在显示和文本模式下使用素数和运算符

Question 1

您可以借助_以下\sump方法进行测试xparse：

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

\makeatletter
\NewDocumentCommand{\sump}{e{_}}
 {%
  \DOTSB
  \mathop{\IfNoValueTF{#1}{\sump@{}}{\sump@{#1}}}%
  \nolimits
 }
\newcommand{\sump@}[1]{\mathpalette\sump@@{#1}}
\newcommand{\sump@@}[2]{%
  \ifx#1\displaystyle
    {\sump@display{#2}}%
  \else
    \sum@\nolimits'_{#2}%
  \fi
}
\newcommand{\sump@display}[1]{%
  \sbox\z@{$\m@th\displaystyle\sum@\nolimits'$}%
  \sbox\tw@{$\m@th\displaystyle\sum@\limits_{#1}$}%
  \sbox\@tempboxa{$\m@th\displaystyle'$}
  \mathop{\sum@\nolimits' \kern-\wd\@tempboxa}\limits_{#1}%
  \ifdim\wd\z@>\wd\tw@
    \kern\dimexpr\wd\z@-\wd\tw@\relax
  \fi
}
\makeatother

\begin{document}

\begin{align}
&\sum_{n} a_n &&\textstyle \sum_{n\ge0} a_n \\
&\sump_{n} a_n &&\textstyle \sump_{n\ge0} a_n \\
&\sum_{n\ge0} a_n &&\textstyle \sum_{n\ge0} a_n \\
&\sump_{n\ge0} a_n &&\textstyle \sump_{n\ge0} a_n \\
&\sum_{1\le n\le 32} a_n &&\textstyle \sum_{1\le n\le 32} a_n \\
&\sump_{1\le n\le 32} a_n &&\textstyle \sump_{1\le n\le 32} a_n
\end{align}

\end{document}

在“非显示”情况下，只需排版\sum'_{#1}，在显示情况下，我测量素数，以便调整字距并生成以求和符号为中心的下标。然后，我测量带素数的和和带下标的和，以决定是否需要进行一些字距调整，如果“带素数的和”的大小超过“带下标的和”的大小，则需要进行字距调整。

Answer

您可以借助_以下\sump方法进行测试xparse：

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

\makeatletter
\NewDocumentCommand{\sump}{e{_}}
 {%
  \DOTSB
  \mathop{\IfNoValueTF{#1}{\sump@{}}{\sump@{#1}}}%
  \nolimits
 }
\newcommand{\sump@}[1]{\mathpalette\sump@@{#1}}
\newcommand{\sump@@}[2]{%
  \ifx#1\displaystyle
    {\sump@display{#2}}%
  \else
    \sum@\nolimits'_{#2}%
  \fi
}
\newcommand{\sump@display}[1]{%
  \sbox\z@{$\m@th\displaystyle\sum@\nolimits'$}%
  \sbox\tw@{$\m@th\displaystyle\sum@\limits_{#1}$}%
  \sbox\@tempboxa{$\m@th\displaystyle'$}
  \mathop{\sum@\nolimits' \kern-\wd\@tempboxa}\limits_{#1}%
  \ifdim\wd\z@>\wd\tw@
    \kern\dimexpr\wd\z@-\wd\tw@\relax
  \fi
}
\makeatother

\begin{document}

\begin{align}
&\sum_{n} a_n &&\textstyle \sum_{n\ge0} a_n \\
&\sump_{n} a_n &&\textstyle \sump_{n\ge0} a_n \\
&\sum_{n\ge0} a_n &&\textstyle \sum_{n\ge0} a_n \\
&\sump_{n\ge0} a_n &&\textstyle \sump_{n\ge0} a_n \\
&\sum_{1\le n\le 32} a_n &&\textstyle \sum_{1\le n\le 32} a_n \\
&\sump_{1\le n\le 32} a_n &&\textstyle \sump_{1\le n\le 32} a_n
\end{align}

\end{document}

在“非显示”情况下，只需排版\sum'_{#1}，在显示情况下，我测量素数，以便调整字距并生成以求和符号为中心的下标。然后，我测量带素数的和和带下标的和，以决定是否需要进行一些字距调整，如果“带素数的和”的大小超过“带下标的和”的大小，则需要进行字距调整。

Question 2

正如评论中所说，这个问题是练习 18.44 的主要主题TeXbook. Barbara Beeton 已经建议这是 Knuth 在附录 A 中提出的两个解决方案之一的变体，尽管她的代码应该或多或少地修改为：

\DeclareRobustCommand*{\sump}{%
    \mathop{{\sum}^{\mathrlap{\prime}}}%
}

这是一个完整的、可编译的示例，它还提请注意该解决方案可能存在的缺陷：

% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly 
                                 % declare the paper format.

