ICS/hw5/hw5.tex

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\title{Howework 5 -- Introduction to Computational Science}

\author{Claudio Maggioni}

\begin{document} \maketitle
\section*{Question 1}
Given the definition of degree of exactness being the highest polynomial degree
$n$ at which a quadrature, for every polynomial of degree $n$, produces exactly
the same polynomial, these are the proofs.

\subsection*{Midpoint rule}
All polynomials of degree 1 can be expressed as:

\[p_1(x) = a_1 \cdot x + a_0\]

Therefore their integral is:

\[\int_0^1 a_1 \cdot x + a_0 dx = \frac{a_1}{2} + a_0\]

The midpoint rule for $p_1(x)$ is

\[f\left(\frac{1}{2}\right) \cdot 1 = \frac{a_1}{2} + a_0 = \int_0^1 a_1 \cdot x
+ a_0 dx\]

Therefore the midpoint rule has a degree of exactness of at least 1.

It is easy to show that the degree of exactness is not higher than 1 by
considering the degree 2
polynomial $x^2$, which has an integral in $[0, 1]$ of $\frac{1}{3}$ but a midpoint rule
quadrature of $\frac{1}{4}$.

\subsection*{Trapezoidal rule}
The proof is similar to the one for the midpoint rule, but with this quadrature
for degree 1 polynomials:

\[\frac{f(0)}{2} + \frac{f(1)}{2} = \frac{a_0 + a_1 + a_0}{2} = \frac{a_1}{2} +
a_0\]

Which is again equal to the general integral for these polynomials.

Again $x^2$ is a degree 2 polynomial with integral $\frac{1}{3}$ but a midpoint quadrature
of $\frac{0 + 1}{2} = \frac{1}{2}$, thus bounding the degree of exactness to 1.

\subsection*{Simpson rule}
The proof is again similar, but for degree 3 polynomials which can all be
written as:

\[p_3(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0\]

The integral is:

\[\int_0^1p_3(x) dx = \frac{a_3}{4} + \frac{a_2}{3} + \frac{a_1}{2} + a_0\]

The Simpson rule gives:

\[\frac{1}{6} \cdot f(0) + \frac{4}{6}\cdot  f\left(\frac{1}{2}\right) + \frac{1}{6} \cdot f(1) =
\frac{1}{6} a_0 + \frac{4}{6} \left(\frac{a_3}{8} + \frac{a_2}{4} +
\frac{a_1}{2} + a_0\right) + \]\[\frac{1}{6} \left(a_3 + a_2 + a_1 + a_0\right) =
\frac{a_3}{4} + \frac{a_2}{3} + \frac{a_1}{2} + a_0 = \int_0^1 p_3(x)dx\]

Which tells us that the degree of exactness is at least 1.

We can bound the degree of exactness to 3 with the 4th degree polynomial $x^4$
which has integral in $[0, 1]$ of $\frac{1}{5}$ but has a quadrature of
$\frac{1}{6} \cdot 0 + \frac{2}{3} \cdot \frac{1}{16} + \frac{1}{6} \cdot 1 =
\frac{5}{24}$.

\section*{Question 2}
The algebraic solution is:

\[\int_0^1 1 - 4(x - 0.5)^2 dx = 1 - 4 \cdot \int_0^1 x^2 - x + \frac14 =
1 - 4 \left(\frac{4-6+3}{12}\right) = 1 - \frac13 = \frac23\]

The solution using quadrature is:

\[Q = \frac{1}{2} (f(0) + f(1)) = \frac{1}{2}(0 + 0) = 0 \qquad (x, h) =
\left(\frac{1}{2}, \frac{1}{2}\right) \qquad \epsilon = \frac{1}{10}\]
\[E\left(\frac{1}{2}, \frac12\right) = f\left(\frac12\right) - \frac12\left(f(0)
+ f(1)\right) = 1 > \frac1{10} \qquad Q = 0 + \frac12 \cdot 1 = \frac12\]

\[E\left(\frac{1}{4}, \frac14\right) = f\left(\frac14\right) - \frac12\left(f(0)
+ f\left(\frac12\right)\right) = \frac34 - \frac12 = \frac14 > \frac1{10} \qquad
Q = \frac12 + \frac14 \cdot \frac14 = \frac9{16}\]

\[E\left(\frac{3}{4}, \frac14\right) = f\left(\frac34\right) -
\frac12\left(f\left(\frac12\right)
+ f(1)\right) = \frac14 > \frac1{10} \qquad
Q = \frac9{16} + \frac14 \cdot \frac14 = \frac{10}{16}\]

