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5_subspaces.tex
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\documentclass[aspectratio=169]{beamer}
%\documentclass[aspectratio=169,handout]{beamer}
\def\stylepath{../styles}
\usepackage{\stylepath/com303}
\usepackage{pst-3dplot}
%\setbeameroption{show only notes}\def\logoEPFL{}
\setbeameroption{show notes}
\begin{document}
\begin{frame} \frametitle{Vector subspace}
\begin{center}
a subset of vectors \textit{closed} under addition and scalar multiplication
\end{center}
\end{frame}
\begin{frame}
\frametitle{Example in Euclidean Space}
intuition: $\mathbb{R}^2 \subset \mathbb{R}^{3}$
\begin{center}
\psset{unit=3cm}
\begin{pspicture}(-1.5,-.8)(1,1)
\psset{Alpha=40,linewidth=1pt}
\pstThreeDCoor[linecolor=black,linewidth=1pt,%
xMax=1.5,yMax=1.5,zMax=1,%
xMin=-.2,yMin=-.2,zMin=-.2,%
nameX=$\mathbf{e}^{(0)}$,
nameY=$\mathbf{e}^{(1)}$,
nameZ=$\mathbf{e}^{(2)}$]
\pstThreeDLine[linecolor=gray,linewidth=1.8pt]{->}(0,0,0)(1,.3,0)
\pstThreeDLine[linecolor=gray,linewidth=1.8pt]{->}(0,0,0)(1,.7,0)
\pstThreeDLine[linecolor=blue,linewidth=1.8pt]{->}(1,.3,0)(1,.7,0)
\pstThreeDPut(1.25,.45,0){$\mathbf{x+y}$}
\end{pspicture}
\end{center}
\end{frame}
\begin{frame} \frametitle{Subspace of symmetric functions over $L_2[-1,1]$}
\begin{center}
\begin{figure}
\begin{dspPlot}[xticks=1,xout=true,yticks=2,sidegap=0]{-1, 1}{-2.2, 2.2}
\moocStyle
\only<1-2>{\dspFunc{x 180 mul cos}}
\only<2>{\dspFunc[linecolor=orange]{x 180 mul 5 mul cos}}
\only<3>{
\dspFunc[linecolor=lightgray]{x 180 mul cos}
\dspFunc[linecolor=lightgray]{x 180 mul 5 mul cos}
\dspFunc[linecolor=blue]{x 180 mul cos x 180 mul 5 mul cos add}}
\end{dspPlot}
\end{figure}
\only<1>{$\mathbf{x} = \cos(\pi t)$}
\only<2>{$\mathbf{y} = \cos(5\pi t)$}
\only<3>{$\mathbf{x} + \mathbf{y}$, symmetric}
\end{center}
\end{frame}
\begin{frame} \frametitle{Subspaces have their own basis}
\centering
\[
\mathbf{e}^{(0)} = \myvector{1 \\ 0 \\ 0} \qquad
\mathbf{e}^{(1)} = \myvector{0 \\ 1 \\ 0}
\]
basis vector for the plane in $\mathbb{R}^3$
\end{frame}
\begin{frame} \frametitle{Approximation}
\begin{columns}
\begin{column}{.4\paperwidth}
Problem:
\begin{itemize}[<+->]
\item vector $\mathbf{x} \in V$
\item subspace $S \subseteq V$
\vspace{1em}
\item approximate $\mathbf{x}$ with $\hat{\mathbf{x}} \in S$
\end{itemize}
\end{column}
\begin{column}{.5\paperwidth}
\centering
\psset{unit=3cm}
\begin{pspicture}(-1.5,-.8)(1.5,1.8)
\psset{Alpha=40,linewidth=1pt}
\pstThreeDCoor[linecolor=black,linewidth=1pt,%
xMax=1.5,yMax=1.5,zMax=1.5,%
xMin=-.2,yMin=-.2,zMin=-.2,%
nameX=$\mathbf{e}^{(0)}$,
nameY=$\mathbf{e}^{(1)}$,
nameZ=$\mathbf{e}^{(2)}$]
\pstThreeDLine[linewidth=1.8pt]{->}(0,0,0)(1,.5,2)\pstThreeDPut(1,.5,2.15){$\mathbf{x}$}
\only<2->{%
\pstThreeDLine[linecolor=darkred]{->}(0,0,0)(1.5,0,0)
\pstThreeDLine[linecolor=darkred]{->}(0,0,0)(0,1.5,0)}
\only<3->{
\pstThreeDLine[linestyle=dashed,linewidth=0.5pt](0,.5,0)(1,.5,0)
\pstThreeDLine[linestyle=dashed,linewidth=0.5pt](1,.5,0)(1,0,0)
