Merge branch 'master' of git.maggioni.xyz:maggicl/ICS

hw4/assignment4.m

@@ -1,7 +1,7 @@
 %% Assignment 4
 % Name: Claudio Maggioni
 %
-% Date: 19/3/2019
+% Date: 2020-05-17
 %
 % 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
@@ -11,11 +11,13 @@
 % 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)
+%  >> options.format = 'pdf'; options.outputDir = pwd; publish('assignment4.m', options)
 %
 
 %% Problem 3
-clear;
+clear all;
+clc;
+clf reset;
 
 n=10;
 f = @(x) (exp(-(x^2)/2)/sqrt(2*pi));
@@ -36,12 +38,21 @@ p_10 = computePolypoints(f, xe, x_10, 10);
 p_5c = computePolypoints(f, xe, x_5c, 5);
 p_10c = computePolypoints(f, xe, x_10c, 10);
 
-figure;
-subplot(1,2,1);
 plot(xe, p_5, xe, p_10, xe, y);
-subplot(1,2,2);
+figure;
 plot(xe, p_5c, xe, p_10c, xe, y);
 
+%% Question 6
+
+y = [-0.0044; -0.0771; -0.2001; -0.3521; -0.3520; 0; 0.5741; 0.8673; ...
+    0.5741; 0; 0.3520; -0.3521; 0.2001; -0.0771; -0.0213; -0.0044];
+figure;
+alpha = b3interpolate(y);
+x = (0:0.01:16)';
+y_c = spline_curve(alpha, x);
+plot(x, y_c);
+
+%% Question 3 (continued)
 function x = computeEquidistantXs(n)
     x = zeros(2*n+1,1);
     for i = 1:2*n+1
@@ -88,3 +99,46 @@ function p = HornerNewton(N, x, xe)
         p = p .* (xe - x(i)) + N(i);
     end
 end
+
+%% Problem 6 (continued)
+
+% assuming x_i = i - 1
+function [alpha] = b3interpolate(y)
+    n = size(y, 1);
+    A = zeros(n+2);
+    B = [y; 0; 0];
+    for x_i = 0:n-1 
+        for j = -1:n
+            %fprintf("A(%d, %d) = B3(%d - %d)\n", x_i + 1, j+2, x_i, j);
+            A(x_i + 1, j + 2) = B3(x_i - j);
+        end
+    end
+    A(n+1, 1:3) = [1 -2 1];
+    A(n+2, n-1:n+1) = [1 -2 1];
+    alpha = A \ B;
+end
+
+function [v] = spline_curve(alpha, x)
+    n = size(alpha, 1) - 2;
+    v = zeros(size(x));
+    for i = 1:size(x,1)
+        for j = -1:n
+            v(i) = v(i) + alpha(j + 2) * B3(x(i) - j);
+        end
+    end
+end
+
+function [y] = B3(x)
+y = zeros (size (x));
+i1 = find (-2 < x & x < -1);
+i2 = find (-1 <= x & x < 1);
+i3 = find (1 <= x & x < 2);
+
+y(i1) = 0.5 * (x(i1) + 2) .^ 3;
+y(i2) = 0.5 * (3 * abs (x(i2)) .^ 3 - 6 * x(i2) .^ 2 + 4);
+y(i3) = 0.5 * (2 - x(i3)) .^ 3;
+y = y / 3;
+end
+ 
+
+

hw4/hw4.tex

@@ -89,7 +89,7 @@ $$y = y (x - x_0) + a_0 = -\frac{23}{6} \cdot 0.5 + 0 = -\frac{23}{12}$$
 \section*{Question 4}
 \subsection*{Point a)}
 
-The node coordinates to which fix the quadratic spline are $(1/2, y_0), (1, y_1),(3/2, y_2),(2, y_3)$.
+The node coordinates to which fix the quadratic spline are $(1/2, y_0), (3/2, y_1),(5/2, y_2),(7/2, y_3)$.
 
