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\title{Math 165: Honors Calculus I\\
Assignment 9 Solutions}

\begin{document}

\maketitle


\problem{1} p.45, \#13

a) $b^p - a^p = (b - a)\sum_{k=0}^{p-1} b^{p-k-1}a^k$.

\begin{proof}
\begin{eqnarray*}
(b - a)\sum_{k=0}^{p-1} b^{p-k-1}a^k
   &=& \sum_{k=0}^{p-1} b b^{p-k-1}a^k - a b^{p-k-1}a^k\\
   &=& \sum_{k=0}^{p-1} b^{p-k}a^k - b^{p-(k+1)}a^{k+1}\\
   &=& b^{p-0}a^0 - b^{p-(p-1+1)}a^{p-1+1}\quad \mbox{[telescoping sum]}\\
   &=& b^p - a^p\\
\end{eqnarray*}
\end{proof}

b)  $n^p < \frac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^p$

\begin{proof}
Use the formula from a) for the case $p+1$ and plug in $b = n+1$ and
$a = n$.  This gives
  \[
  (n+1)^{p+1} - n^{p+1}=(n+1-n)\sum_{k=0}^{p}(n+1)^{p-k}n^{k}
                       =\sum_{k=0}^{p}(n+1)^{p-k}n^{k}
  \]
Apply the inequality $n^k < (n+1)^k$ for $k=1,\ldots,p$ on the right
hand side to get
  \[
  (n+1)^{p+1} - n^{p+1} < \sum_{k=0}^{p}(n+1)^{p-k}(n+1)^{k} 
                = \sum_{k=0}^{p}(n+1)^p
  \]
The right hand side consists of $(p+1)$ copies of $(n+1)^p$ so we get
  \[
  (n+1)^{p+1} - n^{p+1} < (p+1) (n+1)^p
  \]

The other inequality is similar. Apply the inequality $(n+1)^k > n^k$
for $k=1,\ldots,p$ on the right hand side to get
  \begin{eqnarray*}
  (n+1)^{p+1} - n^{p+1}&=&\sum_{k=0}^{p}(n+1)^{p-k}n^{k}\\
                       &>&\sum_{k=0}^{p}n^{p-k}n^{k} = \sum_{k=0}^{p}n^{p}\\
                       &=&(p+1) n^p\\
  \end{eqnarray*}
\end{proof}

c) $\sum_{k=1}^{n-1} k^p < \frac{n^{p+1}}{p+1} < \sum_{k=1}^n k^p$ (*)

\begin{proof}
Use induction on $n$. The formula is true for $n = 1$:
  \[
  0 = \sum_{k=1}^0 k^p < \frac{1^{p+1}}{p+1} < \sum_{k=1}^1 k^p
  \]
We assume the inequalities (*) are true for $n$, and prove they hold for $n+1$.
The left hand side of (*) for $n+1$ looks like:
  \begin{eqnarray*}
  \sum_{k=1}^{n} k^p&=& \bigg(\sum_{k=1}^{n-1} k^p\bigg) + n^p\\
        &<& \frac{n^{p+1}}{p+1} + n^p \quad \mbox{[induction hypothesis]}\\
        &<& \frac{(n+1)^{p+1}}{p+1} \quad \mbox{[by part b)]}\\
  \end{eqnarray*}
so the left inequality is true for $n+1$. The right hand side of (*) for
$n+1$ is:
  \begin{eqnarray*}
  \sum_{k=1}^{n+1} k^p&=& \bigg(\sum_{k=1}^n k^p\bigg) + (n+1)^p\\
      &>& \frac{n^{p+1}}{p+1} + (n+1)^p \quad \mbox{[induction hypothesis]}\\
      &>& \frac{(n+1)^{p+1}}{p+1} \quad \mbox{[by part b)]}\\
  \end{eqnarray*}
\end{proof}

\problem{2} Let $A$ and $B$ be non-empty sets of real numbers bounded
above and below, respectively.

a) If $k < \sup A$ then there is a number $a\in A$ such that $k < a$.

\begin{proof} Suppose statement a) is false. Then there is some $k < \sup
A$ such that $k \ge a$ for all $a \in A$. Therefore, $k$ is an upper bound for
$A$. Since $\sup A$ is the \emph{least} upper bound for $A$, it must be
less than $k$: $\sup A \le k$.  But this contradicts $k < \sup
A$. Therefore statement a) must be true. \end{proof}

b) If $h > \inf B$ then there is a number $b \in B$ such that $b < h$.

