Half-range sine and cosine series: Worked examples

4.7.6 · D3 · Maths › Partial Differential Equations › Half-range sine and cosine series

Yeh page ek shooting range hai. Parent note ne tumhe do formulas diye; yahan hum unpe har tarah ka function fire karte hain taaki koi bhi exam problem surprise na kar sake. Kisi bhi example se pehle, hum saare cases ka ek map banate hain — phir unhe ek-ek karke knock out karte hain.

Do symbols jinhe hum baar baar use karenge, plain words mein define kiye hain taaki kuch bhi assumed na ho:

The scenario matrix

Har half-range problem jo tumhe mili sakti hai, woh in cells mein se kisi ek mein aati hai. Last column us example ka naam batata hai jo use clear karta hai. Figure s01 is poori page ka dil draw karta hai: wahi function $f (x) = x$ do tarafon se mirror kiya gaya hai. Abhi dekho — left panel (coral solid + lavender dashed) odd mirror dikhata hai, jisme right end par ek violent jump hai (ek sawtooth); right panel (coral solid + mint dashed) even mirror $∣ x ∣$ dikhata hai, jisme sirf ek gentle corner hai. Wahi ek contrast — jump vs corner — isliye sine coefficients dheere marte hain ( $1/ n$ ) aur cosine coefficients jaldi marte hain ( $1/ n^{2}$ ). Jab bhi Cells A, B aur H padho, is picture ko dekhte rehna.

#	Case class	Tricky kya hai	Cleared by
A	Straight line $f = x$ , sine	odd extension ka end par jump hai → slow $1/ n$ decay	Ex 1
B	Straight line $f = x$ , cosine	even extension ek corner hai → faster $1/ n^{2}$ ; $a_{0}$ chahiye	Ex 2
C	Constant $f = c$ , sine vs cosine	cosine trivial hai; sine ek subtle non-zero series hai	Ex 3
D	Piecewise (split-domain) function	har piece par alag integrate karo (sine aur cosine)	Ex 4, Ex 4b
E	Function jo ends par vanish nahi karta	endpoint value ≠ 0 → midpoint (Dirichlet) surprise	Ex 5
F	Even-power polynomial $f = x^{2}$ , cosine	double integration by parts; $a_{0} \neq = 0$	Ex 6
G	Real-world word problem (heated/insulated rod)	physics translate karo → sine ya cosine BC chuno	Ex 7
H	Exam-twist / limiting behaviour	$n \to \infty$ decay rate + endpoint convergence value	Ex 8

Hum do degenerate inputs bhi cover karte hain jo har topic mein handle karne padti hain:

$n = 0$ (average term) — Ex 2, 4b, 6, 7 mein aata hai.
Endpoint values $x = 0$ aur $x = L$ jahan series $f$ ke barabar nahi ho sakti — Ex 5, 8.

Figure — Half-range sine and cosine series

The examples

Cell A — straight line, sine series

Worked example Example 1 —

$f (x) = x$ on $[0, π]$ , half-range sine Pehle forecast: ek rising line ka odd mirror ek sawtooth hai (upar jaata ramp, phir achanak girta hai) — bilkul figure s01 ka left panel. Andaza lagao: coefficients fast marenye ya slow?

Yahan $L = π$ hai, isliye $\frac{nπ x}{L} = n x$ aur $b_{n} = \frac{2}{π} \int_{0}^{π} x sin (n x) d x$ .

Integration by parts karo. Yeh step kyun? Hamare paas polynomial $\times$ trig hai; parts $x$ ko trade kar deta hai, ek pure trig integral chhod ke jo hum finish kar sakte hain. $\int_{0}^{π} x sin (n x) d x = [- \frac{x c o s n x}{n}]_{0}^{π} + \frac{1}{n} \int_{0}^{π} cos n x d x = - \frac{π c o s nπ}{n} + 0.$
Bacha hua integral kill karo. Zero kyun? $\int_{0}^{π} cos n x d x = \frac{s i n nπ}{n} = 0$ kyunki $sin (nπ) = 0$ har integer $n$ ke liye.
$cos nπ = (- 1)^{n}$ use karo. Kyun? $π$ ke whole multiples par cosine $+ 1, - 1, + 1, \dots$ flip karta rehta hai. $b_{n} = \frac{2}{π} (- \frac{π ( - 1 ) ^{n}}{n}) = \frac{2 ( - 1 ) ^{n + 1}}{n} .$ $x = 2 n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n} sin (n x), 0 < x < π .$ Verify karo: $x = \frac{π}{2}$ plug karo: RHS $= 2 (1 - \frac{1}{3} + \frac{1}{5} - \dots) = 2 \cdot \frac{π}{4} = \frac{π}{2} = f (\frac{π}{2}) .$ ✓ Decay $\sim 1/ n$ hai — slow, figure s01 ke sawtooth jump se match karta hai. Forecast confirmed.

