75 lines
2.5 KiB
TeX
75 lines
2.5 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 5 -- 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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Given the definition of degree of exactness being the highest polynomial degree
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$n$ at which a quadrature, for every polynomial of degree $n$, produces exactly
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the same polynomial, these are the proofs.
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\subsection*{Midpoint rule}
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All polynomials of degree 1 can be expressed as:
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\[p_1(x) = a_1 \cdot x + a_0\]
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Therefore their integral is:
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\[\int_0^1 a_1 \cdot x + a_0 dx = \frac{a_1}{2} + a_0\]
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The midpoint rule for $p_1(x)$ is
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\[f\left(\frac{1}{2}\right) \cdot 1 = \frac{a_1}{2} + a_0 = \int_0^1 a_1 \cdot x
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+ a_0 dx\]
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Therefore the midpoint rule has a degree of exactness of at least 1.
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It is easy to show that the degree of exactness is not higher than 1 by
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considering the degree 2
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polynomial $x^2$, which has an integral in $[0, 1]$ of $\frac{1}{3}$ but a midpoint rule
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quadrature of $\frac{1}{4}$.
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\subsection*{Trapezoidal rule}
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The proof is similar to the one for the midpoint rule, but with this quadrature
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for degree 1 polynomials:
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\[\frac{f(0)}{2} + \frac{f(1)}{2} = \frac{a_0 + a_1 + a_0}{2} = \frac{a_1}{2} +
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a_0\]
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Which is again equal to the general integral for these polynomials.
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Again $x^2$ is a degree 2 polynomial with integral $\frac{1}{3}$ but a midpoint quadrature
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of $\frac{0 + 1}{2} = \frac{1}{2}$, thus bounding the degree of exactness to 1.
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\subsection*{Simpson rule}
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The proof is again similar, but for degree 3 polynomials which can all be
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written as:
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\[p_3(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0\]
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The integral is:
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\[\int_0^1p_3(x) dx = \frac{a_3}{4} + \frac{a_2}{3} + \frac{a_1}{2} + a_0\]
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The Simpson rule gives:
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\[\frac{1}{6} \cdot f(0) + \frac{4}{6}\cdot f\left(\frac{1}{2}\right) + \frac{1}{6} \cdot f(1) =
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\frac{1}{6} a_0 + \frac{4}{6} \left(\frac{a_3}{8} + \frac{a_2}{4} +
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\frac{a_1}{2} + a_0\right) + \]\[\frac{1}{6} \left(a_3 + a_2 + a_1 + a_0\right) =
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\frac{a_3}{4} + \frac{a_2}{3} + \frac{a_1}{2} + a_0 = \int_0^1 p_3(x)dx\]
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Which tells us that the degree of exactness is at least 1.
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We can bound the degree of exactness to 3 with the 4th degree polynomial $x^4$
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which has integral in $[0, 1]$ of $\frac{1}{5}$ but has a quadrature of
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$\frac{1}{6} \cdot 0 + \frac{2}{3} \cdot \frac{1}{16} + \frac{1}{6} \cdot 1 =
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\frac{5}{24}$.
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\end{document}
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