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Verify Gamma function, not a homework problem

  1. Apr 18, 2010 #1
    I want to verify the procedure of finding .[tex]\Gamma(n+p+1)[/tex]. 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 [tex]\Gamma(n+p+1) = \frac{\Gamma(n+p+2)}{ (n+p+1) }[/tex] as the bases to expand the equations below.



    I start using this identity of p=1/2

    [tex]\Gamma(n+\frac{1}{2}+1) = ((n-1)+\frac{1}{2}+1) \Gamma[(n-1)+\frac{1}{2}+1] = ((n-n)+\frac{1}{2}+1)((n-n+1)+\frac{1}{2}+1).........[(n-1)+\frac{1}{2}+1]\Gamma(0+\frac{1}{2}+1)[/tex]

    [tex] \Rightarrow \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})[/tex]

    [tex]\Gamma(\frac{3}{2})=\frac{1}{2}\Gamma (\frac{1}{2})[/tex]

    [tex] \Rightarrow \Gamma(n+\frac{1}{2}+1) = (\frac{1}{2})(\frac{3}{2})(\frac{5}{2}).......(\frac{2n+1}{2})\Gamma(\frac{1}{2}) = \frac{(2n+1)!\sqrt{\pi}}{2^{2n+1} n!}[/tex]





    For p=-1/2 I use the result from above and add to it:
    [tex]\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}) } [/tex]

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



    For p=-3/2
    [tex]\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) }[/tex]


    [tex]\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})} [/tex]

    [tex]\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)}[/tex]




    For p=-5/2
    [tex]\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) }[/tex]


    [tex]\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})}[/tex]

    [tex]\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)}[/tex]


    Please tell me whether I did this correct or not. There is not example or homework of this in a few of my books but we do use it in Bessel equation expansion. I just work out this formulas.
     
    Last edited: Apr 18, 2010
  2. jcsd
  3. Apr 19, 2010 #2
    Last edited: Apr 19, 2010
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