Fourier series — motivation from periodic functions: Worked examples

This page is the practice arena for Fourier series. The parent note derived the recipe; here we run it on every kind of function you can be handed — even, odd, neither, constant, discontinuous, with a period that isn't the friendly $2 π$ , and a real heat-flow word problem. By the end there is no case class that can surprise you.

The framework we keep reusing (self-contained recap)

Every example below plugs into the same three formulas from the parent note. State them once here so this page stands alone.

The scenario matrix

The decision-tree has three symmetry classes (even / odd / neither) crossed with two continuity classes (continuous / has a jump) crossed with the period ( $2 π$ or general $2 L$ ). The eight examples below hit each reachable branch at least once; the last column names which branch.

Cell	Symmetry	Period $2 L$	Continuity	Twist	Example
A	Odd	$2 π$ ( $L = π$ )	Jump	classic odd+jump	Ex 1: square wave
B	Even	$2 π$ ( $L = π$ )	Continuous (corner)	corner, no jump	Ex 2: $∣ x ∣$
C	Neither	$2 π$ ( $L = π$ )	Jump	both $a_{n}$ and $b_{n}$ nonzero	Ex 3: $f (x) = x + 1$ on $(- π, π)$
D	Even	General $2 L \neq = 2 π$	Continuous	keep $\frac{nπ x}{L}$	Ex 4: $x^{2}$ on $(- 2, 2)$
E	Constant	any	Continuous	degenerate: no waves	Ex 5: $f = 3$
F	Odd	$2 π$	Jump	value at the jump (not new coeffs)	Ex 6: square-wave series at $x = 0$
G	Half-range	$L$ only given	choose extension	exam twist	Ex 7: sine vs cosine series of $f = 1$
H	Physics	$2 π$	Jump (block)	word problem, units	Ex 8: initial heat profile

Example 1 — Cell A: odd, jump, the square wave

Worked example Square wave

$f (x) = {- 1 + 1 - π < x < 0 0 < x < π$ , period $2 π$ Forecast: Before reading — which of $a_{n}, b_{n}$ are zero, and will the answer contain all $n$ or only some?

Step 1. $f (- x) = - f (x)$ , so $f$ is odd. Hence $a_{n} = 0$ for all $n$ . Why this step? Odd $\times$ cosine (even) is odd, and an odd function integrated over the symmetric interval $[- π, π]$ gives $0$ . This deletes every cosine integral before we start.

Step 2. $b_{n} = \frac{1}{π} \int_{- π}^{π} f (x) sin n x d x = \frac{2}{π} \int_{0}^{π} sin n x d x$ . Why this step? $f sin n x$ is odd $\times$ odd $=$ even, symmetric about $0$ , so the integral over $[- π, π]$ is twice the integral over $[0, π]$ , where $f = + 1$ .

Step 3. $\int_{0}^{π} sin n x d x = [\frac{- c o s n x}{n}]_{0}^{π} = \frac{1 - cos nπ}{n} = \frac{1 - ( - 1 ) ^{n}}{n}$ . Why this step? $cos nπ = (- 1)^{n}$ : it is $+ 1$ for even $n$ (giving $0$ ) and $- 1$ for odd $n$ (giving $2/ n$ ). Only odd harmonics survive.

Step 4. So $b_{n} = \frac{2}{π} \cdot \frac{1 - ( - 1 ) ^{n}}{n}$ , i.e. $b_{1} = \frac{4}{π}, b_{3} = \frac{4}{3 π}, \dots$ $f (x) = \frac{4}{π} (sin x + \frac{s i n 3 x}{3} + \frac{s i n 5 x}{5} + \dots)$

Verify: $b_{1} = \frac{4}{π} \approx 1.2732$ , $b_{2} = 0$ , $b_{3} = \frac{4}{3 π} \approx 0.4244$ . This is exactly the parent's result — consistent.

The figure below carries the geometry of this convergence.

Figure — Fourier series — motivation from periodic functions

Example 2 — Cell B: even, continuous corner

Worked example Triangle-like

$f (x) = ∣ x ∣$ on $(- π, π)$ , period $2 π$ Forecast: $∣ x ∣$ is even and has no jump (just a sharp corner at $0$ ). Guess: which coefficients vanish, and will the series converge fast or slow?

Step 1. $f (- x) = ∣ - x ∣ = ∣ x ∣ = f (x)$ — even. So $b_{n} = 0$ . Why this step? Even $\times$ sine (odd) is odd $\Rightarrow$ integrates to $0$ . Only cosines remain.

