# Good typesetting practice for long equations

I have the following (quite horrible) equation in my thesis:

\begin{align*}
\frac{\partial^2}{\partial t_1^2} f(t_0,t_1) =
( \delta+2t_0+2t_1)^{\alpha( w-t_0+t_1 )-1} \cdot \bigl(
\frac{\partial^2}{\partial t_1^2}\alpha(w-t_0+t_1) \cdot ( \delta+2t_0+2t_1) \cdot  \log ( \delta+2t_0+2t_1) +\\
\alpha'(w-t_0+t_1) \cdot 2 \cdot  \log ( \delta+2t_0+2t_1)+
\alpha'(w-t_0+t_1) \cdot ( \delta+2t_0+2t_1) \cdot  \frac{2}{\delta+2t_0+2t_1} +\\
2 \frac{\partial}{\partial t_1} \alpha( w-t_0+t_1 ) \bigr) +
( \delta+2t_0+2t_1)^{\alpha( w-t_0+t_1 )-2}\cdot\\
\bigl( \frac{\partial}{\partial t_1} \alpha(w-t_0+t_1) \cdot ( \delta+2t_0+2t_1) \cdot \log ( \delta+2t_0+2t_1) + (\alpha (w-t_0+t_1) -2) \bigr) \cdot \\
\bigl( \alpha'(w -t_0+t_1) \cdot ( \delta+2t_0+2t_1) \cdot  \log ( \delta+2t_0+2t_1) +
2\alpha( w-t_0+t_1)\bigr) = \\
( \delta+2t_0+2t_1)^{\alpha( w-t_0+t_1 )-1} \cdot \Bigl(
\frac{\partial^2}{\partial t_1^2}\alpha(w -t_0+t_1) \cdot ( \delta+2t_0+2t_1) \cdot  \log ( \delta+2t_0+2t_1) +\\
2 \cdot \alpha'(w-t_0+t_1)  \cdot  \bigl( 2 + \log ( \delta+2t_0+2t_1) \bigr) \Bigr) +
( \delta+2t_0+2t_1)^{\alpha( w-t_0+t_1)-2} \cdot \Bigl( \\
\alpha '(w-t_0+t_1) \cdot
(\delta + 2t_0+2t_1) \cdot \log (\delta + 2t_0+2t_1) +
\bigl(\alpha (w-t_0+t_1) -2) \bigr) \cdot
\bigl(   \\
\alpha'(w-t_0+t_1) \cdot ( \delta+2t_0+2t_1) \cdot  \log ( \delta+2t_0+2t_1) +2\alpha( w-t_0+t_1)\bigr) \Bigr)  < 0
\end{align*}


Using this exact piece of code, without any special formatting commands such as & or 2mm] the resulting mathematical text is quite unreadable: How would you format such equations in LaTeX and what would you say is good practice when typesetting such large equations? • Either keep your align* or use a split inside a display-math environment, but add breaks in places unlikely to throw your reader off. Break lines before plus signs, but after multiplication signs. For the latter, I think \times is easier to parse than \cdot, here. Also, use \left(, \right) for an automatic hierarchy in delimiter size; that will help your reader parse your equation. – jub0bs Apr 14 '13 at 11:30 • You want to prove that this second derivative is negative, right? I guess there's no good typesetting answer to your question, but my mathematical answer is: try and give the proof more structure. – Hendrik Vogt Apr 15 '13 at 12:12 ## 4 Answers I'd try to make the equation smaller by grouping parts: • Don't use \cdot where it's not necessary. I use it only for scalar products of vectors and for numbers, but not for symbolic factors or before parentheses. • Derivatives are often written as \partial_{t_1} instead of \frac{\partial}{\partial t_1}. This can save some space. • Introducing substitutions can be helpful. In your code (\delta+2t_0+2t_1) appears quite often and it could be replaced by a new symbol which will be defined before or after the equation • Align the equation at least on all equal signs: &= • Other line breaks may be before + signs to "group" summands (this shows that the equation consists of similar parts that are added together) breaking before not after operators and defining names for the subterms \documentclass{article} \usepackage{amsmath} \begin{document} \begin{align*} \frac{\partial^2}{\partial t_1^2} f(t_0,t_1) &= b^{a-1} \cdot \bigl( \frac{\partial^2}{\partial t_1^2}a \cdot b \cdot \log ( b) + a' \cdot 2 \cdot \log ( b)+ a' \cdot b \cdot \frac{2}{b} + 2 \frac{\partial}{\partial t_1} a \bigr) \\ &\quad+ b^{a-2}\cdot \bigl( \frac{\partial}{\partial t_1}a \cdot b \cdot \log ( b) + (a -2) \bigr) \cdot \bigl( a' \cdot b \cdot \log ( b) + 2a\bigr)\\ & = b^{a-1} \cdot \Bigl( \frac{\partial^2}{\partial t_1^2}a \cdot b \cdot \log ( b) + 2 \cdot a' \cdot \bigl( 2 + \log ( b) \bigr) \Bigr)\\ &\quad + b^{a-2} \cdot \bigl(a' \cdot c \cdot \log (c) + \bigl(a -2) \bigr) \cdot \bigl(a' \cdot b \cdot \log ( b) +2a)\bigr)\bigr)\\ &< 0 \end{align*} where:\\ a=\alpha( w-t_0+t_1 )\\ a'=\alpha'(w-t_0+t_1)\\ b=\delta+2t_0+2t_1\\ c=\delta + 2t_0+2t_1 \end{document}  • I think adding a \qquad after the first and third line breaks would help the parsing. – jub0bs Apr 14 '13 at 11:35 • @FooBar ' is equivalent to ^\prime. – jub0bs Apr 14 '13 at 11:36 • @Jubobs always? – Foo Bar Apr 14 '13 at 11:37 • @FooBar yes (unless you define it not to be) – David Carlisle Apr 14 '13 at 11:38 • @David Carlisle, Thanks, didn't knew that. – Foo Bar Apr 14 '13 at 11:39 Actually, I would like to start answering with a question: Is it very informative to display an equation that long? I would try to identify parts in your equation, and write something like \[a (A + B + C) < 0
where
$a = ...$
and
\begin{align}
A &= ... \\
B &= ... \\
C &= ...
\end{align}


