Has anyone seen this problem before?

  • Thread starter ZioX
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  • #1
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Main Question or Discussion Point

Cute solution.

[tex]\lim_{n\to+\infty}e^{-n}\sum_{i=n}^{+\infty}\frac{n^i}{i!}[/tex]
 

Answers and Replies

  • #2
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Has anyone? Is anyone working on this? I can give you a hint if it's needed. I thought this was really cool when I saw it the other day and I figured I would share it. Wasn't sure how well known this problem is.
 
  • #3
StatusX
Homework Helper
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Why don't you answer the (my) question in the math Q&A topic in the general math section, then you could post this as a problem.
 
  • #4
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that is the formula for e approximation. i did that in programming, but not approaching infinity, it should be real number.. i might be wrong.
 
  • #5
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that is the formula for e approximation. i did that in programming, but not approaching infinity, it should be real number.. i might be wrong.
Sum begins at n, which is going to infinity.
 
  • #6
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the series is just e^n- s, where s is the partial sum of formal power series e^n truncated when i=n

so the limit becomes
lim n-> inf
1-e^{-n)*s

s goes to e^{n} and the whole thing goes to zero...?
 
  • #7
Gib Z
Homework Helper
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Yea i get the same answer as tim, The whole thing goes to zero.

Edit: Working-

[tex]\lim_{n\to\infty}e^{-n}\sum_{i=n}^{\infty}\frac{n^i}{i!}[/tex]

[tex]\lim_{n\to\infty}e^{-n}(e^n - \sum_{i=0}^n \frac{x^n}{n!})[/tex]

[tex]\lim_{n\to\infty}(1 - \frac{\sum_{i=0}^n \frac{x^n}{n!}}{e^n})[/tex]

[tex]1 - \frac{\lim_{n\to\infty}\sum_{i=0}^n \frac{x^n}{n!}}{\lim_{n\to\infty} e^n}[/tex]

[tex]1 - \frac{\sum_{n=0}^{\infty} \frac{x^n}{n!}}{e^n}[/tex]

[tex]1 - \frac{e^n}{e^n}[/tex]

Which is zero. I think i included enough obvious steps for everyone to follow :)
 
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  • #8
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Woah! Are you sure you can go here, man?

[tex]1 - \frac{\sum_{n=0}^{\infty} \frac{x^n}{n!}}{e^n}[/tex]

I can give you guys a hint: I could've posted this in the probability forum.
 
  • #9
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if you graph this, this equation the domain is approximately (-inf, inf), while its range is the same which is infinity. ziox is right it's not zero, but if you take n as real number the equation solves the e approximation in e^n.
 
  • #10
ssd
268
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[tex]\lim_{n\to\infty}e^{-n}\sum_{i=n}^{\infty}\frac{n^i}{i!}[/tex]

[tex]\lim_{n\to\infty}e^{-n}(e^n - \sum_{i=0}^n \frac{x^n}{n!})[/tex]
First correction
[tex]\lim_{n\to\infty}e^{-n}\sum_{i=n}^{\infty}\frac{n^i}{i!}[/tex]

[tex]\lim_{n\to\infty}e^{-n}(e^n - \sum_{i=0}^{n-1} \frac{n^i}{i!})[/tex]
 
  • #11
ssd
268
6
By the way, is the limit = 0.5 = P[Z>0], Z~N(0,1).
 
Last edited:
  • #12
370
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Yes.

CLT for the win.

My statistician friend gave me the problem, and I hacked away at it for a good 5 hours and finally got a very non-trivial solution.
 
Last edited:

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