\documentclass[11pt]{article} \usepackage{amsthm,amsmath,amsfonts,amssymb} \usepackage[ruled,noline,noend]{algorithm2e} \usepackage[margin=1in]{geometry} \usepackage{enumerate} %%% Update the following commands with your name, ID, and acknowledgments. \newcommand{\YourName}{Tareef Dedhar} \newcommand{\YourStudentID}{20621325} \newcommand{\YourAck}{This question was completed only with the teachings from CS courses here at uWaterloo.} \newenvironment{solution}{\paragraph{Solution.}}{} \title{ \vspace{-0.8in} \normalsize \begin{flushright} \YourName \\ \YourStudentID \end{flushright} \rule{\textwidth}{.5pt}\\[0.4cm] \huge Assignment 1: Question 1 \\ \rule{\textwidth}{2pt} \vspace{-1in} } \date{} \begin{document} \maketitle \paragraph{Acknowledgments.} \YourAck %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section*{Asymptotics} Prove or disprove each of the following statements. \begin{enumerate}[(a)] \item For any constant $b > 0$, the function $f : n \mapsto 1 + b + b^2 + b^3 + \cdots + b^n$ satisfies \[ f(n) = \begin{cases} \Theta(b^n) & \mbox{if } b > 1 \\ \Theta(1) & \mbox{if } b \le 1. \end{cases} \] \begin{solution} Since f(n) is a geometric series, we can express it as $\sum_{i=1}^{n} {b^i}$ = $\frac{1 - {b^N}}{1 - b}$ = $\frac{{b^n} - 1}{b - 1}$ Using this:\\ When b=1: $\lim{n\to\infty}{\dfrac{\tfrac{{b^n} - 1}{b - 1}}{b^n}}$=$\frac{{b^n} - 1}{{b^{n+1}} - b}$, which, by :'Hopital's rule = $1 \div b$. Since $0 \lt 1 \div b \lt \infty$ , we have a \theta bound. \end{solution} \item For every pair of functions $f, g : \mathbb{Z}^+ \to \mathbb{R}^{\ge 1}$ that satisfy $f = \Theta(g)$, the functions $F : n \mapsto 2^{f(n)}$ and $G : n \mapsto 2^{g(n)}$ also satisfy $F = \Theta(G)$. \begin{solution} (ENTER YOUR SOLUTION HERE.) \end{solution} \item For every pair of functions $f, g : \mathbb{Z}^+ \to \mathbb{R}^{\ge 1}$ that satisfy $f = o(g)$, the functions $F : n \mapsto 2^{f(n)}$ and $G : n \mapsto 2^{g(n)}$ also satisfy $F = o(G)$. \begin{solution} (ENTER YOUR SOLUTION HERE.) \end{solution} \end{enumerate} \end{document}