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This is the equation:

\begin{equation}
U^*=\arg \underset{U}{\max } \sum _{m=1}^M  \omega _m \left[\ln \left(P_{\Lambda _m} (X(U,m)|C)\right)- \omega _{\text{KLD}} D_{\Lambda_m}\left(C(U),C\right)\right]
\end{equation}

I try to use split but it has error:

\begin{equation}
\begin{split}
U^*=\arg \underset{U}{\max } \sum _{m=1}^M  & \omega _m \left[\ln \left(P_{\Lambda _m} (X(U,m)|C)\right)- \\
 &\quad \omega _{\text{KLD}} D_{\Lambda_m}\left(C(U),C\right)\right]
\end{split}
\end{equation}

enter image description here

I follow How to break a long equation?, but it seems not work for me.

How to make it effect?

partida
  • 101
  • 2
  • See https://tex.stackexchange.com/q/21290/15925 – Andrew Swann Jan 18 '18 at 10:02
  • This works for me.
    \begin{equation}\label{eq1} \begin{split} U^*=\arg \underset{U}{\max } \sum {m=1}^M & \omega _m \left[\ln \left(P{\Lambda m} (X(U,m)|C)\right)- \right. \ & \left. {} \omega _{\text{KLD}} D{\Lambda_m}\left(C(U),C\right)\vphantom{\ln}\right] \end{split} \end{equation}     – partida Jan 23 '18 at 07:06
  • Yes, that works but you have to manually choose the right vphantom(s). – Andrew Swann Jan 23 '18 at 07:09

0 Answers0