In December of 2009, I picked up a book by Steven Strogatz called The Calculus of Friendship.0 Reading the book, I found myself touched by the story of the author’s life-long relationship with his high school Calculus teacher, affectionately called Joff. They corresponded about interesting math problems for years as Strogatz went to college, graduate school, and then became a professor of mathematics. Later in life, their missives, phone calls, and visits expanded to include more than math, reaching out to the relationships in their lives. I was inspired by the math and moved by the friendship. Tears well in my eyes even now, more than two years since reading the book.
There is a section on Fourier series in the Strogatz book. Fourier series are part of analysis (the field of mathematics where calculus lives) useful for engineering problems involving heat, sound, and electricity. Fourier series are named for Joseph Fourier, a French mathematician who lived and worked at the turn of the 19th century (late 17, early 1800s). Fourier helped grow the field of analysis from the rich soil of calculus. Fourier put his series to work sorting out how heat moves through solid objects, in the form of the heat equation. Fourier’s work stood on the shoulders of great 18th century mathematicians, including Euler, Lagrange, and Laplace.
Fourier also was a member of the Revolutionary Committee during the French Revolution. Like many others, Fourier was arrested and imprisoned, likely forced to come to terms with the possibility of literally losing his brilliant mind to the Reign of Terror. Luckily, for Fourier, Robespierre lost his first. Thankfully, Fourier survived the Reign of Terror and the Napoleonic Wars and was able to produce the groundbreaking series solution to the heat equation.1
While I am willing to meander into Fourier’s past, I am not going to venture very far into the workings of his series in this post. I am, however, going to explore one idea used in solving one of the classic Fourier series problems. The idea is from calculus.
This idea is put to work in solving the problem of decomposing a square wave into a sum of sine waves. “Decomposing” is a sort of rewriting. With Fourier, the square wave can be expressed in a nice, clean package involving Σ (for sum) and sin (for sine). Here is a square wave:2
We saw a sine wave in one of my first posts. Here is a sine wave superimposed on the square wave:
Already, the two graphs share infinitely many points, the intersections at the x-intercepts happening every 3 or so units. (Yes, yes. Those intersections occur every π units.) Of course, the infinity of shared points assumes the patterns continue just as we see them in the small window of the image. They do. This is a simple window into the larger world of the two graphs.
Now, we have the classic problem of building a square-shaped graph out of the sexy curves of a sinusoid.3 The secret is to use more than one sinusoid. The more sinusoids you use, the better the approximation.
When attempting to model a particular curve (for us, the square wave), Fourier analysis offers a process for finding the sinusoids to use. I want to play with just one of the ideas used in the process. The analytical methods of writing the square wave as a sum of sine waves includes a step where the sum of sines is multiplied by another sine. Early in the process, one is faced with a sin2x.
This equation alone, without the context of Fourier series, is pretty remarkable (hence all of these remarks). In math-jargony English, the equation reads, “The integral from negative pi to pi of the sine of x (pause) squared with respect to x equals pi.” The s-like curve is the integral symbol. It is a stretched out “s”, and it represents a sum. Integrals, and their related process called “integration”, provide a means for finding the accumulation of change. What I mean is, if you have a function for the velocity of an object (an object’s speed and the direction it is moving), the integral can be used to find out how far the object travels in a given amount of time. This is very similar to the notion d = rt.4
The integral not only gives us the means to find accumulating change, it also can be used to find area. This is one of the amazing connections of calculus. The value of the integral of sin2x with respect to x from −π to π is also the area between the curve and the x-axis, the gray portion in the following image.
The technology, if you will, of the integral answers one of the two fundamental questions of calculus.5 There are analytical methods using algebra, trigonometry, and calculus for finding the area “under the curve” of sin2x between −π and π.6 In his book, Strogatz shares a simple geometric method for finding the value of the integral from a letter his high school math teacher sent. This method is what is featured in chalk.
The graph of sin2x between −π to π fits nicely into a box 1 unit tall and 2π units wide. The box has an area of 2π units2. (Remember: the area of a rectangle is the product of its length and width.) The graph of sin2x doesn’t just fit nicely, in fact, it takes up exactly half of the area of the box. How do we know? There is a trigonometric identity at play.
