55 lines
2.4 KiB
TeX
55 lines
2.4 KiB
TeX
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% 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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$$\frac{\max_{x \in [-1,1]} |f^{(n+1)}|}{5!} =
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\frac{\max_{x \in [-1,1]}|\frac{7680}{|2x+3|^6}|}{120} =
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\max_{x \in [-1,1]}\frac{64}{|2x+3|^6} = 64$$
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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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$$\max_{x \in [-1,1]} |f(x) - p(x)| \leq \frac{\max_{x \in [-1,1]} |f^{(n+1)}|}{5!} \max_{x \in [-1,1]} \left|(x - 1)\left(x - \frac{1}{2}\right)
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\left(x + \frac{1}{2}\right)(x+1)\right| =$$$$= 64 \cdot \frac{1}{4} = 8 \leq 8$$
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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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\end{document}
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