author

松鼠

date

2022-11-10

editdate

2022-11-28

formatted-for

salenzo.readthedocs.io

实验性的数学公式支持 / 在网页上渲染数学公式的另类解决方案

为了免去MathJax丑陋又缓慢的“正在渲染数学”提示，切实提升页面加载速度，同时调整公式的字体以免MathJax默认的TeX字体与页面的其余部分格格不入，仅使用MathJax将LaTeX公式转换到MathML。虽然MathML标准已经有好些年头了，但是浏览器支持仍旧一塌糊涂，不少屑东西明确表示拒绝支持。我想，借助现今更为先进的CSS排版功能，能否不用JavaScript测量文本尺寸就渲染出漂亮——至少是不难看——的数学公式呢？

$$\left\{\frac{1}{2}\right(\genfrac{}{}{0ex}{}{\mathit{n}}{\mathit{k}}),2\times (-1)\}$$

这可是纯CSS实现的可伸缩大括号哦！这些字符是各种正文字体通用的，因为用CSS也画不出更具特色的符号了。

MathJax的文档写得真是烂爆了……

支持下列标签：

• [ ] <math>
• [ ] <merror>
• [ ] <mfrac>
• [ ] <mi>
• [ ] <mmultiscripts>
• [ ] <mn>
• [ ] <mo>
• [ ] <mover>
• [ ] <mpadded>
• [ ] <mphantom>
• [ ] <mprescripts>
• [ ] <mroot>
• [ ] <mrow>
• [ ] <ms>
• [ ] <mspace>
• [ ] <msqrt>
• [ ] <mstyle>
• [ ] <msub>
• [ ] <msubsup>
• [ ] <msup>
• [ ] <mtable>
• [ ] <mtd>
• [ ] <mtext>
• [ ] <mtr>
• [ ] <munder>
• [ ] <munderover>
• [ ] <none>

绝对不会支持下列玩意：

• dir="rtl"
• <annotation>
• <annotation-xml>
• <maction>
• <semantics>

$1+1=2⨌{\displaystyle \frac{1}{4}}\in ℝ,\Re$ is true. $\mathrm{\backslash undef}=1d\mathit{x}$.

This mathematical formula with a big summation and the number pi $${\sum }_{\mathit{n}=1}^{+\infty }\frac{1}{{\mathit{n}}^2}=\frac{{\pi }^2}{6}$$ is easy to prove.

This mathematical formula with a big summation and the number pi ${\sum }_{\mathit{n}=1}^{+\infty }\frac{1}{{\mathit{n}}^2}=\frac{{\pi }^2}{6}$ is easy to prove.

