Added Euler's handout

Advanced/Euler's Number/main.tex

@@ -0,0 +1,418 @@
+% https://git.betalupi.com/Mark/latex
+% use [nosolutions] flag to hide solutions.
+% use [solutions] flag to show solutions.
+% Last built with version 1.1.0
+\documentclass[
+	singlenumbering,
+	nosolutions
+]{ormc_handout}
+
+\usepackage{amsmath}
+\usepackage{amssymb}
+
+\usepackage{multicol}
+\usepackage{tikz}
+\usepackage{graphicx}
+\graphicspath{ {.} }
+
+\newcommand{\qe}{\stackrel{?}{=}}
+\newcommand{\qgt}{\stackrel{?}{>}}
+\newcommand{\qlt}{\stackrel{?}{<}}
+
+
+\begin{document}
+
+	\maketitle
+		<Advanced 2>
+		<Fall 2022>
+		{Euler's Number}
+		{
+			By Oleg Gleizer and Olga Radko. \\
+			Prepared by Mark on \today
+		}
+
+
+	\paragraph{}
+	The goal of this mini-course is to construct Euler's number, one of the most important constants in mathematics, physics, economics, and finance. Make sure you fully understand all definitions before trying to solve problems that use them.
+
+	\section{Compound Interest}
+
+	Let $P$ be the primary capital invested at a constant rate $r$ compounded annually. Let $V(t)$ be the value of the investment in $t$ years.
+
+	\problem{}
+	Derive the formula for $V(t)$. \\
+	Derive the formula for $V(t)$ if the annual rate $r$ is compounded monthly. \\
+	Derive the formula for $V(t)$ if the annual rate $r$ is compounded $n$ times a year, $n \in \mathbb{N}$. \\
+
+	\begin{solution}
+		$V(t) = P(1 + \frac{r}{n})^{nt}$
+	\end{solution}
+
+	Next, let's try to understand how we can compound interest continuously.
+
+	\vfill
+
+	\section{Limits}
+
+	\definition{}
+	Let $[a_1, a_2, a_3, ... ]$, or alternately, $[a_n]^\infty_{n=1}$ (abbreviated $[a_n]$ in this handout) be a sequence of real numbers. \\
+
+	Any number $u \in \mathbb{R}$ is called an \textit{upper bound} of $[a_n]$ if $a_n \leq u$ for all $u$. \\
+	We say a sequence $[a_n]$ with an an upper bound is \textit{bounded from above}.
+
+	\definition{}
+	We say a sequence of real numbers $[a_n]^\infty_{n=1}$ is \textit{monotonically increasing} if $m < n \implies a_m < a_n$.
+
+	\vfill
+
+	\theorem{}<limexists>
+	A monotonically increasing sequence that is bounded from above has a unique limit. \\
+
+	Intuitively, the limit of a sequence is a point to which the sequence gets closer and closer to. Consider the following examples: $[\frac{1}{1}, \frac{1}{2}, \frac{1}{4}, \frac{1}{4}, ...]$, which has the limit $0$, and the sequence $[3, 3.1, 3.14, ...]$ which approaches $\pi$.
+
+	\vfill
+
+	\problem{}
+	Let $a_1 = 2$, and $a_{n+1} = 2 + \sqrt{a_n}$. \\
+
+	Show that the sequence $[a_n]$ is monotonically increasing and bounded above. Find its limit.
+
+	\begin{solution}
+		\textbf{$[a_n]$ is monotonically increasing:} Show this by induction. \\
+		\textbf{$[a_n]$ is bounded:} Induction again. Show that $\sqrt{a_n} < 2$. \\
+
+		\textbf{$\lim_{n \to\infty}a_n = 4$:} Use the fact that $\lim_{n\to\infty} a_n = \lim_{n\to\infty}a_{n+1}$, the hit the resulting equation with some algebra. Note that $1$ cannot be a solution, since $a_n \geq 2\ \forall n$.
+
+
+	\end{solution}
+
+	\vfill
+
+	\problem{}
+	Show that both assumptions of \ref{limexists} are necessary: \\
+
+	Find an example of a monotonically increasing sequence that does not have a limit, \\
+	and of a bounded sequence that does not have a limit.
+
+	\vfill
+
+	\definition{}
+	Let us provide a formal definition for a ``limit''. If you're stuck trying to prove something with this formal definition, give an informal explanation and come back to the problem later. \\
+
+	The number $A$ is the \textit{limit} of the sequence $[a_n]$ if for every positive real number $\epsilon < 0$, we can find a natural number N so that for all $n \geq N$, $|a_n - A| < \epsilon$. \\
+	If a sequence $[a_n]$ has a limit, we write $\lim_{n \to\infty}(a_n)$.
+
+	\vfill
+
+
+	\problem{}
+	Show that $0$ is the limit of $a_n = -\frac{1}{n}$, where $n \geq 1$. \\
+	Show that $\pi$ is the limit of the sequence $[3, 3.14, 3.141, ...]$.
+
+	\begin{solution}
+		$\lim_{n\to\infty}a_n = 0$ means that $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ so that $|a_n| < \epsilon\ \forall n < N$. \\
+
+		We know that for any $\epsilon$, $\exists p$ so that $\frac{1}{p} = \epsilon$. This is the \textit{Archemedian Property}. \\
+		It is now clear that $\lim_{n\to\infty}a_n = 0$
+
+		\boxlinehack
+
+		$\lim_{n\to\infty}a_n = \pi$ means that $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ so that $|a_n - \pi| < \epsilon\ \forall n < N$. \\
+
+		Looking at the definition of this sequence, we get a similar ``Archemedian Property'': \\
+		$\forall \epsilon$, $\exists m$ so that $|a_m - \pi| < \epsilon$.
+	\end{solution}
+
+	\vfill
+
+	\problem{}
+	Show that if a sequence $[a_n]$ has a limit, that limit is unique. \\
+	This can be done by showing that if $A, B$ are both limits of $[a_n]$, $A$ and $B$ must be equal to $A$.
+
+	\begin{solution}
+		If both $A$ and $B$ are limits of $[a_n]$, we have the following: \\
+		$\forall \epsilon > 0$, $\exists N_A \in \mathbb{N}$ so that $|a_n - A| < \epsilon\ \forall n < N_A$. \\
+		$\forall \epsilon > 0$, $\exists N_B \in \mathbb{N}$ so that $|a_n - B| < \epsilon\ \forall n < N_B$. \\
+
+		Let $N = \max(N_A, N_B)$. \\
+		Then, $|a_n - A| + |a_n - B| < 2\epsilon\ \forall n > N$, \\
+		which can be writen as $|a_n - A| + |B - a_n| < 2\epsilon\ \forall n > N$. \\
+
+		By the triangle inequality, we have \\
+		$|a_n - A + B - a_n| \leq |a_n - A| + |B - a_n|$, \\
+		And since $|a_n - A + B - a_n| = |B - A|$, we have \\
+		$|B - A| < 2\epsilon\ \forall n > N$. \\
+
+		This should be true for all $\epsilon > 0$. \\
+		Let's set $\epsilon = \frac{|B-A|}{4}$, which is greater than zero iff $A \neq B$ \\
+		Then, $|B-A| < \frac{|B-A|}{2}$, which is impossible\textsuperscript{*}. \\
+
+		Therefore, if both $A$ and $B$ are limits of $[a_n]$, $A$ and $B$ must be equal.
+
+		\boxlinehack
+
+		\textsuperscript{*}Note that we can also set $\epsilon = \frac{|B-A|}{2}$, which gives $|B-A| < |B-A|$. This is also false, and the proof still works. However, the extra $\frac{\hspace{2em}}{2}$ gives a clearer explanation.
+
+	\end{solution}
+
+	\vfill
+	\pagebreak
+
+
+	\section{Defining $e$}
+
+	\problem{}
+	Recall and prove the binomial theorem.
+
+	\begin{solution}
+		The binomial theorem:
+		$$(a + b)^n = \sum_{i=0}^{n} \binom{n}{i}a^ib^{n-i}$$
+
+		Prove this by induction, or however else you can.
