Fourier series — motivation from periodic functions: Worked examples

4.7.3 · D3 · Maths › Partial Differential Equations › Fourier series — motivation from periodic functions

Yeh page Fourier series ki practice arena hai. Parent note ne recipe derive ki thi; yahan hum isse har tarah ke function par run karte hain jo tum ko diye ja sakte hain — even, odd, neither, constant, discontinuous, aisi period jo friendly $2 π$ nahi hai, aur ek real heat-flow word problem. Is page ke end tak koi bhi case class tumhe surprise nahi kar sakti.

Woh framework jo hum baar baar use karte hain (self-contained recap)

Neeche ke har example ko same teen formulas mein daalte hain jo parent note se hain. Unhe ek baar yahan state karte hain taaki yeh page apne aap mein complete ho.

Scenario matrix

Decision-tree mein teen symmetry classes hain (even / odd / neither) jo do continuity classes (continuous / jump wali) aur period ( $2 π$ ya general $2 L$ ) se cross hoti hain. Neeche ke aath examples har reachable branch ko kam se kam ek baar cover karte hain; last column mein woh branch likhi hai.

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	dono $a_{n}$ aur $b_{n}$ nonzero	Ex 3: $f (x) = x + 1$ on $(- π, π)$
D	Even	General $2 L \neq = 2 π$	Continuous	$\frac{nπ x}{L}$ rakho	Ex 4: $x^{2}$ on $(- 2, 2)$
E	Constant	koi bhi	Continuous	degenerate: no waves	Ex 5: $f = 3$
F	Odd	$2 π$	Jump	jump par value (naye coeffs nahi)	Ex 6: square-wave series at $x = 0$
G	Half-range	sirf $L$ diya	extension choose karo	exam twist	Ex 7: $f = 1$ ki sine vs cosine series
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: Padhne se pehle — $a_{n}, b_{n}$ mein se kaun zero hoga, aur answer mein sab $n$ honge ya sirf kuch?

Step 1. $f (- x) = - f (x)$ , toh $f$ odd hai. Hence $a_{n} = 0$ for all $n$ . Yeh step kyun? Odd $\times$ cosine (even) odd hota hai, aur symmetric interval $[- π, π]$ par odd function ka integral $0$ hota hai. Isse shuru hone se pehle hi saare cosine integrals delete ho jaate hain.

Step 2. $b_{n} = \frac{1}{π} \int_{- π}^{π} f (x) sin n x d x = \frac{2}{π} \int_{0}^{π} sin n x d x$ . Yeh step kyun? $f sin n x$ hai odd $\times$ odd $=$ even, $0$ ke baare mein symmetric, toh $[- π, π]$ par integral, $[0, π]$ par integral ka do guna hai, jahan $f = + 1$ hai.

Step 3. $\int_{0}^{π} sin n x d x = [\frac{- c o s n x}{n}]_{0}^{π} = \frac{1 - cos nπ}{n} = \frac{1 - ( - 1 ) ^{n}}{n}$ . Yeh step kyun? $cos nπ = (- 1)^{n}$ : even $n$ ke liye $+ 1$ hai (zero deta hai) aur odd $n$ ke liye $- 1$ hai ( $2/ n$ deta hai). Sirf odd harmonics bachte hain.

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

Verify: $b_{1} = \frac{4}{π} \approx 1.2732$ , $b_{2} = 0$ , $b_{3} = \frac{4}{3 π} \approx 0.4244$ . Yeh exactly parent ka result hai — 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: $∣ x ∣$ even hai aur iska koi jump nahi hai (bas $0$ par ek sharp corner). Guess: kaun se coefficients vanish honge, aur series fast ya slow converge karegi?

Step 1. $f (- x) = ∣ - x ∣ = ∣ x ∣ = f (x)$ — even. Toh $b_{n} = 0$ . Yeh step kyun? Even $\times$ sine (odd) odd hota hai $\Rightarrow$ integrate karke $0$ deta hai. Sirf cosines bachte hain.

