Recipe

Compute the Fourier coefficients of a periodic function

Use when: you have a periodic function with period $T$ and need $a_0$ , $a_n$ , and $b_n$ for its Fourier series

Goal: explicit formulas or numbers for $a_0$ , $a_n$ , and $b_n$ that determine the function's Fourier series

Taught in Deriving the coefficient formulas from orthogonality

Quick recipe

Identify the period and one full interval to integrate over
Compute $a_0$ from the mean-value formula
Compute $a_n$ by integrating against $\cos(n\omega_0x)$
Compute $b_n$ by integrating against $\sin(n\omega_0x)$

1
Read off the period $T$ , then pick any one period-length window to integrate over, usually the one where $f$ 's own formula is stated most simply.

Example

The rectangular pulse has period $T=2\pi$ ; its own definition is stated on $(-\pi,\pi)$ , so that's the window used below.

Why: every coefficient formula integrates over exactly one period, and a window of the wrong length changes the constant out front, not just the arithmetic.
2
Compute $a_0=\dfrac2T\displaystyle\int f(x)\,dx$ over the chosen period; $a_0/2$ is $f$ 's average value over that period, a fast sanity check.

Example

$a_0=\dfrac1\pi\displaystyle\int_{-\pi/2}^{\pi/2}1\,dx=1$ , so the constant term is $a_0/2=1/2$ , exactly the pulse's average, since it's $1$ on half its period and $0$ on the other half.

Why: setting $n=0$ in the $a_n$ formula makes $\cos(0\cdot x)=1$ , so $a_0$ is the same formula evaluated once at $n=0$ , not a separate idea.
3
Compute $a_n=\dfrac2T\displaystyle\int f(x)\cos(n\omega_0x)\,dx$ , then look for a pattern across $n$ rather than stopping after one value.

Example

$a_n=\dfrac2\pi\displaystyle\int_{-\pi/2}^{\pi/2}\cos(nx)\,dx=\dfrac{4}{n\pi}\sin\!\left(\dfrac{n\pi}2\right)$ , giving $a_1=\dfrac4\pi$ , $a_2=0$ , $a_3=-\dfrac{4}{3\pi}$ .
- $f$ is even — $b_n=0$ for every $n$ , so Step 4 needs no work at all.
- $f$ is odd — $a_0=0$ and $a_n=0$ for every $n$ , so only Step 4 has any content.
Why: multiplying by $\cos(n\omega_0x)$ and integrating isolates $a_n$ because orthogonality sends every other frequency's cross term to $0$ , leaving only the one that matches.
4
Compute $b_n=\dfrac2T\displaystyle\int f(x)\sin(n\omega_0x)\,dx$ , the same idea run with $\sin$ in place of $\cos$ .

Example

$b_1=\dfrac1\pi\displaystyle\int_{-\pi/2}^{\pi/2}\sin(x)\,dx=\dfrac1\pi\big({-}\cos(\pi/2)+\cos({-}\pi/2)\big)=0$ , straight from the integral, before appealing to any symmetry shortcut.

Why: the same orthogonality argument isolates $b_n$ instead of $a_n$ , using $\sin(n\omega_0x)$ as the matching frequency.

Common mistakes

Integrating over a window that isn't exactly one full period, which changes the constant $2/T$ out front.
Forgetting the factor of $\tfrac12$ that turns $a_0$ into the series' actual constant term $a_0/2$ .
Computing $a_n$ or $b_n$ for a single value of $n$ and treating it as the general formula instead of finding the pattern.
Not checking evenness or oddness first, which would have predicted that half the coefficients are automatically zero.