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Partial exponential sum

  1. Dec 15, 2008 #1
    It is known that
    \[\sum\limits_{k = 0}^\infty {\frac{{N^k }}
    {{k!}}} = e^N

    My question is

    \[\sum\limits_{k = 0}^M {\frac{{N^k }}
    {{k!}}} = ?
    where $M\leq N$ an integer.

    This is not an homework
  2. jcsd
  3. Dec 15, 2008 #2


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    Gold Member

    recheck your code please
  4. Dec 15, 2008 #3
    I do not understand the question. Whether [itex]M \leq N[/itex] or not, as long as M is not infinity, all you have is a Taylor polynomial approximation of eN. The larger M is, the better your approximation.

    Your LaTeX is fine, but the forum requires you to put code in between [ tex ][ /tex ] tags (no spaces) or [ itex ][ /itex ] tags (no spaces) for inline TeX.
  5. Dec 15, 2008 #4
    You should use [tex ][/tex] tags.
    What kind of answer are you looking for? There isn't really a simpler expression, and why the restriction M ≤ N?
  6. Dec 15, 2008 #5
    My empiric results show that if M=N the answer is (e^N)/2. I am looking for any asymptotic approach that gives the solution as M=N.
  7. Dec 15, 2008 #6
    Yes, it does seem to approach that (although of course it's not exact).

    Interestingly, according to Mathematica,
    [tex]\sum_{k = 0}^{n} \frac{x^k}{k!} = e^x \frac{\Gamma(n + 1, x)}{\Gamma(n + 1)}[/tex]
    where [tex]\Gamma(a, x)[/tex] is the incomplete gamma function
    [tex]\Gamma(a, x) = \int_x^\infty t^{a - 1} e^t \,dt[/tex]
    and [tex]\Gamma(a) = \Gamma(a, 0)[/tex].
  8. Dec 15, 2008 #7
    It is ok. I know the proof of this but still i need some approximations
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