Step 2. $a_{0} = \frac{1}{π} \int_{- π}^{π} ∣ x ∣ d x = \frac{2}{π} \int_{0}^{π} x d x = \frac{2}{π} \cdot \frac{π ^{2}}{2} = π$ . Yeh step kyun? $a_{0} /2$ average height hai. $∣ x ∣$ ka average $π /2$ hai, aur wakai $a_{0} /2 = π /2$ hai. Ek built-in sanity check.

Step 3. $a_{n} = \frac{2}{π} \int_{0}^{π} x cos n x d x$ . Integration by parts karo: $\int_{0}^{π} x cos n x d x = [\frac{x s i n n x}{n}]_{0}^{π} - \int_{0}^{π} \frac{s i n n x}{n} d x = 0 + [\frac{c o s n x}{n ^{2}}]_{0}^{π} = \frac{( - 1 ) ^{n} - 1}{n ^{2}} .$ Yeh step kyun? Integration by parts sahi tool hai kyunki integrand polynomial $\times$ trig hai — $x$ differentiate karne se polynomial khatam ho jaata hai, aur hum pure trig integral ki taraf aate hain. (Boundary term $\frac{x s i n n x}{n}$ dono ends par vanish karta hai kyunki $sin nπ = 0$ aur lower limit par $x = 0$ hai.)

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

Verify: $a_{1} = - \frac{4}{π} \approx - 1.2732$ , $a_{2} = 0$ , $a_{3} = - \frac{4}{9 π} \approx - 0.1415$ . Kyunki coefficients $1/ n^{2}$ ki tarah shrink karte hain (jump square wave ke $1/ n$ ke comparison mein), yeh faster converge karta hai — corner, jump se milder hai. $x = 0$ plug karo: $0 = \frac{π}{2} - \frac{4}{π} \sum_{odd} \frac{1}{n ^{2}}$ , jisse $\sum_{odd} \frac{1}{n ^{2}} = \frac{π ^{2}}{8}$ milta hai, jo ek known identity hai.

Example 3 — Cell C: neither even nor odd (aur jump bhi hai)

Worked example

$f (x) = x + 1$ on $(- π, π)$ , period $2 π$ Forecast: Constant $1$ add karne se odd symmetry toot jaati hai: $f (- x) = - x + 1 \neq = \pm f (x)$ . Toh yeh neither even hai na odd — aur $\pm π$ par jump bhi hai. Guess: ab kaun se coefficients bachenge?

Step 1. $f = x + 1$ ko split karo. Constant $1$ sirf $a_{0}$ mein contribute karta hai; linear $x$ odd hai aur sirf $b_{n}$ mein contribute karta hai. Yeh step kyun? Fourier coefficients $f$ mein linear hain, toh har piece compute karo aur add karo. Yeh $x + 1$ ko poora attack karne se kaafi aasan hai.

Step 2. Constant se: $a_{0} = \frac{1}{π} \int_{- π}^{π} 1 d x = 2$ , aur baaki saare $a_{n} = \frac{1}{π} \int_{- π}^{π} 1 \cdot cos n x d x = 0$ . Yeh step kyun? $\int_{- π}^{π} cos n x d x = \frac{s i n nπ}{n} \cdot 2$ -type terms vanish karte hain kyunki $sin nπ = 0$ ; ek constant ko koi non-constant cosine nahi chahiye.

Step 3. Linear part se (sawtooth work reuse karte hue): $b_{n} = \frac{1}{π} \int_{- π}^{π} x sin n x d x = \frac{2 ( - 1 ) ^{n + 1}}{n}$ . Yeh step kyun? $x$ odd hai, $sin n x$ odd hai, product even hai; by parts $\int_{- π}^{π} x sin n x d x = 2 \int_{0}^{π} x sin n x d x = \frac{- 2 π ( - 1 ) ^{n}}{n}$ , aur $π$ se divide karne par $\frac{2 ( - 1 ) ^{n + 1}}{n}$ milta hai.