\pstThreeDLine[linestyle=dashed,linewidth=0.5pt](1,.5,0)(1,.5,2)
\pstThreeDLine[linecolor=gray,linewidth=1.8pt]{->}(0,0,0)(1,.5,0)
\pstThreeDPut(1.15,.65,0){$\hat{\mathbf{x}}$}}
\end{pspicture}
\end{column}
\end{columns}
\end{frame}
\begin{frame} \frametitle{Least-Squares Approximation}
\begin{itemize}[<+->]
\item $\{ \mathbf{s}^{(k)} \}_{k =0,1,\ldots, K-1}$ orthonormal basis for $S$
\item orthogonal projection:
\[
\hat{\mathbf{x}} = \sum_{k=0}^{K-1} \langle \mathbf{s}^{(k)}, \mathbf{x} \rangle \, \mathbf{s}^{(k)}
\]
\end{itemize}
\vspace{2em}
\centering
\uncover<3->{orthogonal projection is the ``best'' approximation over $S$}
\end{frame}
\begin{frame} \frametitle{Least-Squares Approximation}
\begin{itemize}[<+->]
\item orthogonal projection has minimum-norm error:
\[
\arg\min_{\mathbf{y} \in S} \| \mathbf{x} - \mathbf{y} \| = \hat{\mathbf{x}}
\]
\vspace{1em}
\item error is orthogonal to approximation:
\[
\langle \mathbf{x} - \hat{\mathbf{x}}, \, \hat{\mathbf{x}} \rangle = 0
\]
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Least Squares Approximation}
\begin{center}
\psset{unit=10mm}
\begin{pspicture}(-1,0)(8,6)
\psline{->}(5,4)\rput[bl]{0}(5,4){~$\mathbf{x}$}
\psline{->}(10,2)\rput[tl]{0}(10,2){~$\mathbf{s}$}
\only<2->{\pscircle[linecolor=green,linewidth=0.5pt](5,4){0.49}}
\only<3->{\pscircle[linecolor=green,linewidth=0.5pt](5,4){0.98}}
\only<4->{\pscircle[linecolor=green,linewidth=0.5pt](5,4){1.47}}
\only<5->{\pscircle[linecolor=green,linewidth=0.5pt](5,4){1.96}}
\only<6->{%
\pscircle[linecolor=green,linewidth=0.5pt](5,4){2.45}
\pscircle[linecolor=green,linewidth=0.5pt](5,4){2.94}
\psline[linecolor=red]{->}(5.57,1.11)(5,4)\rput[bl]{0}(5.3,3){~$\mathbf{x}-\hat{\mathbf{x}}$}
\psline[linecolor=blue]{->}(5.57,1.11)\rput[tl]{0}(3,0.2){~$\hat{\mathbf{x}}$}
}
\end{pspicture}
\end{center}
\end{frame}
\begin{frame} \frametitle{Example: polynomial approximation}
\begin{itemize}[<+->]
\item vector space $P_N[-1,1] \subset L_2[-1,1]$
\item $\mathbf{p} = a_0 + a_1 t + \ldots + a_{N-1} t^{N-1}$
\item a self-evident, naive basis: $\mathbf{s}^{(k)} = t^k, \quad k = 0, 1, \ldots, N-1$
\item naive basis is not orthonormal
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Example: polynomial approximation}
\centering
goal: approximate $\mathbf{x} = \sin t \in L_2[-1,1]$ over $P_3[-1,1]$
\vspace{1.5em}
\begin{itemize}
\item<2-> build orthonormal basis from naive basis
\item<3-> project $\mathbf{x}$ over the orthonormal basis
\item<4-> compute approximation error
\item<5-> compare error to Taylor approximation (well known but not optimal over the interval)
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Building an orthonormal basis}
\begin{center}
Gram-Schmidt orthonormalization procedure:
\begin{figure}
\begin{dspBlocks}{1}{0.2}
$\{ \mathbf{s}^{(k)} \}$~~ & $~~\{ \mathbf{u}^{(k)} \}$ \\
original set & orthonormal set \\
\BDConnH{1}{1}{2}{}
\end{dspBlocks}
\end{figure}
\end{center}
\pause
Algorithmic procedure: at each step $k$
