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{\LARGE
Math 1650  \hfill Project: ``A Piece of $\pi$'' \\
}
\end{center}

This project will introduce to various techniques that mathematicians
used to calculate $\pi$ before the advent of computers. All of these
techniques involve the creative use of trig identities. 

Although you have a month to do this project, I {\it strongly} urge you 
not to put this off until the last 
minute, as the project does require some creative thinking.

The grading for the project will be 80\% correctness, 20\% clarity
of presentation. Collaboration with your classmates is permitted and
encouraged; however, you may {\it not} discuss the project with 
anyone else (e.g., the seniors). On the front page on your write-up,
you must describe all such collaboration.

\vskip 0.1in

\noindent {\bf 1. Archimedes (c. 250 B.C.).}

Consider regular $n$-sided polygons (called $n$-gons for short) which
are either inscribed in or circumscribed about a circle with diameter 1. 
Then the perimeters of the inscribed $n$-gons must be less than the 
circumference of the circle, which is $\pi$. Likewise, the perimeters
of the circumscribed $n$-gons must be greater than $\pi$.
As $n$ increases, then the perimeters of the $n$-gons will approach $\pi$.

Notice that the perimeter $p_n$ of an inscribed regular $n$-gon is equal
to $nx$, where $x$ is the third side of a triangle with two sides of
length $1/2$ and an included (central) angle of $2\pi/n$. Likewise, the 
perimeter $P_n$ of a circumscribed $n$-gon is equal to $ny$, where $y$ is 
the base of an isoceles triangle with altitude is equal to $1/2$ and
central angle $2\pi/n$.

\begin{figure}[h]
\centerline{\psfig{file=inscribe.ps,width=3.0in} \hfill
            \psfig{file=circum.ps,width=3.106in}}
\end{figure}

(a) Use the law of cosines and a double-angle identity to show that
\begin{displaymath}
x^2 = \sin^2 \myfrac{\pi}{n},
\end{displaymath}
and hence conclude that the perimeter of an inscribed $n$-gon is
given by
\begin{displaymath}
p_n = n \sin \myfrac{\pi}{n}.
\end{displaymath}

(b) Show that $y = n \tan (\pi/n)$,
and hence conclude that the perimeter of a circumscribed $n$-gon is
\begin{displaymath}
P_n = n \tan \myfrac{\pi}{n}.
\end{displaymath}

\vskip 0.3in

(c) Without using a calculator, fill in the second and third columns of the 
following table. You will need to use the half-angle trig identities
to obtain $P_{12}$ and $P_{24}$. Copy the following table, with the
values filled in, onto your write-up.

