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I recall my teacher mentioned that you can only use the Stirling formula with Maxwell-Boltzmann statistics, not in FD or BE statistics, because of large numbers. How can you then approximate $\ln(x!)$ with FD or BE?

Example approximation with Bose-Einstein statistics

The probability function for Bose-Einstein with $N_i$ particles with $M-1$ walls is

$$P=\Pi_i \frac{(N_i+M-1)!}{N_i!(M-1)!}$$

Now my lecture slides for the physics 2062 in Aalto University claims (p.146 here)

$$\ln P \approx \sum_i \left[\left(N_i+M-1\right)\ln\left(N_i+M-1\right)-(N_i+M-1)-N_i \ln(N_i)+N_i\right]-\ln\left(\left(M-1\right)!\right)$$

where the premise is $\ln(n!)\approx n \ln(n)-n$ but a small err in the constant term, more in the comment.

Now Stirling formula is not precise with small amount of particles but the teacher still uses it with both fermions and bosons. I am uneasy about this: more accurate approximation with smaller systems of particles?

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Out of curiosity, did you ask your teacher? –  Jonas Meyer Jan 8 '13 at 4:54
    
@JonasMeyer Better get peer-previewed answers in Math SE, I like it. –  hhh Jan 8 '13 at 5:04
    
I think there is a small mistake in the step of the example: should it be $i \ln((M-1)!)$ instead (not just $\ln((M-1)!)$? It matters none in the following step when differentiating because of being just a large number but confusing. –  hhh Jan 8 '13 at 5:32
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+1 to the question. Quickly reading through the manuscript left me with the impression that in all the cases, whenever the approximate formula $\ln(n!)\approx n\ln n-n$ was used, the numbers $n$ where large enough not to be cause for concern. After all, even though this is quantum physics, the number of particles in the system is very large. The multiplicative constant $\sqrt{2\pi}$ of the factorial does not affect logarithm much, and will drop out when you take partial derivatives. See the derivation of MB on page 133. –  Jyrki Lahtonen Jan 8 '13 at 6:00
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You can derive better Stirling-like approximations of the form $$n! = \sqrt{2 \pi n} \left(\dfrac{n}{e} \right)^n \left(1 + \dfrac{a_1}n + \dfrac{a_2}{n^2} + \dfrac{a_3}{n^3} + \cdots \right)$$ using Abel summation technique (For instance, see here), where $$a_1 = \dfrac1{12}, a_2 = \dfrac1{288}, a_3 = -\dfrac{139}{51740}, a_4 = - \dfrac{571}{2488320}, \ldots$$ The hard part in Stirling's formula is deriving the $\sqrt{2 \pi}$ leading coefficient. The remaining coefficients $a_k$'s are not hard to obtain.

Posting this as an answer since it was a bit too long for a comment.

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