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\title{Math 165: Honors Calculus I\\
Exam I Solutions}
\date{September 30, 1999}

\begin{document}

\maketitle

\bigskip

\begin{enumerate}

\item Let $a\in\R$ be an arbitrary real number.

  \begin{enumerate}[a)]

  \item Define $-a$.

\emph{Definition.} $-a$ is the unique number $x$ such that $a + x = 0$. It
exists by Axiom~5.  It is unique because if $x'$ is another such number
then $a + x = 0 = a + x'$ and adding $x$ to both sides gives $(x+a)
+ x = (x+a) + x'$, which simplifies to $0 + x = 0 + x'$ and $x = x'$
using Axioms~1, 2 and~4.

  \item Prove that $0\cdot a = 0$.

\emph{Proof.}
By Axiom~4, $0 + 0 = 0$, multiplying by~$a$ gives
$(0 + 0)a= 0\cdot a$, and by Axiom~3,
$0\cdot a + 0\cdot a = 0\cdot a$. Now adding $-(0\cdot a)$ to both sides
gives $-(0\cdot a) + (0\cdot a + 0\cdot a) = -(0\cdot a) + 0\cdot a$ which
simplifies to $0 + 0\cdot a = 0$ and $0\cdot a = 0$ using Axioms~1, 2
and~4, and the definition of $-(0\cdot a)$.

  \item Prove that $(-1)\cdot a = -a$.

\medskip
\emph{Proof.}
By definition of $-1$, $1 + (-1) = 0$. Multiplying by $a$ gives $(1 +
(-1))a = 0\cdot a$, and by Axiom~3, $1\cdot a + (-1)\cdot a = 0\cdot
a$. Using part~b) and Axiom~4, this simplifies to $a + (-1)\cdot a = 0$.
Since $-a$ is the unique number $x$ satisfying $a + x = 0$ (see part a)),
we conclude that $(-1)\cdot a = -a$.

  \end{enumerate}

\problemspace
\item \begin{enumerate}[a)]
  \item Define the positive integers,~$\N$.

\medskip
\emph{Definition.}
$n \in \N$ if and only if $n$ is in every inductive set. A set $S$
is inductive if $1\in S$, and $x+1\in S$ whenever $x\in S$.

\medskip
  \item Prove that there is no upper bound for~$\N$.

\medskip
\emph{Proof.}
If $\N$ has an upper bound, then it has a least upper bound $b = \sup(\N)$ by
Axiom~10. Since $b-1$ cannot be an upper bound of $\N$, there must be some
$n\in \N$ such that $b-1 < n$. But then $n+1\in\N$ and $b < n+1$,
contradicting the fact that $b$ is an upper bound of~$\N$. Therefore, there
can be no upper bound of~$\N$.

\medskip
  \item Let $a > 0$ and let $T = \Big\{a\Big(1 +
  \frac{1}{n}\Big)\,|\,n\in\N\Big\}$.  Prove that $ a = \inf(T)$.

\medskip
\emph{Proof.}
Since $a/n>0$ for all $n\in\N$, $a < a+a/n$. Thus, $a$ is a lower bound
of~$T$.  Suppose $b$ is another lower bound of $T$ and $b > a$. Then $a
< b \le a+a/n$ for all $n\in\N$. This implies that $0 < b-a \le a/n$
and $n \le a/(b-a)$ for all $n\in\N$. But then $a/(b-a)$ is an upper
bound for $\N$ contradicting part b). Therefore, $b \le a$ and $a$
must be the greatest lower bound of~$T$, $a = \inf(T)$.

\end{enumerate}

\problemspace
\item 
  \begin{enumerate}[a)]
  \item Define $\sum_{k=1}^n a_k$.

\medskip
\emph{Definition.}
$\sum_{k=1}^1 a_k$ is defined to be $a_1$ and if $\sum_{k=1}^n a_k$ has
been defined for a positive integer $n$, then  $\sum_{k=1}^{n+1} a_k$ is
defined to be $\Big(\sum_{k=1}^n a_k\Big) + a_{n+1}$.
\medskip

  \item Use induction to prove that for any positive integer~$n$,
  $\sum_{k=1}^n (k-1)^3 < \frac{n^4}{4}$.

\medskip
\emph{Proof.}
  \begin{enumerate}[i)]

  \item The inequality holds for $n=1$ since $\sum_{k=1}^1 (k-1)^3 =
  (1-1)^3 = 0 < \frac{1^4}{4}$.