\vskip 0.25in

\begin{tabular}{|c||c|c||c|c|}
\hline
~~~~$n$~~~~ & ~~~~$p_n$ (exact)~~~~ & ~~~~$P_n$ (exact)~~~~ & ~~~~$p_n$ (approximate)~~~~ & ~~~~$P_n$ (approximate)~~~~ \\
\hline
              &                         &           & &                    \\
      4       &                         &           & &                    \\
              &                         &           & &                    \\
\hline
              &                         &           & &                    \\
      6       &                         &           & &                    \\
              &                         &           & &                    \\
\hline
              &                         &           & &                    \\
      8       &                         &           & &                    \\
              &                         &           & &                    \\
\hline
              &                         &           & &                    \\
      12      &                         &           & &                    \\
              &                         &           & &                    \\
\hline
              &                         &           & &                    \\
      24      &                         &           & &                    \\
              &                         &           & &                    \\
\hline
\end{tabular}

\vskip 0.25in

(c) Using a calculator, evaluate each of these approximations to $\pi$.
Not surprisingly, observe that
\begin{equation}
p_n = n \sin \myfrac{\pi}{n} < \pi < n \tan \myfrac{\pi}{n} = P_n.
\label{bounds1}
\end{equation}

\vfill

Note: Archimedes used this method with inscribed and circumscribed
$96-$gons to evaluate $\pi$ to three decimal places. However, this 
method converges quite slowly, as you can check on
your calculator: $p_{1000}$ only gives $\pi$ to five decimal places, 
while $p_{10000}$ is accurate to seven.

\newpage

\noindent{\bf 2. Francois Vieta (1593).}

(a) Use a double-angle identity repeatedly to show that
\begin{eqnarray*}
\sin x &=& 2 \cos \myfrac{x}{2} \sin \myfrac{x}{2} \\
       &=& 4 \cos \myfrac{x}{2} \cos \myfrac{x}{4} \sin \myfrac{x}{4} \\
       &=& 8 \cos \myfrac{x}{2} \cos \myfrac{x}{4} \cos \myfrac{x}{8} 
                 \sin \myfrac{x}{8} \\
       &=& 2^n \cos \myfrac{x}{2} \cos \myfrac{x}{4} \dots
               \cos \myfrac{x}{2^{n-1}} \cos \myfrac{x}{2^n}
               \sin \myfrac{x}{2^n} 
\end{eqnarray*}

(b) Recall that, when $\theta$ is measured in radians, 
   $\sin \theta \approx \theta$ for small $\theta$. Let $\theta = x/2^n$,
   and conclude that
\begin{displaymath}
\frac{\sin x}{x} \approx \cos \myfrac{x}{2} \cos \myfrac{x}{4} \dots
                        \cos \myfrac{x}{2^{n-1}} \cos \myfrac{x}{2^n}.
\end{displaymath}

(c) Substitute $x = \pi/2$ into (b) to obtain the following approximations:

\begin{eqnarray}
\frac{2}{\pi} &\approx& \cos \frac{\pi}{4} \cos \frac{\pi}{8} \nonumber \\
\frac{2}{\pi} &\approx& \cos \frac{\pi}{4} \cos \frac{\pi}{8}
                        \cos \frac{\pi}{16} \nonumber \\
\frac{2}{\pi} &\approx& \cos \frac{\pi}{4} \cos \frac{\pi}{8}
                        \cos \frac{\pi}{16} \cos \frac{\pi}{32} \label{Vieta}
\end{eqnarray}
These approximations become quite accurate as $n$, the number of terms, 
increases.

\vskip 0.1in

(d) Use $\cos(\pi/4) = \sqrt{2}/2$ and a half-angle identity to show that
\begin{eqnarray*}
\cos \frac{\pi}{8}  &=& \frac{\sqrt{2+\sqrt{2}}}{2} \\
\cos \frac{\pi}{16} &=& \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2} \\
\cos \frac{\pi}{32} &=& \frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2} \\
\end{eqnarray*}

(e) Use your calculator and the answers to (c) and (d) to obtain three
successive approximations of $\pi$.

\vfill

Note: As you can check on your calculator, $\pi$ can be obtained to four 
decimal places by going up to $\pi/256$. Vieta used this method with 
{\it thirty terms} to evaluate $\pi$ to seventeen decimal places: 
\begin{displaymath}
\pi \approx 3.14159265358979323.
\end{displaymath}
This was a fairly amazing accomplishment since he obtained all of 
those square roots {\it by hand}.

\newpage

\noindent {\bf 3. Willebrod Snell (1621) and Christian Huygens (1654).}

\begin{figure}[h]
\centerline{\psfig{file=slower.ps,width=4.5in}}
\end{figure}

(a) In the above figure, show that
\begin{displaymath}
\tan \alpha = \frac{\sin \theta}{2 + \cos \theta}.
\end{displaymath}

(b) Since $x = 3 r \tan \alpha$, show that
\begin{displaymath}
x = \frac{3 r \sin \theta}{2 + \cos \theta}.
\end{displaymath}

(c) We now make the approximation that $x \approx \widehat{BE}$. Using the
formula $\widehat{BE} = r\theta$, show that
\begin{displaymath}
\theta \approx \frac{3 \sin \theta}{2 + \cos \theta}.
