Visual walkthrough — Fourier series — motivation from periodic functions

4.7.3 · D2 · Maths › Partial Differential Equations › Fourier series — motivation from periodic functions

Hum ek aisi function $f$ ke saath kaam karte hain jo repeat karti hai. Kisi bhi formula se pehle, yeh pin down karte hain ki "repeat" ka matlab aakhir hai kya.

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Step 1 — "Periodic" dikhta kaisa hai

KYA HAI. Ek function ek machine hai: usse ek number $x$ (line par ek position) do, woh height $f(x)$ return karta hai. Ek periodic function woh hai jiska graph ek stencil hai — use sideways slide karo aur woh exactly apne upar aa jaata hai. Woh slide distance jis par woh land karta hai period $T$ hai: symbols mein, $f(x+T)=f(x)$ — "input ko $T$ se shift karo, output unchanged rehta hai."

KYUN. Neeche ki saari cheez ek period par jeeti hai. Agar shape endlessly repeat karti hai, toh hume sirf ek tile samajhni hoti hai; baaki copies hain. Yahi akela reason hai ki hum baad mein sirf ek width par integrate karte hain.

PICTURE. Neeche, wahi tile (blue) baar baar stamp ki gayi hai. Orange bracket ek period $T$ hai. Dhyan do ki $x$ par aur $x+T$ par height (red dots) barabar hai — yahi definition hai.

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Step 2 — Building bricks: waves jo tile mein fit hote hain

KYA HAI. Hamare bricks pure waves $\cos \frac{n\pi x}{L}$ aur $\sin \frac{n\pi x}{L}$ hain, har whole number $n=1,2,3,\dots$ ke liye. Andar wala piece $\frac{n\pi x}{L}$ ek speed dial ki tarah padho: $n$ count karta hai ki wave $2L$ width ki ek tile mein kitne complete wiggles pack karti hai.

YEH bricks kyun, seedhi lines kyun nahi? Kyunki har ek already period $2L$ se repeat karta hai: $x$ ko $2L$ se badhao aur angle $\frac{n\pi x}{L}$, $2n\pi$ se jump karta hai — puri turns ki whole number — toh wave waahan wापस aa jaati hai jahan thi. Un cheezaon ka sum jo har ek $2L$ mein repeat karti hain automatically $2L$ mein repeat karta hai — bricks wall se match karte hain.

PICTURE. Teen bricks: $n=1$ (ek wiggle), $n=2$ (do), $n=3$ (teen). Sab apna period tile ke edges par ek saath start aur khatam karte hain (dashed).

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Step 3 — Vectors se idea: "dot karke extract karo"

KYA HAI. Maano ek 2-D arrow hai $\vec{v}=a\hat{i}+b\hat{j}$. Sirf number $a$ nikaalने ke liye, $\hat{i}$ se dot karo: $\vec{v}\cdot \hat{i}=a$, kyunki $\hat{i}\cdot \hat{i}=1$ lekin $\hat{j}\cdot \hat{i}=0$. Perpendicular partner kuch contribute nahi karta.

Vectors kyun laaye? Kyunki haari series ka exactly same shape hai: $f$ ek arrow hai, bricks axis directions hain, aur $a_n,b_n$ uske coordinates hain. Agar hum functions ke liye ek "dot product" dhundh sakein jismein alag bricks perpendicular hon, toh wahi extraction trick kaam karti hai.

PICTURE. Left: vector case — $\hat{i}$ par project karna $\hat{j}$ part ko ignore karta hai. Right: analogy — $f$ arrow hai, cosine-1 aur sine-1 do perpendicular axes hain.

Function ka "dot product" ek period par integral hai: $⟨g,h⟩={\int }_{-L}^{L}g(x)h(x)dx$. (Integral kyun? Yeh tile ke across height-by-height product sum karta hai, exactly jaise dot component-by-component sum karta hai.) Dekho Orthogonality of Functions.

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Step 4 — Alag bricks perpendicular kyun hain (orthogonality)

KYA HAI. Hum check karte hain: do alag bricks multiply karo aur ek tile par integrate karo. Jawab zero hai. Ek brick ko itself se multiply karke integrate karo: tumhe ek non-zero number milta hai ($L$).

