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77
Advanced/Euler's Number/main.tex
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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering,
unfinished
]{../../resources/ormc_handout}
\usepackage{../../resources/macros}
\usepackage{multicol}
\usepackage{tikz}
\usepackage{graphicx}
\graphicspath{ {.} }
\uptitlel{Advanced 2}
\uptitler{Fall 2022}
\title{Euler's Number}
\subtitle{
Prepared by \githref{Mark} on \today. \\
Based on a handout by Oleg Gleizer and Olga Radko
}
\begin{document}
\maketitle
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 an amount of primary capital which is invested at an annual rate $r$. \par
Let $V(t)$ be the value of the investment in $t$ years.
\problem{}
Derive the formula for $V(t)$ if the annual rate $r$ is compounded yearly. \par
Derive the formula for $V(t)$ if the annual rate $r$ is compounded monthly. \par
Derive the formula for $V(t)$ if the annual rate $r$ is compounded $n$ times a year, $n \in \mathbb{N}$. \par
\hint{\say{Compound monthly} means that $\frac{1}{12}$ of $r$ is applied every month.}
\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
\pagebreak
\input{parts/1 limits.tex}
\input{parts/2 e.tex}
\input{parts/3 more e.tex}
\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?
\vfill
\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?
\vfill
\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?
\vfill
\pagebreak
\end{document}

122
Advanced/Euler's Number/parts/1 limits.tex
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\section{Limits}
\definition{Boundedness}
Let $a_1, a_2, a_3, ... $ (abbreviated $a_n$ in this handout) be a sequence of real numbers. \par
We say a number $u$ is an \textit{upper bound} of $a_n$ if $a_i \leq u ~ \forall i$. \par
If $a_n$ has an upper bound, we could say it is \textit{bounded from above}.
\definition{Monotonically increasing}
We say a sequence of real numbers $a_n$ is \textit{monotonically increasing} if $m < n \implies a_m < a_n$.
\definition{Limits (informal)}
The \textit{limit} of a sequence is a point to which the sequence gets \say{closer} to. \par
Not all sequences have limits. \par
Don't let this hand-wavy definition bother you too much, we'll see a proper definition soon.
\theorem{}<limexists>
Any monotonically increasing sequence that is bounded from above has a unique limit. \par
\vfill
\problem{}
Let $a_1 = 2$, and $a_{n+1} = 2 + \sqrt{a_n}$. \par
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. \par
\textbf{$a_n$ is bounded:} Induction again. Show that $\sqrt{a_n} < 2$. \par
\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: \par
Find an example of a monotonically increasing sequence that does not have a limit, \par
and of a bounded sequence that does not have a limit.
\vfill
\pagebreak
\definition{Limits (formal)}
Let $a_n$ be a sequence.
$L$ is the \textit{limit} of this sequence if for any $\varepsilon < 0$, \par
we can find an $N$ so that $|a_n - L| < \varepsilon ~~ \forall n \geq N$.
\vfill
\problem{}
Show that $0$ is the limit of $a_n = \frac{1}{n}$, where $n \geq 1$. \par
Show that $\pi$ is the limit of the sequence $3,~ 3.1,~ 3.14,~ ...$.
\begin{solution}
$\lim_{n\to\infty}a_n = 0$ means that $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ so that $|a_n| < \epsilon\ \forall n < N$. \par
We know that for any $\epsilon$, $\exists p$ so that $\frac{1}{p} = \epsilon$. This is the \textit{Archemedian Property}. \par
It is now clear that $\lim_{n\to\infty}a_n = 0$
\linehack{}
$\lim_{n\to\infty}a_n = \pi$ means that $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ so that $|a_n - \pi| < \epsilon\ \forall n < N$. \par
Looking at the definition of this sequence, we get a similar ``Archemedian Property'': \par
$\forall \epsilon$, $\exists m$ so that $|a_m - \pi| < \epsilon$.
\end{solution}
\vfill
\problem{}
Show that if a sequence $a_n$ has a limit, that limit is unique. \par
\hint{Show that if $A, B$ are both limits of $a_n$, $A$ and $B$ must be equal.}
\begin{solution}
If both $A$ and $B$ are limits of $[a_n]$, we have the following: \par
$\forall \epsilon > 0$, $\exists N_A \in \mathbb{N}$ so that $|a_n - A| < \epsilon\ \forall n < N_A$. \par
$\forall \epsilon > 0$, $\exists N_B \in \mathbb{N}$ so that $|a_n - B| < \epsilon\ \forall n < N_B$. \par
Let $N = \max(N_A, N_B)$. \par
Then, $|a_n - A| + |a_n - B| < 2\epsilon\ \forall n > N$, \par
which can be written as $|a_n - A| + |B - a_n| < 2\epsilon\ \forall n > N$. \par
By the triangle inequality, we have \par
$|a_n - A + B - a_n| \leq |a_n - A| + |B - a_n|$, \par
And since $|a_n - A + B - a_n| = |B - A|$, we have \par
$|B - A| < 2\epsilon\ \forall n > N$. \par
This should be true for all $\epsilon > 0$. \par
Let's set $\epsilon = \frac{|B-A|}{4}$, which is greater than zero iff $A \neq B$ \par
Then, $|B-A| < \frac{|B-A|}{2}$, which is impossible\textsuperscript{*}. \par
Therefore, if both $A$ and $B$ are limits of $[a_n]$, $A$ and $B$ must be equal.
\linehack{}
\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