\end{displaymath}

(d) Let $\theta = \pi/n$ to obtain the following approximation to $\pi$:
\begin{displaymath}
\pi \approx \frac{3 n \sin(\pi/n)}{2 + \cos(\pi/n)}.
\end{displaymath}

\begin{figure}[h]
\centerline{\psfig{file=supper.ps,width=4.5in}}
\end{figure}

\vskip 0.3in

(e) In the following figure, use simple angle formulas from geometry
(e.g., isoceles triangles, exterior angles) to show that $\theta = 3\beta$.

(f) Show that
\begin{displaymath}
y = r \left( 1 + 2\cos \frac{\theta}{3} \right) \tan \frac{\theta}{3}.
\end{displaymath}

(g) Again using the approximation $y \approx \widehat{BE}$ and
the formula for the length of a circular arc, show that
\begin{displaymath}
\theta = \left( 1 + 2\cos \frac{\theta}{3} \right) \tan \frac{\theta}{3}.
\end{displaymath}

(h) Let $\theta = \pi/n$ to obtain the following approximation to $\pi$:
\begin{displaymath}
\pi \approx n\left( 1 + 2\cos \frac{\pi}{3n} \right) \tan \frac{\pi}{3n}.
\end{displaymath}

\vskip 0.3in

Although we do not prove this here, the approximations (d) and (h) are in
fact {\it bounds} on $\pi$:
\begin{equation}
\frac{3 n \sin(\pi/n)}{2 + \cos(\pi/n)} < \pi < 
n\left( 1 + 2\cos \frac{\pi}{3n} \right) \tan \frac{\pi}{3n}.
\label{bounds2}
\end{equation}
We will now show that these bounds are {\it tighter} than the
bounds (\ref{bounds1}); that is,
\begin{displaymath}
n \sin \myfrac{\pi}{n} <  \frac{3 n \sin(\pi/n)}{2 + \cos(\pi/n)} < \pi
< n\left( 1 + 2\cos \frac{\pi}{3n} \right) \tan \frac{\pi}{3n}
< n \tan \myfrac{\pi}{n} 
\end{displaymath}

(i) Show that 
\begin{displaymath}
n \sin \myfrac{\pi}{n} <  \frac{3 n \sin(\pi/n)}{2 + \cos(\pi/n)}.
\end{displaymath}
{\it Hint:} $\cos(\pi/n) < 1$ for all positive integers $n \ge 2$.

\vskip 0.1in

(j) To show that the Snell-Huygens upper bound is less than Archimedes'
upper bound, first prove the trig identity
\begin{eqnarray*}
\tan 3x &=& \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} \\
        &=& 3 \tan x + \frac{8 \tan^3 x}{1 - 3\tan^2 x}
\end{eqnarray*}

(k) Substitute $x = \pi/(3n)$ into the above identity, and show that
the second term is positive for $n \ge 3$. Conclude that
\begin{displaymath}
n\left( 1 + 2\cos \frac{\pi}{3n} \right) \tan \frac{\pi}{3n}
< 3 n \tan \frac{\pi}{3n} < n \tan \myfrac{\pi}{n} 
\end{displaymath}

(l) {\bf Beauty contest.} Evaluate the approximations (\ref{bounds1}),
(\ref{Vieta}) and (\ref{bounds2}) up to terms involving $\pi/128$.
That is, substitute $n=128$ into (\ref{bounds1}) and (\ref{bounds2}),
and evaluate a product similar to (\ref{Vieta}) up to and including
$\cos(\pi/128)$. Which method provides the best estimate of $\pi$?

\newpage

\noindent {\bf 4. John Machin (1706).}

(a) Use a double-angle identity twice to prove the following identity:
\begin{displaymath}
\tan 4\theta = \frac{4\tan \theta - 4\tan^3 \theta}
                    {1 - 6\tan^2 \theta + \tan^4 \theta}.
\end{displaymath}

(b) Use (a) to show that 
\begin{displaymath}
\tan\left[4 \tan^{-1}\myfrac{1}{5}\right] = \frac{240}{238}.
\end{displaymath}

(c) Prove the following identity:
\begin{displaymath}
\tan\left(\frac{\pi}{4} + x\right) = \frac{1+\tan x}{1-\tan x}.
\end{displaymath}

(d) Use (c) to find the solution of the equation
\begin{displaymath}
\tan\left(\frac{\pi}{4} + x\right) = \frac{240}{238}.
\end{displaymath}

(e) Use (b) and (d) to show that
\begin{displaymath}
\pi = 16 \tan^{-1} \myfrac{1}{5} - 4 \tan^{-1} \myfrac{1}{239}.
\end{displaymath}

(f) In calculus, we will show that, for small $x$, $\tan^{-1} x$ can be
accurately approximated by
\begin{displaymath}
\tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} 
               + \frac{x^9}{9} - \frac{x^{11}}{11} \dots
\end{displaymath}
We will use this result to approximate $\pi$. Use the first two terms of
the above expression to approximate $\tan^{-1}(\frac{1}{239})$, and all
six terms to approximate $\tan^{-1}(\frac{1}{5})$. You may use a calculator
if you wish, but observe that this calculation could be done without
a calculator with sufficient patience. 

\vskip 0.1in

(g) Use (e) and (f) to approximate $\pi$ to seven decimal places.

\vfill

With sufficient creativity, other trig identities can be exploited to obtain 
approximations to $\pi$. Before the advent of computers, the record was held 
by William Shanks in 1829, who used the series in (f), expanded to {\it over 
two hundred} terms, to evaluate $\pi$ (or so he thought) to
707 decimal places. Unfortunately, it turned out there was an error in
the 528th decimal place, so that all subsequent digits were wrong. At
present and with the help of supercomputers, $\pi$ is known to several
billion decimal places. The first two million places were published about
fifteen years ago in what must be the world's most boring 800-page book.
\end{document}