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I need to understand what the summation below mean. I specifically do not understand the $i<j$ below the summation.

$$\sum_{\substack{ i<j}}^{N} d_{ij}$$

Insight about the equation There are $N$ vectors and $d_{ij}$ is the Euclidean distance between two vectors $i,j \in N $

So if $N=5$ how would the summation work. I would be grateful if it could be explained in numbers so I exactly know how the summation happen.

Additional Information: I encountered this notation in this paper. Looking at your answers and the content of the paper I think the meaning intended is

$$\sum_{j = 1}^{N} \sum_{i<j} d_{ij} $$

Final edit: There are a number of answers which fully explains the notation. Since I have to choose one answer as the best I had to choose the one which explains the notation in detail and thank you all for answering.

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    $\begingroup$ Personally, I think that notation stinks. The $N$ is clearly intended to be like the $N$ in $\sum_{i=1}^N$, but the $i<j$ below the summation sign defeats that interpretation. It would have been better to leave it as $\sum_{i<j} d_{ij}$ and let $N$ come from the context. $\endgroup$ – Harald Hanche-Olsen Aug 6 '13 at 10:04
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    $\begingroup$ Note that in the linked paper the summation the notation $\sum\limits_{i<j}$ occurs in most cases without the upper $N$; even in the one case where the $N$ appears (in formula$~(1)$), there is another similar summation (in the denominator just before) where it doesn't occur, while that summation has precisely the same interpretation (double sum over $i$ and $j$, restricted to $i<j$). This suggests that this one occurrence is a simple typographical error. $\endgroup$ – Marc van Leeuwen Aug 6 '13 at 15:25
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This is a bad notation: a summation expresssion should either involve one or more relations that the summation index has to satisfy, or specify the lower and upper bound of summation over a contiguous interval, according to the format $\sum_{i=a}^b\ldots$, but the two cannot be mixed; notably if the relation $i<j$ is used below, there is no possible interpretation for the $N$ above as upper bound of an interval.

Even without the misplaced upper $N$, the notation could mean two different things, depending on the context. As you wrote it neither $i$ nor $j$ seem to have been introduced, and this might suggest that both are to be varied in the summation. However it is far more common to use a single relation only to introduce a single index of summation, necessarily the first one mentioned (here $i$, rather than $j$). Then the summation would only be over $i$, from some unspecified lower bound (presumably$~1$, but it could be$~0$ in certain contexts) up to $j-1$, where the value of $j$ is assumed to be already fixed at that point.

Assuming that the former interpretation was meant (double summation) it would be much better to write it as $$ \sum_{j=1}^N \sum_{i<j} d_{i,j}, $$ which makes the double summation obvious; the possibility of this better alternative makes it unlikely to me that the former interpretation was actually meant. The latter interpretation mainly suffers from the impossibility to give any meaning to the upper bound $N$ (since $i$ is already bounded by $j$); nevertheless it would be my bet that this was the intended interpretation.

I'll finish with one more alternative for the double sum expression (the sum of all distances between unordered pairs of distinct points): $$ \sum_{1\leq i<j\leq N} d_{i,j}; $$ here the double-sum aspect of the notation is suggested by the presence of more than one relation (in fact three) in the expression under the summation sign.

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  • $\begingroup$ Thank you for this detailed explanation. By going through your interpretation the first meaning is the one relevant to me. This is the paper with this notation link which prompted the question. $\endgroup$ – Synex Aug 6 '13 at 10:19
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It's shorthand for $$ \sum_{j=1}^N \sum_{i=1}^{j-1}d_{ij}$$

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    $\begingroup$ Alternatively, it is shorthand for: $$ \sum_{i=1}^N \sum_{j=i+1}^N d_{ij} $$ $\endgroup$ – Adriano Aug 6 '13 at 9:56
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I think it should be written as

$$\sum_{1\leq i < j \leq N} {d_{ij}}. $$

This way, it's clear that you sum over all $i,j$ satisfying the inequality $1\leq i < j \leq N$.

If $N=5$, this is equal to

$$d_{12} + d_{13} + d_{14} + d_{15} + d_{23} + d_{24} + d_{25} + d_{34} + d_{35} + d_{45}.$$

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$d_{12}+d_{13}+d_{14}+d_{15}+d_{23}+d_{24}+d_{25}+d_{34}+d_{35}+d_{45}$

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  • $\begingroup$ That's the rightest. The notation means "go through every combination of $i$ and $j$ values where they're both $\leq N$ and $i<j$" $\endgroup$ – Džuris Aug 6 '13 at 9:41

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