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+% 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}
+\usepackage[utf8]{inputenc}
+\usepackage[margin=2cm]{geometry}
+\usepackage{amstext}
+\usepackage{array}
+
+\newcommand{\lra}{\Leftrightarrow}
+\newcolumntype{L}{>{$}l<{$}}
+
+\title{Midterm -- Introduction to Computational Science}
+\author{Claudio Maggioni}
+
+\begin{document}
+\maketitle
+\section*{Question 1}
+\subsection*{Point a)}
+$$7.125_{10} = (1 + 2^{-1} + 2^{-2} + 2^{-5}) * {2^2}_{10} = 0 | 1100 1000 0000 | 110_F$$
+$$0.8_{10} = (1 + 2^{-1} + 2^{-4} + 2^{-5}+ 2^{-8} + 2^{-9} + 2^{-12}) * {2^{-1}}_{10} \approx 0 | 100110011001|011_F$$
+$$0.046875_{10} = (2^{-2} + 2^{-3}) * {2^{-3}}_{10} = 0 | 0011 0000 0000|000_F$$
+
+For the last conversion, we assume that the denormalized mode of this floating point representation implicitly includes an exponent of $-3$ (this makes the first bit in the mantissa of a denormalized number weigh $2^{-3}$). This behaviour is akin to the denormalized implementation of IEEE 754 floating point numbers.
+
+\subsection*{Point b)}
+$$1|011010111000|110_F = -(1+2^{-2} +2^{-3} +2^{-5} +2^{-7} +2^{-8} + 2^{-9}) * {2^{2}}_{10} \approx
+-5.6796875$$
+$$1|101010101010|010_F = -(1+2^{-1} +2^{-3} +2^{-5} +2^{-7} +2^{-9} + 2^{-11}) * {2^{-2}}_{10} \approx
+-0.4166259766$$
+
+\subsection*{Point c)}
+$$1|0000 0000 0000|001_F = 2^{-3}_{10} = 0.125_{10}$$
+
+\subsection*{Point d)}
+$$1|1111 1111 1111|111_F = (1 + 2^{-1}+ 2^{-2} + 2^{-3} +2^{-4} + 2^{-5} +2^{-6} + 2^{-7} + 2^{-8} + 2^{-9}+ 2^{-10}+ 2^{-11}+ 2^{-12}) * {2^3}_{10} = $$
+$$ = 15.998046875_{10}$$
+
+\subsection*{Point e)}
+With 12 independent binary choices (bits to flip), there are $2^{12}$ different denormalized numbers in this encoding.
+
+\subsection*{Point f)}
+With 12 independent binary choices (bits to flip) and 3 extra bits for the exponent, there are $2^{15}-1$ different denormalized numbers in this encoding. We subtract 1 in order to be consistent with the assumption in point \textit{a)}, since $0.125_{10}$ would be representable both as $1|0000 0000 0000|001_F$ and
+as $1|1000 0000 0000|000_F$
+
+\section*{Question 2}
+
+\subsection*{Point a)}
+$$ \sqrt[3]{1 + x} - 1 =  (\sqrt[3]{1 + x} - 1) \cdot
+ \frac{ \sqrt[3]{1 + x} + 1}{ \sqrt[3]{1 + x} + 1} = \frac{\sqrt[3]{(1 + x)^2} - 1}{\sqrt[3]{1 + x} + 1} =
+\frac{\sqrt[3]{(1 + x)^2} - 1}{\sqrt[3]{1 + x} + 1} \cdot \frac{\sqrt[3]{(1 + x)^2} + 1}{\sqrt[3]{(1 + x)^2} + 1}  = $$
