Reducing the white spacing

\documentclass[11pt, a4paper]{report}
\usepackage{bm}
\usepackage{amsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs}
\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}
\newcommand{\iu}{{i\mkern1mu}}
\begin{align*}
\setlength\extrarowheight{3pt}
\noindent\begin{tabular}{c | c c c c c }
& $0$ & $1$ & $2$ & $3$ & $4$\\
\cline{1-6}
$\chi_0$ & $1$ & $1$  &  $1$   & $1$   & $1$\\
$\chi_1$ & $1$ & $a$  &  $a^2$ & $a^3$ & $a^4$\\
$\chi_2$ & $1$ & $a^2$ & $a^4$ & $a$   & $a^3$\\
$\chi_3$ & $1$ & $a^3$ & $a$   & $a^4$ & $a^2$\\
$\chi_4$ & $1$ & $a^4$ & $a^3$ & $a^2$ & $a$\\
\end{tabular}
\end{align*}

with $a = \exp\{\frac{2\pi\iu}{5}\}$ hence $a^5=1$ with $|G|=5$.

Applying the definition of Fourier transform from definition 3.1.2 we have:
\doublespacing{
$\hat{f}(\chi_0)=f(0)+f(1)+f(2)+f(3)+f(4)$\\
$\hat{f}(\chi_1)=f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)$\\
$\hat{f}(\chi_2)=f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)$\\
$\hat{f}(\chi_3)=f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)$\\
$\hat{f}(\chi_4)=f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)$\\
}
Using definition 3.1.3. we can compute the inverse Fourier transform $f(t)$:
\begin{align*}
{f}(0)
&=\frac{1}{5}[ \hat{f}(\chi_0)+\hat{f}(\chi_1)+\hat{f}(\chi_2)+\hat{f}(\chi_3)+\hat{f}(\chi_4)]\\
&\begin{aligned}[t]
{}={}&\frac{1}{5}[f(0)+f(1)+f(2)+f(3)+f(4)]\\
{}+{}&\frac{1}{5}[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)]\\
{}+{}&\frac{1}{5}[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)]\\
{}+{}&\frac{1}{5}[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)]\\
{}+{}&\frac{1}{5}[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)]
\end{aligned}
\\
&\begin{aligned}[t]
{}={}&f(0)\\
{}+{}&\frac{f(1)}{5}[1+a+a^2+a^3+a^4]\\
{}+{}&\frac{f(2)}{5}[1+a+a^2+a^3+a^4]\\
{}+{}&\frac{f(3)}{5}[1+a+a^2+a^3+a^4]\\
{}+{}&\frac{f(4)}{5}[1+a+a^2+a^3+a^4]\\
{}={}&f(0)
\end{aligned}
\end{align*}
Similarly
\begin{align*}
{f}(1)
&= \frac{1}{5}[\hat{f}(\chi_0)+\frac{1}{a}\hat{f}(\chi_1)+\frac{1}{a^2}\hat{f}(\chi_2)+\frac{1}{a^3}\hat{f}(\chi_3)+\frac{1}{a^4}\hat{f}(\chi_4)]\\
&\begin{aligned}[t]
{}={}&f(1)
\end{aligned}
\end{align*}
\begin{align*}
{f}(2)
&= \frac{1}{5}[\hat{f}(\chi_0)+a^2\hat{f}(\chi_1)+a^4\hat{f}(\chi_2)+a\hat{f}(\chi_3)+a^3\hat{f}(\chi_4)]\\
&\begin{aligned}[t]
{}={}&f(2)
\end{aligned}
\end{align*}
\begin{align*}
{f}(3)
&= \frac{1}{5}[\hat{f}(\chi_0)+a^3\hat{f}(\chi_1)+a\hat{f}(\chi_2)+a^4\hat{f}(\chi_3)+a^2\hat{f}(\chi_4)]\\
&\begin{aligned}[t]
{}={}&f(3)
\end{aligned}
\end{align*}
\begin{align*}
{f}(4)
&= \frac{1}{5}[\hat{f}(\chi_0)+a^4\hat{f}(\chi_1)+a^3\hat{f}(\chi_2)+a^2\hat{f}(\chi_3)+a\hat{f}(\chi_4)]\\
&\begin{aligned}[t]
{}={}&f(4)
\end{aligned}
\end{align*}
\end{document}


How can I reduce the spacing in this? Where it says "using definition....." I would like to move this up to $\hat{f}(\chi_4)$. Also I want to reduce the spacing of $f(1) = ... = f(1)$ and $f(2) = ... = f(2)$ etc.

