template and my start on the lab down
This commit is contained in:
Mars Ultor
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**/*.aux
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**/*.fdb_latexmk
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**/*.fls
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**/*.log
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**/*.synctex.gz
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\documentclass{article}
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\usepackage{fancyhdr}
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\usepackage{extramarks}
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\usepackage{amsmath}
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\usepackage{amsthm}
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\usepackage{amsfonts}
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\usepackage{tikz}
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\usepackage[plain]{algorithm}
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\usepackage{algpseudocode}
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\usetikzlibrary{automata,positioning}
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%
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% Basic Document Settings
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%
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\topmargin=-0.45in
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\evensidemargin=0in
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\oddsidemargin=0in
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\textwidth=6.5in
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\textheight=9.0in
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\headsep=0.25in
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\linespread{1.1}
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\pagestyle{fancy}
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\lhead{\hmwkAuthorName}
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\chead{\hmwkClass\ (\hmwkClassInstructor\ \hmwkClassTime): \hmwkTitle}
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\rhead{\firstxmark}
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\lfoot{\lastxmark}
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\cfoot{\thepage}
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\renewcommand\headrulewidth{0.4pt}
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\renewcommand\footrulewidth{0.4pt}
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\setlength\parindent{0pt}
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%
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% Create Problem Sections
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%
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\newcommand{\enterProblemHeader}[1]{
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\nobreak\extramarks{}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
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\nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
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}
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\newcommand{\exitProblemHeader}[1]{
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\nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
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\stepcounter{#1}
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\nobreak\extramarks{Problem \arabic{#1}}{}\nobreak{}
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}
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\setcounter{secnumdepth}{0}
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\newcounter{partCounter}
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\newcounter{homeworkProblemCounter}
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\setcounter{homeworkProblemCounter}{1}
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\nobreak\extramarks{Problem \arabic{homeworkProblemCounter}}{}\nobreak{}
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%
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% Homework Problem Environment
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%
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% This environment takes an optional argument. When given, it will adjust the
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% problem counter. This is useful for when the problems given for your
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% assignment aren't sequential. See the last 3 problems of this template for an
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% example.
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%
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\newenvironment{homeworkProblem}[1][-1]{
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\ifnum#1>0
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\setcounter{homeworkProblemCounter}{#1}
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\fi
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\section{Problem \arabic{homeworkProblemCounter}}
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\setcounter{partCounter}{1}
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\enterProblemHeader{homeworkProblemCounter}
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}{
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\exitProblemHeader{homeworkProblemCounter}
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}
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%
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% Homework Details
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% - Title
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% - Due date
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% - Class
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% - Section/Time
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% - Instructor
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% - Author
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%
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\newcommand{\hmwkTitle}{Homework\ \#2}
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\newcommand{\hmwkDueDate}{February 12, 2014}
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\newcommand{\hmwkClass}{Calculus}
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\newcommand{\hmwkClassTime}{Section A}
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\newcommand{\hmwkClassInstructor}{Professor Isaac Newton}
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\newcommand{\hmwkAuthorName}{\textbf{Josh Davis} \and \textbf{Davis Josh}}
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%
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% Title Page
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%
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\title{
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\vspace{2in}
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\textmd{\textbf{\hmwkClass:\ \hmwkTitle}}\\
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\normalsize\vspace{0.1in}\small{Due\ on\ \hmwkDueDate\ at 3:10pm}\\
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\vspace{0.1in}\large{\textit{\hmwkClassInstructor\ \hmwkClassTime}}
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\vspace{3in}
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}
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\author{\hmwkAuthorName}
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\date{}
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\renewcommand{\part}[1]{\textbf{\large Part \Alph{partCounter}}\stepcounter{partCounter}\\}
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%
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% Various Helper Commands
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%
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% Useful for algorithms
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\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
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% For derivatives
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\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
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% For partial derivatives
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\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
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% Integral dx
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\newcommand{\dx}{\mathrm{d}x}
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% Alias for the Solution section header
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\newcommand{\solution}{\textbf{\large Solution}}
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% Probability commands: Expectation, Variance, Covariance, Bias
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\newcommand{\E}{\mathrm{E}}
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\newcommand{\Var}{\mathrm{Var}}
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\newcommand{\Cov}{\mathrm{Cov}}
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\newcommand{\Bias}{\mathrm{Bias}}
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\begin{document}
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\maketitle
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\pagebreak
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\begin{homeworkProblem}
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Give an appropriate positive constant \(c\) such that \(f(n) \leq c \cdot
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g(n)\) for all \(n > 1\).
