Visual walkthrough — Full Fourier series — coefficients derivation

4.7.5 · D2 · Maths › Partial Differential Equations › Full Fourier series — coefficients derivation

Hum zero se build karte hain. Agar koi symbol aata hai, pehle uski picture milegi.

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Step 0 — Vocabulary, pictures ke roop mein

Kisi bhi derivation se pehle, teen words, teen pictures.

Ek function $f(x)$ bas har horizontal position $x$ par ek height hai. Ek ruler ke upar ek wiggly curve imagine karo.

"Periodic with period $2L$" ka matlab hai ki curve har $2L$ distance ke baad repeat karti hai. Yahan $2L$ ek poori copy ki width hai; $L$ us width ka aadha hai, isliye repeating window $x=-L$ (left edge) se $x=+L$ (right edge) tak chalti hai.

The basis waves pure ripples hain: $\cos \frac{n\pi x}{L}$ aur $\sin \frac{n\pi x}{L}$. Poora number $n$ count karta hai ki window $[-L,L]$ ke andar kitne poore up-down wiggles fit hote hain: $n=1$ ek slow wiggle hai, $n=2$ double speed, aur aise hi aage.

Hamara poora kaam: numbers $a_0,a_n,b_n$ nikalna.

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Step 1 — "Integrate" matlab "signed area", aur yeh hamara measuring tool hai

KYA. Hum baar baar ${\int }_{-L}^L(\text{something})dx$ likhenge. Uss symbol ka matlab sirf itna hai: curve aur ruler ke beech ka area, line ke upar ka area positive aur neeche ka negative count karte hue.

YEH TOOL KYUN, KOI AUR KYUN NAHIN. Hum ek aisi machine chahte hain jo ek wave ka average level measure kar sake aur cancellation detect kar sake. Ek pure ripple poori window mein exactly utna hi time line ke upar guzaarta hai jitna neeche, isliye ek poore window par uska signed area $0$ hota hai. Signed area ek perfect "kya yeh wave cancel hoti hai?" detector hai — koi doosra operation itna clean zero nahin deta.

PICTURE. $\sin \frac{\pi x}{L}$ dekho: upar ka mint hump neeche ke coral dip ko cancel kar deta hai, total signed area $=0$. Lekin ek wave times khud, jaise ${\sin }^2$, kabhi negative nahin hoti — yeh poori tarah line ke upar baiThti hai, isliye uska area genuinely positive hai. Yahi farq neeche ki saari cheez ka engine hai.

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Step 2 — Do alag waves "perpendicular" hoti hain (orthogonality, dekha hua)

KYA. Do alag waves lo, unhe point-by-point multiply karke ek nayi curve banao, phir uska signed area lo. Claim: jab bhi do waves alag hoti hain, answer $0$ hota hai.

KYUN. Yahi property hai jo hume ek time par ek coefficient isolate karne deti hai. Agar alag waves cancel nahin karti, toh ek par project karne se baaki sab ka contribution aa jaata — bilkul naakabil. Parent note ${\int }_{-L}^{L}fgdx$ ko "dot product" kehta hai; orthogonality ka matlab hai ki hamare basis waves mutually perpendicular axes hain.

PICTURE. Product $\cos \frac{\pi x}{L}\cos \frac{2\pi x}{L}$ is tarah wiggle karta hai ki equal purple-above aur purple-below area hota hai — total zero. Compare karo ${\cos }^2\frac{\pi x}{L}$ se (same wave times itself), jiska area genuinely positive $L$ hai.

Sine×cosine ke liye hamesha kyun? Sine odd hai ($x=0$ ke baare mein mirror-flip karke ulta ho jaata hai), cosine even hai (mirror-symmetric). Odd×even odd hota hai, aur ek odd curve ka left area uske right area ko cancel karta hai. Us reflection picture ke liye Even and odd functions dekho, aur general principle ke liye Orthogonality of functions.

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Step 3 — Projection trick, drawn

KYA. Maano $f$ sach mein apni series ke barabar hai. Poori equation ko ek chuni hui wave se multiply karo, maano $\cos \frac{m\pi x}{L}$ se, phir integrate karo.

KYUN. Steps 1–2 guarantee karte hain ki integrate karne ke baad, har term mar jaayegi sivaay us ek ke jo hamari chuni hui wave se match kare. Yeh bilkul waise hai jaise ek vector ka ek coordinate padhna — har perpendicular axis kuch contribute nahin karti.

PICTURE. Terms ki ek infinite row imagine karo. Hum unpar "$\cos m$ filter" daalaate hain. Har mismatched term $0$ ho jaata hai (grey); sirf $\cos m\times \cos m$ term light up hoti hai aur $L$ return karti hai.

${\int }_{-L}^{L}f\cos \frac{m\pi x}{L}dx=\underset{=0}{\underbrace{\frac{a_0}{2}\int \cos \frac{m\pi x}{L}dx}}+{\sum }_n{a}_n\underset{0\text{ unless }n=m,\text{then }L}{\underbrace{\int \cos \frac{n\pi x}{L}\cos \frac{m\pi x}{L}dx}}+{\sum }_n{b}_n\underset{=0}{\underbrace{\int \sin \frac{n\pi x}{L}\cos \frac{m\pi x}{L}dx}}$

• the constant term ::: Step 1 se mar jaata hai (ek akele cosine ka area zero hota hai).
• the sine sum ::: "hamesha $0$" rule se mar jaata hai (sine×cosine).
• the cosine sum ::: ek survivor $n=m$ par, jiska value $a_{m}L$ hai.