146
Advanced/Euler's Number/parts/2 e.tex
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\section{Defining $e$}
\problem{}
Recall and prove the binomial theorem.
\begin{solution}
The binomial theorem:
$$(a + b)^n = \sum_{i=0}^{n} \binom{n}{i}a^ib^{n-i}$$
We usually prove this by induction.
\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, \par
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
\pagebreak
\theorem{Bernoulli's inequality}<bernoulli>
$(x + 1)^n \geq 1 + nx$ for $x \geq -1$ and $n \in \mathbb{N}$
\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:
$$
\bigg(
1 + \frac{1}{n + 1}
\bigg)^{n+1}
>
\bigg(
1 + \frac{1}{n}
\bigg)^n
$$
This 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{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
\pagebreak
\definition{}
\ref{e_n}, \ref{e_n_inc}, and \ref{limexists} tell us that $e_n$ has a limit. \par
Let us define $e$:
$$e = \lim_{n\to\infty}{e_n} = \lim_{n\to\infty}{\bigg(1 + \frac{1}{n} \bigg)^n}$$
\vfill
\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 $a = \frac{n}{r}$ \par
$\lim_{n\to\infty}{a} = \infty$, so $\lim_{n\to\infty}{V(t)} = \lim_{a\to\infty}{V(t)}$ \par
Substituting $a$ for $n$, we get \par
$P(1 + \frac{r}{n})^{nt} = P(1 + \frac{1}{a})^{art}$ \par
And finally, we can evaluate \par
$\lim_{a\to\infty}{P (1 + \frac{1}{a})^{art}} = Pe^{rt}$
\end{solution}
\vfill
\pagebreak

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Advanced/Euler's Number/parts/3 more e.tex
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\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)}}
$ \par
$
= \lim_{n\to\infty}{(\frac{n}{n- 1 })^{-n}}
= \lim_{n\to\infty}{(1 + \frac{1}{n-1})^{-n}}
$ \par
$
= \lim_{n\to\infty}{{(1 + \frac{1}{n-1})^{(n-1)(\frac{n}{n-1})}}^{-1}}
$ \par
$
= \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
\pagebreak
\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 $x > M$. \par
If this is true, we say that $L = \lim_{x\to\infty}{(f(x))}$.
\vfill
\problem{}
Prove the following: \par
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