1

I want to write a big system of equation (about 50 lines) and I found I can use the environment align with the option (? is it an option ?) allowdisplaybreaks[1]. But some of my equations are large... I found some solutions but not with align and I don't know how to combine them.

And if I can add a left brace to this system, my day will be made. (it's optional, but if I can have the three : page breaking, line breaking and left brace...)

Here is my code

\documentclass[a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath}

\allowdisplaybreaks[1]
\begin{document}


\begin{align}
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega')\cdot N_0(\lambda;\Omega')-\phi(\lambda;\Omega')\cdot N_0(\lambda;\Omega')-\phi(\lambda;\Omega')N_0(\lambda;\Omega')-\phi(\lambda;\Omega')\cdot N_0(\lambda;\Omega')-\phi(\lambda;\Omega')\cdot N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
 A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).\\
\end{align}

\end{document}
3
  • Do you really think that a biggggg left brace would help readers?
    – egreg
    Commented Dec 18, 2016 at 21:05
  • I don't know... I can't imagine if I don't see it. But if it's not a left brace, maybe there is other solution to separate two big equations system.
    – Ccile
    Commented Dec 18, 2016 at 21:07
  • 1
    i don't know of any decent way to allow page breaking and at the same time place a left brace around the whole thing. some suitable explanatory prose preceding the system would be preferable, i think. Commented Dec 18, 2016 at 21:52

2 Answers 2

1

There are several possibilities for splitting the large term, I use aligned here

  \documentclass[a4paper]{article}
    \usepackage[utf8]{inputenc}
    \usepackage[T1]{fontenc}
    \usepackage{amsmath}

    \allowdisplaybreaks
    \begin{document}


    \begin{align}
         A_1&\!\begin{aligned}[t]=N_0(&\lambda;\Omega')\\
&-\phi(\lambda;\Omega')\cdot N_0(\lambda;\Omega')\\
&-\phi(\lambda;\Omega')\cdot N_0(\lambda;\Omega')\\
&-\phi(\lambda;\Omega')N_0(\lambda;\Omega')\\
&-\phi(\lambda;\Omega')\cdot N_0(\lambda;\Omega')\\
&-\phi(\lambda;\Omega')\cdot N_0(\lambda;\Omega')\\
&-\phi(\lambda;\Omega'),
\end{aligned}\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
    A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
     A_1&=N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),\\
    A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
    A_3&=\mathcal{N}(\lambda;\omega).\\
    \end{align}

    \end{document}
11
  • And is there an automatic way ?
    – Ccile
    Commented Dec 18, 2016 at 21:20
  • 1
    Mostly tex encourages manual linebreaking of of equations, just as with your outer align, you need to add \\ at semantically meaningful places. It is possible in simple cases to make some kind of automatic line wrapping but it is unlikely to make as good results, and is not provided by amsmath Commented Dec 18, 2016 at 21:30
  • Ok, thanks for your explanation. Any idea for the left brace or something equivalent to separate two big equation systems ?
    – Ccile
    Commented Dec 18, 2016 at 21:32
  • 1
    @Ccile -- align* removes all numbers. if you want to remove the number on just selected lines, use \notag or \nonumber just before the appropriate \\ . Commented Dec 18, 2016 at 21:54
  • 1
    @Ccile yes, it's a personal style choice not really a rule but I think just numbering the lines that you want to call out later helps to highlight them. Commented Dec 18, 2016 at 22:13
1

Here's a different take on using an aligned environment inside an align environment. It differs from David's solution purely in the way the line breaks are chosen.

enter image description here

\documentclass[a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath}
\allowdisplaybreaks
\begin{document}
\begin{align}
A_1&=N_0(\lambda;\Omega')
\!\begin{aligned}[t]
&-\phi(\lambda;\Omega')\cdot N_0(\lambda;\Omega')-\phi(\lambda;\Omega')\cdot N_0(\lambda;\Omega')\\
&-\phi(\lambda;\Omega')\cdot N_0(\lambda;\Omega')-\phi(\lambda;\Omega')\cdot N_0(\lambda;\Omega')\\
&-\phi(\lambda;\Omega')\cdot N_0(\lambda;\Omega')-\phi(\lambda;\Omega'),
\end{aligned}\\
A_2&=\phi(\lambda;\Omega')-\phi(\lambda;\Omega),\\
A_3&=\mathcal{N}(\lambda;\omega).
\end{align}
\end{document}

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