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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}
  • Do you really think that a biggggg left brace would help readers? – egreg Dec 18 '16 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 Dec 18 '16 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. – barbara beeton Dec 18 '16 at 21:52
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}
  • And is there an automatic way ? – Ccile Dec 18 '16 at 21:20
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    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 – David Carlisle Dec 18 '16 at 21:30
  • Ok, thanks for your explanation. Any idea for the left brace or something equivalent to separate two big equation systems ? – Ccile Dec 18 '16 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 \\ . – barbara beeton Dec 18 '16 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. – David Carlisle Dec 18 '16 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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