hw1: done up to 3.2

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Claudio Maggioni (maggicl) 2021-03-21 22:54:15 +01:00
parent 98b7349508
commit 85bb62cced
4 changed files with 14206 additions and 3 deletions

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% Sources:
% https://www.youtube.com/watch?v=91RZYO1cv_o
%% Exercise 3.1 %% Exercise 3.1
% f(x1, x2) = x1^2 + u * x2^2; % f(x1, x2) = x1^2 + u * x2^2;
% [x1 x2] [1 0] [x1] + [0][x1] % 1/2 * [x1 x2] [2 0] [x1] + [0][x1]
% [0 u] [x2] + [0][x2] % [0 2u] [x2] + [0][x2]
% A = [1 0; 0 u]; b = [0; 0] % A = [1 0; 0 u]; b = [0; 0]
@ -40,4 +43,6 @@ for u = 1:10
subplot(4, 5, con); subplot(4, 5, con);
contour(xs, ys, Z, 20); contour(xs, ys, Z, 20);
title(sprintf("Contour for u=%d", u)); title(sprintf("Contour for u=%d", u));
end
end
matlab2tikz('showInfo', false, './ex32.tex')

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hw1/ex32.tex Normal file

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\documentclass{scrartcl} \documentclass{scrartcl}
\usepackage[utf8]{inputenc} \usepackage[utf8]{inputenc}
\usepackage{amsmath} \usepackage{amsmath}
\usepackage{pgfplots}
\pgfplotsset{compat=1.17}
\usepackage{mathrsfs}
\usepackage{hyperref}
\usetikzlibrary{arrows}
\setlength{\parindent}{0cm} \setlength{\parindent}{0cm}
\setlength{\parskip}{0.5\baselineskip} \setlength{\parskip}{0.5\baselineskip}
@ -67,4 +72,25 @@ Not in general. If for example we consider A and b to be only zeros, then $J(x)
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). 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).
\section{Exercise 3}
\subsection{Quadratic form}
$f(x,y)$ can be written in quadratic form in the following way:
\[f(v) = \frac12 \langle\begin{bmatrix}2 & 0\\0 & 2\mu\end{bmatrix}v, v\rangle + \langle\begin{bmatrix}0\\0\end{bmatrix}, x\rangle\]
where:
\[v = \begin{bmatrix}x\\y\end{bmatrix}\]
\subsection{Matlab implementation with \texttt{surf} and \texttt{contour}}
The graphs generated by MATLAB are shown below:
\resizebox{\textwidth}{!}{\input{ex32.tex}}
Isolines get stretched along the y axis as $\mu$ increases. For a large $\mu$, points well far away from the axes could be a
problem since picking search directions and steps using a naive gradient based method iterations will zig-zag to the minimizer reaching it slowly.
\end{document} \end{document}