Step 4 (the twist). Dono ek nonzero average ( $a_{0} /2 = 1$ ) aur sines ka poora set appear karta hai: $x + 1 = 1 + 2 (sin x - \frac{s i n 2 x}{2} + \frac{s i n 3 x}{3} - \dots) .$ $b_{n}$ mein har $n$ present hai, aur ek standalone constant bhi hai — "neither" function ki pehchaan.

Verify: $a_{0} = 2$ (toh average $= 1$ ✓, $\frac{1}{2 π} \int_{- π}^{π} (x + 1) d x = 1$ se match karta hai), $b_{1} = 2$ , $b_{2} = - 1$ , $b_{3} = \frac{2}{3}$ . $x = π /2$ par: $\frac{π}{2} + 1 = 1 + 2 (1 - \frac{1}{3} + \frac{1}{5} - \dots) = 1 + 2 \cdot \frac{π}{4}$ ✓ (phir Leibniz). Jump $x = π$ par series midpoint $\frac{1}{2} [f (π^{-}) + f (- π^{+})] = \frac{1}{2} [(π + 1) + (- π + 1)] = 1$ par converge karti hai.

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

Worked example

$f (x) = x^{2}$ on $(- 2, 2)$ , period $4$ (toh $L = 2$ ) Forecast: Period ab $4$ hai, $2 π$ nahi. Teen formulas mein se kaun si break ho jaati hai agar tum bhool jaate ho aur $\frac{nπ x}{2}$ ki jagah $n x$ likh dete ho?

Step 1. $x^{2}$ even hai $\Rightarrow b_{n} = 0$ . $\frac{nπ x}{L} = \frac{nπ x}{2}$ har jagah use karo. Yeh step kyun? Wave ko period $2 L = 4$ mein poore cycles complete karne chahiye; argument $\frac{nπ x}{L}$ yeh guarantee karta hai. $n x$ likhne par galat period use hogi aur result bakwaas aayega.

Step 2. $a_{0} = \frac{1}{L} \int_{- L}^{L} x^{2} d x = \frac{1}{2} \int_{- 2}^{2} x^{2} d x = \frac{1}{2} \cdot \frac{2 \cdot 2 ^{3}}{3} = \frac{1}{2} \cdot \frac{16}{3} = \frac{8}{3}$ . Toh average $a_{0} /2 = \frac{4}{3}$ hai. ( $[- 2, 2]$ par $x^{2}$ ka average $\frac{1}{4} \int_{- 2}^{2} x^{2} = \frac{4}{3}$ hai ✓.)

Step 3a (parity reduction). $a_{n} = \frac{1}{L} \int_{- L}^{L} x^{2} cos \frac{nπ x}{L} d x$ . Integrand even $\times$ even $=$ even hai, $x = 0$ ke baare mein symmetric: $a_{n} = \frac{1}{L} \int_{- L}^{L} x^{2} cos \frac{nπ x}{L} d x = \frac{2}{L} \int_{0}^{L} x^{2} cos \frac{nπ x}{L} d x .$ Yeh step kyun? Symmetric integral ko aadha fold karne se algebra aadhi ho jaati hai aur limits clean rehti hain ( $0$ se shuru).

Step 3b (pehla integration by parts). $k = \frac{nπ}{L}$ lo. $u = x^{2}, d v = cos (k x) d x$ ke saath: $\int_{0}^{L} x^{2} cos (k x) d x = [\frac{x ^{2} s i n k x}{k}]_{0}^{L} - \int_{0}^{L} \frac{2 x s i n k x}{k} d x .$ Boundary term $\frac{L ^{2} s i n ( k L )}{k} = \frac{L ^{2} s i n ( nπ )}{k} = 0$ hai (kyunki $k L = nπ$ aur $sin nπ = 0$ ). Yeh step kyun? $x^{2}$ differentiate karne par $2 x$ banta hai — ek power khatam, ek aur baaki.