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\begin{enumerate}
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\item \(f(n) = n^2 + n + 1\), \(g(n) = 2n^3\)
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\item \(f(n) = n\sqrt{n} + n^2\), \(g(n) = n^2\)
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\item \(f(n) = n^2 - n + 1\), \(g(n) = n^2 / 2\)
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\end{enumerate}
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\textbf{Solution}
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We solve each solution algebraically to determine a possible constant
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\(c\).
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\\
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\textbf{Part One}
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\[
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\begin{split}
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n^2 + n + 1 &=
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\\
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&\leq n^2 + n^2 + n^2
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\\
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&= 3n^2
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\\
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&\leq c \cdot 2n^3
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\end{split}
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\]
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Thus a valid \(c\) could be when \(c = 2\).
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\\
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\textbf{Part Two}
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\[
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\begin{split}
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n^2 + n\sqrt{n} &=
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\\
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&= n^2 + n^{3/2}
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\\
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&\leq n^2 + n^{4/2}
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\\
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&= n^2 + n^2
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\\
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&= 2n^2
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\\
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&\leq c \cdot n^2
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\end{split}
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\]
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Thus a valid \(c\) is \(c = 2\).
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\\
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\textbf{Part Three}
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\[
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\begin{split}
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n^2 - n + 1 &=
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\\
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&\leq n^2
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\\
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&\leq c \cdot n^2/2
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\end{split}
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\]
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Thus a valid \(c\) is \(c = 2\).
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\end{homeworkProblem}
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\pagebreak
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\begin{homeworkProblem}
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Let \(\Sigma = \{0, 1\}\). Construct a DFA \(A\) that recognizes the
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language that consists of all binary numbers that can be divided by 5.
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\\
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Let the state \(q_k\) indicate the remainder of \(k\) divided by 5. For
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example, the remainder of 2 would correlate to state \(q_2\) because \(7
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\mod 5 = 2\).
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\begin{figure}[h]
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\centering
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\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto]
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\node[state, accepting, initial] (q_0) {$q_0$};
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\node[state] (q_1) [right=of q_0] {$q_1$};
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\node[state] (q_2) [right=of q_1] {$q_2$};
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\node[state] (q_3) [right=of q_2] {$q_3$};
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\node[state] (q_4) [right=of q_3] {$q_4$};
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\path[->]
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(q_0)
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edge [loop above] node {0} (q_0)
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edge node {1} (q_1)
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(q_1)
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edge node {0} (q_2)
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edge [bend right=-30] node {1} (q_3)
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(q_2)
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edge [bend left] node {1} (q_0)
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edge [bend right=-30] node {0} (q_4)
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(q_3)
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edge node {1} (q_2)
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edge [bend left] node {0} (q_1)
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(q_4)
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edge node {0} (q_3)
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edge [loop below] node {1} (q_4);
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\end{tikzpicture}
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\caption{DFA, \(A\), this is really beautiful, ya know?}
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\label{fig:multiple5}
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\end{figure}
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\textbf{Justification}
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\\
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Take a given binary number, \(x\). Since there are only two inputs to our
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state machine, \(x\) can either become \(x0\) or \(x1\). When a 0 comes
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into the state machine, it is the same as taking the binary number and
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multiplying it by two. When a 1 comes into the machine, it is the same as
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multipying by two and adding one.