Cell B — straight line, cosine series

Worked example Example 2 —

$f (x) = x$ on $[0, π]$ , half-range cosine Forecast: $x$ ka even mirror $∣ x ∣$ hai — ek "V" tent, koi jump nahi, sirf ek corner (figure s01 ka right panel). Andaza lagao: Ex 1 se faster decay hogi ya slower?

Average term $a_{0}$ . Alag kyun? $cos (0 \cdot x) = 1$ hai, isliye $n = 0$ term special hai — yeh mean carry karta hai. $a_{0} = \frac{2}{π} \int_{0}^{π} x d x = \frac{2}{π} \cdot \frac{π ^{2}}{2} = π \Rightarrow \frac{a _{0}}{2} = \frac{π}{2} .$ ( $[0, π]$ par $x$ ka average $π /2$ hai. ✓)
$a_{n}$ ke liye parts. Kyun? Same polynomial×trig structure. $\int_{0}^{π} x cos n x d x = [\frac{x s i n n x}{n}]_{0}^{π} - \frac{1}{n} \int_{0}^{π} sin n x d x = 0 + \frac{1}{n ^{2}} [cos n x]_{0}^{π} = \frac{( - 1 ) ^{n} - 1}{n ^{2}} .$
Odd/even $n$ mein split karo. Kyun? $(- 1)^{n} - 1$ even $n$ ke liye $0$ hai, odd $n$ ke liye $- 2$ hai. Isliye $a_{n} = - \frac{4}{π n ^{2}}$ (odd $n$ ke liye), $0$ (even $n$ ke liye). $x = \frac{π}{2} - \frac{4}{π} k = 0 \sum \infty \frac{cos (( 2 k + 1 ) x )}{( 2 k + 1 ) ^{2}}, 0 < x < π .$ Verify karo $x = 0$ par: RHS $= \frac{π}{2} - \frac{4}{π} \sum \frac{1}{( 2 k + 1 ) ^{2}} = \frac{π}{2} - \frac{4}{π} \cdot \frac{π ^{2}}{8} = 0 = f (0)$ . ✓ Decay $\sim 1/ n^{2}$ — faster hai Ex 1 se. Forecast confirmed: smooth corner, jump ko beat karta hai. (Dekho Even and Odd Functions.)

Cell C — constant function

Worked example Example 3 —

$f (x) = c$ (ek constant) on $[0, L]$ , dono series Forecast: ek constant pehle se hi flat hai. Kaun si series trivial hogi, kaun si surprising?

Cosine side. Trivial kyun? Ek constant ka even mirror wahi constant hai — koi wiggle nahi chahiye. $\frac{a _{0}}{2} = \frac{1}{L} \int_{0}^{L} c d x = c, a_{n} = \frac{2}{L} \int_{0}^{L} c cos \frac{nπ x}{L} d x = \frac{2 c}{L} \cdot \frac{L}{nπ} [sin \frac{nπ x}{L}]_{0}^{L} = 0.$ Toh cosine series sirf $f = c$ hai. Ek term, done.

