-1)\Gamma(n+s-1)$ $\displaystyle =\cdots=(n+s)\cdots (s+1)\Gamma(s+1)$. On the other hand, it is not uncommon to define $\displaystyle s!=\Gamma(s+1)$ for non-integer $\displaystyle s$, in which case your expression is trivially true.
    thanks , makes lots of sense now
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