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\title[Archimedes' Quadrature of the Parabola]{Archimedes' Quadrature of the Parabola}

\author
{John B. Little}

\institute[College of the Holy Cross]
{
  Department of Mathematics and Computer Science\\
  College of the Holy Cross\\
}

\date[June 12, 2013]
{June 12, 2013}

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

\begin{frame}
{\bf Outline}

\tableofcontents

\end{frame}

\section{Introduction}

\begin{frame}
{\bf Who was Archimedes?}

\begin{itemize}
\item<1-> Lived ca. 287 - 212 BCE, mostly in Greek city of Syracuse in Sicily
\item<2-> Studied many topics in what we would call mathematics, physics, engineering
(less distinction between them at the time)
\item<3-> We don't know much about his actual life; much of 
his later reputation was based on somewhat dubious anecdotes, e.g. 
the ``eureka moment,'' inventions he was said to have
produced to aid in defense of Syracuse during 
Roman siege in which he was killed, etc. 
\item<4-> Perhaps most telling: we do know he designed a tombstone for himself
illustrating the discovery he wanted most 
to be remembered for (discussed by Plutarch, Cicero)
\end{itemize}
\end{frame}

\begin{frame}
\begin{figure}
	\includegraphics[scale = .5]{ArchimedesTombstone.jpg}
	\caption{Sphere inscribed in cylinder of equal radius}
\end{figure}
$3 V_{sphere} = 2 V_{cyl}$ and $A_{sphere} = A_{cyl}$ (lateral area)
\end{frame}

\begin{frame}
{\bf Surviving works}

\begin{itemize}
\item<1-> {\it On the Equilibrium of Planes} (2 books)
\item<2-> {\it On Floating Bodies} (2 books)
\item<3-> {\it Measurement of a Circle}
\item<4-> {\it On Conoids and Spheroids} 
\item<5-> {\it On Spirals}
\item<6-> {\it On the Sphere and Cylinder} (2 books)
\end{itemize}

\end{frame}

\begin{frame}
{\bf Surviving works, cont.}

\begin{itemize}
\item<1->  {\it Book of Lemmas}, {\it The Sand-Reckoner}, {\it The Cattle-Problem}, 
the {\it Stomachion} 
\item<2->  {\it Quadrature of the Parabola}
\item<3->  {\it The Method of Mechanical Theorems}
\item<4->  {\it The Method} was long known only from references in other works
\item<5->  1906 -- a {\it palimpsest} prayerbook created about 1229 CE
(a reused manuscript) was found by to contain substantial portions 
(a 10th century CE copy from older sources)
\end{itemize}
\end{frame}

\begin{frame}
{\bf A page from the palimpsest}
\begin{figure}
	\includegraphics[scale = .4]{ArchPal.jpg}
\end{figure}
\end{frame}

\begin{frame}
{\bf Scientific context}

\begin{itemize}
\item<1-> Archimedes flourished at height of Hellenistic period; Greek
kingdoms (fragments of Alexander's empire) in Egypt, Syria, Syracuse 
strong patrons of science and mathematics
\item<2-> State of the art in mathematics: 
2nd or 3rd generation {\it after} Euclid, rough contemporary of 
Apollonius (best known for work on conic sections)
\item<3-> Part of an active scientific community in fairly close communication; 
for instance several of the works cited above start with letters to colleagues
\item<4-> Example:  Letter to Eratosthenes (in Alexandria) at start
of {\it The Method} 
\end{itemize}
\end{frame}

\begin{frame}
{\bf Goals for this talk}

\begin{itemize}
\item<1-> Discuss a problem treated by Archimedes in {\it The Method}
and (later?) in {\it Quadrature of the Parabola} and the two different
solutions Archimedes gave
\item<2-> Think about the relation of the methods Archimedes developed
and later innovations in mathematics such as limits, integral
calculus, etc.
\item<3-> Get a glimpse of the way Archimedes thought about what
he was doing
\end{itemize}
\end{frame}

\section{The problem and Archimedes' discovery}

\begin{frame}
{\bf What is the area of a parabolic segment?}
\begin{figure}
	\includegraphics<1>[scale=.27]{para1.png}
        \includegraphics<2->[scale=.27]{para2.png}
\end{figure}
\begin{itemize}
\item<3->Let $C$ have same $x$-coordinate as the midpoint of $AB$
\item<4>The area of the parabolic segment  $= \frac{4}{3} {\rm area}(\Delta ABC)$
\end{itemize}
\end{frame}

\begin{frame}
{\it ``Quadrature of the Parabola''}

\begin{itemize}
\item<1-> In the work in the title of this slide, Archimedes gives
two different arguments for this statement,
\item<2-> One using the ``method of exhaustion'' -- part of standard repertoire
of Euclidean mathematics (see Book 12 of {\it Elements}), based on work of Eudoxus
\item<3-> A recurring theme in his mathematical work (used it as a standard method to obtain areas and volumes, approximate the value of $\pi$, etc.)
\item<4-> One by the novel method also described in 
{\it The Method}
\item<5-> We will look at both of these using modern notation -- both quite interesting!
\end{itemize}
\end{frame}