+$$\frac{(1 + x)\sqrt[3]{1 + x} - 1}{(\sqrt[3]{1 + x} + 1) \cdot (\sqrt[3]{(1 + x)^2} + 1)} $$
+\subsection*{Point b)}
+$$ \frac{1 - cos(x)}{sin(x)} = \frac{sin^2(x)cos^2(x) - cos(x)}{sin(x)} \cdot \frac{sin(x)}{cos(x)} \cdot \frac{cos(x)}{sin(x)} = (sin^2(x)cos(x) - 1)\cdot\frac{cos(x)}{sin(x)}$$ 
+\subsection*{Point c)}
+$$ \frac{1}{1-\sqrt{x^2-1}} = \frac{x}{x^2-\sqrt{x^4-x^2}}$$
+\subsection*{Point d)}
+$$ x^3\cdot\left(\frac{x}{x^2-1}-\frac{1}{x}\right) = x^3\cdot\left(\frac{x^2-x^2+1}{x^3-x}\right) =
+\frac{x^2}{x^2-1}$$
+\subsection*{Point e)}
+$$ \frac{1}{x} - \frac{1}{x+1} = \frac{x + 1 - x}{x^2 + x} = \frac{1}{x^2 + x}$$
+
+\section*{Question 3}
+\subsection*{Point a)}
+First we point out that:
+$$ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \stackrel{k := -h}{=} \lim_{k \to 0} \frac{f(x - (-k)) - f(x)}{-k} = $$
+$$ =  \lim_{k \to 0} \frac{f(x) - f(x + k)}{k} = - \lim_{h \to 0} \frac{f(x) - f(x - h)}{h} $$ 
+ 
+Then, we find an equivalent way to represent $f'(x)$:
+$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = 2 \cdot \lim_{h \to 0} \frac{f(x + h) - f(x)}{2h} =$$
+$$ = \lim_{h \to 0} \frac{f(x + h) - f(x)}{2h} + \left(- \lim_{h \to 0} \frac{f(x) - f(x - h)}{2h}\right) =  \lim_{h \to 0} \frac{f(x + h) - f(x - h)}{2h} $$
+
+Then we consider the \textit{epsilon-delta} definition of limits for the last limit:
+$$ \forall \epsilon > 0 \exists \delta > 0 | \forall h > 0, \text{if } 0 < h < \delta \Rightarrow$$
+$$\left| \frac{f(x + h) - f(x - h)}{2h} - f'(x)\right| = \left|f'(x) - \frac{f(x + h) - f(x - h)}{2h}\right| < \epsilon$$
+
+I GIVE UP :(
+
+\section*{Question 4}
+\subsection*{Point a)}
+\[
+A_1 = 
+ \begin{bmatrix}
+   1 & 1  & 1  & 1  & 1 \\
+   2 & 4  & 4  & 4  & 4 \\
+   3 & 7  & 10 & 10 & 10 \\
+   4 & 10 & 16 & 20 & 20 \\
+   5 & 13 & 22 & 30 & 35 \\
+ \end{bmatrix},
+ b = 
+ \begin{bmatrix}
+ 1 \\ 2 \\ 3 \\ 4 \\ 5 \\
+ \end{bmatrix}
+\]
+\[
+l_1 = 
+  \begin{bmatrix}
+ 1 \\ 2 \\ 3 \\ 4 \\ 5 \\
+ \end{bmatrix},
+u_1 = 
+  \begin{bmatrix}
+ 1 & 1 & 1 & 1 & 1 \\
+ \end{bmatrix},
+ A_2
+ \begin{bmatrix}
+   0 & 0  & 0  & 0  & 0 \\
+   0 & 2  & 2  & 2  & 2 \\
+   0 & 4  & 7  & 7  & 7 \\
+   0 & 6  & 12 & 16 & 16 \\
+   0 & 8  & 17 & 25 & 30 \\
+ \end{bmatrix}
+\]
+\[
+l_2 = 
+  \begin{bmatrix}
+ 0 \\ 1 \\ 2 \\ 3 \\ 4 \\
+ \end{bmatrix},
+u_2 = 
+  \begin{bmatrix}
+ 0 & 2 & 2 & 2 & 2 \\
+ \end{bmatrix},
+ A_3
+ \begin{bmatrix}
+   0 & 0  & 0  & 0  & 0 \\
+   0 & 0  & 0  & 0  & 0 \\
+   0 & 0  & 3  & 3  & 3 \\
+   0 & 0  & 6 & 10 & 10 \\
+   0 & 0  & 9 & 17 & 22 \\
+ \end{bmatrix}
+\]
