Euler edits
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\uptitler{Fall 2022}
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\title{Euler's Number}
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\subtitle{
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By Oleg Gleizer and Olga Radko. \\
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Prepared by Mark on \today
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Prepared by 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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@ -31,12 +31,14 @@
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\section{Compound Interest}
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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.
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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 $V(t)$. \\
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Derive the formula for $V(t)$ if the annual rate $r$ is compounded monthly. \\
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Derive the formula for $V(t)$ if the annual rate $r$ is compounded $n$ times a year, $n \in \mathbb{N}$. \\
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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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@ -44,369 +46,33 @@
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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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\section{Limits}
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\definition{}
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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. \\
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Any number $u \in \mathbb{R}$ is called an \textit{upper bound} of $[a_n]$ if $a_n \leq u$ for all $u$. \\
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We say a sequence $[a_n]$ with an an upper bound is \textit{bounded from above}.
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\definition{}
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We say a sequence of real numbers $[a_n]^\infty_{n=1}$ is \textit{monotonically increasing} if $m < n \implies a_m < a_n$.
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\vfill
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\theorem{}<limexists>
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A monotonically increasing sequence that is bounded from above has a unique limit. \\
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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 $\pi$.
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\vfill
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\problem{}
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Let $a_1 = 2$, and $a_{n+1} = 2 + \sqrt{a_n}$. \\
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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. \\
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\textbf{$[a_n]$ is bounded:} Induction again. Show that $\sqrt{a_n} < 2$. \\
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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: \\
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Find an example of a monotonically increasing sequence that does not have a limit, \\
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and of a bounded sequence that does not have a limit.
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\vfill
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\definition{}
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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. \\
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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 $n \geq N$, $|a_n - A| < \epsilon$. \\
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If a sequence $[a_n]$ has a limit, we write $\lim_{n \to\infty}(a_n)$.
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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 $n \geq 1$. \\
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Show that $\pi$ is the limit of the sequence $[3, 3.14, 3.141, ...]$.
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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 $|a_n| < \epsilon\ \forall n < N$. \\
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We know that for any $\epsilon$, $\exists p$ so that $\frac{1}{p} = \epsilon$. This is the \textit{Archemedian Property}. \\
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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 $|a_n - \pi| < \epsilon\ \forall n < N$. \\
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Looking at the definition of this sequence, we get a similar ``Archemedian Property'': \\
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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. \\
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This can be done by showing that if $A, B$ are both limits of $[a_n]$, $A$ and $B$ must be equal to $A$.
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\begin{solution}
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If both $A$ and $B$ are limits of $[a_n]$, we have the following: \\
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$\forall \epsilon > 0$, $\exists N_A \in \mathbb{N}$ so that $|a_n - A| < \epsilon\ \forall n < N_A$. \\
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$\forall \epsilon > 0$, $\exists N_B \in \mathbb{N}$ so that $|a_n - B| < \epsilon\ \forall n < N_B$. \\
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Let $N = \max(N_A, N_B)$. \\
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Then, $|a_n - A| + |a_n - B| < 2\epsilon\ \forall n > N$, \\
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which can be writen as $|a_n - A| + |B - a_n| < 2\epsilon\ \forall n > N$. \\
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By the triangle inequality, we have \\
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$|a_n - A + B - a_n| \leq |a_n - A| + |B - a_n|$, \\
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And since $|a_n - A + B - a_n| = |B - A|$, we have \\
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$|B - A| < 2\epsilon\ \forall n > N$. \\
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This should be true for all $\epsilon > 0$. \\
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Let's set $\epsilon = \frac{|B-A|}{4}$, which is greater than zero iff $A \neq B$ \\
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Then, $|B-A| < \frac{|B-A|}{2}$, which is impossible\textsuperscript{*}. \\
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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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\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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\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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Prove this by induction, or however else you can.
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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, \\
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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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\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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\vfill
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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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Is 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{theorem: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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\definition{}
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\ref{e_n}, \ref{e_n_inc}, and \ref{limexists} tell us that $[e_n]$ has a limit. \\
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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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\section{Continuously compounding interest}
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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 $a = \frac{n}{r}$ \\
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$\lim_{n\to\infty}{a} = \infty$, so $\lim_{n\to\infty}{V(t)} = \lim_{a\to\infty}{V(t)}$ \\
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Substituting $a$ for $n$, we get \\
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$P(1 + \frac{r}{n})^{nt} = P(1 + \frac{1}{a})^{art}$ \\
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And finally, we can evaluate \\
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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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\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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$ \\
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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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$ \\
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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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$ \\
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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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\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 $x > M$. \\
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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: \\
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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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\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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122
Advanced/Euler's Number/parts/1 limits.tex
Normal file
122
Advanced/Euler's Number/parts/1 limits.tex
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@ -0,0 +1,122 @@
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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
|
||||
|
||||
|
||||
|
||||
\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 writen 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
Normal file
146
Advanced/Euler's Number/parts/2 e.tex
Normal file
@ -0,0 +1,146 @@
|
||||
\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
|
||||
126
Advanced/Euler's Number/parts/3 more e.tex
Normal file
126
Advanced/Euler's Number/parts/3 more e.tex
Normal file
@ -0,0 +1,126 @@
|
||||
\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
|
||||
Loading…
x
Reference in New Issue
Block a user