diff --git a/Advanced/Euler's Number/main.tex b/Advanced/Euler's Number/main.tex index 6d6bc53..a50948a 100755 --- a/Advanced/Euler's Number/main.tex +++ b/Advanced/Euler's Number/main.tex @@ -18,8 +18,8 @@ \uptitler{Fall 2022} \title{Euler's Number} \subtitle{ - By Oleg Gleizer and Olga Radko. \\ - Prepared by Mark on \today + Prepared by Mark on \today. \\ + Based on a handout by Oleg Gleizer and Olga Radko } @@ -31,12 +31,14 @@ \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. + 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 $V(t)$. \\ - 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, $n \in \mathbb{N}$. \\ + 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} $V(t) = P(1 + \frac{r}{n})^{nt}$ @@ -44,369 +46,33 @@ 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{} - 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 $\lim_{n\to\infty}a_n = 0$ - - \linehack{} - - $\lim_{n\to\infty}a_n = \pi$ means that $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ so that $|a_n - \pi| < \epsilon\ \forall n < N$. \\ - - 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. - - \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 - \section{Defining $e$} - \problem{} - Recall and prove the binomial theorem. + \input{parts/1 limits.tex} + \input{parts/2 e.tex} + \input{parts/3 more e.tex} - \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{} - 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} - $(x + 1)^n \geq 1 + nx$ for $x \geq -1$ and $n \in \mathbb{N}$ - - \vfill - - \problem{} - Use induction to prove \ref{bernoulli} - - \vfill - - \problem{} - 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{} - 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? + \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} diff --git a/Advanced/Euler's Number/parts/1 limits.tex b/Advanced/Euler's Number/parts/1 limits.tex new file mode 100644 index 0000000..030edba --- /dev/null +++ b/Advanced/Euler's Number/parts/1 limits.tex @@ -0,0 +1,122 @@ +\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{} +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 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 \ No newline at end of file diff --git a/Advanced/Euler's Number/parts/2 e.tex b/Advanced/Euler's Number/parts/2 e.tex new file mode 100644 index 0000000..1119d1d --- /dev/null +++ b/Advanced/Euler's Number/parts/2 e.tex @@ -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{} +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} +$(x + 1)^n \geq 1 + nx$ for $x \geq -1$ and $n \in \mathbb{N}$ + +\problem{} +Use induction to prove \ref{bernoulli} + +\vfill + + + +\problem{} +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 \ No newline at end of file diff --git a/Advanced/Euler's Number/parts/3 more e.tex b/Advanced/Euler's Number/parts/3 more e.tex new file mode 100644 index 0000000..c40e3fe --- /dev/null +++ b/Advanced/Euler's Number/parts/3 more e.tex @@ -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{} +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 \ No newline at end of file