4.7.6
Intuition The Big PictureJab hum kisi PDE (jaise heat ya wave equation) ko ek finite rod 0 ≤ x ≤ L us interval par kya ho raha hai. Lekin Fourier series periodic functions ke liye bani hain ( − L , L ) [ 0 , L ] [ − L , 0 ] khud values invent karte hain taaki woh ya toh odd ban jaaye (jisse pure sine series milti hai) ya even (jisse pure cosine series milti hai). Hum yeh choose karne ke liye free hain kyunki kisine nahi bataya ki x < 0 humara half range hai.
Definition Half-range expansionEk function f ( x ) sirf [ 0 , L ]
Half-range sine series : f odd function ki tarah extend karo ⇒ sin Half-range cosine series : f even function ki tarah extend karo ⇒ cos
"Half-range" = hum ek full period ( − L , L ) aadha jaante hain, yaani [ 0 , L ]
Do choices kyun? Kyunki PDE ki boundary conditions decide karti hain kaun si useful hai:
Dirichlet BC (u = 0 → x = 0 , L → sine .
Neumann BC (zero derivative / insulated ends) → → cosine .
Shuru karo full Fourier series se ek function ki jo period 2 L ( − L , L )
f ( x ) = 2 a 0 + n = 1 ∑ ∞ [ a n cos L nπ x + b n sin L nπ x ] with
a n = L 1 ∫ − L L f ( x ) cos L nπ x d x , b n = L 1 ∫ − L L f ( x ) sin L nπ x d x .
Intuition Odd kyun cosines ko khatam kar deta haicos even function hai; sin odd hai. Agar hum f odd force karen, toh
f ( x ) cos ( ⋅ ) odd , aur ek odd function ka integral symmetric interval
[ − L , L ] zero hota hai. Toh har a n = 0 x = 0 x = L
Ek odd f f ( x ) sin L nπ x even , isliye
b n = L 1 ∫ − L L f sin L nπ x d x = L 2 ∫ 0 L f ( x ) sin L nπ x d x . Ek even f f ( x ) sin ( ⋅ ) odd ⇒ b n = 0 f ( x ) cos ( ⋅ ) even , toh hum half-integral ko double kar dete hain:
a n = L 2 ∫ 0 L f ( x ) cos L nπ x d x , a 0 = L 2 ∫ 0 L f ( x ) d x .
Worked example Example 1 —f ( x ) = x [ 0 , π ] L = π b n = π 2 ∫ 0 π x sin ( n x ) d x
Step — integrate by parts. Kyun? Polynomial × trig ka product; IBP polynomial ki power kam karta hai.
∫ 0 π x sin ( n x ) d x = [ − n x c o s n x ] 0 π + n 1 ∫ 0 π cos n x d x = − n π c o s nπ + 0. Doosra term zero kyun ho jaata hai? ∫ 0 π cos n x d x = n s i n nπ = 0 n
Kyunki cos nπ = ( − 1 ) n b n = π 2 ⋅ ( − n π ( − 1 ) n ) = n − 2 ( − 1 ) n = n 2 ( − 1 ) n + 1 . x = 2 n = 1 ∑ ∞ n ( − 1 ) n + 1 sin ( n x ) , 0 < x < π . Pehle forecast karo, phir verify: x = π /2 = 2 ( 1 − 3 1 + 5 1 − ⋯ ) = 2 ⋅ 4 π = 2 π f ( π /2 ) = π /2
Worked example Example 2 —f ( x ) = x [ 0 , π ] a 0 = π 2 ∫ 0 π x d x = π 2 ⋅ 2 π 2 = π . 2 a 0 = 2 π x [ 0 , π ]
a n = π 2 ∫ 0 π x cos ( n x ) d x IBP (kyun? same wajah jaise pehle) :
∫ 0 π x cos n x d x = [ n x s i n n x ] 0 π − n 1 ∫ 0 π sin n x d x = 0 + n 2 1 [ cos n x ] 0 π = n 2 ( − 1 ) n − 1 . Pehla bracket zero kyun? sin nπ = 0 a n = π 2 ⋅ n 2 ( − 1 ) n − 1 n 0 n − π n 2 4 x = 2 π − π 4 k = 0 ∑ ∞ ( 2 k + 1 ) 2 cos (( 2 k + 1 ) x ) , 0 < x < π . Verify x = 0 = 2 π − π 4 ∑ ( 2 k + 1 ) 2 1 = 2 π − π 4 ⋅ 8 π 2 = 0 = f ( 0 )
Worked example Example 3 — Cosine series faster kyun converge karti haiUpar sine coefficients ∼ n 1 ∼ n 2 1 Kyun? f ( x ) = x ∣ x ∣ x = π jump hai (sawtooth). Smoother extension ⇒ Take-away: pehle BCs match karo, lekin note karo ki cosine series aksar better converge karti hai jab f
Common mistake "Main bas standard Fourier formula use kar lunga∫ − L L Kyun sahi lagta hai: yahi formula yaad kiya tha tumne. Trap: f [ 0 , L ] [ − L , 0 ] khud koi extension choose naho karte.
