Fourier Sine and Cosine Series, calculus lesson

A guitar string is fixed at both ends and only exists between them. To describe its shape with waves you have a choice to make, because nothing outside the string tells you what the function should be there. Filling in the missing half by mirror symmetry gives a series with only sines or only cosines, and the two choices behave quite differently.

Key terms

Odd and even extensions

Definition Let $f$ be defined on $(0, L)$ with $L > 0$ . The odd extension $f_o$ is the function with period $2L$ given on $(-L, L)$ by

f_o(x) = \begin{cases} f(x), & 0 < x < L, \\ 0, & x = 0, \\ -f(-x), & -L < x < 0, \end{cases}

and $f_o(\pm L) = 0$ . If $f$ is defined on $[0, L]$ , the even extension $f_e$ is the function with period $2L$ given by $f_e(x) = f(\lvert x \rvert)$ for $-L \le x \le L$ . This is consistent because $f_e(-L) = f_e(L) = f(L)$ .

In plain words For the odd extension you copy the graph into the left half turned upside down. For the even extension you copy it as a mirror image. Then you repeat the picture forever in both directions.

Example Take $f(x) = x$ on $(0, 1)$ . The odd extension is $x$ on $(-1, 1)$ repeated with period $2$ , a sawtooth wave that jumps from $1$ down to $-1$ at every odd integer. The even extension is $\lvert x \rvert$ on $[-1, 1]$ repeated with period $2$ , a triangle wave with no jumps at all.

Half-range expansions

Definition Let $f$ be integrable on $[0, L]$ . A half-range expansion of $f$ is the Fourier series of either its odd or its even extension. It has period $2L$ , its coefficients are computed only from the values of $f$ on $[0, L]$ , and it contains only sine terms (odd extension) or only cosine terms (even extension).

Recall that an integrable function with period $2L$ has the Fourier series

\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),

whose coefficients are

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

In plain words You only know the function on half of a period. You choose how to fill in the other half, and the choice decides which kind of wave appears.

Example The odd extension of $f$ is odd, so $f_o(x)\cos\frac{n\pi x}{L}$ is odd and every $a_n$ is $0$ . The product $f_o(x)\sin\frac{n\pi x}{L}$ is even, so

b_n = \frac{1}{L}\int_{-L}^{L} f_o(x)\sin\frac{n\pi x}{L}\,dx = \frac{2}{L}\int_{0}^{L} f(x)\sin\frac{n\pi x}{L}\,dx.

The same symmetry argument for the even extension kills every $b_n$ and doubles the integral over $[0, L]$ for $a_n$ .

Sine series

Definition The Fourier sine series of an integrable function $f$ on $[0, L]$ is

\sum_{n=1}^{\infty} b_n \sin\frac{n\pi x}{L}, \qquad b_n = \frac{2}{L}\int_{0}^{L} f(x)\sin\frac{n\pi x}{L}\,dx.

It is the Fourier series of the odd extension of $f$ .

In plain words You build $f$ from the shapes $\sin\frac{\pi x}{L}$ , $\sin\frac{2\pi x}{L}$ , and so on, each of which is zero at both ends of the interval.

Example For $f(x) = x$ on $[0, \pi]$ , integration by parts gives

b_n = \frac{2}{\pi}\int_{0}^{\pi} x\sin(nx)\,dx = \frac{2}{\pi}\left[-\frac{x\cos(nx)}{n} + \frac{\sin(nx)}{n^2}\right]_0^{\pi} = \frac{2(-1)^{n+1}}{n}.

So on $(0, \pi)$ the sine series of $x$ is

x \sim 2\sin x - \sin(2x) + \frac{2}{3}\sin(3x) - \frac{1}{2}\sin(4x) + \cdots

Cosine series

Definition The Fourier cosine series of an integrable function $f$ on $[0, L]$ is

\frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\frac{n\pi x}{L}, \qquad a_n = \frac{2}{L}\int_{0}^{L} f(x)\cos\frac{n\pi x}{L}\,dx \quad (n \ge 0).

It is the Fourier series of the even extension of $f$ . The constant term $\frac{a_0}{2}$ is the average value of $f$ on $[0, L]$ .

In plain words You build $f$ from a constant and the shapes $\cos\frac{n\pi x}{L}$ , which all have zero slope at both ends.

