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In the multi-dimensional Ito formula, this notation pops up:$$\sum_{i,j = 1}^n$$ What does that even mean? Do we take $i = 1$, then $j = 1, ....,n$, then $i =2$ and $j = 1,...,n$, and so on? Or do we take both $i$ and $j$ equal to 1, and then both $i$ and $j$ equal to 2, and so on? If it is the latter, then why don't they just write $\sum_{k=1}^n$, and then everywhere inside the summation, exchange both $j$ and $i$ for $k$? If it is the former, then surely the double summation notation makes the point clearer and is also better suited for calculation (i.e, if you want to take a constant in the inner sum out into the outer).

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  • $\begingroup$ en.wikipedia.org/wiki/Summation#Capital-sigma_notation $\endgroup$
    – user9464
    Commented Jan 25, 2017 at 3:16
  • $\begingroup$ There is always a sliding scale between clarity and conciseness. For two variables a double sum might be fine, but how would you handle the case of, say, five or more variables? (Not as uncommon as one would think.) $\endgroup$
    – mlk
    Commented Jan 25, 2017 at 15:17

2 Answers 2

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It has the following meaning $$\sum_{i,j=1}^n=\sum_{i=1}^n\sum_{j=1}^n=\sum_{j=1}^n\sum_{i=1}^n$$ Since the sums are finite, the order of summation does not matter. This notation is convenient because it takes up less space than writing it out.

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It's the summation of a square array. Here's a concrete example: $$\sum_{i, j = 1}^5 \gcd(i, j).$$ You could start with $\gcd(1, 1)$, then $\gcd(1, 2)$, $\gcd(1, 3)$, and so on and so forth until reaching $\gcd(5, 5)$. But like Moop said, addition is commutative, so you could fix $j$ while you iterate $i$, or you could proceed in diagonals, or you could even go through it randomly (of course keeping track of what you've already calculated).

I personally would prefer $$\sum_{i = 1}^5 \sum_{j = 1}^5 \gcd(i, j)$$ because you can more easily change that to a nonsquare rectangular array if you need to. But it could be a parsing speed bump if someone gets temporarily confused and thinks there is multiplication involved.

Or do we take both $i$ and $j$ equal to 1, and then both $i$ and $j$ equal to 2, and so on?

As you've already noted, that would be inefficient, since $i$ and $j$ could be collapsed to a single variable, like you said. it's the kind of mistake I would not expect in a book or a peer-reviewed journal article. But on a post on this website? It could happen, I presume we're all human here.

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