Fourier series — motivation from periodic functions

WHY do we even want this?

WHAT is a periodic function?

The simplest periodic functions are $\cos(n\omega x)$ and $\sin(n\omega x)$ . All of them share the period $T$ (they fit a whole number $n$ of cycles inside one period). So they are perfect Lego bricks for anything of period $T$ .

HOW do we build the series? (Derivation from scratch)

We claim that a periodic function $f$ of period $2L$ (so $\omega = \pi/L$ ) can be written:

$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\Big( a_n\cos\tfrac{n\pi x}{L} + b_n\sin\tfrac{n\pi x}{L}\Big)$

The whole problem is: what are the numbers $a_n, b_n$ ? We derive them, we don't memorise them.

Step 1 — The orthogonality trick (WHY it works)

The integrals we need (all over $-L$ to $L$ ):

$\int_{-L}^{L}\cos\tfrac{n\pi x}{L}\cos\tfrac{m\pi x}{L}\,dx = \begin{cases}0 & n\neq m\\ L & n=m\neq0\\ 2L & n=m=0\end{cases}$

$\int_{-L}^{L}\sin\tfrac{n\pi x}{L}\sin\tfrac{m\pi x}{L}\,dx = \begin{cases}0 & n\neq m\\ L & n=m\end{cases}$

$\int_{-L}^{L}\sin\tfrac{n\pi x}{L}\cos\tfrac{m\pi x}{L}\,dx = 0 \quad\text{(always)}$

Why are these true? Use product-to-sum, e.g. $\cos A\cos B = \tfrac12[\cos(A-B)+\cos(A+B)]$ . Each piece is $\cos(k\pi x/L)$ with integer $k$ ; integrating a full number of cycles over $[-L,L]$ gives $0$ , unless $k=0$ (a constant), which survives.

Step 2 — Extract $a_m$ (HOW)

Multiply the series by $\cos(m\pi x/L)$ and integrate $-L$ to $L$ :

$\int_{-L}^{L} f(x)\cos\tfrac{m\pi x}{L}\,dx = \underbrace{\frac{a_0}{2}\int\cos\tfrac{m\pi x}{L}dx}_{=0\ (m\geq1)} + \sum_n a_n\!\!\underbrace{\int\cos\tfrac{n\pi x}{L}\cos\tfrac{m\pi x}{L}dx}_{=L\ \text{only if }n=m} + \sum_n b_n\underbrace{(\dots)}_{=0}$

Why this step? Orthogonality kills every term except the single $n=m$ cosine term, leaving $a_m \cdot L$ . So:

$\boxed{\,a_m = \frac{1}{L}\int_{-L}^{L} f(x)\cos\tfrac{m\pi x}{L}\,dx\,}$

The $a_0/2$ choice makes $a_0 = \frac1L\int_{-L}^L f\,dx$ = twice the average, so $a_0/2$ = the average value. Neat.

Step 3 — Extract $b_m$

Same trick with $\sin(m\pi x/L)$ :

$\boxed{\,b_m = \frac{1}{L}\int_{-L}^{L} f(x)\sin\tfrac{m\pi x}{L}\,dx\,}$

Figure — Fourier series — motivation from periodic functions

Symmetry shortcut (the 80/20 win)

Worked Examples

Worked example Square wave (odd) — the classic

$f(x)=\begin{cases}-1 & -\pi<x<0\\ +1 & 0<x<\pi\end{cases}$ , period $2\pi$ (so $L=\pi$ ).

Step 1: $f$ is odd ⟹ $a_n=0$ . Why? Saves us all cosine integrals.

Step 2: $b_n=\frac1\pi\int_{-\pi}^{\pi} f\sin nx\,dx = \frac2\pi\int_0^\pi \sin nx\,dx$ . Why the 2 and the $0$ to $\pi$ ? Integrand (odd·odd = even) is symmetric, so double the right half.

Step 3: $\int_0^\pi\sin nx\,dx = \frac{1-\cos n\pi}{n}=\frac{1-(-1)^n}{n}$ . Why? This is $0$ for even $n$ , $2/n$ for odd $n$ .

Result: $b_n=\dfrac{2}{\pi}\cdot\dfrac{1-(-1)^n}{n}$ , so $f(x)=\frac{4}{\pi}\Big(\sin x+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+\dots\Big)$ Only odd harmonics — the diagram shows how adding more terms sharpens the square.

Forecast-then-Verify

Recall Forecast before reading on

Q: For an even triangle wave, before computing — which coefficients are zero, and why?

A: All $b_n=0$ , because the function is even and sines are odd; only cosines (which match even symmetry) appear. Verified by the symmetry rule.

Common Mistakes (Steel-manned)

Flashcards

What does it mean for

f

to be periodic with period

T

f(x+T)=f(x)

for all

x

; smallest such

T>0

is the fundamental period.