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\\
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Using this knowledge, we can construct a transition table that tell us
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where to go:
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\begin{table}[ht]
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\centering
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\begin{tabular}{c || c | c | c | c | c}
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& \(x \mod 5 = 0\)
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& \(x \mod 5 = 1\)
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& \(x \mod 5 = 2\)
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& \(x \mod 5 = 3\)
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& \(x \mod 5 = 4\)
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\\
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\hline
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\(x0\) & 0 & 2 & 4 & 1 & 3 \\
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\(x1\) & 1 & 3 & 0 & 2 & 4 \\
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\end{tabular}
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\end{table}
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Therefore on state \(q_0\) or (\(x \mod 5 = 0\)), a transition line should
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go to state \(q_0\) for the input 0 and a line should go to state \(q_1\)
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for input 1. Continuing this gives us the Figure~\ref{fig:multiple5}.
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\end{homeworkProblem}
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\begin{homeworkProblem}
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Write part of \alg{Quick-Sort($list, start, end$)}
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\begin{algorithm}[]
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\begin{algorithmic}[1]
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\Function{Quick-Sort}{$list, start, end$}
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\If{$start \geq end$}
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\State{} \Return{}
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\EndIf{}
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\State{} $mid \gets \Call{Partition}{list, start, end}$
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\State{} \Call{Quick-Sort}{$list, start, mid - 1$}
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\State{} \Call{Quick-Sort}{$list, mid + 1, end$}
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\EndFunction{}
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\end{algorithmic}
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\caption{Start of QuickSort}
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\end{algorithm}
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\end{homeworkProblem}
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\pagebreak
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\begin{homeworkProblem}
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Suppose we would like to fit a straight line through the origin, i.e.,
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\(Y_i = \beta_1 x_i + e_i\) with \(i = 1, \ldots, n\), \(\E [e_i] = 0\),
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and \(\Var [e_i] = \sigma^2_e\) and \(\Cov[e_i, e_j] = 0, \forall i \neq
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j\).
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\\
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\part
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Find the least squares esimator for \(\hat{\beta_1}\) for the slope
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\(\beta_1\).
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\\
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\solution
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To find the least squares estimator, we should minimize our Residual Sum
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of Squares, RSS:
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\[
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\begin{split}
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RSS &= \sum_{i = 1}^{n} {(Y_i - \hat{Y_i})}^2
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\\
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&= \sum_{i = 1}^{n} {(Y_i - \hat{\beta_1} x_i)}^2
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\end{split}
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\]
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By taking the partial derivative in respect to \(\hat{\beta_1}\), we get:
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\[
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\pderiv{
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\hat{\beta_1}
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}{RSS}
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= -2 \sum_{i = 1}^{n} {x_i (Y_i - \hat{\beta_1} x_i)}
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= 0
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\]
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This gives us:
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\[
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\begin{split}
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\sum_{i = 1}^{n} {x_i (Y_i - \hat{\beta_1} x_i)}
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&= \sum_{i = 1}^{n} {x_i Y_i} - \sum_{i = 1}^{n} \hat{\beta_1} x_i^2
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\\
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&= \sum_{i = 1}^{n} {x_i Y_i} - \hat{\beta_1}\sum_{i = 1}^{n} x_i^2
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\end{split}
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\]
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Solving for \(\hat{\beta_1}\) gives the final estimator for \(\beta_1\):
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\[
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\begin{split}
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\hat{\beta_1}
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&= \frac{
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\sum {x_i Y_i}
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}{
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\sum x_i^2
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}
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\end{split}
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||||
\]
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\pagebreak
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\part
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Calculate the bias and the variance for the estimated slope
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\(\hat{\beta_1}\).
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\\
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\solution
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For the bias, we need to calculate the expected value
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\(\E[\hat{\beta_1}]\):
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\[
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\begin{split}
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\E[\hat{\beta_1}]
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&= \E \left[ \frac{
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||||
\sum {x_i Y_i}
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||||
}{
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||||
\sum x_i^2
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||||
}\right]
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||||
\\
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||||
&= \frac{
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||||
\sum {x_i \E[Y_i]}
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||||
}{
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\sum x_i^2
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||||
}
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||||
\\
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||||
&= \frac{
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||||
\sum {x_i (\beta_1 x_i)}
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||||
}{
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||||
\sum x_i^2
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||||
}
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||||
\\
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||||
&= \frac{
|
||||
\sum {x_i^2 \beta_1}
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||||
}{
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||||
\sum x_i^2
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||||
}
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||||
\\
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||||
&= \beta_1 \frac{
|
||||
\sum {x_i^2 \beta_1}
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||||
}{
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||||
\sum x_i^2
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||||
}
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||||
\\
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||||
&= \beta_1
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||||
\end{split}
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||||
\]
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||||
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||||
Thus since our estimator's expected value is \(\beta_1\), we can conclude
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||||
that the bias of our estimator is 0.