Edit: I've attached all 3 pages so you guys get the full picture of what the issue is.

• Unrelated to the issue, but switching from tabular to array you can remove all the repeated $signs. – leandriis May 1 at 16:37 2 Answers With this simpler code, it can all fit on a single page. I loaded nccmath for its medium-sized fractions, which look better for coefficients, in my opinion: \documentclass[11pt, a4paper]{report} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{bm} \usepackage{nccmath} \usepackage{amsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs} \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} \newcommand{\iu}{{i\mkern1mu}} $\setlength\extrarowheight{3pt} \begin{array}{c | c c c c c } & 0 & 1 & 2 & 3 & 4\\ \cline{1-6} \chi_0 & 1 & 1 & 1 & 1 & 1\\ \chi_1 & 1 & a & a^2 & a^3 & a^4\\ \chi_2 & 1 & a^2 & a^4 & a & a^3\\ \chi_3 & 1 & a^3 & a & a^4 & a^2\\ \chi_4 & 1 & a^4 & a^3 & a^2 & a\\ \end{array}$ with$a = \exp\bigl\{\frac{2\pi \iu}{5}\bigr\}$, hence$a^5=1$with$|G|=5. Applying the definition of Fourier transform from definition 3.1.2 we have: \begin{fleqn} \begin{align*} \hat{f}(\chi_0) & =f(0)+f(1)+f(2)+f(3)+f(4) \\ \hat{f}(\chi_1) & =f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4) \\ \hat{f}(\chi_2) & =f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4) \\ \hat{f}(\chi_3) & =f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4) \\ \hat{f}(\chi_4) & =f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4) \end{align*} \end{fleqn} Using definition 3.1.3. we can compute the inverse Fourier transformf(t): \allowdisplaybreaks \begin{align*} {f}(0) &=\mfrac{1}{5}\bigl[ \hat{f}(\chi_0)+\hat{f}(\chi_1)+\hat{f}(\chi_2)+\hat{f}(\chi_3)+\hat{f}(\chi_4)\bigr]\\ & = \begin{aligned}[t] &\mfrac{1}{5}\bigl[f(0)+f(1)+f(2)+f(3)+f(4)]\\ & + \mfrac{1}{5}\bigl[f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4)\bigr]\\ & + \mfrac{1}{5}\bigl[f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4)\bigr]\\ & + \mfrac{1}{5}\bigl[f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4)\bigr]\\ & + \mfrac{1}{5}\bigl[f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4)\bigr] \end{aligned}\\ & =f(0) \begin{aligned}[t] & + \mfrac{f(1)}{5}[1+a+a^2+a^3+a^4]\\ & + \mfrac{f(2)}{5}[1+a+a^2+a^3+a^4]\\ & + \mfrac{f(3)}{5}[1+a+a^2+a^3+a^4]\\ & + \mfrac{f(4)}{5}[1+a+a^2+a^3+a^4] \end{aligned}\\ & = f(0) \shortintertext{Similarly:} {f}(1) &= \mfrac{1}{5}\Bigl[\hat{f}(\chi_0)+\mfrac{1}{a}\hat{f}(\chi_1)+\mfrac{1}{a^2}\hat{f}(\chi_2)+\mfrac{1}{a^3}\hat{f}(\chi_3)+\mfrac{1}{a^4}\hat{f}(\chi_4)\Bigr]\\ & = f(1) \\[1.5ex] f(2) &= \mfrac{1}{5}\bigl[\hat{f}(\chi_0)+a^2\hat{f}(\chi_1)+a^4\hat{f}(\chi_2)+a\hat{f}(\chi_3)+a^3\hat{f}(\chi_4)\bigr] \\ & = f(2) \\[1.5ex] f(3) &= \mfrac{1}{5}\bigl[\hat{f}(\chi_0)+a^3\hat{f}(\chi_1)+a\hat{f}(\chi_2)+a^4\hat{f}(\chi_3)+a^2\hat{f}(\chi_4)\bigr] \\ & = f(3) \\[1.5ex] f(4) &= \mfrac{1}{5}\bigl[\hat{f}(\chi_0)+a^4\hat{f}(\chi_1)+a^3\hat{f}(\chi_2)+a^2\hat{f}(\chi_3)+a\hat{f}(\chi_4)\bigr] \\ & = f(4) \end{align*} \end{document}  • this is nice however you missed out the remaining section forf(0)aha! – Maths May 2 at 11:03 • Oh! yes. I'll fix it in a moment – Bernard May 2 at 11:08 • I had to slightly modify the code to make it fit on a single page (replaced \intertext with shortintertext, and loading nccmath before mathtools to make it work). – Bernard May 2 at 11:24 • there's still some code of f(0) missing aha. its where I group f(1) ... f(4) as f(1)[1+a+...+a^4] etc. Also, I don't mind if it runs over two pages. I don't want the text to be squashed, all I wanted is to make use of the empty white space :) – Maths May 2 at 11:26 • Refer to my code in the question, you'll see the part you missed :) – Maths May 2 at 11:35 You should avoid \\ on the last line of alignments. Perhaps the following is closer to what you want: \documentclass[11pt, a4paper]{report} \usepackage{amsmath,array} \begin{document} \newcommand{\iu}{{i\mkern1mu}} \begin{equation*} \setlength\extrarowheight{3pt} \begin{tabular}{c | c c c c c } &0$&$1$&$2$&$3$&$4$\\ \cline{1-6}$\chi_0$&$1$&$1$&$1$&$1$&$1$\\$\chi_1$&$1$&$a$&$a^2$&$a^3$&$a^4$\\$\chi_2$&$1$&$a^2$&$a^4$&$a$&$a^3$\\$\chi_3$&$1$&$a^3$&$a$&$a^4$&$a^2$\\$\chi_4$&$1$&$a^4$&$a^3$&$a^2$&$a$\\ \end{tabular} \end{equation*} with$a = \exp\{\frac{2\pi\iu}{5}\}$hence$a^5=1$with$|G|=5. Applying the definition of Fourier transform from Definition~3.1.2 we have: \begin{align*} \hat{f}(\chi_0) &=f(0)+f(1)+f(2)+f(3)+f(4),\\ \hat{f}(\chi_1) &=f(0)+af(1)+a^2f(2)+a^3f(3)+a^4f(4),\\ \hat{f}(\chi_2) &=f(0)+a^2f(1)+a^4f(2)+af(3)+a^3f(4),\\ \hat{f}(\chi_3) &=f(0)+a^3f(1)+af(2)+a^4f(3)+a^2f(4),\\ \hat{f}(\chi_4) &=f(0)+a^4f(1)+a^3f(2)+a^2f(3)+af(4). \end{align*} Using Definition~3.1.3 we can compute the inverse Fourier transformf(t)\$:
\begin{align*}
f(0)
&=\frac{1}{5}[ \hat{f}(\chi_0) + \hat{f}(\chi_1) + \hat{f}(\chi_2) +
\hat{f}(\chi_3) + \hat{f}(\chi_4)]\\
&=\frac{1}{5}[f(0)+f(1)+f(2)+f(3)+f(4)]\\
\\
&= f(0)\\
&=f(0).
\end{align*}
Similarly
\begin{align*}
f(1)
&= \frac{1}{5}\Bigl[\hat{f}(\chi_0) + \frac{1}{a}\hat{f}(\chi_1) +
\frac{1}{a^2}\hat{f}(\chi_2) + \frac{1}{a^3}\hat{f}(\chi_3) +
\frac{1}{a^4}\hat{f}(\chi_4)\Bigr]\\
&=f(1),\\
f(2)
&= \frac{1}{5}[\hat{f}(\chi_0) + a^2\hat{f}(\chi_1) +
a^4\hat{f}(\chi_2) + a\hat{f}(\chi_3) + a^3\hat{f}(\chi_4)]\\
&=f(2), \\
f(3)
&= \frac{1}{5}[\hat{f}(\chi_0) + a^3\hat{f}(\chi_1) +
a\hat{f}(\chi_2) + a^4\hat{f}(\chi_3) + a^2\hat{f}(\chi_4)]\\
&=f(3),\\
f(4)
&= \frac{1}{5}[\hat{f}(\chi_0) + a^4\hat{f}(\chi_1) +
a^3\hat{f}(\chi_2) + a^2\hat{f}(\chi_3) + a\hat{f}(\chi_4)]\\
& =f(4).
\end{align*}
\end{document}

• why did you push f(0) outwards? it wasn't necessary. but thanks for your solution – Maths May 1 at 15:42
• The +'s should not be under the =, but to the right of it as they belong to that side of the equation. Whether you want to indent by \qquad as I did, or the smaller \quad is a matter of taste. – Andrew Swann May 1 at 18:28