如何绘制这样的图像？数学和文字还可以，但是水平线和垂直线真的让我很困扰

Question

如何绘制这样的图像？数学和文本还可以，但是水平和垂直线确实困扰着我。

Answer 1

我建议tikzmark这样做。你必须运行三次。

\documentclass[fleqn]{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{tikzmark}
\begin{document}
\[ \qquad\qquad\tikzmarknode[inner sep=1pt]{f}{f(x)}~\tikzmarknode[inner sep=1pt]{eq}{=}~a_0
+\sum\limits_{n=1}^\infty \left(a_n \cos(n\,x)+b_n \cos(n\,x)\right)\]
\medskip
\begin{tabular}{p{2.5cm}l}
\tikzmarknode[inner sep=1pt]{b}{bounded} & \tikzmarknode[inner sep=1pt]{p}{pointwise convergence}\\
integrable & uniform convergence\\
\dots & \dots \\
\end{tabular}
\begin{tikzpicture}[overlay,remember picture]
\draw ([yshift=0.5ex]f.south west) |- (f.south east) coordinate[pos=0.75] (f1)
 -- ++ (0,0.5ex);
\draw ([yshift=0.5ex]eq.south west) |- (eq.south east) coordinate[pos=0.75] (eq1)
 -- ++ (0,0.5ex);
\draw ([yshift=-0.5ex]b.north west) |- (b.north east) coordinate[pos=0.75] (b1)
 -- ++ (0,-0.5ex);
\draw ([yshift=-0.5ex]p.north west) |- (p.north east) coordinate[pos=0.75] (p1)
 -- ++ (0,-0.5ex);
\draw (f1) -- ++ (0,-1ex) |- ([yshift=1ex]b1) -- (b1);
\draw (eq1) -- ++ (0,-1ex) |- ([yshift=1ex]p1) -- (p1);
\end{tikzpicture}
\end{document}

或者

\documentclass[fleqn]{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{tikzmark}
\begin{document}
\[ \qquad\qquad\tikzmarknode[inner sep=1pt]{f}{f(x)}~\tikzmarknode[inner sep=1pt]{eq}{=}~a_0
+\sum\limits_{n=1}^\infty \left(a_n \cos(n\,x)+b_n \cos(n\,x)\right)\]
\medskip
\begin{tabular}{p{2.5cm}l}
\tikzmarknode[inner sep=1pt]{b}{bounded} & \tikzmarknode[inner sep=1pt]{p}{pointwise convergence}\\
integrable & uniform convergence\\
\dots & \dots \\
\end{tabular}
\begin{tikzpicture}[overlay,remember picture,semithick]
\draw ([yshift=0.5ex]f.south west) |- (f.south east) coordinate[pos=0.75] (f1)
 -- ++ (0,0.5ex);
\draw ([yshift=0.5ex]f.south-|eq.west) |- (f.south-|eq.east) coordinate[pos=0.75] (eq1)
 -- ++ (0,0.5ex);
\draw ([yshift=-0.5ex]b.north west) |- (b.north east) coordinate[pos=0.75] (b1)
 -- ++ (0,-0.5ex);
\draw ([yshift=-0.5ex]p.north west) |- (p.north east) coordinate[pos=0.75] (p1)
 -- ++ (0,-0.5ex);
\draw (f1) -- ++ (0,-1ex) |- ([yshift=1ex]b1) -- (b1);
\draw (eq1) -- ++ (0,-1ex) |- ([yshift=1ex]p1) -- (p1);
\end{tikzpicture}
\end{document}

Answer 2

在以下人员的大力帮助下remember picture：

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,positioning}
\begin{document}
\[\tikz[baseline,remember picture]
 \node[inner xsep=0pt,minimum height=.6cm,anchor=base] (f) {$f(x)$};\;
\tikz[baseline,remember picture]
 \node[inner xsep=0pt,minimum height=.6cm,anchor=base] (e)
 {$=$\vphantom{$f(x)$}};\;a_0+\sum_{n=1}^\infty a_n\cos(nx)+b_n\sin(nx)\]

\begin{tikzpicture}[overlay,remember picture]
\draw (f.south west)|-($(f.south east)+(0,-.1)$)--(f.south east);
\draw ($(f.south)+(0,-.1)$)--++(0,-.3)-|++(-1,-.3) node[below,align=left] 
{bounded\\integrable\\continuous\\differentiable\\$f'$ continuous};
\draw (e.south west)|-($(e.south east)+(0,-.1)$)--(e.south east);
\draw ($(e.south)+(0,-.1)$)--++(0,-.3)-|++(1,-.3) 
node[below right=0pt and -5ex,align=left] 
{pointwise convergence\\uniform convergence\\$L^2$ convergence\\Cesaro mean convergence};
\end{tikzpicture}
\end{document}

---

然而，由于选项overlay，Ti钾Z 图片与普通文档配合得不太好。

一个棘手的解决方案是\vspace{}：

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,positioning}
\usepackage{lipsum}
\begin{document}
\lipsum[1]
\[\tikz[baseline,remember picture]
 \node[inner xsep=0pt,minimum height=.6cm,anchor=base] (f) {$f(x)$};\;
\tikz[baseline,remember picture]
 \node[inner xsep=0pt,minimum height=.6cm,anchor=base] (e)
 {$=$\vphantom{$f(x)$}};\;a_0+\sum_{n=1}^\infty a_n\cos(nx)+b_n\sin(nx)\]

\begin{tikzpicture}[overlay,remember picture]
\draw (f.south west)|-($(f.south east)+(0,-.1)$)--(f.south east);
\draw ($(f.south)+(0,-.1)$)--++(0,-.3)-|++(-1,-.3) node[below,align=left] 
{bounded\\integrable\\continuous\\differentiable\\$f'$ continuous};
\draw (e.south west)|-($(e.south east)+(0,-.1)$)--(e.south east);
\draw ($(e.south)+(0,-.1)$)--++(0,-.3)-|++(1,-.3) 
node[below right=0pt and -5ex,align=left] 
{pointwise convergence\\uniform convergence\\$L^2$ convergence\\Cesaro mean convergence};
\end{tikzpicture}
\vspace{6\baselineskip}

\lipsum[2]
\end{document}

欢迎任何建议。