There’s an interesting book I’m working on called The Fall of Rome: And the End of Civilization, by historian Brian Ward-Perkins. He argues (against a prominent modern school of thought) that Rome did indeed fall rather than merely change, and that European civilization really was wrecked pretty hard with a recovery that by many fairly objective numerical metrics took almost a thousand years to reach its former level of economic complexity and living standards. It wasn’t a fast and sharp break separating the Roman from post-Roman world, but it was in fact a break.
You can make a similar argument in a positive direction about the history of mathematics. Math has existed more or less forever, but I’d say there’s a definite new era that started at roughly the end of the Renaissance about the time Newton and Leibniz independently developed calculus. Simply because calculus was so brilliantly effective at solving problems, subsequent mathematics tended to focus on whatever worked rather than whatever could be rigorously proved. In math this is a dangerous thing. If your reasoning is unclear, you won’t be able to tell under what circumstances your results no longer work. Eventually the sketchier concepts of modern math were examined formally and were put into a logically sound and rigorous framework. But today, we’re going to do this old school and assert a whopper of a supposition with no other support. I’ll tell you when we get there.
First, our Sunday Function. Unlike most other Sunday Functions, I’m not going to write this one down in terms of f(x) = stuff, because this one is just piecewise continuous straight lines. You can easily write it in terms of mx + b for each piece, but it will save space to define it just in terms of the graph. It’s a triangle wave of period 2, and we’re only going to deal with it on the interval [0, 2]. Therefore, here simultaneously is the graph and definition:
Now here comes our bold and utterly unsupported assertion. We know many functions can be written in terms of a power series. What if it were possible to write this in sort of a trigonometric series? In other words, are there constants a0, a1, a2, and so on such that this is true for our particular f(x):
Well, maybe. First we have to figure out what the various coefficients are in this series (which happens to be called a Fourier series, after the mathematician Jean Baptiste Joseph Fourier). First let’s dispatch a0 by means of a handwaving but nonetheless valid argument. Here goes. It’s clear that the average value of our function on this interval is 0 by virtue of its symmetry about the x-axis. It’s also clear that all of the sine terms on the right have an average value of 0 as well for the same reason. But any nonzero constant term will have a nonzero average value. The average of a constant is just that constant, after all. Therefore in order to preserve symmetry the constant term a0 must itself be equal to zero.
The other terms look harder at first. Fortunately there is a very convenient property of the sine function that can help us out with the rest. We call it orthogonality*:
The delta on the right side is the Kronecker delta, which is just a symbol that means “equals 1 if m = n, equals 0 otherwise”. So we have that weird sine series, and we just want to find (for instance) a2. We can just multiply both sides of the series equation by sin(2 pi x) and integrate. Every single one of those infinite terms will integrate out to 0 due to orthogonality, except for the one with n = 2, which is attached to the a2 term. Doing that, we can see that a2 (and in general if we had picked a different numbered “a”) is equal to:
Sweet! We can evaluate those. I won’t bore you by actually doing it (you can try it yourself or take a look at the result). Once you do evaluate them, you can actually plot the first few terms and see what you get. It’s going to turn out that all the even terms will equal zero. Let me give you numerical values for the first few odd terms:
a1 = 0.810569469
a3 = -0.0900632743
a5 = 0.0324227788
Plug those in and plot, with the original overlaid for comparison. Only including the a1 term:
With the first two nonzero terms:
With the first three nonzero terms:
So it sure looks like our wild supposition was justified and our function can legitimately be written in terms of trig functions. As indeed it can be, and as indeed pretty much any well-behaved periodic function can be. (Crucially, you must also include cosines in general. In the interests of simplicity I’ve picked an odd function as our example. These have the property that all the cosine coefficients are zero. Conversely, even functions have all the sine terms equal to zero.) This property of the set of sine/cosine functions is called completeness, and unlike orthogonality it’s way out of Sunday Function’s league to prove from scratch. Way out of my league, to be honest. But that’s what mathematicians are for. They’ve showed that a very broad class of different types of orthogonal functions are complete, and showing that a particular set of functions is in that broad class isn’t actually so hard. We might even do it one of these days.
*Here I’m writing the statement of orthogonality in a slightly nonstandard way. More usually it’s written with respect to the interval from [-pi, pi] or [0, 2*pi]. Our problem is on the interval [0, 2], so just like on the cooking shows I’m pulling this equation out of the oven “pre-cooked”, appropriately scaled to the interval in our problem.