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||||
\\
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||||
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||||
For the variance:
|
||||
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||||
\[
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||||
\begin{split}
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||||
\Var[\hat{\beta_1}]
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||||
&= \Var \left[ \frac{
|
||||
\sum {x_i Y_i}
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||||
}{
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||||
\sum x_i^2
|
||||
}\right]
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||||
\\
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||||
&=
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||||
\frac{
|
||||
\sum {x_i^2}
|
||||
}{
|
||||
\sum x_i^2 \sum x_i^2
|
||||
} \Var[Y_i]
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||||
\\
|
||||
&=
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||||
\frac{
|
||||
\sum {x_i^2}
|
||||
}{
|
||||
\sum x_i^2 \sum x_i^2
|
||||
} \Var[Y_i]
|
||||
\\
|
||||
&=
|
||||
\frac{
|
||||
1
|
||||
}{
|
||||
\sum x_i^2
|
||||
} \Var[Y_i]
|
||||
\\
|
||||
&=
|
||||
\frac{
|
||||
1
|
||||
}{
|
||||
\sum x_i^2
|
||||
} \sigma^2
|
||||
\\
|
||||
&=
|
||||
\frac{
|
||||
\sigma^2
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||||
}{
|
||||
\sum x_i^2
|
||||
}
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
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||||
\begin{homeworkProblem}
|
||||
Prove a polynomial of degree \(k\), \(a_kn^k + a_{k - 1}n^{k - 1} + \hdots
|
||||
+ a_1n^1 + a_0n^0\) is a member of \(\Theta(n^k)\) where \(a_k \hdots a_0\)
|
||||
are nonnegative constants.
|
||||
|
||||
\begin{proof}
|
||||
To prove that \(a_kn^k + a_{k - 1}n^{k - 1} + \hdots + a_1n^1 +
|
||||
a_0n^0\), we must show the following:
|
||||
|
||||
\[
|
||||
\exists c_1 \exists c_2 \forall n \geq n_0,\ {c_1 \cdot g(n) \leq
|
||||
f(n) \leq c_2 \cdot g(n)}
|
||||
\]
|
||||
|
||||
For the first inequality, it is easy to see that it holds because no
|
||||
matter what the constants are, \(n^k \leq a_kn^k + a_{k - 1}n^{k - 1} +
|
||||
\hdots + a_1n^1 + a_0n^0\) even if \(c_1 = 1\) and \(n_0 = 1\). This
|
||||
is because \(n^k \leq c_1 \cdot a_kn^k\) for any nonnegative constant,
|
||||
\(c_1\) and \(a_k\).
|
||||
\\
|
||||
|
||||
Taking the second inequality, we prove it in the following way.
|
||||
By summation, \(\sum\limits_{i=0}^k a_i\) will give us a new constant,
|
||||
\(A\). By taking this value of \(A\), we can then do the following:
|
||||
|
||||
\[
|
||||
\begin{split}
|
||||
a_kn^k + a_{k - 1}n^{k - 1} + \hdots + a_1n^1 + a_0n^0 &=
|
||||
\\
|
||||
&\leq (a_k + a_{k - 1} \hdots a_1 + a_0) \cdot n^k
|
||||
\\
|
||||
&= A \cdot n^k
|
||||
\\
|
||||
&\leq c_2 \cdot n^k
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
where \(n_0 = 1\) and \(c_2 = A\). \(c_2\) is just a constant. Thus the
|
||||
proof is complete.
|
||||
\end{proof}
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
%
|
||||
% Non sequential homework problems
|
||||
%
|
||||
|
||||
% Jump to problem 18
|
||||
\begin{homeworkProblem}[18]
|
||||
Evaluate \(\sum_{k=1}^{5} k^2\) and \(\sum_{k=1}^{5} (k - 1)^2\).
