Full Fourier series — coefficients derivation: Worked examples

4.7.5 · D3 · Maths › Partial Differential Equations › Full Fourier series — coefficients derivation

Neeche sirf yeh teen boxed results use hote hain jo parent note se liye gaye hain. Hum unhe ek baar phir likhte hain taaki koi symbol unexplained na rahe.

Scenario matrix

Har Fourier problem jo tumhe mila ho, in cells mein se kisi ek mein aata hai. Is page ka point yeh hai ki neeche ke saat examples unhe sab cover karte hain.

Cell	Situation	Symmetry shortcut	Kya bachta hai	Example
C1	$f$ odd hai (jaise square wave, ya $x$ )	saare $a_{n} = 0$	sirf $b_{m}$ (sines)	Ex 1
C2	$f$ even hai (jaise $∣ x ∣$ )	saare $b_{m} = 0$	$a_{0}, a_{m}$ (cosines)	Ex 2
C3	$f$ mein koi symmetry nahi (jaise $e^{x}$ , ya shifted $x$ )	koi nahi — teeno karo	$a_{0}, a_{m}, b_{m}$	Ex 3
C4	$f$ ek constant hai (degenerate: sirf zero frequency)	trivial	sirf $a_{0}$	Ex 4
C5	$f$ pehle se ek basis wave hai (limiting / pure tone)	orthogonality	ek single coefficient	Ex 5
C6	Real-world word problem + exam twist (interval centred nahi, relocate karna hoga)	C1–C5 pe reduce karo	depend karta hai	Ex 6
C7	$f$ discontinuous hai (ek jump / square wave) — Gibbs behaviour	symmetry agar ho	$b_{m}$ (odd jump)	Ex 7

Unhi saat cells ka ek visual map — koi bhi problem shuru karne se pehle yahan dekho:

Figure — Full Fourier series — coefficients derivation

Example 1 — Cell C1 (odd function): triangle-slope wave $f (x) = x$ on $[- L, L]$

Origin se guzarti seedhi line aur uska odd flip dekho:

Worked example

$f (x) = x$ ko $[- L, L]$ par sines aur cosines mein rebuild karo. Forecast (pehle guess karo!): $f (x) = x$ origin se guzarti seedhi line hai. $x \to - x$ flip karo aur poora graph ulta ho jaata hai — yeh odd function ki pehchaan hai. Toh predict karo: saare cosine coefficients vanish ho jayenge, sirf sines bachenge. Kya tum yeh bhi expect karte ho ki $a_{0} = 0$ hoga? (Origin ke baare mein symmetric line ka average zero hota hai — haan.)

Step 1 — symmetry classify karo. $f (- x) = - x = - f (x)$ , toh $f$ odd hai → Cell C1. Yeh step kyun? Even and odd functions ke hisab se, odd functions mein $a_{0} = 0$ aur $a_{m} = 0$ automatically hote hain. Isse pen uthane se pehle teeno mein se do integrals delete ho jaate hain.

Step 2 — $b_{m}$ compute karo. Kyunki $f sin$ hai (odd)(odd)=even, hum interval ko fold kar sakte hain: $\int_{- L}^{L} = 2 \int_{0}^{L}$ . $b_{m} = \frac{1}{L} \int_{- L}^{L} x sin \frac{mπ x}{L} d x = \frac{2}{L} \int_{0}^{L} x sin \frac{mπ x}{L} d x .$ Yeh step kyun? Folding se kaam aadha ho jaata hai aur woh symmetry use hoti hai jo humne abhi find ki. Extra factor $2$ ek consequence hai, base $\frac{1}{L}$ nahi badla.

Step 3 — integration by parts. Hamare paas polynomial $\times$ trig hai; $x$ differentiate karne se constant aata hai, toh $u = x, d v = sin \frac{mπ x}{L} d x$ lo: $\int_{0}^{L} x sin \frac{mπ x}{L} d x = [- \frac{Lx}{mπ} cos \frac{mπ x}{L}]_{0}^{L} + \frac{L}{mπ} \int_{0}^{L} cos \frac{mπ x}{L} d x .$ Bacha hua integral evaluate hota hai $\frac{L}{mπ} [sin \frac{mπ x}{L}]_{0}^{L} = \frac{L}{mπ} (sin mπ - 0) = 0$ , kyunki $sin mπ = 0$ har integer $m$ ke liye — yeh zero boundary values se aata hai, kisi averaging argument se nahi. $x = L$ par bracket term deta hai $- \frac{L ^{2}}{mπ} cos mπ = - \frac{L ^{2}}{mπ} (- 1)^{m}$ ; $x = 0$ par yeh $0$ hai. Yeh step kyun? Integration by parts "power times wiggle" ka tool hai — yeh $x$ ko constant se replace kar deta hai.