\begin{frame}
{\bf What's special about this $C$?}

\begin{itemize}
\item<1-> Can show easily with calculus (and less so using the geometric definition of a parabola):  The tangent to the parabola at $C$ is {\it parallel} to $AB$ 
\item<2-> So for this $C$, and this $C$ only, the heights of the triangle
and the parabolic segment are equal
\item<3-> The point $C$ is called the \emph{vertex} of the segment
\end{itemize}
\end{frame}

\begin{frame}
{\bf Sketch of Archimedes' first proof}

\begin{itemize}
\item<1-> Now have two smaller parabolic segments; let their vertices 
be $C_1$ and $C_2$ and construct triangles $\Delta AC_1C$ and $\Delta CC_2B$ 
\end{itemize}
\begin{figure}
	\includegraphics<2>[scale=.4]{para3.png}
\end{figure}
\end{frame}

\begin{frame}
{\bf The first proof, continued}

\begin{itemize}
\item<1-> Archimedes shows that each of those triangles has area $\frac{1}{8}$ the 
area of $\Delta ABC$, so 
$${\rm area}(\Delta AC_1C) + {\rm area}(\Delta CC_2B) = 
\frac{1}{4} {\rm area}(\Delta ABC).$$
\item<2-> We do this construction some finite number $n$ of times and 
add together the areas of all the triangles,
\end{itemize}
\end{frame}

\begin{frame}
{\bf First proof, continued}

\begin{itemize}
\item<1-> letting $A_0$ be the original area, $A_1 = \frac{1}{4} A_0$ 
the area of the first two smaller triangles, 
then $A_2 = \frac{1}{4} A_1 = \frac{1}{4^2} A_0$ 
the area of the next four, etc. 
\item<2-> we have total area
$$S_n = A_0 + \frac{1}{4} A_0 + \frac{1}{4^2} A_0 + \cdots + \frac{1}{4^n} A_0$$
\item<3-> Moreover, $\frac{4}{3} A_0 - S_n = \frac{1}{3} A_n$, so as $n$ increases
without bound, $S_n$ tends to $\frac{4}{3} A_0$
\end{itemize}
\end{frame}

\begin{frame}
{\bf First proof, concluded}

\begin{itemize}
\item<1->  On the other hand, as $n$ increases without bound, 
the triangles constructed above {\it fill out} all of the area
in the parabolic segment, so
\item<2->  Archimedes concludes that the area of the parabolic segment  
$= \frac{4}{3} A_0 = \frac{4}{3} {\rm area}(\Delta ABC)$ as claimed before.
\item<3-> We would derive all this, of course, using the usual properties
of finite and infinite \emph{geometric series}:
$$\text{area of segment} = \sum_{n=0}^\infty \frac{A_0}{4^n} = \frac{A_0}{1 - \frac{1}{4}} = \frac{4A_0}{3}.$$  Archimedes had to do it
all in an \emph{ad hoc} way because that general theory did not exist yet!
\end{itemize}
\end{frame}

\begin{frame}
{\bf Archimedes' second proof}

\begin{itemize}
\item<1-> Now, the remarkable second proof that Archimedes presents both
in {\it Quadrature of the Parabola} and in {\it The Method}
\item<2-> We'll follow the presentation in {\it The Method} (in Heath's translation)
-- this seems (to me) to be a less polished form of the argument and 
even more instructive about what Archimedes was getting at  
\item<3-> Key Idea:  Apply a physical analogy -- will compute the area of the parabolic
segment essentially by considering it as a thin plate, or ``lamina,'' of 
constant density and ``weighing it on a balance'' in an extremely clever way
\end{itemize}
\end{frame}

\begin{frame}
{\bf The construction}

\begin{columns}
\column{.5 \textwidth}
\begin{figure}
\includegraphics<1>[scale=.33]{balance.png}
\end{figure}

\column{.5 \textwidth}
\begin{itemize}
\item \emph{Note:} $\Delta$ in figure $= D$ in text (a Maple issue!)
\item $CF$ is tangent to the parabola at $C$
\item $B$ is the vertex 
\item $O$ is an arbitrary point along the line from $A$ to $C$
\item $ED$, $MO$, $FA$ are parallel to the axis of the parabola
\end{itemize}
\end{columns}
\end{frame}

\begin{frame}
{\bf First observation}

\begin{columns}
\column{.5 \textwidth}
\begin{figure}
\includegraphics<1>[scale=.33]{balance.png}
\end{figure}