Step 3c (doosra integration by parts). Baaki $\int_{0}^{L} 2 x sin (k x) d x$ ke liye, $u = 2 x, d v = sin (k x) d x$ lo: $\int_{0}^{L} 2 x sin (k x) d x = [\frac{- 2 x c o s k x}{k}]_{0}^{L} + \int_{0}^{L} \frac{2 c o s k x}{k} d x = \frac{- 2 L c o s ( nπ )}{k} + \frac{2 s i n ( nπ )}{k ^{2}} = \frac{- 2 L ( - 1 ) ^{n}}{k} .$ Step 3b aur 3c ko saath rakhte hain (3b ka minus sign dhyan mein rakho): $\int_{0}^{L} x^{2} cos (k x) d x = 0 - \frac{1}{k} (\frac{- 2 L ( - 1 ) ^{n}}{k}) = \frac{2 L ( - 1 ) ^{n}}{k ^{2}} = \frac{2 L ( - 1 ) ^{n} L ^{2}}{n ^{2} π ^{2}} = \frac{2 L ^{3} ( - 1 ) ^{n}}{n ^{2} π ^{2}} .$ Yeh step kyun? $2 x$ differentiate ho kar constant ban jaata hai, toh doosra part-step kaam khatam karta hai.

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: $a_{1} = \frac{- 16}{π ^{2}} \approx - 1.6211$ , $a_{2} = \frac{16}{4 π ^{2}} = \frac{4}{π ^{2}} \approx 0.4053$ . $x = 2$ par (jahan $cos \frac{nπ \cdot 2}{2} = cos nπ = (- 1)^{n}$ ) series deti hai $\frac{4}{3} + \frac{16}{π ^{2}} \sum \frac{( - 1 ) ^{n} ( - 1 ) ^{n}}{n ^{2}} = \frac{4}{3} + \frac{16}{π ^{2}} \cdot \frac{π ^{2}}{6} = \frac{4}{3} + \frac{8}{3} = 4 = 2^{2}$ ✓ ( $\sum 1/ n^{2} = π^{2} /6$ use kiya).

Example 5 — Cell E: degenerate constant

Example 6 — Cell F: jump par exactly woh value

Worked example Square-wave series

$x = 0$ par kahan converge karti hai? Forecast: $x = 0$ par square wave undefined hai (woh $- 1$ se $+ 1$ ki taraf jump karta hai). $\frac{4}{π} (sin x + \dots)$ wahan actually kaunsa number output karta hai?

Step 1. $\frac{4}{π} (sin x + \frac{s i n 3 x}{3} + \dots)$ mein $x = 0$ plug karo. Har $sin (n \cdot 0) = 0$ , toh sum $0$ hai. Yeh step kyun? Hum danger point par series ko numerically test karte hain.

Step 2. Convergence theorem kehta hai ki jump par series midpoint ke barabar hoti hai: $\frac{1}{2} [f (0^{+}) + f (0^{-})] = \frac{1}{2} [(+ 1) + (- 1)] = 0$ . Yeh step kyun? Yeh koi coincidence nahi — series hamesha jump par difference split karti hai (jump ke paas overshoot ke liye Gibbs Phenomenon dekho).

Verify: theorem se predicted $0$ $=$ sum se computed $0$ ✓. Lesson: Fourier series mein " $=$ " discontinuities par midpoint equality hai, literal equality nahi.

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

Intuition Half-range expansion kya hoti hai, aur

$\frac{2}{π}$ kahan se aata hai Kabhi kabhi $f$ sirf $(0, π)$ par diya jaata hai — poore $2 π$ -period ka aadha. Fourier machinery use karne ke liye pehle $(- π, 0)$ par missing half invent karni padti hai. Hum isko choose karne ke liye free hain, aur do choices special hain: oddly extend karo (mirror-and-flip, sirf sine series force karta hai) ya evenly (seedha mirror, sirf cosine series force karta hai). Dono taraf se period $2 π$ ban jaata hai, toh $L = π$ .

