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Stirling's approximation for Gamma functions with a negative argument

  1. Jul 19, 2012 #1
    Hi, fellow physicists (to be). This is my first post on the forum, so I hope I get it right. If not so, please let me know :)

    introduction to the problem
    At the moment I am working on my physics bachelor's thesis at the theoretical department of my university (Amsterdam). My thesis focusses on a certain supersymmetric 1D lattice model, on which spinless fermions can be placed. While working in Mathematica, I found an exact expression for the occupation of sub-levels on the chain of length l (l being the number of sites on the lattice). The expression I found (f(l)), however, is a product over indices, which is not very insightful. My supervisor has asked me to rewrite this (f(l)) into a power of l. To do this, he said that I should use Stirling's approximation.

    The problem
    The function I found is
    \begin{equation}
    f(l) = \frac{2}{5} {(-1)}^l \prod_{i=0}^{l-3} \frac{3(l-i)-2}{3(l-i)-1}.
    \end{equation}

    While being not skilled in rewriting this, Mathematica rewrote f(l) in terms of Gamma functions. leaving out the pre-factors, I am left with
    \begin{equation}
    f(l)={(-1)}^l \frac{3l-2}{3l-1} \frac{\Gamma(\frac{4}{3}-l)}{\Gamma(\frac{5}{3}-l)}.
    \end{equation}

    On this expression I wanted to use Sterling's approximation for Gamma functions:
    \begin{equation}
    \Gamma(z)=\sqrt{\frac{2π}{z}}(\frac{z}{e})^{z}.
    \end{equation}
    The trouble that arises is that this approximation is only valid for positive arguments z of Gamma.

    Attempt to Solve
    To solve this, I tried to use the recursion relation

    \begin{equation}
    \Gamma(z)=\frac{\Gamma(z+1)}{z}
    \end{equation}

    If I use this recursion relation, however, I'm back at the indexed product I started with.

    Does anyone on this forum know of a different approach, some sparks of creativity or otherwise good tips? All help is very welcome.

    greetings,

    Ludo
     
  2. jcsd
  3. Jul 19, 2012 #2
    Re: Sterling's approximation for Gamma functions with a negative argument

    I misspelled 'Stirling' in the title, can an admin maybe change this?
     
  4. Jul 19, 2012 #3
    Re: Sterling's approximation for Gamma functions with a negative argument

    There definitely exists an expression which is more handy than the one Mathematica gave you... If you just write out the product, you get
    [tex] f(l) = \frac{2}{5} (-1)^l \frac{7 \cdot 10 \cdot 13 \cdot ... \cdot (3l-2)}{8 \cdot 11 \cdot 14 \cdot ... \cdot (3l-1)} = (-1)^l \frac{\Gamma(2/3) \Gamma(l+1/3)}{\Gamma(1/3) \Gamma(l+2/3)}, [/tex]
    where the last step is from wolfram alpha :) It's difficult to estimate if the two expressions are the same without knowing the prefactors, but this certainly looks more appealing in terms of using Stirling.
     
  5. Jul 19, 2012 #4

    gabbagabbahey

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    Re: Sterling's approximation for Gamma functions with a negative argument

    You can also get your expression into the same form as the one clamtrox posted by use Euler's reflection formula:

    [tex] \Gamma(1-z) \Gamma(z)=\frac{ \pi }{ \sin( \pi z ) }[/tex]
     
  6. Jul 19, 2012 #5
    Re: Sterling's approximation for Gamma functions with a negative argument

    Thanks, Clamtrox and gabbagabbahey (Gabba as in the music genre?) for the useful comments. With a positive-argument Gamma function, my world is a lot brighter :)
     
  7. Jul 19, 2012 #6

    gabbagabbahey

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    Re: Sterling's approximation for Gamma functions with a negative argument

    You're welcome! (And my handle is taken from the lyrics to a Ramones track)
     
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