Yeh sach KYUN hona chahiye — mechanism. Product-to-sum identity use karo: $\cos A\cos B=\frac{1}{2}[\cos (A-B)+\cos (A+B)].$ $A=\frac{n\pi x}{L}$ aur $B=\frac{m\pi x}{L}$ ke saath, right-hand ka har piece $\cos \frac{(n\pm m)\pi x}{L}$ hai — ek aur pure wave with integer speed $k=n\pm m$. Ek pure wave ko uske cycles ki poori number par integrate karna axis ke upar aur neeche equal area sweep karta hai: woh cancel hokar zero ho jaate hain. Iska ek hi bachne ka rasta hai jab speed $k=0$ ho (yaani $A-B$ term mein $n=m$): tab $\cos 0=1$, height $1$ ki flat line, jiska width $2L$ par integral $2L$ hai, aur $\frac{1}{2}$ use $L$ banaa deta hai.

PICTURE. Upar: $\cos 1\cdot \cos 2$ — upar shaded area (blue) exactly neeche area (orange) se match karta hai: net $0$. Neeche: ${\cos }^{2}1$ poora $\ge 0$ rehta hai, toh uska area bachta hai.

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Step 5 — "$\cos \frac{m\pi x}{L}$ se dot karke" $a_m$ extract karna

KYA HAI. Poori series lo, dono sides ko ek chosen cosine brick $\cos \frac{m\pi x}{L}$ se multiply karo, aur $-L$ se $L$ tak integrate karo.

KYUN. Step 4 ne promise kiya tha ki yeh "dot" har galat term ko vanish kar dega. Tabahi dekho:

${\int }_{-L}^{L}f(x)\cos \frac{m\pi x}{L}dx=\underset{=0(m\ge 1,\text{full cycles cancel})}{\underbrace{\frac{a_0}{2}{\int }_{-L}^L\cos \frac{m\pi x}{L}dx}}+{\sum }_n{a}_n\underset{=L\text{only when }n=m,\text{else }0}{\underbrace{{\int }_{-L}^L\cos \frac{n\pi x}{L}\cos \frac{m\pi x}{L}dx}}+{\sum }_n{b}_n\underset{=0\text{always}}{\underbrace{{\int }_{-L}^L\sin \frac{n\pi x}{L}\cos \frac{m\pi x}{L}dx}}$

Sab kuch mar jaata hai siwaaye ek term $n=m$ ke, jo $a_m\cdot L$ chhodta hai. $L$ se divide karo:

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

Term-by-term: $\frac{1}{L}$, self-overlap ke $L$ ko undo karta hai; $f(x)$ given shape hai; $\cos \frac{m\pi x}{L}$ woh probe hai jo $m$-th cosine dial select karta hai.

PICTURE. Left par infinite tangle; right par "multiply-and-integrate" ki sieve exactly ek marble ($a_m$) ko through jaane deti hai.

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Step 6 — "$\sin \frac{m\pi x}{L}$ se dot karke" $b_m$ extract karna

KYA HAI. Same move, sine probe. Series ko $\sin \frac{m\pi x}{L}$ se multiply karo, tile par integrate karo.

KYUN. Sine–cosine cross fact se ($=0$ always) saare cosine terms aur baseline vanish ho jaate hain; sine orthogonality se sirf $n=m$ wala sine value $L$ ke saath bachta hai:

${\int }_{-L}^{L}f(x)\sin \frac{m\pi x}{L}dx=b_m\cdot L⟹\boxed{b_m=\frac{1}{L}{\int }_{-L}^{L}f(x)\sin \frac{m\pi x}{L}dx}$

PICTURE. Wahi sieve, sine-coloured, sirf $b_m$ ko pakadti hai.

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Step 7 — Degenerate cases (reader ko kabhi stranded mat chhoddo)

KYA AUR KYUN — teen edges jinhe formula survive karna chahiye:

1. $f$ even hai ($f(-x)=f(x)$, $x=0$ ke baare mein mirror-symmetric). Tab $f\cdot \sin$ hai even·odd = odd, toh har $b_m=0$. Sirf cosines bachte hain — cosines khud even hain, toh woh ek even shape mein fit hote hain. Aadhe integrals free hain.
2. $f$ odd hai ($f(-x)=-f(x)$, origin ke baare mein half-turn mein spin karta hai). Tab $f\cdot \cos$ hai odd·even = odd, toh har $a_m=0$ hai, including $a_0$: ek odd function ka average $0$ hota hai. Sirf sines rehte hain.
3. $f$ flat constant hai $f(x)=c$. Yeh even hai, toh saare $b_m=0$; aur kisi bhi wave $n\ge 1$ ke saath net cycles zero hain, toh woh $a_m=0$ bhi. Sirf $a_0/2=c$ bachta hai — ek constant apna khud ka average hai, koi waves nahi chahiye.