Step 2. $a_{0} = \frac{1}{π} \int_{- π}^{π} ∣ x ∣ d x = \frac{2}{π} \int_{0}^{π} x d x = \frac{2}{π} \cdot \frac{π ^{2}}{2} = π$ . Why this step? $a_{0} /2$ is the average height. Average of $∣ x ∣$ is $π /2$ , and indeed $a_{0} /2 = π /2$ . A built-in sanity check.

Step 3. $a_{n} = \frac{2}{π} \int_{0}^{π} x cos n x d x$ . Integrate by parts: $\int_{0}^{π} x cos n x d x = [\frac{x s i n n x}{n}]_{0}^{π} - \int_{0}^{π} \frac{s i n n x}{n} d x = 0 + [\frac{c o s n x}{n ^{2}}]_{0}^{π} = \frac{( - 1 ) ^{n} - 1}{n ^{2}} .$ Why this step? Integration by parts is the right tool because the integrand is a polynomial $\times$ trig — differentiating $x$ kills the polynomial, moving us toward a pure trig integral we can do. (The boundary term $\frac{x s i n n x}{n}$ vanishes at both ends because $sin nπ = 0$ and $x = 0$ at the lower limit.)

Step 4. $a_{n} = \frac{2}{π} \cdot \frac{( - 1 ) ^{n} - 1}{n ^{2}}$ : zero for even $n$ , and $- \frac{4}{π n ^{2}}$ for odd $n$ . $∣ x ∣ = \frac{π}{2} - \frac{4}{π} (cos x + \frac{c o s 3 x}{9} + \frac{c o s 5 x}{25} + \dots)$

Verify: $a_{1} = - \frac{4}{π} \approx - 1.2732$ , $a_{2} = 0$ , $a_{3} = - \frac{4}{9 π} \approx - 0.1415$ . Because coefficients shrink like $1/ n^{2}$ (vs $1/ n$ for the jump square wave), this converges faster — the corner is milder than a jump. Plug $x = 0$ : $0 = \frac{π}{2} - \frac{4}{π} \sum_{odd} \frac{1}{n ^{2}}$ , giving $\sum_{odd} \frac{1}{n ^{2}} = \frac{π ^{2}}{8}$ , a known identity.

Example 3 — Cell C: neither even nor odd (and it jumps)

Worked example

$f (x) = x + 1$ on $(- π, π)$ , period $2 π$ Forecast: Adding the constant $1$ to $x$ breaks odd symmetry: $f (- x) = - x + 1 \neq = \pm f (x)$ . So this is neither even nor odd — and it still jumps at $\pm π$ . Guess: which coefficients survive now?

Step 1. Split $f = x + 1$ . The constant $1$ contributes only to $a_{0}$ ; the linear $x$ is odd and contributes only to $b_{n}$ . Why this step? Fourier coefficients are linear in $f$ , so we compute each piece and add. This is far easier than attacking $x + 1$ whole.

Step 2. From the constant: $a_{0} = \frac{1}{π} \int_{- π}^{π} 1 d x = 2$ , and all other $a_{n} = \frac{1}{π} \int_{- π}^{π} 1 \cdot cos n x d x = 0$ . Why this step? $\int_{- π}^{π} cos n x d x = \frac{s i n nπ}{n} \cdot 2$ -type terms vanish because $sin nπ = 0$ ; a constant needs no non-constant cosine.

Step 3. From the linear part (reusing the sawtooth work): $b_{n} = \frac{1}{π} \int_{- π}^{π} x sin n x d x = \frac{2 ( - 1 ) ^{n + 1}}{n}$ . Why this step? $x$ is odd, $sin n x$ is odd, product even; by parts $\int_{- π}^{π} x sin n x d x = 2 \int_{0}^{π} x sin n x d x = \frac{- 2 π ( - 1 ) ^{n}}{n}$ , and dividing by $π$ gives $\frac{2 ( - 1 ) ^{n + 1}}{n}$ .

Step 4 (the twist). Both a nonzero average ( $a_{0} /2 = 1$ ) and a full set of sines appear: $x + 1 = 1 + 2 (sin x - \frac{s i n 2 x}{2} + \frac{s i n 3 x}{3} - \dots) .$ Every $n$ is present in $b_{n}$ , and there is a standalone constant — the signature of a "neither" function.

