hw4: done ex1

hw4/assignment4.m

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+%% Assignment 2
+% Name: Claudio Maggioni
+%
+% Date: 19/3/2019
+%
+% This is a template file for the first assignment to get started with running
+% and publishing code in Matlab.  Each problem has its own section (delineated
+% by |%%|) and can be run in isolation by clicking into the particular section
+% and pressing |Ctrl| + |Enter| (evaluate current section).
+%
+% To generate a pdf for submission in your current directory, use the following
+% three lines of code at the command window:
+%
+%  >> options.format = 'pdf'; options.outputDir = pwd; publish('assignment3.m', options)
+%
+
+%% Problem 3
+format long
+A = [4 3 2 1; 8 8 5 2; 16 12 10 5; 32 24 20 11];
+[L,U,P] = pivotedOuterProductLU(A);
+display(L);
+display(U);
+display(P);
+
+%% Problem 6
+[X,Y] = meshgrid((0:2000)/2000); 
+A = exp(-sqrt((X-Y).^2));
+L = outerProductCholesky(A);
+disp(norm(L*L'-A, 'fro'));
+
+%% Problem 3 (continued)
+
+function [L,U,P] = pivotedOuterProductLU(A)
+    dimensions = size(A);
+    n = dimensions(1);
+    
+    p = 1:n;
+    L = zeros(n);
+    U = zeros(n);
+
+    for i = 1:n
+        values = A(:,i);
+        values(values == 0) = -Inf;
+        [~, p_k] = max(values);
+        k = find(p == p_k);
+        
+        if k == -Inf
+            disp("Matrix is singular");
+            L = [];
+            U = [];
+            P = [];
+            return
+        end
+        
+        p(k) = p(i);
+        p(i) = p_k;
+        
+        L(:,i) = A(:,i) / A(p(i),i);
+        U(i,:) = A(p(i),:);
+        A = A - L(:,i) * U(i,:);
+    end
+    
+    I = eye(n);
+    P = zeros(n);
+    for i = 1:n
+        P(:,i) = I(:,p(i));
+    end
+    P = transpose(P);
+    L = P * L;   
+end
+
+%% Problem 6 (continued)
+
+function [L] = outerProductCholesky(A)
+    if ~all(eig(A) >= 0)
+        disp("matrix is not positive definite");
+        L = [];
+        return;
+    end
+    dimensions = size(A);
+    n = dimensions(1);
+    L = zeros(n);
+    
+    for i = 1:n
+        L(:, i) = A(:, i) / sqrt(A(i,i));
+        A = A - L(:, i) * transpose(L(:, i));
+    end
+end

hw4/hw4.tex

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+% vim: set ts=2 sw=2 et tw=80:
+
+\documentclass[12pt,a4paper]{article}
+
+\usepackage[utf8]{inputenc} \usepackage[margin=2cm]{geometry}
+\usepackage{amstext} \usepackage{amsmath} \usepackage{array}
+\newcommand{\lra}{\Leftrightarrow}
+
+\title{Howework 4 -- Introduction to Computational Science}
+
+\author{Claudio Maggioni}
+
+\begin{document} \maketitle 
+\section*{Question 1}
+$$L_0(x) = \prod_{j = 0, j \neq 0}^n \frac{x - x_j}{x_i - x_j} = 
+\frac{x - (-0.5)}{(-1) - (-0.5)} \cdot \frac{x - 0.5}{(-1) - 0.5} \cdot 
+\frac{x - 1}{(-1) - 1} =  -\frac{2}{3}x^3 + \frac{2}{3}x^2 
++\frac{1}{6} x - \frac{1}{6}$$
+
+$$L_1(x) = \prod_{j = 0, j \neq 1}^n \frac{x - x_j}{x_i - x_j} = 
+\frac{x - (-1)}{(-0.5) - (-1)} \cdot \frac{x - 0.5}{(-0.5) - 0.5} \cdot 
+\frac{x - 1}{(-0.5) - 1} = \frac{4}{3}x^3 - \frac{2}{3}x^2 - \frac{4}{3}x + \frac{2}{3}$$
+
+$$L_2(x) = \prod_{j = 0, j \neq 2}^n \frac{x - x_j}{x_i - x_j} = 
+\frac{x - (-1)}{0.5 - (-1)} \cdot \frac{x - (-0.5)}{0.5 - (-0.5)} \cdot 
+\frac{x - 1}{0.5 - 1} = -\frac{4}{3}x^3 - \frac{2}{3}x^2 + \frac{4}{3}x + \frac{2}{3}$$
+
+$$L_3(x) = \prod_{j = 0, j \neq 3}^n \frac{x - x_j}{x_i - x_j} = 
+\frac{x - (-1)}{1 - (-1)} \cdot \frac{x - (-0.5)}{1 - (-0.5)} \cdot 
+\frac{x - 0.5}{1 - 0.5} = \frac{2}{3}x^3 + \frac{2}{3}x^2 -\frac{1}{6}x - \frac{1}{6}$$
+
+$$p(x) = \sum_{i=0}^n y_i L_i(x) =
+2 \cdot \left(-\frac{2}{3}x^3 + \frac{2}{3}x^2 
++\frac{1}{6} x - \frac{1}{6}\right) +
+1 \cdot \left(\frac{4}{3}x^3 - \frac{2}{3}x^2 - \frac{4}{3}x + \frac{2}{3}\right) +
+$$$$
+0.5 \cdot \left(-\frac{4}{3}x^3 - \frac{2}{3}x^2 + \frac{4}{3}x + \frac{2}{3}\right) + 
+0.4 \cdot \left(\frac{2}{3}x^3 + \frac{2}{3}x^2 -\frac{1}{6}x - \frac{1}{6}\right) = -\frac{2}{5}x^3 + \frac{3}{5}x^2 - \frac{2}{5}x + \frac{3}{5}$$
+
+$$\frac{\max_{x \in [-1,1]} |f^{(n+1)}|}{5!} = 
+\frac{\max_{x \in [-1,1]}|\frac{7680}{|2x+3|^6}|}{120} =
+\max_{x \in [-1,1]}\frac{64}{|2x+3|^6} = 64$$
+
+$$\max_{x \in [-1,1]} \left|(x - 1)\left(x - \frac{1}{2}\right)
+\left(x + \frac{1}{2}\right)(x+1)\right| = 
+\max_{x \in [-1,1]} \left| x^4 - \frac{5}{4}x^2 + \frac{1}{4}\right| = \frac{1}{4}$$
+
+$$\max_{x \in [-1,1]} |f(x) - p(x)| \leq \frac{\max_{x \in [-1,1]} |f^{(n+1)}|}{5!} \max_{x \in [-1,1]} \left|(x - 1)\left(x - \frac{1}{2}\right)
+\left(x + \frac{1}{2}\right)(x+1)\right| =$$$$= 64 \cdot \frac{1}{4} = 8 \leq 8$$
+
+The statement above is true so p satisfies the error estimate:
+
+$$\max_{x \in [-1,1]} |f(x) - p(x)| \leq 8$$
+\end{document}