Computing \(\pi\) and the February 2018 DMMC Problem
Computing \(\pi\) and the February 2018 Problem
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How the pentagon problem relates to computing \(\boldsymbol{\pi}\)
A. A famous infinite series that can produce expnasions for \(\pi\) is:
\[
\tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots
\quad (-1 \leq x \leq 1)
\]
(known as the Madhava-Gregory-Lebiniz series). Substituting \(x=1\) above,
we get the classic Leibniz formula for \(\pi\)
(really for \(\pi/4\), but we can simply multiply by \(4\) to get \(\pi\)):
\[
\frac{\pi}{4} = \tan^{-1} 1 =
1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots
\]
Although it is a beautiful simple expression for \(\pi\), this
conditionally convergent series converges extremely slowly and
is of no practical use in computing \(\pi\).
B. Now revisit the solution to the pentagon problem, and note the angles \(p\), \(q\), \(r\), \(s\), \(t\), and \(u\), given by:
\[
s = \tan^{-1}\frac{1}{2},\;
q = t = \tan^{-1}\frac{1}{3},\;
p = \tan^{-1}\frac{1}{5},\;
r = \tan^{-1}\frac{1}{7},\;
u = \tan^{-1}\frac{1}{5}.
\]
The key lemma there gave
\[
\frac{\pi}{4} = s + q
= \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3}.
\]
The series for expansion for \(\tan^{-1}x\) then yields the series (pair):
\begin{align*}
\label{eq:pia}
\frac{\pi}{4}
&= \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3}\\
\nonumber{}
&= \left(\frac{1}{2^1} - \frac{1}{3 \cdot 2^3} + \frac{1}{5 \cdot 2^5}
- \cdots\right)
\;+\;
\left(
\frac{1}{3^1} - \frac{1}{3 \cdot 3^3} + \frac{1}{5 \cdot 3^5}
- \cdots
\right),
\end{align*}
which converges much more quickly (at the order of geometric series),
making this formula practical for computing \(\pi\).
C. The proof in the pentagon problem also noted that \(\tan(r + t) = 1/2\), or \(s = \tan^{-1} 1/2 = r + t = r + q\), and so
\[
\frac{\pi}{4} = s + q
= 2q + r
= 2\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{7}.
\]
The \(\tan^{-1}x\) series expansion then yields:
\begin{align*}
\label{eq:pib}
\frac{\pi}{4}
&= 2\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{7}\\
\nonumber{}
&=
2\left(
\frac{1}{3^1} - \frac{1}{3 \cdot 3^3} + \frac{1}{5 \cdot 3^5}
- \cdots
\right)
\;+\;
\left(
\frac{1}{7^1} - \frac{1}{3 \cdot 7^3} + \frac{1}{5 \cdot 7^5}
- \cdots
\right).
\end{align*}
This will converge even faster than the previous series.
In the late 1700's, Vega used this series to compute \(\pi\) to about
140 places.
D. Finally, we also had \(\tan(p+u) = 1/3\), or \(q = t = p + u\) and \(r + t + p + u = \pi/4\), giving us \(\pi/4 = r + (p+u) + p + u\), i.e. \(\pi/4 = 2p + r + 2u\), or:
\[
\frac{\pi}{4} = 2\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7}
+ 2 \tan^{-1} \frac{1}{8}.
\]
Again, applying the series expansion for \(\tan^{-1} x\), we get:
\begin{align*}
\label{eq:pic}
\frac{\pi}{4}
&= 2\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7}
+ 2 \tan^{-1} \frac{1}{8}\\
\nonumber{}
&=
2\left(
\frac{1}{5^1} - \frac{1}{3 \cdot 5^3} + \frac{1}{5 \cdot 5^5} - \cdots
\right)
\;+\;
\frac{1}{7^1} - \frac{1}{3 \cdot 7^3} + \frac{1}{5 \cdot 7^5} - \cdots
\;+\;
2\left(
\frac{1}{8^1} - \frac{1}{3 \cdot 8^3} + \frac{1}{5 \cdot 8^5} - \cdots
\right),
\end{align*}
which will converge even more rapidly! Just computing one term of each series
gives an approximation error of about \(0.007\)
(less than 1 percent). Computing two terms drops
the error to \(0.0001\), and computing four terms drops it
to \(0.0000001\).
Before electronic computers, these methods (e.g. Machin's formula) were used heavily to compute \(\pi\).
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