# Sterling approximation of Beta Function

Tags:
1. Nov 1, 2016

### dykuma

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I think this problem is probably a lot simpler than I am making it out to be. However, when I apply sterling's approx., I get a very nasty equation that does not simplify easily.

One of the biggest problems I have though is dealing with that n^x term. it seems to me that as n approaches infinity, the outcome depends heavily on whether x is a positive or negative integer, or if its zero.

2. Nov 1, 2016

### stevendaryl

Staff Emeritus
Rearrange it to:

$n^x B(x,n) =\Gamma(x) n^x \frac{\Gamma(n)}{\Gamma(n+x)}$

So what you're really interested in is the limit of

$n^x \frac{\Gamma(n)}{\Gamma(n+x)}$

Use Sterling's approximation:

$\Gamma(n) \approx n^{n+\frac{1}{2}} \sqrt{2 \pi} e^{-n}$
$\Gamma(n+x) \approx (n+x)^{n+x+\frac{1}{2}} \sqrt{2 \pi} e^{-(n+x)}$

Write $(n+x)^{n+x+\frac{1}{2}}$ as $(n(1+\frac{x}{n}))^{n+x +\frac{1}{2}} = n^{n+x+\frac{1}{2}} (1+\frac{x}{n})^{n+x+\frac{1}{2}}$
$= n^{n+\frac{1}{2}} n^x (1+\frac{x}{n})^{n} (1+\frac{x}{n})^{x+\frac{1}{2}}$

So
$\Gamma(n+x) \approx n^{n+\frac{1}{2}} n^x (1+\frac{x}{n})^{n} (1+\frac{x}{n})^{x+\frac{1}{2}} \sqrt{2 \pi} e^{-(n+x)}$

Now, you can just use a fact about $e$: $e^x = lim_{n \rightarrow \infty} (1+\frac{x}{n})^n$

3. Nov 1, 2016

Suggest using the simplified Stirling formula without the $\sqrt{2 \pi n}$.Then the result follows immediately with $ln(n+x-1)=ln(n)$ in the limit that n approaches infinity... editing...You can include the $\sqrt{2 \pi n}$ but in the limit that n approaches infinity, this puts the same factor in the numerator that it does in the denominator.

Last edited: Nov 1, 2016
4. Nov 1, 2016

### dykuma

Thanks to both of you! The key to solving that problem noticing the what Steven showed, that
Then the entire problem simplifies quite nicely.