\begin{align*}
\sum_{j=1}^{m_0-1}\,(-1)^{j-1}\hspace{-3ex}
\sum_{\substack{ k_{m_0 +1} \leq k_{m_0} \leq \cdots \leq k_{j+2} \leq k_1 \leq \boldsymbol{\ell} \\ k_2 < \cdots < k_j < k_{j+1} \leq k_1 }}
= \sum_{ k_{m_0 +1} \leq k_{m_0} \leq \cdots \leq k_2 \leq k_1 \leq
\boldsymbol{\ell}, } \tag(3) \label{eq:3}
\end{align*}
which is very close to our goal, equation \eqref{eq:2}.
Comparing equation \eqref{eq:3} with equation \eqref{eq:2}, we discover that both have the identical 1\ts{st} through $(m_0 -2)$\ts{th} terms ($1 \leq j \leq m_0-2$).
The only difference of the two is that the $(m_0 -1)$\ts{th} term in equation \eqref{eq:3} is ...
\end{document}
我尝试在等式旁边放置一个标签,但结果却是这样的:
标签显示不正确。此外, 也以相同的错误方式显示。(我省略了和 的\eqref{eq:3}
代码。)eq:2
eq:1
答案1
\tag{<tag>}
接受强制参数。如果不提供括号,则接受下一个令牌作为论据。
在您的情况下,\tag(3)
不使用建议的格式\tag{<tag>}
,因此它变为\tag{(}3)
,用对方程进行编号(
并留3)
在输入流中。您正在寻找\tag{3}
:
\documentclass{article}
\usepackage{amsmath,amssymb}
\begin{document}
\begin{align*}
\sum_{j = 1}^{m_0 - 1}\,(-1)^{j - 1}\hspace{-3ex}
\sum_{
\substack
{ k_{m_0 + 1} \leq k_{m_0} \leq \cdots \leq k_{j + 2} \leq k_1 \leq \boldsymbol{\ell} \\
k_2 < \cdots < k_j < k_{j+1} \leq k_1
}
}
= \sum_{ k_{m_0 + 1} \leq k_{m_0} \leq \cdots \leq k_2 \leq k_1 \leq \boldsymbol{\ell} }, \tag{3} \label{eq:3}
\end{align*}
which is very close to our goal, equation~(2). Comparing equation~\eqref{eq:3} with equation~(2), we discover that both have the identical
1\textsuperscript{st} through $(m_0 - 2)$\textsuperscript{th} terms ($1 \leq j \leq m_0 - 2$). The only difference of the two is that the
$(m_0 - 1)$\textsuperscript{th} term in equation \eqref{eq:3} is \ldots
\end{document}