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Kainui
 one year ago
I call this problem "the big kid's club":
\[\Large \int_0^\infty (1+\frac{x}{n})^n e^{x}dx + \int_0^n (1\frac{x}{n})^n e^x dx\]
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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Kainui
 one year ago
Best ResponseYou've already chosen the best response.3WHOOPS! 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
 one year ago
Best ResponseYou've already chosen the best response.3I just went through it again to make sure I didn't make any mistakes in coming up with this thing.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0Binomial theorem? I don't think it is hard, but it is very tedious...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{0}^{n}[(1+\frac{x}{n})^n e^{x}+(1\frac{x}{n})^n e^{x}] dx+\int_n^\infty (1+\frac{x}{n})^n e^{x}dx\] I wonder if we can do something with this

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3I 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
 one year ago
Best ResponseYou've already chosen the best response.3@freckles That looks like it might be promising. :D

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Well... It's on the right track, considering the bounds of the integrals, I'll say that much.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1did you get this from playing around with some stirling type stuff?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\int\limits_0^n n^{n}e^{x}(e^{2x}(nx)^n+(n+x)^n) dx+\int\limits_n^\infty (1+\frac{x}{n})^n e^{x}\] like some of the factors in the first integrand look similar to some factors in integrand on that one stirling page

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Yeah, I did actually, specifically the formula in here they have for the integral to n! \[n! = \int_0^\infty x^ne^{x}dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So what language is this called I know its not English...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i knew it was something to gamma :\

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Haha. 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
 one year ago
Best ResponseYou've already chosen the best response.0@Kainui we can use Integration by parts right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.1i think n is a positive integer

freckles
 one year ago
Best ResponseYou've already chosen the best response.1though he said something about if you find a formula for n being a negative integer he would be really impressed

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i already know the solution so i wont comment

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.9let \(x=u \) 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
 one year ago
Best ResponseYou've already chosen the best response.1wow you made it look not so terrible

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.9Haha I hope it is right, didn't verify if it really works...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 it is beautiful and not crazy!! :)

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.0@Kainui Brilliant i was waiting to hear why u called it that more than the way to solve xD

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Just 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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