\usepackage[T1]{fontenc}         % Not always necessary, but recommended.
% End of standard header.  What follows pertains to the problem at hand.

\usepackage{mathtools}

\newcommand*{\sump}{} % check that the name can be used
\DeclareRobustCommand*{\sump}{%
    \mathop{{\sum}^{\mathrlap{\prime}}}%
}



\begin{document}

First in-line:
\( \sump_{i\in I}a_{i} \ne \sum_{i\in I}a_{i} \).
Then in display:
\[
    \sump_{i\in I}a_{i} \ne \sum_{i\in I}a_{i}
    \mbox{.}
\]

You can also say, for instance,
\( \sump\limits_{i\in I}a_{i} \ne \sum\limits_{i\in I}a_{i} \)
and
\[
    \sump\nolimits_{i\in I}a_{i} \ne \sum\nolimits_{i\in I}a_{i}
    \mbox{,}
\]
respectively.

There's a caveat, though: because the prime is not taken into account
when computing the width of the \verb|\sump| symbol, it may bump into
adjacent characters:
\[
    \sump \biggl[\frac{a+b}{c+d}-\frac{x+y}{x-y}\biggr]
        \ne \sum \biggl[\frac{a+b}{c+d}-\frac{x+y}{x-y}\biggr]
\]
The clash is more problematic in in-line math:
\( \sump[a+b] \ne \sum[a+b] \).
However, in practice this is not going to be a problem if you stick to
using the \verb|\sump| symbol only with a (non-empty) subscript.

\end{document}

然而，如果我们必须加载该mathtools包，而这又需要amsmath，我认为我们也应该支持后者的 [ no]sumlimits选项，以及它的\dots功能：

% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly 
                                 % declare the paper format.

\usepackage[T1]{fontenc}         % Not always necessary, but recommended.
% End of standard header.  What follows pertains to the problem at hand.

% Try decommenting the following line:
% \usepackage[nosumlimits]{amsmath}
\usepackage{mathtools}

\makeatletter

% Cannot directly use "\DeclareRobustCommand" with "\DOTSB":
\@ifdefinable\sump@{\DeclareRobustCommand*\sump@{%
    \mathop{{\sum}^{\mathrlap{\prime}}}%
}}
\newcommand*\sump{%
    \DOTSB\sump@\slimits@ % ...or move "\slimits@" into "\sump@"
}

\makeatother



\begin{document}

First in-line:
\( \sump_{i\in I}a_{i} \ne \sum_{i\in I}a_{i} \).
Then in display:
\begin{equation}
    \sump_{i\in I}a_{i} \ne \sum_{i\in I}a_{i}
    \mbox{.}
    \label{eq:displaylimits}
\end{equation}

You can also say, for instance,
\( \sump\limits_{i\in I}a_{i} \ne \sum\limits_{i\in I}a_{i} \)
and
\[
    \sump\nolimits_{i\in I}a_{i} \ne \sum\nolimits_{i\in I}a_{i}
    \mbox{,}
\]
respectively.  And see what happens in the ``default'' case (that is,
in~\eqref{eq:displaylimits}) if you load the \textsf{amsmath} package
with the \texttt{nosumlimits} option.

Finally, note that
\( \sump_{i_{1}}\dots\sump_{i_{p}} x_{i_{1}}\otimes\dots\otimes x_{i_{p}} \)
works exactly in the same way as
\( \sum_{i_{1}}\dots\sum_{i_{p}} x_{i_{1}}\otimes\dots\otimes x_{i_{p}} \)
(as one would expect).  Let's repeat it in display:
\[
    \sump_{i_{1}}\dots\sump_{i_{p}} x_{i_{1}}\otimes\dots\otimes x_{i_{p}}
    \ne
    \sum_{i_{1}}\dots\sum_{i_{p}} x_{i_{1}}\otimes\dots\otimes x_{i_{p}}
\]
(this way the formula is more readable!).

\typeout{A test for robustness: \sump}
\typeout{Compare the above with \sum}

\end{document}

请注意，上述代码（第二个 MWE）在编译期间写入了几条消息，这些消息说明了如何将命令\sump写出到辅助文件中，并表明其在这方面的行为与的行为相似\sum。

Answer

正如评论中所说，这个问题是练习 18.44 的主要主题TeXbook. Barbara Beeton 已经建议这是 Knuth 在附录 A 中提出的两个解决方案之一的变体，尽管她的代码应该或多或少地修改为：

\DeclareRobustCommand*{\sump}{%
    \mathop{{\sum}^{\mathrlap{\prime}}}%
}

这是一个完整的、可编译的示例，它还提请注意该解决方案可能存在的缺陷：

% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly 
                                 % declare the paper format.

