Fourier Series

The idea

Any reasonably well-behaved periodic function can be rebuilt out of sines and cosines. For a function $f(x)$ with period $2\pi$ :

f(x) = \dfrac{a_0}{2} + \sum_{n=1}^{\infty}\left(a_n\cos nx + b_n\sin nx\right)

where the coefficients are found by integrating over one full period:

a_0 = \dfrac{1}{\pi}\int_{-\pi}^{\pi} f(x)\,dx, \qquad a_n = \dfrac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos nx\,dx, \qquad b_n = \dfrac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin nx\,dx

A shortcut for odd and even functions

If $f(x)$ is odd (symmetric about the origin), every $a_n$ (including $a_0$ ) is automatically zero, since cosine is even and an odd-times-even integrand over a symmetric interval vanishes. Only the sine terms survive.

Worked example: the Fourier series of f(x) = x

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

Solution: $f(x)=x$ is odd, so $a_0 = 0$ and every $a_n = 0$ . Only $b_n$ needs computing, and since $x\sin nx$ is even, the integral over $(-\pi,\pi)$ is twice the integral over $(0,\pi)$ :

b_n = \dfrac{2}{\pi}\int_0^{\pi} x\sin nx\,dx

Integrating by parts (u = x, dv = sin nx dx):

\int_0^\pi x\sin nx\,dx = \left[-\dfrac{x\cos nx}{n}\right]_0^\pi + \dfrac{1}{n}\int_0^\pi \cos nx\,dx = -\dfrac{\pi\cos n\pi}{n} + \dfrac{1}{n}\left[\dfrac{\sin nx}{n}\right]_0^\pi

The second term vanishes since $\sin n\pi = 0$ for every integer $n$ , and $\cos n\pi = (-1)^n$ :

\int_0^\pi x\sin nx\,dx = -\dfrac{\pi(-1)^n}{n} = \dfrac{\pi(-1)^{n+1}}{n}

So:

b_n = \dfrac{2}{\pi}\cdot\dfrac{\pi(-1)^{n+1}}{n} = \dfrac{2(-1)^{n+1}}{n}

giving $b_1=2$ , $b_2=-1$ , $b_3=\tfrac{2}{3}$ , and so on, so the full series is:

x = 2\sin x - \sin 2x + \dfrac{2}{3}\sin 3x - \dfrac{1}{2}\sin 4x + \cdots = \sum_{n=1}^{\infty} \dfrac{2(-1)^{n+1}}{n}\sin nx

Fourier Series

The idea

Any reasonably well-behaved periodic function can be rebuilt out of sines and cosines. For a function $f(x)$ with period $2\pi$ :

f(x) = \dfrac{a_0}{2} + \sum_{n=1}^{\infty}\left(a_n\cos nx + b_n\sin nx\right)

where the coefficients are found by integrating over one full period:

a_0 = \dfrac{1}{\pi}\int_{-\pi}^{\pi} f(x)\,dx, \qquad a_n = \dfrac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos nx\,dx, \qquad b_n = \dfrac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin nx\,dx

A shortcut for odd and even functions

Worked example: the Fourier series of f(x) = x

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

Solution: $f(x)=x$ is odd, so $a_0 = 0$ and every $a_n = 0$ . Only $b_n$ needs computing, and since $x\sin nx$ is even, the integral over $(-\pi,\pi)$ is twice the integral over $(0,\pi)$ :

b_n = \dfrac{2}{\pi}\int_0^{\pi} x\sin nx\,dx

Integrating by parts (u = x, dv = sin nx dx):

\int_0^\pi x\sin nx\,dx = \left[-\dfrac{x\cos nx}{n}\right]_0^\pi + \dfrac{1}{n}\int_0^\pi \cos nx\,dx = -\dfrac{\pi\cos n\pi}{n} + \dfrac{1}{n}\left[\dfrac{\sin nx}{n}\right]_0^\pi

The second term vanishes since $\sin n\pi = 0$ for every integer $n$ , and $\cos n\pi = (-1)^n$ :

\int_0^\pi x\sin nx\,dx = -\dfrac{\pi(-1)^n}{n} = \dfrac{\pi(-1)^{n+1}}{n}

So:

b_n = \dfrac{2}{\pi}\cdot\dfrac{\pi(-1)^{n+1}}{n} = \dfrac{2(-1)^{n+1}}{n}

giving $b_1=2$ , $b_2=-1$ , $b_3=\tfrac{2}{3}$ , and so on, so the full series is:

x = 2\sin x - \sin 2x + \dfrac{2}{3}\sin 3x - \dfrac{1}{2}\sin 4x + \cdots = \sum_{n=1}^{\infty} \dfrac{2(-1)^{n+1}}{n}\sin nx

Fourier Series

The idea

A shortcut for odd and even functions

Worked example: the Fourier series of f(x) = x

Continue learning

Fourier Series

The idea

A shortcut for odd and even functions

Worked example: the Fourier series of f(x) = x

Continue learning