309 lines
9.6 KiB
TeX
309 lines
9.6 KiB
TeX
% vim: set ts=2 sw=2 et tw=80:
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\documentclass[12pt,a4paper]{article}
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\usepackage[utf8]{inputenc} \usepackage[margin=2cm]{geometry}
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\usepackage{amstext} \usepackage{amsmath} \usepackage{array}
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\newcommand{\lra}{\Leftrightarrow}
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\title{Howework 4 -- Introduction to Computational Science}
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\author{Claudio Maggioni}
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\begin{document} \maketitle
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\section*{Question 1}
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$$L_0(x) = \prod_{j = 0, j \neq 0}^n \frac{x - x_j}{x_i - x_j} =
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\frac{x - (-0.5)}{(-1) - (-0.5)} \cdot \frac{x - 0.5}{(-1) - 0.5} \cdot
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\frac{x - 1}{(-1) - 1} = -\frac{2}{3}x^3 + \frac{2}{3}x^2
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+\frac{1}{6} x - \frac{1}{6}$$
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$$L_1(x) = \prod_{j = 0, j \neq 1}^n \frac{x - x_j}{x_i - x_j} =
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\frac{x - (-1)}{(-0.5) - (-1)} \cdot \frac{x - 0.5}{(-0.5) - 0.5} \cdot
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\frac{x - 1}{(-0.5) - 1} = \frac{4}{3}x^3 - \frac{2}{3}x^2 - \frac{4}{3}x + \frac{2}{3}$$
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$$L_2(x) = \prod_{j = 0, j \neq 2}^n \frac{x - x_j}{x_i - x_j} =
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\frac{x - (-1)}{0.5 - (-1)} \cdot \frac{x - (-0.5)}{0.5 - (-0.5)} \cdot
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\frac{x - 1}{0.5 - 1} = -\frac{4}{3}x^3 - \frac{2}{3}x^2 + \frac{4}{3}x + \frac{2}{3}$$
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$$L_3(x) = \prod_{j = 0, j \neq 3}^n \frac{x - x_j}{x_i - x_j} =
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\frac{x - (-1)}{1 - (-1)} \cdot \frac{x - (-0.5)}{1 - (-0.5)} \cdot
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\frac{x - 0.5}{1 - 0.5} = \frac{2}{3}x^3 + \frac{2}{3}x^2 -\frac{1}{6}x - \frac{1}{6}$$
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$$p(x) = \sum_{i=0}^n y_i L_i(x) =
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2 \cdot \left(-\frac{2}{3}x^3 + \frac{2}{3}x^2
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+\frac{1}{6} x - \frac{1}{6}\right) +
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1 \cdot \left(\frac{4}{3}x^3 - \frac{2}{3}x^2 - \frac{4}{3}x + \frac{2}{3}\right) +
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$$$$
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0.5 \cdot \left(-\frac{4}{3}x^3 - \frac{2}{3}x^2 + \frac{4}{3}x + \frac{2}{3}\right) +
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0.4 \cdot \left(\frac{2}{3}x^3 + \frac{2}{3}x^2 -\frac{1}{6}x - \frac{1}{6}\right) = -\frac{2}{5}x^3 + \frac{3}{5}x^2 - \frac{2}{5}x + \frac{3}{5}$$
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$$\frac{\max_{x \in [-1,1]} |f^{(n+1)}|}{4!} =
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\frac{\max_{x \in [-1,1]}|\frac{768}{|2x+3|^6}|}{24} =
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\max_{x \in [-1,1]}\frac{32}{|2x+3|^6} = 32$$
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$$\max_{x \in [-1,1]} \left|(x - 1)\left(x - \frac{1}{2}\right)
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\left(x + \frac{1}{2}\right)(x+1)\right| =
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\max_{x \in [-1,1]} \left| x^4 - \frac{5}{4}x^2 + \frac{1}{4}\right| = \frac{1}{4}$$
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$$\max_{x \in [-1,1]} |f(x) - p(x)| \leq \frac{\max_{x \in [-1,1]} |f^{(n+1)}|}{4!} \max_{x \in [-1,1]} \left|(x - 1)\left(x - \frac{1}{2}\right)
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\left(x + \frac{1}{2}\right)(x+1)\right| =$$$$= 32 \cdot \frac{1}{4} = 8 \leq 8$$
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The statement above is true so p satisfies the error estimate:
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$$\max_{x \in [-1,1]} |f(x) - p(x)| \leq 8$$
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\section*{Question 2}
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We first use the Lagrange method:
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$$L_1(x) = \prod_{j = 0, j \neq 1}^2 \frac{x - x_j}{x_i - x_j} =
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\frac{x-0}{1-0} \frac{x-3}{1-3} = -\frac{1}{2}x^2 + \frac{3}{2}x$$
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$$L_2(x) = \prod_{j = 0, j \neq 2}^2 \frac{x - x_j}{x_i - x_j} =
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\frac{x-0}{3-0} \frac{x-1}{3-1} = \frac{1}{6}x^2 - \frac{1}{6}x $$
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$$p(x) = (-3) \cdot \left(-\frac{1}{2}x^2 + \frac{3}{2}x\right) +
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1 \cdot \left(\frac{1}{6}x^2 - \frac{1}{6}x\right) =
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\frac{5}{3}x^2 - \frac{14}{3}x$$
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Then we use the Newtonian method:
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$$a_0 = f[0] = 0, \hspace{2cm} f[1] = -3 \hspace{2cm} f[3] = 1$$
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$$a_1 = f[0,1] = \frac{-3-0}{1-0} = -3, \hspace{2cm} f[1,3] = \frac{1-(-3)}{3-1} = 2 $$
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$$a_2 = f[0,1,3] = \frac{2-(-3)}{3-0} = \frac{5}{3}$$
