4

I would like to have a few tabs in a cases environment so that the "for"s and the "u"s in my example (and in the best case the "j"s as well) are on one line, respectively. I tried different environments for equations but I haven't found a solution. The quad-distances don't satisfy me...

\begin{align*}
    \phi_{C}(u)= 
        \begin{cases}
            (j-1)\cdot \lambda & \text{for } d_{j-1}\leq u \leq d_{j}-\lambda \qquad j=1,\ldots ,r\\
            j\cdot \lambda +u-d_{j} & \text{for } d_{j}-\lambda\leq u\leq d_{j} \qquad \qquad j=1,\ldots ,r-1 \qquad  \text{with } r\geq 2 \\
            r\cdot \lambda +u-d_{r} & \text{for } d_{r}-\lambda\leq u
    \end{cases}
\end{align*}
4

We can define a multicases environment modeled on cases that accepts as argument the number of conditions:

\documentclass{article}
\usepackage{amsmath}

\makeatletter
\newenvironment{multicases}[1]
  {\let\@ifnextchar\new@ifnextchar
   \left\lbrace\def\arraystretch{1.2}%
   \array{@{}l*{#1}{@{\quad}l}@{}}}
  {\endarray\right.\kern-\nulldelimiterspace}
\makeatother

\begin{document}
\begin{equation*}
\phi_{C}(u)= 
   \begin{multicases}{2}
   (j-1)\cdot \lambda      & \text{for } d_{j-1}\leq u \leq d_{j}-\lambda & j=1,\dots ,r\\
   j\cdot \lambda +u-d_{j} & \text{for } d_{j}-\lambda\leq u\leq d_{j}    & j=1,\dots ,r-1 \text{ with } r\geq 2 \\
   r\cdot \lambda +u-d_{r} & \text{for } d_{r}-\lambda\leq u
   \end{multicases}
\end{equation*}
\end{document}

enter image description here

Use \dots and not \ldots. It's possibly improved by adding parentheses around the additional conditions:

enter image description here

| improve this answer | |
3

Easy with alignat and empheq to replace the cases environment. Empheq loads mathtools, hence amsmath:

\documentclass[a4paper, 11pt]{book}
\usepackage[utf8]{inputenc}
\usepackage[showframe]{geometry}
\usepackage{fourier}

\usepackage[overload]{empheq}

\begin{document}

\begin{alignat*}{4}[left = {ϕ_{C}(u)=\empheqlbrace}]
    & (j-1) · λ & \quad & \text{for\enspace } & d_{j-1} & \leq u \leq d_{j}-λ & \qquad & j=1,\ldots ,r \\
    & j · λ +u-d_{j} & & \text{for } & d_{j}-λ & \leq u\leq d_{j} & & j=1,\ldots ,r-1 \quad \text{with } r\geq 2 \\
  & r · λ +u-d_{r} & & \text{for } & d_{r}-λ & \leq u
\end{alignat*}

\end{document} 

enter image description here

| improve this answer | |
0

Isn't an elegant response, but it works.

\begin{align*}
  \phi_{C}(u)= 
    \begin{cases}
        (j-1)\cdot \lambda & \text{for } \;\,d_{j-1}\;\leq u \leq    d_{j}-\lambda \qquad j=1,\ldots ,r\\
        j\cdot \lambda +u-d_{j} & \text{for } d_{j}-\lambda\leq u\leq d_{j} \qquad \quad\;\;\, j=1,\ldots ,r-1 \qquad  \text{with } r\geq 2 \\
        r\cdot \lambda +u-d_{r} & \text{for } d_{r}-\lambda\leq u
    \end{cases}
\end{align*}
| improve this answer | |

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.