Fit inside enumerate

Question

\begin{enumerate}
  \item $\mathcal{D}(m)=0$ for $m\in\mathbb{Z}$,

  \item $\mathcal{D}(x^{-1})=-x^{-1}\sigma(x^{1})\mathcal{D}(x)$,

  \item $\mathcal{D}(x^{n})=\left( {\sigma ({x^{n - 1}}) + \sigma ({x^{n - 2}})x + ... + \sigma (x){x^{n - 2}} + {x^{n - 1}}} \right)\mathcal{D}(x)$,

  \item $\mathcal{D}(x^{-n})=- \left( {\sigma ({x^-n}){x^-1} + \sigma ({x^{-(n - 1)}}){x^-2} + ... + \sigma ({x^-2}){x^{-(n - 1)}} + \sigma ({x^-1}){x^-n}} \right)\mathcal{D}(x)$ for $n\geq1$.
\end{enumerate}

How can I fit the last equation in article'paper size?

Could you please make it a fully compilable example? It is helping solvers a lot to actually start experimenting with the code! — Malipivo, Apr 12 '14 at 13:09
One option is to use the \sum notation: \item $\mathcal{D}(x^{-n})= -\mathcal{D}(x)\sum_{i=1}^{n}\sigma ({x^{-n+i-1}}){x^{i}}$, for $n\geq1$. — Gonzalo Medina, Apr 12 '14 at 13:44
LaTeX won't break the equation, since the breaking point doesn't lie with an operator. You may have to rewrite the equation (perhaps using a different notation) or manually break it. — Werner, Apr 12 '14 at 13:46
Thank u for advice Gonzalo Medina but it was thougt. My post is only a sample for this porblem. — Cade, Apr 12 '14 at 16:39

score 1 · Accepted Answer · edited Jun 10 '20 at 12:32

My recommendation would be to use the inner aligned environment and make the equation over several lines:

enter image description here

Notes:

Use x^{-1} instead of x^-1.
Use \dotsb instead of ... as per Difference of the \dots*.
The showframe package was used just to show the page margins. It is not needed in you actual use case.

You could also consider aligning the left hand side of the equations. The MWE below produces a right alignment:

enter image description here

But by replacing the [r] with a [l] in the \MakeBox you get:

enter image description here

I think I prefer the [r] aligned version.

Code: `aligned`

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{showframe}
\begin{document}
\begin{enumerate}
  \item $\mathcal{D}(m)=0$ for $m\in\mathbb{Z}$,
\item $\mathcal{D}(x^{-1})=-x^{-1}\sigma(x^{1})\mathcal{D}(x)$,
\item $\mathcal{D}(x^{n})=\left( {\sigma ({x^{n - 1}}) + \sigma ({x^{n - 2}})x + ... + \sigma (x){x^{n - 2}} + {x^{n - 1}}} \right)\mathcal{D}(x)$,
\item $\begin{aligned}[t]\mathcal{D}(x^{-n}) 
          &=- \Big( \sigma ({x^-n}){x^{-1}} + \sigma ({x^{-(n - 1)}}){x^-2} + \dotsb \
          &\qquad+ \sigma ({x^-2}){x^{-(n - 1)}} + \sigma ({x^{-1}}){x^-n} \Big)\mathcal{D}(x)
           \text{ for } n\geq1.
          \end{aligned}$
\end{enumerate}
\end{document}

Code: `aligned` with `\MakeBox`:

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{calc}
\usepackage{showframe}
\newcommand*{\MakeBox}[1]{\makebox[\widthof{$\mathcal{D}(x^{-n})$}][r]{$#1$}}%
\begin{document}
\begin{enumerate}
  \item $\MakeBox{\mathcal{D}(m)}=0$ for $m\in\mathbb{Z}$,
\item $\MakeBox{\mathcal{D}(x^{-1})}=-x^{-1}\sigma(x^{1})\mathcal{D}(x)$,
\item $\MakeBox{\mathcal{D}(x^{n})}=\left( {\sigma ({x^{n - 1}}) + \sigma ({x^{n - 2}})x + ... + \sigma (x){x^{n - 2}} + {x^{n - 1}}} \right)\mathcal{D}(x)$,
\item $\begin{aligned}[t]\mathcal{D}(x^{-n}) 
          &=- \Big( \sigma ({x^-n}){x^{-1}} + \sigma ({x^{-(n - 1)}}){x^-2} + \dotsb \
          &\qquad+ \sigma ({x^-2}){x^{-(n - 1)}} + \sigma ({x^{-1}}){x^-n} \Big)\mathcal{D}(x)
           \text{ for } n\geq1.
          \end{aligned}$
\end{enumerate}
\end{document}

Fit inside enumerate

1 Answers1

Notes:

Code: aligned

Code: aligned with \MakeBox:

Code: `aligned`

Code: `aligned` with `\MakeBox`: