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\title{ESP Workshop, Worksheet \#3}
\date{\today}
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\begin{center}
{\bf \Large ESP Workshop, Worksheet \#3}

{\bf \large Thursday \today}

{\bf \large AI: Eric Katerman}
\end{center}
\vspace{.15in}
\begin{enumerate}
\item Consider the sequence $\{1,-1/2,1/3,-1/4,\ldots \}$.
\begin{enumerate}
\item Find a formula for the general term $a_n$ of the sequence.
\item Sketch the sequence both as points on a number line and as a
  graph of a function whose domain is the set of positive integers.
\item Find the limit of this sequence as $n \rightarrow \infty$.
\item Repeat parts (a) -- (c) with the sequence
  $\{2/9,6/27,24/81,120/243,\ldots \}$
\end{enumerate}

\item Now we use the Squeeze theorem...

\begin{enumerate}
\item Derive $\sin (\pi /4) = \sqrt{2}/2$ using an argument involving
  isosceles triangles. Also show that $\sin (\pi /6) = 1/2$
  using equilateral triangles.
\item Consider the sequence $a_n = \sin (n\pi /6), n\geq 0$. Write out
  the first thirteen terms. What is $a_{100}$?
\item Does this sequence converge? What about the sequence $b_n = a_n
  /(\ln n)$? Justify your answers, and sketch $a_n$ and $b_n$ as graphs (with
  domain the positive integers).
\item Find
\begin{displaymath}
  \lim _{n\rightarrow \infty} \frac{n + 2\cos n}{n}
\end{displaymath}
and sketch the graph to see what's going on.
\end{enumerate}

\item Let $f(x) = \lfloor x \rfloor$. This is called the ``floor''
  function, and it returns the largest integer less than or equal to
  $x$. Also, let $g:\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}$ be
  defined by $g(x) = (f(x))!$. This function $g$ \textit{extends} the
  factorial function, which is only defined for non-negative integers:
  $n!: \mathbb{Z}_{\geq 0} \rightarrow \mathbb{Z},\ n! =
  n(n-1)\cdots 2\cdot 1$.

\begin{enumerate}
  
\item What's the limit of the sequence $\{1.9, 1.99, 1.999, 1.9999,
  \ldots \}$?

\item Draw the graph of $y=f(x)$ for $x$ in the range $-3 \leq x \leq
  3$. What is $f(-1.5)$? What about $f(1.9999)$?  And
  $f(1.\overline{9})$?

\item Draw the graph of $y=g(x)$ for $x$ in the range $0\leq x \leq 4$.

Let's define a new function, called $\Gamma(x)$, like this:
\begin{displaymath}
\Gamma(x)=\int_0^\infty e^{-t}t^{x-1}\ dt,\qquad x>0.
\end{displaymath}

\item Use integration by parts to prove that $\Gamma(x+1)=x\Gamma(x)$.

\item Show that $\Gamma(1)=1$.  Conclude that $\Gamma(n)=(n-1)!$ for
  all natural numbers $n$.
  
  The gamma function is very special: not only does it \textit{extend}
  the factorial function (like $f$), but it is actually
  \textit{differentiable} on the positive real line! Can you think of
  a continuous function that extends the sequence $a_n = (-1)^n$ to
  $\mathbb{R}$? What about a differentiable one?
  
\end{enumerate}

\item To construct the \textbf{snowflake curve}, start with an
  equilateral triangle with sides of length 1. Step 1 in the
  construction is to divide each side into three equal parts,
  construct an equilateral triangle on the middle part, and then
  delete the middle part. Step 2 is to repeat Step 1 for each side of
  the resulting polygon. This process is repeated at each succeeding
  step. The snowflake curve is the curve that results from repeating
  this process indefinitely.
\begin{enumerate}
  
\item Draw the first few stages of this construction.

\item Let $s_n, l_n,$ and $p_n$ represent the number of sides, the
  length of a side, and the total length of the $n$th approximating
  curve (the curve obtained after Step $n$ of the construction),
  respectively. Find formulas for $s_n, l_n,$ and $p_n$.
  
\item Show that $p_n \rightarrow \infty$ as $n\rightarrow \infty$, so
  the snowflake curve has infinite perimeter!
  
\item Give an easy argument to show that the snowflake curve have
  finite area. Can you compute its area explicitly?
\end{enumerate}

\item We say that the set of all real numbers $\mathbb{R}$ is
  \textbf{complete} because every non-empty set of real numbers with
  an upper bound \textit{in $\mathbb{R}$} has a least upper bound
  \textit{in $\mathbb{R}$}, so there are no ``holes'' in $\mathbb{R}$.
  We will show that the set of all rational numbers
\begin{displaymath}
\mathbb{Q} = \left\{ \frac{r}{s} : r,s\ \textrm{are integers} \right \}
\end{displaymath}
does not have this property. Here's the strategy: we will find a real
number $x$ that is not rational, and then we will construct a subset
of $\mathbb{Q}$ whose least upper bound \textit{in $\mathbb{R}$} is
$x$.

\begin{enumerate}
\item Suppose that $\sqrt{2}$ is rational, i.e. suppose $\sqrt{2} =
  a/b$ for some integers $a, b$. Also suppose that $a/b$ IS IN LOWEST
  TERMS. What does $2b^2$ equal?
  
\item Is $a^2$ odd or even? (If you can't tell, take another look at
  part (a).)
\item Is $b^2$ odd or even? Why does this prove that $\sqrt{2}$ is not
  rational?

\item Let $S$ be the set of all rational numbers less than $\sqrt{2}$, i.e.
\begin{displaymath}
S = \{ x\in \mathbb{Q} : x < \sqrt{2} \}
\end{displaymath}
Use the fact that between any two distinct real numbers $x,y \in
\mathbb{R}$, $x\neq y$, there exists a rational number $r/s \in
\mathbb{Q}$ (i.e.  $x < r/s < y$) to show that $S$ does not have a
least upper bound in $\mathbb{Q}$.

\item (Challenge.) Can you prove the fact we used in the last part?

\end{enumerate}

\end{enumerate}

\end{document}