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This is the fomula (there might be some terms added, it is just boiled down to these two sums): $$\sum_{i = 0}^{N} \Big(\ Term_i * \sum_{j = 0}^{i - 1} x_j\ \Big)$$ The outer sum will iterate from $0$ to $N$ and the inner one from $0$ to $i - 1$. It has a recursive character. Is the inner sum valid on the first term ($i = 0 $ and $j = -1$)?

Or do I have to rewrite the formula this way: $$Term_0 + \sum_{i = 1}^{N} \Big(\ Term_i * \sum_{j = 0}^{i - 1} x_j\ \Big)$$

This is all about how to write down the formula in a correct manner.

Buni
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  • Depends on what you want to do with initial term. Although upper one makes no sense when i=0 – evil999man Apr 11 '14 at 11:52
  • When $i=0$, the inner sum is empty. This is not a problem. – Gerry Myerson Apr 11 '14 at 12:05
  • But then $j \gt i$ will occour, is this valid? – Buni Apr 11 '14 at 12:06
  • More general: is there a NEED of the starting index $j$ to be less or equal to the ending one? – Buni Apr 11 '14 at 12:13
  • You can define $\sum_{j=0}^{-1} x_j = 0$ or just shift the indeces of $x_j$ so $x_0 = 0$ than you can use $\sum_{j=0}^{i} x_{j}$. or you use the second term which is easier to understand anyway! – mjb4 Apr 11 '14 at 14:07
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    $j>i$ does not occur. The sum is over all $j$ that satisfy $j\ge0$ and, simultaneously, $j\le-1$. There are no such $j$, so the sum is over the empty set --- it's an empty sum --- this is not a problem. – Gerry Myerson Apr 13 '14 at 04:49
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    Thank you @GerryMyerson, I'd like to accept this one as answer, if you post it as one. Also kudos to the other commentators, I apprecciate this! – Buni Apr 14 '14 at 08:27
  • Will do, if I get the time --- but it's OK with me if you want to post it as an answer, now that you understand what's going on. – Gerry Myerson Apr 14 '14 at 14:59

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When $i=0$, the inner sum is taken over all $j$ that satisfy $j\ge 0$ and, simultaneously, $j\le-1$. There are no such $j$, so the sum is over the empty set. The empty sum contributes nothing, i.e., $0$. -- Gerry Myerson (slightly edited)

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