2

enter image description here

    \documentclass[twocolumn]{article}
\usepackage{graphicx} 
\usepackage{cuted}
\usepackage{amsmath}



\begin{document}
\maketitle

\section{Introduction}
\subsection{Equations}
In this section, we will explore how to compose lengthy equations intended to be formatted within a single column on a two-column paper.\\
\newline

\begin{strip}
    \begin{equation}
        A = Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \left(1 - \frac{r}{2}\right)- \frac{-Q^{u} \left(1 - r \cdot c\right)}{r \cdot c} \left(1 - \frac{r}{2}\right) - Q_{\text{rand}}^{u} \left(\frac{r}{2} - 1\right)
    \end{equation}
\end{strip}

\begin{strip}
   \begin{equation}\label{eq13}
        A = Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \left(1 - \frac{r}{2}\right)- \frac{-Q^{u} \left(1 - r \cdot c\right)}{r \cdot c} \left(1 - \frac{r}{2}\right) - Q_{\text{rand}}^{u} \left(\frac{r}{2} - 1\right)
    \end{equation}
\end{strip}

\begin{strip}
   \begin{equation}\label{eq13}
        A = Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \left(1 - \frac{r}{2}\right)- \frac{-Q^{u} \left(1 - r \cdot c\right)}{r \cdot c} \left(1 - \frac{r}{2}\right) - Q_{\text{rand}}^{u} \left(\frac{r}{2} - 1\right) 
    \end{equation}
\end{strip}

\end{document}
3
  • 1
    I get no pdf output at all from that test file Commented Oct 1, 2023 at 13:49
  • you need x (or anything) before \end{document} to get output (which then does show the 5) Commented Oct 1, 2023 at 13:51
  • I have edited, please check now. @DavidCarlisle
    – Dileep
    Commented Oct 1, 2023 at 13:56

2 Answers 2

2

As @Zarko has already suggested in a comment, do consider placing the three separate equations in a suitably-chosen align or gather environment.

enter image description here

\documentclass[twocolumn]{article}
\usepackage{cuted,amsmath,lipsum}

\begin{document}

\section{Introduction}
\subsection{Equations}
\lipsum[1][1-9] % filler text

\begin{strip}
\begin{align}
A &= Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \bigl(1 - \tfrac{r}{2}\bigr)
     - \frac{-Q^{u} (1 - r \cdot c)}{r \cdot c} \bigl(1 - \tfrac{r}{2}\bigr) 
     - Q_{\textrm{rand}}^{u} \bigl(\tfrac{r}{2} - 1\bigr)
     \\
A &= Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \bigl(1 - \tfrac{r}{2}\bigr)
    - \frac{-Q^{u} (1 - r \cdot c)}{r \cdot c} \bigl(1 - \tfrac{r}{2}\bigr) 
    - Q_{\textrm{rand}}^{u} \bigl(\tfrac{r}{2} - 1\bigr) 
    \label{eq23} \\
A &= Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \bigl(1 - \tfrac{r}{2}\bigr)
    - \frac{-Q^{u} (1 - r \cdot c)}{r \cdot c} \bigl(1 - \tfrac{r}{2}\bigr) 
    - Q_{\textrm{rand}}^{u} \bigl(\tfrac{r}{2} - 1\bigr) 
    \label{eq33}
\end{align}
\end{strip}

\null % insert an invisible marker

\end{document}
1
  • 2
    this is what the OP should do but doesn't actually answer the more interesting question of where the heck the 5 comes from if you use adjacent strip. I spent a few minutes with \tracingall earlier (without success:-) Commented Oct 1, 2023 at 16:23
0

Alright, I found the answer. In above case we add all three equations within one command of \strip.

\begin{strip}

\begin{equation}
        A = Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \left(1 - \frac{r}{2}\right)- \frac{-Q^{u} \left(1 - r \cdot c\right)}{r \cdot c} \left(1 - \frac{r}{2}\right) - Q_{\text{rand}}^{u} \left(\frac{r}{2} - 1\right)

\begin{equation}\label{eq13}
 A = Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \left(1 - \frac{r}{2}\right)- \frac{-Q^{u} \left(1 - r \cdot c\right)}{r \cdot c} \left(1 - \frac{r}{2}\right) - Q_{\text{rand}}^{u} \left(\frac{r}{2} - 1\right)
\end{equation}

\begin{equation}\label{eq13}
        A = Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \left(1 - \frac{r}{2}\right)- \frac{-Q^{u} \left(1 - r \cdot c\right)}{r \cdot c} \left(1 - \frac{r}{2}\right) - Q_{\text{rand}}^{u} \left(\frac{r}{2} - 1\right) 
    \end{equation}

\end{strip}
4
  • 1
    It will far better to merge your equation in gather math environment ...
    – Zarko
    Commented Oct 1, 2023 at 14:17
  • Tell me more about it. :)
    – Dileep
    Commented Oct 1, 2023 at 14:19
  • The first equation is missing an \end{equation} directive.
    – Mico
    Commented Oct 1, 2023 at 15:14
  • @Mico Yes, it's typo.
    – Dileep
    Commented Oct 2, 2023 at 5:35

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