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

For a general period $2L$ on $(-L, L)$ , replace $nx$ by $\dfrac{n\pi x}{L}$ everywhere and $\dfrac{1}{\pi}$ by $\dfrac{1}{L}$ in front of each integral.

Shortcuts 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.

If $f(x)$ is even (symmetric about the $y$ -axis), it's the other way round: every $b_n = 0$ , and only $a_0$ and the cosine terms survive. In both cases the remaining integrals can be taken over $(0, \pi)$ and doubled.

What survives for odd and even functions on (−π, π)

Function	Zero coefficients	Series contains
Odd, e.g. $x$ , $x^3$ , $\sin x$	$a_0$ and every $a_n$	Sine terms only
Even, e.g. $x^2$ , $\|x\|$ , $\cos x$	Every $b_n$	Constant and cosine terms only
Neither	None in general	Both

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

Worked example: the Fourier series of f(x) = x²

Find the Fourier series of $f(x) = x^2$ on $(-\pi, \pi)$ , and use it to find $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2}$ .

Solution: $x^2$ is even, so every $b_n = 0$ . For the constant term:

a_0 = \dfrac{2}{\pi}\int_0^\pi x^2\,dx = \dfrac{2}{\pi}\cdot\dfrac{\pi^3}{3} = \dfrac{2\pi^2}{3}, \qquad \dfrac{a_0}{2} = \dfrac{\pi^2}{3}

For $a_n$ , integrate by parts twice. The boundary terms leave only the part with $\cos n\pi = (-1)^n$ :

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

x^2 = \dfrac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^2}\cos nx = \dfrac{\pi^2}{3} - 4\cos x + \cos 2x - \dfrac{4}{9}\cos 3x + \cdots

Now put $x = \pi$ . Since $\cos n\pi = (-1)^n$ , every term becomes $\dfrac{4}{n^2}$ :

\pi^2 = \dfrac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\dfrac{1}{n^2} \;\Rightarrow\; \sum_{n=1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}

What the series does at a jump

Under the usual (Dirichlet) conditions, the series equals $f(x)$ wherever $f$ is continuous. At a jump it converges to the average of the values on either side.

The series for $x$ shows this. Repeated with period $2\pi$ , the function jumps from $\pi$ down to $-\pi$ at $x = \pi$ . Every term $\sin n\pi$ is 0 there, so the series gives $0 = \dfrac{\pi + (-\pi)}{2}$ , the midpoint of the jump.

That is the value at the jump. Just beside it, the partial sums overshoot by about 9% of the jump however many terms you add, the Gibbs phenomenon, which has its own worked post.

Common mistakes

Mixing the two conventions for the constant. With $a_0 = \dfrac{1}{\pi}\displaystyle\int f\,dx$ , the series starts with $\dfrac{a_0}{2}$ . Some books define $a_0$ with $\dfrac{1}{2\pi}$ and start with $a_0$ . Use one convention throughout.
Computing coefficients you already know are zero. Check odd or even first: it removes half the work.
Forgetting $\cos n\pi = (-1)^n$ and $\sin n\pi = 0$ . Almost every boundary term simplifies through these two facts.
Keeping $nx$ for a period other than $2\pi$ . On $(-L, L)$ the arguments become $\dfrac{n\pi x}{L}$ .
Claiming the series equals $f$ at a discontinuity. It gives the average of the two one-sided values there.

Where this comes up in exams

Fourier series are listed in the GATE 2027 EE syllabus under both Engineering Mathematics and Signals and Systems, and under Signals and Systems in the GATE 2027 EC and IN syllabi.

Checked against GATE 2027 EE syllabus, GATE 2027 IN syllabus and GATE 2027 EC syllabus. Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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