What single property of sines/cosines lets us extract Fourier coefficients?

Orthogonality: integrals of products of distinct sines/cosines over one period are zero.

Formula for

a_n

(period

2L

a_n=\frac1L\int_{-L}^{L} f(x)\cos\frac{n\pi x}{L}\,dx

Formula for

b_n

(period

2L

b_n=\frac1L\int_{-L}^{L} f(x)\sin\frac{n\pi x}{L}\,dx

Why is the constant term written

a_0/2

So that

a_0

uses the same

\frac1L

formula as other

a_n

; then

a_0/2

equals the average of

f

f

is even, which coefficients vanish?

All

b_n=0

(only cosines survive).

f

is odd, which coefficients vanish?

All

a_n=0

(only sines survive).

What does the Fourier series converge to at a jump discontinuity?

The midpoint

\frac12[f(x^+)+f(x^-)]

What is the Gibbs phenomenon?

Persistent ~9% overshoot of the partial sums near a jump that doesn't vanish as terms increase.

Fourier series of the square wave (±1, period

2\pi

\frac4\pi\sum_{\text{odd }n}\frac{\sin nx}{n}

— odd harmonics only.

Why do PDE solvers need Fourier series?

Separation of variables gives sine/cosine modes; matching an arbitrary initial condition requires expanding it in those modes.

Recall Feynman: explain to a 12-year-old

Imagine you have a weird wiggly line that repeats forever, like a heartbeat monitor. You also have a box of "pure musical notes," each a smooth wave. Fourier's amazing discovery: you can recreate that weird wiggle perfectly by playing the right mix of those pure notes — some loud, some soft, some high-pitched, some low. The "recipe card" for the mix is found by a clever sliding-and-adding trick (integration) that asks "how much of this note is hiding in my wiggle?" Because the notes never get confused with each other (orthogonality), each question gives one clean answer.

Connections

Separation of Variables — produces the sine/cosine modes Fourier series assembles.
Heat Equation — initial temperature profile expanded as a Fourier (sine) series.
Wave Equation — plucked-string shape decomposed into harmonics.
Orthogonality of Functions — the inner-product structure $\langle f,g\rangle=\int fg$ .
Even and Odd Functions — symmetry shortcuts for coefficients.
Fourier Transform — limit of Fourier series as period $\to\infty$ .
Gibbs Phenomenon — convergence behaviour at jumps.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho tumhare paas koi bhi repeating (periodic) function hai — jaise square wave ya heartbeat pattern. Fourier ka kamaal yeh hai ki is tedhe-medhe shape ko tum pure waves yaani $\sin$ aur $\cos$ ko mila-jhula ke banaa sakte ho. Har wave ka apna "kitna chahiye" amount hota hai — usi ko hum coefficient $a_n$ aur $b_n$ bolte hain. Yeh waves Lego blocks ki tarah hain, aur Fourier series unka recipe card hai.

Coefficients nikalne ka raaz hai orthogonality. Matlab agar tum do alag waves ko multiply karke ek period par integrate karo, answer zero aata hai. Isi wajah se jab hum poori series ko $\cos(m\pi x/L)$ se multiply karke integrate karte hain, toh saare terms gayab ho jaate hain, sirf ek bachta hai — aur wahi humein $a_m$ de deta hai. Bilkul vectors mein $\hat i\cdot\hat j=0$ jaisa funda hai.

Yeh PDEs mein kyun zaroori hai? Jab hum heat ya wave equation ko separation of variables se solve karte hain, solutions $\sin(n\pi x/L)$ wale forms mein aate hain. Lekin initial condition $f(x)$ koi bhi shape ho sakta hai. Toh us shape ko inhi sine/cosine ke sum mein todna padta hai — yahi Fourier series karti hai. Bina iske PDE ka initial condition match hi nahi hoga.

Do shortcuts yaad rakho: agar function even hai toh sirf cosines ( $b_n=0$ ), aur odd hai toh sirf sines ( $a_n=0$ ) — aadha kaam bach gaya. Aur ek warning: jahan function mein sudden jump hai, wahan series exact value nahi, balki midpoint deti hai, aur thoda overshoot karti hai jise Gibbs phenomenon bolte hain. Yeh galti exam mein bahut log karte hain, toh dhyan rakhna.

Fourier series — motivation from periodic functions

WHY do we even want this?

HOW do we build the series? (Derivation from scratch)

Step 1 — The orthogonality trick (WHY it works)

Step 2 — Extract $a_m$ (HOW)

Step 3 — Extract $b_m$

Symmetry shortcut (the 80/20 win)

Worked Examples

Forecast-then-Verify

Common Mistakes (Steel-manned)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Partial Differential Equations

Connections