12

I am using the IEEEtran document class to generate my PDF file so that it has two columns, but now I get a long equation and I cannot put it into just one of the two columns. Thus I used \onecolumn at the beginning of my equation and \twocolumn at the end of my equation like this:

\documentclass[journal]{IEEEtran}
\begin{document}
...
\onecolumn
\begin{align}
    T\triangleq{} &
    \int_{0}^{\infty}
    \frac
    {\gamma^{k+m_2-1}\exp\Bigl(-\frac{m_2\gamma}{\bar\gamma_2}\Bigr)}
    {1+\Bigl(\frac{(m_1+s\bar\gamma_1)\gamma}{cm_1}\Bigr)^{m_1}(1+\frac{1}{c}\gamma_2)
        ^{\lceil m_1 \rceil -m_1}}
    \,\mathrm{d}\gamma
    \\
    \nonumber
    ={}& C\int_{0}^{\infty}\gamma^{k+m_2-1}\exp\Biggl(-\frac{m_2\gamma}{\bar\gamma_2}\Biggr)
    \\
    &\times H_{1,1}^{1,1}
    \left[
    \frac{(m_1+s\bar\gamma_1)\gamma}{cm_1}
    \;\middle|\;
    \begin{gathered} (1-m_1,1) \\ (0,1)\end{gathered}
    \right]
    H_{1,1}^{1,1}
    \left[
    \frac{\gamma}{c}
    \;\middle|\;
    t \begin{gathered} (1-\lceil m_1 \rceil +m_1,1) \\ (0,1)\end{gathered}
    \right]
    \mathrm{d}\gamma
    \\
    ={}& C\Biggl(\frac{\bar\gamma_2}{m_2}\Biggr)^{k+m_2}
    H_{1,\,[1:\,1],\,0,\,[1:\,1]}^{1,\,1,\,1,\,1,\,1}
    \left[
    \begin{gathered}
        \frac{(m_1+s\bar{\gamma}^{}_{1})\bar{\gamma}^{}_{2}}{c\,m_1m_2} \\
        \frac{\bar{\gamma}^{}_{2}}{c\,m_2}
    \end{gathered}
    \;\middle|\;
    \begin{gathered}
        (k+m_2,1) \\
        (1-m_1,1);(1-\lceil{m_1}\rceil+m_1,) \\
        - \\
        (0,1);(0,1)
    \end{gathered}
    \right]
    \\
    ={}& C\Biggl(\frac{\bar\gamma_2}{m_2}\Biggr)^{k+m_2}
    G_{1,\,[1:\,1],\,0,\,[1:\,1]}^{1,\,1,\,1,\,1,\,1}
    \left[
    \begin{gathered}
        \frac{(m_1+s\bar{\gamma}^{}_{1})\bar{\gamma}^{}_{2}}{c\,m_1m_2}\\
        \frac{\bar{\gamma}_{_2}}{c\,m_2}
    \end{gathered}
    \;\middle|\;
    \begin{gathered}
        k+m_2\\1-m_1;1-\lceil{m_1}\rceil+m_1 \\
        - \\
        0;0\end{gathered}
    \right].
\end{align}
\twocolumn

which will generate an equation in the middle of the paper like this: The long equation I want to break But I was told that the format is not suitable for publication, and I have to put this long equation spanning two columns. What's the solution?

3 Answers 3

21

We can make the whole group fit one column if we break some equations and use the \medmath command (from nccmath) which reduces the formulae size by ~20%.

I add another solution based on the strip environment from cuted (a component of the sttools bundle) which allows to have full width formulae in a two-column environment, but in contrast to table*, is inserted at the point where it is called.

