Half-range sine and cosine series: Question bank — Hinglish deep dive

4.7.6 · D5 · Maths › Partial Differential Equations › Half-range sine and cosine series

Shuru karne se pehle, teen foundations ko refresh karo jis par ye page tika hai: Even and Odd Functions (wo symmetry jo aadhe coefficients delete kar deti hai), Dirichlet Conditions (series jump par kya karta hai), aur Fourier Series — full range jise hum specialise karte hain.

Sahi hai ya galat — justify karo

"Extension" ka matlab hai wo invented values jo hum $[- L, 0]$ par rakhte hain taaki sirf $[0, L]$ par jaani hui function ek full-period function ban sake $(- L, L)$ par. Har item ke liye yahi picture yaad rakho.

Ek hi

f

ki half-range sine series aur half-range cosine series

(0, L)

par agree karti hain

Sahi — dono open interval

0 < x < L

par wohi

f (x)

reproduce karte hain; fark sirf

[0, L]

ke bahar invented values mein hai (ek odd, ek even), isliye physical interval par dono identical function represent karte hain.

Half-range sine series

f (x) = 5

(ek nonzero constant) ko

[0, L]

ke har point par exactly represent kar sakti hai

Galat — odd extension series ko

x = 0

par

0

force karta hai (odd functions mein

f (0) = 0

hota hai), aur periodic extension

x = 0, L

par jump karti hai, isliye sine series sirf open

(0, L)

par

5

ke barabar hoti hai, ends par midpoint par converge karti hai.

Kyunki sine series ke liye

a_{n} = 0

hota hai, sine series mein koi constant (average) term nahi hoti

Sahi — odd extension origin ke baare mein antisymmetric hai, isliye ek full period par uska average

0

hai; simply koi

\frac{a _{0}}{2}

carry karne ko hai hi nahi.

Agar do functions ke values

[0, L]

par same hain toh unke half-range sine coefficients bhi same honge

Sahi —

b_{n} = \frac{2}{L} \int_{0}^{L} f sin \frac{nπ x}{L} d x

sirf

[0, L]

par

f

par depend karta hai; kahin aur kya hota hai ye irrelevant hai kyunki hum wahan kabhi integrate nahi karte.

Half-range expansion mein extended function ki period

L

hoti hai

Galat — period

2 L

hoti hai; hum us period ka sirf aadha hi jaante hain (yaani

[0, L]

), isliye ise "half-range" kaha jaata hai.

Even extension ke liye, sine coefficients

b_{n}

zero ho jaate hain kyunki

f (x) sin \frac{nπ x}{L}

odd hoti hai

Sahi — (even)

\times

(odd)

=

odd, aur ek odd function ko symmetric interval

[- L, L]

par integrate karne par

0

milta hai, jo har

b_{n}

ko khatam kar deta hai.

Half-range cosine series hamesha same

f

ki sine series se kam se kam utni hi tezi se converge karti hai

Saamanya taur par Galat — ye tabhi tezi se converge karti hai jab even extension odd extension se smoother ho; agar

f

ends par already zero ho jaata hai, toh odd extension continuous wala ho sakta hai aur jeet sakta hai.

Error dhundho

Neeche diye har statement mein ek kharabi hai jo ek dhyan se padhne wala student pakad sakta hai. Use dhundho aur batao kyun ye galat hai.

f (x) = x

[0, π]

ke sine coefficients

b_{n} \sim 1/ n^{2}

hain, isliye series tezi se converge karti hai."

Galat decay rate hai —

x

ka odd extension ek sawtooth with a jump hai

x = π

par, aur jump discontinuities

b_{n} \sim 1/ n

dete hain, jo slow rate hai.

1/ n^{2}

rate cosine series ka hai (jiska even extension

∣ x ∣

continuous hai).

"Sine series ke liye main

b_{n} = \frac{1}{L} \int_{- L}^{L} f sin \frac{nπ x}{L} d x

standard formula se use karta hun."

