# How to wrap a long equation in Latex

I have a very big equation and using split or multiline is not helping. How do i wrap it and how do i make sure thatRHS of the eqn is always on the right side of equal to ('=') even when line is changed

\begin{equation*}
\begin{split}
m_{12}(\{ \emph{Red},\emph{Blue} \})= K'* [m_{1}(\{ \emph{Red},\emph{Blue} \})*m_{2}(\{    \emph{Red},\emph{Blue} \})+m_{1}(\{ \emph{Red},\emph{Blue} \})*m_{2}(\{ \emph{Red},\emph{Blue},\emph{Green} \})+m_{2}(\{ \emph{Red},\emph{Blue} \})*m_{1}(\{ \emph{Red},\emph{Blue},\emph{Green} \})]
\end{split}
\end{equation*}


Use aligned:

\begin{equation*}
\begin{aligned}
m_{12}(\{ \emph{Red},\emph{Blue} \})
= K' * \bigl[ & m_{1}(\{ \emph{Red},\emph{Blue} \})*m_{2}(\{    \emph{Red},\emph{Blue} \}) \\
+ & m_{1}(\{ \emph{Red},\emph{Blue} \})*m_{2}(\{ \emph{Red},\emph{Blue},\emph{Green} \}) \\
+ & m_{2}(\{ \emph{Red},\emph{Blue} \})*m_{1}(\{ \emph{Red},\emph{Blue},\emph{Green} \})\bigr]
\end{aligned}
\end{equation*}

• I'd use \mathit rather than \emph May 3 '13 at 20:06
• @egreg The italic font is strange anyway, I just copied from the question.
– Alex
May 3 '13 at 20:08

Have you tried using the align environment? \documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{align*}
m_{12}(\{Red,Blue\})&=K'*[m_1 (\{Red,Blue\})*m_2(\{Red,Blue\})\\

• using \mathit{Red,Blue} instead of just the words would result in a nicer appearance with respect to intra-word kerning. (the alignment is very nice, however.) May 4 '13 at 13:11
• Any easy way to line up the m_1 on the first two lines?
$= -2 \displaystyle \left[ \frac{e^{\delta \sin (\frac{\pi}{2}-\phi)}}{\delta} +e^{\delta \sin (\frac{\pi}{2}-\phi)}\sum\limits_{n=1}^{\infty}T_n \cdot \sum \limits_{\alpha^{/}=0}^{2n-1} {2n(2n-1)(2n-2)...\times 1} \times \frac {(-\sin (\frac{\pi}{2}-\phi))^{2n-\alpha^{'}}}{\delta ^{\alpha^{'} +1}} -\displaystyle \left\{ -\frac{e^{\delta \sin (\frac{\pi}{2}-\phi)}}{\delta} +e^{-\delta \sin (\frac{\pi}{2}-\phi)}\sum\limits_{n=1}^{\infty}T_n \cdot \sum\limits_{\alpha^{/}=0}^{2n-1} {2n(2n-1)(2n-2)...\times 1} \times \frac {(-\sin (-(\frac{\pi}{2}-\phi))^{2n-\alpha^{'}})}{\delta ^{\alpha^{'} +1}} \right \} \right]$