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Calculating a double summation

  1. Jul 10, 2011 #1

    Can every body solve this problem in terms of [itex] a [/itex] and [itex] b [/itex].

    [itex]e^{-(a+b)}\sum _{n=0}^{\infty} \frac{a^n}{n!} \sum_{m=0}^{n}\frac{b^m}{m!}[/itex]

    Thanks in advance for your participation.
  2. jcsd
  3. Jul 10, 2011 #2


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    hi sabbagh80! :smile:

    hmm … looks a bit like a binomial expansion, doesn't it? :rolleyes:

    it's over all pairs (m,n) with m ≤ n

    try rearranging the ∑∑ into a ∑∑ over different variables :wink:
  4. Jul 10, 2011 #3
    I mean that is it possible to further simplify this expression or maybe rewrite it by a more known functions. Because computing this expression in a long loop is time consuming process. isn't it?
  5. Jul 10, 2011 #4


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    yes … try rearranging the ∑∑ into a ∑∑ over different variables :smile:
  6. Jul 15, 2011 #5
    Are you meaning
    *. e^−(a+b) * ( ∑a^n/n!) (∑b^m/m!)
    **. e^−(a+b) * ∑ (a^n/n! (∑b^m/m!) )

    For *, we can evaluate each sum separately, then multiply the result. Hint: try a power series formula sheet. * simplifies beautifully.

    I suspect you mean **, where the b_m/m! sum gets evaluated first. Then multiplied by a single "a" term.

    Arggh, almost had it until I realized the sum b^m/m! is not an infinite number of terms, but n terms.
  7. Jul 15, 2011 #6
    Can everybody show me how could I do the above suggestion?
  8. Jul 15, 2011 #7
    What is your meaning, I am confused!
  9. Jul 15, 2011 #8
    Let me first say that even with tinytim's suggestion, I still haven't figured this out!

    Anyway, the idea is straightforward enough. If we sketch out the area over which you are summing with m on the horizontal axis and n on the vertical, then we see the sum is taken over all pairs (m,n) on or above the line m=n (or the line y=x if that is any clearer). Now at the moment, your sum is calculated by picking a value for n and then summing the first n terms of bm. So, using our picture, you're summing horizontally. However, you could also sum vertically, or by diagonals parallel to the line m=n or by diagonals parallel to m=-n.

    That said, I'm not sure which method will make things easier, though tinytim's comment makes me think it might be the last one I mentioned...
  10. Jul 15, 2011 #9


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    hi spamiam! :smile:
    yup …

    instead of summing over all values of m and n,

    sum over all values of m and m+n :wink:
  11. Jul 17, 2011 #10
    ok. didn`t see the `n`.
    Last edited: Jul 17, 2011
  12. Jul 17, 2011 #11
    It is a cute joke!
    m is from 0 to n not to infinity.
    by the way, thanks for your participation.
  13. Jul 19, 2011 #12

    The outline of the derivation is as follows:
    First, check that the double sum converges!

    Next, the summation from m=0 to m=n. Rewrite this as the difference of the summation from m=0 to ∞ and the summation from m=n+1 to ∞. The first member of this difference is eᵇ. For the 2nd member set p:=m-(n+1), where p runs from zero to ∞.

    We may differentiate this 2nd member (n+1) times (= eᵇ). Then (indefinite) integrate this (n+1) times to get,

    (*)- eᵇ + c₁bⁿ/n! + c₂bⁿ⁻¹/(n-1)! +...+cⁿ⁺¹,

    where the `c` are integration constants which may all be set equal to unity.

    Last, multiply Σaⁿ/n! with each term of (*) and sum each resulting term to get the result.

    NOTE: Nobody cares about making computations such as this by hand anymore. They do this by computer.

    Addendum: See, for example, the work of Doron Zeilberger.
    Last edited: Jul 19, 2011
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