Visual walkthrough — Half-range sine and cosine series

4.7.6 · D2 · Maths › Partial Differential Equations › Half-range sine and cosine series

Hum, bilkul zero se, yeh ek result derive karenge:

$$b_n=\frac{2}{L}{\int }_0^{L}f(x)\sin \frac{n\pi x}{L}dx$$

aur isme har symbol ko samjhenge.

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Step 1 — Hume sirf aadhi kahani pata hai

KYA. Hume ek function $f(x)$ diya gaya hai jo sirf $x$ ke liye defined hai, $0$ aur $L$ ke beech mein. Socho ek metal rod ke saath temperature reading — rod left wall ($x=0$) se shuru hoti hai aur right wall ($x=L$) par khatam hoti hai. $0$ ke baaye, kuch bhi nahi hai — na koi rule, na koi values.

YEH KYUN MATTER KARTA HAI. Fourier machine (woh tool jo kisi bhi repeating wiggle ko sines aur cosines ke sum mein likhti hai) ko ek full symmetric interval $[-L,L]$ par function chahiye, taaki woh left aur right side compare kar sake. Hamare paas sirf right side $[0,L]$ hai. To machine jo input chahti hai uska exactly aadha hume missing hai.

PICTURE. Solid burnt-orange curve wohi hai jo hume di gayi thi; left mein grey shaded region "unknown zone" hai — bilkul blank.

Abhi tak ke symbols:

• $x$ ::: rod ke saath position, $0$ se $L$ tak ka ek number.
• $L$ ::: us interval ki length jo hume pata hai (yahan right wall bhi).
• $f(x)$ ::: diye gaye values, sirf right side par draw kiye gaye.

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Step 2 — Hum left half KHUD BANANE ke liye FREE hain

KYA. Kyunki kisi ne nahi bataya ki $x<0$ ke liye kya hota hai, hum missing values choose kar sakte hain. Do natural choices hain. Is walkthrough mein hum odd choice lete hain: curve ko origin se reflect karo — ek saath left–right aur ulta flip karo.

YEH CHOICE KYUN. Ek odd function woh hoti hai jahan $f(-x)=-f(x)$: distance $x$ par right side par jo bhi height hai, left side par same distance par uski negative rakh do. Is particular flip ki ek magic property hai (Step 4 mein reveal hogi) jo har cosine ko khatam kar deti hai. Yeh curve ko origin par zero se guzarne par bhi force karta hai — fixed walls wali problems ke liye convenient.

PICTURE. Original (burnt orange) wahi rehta hai; left par nayi plum curve point-flipped copy hai. Origin par chhota rotation arrow notice karo — odd extension ek $180^{\circ }$ spin hai point $(0,0)$ ke baare mein.

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Step 3 — Ab full Fourier machine apply hoti hai

KYA. $[-L,L]$ par values hone se, hum standard full Fourier series likh sakte hain. Yeh kehta hai: koi bhi well-behaved periodic wiggle ek constant plus cosines ka dhera plus sines ka dhera hota hai.

$$f(x)=\underset{\text{average height}}{\underbrace{\frac{a_0}{2}}}+\sum _{n=1}^{\infty }[\underset{\begin{array}{c}\text{how much}\\ \text{of }n\text{-th cosine}\end{array}}{\underbrace{a_n}}\underset{\begin{array}{c}\text{even wave,}\\ n\text{ humps}\end{array}}{\underbrace{\cos \frac{n\pi x}{L}}}+\underset{\begin{array}{c}\text{how much}\\ \text{of }n\text{-th sine}\end{array}}{\underbrace{b_n}}\underset{\begin{array}{c}\text{odd wave,}\\ n\text{ humps}\end{array}}{\underbrace{\sin \frac{n\pi x}{L}}}]$$

YEH PIECES KYUN. $\frac{n\pi x}{L}$ ke $\cos$ aur $\sin$ wahi pure waves hain jo interval mein $n$ humps fit karti hain aur period $2L$ se repeat karti hain. Numbers $a_n,b_n$ "recipe amounts" hain — har wave kitni loud bajti hai. Coefficients integrals se aate hain (woh tool jo overlap measure karta hai):

$$a_n=\frac{1}{L}{\int }_{-L}^{L}f(x)\cos \frac{n\pi x}{L}dx,b_n=\frac{1}{L}{\int }_{-L}^{L}f(x)\sin \frac{n\pi x}{L}dx.$$

YAHAN INTEGRAL KYUN? Integral ek continuous sum hai — yeh $f\times$(ek test wave) ka product poore interval par add karta hai. Agar $f$ us test wave ki taraf jhukta hai, to products pile up hote hain positive aur coefficient bada hota hai; agar disagree karein, positives aur negatives cancel ho jaate hain. Yeh ek similarity meter hai. Dekho Fourier Series — full range.