Sine side. Non-trivial kyun? $c$ ka odd mirror ek square wave hai (upar $+ c$ , neeche $- c$ ) — wiggles se bhara hua. $b_{n} = \frac{2}{L} \int_{0}^{L} c sin \frac{nπ x}{L} d x = \frac{2 c}{L} \cdot \frac{L}{nπ} [- cos \frac{nπ x}{L}]_{0}^{L} = \frac{2 c}{nπ} (1 - cos nπ) = \frac{2 c}{nπ} (1 - (- 1)^{n}) .$ Yeh odd $n$ ke liye $\frac{4 c}{nπ}$ hai, even $n$ ke liye $0$ . $c = \frac{4 c}{π} k = 0 \sum \infty \frac{1}{2 k + 1} sin \frac{( 2 k + 1 ) π x}{L}, 0 < x < L .$ Verify karo ( $c = 1, L = π, x = \frac{π}{2}$ lete hain): RHS $= \frac{4}{π} (1 - \frac{1}{3} + \frac{1}{5} - \dots) = \frac{4}{π} \cdot \frac{π}{4} = 1 = c$ . ✓ Ek flat line ek infinite pile of sines hide karti hai kyunki humne ends par jump force kiya.

Cell D — piecewise function (dono series)

Worked example Example 4 — piecewise

$f$ , half-range sine, $L = 2$ $f (x) = {x, 2 - x, 0 \leq x \leq 1 1 \leq x \leq 2 (ek symmetric "tent") .$ Forecast: yeh pehle badhta hai phir girta hai, $x = 1$ par peak karta hai. Sirf odd multiples of the half survive honge — guess karo kyun.

Corner par integral split karo. Kyun? $f$ ka formula $x = 1$ par change hota hai, isliye har piece ko apne rule se integrate karo. $k_{n} = \frac{nπ}{2}$ shorthand ke liye likhte hain. $b_{n} = \frac{2}{2} I_{1} \int_{0}^{1} x sin \frac{nπ x}{2} d x + I_{2} \int_{1}^{2} (2 - x) sin \frac{nπ x}{2} d x .$
Piece $I_{1}$ par parts. Kyun? polynomial×trig. $u = x, d v = sin (k_{n} x) d x$ ke saath: $I_{1} = [- \frac{x c o s k _{n} x}{k _{n}}]_{0}^{1} + \frac{1}{k _{n}} \int_{0}^{1} cos k_{n} x d x = - \frac{c o s k _{n}}{k _{n}} + \frac{s i n k _{n}}{k _{n}^{2}} .$
Piece $I_{2}$ par parts. Wahi tool kyun? Ab bhi polynomial×trig hai, ab $u = 2 - x$ ke saath (isliye $d u = - d x$ ): $I_{2} = [- \frac{( 2 - x ) c o s k _{n} x}{k _{n}}]_{1}^{2} - \frac{1}{k _{n}} \int_{1}^{2} cos k_{n} x d x = \frac{c o s k _{n}}{k _{n}} - \frac{s i n 2 k _{n} - s i n k _{n}}{k _{n}^{2}} .$
Add karo aur dekho boundary terms cancel ho jaate hain. Kyun collapse hote hain? Dono $\frac{cos k _{n}}{k _{n}}$ terms (corner $x = 1$ se) equal aur opposite hain, isliye woh cancel ho jaate hain; aur $sin 2 k_{n} = sin (nπ) = 0$ hai. Sirf do $sin k_{n} / k_{n}^{2}$ interior pieces bachti hain: $I_{1} + I_{2} = \frac{s i n k _{n}}{k _{n}^{2}} + \frac{s i n k _{n}}{k _{n}^{2}} = \frac{2 s i n k _{n}}{k _{n}^{2}} = \frac{2 s i n \frac{nπ}{2}}{( nπ /2 ) ^{2}} = \frac{8 s i n \frac{nπ}{2}}{n ^{2} π ^{2}} .$
Prefactor $\frac{2}{2} = 1$ se multiply karo. 1 kyun? Yahan $\frac{2}{L} = \frac{2}{2} = 1$ hai, isliye $b_{n} = I_{1} + I_{2}$ directly: $b_{n} = \frac{8}{n ^{2} π ^{2}} sin \frac{nπ}{2} .$
$sin \frac{nπ}{2}$ padho. Yeh even $n$ ke liye $0$ hai aur odd $n$ ke liye $\pm 1$ — isliye sirf odd harmonics aate hain (tent apne middle ke baare mein symmetric hai). $f (x) = \frac{8}{π ^{2}} k = 0 \sum \infty \frac{( - 1 ) ^{k}}{( 2 k + 1 ) ^{2}} sin \frac{( 2 k + 1 ) π x}{2}, 0 < x < 2.$ Verify karo peak $x = 1$ par (should give $f (1) = 1$ ): $sin \frac{( 2 k + 1 ) π}{2} = (- 1)^{k}$ , isliye RHS $= \frac{8}{π ^{2}} \sum \frac{( - 1 ) ^{k} ( - 1 ) ^{k}}{( 2 k + 1 ) ^{2}} = \frac{8}{π ^{2}} \cdot \frac{π ^{2}}{8} = 1$ . ✓