\begin{enumerate}[<+->]
\item $\displaystyle \mathbf{p}^{(k)} = \mathbf{s}^{(k)} - \sum_{n=0}^{k-1} \langle \mathbf{u}^{(n)}, \mathbf{s}^{(k)} \rangle \, \mathbf{u}^{(n)}$
\item $\mathbf{u}^{(k)} = \mathbf{p}^{(k)}/ \|\mathbf{p}^{(k)}\|$
\end{enumerate}
\end{frame}
\begin{frame} \frametitle{Building an orthonormal basis}
\begin{center}
\psset{unit=10mm}
\begin{pspicture}(-1,0)(8,6)
{
\only<2->{\psset{linecolor=lightgray}}
\psline{->}(5,4)\rput[bl]{0}(5,4){~$\mathbf{s}^{(1)}$}
\psline{->}(10,2)\rput[tl]{0}(10,2){~$\mathbf{s}^{(0)}$}
}
\only<2>{
\psline[linecolor=red]{->}(10,2)
\rput[bl]{0}(10,2){~$\mathbf{p}^{(0)}$}}
\only<3->{
\psline[linecolor=blue]{->}(3.9223, 0.7844)
\rput[tl]{0}(3.9223, 0.7844){\color{blue}~$\mathbf{u}^{(0)}$}}
\only<4-5>{
\psline[linecolor=gray,linestyle=dashed](5.57,1.11)(5,4)}
\only<4>{
\psline[linecolor=red!50]{->}(5.57,1.11)
\rput[tl]{0}(5.57,1.11){\color{red!50}$\langle \mathbf{u}^{(0)}, \mathbf{s}^{(1)} \rangle \mathbf{u}^{(0)}$}}
\only<5>{
\psline[linecolor=red!50]{->}(5,4)(-0.57,2.89)
\psline[linecolor=red]{->}(-0.57,2.89)
\rput[bl]{0}(-0.57,2.89){~$\mathbf{p}^{(1)}$}}
\only<6->{
\psline[linecolor=blue]{->}(-0.7740, 3.9244)
\rput[br]{0}(-0.7740, 3.9244){\color{blue}~$\mathbf{u}^{(1)}$}}
\end{pspicture}
\end{center}
\end{frame}
\begin{frame} \frametitle{Building an orthonormal basis}
\begin{center}
Gram-Schmidt orthonormalization of the naive basis: $\{ \mathbf{s}^{(k)} \} \rightarrow \{ \mathbf{u}^{(k)} \}$
\end{center}
\begin{columns}
\begin{column}{.5\paperwidth}
\begin{itemize}[<+->]
\item $\mathbf{s}^{(0)} = 1$
\begin{itemize}
\item $\mathbf{p}^{(0)} = \mathbf{s}^{(0)} = 1$
\item $\|\mathbf{p}^{(0)}\|^2 = 2$
\item $\mathbf{u}^{(0)} = \mathbf{p}^{(0)}/\|\mathbf{p}^{(0)}\| = \sqrt{1/2}$
\end{itemize}
\item $\mathbf{s}^{(1)} = t$
\begin{itemize}
\item $\langle \mathbf{u}^{(0)}, \mathbf{s}^{(1)} \rangle = \int_{-1}^{1}t/\sqrt{2} = 0$
\item $\mathbf{p}^{(1)} = \mathbf{s}^{(1)} = t$
\item $\|\mathbf{p}^{(1)}\|^2 = 2/3$
\item $\mathbf{u}^{(1)} = \sqrt{3/2}\,t$
\end{itemize}
\end{itemize}
\end{column}
\begin{column}{.5\paperwidth}
\begin{itemize}[<+->]
\item $\mathbf{s}^{(2)} = t^2$
\begin{itemize}
\item $\langle \mathbf{u}^{(0)}, \mathbf{s}^{(2)} \rangle = \int_{-1}^{1}t^2/\sqrt{2} = 2/3\sqrt{2}$
\item $\langle \mathbf{u}^{(1)}, \mathbf{s}^{(2)} \rangle = \int_{-1}^{1}t^3/\sqrt{2} = 0$
\item $\mathbf{p}^{(2)} = \mathbf{s}^{(2)} - (2/3\sqrt{2})\mathbf{u}^{(0)} = t^2 - 1/3$
\item $\|\mathbf{p}^{(2)}\|^2 = 8/45$
\item $\mathbf{u}^{(2)} = \sqrt{5/8}(3t^2-1)$
\end{itemize}
\end{itemize}
\end{column}
\end{columns}
\end{frame}
\begin{frame} \frametitle{Legendre polynomials}
\begin{center}
The Gram-Schmidt algorithm leads to an orthonormal basis for $P_N([-1,1])$
\end{center}
\begin{align*}
\mathbf{u}^{(0)} &= \sqrt{1/2} \\
\mathbf{u}^{(1)} &= \sqrt{3/2}\,t \\
\mathbf{u}^{(2)} &= \sqrt{5/8}(3t^2-1) \\
\mathbf{u}^{(3)} &= \ldots
\end{align*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%
%\begin{align*}
