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template and my start on the lab down

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**/*.aux
**/*.fdb_latexmk
**/*.fls
**/*.log
**/*.synctex.gz

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\documentclass{article}
\usepackage{fancyhdr}
\usepackage{extramarks}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{tikz}
\usepackage[plain]{algorithm}
\usepackage{algpseudocode}
\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\ \#2}
\newcommand{\hmwkDueDate}{February 12, 2014}
\newcommand{\hmwkClass}{Calculus}
\newcommand{\hmwkClassTime}{Section A}
\newcommand{\hmwkClassInstructor}{Professor Isaac Newton}
\newcommand{\hmwkAuthorName}{\textbf{Josh Davis} \and \textbf{Davis Josh}}
%
% 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{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{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}