Andrew Glassner’s Notebook http://www.research.microsoft.com/glassner Fourier Polygons ____________________________________ Andrew olygons are everywhere. They’re in our cereal, on Napoleon’s Theorem: Given any triangle ∆ABC, Glassner Pour kitchen floors, and in the constellations in the erect equilateral triangles on each side (all facing skies. One place I didn’t expect to see polygons is in the inward or all facing outward), and connect the Microsoft Fourier transform, but I found them there as well. centroids of each of these triangles. The resulting Research The Fourier transform is an indispensable tool in sig- triangle will be equilateral. nal processing. In computer graphics, it helps us under- stand and cure problems as diverse as jaggies on the You may recall that the centroid of a triangle is the edge of polygons, blocky looking textures, and animat- arithmetic average of the three vertices. Figure 1 shows ed objects that appear to jump erratically as they move an example of this theorem in action. I’ll call the origi- across the screen. nal three points V = (v0, v1, v2), the three points at the My friend and colleague Alvy Ray Smith recently wrote tips of the new triangles T, and the three points that a memo that demonstrated a surprising interpretation of make up the Napoleon triangle N. the Fourier transform. He showed how in some circum- You can prove this theorem in lots of ways. If you like stances the Fourier transform looks like nothing more working things out for yourself you may want to take a than operations on regular polygons. This column is shot at it before reading on. I’ve seen proofs that are about that fascinating insight. I’ll start off with a warm- strictly geometric, strictly algebraic, and various com- up in using complex numbers to do geometry and then binations of the two. move on to the Fourier series, building up to a discussion The approach I’ll use here is based on representing of the new interpretation. If you’re unfamiliar with com- each vertex of the triangle as a complex number. We’ll plex numbers, I invite you to check out the sidebar “A carry out the construction with the complex interpre- Quick Refresher on Complex Numbers.” I’ll introduce the tation and then prove that it’s right. Fourier series in the main body of the text. If it’s new to Figure 2 shows the edge formed by v1 and v2. Although you, don’t worry—you’ll see it’s actually pretty simple by I chose this edge at random, they all work the same way. the time we get there. And the payoff is worth the journey. Figure 2 shows the geometry resulting from building an equilateral triangle on these points and then finding the Napoleon’s Theorem centroid. First find point t0 at the apex of the triangle, Let’s begin with a lovely little theorem in elementary and drop the median down from t0 to vm12, the midpoint geometry: of v1 and v2. Next draw the median from v1 to the mid- point of v2 and t0. The intersection of these two medians denotes the centroid, marked by the complex point n0. It doesn’t matter which two medians we use in this con- struction, because all three meet at n0. t 1 A demonstration of Napoleon’s 1 We can save time by observing that we don’t have to Theorem. The original triangle, V, shown in blue, consists of points n v2 1 t0 (v0, v1, v2). On each edge, we build an equilateral triangle facing out- n v0 wards. The new points, T, at the tip 0 t0 of these triangles are (t0, t1, t2). The 2 Finding the centroids of the three new triangles centroid n0 at (n0, n1, n2), are joined with heavy the intersection v black lines to form the Napoleon 1 n2 of two medians. triangle, N, which is equilateral. n0 v1 v2 vm12 t2 84 January/February 1999 0272-1716/99/$10.00 © 1999 IEEE . A Quick Refresher on Complex Numbers Complex numbers arose naturally as people thought A remarkable property of complex numbers is revealed about how to find some value of x that would satisfy when we compare the series expansions for sine and cosine x2 + 1 = 0. It’s a logical step in the sequence of trying to with the series of expansion of powers of e, Euler’s constant. solve equations. Many books derive these expansions. They work out to be: To begin with, consider the equation x + 3 = 5. Only one value of x will work (that is, only one solution exists): x = 2. x2 x3 x 4 x5 x6 e x = 1 + x + + + + + + L This x is an integer. Now consider 3x = 5. The solution, 2 3 4 5 6 x = 5/3, is a real number (that is, not an integer) and a x2 x 4 x6 x8 rational number, which is specified as the ratio of two cos x = 1 − + − + − L 2 4 6 8 integers. Getting more ambitious, suppose we want to 3 5 7 9 2 x x x x solve x = 3. Now the only solution is x =√3 , which is also sin x = x − + − + − L real, but irrational. Mathematically, irrational doesn’t mean 3 5 7 9 emotionally unstable, but simply nonrational (that is, √3 cannot be expressed as the ratio of two integers). Some If we plug in i for x, we get the very useful identity common irrational numbers are √2, √3, e, and π. Consider the equation x2 + 1 = 0. If we just grind out a ei = cos 1 + i sin 1 solution, we find x =√−1. Whatever this may mean, it’s been given the label i, so i=√−1 (engineers often use the symbol j Combining this with de Moivre’s theorem, which tell us θ for this; the right interpretation is usually clear from (cos θ + i sin θ) = (cos 1 + i sin 1) , we find the important context). The question of just what √−1 ”means” has been identity, debated for a very long time. Regardless of the θ philosophical interpretation, it’s clearly a very convenient ei = cos θ + i sin θ computational entity. We can combine a real and an imaginary number into a single new number z as z = a + bi, which relates imaginary powers of e with complex where a and b are real numbers (rational or irrational), and numbers or simply points on the complex plane. Notice i=√−1. We say z is a complex number. Note that either part that the magnitude of these points is 1, and the phase is of z may be zero, so the complex numbers include all the simply θ. So if we know a complex number has magnitude pure reals and imaginaries. r and phase θ, we may write it as r(cos θ + i sin θ) or much θ What are the powers of i? Since anything raised to the 0 more compactly as rei . power is defined as 1, i0 = 1 and anything to the first power This expression, r(cos θ + i sin θ), is called a complex is 1, so i1 = i. By definition, i2 = −1. Building on these starting sinusoid. If we plot its coordinates on the complex plane as points, we find i3 = −i, i4 = 1, i5 = i, i6 = −1, and so on. The θ moves from 0 to 2π, we get one complete circle of radius pattern repeats indefinitely. r. As θ grows beyond 2π, the sinusoid simply wraps around Algebra with complex numbers is straightforward. To add, the circle again and again. Suppose we have two of these: iθ i3θ just add components. If w = a + bi and z = c + di, then w + z s1 = e and s3 = e . Clearly s3 spins around the unit circle = (a + c) + (b + d)i. To multiply, treat the complex numbers three times faster than s1. Just as Euclidean space builds on as factors: (a + bi)(c + di) = ac + bci + adi + bdi2 = (ac − bd) + the idea that the x-, y-, and z-axes are orthogonal (or at (bc + ad)i, using i2 = −1. It’s natural to try to get some kind of right angles) to one another, so does the Fourier transform visual take on complex numbers. Since real numbers can be makes use of the fact that these two functions are also plotted along an axis, we can try to put them together with orthogonal (though in a more abstract way). imaginary numbers along another axis. The plus sign in the Writing complex numbers as points in the complex plane iθ complex number a + bi seems to invite the idea of replacing gives us another way to see multiplication. If z1 = re and z2 iϕ i(θ+ϕ) it with a comma, giving the 2D point (a, b). = se , then z1z2 = (rs)e . In words, this tells us that when In 1797 Caspar Wessel did just this. He placed the reals on two complex numbers are multiplied, we multiply their the x-axis of a normal 2D coordinate system and the purely magnitudes and add their phases. If the magnitude of a imaginary numbers on the y-axis, creating what was later complex number z is 1, then multiplying by z is equivalent called an Argand diagram. The complex number z = a + bi is to a rotation counterclockwise by the phase of z. θ represented by the point (a, b). A purely real number r sits on To wrap up with a bang, let’s return to ei = cos θ + i sin θ the x-axis and may be thought of as r + 0i.