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Step 4 — $a_m$ padh lo

KYA. Right side par sirf $a_{m}L$ bacha hai, isliye equation ab solve karna trivial hai.

KYUN. Ulajha hua infinite sum ek single unknown times $L$ tak simat gaya hai. $L$ se divide karo aur kaam khatam.

PICTURE. Ek bar ko chhodkar saari bars zero ho gayi hain; akeli surviving bar ki height $a_{m}L$ hai. Left side par jo area measure kiya woh wahi bar hai.

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Step 5 — $b_m$ padh lo (same trick, sine filter)

KYA. Steps 3–4 repeat karo lekin $\sin \frac{m\pi x}{L}$ se multiply karo is baar.

KYUN. Conceptually kuch nahin badla: ab sine filter saari cosines (odd×even) aur saari mismatched sines ko kill kar deta hai, sirf akela $b_{m}L$ bachta hai.

PICTURE. Step 4 ka mirror image: sine filter exactly ek sine bar pick karta hai.

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Step 6 — Constant term $a_0$, aur humne $\frac{1}{2}$ kyun chhupaya

KYA. Flat filter use karo — $1$ se multiply karo — aur integrate karo.

KYUN. Flat line hi akeli "wave" hai jo ek period mein cancel nahin hoti (Step 1), isliye $f$ ko integrate karne se seedha uska sea level milta hai.

PICTURE. Har wiggle cancel hokar zero ho jaati hai; sirf height $\frac{a_0}{2}$ aur width $2L$ ka constant slab bachta hai, area $a_{0}L$.

${\int }_{-L}^{L}fdx=\frac{a_0}{2}\underset{2L}{\underbrace{{\int }_{-L}^{L}1dx}}+0+0=a_{0}L⟹\boxed{a_0=\frac{1}{L}{\int }_{-L}^{L}f(x)dx}$

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Step 7 — Degenerate aur edge cases (koi gap mat chhodho)

Chaar scenarios jo reader eventually milenge:

(a) $f$ odd hai ($x=0$ ke paar ulta flip hota hai). Tab $f\cos$ odd hai → saare $a_n=0$, including $a_0$. Sirf sines bachti hain. Square wave aur $f(x)=x$ exactly yahi karte hain.

(b) $f$ even hai (mirror-symmetric). Tab $f\sin$ odd hai → saare $b_n=0$. Sirf cosines bachti hain. Yeh cosine half-range situation hai.

(c) $f$ khud ek basis wave hai, jaise $f=\cos \frac{3\pi x}{L}$. Tab filter exactly $m=3$ par match karta hai: $a_3=\frac{1}{L}{\int }_{-L}^L{\cos }^2\frac{3\pi x}{L}dx=\frac{1}{L}\cdot L=1$, baaki sab $0$. Ek building block ki series bas wahi block hoti hai.

(d) $f$ mein jump hai (jaise $x=0$ par square wave). Coefficients phir bhi exist karte hain (integrals finite hain), lekin series wahan jump ke midpoint par converge karti hai. Yahi Convergence of Fourier series (Dirichlet conditions) ka content hai.

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Ek-picture summary

Is poore page ki ek sentence hai: $f$ ko ek wave se multiply karo, signed area lo, $L$ se divide karo — orthogonality guarantee karti hai ki sirf matching amount bachega.

Recall Poore walkthrough ki Feynman-style retelling

Tumhe ek gaana diya gaya hai jo kai tuning forks se mix hai, har ek ek steady note hum raha hai (hamaari waves). Tum jaanna chahte ho ki har fork kitni loud thi (the coefficients). Yeh hai woh trick jo tumne abhi pictures mein dekhi: ek matching fork gaane ke paas pakdo aur ek poore beat par "running average" lo. Har doosri fork ka note utna hi time upar push karne mein guzaarta hai jitna neeche, isliye average hokar gayab ho jaata hai. Sirf woh fork jo tumhari se match karti hai khud ko reinforce karti hai aur ek clean, positive amount chhodti hai — woh amount, $\frac{1}{L}$ se scale karke, uska coefficient hai. Har fork ke saath ek baar karo (har cosine, har sine, aur average ke liye silent "flat" fork) aur tumne gaane ka har ingredient recover kar liya. Bas yahi ek Fourier coefficient hai: ek pure tone ka volume, cancellation se measure kiya gaya.

Recall Quick self-test

$\cos \frac{m\pi x}{L}$ se multiply karne par $a_m$ isolate kyun hota hai? ::: Orthogonality: har doosri wave times $\cos m$ integrate hokar $0$ deti hai, sirf $\cos m\times \cos m$ bachta hai (jo $L$ deta hai). $f$ ka average $a_0$ nahin balki $\frac{a_0}{2}$ kyun hai? ::: Constant term likha $\frac{a_0}{2}$ jaata hai; ek period ka average us constant ke barabar hota hai, isliye average $=\frac{a_0}{2}$. Agar $f$ odd hai, toh kaun se coefficients gayab ho jaate hain? ::: Saare cosine coefficients $a_n$ ($a_0$ including); sirf sines bachti hain.

Related next steps: Complex (exponential) Fourier series (same trick, ek wave $e^{in\pi x/L}$), Parseval's theorem (orthogonality ka "energy" version), aur Separation of variables for the heat equation (jahan yeh coefficients initial data bante hain).