2

I am trying to align an equation within the cases environment like so:

\begin{equation}
    R_i=
    \begin{cases}
        2G_{\frac{i}{2} - 1}  &= \frac{i^2 - 2i}{4}, & \text{if}\ 2 | i \\
        2G_{\frac{i - 1}{2} - 1} + \frac{i-1}{2}  &= \frac{i^2 - 2i + 1}{4}, & \text{otherwise}.
    \end{cases}
\end{equation}

Unfortunately this errors with

Extra alignment tab has been changed to \cr. [...\frac{i}{2} - 1} & = \frac{i^2 - 2i}{4}, &]

How would I go about having two alignment tabs in this case?

it works without the extra &

\begin{equation}
    R_i=
    \begin{cases}
        2G_{\frac{i}{2} - 1}  = \frac{i^2 - 2i}{4}, & \text{if}\ 2 | i \\
        2G_{\frac{i - 1}{2} - 1} + \frac{i-1}{2}  = \frac{i^2 - 2i + 1}{4}, & \text{otherwise}.
    \end{cases}
\end{equation}

enter image description here

  • Do you really need alignment at =? – egreg Feb 1 '18 at 16:27
5

You can use aligned or alignedat. But, in my opinion, a plain dcases (requires mathtools) is the best choice.

\documentclass{article}
\usepackage{amsmath,mathtools}

\begin{document}

\begin{equation}
R_i=
\begin{cases}
  \begin{aligned}
  2G_{\frac{i}{2} - 1}                   &= \frac{i^2-2i}{4},   && \text{if $2\mid i$}
  \\
  2G_{\frac{i-1}{2} - 1} + \frac{i-1}{2} &= \frac{i^2-2i+1}{4}, && \text{otherwise}.
  \end{aligned}
\end{cases}
\end{equation}

\begin{equation}
R_i=
\begin{cases}
  \begin{alignedat}{3}
  &2G_{\frac{i}{2} - 1}                   &&= \frac{i^2-2i}{4},
  &\quad& \text{if $2\mid i$}
  \\
  &2G_{\frac{i-1}{2} - 1} + \frac{i-1}{2} &&= \frac{i^2-2i+1}{4},
  &\quad& \text{otherwise}.
  \end{alignedat}
\end{cases}
\end{equation}

\begin{equation}
R_i=
\begin{dcases}
2G_{\frac{i}{2} - 1} = \frac{i^2-2i}{4},                     & \text{if $2\mid i$}
\\
2G_{\frac{i-1}{2} - 1} + \frac{i-1}{2} = \frac{i^2-2i+1}{4}, & \text{otherwise}.
\end{dcases}
\end{equation}

\end{document}

enter image description here

For divisibility, it's better to use 2\mid i rather than 2|i. I'd use \text{even $i$} and \text{odd $i$} instead.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.