I am having the following equation:

         \sum^{n}_{\stackrel{i=1}{(j,i)\in Q}} \left\{ w^{1}_{ijR}+\sum^{m=1}_{\stackrel{(m,i)\in Q}{m\neq j}} \left[ (w^{2}_{ijmR}+w^{2}_{ijm,R-1})  + \sum^{l=1}_{\stackrel{(l,i)\in Q}{i\neq j\neq m}} (w^{3}_{ijmlR}+w^{3}_{ijml,R-1}+w^{3}_{ijml,R-2}) \right]                         \right\}+     \sum^{n}_{\stackrel{j=1}{(l,i)\in Q}} \left\{ w^{1}_{jiR}+ \sum^{m=1}_{\stackrel{(m,j)\in Q}{m\neq i}}         \left[ (w^{2}_{jimR}+w^{2}_{jmi,R-1}) +\sum^{l=1}_{\stackrel{(l,j)\in Q}{m\neq i\neq l}}(w^{3}_{jimlR}+w^{3}_{jmil,R-1}+w^{3}_{jmli,R-2})   \right]  \right\}=1    ~ ~ ~ i=1,...,n  ~ ~ ~ R=1,...,n-1

How can I split this over three lines??


Here is a solution, if you don't need margin notes, with two variants for the second equation. A small remark: for multiline indices, you should use \substack, not \stackrel which is designed to add something over a relation symbol, so that the two rows use the same font size. I also used smaller delimiters.



          &\smashoperator{ \sumⁿ_{\substack{i=1\\(j,i) ∈ Q}}}\, \Biggl\{ w^{1}_{ijR}+\sum^{m=1}_{\substack{(m,i) ∈ Q\\m ≠ j}} \Biggl[ (w^{2}_{ijmR}+w^{2}_{ijm,R-1}) + \smashoperator{\sum^{l=1}_{\substack{(l,i) ∈ Q\\i ≠ j ≠ m}}} (w^{3}_{ijmlR}+w^{3}_{ijml,R-1}+w^{3}_{ijml,R-2}) \Biggr] \Biggr\} \\
          + & \smashoperator{\sumⁿ_{\substack{j=1\\(l,i) ∈ Q}}} \Biggl\{ w^{1}_{jiR}+ \sum^{m=1}_{\substack{(m,j) ∈ Q\\ m ≠ i}} \Biggl[ (w^{2}_{jimR}+w^{2}_{jmi,R-1}) + \smashoperator{\sum^{l=1}_{\substack{(l,j) ∈ Q\\ m ≠ i ≠ l}}}(w^{3}_{jimlR}+w^{3}_{jmil,R-1}+w^{3}_{jmli,R-2}) \Biggr] \Biggr\}
    \end{aligned} \\
         =1 ,\qquad i=1,...,n,\quad R=1,...,n-1

 \begin{multline} \smashoperator{\sum^{n}_{\substack{j=1\\ (j,i)\in Q}}} \Biggl\{w^{1}_{ijR}+w^{1}_{ij,R+1}+\smashoperator{\sum^{n}_{\substack{m=1 \\ (m,i)\in Q \\ m\neq j}}} \,\biggl[w^{2}_{ijm,R-1}+w^{2}_{ijmR}+w^{2}_{ijm,R+1} \\[-6ex]
  + \smashoperator{\sum^{n}_{\substack{l=1 \\ (l,i)\in Q \\ i\neq j\neq m}}}(w^{3}_{ijml,R-2}+w^{3}_{ijmlR}+w^{3}_{ijml,R+2}) \biggr] \Biggr\} \leq 1
\end{multline} \\

  \smashoperator{\sum^{n}_{\substack{j=1\\ (j,i)\in Q}}} \Biggl\{w^{1}_{ijR}+w^{1}_{ij,R+1}+\smashoperator{\sum^{n}_{\substack{m=1 \\ (m,i)\in Q \\ m\neq j}}} \,\biggl[w^{2}_{ijm,R-1} & + w^{2}_{ijmR}+w^{2}_{ijm,R+1} \\[-5.5ex]
  & + \smashoperator{\sum^{n}_{\substack{l=1 \\ (l,i)\in Q \\ i\neq j\neq m}}}(w^{3}_{ijml,R-2}+w^{3}_{ijmlR}+w^{3}_{ijml,R+2}) \biggr] \Biggr\} \leq 1


enter image description here

  • many thanks but this is not working with this equation: – mary Aug 11 '18 at 14:53
  • @mary: What do you mean with ‘not working’? What happens exactly? – Bernard Aug 11 '18 at 16:19
  • I mean erreur latex – mary Aug 11 '18 at 16:28
  • Could you post the error message? Are the required packages installed? – Bernard Aug 11 '18 at 16:32
  • this is the equation: \begin{multline} \sum^{n}_{\stackrel{j=1}{(j,i)\in Q}}\left\{w^{1}_{ijR}+w^{1}_{ij,R+1}+\sum^{n}_{\stackrel{m=1}{\stackrel{(m,i)\in Q}{m\neq j}}} \left[w^{2}_{ijm,R-1}+w^{2}_{ijmR}+w^{2}_{ijm,R+1}+ \\ \sum^{n}_{\stackrel{l=1}{\stackrel{(l,i)\in Q}{i\neq j\neq m}}}(w^{3}_{ijml,R-2}+w^{3}_{ijmlR}+w^{3}_{ijml,R+2}) \right] \right\} \leq 1 – mary Aug 11 '18 at 16:33
  • assumed, that the page borders widths are 25mm
  • with use of multlined math environment from the mathtools package:

enter image description here

(red lines indicate text bordes)

%---------------- show page layout. don't use in a real document!

\sum^{n}_{\stackrel{i=1}{(j,i)\in Q}} 
\left\{ w^{1}_{ijR}+\smashoperator[l]{\sum^{m=1}_{\stackrel{(m,i)\in Q}{m\neq j}}} 
\left[  \Bigl(w^{2}_{ijmR}+w^{2}_{ijm,R-1}\Bigr)  + 
                    \smashoperator[l]{\sum^{l=1}_{\stackrel{(l,i)\in Q}{i\neq j\neq m}}}
        \Bigl(w^{3}_{ijmlR}+w^{3}_{ijml,R-1}+w^{3}_{ijml,R-2} \Bigr)
\right\}+  \\   
\sum^{n}_{\stackrel{j=1}{(l,i)\in Q}} 
\left\{ w^{1}_{jiR}+ \smashoperator[l]{\sum^{m=1}_{\stackrel{(m,j)\in Q}{m\neq i}}}         
\left[  \Bigl(w^{2}_{jimR}+w^{2}_{jmi,R-1}\Bigr) +
                    \smashoperator[l]{\sum^{l=1}_{\stackrel{(l,j)\in Q}{m\neq i\neq l}}}
 i=1,...,n ;\ R=1,...,n-1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.