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$$p(x) = \left(\frac{5}{3}(x-1)-3\right) x + 0 = \frac{5}{3}x^2 - \frac{14}{3}x$$
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The interpolating polynomials are indeed equal.
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Now we use the Horner method to compute $p(0.5)$:
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$$y=a_n = a_2 = \frac{5}{3}$$
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$$i = 1$$
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$$y = y (x - x_1) + a_1 = \frac{5}{3} (0.5 - 1) + a_1 - 3 = -\frac{5}{6} - 3 = -\frac{23}{6}$$
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$$i = 0$$
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$$y = y (x - x_0) + a_0 = -\frac{23}{6} \cdot 0.5 + 0 = -\frac{23}{12}$$
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\section*{Question 4}
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\subsection*{Point a)}
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The node coordinates to which fix the quadratic spline are $(1/2, y_0), (1, y_1),(3/2, y_2),(2, y_3)$.
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Then, we can start formulating the equations for the linear system:
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$$y_0 = s\left(\frac{1}{2}\right) = a_{-1}B_2\left(\frac{1}{2} + 1\right) + a_{0}B_2\left(\frac{1}{2}\right) + a_{1}B_2\left(\frac{1}{2} - 1\right) + $$$$
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+ a_{2}B_2\left(\frac{1}{2} - 2\right)
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+ a_{-1}B_2\left(\frac{1}{2} - 3\right)
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+ a_{0}B_2\left(\frac{1}{2} - 4\right)
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+ a_{1}B_2\left(\frac{1}{2} - 5\right)=$$$$
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a_{-1} \cdot 0 + a_{0} \cdot \frac{1}{2} + a_{1} \cdot \frac{1}{2} + a_{2} \cdot 0 + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = \frac{a_0 + a_1}{2}$$
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$$y_1 = s\left(1\right) = a_{-1}B_2\left(1 + 1\right) + a_{0}B_2\left(1\right) + a_{1}B_2\left(1 - 1\right) + $$$$
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+ a_{2}B_2\left(1 - 2\right)
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+ a_{-1}B_2\left(1 - 3\right)
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+ a_{0}B_2\left(1 - 4\right)
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+ a_{1}B_2\left(1 - 5\right)=$$$$
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a_{-1} \cdot 0 + a_{0} \cdot \left(\frac{1}{8}\right) + a_{1} \cdot \frac{3}{4} + a_{2} \cdot \left(\frac{1}{8}\right) + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 =
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\frac{1}{8} a_{0} + \frac{3}{4} a_{1} + \frac{1}{8} a_{2}$$
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$$y_2 = s\left(\frac{3}{2}\right) = a_{-1}B_2\left(\frac{3}{2} + 1\right) + a_{0}B_2\left(\frac{3}{2}\right) + a_{1}B_2\left(\frac{3}{2} - 1\right) + $$$$
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+ a_{2}B_2\left(\frac{3}{2} - 2\right)
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+ a_{-1}B_2\left(\frac{3}{2} - 3\right)
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+ a_{0}B_2\left(\frac{3}{2} - 4\right)
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+ a_{1}B_2\left(\frac{3}{2} - 5\right)=$$$$
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a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot \frac{1}{2} + a_{2} \cdot \frac{1}{2} + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = \frac{a_1 + a_2}{2}$$
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$$y_3 = s\left(2\right) = a_{-1}B_2\left(2 + 1\right) + a_{0}B_2\left(2\right) + a_{1}B_2\left(2 - 1\right) + $$$$
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+ a_{2}B_2\left(2 - 2\right)
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+ a_{-1}B_2\left(2 - 3\right)
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+ a_{0}B_2\left(2 - 4\right)
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+ a_{1}B_2\left(2 - 5\right)=$$$$
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a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot \left(\frac{1}{8}\right) + a_{2} \cdot \frac{3}{4} + a_{-1} \cdot \left(\frac{1}{8}\right) + a_{0} \cdot 0 + a_{1} \cdot 0 =
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\frac{1}{8} a_{1} + \frac{3}{4} a_{2} + \frac{1}{8} a_{-1}$$
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The linear system in matrix form is:
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\[\frac{1}{8} \cdot
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\begin{bmatrix}
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4&4&0&0\\
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1&6&1&0\\
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0&4&4&0\\
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0&1&6&1\\
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\end{bmatrix}
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\begin{bmatrix}
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a_0\\a_1\\a_2\\a_{-1}\\
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\end{bmatrix}
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=
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\begin{bmatrix}
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y_0\\y_1\\y_2\\y_{3}\\
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\end{bmatrix}\]
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The determinant of that matrix is 0 so the matrix is singular.
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\subsection*{Point b)}
