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Computing the sum of a particular series.

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Capture.PNG

    3. The attempt at a solution

    Alright, so, I'm clueless about doing this one. I do know that it's extremely similar to e^x

    [itex]\sum_{n->0}^{\infty}\frac{x^n}{n!}[/itex]

    But really, that means nothing! Usually there's another function/series I can compare and then integrate, in this case I don't have any idea on how to procceed.

    Can someone tip me in the right direction, please?

    Thanks in advance!
     
  2. jcsd
  3. Jan 29, 2013 #2

    mfb

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    Staff: Mentor

    Those quotation marks are "n"?

    ##\sum_{n=0}^\infty \frac{(x-1)^n}{(n+2)!} = \sum_{n=2}^\infty \frac{(x-1)^{n-2}}{n!}
    = \frac{1}{(x-1)^2} \sum_{n=2}^\infty \frac{(x-1)^n}{n!}##
    Just 2 steps left to get an exponential function.
     
  4. Jan 29, 2013 #3
    You can try to use the Poisson distribution find the solution from there.

    In other words,

    multiply your summation with the e^(-x)/(e^(-x)). Move the denominator out. We know that the summation(e^(-x) * x^n/ n!) must be 1 since it is a distribution (Poisson). So the answer is 1/(e^-x).
     
  5. Jan 29, 2013 #4
    I think they're "n", I'm not absolutely sure, but since the ebook was scanned with auto recognition, it's very possible that when scanned the "n" was with a poor print, and it recognized as ", it's the only explanation I could come up with.

    Regarding the answer, whoa, I never heard of shifting the start of the sum to do that, it was very clever, so the answer would be:
    ##
    \frac{1}{(x-1)^2} \sum_{n=2}^\infty \frac{(x-1)^n}{n!} = \frac{1}{(x-1)^2}*(e^{(x-1)} - \frac{x-1}{6} - \frac{1}{2}) = \frac{-ex+6e^x-2e}{6e(x-1)^2}##

    Is this correct?

    I haven't learned Poisson distribution yet, I'm going to see if I find this further in the book.
     
  6. Jan 30, 2013 #5

    mfb

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    I think the denominators for (x-1) and 1 are wrong.
    And I would not split that factor of e at e^(x-1), that looks weird (at least to me).
     
  7. Jan 30, 2013 #6
    Ohh, I think I get it, I used the wrong term: (N+2)! , I should have used N!

    So in fact:
    ##
    \frac{1}{(x-1)^2} \sum_{n=2}^\infty \frac{(x-1)^n}{n!} = \frac{1}{(x-1)^2}*(e^{(x-1)} - \frac{x-1}{1!} - \frac{1}{0!}) = \frac{e^x - ex}{e(x-1)^2}##

    Perfect!Thanks!

    Why do you say it looks weird? You have a more elegant solution in mind?
     
  8. Jan 30, 2013 #7

    mfb

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    Well, that is just a matter of taste (and you have less e hanging around now), but I would write it as $$\frac{e^{x-1}-x}{(x-1)^2}$$
     
  9. Jan 30, 2013 #8
    Hmm indeed.

    Well, thanks you for all your help so far mfb! That was great.
     
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