\documentclass[journal]{IEEEtran}
\usepackage{mathtools,amssymb,lipsum, nccmath}
\DeclarePairedDelimiter{\ceil}{\lceil}{\rceil}
\usepackage{cuted}
\setlength\stripsep{3pt plus 1pt minus 1pt}

\begin{document}

\lipsum[1-2] % filler text

\begin{align}
  & T
  \triangleq
  \int_{0}^{\infty}
  \frac
  {\gamma^{k+m_2-1}\exp\Bigl(-\mfrac{m_2\gamma}{\bar\gamma_2}\Bigr)}
  {1+\Bigl(\mfrac{(m_1+s\bar\gamma_1)\gamma}{cm_1}\Bigr)^{\mathstrut m_1}
  \Bigl(1+\mfrac{1}{c}\gamma_2\Bigr)^{\ceil{m_1}-m_1}}
  \,\mathrm{d}\gamma
  \\[1ex]
    & =\!\begin{aligned}[t]\medmath{C\!\int_{0}^{\infty}\!\!\gamma^{k+m_2-1}} & \medmath{\exp\Bigl(\!-\frac{m_2\gamma}{\bar\gamma_2}\!\Bigr)
  H_{1,1}^{1,1}
  \mathrlap{\left[
  \frac{(m_1\!+s\bar\gamma_1)\gamma}{cm_1}
  \middle\vert
  \begin{gathered} (1\!- m_1,1) \\ (0,1)\end{gathered}
  \right]}}\\
  & {}\times\medmath{H_{1,1}^{1,1}
  \left[
  \frac{\gamma}{c}
  \;\middle|\;
  t \begin{gathered} (1-\ceil{m_1} +m_1,1) \\ (0,1)\end{gathered}
  \right]
  \mathrm{d}\gamma}
  \end{aligned}
  \\
  & =\!\begin{aligned}[t] \medmath{
  C\biggl(\frac{\bar\gamma_2}{m_2}\biggr)^{k+m_2}} & \medmath{H_{1,\,[1:1],\,0,\,[1:1]} ^{1,\,1,\,1,\,1,\,1}}\times {}\\[-1ex]
  \MoveEqLeft\medmath{\times\left[
  \begin{gathered}
  \frac{(m_1\!+s\bar{\gamma}^{}_{1})\bar{\gamma}^{}_{2}}{c\,m_1m_2} \\
  \frac{\bar{\gamma}^{}_{2}}{c\,m_2}
  \end{gathered}
  \middle\vert
  \begin{gathered}
  (k+m_2,1) \\
  (1\!- m_1,1);(1\!-\!\ceil{m_1}\!+ m_1,) \\
  - \\
  (0,1);(0,1)
  \end{gathered}
  \right]}
  \end{aligned}
  \\[1ex]
  & =\!\begin{aligned}[t] \medmath{C\biggl(\frac{\bar\gamma_2}{m_2}\biggr)^{k+m_2}}
  & \medmath{G_{1,\,[1:1],\,0,\,[1:1]}^{1,\,1,\,1,\,1,\,1}}\times{}\\[-1ex]
  \MoveEqLeft\times \medmath{\left[
  \begin{gathered}
  \frac{(m_1+s\bar{\gamma}^{}_{1})\bar{\gamma}^{}_{2}}{c\,m_1m_2}\\
  \frac{\bar{\gamma}_{_2}}{c\,m_2}
  \end{gathered}
  \;\middle|\;
  \begin{gathered}
  k+m_2\\
  1-m_1;1-\ceil{m_1}+m_1 \\
  - \\
  0;0
  \end{gathered}
  \right]}.
  \end{aligned}
\end{align}

\lipsum[3-6] % more filler text
\end{document} 

enter image description here

\documentclass[journal]{IEEEtran}
\usepackage{mathtools,amssymb,lipsum}
\DeclarePairedDelimiter{\ceil}{\lceil}{\rceil}
\usepackage{cuted}
\setlength\stripsep{3pt plus 1pt minus 1pt}