Coefficient formula mein $\frac{2}{π}$ kyun aata hai: general $b_{n} = \frac{1}{L} \int_{- L}^{L} f sin \frac{nπ x}{L} d x$ se $L = π$ ke saath shuru karo, milta hai $b_{n} = \frac{1}{π} \int_{- π}^{π} f sin n x d x$ . Odd extension ke liye integrand $f sin n x$ even hai (odd $\times$ odd), $0$ ke baare mein symmetric, toh $[- π, π]$ integral $[0, π]$ integral ka do guna hai — aur $\frac{1}{π}$ ban jaata hai $\frac{2}{π}$ : $b_{n} = \frac{2}{π} \int_{0}^{π} f (x) sin n x d x .$ Bilkul waise hi, even extension ke liye $f cos n x$ even hai, toh $a_{n} = \frac{2}{π} \int_{0}^{π} f (x) cos n x d x$ . $\frac{2}{π}$ koi nayi baat nahi — yeh standard formula hai jo symmetry se aadha fold hua hai.

Worked example

$f (x) = 1$ on $(0, π)$ ko (a) sine series, (b) cosine series mein expand karo Forecast: Same function, same interval — phir bhi tum do alag valid Fourier series nikaal sakte ho. Kaise? Kyunki tum choose karte ho ki $f$ ko $(- π, 0)$ par kaise extend karo.

Step 1 (sine series — odd extension). $f$ ko oddly reflect karo: $f (- x) = - 1$ on $(- π, 0)$ . Toh $a_{n} = 0$ aur, upar ke half-range formula se, $b_{n} = \frac{2}{π} \int_{0}^{π} 1 \cdot sin n x d x = \frac{2}{π} [\frac{- c o s n x}{n}]_{0}^{π} = \frac{2}{π} \cdot \frac{1 - c o s nπ}{n} = \frac{2}{π} \cdot \frac{1 - ( - 1 ) ^{n}}{n} .$ Yeh step kyun? Odd extension sine-only series force karti hai; yeh exactly wahi hai jo PDE demands karta hai jab $f = 0$ dono ends par ho (Dirichlet boundary).

Step 2 (sine coefficients evaluate karo). $b_{n}$ even $n$ ke liye $0$ hai aur odd $n$ ke liye $\frac{4}{π n}$ hai, toh $1 = \frac{4}{π} (sin x + \frac{s i n 3 x}{3} + \frac{s i n 5 x}{5} + \dots) (0 < x < π) .$ Yeh step kyun? Constant $1$ ki odd extension hi $\pm 1$ square wave hai, toh coefficients exactly Ex 1 se match karte hain — ek satisfying consistency check.

Step 3 (cosine series — even extension). $f$ ko evenly reflect karo: $f (- x) = + 1$ . Toh $b_{n} = 0$ , aur $a_{0} = \frac{2}{π} \int_{0}^{π} 1 d x = 2, a_{n} = \frac{2}{π} \int_{0}^{π} cos n x d x = \frac{2}{π} \cdot \frac{s i n nπ}{n} = 0 (n \geq 1) .$ Yeh step kyun? Constant ka even extension sirf constant hi hota hai — flat line ko average ke alaawa koi cosine nahi chahiye.

Step 4 (cosine series likho). $1 = \frac{a _{0}}{2} = 1 (0 < x < π) .$ Dono series bilkul alag lagti hain (ek infinite sine sum vs. ek single constant) phir bhi $(0, π)$ par same $f$ represent karti hain — yahi exam twist hai.

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 ✓. Dono given half-interval par $1$ ke barabar hain.

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

Worked example

$π$ m lambi ek rod ka temperature $u (x, 0) = 100$ °C middle third $(π /3, 2 π /3)$ mein aur baaki jagah $0$ °C se shuru hota hai. Initial profile ke sine coefficients nikalo. Forecast: Hot block middle mein symmetrically baithta hai, toh profile rod ke centre ke baare mein symmetric hai phir bhi discontinuous hai (block edges par jump karti hai) — "even+jump" branch. Heat Equation par Separation of Variables modes $sin (n x)$ deta hai; humein yeh blocky initial temperature unme expand karni hai. Kaun se coefficients dominate karenge?

