\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsfonts}
\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{enumerate}
\usepackage{hyperref}
\hypersetup{
    colorlinks,
    citecolor=black,
    filecolor=black,
    linkcolor=black,
    urlcolor=blue
}

\theoremstyle{definition}
\newtheorem{definition}{Def}[section]
\newtheorem{theorem}[definition]{Thm}
\newtheorem{innersolution}{}
\newenvironment{solution}[1]
{\renewcommand\theinnersolution{#1}\innersolution}
{\endinnersolution}


\title{FFT: Fast Fourier Transform}
\author{arnaucube}
\date{August 2022}

\begin{document}

\maketitle

\begin{abstract}
  Usually while reading papers and books I take handwritten notes, this document contains some of them re-written to $LaTeX$.

  The notes are not complete, don't include all the steps neither all the proofs. I use these notes to revisit the concepts after some time of reading the topic.

  This document are notes done while reading about the topic from \cite{gstrang}, \cite{tpornin}, \cite{rfateman}.
\end{abstract}

\tableofcontents

\section{Discrete \& Fast Fourier Transform}

\subsection{Discrete Fourier Transform (DFT)}

Continuous:

$$
x(f) = \int_{-\infty}^{\infty} x(t) e^{-2 \pi f t} dt
$$

Discrete:
The $k^{th}$ frequency, evaluating at $n$ of $N$ samples.
$$
\hat{f_k} = \sum_{n=0}^{n-1} f_n e^{\frac{-j \pi kn}{N}}
$$

where we can group under $b_n = \frac{\pi kn}{N}$. The previous expression can be expanded into:
$$
x_k = x_0 e^{-b_0 j} + x_1 e^{-b_1 j} + ... + x_n e^{-b_n j}
$$

By the \emph{Euler's formula} we have $e^{jx} = cos(x) + j\cdot sin(x)$, and using it in the previous $x_k$, we obtain

$$
x_k = x_0 [cos(-b_0) + j \cdot sin(-b_0)] + \ldots
$$

Using $\hat{f_k}$ we obtained
$$
\{f_0, f_1, \ldots, f_N\} \xrightarrow{DFT} \{ \hat{f_0}, \hat{f_1}, \ldots, \hat{f_N} \}
$$

To reverse the $\hat{f_k}$ back to $f_k$:
$$
f_k = \left( \sum_{n=0}^{n-1} \hat{f_n} e^{\frac{-j \pi kn}{N}} \right) \cdot \frac{1}{N}
$$


$$
DFT =
\begin{pmatrix}
\hat{f_0}\\
\hat{f_1}\\
\hat{f_2}\\
\vdots\\
\hat{f_n}\\
\end{pmatrix}=
\begin{pmatrix}
1 & 1 & 1 & \ldots & 1 \\
1 & w_n & w_n^2 & \ldots & w_n^{N-1} \\
1 & w_n^2 & w_n^4 & \ldots & w_n^{2(N-1)} \\
\vdots & \vdots & \vdots & & \vdots\\
1 & w_n^{n-1} & w_n^{2(n-1)} & \ldots & w_n^{(N-1)^2} \\
\end{pmatrix}
\begin{pmatrix}
f_0 \\ f_1 \\ f_2 \\ \vdots \\ f_n
\end{pmatrix}
$$

\subsection{Fast Fourier Transform (FFT)}
While DFT is $O(n)$, FFT is $O(n \space log(n))$

Here you can find a simple implementation of the these concepts in Rust: \href{https://github.com/arnaucube/fft-rs}{arnaucube/fft-rs} \cite{fftrs}


\section{FFT over finite fields, roots of unity, and polynomial multiplication}

FFT is very useful when working with polynomials. [TODO poly multiplication]

An implementation of the FFT over finite fields using the Vandermonde matrix approach can be found at \cite{fftsage}.