|
||||
\end{homeworkProblem}
|
||||
|
||||
% Continue counting to 19
|
||||
\begin{homeworkProblem}
|
||||
Find the derivative of \(f(x) = x^4 + 3x^2 - 2\)
|
||||
\end{homeworkProblem}
|
||||
|
||||
% Go back to where we left off
|
||||
\begin{homeworkProblem}[6]
|
||||
Evaluate the integrals
|
||||
\(\int_0^1 (1 - x^2) \dx\)
|
||||
and
|
||||
\(\int_1^{\infty} \frac{1}{x^2} \dx\).
|
||||
\end{homeworkProblem}
|
||||
|
||||
\end{document}
|
||||
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|
||||
\documentclass{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{comment}
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ (\hmwkClassInstructor\ \hmwkClassTime): \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Problem \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Problem \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
%
|
||||
% Homework Problem Environment
|
||||
%
|
||||
% This environment takes an optional argument. When given, it will adjust the
|
||||
% problem counter. This is useful for when the problems given for your
|
||||
% assignment aren't sequential. See the last 3 problems of this template for an
|
||||
% example.
|
||||
%
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Problem \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
%
|
||||
% Homework Details
|
||||
% - Title
|
||||
% - Due date
|
||||
% - Class
|
||||
% - Section/Time
|
||||
% - Instructor
|
||||
% - Author
|
||||
%
|
||||
|
||||
\newcommand{\hmwkTitle}{Homework\ \#1}
|
||||
\newcommand{\hmwkDueDate}{September 21, 2025}
|
||||
\newcommand{\hmwkClass}{Calc III}
|
||||
\newcommand{\hmwkClassTime}{$6^{\text{th}}$ Hour}}
|
||||
\newcommand{\hmwkClassInstructor}{Professor Foresee}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Krishna A.}}
|
||||
|
||||
%
|
||||
% Title Page
|
||||
%
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass:\ \hmwkTitle}}\\
|
||||
\normalsize\vspace{0.1in}\small{Due\ on\ \hmwkDueDate\ at 3:10pm}\\
|
||||
\vspace{0.1in}\large{\textit{\hmwkClassInstructor\ \hmwkClassTime}}
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName}
|
||||
\date{}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \Alph{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large Solution}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{comment}
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Give an appropriate positive constant \(c\) such that \(f(n) \leq c \cdot
|
||||
g(n)\) for all \(n > 1\).
|
||||
|
||||
\begin{enumerate}
|
||||
\item \(f(n) = n^2 + n + 1\), \(g(n) = 2n^3\)
|
||||
\item \(f(n) = n\sqrt{n} + n^2\), \(g(n) = n^2\)
|
||||
\item \(f(n) = n^2 - n + 1\), \(g(n) = n^2 / 2\)
|
||||
\end{enumerate}
|
||||
|
||||
\textbf{Solution}
|
||||
|
||||
We solve each solution algebraically to determine a possible constant
|
||||
\(c\).
|
||||
\\
|
||||
|
||||
\textbf{Part One}
|
||||
|
||||
\[
|
||||
\begin{split}
|
||||
n^2 + n + 1 &=
|
||||
\\
|
||||
&\leq n^2 + n^2 + n^2
|
||||
\\
|
||||
&= 3n^2
|
||||
\\
|
||||
&\leq c \cdot 2n^3
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
Thus a valid \(c\) could be when \(c = 2\).
|
||||
\\
|
||||
|
||||
\textbf{Part Two}
|
||||
|
||||
\[
|
||||
\begin{split}
|
||||
n^2 + n\sqrt{n} &=
|
||||
\\
|
||||
&= n^2 + n^{3/2}
|
||||
\\
|
||||
&\leq n^2 + n^{4/2}
|
||||
\\
|
||||
&= n^2 + n^2
|
||||
\\
|
||||
&= 2n^2
|
||||
\\
|
||||
&\leq c \cdot n^2
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
Thus a valid \(c\) is \(c = 2\).
|
||||
\\
|
||||
|
||||
\textbf{Part Three}
|
||||
|
||||
\[
|
||||
\begin{split}
|
||||
n^2 - n + 1 &=
|
||||
\\
|
||||
&\leq n^2
|
||||
\\
|
||||
&\leq c \cdot n^2/2
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
Thus a valid \(c\) is \(c = 2\).
