有没有办法让默认字体设置中的文本变得更小？

Question

因此，我使用默认边距设置编写了一个文档，并使用不同的边距设置（更宽）编写了另一个文档。我需要编译这两个文档，所以现在有一个明显的问题。让我告诉你我的意思：

\documentclass[11pt, a4paper]{report}
\usepackage{bm}
\usepackage{amsfonts, graphicx, verbatim, amsmath,amssymb, amsthm}
\usepackage{color}
\usepackage{array}
\usepackage{setspace}% if you must (for double spacing thesis)
\usepackage{fancyhdr}
\usepackage{enumitem}
\usepackage{tikz}
\usepackage{parskip}
\usepackage{lipsum}
\usepackage{floatrow}
\begin{document}
Computation of the Inverse Fourier transform from definition 3.1.3:
\[
f_{(0,0,0)} = \frac{1}{8}[f_{(0,0,0)}+f_{(0,0,0)}+f_{(0,0,0)}+f_{(0,0,0)}+f_{(0,0,0)}+f_{(0,0,0)}+f_{(0,0,0)}+f_{(0,0,0)}] = f_{(0,0,0)}
\]
\[
f_{(0,0,1)} = \frac{1}{8}[f_{(0,0,1)}+f_{(0,0,1)}+f_{(0,0,1)}+f_{(0,0,1)}-f_{(0,0,1)}\cdot {(-1)}-f_{(0,0,1)}\cdot {(-1)}-f_{(0,0,1)}\cdot {(-1)}-f_{(0,0,1)}\cdot {(-1)}] = f_{(0,0,1)}
\]
\[
f_{(0,1,0)} = \frac{1}{8}[f_{(0,1,0)}+f_{(0,1,0)}-f_{(0,1,0)}\cdot{(-1)}-f_{(0,1,0)}\cdot{(-1)}+f_{(0,1,0)}+f_{(0,1,0)}-f_{(0,1,0)}\cdot{(-1)}-f_{(0,1,0)}\cdot{(-1)}] = f_{(0,1,0)}
\]
\[
f_{(0,1,1)} = \frac{1}{8}[f_{(0,1,1)}+f_{(0,1,1)}-f_{(0,1,1)}\cdot{(-1)}-f_{(0,1,1)}\cdot{(-1)}-f_{(0,1,1)}\cdot{(-1)}-f_{(0,1,1)}\cdot{(-1)}+f_{(0,1,1)}+f_{(0,1,1)}] = f_{(0,1,1)}
\]
\[
f_{(1,0,0)} = \frac{1}{8}[f_{(1,0,0)}-f_{(1,0,0)}\cdot{(-1)}+f_{(1,0,0)}-f_{(1,0,0)}\cdot{(-1)}+f_{(1,0,0)}-f_{(1,0,0)}\cdot{(-1)}+f_{(1,0,0)}-f_{(1,0,0)}\cdot{(-1)}] = f_{(1,0,0)}
\]
\[
f_{(1,0,1)} = \frac{1}{8}[f_{(1,0,1)}-f_{(1,0,1)}\cdot{(-1)}+f_{(1,0,1)}-f_{(1,0,1)}\cdot{(-1)}-f_{(1,0,1)}\cdot{(-1)}+f_{(1,0,1)}-f_{(1,0,1)}\cdot{(-1)}+f_{(1,0,1)}] = f_{(1,0,1)}
\]
\[
f_{(1,1,0)} = \frac{1}{8}[f_{(1,1,0)}-f_{(1,1,0)}\cdot{(-1)}-f_{(1,1,0)}\cdot{(-1)}+f_{(1,1,0)}+f_{(1,1,0)}-f_{(1,1,0)}\cdot{(-1)}-f_{(1,1,0)}\cdot{(-1)}+f_{(1,1,0)}] = f_{(1,1,0)}
\]
\[
f_{(1,1,1)} = \frac{1}{8}[f_{(1,1,1)}-f_{(1,1,1)}\cdot{(-1)}-f_{(1,1,1)}\cdot{(-1)}+f_{(1,1,1)}-f_{(1,1,1)}\cdot{(-1)}+f_{(1,1,1)}+f_{(1,1,1)}-f_{(1,1,1)}\cdot{(-1)}] = f_{(1,1,1)}
\]
\end{document}