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The node coordinates to which fix the quadratic spline are $(0, y_0), (1, y_1),(2, y_2),(3, y_3)$.
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Then, we can start formulating the equations for the linear system:
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$$y_0 = s(0) = a_{-1}B_2\left(1\right) + a_{0}B_2\left(0\right) + a_{1}B_2\left(- 1\right) + $$$$
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+ a_{2}B_2\left(- 2\right)
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+ a_{-1}B_2\left(- 3\right)
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+ a_{0}B_2\left(- 4\right)
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+ a_{1}B_2\left(- 5\right)=$$$$
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a_{-1} \cdot \frac{1}{8} + a_{0} \cdot \frac{3}{4} + a_{1} \cdot \frac{1}{8} + a_{2} \cdot 0 + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 =
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\frac{1}{8} a_{-1} + \frac{3}{4} a_{0} + \frac{1}{8} a_{1}$$
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$$y_1 = s\left(1\right) = \frac{1}{8} a_{0} + \frac{3}{4} a_{1} + \frac{1}{8} a_{2}$$
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$$y_2 = s\left(2\right) = \frac{1}{8} a_{1} + \frac{3}{4} a_{2} + \frac{1}{8} a_{-1}$$
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$$y_3 = s\left(3\right) = a_{-1}B_2\left(3 + 1\right) + a_{0}B_2\left(3\right) + a_{1}B_2\left(3 - 1\right) + $$$$
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+ a_{2}B_2\left(3 - 2\right)
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+ a_{-1}B_2\left(3 - 3\right)
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+ a_{0}B_2\left(3 - 4\right)
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+ a_{1}B_2\left(3 - 5\right)=$$$$
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a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 + a_{2} \cdot \frac{1}{8} + a_{-1} \cdot \frac{3}{4} + a_{0} \cdot \frac{1}{8} + a_{1} \cdot 0 =
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\frac{1}{8} a_{2} + \frac{3}{4} a_{-1} + \frac{1}{8} a_{0}$$
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The linear system in matrix form is:
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\[\frac{1}{8} \cdot
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\begin{bmatrix}
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6&1&0&1\\
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1&6&1&0\\
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0&1&6&1\\
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1&0&1&6\\
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\end{bmatrix}
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\begin{bmatrix}
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a_0\\a_1\\a_2\\a_{-1}\\
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\end{bmatrix}
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=
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\begin{bmatrix}
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y_0\\y_1\\y_2\\y_{3}\\
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\end{bmatrix}\]
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\subsection*{Question 5}
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$$1 = y_0 = s(x_0) = 0 = a_{-1}B_3(1) +a_0B_3(0)
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+a_1B_3(-1)
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+a_2B_3(-2)
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+a_3B_3(-3) = \frac{1}{6}a_{-1} + \frac{2}{3}a_0 + \frac{1}{6}a_1$$
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$$5 = y_1 = s(x_1) = 1 = a_{-1}B_3(2) +a_0B_3(1)
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+a_1B_3(0)
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+a_2B_3(-1)
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+a_3B_3(-2) = \frac{1}{6}a_{0} + \frac{2}{3}a_1 + \frac{1}{6}a_2$$
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$$1 = y_2 = s(x_2) = 2 = a_{-1}B_3(3) +a_0B_3(2)
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+a_1B_3(1)
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+a_2B_3(0)
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+a_3B_3(-1) = \frac{1}{6}a_{1} + \frac{2}{3}a_2 + \frac{1}{6}a_3$$
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$$s''(0) = a_{-1}B''_3(1) +a_0B''_3(0)
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+a_1B''_3(-1)
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+a_2B''_3(-2)
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+a_3B''_3(-3) = a_{-1} - 2 a_0 + a_{1}$$
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$$s''(2) = a_{-1}B''_3(3) +a_0B''_3(2)
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+a_1B''_3(1)
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+a_2B''_3(-0)
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+a_3B''_3(-1) = a_{1} - 2 a_2 + a_{3}$$
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\[\frac{1}{6} \cdot
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\begin{bmatrix}
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1&-2&1&0&0\\
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1&4&1&0&0\\
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0&1&4&1&0\\
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0&0&1&4&1\\
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0&0&1&-2&1\\
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\end{bmatrix}
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\begin{bmatrix}
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a_{-1}\\a_0\\a_1\\a_2\\a_3\\
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\end{bmatrix}
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=
