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

  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} \})]

Use 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]
  • 6
    I'd use \mathit rather than \emph
    – egreg
    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?

enter image description here



  m_{12}(\{Red,Blue\})&=K'*[m_1 (\{Red,Blue\})*m_2(\{Red,Blue\})\\
                      &\qquad +m_1(\{Red,Blue\})*m_2(\{Red,Blue,Green\})\\
                      &\qquad +m_2(\{Red,Blue\})*m_1(\{Red,Blue,Green\})]

  • 2
    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?
    – Eric
    Dec 27 '19 at 11:03

How do I make alignment like the one above for equation with many brackets and braces?

\[ = -2 \displaystyle \left[ \frac{e^{\delta \sin (\frac{\pi}{2}-\phi)}}{\delta} 
 +e^{\delta \sin (\frac{\pi}{2}-\phi)}\sum\limits_{n=1}^{\infty}T_n
 \cdot \sum \limits_{\alpha^{/}=0}^{2n-1} {2n(2n-1)(2n-2)...\times 1} \times 
 \frac {(-\sin (\frac{\pi}{2}-\phi))^{2n-\alpha^{'}}}{\delta ^{\alpha^{'} +1}} 
 -\displaystyle \left\{ -\frac{e^{\delta \sin (\frac{\pi}{2}-\phi)}}{\delta} 
 +e^{-\delta \sin (\frac{\pi}{2}-\phi)}\sum\limits_{n=1}^{\infty}T_n 
 \cdot \sum\limits_{\alpha^{/}=0}^{2n-1} {2n(2n-1)(2n-2)...\times 1} \times 
 \frac {(-\sin (-(\frac{\pi}{2}-\phi))^{2n-\alpha^{'}})}{\delta ^{\alpha^{'} +1}}  \right \} \right]\]  

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