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需要帮助来创建一个更灵活的 \cancel 命令，该命令可以在对齐环境中工作

tag-icon • tag-icon tikz-pgf align cancel
需要帮助来创建一个更灵活的 \cancel 命令，该命令可以在对齐环境中工作

请考虑下面的片段。

我想取消方程（1）的两个中间项，并将方程（2）中括号之间的表达式取消为 1。

如果我使用包中的\cancel和，这会导致对角线起始太低结束太高，因为表达式非常长。对于小表达式，结果很好。\canceltocancel

针对此问题的一些解决方案斜线开始太低，结束太高。然而，这些解决方案似乎只适用于内联方程。例如，Speravir 给出的解决方案看起来不错，但是如果我在对齐环境中使用它，它会改变被取消的术语的排版，使其与其他术语不同，如下所示，我想要取消的第二个术语。

问题：

  1. 是否有可能调整上述解决方案以在对齐环境中工作？

  2. 是否有可能创建一个等效的 \hcancelto 命令，用于公式 (2)？

谢谢！

附言：为了与我的文档保持一致，我yathesis在代码片段中使用了文档类（用法语写论文）。没有yathesis安装该类的人可以使用书籍或文章类。

\documentclass[mainlanguage=english,babel={main=english,french,brazil},sepcorpaffilfrench={,~},sepcorpaffilenglish={,~},version=inprogress]{yathesis}
%\documentclass{book}
%\documentclass{article}
%
% If using XeLaTex or LuaLaTex
%\usepackage{fontspec}
%
% If using LaTex or PdfLaTex
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
%
\usepackage{lmodern} % latin modern font
\usepackage{amsmath}
\usepackage{cancel} % draw diagonal lines  ("cancelling" a term)
%
%%% Code from https://tex.stackexchange.com/a/156581/95438 %%%
%
\usepackage{keycommand}
% Patch by Joseph Wright ("bug in the definition of \ifcommandkey (2010/04/27 v3.1415)"),
% https://tex.stackexchange.com/a/35794
\begingroup
\makeatletter
\catcode`\/=8 %
\@firstofone
{
    \endgroup
    \renewcommand{\ifcommandkey}[1]{%
        \csname @\expandafter \expandafter \expandafter
        \expandafter \expandafter \expandafter \expandafter
        \kcmd@nbk \commandkey {#1}//{first}{second}//oftwo\endcsname
    }
}
%--------%
\usepackage{tikz}
\usetikzlibrary{calc}
\newkeycommand{\hcancel}[hshiftstart=0pt,vshiftstart=0pt,hshiftend=0pt,vshiftend=0pt,color=red][1]{%
    \tikz[baseline=(tocancel.base)]{
        \node[inner sep=0pt,outer sep=0pt] (tocancel) {#1};
        \draw[\commandkey{color}] ($(tocancel.south west)+(\commandkey{hshiftstart},\commandkey{vshiftstart})$) --
        ($(tocancel.north east)+(\commandkey{hshiftend},\commandkey{vshiftend})$);
    }%
}%
%
%%% End of code %%%
%
\begin{document}
%
\begin{align}
\begin{split}
&= \mp \cos^{2} \left[ 2\pi\left(  \frac{k}{2}t^{2} + f_{1}t + \varphi\right)\right] \int_{0}^{t} s_{i}\left( \tau\right) \sin \left[ 2\pi\left( kt + f_{1}\right)\tau\right] \, d\tau \\
&\mathrel{\phantom{=}} \cancel{\pm \frac{1}{2} \sin \left[ 4\pi\left(  \frac{k}{2}t^{2} + f_{1}t + \varphi\right)\right] \int_{0}^{t} s_{i}\left( \tau\right) \cos \left[ 2\pi\left( kt + f_{1}\right)\tau\right] \, d\tau} \\
&\mathrel{\phantom{=}} \hcancel{$\mp \frac{1}{2} \sin \left[ 4\pi\left(  \frac{k}{2}t^{2} + f_{1}t + \varphi\right)\right] \int_{0}^{t} s_{i}\left( \tau\right) \cos \left[ 2\pi\left( kt + f_{1}\right)\tau\right] \, d\tau$} \\
&\mathrel{\phantom{=}} \mp \sin^{2} \left[ 2\pi\left(  \frac{k}{2}t^{2} + f_{1}t + \varphi\right)\right] \int_{0}^{t} s_{i}\left( \tau\right) \sin \left[ 2\pi\left( kt + f_{1}\right)\tau\right] \, d\tau
\end{split} \\
\begin{split}
&= \mp \left\lbrace \cancelto{1}{\cos^{2} \left[ 2\pi\left(  \frac{k}{2}t^{2} + f_{1}t + \varphi\right)\right] + \sin^{2} \left[ 2\pi\left(  \frac{k}{2}t^{2} + f_{1}t + \varphi\right)\right]} \right\rbrace \\
&\mathrel{\phantom{=}} \int_{0}^{t} s_{i}\left( \tau\right) \sin \left[ 2\pi\left( kt + f_{1}\right)\tau\right] \, d\tau
\end{split} \\
&= \mp \int_{0}^{t} s_{i}\left( \tau\right) \sin \left[ 2\pi\left( kt + f_{1}\right)\tau\right] \, d\tau
\end{align}
%
\end{document}