 Then, we can start formulating the equations for the linear system:
 
@@ -100,40 +100,38 @@ $$y_0 = s\left(\frac{1}{2}\right) = a_{-1}B_2\left(\frac{1}{2} + 1\right) + a_{0
 + a_{1}B_2\left(\frac{1}{2} - 5\right)=$$$$
 a_{-1} \cdot 0 + a_{0} \cdot \frac{1}{2} + a_{1} \cdot \frac{1}{2} + a_{2} \cdot 0 + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = \frac{a_0 + a_1}{2}$$ 
 
-$$y_1 = s\left(1\right) = a_{-1}B_2\left(1 + 1\right) + a_{0}B_2\left(1\right) + a_{1}B_2\left(1 - 1\right) + $$$$
-+ a_{2}B_2\left(1 - 2\right)
-+ a_{-1}B_2\left(1 - 3\right)
-+ a_{0}B_2\left(1 - 4\right)
-+ a_{1}B_2\left(1 - 5\right)=$$$$
-a_{-1} \cdot 0 + a_{0} \cdot \left(\frac{1}{8}\right) + a_{1} \cdot \frac{3}{4} + a_{2} \cdot \left(\frac{1}{8}\right) + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = 
-\frac{1}{8} a_{0} + \frac{3}{4} a_{1} + \frac{1}{8} a_{2}$$ 
-
-$$y_2 = s\left(\frac{3}{2}\right) = a_{-1}B_2\left(\frac{3}{2} + 1\right) + a_{0}B_2\left(\frac{3}{2}\right) + a_{1}B_2\left(\frac{3}{2} - 1\right) + $$$$
+$$y_1 = s\left(\frac{3}{2}\right) = a_{-1}B_2\left(\frac{3}{2} + 1\right) + a_{0}B_2\left(\frac{3}{2}\right) + a_{1}B_2\left(\frac{3}{2} - 1\right) + $$$$
 + a_{2}B_2\left(\frac{3}{2} - 2\right)
 + a_{-1}B_2\left(\frac{3}{2} - 3\right)
 + a_{0}B_2\left(\frac{3}{2} - 4\right)
 + a_{1}B_2\left(\frac{3}{2} - 5\right)=$$$$
 a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot \frac{1}{2} + a_{2} \cdot \frac{1}{2} + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = \frac{a_1 + a_2}{2}$$ 
 
-$$y_3 = s\left(2\right) = a_{-1}B_2\left(2 + 1\right) + a_{0}B_2\left(2\right) + a_{1}B_2\left(2 - 1\right) + $$$$
-+ a_{2}B_2\left(2 - 2\right)
-+ a_{-1}B_2\left(2 - 3\right)
-+ a_{0}B_2\left(2 - 4\right)
-+ a_{1}B_2\left(2 - 5\right)=$$$$
-a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot \left(\frac{1}{8}\right) + a_{2} \cdot \frac{3}{4} + a_{-1} \cdot \left(\frac{1}{8}\right) + a_{0} \cdot 0 + a_{1} \cdot 0 = 
-\frac{1}{8} a_{1} + \frac{3}{4} a_{2} + \frac{1}{8} a_{-1}$$
+$$y_2 = s\left(\frac{5}{2}\right) = a_{-1}B_2\left(\frac{5}{2} + 1\right) + a_{0}B_2\left(\frac{5}{2}\right) + a_{1}B_2\left(\frac{5}{2} - 1\right) + $$$$
++ a_{2}B_2\left(\frac{5}{2} - 2\right)
++ a_{-1}B_2\left(\frac{5}{2} - 3\right)
++ a_{0}B_2\left(\frac{5}{2} - 4\right)
++ a_{1}B_2\left(\frac{5}{2} - 5\right)=$$$$
+a_{-1} \cdot 0 + a_{0} \cdot 0 +  a_{1} \cdot 0 + a_{2} \cdot \frac{1}{2} + a_{-1} \cdot \frac{1}{2} + a_{0} \cdot 0 + a_{1} \cdot 0 = \frac{a_2 + a_{-1}}{2}$$ 
+
+$$y_1 = s\left(\frac{7}{2}\right) = a_{-1}B_2\left(\frac{7}{2} + 1\right) + a_{0}B_2\left(\frac{7}{2}\right) + a_{1}B_2\left(\frac{7}{2} - 1\right) + $$$$
++ a_{2}B_2\left(\frac{7}{2} - 2\right)
++ a_{-1}B_2\left(\frac{7}{2} - 3\right)
++ a_{0}B_2\left(\frac{7}{2} - 4\right)
++ a_{1}B_2\left(\frac{7}{2} - 5\right)=$$$$
+a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 + a_{2} \cdot 0 + a_{-1} \cdot \frac{1}{2} + a_{0} \cdot \frac{1}{2} + a_{1} \cdot 0  = \frac{a_{-1} + a_0}{2}$$ 
 
