\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}
2 Answers
As @Zarko has already suggested in a comment, do consider placing the three separate equations in a suitably-chosen align
or gather
environment.
\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}
-
2this 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:-) Oct 1 at 16:23
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}
-
1
-
-
-
x
(or anything) before\end{document}
to get output (which then does show the 5)