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Let \(\Sigma = \{0, 1\}\). Construct a DFA \(A\) that recognizes the
|
||||
language that consists of all binary numbers that can be divided by 5.
|
||||
\\
|
||||
|
||||
Let the state \(q_k\) indicate the remainder of \(k\) divided by 5. For
|
||||
example, the remainder of 2 would correlate to state \(q_2\) because \(7
|
||||
\mod 5 = 2\).
|
||||
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto]
|
||||
\node[state, accepting, initial] (q_0) {$q_0$};
|
||||
\node[state] (q_1) [right=of q_0] {$q_1$};
|
||||
\node[state] (q_2) [right=of q_1] {$q_2$};
|
||||
\node[state] (q_3) [right=of q_2] {$q_3$};
|
||||
\node[state] (q_4) [right=of q_3] {$q_4$};
|
||||
\path[->]
|
||||
(q_0)
|
||||
edge [loop above] node {0} (q_0)
|
||||
edge node {1} (q_1)
|
||||
(q_1)
|
||||
edge node {0} (q_2)
|
||||
edge [bend right=-30] node {1} (q_3)
|
||||
(q_2)
|
||||
edge [bend left] node {1} (q_0)
|
||||
edge [bend right=-30] node {0} (q_4)
|
||||
(q_3)
|
||||
edge node {1} (q_2)
|
||||
edge [bend left] node {0} (q_1)
|
||||
(q_4)
|
||||
edge node {0} (q_3)
|
||||
edge [loop below] node {1} (q_4);
|
||||
\end{tikzpicture}
|
||||
\caption{DFA, \(A\), this is really beautiful, ya know?}
|
||||
\label{fig:multiple5}
|
||||
\end{figure}
|
||||
|
||||
\textbf{Justification}
|
||||
\\
|
||||
|
||||
Take a given binary number, \(x\). Since there are only two inputs to our
|
||||
state machine, \(x\) can either become \(x0\) or \(x1\). When a 0 comes
|
||||
into the state machine, it is the same as taking the binary number and
|
||||
multiplying it by two. When a 1 comes into the machine, it is the same as
|
||||
multipying by two and adding one.
|
||||
\\
|
||||
|
||||
Using this knowledge, we can construct a transition table that tell us
|
||||
where to go:
|
||||
|
||||
\begin{table}[ht]
|
||||
\centering
|
||||
\begin{tabular}{c || c | c | c | c | c}
|
||||
& \(x \mod 5 = 0\)
|
||||
& \(x \mod 5 = 1\)
|
||||
& \(x \mod 5 = 2\)
|
||||
& \(x \mod 5 = 3\)
|
||||
& \(x \mod 5 = 4\)
|
||||
\\
|
||||
\hline
|
||||
\(x0\) & 0 & 2 & 4 & 1 & 3 \\
|
||||
\(x1\) & 1 & 3 & 0 & 2 & 4 \\
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
|
||||
Therefore on state \(q_0\) or (\(x \mod 5 = 0\)), a transition line should
|
||||
go to state \(q_0\) for the input 0 and a line should go to state \(q_1\)
|
||||
for input 1. Continuing this gives us the Figure~\ref{fig:multiple5}.
|
||||
\end{homeworkProblem}
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Write part of \alg{Quick-Sort($list, start, end$)}
|
||||
|
||||
\begin{algorithm}[]
|
||||
\begin{algorithmic}[1]
|
||||
\Function{Quick-Sort}{$list, start, end$}
|
||||
\If{$start \geq end$}
|
||||
\State{} \Return{}
|
||||
\EndIf{}
|
||||
\State{} $mid \gets \Call{Partition}{list, start, end}$
|
||||
\State{} \Call{Quick-Sort}{$list, start, mid - 1$}
|
||||
\State{} \Call{Quick-Sort}{$list, mid + 1, end$}
|
||||
\EndFunction{}
|
||||
\end{algorithmic}
|
||||
\caption{Start of QuickSort}
|
||||
\end{algorithm}
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Suppose we would like to fit a straight line through the origin, i.e.,
|
||||
\(Y_i = \beta_1 x_i + e_i\) with \(i = 1, \ldots, n\), \(\E [e_i] = 0\),
|
||||
and \(\Var [e_i] = \sigma^2_e\) and \(\Cov[e_i, e_j] = 0, \forall i \neq
|
||||
j\).