Example For $f(x) = x$ on $[0, \pi]$ the constant term is

\frac{a_0}{2} = \frac{1}{\pi}\int_0^{\pi} x\,dx = \frac{\pi}{2},

and for $n \ge 1$ integration by parts gives

a_n = \frac{2}{\pi}\int_{0}^{\pi} x\cos(nx)\,dx = \frac{2}{\pi}\cdot\frac{(-1)^n - 1}{n^2},

which is $-\frac{4}{\pi n^2}$ for odd $n$ and $0$ for even $n$ . So the cosine series of $x$ on $[0, \pi]$ is

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

Endpoint behavior

Definition Let $f$ be piecewise smooth on $[0, L]$ , meaning $f$ and $f'$ are piecewise continuous there. Then for $0 < x < L$ both half-range series converge to

\frac{f(x^+) + f(x^-)}{2},

which is $f(x)$ wherever $f$ is continuous. At the ends the sine series converges to $0$ at both $x = 0$ and $x = L$ , whatever the values of $f$ there, while the cosine series converges to $f(0^+)$ at $x = 0$ and to $f(L^-)$ at $x = L$ .

In plain words Every sine term vanishes at the ends, so the sine series is stuck at $0$ there. The even extension has no jump at the ends, so the cosine series lands on the true end values.

Example For $f(x) = x$ on $[0, \pi]$ , the sine series at $x = \pi$ is $\sum b_n \sin(n\pi) = 0$ , the average of the jump from $\pi$ to $-\pi$ in the sawtooth. The cosine series at $x = 0$ is

\frac{\pi}{2} - \frac{4}{\pi}\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2} = \frac{\pi}{2} - \frac{4}{\pi}\cdot\frac{\pi^2}{8} = 0 = f(0).

The second step uses the known sum

\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2} = \frac{\pi^2}{8}.

Approximating periodic signals

Definition For an integrable function $f$ with period $2L$ , the partial sum

S_N(x) = \frac{a_0}{2} + \sum_{n=1}^{N}\left(a_n\cos\frac{n\pi x}{L} + b_n\sin\frac{n\pi x}{L}\right)

is an approximation built from finitely many frequencies. If $f$ is continuous on $\mathbb{R}$ and $f'$ is piecewise continuous, then $\sum (\lvert a_n \rvert + \lvert b_n \rvert)$ converges and $S_N \to f$ uniformly on $\mathbb{R}$ . If $f$ has a jump, $S_N$ cannot converge uniformly near it, and the overshoot of the Gibbs phenomenon persists. For a half-range expansion this means the cosine series of a continuous piecewise smooth $f$ always converges uniformly on $[0, L]$ , while the sine series does so only when also $f(0) = f(L) = 0$ .

In plain words A signal with no jumps is captured well by a handful of frequencies. A jump forces you to keep many terms and still leaves a ripple next to the jump.

Example For $f(x) = x$ on $[0, \pi]$ , the sine coefficients shrink like $\frac{2}{n}$ and the cosine coefficients like $\frac{4}{\pi n^2}$ . The largest error of the cosine partial sum on $[0, \pi]$ occurs at the ends.

Terms up to $n = N$	Largest error of cosine sum on $[0, \pi]$
$5$	$0.105$
$20$	$0.032$
$100$	$0.0064$

The sine partial sums equal $0$ at $x = \pi$ while $f(\pi) = \pi$ , and they are continuous, so their largest error on $(0, \pi)$ comes arbitrarily close to $\pi$ for every $N$ .

Check yourself

Find the sine series of $f(x) = 1$ on $(0, \pi)$ .

Show the answer

The coefficients are

b_n = \frac{2}{\pi}\int_0^{\pi}\sin(nx)\,dx = \frac{2}{\pi}\cdot\frac{1 - (-1)^n}{n},

which is $\frac{4}{\pi n}$ for odd $n$ and $0$ for even $n$ . So on $(0, \pi)$

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

What is the cosine series of $f(x) = 1$ on $[0, \pi]$ ?

Show the answer

Just the constant $1$ . The even extension is the constant function $1$ , so $a_0 = 2$ and $a_n = 0$ for $n \ge 1$ .

To what value does the sine series of $f(x) = x^2$ on $[0, 2]$ converge at $x = 2$ ? What about the cosine series?

Show the answer

The sine series converges to $0$ at $x = 2$ , since every term $\sin(n\pi)$ is $0$ . The cosine series converges to $f(2) = 4$ , because the even extension is continuous there.

Fourier Sine and Cosine Series

Key terms

Odd and even extensions

Half-range expansions

Sine series

Cosine series

Endpoint behavior

Approximating periodic signals

Check yourself

How this lesson connects

Builds on