\documentclass{amsart}

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\title{Department of Mathematics}
\author{Solutions for the Auman Prize\\
2000}

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\begin{document}

\maketitle

\bigskip

\begin{enumerate}
\item
Let $f(x)$ be differentiable at $x=a$, and $f(a)\ne0$.
\begin{eqnarray*}
 \lim_{n\to\infty}\left[\frac{f(a+1/n)}{f(a)}\right]^n
 &=& \lim_{x\to0^+}\left[\frac{f(a+x)}{f(a)}\right]^{1/x}\\
 &=& \lim_{x\to0^+}\exp\big([\log f(a+x) - \log f(a)]/x\big)\\
 &=& \exp\big(\lim_{x\to0^+}[\log f(a+x) - \log f(a)]/x\big)\\
 &=& \exp\big(\lim_{x\to0^+}[f'(a+x)/f(a+x)\big)
      \quad\textsc{[L'H\^opital]}\\
 &=& \exp\big(f'(a)/f(a)\big)
\end{eqnarray*}

\bigskip
\item
Prove that there is no continuous function $f$ with domain the closed
interval $[0,1]$ and range the open interval $(0,1)$.

\bigskip
\emph{Proof.}
Assume $f$ is continuous on $[0,1]$.  By the \textsc{Extreme Value
Theorem} there exist $a,b\in[0,1]$ such that $f(a) \le f(x) \le f(b)$ for
all $x\in[0,1]$. By the \textsc{Intermediate Value Theorem}, the range
of $f$ must include all numbers between $f(a)$ and $f(b)$ and therefore
equals the closed interval $[f(a),f(b)]$.  No closed interval can
equal an open interval. If $[f(a),f(b)] = (0,1)$, for example,
we would have $0<f(a)<1$. But then $0<f(a)/2<f(a)$, so $f(a)/2\in(0,1)$
but $f(a)/2\notin[f(a),f(b)]$, a contradiction.  Therefore, the range
of $f$ cannot be $(0,1)$.

\bigskip
\item
\begin{eqnarray*}
\lim_{x\to\infty}x\int_0^x e^{t^2-x^2}dt
 &=& \lim_{x\to\infty}xe^{-x^2}\int_0^x e^{t^2}dt\\
 &=& \lim_{x\to\infty}\frac{\int_0^x e^{t^2}dt}{e^{x^2}/x}\\
 &=& \lim_{x\to\infty}\frac{e^{x^2}}{2e^{x^2}-e^{x^2}/x^2}
      \quad\textsc{[L'H\^opital]}\\
 &=& \lim_{x\to\infty}\frac{1}{2-1/x^2} = \frac{1}{2}
\end{eqnarray*}

\bigskip
\item
Determine how far a stack of $n$ identical books can be made to
project over the edge of a table by stacking them one on top of the
other. Determine the theoretical limit of this overhang as $n\to\infty$.

\bigskip
\emph{Solution.}
Let's take the length of the book to be one unit and label the books $b_1$
through $b_n$ from the top down.  In order to stack the books so that
each book extends as far as possible from the edge of the book below it
without falling, we must arrange for the center of gravity, $c_k$, of books
$b_1$ through $b_{k-1}$ together to lie at the right edge of book $b_k$.
Calculating distances from the left edge of $b_k$, we see that $c_k$
is the weighted average of the center of gravity of $b_k$---a distance of
1/2 and weight 1---and the center of gravity of $b_1$ through $b_{k-1}$
together---a distance of 1 (the right edge of $b_k$) and weight $k-1$:
  \[ c_k = \frac{1\cdot\frac{1}{2} + (k-1)\cdot 1}{k} = 1 - \frac{1}{2k} \]
This means that the distance $b_k$ projects from the book below it is
$1-c_k = 1/(2k)$. Therefore, the total distance the books project over
the edge of the table together is:
  \[ \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n} = \frac{1}{2}\big(1
  + \frac{1}{2} + \cdots + \frac{1}{n}\big) \approx \frac{1}{2}\log(n) \]
In fact, comparing areas of rectangles above and below the graph of $1/x$,
we find that
  \[ \log(n+1) < 1 + \frac{1}{2} + \cdots + \frac{1}{n}< \log(n) + 1 \]
Since $\log(n)\to\infty$ as $n\to\infty$, there is no \emph{theoretical}
limit to the overhang, although there are obvious practical limitations.

\bigskip
\item Show if there are $n\ge2$ people in a room, then two of them know
the same number of people among those present. (Assume that if $A$ knows
$B$, then $B$ knows $A$).

\bigskip
\emph{Solution.}
If the statement is false there must be a group of $n$ people, each
of whom knows a different number of people in the group.  Since any
list of $n$ distinct integers, each between 0 and $n-1$, must contain
all of the numbers from 0 through $n-1$ exactly once, it follows that
there must be one person in the group who knows nobody, and one person
who knows everybody. But then the person who knows nobody would know the
person who knows everybody, a contradiction.  Therefore, no such group
of people exists.

\end{enumerate}

\end{document}