Worked example Example 4b —

wahi piecewise tent, half-range cosine, $L = 2$ Yeh natural counterpart hai: identical tent ko cosine series ke roop mein expand karo (jo ek insulated rod ko chahiye hoga). Forecast: cosine ek constant term rakhta hai = tent ki average height. Tent $1$ tak jaati hai phir girti hai, isliye uska average $\frac{1}{2}$ hona chahiye. Predict karo $\frac{a _{0}}{2} = \frac{1}{2}$ .

Average term $a_{0}$ . Pehle kyun? $n = 0$ mean carry karta hai, do pieces mein split hoga. $a_{0} = \frac{2}{2} [\int_{0}^{1} x d x + \int_{1}^{2} (2 - x) d x] = [\frac{1}{2} + \frac{1}{2}] = 1 \Rightarrow \frac{a _{0}}{2} = \frac{1}{2} .$ (Forecast se match karta hai ✓.)
$a_{n}$ do pieces mein. Phir split kyun? Same reason — formula $x = 1$ par change hota hai. $k_{n} = \frac{nπ}{2}$ likhte hain; note karo $cos 2 k_{n} = cos (nπ) = (- 1)^{n}$ aur $sin 2 k_{n} = sin (nπ) = 0$ . Ex 4 jaise $u$ -choices ke saath: $\int_{0}^{1} x cos k_{n} x d x = [\frac{x s i n k _{n} x}{k _{n}}]_{0}^{1} - \frac{1}{k _{n}} \int_{0}^{1} sin k_{n} x d x = \frac{s i n k _{n}}{k _{n}} + \frac{c o s k _{n} - 1}{k _{n}^{2}},$ $\int_{1}^{2} (2 - x) cos k_{n} x d x = [\frac{( 2 - x ) s i n k _{n} x}{k _{n}}]_{1}^{2} + \frac{1}{k _{n}} \int_{1}^{2} sin k_{n} x d x = - \frac{s i n k _{n}}{k _{n}} + \frac{c o s k _{n} - c o s 2 k _{n}}{k _{n}^{2}} .$
Add karo — $sin k_{n} / k_{n}$ terms cancel ho jaate hain aur $cos 2 k_{n}$ enter karta hai. Kyun? Do $\pm sin k_{n} / k_{n}$ boundary terms (corner $x = 1$ ) equal aur opposite hain, isliye cancel ho jaate hain. Bachne wale numerators hain $(cos k_{n} - 1)$ pehli piece se aur $(cos k_{n} - cos 2 k_{n})$ doosri se. Unhe add karo aur $cos 2 k_{n} = (- 1)^{n}$ use karo: $a_{n} = \frac{( c o s k _{n} - 1 ) + ( c o s k _{n} - ( - 1 ) ^{n} )}{k _{n}^{2}} = \frac{2 c o s k _{n} - 1 - ( - 1 ) ^{n}}{k _{n}^{2}} .$
$- 1 - (- 1)^{n}$ simplify karo. Kyun? Even $n$ ke liye, $- 1 - (- 1)^{n} = - 2$ hai, numerator deta hai $2 cos k_{n} - 2 = 2 (cos \frac{nπ}{2} - 1)$ . Odd $n$ ke liye, $- 1 - (- 1)^{n} = 0$ aur $cos k_{n} = cos \frac{nπ}{2} = 0$ hai, numerator deta hai $0 - 0 = 0 = 2 (cos \frac{nπ}{2} - 1)$ (kyunki $cos \frac{nπ}{2} = 0$ se $2 (0 - 1) = - 2$ ... seedha case by case compute karte hain). Sab cases mein numerator $2 (cos \frac{nπ}{2} - 1)$ ke barabar hai ek baar verify karne par — lekin us shortcut par trust karne ke bajaye, neeche teen residue classes directly padho.
$n$ ke teen classes padho. Teen kyun? $cos \frac{nπ}{2}$ aur $(- 1)^{n}$ milke har $4$ mein repeat karte hain. $k_{n}^{2} = (nπ /2)^{2} = n^{2} π^{2} /4$ use karo toh ek factor $4$ aata hai:

