I would like to know if it is possible to force the equation to be closer to the operation sum? When there are a lot of subscript under the sum operator, it tends to pouch the equation to the right, and this is not very aesthetic.

\sum_{n\in N_{-i}}\sum_{\substack{j \in N_{-i} \\ j \neq n}}v_j

1 Answer 1


The first part of following code shows the problem you mentioned and the "standard" possible solutions for a single sum: using a \makebox or using \mathclap (from the mathtools package).

Next I present your concrete example and several variations; now, only using \mathclap makes things worst since the subscripts overlap; using \mathclap and adding some space between the sums could be an option, but I think the best solution would be the last one in which the amount of symbols used in the subscripts has been reduced:




No special treatment (ugly):
\sum_{1\leq i < j < k \leq n}a_{ijk}

Using a \verb!\makebox!:
\sum_{\makebox[0pt]{$\scriptstyle 1\leq i < j < k \leq n$}}a_{ijk}

Using \verb!\mathclap! (requires the \texttt{mathtools} package):
\sum_{\mathclap{1\leq i < j < k \leq n}}a_{ijk}

Your concrete example (ugly): 
\sum_{n \in N_{-i}}\sum_{\substack{j \in N_{-i} \\ j \neq n}}v_j

Your concrete example using \verb!\mathclap! (uglier since scripts overlap):
\sum_{n \in N_{-i}}\sum_{\mathclap{\substack{j \in N_{-i} \\ j \neq n}}}v_j

Your concrete example using \verb!\mathclap! and some space between the sums (a little better?):
\sum_{n \in N_{-i}}\mkern13mu\sum_{\mathclap{\substack{j \in N_{-i} \\ j \neq n}}}v_j

Your concrete example reformulated (better, when possible):
\sum_{n}\sum_{ j \neq n}v_j,
where $n$ and $j$ run over $N_{-i}$.


output of code example

  • I wish it was that simple, sadely the notation \in N_{-i} is necessary in order to differenciate from n \in N Aug 18, 2012 at 1:22
  • Thank you very much for your help, though. It was very well written and detailed. I will remember those function if I ever encounter this problem. sadely, it seem that for this particular one, there isnt one. Aug 18, 2012 at 1:23
  • Actually, by applying it to my non-double sum terms, I was able to gain enough space, which was my initial problem and rose the question. So this was indeed useful :) Aug 18, 2012 at 1:31

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