Fourier Coefficients and Fourier Series, calculus lesson

A chord played on a piano is one wave, yet your ear hears separate notes inside it. Fourier series do the same thing with any periodic function. They split it into a constant plus harmonics, and the orthogonality of sines and cosines tells you exactly how much of each harmonic is present.

Key terms

Fourier series

Definition Let $f$ be periodic with period $2\pi$ and integrable on $[-\pi, \pi]$ . The Fourier series of $f$ is the trigonometric series

f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} \bigl(a_n \cos(nx) + b_n \sin(nx)\bigr),

with the coefficients $a_n$ and $b_n$ defined below. The symbol $\sim$ means the series is attached to $f$ . Whether and where it converges to $f(x)$ is a separate question.

In plain words It is a recipe for rebuilding $f$ from a constant and the harmonics $\cos(nx)$ and $\sin(nx)$ , each with its own weight.

Example The square wave equal to $-1$ on $(-\pi, 0)$ and $1$ on $(0, \pi)$ , repeated with period $2\pi$ , has Fourier series

\frac{4}{\pi}\left(\sin x + \frac{\sin(3x)}{3} + \frac{\sin(5x)}{5} + \cdots\right).

The coefficients are computed in the next section.

Sine and cosine coefficients

Definition The Fourier coefficients of a $2\pi$ -periodic $f$ that is integrable on $[-\pi, \pi]$ are

a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \quad (n \ge 0), \qquad b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \quad (n \ge 1).

These formulas come from orthogonality. If $f$ equaled a uniformly convergent trigonometric series, multiplying by $\cos(mx)$ and integrating term by term would kill every term except the one with $a_m$ , leaving

\int_{-\pi}^{\pi} f(x) \cos(mx) \, dx = \pi a_m

for $m \ge 1$ , and $2\pi$ times the constant term for $m = 0$ .

In plain words To find how much of $\cos(nx)$ is in $f$ , take the inner product of $f$ with $\cos(nx)$ and divide by $\pi$ . Same for sine.

Example For the square wave every $a_n$ is $0$ , because $f(x)\cos(nx)$ is odd. For $b_n$ ,

b_n = \frac{1}{\pi}\left(\int_{-\pi}^{0} -\sin(nx) \, dx + \int_{0}^{\pi} \sin(nx) \, dx\right) = \frac{2}{\pi} \cdot \frac{1 - \cos(n\pi)}{n} = \frac{2\bigl(1 - (-1)^n\bigr)}{n\pi}.

So $b_n = \frac{4}{n\pi}$ for odd $n$ and $b_n = 0$ for even $n$ .

Constant term

Definition The constant term of the Fourier series is

\frac{a_0}{2} = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx,

the average value of $f$ over one period. The factor $\frac{1}{2}$ is there so that the single formula for $a_n$ also covers $n = 0$ .

In plain words The constant term is the level the function oscillates around.

Example For $f(x) = x^2$ on $[-\pi, \pi]$ , extended with period $2\pi$ ,

\frac{a_0}{2} = \frac{1}{2\pi} \int_{-\pi}^{\pi} x^2 \, dx = \frac{1}{2\pi} \cdot \frac{2\pi^3}{3} = \frac{\pi^2}{3}.

Symmetry

Definition Let $f$ be $2\pi$ -periodic and integrable on $[-\pi, \pi]$ .

If $f$ is even, then $b_n = 0$ for all $n \ge 1$ and

a_n = \frac{2}{\pi} \int_0^{\pi} f(x) \cos(nx) \, dx \qquad (n \ge 0),

so the series has cosines only. If $f$ is odd, then $a_n = 0$ for all $n \ge 0$ and

b_n = \frac{2}{\pi} \int_0^{\pi} f(x) \sin(nx) \, dx \qquad (n \ge 1),

so the series has sines only.

This follows because an odd function integrates to $0$ over $[-\pi, \pi]$ , and an even one integrates to twice its integral over $[0, \pi]$ .

In plain words Check symmetry first. It can cut the work in half and tells you which coefficients vanish without computing them.

Example The function $f(x) = x^2$ is even. Integrating by parts twice,

a_n = \frac{2}{\pi} \int_0^{\pi} x^2 \cos(nx) \, dx = \frac{4(-1)^n}{n^2}, \qquad n \ge 1,

so on $[-\pi, \pi]$

x^2 \sim \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos(nx).

Arbitrary period

Definition If $f$ has period $2L$ with $L > 0$ and is integrable on $[-L, L]$ , its Fourier series is

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, \qquad b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\frac{n\pi x}{L} \, dx.