Step 4 — assemble karo. $b_{m} = \frac{2}{L} \cdot (- \frac{L ^{2}}{mπ} (- 1)^{m}) = \frac{2 L ( - 1 ) ^{m + 1}}{mπ} .$ Ab series kyun likh sakte hain: humne measure kar liya hai ki har sine wave ka kitna hissa $f$ ke andar hai ( $b_{m}$ ), aur saare cosine amounts zero hain. Exactly unhi amounts ko re-stack karne se $f$ rebuild hota hai (upar "matching terms" box dekho), toh $x = \frac{2 L}{π} m = 1 \sum \infty \frac{( - 1 ) ^{m + 1}}{m} sin \frac{mπ x}{L} on (- L, L) .$

Verify: $L = π$ set karo: $b_{m} = \frac{2 ( - 1 ) ^{m + 1}}{m}$ , parent note ke Example 2 se match karta hai. Numerically $b_{1} = 2, b_{2} = - 1, b_{3} = \frac{2}{3}$ . ✔

Partial sums sach mein line ke paas aate hain (endpoints baad mein discuss honge):

Example 2 — Cell C2 (even function): $f (x) = ∣ x ∣$ on $[- π, π]$

V-shape vertical axis ke across apna khud ka mirror image hai:

Worked example

$f (x) = ∣ x ∣$ ko $[- π, π]$ par rebuild karo (toh $L = π$ ). Forecast: $∣ x ∣$ ek V-shape hai, vertical axis ke across mirror-symmetric. Yeh even hai. Predict karo: saare sine coefficients vanish ho jayenge, aur kyunki graph poori tarah axis ke upar hai, uska average $\frac{a _{0}}{2}$ positive hoga (is baar zero nahi!).

Step 1 — classify karo. $f (- x) = ∣ - x ∣ = ∣ x ∣ = f (x)$ : even → Cell C2, toh saare $b_{m} = 0$ . Yeh step kyun? Even $\times$ odd sine = odd, integrate karke $0$ milta hai. Aadha kaam khatam.

Step 2 — $a_{0}$ . Fold karo ( $f$ even hai): $a_{0} = \frac{1}{π} \int_{- π}^{π} ∣ x ∣ d x = \frac{2}{π} \int_{0}^{π} x d x = \frac{2}{π} \cdot \frac{π ^{2}}{2} = π .$ Toh average value hai $\frac{a _{0}}{2} = \frac{π}{2}$ — positive, jaise forecast tha. Yeh step kyun? Constant term wave ke DC level ke barabar hona chahiye.

Step 3 — $a_{m}$ . $L = π$ ke saath cosine hai $cos m x$ ; $f cos$ even hai, fold karo aur parts ( $u = x$ ): $a_{m} = \frac{2}{π} \int_{0}^{π} x cos m x d x = \frac{2}{π} ([\frac{x s i n m x}{m}]_{0}^{π} - \frac{1}{m} \int_{0}^{π} sin m x d x) .$ Boundary term vanish hota hai ( $sin mπ = 0$ ). Bacha hua integral: $\int_{0}^{π} sin m x d x = \frac{1 - c o s mπ}{m} = \frac{1 - ( - 1 ) ^{m}}{m}$ . Toh $a_{m} = \frac{2}{π} \cdot (- \frac{1}{m}) \cdot \frac{1 - ( - 1 ) ^{m}}{m} = - \frac{2}{π m ^{2}} (1 - (- 1)^{m}) .$ Yeh step kyun? Ex 1 jaisa hi parts logic; yahan $x$ ko $cos$ ke against differentiate karne se $sin$ boundary term aata hai jo khatam ho jaata hai.