Verify: $a_{0} = 2$ (so average $= 1$ ✓, matches $\frac{1}{2 π} \int_{- π}^{π} (x + 1) d x = 1$ ), $b_{1} = 2$ , $b_{2} = - 1$ , $b_{3} = \frac{2}{3}$ . At $x = π /2$ : $\frac{π}{2} + 1 = 1 + 2 (1 - \frac{1}{3} + \frac{1}{5} - \dots) = 1 + 2 \cdot \frac{π}{4}$ ✓ (Leibniz again). At the jump $x = π$ the series converges to the midpoint $\frac{1}{2} [f (π^{-}) + f (- π^{+})] = \frac{1}{2} [(π + 1) + (- π + 1)] = 1$ .

Example 4 — Cell D: even, general period $2 L \neq = 2 π$

Worked example

$f (x) = x^{2}$ on $(- 2, 2)$ , period $4$ (so $L = 2$ ) Forecast: The period is now $4$ , not $2 π$ . Which of the three formulas break if you forget and write $n x$ instead of $\frac{nπ x}{2}$ ?

Step 1. $x^{2}$ is even $\Rightarrow b_{n} = 0$ . Use $\frac{nπ x}{L} = \frac{nπ x}{2}$ everywhere. Why this step? The wave must complete whole cycles in the period $2 L = 4$ ; the argument $\frac{nπ x}{L}$ guarantees that. Writing $n x$ would use the wrong period and give nonsense.

Step 2. $a_{0} = \frac{1}{L} \int_{- L}^{L} x^{2} d x = \frac{1}{2} \int_{- 2}^{2} x^{2} d x = \frac{1}{2} \cdot \frac{2 \cdot 2 ^{3}}{3} = \frac{1}{2} \cdot \frac{16}{3} = \frac{8}{3}$ . So the average is $a_{0} /2 = \frac{4}{3}$ . (Average of $x^{2}$ on $[- 2, 2]$ is $\frac{1}{4} \int_{- 2}^{2} x^{2} = \frac{4}{3}$ ✓.)

Step 3a (parity reduction). $a_{n} = \frac{1}{L} \int_{- L}^{L} x^{2} cos \frac{nπ x}{L} d x$ . The integrand is even $\times$ even $=$ even, so it is symmetric about $x = 0$ : $a_{n} = \frac{1}{L} \int_{- L}^{L} x^{2} cos \frac{nπ x}{L} d x = \frac{2}{L} \int_{0}^{L} x^{2} cos \frac{nπ x}{L} d x .$ Why this step? Folding the symmetric integral in half halves the algebra and keeps the limits clean (starting from $0$ ).

Step 3b (first integration by parts). Let $k = \frac{nπ}{L}$ . With $u = x^{2}, d v = cos (k x) d x$ : $\int_{0}^{L} x^{2} cos (k x) d x = [\frac{x ^{2} s i n k x}{k}]_{0}^{L} - \int_{0}^{L} \frac{2 x s i n k x}{k} d x .$ The boundary term is $\frac{L ^{2} s i n ( k L )}{k} = \frac{L ^{2} s i n ( nπ )}{k} = 0$ (since $k L = nπ$ and $sin nπ = 0$ ). Why this step? Differentiating $x^{2}$ drops it to $2 x$ — one power gone, one to go.

Step 3c (second integration by parts). For the remaining $\int_{0}^{L} 2 x sin (k x) d x$ , take $u = 2 x, d v = sin (k x) d x$ : $\int_{0}^{L} 2 x sin (k x) d x = [\frac{- 2 x c o s k x}{k}]_{0}^{L} + \int_{0}^{L} \frac{2 c o s k x}{k} d x = \frac{- 2 L c o s ( nπ )}{k} + \frac{2 s i n ( nπ )}{k ^{2}} = \frac{- 2 L ( - 1 ) ^{n}}{k} .$ Putting Step 3b and 3c together (mind the minus sign in 3b): $\int_{0}^{L} x^{2} cos (k x) d x = 0 - \frac{1}{k} (\frac{- 2 L ( - 1 ) ^{n}}{k}) = \frac{2 L ( - 1 ) ^{n}}{k ^{2}} = \frac{2 L ( - 1 ) ^{n} L ^{2}}{n ^{2} π ^{2}} = \frac{2 L ^{3} ( - 1 ) ^{n}}{n ^{2} π ^{2}} .$ Why this step? $2 x$ differentiates to a constant, so the second part-step finishes the job.