$n \equiv 0 (mod 4)$ : $cos k_{n} = 1$ , $(- 1)^{n} = 1$ → numerator $2 (1) - 1 - 1 = 0$ → $a_{n} = 0$ .
$n \equiv 2 (mod 4)$ : $cos k_{n} = - 1$ , $(- 1)^{n} = 1$ → numerator $2 (- 1) - 1 - 1 = - 4$ → $a_{n} = \frac{- 4}{n ^{2} π ^{2} /4} = \frac{- 16}{n ^{2} π ^{2}}$ .
$n$ odd: $cos k_{n} = 0$ , $(- 1)^{n} = - 1$ → numerator $0 - 1 + 1 = 0$ → $a_{n} = 0$ .

Toh sirf $n \equiv 2 (mod 4)$ survive karte hain. Notice karo yeh teen lines exactly wahi hain jo $\frac{8 ( cos \frac{nπ}{2} - 1 )}{n ^{2} π ^{2}}$ produce karta hai ( $0, - \frac{16}{n ^{2} π ^{2}}, 0$ respectively), isliye hum compact form likh sakte hain: $f (x) = \frac{1}{2} + n = 1 \sum \infty \frac{8 ( cos \frac{nπ}{2} - 1 )}{n ^{2} π ^{2}} cos \frac{nπ x}{2}, 0 < x < 2.$ Verify karo $x = 1$ par (tent peak, cosine even-mirror continuous hai, should give $1$ ): sirf $n \equiv 2 (mod 4)$ contribute karte hain, $cos \frac{nπ \cdot 1}{2} = cos \frac{nπ}{2} = - 1$ wahan hai, aur numerically sum karne par $1.000$ milta hai. ✓ Ex 4 se compare karo — same function, do legitimate series; choice boundary condition se dictate hoti hai, function se nahi.

Cell E — function jo ends par vanish nahi karta

Worked example Example 5 —

$f (x) = 1$ on $[0, π]$ , sine series, endpoint surprise Yeh Cell C hai $c = 1, L = π$ ke saath, lekin ab hum endpoints interrogate karte hain — woh case jo sabko trip karta hai.

Ex 3 se: $1 = \frac{4}{π} k = 0 \sum \infty \frac{sin (( 2 k + 1 ) x )}{2 k + 1}$ for $0 < x < π$ .

Poocho series $x = 0$ par kya deti hai. Kyun? Kyunki $f (0) = 1$ hai, lekin har $sin (n \cdot 0) = 0$ hai, isliye series 0 sum karta hai, 1 nahi. Contradiction?
Dirichlet ke midpoint rule se resolve karo. Yeh rule exist kyun karta hai? Periodic extension ke ek jump par, ek Fourier series dono sides ke average par converge karti hai. Odd square-wave $0$ ke just left se $- 1$ se $+ 1$ pe jump karta hai (just right), isliye series $\frac{- 1 + 1}{2} = 0$ deta hai. ✓ Yeh bug nahi hai — yeh rule hai. (Dekho Dirichlet Conditions.)
$x = π$ par bhi same. $+ 1$ se $- 1$ tak jump, midpoint $= 0$ ; series $0$ deta hai, $f (π) = 1$ nahi.

Inside point $x = \frac{π}{2}$ verify karo: RHS $= \frac{4}{π} (1 - \frac{1}{3} + \frac{1}{5} - \dots) = 1$ . ✓ Take-away: series equalities open interval $0 < x < π$ par likho; ends ko midpoint rule se check karo.

Cell F — even-power polynomial

Worked example Example 6 —

$f (x) = x^{2}$ on $[0, π]$ , half-range cosine Forecast: ek parabola ka even mirror smooth hota hai (koi corner value jump nahi), isliye fast decay aur ek hefty average expect karo.

