hw1: done up to 3.2
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% Sources:
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% https://www.youtube.com/watch?v=91RZYO1cv_o
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%% Exercise 3.1
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% f(x1, x2) = x1^2 + u * x2^2;
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% [x1 x2] [1 0] [x1] + [0][x1]
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% [0 u] [x2] + [0][x2]
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% 1/2 * [x1 x2] [2 0] [x1] + [0][x1]
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% [0 2u] [x2] + [0][x2]
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% A = [1 0; 0 u]; b = [0; 0]
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@ -40,4 +43,6 @@ for u = 1:10
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subplot(4, 5, con);
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contour(xs, ys, Z, 20);
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title(sprintf("Contour for u=%d", u));
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end
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matlab2tikz('showInfo', false, './ex32.tex')
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hw1/ex32.tex
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hw1/ex32.tex
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hw1/main.pdf
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hw1/main.tex
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hw1/main.tex
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\documentclass{scrartcl}
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\usepackage[utf8]{inputenc}
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\usepackage{amsmath}
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\usepackage{pgfplots}
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\pgfplotsset{compat=1.17}
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\usepackage{mathrsfs}
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\usepackage{hyperref}
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\usetikzlibrary{arrows}
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\setlength{\parindent}{0cm}
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\setlength{\parskip}{0.5\baselineskip}
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@ -67,4 +72,25 @@ Not in general. If for example we consider A and b to be only zeros, then $J(x)
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However, for if $A$ would be guaranteed to have full rank, the minimizer would be unique because the first order necessary condition would hold only for one value $x^*$. This is because the linear system $Ax^* = b$ would have one and only one solution (due to $A$ being full rank).
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\section{Exercise 3}
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\subsection{Quadratic form}
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$f(x,y)$ can be written in quadratic form in the following way:
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\[f(v) = \frac12 \langle\begin{bmatrix}2 & 0\\0 & 2\mu\end{bmatrix}v, v\rangle + \langle\begin{bmatrix}0\\0\end{bmatrix}, x\rangle\]
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where:
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\[v = \begin{bmatrix}x\\y\end{bmatrix}\]
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\subsection{Matlab implementation with \texttt{surf} and \texttt{contour}}
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The graphs generated by MATLAB are shown below:
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\resizebox{\textwidth}{!}{\input{ex32.tex}}
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Isolines get stretched along the y axis as $\mu$ increases. For a large $\mu$, points well far away from the axes could be a
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problem since picking search directions and steps using a naive gradient based method iterations will zig-zag to the minimizer reaching it slowly.
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\end{document}
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