我该如何修复涉及调整大小的方程式中的错误？

Question

我该如何调整

\begin{align}
\resizebox{.9\hsize}{!}{$\displaystyle{\inf \lambda^{*}_{A}(S)=
\left\{\frac{\sum\limits_{k=1}^{m}\ell(I_{k,\epsilon})
\bigg[1-\mu(A)(1-\mu(S\capI_{k,\epsilon}))\bigg]}{\sum\limits_{k=1}^{n}\ell(J_{k,\epsilon})
\bigg[1-\mu(A)(1-\mu(A\cap J_{k,\epsilon}))\bigg]}:
S\subseteq\bigcup\limits_{k=1}^{m}I_{k,\epsilon},
A\subseteq\bigcup\limits_{k=1}^{n}J_{k,\epsilon} \right\}}$}
\end{align}\\

以避免错误。目前，我的预览页面是空白的。

以下是我使用的所有软件包

\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{hyperref}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{cases}
\usepackage{graphicx}
\DeclareMathOperator{\sign}{sign}
\makeatletter
\newcommand{\vast}{\bBigg@{4.5}}
\newcommand{\Vast}{\bBigg@{5}}
\makeatother

Answer 1

调整材料大小是危险的；当您确定文本是最终形式时，将问题留到最后的修订。

关于您输入的几点说明：\bigg应该是\biggl或\biggr（对于左边、开幕和正确的，关闭），但这里它太大了：一个简单的\bigl–\bigr对就可以更好地完成这项工作。

为了应用\small或者\footnotesize最好在equation环境里面使用一个盒子，否则显示器周围的线间距可能会不均匀。

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{cases}
\usepackage{graphicx,varwidth}
\usepackage{hyperref}

\usepackage{lipsum}% just for the example

\DeclareMathOperator{\sign}{sign}
\makeatletter
\newcommand{\vast}{\bBigg@{4.5}}
\newcommand{\Vast}{\bBigg@{5}}
\makeatother

\begin{document}

Equation\eqref{small} has been typeset in \verb|\small| size; depending
on your text width you may not need the \verb|\hspace| instructions,
which are needed here to avoid the equation number is moved down.
In equation\eqref{footnotesize}, there should be no need for tricks.

\lipsum[1][1-6]
\begin{equation}\label{small}
\hspace{-0.5em}
\mbox{\small$\displaystyle
  \inf \lambda^{*}_{A}(S)=
  \left\{
    \frac{
      \sum\limits_{k=1}^{m}\ell(I_{k,\epsilon})
      \bigl[1-\mu(A)(1-\mu(S\cap I_{k,\epsilon}))\bigr]
    }{
      \sum\limits_{k=1}^{n}\ell(J_{k,\epsilon})
      \bigl[1-\mu(A)(1-\mu(A\cap J_{k,\epsilon}))\bigr]
    }
    :
    S\subseteq\bigcup\limits_{k=1}^{m}I_{k,\epsilon},
    A\subseteq\bigcup\limits_{k=1}^{n}J_{k,\epsilon}
  \right\}
$}
\hspace{10000pt minus 1fil}
\end{equation}
\lipsum[1][1-6]
\begin{equation}\label{footnotesize}
\mbox{\footnotesize$\displaystyle
  \inf \lambda^{*}_{A}(S)=
  \left\{
    \frac{
      \sum\limits_{k=1}^{m}\ell(I_{k,\epsilon})
      \bigl[1-\mu(A)(1-\mu(S\cap I_{k,\epsilon}))\bigr]
    }{
      \sum\limits_{k=1}^{n}\ell(J_{k,\epsilon})
      \bigl[1-\mu(A)(1-\mu(A\cap J_{k,\epsilon}))\bigr]
    }
    :
    S\subseteq\bigcup\limits_{k=1}^{m}I_{k,\epsilon},
    A\subseteq\bigcup\limits_{k=1}^{n}J_{k,\epsilon}
  \right\}
$}
\end{equation}
\lipsum[1][1-6]

\end{document}

Answer 2

你的例子是

\documentclass{article}

\usepackage{amsmath,graphicx}

\begin{document}
\begin{align}
\resizebox{.9\hsize}{!}{$\displaystyle{\inf \lambda^{*}_{A}(S)=
\left\{\frac{\sum\limits_{k=1}^{m}\ell(I_{k,\epsilon})
\bigg[1-\mu(A)(1-\mu(S\capI_{k,\epsilon}))\bigg]}{\sum\limits_{k=1}^{n}\ell(J_{k,\epsilon})
\bigg[1-\mu(A)(1-\mu(A\cap J_{k,\epsilon}))\bigg]}:
S\subseteq\bigcup\limits_{k=1}^{m}I_{k,\epsilon},
A\subseteq\bigcup\limits_{k=1}^{n}J_{k,\epsilon} \right\}}$}
\end{align}\\
\end{document}

产生

! Undefined control sequence.
<argument> ...n }) \bigg [1-\mu (A)(1-\mu (S\capI 
                                                  _{k,\epsilon }))\bigg ]
l.13 \end{align}
                \\
? 

Underfull \hbox (badness 10000) in paragraph at lines 13--14

因此唯一的错误\capI应该是，并且放错位置后\cap I会发出警告。\\\end{align}

但是您应该使用\bigl[而\bigr]不是\big 并且align在这里什么也不做，因为您没有指定对齐。

\documentclass[a4paper]{article}

\usepackage{amsmath,graphicx}

\begin{document}
\begin{multline}
\inf \lambda^{*}_{A}(S)=\\
\left\{\frac{\sum\limits_{k=1}^{m}\ell(I_{k,\epsilon})
\biggl[1-\mu(A)(1-\mu(S\cap I_{k,\epsilon}))\biggr]}{\sum\limits_{k=1}^{n}\ell(J_{k,\epsilon})
\biggl[1-\mu(A)(1-\mu(A\cap J_{k,\epsilon}))\biggr]}:
S\subseteq\bigcup\limits_{k=1}^{m}I_{k,\epsilon},
A\subseteq\bigcup\limits_{k=1}^{n}J_{k,\epsilon} \right\}
\end{multline}

Or if you really must fit on one line
{\small\begin{equation}
\inf \lambda^{*}_{A}(S)=
\left\{\frac{\sum\limits_{k=1}^{m}\ell(I_{k,\epsilon})
\bigl[1{-}\mu(A)(1{-}\mu(S\cap I_{k,\epsilon}))\bigr]}{\sum\limits_{k=1}^{n}\ell(J_{k,\epsilon})
\bigl[1{-}\mu(A)(1{-}\mu(A\cap J_{k,\epsilon}))\bigr]}:
S\subseteq\bigcup\limits_{k=1}^{m}I_{k,\epsilon},
A\subseteq\bigcup\limits_{k=1}^{n}J_{k,\epsilon} \right\}
\end{equation}}
\end{document}