Euler edits

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