  \item Assume the inequality holds for $n$. Then
  \begin{eqnarray*}
  \sum_{k=1}^{n+1} (k-1)^3 &=& \Big(\sum_{k=1}^n (k-1)^3\Big) + n^3 
    < \frac{n^4}{4} + n^3 = \frac{1}{4}(n^4 + 4n^3) \\
    &<& \frac{1}{4}(n^4 + 4n^3 + 6n^2 + 4n + 1) = \frac{(n+1)^4}{4}
  \end{eqnarray*}

  \end{enumerate}

  \item  Use telescoping sums to calculate $\sum_{k=1}^n \frac{2k-1}{2^kk!}$

\medskip
\emph{Solution.}
  \begin{eqnarray*}
  \sum_{k=1}^n \frac{2k-1}{2^kk!}
  &=& \sum_{k=1}^n \frac{2k}{2^kk!} - \frac{1}{2^k k!}
  = \sum_{k=1}^n \frac{1}{2^{k-1}(k-1)!} - \frac{1}{2^k k!} \\
  &=& \frac{1}{2^0 0!} - \frac{1}{2^n n!} = 1 - \frac{1}{2^n n!} \\
  \end{eqnarray*}

  \end{enumerate}

\problemspace
\item 
Find the coefficient of $x^{81}$ in $(x^3 - 7)^{30}$
(give its prime factorization).

\medskip
\emph{Solution.}
By the binomial theorem, the terms in the expansion of  $(x^3 - 7)^{30}$
have the form $\choose{30}{k} (x^3)^k (-7)^{30-k}$. The term containing
$x^{81}$ corresponds to~$k=27$. Therefore, the coefficient of $x^{81}$ is
\[ \choose{30}{27} (-7)^3 = \frac{30\cdot29\cdot28}{3\cdot2\cdot1} (-7)^3 
= - 4\cdot5\cdot7^4\cdot29 \]

\problemspace
\item Give precise definitions of the following.
  \begin{enumerate}[a)]
  \item A step function $s$ on $[a,b]$.

\medskip
\emph{Definition.}
A function $s(x)$ is a step function on $[a,b]$ if there is a partition
$P = \{x_0, x_1, \dots, x_n\}$ of $[a,b]$ such that $s(x)$ is constant
on the open subintervals of $P$: $s(x) = s_k$ for $x\in (x_{k-1},x_k)$,
$k=1,\dots,n$.

  \item $\int_a^b s(x)\,dx$ where $s$ is a step function on~$[a,b]$.

\medskip
\emph{Definition.}
Using the notation of part a), $\int_a^b s(x)\,dx$ is defined to be
$\sum_{k=1}^n s_k(x_k - x_{k-1})$.

  \item The lower integral, $\underline{I}(f)$, and upper integral,
  $\overline{I}(f)$, of a bounded function $f$ on $[a,b]$.

\medskip
\emph{Definition.}
Let $S$ be the set of numbers $\int_a^b s(x)\,dx$ where $s$ runs through
all the step functions on $[a,b]$ below $f$ and let $T$ be the set of
numbers $\int_a^b t(x)\,dx$ where $t$ runs through all the step functions
on $[a,b]$ above~$f$.  Since $f$ is bounded on $[a,b]$, the sets $S$
and $T$ are non-empty and bounded from above and below, respectively.
By Axiom~10, the supremum of $S$ and the infimum of $T$ exist.  The lower
integral of $f$ is defined to be $\underline{I}(f) = \sup(S)$, and the
upper integral of $f$ is defined to be $\underline{I}(f) = \inf(T)$.

\medskip

  \item $\int_a^b f(x)\,dx$ where $f$ is a bounded function on $[a,b]$.

\medskip
\emph{Definition.}
If there is \emph{exactly one} number $I$ satisfying $\int_a^b s(x)\,dx
\le I \le \int_a^b t(x)\,dx$ for all step functions $s$, $t$ such that
$s(x) \le f(x) \le t(x)$ for $x\in[a,b]$, then $\int_a^b f(x)\,dx$
is defined to be this number~$I$.  Alternately, if $\underline{I}(f) =
\overline{I}(f) = I$, where $\underline{I}(f)$ and $\overline{I}(f)$
are defined as in part c), then $\int_a^b f(x)\,dx = I$.

  \end{enumerate}

\problemspace
\item Give precise statements of the following theorems for integrals.
  \begin{enumerate}[a)]
  \item \textsc{Linearity with Respect to the Integrand}.

