Stirling's approximation for Gamma functions with a negative argument

[L. de Pudo]

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

 

[L. de Pudo]

I misspelled 'Stirling' in the title, can an admin maybe change this?

 

[clamtrox]

There definitely exists an expression which is more handy than the one Mathematica gave you... If you just write out the product, you get
$$f(l) = \frac{2}{5} (-1)^l \frac{7 \cdot 10 \cdot 13 \cdot ... \cdot (3l-2)}{8 \cdot 11 \cdot 14 \cdot ... \cdot (3l-1)} = (-1)^l \frac{\Gamma(2/3) \Gamma(l+1/3)}{\Gamma(1/3) \Gamma(l+2/3)},$$
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.

 

[gabbagabbahey]

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

$$\Gamma(1-z) \Gamma(z)=\frac{ \pi }{ \sin( \pi z ) }$$

 

[L. de Pudo]

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 :)

 

[gabbagabbahey]

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