Kainui
  • Kainui
I call this problem "the big kid's club": \[\Large \int_0^\infty (1+\frac{x}{n})^n e^{-x}dx + \int_0^n (1-\frac{x}{n})^n e^x dx\]
Calculus1
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
sweet baby jesus
IrishBoy123
  • IrishBoy123
.
anonymous
  • anonymous
is this calculus

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Kainui
  • Kainui
WHOOPS! Slight mistake in the signs ahehehe... \[\Large \int_0^\infty (1+\frac{x}{n})^n e^{-x}dx + \int_0^n (1-\frac{x}{n})^n e^x dx\]
Kainui
  • Kainui
I just went through it again to make sure I didn't make any mistakes in coming up with this thing.
geerky42
  • geerky42
Binomial theorem? I don't think it is hard, but it is very tedious...
freckles
  • freckles
\[\int\limits_{0}^{n}[(1+\frac{x}{n})^n e^{-x}+(1-\frac{x}{n})^n e^{x}] dx+\int_n^\infty (1+\frac{x}{n})^n e^{-x}dx\] I wonder if we can do something with this
Kainui
  • Kainui
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.
Kainui
  • Kainui
@freckles That looks like it might be promising. :D
Kainui
  • Kainui
Well... It's on the right track, considering the bounds of the integrals, I'll say that much.
freckles
  • freckles
did you get this from playing around with some stirling type stuff?
freckles
  • freckles
\[\int\limits_0^n n^{-n}e^{-x}(e^{2x}(n-x)^n+(n+x)^n) dx+\int\limits_n^\infty (1+\frac{x}{n})^n e^{-x}\] like some of the factors in the first integrand look similar to some factors in integrand on that one stirling page
Kainui
  • Kainui
Yeah, I did actually, specifically the formula in here they have for the integral to n! \[n! = \int_0^\infty x^ne^{-x}dx\]
anonymous
  • anonymous
So what language is this called I know its not English...
anonymous
  • anonymous
i knew it was something to gamma :-\
Kainui
  • Kainui
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
dinamix
  • dinamix
@Kainui we can use Integration by parts right ?
anonymous
  • anonymous
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
freckles
  • freckles
i think n is a positive integer
freckles
  • freckles
though he said something about if you find a formula for n being a negative integer he would be really impressed
dan815
  • dan815
i already know the solution so i wont comment
ganeshie8
  • ganeshie8
let \(x=-u \) for the first integral, \[\begin{align} &\int_0^\infty (1+\frac{x}{n})^n e^{-x}dx + \int_0^n (1-\frac{x}{n})^n e^x dx\\~\\ &=\int_{-\infty}^0 (1-\frac{u}{n})^n e^{u}du + \int_0^n (1-\frac{x}{n})^n e^x dx\\~\\ &=\int_{-\infty}^n (1-\frac{x}{n})^n e^{x}dx \\~\\ &=ne^n\int_0^{\infty}t^n e^{-nt}dt \\~\\ &=e^n\dfrac{\Gamma(n+1)}{n^n} \end{align}\]
freckles
  • freckles
wow you made it look not so terrible
ganeshie8
  • ganeshie8
Haha I hope it is right, didn't verify if it really works...
freckles
  • freckles
works for n=2
anonymous
  • anonymous
@ganeshie8 it is beautiful and not crazy!! :)
Kainui
  • Kainui
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.
anonymous
  • anonymous
@Kainui Brilliant i was waiting to hear why u called it that more than the way to solve xD
SolomonZelman
  • SolomonZelman
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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