Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am reading an article and in one section it uses stirling's approximation. I decided to do the math and check if it's ok, but I got a different result than the one in the article.

A screenshot of the use of stirling's approximation

Where r,p are natural numbers.

I used the "often written" part in Wikipidia (the one with the 'e' in it)

How did the article got this result ?

share|improve this question
    
We need a direction for a limit I guess. Say, $r$ is fixed but $p \to \infty$. Is that what you want? To tell, we need to know more than just "an article". –  GEdgar Mar 25 '12 at 16:47
    
@GEdgar- you are right, sorry! r is fixed and p tends to infinity –  Belgi Mar 25 '12 at 17:13
add comment

1 Answer

up vote 2 down vote accepted

The result mentioned in the paper you are reading holds in the sense that, for every fixed $r\gt2$, $$ \lim\limits_{p\to\infty}\frac1p\log{(r-1)p\choose p}=\log c(r),\quad \text{with}\quad c(r)=\frac{(r-1)^{r-1}}{(r-2)^{r-2}}. $$ In other words, when $p\to\infty$, $$ {(r-1)p\choose p}=c(r)^{p+o(p)}. $$ Stirling's approximation (which you link to) yields the (stronger, non logarithmic) equivalent $$ {(r-1)p\choose p}=c(r)^p\cdot\frac1{\sqrt{p}}\cdot\sqrt{2\pi}\cdot\sqrt\frac{r-1}{r-2}\cdot(1+o(1)). $$

share|improve this answer
    
What is the base for the log ? if it's p I understand the result in the article. what did you use for your approximation (what did you substitute n! for ?) –  Belgi Mar 25 '12 at 17:12
    
Thank you for the extra explanation! are you sure that it's a small o and not O ? –  Belgi Mar 25 '12 at 17:48
    
Yes. $ $ $ $ $ $ –  Did Mar 25 '12 at 17:53
    
very interesting. Do you have a reference to the version of Stirling's approximation you are using, or maybe you have another reason for this ? (all versions I know are with big O...) –  Belgi Mar 25 '12 at 17:55
    
Yes: the second formula on the WP page you linked to says that $n!=\sqrt{2\pi n}\cdot(n/\mathrm e)^n\cdot(1+o(1))$. You might want to check again your use of Bachmann-Landau notations. –  Did Mar 25 '12 at 18:00
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.