Step 4 — cases padho. $1 - (- 1)^{m} = 0$ even $m$ ke liye, $= 2$ odd $m$ ke liye: $a_{m} = ⎩ ⎨ ⎧ - \frac{4}{π m ^{2}} 0 m odd m even$ Ab series kyun likh sakte hain: hum jaante hain ki $f$ ke andar har cosine wave ka exact amount kya hai ( $a_{0}$ aur har $a_{m}$ ), aur saare sine amounts zero hain. Un measured amounts ko re-stack karne se V-shape rebuild hoti hai: $∣ x ∣ = \frac{π}{2} - \frac{4}{π} k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{2}} cos ((2 k + 1) x) .$

Verify: $x = 0$ rakho: LHS $= 0$ , toh $\frac{π}{2} = \frac{4}{π} \sum \frac{1}{( 2 k + 1 ) ^{2}} \Rightarrow \sum_{k \geq 0} \frac{1}{( 2 k + 1 ) ^{2}} = \frac{π ^{2}}{8}$ — ek famous identity. ✔ Aur $a_{1} = - \frac{4}{π}, a_{2} = 0, a_{3} = - \frac{4}{9 π}$ . ✔

Example 3 — Cell C3 (koi symmetry nahi): $f (x) = e^{x}$ on $[- 1, 1]$

Worked example

$f (x) = e^{x}$ ko $[- 1, 1]$ par rebuild karo (toh $L = 1$ ). Forecast: $e^{x}$ na toh apna mirror image hai na apna flip, toh koi shortcut nahi — hume $a_{0}$ , saare $a_{m}$ , aur saare $b_{m}$ nikalne honge. Yeh general case hai jo tumhe honest rakhta hai.

Step 1 — classify karo. $e^{- x} \neq = e^{x}$ aur $\neq = - e^{x}$ : na even na odd → Cell C3. Yeh step kyun? "No symmetry" confirm karna batata hai ki koi bhi coefficient free mein vanish nahi hoga.

Step 2 — $a_{0}$ . $L = 1$ ke saath: $a_{0} = \int_{- 1}^{1} e^{x} d x = e - e^{- 1} = 2 sinh 1$ (upar wali definition use karke). Yeh step kyun? Seedha integral; average level set karta hai $\frac{a _{0}}{2} = sinh 1$ .

Step 3 — $a_{m}$ do baar integration by parts se (koi black box nahi). Hume chahiye $I_{c} = \int e^{x} cos (mπ x) d x$ . Parts ek baar apply karo $u = e^{x}, d v = cos (mπ x) d x$ ke saath: $I_{c} = \frac{e ^{x} s i n ( mπ x )}{mπ} - \frac{1}{mπ} \int e^{x} sin (mπ x) d x .$ Bacha hua integral $I_{s} = \int e^{x} sin (mπ x) d x$ same type ka hai, toh parts phir apply karo ( $u = e^{x}, d v = sin (mπ x) d x$ ): $I_{s} = - \frac{e ^{x} c o s ( mπ x )}{mπ} + \frac{1}{mπ} \int e^{x} cos (mπ x) d x = - \frac{e ^{x} c o s ( mπ x )}{mπ} + \frac{1}{mπ} I_{c} .$ Yeh step kyun? Har baar jab hum trig integrate karte hain aur $e^{x}$ differentiate karte hain, same shape wapas aata hai — toh integral khud pe feed karta hai, aur hum ise forever integrate karne ke bajaay algebraically solve kar sakte hain. $I_{s}$ ko $I_{c}$ mein substitute karo: $I_{c} = \frac{e ^{x} s i n ( mπ x )}{mπ} - \frac{1}{mπ} (- \frac{e ^{x} c o s ( mπ x )}{mπ} + \frac{1}{mπ} I_{c}) \Rightarrow I_{c} (1 + \frac{1}{m ^{2} π ^{2}}) = \frac{e ^{x} ( c o s mπ x + mπ s i n mπ x )}{m ^{2} π ^{2}} .$ Bracket clear karne se closed form milta hai $I_{c} = \frac{e ^{x} ( c o s mπ x + mπ s i n mπ x )}{1 + m ^{2} π ^{2}} .$ Ab $[- 1, 1]$ par evaluate karo: $x = \pm 1$ par, $sin (\pm mπ) = 0$ aur $cos (\pm mπ) = (- 1)^{m}$ , toh $a_{m} = [I_{c}]_{- 1}^{1} = \frac{( - 1 ) ^{m}}{1 + m ^{2} π ^{2}} (e - e^{- 1}) = \frac{2 ( - 1 ) ^{m} s i n h 1}{1 + m ^{2} π ^{2}} .$

