Stirling's approximation/Multiplicity

[JaWiB]

Homework Statement

For a single large two-state paramagnet, the multiplicity function is very sharply peaked about $$N_{\uparrow} = N/2$$

a) Use Stirling's approximation to estimate the height of the peak in the multiplicity function.

b) Use the methods in this section to derive a formula for the multiplicity function in the vicinity of the peak, in terms of $$x \equiv N_{\uparrow} - (N/2)$$ Check that your formula agrees with your answer to part (a) when x = 0.

c) How wide is the peak in the multiplicity function?

Homework Equations

Stirling's approximation: $$ln(N!) = N ln(N) - N$$

# of microstates = $$\frac{N!}{N_{\uparrow}!N_{\downarrow}!}$$

The Attempt at a Solution

The first question seems simple enough; I'll just plug in N/2:
$$ln \Omega = ln(N!) - ln((N/2)!) - ln((N/2)!)$$
$$ln \Omega \approx Nln(N) - N - 2((N/2)ln(N/2)-N/2) = Nln(N)-Nln(N/2) = Nln(2)$$

So the answer to (a) is 2^N. This seems problematic though because I think 2^N should be the total number of microstates for all macrostates, not just the macrostate where half the dipoles are "up"...Do I need to include the $$\sqrt{2{\pi}N}$$ term in Stirling's approximation?

For (b), I managed to get the formula:
$$\Omega = N^N[(N/2+x)(N/2-x)]^{-N/2}(\frac{N/2-x}{N/2+x})^x$$
which does give me 2^N where x=0, but I expected to get a gaussian curve, especially considering part (c). But I can't see how I can get an exponential in the final answer because the exponentials seem to cancel out whether or not the square root term is in there.

What am I missing here? Am I on the right track at all?

 

[JaWiB]

So it turns out this is fairly involved but not too complicated. All that's required is using the Taylor series expansion for the natural logarithm and neglecting terms because x<<N (since N is large and the question asks for the behavior "in the vicinity" of the peak)