# How to break a long expression in the denominator of a fraction?

I used the split command to break my equation but I receive the error:

! Missing } inserted.
<inserted text>
}
l. \end {split}


The code in .tex file is as follows:

$$\label{eq:5} \begin{split} A =1-&\frac{h_{FS}p_{I}p_{DN}p_{FS} +h_{F}p_{I}p_{DN}p_{F} + h_{UNC}p_{I}p_{UN}}{h_{G}+h_{V}+p_{I}(h_{I}+h_{DMC}p_{DM}+h_{UNC}p_{UN}+h_{UMC}p_{UM}+h_{DNC}p_{DN}+h_{FS}p_{DN}p_{FS} +\\ &h_{GD}p_{DN}p_{GD} + h_{F}p_{DN}p_{F})} \end {split}$$


Actually, I can see the produced equation in the PDF and it's
correct but this error exists. I want to submit my source file to a journal, but the PDF cannot be generated in the submission process. Any help is really appreciated.

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$$\label{eq:5} \begin{split} A =1-&\frac{h_{FS}p_{I}p_{DN}p_{FS} +h_{F}p_{I}p_{DN}p_{F} + h_{UNC}p_{I}p_{UN}}{h_{G}+h_{V}+p_{I}(h_{I}+h_{DMC}p_{DM}+h_{UNC}p_{UN}+h_{UMC}p‌​_{UM}+h_{DNC}p_{DN}+h_{FS}p_{DN}p_{FS} +\\ &h_{GD}p_{DN}p_{GD} + h_{F}p_{DN}p_{F})} \end {split}$$ –  Mary Sep 27 '12 at 2:20
You can't split the second part of a \frac on two line unless you put it inside something like a \parbox –  Peter Grill Sep 27 '12 at 2:23

You could use the \splitfrac instruction of the mathtools package to split the long denominator into two lines. Note that you needn't employ a split environment in this case. I've increased the size of the round parentheses used in the denominator (via \big instructions) to give them a bit more visibility.

\documentclass{article}
\usepackage{mathtools} % mathtools builds on and extends amsmath package
\begin{document}

$$\label{eq:5} A = 1-\frac{h_{FS}p_{I}p_{DN}p_{FS} +h_{F}p_{I}p_{DN}p_{F} + h_{UNC}p_{I}p_{UN}}{ \splitfrac{h_{G}+h_{V}+p_{I}\big( h_{I}+h_{DMC}p_{DM}+h_{UNC}p_{UN}+h_{UMC}p_{UM}} {+h_{DNC}p_{DN}+h_{FS}p_{DN}p_{FS} +h_{GD}p_{DN}p_{GD} + h_{F}p_{DN}p_{F}\big)}}$$
\end{document}

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The other answers give you what you wanted (+1 to both), but for what it's worth, I would consider tackling this using local definitions to make it a little easier to read

\documentclass{article}
\usepackage{mathtools} % mathtools builds on and extends amsmath package
\begin{document}

$$\label{eq:5} A = 1-\frac{f(h,p)}{g(h,p)}$$
where
\begin{align*}
f(h,p) & =h_{FS}p_{I}p_{DN}p_{FS} +h_{F}p_{I}p_{DN}p_{F} + h_{UNC}p_{I}p_{UN}                                    \\
g(h,p) & =h_{G}+h_{V}+p_{I}\left( h_{I}+h_{DMC}p_{DM}+h_{UNC}p_{UN}+h_{UMC}p_{UM}\right.                         \\
& \phantom{ {}= } + \left. h_{DNC}p_{DN}+h_{FS}p_{DN}p_{FS} +h_{GD}p_{DN}p_{GD} + h_{F}p_{DN}p_{F}\right)
\end{align*}
\end{document}

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In the present case, the \left( ... \right. and \left. ... \right) don't actually increase the size of the parentheses. It may be better to use \big statements instead. –  Mico Sep 27 '12 at 13:47
@Mico thanks for the feedback; I view \big as a bit too manual though, perhaps that's just me :) –  cmhughes Sep 27 '12 at 15:58
Great answer, Chris! :) I think this is the cleanest way to tackle this expression. :) –  Paulo Cereda Sep 28 '12 at 1:08
@PauloCereda thanks a lot :) I always appreciate your positivity :) –  cmhughes Sep 28 '12 at 1:26

If you place the denominator of the \frac with a \parbox then you can have a new line in it:

## Code:

\documentclass{article}
\usepackage{amsmath}

\begin{document}
$$\label{eq:5} A =1-\frac{h_{FS}p_{I}p_{DN}p_{FS} +h_{F}p_{I}p_{DN}p_{F} + h_{UNC}p_{I}p_{UN}}{\parbox{4.2in}{h_{G}+h_{V}+p_{I}(h_{I}+h_{DMC}p_{DM}+h_{UNC}p_{UN}+h_{UMC}p_{UM}+h_{DNC}p_{DN} + \\ \hspace*{2.1cm}h_{FS}p_{DN}p_{FS} + h_{GD}p_{DN}p_{GD} + h_{F}p_{DN}p_{F})}}$$
\end{document}

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Here's a TeXy way (amsmath will nag though if you're using it, IIRC.):

A = 1-{h_{FS}p_Ip_{DN}p_{FS} +h_Fp_Ip_{DN}p_F + h_{UNC}p_Ip_{UN}
\over\displaystyle{
h_G+h_V+p_I\big( h_I+h_{DMC}p_{DM}+h_{UNC}p_{UN}+h_{UMC}p_{UM}

Even if it doesn't change anything in this case, I'd use \bigl and \bigr instead of \big. –  Hendrik Vogt Sep 28 '12 at 7:54