%typos corrected 4/9/98 ARM
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\begin{document}

\suppressfloats[t]

\handout{notes-books}{Notes on Book Stacking}{March 16, 1998}

We consider a pile of unit-length books, stacked one on another and placed
on a table.  The center of mass of the whole stack must lie over the table
or it would pivot over the edge.  In fact, if the center of mass of the
stack on top of the $i$th book was not positioned on some vertical line
through the $i$th book, then the stack would pivot over the edge of that
book, as shown in Figure~1.  On the other hand, the stack will be
physically \emph{stable} providing that for every $i$, the center of mass
of the top $i-1$ books lies somewhere vertically above book $i$, and the
center of mass of the whole stack lies vertically over the table.

\begin{figure}[h]
\centerline{\psfig{file=figures/bookstack-1.ps,width=.70\textwidth}}
\caption{}
\end{figure}


Define the \emph{overhang} of a stack of books to be the distance from its
center of mass to the end of the book furthest to the right of the stack.
When the stack is placed so its center of mass is at the edge of the
table, which is as far to the right as the stack can be placed without
tipping, the overhang of the stack measures exactly how far the rightmost
book extends past the table edge, as shown in Figure~2.  So we want a
formula for the maximum possible overhang, $B_n$, achievable with a stack
of $n$~books.

\begin{figure}
\centerline{\psfig{file=figures/bookstack-2.ps,width=.4\textwidth}}
\caption{}
\end{figure}

For example, with one book, the center of mass is~1/2 a book length from
the righthand edge of the book, so the overhang is~1/2.  That is,
\[
    B_1 = \frac{1}{2}.
\]
A ``stack'' of one book is stable as long as the center of mass of the
book is over the table.  So by placing the center of mass at the edge of
the table, we can have the righthand edge stick out by the overhang, as
shown in Figure~3.

\begin{figure}
\centerline{\psfig{file=figures/bookstack-3.ps,width=.40\textwidth}}
\caption{}
\end{figure}

Now suppose we have a stack of $n+1$ books with maximum overhang.  If the
overhang of the stack of $n$ books on top of the the bottom book was not
maximum, we could replace it with a stack of $n$ books with larger
overhang, positioning the new stack so its center of mass was at the same
position as the original stack.  In other words, the maximum overhang,
$B_{n+1}$, of the stack of $n+1$ books is obtained by placing a stable
stack of $n$ books with maximum overhang, $B_n$, somewhere on top of the
bottom book.

Suppose we place the stable stack of $n$ books on the $(n+1)$st bottom
book so that the stack's center of mass is distance~$x$ to the right of
the left edge of the bottom book, as shown in Figure~4.  The center of
mass of the top $n$ books must lie above the bottom book for the stack of
all $n+1$ books to be stable, so we must have $0 \le x \le 1$.

\begin{figure}
\centerline{\psfig{file=figures/bookstack-4.ps,width=.4\textwidth}}
\caption{}
\end{figure}

Now the center of mass of the stack of $n+1$ books is distance $(1\cdot
1/2 + nx)/(n+1)$ to the right of the left edge of the bottom book, so the
overhang of the stack of $n+1$ books has increased by $x - (1\cdot 1/2 +
nx)/(n+1) = (2x-1)/2(n+1)$.  This increase is maximized when $x$ takes its
maximum value 1.  So the maximum increase in overhang is $1/2(n+1)$, and
occurs when the center of mass of the top $n$ books is placed exactly at
the righthand edge of the bottom book, as shown in Figure~5.

\begin{figure}
\centerline{\psfig{file=figures/bookstack-5.ps,width=.80\textwidth}}
\caption{}
\end{figure}

So,
\[
    B_{n+1} = B_n + \frac{1}{2(n+1)}
\]
and we have
\begin{align*}
    B_{n+1} & = \frac{1}{2(n+1)} + \frac{1}{2n} + \frac{1}{2(n-1)} + \cdots +
 \frac{1}{2\cdot 2} + 
 \frac{1}{2 \cdot 1}\\
       & = \frac{1}{2}\sum_{i=0}^{n} \frac{1}{(n+1) -i}\\
       & = \frac{1}{2}\sum_{i=1}^{n+1} \frac{1}{i}.
\end{align*}

Let $H_n$ be the value of $\sum_{i=1}^n (1/i)$, so $B_n = H_n/2$.
$H_n$~is called the $n$th \emph{Harmonic number}.  In lecture we used the
integral method to conclude that $H_n \sim \ln n$.  In particular, $H_n$
grows slowly, but unboundedly with~$n$.  So we can get books to overhang
any distance past the edge of the table by piling them high enough!

\end{document}
