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I have seen notation like this, commonly in regard to matrix multiplication:

$$ b_i = \sum_{j \ne i} A_{ij} x_j $$

So this is a matrix multiplication that excludes the diagonal ($j=i$).

I'm kind of confused when it's ok to just leave off specification of the upper limit or lower limit from the sum. From here, I'm assuming if you have an exclusion, you're supposed to sum over all elements not excluded?

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It's common to omit when understood. The fact there's an $i$ on the LHS, makes it clear that the sum on RHS is over $j$ (and $i$ is fixed), where $j \in [a, b]$ for some $a,b$ inferred from the context. –  user2468 Aug 22 '12 at 23:07
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Speaking of summation notations, you can also have things like $\large\sum\limits_{d \mid n} d,$ or any other predicate. –  user2468 Aug 23 '12 at 1:53

2 Answers 2

In the above, $i$ is fixed and $j$ varies, so sum over all the $j$'s that are not $i$'s. It is common to omit complete specification when the context makes it clear.

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The notation you give is fairly common, and Jasper explains its meaning. If you wanted to write the same sum in a more explicit (albeit heavy-handed) way, you could write something like $$ b_i = \sum_{j \in J_i} A_{ij}x_j $$ where $J_i = \{ j \in \mathbb{N} \mid 1 \leq j \leq n,\ j \neq i\}$. This notation means to sum over every value of $j$ belonging to the set $J_i$. The subscript $i$ is a way to denote that you will have a different "$J$" set for each choice of $i$. In other words, each "$J$" set depends on $i$.

As remarked in the comments, there are ways to denote this sum that are easier on the eye. The point of my answer is that every summation is just a sum over some indexing set. If all else fails, you can always say "sum over indices belonging to some set" and then describe to the reader what that set is.

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Yow. I'd just write $\sum\limits_{\substack{1\le j\le n,\\j\ne i}} A_{ij} x_j$. –  Rahul Aug 22 '12 at 23:23
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@RahulNarain I would write precisely what the OP wrote. The point of my answer wasn't to convey the most succinct way to denote the sum, but to make explicit what is meant by "$j \neq i$" in the summation. –  Austin Mohr Aug 23 '12 at 1:11

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