idku
  • idku
What is going on here? (Stirling's approximation of n!)
Mathematics
chestercat
  • chestercat
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e.mccormick
  • e.mccormick
Inquiring minds want to know.... what is the rest of this question?
idku
  • idku
Ok, there Stirling's approximation of the factorial \(\large \displaystyle n! {\LARGE\text{ ~}} ~\sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n\) and I take two limits to test how reasonable it is, and this is what I get: \(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~-~n! \right]=-\infty\) \(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \frac{\sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~}{~n!} \right]=1\)
idku
  • idku
I can see Stirling's is smaller from the first limit, but why is then 2nd limit =1? (btw, the source is wolfram)

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anonymous
  • anonymous
i'm not sure you're first limit is correct.
idku
  • idku
http://www.wolframalpha.com/input/?i=Lim+n%E2%86%92%E2%88%9E+%28%E2%88%9A%282%CF%80n%29%E2%80%A2%28%28n%2Fe%29%5En%29-%28n%21%29%29
idku
  • idku
wolfram said so
idku
  • idku
And if you subtract Stirling's approximation from factorial: n! - Stirling's approximation , then the limit as n-> infinity , = infinity.
idku
  • idku
http://www.wolframalpha.com/input/?i=Lim+n%E2%86%92%E2%88%9E+%28%28n%21%29-%E2%88%9A%282%CF%80n%29%E2%80%A2%28%28n%2Fe%29%5En%29%29
ikram002p
  • ikram002p
so basically n! is increasing faster than (n)^{n+1/2} :)
ikram002p
  • ikram002p
@ganeshie8 check this
idku
  • idku
Yes, basically plain factorial is increasing faster than stirlings. I guess for limits as n-> infinity, I can only use stirling in a denominator if I want to shjow convergence.
idku
  • idku
show*
ganeshie8
  • ganeshie8
whats the question
idku
  • idku
My question is why this limit is (- infinity) \(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~-~n! \right]=-\infty\) and this limit is 1 for some reason \(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \frac{\sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~}{~n!} \right]=1\)
ganeshie8
  • ganeshie8
let me ask you a question
ganeshie8
  • ganeshie8
\(\lim\limits_{n\to\infty} (2n-n) = ?\) \(\lim\limits_{n\to\infty} \dfrac{2n}{n} = ?\)
idku
  • idku
Yes, yes I think I found why 1000-100 10000-1000 100000-10000 1×10^(n+1)-1×10^n = ∞ but if you divide it is just 10 for every one of them
ikram002p
  • ikram002p
this what u need to explain http://prntscr.com/8977k3
ganeshie8
  • ganeshie8
right, good question to spend time on !
idku
  • idku
\(\large \displaystyle \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~~~?~~~~n! \) \(\large \displaystyle \sqrt{2\pi n}\cdot n^n~~~?~~~~e^nn! \) I guess like this. \(\large \displaystyle \sqrt{2\pi}\cdot n^{n+0.5}~~~?~~~~e^nn! \) we can disregard coefficient 2pi for this, perhaps. \(\large \displaystyle n^{n+0.5}~~~?~~~~e^nn! \)
idku
  • idku
so two growths verses one, but still kind of strange I thought that n^n is the boss
ikram002p
  • ikram002p
well in the long time no one can beats factorial :3
idku
  • idku
No, we do know that n^n>n!
ikram002p
  • ikram002p
what about e^n ?
idku
  • idku
yes, so with e^n together factorial is greater, and my logic of n^n being the boss doesn't apply
idku
  • idku
Well, \[\frac{e}{n}\times\frac{2e}{n}\times\frac{3e}{n}\times\frac{4e}{n}\times...\frac{(n-1)e^{n-1}}{n}\times\frac{e^nn}{n}\] so that e^n * n! can be in a sense intuitively larger if you think about it...
idku
  • idku
oh, my bad
idku
  • idku
wait, I am mixinging something up
idku
  • idku
\[\frac{n!~e^n}{n^n}=\frac{e}{n}\times\frac{2e}{n}\times\frac{3e}{n}\times\frac{4e}{n}\times...\frac{(n-1)e}{n}\times\frac{en}{n}\] just this.
idku
  • idku
i don't know intuitively n^n still rules, but it will take me a while to digest that and actually think about the truth and make some sense of it. Thanks guys!
ganeshie8
  • ganeshie8
|dw:1440615166212:dw|
idku
  • idku
well, that is very beginning tho...... and the difference will grow infinitely...
idku
  • idku
in any case.... I guess will just take the fact and some reasoning without complete comprehension. THere are times when we humans have to do that.
idku
  • idku
Tnx for your input ganeshie and ikram

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