• $\mathit{A}A𝔄𝔸$
• ${\sum }_{\mathit{A}}{\displaystyle \sum {\mathrm{\backslash nolimits}}_{\mathit{A}}}{\sum }_{\mathit{A}}$
• Fill the blank: $\sqrt{2\boxed{}-1}=3$
• $$\left|\begin{array}{cc}\mathit{x}& \left(\genfrac{}{}{0ex}{}{5}{3}\right)\\ \sqrt{7+2}& \mathit{y}\end{array}\right|$$
• $\mathit{y}={\mathit{x}}^2\text{ if }\mathit{x}\ge 1\text{ and }2\text{ otherwise.}$
• $\cos ,\mathit{c},c$
• $3.141592653589793$
• $1+2+3+4+5$
• ${\displaystyle {\sum }_5{\sum }_6}{\sum }_5\sum _7$
• $\frac{\mathrm{\backslash rule}[-50\mathit{p}\mathit{x}]10\mathit{p}\mathit{x}100\mathit{p}\mathit{x}}{\mathrm{\backslash rule}[-25\mathit{p}\mathit{x}]10\mathit{p}\mathit{x}25\mathit{p}\mathit{x}}\uparrow \uparrow$ (no asymmetric delimiters in TeX)
• $1\frac{2\mathrm{\backslash vrule}\mathit{w}\mathit{i}\mathit{d}\mathit{t}\mathit{h}\mathrm{0pt}\mathit{h}\mathit{e}\mathit{i}\mathit{g}\mathit{h}\mathit{t}\mathrm{0pt}\mathit{d}\mathit{e}\mathit{p}\mathit{t}\mathit{h}\mathrm{1em}}{3\mathrm{\backslash vrule}\mathit{w}\mathit{i}\mathit{d}\mathit{t}\mathit{h}\mathrm{0pt}\mathit{h}\mathit{e}\mathit{i}\mathit{g}\mathit{h}\mathit{t}\mathrm{2em}\mathit{d}\mathit{e}\mathit{p}\mathit{t}\mathit{h}\mathrm{0pt}}$
• $\mathit{s}=\mathtt{"}\mathtt{h}\mathtt{e}\mathtt{l}\mathtt{l}\mathtt{o}\mathtt{w}\mathtt{o}\mathtt{r}\mathtt{l}\mathtt{d}\mathtt{"}$
• ${\left(\frac{1+\frac{2}{3}}{4}\right)}^5$
• $0+{\displaystyle \frac{1}{2}}-\frac{1}{2}+\frac{1}{234}-\left(\genfrac{}{}{0ex}{}{123}{4}\right)$
• $\sqrt[3]{\sqrt{\frac{1}{2}+4}}+0$
• ${*}_{\mathit{A}}\underset{\mathit{B}}{*}\underset{\mathit{C}}{*}{*}_{\mathit{D}}$
• $\frac{\frac{1}{\mathit{\}}}3}{}$
• $1\mathrm{\backslash colorbox}\mathit{l}\mathit{i}\mathit{g}\mathit{h}\mathit{t}\mathit{b}\mathit{l}\mathit{u}\mathit{e}\frac{23456}{78}9+1\mathrm{\backslash colorbox}\mathit{l}\mathit{i}\mathit{g}\mathit{h}\mathit{t}\mathit{b}\mathit{l}\mathit{u}\mathit{e}\frac{23456}{78}9$
• $\frac{\mathit{x}+\mathit{y}+\mathit{z}}{\mathit{x}\phantom{+\mathit{y}}+\mathit{z}}$
• $1_2+3^4+5_6^7$
• $\underset{2}{1}+\stackrel{4}{3}+\stackrel{7}{\underset{6}{5}}+\underset{9}{\hat{8}}+\stackrel{13}{\underbrace{11}}$
• ${}_6\mathrm{\backslash prescript}99981_2^3{}^5$
• $$\frac{\mathit{A}}{2}=\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right)$$
• $$\left\{\frac{1}{2}\right\}$$
• $${\left[}^{\begin{array}{ccc}12& 34& 56\\ 7& 8& 9\end{array}}$$
• $\exp 𝐀\in {\mathfrak{g}\mathfrak{l}}_{\mathit{n}}(ℝ)$
• ${\mathit{x}}^{\mathit{n}}+{\mathit{y}}^{\mathit{n}}={\mathit{z}}^{\mathit{n}}$
• $\sqrt{{\sum }_{\mathit{n}=1}^{+\infty }\frac{10}{{\mathit{n}}^4}=\frac{{\pi }^2}{3}}$ VS ${\displaystyle {\sum }_{\mathit{n}=1}^{+\infty }\frac{10}{{\mathit{n}}^4}=\frac{{\pi }^2}{3}}$
• $\sqrt{{\mathit{x}}^2}\ne {\mathit{x}}^2$
• ${\mathit{A}}^{{\mathit{A}}^{\mathit{A}}}+\sqrt[\mathit{A}]{\mathit{A}}+{\displaystyle \frac{\mathit{A}+\frac{\mathit{A}}{\mathit{A}}}{\mathit{A}}}$

\mathbf{Ab}^\mathsf{T}