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Kainui

  • one year ago

I call this problem "the big kid's club": \[\Large \int_0^\infty (1+\frac{x}{n})^n e^{-x}dx + \int_0^n (1-\frac{x}{n})^n e^x dx\]

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  1. anonymous
    • one year ago
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    sweet baby jesus

  2. IrishBoy123
    • one year ago
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    .

  3. anonymous
    • one year ago
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    is this calculus

  4. Kainui
    • one year ago
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    WHOOPS! Slight mistake in the signs ahehehe... \[\Large \int_0^\infty (1+\frac{x}{n})^n e^{-x}dx + \int_0^n (1-\frac{x}{n})^n e^x dx\]

  5. Kainui
    • one year ago
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    I just went through it again to make sure I didn't make any mistakes in coming up with this thing.

  6. geerky42
    • one year ago
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    Binomial theorem? I don't think it is hard, but it is very tedious...

  7. freckles
    • one year ago
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    \[\int\limits_{0}^{n}[(1+\frac{x}{n})^n e^{-x}+(1-\frac{x}{n})^n e^{x}] dx+\int_n^\infty (1+\frac{x}{n})^n e^{-x}dx\] I wonder if we can do something with this

  8. Kainui
    • one year ago
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    I didn't use the binomial theorem to solve this, but you're welcome to try. I guess I should also mention that you can keep n as a positive integer if you like, but if you solve it more generally I will be impressed cause I don't know how to do that.

  9. Kainui
    • one year ago
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    @freckles That looks like it might be promising. :D

  10. Kainui
    • one year ago
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    Well... It's on the right track, considering the bounds of the integrals, I'll say that much.

  11. freckles
    • one year ago
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    did you get this from playing around with some stirling type stuff?

  12. freckles
    • one year ago
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    \[\int\limits_0^n n^{-n}e^{-x}(e^{2x}(n-x)^n+(n+x)^n) dx+\int\limits_n^\infty (1+\frac{x}{n})^n e^{-x}\] like some of the factors in the first integrand look similar to some factors in integrand on that one stirling page

  13. Kainui
    • one year ago
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    Yeah, I did actually, specifically the formula in here they have for the integral to n! \[n! = \int_0^\infty x^ne^{-x}dx\]

  14. anonymous
    • one year ago
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    So what language is this called I know its not English...

  15. anonymous
    • one year ago
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    i knew it was something to gamma :-\

  16. Kainui
    • one year ago
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    Haha. I'm going to go down and swim in the lake since it's a beautiful day, in about 30 minutes to an hour I'll be back to give the answer I think or hints... or hopefully even congratulate whoever gets it :P

  17. dinamix
    • one year ago
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    @Kainui we can use Integration by parts right ?

  18. anonymous
    • one year ago
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    i am able to reduce it to one integral.. is there any constraint to 'n' coz if the limit tends to infinite then um getting 1^infinite form

  19. freckles
    • one year ago
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    i think n is a positive integer

  20. freckles
    • one year ago
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    though he said something about if you find a formula for n being a negative integer he would be really impressed

  21. dan815
    • one year ago
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    i already know the solution so i wont comment

  22. ganeshie8
    • one year ago
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    let \(x=-u \) for the first integral, \[\begin{align} &\int_0^\infty (1+\frac{x}{n})^n e^{-x}dx + \int_0^n (1-\frac{x}{n})^n e^x dx\\~\\ &=\int_{-\infty}^0 (1-\frac{u}{n})^n e^{u}du + \int_0^n (1-\frac{x}{n})^n e^x dx\\~\\ &=\int_{-\infty}^n (1-\frac{x}{n})^n e^{x}dx \\~\\ &=ne^n\int_0^{\infty}t^n e^{-nt}dt \\~\\ &=e^n\dfrac{\Gamma(n+1)}{n^n} \end{align}\]

  23. freckles
    • one year ago
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    wow you made it look not so terrible

  24. ganeshie8
    • one year ago
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    Haha I hope it is right, didn't verify if it really works...

  25. freckles
    • one year ago
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    works for n=2

  26. anonymous
    • one year ago
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    @ganeshie8 it is beautiful and not crazy!! :)

  27. Kainui
    • one year ago
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    Yeah that's exactly right! This is why I called it "the big kid's club" because \[\huge \frac{e^n n!}{n^n}\] \(e^n\), \(n!\), \(n^n\) are 3 popular big functions, all in one place haha.

  28. anonymous
    • one year ago
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    @Kainui Brilliant i was waiting to hear why u called it that more than the way to solve xD

  29. SolomonZelman
    • one year ago
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    Just a little addition, that according to Stirling's approximation, \(\large\color{black}{ \displaystyle n! {\LARGE\text{ ~}}\sqrt{2\pi n}\frac{n^n}{e^n} }\) So, when we apply that: \(\large\color{black}{ \displaystyle \frac{\color{red}{n!}e^n}{n^n} =}\)\(\large\color{black}{ \displaystyle \frac{\color{red}{\sqrt{2\pi n}\frac{n^n}{e^n}}e^n}{n^n}= }\)\(\large\color{black}{ \displaystyle \frac{\color{red}{\sqrt{2\pi n}~n^n}}{n^n} =}\)\(\large\color{black}{ \displaystyle \sqrt{2\pi n}}\) But, considering the fact that Factorial is greater than its Stirling's approximation of it, (for all whole number x), your value is really a little greater than \(\large \color{black}{\sqrt{2 \pi n}}\) .

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