Stirling's approximation for Gamma functions with a negative argument

Click For Summary
The discussion revolves around using Stirling's approximation for Gamma functions with negative arguments in the context of a physics bachelor's thesis focused on a supersymmetric 1D lattice model. The original expression for the occupation of sub-levels, derived in Mathematica, is transformed into a product involving Gamma functions, which poses a challenge since Stirling's approximation is only valid for positive arguments. Contributors suggest rewriting the expression using known identities, such as the recursion relation for Gamma functions and Euler's reflection formula, to facilitate the application of Stirling's approximation. A more manageable expression is proposed, which could potentially simplify the analysis. The thread highlights the collaborative effort to find a solution to the mathematical challenge presented.
L. de Pudo
Messages
3
Reaction score
0
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
 
Physics news on Phys.org


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


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.
 


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 ) }
 


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 :)
 


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

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
0
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
Replies
95
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
12
Views
3K