Step 4 — $b_{m}$ usi $I_{s}$ se. Pair ko $I_{s}$ ke liye solve karne par similarly milta hai $I_{s} = \frac{e ^{x} ( s i n mπ x - mπ c o s mπ x )}{1 + m ^{2} π ^{2}} .$ $sin$ boundary terms vanish ho jaate hain, bachhta hai $b_{m} = [I_{s}]_{- 1}^{1} = \frac{- mπ ( - 1 ) ^{m}}{1 + m ^{2} π ^{2}} (e - e^{- 1}) = \frac{- 2 mπ ( - 1 ) ^{m} s i n h 1}{1 + m ^{2} π ^{2}} .$ Yeh step kyun? Same self-feeding pair; note karo ki $b_{m}$ mein extra $mπ$ hai, toh sine amplitudes yahan slower decay karti hain.

Verify: Numerically $sinh 1 \approx 1.1752$ . Toh $a_{0} \approx 2.3504$ , $a_{1} = \frac{- 2 s i n h 1}{1 + π ^{2}} \approx - 0.2160$ , $b_{1} = \frac{2 π s i n h 1}{1 + π ^{2}} \approx 0.6785$ . Teeno families nonzero hain, jaise no-symmetry cell demand karta hai. ✔

Example 4 — Cell C4 (degenerate constant): $f (x) = 5$ on $[- L, L]$

Worked example Constant

$f (x) = 5$ ko rebuild karo. Forecast: Ek flat line mein bilkul koi wiggle nahi hoti. Toh har genuine harmonic ( $m \geq 1$ ) zero hona chahiye, aur sirf "zero-frequency" term $a_{0}$ nonzero ho sakta hai. Yeh matrix ka degenerate edge case hai.

Step 1 — $a_{0}$ . $a_{0} = \frac{1}{L} \int_{- L}^{L} 5 d x = \frac{1}{L} \cdot 5 \cdot 2 L = 10$ . Yeh step kyun? Constant term $\frac{a _{0}}{2} = 5$ function ke barabar hona chahiye. ✔

Step 2 — $m \geq 1$ ke liye $a_{m}, b_{m}$ . $a_{m} = \frac{5}{L} \int_{- L}^{L} cos \frac{mπ x}{L} d x = 0$ aur $b_{m} = \frac{5}{L} \int_{- L}^{L} sin \frac{mπ x}{L} d x = 0$ , kyunki har integral endpoints par sines/cosines ka ek aisa difference deta hai jo cancel ho jaata hai. Yeh step kyun? Yeh constant $1$ ke saath har wave ki orthogonality hai — DC part har wiggle ke "perpendicular" hota hai.

Verify: Series $= \frac{a _{0}}{2} = 5 = f (x)$ exactly, baaki saare terms zero. ✔ Degenerate case confirm: ek constant apna khud ka Fourier series hai.

Example 5 — Cell C5 (limiting: pehle se ek basis wave hai) $f (x) = sin \frac{2 π x}{L} + 3 cos \frac{5 π x}{L}$

Worked example Ek aisi function rebuild karo jo

pehle se hi do basis waves ka sum hai. Forecast: Orthogonality of functions ke hisab se, machine ko simply woh coefficients wapas dene chahiye jo tum formula se padh sakte ho: $b_{2} = 1$ , $a_{5} = 3$ , baaki sab $0$ . Agar usne kuch aur kiya, toh orthogonality tooti hogi.

Step 1 — target coefficients padho. $\sum [a_{n} cos \frac{nπ x}{L} + b_{n} sin \frac{nπ x}{L}]$ se compare karne par, sirf nonzero pieces $n = 2$ (ek sine, amplitude $1$ ) aur $n = 5$ (ek cosine, amplitude $3$ ) par hain. Yeh step kyun? Agar answer obvious hai, toh ise integral machinery ko test karne ke liye use karo, blindly trust karne ke bajaay.

Step 2 — integrate karke $b_{2}$ confirm karo. $b_{2} = \frac{1}{L} \int_{- L}^{L} (sin \frac{2 π x}{L} + 3 cos \frac{5 π x}{L}) sin \frac{2 π x}{L} d x .$ Cross term $\int cos \frac{5 π x}{L} sin \frac{2 π x}{L} d x = 0$ (sine–cosine orthogonality, hamesha). Self term $\int sin^{2} \frac{2 π x}{L} d x = L$ . Toh $b_{2} = \frac{1}{L} \cdot L = 1$ . ✔ Yeh step kyun? Yeh orthogonality in action hai: $sin \frac{2 π x}{L}$ se multiply karne par $n = 2$ wala sine isolate hota hai aur baaki sab silence ho jaate hain.

