amsmath align 環境の視覚的な方程式リンク

Question

方程式を「視覚的にリンク」したいのですがalign

一枚の写真は多くの言葉に値するので、私が言いたいことを示すために MS ペイントで作成した例を以下に示します。

これを実現するにはどうすればよいでしょうか? テスト目的の 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

私のコメントによる解決策:

\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}

出力：

線は各行（各数式行）の中央から始まるため、\equiv記号に合わせて中央に調整する必要がある場合があることに注意してください。

ご興味があれば、後でこれを自動化できるかもしれません。(yshift=2mm描画コマンドの tikzmark の文字の前のオプションで手動で修正できます)