+	\end{solution}
+
+	\vfill
+
+	\problem{}<e_n>
+	Prove the following:
+
+	$$ \bigg(1 + \frac{1}{n} \bigg)^n < 3 - \frac{1}{n}\ \text{for all } n = 3, 4, 5, ...$$
+
+	This proves that that the sequence $e_n = (1 + \frac{1}{n})^n$, $n = 1, 2, ...$ is bounded from above, \\
+	since $e_n < 3\ \forall n \in \mathbb{N}$.
+
+	\begin{solution}
+		$$
+			\bigg(1 + \frac{1}{n}\bigg)^n =
+			\sum_{k=0}^n \binom{n}{k}\frac{1}{n^i} =
+			2 + \sum_{k=2}^n \binom{n}{k}\frac{1}{n^i} =
+			2 + \sum_{k=2}^n \frac{1}{k!}\frac{(n)(n-1)...(n-k+1)}{n^k} <
+			2 + \sum_{k=2}^n \frac{1}{k!}
+		$$
+
+		$$
+			2 + \sum_{k=2}^n \frac{1}{k!} <
+			2 + \sum_{k=2}^n \frac{1}{k(k-1)} =
+			2 + 1 - \frac{1}{n} = 3 - \frac{1}{n}
+		$$
+	\end{solution}
+
+	\vfill
+
+	\theorem{Bernoulli's inequality}<bernoulli>
+	$(x + 1)^n \geq 1 + nx$ for $x \geq -1$ and $n \in \mathbb{N}$
+
+	\vfill
+
+	\problem{}
+	Use induction to prove \ref{bernoulli}
+
+	\vfill
+
+	\problem{}<e_n_inc>
+	Use \ref{bernoulli} to prove that the $[e_n]$ defined in \ref{e_n} is monotonically increasing.
+
+	\begin{solution}
+		We want to show that the following is true:
+			$$
+				\bigg(
+					1 + \frac{1}{n + 1}
+				\bigg)^{n+1}
+				>
+				\bigg(
+					1 + \frac{1}{n}
+				\bigg)^n
+			$$
+
+		Is inequality is equivalent to
+			$$
+				\Bigg(
+					\frac{
+						1 + \frac{1}{n+1}
+					}{
+						1 + \frac{1}{n}
+					}
+				\Bigg)^{n+1}
+				>
+				\bigg(
+					1 + \frac{1}{n}
+				\bigg)^{-1}
+				= \frac{1}{1 + \frac{1}{n}}
+				= \frac{n}{n+1}
+				= 1 - \frac{1}{n+1}
+			$$
+
+		Also,
+			$$
+				\frac{
+					1 + \frac{1}{n+1}
+				} {
+					1 + \frac{1}{n}
+				}
+				= 1 - \frac{1}{(n + 1)^2}
+			$$
+
+		\ref{theorem:bernoulli} tells us that
+			$$
+				\bigg(
+					1 - \frac{1}{(n+1)^2}
+				\bigg) ^ {n+1}
+				= 1 - \frac{n+1}{(n+1)^2}
+				= 1 - \frac{1}{n+1}
+			$$
+
+		Since this is equivalent to our original inequality, we are done.
+
+	\end{solution}
+
+	\vfill
+
+	\definition{}
+	\ref{e_n}, \ref{e_n_inc}, and \ref{limexists} tell us that $[e_n]$ has a limit. \\
+	Let us define $e$:
+	$$e = \lim_{n\to\infty}{e_n} = \lim_{n\to\infty}{\bigg(1 + \frac{1}{n} \bigg)^n}$$
+
+	\vfill
+
+	\section{Continuously compounding interest}
+
+	\problem{}
+	Derive the formula for $V(t)$ if the annual rate $r$ is compounded continuously.
+
+	\begin{solution}
+		For a rate $r$ compounded $n$ times per year, we have $V(t) = P(1 + \frac{r}{n})^{nt}$.
