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## Infinite Series Approximations for Trigonometric functions from 14th century India

The infinite series approximations that have been used for many years to calculate the values of trigonometric functions have traditionally been attributed to Brook Taylor and Colin Maclaurin, European mathematicians of the early 18th century who were building on the work of Newton, Leibniz, James Gregory and Isaac Barrow among others.

However, I recently discovered that they were not the first to use these techniques.  As author George Gheverghese Joseph points out in the previous link, the work of Newton and Leibniz was tremendous, however the Indian development of infinite series approximations for trigonometric functions was equally amazing and important.  In addition, it came nearly 300 years before the European development of these techniques.

Madhava of Sangamagrama is generally recognized of the founder of the Kerala school of mathematics and astronomy in what is today the state of Kerala in southwest India.  The work of the mathematicians of the Kerala school was based on a desire for accurate trigonometric values for use in navigation.

Madhava lived in the late 1300s and early 1400s and most of his original work has been lost.  However, he is mentioned frequently in the surviving work of later mathematicians from the Kerala school.  Madhava is credited with power series calculations for the sine, cosine, tangent and arctangent, and like Leibniz, he used the arctangent power series to approximate the value of $\pi$ to 13 decimal places.

Victor Katz’ A History of Mathematics (Brief Edition) has a wonderful and detailed derivation of the Kerala school trigonometric series, with diagrams showing how they used the relationships between the angles, radii, chords and arcs in a circle to arrive at these amazing calculations.

Katz also published this derivation in a paper for the MAA (Mathematics Magazine, vol. 68, n. 3, June 1995, pp. 163-174)

The derivation for the infinite series begins on page 169 (pg. 7 in the pdf).

I’ve just begun to unpack this derivation and will post a step-by-step explanation of Katz’ work in the “near” future.

## Deriving the Area of a Circle

During the 1998-99 school year I was teaching math at Ellsworth High School in Maine.  I got to talking with one of the other math teachers there about deriving the area of a circle.  We started approaching the idea from a similar perspective as Archimedes had 2300 years ago in his work with circles and $\pi$.  We inscribed a polygon in a circle and then took the limit of the area of the polygon as the number of sides ($n$) approached infinity.  For the area of the polygon, we divided it into $n$ triangles and used the area formula $A=\frac {1}{2} ab\sin C$.

In this diagram, both $a$ and $b$ are radii of the circle and are labeled as $R$.  The central angle $\theta$ which will be used to find the area of the triangle is actually equal to $\frac {360^\circ}{n}$ or (and this will be VERY important later) $\frac {2\pi}{n}$.

So the area of the polygon is $n$ times the area of the triangle, or:

$n*\frac {1}{2} R*R*\sin \frac{2 \pi}{n}$

For this to be the area of the circle we want to take the limit of the previous expression as $n \to \infty$.

$A=\lim_{n \to \infty} n*\frac {1}{2} R*R*\sin \frac {2 \pi}{n}$

Since the $\frac {1}{2}*R*R$ don’t involve $n$, we can pull them out from the limit.

$A=\frac {1}{2} R^2*\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}$

Now comes the calculus – we were just talking about the derivative of the sine function in Calculus I last week, which is what made me think about this.  Remember that:

$\lim_{x \to 0} \frac{\sin x}{x}=1$

So, we need to establish this limit somewhere in our derivation and then replace it with $1$.  The most likely place for this would be in here:

$\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}$

but we need to have:

$\lim_{\frac{2 \pi}{n} \to 0} \frac{\sin \frac {2 \pi}{n}}{\frac{2 \pi}{n}}$

OK – so, let’s multiply our origninal expression by $\frac{2 \pi}{2 \pi}$.

$\lim_{n \to \infty} \frac{2 \pi}{2 \pi}*n*\sin \frac {2 \pi}{n}$

or

$\lim_{n \to \infty} 2 \pi*\frac{n}{2 \pi}*\sin \frac {2 \pi}{n}$

which is equivalent to

$\lim_{n \to \infty} 2 \pi*\frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$

Let’s bring back the whole expression:

$A=\frac {1}{2} R^2*\lim_{n \to \infty} n*\sin \frac {2 \pi}{n}$

$A=\frac {1}{2} R^2*\lim_{n \to \infty} 2 \pi*\frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$

$A=\frac {1}{2} R^2*2 \pi \lim_{n \to \infty} \frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$

as $n \to \infty$,

$\frac{2 \pi}{n} \to 0$

$A=\frac {1}{2} R^2*2 \pi \lim_{\frac{2 \pi}{n} \to 0} \frac{\sin \frac {2 \pi}{n}}{\frac {2 \pi}{n}}$

$A=\frac {1}{2} R^2*2 \pi *1$

or

$A=\pi R^2$

When we were working through this the first time, we kept the $360^\circ$ in the formula and ended up with $A=180^\circ R^2$, which didn’t make any sense until we realized that $180^\circ$ is $\pi$ radians!

Archimedes’ conclusion was that the area of the circle was equal to one-half the radius times the circumference:

$\frac{1}{2} R*2\pi R = \pi R^2$

or, since the mathematicians in the Greek tradition like Archimedes thought in terms of geometry – the area of a circle is the same as the area of triangle with height equal to the radius of the circle and base equal to the circumference of the circle!