Step 3d. Therefore $a_{n} = \frac{2}{L} \cdot \frac{2 L ^{3} ( - 1 ) ^{n}}{n ^{2} π ^{2}} = \frac{4 L ^{2} ( - 1 ) ^{n}}{n ^{2} π ^{2}} = \frac{16 ( - 1 ) ^{n}}{n ^{2} π ^{2}} (L = 2) .$

Step 4. $x^{2} = \frac{4}{3} + \frac{16}{π ^{2}} \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n}}{n ^{2}} cos \frac{nπ x}{2} .$

Verify: $a_{1} = \frac{- 16}{π ^{2}} \approx - 1.6211$ , $a_{2} = \frac{16}{4 π ^{2}} = \frac{4}{π ^{2}} \approx 0.4053$ . At $x = 2$ (where $cos \frac{nπ \cdot 2}{2} = cos nπ = (- 1)^{n}$ ) the series gives $\frac{4}{3} + \frac{16}{π ^{2}} \sum \frac{( - 1 ) ^{n} ( - 1 ) ^{n}}{n ^{2}} = \frac{4}{3} + \frac{16}{π ^{2}} \cdot \frac{π ^{2}}{6} = \frac{4}{3} + \frac{8}{3} = 4 = 2^{2}$ ✓ (used $\sum 1/ n^{2} = π^{2} /6$ ).

Example 5 — Cell E: the degenerate constant

Example 6 — Cell F: the value exactly at a jump

Example 7 — Cell G: half-range twist (exam classic)

Intuition What a half-range expansion is, and where the

$\frac{2}{π}$ comes from Sometimes $f$ is given only on $(0, π)$ — half of a full $2 π$ -period. To use the Fourier machinery we must first invent the missing half on $(- π, 0)$ . We are free to choose it, and two choices are special: extend oddly (mirror-and-flip, forcing a sine-only series) or evenly (mirror straight across, forcing a cosine-only series). Either way the period becomes $2 π$ , so $L = π$ .

Why the coefficient formula gains a $\frac{2}{π}$ : start from the general $b_{n} = \frac{1}{L} \int_{- L}^{L} f sin \frac{nπ x}{L} d x$ with $L = π$ , giving $b_{n} = \frac{1}{π} \int_{- π}^{π} f sin n x d x$ . For the odd extension the integrand $f sin n x$ is even (odd $\times$ odd), symmetric about $0$ , so the $[- π, π]$ integral is twice the $[0, π]$ integral — and the $\frac{1}{π}$ becomes $\frac{2}{π}$ : $b_{n} = \frac{2}{π} \int_{0}^{π} f (x) sin n x d x .$ Identically, for the even extension $f cos n x$ is even, so $a_{n} = \frac{2}{π} \int_{0}^{π} f (x) cos n x d x$ . The $\frac{2}{π}$ is nothing new — it is the standard formula folded in half by symmetry.

Worked example Expand

$f (x) = 1$ on $(0, π)$ as (a) a sine series, (b) a cosine series Forecast: Same function, same interval — yet you can get two different valid Fourier series. How? Because you choose how to extend $f$ to $(- π, 0)$ .

Step 1 (sine series — odd extension). Reflect $f$ oddly: $f (- x) = - 1$ on $(- π, 0)$ . Then $a_{n} = 0$ and, by the half-range formula above, $b_{n} = \frac{2}{π} \int_{0}^{π} 1 \cdot sin n x d x = \frac{2}{π} [\frac{- c o s n x}{n}]_{0}^{π} = \frac{2}{π} \cdot \frac{1 - c o s nπ}{n} = \frac{2}{π} \cdot \frac{1 - ( - 1 ) ^{n}}{n} .$ Why this step? An odd extension forces a sine-only series; this is exactly what a PDE with $f = 0$ at both ends (Dirichlet boundary) demands.

Step 2 (evaluate the sine coefficients). $b_{n}$ is $0$ for even $n$ and $\frac{4}{π n}$ for odd $n$ , so $1 = \frac{4}{π} (sin x + \frac{s i n 3 x}{3} + \frac{s i n 5 x}{5} + \dots) (0 < x < π) .$ Why this step? The odd extension of the constant $1$ is the $\pm 1$ square wave, so the coefficients match Ex 1 exactly — a satisfying consistency check.

Step 3 (cosine series — even extension). Reflect $f$ evenly: $f (- x) = + 1$ . Then $b_{n} = 0$ , and $a_{0} = \frac{2}{π} \int_{0}^{π} 1 d x = 2, a_{n} = \frac{2}{π} \int_{0}^{π} cos n x d x = \frac{2}{π} \cdot \frac{s i n nπ}{n} = 0 (n \geq 1) .$ Why this step? The even extension of a constant is just the constant again — a flat line needs no cosines beyond its average.

