hw2 done up to ex4

hw2/hw2.tex

@@ -1,66 +1,90 @@
+% vim: set ts=2 sw=2 et tw=80:
+
 \documentclass[12pt,a4paper]{article}
 
-\usepackage[utf8]{inputenc}
+\usepackage[utf8]{inputenc} \usepackage[margin=2cm]{geometry}
-\usepackage[margin=2cm]{geometry}
+\usepackage{amstext} \usepackage{amsmath} \usepackage{array}
-\usepackage{amstext}
+\newcommand{\lra}{\Leftrightarrow}
-\usepackage{array}
 
 \title{Howework 2 -- Introduction to Computational Science}
+
 \author{Claudio Maggioni}
 
-\begin{document}
+\begin{document} \maketitle \section*{Question 1} The solutions assume that the
-\maketitle
+sign bit $1$ is negative and $0$ is positive.
-\section*{Question 1}
-The solutions assume that the sign bit $1$ is negative and $0$ is positive.
 
-\subsection*{Point a}
+\subsection*{Point a} \begin{itemize} \item 13 is equal to $0 1010 0000 000
-\begin{itemize}
+1011$; \item 42.125 is equal to $0 0101 0001 000 1101$; \item 0.8 is equal to $0
-\item 13 is equal to $0 1010 0000 000 1011$;
+1 1001 1001 100 0011$. 0.78 is approximated to 0.7998046875; \end{itemize}
-\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}
+\subsection*{Point b} $1 0110 1011 100 1101$ is $(-1) *
-$1 0110 1011 100 1101$ is $(-1) * (0.25+0.125+0.03125+0.0078125+0.00320625+0.001953125) *2^5$,
+(0.25+0.125+0.03125+0.0078125+0.00320625+0.001953125) *2^5$, which is equal to
-which is equal to $-13.4151$.
+$-13.4151$.
+
+\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}
-$x_{max}$ 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}
+\[ C= \begin{bmatrix} 2 & 3 \\ 1.5 & -1 \\ \end{bmatrix} \begin{bmatrix} 2 & 1.5
-\subsection*{Point a}
+  \\ 3 & -1 \\ \end{bmatrix} = \begin{bmatrix} 4 + 9 & 3 - 3 \\ 3 - 3 & 2.25 + 1
-$$\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} +
+    \\ \end{bmatrix} = \begin{bmatrix} 13 & 0 \\ 0 & 3.25 \\ \end{bmatrix} \]
-\frac{y}{x + y}\frac{\Delta y}{y}$$
 
-\subsection*{Point b}
+\[ det(C - \lambda i) = det\left( \begin{bmatrix} 13 - \lambda & 0 \\ 0 & 3.25 -
-$$\frac{(x + \Delta x) - (y + \Delta y) - (x - y)}{x - y} = \frac{\Delta x}{x - y} - \frac{\Delta y}{x - y} = \frac{x}{x - y}\frac{\Delta x}{x} -
+\lambda\\ \end{bmatrix} \right) = (42.25 - 13 \lambda -3.25 \lambda + \lambda^2)
-\frac{y}{x - y}\frac{\Delta y}{y}$$
+\\ = \lambda^2 - 16.25 \lambda + 42.25 \]
 
-\subsection*{Point c}
+$$det(C - \lambda i) = 0 \lra = \lambda^2 - 16.25 \lambda + 42.25 = 0 \lra
-$$\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
+\lambda = 3.25 \lor \lambda = 13$$
-\frac{y \Delta x + x \Delta y}{x y} = \frac{\Delta x}{y} + \frac{\Delta y}{x}$$
 
-\subsection*{Point d}
+$$||A||_2 = \sqrt{\lambda_{max}(AA^T)} = \sqrt{13} \approx 3.606$$
-$$\frac{((x + \Delta x) / (y + \Delta y)) - (x / y)}{x / y}
-= \frac{\frac{(x + \Delta x) y}{(y + \Delta y) y} - \frac{(y + \Delta y)x}{(y + \Delta y)y}}{x / y}
-= \frac{y \Delta x  - x \Delta y}{x (y + \Delta y)}
-= \frac{y \Delta x}{x(y + \Delta y)} - \frac{\Delta y}{y + \Delta y} = $$
-$$ = \left(\frac{x(y + \Delta y)}{y \Delta x}\right)^{-1} - \left(\frac{y + \Delta y}{\Delta y}\right)^{-1}
-= \left(\frac{x}{\Delta x} - \frac{y \Delta x}{x \Delta y}\right)^{-1} - \left(\frac{y}{\Delta y} + 1\right)^{-1} = $$
-$$ = \left(\frac{x}{\Delta x} \left(1 - \frac{\Delta y}{y}\right)\right)^{-1} - \left(\frac{y}{\Delta y} + 1\right)^{-1}
-\approx \left(\frac{x}{\Delta x}\right)^{-1} - \left(\frac{y}{\Delta y}\right)^{-1}
-= \frac{\Delta x}{x} - \frac{\Delta y}{y}
-$$
-
-\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.
 \end{document}