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idku
 one year ago
What is going on here? (Stirling's approximation of n!)
idku
 one year ago
What is going on here? (Stirling's approximation of n!)

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e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0Inquiring minds want to know.... what is the rest of this question?

idku
 one year ago
Best ResponseYou've already chosen the best response.1Ok, 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
 one year ago
Best ResponseYou've already chosen the best response.1I can see Stirling's is smaller from the first limit, but why is then 2nd limit =1? (btw, the source is wolfram)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm not sure you're first limit is correct.

idku
 one year ago
Best ResponseYou've already chosen the best response.1And if you subtract Stirling's approximation from factorial: n!  Stirling's approximation , then the limit as n> infinity , = infinity.

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1so basically n! is increasing faster than (n)^{n+1/2} :)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 check this

idku
 one year ago
Best ResponseYou've already chosen the best response.1Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.1My 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
 one year ago
Best ResponseYou've already chosen the best response.2let me ask you a question

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(\lim\limits_{n\to\infty} (2nn) = ?\) \(\lim\limits_{n\to\infty} \dfrac{2n}{n} = ?\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1Yes, yes I think I found why 1000100 100001000 10000010000 1×10^(n+1)1×10^n = ∞ but if you divide it is just 10 for every one of them

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1this what u need to explain http://prntscr.com/8977k3

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2right, good question to spend time on !

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \displaystyle \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~~~?~~~~n! \) \(\large \displaystyle \sqrt{2\pi n}\cdot n^n~~~?~~~~e^nn! \) 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
 one year ago
Best ResponseYou've already chosen the best response.1so two growths verses one, but still kind of strange I thought that n^n is the boss

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1well in the long time no one can beats factorial :3

idku
 one year ago
Best ResponseYou've already chosen the best response.1No, we do know that n^n>n!

idku
 one year ago
Best ResponseYou've already chosen the best response.1yes, so with e^n together factorial is greater, and my logic of n^n being the boss doesn't apply

idku
 one year ago
Best ResponseYou've already chosen the best response.1Well, \[\frac{e}{n}\times\frac{2e}{n}\times\frac{3e}{n}\times\frac{4e}{n}\times...\frac{(n1)e^{n1}}{n}\times\frac{e^nn}{n}\] so that e^n * n! can be in a sense intuitively larger if you think about it...

idku
 one year ago
Best ResponseYou've already chosen the best response.1wait, I am mixinging something up

idku
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{n!~e^n}{n^n}=\frac{e}{n}\times\frac{2e}{n}\times\frac{3e}{n}\times\frac{4e}{n}\times...\frac{(n1)e}{n}\times\frac{en}{n}\] just this.

idku
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1440615166212:dw

idku
 one year ago
Best ResponseYou've already chosen the best response.1well, that is very beginning tho...... and the difference will grow infinitely...

idku
 one year ago
Best ResponseYou've already chosen the best response.1in 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
 one year ago
Best ResponseYou've already chosen the best response.1Tnx for your input ganeshie and ikram
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