% vim: set ts=2 sw=2 et tw=80:

\documentclass[12pt,a4paper]{article}

\usepackage[utf8]{inputenc} \usepackage[margin=2cm]{geometry}
\usepackage{amstext} \usepackage{amsmath} \usepackage{array}
\newcommand{\lra}{\Leftrightarrow}

\title{Howework 2 -- Introduction to Computational Science}

\author{Claudio Maggioni}

\begin{document} \maketitle \section*{Question 1} The solutions assume that the
sign bit $1$ is negative and $0$ is positive.

\subsection*{Point a} \begin{itemize} \item 13 is equal to $0 1010 0000 000
1011$; \item 42.125 is equal to $0 0101 0001 000 1101$; \item 0.8 is equal to $0
1 1001 1001 100 0011$. 0.78 is approximated to 0.7998046875; \end{itemize}

\subsection*{Point b} $1 0110 1011 100 1101$ is $(-1) *
(0.25+0.125+0.03125+0.0078125+0.00320625+0.001953125) *2^5$, which is equal to
$-13.4151$.

\subsection*{Point c} $x_{max}$ is $0 1111 1111 111 1111$, equal to $255.9375$.
Since denormalized numbers do not belong to this representation (since the
exponent $0000$ cannot be used for valid numbers other than 0) $x_{min}$ is $0
0000 0000 000 0001$, equal to $0.0078125$.

\section*{Question 2} \subsection*{Point a} $$\frac{(x + \Delta x) + (y + \Delta
y) - (x + y)}{x + y} = \frac{\Delta x}{x + y} + \frac{\Delta y}{x + y} =
\frac{x}{x + y}\frac{\Delta x}{x} + \frac{y}{x + y}\frac{\Delta y}{y}$$

\subsection*{Point b} $$\frac{(x + \Delta x) - (y + \Delta y) - (x - y)}{x - y}
= \frac{\Delta x}{x - y} - \frac{\Delta y}{x - y} = \frac{x}{x - y}\frac{\Delta
x}{x} - \frac{y}{x - y}\frac{\Delta y}{y}$$

\subsection*{Point c} $$\frac{((x + \Delta x) (y + \Delta y)) - (x y)}{x y} =
\frac{y \Delta x + x \Delta y + \Delta x \Delta y}{x y} \approx \frac{y \Delta x
+ x \Delta y}{x y} = \frac{\Delta x}{y} + \frac{\Delta y}{x}$$

\subsection*{Point d} $$\frac{((x + \Delta x) / (y + \Delta y)) - (x / y)}{x /
y} = \frac{\frac{(x + \Delta x) y}{(y + \Delta y) y} - \frac{(y + \Delta y)x}{(y
+ \Delta y)y}}{x / y} = \frac{y \Delta x  - x \Delta y}{x (y + \Delta y)} =
\frac{y \Delta x}{x(y + \Delta y)} - \frac{\Delta y}{y + \Delta y} = $$ $$ =
\left(\frac{x(y + \Delta y)}{y \Delta x}\right)^{-1} - \left(\frac{y + \Delta
y}{\Delta y}\right)^{-1} = \left(\frac{x}{\Delta x} - \frac{y \Delta x}{x \Delta
y}\right)^{-1} - \left(\frac{y}{\Delta y} + 1\right)^{-1} = $$ $$ =
\left(\frac{x}{\Delta x} \left(1 - \frac{\Delta y}{y}\right)\right)^{-1} -
\left(\frac{y}{\Delta y} + 1\right)^{-1} \approx \left(\frac{x}{\Delta
x}\right)^{-1} - \left(\frac{y}{\Delta y}\right)^{-1} = \frac{\Delta x}{x} -
\frac{\Delta y}{y} $$

\subsection*{Point e} Division and multiplication may suffer from cancellation.

\section*{Exercise 3} \subsection*{Point d} The error at first keeps getting
exponentially smaller due to a better approximation of $h$ when computing the
derivative (i.e. $h$ is exponentially nearer to 0), but at $10^{-9}$ this trend
almost becomes the opposite due to loss of significant digits when subtracting
from $e^{x+h}$ $e^x$ and amplifiying this error by effectively multiplying that
with exponentially increasing powers of 10.

