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Math Help - nested summation

  1. #1
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    nested summation

    Is there any way to calculate this efficiently?
    <br />
\sum_{i=min_N}^{max_N}\left\{<br />
\sum_{j=min_{N-1}}^{i-1}\left\{<br />
...\left\{<br />
\sum_{q=min_2}^{p-1} 1<br />
...<br />
\right\}<br />
\right\}<br />
\right\}<br />
    It's only N-1 nested summations where
    the top limit of the next summation, is the iterator variable of the previous.
    Thanx in advance
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  2. #2
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by phsoftnet View Post
    Is there any way to calculate this efficiently?
    <br />
\sum_{i=min_N}^{max_N}\left\{<br />
\sum_{j=min_{N-1}}^{i-1}\left\{<br />
...\left\{<br />
\sum_{q=min_2}^{p-1} 1<br />
...<br />
\right\}<br />
\right\}<br />
\right\}<br />
    It's only N-1 nested summations where
    the top limit of the next summation, is the iterator variable of the previous.
    Thanx in advance
    I hope you are asking the following formula (this is what my signature reads)
    \sum\limits_{i=a}^{b-1}\sum\limits_{j=a}^{i-1}f(j)=\sum\limits_{j=a}^{b-1}\sum\limits_{i=j+1}^{b-1}f(j)
    ................_ =\sum\limits_{i=a}^{b-1}[b-(i+1)]f(i).
    You may easiliy generalize this as follows:
    \sum\limits_{i_{n}=a}^{b-1}\sum\limits_{i_{n-1}=a}^{i_{n}-1}\cdots\sum\limits_{i_{1}=a}^{i_{2}-1}f(i_{1})=\sum\limits_{i=a}^{b-1}\frac{[b-(i+1)]^{(n-1)}}{(n-1)!}f(i),
    where ~^{(\cdot)} stands for the factorial function, i.e. n^{(k)}:=n(n-1)\cdots(n-k+1).
    You may also refer to discrete Taylor's formula.
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