this makes it much easier to read it, and you can maybe give also an exlanation to every term.

Try using the breqn package. Begin with usepackage{breqn}, then replace the align* environment with dmath*. Then remove all the manual linebreaks \\, because breqn does the line-breaking and aligning automatically. Also you can replace \bigl and \bigr with \left and \right, because breqn allows line breaks within a \left-\right pair.

\documentclass{article}
\usepackage{breqn}  % from the "mh" bundle

\begin{document}

\begin{dmath*}
\frac{\partial^2}{\partial t_1^2} f(t_0,t_1) =
( \delta+2t_0+2t_1)^{\alpha( w-t_0+t_1 )-1} \cdot \left(
\frac{\partial^2}{\partial t_1^2}\alpha(w-t_0+t_1) \cdot ( \delta+2t_0+2t_1) \cdot
\log ( \delta+2t_0+2t_1) +
\alpha'(w-t_0+t_1) \cdot 2 \cdot  \log ( \delta+2t_0+2t_1)+
\alpha'(w-t_0+t_1) \cdot ( \delta+2t_0+2t_1) \cdot  \frac{2}{\delta+2t_0+2t_1} +
2 \frac{\partial}{\partial t_1} \alpha( w-t_0+t_1 ) \right) +
( \delta+2t_0+2t_1)^{\alpha( w-t_0+t_1 )-2}\cdot
\left( \frac{\partial}{\partial t_1} \alpha(w-t_0+t_1) \cdot ( \delta+2t_0+2t_1)
\cdot \log ( \delta+2t_0+2t_1) + (\alpha (w-t_0+t_1) -2) \right) \cdot
\left( \alpha'(w -t_0+t_1) \cdot ( \delta+2t_0+2t_1) \cdot  \log ( \delta+2t_0+2t_1) +
2\alpha( w-t_0+t_1)\right) =
( \delta+2t_0+2t_1)^{\alpha( w-t_0+t_1 )-1} \cdot \left(
\frac{\partial^2}{\partial t_1^2}\alpha(w -t_0+t_1) \cdot ( \delta+2t_0+2t_1) \cdot
\log ( \delta+2t_0+2t_1) +
2 \cdot \alpha'(w-t_0+t_1)  \cdot  \left( 2 + \log ( \delta+2t_0+2t_1) \right) \right)
+ ( \delta+2t_0+2t_1)^{\alpha( w-t_0+t_1)-2} \cdot \Bigl(
\alpha '(w-t_0+t_1) \cdot
(\delta + 2t_0+2t_1) \cdot \log (\delta + 2t_0+2t_1) +
\left(\alpha (w-t_0+t_1) -2 \right) \cdot
\left(
\alpha'(w-t_0+t_1) \cdot ( \delta+2t_0+2t_1) \cdot  \log ( \delta+2t_0+2t_1) +2\alpha(
w-t_0+t_1)\right) \Bigr)  < 0
\end{dmath*}
\end{document}