\begin{document}

\lipsum[1-2] % filler text

\begin{strip}
  \begin{align}
    T & \triangleq
    \int_{0}^{\infty}
    \frac
    {\gamma^{k+m_2-1}\exp\Bigl(-\frac{m_2\gamma}{\bar\gamma_2}\Bigr)}
    {1+\Bigl(\frac{(m_1+s\bar\gamma_1)\gamma}{cm_1}\Bigr)^{\mathstrut m_1}
    \bigl(1+\frac{1}{c}\gamma_2\bigr)^{\ceil{m_1}-m_1}}
    \,\mathrm{d}\gamma
    \\
      & = C\int_{0}^{\infty}\gamma^{k+m_2-1}\exp\Bigl(-\frac{m_2\gamma}{\bar\gamma_2}\Bigr)
    H_{1,1}^{1,1}
    \left[
    \frac{(m_1+s\bar\gamma_1)\gamma}{cm_1}
    \;\middle|\;
    \begin{gathered} (1-m_1,1) \\ (0,1)\end{gathered}
    \right]
    H_{1,1}^{1,1}
    \left[
    \frac{\gamma}{c}
    \;\middle|\;
    t \begin{gathered} (1-\ceil{m_1} +m_1,1) \\ (0,1)\end{gathered}
    \right]
    \mathrm{d}\gamma
    \\
      & = C\biggl(\frac{\bar\gamma_2}{m_2}\biggr)^{k+m_2}
    H_{1,\,[1:\,1],\,0,\,[1:\,1]}^{1,\,1,\,1,\,1,\,1}
    \left[
    \begin{gathered}
    \frac{(m_1+s\bar{\gamma}^{}_{1})\bar{\gamma}^{}_{2}}{c\,m_1m_2} \\
    \frac{\bar{\gamma}^{}_{2}}{c\,m_2}
    \end{gathered}
    \;\middle|\;
    \begin{gathered}
    (k+m_2,1) \\
    (1-m_1,1);(1-\ceil{m_1}+m_1,) \\
    - \\
    (0,1);(0,1)
    \end{gathered}
    \right]
    \\
      & = C\biggl(\frac{\bar\gamma_2}{m_2}\biggr)^{k+m_2}
    G_{1,\,[1:\,1],\,0,\,[1:\,1]}^{1,\,1,\,1,\,1,\,1}
    \left[
    \begin{gathered}
    \frac{(m_1+s\bar{\gamma}^{}_{1})\bar{\gamma}^{}_{2}}{c\,m_1m_2}\\
    \frac{\bar{\gamma}_{_2}}{c\,m_2}
    \end{gathered}
    \;\middle|\;
    \begin{gathered}
    k+m_2\\
    1-m_1;1-\ceil{m_1}+m_1 \\
    - \\
    0;0
    \end{gathered}
    \right].
  \end{align}
\end{strip}
\lipsum[3-6] % more filler text

\end{document} 

enter image description here

3
  • +1. I wasn't sure if the termH_{1,\,[1:1],\,0,\,[1:1]} ^{1,\,1,\,1,\,1,\,1} and the subsequent large expression in square brackets are connected multiplicatively or not; that's why I stayed away from inserting line breaks and \times symbols in those spots in my answer...
    – Mico
    Jul 13, 2015 at 13:59
  • Thanks for your reply, and the second method is what I needed. :-) Jul 13, 2015 at 15:05
  • @Mico: I wasn't so sure too, except by elimination ;o)
    – Bernard
    Jul 13, 2015 at 18:54
8

My understanding of the equations you're trying to display isn't sufficient to judge if it's sensible to introduce additional line breaks in order to make them fit in a single column of the two-column document layout that's provided by the IEEEtran class.

If it doesn't really make sense to introduce additional line breaks, I would suggest that you place the entire align environment inside a double-width table* environment; you'll need to choose a suitable caption, of course. Note that table* environments can only occur at the top of a page.

enter image description here

\documentclass[journal]{IEEEtran}
\usepackage{mathtools,amssymb,lipsum,caption}
\DeclarePairedDelimiter{\ceil}{\lceil}{\rceil}
\begin{document}

