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Hypergeometic function. Questions.

  1. Dec 1, 2015 #1
    Legendre polynomial is defined as
    ##P_n(x)=_2F_1(-n,n+1;1,\frac{1-x}{2}) ##
    Pochammer symbols are defined as ##(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}##. If I undestand well
    [tex]P_n(x)=_2F_1(-n,n+1;1,\frac{1-x}{2}) =\sum^{\infty}_{k=0}\frac{(-n)_k(n+1)_k}{(1)_kk!}x^k [/tex]
    I am not sure what happens with
    [tex] (-n)_k=\frac{\Gamma(-n+k)}{\Gamma(-n)} [/tex]
    because ##\Gamma## diverge for negative integers.
     
  2. jcsd
  3. Dec 1, 2015 #2

    Geofleur

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    Even though ## \Gamma ## diverges for negative integers, the ratio of gamma functions evaluated at different integer arguments does not necessarily diverge. First, note that ## \Gamma(n) = (n-1)! ##. Now write out the terms as follows:

    ## \frac{\Gamma(-n+k)}{\Gamma(-n)} = \frac{(k-n-1)!}{(-n-1)!} = \frac{(k-n-1)(k-n-2)\cdots (k-n-(k-1))(k-n-k)(k-n-(k+1))\cdots }{(-n-1)(-n-2)(-n-3)\cdots}.##

    Therefore, we have

    ## \frac{\Gamma(-n+k)}{\Gamma(-n)} = \frac{(k-n-1)(k-n-2)\cdots (-n+1)(-n)(-n-1)\cdots }{(-n-1)(-n-2)(-n-3)\cdots} = (k-n-1)(k-n-2)\cdots(-n+1)(-n)##,

    and this last result is finite.
     
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