From Archimedes to Infinity

 Circles, Ellipses, and Elliptic Integrals

Archimedes

In his work Κύκλου Μέτρησις (Measurement of a Circle), Archimedes calculates successive approximations to

which is equal to , by using reduction formulas relating the perimeters of regular inscribed and circumscribed -gons to those of the corresponding regular -gons of the circle of radius one.

Why does the integral   appear?

Consider the upper semicircle

To compute its arc length, partition the interval  into subintervals

and approximate the semicircle by the polygonal path joining the points

for .

By the Pythagorean Theorem, the length of the -th segment is

where  and . Assuming a uniform partition,  is constant, and we may write

By the Mean Value Theorem, there exists a point

such that

Hence the length of the segment becomes

Summing the lengths of all segments gives the approximation

which is the Riemann sum for

Since

the arc length of the upper semicircle is

The integrand

is continuous on  but it blows up at . So we interpret the integral as an improper integral:

For , the integrand is continuous and

Therefore,

Since  is continuous at

Hence

Greek mathematicians carried out such computations using only a ruler and a compass.

What follows is not the method Archimedes himself used to approximate the circumference of the unit circle. Rather, it is an analytical reconstruction of the ideas underlying his procedure.

 

The reduction formula that relates the circumscribed -gons to the corresponding -gons follows directly from the figure above. Indeed,

and

Since, in a triangle, the bisector of an interior angle divides the opposite side into segments whose lengths are proportional to the lengths of the adjacent sides (see Basic Geometry in The Riemann Hypothesis Revealed), we have

 

From this we get:

or

For the reduction formula that relates the inscribed -gons to the corresponding -gons (see the following figure) we have

and

Triangles ,  and  are obviously similar. Therefore

Adding yields

Using Pythagoras’ Theorem

Since ,  and , we obtain

For convenience set  and . So

or

 

Thus, the side lengths of the circumscribed and inscribed polygons satisfy the reduction formulas

and

for .

Now let

be the perimeters of the inscribed and circumscribed polygons.

Obviously

If  is the apothem corresponding to the side of the regular -gon then from Pythagoras’ Theorem

From

we get

hence

Thus,

Additionally,

and for any  there exists an -gon such that

The side length of the inscribed -gon is  and . Also, the side length of the circumscribed -gon is . Therefore (this can be proved purely from the geometry of the inscribed and circumscribed polygons, with no trigonometry at all)

and

So

Since

the squeeze theorem implies

This is essentially Archimedes’ use of the Squeeze Theorem. Consequently,

 The Man Who Knew Infinity

In the movie The Man Who Knew Infinity (2015), there is a memorable classroom scene at Cambridge in which Ramanujan, having neglected to take notes during a lecture, is challenged by Professor Howard to contribute to the evaluation the integral:

This integral is known as the complete elliptic integral of the first kind. Elliptic integrals owe their name to their historical appearance in the problem of determining the arc length of an ellipse, although they arise naturally in many other areas of mathematics and physics.

An ellipse is defined by the major axis of length  and the minor axis of length . The exact perimeter of an ellipse is given by the integral

where  is the eccentricity of the ellipse.

The integral

is known as the complete elliptic integral of the second kind.

To see why

we start with a natural parametrization

because

Use the arc-length formula for a parametrized curve over the first quadrant

Here,

Therefore

Factoring out :

But

so

Thus,

Infinite Sums

In the movie Ramanujan immediately writes on the blackboard:

astonishing both the professor and the class. The scene is intended to showcase the exceptional mathematical insight for which Ramanujan became legendary.

To prove this result, we follow the approach presented in (Almost) Impossible Integrals, Sums, and Series by Cornel Ioan Vălean. We begin with the following auxiliary integral:

We have:

Now set

Set

Now take

Set  in . Then

Thus,

Expand

so

Now use the well-known Wallis’ integral

It follows that

 

To find a similar formula for  we write the integrand as

and use the binomial series expansion for  with , :

where .

Set . Then

Assuming term wise integration is justified (it is ), we get

We have

Multiplying by

so

because

Thus,

Apply this formula to find the perimeter of the ellipsis:

Epilogue

When the subject is rich in geometry, I find myself especially excited. Even more so when I can build my ideas around the geometric intuition of Archimedes. There is something deeply satisfying about seeing an abstract mathematical argument come alive through geometry.

But there is another important element: the reader and I need to have some common ground. For my essays, that common ground is provided by my book, The Riemann Hypothesis Revealed. Everything I use in these essays can, of course, be found elsewhere, and I make no claim otherwise. But if you are looking for a single book that brings together all the mathematical tools and background needed for these ideas, this is the book for you.

The Riemann Hypothesis Revealed is available through Amazon marketplaces worldwide.