\column{.5 \textwidth}
\begin{itemize}
\item By construction of the vertex $B$, 
$\text{area}(\Delta ADB) = \text{area}(\Delta CDB)$
\item ``Well-known'' property of parabolas (at least in Archimedes' time!):  $BE = BD$, 
$NM = NO$, $KA = KF$
\item $\therefore \text{area}(\Delta EDC) = \text{area}(\Delta ABC)$ and by 
similar triangles,
$$\text{area}(\Delta AFC) = 4\cdot\text{area}(ABC) $$
\end{itemize}
\end{columns}
\end{frame}


\begin{frame}
{\bf Next steps}

\begin{columns}
\column{.5 \textwidth}
\begin{figure}
\includegraphics<1>[scale=.33]{balance.png}
\end{figure}

\column{.5 \textwidth}
\begin{itemize}
\item Produce $CK$ to $H$ making $CK = KH$
\item Consider $CH$ ``as the bar of a balance'' with $K$ at 
fulcrum
\item Known facts about triangles (proved by Archimedes earlier):  
The {\it centroid} of $\Delta AFC$ is located at a point $W$ along $CK$
with $CK = 3\cdot KW$.
\end{itemize}
\end{columns}
\end{frame}

\begin{frame}
{\bf ``Dissection'' step}

\begin{columns}
\column{.5 \textwidth}
\begin{figure}
\includegraphics<1>[scale=.33]{balance.png}
\end{figure}

\column{.5 \textwidth}
\begin{itemize}
\item Consider $PO$ (think of a thin strip approximating the 
part of the parabolic segment with some thickness $\Delta x$)
\item Properties of parabolas and similar triangles imply
$MO:PO = AC:AO = CK:KN = HK:KN$
\item Place segment $TG = PO$ with midpoint at $H$.
\item Then Archimedes' {\it law of the lever} (fulcrum at $K$) 
says $TG$ balances $MO$ 
\end{itemize}
\end{columns}
\end{frame}

\begin{frame}
{\bf Conclusion of the proof}

\begin{columns}
\column{.5 \textwidth}
\begin{figure}
\includegraphics<1>[scale=.33]{balance.png}
\end{figure}

\column{.5 \textwidth}
\begin{itemize}
\item Do this for all ``strips'' like $PO$.  We get that $\Delta ACF$ exactly 
balances the whole collection of vertical strips making up the parabolic segment.  
\item By the law of the lever again, 
$\text{area}(\Delta ACF):\text{area of segment} = HK:KW = 3:1$
\item Since $\text{area}(\Delta ACF) = 4\cdot \text{area}(\Delta ABC)$,
the proof is complete.  
\end{itemize}
\end{columns}
\end{frame}

\section{Some conclusions}

\begin{frame}
{\bf From the introductory letter of {\it The Method} to Eratosthenes}

\vskip 10pt
``Seeing moreover in you, as I say, an earnest student, a man of considerable
eminence in philosophy, and an admirer of mathematical inquiry, I thought fit
to write out for you and explain in detail ... a certain method by which it will be 
possible for you to get a start to enable you to investigate some of the 
problems in mathematics by means of mechanics.  This procedure is, I am
persuaded, no less useful even for the proof of the theorems themselves; for 
certain things first became clear to me by a mechanical method, 
although they had to be demonstrated by geometry afterwards because 
their investigation by the said method
did not furnish an actual demonstration.''

Translation by T.L. Heath
\end{frame}

\begin{frame}
{\bf What does this tell us?}

\begin{itemize}
\item<1-> Can certainly say that Archimedes anticipated many ideas of \emph{integral calculus} in this work
\item<2-> But he was also \emph{careful} and realized that he didn't have
a complete justification for the idea of ``balancing'' an area with a 
collection of line segments   
\item<3-> So he also worked out the first ``method of exhaustion'' proof
to supply an argument that would satisfy the mathematicians
of his day
\end{itemize} 
\end{frame}

\begin{frame}
{\bf Recognized in antiquity as a master}

\vskip 10pt
``In weightiness of matter and elegance of style, no classical
mathematics treatise surpasses the works of Archimedes. This
was recognized in antiquity; thus Plutarch says of Archimedes'
works:

`It is not possible to find in all geometry more difficult
and intricate questions, or more simple and lucid
explanations. Some ascribe this to his genius; while
others think that incredible effort and toil produced
these, to all appearances, easy and unlaboured
results.' ''

\vskip 10pt
A. Aaboe, \emph{Episodes from the early history of mathematics}
\end{frame}


\begin{frame}
{\bf Archimedes' place in mathematical history}

\begin{itemize}
\item<1-> It should be clear from these examples that Archimedes was one 
of the greatest mathematicians in all of our known history -- his work shows extremely strong originality, masterful use of an intuitive physical analogy, and technical power
\item<2-> But he was so far ahead of his time that within 100 years or
so of his death very few, if any, people were reading him with complete understanding
\item<3-> Almost miraculous that this has survived
\item<4-> Also extremely poignant to think what might have been if others
had been better prepared to follow his lead at the time!
\end{itemize}
\end{frame}

\end{document}

