Removed Euler's handout
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% use [nosolutions] flag to hide solutions.
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% use [solutions] flag to show solutions.
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\documentclass[
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solutions,
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singlenumbering,
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unfinished
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]{../../resources/ormc_handout}
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\usepackage{../../resources/macros}
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\usepackage{multicol}
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\usepackage{tikz}
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\usepackage{graphicx}
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\graphicspath{ {.} }
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\uptitlel{Advanced 2}
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\uptitler{Fall 2022}
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\title{Euler's Number}
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\subtitle{
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Prepared by \githref{Mark} on \today. \\
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Based on a handout by Oleg Gleizer and Olga Radko
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}
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\begin{document}
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\maketitle
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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.
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\section{Compound Interest}
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Let $P$ be an amount of primary capital which is invested at an annual rate $r$. \par
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Let $V(t)$ be the value of the investment in $t$ years.
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\problem{}
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Derive the formula for $V(t)$ if the annual rate $r$ is compounded yearly. \par
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Derive the formula for $V(t)$ if the annual rate $r$ is compounded monthly. \par
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Derive the formula for $V(t)$ if the annual rate $r$ is compounded $n$ times a year, $n \in \mathbb{N}$. \par
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\hint{\say{Compound monthly} means that $\frac{1}{12}$ of $r$ is applied every month.}
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\begin{solution}
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$V(t) = P(1 + \frac{r}{n})^{nt}$
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\end{solution}
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Next, let's try to understand how we can compound interest continuously.
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\vfill
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\pagebreak
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\input{parts/1 limits.tex}
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\input{parts/2 e.tex}
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\input{parts/3 more e.tex}
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\section{$e$ and Probability}
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\problem{}
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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?
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\vfill
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\problem{}
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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?
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\vfill
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\problem{}
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$n$ 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?
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\vfill
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\pagebreak
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\end{document}
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\section{Limits}
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\definition{Boundedness}
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Let $a_1, a_2, a_3, ... $ (abbreviated $a_n$ in this handout) be a sequence of real numbers. \par
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We say a number $u$ is an \textit{upper bound} of $a_n$ if $a_i \leq u ~ \forall i$. \par
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If $a_n$ has an upper bound, we could say it is \textit{bounded from above}.
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\definition{Monotonically increasing}
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We say a sequence of real numbers $a_n$ is \textit{monotonically increasing} if $m < n \implies a_m < a_n$.
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\definition{Limits (informal)}
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The \textit{limit} of a sequence is a point to which the sequence gets \say{closer} to. \par
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Not all sequences have limits. \par
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Don't let this hand-wavy definition bother you too much, we'll see a proper definition soon.
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\theorem{}<limexists>
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Any monotonically increasing sequence that is bounded from above has a unique limit. \par
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\vfill
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\problem{}
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Let $a_1 = 2$, and $a_{n+1} = 2 + \sqrt{a_n}$. \par
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Show that the sequence $a_n$ is monotonically increasing and bounded above. Find its limit.
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\begin{solution}
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\textbf{$a_n$ is monotonically increasing:} Show this by induction. \par
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\textbf{$a_n$ is bounded:} Induction again. Show that $\sqrt{a_n} < 2$. \par
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\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 $a_n \geq 2\ \forall n$.
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\end{solution}
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\vfill
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\problem{}
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Show that both assumptions of \ref{limexists} are necessary: \par
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Find an example of a monotonically increasing sequence that does not have a limit, \par
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and of a bounded sequence that does not have a limit.
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\vfill
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\pagebreak
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\definition{Limits (formal)}
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Let $a_n$ be a sequence.
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$L$ is the \textit{limit} of this sequence if for any $\varepsilon < 0$, \par
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we can find an $N$ so that $|a_n - L| < \varepsilon ~~ \forall n \geq N$.
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\vfill
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\problem{}
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Show that $0$ is the limit of $a_n = \frac{1}{n}$, where $n \geq 1$. \par
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Show that $\pi$ is the limit of the sequence $3,~ 3.1,~ 3.14,~ ...$.
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\begin{solution}
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$\lim_{n\to\infty}a_n = 0$ means that $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ so that $|a_n| < \epsilon\ \forall n < N$. \par
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We know that for any $\epsilon$, $\exists p$ so that $\frac{1}{p} = \epsilon$. This is the \textit{Archemedian Property}. \par
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It is now clear that $\lim_{n\to\infty}a_n = 0$
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\linehack{}
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$\lim_{n\to\infty}a_n = \pi$ means that $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ so that $|a_n - \pi| < \epsilon\ \forall n < N$. \par
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Looking at the definition of this sequence, we get a similar ``Archemedian Property'': \par
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$\forall \epsilon$, $\exists m$ so that $|a_m - \pi| < \epsilon$.
