# Is there a way to make the text smaller inside a default font setting?

so I've written one document using the default margin settings and another using different margin settings(much wider). I needed to compile both documents so there's an obvious issue now. Let me show you what I mean:

\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}


I want to use the default margins which means making the code above font smaller or are there better solutions?

Thank you.

EDIT: I have rechecked the output and making the font smaller wont be useful. I think I should break the line using align but I don't know how to use it properly.

\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*}

• I am not sure I understand. What's it with the two documents you wrote? I only see one. You can change the fontsize e.g. with \footnotesize, but that's probably not what you're asking, is it? – sheß Apr 1 at 11:04
• You should not change the font sizes, you are already using amsmath, so why aren't you using the environments from it? I'd use align* and introduce line breaks on each of the original lines – daleif Apr 1 at 11:04
• @daleif so it almost fits the default margins but it is off by a word or two so I don't want to break the line as it wouldn't look as neat. Ok I rechecked, its off by quite a lot, not what I aforementioned. – Maths Apr 1 at 11:08
• For me those lines as off my up to 150pt. Remember, this is not about what is neat, it is about readability, consistency and good design. Changing the font size because something does not fit, or does not look neat, is not a good design. – daleif Apr 1 at 11:25
• Regarding tables, quoting Robert Bringhurst (author of the Elements of Typographic style): "Tables are notoriously time-consuming to typeset, but the problems posed are often editorial as much as typographic. If the table is not planned in a readable form to begin with, the typographer can render it readable only by rewriting or redesigning it from scratch." – daleif Apr 1 at 11:38

Here is probably how I'd typeset it. Note, there is a thought behind where each line is broken, highlighting the ending =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}


• I cannot thank you enough for the solution provided. +1 – Maths Apr 1 at 11:45
• Now If I may ask, how can I apply a similar trick to the tables in the question? – Maths Apr 1 at 11:47
• Please post another question instead of highjacking, and note that tabulars does not belong in math mode, use array instead. – daleif Apr 1 at 11:56