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\begin{bmatrix}
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0\\y_0\\y_1\\y_2\\0\\
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\end{bmatrix}\]
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We then use Gaussian \textit{ellimination} to solve the system.
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\[
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\begin{array}{@{}ccccc|c@{}}
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1&-2&1&0&0&0\\
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1&4&1&0&0&6\\
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0&1&4&1&0&30\\
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0&0&1&4&1&6\\
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0&0&1&-2&1&0\\
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\end{array}
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\qquad
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\begin{array}{@{}ccccc|c@{}}
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1&-2&1&0&0&0\\
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0&6&0&0&0&6\\
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0&1&4&1&0&30\\
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0&0&1&4&1&6\\
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0&0&1&-2&1&0\\
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\end{array}
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\qquad
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\begin{array}{@{}ccccc|c@{}}
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1&-2&1&0&0&0\\
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0&1&4&1&0&30\\
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0&6&0&0&0&6\\
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0&0&1&4&1&6\\
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0&0&1&-2&1&0\\
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\end{array}
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\]
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\[
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\begin{array}{@{}ccccc|c@{}}
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1&0&9&2&0&60\\
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0&1&4&1&0&30\\
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0&0&-24&-6&0&-174\\
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0&0&1&4&1&6\\
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0&0&1&-2&1&0\\
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\end{array}
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\qquad
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\begin{array}{@{}ccccc|c@{}}
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1&0&9&2&0&60\\
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0&1&4&1&0&30\\
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0&0&1&4&1&6\\
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0&0&-24&-6&0&-174\\
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0&0&1&-2&1&0\\
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\end{array}
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\qquad
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\begin{array}{@{}ccccc|c@{}}
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1&0&0&-34&-9&6\\
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0&1&0&-15&-4&6\\
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0&0&1&4&1&6\\
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0&0&0&90&24&-30\\
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0&0&0&6&0&-6\\
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\end{array}
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\]
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\[
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\begin{array}{@{}ccccc|c@{}}
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1&0&0&-34&-9&6\\
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0&1&0&-15&-4&6\\
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0&0&1&4&1&6\\
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0&0&0&1&4/15&-1/3\\
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0&0&0&6&0&-6\\
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\end{array}
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\qquad
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\begin{array}{@{}ccccc|c@{}}
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1 & 0 & 0 & 0 & 1/15 & -16/3 \\
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0 & 1 & 0 & 0 & 0 & 1 \\
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0 & 0 & 1 & 0 & -1/15 & 22/3 \\
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0 & 0 & 0 & 1 & 4/15 & -1/3 \\
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0 & 0 & 0 & 0 & 8/5 & -8 \\
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\end{array}
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\qquad
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\begin{array}{@{}ccccc|c@{}}
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1 & 0 & 0 & 0 & 0 & -5 \\
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0 & 1 & 0 & 0 & 0 & 1 \\
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0 & 0 & 1 & 0 & 0 & 7 \\
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0 & 0 & 0 & 1 & 0 & 1 \\
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0 & 0 & 0 & 0 & 1 & -5 \\
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\end{array}
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\]
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Therefore, the coefficients are $a_{-1} = a_3 = -5$, $a_0 = a_2 = 1$, and $a_1 = 7$.
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\end{document}
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