 The linear system in matrix form is:
 
-\[\frac{1}{8} \cdot
+\[\frac{1}{2} \cdot
 \begin{bmatrix}
-4&4&0&0\\
-1&6&1&0\\
-0&4&4&0\\
-0&1&6&1\\
+0&1&1&0\\
+0&0&1&1\\
+1&0&0&1\\
+1&1&0&0\\
 \end{bmatrix}
 \begin{bmatrix}
-a_0\\a_1\\a_2\\a_{-1}\\
+a_{-1}\\a_0\\a_1\\a_2\\
 \end{bmatrix}
 =
 \begin{bmatrix}
@@ -155,9 +153,21 @@ $$y_0 = s(0) = a_{-1}B_2\left(1\right) + a_{0}B_2\left(0\right) + a_{1}B_2\left(
 a_{-1} \cdot \frac{1}{8} + a_{0} \cdot \frac{3}{4} + a_{1} \cdot \frac{1}{8} + a_{2} \cdot 0 + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = 
 \frac{1}{8} a_{-1} + \frac{3}{4} a_{0} + \frac{1}{8} a_{1}$$
 
-$$y_1 = s\left(1\right) = \frac{1}{8} a_{0} + \frac{3}{4} a_{1} + \frac{1}{8} a_{2}$$ 
+$$y_1 = s\left(1\right) = a_{-1}B_2\left(1 + 1\right) + a_{0}B_2\left(1\right) + a_{1}B_2\left(1 - 1\right) + $$$$
++ a_{2}B_2\left(1 - 2\right)
++ a_{-1}B_2\left(1 - 3\right)
++ a_{0}B_2\left(1 - 4\right)
++ a_{1}B_2\left(1 - 5\right)=$$$$
+a_{-1} \cdot 0 + a_{0} \cdot \left(\frac{1}{8}\right) + a_{1} \cdot \frac{3}{4} + a_{2} \cdot \left(\frac{1}{8}\right) + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = 
+\frac{1}{8} a_{0} + \frac{3}{4} a_{1} + \frac{1}{8} a_{2}$$ 
 
-$$y_2 = s\left(2\right) = \frac{1}{8} a_{1} + \frac{3}{4} a_{2} + \frac{1}{8} a_{-1}$$
+$$y_2 = s\left(2\right) = a_{-1}B_2\left(2 + 1\right) + a_{0}B_2\left(2\right) + a_{1}B_2\left(2 - 1\right) + $$$$
++ a_{2}B_2\left(2 - 2\right)
++ a_{-1}B_2\left(2 - 3\right)
++ a_{0}B_2\left(2 - 4\right)
++ a_{1}B_2\left(2 - 5\right)=$$$$
+a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot \left(\frac{1}{8}\right) + a_{2} \cdot \frac{3}{4} + a_{-1} \cdot \left(\frac{1}{8}\right) + a_{0} \cdot 0 + a_{1} \cdot 0 = 
+\frac{1}{8} a_{1} + \frac{3}{4} a_{2} + \frac{1}{8} a_{-1}$$
 
 $$y_3 = s\left(3\right) = a_{-1}B_2\left(3 + 1\right) + a_{0}B_2\left(3\right) + a_{1}B_2\left(3 - 1\right) + $$$$
 + a_{2}B_2\left(3 - 2\right)
@@ -184,7 +194,7 @@ a_0\\a_1\\a_2\\a_{-1}\\
 y_0\\y_1\\y_2\\y_{3}\\
 \end{bmatrix}\]
 
-\subsection*{Question 5}
+\section*{Question 5}
 
 $$1 = y_0 = s(x_0) = 0 = a_{-1}B_3(1) +a_0B_3(0)
 +a_1B_3(-1)