Step 3 — $a_{5}$ confirm karo. $cos \frac{5 π x}{L}$ se multiply karo: sine term khatam (orthogonal), $3 cos^{2} \frac{5 π x}{L}$ se milta hai $\frac{1}{L} \cdot 3 L = 3$ . Toh $a_{5} = 3$ . ✔

Step 4 — baaki sab. Kisi bhi doosre $m$ ke liye, dono self-terms $n \neq = m$ par hain, toh orthogonality $0$ deta hai.

Verify: Nonzero coefficients: $b_{2} = 1, a_{5} = 3$ , baaki sab $0$ . Series exactly $f$ reproduce karta hai — ek basis function apna khud ka series hai. ✔

Example 6 — Cell C6 (word problem + exam twist): off-centre interval par ek heater

Full period ek full period hoti hai chahe window kahaan se shuru ho — neeche orange aur teal windows identical coefficients capture karti hain:

Worked example Length

$L$ ki ek metal rod ka starting temperature $u_{0} (x) = x$ hai $0 < x < L$ ke liye. Separation of variables se heat equation solve karne ke liye hume ise $[0, L]$ par sine series ke roop mein chahiye. Lekin maan lo exam $g (x) = x$ shifted interval $[a, a + 2 L]$ par deta hai jahan $a \neq = - L$ , aur period $2 L$ ke saath full series maangta hai. Dikhao ki yeh compatible hain.

Forecast: Fourier coefficients sirf ek full period par depend karte hain, na ki tum kahaan se integrate shuru karte ho. Toh window ko $[- L, L]$ se $[a, a + 2 L]$ slide karne par same $a_{m}, b_{m}$ milne chahiye — agar tum $g$ ko periodically extend karo. "Off-centre" twist ek trap hai: yeh kuch nahi badalta.

Step 1 — shift-invariance principle. Kisi bhi $2 L$ -periodic integrand $h$ ke liye, $\int_{a}^{a + 2 L} h (x) d x = \int_{- L}^{L} h (x) d x .$ Yeh step kyun? Periodicity ka matlab hai ki jo piece right end se drop hoti hai woh left end par wapas aati hai — kisi bhi full period par total fixed hota hai. Toh tum hamesha $[- L, L]$ par re-centre kar sakte ho.

Step 2 — Cell C1 par reduce karo. Re-centre hone ke baad, $g (x) = x$ on $[- L, L]$ exactly Example 1 hai, ek odd function → saare $a_{m} = 0$ aur $b_{m} = \frac{2 L ( - 1 ) ^{m + 1}}{mπ}$ . Yeh step kyun? Scary shifted interval ek aisi case par collapse ho jaata hai jo humne pehle se solve kar li hai.

Step 3 — half-range connection, poori tarah through. Rod problem ko $[0, L]$ par sine-only series chahiye. Half-range sine expansion ek recipe se define hoti hai: $u_{0}$ ko sirf $[0, L]$ par lo, ise odd function ki tarah $[- L, L]$ par reflect karo, yaani $u_{0}^{odd} (x) = u_{0} (x)$ $x > 0$ ke liye aur $u_{0}^{odd} (- x) = - u_{0} (x)$ , phir full-series machinery chalao. Kyunki extension odd hai, saare $a_{m} = 0$ , aur surviving sine coefficients hain $b_{m} = \frac{1}{L} \int_{- L}^{L} u_{0}^{odd} (x) sin \frac{mπ x}{L} d x .$ Integrand $u_{0}^{odd} sin$ hai (odd)(odd)=even, toh fold kar sakte hain: $b_{m} = \frac{2}{L} \int_{0}^{L} u_{0}^{odd} (x) sin \frac{mπ x}{L} d x = \frac{2}{L} \int_{0}^{L} u_{0} (x) sin \frac{mπ x}{L} d x,$ aur $[0, L]$ par extension $u_{0}$ khud ke barabar hai, toh odd reflection final formula se gayab ho gayi. Woh last expression hi half-range sine formula hai. Ab $u_{0} (x) = x$ specialize karo: $b_{m} = \frac{2}{L} \int_{0}^{L} x sin \frac{mπ x}{L} d x = \frac{2 L ( - 1 ) ^{m + 1}}{mπ},$ jo exactly Step 2 (aur Example 1) ka coefficient hai. Toh off-centre full series, re-centred full series, aur half-range sine series teeno same $b_{m}$ dete hain. Yeh step kyun? Yeh loop close karta hai: full-series derivation aur half-range recipe do alag rules nahi hain balki ek hi hain — ek odd function ki full series pure sine hoti hai, aur $\frac{2}{L} \int_{0}^{L}$ sirf folded $\frac{1}{L} \int_{- L}^{L}$ hai.