Nahi kar sakte —

f

[- L, 0]

par undefined hai, isliye us integral ka wahan koi integrand nahi hai. Sahi tarika ye hai ki odd extension choose karo, jo

\frac{1}{L} \int_{- L}^{L}

\frac{2}{L} \int_{0}^{L}

mein badal deta hai; factor

2

us choice ka fingerprint hai.

f (x) = x

[0, π]

ki cosine series

\sum a_{n} cos n x

hai jisme koi alag constant term nahi hai."

\frac{a _{0}}{2} = \frac{π}{2}

term miss ho gayi —

x

ka even extension

∣ x ∣

hai, jiska ek period par average nonzero hai, isliye constant term zaroori hai; ise drop karne se poori series

π /2

neeche shift ho jaati hai.

x = π

par

x

[0, π]

ki sine series

π

ke barabar hai, jo

f (π) = π

se match karti hai."

Nahi — periodic odd extension

+ π

se (

π

ke bilkul baayin taraf)

- π

par (

π

ke bilkul daayin taraf) jump karti hai, isliye midpoint rule se series

\frac{1}{2} (π + (- π)) = 0

par converge karti hai,

π

par nahi.

"Kyunki sine series fixed ends ke liye suit karti hai, main ek insulated rod ke liye sine series use karunga."

Ulta hai — insulated (zero-slope) ends Neumann conditions hain, jinhe ends par zero derivative wale functions chahiye.

cos \frac{nπ x}{L}

ka slope

x = 0, L

par zero hota hai, isliye cosine match karta hai; sine zero-value (Dirichlet) ends se match karti hai.

"Integral ko double karna

(\frac{2}{L} \int_{0}^{L})

sirf convention hai; plain

\frac{1}{L} \int_{0}^{L}

bhi kaam karega."

Nahi — factor

2

ek symmetric integrand ko fold karne se aata hai: even integrand ke liye

\int_{- L}^{L} = 2 \int_{0}^{L}

\frac{1}{L} \int_{0}^{L}

use karne se har coefficient aadha ho jaega aur series

f

nahi balki

f /2

reconstruct karegi.

"Kyun" wale questions

Kyun

f

ko odd force karne se har cosine coefficient delete ho jaata hai?

Kyunki

cos

even hai, isliye (odd

f

)

\times

(even

cos

) ek odd integrand hai, aur koi bhi odd function symmetric

[- L, L]

par integrate karne se

0

milta hai — sirf sine (odd

\times

odd

=

even) terms bachte hain.

Hum extension choose karne ke liye free kyun hain?

PDE ki boundary conditions sirf physical interval

[0, L]

par

f

specify karti hain;

x < 0

ke liye kuch bhi constrain nahi hai, isliye wahan values hum invent karte hain — hum wo symmetry chunte hain jo series ko BCs se match karaaye.

Extension ki smoothness coefficients ki decay rate ko kyun control karti hai?

Periodic extension mein ek jump slow

1/ n

decay force karta hai, continuous-but-kinked extension

1/ n^{2}

deta hai, aur har ek extra continuous derivative

1/ n

ka ek aur power kamaata hai — Fourier coefficients extension ki sabse buri discontinuity ko "feel" karte hain.

Sine series automatically

u (0) = u (L) = 0

kyun satisfy karti hai?

Har basis function

sin \frac{nπ x}{L}

x = 0

aur

x = L

par zero hai (kyunki

sin nπ = 0

), isliye unka koi bhi sum wahan zero hoga — exactly wahi jo fixed-end (Dirichlet) problems maangti hain.

Cosine series mein

\frac{a _{0}}{2}

term kyun hoti hai jabki sine series mein nahi?

\frac{a _{0}}{2} = \frac{1}{L} \int_{0}^{L} f d x

f

ki average value hai; even extension ye average rakhta hai, lekin odd extension antisymmetry ki wajah se average

0

rakhta hai, isliye uski constant term absent hai.

