Taylor series build a function from powers of $x$. Fourier series build a periodic function from sines and cosines, the eigenfunctions of the previous lesson. They describe sound, heat and signals, and they solve the heat and wave equations. Computing coefficients uses integration by parts again and again.
Key idea
For $f$ with period $2L$, $$f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty}\left(a_n\cos\frac{n\pi x}{L} + b_n\sin\frac{n\pi x}{L}\right),$$ $$a_n = \frac{1}{L}\int_{-L}^{L} f(x)\cos\frac{n\pi x}{L}\,dx,$$ $$b_n = \frac{1}{L}\int_{-L}^{L} f(x)\sin\frac{n\pi x}{L}\,dx.$$ Here $L$ is half the period. If $f$ is odd, every $a_n = 0$; if $f$ is even, every $b_n = 0$. At a jump, the series converges to the average of the two one-sided values.
Why it works
The building blocks are orthogonal on $[-L, L]$. Using $\cos A\cos B$ $= \frac{1}{2}\left(\cos(A - B) + \cos(A + B)\right)$, the integral $\int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}\,dx$ is a sum of cosines over whole periods, so it is 0 when $m \ne n$ and $L$ when $m = n \ge 1$. The same holds for sines, and every sine-cosine product integrates to 0. Now multiply the series by $\cos\frac{m\pi x}{L}$ and integrate: every term dies except $a_m\cdot L$, which gives the formula for $a_m$. For $n = 0$, $\int_{-L}^{L} 1\,dx = 2L$, which is why the constant term is written $\frac{a_0}{2}$: then one formula covers every $a_n$.
Worked example: A square wave
Find the Fourier series of $f = -1$ on $(-\pi, 0)$ and $f = 1$ on $(0, \pi)$, period $2\pi$.
- $f$ is odd, so $a_n = 0$. Here $L = \pi$.
- $b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f\sin nx\,dx$ $= \frac{2}{\pi}\int_0^{\pi} \sin nx\,dx$ $= \frac{2}{n\pi}(1 - \cos n\pi)$.
- $\cos n\pi = (-1)^{n}$: $b_n = \frac{4}{n\pi}$ for odd $n$ and 0 for even $n$.
Answer: $f(x) \sim \frac{4}{\pi}\left(\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \dots\right)$
Worked example: A sawtooth
Find $b_n$ for $f(x) = x$ on $(-1, 1)$, period 2.
- $L = 1$ and $f$ is odd, so $b_n = 2\int_0^{1} x\sin n\pi x\,dx$.
- By parts: $\int_0^{1} x\sin n\pi x\,dx$ $= \left[-\frac{x\cos n\pi x}{n\pi}\right]_0^{1}$ $+ \frac{1}{n\pi}\int_0^{1}\cos n\pi x\,dx$ $= -\frac{(-1)^{n}}{n\pi} + 0$.
Answer: $b_n = \frac{2(-1)^{n+1}}{n\pi}$
Worked example: An even function
Find the series of $f(x) = x^{2}$ on $(-\pi, \pi)$.
- $f$ is even: $b_n = 0$. $a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi} x^{2}\,dx = \frac{2\pi^{2}}{3}$, so $\frac{a_0}{2} = \frac{\pi^{2}}{3}$.
- Integrating by parts twice, $a_n = \frac{2}{\pi}\int_0^{\pi} x^{2}\cos nx\,dx$ $= \frac{4(-1)^{n}}{n^{2}}$.
Answer: $x^{2} \sim \frac{\pi^{2}}{3} - 4\cos x + \cos 2x$ $- \frac{4}{9}\cos 3x + \dots$; at $x = \pi$ this gives $\sum \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$
Common mistake
Using the wrong interval length. $L$ is half the period: for period $2\pi$, $L = \pi$, and for period 6, $L = 3$. The factor in front is $\frac{1}{L}$ (or $\frac{2}{L}$ over half the interval for odd or even functions), and the same $L$ appears inside $\frac{n\pi x}{L}$. Using $\frac{1}{2L}$, or $\frac{1}{\pi}$ when the period is not $2\pi$, scales every coefficient wrongly.