Conclusion. Exam twist (shifted interval) aur physics goal (half-range sine on $[0, L]$ ) dono Cell C1 par reduce hote hain. Temperature profile expand hota hai: $u_{0} (x) = x = \frac{2 L}{π} \sum_{m = 1}^{\infty} \frac{( - 1 ) ^{m + 1}}{m} sin \frac{mπ x}{L} on (0, L) .$

Verify: $L = π$ ke saath: $b_{m} = \frac{2 ( - 1 ) ^{m + 1}}{m}$ , toh $b_{1} = 2, b_{2} = - 1, b_{3} = \frac{2}{3}$ — same numbers jo Example 1 ke check mein the, starting endpoint $a$ ki parwah kiye bina. ✔

Example 7 — Cell C7 (discontinuous jump): square wave, endpoints aur Gibbs

Square wave $- 1$ se $+ 1$ ki taraf leap karta hai; dekho ki series exactly jump par kya value leta hai, aur uske paas ka stubborn overshoot:

Worked example Square wave

$f (x) = {- 1 + 1 - L < x < 0 0 < x < L$ rebuild karo aur describe karo ki jump par aur endpoints $x = \pm L$ par kya hota hai. Forecast: Graph $x = 0$ par $- 1$ se $+ 1$ ki taraf leap karta hai — ek genuine discontinuity. Yeh odd bhi hai ( $x \to - x$ flip karo aur value ka sign badal jaata hai), toh sirf sines predict karo. Lekin ek nayi question aati hai: series exactly jump par aur ends par kya karti hai, aur kya woh unke paas overshoot karti hai?

Step 1 — classify karo. $f (- x) = - f (x)$ : odd → Cell C7 with odd symmetry, toh saare $a_{m} = 0$ . Yeh step kyun? Odd cosines kill kar deta hai; jump isse nahi badalta.

Step 2 — $b_{m}$ compute karo. Kyunki $f sin$ even hai, fold karo: $b_{m} = \frac{2}{L} \int_{0}^{L} (1) sin \frac{mπ x}{L} d x = \frac{2}{L} \cdot \frac{L}{mπ} [- cos \frac{mπ x}{L}]_{0}^{L} = \frac{2}{mπ} (1 - cos mπ) .$ $cos mπ = (- 1)^{m}$ ke saath: $b_{m} = \frac{2}{mπ} (1 - (- 1)^{m}) = {\frac{4}{mπ} 0 m odd m even$ Yeh step kyun? Ex 1 jaisa hi folding + endpoint evaluation; odd terms saara weight uthate hain. Toh $f (x) = \frac{4}{π} k = 0 \sum \infty \frac{1}{2 k + 1} sin \frac{( 2 k + 1 ) π x}{L}$ .

Step 3 — interior jump $x = 0$ par value. Har $sin 0 = 0$ hai, toh series wahan $0$ sum karti hai. Yeh exactly jump ka midpoint hai, $\frac{( - 1 ) + ( + 1 )}{2} = 0$ . Dirichlet's theorem ke hisab se, ek Fourier series hamesha jump par left aur right limits ke average par converge karti hai — kisi ek side par nahi. Yeh step kyun? Yeh edge-case question ka jawab hai "break par series kya hai?" jo smooth examples mein kabhi nahi uthti.

Step 4 — endpoints $x = \pm L$ par value (doosra jump). Is square wave ka $2 L$ -periodic extension phir se $x = \pm L$ par jump karta hai, jahan right value ( $+ 1$ , $L$ ke just left) wrap-around value ( $- 1$ , $L$ ke just right) se milti hai. $x = L$ par har $sin \frac{( 2 k + 1 ) π L}{L} = sin ((2 k + 1) π) = 0$ hai, toh series $0$ ke barabar hoti hai — phir se $+ 1$ aur $- 1$ ka midpoint. Toh endpoints usi Dirichlet midpoint rule ka paalan karte hain jaise koi bhi interior jump. Yeh step kyun? "Full" series ke liye $\pm L$ endpoints periodicity se glued hote hain; agar $f (- L^{+}) \neq = f (L^{-})$ toh woh ek hidden discontinuity banaate hain, aur series unke average par land karti hai — ek case jo students aksar bhool jaate hain.