我想使用默认边距，这意味着使上面的代码字体更小，或者有更好的解决方案？

谢谢。

编辑：我重新检查了输出，发现将字体缩小是没有用的。我认为我应该使用对齐来换行，但我不知道如何正确使用它。

\begin{align*}
\setlength\extrarowheight{3pt}
\noindent\begin{tabular}{c | c c c c c c c c }
  +  & $(0,0,0)$ & $(0,0,1)$ & $(0,1,0)$ & $(0,1,1)$ & $(1,0,0)$ & $(1,0,1)$ & $(1,1,0)$ & $(1,1,1)$\\
    \cline{1-9}
   $(0,0,0)$ & $(0,0,0)$ & $(0,0,1)$ & $(0,1,0)$ & $(0,1,1)$ & $(1,0,0)$ & $(1,0,1)$ & $(1,1,0)$ & $(1,1,1)$\\
    $(0,0,1)$ & $(0,0,1)$ & $(0,0,0)$ & $(0,1,1)$ & $(0,1,0)$ & $(1,0,1)$ & $(1,0,0)$ & $(1,1,1)$ & $(1,1,0)$\\
    $(0,1,0)$ & $(0,1,0)$ & $(0,1,1)$ & $(0,0,0)$ & $(0,0,1)$ & $(1,1,0)$ & $(1,1,1)$ & $(1,0,0)$ & $(1,0,1)$\\
    $(0,1,1)$ & $(0,1,1)$ & $(0,1,0)$ & $(0,0,1)$ & $(0,0,0)$ & $(1,1,1)$ & $(1,1,0)$ & $(1,0,1)$ & $(1,0,0)$\\
    $(1,0,0)$ & $(1,0,0)$ & $(1,0,1)$ & $(1,1,0)$ & $(1,1,1)$ & $(0,0,0)$ & $(0,0,1)$ & $(0,1,0)$ & $(0,1,1)$\\
    $(1,0,1)$ & $(1,0,1)$ & $(1,0,0)$ & $(1,1,1)$ & $(1,1,0)$ & $(0,0,1)$ & $(0,0,0)$ & $(0,1,1)$ & $(0,1,0)$\\
    $(1,1,0)$ & $(1,1,0)$ & $(1,1,1)$ & $(1,0,0)$ & $(1,0,1)$ & $(0,1,0)$ & $(0,1,1)$ & $(0,0,0)$ & $(0,0,1)$\\
    $(1,1,1)$ & $(1,1,1)$ & $(1,1,0)$ & $(1,0,1)$ & $(1,0,0)$ & $(0,1,1)$ & $(0,1,0)$ & $(0,0,1)$ & $(0,0,0)$\\
\end{tabular}
\end{align*}

The character table:

\begin{align*}
\setlength\extrarowheight{3pt}
\noindent\begin{tabular}{c | c c c c c c c c }
  +  & $(0,0,0)$ & $(0,0,1)$ & $(0,1,0)$ & $(0,1,1)$ & $(1,0,0)$ & $(1,0,1)$ & $(1,1,0)$ & $(1,1,1)$\\
    \cline{1-9}
    $\chi_{(0,0,0)}$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$\\
    $\chi_{(0,0,1)}$ & $1$ & $1$ & $1$ & $1$ & $-1$ & $-1$ & $-1$ & $-1$\\
    $\chi_{(0,1,0)}$ & $1$ & $1$ & $-1$ & $-1$ & $1$ & $1$ & $-1$ & $-1$\\
    $\chi_{(0,1,1)}$ & $1$ & $1$ & $-1$ & $-1$ & $-1$ & $-1$ & $1$ & $1$\\
    $\chi_{(1,0,0)}$ & $1$ & $-1$ & $1$ & $-1$ & $1$ & $-1$ & $1$ & $-1$\\
    $\chi_{(1,0,1)}$ & $1$ & $-1$ & $1$ & $-1$ & $-1$ & $1$ & $-1$ & $1$\\
    $\chi_{(1,1,0)}$ & $1$ & $-1$ & $-1$ & $1$ & $1$ & $-1$ & $-1$ & $1$\\
    $\chi_{(1,1,1)}$ & $1$ & $-1$ & $-1$ & $1$ & $-1$ & $1$ & $1$ & $-1$\\
\end{tabular}
\end{align*}

Answer 1

我可能是这样排版的。注意，每一行的断行背后都有一个想法，突出结尾=f...

