绘制括号多项式的规则

Question 1

您应该为图片定义宏；这是一种可能性。

\documentclass{article}
\usepackage[utf8]{inputenc}

\usepackage{mathtools}
\usepackage{amssymb}
\usepackage{tikz}
\usetikzlibrary{shapes,snakes}

\newcommand{\KP}[1]{%
  \begin{tikzpicture}[baseline=-\dimexpr\fontdimen22\textfont2\relax]
  #1
  \end{tikzpicture}%
}
\newcommand{\KPA}{%
  \KP{\filldraw[color=gray, fill=none, thick] circle (0.3);}%
}
\newcommand{\KPB}{%
  \KP{
    \draw[color=gray,thick] (-0.3,0.3) -- (0.3,-0.3);
    \draw[color=gray,thick] (-0.3,-0.3) -- (-0.05,-0.05);
    \draw[color=gray,thick] (0.05,0.05) -- (0.3,0.3);
  }%
}
\newcommand{\KPC}{%
  \KP{%
    \draw[color=gray,thick] (-0.3,0.3) .. controls (0,-0.05) .. (0.3,0.3);
    \draw[color=gray,thick] (-0.3,-0.3) .. controls (0,0.05) .. (0.3,-0.3);
  }%
}
\newcommand{\KPD}{%
  \KP{%
    \draw[color=gray,thick] (-0.3,-0.3) .. controls (0.05,0) .. (-0.3,0.3);
    \draw[color=gray,thick] (0.3,-0.3) .. controls (-0.05,0) .. (0.3,0.3);
  }%
}


\begin{document}

\begin{enumerate}
%the first rule
\item
  $\left\langle\KPA\right\rangle=1$

%the second rule
\item
  $\left\langle L \cup \KPA\right\rangle=(-A^{2}-A^{-2})\langle L\rangle$

%the third rule
\item
  $\left\langle\KPB\right\rangle=
  A\left\langle\KPC\right\rangle + A^{-1} \left\langle \KPD \right\rangle$

\end{enumerate}

\end{document}

一些注释。

绝不问题\everymath{\displaystyle}。曾经。
该commath软件包提供了大量有缺陷的命令。请避免使用它。

Answer

您应该为图片定义宏；这是一种可能性。

\documentclass{article}
\usepackage[utf8]{inputenc}

\usepackage{mathtools}
\usepackage{amssymb}
\usepackage{tikz}
\usetikzlibrary{shapes,snakes}

\newcommand{\KP}[1]{%
  \begin{tikzpicture}[baseline=-\dimexpr\fontdimen22\textfont2\relax]
  #1
  \end{tikzpicture}%
}
\newcommand{\KPA}{%
  \KP{\filldraw[color=gray, fill=none, thick] circle (0.3);}%
}
\newcommand{\KPB}{%
  \KP{
    \draw[color=gray,thick] (-0.3,0.3) -- (0.3,-0.3);
    \draw[color=gray,thick] (-0.3,-0.3) -- (-0.05,-0.05);
    \draw[color=gray,thick] (0.05,0.05) -- (0.3,0.3);
  }%
}
\newcommand{\KPC}{%
  \KP{%
    \draw[color=gray,thick] (-0.3,0.3) .. controls (0,-0.05) .. (0.3,0.3);
    \draw[color=gray,thick] (-0.3,-0.3) .. controls (0,0.05) .. (0.3,-0.3);
  }%
}
\newcommand{\KPD}{%
  \KP{%
    \draw[color=gray,thick] (-0.3,-0.3) .. controls (0.05,0) .. (-0.3,0.3);
    \draw[color=gray,thick] (0.3,-0.3) .. controls (-0.05,0) .. (0.3,0.3);
  }%
}


\begin{document}

\begin{enumerate}
%the first rule
\item
  $\left\langle\KPA\right\rangle=1$

%the second rule
\item
  $\left\langle L \cup \KPA\right\rangle=(-A^{2}-A^{-2})\langle L\rangle$

%the third rule
\item
  $\left\langle\KPB\right\rangle=
  A\left\langle\KPC\right\rangle + A^{-1} \left\langle \KPD \right\rangle$

\end{enumerate}

\end{document}

一些注释。

绝不问题\everymath{\displaystyle}。曾经。
该commath软件包提供了大量有缺陷的命令。请避免使用它。

Question 2

已经给出的两个答案都有一个明显的改进，即使用数学样式使符号可扩展。即使是@egreg的答案（比@LoopSpaces的答案更灵活）在这方面也有缺陷，因为它静态地引用了“文本大小”（它使用了\textfont2）。

这些增强只是例行公事，但由于已经过去了 24 小时，我会认为这两位作者没有费心提交它们（它们也相当无聊！），并提出自己承担这项任务。

除了从缩小到\textfont和\scriptfont缩小到\scriptscriptfont尺寸之外，我还决定让符号在显示时以“大”的形式出现，就像“大”运算符（\sum、\int、\bigoplus、...）。诚然，这可能是值得怀疑的。

这是代码：

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

\usepackage[T1]{fontenc}         % Not necessary, but recommended.

\usepackage[ascii]{inputenc}     % Just to check that the source is still pure,
                                 % 7-bit-clean ASCII when you execute it, as it
                                 % was when I wrote it.
% End of standard header.  What follows pertains to the problem at hand.