输出：

在此处输入图片描述

答案1

我有一个建议。最新版本的tikzmark有一个新命令，\tikzmarkmode，它有许多令人惊叹的功能，其中之一就是它可以检测您所处的模式。因此，$如果您使用它，您不必添加任何内容，也不必担心内联方程式与显示样式。只需一张简单的路径图即可实现取消。

\documentclass[mainlanguage=english,babel={main=english,french,brazil},sepcorpaffilfrench={,~},sepcorpaffilenglish={,~},version=inprogress]{yathesis}
%\documentclass{book}
%\documentclass{article}
%
% If using XeLaTex or LuaLaTex
%\usepackage{fontspec}
%
% If using LaTex or PdfLaTex
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
%
\usepackage{lmodern} % latin modern font
\usepackage{amsmath}
\usepackage{cancel} % draw diagonal lines  ("cancelling" a term)
%
%%% Code from https://tex.stackexchange.com/a/156581/95438 %%%
%
\usepackage{keycommand}
% Patch by Joseph Wright ("bug in the definition of \ifcommandkey (2010/04/27 v3.1415)"),
% https://tex.stackexchange.com/a/35794
\begingroup
\makeatletter
\catcode`\/=8 %
\@firstofone
{
    \endgroup
    \renewcommand{\ifcommandkey}[1]{%
        \csname @\expandafter \expandafter \expandafter
        \expandafter \expandafter \expandafter \expandafter
        \kcmd@nbk \commandkey {#1}//{first}{second}//oftwo\endcsname
    }
}
%--------%
\usepackage{tikz}
\usetikzlibrary{calc,tikzmark}
\newkeycommand{\hcancel}[hshiftstart=0pt,vshiftstart=0pt,hshiftend=0pt,vshiftend=0pt,color=red][1]{%
    \tikz[baseline=(tocancel.base)]{
        \node[inner sep=0pt,outer sep=0pt] (tocancel) {#1};
        \draw[\commandkey{color}] ($(tocancel.south west)+(\commandkey{hshiftstart},\commandkey{vshiftstart})$) --
        ($(tocancel.north east)+(\commandkey{hshiftend},\commandkey{vshiftend})$);
    }%
}%
\tikzset{cancel/.style={path picture={
\draw[#1] (path picture bounding box.south west) -- 
(path picture bounding box.north east);
}}}
%
%%% End of code %%%
%
\begin{document}
%
\begin{align}
\begin{split}
&= \mp \cos^{2} \left[ 2\pi\left(  \frac{k}{2}t^{2} + f_{1}t + \varphi\right)\right] \int_{0}^{t} s_{i}\left( \tau\right) \sin \left[ 2\pi\left( kt + f_{1}\right)\tau\right] \, d\tau \\
&\mathrel{\phantom{=}} \cancel{\pm \frac{1}{2} \sin \left[ 4\pi\left(  \frac{k}{2}t^{2} + f_{1}t + \varphi\right)\right] \int_{0}^{t} s_{i}\left( \tau\right) \cos \left[ 2\pi\left( kt + f_{1}\right)\tau\right] \, d\tau} \\
&\mathrel{\phantom{=}} \tikzmarknode[cancel]{cc}{\mp \frac{1}{2} \sin \left[ 4\pi\left(  \frac{k}{2}t^{2} + f_{1}t + \varphi\right)\right] \int_{0}^{t} s_{i}\left( \tau\right) \cos \left[ 2\pi\left( kt + f_{1}\right)\tau\right] \, d\tau} \\
&\mathrel{\phantom{=}} \mp \sin^{2} \left[ 2\pi\left(  \frac{k}{2}t^{2} + f_{1}t + \varphi\right)\right] \int_{0}^{t} s_{i}\left( \tau\right) \sin \left[ 2\pi\left( kt + f_{1}\right)\tau\right] \, d\tau
\end{split} \\
\begin{split}
&= \mp \left\lbrace \cancelto{1}{\cos^{2} \left[ 2\pi\left(  \frac{k}{2}t^{2} + f_{1}t + \varphi\right)\right] + \sin^{2} \left[ 2\pi\left(  \frac{k}{2}t^{2} + f_{1}t + \varphi\right)\right]} \right\rbrace \\
&\mathrel{\phantom{=}} \int_{0}^{t} s_{i}\left( \tau\right) \sin \left[ 2\pi\left( kt + f_{1}\right)\tau\right] \, d\tau
\end{split} \\
&= \tikzmarknode[cancel=red]{cc}{\mp \int_{0}^{t} s_{i}\left( \tau\right) \sin
\left[ 2\pi\left( kt + f_{1}\right)\tau\right] \, d\tau}
\end{align}
%
\end{document}

在此处输入图片描述

离题了，但我会把d积分中出现的微分排版成直立的。

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