Add DCT least squares proof to the supplement

parent 67ef2697
......@@ -28,14 +28,54 @@
\maketitle
\section{Proof of the DCT Least Squares Theorem}
\section{Proof of the DCT Least Squares Approximation Theorem}
\begin{theorem}[DCT Least Squares Approximation Theorem]
Given a set of $N$ samples of a signal $X = \{x_0, ... x_N\}$, let $Y = \{y_0, ... y_N\}$ be the DCT coefficients of $X$. Then, for any $1 \leq m \leq N$, the approximation
\begin{equation}
p_m(t) = \frac{1}{\sqrt{n}}y_o + \sqrt{\frac{2}{n}}\sum_{k=1}^{m} y_k\cos\left(\frac{k(2t + 1)\pi}{2n}\right)
\label{eq:dct1d}
\end{equation}
of $X$ minimizes the least squared error
\begin{equation}
e_m = \sum_{i=0}^{n} (p_m(i) - x_i)^2
\end{equation}
\label{thm:dctls}
\end{theorem}
\begin{proof}
First consider that since Equation \ref{eq:dct1d} represents the Discrete Cosine Transform, which is a Linear map, we can write the least squares error as
\begin{equation}
D^T_my = x
\end{equation}
where $D_m$ is formed from the first $m$ rows of the DCT matrix, $y$ is a row vector of the DCT coefficients, and $x$ is a row vector of the original samples.
To solve for the least squares solution, we use the the normal equations, that is we solve
\begin{equation}
D_mD^T_my = D_mx
\end{equation}
and since the DCT is an orthonormal transformation, the rows of $D_m$ are orthogonal, so $D_mD^T_m = I$. Therefore
\begin{equation}
y = D_mx
\end{equation}
Since there is no contradiction, the least squares solution must use the first $m$ DCT coefficients.
\end{proof}
\section{Proof of the DCT Mean-Variance Theorem}
\begin{theorem}[DCT Mean-Variance Theorem]
Given a set of samples of a signal $X$ such that $\e[X] = 0$, let $Y$ be the DCT coefficients of $X$. Then
\begin{equation}
\var[X] = \e[Y^2]
\end{equation}
\end{theorem}
\begin{proof}
\end{proof}
\section{Algorithms}
\begin{algorithm}
\caption{Direct Convolution Explosion. $K$ is an initial filter, $m, n$ are the input and output channels, $h, w$ are the image height and width, $s$ is the stride, $\star_s$ denotes the discrete convolution with stride $s$}
\caption{Convolution Explosion. $K$ is an initial filter, $m, n$ are the input and output channels, $h, w$ are the image height and width, $s$ is the stride, $\star_s$ denotes the discrete convolution with stride $s$}
\label{alg:dce}
\begin{algorithmic}
\Function{Explode}{$K, m, n, h, w, s$}
......@@ -51,7 +91,7 @@
\end{algorithm}
\begin{algorithm}
\caption{Automated Spatial Masking for ReLu. $F$ is a DCT domain block, $\phi$ is the desired maximum spatial frequencies, $N$ is the block size.}
\caption{Approximated Spatial Masking for ReLu. $F$ is a DCT domain block, $\phi$ is the desired maximum spatial frequencies, $N$ is the block size.}
\label{alg:asmr}
\begin{algorithmic}
\Function{ReLu}{$F, \phi, N$}
......@@ -84,4 +124,10 @@
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Batch Normalization}
\label{alg:bn}
\end{algorithm}
\end{document}
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