Assignment 1 except 1.2d, 1.4, 1.5

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+\documentclass[10pt,a4paper]{article}
+
+\usepackage[margin=2cm]{geometry}
+
+\usepackage{enumitem}
+\setenumerate[1]{label=1.\arabic*.}
+\setenumerate[2]{label=\textit{\alph*})}
+
+% textcomp package is not available everywhere, and we only need the Copyright symbol
+% taken from http://tex.stackexchange.com/a/1677/23992
+\DeclareTextCommandDefault{\textregistered}{\textcircled{\check@mathfonts\fontsize\sf@size\z@\math@fontsfalse\selectfont R}}
+
+\usepackage{fancyhdr}
+\renewcommand{\headrulewidth}{0pt}
+\renewcommand{\footrulewidth}{0pt}
+\fancyhead{}
+\fancyfoot[C]{Copyright {\textcopyright} 2015 Camil Staps}
+\pagestyle{fancy}
+
+\usepackage{amsmath}
+\usepackage{amsthm}
+\usepackage{amsfonts}
+\DeclareMathAlphabet{\pazocal}{OMS}{zplm}{m}{n}
+
+\parindent0pt
+
+\title{Algoritmen en Datastructuren - assignment 1}
+\author{Camil Staps\\\small{s4498062}}
+
+\begin{document}
+
+\maketitle
+\thispagestyle{fancy}
+
+\begin{enumerate}
+    \item Essentially we need to equate these expressions to $60\cdot10^6$ and solve for $n$.
+
+        \begin{enumerate}
+            \item $n^2 = 60 \cdot 10^6$; $n = \sqrt{60 \cdot 10^6} \approx 7745.97$.
+            \item $n\log n = 60 \cdot 10^6$; $n \approx 8.65 \cdot 10^6$.\footnote{Assuming the base 10 logarithm.}
+            \item $2^n = 60 \cdot 10^6$; $n = \log_2(60\cdot10^6) \approx 25.84$.
+            \item $n\sqrt n = 60 \cdot 10^6$; $n = \left(60\cdot10^6\right)^{\frac23} \approx 153261.89$.
+            \item $n^{100} = 60 \cdot 10^6$; $n = \sqrt[100]{60\cdot10^6} \approx 1.20$.
+            \item $4^n = 60 \cdot 10^6$; $n = \log_4(60\cdot10^6) \approx 12.92$.
+            \item $n = 60 \cdot 10^6$.
+        \end{enumerate}
+
+    \item \begin{enumerate}
+            \item Yes.
+            \item Yes.
+            \item No.
+            \item %Todo
+            \item Yes.
+            \item Yes.
+            \item No.
+            \item Yes.
+            \item No.
+            \item Yes.
+            \item No.
+        \end{enumerate}
+
+    \item \begin{enumerate}
+            \item \begin{enumerate}[label=\textit{\alph*})]
+                    \item Take $c=10,n_0=1$. For $n\geq n_0 = 1$ we know $9n\geq9>1$, therefore $cn = 10n > n + 1$ for all $n\geq n_0$ and thus $n+1 \in \pazocal{O}(n)$.
+                    \item Take $c=3,n_0=0$. Then $3n>2n$ for all $n\geq n_0$, therefore $2n \in \pazocal{O}(n)$.
+                \end{enumerate}
+            \item We take $c=1,n_0=1$. Claim: $\forall_{n\geq n_0=1}\left[n < c\cdot2^n = 2^n\right]$.
+                \begin{proof}
+                    We prove this with induction over $n$. 
+
+                    \textbf{Base case}: Let $n=1$, then $n = 1 < 2 = 2^1 = 2^n$.
+
+                    \textbf{Inductive step}: assume that the claim holds for a certain $n=k\geq n_0=1$ (inductive hypothesis). Now, we know $k+1 - k = 1 < 2^k = 2^{k+1} - 2^k$. Therefore, also $k+1 < 2^{k+1}$, i.e. the claim holds for $k+1$ as well.
+
+                    From the principle of induction it now follows that $\forall_{n\geq n_0=1}\left[n < c\cdot2^n = 2^n\right]$.
+                \end{proof}
+        \end{enumerate}
+
+    \item %Todo
+
+    \item %Todo
+\end{enumerate}
+
+\end{document}
+