Average term. Pehle kyun? Phir se $n = 0$ special hai. $a_{0} = \frac{2}{π} \int_{0}^{π} x^{2} d x = \frac{2}{π} \cdot \frac{π ^{3}}{3} = \frac{2 π ^{2}}{3} \Rightarrow \frac{a _{0}}{2} = \frac{π ^{2}}{3} .$
$a_{n}$ ke liye parts do baar. Do baar kyun? Har parts polynomial power ko ek se ghata deta hai; $x^{2}$ ke do rounds chahiye. $\int_{0}^{π} x^{2} cos n x d x = [\frac{x ^{2} s i n n x}{n}]_{0}^{π} - \frac{2}{n} \int_{0}^{π} x sin n x d x = 0 - \frac{2}{n} (- \frac{π ( - 1 ) ^{n}}{n}) = \frac{2 π ( - 1 ) ^{n}}{n ^{2}} .$
Assemble karo. $a_{n} = \frac{2}{π} \cdot \frac{2 π ( - 1 ) ^{n}}{n ^{2}} = \frac{4 ( - 1 ) ^{n}}{n ^{2}} .$ $x^{2} = \frac{π ^{2}}{3} + 4 n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n ^{2}} cos (n x), 0 < x < π .$ Verify karo $x = π$ par (parabola even-mirror continuous hai, isliye $π^{2}$ dena chahiye): RHS $= \frac{π ^{2}}{3} + 4 \sum \frac{( - 1 ) ^{n} c o s ( nπ )}{n ^{2}} = \frac{π ^{2}}{3} + 4 \sum \frac{( - 1 ) ^{n} ( - 1 ) ^{n}}{n ^{2}} = \frac{π ^{2}}{3} + 4 \cdot \frac{π ^{2}}{6} = π^{2}$ . ✓ ( $\sum 1/ n^{2} = π^{2} /6$ use kiya.)

Cell G — real-world word problem

Worked example Example 7 — insulated rod (word problem → series chuno)

"Ek rod $0 \leq x \leq L$ , $L = π$ , ke dono ends insulated hain (koi heat bahar nahi jaati: ends par zero slope). Uska initial temperature profile $f (x) = x$ hai. Initial state ko kaun si half-range series describe karti hai, aur uska steady-state (eventual uniform) temperature kya hai?"

Forecast: insulated = ends par zero slope = Neumann BC → hamein ends par flat functions chahiye → cosine. Final temperature guess karo.

Physics ko series mein translate karo. Cosine kyun? Cosine waves ki $x = 0, π$ par zero derivative hoti hai, exactly insulated ends match karta hai (dekho Neumann Boundary Conditions). Sine galat hoga (uski slope ends par non-zero hoti hai). Toh Ex 2 reuse karo: $x = \frac{π}{2} - \frac{4}{π} \sum_{k = 0}^{\infty} \frac{c o s (( 2 k + 1 ) x )}{( 2 k + 1 ) ^{2}} .$
Steady state dhundho. $a_{0} /2$ term kyun? Heat Equation — separation of variables mein, har $cos (n x)$ mode time $\to \infty$ ke saath zero par decay hota hai; sirf constant $\frac{a _{0}}{2}$ bachta hai. Ek insulated rod total heat conserve karta hai, isliye yeh average temperature par settle hota hai. $T_{\infty} = \frac{a _{0}}{2} = \frac{1}{π} \int_{0}^{π} x d x = \frac{π}{2} .$ Physics se verify karo: total heat $\propto \int_{0}^{π} x d x = π^{2} /2$ ; length $π$ par uniformly spread karne par $\frac{π ^{2} /2}{π} = \frac{π}{2}$ milta hai. ✓ Match karta hai. Cosine series ka constant term hi steady state hai — ek beautiful consistency.

Cell H — exam-twist: decay rate & limiting behaviour

Worked example Example 8 — "which converges faster and to what" twist

Exam wording: "For $f (x) = x$ on $[0, π]$ , (i) sine vs cosine coefficients ka large- $n$ decay batao, (ii) dhundho ki har series $x = π$ par kya converge karti hai."

Forecast: tumne dono series dekhi hain (Ex 1, 2) aur figure s01 mein unke do mirrors. Aage padhne se pehle answer do.

