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idku

  • one year ago

What is going on here? (Stirling's approximation of n!)

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  1. e.mccormick
    • one year ago
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    Inquiring minds want to know.... what is the rest of this question?

  2. idku
    • one year ago
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    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: \(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~-~n! \right]=-\infty\) \(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \frac{\sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~}{~n!} \right]=1\)

  3. idku
    • one year ago
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    I can see Stirling's is smaller from the first limit, but why is then 2nd limit =1? (btw, the source is wolfram)

  4. anonymous
    • one year ago
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    i'm not sure you're first limit is correct.

  5. idku
    • one year ago
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    wolfram said so

  6. idku
    • one year ago
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    And if you subtract Stirling's approximation from factorial: n! - Stirling's approximation , then the limit as n-> infinity , = infinity.

  7. ikram002p
    • one year ago
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    so basically n! is increasing faster than (n)^{n+1/2} :)

  8. ikram002p
    • one year ago
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    @ganeshie8 check this

  9. idku
    • one year ago
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    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.

  10. idku
    • one year ago
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    show*

  11. ganeshie8
    • one year ago
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    whats the question

  12. idku
    • one year ago
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    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 \(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \frac{\sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~}{~n!} \right]=1\)

  13. ganeshie8
    • one year ago
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    let me ask you a question

  14. ganeshie8
    • one year ago
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    \(\lim\limits_{n\to\infty} (2n-n) = ?\) \(\lim\limits_{n\to\infty} \dfrac{2n}{n} = ?\)

  15. idku
    • one year ago
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    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

  16. ikram002p
    • one year ago
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    this what u need to explain http://prntscr.com/8977k3

  17. ganeshie8
    • one year ago
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    right, good question to spend time on !

  18. idku
    • one year ago
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    \(\large \displaystyle \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~~~?~~~~n! \) \(\large \displaystyle \sqrt{2\pi n}\cdot n^n~~~?~~~~e^nn! \) I guess like this. \(\large \displaystyle \sqrt{2\pi}\cdot n^{n+0.5}~~~?~~~~e^nn! \) we can disregard coefficient 2pi for this, perhaps. \(\large \displaystyle n^{n+0.5}~~~?~~~~e^nn! \)

  19. idku
    • one year ago
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    so two growths verses one, but still kind of strange I thought that n^n is the boss

  20. ikram002p
    • one year ago
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    well in the long time no one can beats factorial :3

  21. idku
    • one year ago
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    No, we do know that n^n>n!

  22. ikram002p
    • one year ago
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    what about e^n ?

  23. idku
    • one year ago
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    yes, so with e^n together factorial is greater, and my logic of n^n being the boss doesn't apply

  24. idku
    • one year ago
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    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...

  25. idku
    • one year ago
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    oh, my bad

  26. idku
    • one year ago
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    wait, I am mixinging something up

  27. idku
    • one year ago
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    \[\frac{n!~e^n}{n^n}=\frac{e}{n}\times\frac{2e}{n}\times\frac{3e}{n}\times\frac{4e}{n}\times...\frac{(n-1)e}{n}\times\frac{en}{n}\] just this.

  28. idku
    • one year ago
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    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!

  29. ganeshie8
    • one year ago
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    |dw:1440615166212:dw|

  30. idku
    • one year ago
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    well, that is very beginning tho...... and the difference will grow infinitely...

  31. idku
    • one year ago
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    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.

  32. idku
    • one year ago
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    Tnx for your input ganeshie and ikram

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