amsmath 對齊環境中的視覺方程式鏈接

Question

我想「直觀地連結」中的方程align

由於一張圖片勝過很多文字，這裡是我在 MS Paint 中製作的一個範例來說明我的意思：

怎樣才能實現這目標呢？用於測試目的的 MWE：

\documentclass{article}
\usepackage{amsmath}
\allowdisplaybreaks

\begin{document}
    First, let us solve the following recursion formula:
        $$ F_{n + 1} = \alpha F_{n} + \beta$$
        \begin{align*}
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t}
        \end{align*}
\end{document}

Answer 1

根據我的評論的解決方案：

\documentclass{article}
\usepackage{amsmath}
\allowdisplaybreaks
\usepackage{tikz}
\def\tikzmark#1{\begin{tikzpicture}[remember picture]\coordinate(#1);\end{tikzpicture}}
\begin{document}
 $$ F_{n + 1} = \alpha F_{n} + \beta$$
        \begin{align*}
            &\tikzmark{A}\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\tikzmark{C}\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
          &\tikzmark{D}\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
      &\tikzmark{B}\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t}
        \end{align*}
\begin{tikzpicture}[remember picture,overlay]
  \draw[-,red] (A)--([xshift=-0.6cm]A)|-(B);
   \draw[-,blue] (C)--([xshift=-0.4cm]C)|-(D);
\end{tikzpicture}
\end{document}

輸出：

請注意，這些線從每行（每條數學線）的中心開始，可能需要調整為以\equiv符號為中心。

如果您有興趣的話，也許我可以稍後將其自動化。 （yshift=2mmdraw指令中tikzmark字母前的選項可以手動修復）