Decay rates. Alag kyun hote hain? Coefficient decay periodic extension ki smoothness se set hoti hai (figure s01). Sine → odd extension = sawtooth with a jump → $b_{n} \sim 1/ n$ . Cosine → even extension $= ∣ x ∣$ , continuous with a corner → $a_{n} \sim 1/ n^{2}$ . Smoothness ka ek extra order, $n$ ke decay mein ek extra power buy karta hai.
$x = π$ par sine series. $π$ kyun nahi? Sawtooth ka $+ π$ se $- π$ tak jump; midpoint $= 0$ . Check karo: har $sin (nπ) = 0$ hota hai, isliye series literally $0$ sum karta hai. ✓
$x = π$ par cosine series — sign unpack karo. $π$ kyun? Even extension wahan continuous hai (koi jump nahi), isliye series ko $f (π) = π$ equal karna chahiye. Explicitly arithmetic verify karte hain. Cosine series (Ex 2) hai $x = \frac{π}{2} - \frac{4}{π} \sum_{k = 0}^{\infty} \frac{c o s (( 2 k + 1 ) x )}{( 2 k + 1 ) ^{2}} .$ $x = π$ par har term ko $cos ((2 k + 1) π)$ chahiye. Yeh $- 1$ kyun hai? Kyunki $2 k + 1$ odd hai, aur $π$ ke odd multiple par cosine $- 1$ hota hai: $cos ((2 k + 1) π) = (- 1)^{2 k + 1} = - 1$ . Toh har term $\frac{- 1}{( 2 k + 1 ) ^{2}}$ ban jaata hai, aur aage ka minus use $+$ mein flip kar deta hai: $x_{x = π} = \frac{π}{2} - \frac{4}{π} \sum_{k = 0}^{\infty} \frac{- 1}{( 2 k + 1 ) ^{2}} = \frac{π}{2} + \frac{4}{π} \sum_{k = 0}^{\infty} \frac{1}{( 2 k + 1 ) ^{2}} = \frac{π}{2} + \frac{4}{π} \cdot \frac{π ^{2}}{8} = \frac{π}{2} + \frac{π}{2} = π .$ (Humne $\sum_{k \geq 0} \frac{1}{( 2 k + 1 ) ^{2}} = \frac{π ^{2}}{8}$ use kiya.) Toh cosine series exactly $π = f (π)$ deti hai. ✓

Verify karo: cosine endpoint value $= π$ (upar dikhaya), sine endpoint $= 0$ (har $sin (nπ) = 0$ ). ✓ Exams ke liye take-away: smoother extension → faster decay; jump par answer midpoint hota hai.

Active recall

Recall Answers cover karo aur fire karo

Insulated rod ke liye kaun sa cell chuno, aur kyun? ::: Cosine (Neumann, ends par zero slope). Ek constant $c$ ki trivial cosine series hoti hai lekin non-trivial sine series — kyun? ::: Even mirror of $c$ sirf $c$ hai; odd mirror ek square wave hai jo sine wiggles se bhari hai. $x$ ki sine series $x = π$ par kya sum karti hai, aur kyun? ::: $0$ , sawtooth jump ka midpoint (Dirichlet rule). Kaun sa faster decay karta hai, sawtooth ka $b_{n}$ ya $∣ x ∣$ ka $a_{n}$ , aur kyun? ::: $a_{n} \sim 1/ n^{2}$ faster hai $b_{n} \sim 1/ n$ se; corner, jump se smoother hota hai. Same tent, sine vs cosine — kis mein constant term hota hai, aur woh kya hai? ::: Cosine mein; uska $a_{0} /2 = \frac{1}{2}$ hai (tent ki average height).

Connections

Parent topic — formulas jinhein yeh page stress-test karta hai.
Even and Odd Functions — odd→sine, even→cosine kyun.
Dirichlet Conditions — Ex 5 & 8 mein use hua midpoint rule.
Heat Equation — separation of variables — Ex 7 mein steady state.
Neumann Boundary Conditions — Ex 7 mein insulated-end pairing.
Fourier Series — full range — general theory jise yeh specialize karta hai.
Wave Equation on a finite string — fixed ends ke liye sine series.