Periodic and Orthogonal Functions

Two functions are orthogonal when you multiply them together, add up the result over an interval, and get zero, and waves of different pitches turn out to be orthogonal over a full cycle. ("Orthogonal" comes from the Greek for "right-angled": it's the function version of two vectors being perpendicular, with a dot product of zero.) That's what lets you take a mixed-up signal and pull out each frequency in it separately. The integral $\int_{-L}^{L}\sin\frac{n\pi x}{L}\,\sin\frac{m\pi x}{L}\,dx$ looks like a lot, but it's just three simple steps: take two waves (one with $n$ bumps, one with $m$), multiply them point by point, and total up the signed area over one period (that's what $\int_{-L}^{L}$ does). The answer is $0$ when the waves are different and $L$ when they're the same. That one fact is the key to everything else in this unit, so bear with it.

The eigenfunctions from the last lesson, $\sin\frac{n\pi x}{L}$, have a property that makes the rest of this unit work. Before we get to it, one quick definition.

Periodic functions repeat: $f(x + T) = f(x)$. $\sin\frac{n\pi x}{L}$ has period $\frac{2L}{n}$, so all of them repeat every $2L$.

Orthogonality. Think of functions as vectors with infinitely many components and $\int fg\,dx$ as the dot product. Then $\sin x, \sin 2x, \cos x, \dots$ are perpendicular to each other, like $\vec i, \vec j, \vec k$.

Think about why the integral of $\sin\frac{n\pi x}{L}\sin\frac{m\pi x}{L}$ comes out to zero. The product-to-sum identity rewrites it as $\frac12\left[\cos\frac{(n - m)\pi x}{L} - \cos\frac{(n + m)\pi x}{L}\right]$. When $n \ne m$, both cosines complete whole cycles over $[-L, L]$, so each one's positive and negative parts cancel and the integral is $0$. When $n = m$, the first cosine is $\cos 0 = 1$, which doesn't cancel, and it leaves $\frac12\cdot 2L = L$.

A common mistake. It's tempting to think $\int_{-L}^{L}\sin^2\frac{n\pi x}{L}\,dx$ is zero too, because a sine averages out to zero over a period. It isn't: $\sin^2$ is never negative, so nothing cancels. Its average is $\frac12$, and over an interval of length $2L$ that gives $L$.

The payoff. Remember how you find the $\vec j$ component of a vector: dot it with $\vec j$, and the $\vec i$ and $\vec k$ parts drop out. Functions work the same way. To find how much of $\sin\frac{m\pi x}{L}$ is in $f$, take the "dot product" of $f$ with it: every other term integrates to zero. That's how Fourier coefficients are computed, starting next lesson. There's an animation further down that shows the cancellation happening for $\sin 2x$ and $\sin 3x$.

Orthogonality is why a radio can pick one station out of the air, why JPEG and MP3 compression work, and how noise-cancelling headphones isolate frequencies.