\medskip
\emph{Theorem.}
If $f$ and $g$ are integrable on $[a,b]$, then for any constants
$c_1$, $c_2\in\R$, the function $c_1 f + c_2 g$ is also integrable on
$[a,b]$ and
  \[
  \int_a^b c_1f(x)+c_2g(x)\,dx = c_1\int_a^b f(x)\,dx + c_2\int_a^b g(x)\,dx
  \]

  \item \textsc{Additivity with Respect to the Interval of
  Integration}.

\medskip
\emph{Theorem.}
Let $a \le c \le b$. Then $f$ is integrable on $[a,b]$ if and only if $f$
is integrable on $[a,c]$ and on $[c,b]$. Moreover,
  \[ \int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx \]

  \item \textsc{Comparison Theorem}

\medskip
\emph{Theorem.}
If $f$ and $g$ are integrable on $[a,b]$ and $f(x) \le g(x)$ for all
$x\in[a,b]$, then \[ \int_a^b f(x)\,dx \le \int_a^b g(x)\,dx \]

  \end{enumerate}

\problemspace
  \item  Evaluate $\int_{-2}^2 3 - |x^2-1| \,dx$.

\medskip
\emph{Solution.}
$x^2-1 \ge 0$ if and only if $x\le -1$ or $x \ge 1$. Using additivity with
respect to the interval, the definition of absolute value, and the formula
$\int_a^b x^n\,dx = (b^{n+1}-a^{n+1})/(n+1)$, we get
\begin{eqnarray*}
  \int_{-2}^2 3 - |x^2-1| \,dx
    &=& \int_{-2}^{-1} 3 - (x^2-1) \,dx + 
        \int_{-1}^1 3 + (x^2-1) \,dx  + \int_{1}^2 3 - (x^2-1) \,dx \\
    &=& (4 - \frac{1}{3}7) + (4 + \frac{1}{3}2) + (4 - \frac{1}{3}7) = 8
\end{eqnarray*}

\problemspace
\item Let $b > 0$. Show that $\int_0^b x\,dx = \frac{1}{2}b^2$ by
considering step functions constant on subintervals of equal length.

\medskip
\emph{Solution.}
The partition points for $n$ subintervals of equal length are given by
$x_k = k\frac{b}{n}$, $k = 0,\dots,n$. Define step functions by
\[ \begin{array}{rclcll}
 s(x) &=& s_k &=& (k-1)\frac{b}{n}, &\qquad x_{k-1}\le x < x_k\\
 t(x) &=& t_k  &=& k\frac{b}{n}, &\qquad x_{k-1} < x \le x_k
\end{array} \]
for $k=1,\dots,n$ (also define $s(b) = b$ and $t(0) = 0$). Since $f(x) = x$
is increasing on $[0,b]$, $s(x) \le f(x) \le t(x)$ for all $x\in[0,b]$.

The integrals of these step functions can be easily  calculated since
$(x_k - x_{k-1}) = \frac{b}{n}$:
 \[
 \int_0^b s(x)\,dx = \sum_{k=1}^n s_k(x_k - x_{k-1}) 
 = \sum_{k=1}^n (k-1)\frac{b}{n}\Big(\frac{b}{n}\Big)
 = \frac{b^2}{n^2}\sum_{k=1}^n (k-1)
 = \frac{b^2}{n^2}\frac{n(n-1)}{2}
 = \frac{1}{2}b^2\Big(1 - \frac{1}{n}\Big)
 \]
Similarly,
 \[
 \int_0^b t(x)\,dx = \sum_{k=1}^n t_k(x_k - x_{k-1}) 
 = \sum_{k=1}^n k\frac{b}{n}\Big(\frac{b}{n}\Big)
 = \frac{b^2}{n^2}\sum_{k=1}^n k
 = \frac{b^2}{n^2}\frac{n(n+1)}{2}
 = \frac{1}{2}b^2\Big(1 + \frac{1}{n}\Big)
 \]
There is only one number $I$ that satisfies
\[ \frac{1}{2}b^2\Big(1 - \frac{1}{n}\Big) \le I \le
   \frac{1}{2}b^2\Big(1 + \frac{1}{n}\Big) \]
for all $n\in \N$, namely, $I = \frac{1}{2}b^2$ (see for example,
Problem~2c). We conclude that $\int_0^1 x\,dx = \frac{1}{2}b^2$.

\end{enumerate}

\end{document}