Step 4 (write the cosine series). $1 = \frac{a _{0}}{2} = 1 (0 < x < π) .$ The two series look utterly different (an infinite sine sum vs. a single constant) yet represent the same $f$ on $(0, π)$ — that is the exam twist.

Verify: sine version at $x = π /2$ : $\frac{4}{π} (1 - \frac{1}{3} + \frac{1}{5} - \dots) = \frac{4}{π} \cdot \frac{π}{4} = 1$ ✓. Cosine version $= 1$ trivially ✓. Both equal $1$ on the given half-interval.

Example 8 — Cell H: physics word problem (even block, with a jump)

Worked example A rod of length

$π$ m starts at temperature $u (x, 0) = 100$ °C in the middle third $(π /3, 2 π /3)$ and $0$ °C elsewhere. Find the sine coefficients of the initial profile. Forecast: The hot block sits symmetrically in the middle, so the profile is symmetric about the rod's centre yet discontinuous (jumps at the block edges) — the "even+jump" branch. Separation of Variables on the Heat Equation gives modes $sin (n x)$ ; we must expand this blocky initial temperature in them. Which coefficients dominate?

Step 1. Model: $u (x, 0) = {1000 π /3 < x < 2 π /3 else$ on $(0, π)$ . Use the sine (odd, half-range) expansion because the rod ends are held at $0$ °C, i.e. $u (0) = u (π) = 0$ . Why sine? The boundary conditions kill cosines; only $sin (n x)$ satisfies $u (0) = u (π) = 0$ . We reuse the half-range formula $b_{n} = \frac{2}{π} \int_{0}^{π} u sin n x d x$ from Ex 7.

Step 2. $b_{n} = \frac{2}{π} \int_{π /3}^{2 π /3} 100 sin n x d x = \frac{200}{π} [\frac{- cos n x}{n}]_{π /3}^{2 π /3} = \frac{200}{nπ} (cos \frac{nπ}{3} - cos \frac{2 nπ}{3}) .$ Why this step? Integrate only where $u = 100$ ; elsewhere the integrand is $0$ .

Step 3. Compute the amplitudes (°C): with $n = 1$ , $cos \frac{π}{3} - cos \frac{2 π}{3} = \frac{1}{2} - (- \frac{1}{2}) = 1$ , so $b_{1} = \frac{200}{π} \approx 63.66$ . With $n = 2$ , $cos \frac{2 π}{3} - cos \frac{4 π}{3} = - \frac{1}{2} - (- \frac{1}{2}) = 0$ , so $b_{2} = 0$ . With $n = 3$ , $cos π - cos 2 π = - 1 - 1 = - 2$ , so $b_{3} = \frac{200}{3 π} \cdot (- 2) = \frac{- 400}{3 π} \approx - 42.44$ . Why check units? $b_{n}$ carries the same unit as $u$ , i.e. °C — it is the amplitude of that temperature mode.

Step 4. The initial profile is $u (x, 0) = \sum_{n = 1}^{\infty} b_{n} sin n x$ , and as time runs the Heat Equation multiplies mode $n$ by $e^{- n^{2} t}$ , so high- $n$ modes decay fastest — the sharp block smooths quickly. (Compare the Wave Equation, where each mode instead oscillates in time.)

Verify: $b_{1} \approx 63.66$ °C, $b_{2} = 0$ , $b_{3} \approx - 42.44$ °C. The dominant $b_{1}$ hump is the slowest-decaying, so late-time temperature looks like a single $sin x$ arch — physically sensible.

Recall Scenario checklist before any Fourier problem

Symmetry? ::: Even ⇒ only $a_{n}$ ; odd ⇒ only $b_{n}$ ; neither ⇒ both. Period $2 L = ?$ ::: Always write $\frac{nπ x}{L}$ ; only simplify to $n x$ when $L = π$ . Jump present? ::: Series converges to the midpoint there (Gibbs overshoot nearby). Half-range given? ::: You choose even (cosine) or odd (sine) extension per the boundary conditions; the $\frac{1}{π}$ becomes $\frac{2}{π}$ by symmetry.

Connections

Parent recipe & derivation: Fourier series motivation
Why coefficients isolate: Orthogonality of Functions
Symmetry shortcuts: Even and Odd Functions
Where these series come from: Separation of Variables, applied to the Heat Equation and Wave Equation
The overshoot at jumps: Gibbs Phenomenon
The continuous-spectrum sibling: Fourier Transform