nested summation

**phsoftnet** · September 20th 2008, 09:47 AM

Is there any way to calculate this efficiently?
$ \sum_{i=min_N}^{max_N}\left\{ \sum_{j=min_{N-1}}^{i-1}\left\{ ...\left\{ \sum_{q=min_2}^{p-1} 1 ... \right\} \right\} \right\} $
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

**bkarpuz** · September 20th 2008, 10:52 AM

Originally Posted by phsoftnet

Is there any way to calculate this efficiently?
$ \sum_{i=min_N}^{max_N}\left\{ \sum_{j=min_{N-1}}^{i-1}\left\{ ...\left\{ \sum_{q=min_2}^{p-1} 1 ... \right\} \right\} \right\} $
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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