Step 5 — Gibbs phenomenon. Har jump ke paas partial sums ek fixed fraction se overshoot karte hain (jump height ka lagbhag $9%$ ) chahe kitne bhi terms add karo — ripple narrow hoti hai lekin kabhi shrink nahi hoti. Yeh Gibbs phenomenon hai, jo smooth waves se discontinuity approximate karne ki pehchaan hai. (Example 1 ki figure mein wahi overshoot $x = \pm L$ par appear hota hai, jahan sawtooth line $x$ secretly $+ L$ se $- L$ ki taraf jump karti hai.) Yeh step kyun? Yeh warn karta hai ki jumps par uniform closeness expect mat karo — sirf jump par midpoint par convergence aur dono taraf se door points par $f$ tak convergence.

Verify: $b_{1} = \frac{4}{π}, b_{2} = 0, b_{3} = \frac{4}{3 π}$ ; $x = 0$ aur $x = L$ par series ki value $0$ hai (jump ka midpoint). Saath hi, $L = 1, x = \frac{1}{2}$ set karne par $\frac{4}{π} \sum_{k} \frac{( - 1 ) ^{k}}{2 k + 1} = \frac{4}{π} \cdot \frac{π}{4} = 1$ milta hai, jo $f (\frac{1}{2}) = + 1$ se match karta hai. ✔

Endpoint & discontinuity convergence — "full" series ke liye edge case

Intuition Jump par ya

$x = \pm L$ par series actually kya value leta hai? Ek full Fourier series $x = - L$ aur $x = + L$ ko same point maanti hai (period $2 L$ unhe wrap kar deta hai). Toh do edge cases hain jo smooth examples chhupa dete hain:

Kisi bhi interior jump par $(- L, L)$ ke andar: Dirichlet's theorem ke hisab se, series $\frac{1}{2} (f (x^{-}) + f (x^{+}))$ par converge karti hai, do one-sided limits ka midpoint — Ex 7, $x = 0$ dekho.
Endpoints $x = \pm L$ par: series $\frac{1}{2} (f (- L^{+}) + f (L^{-}))$ par converge karti hai. Agar woh do values disagree karein (jaise square wave ke liye, aur secretly $f (x) = x$ ke liye jahan $+ L$ , $- L$ se milta hai), toh endpoints khud ek jump hain aur series unke average par land karti hai, kisi bhi end value par nahi.
Jahan $f$ continuous hai: series $f (x)$ ke exactly barabar hoti hai, lekin kisi bhi jump ke paas yeh hamesha ~9% overshoot karti hai (Gibbs) — ek aisa overshoot jo kabhi nahi jaata, sirf narrow hota hai.

Recall Case-map self-test

Odd function: kaun se coefficients bachte hain? ::: sirf $b_{m}$ (sines); $a_{0} = a_{m} = 0$ . Even function: kaun se bachte hain? ::: sirf $a_{0}$ aur $a_{m}$ (cosines); $b_{m} = 0$ . Ek constant $c$ : full series? ::: $a_{0} = 2 c$ , baaki sab $0$ ; series sirf $c$ hai. Kya full-period integral ka starting endpoint matter karta hai? ::: Nahi — $2 L$ length ka koi bhi interval same coefficients deta hai. Jump par, series kis value par converge karti hai? ::: midpoint (left aur right limits ka average). $x = \pm L$ par, kya value? ::: $f (- L^{+})$ aur $f (L^{-})$ ka average — jump agar woh differ karein. Gibbs phenomenon kya hai? ::: discontinuity ke paas partial sums ka ek fixed ~9% overshoot jo kabhi nahi jaata. $\sum_{k \geq 0} \frac{1}{( 2 k + 1 ) ^{2}} = ?$ (Ex 2 mein $x = 0$ se) ::: $\frac{π ^{2}}{8}$ .

Related: Convergence of Fourier series (Dirichlet conditions) (kya series actually jumps par $f$ ke barabar hoti hai?), Complex (exponential) Fourier series (same coefficients repackaged), Parseval's theorem (upar ke coefficients par energy check).