\usepackage[T1]{fontenc}         % Not always necessary, but recommended.
% End of standard header.  What follows pertains to the problem at hand.

\usepackage{mathtools}

\newcommand*{\sump}{} % check that the name can be used
\DeclareRobustCommand*{\sump}{%
    \mathop{{\sum}^{\mathrlap{\prime}}}%
}



\begin{document}

First in-line:
\( \sump_{i\in I}a_{i} \ne \sum_{i\in I}a_{i} \).
Then in display:
\[
    \sump_{i\in I}a_{i} \ne \sum_{i\in I}a_{i}
    \mbox{.}
\]

You can also say, for instance,
\( \sump\limits_{i\in I}a_{i} \ne \sum\limits_{i\in I}a_{i} \)
and
\[
    \sump\nolimits_{i\in I}a_{i} \ne \sum\nolimits_{i\in I}a_{i}
    \mbox{,}
\]
respectively.

There's a caveat, though: because the prime is not taken into account
when computing the width of the \verb|\sump| symbol, it may bump into
adjacent characters:
\[
    \sump \biggl[\frac{a+b}{c+d}-\frac{x+y}{x-y}\biggr]
        \ne \sum \biggl[\frac{a+b}{c+d}-\frac{x+y}{x-y}\biggr]
\]
The clash is more problematic in in-line math:
\( \sump[a+b] \ne \sum[a+b] \).
However, in practice this is not going to be a problem if you stick to
using the \verb|\sump| symbol only with a (non-empty) subscript.

\end{document}

然而，如果我们必须加载该mathtools包，而这又需要amsmath，我认为我们也应该支持后者的 [ no]sumlimits选项，以及它的\dots功能：

% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly 
                                 % declare the paper format.

\usepackage[T1]{fontenc}         % Not always necessary, but recommended.
% End of standard header.  What follows pertains to the problem at hand.

% Try decommenting the following line:
% \usepackage[nosumlimits]{amsmath}
\usepackage{mathtools}

\makeatletter

% Cannot directly use "\DeclareRobustCommand" with "\DOTSB":
\@ifdefinable\sump@{\DeclareRobustCommand*\sump@{%
    \mathop{{\sum}^{\mathrlap{\prime}}}%
}}
\newcommand*\sump{%
    \DOTSB\sump@\slimits@ % ...or move "\slimits@" into "\sump@"
}

\makeatother



\begin{document}

First in-line:
\( \sump_{i\in I}a_{i} \ne \sum_{i\in I}a_{i} \).
Then in display:
\begin{equation}
    \sump_{i\in I}a_{i} \ne \sum_{i\in I}a_{i}
    \mbox{.}
    \label{eq:displaylimits}
\end{equation}

You can also say, for instance,
\( \sump\limits_{i\in I}a_{i} \ne \sum\limits_{i\in I}a_{i} \)
and
\[
    \sump\nolimits_{i\in I}a_{i} \ne \sum\nolimits_{i\in I}a_{i}
    \mbox{,}
\]
respectively.  And see what happens in the ``default'' case (that is,
in~\eqref{eq:displaylimits}) if you load the \textsf{amsmath} package
with the \texttt{nosumlimits} option.

Finally, note that
\( \sump_{i_{1}}\dots\sump_{i_{p}} x_{i_{1}}\otimes\dots\otimes x_{i_{p}} \)
works exactly in the same way as
\( \sum_{i_{1}}\dots\sum_{i_{p}} x_{i_{1}}\otimes\dots\otimes x_{i_{p}} \)
(as one would expect).  Let's repeat it in display:
\[
    \sump_{i_{1}}\dots\sump_{i_{p}} x_{i_{1}}\otimes\dots\otimes x_{i_{p}}
    \ne
    \sum_{i_{1}}\dots\sum_{i_{p}} x_{i_{1}}\otimes\dots\otimes x_{i_{p}}
\]
(this way the formula is more readable!).

\typeout{A test for robustness: \sump}
\typeout{Compare the above with \sum}

\end{document}

请注意，上述代码（第二个 MWE）在编译期间写入了几条消息，这些消息说明了如何将命令\sump写出到辅助文件中，并表明其在这方面的行为与的行为相似\sum。

在显示和文本模式下使用素数和运算符

答案1

答案2

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