|
||||
\\
|
||||
|
||||
\part
|
||||
|
||||
Find the least squares esimator for \(\hat{\beta_1}\) for the slope
|
||||
\(\beta_1\).
|
||||
\\
|
||||
|
||||
\solution
|
||||
|
||||
To find the least squares estimator, we should minimize our Residual Sum
|
||||
of Squares, RSS:
|
||||
|
||||
\[
|
||||
\begin{split}
|
||||
RSS &= \sum_{i = 1}^{n} {(Y_i - \hat{Y_i})}^2
|
||||
\\
|
||||
&= \sum_{i = 1}^{n} {(Y_i - \hat{\beta_1} x_i)}^2
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
By taking the partial derivative in respect to \(\hat{\beta_1}\), we get:
|
||||
|
||||
\[
|
||||
\pderiv{
|
||||
\hat{\beta_1}
|
||||
}{RSS}
|
||||
= -2 \sum_{i = 1}^{n} {x_i (Y_i - \hat{\beta_1} x_i)}
|
||||
= 0
|
||||
\]
|
||||
|
||||
This gives us:
|
||||
|
||||
\[
|
||||
\begin{split}
|
||||
\sum_{i = 1}^{n} {x_i (Y_i - \hat{\beta_1} x_i)}
|
||||
&= \sum_{i = 1}^{n} {x_i Y_i} - \sum_{i = 1}^{n} \hat{\beta_1} x_i^2
|
||||
\\
|
||||
&= \sum_{i = 1}^{n} {x_i Y_i} - \hat{\beta_1}\sum_{i = 1}^{n} x_i^2
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
Solving for \(\hat{\beta_1}\) gives the final estimator for \(\beta_1\):
|
||||
|
||||
\[
|
||||
\begin{split}
|
||||
\hat{\beta_1}
|
||||
&= \frac{
|
||||
\sum {x_i Y_i}
|
||||
}{
|
||||
\sum x_i^2
|
||||
}
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
Calculate the bias and the variance for the estimated slope
|
||||
\(\hat{\beta_1}\).
|
||||
\\
|
||||
|
||||
\solution
|
||||
|
||||
For the bias, we need to calculate the expected value
|
||||
\(\E[\hat{\beta_1}]\):
|
||||
|
||||
\[
|
||||
\begin{split}
|
||||
\E[\hat{\beta_1}]
|
||||
&= \E \left[ \frac{
|
||||
\sum {x_i Y_i}
|
||||
}{
|
||||
\sum x_i^2
|
||||
}\right]
|
||||
\\
|
||||
&= \frac{
|
||||
\sum {x_i \E[Y_i]}
|
||||
}{
|
||||
\sum x_i^2
|
||||
}
|
||||
\\
|
||||
&= \frac{
|
||||
\sum {x_i (\beta_1 x_i)}
|
||||
}{
|
||||
\sum x_i^2
|
||||
}
|
||||
\\
|
||||
&= \frac{
|
||||
\sum {x_i^2 \beta_1}
|
||||
}{
|
||||
\sum x_i^2
|
||||
}
|
||||
\\
|
||||
&= \beta_1 \frac{
|
||||
\sum {x_i^2 \beta_1}
|
||||
}{
|
||||
\sum x_i^2
|
||||
}
|
||||
\\
|
||||
&= \beta_1
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
Thus since our estimator's expected value is \(\beta_1\), we can conclude
|
||||
that the bias of our estimator is 0.