PICTURE. Bar chart "recipe" dikhata hai: kuch cosine bars (teal) aur kuch sine bars (plum), har ek ki height $a_n$ ya $b_n$. Abhi hume nahi pata kaun se zero hain — yeh Step 4 ka kaam hai.

Introduce kiye gaye symbols:

• $n$ ::: ek counting number $1,2,3,\dots$ — wave mein kitne humps hain.
• $\frac{n\pi x}{L}$ ::: $\sin /\cos$ mein diya jaane wala angle, scale kiya gaya taaki $[0,L]$ mein $n$ full humps fit hon.
• $a_0/2$ ::: interval par $f$ ki average height.

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Step 4 — Cosines ko marte hue dekho (odd × even = odd)

KYA. Cosine coefficient integrand lo: $f(x)\cos \frac{n\pi x}{L}$. Hamara $f$ odd hai (humne aise banaya). Cosine even hai ($\cos (-\theta )=\cos \theta$ — iska graph vertical axis ke across mirror hai). Ek odd cheez aur ek even cheez ka product odd hota hai.

YEH USE KYUN MARTA HAI. Symmetric interval $[-L,L]$ par ek odd function integrate karo aur tumhe exactly zero milega: right par har positive bump, left par ek equal negative bump se match hoti hai. Woh pairs mein cancel ho jaate hain. To

$$a_n=\frac{1}{L}{\int }_{-L}^L\underset{\text{odd}}{\underbrace{f(x)}}\underset{\text{even}}{\underbrace{\cos \frac{n\pi x}{L}}}dx=\frac{1}{L}{\int }_{-L}^L(\text{odd})dx=0.$$

Har cosine amount zero hai. Step 3 ke teal bars sab gayab ho jaate hain.

PICTURE. Shaded area ka left half (plum, negative) right half (orange, positive) ka exact mirror-and-flip hai. Dono signed areas size mein equal hain, sign mein opposite — total area $=0$.

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Step 5 — Sines ko double hote dekho (odd × odd = even)

KYA. Ab sine coefficient integrand: $f(x)\sin \frac{n\pi x}{L}$. Sine odd hai ($\sin (-\theta )=-\sin \theta$ — left flip karo to down flip milta hai). Odd $\times$ odd $=$ even.

YEH DOUBLE KYUN HOTA HAI. Ek even function ($g(-x)=g(x)$) ka left half right half ki perfect mirror copy hoti hai — same sign, same size. To $-L$ se $0$ tak ka area equals karta hai $0$ se $L$ tak ka area. Poore symmetric interval par integrate karne ki jagah, right half par integrate karo aur two se multiply karo:

$$b_n=\frac{1}{L}{\int }_{-L}^L\underset{\text{even}}{\underbrace{f(x)\sin \frac{n\pi x}{L}}}dx=\frac{1}{L}\cdot 2{\int }_0^{L}f(x)\sin \frac{n\pi x}{L}dx=\frac{2}{L}{\int }_0^{L}f(x)\sin \frac{n\pi x}{L}dx.$$

PICTURE. Dono shaded halves ab same sign mein hain (dono orange, dono axis ke upar is example mein). Unke areas identical hain — left right ka mirror hai — to total ek half ka double hai. $2$ ka woh factor literally area ki doosri copy hai.

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Step 6 — Average kahaan gaya ($a_0$ missing kyun hai)

KYA. Full series mein ek constant term $a_0/2$ tha. Uska kya hua? Constant sirf $n=0$ cosine hai (kyunki $\cos 0=1$). Uska integrand $f(x)\cdot 1=f(x)$ hai, jo odd hai. $[-L,L]$ par yeh bhi zero integrate hota hai.