\documentclass[11pt, a4paper]{report}
\usepackage{amsmath}
\begin{document}
Computation of the Inverse Fourier transform from definition 3.1.3:
\begin{align*}
f_{(0,0,0)} &= \frac{1}{8}
\begin{aligned}[t]
  \bigl[&f_{(0,0,0)}+f_{(0,0,0)}+f_{(0,0,0)}
  \\
  &+f_{(0,0,0)}
  +f_{(0,0,0)}
  +f_{(0,0,0)}+f_{(0,0,0)}+f_{(0,0,0)}\bigr]
  = f_{(0,0,0)}
\end{aligned}
\\
f_{(0,0,1)} &= \frac{1}{8}
\begin{aligned}[t]
  \bigl[&f_{(0,0,1)}+f_{(0,0,1)}+f_{(0,0,1)}+f_{(0,0,1)}-f_{(0,0,1)}\cdot
  {(-1)}\\
  &-f_{(0,0,1)}\cdot {(-1)}-f_{(0,0,1)}\cdot
  {(-1)}-f_{(0,0,1)}\cdot {(-1)}\bigr] = f_{(0,0,1)}
\end{aligned}
\\
f_{(0,1,0)} &= \frac{1}{8}
\begin{aligned}[t]
  \bigl[&f_{(0,1,0)}+f_{(0,1,0)}-f_{(0,1,0)}\cdot{(-1)}-f_{(0,1,0)}\cdot{(-1)}
  \\
  &+f_{(0,1,0)}
  +f_{(0,1,0)}-f_{(0,1,0)}\cdot{(-1)}-f_{(0,1,0)}\cdot{(-1)}\bigr] =
  f_{(0,1,0)}
\end{aligned}
\\
f_{(0,1,1)} &= \frac{1}{8}
\begin{aligned}[t]
  \bigl[&f_{(0,1,1)}+f_{(0,1,1)}-f_{(0,1,1)}\cdot{(-1)}-f_{(0,1,1)}\cdot{(-1)}
  \\
  &-f_{(0,1,1)}\cdot{(-1)}
  -f_{(0,1,1)}\cdot{(-1)}+f_{(0,1,1)}+f_{(0,1,1)}\bigr]
  = f_{(0,1,1)}
\end{aligned}
\\
f_{(1,0,0)} &= \frac{1}{8}
\begin{aligned}[t]
  \bigl[&f_{(1,0,0)}-f_{(1,0,0)}\cdot{(-1)}+f_{(1,0,0)}-f_{(1,0,0)}\cdot{(-1)}
  \\
  &+f_{(1,0,0)}
  -f_{(1,0,0)}\cdot{(-1)}
  +f_{(1,0,0)}-f_{(1,0,0)}\cdot{(-1)}\bigr]
  = f_{(1,0,0)}
\end{aligned}
\\
f_{(1,0,1)} &= \frac{1}{8}
\begin{aligned}[t]
  \bigl[&f_{(1,0,1)}-f_{(1,0,1)}\cdot{(-1)}+f_{(1,0,1)}-f_{(1,0,1)}\cdot{(-1)}
  \\
  &-f_{(1,0,1)}\cdot{(-1)}+f_{(1,0,1)}-f_{(1,0,1)}\cdot{(-1)}+f_{(1,0,1)}\bigr]
  = f_{(1,0,1)}
\end{aligned}
\\
f_{(1,1,0)} &= \frac{1}{8}
\begin{aligned}[t]
  \bigl[&f_{(1,1,0)}-f_{(1,1,0)}\cdot{(-1)}-f_{(1,1,0)}\cdot{(-1)}+f_{(1,1,0)}
  \\
  &+f_{(1,1,0)}
  -f_{(1,1,0)}\cdot{(-1)}-f_{(1,1,0)}\cdot{(-1)}+f_{(1,1,0)}\bigr]
  = f_{(1,1,0)}
\end{aligned}
\\
f_{(1,1,1)} &= \frac{1}{8}
\begin{aligned}[t]
  \bigl[&f_{(1,1,1)}-f_{(1,1,1)}\cdot{(-1)}-f_{(1,1,1)}\cdot{(-1)}+f_{(1,1,1)}
  \\
  &
  -f_{(1,1,1)}\cdot{(-1)}
  +f_{(1,1,1)}+f_{(1,1,1)}-f_{(1,1,1)}\cdot{(-1)}\bigr]
  = f_{(1,1,1)}
\end{aligned}
\end{align*}
\end{document}