\subsection{Intro}
Let $A(x)$ be a polynomial of degree $n-1$,

$$
A(x) = a_0 + a_1 \cdot x + a_2 \cdot x^2 + \cdots + a_{n-1} \cdot x^{n-1} = \sum_{i=0}^{n-1} a_i \cdot x^i
$$

We can represent $A(x)$ in its evaluation form,

$$
(x_0, A(x_0)), (x_1, A(x_1)), \cdots, (x_{n-1}, A(x_{n-1})) = (x_i, A(x_i))
$$


We can evaluate A(x) at n given points $(x_0, x_1, ..., x_{n-1}$):

$$
\begin{pmatrix}
A(x_0)\\ A(x_1)\\ A(x_2)\\ \vdots\\ A(x_{n-1})
\end{pmatrix}=
\begin{pmatrix}
x_0^0 & x_0^1 & x_0^2 & \ldots & x_0^{n-1} \\
x_1^0 & x_1^1 & x_1^2 & \ldots & x_1^{n-1} \\
x_2^0 & x_2^1 & x_2^2 & \ldots & x_2^{n-1} \\
\vdots & \vdots & \vdots & & \vdots\\
x_{n-1}^0 & x_{n-1}^1 & x_{n-1}^2 & \ldots & x_{n-1}^{n-1} \\
\end{pmatrix}
\begin{pmatrix}
a_0 \\ a_1 \\ a_2 \\ \vdots \\ a_{n-1}
\end{pmatrix}
$$

This is known by the Vandermonde matrix.

But this will not be too efficient. Instead of random $x_i$ values, we use \emph{roots of unity}, where $\omega_n^n = 1$. We denote $\omega$ as a primitive $n^{th}$ root of unity:

$$
\begin{pmatrix}
A(1)\\ A(\omega)\\ A(\omega^2)\\ \vdots\\ A(\omega^{n-1})
\end{pmatrix}=
\begin{pmatrix}
1 & 1 & 1 & \ldots & 1 \\
1 & \omega & \omega^2 & \ldots & \omega^{n-1} \\
1 & \omega^2 & \omega^4 & \ldots & \omega^{2(n-1)} \\
\vdots & \vdots & \vdots & & \vdots\\
1 & \omega^{n-1} & \omega^{2(n-1)} & \ldots & \omega^{(n-1)^2} \\
\end{pmatrix}
\begin{pmatrix}
a_0 \\ a_1 \\ a_2 \\ \vdots \\ a_{n-1}
\end{pmatrix}
$$

Which we can see as

$$
\hat{A} = F_n \cdot A
$$

This matches our system of equations:

\begin{itemize}
  \item at $x=0$, $a_0 + a_1 + \cdots + a_{n-1} = A_0 = A(1)$
  \item at $x=1$, $a_0 \cdot 1 + a_1 \cdot \omega + a_2 \cdot \omega^2 + \cdots + a_{n-1} \cdot \omega^{n-1} = A_1 = A(\omega)$
  \item at $x=2$, $a_0 \cdot 1 + a_1 \cdot \omega^2 + a_2 \cdot \omega^4 + \cdots + a_{n-1} \cdot \omega^{2(n-1)} = A_2 = A(\omega^2)$
  \item $\cdots$
  \item at $x=n-1$, $a_0 \cdot 1 + a_1 \cdot \omega^{n-1} + a_2 \cdot \omega^{2(n-1)} + \cdots + a_{n-1} \cdot \omega^{(n-1)(n-1)} = A_2 = A(\omega^{n-1})$
\end{itemize}

We denote the $F_n$ as the Fourier matrix, with $j$ rows and $k$ columns, where each entry can be expressed as $F_{jk} = \omega^{jk}$.

To find the $a_i$ values, we use the inverted $F_n = F_n^{-1}$

\subsection{Roots of unity}
todo

\subsection{FFT over finite fields}
todo

\subsection{Polynomial multiplication with FFT}
todo


\bibliography{fft-notes.bib}
\bibliographystyle{unsrt}

\end{document}