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Math Help - Gamma function

  1. #1
    Junior Member
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    Gamma function

    I want to verify the procedure of finding . \Gamma(n+p+1). with p =-ve. This is usually found in Bessel's equation. It is well talked about if p=+ve. But books I have don't even talking about in general how to find the series representation when p=-ve. I worked this out and I want to verify with you guys whether I am correct for some simple numbers of p= -1/2, -3/2,-5/2.
    I use \Gamma(n+p+1) = \frac{\Gamma(n+p+2)}{ (n+p+1) } as the bases to expand the equations below.





    I start using this identity of p=1/2

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

    \Gamma(n+\frac{1}{2}+1) =((n-1)+\frac{1}{2}+1)((n-2)+\frac{1}{2}+1)....((n-n)+\frac{1}{2}+1)\Gamma(0+\frac{1}{2}+1) = \frac{1}{2} \sqrt{\pi}

    \Gamma(n+\frac{1}{2}+1) = (\frac{0+1+2}{2})(\frac{2+1+2}{2})(\frac{4+1+2}{2}  )...(\frac{(2n-2)+1+2}{2})\Gamma(\frac{0+1+2}{2})


     \Rightarrow \Gamma(n+\frac{1}{2}+1) = (\frac{1}{2})(\frac{3}{2})(\frac{5}{2}).......(\fr  ac{2n+1}{2})\Gamma(\frac{1}{2}) = \frac{(2n+1)!\sqrt{\pi}}{2^{2n+1} n!}





    For p=-1/2, I use the result from above and add on to it:
    \Gamma(n-\frac{1}{2}+1) = \frac{\Gamma(n-\frac{1}{2}+2)}{ (n-\frac{1}{2}+1) } = \frac{\Gamma(n+\frac{1}{2}+1)}{ (n-\frac{1}{2}+1) } = \frac{(2n+1)!\sqrt{\pi}}{2^{2n+1}n! (n-\frac{1}{2}+1) } = \frac{(2n+1)!\sqrt{\pi}}{2^{2n+1}n! (\frac{2n-1+2}{2}) }

    \Rightarrow \Gamma(n-\frac{1}{2}+1) = \frac{(2n+1)!\sqrt{\pi}}{2^{2n}n! (2n+1) } = \frac{(2n)!\sqrt{\pi}}{2^{2n}n! }



    For p=-3/2
    \Gamma(n-\frac{3}{2}+1) = \frac{\Gamma(n-\frac{3}{2}+2)}{ (n-\frac{3}{2}+1) } = \frac{\Gamma(n-\frac{1}{2}+1)}{ (n-\frac{3}{2}+1) } = \frac{\Gamma(n+\frac{1}{2}+1)}{ (n-\frac{3}{2}+1) (n-\frac{1}{2}+1) }


    \Rightarrow \Gamma(n-\frac{3}{2}+1) = \frac{(2n+1)!\sqrt{\pi}}{2^{2n+1}n! (n-\frac{3}{2}+1)(n-\frac{1}{2}+1) } = \frac{(2n+1)!\sqrt{\pi}}{2^{2n+1}n! (\frac{2n-3+2}{2}) (\frac{2n-1+2}{2})}

    \Rightarrow \Gamma(n-\frac{3}{2}+1) = \frac{(2n+1)!\sqrt{\pi}}{2^{2n-1}n! (2n-1)(2n+1) } = \frac{(2n)!\sqrt{\pi}}{2^{2n-1}n! (2n-1)}




    For p=-5/2
    \Gamma(n-\frac{5}{2}+1) = \frac{\Gamma(n-\frac{3}{2}+1)}{ (n-\frac{5}{2}+1) } = \frac{\Gamma(n+\frac{1}{2}+1)}{ (n-\frac{5}{2}+1) (n-\frac{3}{2}+1) (n-\frac{1}{2}+1) }


    \Gamma(n-\frac{5}{2}+1) = \frac{(2n+1)!\sqrt{\pi}}{2^{2n+1}n! (\frac{2n-3}{2}) (\frac{2n-1}{2})(\frac{2n+1}{2})}

    \Rightarrow \Gamma(n-\frac{5}{2}+1) = \frac{(2n+1)!\sqrt{\pi}}{2^{2n-2}n! (2n-3)(2n-1) (2n+1)} = \frac{(2n)!\sqrt{\pi}}{2^{2n-2}n! (2n-3)(2n-1)}


    Please tell me whether I did this correct or not.
    Last edited by yungman; April 18th 2010 at 04:47 PM.
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  2. #2
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    Anyone please?
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