\usepackage{amsmath}

% Old uncle Gustavo prefers to stick to the "picture" environment:
\usepackage{pict2e}
% \usepackage{xcolor} % I'd leave it out

%--------------------------------------------------------------%
\makeatletter

% We define separate versions (large and small) for the frames:
\newcommand*\@KP@Large@frame[2]{%
    \setlength\unitlength{\fontdimen 22 #1\tw@}%
    \vrule \@width\z@ \@height 4\unitlength \@depth\tw@\unitlength
    \begin{picture}(6,2)(-3,-1)%
        \def\@KP@Radius     {3}%
        \def\@KP@Hole@radius{.5}% The same value seem adequate for both...
        \def\@KP@Diameter   {6}%
        #2%
    \end{picture}%
}
\newcommand*\@KP@Small@frame[2]{%
    \setlength\unitlength{\fontdimen 22 #1\tw@}%
    \vrule \@width\z@ \@height \thr@@\unitlength \@depth\@ne\unitlength
    \begin{picture}(4,2)(-2,-1)%
        \def\@KP@Radius     {2}%
        \def\@KP@Hole@radius{.5}% ... but let it be customizable too.
        \def\@KP@Diameter   {4}%
        #2%
    \end{picture}%
}

% On the other hand, for the commands that draw the four different shapes, it
% seems that all differences between the small variant and the large one can be
% confined in the following three macros (here, we just declare their name):
\newcommand*\@KP@Radius     {}
\newcommand*\@KP@Hole@radius{}
\newcommand*\@KP@Diameter   {}
%
% The four shapes:
\newcommand*\@KP@Shape@A{%
    \put(0,0){\circle{\@KP@Diameter}}%
}
\newcommand*\@KP@Shape@B{%
    \Line(-\@KP@Radius,\@KP@Radius )(\@KP@Radius,-\@KP@Radius)%
    \Line(-\@KP@Radius,-\@KP@Radius)(-\@KP@Hole@radius,-\@KP@Hole@radius)%
    \Line(\@KP@Radius ,\@KP@Radius )(\@KP@Hole@radius ,\@KP@Hole@radius )%
}
\newcommand*\@KP@Shape@C{%
    \cbezier(-\@KP@Radius,\@KP@Radius )(0,0)(0,0)(\@KP@Radius,\@KP@Radius )%
    \cbezier(-\@KP@Radius,-\@KP@Radius)(0,0)(0,0)(\@KP@Radius,-\@KP@Radius)%
}
\newcommand*\@KP@Shape@D{%
    \cbezier(-\@KP@Radius,-\@KP@Radius)(0,0)(0,0)(-\@KP@Radius,\@KP@Radius)%
    \cbezier(\@KP@Radius ,-\@KP@Radius)(0,0)(0,0)(\@KP@Radius ,\@KP@Radius)%
}

\newcommand*\@KP@Atomic@mathpalette[1]{%
    \mathinner{% or "\mathord"?
        % Note that a new level of grouping has just been entered (p. 290).
        % \color{gray}% not used, for now
        \mathchoice{%
            \linethickness{.6\p@}% Tip: use thicker lines if you decide to 
                                 % revert to using gray.
            \@KP@Large@frame \textfont {#1}%
        }{%
            \linethickness{.4\p@}% adjustable
            \@KP@Small@frame \textfont {#1}%
        }{%
            \linethickness{.3\p@}% adjustable
            \@KP@Small@frame \scriptfont {#1}%
        }{%
            \linethickness{.2\p@}% adjustable
            \@KP@Small@frame \scriptscriptfont {#1}%
        }%
    }%
}

% User-level commands:
\newcommand*\KPA{\@KP@Atomic@mathpalette \@KP@Shape@A}
\newcommand*\KPB{\@KP@Atomic@mathpalette \@KP@Shape@B}
\newcommand*\KPC{\@KP@Atomic@mathpalette \@KP@Shape@C}
\newcommand*\KPD{\@KP@Atomic@mathpalette \@KP@Shape@D}

\makeatother
%--------------------------------------------------------------%



\begin{document}

The Kauffmann Polynomial is defined by:
\begin{enumerate}
    \item
        Trivial link: \( \left<\KPA\right> = 1 \).

    \item
        Disjoint union etc.:
        \( \left<L\cup\KPA\right> = (-A^{2}-A^{-2})\langle L\rangle \).

    \item
        Skein relation:
        \( \left<\KPB\right> = A\left<\KPC\right> + A^{-1}\left<\KPD\right> \).
\end{enumerate}
Now the same three relations in display:
%
\begin{align}
    \left<\KPA\right> &= 1
        &&\text{trivial link} \label{eq:trivial} \\
    \left<L\cup\KPA\right> &= (-A^{2}-A^{-2})\langle L\rangle
        &&\text{disjoint union} \label{eq:union} \\
    \left<\KPB\right> &= A\left<\KPC\right> + A^{-1}\left<\KPD\right>
        &&\text{skein relation} \label{eq:skein}
\end{align}
%
From \eqref{eq:trivial} and~\eqref{eq:skein} I infere that
\( \frac{\left<\KPB\right>}{A\left<\KPC\right>
+ A^{-1}\left<\KPD\right>} = \left<\KPA\right> \);
let us repeat it once again:
%
\begin{equation*}
    \frac{\left<\KPB\right>}{A\left<\KPC\right> + A^{-1}\left<\KPD\right>}
        = \left<\KPA\right>
\end{equation*}
%

Also work: \( 2^{\left<\KPB\right>} \) and \( 2^{2^{\left<\KPB\right>}} \).

\end{document}

它产生的输出如下：

Answer