|
||||
\\
|
||||
|
||||
For the variance:
|
||||
|
||||
\[
|
||||
\begin{split}
|
||||
\Var[\hat{\beta_1}]
|
||||
&= \Var \left[ \frac{
|
||||
\sum {x_i Y_i}
|
||||
}{
|
||||
\sum x_i^2
|
||||
}\right]
|
||||
\\
|
||||
&=
|
||||
\frac{
|
||||
\sum {x_i^2}
|
||||
}{
|
||||
\sum x_i^2 \sum x_i^2
|
||||
} \Var[Y_i]
|
||||
\\
|
||||
&=
|
||||
\frac{
|
||||
\sum {x_i^2}
|
||||
}{
|
||||
\sum x_i^2 \sum x_i^2
|
||||
} \Var[Y_i]
|
||||
\\
|
||||
&=
|
||||
\frac{
|
||||
1
|
||||
}{
|
||||
\sum x_i^2
|
||||
} \Var[Y_i]
|
||||
\\
|
||||
&=
|
||||
\frac{
|
||||
1
|
||||
}{
|
||||
\sum x_i^2
|
||||
} \sigma^2
|
||||
\\
|
||||
&=
|
||||
\frac{
|
||||
\sigma^2
|
||||
}{
|
||||
\sum x_i^2
|
||||
}
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Prove a polynomial of degree \(k\), \(a_kn^k + a_{k - 1}n^{k - 1} + \hdots
|
||||
+ a_1n^1 + a_0n^0\) is a member of \(\Theta(n^k)\) where \(a_k \hdots a_0\)
|
||||
are nonnegative constants.
|
||||
|
||||
\begin{proof}
|
||||
To prove that \(a_kn^k + a_{k - 1}n^{k - 1} + \hdots + a_1n^1 +
|
||||
a_0n^0\), we must show the following:
|
||||
|
||||
\[
|
||||
\exists c_1 \exists c_2 \forall n \geq n_0,\ {c_1 \cdot g(n) \leq
|
||||
f(n) \leq c_2 \cdot g(n)}
|
||||
\]
|
||||
|
||||
For the first inequality, it is easy to see that it holds because no
|
||||
matter what the constants are, \(n^k \leq a_kn^k + a_{k - 1}n^{k - 1} +
|
||||
\hdots + a_1n^1 + a_0n^0\) even if \(c_1 = 1\) and \(n_0 = 1\). This
|
||||
is because \(n^k \leq c_1 \cdot a_kn^k\) for any nonnegative constant,
|
||||
\(c_1\) and \(a_k\).
|
||||
\\
|
||||
|
||||
Taking the second inequality, we prove it in the following way.
|
||||
By summation, \(\sum\limits_{i=0}^k a_i\) will give us a new constant,
|
||||
\(A\). By taking this value of \(A\), we can then do the following:
|
||||
|
||||
\[
|
||||
\begin{split}
|
||||
a_kn^k + a_{k - 1}n^{k - 1} + \hdots + a_1n^1 + a_0n^0 &=
|
||||
\\
|
||||
&\leq (a_k + a_{k - 1} \hdots a_1 + a_0) \cdot n^k
|
||||
\\
|
||||
&= A \cdot n^k
|
||||
\\
|
||||
&\leq c_2 \cdot n^k
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
where \(n_0 = 1\) and \(c_2 = A\). \(c_2\) is just a constant. Thus the
|
||||
proof is complete.
|
||||
\end{proof}
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
%
|
||||
% Non sequential homework problems
|
||||
%
|
||||
|
||||
% Jump to problem 18
|
||||
\begin{homeworkProblem}[18]
|
||||
Evaluate \(\sum_{k=1}^{5} k^2\) and \(\sum_{k=1}^{5} (k - 1)^2\).
|
||||
\end{homeworkProblem}
|
||||
|
||||
% Continue counting to 19
|
||||
\begin{homeworkProblem}
|
||||
Find the derivative of \(f(x) = x^4 + 3x^2 - 2\)
|
||||
\end{homeworkProblem}
|
||||
|
||||
% Go back to where we left off
|
||||
\begin{homeworkProblem}[6]
|
||||
Evaluate the integrals
|
||||
\(\int_0^1 (1 - x^2) \dx\)
|
||||
and
|
||||
\(\int_1^{\infty} \frac{1}{x^2} \dx\).
|
||||
\end{homeworkProblem}
|
||||
\end{comment}
|
||||
\end{document}
|
||||
Reference in New Issue
Block a user