升级到MathJax v3后方程移至最右边

Question

我有一个基于 minimal-mistakes Jekyll 主题构建的个人博客。一些帖子使用由 MathJax 解析的 LaTex。我一直在使用 MathJax v2 并进行以下配置：

MathJax.Hub.Config({

  showProcessingMessages: false,
     messageStyle: "none",
     tex2jax: {
       inlineMath: [ ['$','$'], ["\\(","\\)"] ],
       displayMath: [ ['$$','$$'], ["\\[","\\]"] ],
       processEscapes: true
     },
     TeX: {
       MultLineWidth: "100%",
       equationNumbers: { autoNumber: "AMS" }
     },
     "HTML-CSS": { fonts: ["Latin-Modern"] }

});

MathJax.js?config=TeX-AMS-MML_HTMLorMML

这是一个方程及其输出。

$$
E[X] = \sum_{i=1}^{k-1}\sum_{j=i+1}^{k} X_{ij}Pr[\text{i and j have the same birthday}]
$$
$$ Pr[\text{i and j have unique birthdays}] = 365/365 * 364/365 $$ ($$ i $$ may have been born on any of the $$ 365 $$ days, and $$ j $$ on any of the remaining $$ 364 $$ days).
$$
\therefore Pr[\text{i and j have the same birthday}] = 1 - \frac{364}{365} = \frac{1}{365} \\
\begin{equation*}
\begin{aligned}
  E[X] & = \frac{1}{365} * \sum_{i=1}^{k-1}\sum_{j=i+1}^{k} X_{ij} \\
   & = \frac{1}{365} * \sum_{i=1}^{k-1} (k - i - 1 + 1) \\
   & = \frac{1}{365} * \sum_{i=1}^{k-1} (k - i) \\
   & = \frac{1}{365} * (\sum_{i=1}^{k-1} k - \mathop{\sum_{i=1}^{k-1}} i) \\
   & = \frac{1}{365} * (k(k - 1) - (1 + 2 +...+ k - 1)) \\
   & = \frac{1}{365} * (k(k - 1) - \frac{k(k - 1)}{2}) \\
   & = \frac{k(k - 1)}{(365 * 2)}
\end{aligned}
\end{equation*}
$$

我将MaxJax升级到v3，并使用MathJax提供的转换工具获取v3配置。

v3 配置：

window.MathJax = {
  tex: {
    inlineMath: [ ['$','$'], ["\\(","\\)"] ],
    displayMath: [ ['$$','$$'], ["\\[","\\]"] ],
    processEscapes: true,
    multlineWidth: "100%",
    tags: "ams"
  },
  options: {
    ignoreHtmlClass: 'tex2jax_ignore',
    processHtmlClass: 'tex2jax_process'
  }
};

如您所见，方程移至最右侧，并覆盖在其上方的文本之上。我该如何解决这个问题？

Answer 1

我实际上升级到了 v4.0.0-beta.7，以获得更好的验证，以及换行符支持。

对于上面的问题，我最终得到了以下结果。感谢 Davide P. Cervone 查看我在 MathJax GitHub 存储库上打开的ticket。

$$
\\ X = \begin{cases}
  1 & \text{if the pair has the same birthday} \\
  0 & \text{otherwise}
\end{cases} \\
\begin{aligned}
E[X] = \sum_{i=1}^{k-1}\sum_{j=i+1}^{k} X_{ij}Pr[\text{i and j have the same birthday}]
\end{aligned} \\
Pr[\text{i and j have unique birthdays}] = 365/365 * 364/365 $$ ($$ i $$ may have been born on any of the $$ 365 $$ days, and $$ j $$ on any of the remaining $$ 364 $$ days).
$$\\
\therefore Pr[\text{i and j have the same birthday}] = 1 - \frac{364}{365} = \frac{1}{365}
$$
$$
\begin{equation*}
\begin{aligned}
  E[X] & = \frac{1}{365} * \sum_{i=1}^{k-1}\sum_{j=i+1}^{k} X_{ij} \\
   & = \frac{1}{365} * \sum_{i=1}^{k-1} (k - i - 1 + 1) \\
   & = \frac{1}{365} * \sum_{i=1}^{k-1} (k - i) \\
   & = \frac{1}{365} * (\sum_{i=1}^{k-1} k - \mathop{\sum_{i=1}^{k-1}} i) \\
   & = \frac{1}{365} * (k(k - 1) - (1 + 2 +...+ k - 1)) \\
   & = \frac{1}{365} * (k(k - 1) - \frac{k(k - 1)}{2}) \\
   & = \frac{k(k - 1)}{(365 * 2)}
\end{aligned}
\end{equation*}
$$

可以在我的博客的此页面上看到实时渲染。