Add DCT least squares proof to the supplement

paper/supplement.tex

@@ -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}