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\end{solution}
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\vfill
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\problem{}
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Show that if a sequence $a_n$ has a limit, that limit is unique. \par
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\hint{Show that if $A, B$ are both limits of $a_n$, $A$ and $B$ must be equal.}
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\begin{solution}
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If both $A$ and $B$ are limits of $[a_n]$, we have the following: \par
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$\forall \epsilon > 0$, $\exists N_A \in \mathbb{N}$ so that $|a_n - A| < \epsilon\ \forall n < N_A$. \par
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$\forall \epsilon > 0$, $\exists N_B \in \mathbb{N}$ so that $|a_n - B| < \epsilon\ \forall n < N_B$. \par
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Let $N = \max(N_A, N_B)$. \par
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Then, $|a_n - A| + |a_n - B| < 2\epsilon\ \forall n > N$, \par
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which can be written as $|a_n - A| + |B - a_n| < 2\epsilon\ \forall n > N$. \par
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By the triangle inequality, we have \par
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$|a_n - A + B - a_n| \leq |a_n - A| + |B - a_n|$, \par
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And since $|a_n - A + B - a_n| = |B - A|$, we have \par
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$|B - A| < 2\epsilon\ \forall n > N$. \par
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This should be true for all $\epsilon > 0$. \par
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Let's set $\epsilon = \frac{|B-A|}{4}$, which is greater than zero iff $A \neq B$ \par
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Then, $|B-A| < \frac{|B-A|}{2}$, which is impossible\textsuperscript{*}. \par
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Therefore, if both $A$ and $B$ are limits of $[a_n]$, $A$ and $B$ must be equal.
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\linehack{}
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\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.
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\end{solution}
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\vfill
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\pagebreak
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\section{Defining $e$}
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\problem{}
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Recall and prove the binomial theorem.
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\begin{solution}
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The binomial theorem:
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$$(a + b)^n = \sum_{i=0}^{n} \binom{n}{i}a^ib^{n-i}$$
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We usually prove this by induction.
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\end{solution}
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\vfill
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\problem{}<e_n>
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Prove the following:
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$$ \bigg(1 + \frac{1}{n} \bigg)^n < 3 - \frac{1}{n}\ \text{for all } n = 3, 4, 5, ...$$
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This proves that that the sequence $e_n = (1 + \frac{1}{n})^n$, $n = 1, 2, ...$ is bounded from above, \par
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since $e_n < 3\ \forall n \in \mathbb{N}$.
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\begin{solution}
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$$
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\bigg(1 + \frac{1}{n}\bigg)^n =
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\sum_{k=0}^n \binom{n}{k}\frac{1}{n^i} =
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2 + \sum_{k=2}^n \binom{n}{k}\frac{1}{n^i} =
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2 + \sum_{k=2}^n \frac{1}{k!}\frac{(n)(n-1)...(n-k+1)}{n^k} <
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2 + \sum_{k=2}^n \frac{1}{k!}
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$$
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$$
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2 + \sum_{k=2}^n \frac{1}{k!} <
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2 + \sum_{k=2}^n \frac{1}{k(k-1)} =
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2 + 1 - \frac{1}{n} = 3 - \frac{1}{n}
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$$
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\end{solution}
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\vfill
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\pagebreak
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\theorem{Bernoulli's inequality}<bernoulli>
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$(x + 1)^n \geq 1 + nx$ for $x \geq -1$ and $n \in \mathbb{N}$
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\problem{}
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Use induction to prove \ref{bernoulli}
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\vfill
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\problem{}<e_n_inc>
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Use \ref{bernoulli} to prove that the $e_n$ defined in \ref{e_n} is monotonically increasing.