\[E\left(\frac18, \frac18\right) = f\left(\frac18\right) -
\frac12\left(f\left(0\right)
+ f\left(\frac14\right)\right) = \frac7{16} - \frac12 \cdot \frac34 = \frac1{16} < \frac1{10}\]
\[E\left(\frac38, \frac18\right) = f\left(\frac38\right) -
\frac12\left(f\left(\frac14\right)
+ f\left(\frac12\right)\right) = \frac{15}{16} - \frac12 \cdot \frac74 = \frac1{16} < \frac1{10}\]
\[E\left(\frac58, \frac18\right) = f\left(\frac58\right) -
\frac12\left(f\left(\frac12\right)
+ f\left(\frac34\right)\right) = \frac{15}{16} - \frac12 \cdot \frac74 = \frac1{16} < \frac1{10}\]
\[E\left(\frac78, \frac18\right) = f\left(\frac78\right) -
\frac12\left(f\left(\frac34\right)
+ f\left(1\right)\right) = \frac7{16} - \frac12 \cdot \frac34 = \frac1{16} < \frac1{10}\]

Thus the solution using quadrature is $\frac{5}{8}$.

\section*{Question 4}

\[\begin{bmatrix} x_0^2 & x_0 & 1 \\x_1^2 & x_1 & 1 \\x_2^2 & x_2 & 1 \\x_3^2 &
  x_3 & 1 \end{bmatrix} \begin{bmatrix}a \\b\\c\\\end{bmatrix} = \begin{bmatrix}
y_0 \\y_1\\y_2\\y_3\\\end{bmatrix}\]

\[\begin{bmatrix} x_0^2 & x_1^2 & x_2^2 & x_3^2 \\ x_0 & x_1 & x_2 & x_3 \\
1&1&1&1\end{bmatrix} \begin{bmatrix} x_0^2 & x_0 & 1 \\x_1^2 & x_1 & 1 \\x_2^2 & x_2 & 1 \\x_3^2 &
  x_3 & 1 \end{bmatrix} \begin{bmatrix}a \\b\\c\\\end{bmatrix} = \begin{bmatrix} x_0^2 & x_1^2 & x_2^2 & x_3^2 \\ x_0 & x_1 & x_2 & x_3 \\
  1&1&1&1\end{bmatrix} \begin{bmatrix} y_0 \\y_1\\y_2\\y_3\\\end{bmatrix}\]

\[\begin{bmatrix}18&8&6\\8&6&2\\6&2&4\\\end{bmatrix}\begin{bmatrix}a\\b\\c\\\end{bmatrix}=\begin{bmatrix}2\\0\\2\\\end{bmatrix}\]

We now use Gaussian \textit{ellimination} to solve the system:

\[
  \begin{array}{@{}ccc|c@{}}
  18&8&6&2\\
  8&6&2&0\\
  6&2&4&2\\
  \end{array}
  \qquad
  \begin{array}{@{}ccc|c@{}}
  1&\frac49	&\frac13 &	\frac19\\
  8&	6&	2	&0\\
  6	&2	&4	&2\\
  \end{array}
  \qquad
  \begin{array}{@{}ccc|c@{}}
    1&	\frac49&	\frac13&	\frac19\\
    0&	\frac{22}9&	\frac{-2}3&	\frac{-8}9\\
    0&	\frac{-2}3&	2&	\frac43\\
  \end{array}
  \qquad
  \begin{array}{@{}ccc|c@{}}
    1&	\frac49&	\frac13&	\frac19\\
    0	&1	&\frac{-3}{11} & \frac{-4}{11}\\
    0	& \frac{-2}{3} & 2	& \frac43\\
  \end{array}
\]\[
  \begin{array}{@{}ccc|c@{}}
    1&	0&	\frac5{11}	&\frac3{11}\\
    0	&1	&\frac{-3}{11}&\frac{-4}{11}\\
    0	&0	&\frac{20}{11}&\frac{12}{11}\\
 \end{array}
  \qquad
  \begin{array}{@{}ccc|c@{}}
    1&	0	&\frac5{11}	& \frac{3}{11}\\
    0	&1	& \frac{-3}{11} &	\frac{-4}{11}\\
    0&	0&	1	&\frac35\\
 \end{array}
  \qquad
  \begin{array}{@{}ccc|c@{}}
    1	&0	&0	&0\\
    0&	1	&0	&\frac{-1}5\\
    0&	0&	1	&\frac35\\
 \end{array}
  \qquad
  \begin{bmatrix}a\\b\\c\\\end{bmatrix}=\begin{bmatrix}0\\-\frac15\\\frac35\\\end{bmatrix}
\]
\end{document}