\lipsum[1-2] % filler text

\begin{table*}
\caption{Derivation of Result XYZ}
\centering
\begin{minipage}{0.75\textwidth}
\begin{align}
    T&\triangleq
    \int_{0}^{\infty}
    \frac
    {\gamma^{k+m_2-1}\exp\Bigl(-\frac{m_2\gamma}{\bar\gamma_2}\Bigr)}
    {1+\Bigl(\frac{(m_1+s\bar\gamma_1)\gamma}{cm_1}\Bigr)^{\mathstrut m_1}
    \bigl(1+\frac{1}{c}\gamma_2\bigr)^{\ceil{m_1}-m_1}}
    \,\mathrm{d}\gamma
    \\
    &=  \nonumber C\int_{0}^{\infty}\gamma^{k+m_2-1}\exp\Bigl(-\frac{m_2\gamma}{\bar\gamma_2}\Bigr)
    \\
    &\qquad\qquad \times 
    H_{1,1}^{1,1}
    \left[
    \frac{(m_1+s\bar\gamma_1)\gamma}{cm_1}
    \;\middle|\;
    \begin{gathered} (1-m_1,1) \\ (0,1)\end{gathered}
    \right]
    H_{1,1}^{1,1}
    \left[
    \frac{\gamma}{c}
    \;\middle|\;
    t \begin{gathered} (1-\ceil{m_1} +m_1,1) \\ (0,1)\end{gathered}
    \right]
    \mathrm{d}\gamma
    \\
    &= C\biggl(\frac{\bar\gamma_2}{m_2}\biggr)^{k+m_2}
    H_{1,\,[1:\,1],\,0,\,[1:\,1]}^{1,\,1,\,1,\,1,\,1}
    \left[
    \begin{gathered}
        \frac{(m_1+s\bar{\gamma}^{}_{1})\bar{\gamma}^{}_{2}}{c\,m_1m_2} \\
        \frac{\bar{\gamma}^{}_{2}}{c\,m_2}
    \end{gathered}
    \;\middle|\;
    \begin{gathered}
        (k+m_2,1) \\
        (1-m_1,1);(1-\ceil{m_1}+m_1,) \\
        - \\
        (0,1);(0,1)
    \end{gathered}
    \right]
    \\
    &= C\biggl(\frac{\bar\gamma_2}{m_2}\biggr)^{k+m_2}
    G_{1,\,[1:\,1],\,0,\,[1:\,1]}^{1,\,1,\,1,\,1,\,1}
    \left[
    \begin{gathered}
        \frac{(m_1+s\bar{\gamma}^{}_{1})\bar{\gamma}^{}_{2}}{c\,m_1m_2}\\
        \frac{\bar{\gamma}_{_2}}{c\,m_2}
    \end{gathered}
    \;\middle|\;
    \begin{gathered}
        k+m_2\\
        1-m_1;1-\ceil{m_1}+m_1 \\
        - \\
        0;0
    \end{gathered}
    \right].
\end{align}
\medskip
\hrule
\end{minipage}
\end{table*}
%\twocolumn
\lipsum[3-20] % more filler text
\end{document}
1
  • Thanks for your reply :-) But I think maybe Bernard's answer is closer to what I want, sorry for that. Jul 13, 2015 at 15:06
0

Finally:

\documentclass[journal]{IEEEtran}
\usepackage{mathtools,amssymb,lipsum,caption}
\DeclarePairedDelimiter{\ceil}{\lceil}{\rceil}

\usepackage{glossaries}
\makeglossaries
\newacronym{rog}{ROG}{radius of gyration}

\begin{document}

\lipsum[1-2] % filler text

\begin{table*}
\caption{Derivation of Result XYZ}
\centering
\begin{minipage}{0.75\textwidth}
\begin{align}
    \text{mean} &={\frac {1}{n}}\sum _{i=1}^{n}x_{i} \\
    \text{weightedMean} &= \frac{\sum_{i_1}^{n} w_i * x_i}{\sum_{i_1}^{n} w_i} \\
    \text{centroid} &= \left[ mean(x_{long}), mean(y_{lat}) \right] \\
    \text{distanceHaversine} &= 2r \arcsin{\sqrt{\sin^2{\left(\frac{y_{lat_1}-y_{lat_2}}{2} \right)}+ \cos(y_{lat_1}) \cos{y_{lat_2}} \sin^2{\left(\frac{x_{long_1}-x_{long_2}}{2} \right)}}} \\

\end{align}
\medskip
\hrule
\end{minipage}
\end{table*}
%\twocolumn
\lipsum[3-20] % more filler text
\end{document}
1
  • This fixes the two-column layout problem Apr 23, 2020 at 10:53

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