If one takes a circle and compare the length of its circumference to that of its diameter, you get  $$\pi$$. This constant has a surprising number of applications outside geometry, notably in infinite series of simple fractions. What sequences come out as functions of $$\pi$$, and how on earth can we derive them? These questions have bothered mathematicians for thousands of years.

First, a brief history of $$\pi$$. Its existence has been known for thousands of years: In the Rhind Papyrus (which is estimated to be over 3.5 thousand years old), $$\pi$$ is said to equal $$4(\frac{8}{9})^2$$, which roughly equals $$3.16$$1.

Many mathematicians managed to approximate $$\pi$$ using many-sided polygons and the circles inscribed or circumscribed around them. The ancient Greek Archimedes derived the inequality $$\frac{223}{71} < \pi < \frac{22}{7}$$, using a regular polygon with a whopping 96 sides[1]. In the early 17th century, the German mathematician Ludolph Van Ceulen used a polygon with $$2^{62}$$ sides to calculate $$\pi$$ to an accuracy of 35 decimal places2.

The understanding of $$\pi$$ hasn’t necessarily advanced with time, however. In 1897, an amateur mathematician introduced a bill to the Indiana State Legislature that would set $$\pi$$ as being exactly $$3.2$$ 3. This mathematician also happened to have devised “proofs” of the trisection of the angle and the doubling of the cube. Fortunately, the bill was shot down thanks to a visiting professor from Purdue University.

Our first method of deriving $$\pi$$ is known as Leibniz’s formula. It states that

$$\frac{\pi}{4} = 1 − \frac{1}{3} + \frac{1}{5} −\frac{1}{7} + \dots$$

This happens to be a special case of the power series for arctan:

$$arctan(x) = \frac{x^1}{1} −\frac{x^3}{3} + \frac{x^5}{5} − \frac{x^7}{7} + \dots$$

But why is this true?

When we take the derivative of both sides of the equation, we get that $$\frac{1}{1+x^2}=1-x^2+x^4-x^6+ \dots$$

Multiplying both sides by $$1 + x^2$$ yields

$$1 = 1 + x^2 − x^2 − x^4 + x^4 + x^6 −x^6 −x^8 + \dots$$

The right-hand side of the equation telescopes, so we see that the derivatives of the two sides are in fact equal. Since $$arctan (0) = 0$$, the two sides must be the same (and not offset by some constant). When we plug $$x=0$$ into this expansion, we get the desired equation.

If we take the first 10 terms of Leibniz’s formula and multiply by 4, we get an approximate value of 3.04; evidently, this is not a series that converges very quickly.

Another way to write out $$\frac{\pi}{4}$$ is as the following fraction:

$$\dfrac 4 \pi = 1 + \cfrac {1^2} {2 + \cfrac {3^2} {2 + \cfrac {5^2} {2 + \cfrac {7^2} {2 + \cfrac {9^2} {2 + \cdots} } } } }$$

This is known as Brouncker’s Formula4. It was first presented in 1655 by William Brouncker, the first president of the Royal Society. He derived it from Wallis’s Product (Brouncker’s Formula can also be derived from plugging $$x=1$$ into an expansion of $$arctan(x)$$). Wallis’s Product is as follows:

$$\frac{4}{\pi} = \frac{3}{2} \cdot \frac{3}{4} \cdot \frac{5}{4} \cdot \frac{5}{6} \cdot \frac{7}{6} \cdot \frac{7}{8} \cdot \dots$$

It was derived by John Wallis and published (along with Brouncker’s Formula) in Wallis’s massive and impressive work, Arithmetica Infinitorum. Interestingly enough, Brouncker derived a whole series of theorems from Wallis’s formula: The $$2$$’s in Brouncker’s Formula can be replaced with any even integer not divisible by $$4$$. The left-hand side must be compensated with corresponding terms from Wallis’s formula. For example, $$5 \cdot \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{3}{4} \cdot \frac{5}{4} \cdot \dfrac 4 \pi = 11 + \cfrac {1^2} {22 + \cfrac {3^2} {22 + \cfrac {5^2} {22 + \cfrac {7^2} {22 + \cfrac {9^2} {22 + \cdots} } } } }$$

If we take the product of the first 10 terms of Wallis’s Product, we can approximate $$\pi$$ as being roughly 3.275, which, like Leibniz’s formula, isn’t very accurate.

In the modern-day, we calculate $$\pi$$ using far more efficient formulas. Around 1910, the Indian mathematician Srinivasa Ramanujan found the following formula for $$\pi$$5:

$$\frac{1}{\pi} = \frac{2 \sqrt 2}{99^2} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{26390k+1103}{396^{4k}}$$

This formula converges insanely fast; even when we take only one term from the series, we can approximate $$\pi$$ as $$3.14159273 \dots$$. Ramanujan’s formula was later improved upon by the Chudnovsky brothers, who calculated over 2 billion digits of $$\pi$$ in the 1990s6. In 2019, Google employee Emma Haruka Iwao calculated 31.4 trillion digits of $$\pi$$, setting a world record. She used y-cruncher, a program created specifically to calculate the digits of $$\pi$$7. For those interested, y-cruncher can be downloaded for free at http://www.numberworld.org/y-cruncher/#Download.

But even though we’ve calculated this number to ridiculous lengths, the digits still go on, and so does the quest to find them.

1. O’Connor, J. J., & Robertson, E. F. (2001, August). Pi history. Retrieved from http://www-history.mcs.st-and.ac.uk/HistTopics/Pi_through_the_ages.html
2. O’Connor, J. J., & Robertson, E. F. (2009, April). Ludolph Van Ceulen. Retrieved from https://www-history.mcs.st-andrews.ac.uk/Biographies/Van_Ceulen.html
3. Goins, E. H. (2013). In Celebration of Pi Day: The History of the Indiana Pi Bill. Retrieved from https://www.math.purdue.edu/people/bio/egoins/Indiana%20Pi%20Bill.html
4. Osler, T. J. (2010). Lord Brouncker’s forgotten sequence of continued fractions for pi. International Journal of Mathematical Education in Science and Technology, 41(1), 105-110. doi:10.1080/00207390903189195