
Me gustaría "vincular visualmente" ecuaciones en elalign
Dado que una imagen vale más que muchas palabras, aquí hay un ejemplo que hice en MS Paint para mostrar lo que quiero decir:
¿Cómo se podría lograr esto? Un MWE para fines de prueba:
\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}
Respuesta1
Una solución según mi comentario:
\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}
Producción:
Tenga en cuenta que las líneas comienzan desde el centro de cada fila (cada línea matemática) y es posible que deba ajustarse para que queden centradas con \equiv
el símbolo.
Quizás pueda automatizar esto más tarde si está interesado. (Una yshift=2mm
opción antes de la letra del tikzmark en el comando de dibujo puede solucionarlo manualmente)