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Has anyone seen this problem before?

  1. Mar 3, 2007 #1
    Cute solution.

  2. jcsd
  3. Mar 4, 2007 #2
    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.
  4. Mar 4, 2007 #3


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    Homework Helper

    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.
  5. Mar 4, 2007 #4
    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.
  6. Mar 4, 2007 #5
    Sum begins at n, which is going to infinity.
  7. Mar 4, 2007 #6
    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

    s goes to e^{n} and the whole thing goes to zero...?
  8. Mar 5, 2007 #7

    Gib Z

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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}(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 :)
    Last edited: Mar 5, 2007
  9. Mar 5, 2007 #8
    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.
  10. Mar 5, 2007 #9
    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.
  11. Mar 5, 2007 #10


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    First correction

    [tex]\lim_{n\to\infty}e^{-n}(e^n - \sum_{i=0}^{n-1} \frac{n^i}{i!})[/tex]
  12. Mar 5, 2007 #11


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    By the way, is the limit = 0.5 = P[Z>0], Z~N(0,1).
    Last edited: Mar 5, 2007
  13. Mar 5, 2007 #12

    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: Mar 5, 2007
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