\documentclass{amsart}

\usepackage{graphics}
\setlength{\parindent}{0pt}
\everymath{\displaystyle}
\setlength{\textheight}{9in}
\setlength{\voffset}{-0.5in}

\newcommand{\N}{\ensuremath{\mathbb{N}}}

\title{Math 165: Honors Calculus I\\
Assignment 10 Solutions}

\newcommand{\problem}[1]{\bigskip\textbf{Problem #1}\ }


\begin{document}

\maketitle

\problem{1}
Let $A$ and $B$ be subsets of real numbers each of which has an
infimum.  Let $C = \{ a + b \, | \, a\in A,\ b \in B \}$. Prove that
$C$ has an infimum and $\inf C = \inf A + \inf B$. (The proof of the
corresponding theorem for supremum, Theorem~I.33~(a), is given on p.27.)

\begin{proof}
Any $c\in C$ has the form $c = a + b$ where $a \in A$ and $b \in B$.
Since $a \ge \inf A$ and $b \ge \inf B$, we get $c = a + b \ge \inf A +
\inf B$, which shows that $\inf A + \inf B$ is a lower bound for~$C$. By
Theorem~I.27, $C$ has an infimum, $\inf C$, which is greater than or equal
to any lower bound of $C$, so $\inf C \ge \inf A + \inf B$.  If $\inf
C$ is strictly greater than $\inf A + \inf B$ then $\inf C = \inf A +
\inf B + h$ for some positive number $h > 0$. By Assignment~9, 2b)
(see also Theorem~I.32), there is an $x \in A$ and a $y\in B$ such that
$x < \inf A + h/2$ and $y < \inf B + h/2$. Since $x+y \in C$,
   \[
   \inf C \le x + y < (\inf A + h/2) + (\inf B + h/2)
   = \inf A + \inf B + h = \inf C
   \]
But $\inf C < \inf C$ is a contradiction! Therefore, $\inf C = \inf A +
\inf B$.
\end{proof}

\problem{2} Define
  \[
  f(x) =\left\{\begin{array}{ll}
        0 &\mbox{if $x = 1/n$ for some $n\in \N$}\\
        1 &\mbox{otherwise}\\
        \end{array}\right.
  \]
Prove that $f(x)$ is integrable on $[0,1]$.

\begin{proof} Fix $n\in \N$ and define a step function $s_n$ by
  \[ s_n(x) = \left\{\begin{array}{ll}
           0    & \mbox{if $0 \le x \le 1/n$}\\
           f(x) & \mbox{if $1/n < x \le 1$}
           \end{array}\right.
  \]
  \[
  \resizebox{1.5in}{!}{\includegraphics{assign10.eps}}
  \]
A partition for $s_n$ is given by the points  $x_0=0$, $x_1=1/n$,
$x_2=1/(n-1)$, \ldots, $x_{n-1} = 1/2$, $x_n = 1$.  Define a second step
function by $t(x) = 1$ for $0 \le x \le 1$. Then
  \[ s_n(x) \le f(x) \le t(x)\ \mbox{for all $0\le x \le1$} \]

The integrals of these steps functions are easy to calculate:
  \begin{eqnarray*}
  \int_0^1 s_n &=& \sum_{k=1}^n s_k(x_k - x_{k-1})
                =  0(x_1 - x_0) + \sum_{k=2}^n 1(x_k - x_{k-1})\\
               &=& x_n - x_1 = 1 - \frac{1}{n}\\
  \int_0^1 t &=& 1
  \end{eqnarray*}
Since $\int_0^1 s \le \underline{I}(f) \le \overline{I}(f) \le \int_0^1 t$
for any step functions satisfying $s(x) \le f(x) \le t(x)$ on $[0,1]$,
we see that
  \[ 1 - \frac{1}{n} \le \underline{I}(f) \le \overline{I}(f) \le 1 \]
and these inequalities must hold for any $n\in\N$. Since $I = 1$ is the
only number that satisfies
  \[ 1 - \frac{1}{n} \le I \le  1 \quad \mbox{for all $n \in \N$} \]
we conclude that $\underline{I}(f) = \overline{I}(f) = 1$. Therefore $f$
is integrable and $\int_0^1 f = 1$.
\end{proof}

\problem{3} Use Theorem 1.14 to find an approximation to the integral $\int_1^2
\frac{1}{x}\,dx$ so that the error is $< .05$.

\bigskip
\emph{Solution.}
The function $f(x) = \frac{1}{x}$ is decreasing on $[1,2]$. The partition
points for $n$ subintervals of equal length are given by $x_k = 1 +
k\frac{1}{n}$, for $k = 1,\ldots,n$.
Let
  \[
  \sigma_n = \frac{1}{n}\sum_{k=1}^nf(x_k)
      = \frac{1}{n}\sum_{k=1}^n\frac{1}{1+k\frac{1}{n}} 
      = \sum_{k=1}^n\frac{1}{n+k}
  \]
By Theorem~1.14,
  \[
  \sigma_n \le \int_1^2 \frac{1}{x}\, dx \le \sigma_n + \varepsilon_n
  \]
where
  \[
   \varepsilon_n = \frac{(2-1)(f(1) - f(2))}{n} =  \frac{1}{2n}
  \]
If we take our approximation to be the midpoint of the interval
$[\sigma_n, \sigma_n + \varepsilon_n]$, namely $\alpha_n = \sigma_n +
\varepsilon_n/2$, then the true value of the integral will lie with
$\pm \varepsilon_n/2$ of $\alpha_n$. We would like this error to be less
than $0.05$:
  \[
  \frac{1}{2}\varepsilon_n < 0.05 \Longleftrightarrow \frac{1}{4n}
  < 0.05 \Longleftrightarrow n > \frac{1}{4\cdot0.05} =  5
  \]
Therefore, the desired approximation is
  \begin{eqnarray*}
  \alpha_6 &=& \sigma_6 + \varepsilon_6/2\\
           &=& \bigg(\frac{1}{6+1} + \frac{1}{6+2} + \frac{1}{6+3} +
               \frac{1}{6+4} + \frac{1}{6+5} + \frac{1}{6+6}\bigg) +
               \frac{1}{2}\Big(\frac{1}{2\cdot6}\Big)\\
           &=& 0.694877
  \end{eqnarray*}
The actual value of this integral is $\log(2) = 0.693147$ and the
error of our approximation is $0.694877 -  0.693147 = 0.00173 < 0.05$.
\end{document}
