Stirling Approximation for a thermodynamic system

1. Nov 1, 2014

Demroz

1. The problem statement, all variables and given/known data
The stirling approximation, J! = √JJ+1/2e-J, is very handy when dealing with numbers larger than about 100. consider the following ratio: the number of ways N particles can be evenly divided between two halves of a room to the number of ways they can be divided with 60% on the right and 40% on the left.

a. Show using the stirling approximation that the ratio is approximately (0.40.40.60.6/5)n

2. Relevant equations

N!/NR!(N - NR)!

J! = √JJ+1/2e-J

3. The attempt at a solution

So i figured out the ratio to be
(0.5n)!(0.4n)!/(0.5n)!

Then by using the stirling approximation I get

0.6n0.6n+1/20.4n0.4n+1/2 / 0.25nn+1

all of my attempts so far have given me that answer which simplifies to 24/25. can you guys point me in the right direction/show me what I'm doing wrong?

2. Nov 1, 2014

haruspex

No, it's $J^Je^{-J}\sqrt{2\pi J}$. For the present problem the $2\pi$ doesn't matter since it will cancel, but the square root you have at the front is definitely wrong.
A typo in there?
Not exactly sure what you mean by that since you've left out all the parentheses, but the 0.25 looks wrong. Please post your working, including all necessary parentheses.

3. Nov 1, 2014

Demroz

Sorry about that, I screwed up a lot of things in typing it out

JJ+0.5e-J√2pi

My work

probability for a 50/50 distribution is

(N!)/((0.5N)!(0.5N)!)

probability for a 40/60 distribution

(N!)/((0.4N)!(0.6N)!)

when you divide 50/50 over 40/60 you get

((0.4N)!(0.6N)!)/((0.5N)!)2

So when I plug in to the stirling approximation (I reworked it from the previous attempt)

((0.4n0.4n)(0.6n0.6n))/((2-n-1)(nn+1))

Which I was able to simplify to:

(0.40.4n0.60.6n)(2n+1)(n-n)

I have no idea where to go from here

4. Nov 1, 2014

haruspex

Agreed.
That's not what I get. Please post all your working.
You can see from the question what form is required. After correcting the algebra so far, can you find a way to extract the n exponent to be outside as in the desired answer:
You will find there is a constant factor that the n exponent does not apply to, but for these sorts of questions you don't have to worry about that. It's the asymptotic form that matters. E.g. 0.1n may be considered an approximation to 1000*0.1n for large n.
But btw, that answer is incorrect. It should be (0.40.40.60.6/0.5)n

5. Nov 1, 2014

Demroz

I found what I did wrong, thank you very much for the help. It was a simple algebra mistake when distributing exponents

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