# aliquote

## < a quantity that can be divided into another a whole number of time />

I am re-reading Melzak’s Companion to Concrete Mathematics, and there’s a section dedicated to $\pi$ (pp. 164–169). There are various formulas to approximate $\pi$ to a given precision, the first being probably the fraction 22/7, from Archimedes. This is only correct to three decimal places, so a better fractional approximation is 355/113 = 3.1415929, which is easy to memorize and probably the one we are first taught in (French) college (in addition to the mnemonic trick — “que j’aime a faire apprendre ce nombre utile aux sages”).

If you’re familiar with the bc program, you will recall that it relies on the arc tangent. This follows from Leibniz’s approximation, which starts with $\arctan x = x - x^3/3 + x^5/5 + \dots$, which yields $\pi/4 = 1 - 1/3 + 1/5 + \dots$ for $x = 1$. From this, there exists a variety of arctan-type formulas for $\pi$, e.g. $\pi/4 = \arctan\tfrac{1}{2} + \arctan\tfrac{1}{3}$ or $\pi/4 = 5\arctan\tfrac{1}{7} + 2\arctan\tfrac{3}{79}$.1

Let’s try it in a Fish shell:

 echo "scale=11; 4*a(1)" | bc -l
3.14159265356


Of note, the scale parameter is quite important when a high precision is required.

Another well-known formula, at least for $\LaTeX$ aficionado, is the following continued fraction, due to Brouncker:

$$\frac{4}{\pi} = 1 + \frac{1^2}{2 + \frac{3^2}{2 + \frac{5^2}{2 + \dots}}}$$

Of course, many more approximations are available. Although Melzak notes that no hyperexponentially fast procedure2 appears to be known for computing $\pi$, there does exist efficient algorithms to compute $\pi$ to n exact figures. A short snippet of Python code is available in the case of the Chudnovsky algorithm, which remains the most efficient algorithm at the time of this writing. Scheme code is available on Programming Praxis. Other iterative algorithms, like Borwein’s algorithm, are also simple to implement in languages that offer support for large integers.

Note that this is only if you are interested in computing $\pi$ to a large number of decimal places since most PLs will provide you with built-in constants for $\pi$ or $\pi/2$. E.g., in C (using clang on macOS) $\pi$ is stored as a constant in math.h: #define M_PI 3.14159265358979323846264338327950288. This file is actually located under the command-line tools directory, that can be located using, e.g., echo "#include <math.h>" | gcc -v -x c -. Racket provides a double-precision flonum for $\pi$, but fractional approximations are used in various place the math library (e.g., 14488038916154245685/4611686018427387904 = 3.141592653589793).

Melzak provides two additional formulas to compute $\pi$ and $e^{-\pi}$, based on the theory of elliptic functions which is far beyond the scope of this short post. Those formulas are:

\begin{align} e^{-\pi} &= b + 2b^5 + 15b^9 +150b^{13} + 1707b^{17} + \dots \newline \pi &= \log\frac{1}{b} - 2b^4 - 13b^8 - \frac{368}{3}b^{12} - \frac{2701}{2}b^{16} + \dots, \end{align}

with $b = \tfrac{1}{2}\frac{\sqrt[\leftroot{-1}\uproot{2}4]{2}-1}{\sqrt[\leftroot{-1}\uproot{2}4]{2}+1} = 0.0432136168629448960219378\dots$ It should be noted that the very first term, $\log\tfrac{1}{b}$, already gives $\pi$ correctly to five decimal places. Using Mathematica, I got the following result:

In:= b = 1/2*(Power[2, (4)^-1] - 1)/(Power[2, (4)^-1] + 1)
In:= N[b, 24]
Out= 0.0432136168629448960219378
In:= N[Log[1/b], 24]
Out= 3.14159962823802109942254
In:= N[Pi, 24]
Out= 3.14159265358979323846264


As a side note, Mathematica also allows us to compute any terms of the continued fraction of $\pi$:3

In:= ContinuedFraction[Pi, 20]
Out= {3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2}
In:= FromContinuedFraction[%]
Out= 14885392687/4738167652


1. Such formulas rely on the following identity, attributed to Dogson: if $qr = 1+p^2$, then $\arctan\tfrac{1}{p} = \arctan\tfrac{1}{p+q} + \arctan\tfrac{1}{p+r}$. This identity was also proposed by Euler, who further demonstrated that $\arctan\tfrac{m}{n} = \frac{mn}{m^2+n^2}\left[ 1 + \frac{2}{3}\frac{m^2}{m^2+n^2} + \frac{2\cdot 4}{3\cdot 5}\left(\frac{m^2}{m^2+n^2}\right)^2 + \dots \right]$. ↩︎

2. A numerical procedure is said to be exponentially fast if for large $n$, $E^n\sim c^n$ for some $c$, $0 < c < 1$. An hyperexponentially fast procedure is one for which $E_n\sim c^{n^{\alpha}}$ for some $c$, $0 < c < 1$ and $\alpha > 1$. ↩︎

3. To compute the continued fraction of a number $x$, use the recurrence $a_0 = x$ and $a_n = (a_{n-1} - \lfloor a_{n_1} \rfloor)^{-1}$; the $n$th term will be $\lfloor a_n \rfloor$. Consider the precision of 20 digits illustrated above, we have: $n=0$, $a_n = 3.1415926535897932385$ and $d_n = 3$; $n=1$, $a_n = 0.1415926535897932385^{-1}=7.06251330593104577$ and $d_n = 7$; $n=2$, $a_n = 06251330593104577^{-1} = 15.9966$ and $d_n=15$; etc. ↩︎