\section*{Exercise 4} \subsection*{Point a} $$||A||_\infty = \max_{1 \leq j \leq
2}{\sum_{j=1}^{2} |a_{i,j}|} = \max{\{5_{j=1}, 2.5_{j=2}\}} = 5$$ $$||A||_1 =
\max_{1 \leq i \leq 2}{\sum_{i=1}^{2} |a_{i,j}|} = \max{\{3.5_{i=1}, 4_{i=2}\}}
= 4$$ $$||A||_F = \left(\sum_{i=1}^{2}\sum_{j=1}^{2}
(a_{i,j})^2\right)^{\frac{1}{2}} = (4 + 9 + 2.25 + 1)^{\frac{1}{2}} \approx
4.031$$

\subsection*{Point b} $$||B||_\infty = \max_{1 \leq j \leq 3}{\sum_{j=1}^{3}
|b_{i,j}|} = \max{\{12_{j=1}, 6_{j=2}, 3_{j=3}\}} = 12$$ $$||B||_1 = \max_{1
\leq i \leq 3}{\sum_{i=1}^{3} |b_{i,j}|} = \max{\{9_{i=1}, 7_{i=2}, 5_{i=3}\}} =
9$$ $$||B||_F = \left(\sum_{i=1}^{3}\sum_{j=1}^{3}
(b_{i,j})^2\right)^{\frac{1}{2}} = (36 + 16 + 4 + 1 + 4 + 9 + 4 + 1 +
0)^{\frac{1}{2}} \approx 8.660$$

\subsection*{Point c}

\[ C= \begin{bmatrix} 2 & 3 \\ 1.5 & -1 \\ \end{bmatrix} \begin{bmatrix} 2 & 1.5
  \\ 3 & -1 \\ \end{bmatrix} = \begin{bmatrix} 4 + 9 & 3 - 3 \\ 3 - 3 & 2.25 + 1
    \\ \end{bmatrix} = \begin{bmatrix} 13 & 0 \\ 0 & 3.25 \\ \end{bmatrix} \]

\[ det(C - \lambda i) = det\left( \begin{bmatrix} 13 - \lambda & 0 \\ 0 & 3.25 -
\lambda\\ \end{bmatrix} \right) = (42.25 - 13 \lambda -3.25 \lambda + \lambda^2)
\\ = \lambda^2 - 16.25 \lambda + 42.25 \]

$$det(C - \lambda i) = 0 \lra = \lambda^2 - 16.25 \lambda + 42.25 = 0 \lra
\lambda = 3.25 \lor \lambda = 13$$

$$||A||_2 = \sqrt{\lambda_{max}(AA^T)} = \sqrt{13} \approx 3.606$$

\section*{Exercise 5}
\subsection*{Point a}
\[
   A^{-1} =
    \frac{1}{(-2) - 4.5}
   \begin{bmatrix}
      -1 & -3 \\
      -1.5 & 2 \\
   \end{bmatrix}
   =
   \begin{bmatrix}
      2/13 & 6/13 \\
      3/13 & -4/13 \\
   \end{bmatrix}
 \]
 
$$||A^{-1}||_\infty = \max_{1 \leq j \leq 2}{\sum_{j=1}^{2} |a_{i,j}|} =
 \max{\{\frac{8}{13}. \frac{7}{13}\}} = \frac{8}{13}$$

$$||A^{-1}||_1 = \max_{1 \leq i \leq 2}{\sum_{i=1}^{2} |a_{i,j}|} =
 \max{\{\frac{5}{13}, \frac{10}{13}\}} = \frac{10}{13}$$

$$||A^{-1}||_F = \left(\sum_{i=1}^{2}\sum_{j=1}^{2}
(a_{i,j})^2\right)^{\frac{1}{2}} = \left(\frac{65}{13^2}\right)^{\frac{1}{2}}
= \sqrt{\frac{5}{13}}$$

$$k_1(A) = \frac{4 * 10}{13} \approx 3.077$$

$$k_\infty(A) = \frac{5 * 8}{13} \approx 3.077$$

$$k_F(A) \approx 8.660 * \sqrt{\frac{5}{13}} \approx 5.371$$

\subsection*{Point b}
$$||A^{-1}||_2 = \sigma_{max}(A^{-1}) = \frac{1}{\sigma_{min}(A)} = \frac{1}{\sqrt{\lambda_{min}}} =\frac{1}{\sqrt{3.25}}$$

$$k_2(A) = \frac{\sqrt{13}}{\sqrt{3.25}} = 2$$

\subsection*{Point c}

\[
  k_\infty
  \begin{bmatrix}
  -4 & 0 & 0 \\
  0 & 5 & 0 \\
  0 & 0 & 3 \\
  \end{bmatrix}
  =
  \left|\left| \begin{bmatrix}
  -4 & 0 & 0 \\
  0 & 5 & 0 \\
  0 & 0 & 3 \\
  \end{bmatrix} \right|\right|_\infty 
  \left|\left| \begin{bmatrix}
  -1/4 & 0 & 0 \\
  0 & 1/5 & 0 \\
  0 & 0 & 1/3 \\
  \end{bmatrix} \right|\right|_\infty 
  = 5 * \frac{1}{3} = \frac{5}{3}
\]

\subsection*{Point d}
The condition of that matrix cannot be computed since that matrix has no inverse since it is not full rank ($M_{2,:} = M_{1,:} * -2$).
\end{document}