YEH VANISH KYUN HONA CHAHIYE. Ek odd extension mein left par axis ke neeche utni hi curve hoti hai jitni right par upar hoti hai; uska overall average zero hota hai. Isliye ek sine series mein koi constant term nahi hota — yeh kisi nonzero background level represent nahi kar sakti. (Agar tumhare problem ko woh background chahiye, to cosine series use karo — parent note mein cosine branch dekho.)

PICTURE. Full odd curve $[-L,L]$ par: axis ke upar orange area exactly neeche wali plum area se match karta hai. Net signed area (average) zero hai.

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Step 7 — Edge cases: endpoints aur jumps

KYA. Do degenerate spots ko apna alag look milna chahiye: origin $x=0$ aur wall $x=L$.

$x=0$ par: har $\sin \frac{n\pi \cdot 0}{L}=\sin 0=0$. To sine series hamesha left end par $0$ return karti hai, chahe $f(0)$ kuch bhi ho. Agar $f(0)\ne 0$, series simply use match nahi kar sakti — woh difference split karti hai (midpoint rule) right se aane wali value ($f(0^{+})$) aur left ki flipped value ($-f(0^{+})$) ke beech, jinka average $0$ hai.

$x=L$ par: periodic extension mein jump ho sakta hai. Agar $f(L)\ne 0$, to $L$ ke just baad ki flipped copy $-f(L)$ par drop karti hai, ek sudden cliff. Wahan series midpoint $\frac{1}{2}(f(L^{-})+(-f(L^{-})))=0$ par converge karti hai, $f(L)$ par nahi.

KYUN CARE KAREIN. Ye exactly woh points hain jahan "$f(x)=\sum \dots$" likhna ek jhooth hai. Honest statement open interval $0<x<L$ par hold karta hai.

PICTURE. $x=L$ par sawtooth jump ka zoom: dono branches (orange $f(L^{-})$ tak upar aa rahi hai, plum $-f(L^{-})$ se shuru ho rahi hai) aur ek plum dot jo midpoint mark karta hai jahan series actually hit karti hai.

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Ek-picture summary

Sab ek saath. Left panel: diya gaya right half → odd-reflected karke left fill kiya gaya. Middle: even cosine ke saath pair karne par cancelling areas milte hain (⇒ $a_n=0$). Right: odd sine ke saath pair karne par matching areas milte hain (⇒ half-integral ka double). Teen moves poori derivation hain.

Recall Feynman: poora walkthrough plain words mein retell karo

Mujhe ek wiggle di gayi thi jo page ke right side par sirf $0$ se $L$ tak draw thi. Fourier machine use karne ke liye — jise dono sides chahiye — maine missing left side khud draw ki, picture ko centre dot ke around $180^{\circ }$ spin karke; isse woh "odd" ban jaati hai. Phir maine machine se uski waves ki recipe maangi. Jab usne measure karne ki koshish ki ki kitna cosine andar tha, left area (ab ulta) right area se perfectly cancel ho gaya, to answer zero tha — cosines bilkul nahi. Jab usne measure kiya ki kitna sine andar tha, left area right area se sign-for-sign match kar gaya, to maine sirf right half measure kiya aur use double kiya. Woh doubling woh chhota "$2$" hai saamne, aur "sirf right half measure karo" woh ${\int }_0^L$ hai. Constant background bhi zero cancel ho gaya, kyunki ek ulta mirror mein line ke neeche utni hi curve hoti hai jitni upar hoti hai. Sirf ek jagah mujhe careful rehna hai woh hai dono ends par: sines hamesha $x=0$ par zero hote hain, aur agar meri wiggle $x=L$ par already zero nahi chhoo rahi thi, to mirror wahan ek cliff banata hai, aur series us cliff ke middle par land karti hai, top par nahi. Isliye main equality sirf $0<x<L$ ke liye likhta hoon. Yahi poori half-range sine story hai.

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Connections

• Parent topic — full statement aur is walkthrough ka cosine twin.
• Even and Odd Functions — reflection symmetry jo Steps 4–6 ko power deti hai.
• Fourier Series — full range — woh machine jo humne Step 3 mein specialise ki.
• Dirichlet Conditions — midpoint rule jo Step 7 mein use hua.
• Heat Equation — separation of variables — jahan yeh sine series use hoti hain.
• Wave Equation on a finite string — fixed ends → yahi exact sine expansion.
• Neumann Boundary Conditions — Step 2 ki choice ka even/cosine alternative.