Commit 8a56afb2 by Max Ehrlich

### 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 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} of $X$ minimizes the least squared error e_m = \sum_{i=0}^{n} (p_m(i) - x_i)^2 \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 D^T_my = x 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 D_mD^T_my = D_mx and since the DCT is an orthonormal transformation, the rows of $D_m$ are orthogonal, so $D_mD^T_m = I$. Therefore y = D_mx 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 \var[X] = \e[Y^2] \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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