This is the foundational Pythagorean trigonometric identity. Given the box is 1 unit tall, a light may start shining out of the darkness if you consider adding the area between the graph of sin2x and 1 to the area “under” sin2x. What you’ll find is you are adding the area enclosed by cos2x to the area enclosed by sin2x. What you’ll get is the area enclosed by the horizontal line y = 1, yielding our good ol’ 2π units2. If you want a more rigorous explanation, check out the footnote below.
0. Strogatz, Steven H, and Don Joffray. The Calculus of Friendship: What a Teacher and a Student Learned About Life While Corresponding About Math. Princeton, N.J: Princeton University Press, 2009. Print.
1. I am curious if and how Fourier’s experience as a twenty-something Frenchman during the French Revolution shaped his mathematical interests. (He was 26 when he was imprisoned.) Fourier published Mémoire sur la propagation de la chaleur dans les corps solides thirteen years after the Reign of Terror. How was Fourier’s thinking during that decade impacted by his experiences during Napolean’s meteoric rise to power? Was his work on the heat equation fueled by military applications? Am I being naïve to even ask?
For more on Fourier, the man or the series, check out:
2. The CG images in this post were made with Apple’s Grapher©. The animation uses several images popped through GIFMAKE.
3. For a while, pop culture would have likely wanted us to go the other way. I am thinking of Ronald Miller’s transformation in Can’t Buy Me Love. Patrick Dempsey’s character, Ronald, falls in love with a “hot” girl, and being a nerd, he thinks he has only one choice – to become cool. Little did Ronald know he could have simply waited 3 years for the 1990s to arrive when nerds would start to become pop-icon kajillionaires. How is that for the transformation of square to sexy?
4.↑ What the creators and developers of calculus gave us with integration is the means to find the distance traveled when the rate, i.e. the velocity, is not constant. Good thing, too, because I would hate if my car only had two speeds, 0 and 30. I am pretty sure I would be killed instantly, if I attempted travel in that car.
Our lives move at ever changing rates. My car, thankfully, can accelerate from 0 to 30 and back down to 0. Of course, that makes the question, “How far does Eric travel if he is moving at 30 miles per hour for 2 hours?” a severe over simplification. The better question is, “How far does Eric travel if his velocity, as a function of time, is 30sin2(t), from time 0 to time 2?” I know this question is more complicated. So is real life.
5. The other question has to do with finding the instantaneous velocity of an object given a function for its displacement, i.e. distance traveled. For an entertaining, accessible, and informative introduction to the two fundamental questions of calculus, check out Edward Burger’s video, The Two Questions of Calculus courtesy of thinkwell. Edward Burger is an award-winning professor of mathematics at Williams College. His deep insight, clear explanations, wit, and colorful shirts & ties inspire me.
6. Buckle up. We are about to do some calculus. In the following, I am assuming you have taken at least a year of high school calculus or a semester of it in college. Also, you will see a method called integration by parts in this solution. This is not the only way to solve this problem. A trigonometric identity for sin2x could be implemented instead. However, for related parts of the larger Fourier series square wave problem, integration by parts is a handy tool. So, let’s use it here.
To begin, unpack the square.
From here, we can follow the general format for integration by parts.
For our integral, let u = sin x and dv = sin x dx. Then, du = cos x dx, v = − cos x.
Since sin(π) = sin(−π) = 0, the first term on the right-hand side is zero.
This means the integrals are equal.
This is a great little trig-calculus identity. It speaks to the geometric interpretation of the integral discussed in the main text of this post. Now, using some fancy trig identity footwork
we can say
Let’s evaluate the first integral on the right, the integral from −π to π of dx. That dx is really 1 dx, so that the function we are integrating is f (x) = 1. Thus, this integral is equivalent to the area “under” the line y = 1 between −π and π. That region is a rectangle with width 1 and length 2π. Hence, the cute little integral of dx on the right equals 2π, which means
We are now at a magical moment for many students of calculus. Many of us trained ourselves to see an operation when we look at an integral, even with definite integrals. The truth is, a definite integral is a number. We are looking for the value of the integral of sin2x between −π and π, the integral on the left and the right. If you are willing to cast this into algebraic terms, we are attempting to solve the following equation for x.
Ask most algebra students how to solve this, and they will say something about getting x “all by itself,” meaning solve for x. To do that, you might ________. Just give yourself a moment to think.
To apply this algebra idea to our integral problem, let’s add that integral of sin squared to both sides of the equation.
I think we all know if 2 of this equals 2 of that, then this equals that. Therefore,