PICTURE. Left: even shape ek cosine ke saath flush baithta hai. Middle: odd shape ek sine se match karta hai, aur uska signed area kisi bhi cosine ke against cancel ho jaata hai. Right: ek constant pure baseline hai.

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Step 8 — Ise kaam karte dekhna: square wave khud ko rebuild karti hai

KYA HAI. Odd square wave lo $f(x)=+1$ on $(0,\pi )$, $-1$ on $(-\pi ,0)$, period $2\pi$ toh $L=\pi$. Odd ⟹ saare $a_n=0$ (Step 7). Sine dials aate hain: $b_n=\frac{2}{\pi }\cdot \frac{1-(-1{)}^n}{n}=\{\begin{array}{ll}\frac{4}{\pi n}& n\text{ odd}\\ 0& n\text{ even}\end{array}\Rightarrow f(x)=\frac{4}{\pi }(\sin x+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+\cdots ).$

KYUN yeh convince karta hai. Har nayi odd brick vertical wall ko steepen karti hai aur plateau ko flatten karti hai — Steps 5–6 mein mile dials exactly woh amounts hain jo pile ko square se agree karaate hain.

PICTURE. 1, 3, aur 15 terms ke partial sums square ki taraf chadhte hue; dhyan do jump par woh chhota Gibbs ear jo kabhi fully shrink nahi karta.

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Ek-picture summary

Left se right padho: ek repeating tile ($T=2L$) → perpendicular wave-axes mein decompose karo → har axis ko multiply-and-integrate se probe karo (orthogonality sab ko maar deta hai siwaaye ek ke) → dials $a_n,b_n$ nikalte hain → bricks ko wापस $f$ mein stack karo.

Recall Feynman retelling — plain words mein wापस bolo

Hamare paas ek shape thi jo har $2L$ mein repeat karti thi. Humne guess kiya ki yeh pure waves ka pile hai, har ek kisi unknown loudness par set. Ek wave ki loudness read karne ke liye, humne vector trick udhaar li: ek vector ka $x$-part paane ke liye $\hat{i}$ se dot karo aur $\hat{j}$-part drop ho jaata hai kyunki woh perpendicular hain. Functions ke liye, "dot" matlab hai dono ko multiply karo aur ek tile par integrate karo, aur yahi miracle hai — do alag waves multiply-and-integrate karke zero dete hain (unka overlap axis ke upar aur neeche cancel ho jaata hai), jabki ek wave apne saath $L$ deti hai. Toh jab hum poore pile ko ek probe wave se multiply karke integrate karte hain, har stranger vanish ho jaata hai aur ek survivor rehta hai, jo hume exactly woh dial de deta hai: $a_m=\frac{1}{L}\int f\cos$, $b_m=\frac{1}{L}\int f\sin$. Even shapes ko sirf cosines chahiye, odd shapes ko sirf sines, constants ko sirf baseline; aur ek cliff-edge jump par pile midpoint par agree karta hai, hamesha ek tiny 9% Gibbs ear pahnta hua.

Recall

$a_m$ extract karna: $f$ ko kis probe se multiply karo, phir kya karo? ::: $\cos \frac{m\pi x}{L}$ se multiply karo aur $-L$ se $L$ tak integrate karo; orthogonality $a_{m}L$ chhodti hai, toh $a_m=\frac{1}{L}{\int }_{-L}^{L}f\cos \frac{m\pi x}{L}dx$. Saare galat terms kyun vanish ho jaate hain? ::: Alag sines/cosines orthogonal hain — unka product ek period par zero integrate karta hai; sirf matching brick value $L$ ke saath bachta hai. Square-wave coefficient $b_n$? ::: $b_n=\frac{2}{\pi }\cdot \frac{1-(-1{)}^n}{n}$: odd $n$ ke liye $\frac{4}{\pi n}$, even $n$ ke liye $0$.