The symmetry rules hold with the factor $\frac{2}{L}$ and integrals over $[0, L]$ in place of $\frac{2}{\pi}$ and $[0, \pi]$ .

In plain words Stretch the harmonics so that they fit the period $2L$ , and divide by $L$ instead of $\pi$ .

Example Let $f(x) = x$ on $(-1, 1)$ with period $2$ , so $L = 1$ . It is odd, and

b_n = 2 \int_0^1 x \sin(n\pi x) \, dx = 2\left(\left[-\frac{x \cos(n\pi x)}{n\pi}\right]_0^1 + \int_0^1 \frac{\cos(n\pi x)}{n\pi} \, dx\right) = \frac{2(-1)^{n+1}}{n\pi}.

So on $(-1, 1)$

x \sim \frac{2}{\pi}\left(\sin(\pi x) - \frac{\sin(2\pi x)}{2} + \frac{\sin(3\pi x)}{3} - \cdots\right).

Partial sums

Definition The $N$ th partial sum of the Fourier series of a $2\pi$ -periodic $f$ is the trigonometric polynomial

S_N(x) = \frac{a_0}{2} + \sum_{n=1}^{N} \bigl(a_n \cos(nx) + b_n \sin(nx)\bigr).

The series converges at $x$ when $S_N(x)$ has a limit as $N \to \infty$ .

In plain words Keep the first $N$ harmonics and drop the rest. Watching $S_N$ as $N$ grows shows the series building up the function.

Example For the square wave at $x = \frac{\pi}{2}$ , where $f = 1$ , the odd sines alternate in sign.

$N$	$S_N(\pi/2)$
$1$	$1.2732$
$3$	$0.8488$
$5$	$1.1035$
$7$	$0.9216$

The values swing above and below $1$ and close in on it.

Spectral interpretation

Definition For $n \ge 1$ the amplitude of the $n$ th harmonic is $A_n = \sqrt{a_n^2 + b_n^2}$ , and for a suitable phase $\varphi_n$

a_n \cos(nx) + b_n \sin(nx) = A_n \cos(nx - \varphi_n).

The sequence $(A_n)$ is the amplitude spectrum of $f$ . For $f$ that is piecewise continuous on $[-\pi, \pi]$ , Parseval’s identity holds,

\frac{1}{\pi} \int_{-\pi}^{\pi} f(x)^2 \, dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} \bigl(a_n^2 + b_n^2\bigr).

In plain words The coefficients say how strongly each frequency is present. Parseval says the total energy of $f$ is the sum of the energies of its harmonics.

Example The square wave has $A_n = \frac{4}{n\pi}$ for odd $n$ and $0$ for even $n$ , so only odd harmonics are present and they fade like $\frac{1}{n}$ . Since $f^2 = 1$ , Parseval gives

2 = \frac{16}{\pi^2}\left(1 + \frac{1}{9} + \frac{1}{25} + \cdots\right), \qquad \text{so} \qquad 1 + \frac{1}{3^2} + \frac{1}{5^2} + \cdots = \frac{\pi^2}{8}.

Check yourself

Find the Fourier series of $f(x) = \lvert x \rvert$ on $[-\pi, \pi]$ , extended with period $2\pi$ .

Show the answer

The function is even, so $b_n = 0$ . The cosine coefficients are

a_0 = \frac{2}{\pi}\int_0^{\pi} x \, dx = \pi

and, for $n \ge 1$ ,

a_n = \frac{2}{\pi}\int_0^{\pi} x \cos(nx) \, dx = \frac{2\bigl((-1)^n - 1\bigr)}{\pi n^2},

which is $-\frac{4}{\pi n^2}$ for odd $n$ and $0$ for even $n$ . So

\lvert x \rvert \sim \frac{\pi}{2} - \frac{4}{\pi}\left(\cos x + \frac{\cos(3x)}{9} + \frac{\cos(5x)}{25} + \cdots\right).

A function with period $4$ equals $0$ on $(-2, 0)$ and $1$ on $(0, 2)$ . What is the constant term of its Fourier series?

Show the answer

The constant term is the average over one period,

\frac{a_0}{2} = \frac{1}{4}\int_{-2}^{2} f(x) \, dx = \frac{2}{4} = \frac{1}{2}.

Which coefficients of an odd $2\pi$ -periodic function must be zero, and why?

Show the answer

All $a_n$ , including $a_0$ . The product $f(x)\cos(nx)$ is odd, so its integral over $[-\pi, \pi]$ is $0$ .

Fourier Coefficients and Fourier Series

Key terms

Fourier series

Sine and cosine coefficients

Constant term

Symmetry

Arbitrary period

Partial sums

Spectral interpretation

Check yourself

How this lesson connects

Builds on

Used later in