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Math Help - Gamma Function

  1. #1
    Member iPod's Avatar
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    Gamma Function

    I have attached the question.

    I have attempted this question from many angles yet I havn't been successful.
    I would like some help on this.
    Thank you
    Attached Thumbnails Attached Thumbnails Gamma Function-gamma-function.png  
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Gamma Function

    I tried proving it with induction.
    First case: n=0 then we need to prove:
    \Gamma\left(\frac{3}{2}\right)=\frac{\sqrt{\pi}}{2  }
    This is true because
    \Gamma\left(\frac{3}{2}\right)=\Gamma\left(\frac{1  }{2}+1\right) =\frac{1}{2}\Gamma\left(\frac{1}{2}\right) =\frac{\sqrt{\pi}}{2}

    Suppose the statement is true for n=k-1 (induction hypothesis), we accept
    \Gamma\left(\frac{2k+1}{2}\right) =\frac{(2k-1)!}{2^{2k-1}(k-1)!}\sqrt{\pi}

    Proving the statement for n=k, we obtain:
    \Gamma\left(\frac{2k+3}{2}\right) =\Gamma\left(\frac{2k+1}{2}+1\right)=\left(\frac{2  k+1}{2}\right)\Gamma\left(\frac{2k+1}{2}\right)
    =\left(\frac{2k+1}{2}\right)\frac{(2k-1)!}{2^{2k-1}(k-1)!}\sqrt{\pi}
    =\frac{(2k+1)(2k-1)!}{2^{2k}(k-1)!}\sqrt{\pi}
    =\frac{(2k+1)(2k)(2k-1)!}{2^{2k}(2k)(k-1)!} (add a factor 2k in the numerator and the denominator)
    =\frac{(2k+1)!}{2^{2k+1}k!}\sqrt{\pi}

    This is what we wanted to prove.
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