Cosine basis naturally insulated (Neumann) ends ke saath kyun pair hoti hai?

\frac{d}{d x} cos \frac{nπ x}{L} = - \frac{nπ}{L} sin \frac{nπ x}{L}

x = 0, L

par zero hai, isliye har cosine ka ends par zero slope hai — "boundary se koi heat flow nahi" condition se match karta hai.

Do bilkul alag dikhne wale extensions

[0, L]

par same answer kyun de sakte hain?

Dono construction se

[0, L]

par

f

reproduce karne ke liye banaye jaate hain; extension sirf us interval ke bahar behaviour decide karta hai, jo physics kabhi nahi dekhti.

Edge cases

"Degenerate" input woh hota hai jahan general rule ko dobara dekhna padta hai — ek constant, ek zero, ek endpoint, ya ek $\to \infty$ limit. Yahan marks jaate hain.

f (x) = 0

ki half-range sine series kya hai?

Har

b_{n} = \frac{2}{L} \int_{0}^{L} 0 d x = 0

, isliye series identically

0

hai — trivial lekin valid consistency check.

f (x) = 1

[0, L]

lo: kya uski cosine series mein koi

cos

terms hain?

Nahi — ek constant ka even extension bhi constant hi rehta hai, isliye

a_{0} /2 = 1

aur saare

a_{n} = 0

for

n \geq 1

; "series" sirf constant

1

hai, pure average.

f (x) = 1

[0, L]

ki sine series ke endpoints par kya hota hai?

Odd extension

x = 0

par

- 1

+ 1

ki taraf jump karta hai aur phir

x = L

par, isliye series dono ends par

0

(midpoint) par converge karti hai, jabki

f = 1

open interval par hai.

n \to \infty

ke saath, Dirichlet conditions satisfy karne wale kisi bhi legitimate

f

ke liye

b_{n}

(ya

a_{n}

) ko kya karna chahiye?

Unhe

0

ki taraf jaana chahiye (Riemann–Lebesgue) — jo coefficients zero nahi hote unka matlab hoga series converge nahi kar sakti; decay ka rate hi wo hai jo smoothness control karti hai.

Agar

f

[- L, L]

par already odd hai lekin aap sirf

[0, L]

feed karte ho aur cosine series lete ho, toh kya milta hai?

Aapko

f

itself ka off

[0, L]

cosine series nahi milta, balki uske even reflection ka milta hai;

(0, L)

par ye phir bhi

f

ke barabar hai, lekin bahar reconstructed function original odd

f

nahi, mirror image hai.

Jahan periodic extension continuous hai, series wahan kya converge karti hai?

Exactly

f (x)

par — midpoint rule sirf jumps par kaam karta hai; jahan extension continuous hai, Dirichlet ka theorem function value par pointwise convergence deta hai.

Ek rod ke liye kaun si half-range series choose karoge jiska ek end fixed ho (

u = 0

) aur ek end insulated (

u_{x} = 0

Na pure sine na pure cosine — mixed boundary conditions ko odd-quarter/eighth-range basis chahiye jaise

sin \frac{( 2 n - 1 ) π x}{2 L}

, jo ek end par zero hoti hai aur doosre par zero slope rakhti hai. (Dekho Neumann Boundary Conditions aur Heat Equation — separation of variables.)

Connections

Half-range sine and cosine series (index 4.7.6) — wo parent note jinhe ye traps drill karte hain.
Even and Odd Functions — wo symmetry jis par upar har "kyun" ka jawab tika hai.
Dirichlet Conditions — saare endpoint edge cases ke peeche midpoint rule.
Fourier Series — full range — jahan se doubled-integral factor of $2$ aata hai.
Heat Equation — separation of variables — Dirichlet ends $\to$ sine.
Wave Equation on a finite string — fixed ends $\to$ sine.
Neumann Boundary Conditions — insulated ends $\to$ cosine.