+\[
+l_3 = 
+  \begin{bmatrix}
+ 0 \\ 0 \\ 1 \\ 2 \\ 3 \\
+ \end{bmatrix},
+u_3 = 
+  \begin{bmatrix}
+ 0 & 0 & 3 & 3 & 3 \\
+ \end{bmatrix},
+ A_4
+ \begin{bmatrix}
+   0 & 0  & 0  & 0  & 0 \\
+   0 & 0  & 0  & 0  & 0 \\
+   0 & 0  & 0  & 0  & 0 \\
+   0 & 0  & 0 & 4 & 4 \\
+   0 & 0  & 0 & 8 & 13 \\
+ \end{bmatrix}
+\]
+\[
+l_4 = 
+  \begin{bmatrix}
+ 0 \\ 0 \\ 0 \\ 1 \\ 2 \\
+ \end{bmatrix},
+u_4 = 
+  \begin{bmatrix}
+ 0 & 0 & 0 & 4 & 4 \\
+ \end{bmatrix},
+ A_5
+ \begin{bmatrix}
+   0 & 0  & 0  & 0  & 0 \\
+   0 & 0  & 0  & 0  & 0 \\
+   0 & 0  & 0  & 0  & 0 \\
+   0 & 0  & 0 & 0 & 0 \\
+   0 & 0  & 0 & 0 & 5 \\
+ \end{bmatrix}
+\]
+\[
+\l_5 = 
+  \begin{bmatrix}
+ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\
+ \end{bmatrix},
+u_5 = 
+  \begin{bmatrix}
+ 0 & 0 & 0 & 0 & 5 \\
+ \end{bmatrix},
+ L = 
+ \begin{bmatrix}
+   1 & 0 & 0 & 0 & 0 \\
+   2 & 1 & 0 & 0 & 0 \\
+   3 & 2 & 1 & 0 & 0 \\
+   4 & 3 & 2 & 1 & 0 \\
+   5 & 4 & 3 & 2 & 1 \\
+ \end{bmatrix},
+  U = 
+ \begin{bmatrix}
+   1 & 1 & 1 & 1 & 1 \\
+   0 & 2 & 2 & 2 & 2 \\
+   0 & 0 & 3 & 3 & 3 \\
+   0 & 0 & 0 & 4 & 4 \\
+   0 & 0 & 0 & 0 & 5 \\
+ \end{bmatrix} 
+ \]
+\subsection*{Point b)}
+\[
+\l_1 = 
+  \begin{bmatrix}
+ 1 \\ 2 \\ 3 \\ 4 \\ 5 \\
+ \end{bmatrix},
+e^{T}_1 = 
+  \begin{bmatrix}
+ 1 & 0 & 0 & 0 & 0 \\
+ \end{bmatrix},
+ L_1 = 
+ \begin{bmatrix}
+   0 & 0 & 0 & 0 & 0 \\
+   -2 & 1 & 0 & 0 & 0 \\
+   -3 & 0 & 1 & 0 & 0 \\
+   -4 & 0 & 0 & 1 & 0 \\
+   -5 & 0 & 0 & 0 & 1 \\
+ \end{bmatrix}
+ \]
+ \[
+\l_2 = 
+  \begin{bmatrix}
+ 0 \\ 1 \\ 2 \\ 3 \\ 4 \\
+ \end{bmatrix},
+e^{T}_2 = 
+  \begin{bmatrix}
+ 0 & 1 & 0 & 0 & 0 \\
+ \end{bmatrix},
+ L_2 = 
+ \begin{bmatrix}
+   1 & 0 & 0 & 0 & 0 \\
+   0 & 0 & 0 & 0 & 0 \\
+   0 & -2 & 1 & 0 & 0 \\
+   0 & -3 & 0 & 1 & 0 \\
+   0 & -4 & 0 & 0 & 1 \\
+ \end{bmatrix}
+ \]
+  \[
+\l_3 = 
+  \begin{bmatrix}
+ 0 \\ 0 \\ 1 \\ 2 \\ 3 \\
+ \end{bmatrix},
+e^{T}_3 = 
+  \begin{bmatrix}
+ 0 & 0 & 1 & 0 & 0 \\
+ \end{bmatrix},
+ L_3 = 
+ \begin{bmatrix}
+   1 & 0 & 0 & 0 & 0 \\
+   0 & 1 & 0 & 0 & 0 \\
+   0 & 0 & 0 & 0 & 0 \\
+   0 & 0 & -2 & 1 & 0 \\
+   0 & 0 & -3 & 0 & 1 \\
+ \end{bmatrix}
+ \]
+   \[
+\l_4 = 
+  \begin{bmatrix}
+ 0 \\ 0 \\ 0 \\ 1 \\ 2 \\
+ \end{bmatrix},
+e^{T}_4 = 
+  \begin{bmatrix}
+ 0 & 0 & 0 & 1 & 0 \\
+ \end{bmatrix},
+ L_4 = 
+ \begin{bmatrix}
+   1 & 0 & 0 & 0 & 0 \\
+   0 & 1 & 0 & 0 & 0 \\
+   0 & 0 & 1 & 0 & 0 \\
+   0 & 0 & 0 & 0 & 0 \\
+   0 & 0 & 0 & -2 & 1 \\
+ \end{bmatrix}
+\] 
+\subsection{Point c)}
+ \end{document}