146 lines
2.8 KiB
TeX
146 lines
2.8 KiB
TeX
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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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