# Raise equation number position from new line

I am trying to write an equation in a multicolumn environment and while there is space on the final line for the equation number, LaTeX is seemingly putting it on a new line.

\documentclass[a4paper, 12pt]{article}
\usepackage{lipsum, amsmath, multicol, geometry}
\geometry{left=20mm, right=20mm, top=20mm, bottom=20mm}
\begin{document}
\begin{multicols}{2}
Using central differences for the first and second order derivative our numerical scheme becomes
\begin{equation}
\begin{aligned}
h_j^{n+1} =  h_j^n &+  \dfrac{\Delta t \left(h_j^n\right)^3}{(\Delta x)^2}
\left(h_{j+1}^n - 2 h_j^n + h_{j-1}^n\right)\\
&+  \dfrac{3\Delta t \left(h_j^n\right)^2}{4(\Delta x)^2}
\left(h_{j+1}^n - h_{j-1}^n\right)
\end{aligned}
\label{eqt:numerical_scheme}
\end{equation}
where we notice something\\
\lipsum
\end{multicols}
\end{document}


gives the following: I would like the equation number to be higher on the final line where there is room. I have tried swapping the lines and various environments, but no luck.

Any help would be appreciated. \documentclass{article}
\usepackage{multicol,amsmath}
\begin{document}

\begin{multicols}{2}
Using central differences for the first and second order derivative our numerical scheme becomes
\begin{gather}
\begin{aligned}
h_j^{n+1} =  h_j^n &+  \dfrac{\Delta t \left(h_j^n\right)^3}{(\Delta x)^2}
\left(h_{j+1}^n - 2 h_j^n + h_{j-1}^n\right)\\
&+  \dfrac{3\Delta t \left(h_j^n\right)^2}{4(\Delta x)^2}
\left(h_{j+1}^n - h_{j-1}^n\right)
\end{aligned}
\label{eqt:numerical_scheme1}
\raisetag{20pt}
\end{gather}
where we notice something

Using central differences for the first and second order derivative our numerical scheme becomes
\begin{align}
h_j^{n+1} =  h_j^n &+  \dfrac{\Delta t \left(h_j^n\right)^3}{(\Delta x)^2}
\left(h_{j+1}^n - 2 h_j^n + h_{j-1}^n\right)\nonumber\\
&+  \dfrac{3\Delta t \left(h_j^n\right)^2}{4(\Delta x)^2}
\left(h_{j+1}^n - h_{j-1}^n\right)
\label{eqt:numerical_scheme2}
\end{align}
where we notice something

\end{multicols}

\end{document}


Please always post full documents, I had to guess a text width to get the effect that you showed. You can use \raisetag (But apparently not in equation so I used a one line gather) or you can use align and just number one line.

• Sorry about the full document, will add it in now (always apprehensive to explode the 'snippit' with what are usually long preambles)! Why did you use \raisetag{20pt}, was 20pt a guess or did you work it out? Thanks! Oct 29 '16 at 21:15
• @oliversm -- 20pt was obviously a guess. if you look carefully, the parentheses in the right-hand column (using align) are lined up, but the ones in the left-hand column aren't. Oct 29 '16 at 21:23
• @barbarabeeton -- Thanks for pointing that out. The solution using aligned seems ideal. Oct 29 '16 at 21:32
• @oliversm well also it depends on the intention if the equation had been smaller the aligned method puts the number vertically centred between the two lines (so iIused 20pt being a bit more than a baseline) it should be a bit more for that effect, if you want it lined with a baseline using aligned of multline is better as the baseline alignment of the number will be exact then Oct 29 '16 at 21:36
• @barbarabeeton ^^ Oct 29 '16 at 21:37

The equation-number-placement issues can be avoided entirely by using a multline environment instead of nested equation and aligned environments. (Vertical alignment of the two + symbols would not appear to be particularly urgent.) \documentclass{article}
\setlength\textwidth{2.75in} % an educated guess...
\usepackage{amsmath}
% a version of \frac that uses \displaystyle for numerator and deminator:
\newcommand\ddfrac{\frac{\displaystyle#1}{\displaystyle#2}}

\begin{document}
Using central differences for the first and second order derivatives
our numerical scheme becomes
\begin{multline}
h_j^{n+1} =  h_j^n + \ddfrac{\Delta t(h_j^n)^3}{(\Delta x)^2}
\bigl(h_{j+1}^n - 2h_j^n + h_{j-1}^n\bigr)\\
+  \ddfrac{3\Delta t(h_j^n)^2}{4(\Delta x)^2}
\bigl(h_{j+1}^n - h_{j-1}^n \bigr) \label{eqt:numerical_scheme2}
\end{multline}
where we notice something
\end{document}


Another option using align:

\documentclass[12pt]{article}
\setlength\textwidth{19.18em}
\usepackage{amsmath,microtype}
\newcommand\ddfrac{\frac{\displaystyle#1}{\displaystyle#2}}

\begin{document}
\noindent differences for the first and second order derivatives
our numerical scheme becomes
\begin{align}
h_j^{n+1} = h_j^n &+ \ddfrac{\Delta t(h_j^n)^3}{(\Delta x)^2}
\bigl(h_{j+1}^n - 2h_j^n + h_{j-1}^n\bigr) \notag \\
& + \ddfrac{3\Delta t(h_j^n)^2}{4(\Delta x)^2}
\bigl(h_{j+1}^n - h_{j-1}^n \bigr) \label{eqt:numerical_scheme2}
\end{align}
where we notice something

\end{document} Here is another way, with the optional argument of aligned and \mathrlap:

\documentclass{article}
\usepackage{multicol,mathtools, lipsum}
\begin{document}

\begin{multicols}{2}
Using central differences for the first and second order derivative our numerical scheme becomes
\begin{equation}
\begin{aligned}[b]
h_j^{n+1} =h_j^n & + \dfrac{Δt \left(h_j^n\right)³}{(Δx)²}
\bigl(h_{j+1}^n - 2 h_j^n +\mathrlap{ h_{j-1}^n\bigr)} \\
& +\dfrac{3Δt \left(h_j^n\right)²}{4(Δx)²}
\left(h_{j+1}^n - h_{j-1}^n\right)
\end{aligned}
\label{eqt:numerical_scheme1}
\end{equation}
where we notice something.

\lipsum

\end{multicols}

\end{document} 