\begin{proof} Suppose statement b) is false. Then there is some $h >
\inf B$ such that $b \ge h$ for all $b \in B$. Therefore, $h$ is a lower
bound for $B$. Since $\inf B$ is the \emph{greatest} lower bound for $B$,
it must be greater than $h$: $\inf B \ge h$. But this contradicts $h >
\inf B$. Therefore, statement b) must be true. \end{proof}

\problem{3} Let $A$ and $B$ be non-empty sets of real numbers with $A$
contained in $B$.

a) If $A$ and $B$ are bounded above, then $\sup A \le \sup B$.

\begin{proof} Let $k = \sup B$ and suppose $k < \sup A$.  By 2a), there is
a number $a \in A$ such that $k < a$. Since $k = \sup B \ge b$ for all $b
\in B$, the number $a$ cannot be in $B$. But this contradicts the assumption
that $B$ contains $A$.  Since assuming $k < \sup A$ leads to a contradiction,
it must be that $\sup A \le k$, i.e., $\sup A \le \sup B$. \end{proof}

b) If $A$ and $B$ are bounded below, then $\inf B \le \inf A$.

\begin{proof} Let $h = \inf B$ and suppose $h > \inf A$.  By 2b), there
is a number $a \in A$ such that $a < h$. Since $h = \inf B \le b$ for
all $b$ in $B$, the number $a$ cannot be in $B$. But this contradicts the
assumption that $B$ contains $A$.  Since assuming $h > \inf A$ leads to a
contradiction, it must be that $h \le \inf A$, i.e., $\inf B \le \inf A$.
\end{proof}

\problem{4} For $0 \le x \le 1$ define
  \[
  f(x) = \left\{\begin{array}{ll}
         1 & \mbox{if $x$ is rational}\\
         0 & \mbox{if $x$ is irrational}\\
         \end{array}\right.
  \]

a) If $s$ and $t$ are step functions satisfying $s(x) \le f(x) \le t(x)$
for all $0 \le x \le 1$, then $s(x) \le 0$ and $t(x) \ge 1$, except
possibly at partition points.

\begin{proof} Let $P = \{x_0, x_1, ..., x_n\}$ be a common refinement
of partitions for $s$ and $t$ so that $s$ and $t$ are constant on the
open subintervals of $P$: $s(x) = s_k$ and $t(x) = t_k$ for $x_{k-1} < x <
x_k$, $k = 1,\ldots, n$.  By assumption, $s_k \le f(x) \le t_k$ for
$x_{k-1} < x < x_k$. Let $q_k$ be a rational number and $r_k$ an irrational
number in the open interval $(x_{k-1}, x_k)$, for $k = 1,\ldots,n$. Then
$s_k \le f(r_k) = 0$ and $1 = f(q_k) \le t_k$ for $k = 1,...,n$. Therefore,
$s(x) \le 0$ and $t(x) \ge 1$, except possibly at partition points.
\end{proof}

b) The lower integral of $f$, $\underline{I}(f) = 0$, and the upper
integral of $f$, $\overline{I}(f) = 1$. Hence $f$ cannot be integrable
by Theorem~1.9.

\begin{proof} Let $s$ and $t$ be step functions as in part~a). Then
  \[
  \int_0^1 s(x)\, dx = \sum_{k=1}^n s_k(x_k - x_{k-1})
                   \le \sum_{k=1}^n 0 (x_k - x_{k-1}) = 0
  \]
and
  \[
  \int_0^1 t(x)\, dx = \sum_{k=1}^n t_k(x_k - x_{k-1)})
                   \ge \sum_{k=1}^n 1 (x_k - x_{k-1}) = 1.
  \]
We conclude that every number in the set
  \[
  S = \bigg\{ \int_0^1 s(x)\, dx\; \big|\; s \le f \bigg\}
  \]
is $\le 0$ and every number in the set
  \[
  T = \bigg\{ \int_0^1 t(x)\, dx\; \big|\; f \le t \bigg\}
  \]
is $\ge 1$. Therefore $\underline{I}(f) = \sup S \le 0$ and
$\overline{I}(f) = \inf T \ge 1$.  Notice also that the constant step
functions $s(x) = 0$ and $t(x) = 1$ satisfy $s(x) \le f(x) \le t(x)$ for
all $0 \le x \le 1$, so $0\in S$ and $1 \in T$. Therefore
$\underline{I}(f) = \sup S = 0$ and $\overline{I}(f) = \inf T = 1$.
\end{proof}

\end{document}