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\begin{solution}
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We want to show that the following is true:
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$$
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\bigg(
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1 + \frac{1}{n + 1}
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\bigg)^{n+1}
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>
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\bigg(
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1 + \frac{1}{n}
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\bigg)^n
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$$
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This inequality is equivalent to
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$$
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\Bigg(
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\frac{
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1 + \frac{1}{n+1}
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}{
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1 + \frac{1}{n}
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}
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\Bigg)^{n+1}
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>
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\bigg(
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1 + \frac{1}{n}
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\bigg)^{-1}
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= \frac{1}{1 + \frac{1}{n}}
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= \frac{n}{n+1}
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= 1 - \frac{1}{n+1}
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$$
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Also,
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$$
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\frac{
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1 + \frac{1}{n+1}
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} {
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1 + \frac{1}{n}
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}
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= 1 - \frac{1}{(n + 1)^2}
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$$
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\ref{bernoulli} tells us that
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$$
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\bigg(
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1 - \frac{1}{(n+1)^2}
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\bigg) ^ {n+1}
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= 1 - \frac{n+1}{(n+1)^2}
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= 1 - \frac{1}{n+1}
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$$
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Since this is equivalent to our original inequality, we are done.
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\end{solution}
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\vfill
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\pagebreak
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\definition{}
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\ref{e_n}, \ref{e_n_inc}, and \ref{limexists} tell us that $e_n$ has a limit. \par
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Let us define $e$:
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$$e = \lim_{n\to\infty}{e_n} = \lim_{n\to\infty}{\bigg(1 + \frac{1}{n} \bigg)^n}$$
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\vfill
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\problem{}
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Derive the formula for $V(t)$ if the annual rate $r$ is compounded continuously.
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\begin{solution}
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For a rate $r$ compounded $n$ times per year, we have $V(t) = P(1 + \frac{r}{n})^{nt}$.
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$\lim_{n\to\infty}{(V(t))} = \lim_{n\to\infty}{P(1 + \frac{r}{n})^{nt}}$
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Let $a = \frac{n}{r}$ \par
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$\lim_{n\to\infty}{a} = \infty$, so $\lim_{n\to\infty}{V(t)} = \lim_{a\to\infty}{V(t)}$ \par
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Substituting $a$ for $n$, we get \par
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$P(1 + \frac{r}{n})^{nt} = P(1 + \frac{1}{a})^{art}$ \par
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And finally, we can evaluate \par
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$\lim_{a\to\infty}{P (1 + \frac{1}{a})^{art}} = Pe^{rt}$
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\end{solution}
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\vfill
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\pagebreak
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@ -1,126 +0,0 @@
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\section{More about $e$}
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\problem{}
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Show that
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$$
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\lim_{n\to\infty}{
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\bigg(
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1 + \frac{1}{n+1}
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\bigg)^n
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= e
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}
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$$
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\vfill
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\problem{}
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Show that
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$$
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\lim_{n\to\infty}{
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\bigg(
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1 + \frac{1}{n}
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\bigg)^{n+1}
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= e
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}
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$$
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\vfill
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\problem{}<inverse_e>
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Show that
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$$
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\lim_{n\to\infty}{
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\bigg(
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1 - \frac{1}{n}
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\bigg)^n
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= \frac{1}{e}
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}
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$$
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\begin{solution}
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$
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\lim_{n\to\infty}{(1 - \frac{1}{n})^n} =
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\lim_{n\to\infty}{(\frac{n-1}{n})^{(-1)(-n)}}
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$ \par
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$
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= \lim_{n\to\infty}{(\frac{n}{n- 1 })^{-n}}
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= \lim_{n\to\infty}{(1 + \frac{1}{n-1})^{-n}}
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$ \par
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$
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= \lim_{n\to\infty}{{(1 + \frac{1}{n-1})^{(n-1)(\frac{n}{n-1})}}^{-1}}
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$ \par
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$
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= \frac{1}{e}
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$
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\end{solution}
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\vfill
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\problem{}
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Show that
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$$
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\lim_{n\to\infty}{
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\bigg(
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1 + \frac{x}{n}
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\bigg)^n
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= e^x
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}
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$$
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Note that \ref{inverse_e} is a special case of this problem.
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\vfill
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\pagebreak
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\theorem{}
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The following important formula is proven in most calculus courses.
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$$e^x = \sum_{n=0}^{\infty}{\frac{x^n}{n!}} = 2 + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
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\vfill
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\problem{}
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What are the first six digits of $e$?
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\begin{solution}
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$e = 2.718\ 281\ 828$
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\end{solution}
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\vfill
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\definition{}
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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
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If this is true, we say that $L = \lim_{x\to\infty}{(f(x))}$.
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\vfill
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\problem{}
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Prove the following: \par
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Hint: If $x > 0$, then $\lfloor x \rfloor \leq x \leq \lceil x \rceil$
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$$ \lim_{x\to\infty}{\bigg(1 + \frac{1}{x}\bigg)^x} = e$$
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\vfill
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\pagebreak
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