% P_N(x) &= \sqrt{\frac{2N+1}{2}}p_N(x) \\
% p_0(x) &= 1\\
% p_1(x) &= x \\
% p_2(x) &= \frac{1}{2}(3x^2-1) \\
% p_3(x) &= \frac{1}{2}(5x^3-3x) \\
% p_4(x) &= \frac{1}{8}(35x^4 - 30x^2 + 3) \\
% p_5(x) &= \frac{1}{8}(63x^5 - 70x^3 + 15x) \\
%\end{align*}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame} \frametitle{Legendre Polynomials}
\begin{center}
\begin{figure}
\begin{dspPlot}[yticks=1,xticks=1,height=5cm,sidegap=0,xout=true]{-1, 1}{-2.3, 2.3}
\begin{dspClip}
\only<1->{\dspFunc[linecolor=black]{ 1 2 div sqrt 1 mul}}%
\only<2->{\dspFunc[linecolor=red]{ 3 2 div sqrt x mul}}%
\only<3->{\dspFunc[linecolor=yellow]{5 2 div sqrt 3 x x mul mul 1 sub mul 2 div}}%
\only<4->{\dspFunc[linecolor=green]{ 7 2 div sqrt 5 x x x mul mul mul -3 x mul add mul 2 div}}%
\only<5->{\dspFunc[linecolor=blue]{ 9 2 div sqrt 35 x x x x mul mul mul mul -30 x x mul mul 3 add add mul 8 div}}%
\only<6->{\dspFunc[linecolor=cyan]{ 11 2 div sqrt 63 x x x x x mul mul mul mul mul -70 x x x mul mul mul 15 x mul add add mul 8 div}}%
\end{dspClip}
\dspCustomTicks{-1 -1 0 0 1 1}%
\end{dspPlot}
\end{figure}
\end{center}
\end{frame}
\begin{frame} \frametitle{Orthogonal projection over $P_3[-1,1]$}
\[
\alpha_k = \langle \mathbf{u}^{(k)}, \mathbf{x} \rangle = \int_{-1}^{1}u_k(t)\,\sin t \, dt
\]
\begin{itemize}[<+->]
\item $\alpha_0 = \langle \sqrt{1/2}, \sin t \rangle = 0$
\item $\alpha_1 = \langle \sqrt{3/2} \, t, \sin t \rangle \approx 0.7377$
\item $\alpha_2 = \langle \sqrt{5/8}(3t^2 - 1), \sin t \rangle = 0$
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Approximation}
\begin{center}
Using the orthogonal projection over $P_3[-1,1]$:
\[
\sin t \rightarrow \alpha_1 \mathbf{u}^{(1)} \approx 0.9035\,t
\]
\vspace{3em}
\uncover<2->{
Using Taylor's series:
\[
\sin t \approx t
\]
}
\end{center}
\end{frame}
\begin{frame} \frametitle{Sine approximation}
\begin{center}
\begin{figure}
\begin{dspPlot}[yticks=1,xticks=1,height=5cm,sidegap=0,xout=true]{-1, 1}{-1.2, 1.2}
\dspFunc[linecolor=blue]{x 3.14 div 180 mul sin}
\dspFunc[linecolor=red,linewidth=1pt]{x }
\dspFunc[linecolor=green,linewidth=1pt]{x 0.9035 mul}
\dspLegend(-0.9,1){blue {$\sin t$} red $t$ green $0.9035t$}
\end{dspPlot}
\end{figure}
\end{center}
\end{frame}
\begin{frame} \frametitle{Approximation error}
\begin{center}
\begin{figure}
%\dspCustomTicks{-1 -1 0 0 1 1}
\begin{dspPlot}[yticks=1,xticks=1,height=5cm,sidegap=0,xout=true]{-1, 1}{0, 0.22}
\dspFunc[linecolor=red]{x 3.14 div 180 mul sin x sub abs}
\dspFunc[linecolor=green]{x 3.14 div 180 mul sin x 0.9035 mul sub abs}
\dspLegend(-0.5,0.2){red {$|\sin t - t|$} green {$|\sin t - 0.9035t|$}}
\end{dspPlot}
\end{figure}
\end{center}
\end{frame}
\begin{frame} \frametitle{Error norm}
\begin{center}
Orthogonal projection over $P_3[-1,1]$:
\[
\| \sin t - \alpha_1\,\mathbf{u}^{(1)} \| \approx 0.0337
\]
\vspace{3em}
\uncover<2->{
Taylor series:
\[
\| \sin t - t \| \approx 0.0857
\]
}
\end{center}
\end{frame}
\end{document}

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