+
+		$\lim_{n\to\infty}{(V(t))} = \lim_{n\to\infty}{P(1 + \frac{r}{n})^{nt}}$
+
+		Let $a = \frac{n}{r}$ \\
+
+		$\lim_{n\to\infty}{a} = \infty$, so $\lim_{n\to\infty}{V(t)} = \lim_{a\to\infty}{V(t)}$ \\
+
+		Substituting $a$ for $n$, we get \\
+		$P(1 + \frac{r}{n})^{nt} = P(1 +  \frac{1}{a})^{art}$ \\
+
+		And finally, we can evaluate \\
+		$\lim_{a\to\infty}{P (1 +  \frac{1}{a})^{art}} = Pe^{rt}$
+	\end{solution}
+
+	\vfill
+	\pagebreak
+
+	\section{More about $e$}
+
+	\problem{}
+	Show that
+	$$
+		\lim_{n\to\infty}{
+			\bigg(
+				1 + \frac{1}{n+1}
+			\bigg)^n
+			= e
+		}
+	$$
+
+	\vfill
+
+	\problem{}
+	Show that
+	$$
+		\lim_{n\to\infty}{
+			\bigg(
+				1 + \frac{1}{n}
+			\bigg)^{n+1}
+			= e
+		}
+	$$
+
+	\vfill
+
+	\problem{}<inverse_e>
+	Show that
+	$$
+		\lim_{n\to\infty}{
+			\bigg(
+				1 - \frac{1}{n}
+			\bigg)^n
+			= \frac{1}{e}
+		}
+	$$
+
+	\begin{solution}
+		$
+			\lim_{n\to\infty}{(1 - \frac{1}{n})^n} =
+			\lim_{n\to\infty}{(\frac{n-1}{n})^{(-1)(-n)}}
+		$ \\
+
+		$
+		= \lim_{n\to\infty}{(\frac{n}{n- 1 })^{-n}}
+		= \lim_{n\to\infty}{(1 + \frac{1}{n-1})^{-n}}
+		$ \\
+
+		$
+		= \lim_{n\to\infty}{{(1 + \frac{1}{n-1})^{(n-1)(\frac{n}{n-1})}}^{-1}}
+		$ \\
+
+		$
+		= \frac{1}{e}
+		$
+	\end{solution}
+
+	\vfill
+
+	\problem{}
+	Show that
+	$$
+		\lim_{n\to\infty}{
+			\bigg(
+				1 + \frac{x}{n}
+			\bigg)^n
+			= e^x
+		}
+	$$
+	Note that \ref{inverse_e} is a special case of this problem.
+
+	\vfill
+
+	\theorem{}
+	The following important formula is proven in most calculus courses.
+
+	$$e^x = \sum_{n=0}^{\infty}{\frac{x^n}{n!}} = 2 + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
+
+	\vfill
+
+	\problem{}
+	What are the first six digits of $e$?
+
+	\begin{solution}
+		$e = 2.718\ 281\ 828$
+	\end{solution}
+
+	\vfill
+
+	\definition{}
+	If $f$ is a function, we say that $L$ is a limit of $f$ at $\infty$ if for every $\epsilon > 0$, we can find an $M \in \mathbb{R}$ so that $|f(x) - L| < \epsilon$ for $x > M$. \\
+	If this is true, we say that  $L = \lim_{x\to\infty}{(f(x))}$.
+
+	\vfill
+
+	\problem{}
+	Prove the following: \\
+	Hint: If $x > 0$, then $\lfloor x \rfloor \leq x \leq \lceil x \rceil$
+
+	$$ \lim_{x\to\infty}{\bigg(1 + \frac{1}{x}\bigg)^x} = e$$
+
+	\vfill
+	\pagebreak
+
+	\section{$e$ and Probability}
+
+	\problem{}
+	A gambler plays a game $n$ times. Each time he plays, his chance of winning is $p$. What are the odds he will win exactly $k$ times?
+
+	\problem{}
+	A gambler plays a game $10,000$ times. Each time he plays, he has a $\frac{1}{10,000}$ chance of winning. What are the odds he loses every time?
+
+	\problem{}
+	$n$ people participate in a gift exchange. Each person puts their name in a hat, then names are drawn at random. For a large $n$, what is the probability that someone will draw their own name?
+
+\end{document}