Fix: yeh choice encode hoti hai factor L 2 half integral ∫ 0 L
Common mistake Cosine series mein2 a 0 Kyun sahi lagta hai: sine series mein koi constant term nahi hoti, symmetry ki wajah se. Trap: even
functions mein generally nonzero average hota hai, jo a 0 /2 = L 1 ∫ 0 L f Fix: hamesha a 0
Common mistake Yeh believe karna ki seriesf endpoints par bhi .
Kyun sahi lagta hai: formula "f ( x ) = ∑ … Trap: periodic extension ke jump par,
series midpoint par converge karti hai (Dirichlet's theorem).
Jaise ki x x = π 0 π Fix: equality 0 < x < L
Recall Quick self-test (hide and answer)
Kaun si extension sirf sine terms deti hai, aur kyun?
Half-range sine series mein b n
Odd function ke liye a n = 0
Kaun sa BC type cosine series ke saath pair karta hai?
Answers: 1. Odd extension; cos f cos = 0 b n = L 2 ∫ 0 L f sin L nπ x d x 2 L f cos [ − L , L ] 0
Recall Feynman: ek 12-saal ke bacche ko explain karoSocho tumhare paas ek gaana hai jo sirf tape ke pehle aadhe hisse mein record hua hai. Ise ek aise machine par chalane ke liye jo poora loop chahti hai, tum doosra aadha khud record karte ho. Ya toh pehle aadhe ko ulta aur seedha neeche flip karke copy kar sakte ho (odd → use bilkul "wave-like" bana deta hai, sirf smooth wiggly sine sounds), ya sirf mirror kar sakte ho (even → sirf cosine sounds, aur isme ek steady background hum ho sakti hai = the average). Dono se ek poora gaana milta hai; tum woh choose karo jo tumhari machine ke rules se match kare.
Mnemonic Pairing yaad rakho"S ine = S idon par zeros S lay karo (function = 0 at ends, odd, fixed walls).
C osine = C alm slope, C onstant term, C losed/insulated ends (even)."
Aur: O dd → E ven → O S/E C": O-Sine, E-Cosine).
Half-range expansion kis interval par defined function ko represent karta hai? Sirf [ 0 , L ] 2 L
f Pure sine series (saare a n = 0
f Pure cosine series (saare b n = 0 a 0 /2
Odd function ke liye cosine terms kyun vanish ho jaate hain? f ( x ) cos L nπ x [ − L , L ] 0
Half-range sine coefficient formula b n = L 2 ∫ 0 L f ( x ) sin L nπ x d x
Half-range cosine coefficient formula a n = L 2 ∫ 0 L f ( x ) cos L nπ x d x a 0 = L 2 ∫ 0 L f d x
a 0 /2 f [ 0 , L ]
Kaun sa boundary condition sine series ke saath pair karta hai? Dirichlet (function = 0 at the ends).
Kaun sa boundary condition cosine series ke saath pair karta hai? Neumann (zero derivative / insulated ends).
Periodic extension ke jump discontinuity par, series kis cheez par converge karti hai? Midpoint (average) of left aur right limits par (Dirichlet's theorem).
x [ 0 , π ] x = 2 ∑ n = 1 ∞ n ( − 1 ) n + 1 sin n x
x [ 0 , π ] x = 2 π − π 4 ∑ k = 0 ∞ ( 2 k + 1 ) 2 c o s (( 2 k + 1 ) x )
Full Fourier series period 2L
Neumann BC insulated ends
