Fourier Series

Main idea

On $[-L, L]$ , almost any periodic function is a constant plus cosines plus sines: the cosines carry its even part and the sines its odd part. Where the function jumps, the series splits the difference and lands on the midpoint.

f(x) = A_0 + \sum_{n=1}^\infty\left(A_n\cos\frac{n\pi x}{L} + B_n\sin\frac{n\pi x}{L}\right)

A_0 = \frac{1}{2L}\int_{-L}^L f\,dx, \qquad {\htmlData{sym=0}{A_n}} = \frac1L\int_{-L}^L f\cos\frac{n\pi x}{L}\,dx, \qquad {\htmlData{sym=1}{B_n}} = \frac1L\int_{-L}^L f\sin\frac{n\pi x}{L}\,dx

What the symbols mean

$A_n$
Cosine coefficients
The even part of f.
$B_n$
Sine coefficients
The odd part of f.

Explanation

A Fourier series is simply a way to write almost any repeating pattern, even one with sharp jumps, as a stack of simple sine and cosine waves. (It's named after Joseph Fourier, and the sine and cosine series from the last two lessons are special cases of it.) It's how engineers break sounds, signals and vibrations into the frequencies they're made of. The formula looks like a lot, but it's just three kinds of pieces added together: $A_0$ (the average level), $A_n\cos\frac{n\pi x}{L}$ (the even, mirror-symmetric waves) and $B_n\sin\frac{n\pi x}{L}$ (the odd, flipped-symmetric waves). Each coefficient formula is the same "multiply by one wave and integrate" trick you've already used. It's the sine series and the cosine series glued together, so bear with it.

The sine series and the cosine series only worked on $[0, L]$ , because you got to choose how to extend $f$ . On the full interval $[-L, L]$ you don't get that choice. A function with no symmetry has both an even part and an odd part, so you need both cosines and sines.

Why $\frac1L$ in front now instead of $\frac2L$ ? You're integrating over all of $[-L, L]$ , where $\int_{-L}^{L}\sin^2\frac{n\pi x}{L}\,dx = L$ (the orthogonality lesson's $n = m$ case), and the cosines work the same way. And the constant term is again the average of $f$ , now over an interval of length $2L$ .

Shortcuts: an odd function has only sine terms; an even function has only cosine terms (plus $A_0$ ). That's because an odd function times an even cosine is odd, and an odd function integrates to $0$ over $[-L, L]$ . The same thing kills the sine terms of an even function.

Convergence. For piecewise smooth $f$ , the series converges to $f(x)$ wherever $f$ is continuous, and to the average of the left and right limits at a jump.

Gibbs phenomenon. Near a jump, partial sums overshoot by about 9% of the jump size. Adding terms narrows the overshoot but never removes it.

A common mistake. It's tempting to think enough terms will eventually get rid of the overshoot, because the series converges. It won't: the spike stays at about 9% and only gets narrower. Convergence still holds at each fixed point, since the spike slides closer and closer to the jump and eventually passes any point you pick. There's a graph with a slider further down where you can watch this happen.

See it

N

7.00

This is the sine series of a square wave, using N terms. Raise N and the sum closes in on the square wave, but the overshoot at each jump never shrinks: the peak stays at about 1.18 instead of 1, which is about 9% of the jump from −1 to 1. It only gets narrower.

Worked examples

Example 1

Find the Fourier series of the square wave $f(x) = -1$ on $(-\pi, 0)$ , $1$ on $(0, \pi)$ .

1
$f$ is odd, so every $A_n$ is $0$ and only the sines remain: $B_n = \frac2\pi\int_0^\pi\sin nx\,dx = \frac{2}{n\pi}\left(1 - (-1)^n\right)$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find $A_0$ for $f(x) = x$ on $[-\pi, \pi]$ .

Problem 2+5 XP

Find $B_1$ for $f(x) = x$ on $[-\pi, \pi]$ .

Problem 3+5 XP

Find $B_1$ for the square wave in the example.

Problem 4+5 XP

What value does the square wave's Fourier series give at the jump $x = 0$ ?

Problem 5+5 XP

About how big is the Gibbs overshoot near a jump?

Problem 6+5 XP

A square-wave voltage is $+5$ V for $0 < t < \pi$ and $-5$ V for $-\pi < t < 0$ , repeating. Find its Fourier coefficients $B_1$ and $B_3$ , in that order.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

On $[-\pi, \pi]$ , $f(x) = x^2$ has the Fourier series $\dfrac{\pi^2}{3} + \displaystyle\sum_{n=1}^\infty\frac{4(-1)^n}{n^2}\cos nx$ . Evaluate it at $x = \pi$ to find the exact value of $1 + \frac14 + \frac19 + \frac1{16} + \cdots$ .

Where it's used

Every sound you hear and every signal your phone sends is analyzed in terms of Fourier series; the Fourier series lab builds a square wave from sines term by term.

Fourier SeriesCalc II · Electrical Engineering

Fourier Cosine Series