hw4: almost done
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%% Assignment 4
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% Name: Claudio Maggioni
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%
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% Date: 19/3/2019
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% Date: 2020-05-17
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%
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% This is a template file for the first assignment to get started with running
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% and publishing code in Matlab. Each problem has its own section (delineated
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@ -11,11 +11,13 @@
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% To generate a pdf for submission in your current directory, use the following
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% three lines of code at the command window:
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%
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% >> options.format = 'pdf'; options.outputDir = pwd; publish('assignment3.m', options)
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% >> options.format = 'pdf'; options.outputDir = pwd; publish('assignment4.m', options)
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%
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%% Problem 3
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clear;
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clear all;
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clc;
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clf reset;
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n=10;
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f = @(x) (exp(-(x^2)/2)/sqrt(2*pi));
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@ -36,12 +38,21 @@ p_10 = computePolypoints(f, xe, x_10, 10);
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p_5c = computePolypoints(f, xe, x_5c, 5);
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p_10c = computePolypoints(f, xe, x_10c, 10);
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figure;
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subplot(1,2,1);
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plot(xe, p_5, xe, p_10, xe, y);
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subplot(1,2,2);
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figure;
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plot(xe, p_5c, xe, p_10c, xe, y);
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%% Question 6
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y = [-0.0044; -0.0771; -0.2001; -0.3521; -0.3520; 0; 0.5741; 0.8673; ...
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0.5741; 0; 0.3520; -0.3521; 0.2001; -0.0771; -0.0213; -0.0044];
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figure;
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alpha = b3interpolate(y);
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x = (0:0.01:16)';
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y_c = spline_curve(alpha, x);
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plot(x, y_c);
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%% Question 3 (continued)
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function x = computeEquidistantXs(n)
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x = zeros(2*n+1,1);
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for i = 1:2*n+1
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@ -88,3 +99,46 @@ function p = HornerNewton(N, x, xe)
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p = p .* (xe - x(i)) + N(i);
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end
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end
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%% Problem 6 (continued)
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% assuming x_i = i - 1
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function [alpha] = b3interpolate(y)
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n = size(y, 1);
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A = zeros(n+2);
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B = [y; 0; 0];
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for x_i = 0:n-1
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for j = -1:n
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%fprintf("A(%d, %d) = B3(%d - %d)\n", x_i + 1, j+2, x_i, j);
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A(x_i + 1, j + 2) = B3(x_i - j);
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end
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end
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A(n+1, 1:3) = [1 -2 1];
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A(n+2, n-1:n+1) = [1 -2 1];
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alpha = A \ B;
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end
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function [v] = spline_curve(alpha, x)
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n = size(alpha, 1) - 2;
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v = zeros(size(x));
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for i = 1:size(x,1)
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for j = -1:n
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v(i) = v(i) + alpha(j + 2) * B3(x(i) - j);
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end
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end
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end
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function [y] = B3(x)
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y = zeros (size (x));
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i1 = find (-2 < x & x < -1);
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i2 = find (-1 <= x & x < 1);
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i3 = find (1 <= x & x < 2);
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y(i1) = 0.5 * (x(i1) + 2) .^ 3;
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y(i2) = 0.5 * (3 * abs (x(i2)) .^ 3 - 6 * x(i2) .^ 2 + 4);
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y(i3) = 0.5 * (2 - x(i3)) .^ 3;
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y = y / 3;
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end
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hw4/assignment4.pdf
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@ -184,7 +184,7 @@ a_0\\a_1\\a_2\\a_{-1}\\
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y_0\\y_1\\y_2\\y_{3}\\
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\end{bmatrix}\]
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\subsection*{Question 5}
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\section*{Question 5}
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$$1 = y_0 = s(x_0) = 0 = a_{-1}B_3(1) +a_0B_3(0)
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+a_1B_3(-1)
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