Step 1. Model: $u (x, 0) = {1000 π /3 < x < 2 π /3 else$ on $(0, π)$ . Sine (odd, half-range) expansion use karo kyunki rod ke ends $0$ °C par held hain, yaani $u (0) = u (π) = 0$ . Sine kyun? Boundary conditions cosines ko kill karte hain; sirf $sin (n x)$ satisfy karta hai $u (0) = u (π) = 0$ . Ex 7 se half-range formula reuse karte hain: $b_{n} = \frac{2}{π} \int_{0}^{π} u sin n x d x$ .

Step 2. $b_{n} = \frac{2}{π} \int_{π /3}^{2 π /3} 100 sin n x d x = \frac{200}{π} [\frac{- cos n x}{n}]_{π /3}^{2 π /3} = \frac{200}{nπ} (cos \frac{nπ}{3} - cos \frac{2 nπ}{3}) .$ Yeh step kyun? Sirf wahan integrate karo jahan $u = 100$ hai; baaki jagah integrand $0$ hai.

Step 3. Amplitudes compute karo (°C): $n = 1$ ke saath, $cos \frac{π}{3} - cos \frac{2 π}{3} = \frac{1}{2} - (- \frac{1}{2}) = 1$ , toh $b_{1} = \frac{200}{π} \approx 63.66$ . $n = 2$ ke saath, $cos \frac{2 π}{3} - cos \frac{4 π}{3} = - \frac{1}{2} - (- \frac{1}{2}) = 0$ , toh $b_{2} = 0$ . $n = 3$ ke saath, $cos π - cos 2 π = - 1 - 1 = - 2$ , toh $b_{3} = \frac{200}{3 π} \cdot (- 2) = \frac{- 400}{3 π} \approx - 42.44$ . Units check kyun? $b_{n}$ same unit carry karta hai jitna $u$ , yaani °C — yeh us temperature mode ki amplitude hai.

Step 4. Initial profile hai $u (x, 0) = \sum_{n = 1}^{\infty} b_{n} sin n x$ , aur jaise time chalti hai Heat Equation mode $n$ ko $e^{- n^{2} t}$ se multiply karta hai, toh high- $n$ modes sabse jaldi decay karte hain — sharp block jaldi smooth ho jaata hai. (Wave Equation se compare karo, jahan har mode time mein oscillate karta hai.)

Verify: $b_{1} \approx 63.66$ °C, $b_{2} = 0$ , $b_{3} \approx - 42.44$ °C. Dominant $b_{1}$ hump sabse slow-decaying hai, toh late-time temperature ek single $sin x$ arch jaisi lagti hai — physically sensible.

Recall Kisi bhi Fourier problem se pehle scenario checklist

Symmetry? ::: Even ⇒ sirf $a_{n}$ ; odd ⇒ sirf $b_{n}$ ; neither ⇒ dono. Period $2 L = ?$ ::: Hamesha $\frac{nπ x}{L}$ likho; sirf tab $n x$ simplify karo jab $L = π$ ho. Jump present hai? ::: Series wahan midpoint par converge karti hai (Gibbs overshoot paas mein). Half-range diya hai? ::: Tum even (cosine) ya odd (sine) extension choose karo boundary conditions ke hisaab se; $\frac{1}{π}$ symmetry se $\frac{2}{π}$ ban jaata hai.

Connections

Parent recipe & derivation: Fourier series motivation
Coefficients isolate kyun hote hain: Orthogonality of Functions
Symmetry shortcuts: Even and Odd Functions
Yeh series kahan se aati hain: Separation of Variables, Heat Equation aur Wave Equation par apply ki gayi
Jumps par overshoot: Gibbs Phenomenon
Continuous-spectrum sibling: Fourier Transform