Binomial Coefficient Stirling's Approximation

Question

I have the following expression, $$\frac{\binom{n}{j}}{\binom{n}{j-l}}$$ I have approximated this and wrote: $$\frac{\binom{n}{j}}{\binom{n}{j-l}}\approx \frac{\frac{n^j}{j!}}{\frac{n^{j-l}}{(j-l)!}}=\frac{n^j(j-l)!}{j!n^{j-l}}=\frac{n^l(j-l)!}{j!}$$ I wonder if above can be further simplify?

Answer 1

The question is a bit misleading, because what you've written isn't a direct result of Stirling's approximation. It's a perfectly reasonable approximation, but it's not Stirling's. Stirling's (at least according to wikipedia) is this:

$n! \approx \sqrt{2\pi n}(\frac{n}{e})^n$

You need additional approximation steps to turn that into what you've got.

Carrying on though regardless, the sensible next step would be to apply the same thing to (j-l)! and j! and you'll end up with something like:

$(\frac{n}{j})^l$

However, I think you're better off expanding first before approximating:

$\binom{n}{j}=\frac{n!}{j!(n-j)!}$

$\binom{n}{j-l}=\frac{n!}{(j-l)!(n-j+l)!}$

So,

$\frac{\binom{n}{j}}{\binom{n}{j-l}}=\frac{(j-l)!(n-j+l)!}{j!(n-j)!}$

That's the numbers from j down to j-l+1 multiplied on the denominator, and the numbers from n-j+l down to n-j+1 multiplied on the numerator. A similar approximation to the one you've used gives:

$(\frac{n-j}{j})^l$

for small l. If j is also small relative to n (as you've assumed it is) then this is also approximately

$(\frac{n}{j})^l$

It would need to be clear which numbers are small relative to which to know which approximations are sensible to apply.

PS also try and avoid the letter el it looks a lot like a one and a one was quite plausible in this context.

Answer 2

Let $k:=j-l$ for convenience. After simplification, Stirling's approximation yields

$$\frac{\displaystyle\binom nj}{\displaystyle\binom nk}\approx\frac{k^{k+1/2}(n-k)^{n-k+1/2}}{j^{j+1/2}(n-j)^{n-j+1/2}}.$$