4.7.4 · Maths › Partial Differential Equations
Ek Fourier series ek infinite sum of waves hoti hai jo function f ( x ) ko rebuild karne ki koshish karti hai. Lekin har function rebuild nahi ho sakta . Dirichlet conditions ek simple checklist hai jo guarantee karti hai ki series actually converge hogi — aur yeh hume exactly batati hai ki series kis value pe settle hogi , chahe ugly jump points pe hi kyun na ho.
Deep idea: ek function jo "zyada wild nahi" (finitely many wiggles aur jumps) hota hai, use hamesha sines aur cosines se reconstruct kiya ja sakta hai .
Ek Fourier series hoti hai
f ( x ) ∼ 2 a 0 + n = 1 ∑ ∞ ( a n cos L nπ x + b n sin L nπ x ) .
Hum pehle coefficients likhte hain (sines/cosines ki orthogonality use karke), lekin unhe likh dena yeh prove nahi karta ki sum f ( x ) ke barabar hai. "∼ " sign ek warning hai: coefficients exist karte hain, lekin kya sum converge hoga, aur kis value pe?
Intuition Convergence fail kyun ho sakti hai
Agar f kisi point ke paas infinitely zyada oscillate kare (jaise sin ( 1/ x ) ), toh smooth waves ka koi bhi finite collection uske saath nahi chal sakta. Agar f ∞ tak blow up ho jaye, toh a n , b n define karne wale integrals exist bhi nahi kar sakte. Dirichlet ki conditions exactly woh rules hain jo in pathologies ko ban karti hain.
Definition Dirichlet conditions
Ek periodic function f ( x ) jiska period 2 L hai, uski ek convergent Fourier series hogi agar ek period mein:
f single-valued aur absolutely integrable hai, yaani ∫ − L L ∣ f ( x ) ∣ d x < ∞ .
f mein sirf finite number of finite discontinuities (jumps) hain, koi infinite discontinuity nahi.
f mein sirf finite number of maxima aur minima hain (finitely many wiggles).
Yeh sufficient conditions hain (strictly necessary nahi) — agar yeh hold karti hain, toh convergence guaranteed hai.
N -va partial sum hai
S N ( x ) = 2 L 1 ∫ − L L f ( x − t ) D N ( t ) d t , D N ( t ) = sin ( 2 L π t ) sin ( ( N + 2 1 ) L π t ) .
Yeh form kyun? a n , b n ke integral formulas ko partial sum mein substitute karo, sum aur integral swap karo, aur cosines ka finite geometric sum collapse hokar D N ban jaata hai. Kernel ka total weight 1 hota hai:
2 L 1 ∫ − L L D N ( t ) d t = 1.
Yeh step kyun important hai: jab N → ∞ , toh D N ek spike ban jaata hai jo t = 0 pe concentrated hota hai, lekin left (t > 0 ) aur right (t < 0 ) dono sides pe equal mass hoti hai. Toh yeh f ko dono sides se equally sample karta hai:
S N ( x ) → 2 1 f ( x − ) + 2 1 f ( x + ) .
Yahi midpoint formula hai — derive kiya gaya, assume nahi kiya.
Worked example Example 1 — Square wave (ek jump function)
Maano f ( x ) = { − 1 + 1 − π < x < 0 0 < x < π , period 2 π .
Dirichlet check: integrable ✓ (bounded), x = 0 pe ek jump (finite) ✓, koi extra wiggles nahi ✓. Converges.
Step — x = 0 pe value? Kyun? Yeh ek jump hai.
S ( 0 ) = 2 f ( 0 + ) + f ( 0 − ) = 2 1 + ( − 1 ) = 0.
Series 0 deti hai, haalaanki f wahan kabhi 0 nahi hoti. Yeh correct kyun hai: symmetric ingredients jump ko split kar dete hain.
Step — x = π /2 pe value? Wahan continuous hai, toh S = f ( π /2 ) = 1 .
Worked example Example 2 —
f ( x ) = x on ( − π , π ) , period 2 π
Check: integrable ✓, jump sirf endpoints x = ± π pe jahan yeh π se − π pe wrap karta hai ✓, monotonic hai toh finite extrema ✓. Converges.
Step — x = π pe value? Kyun? Period wrap ek jump create karta hai: f ( π − ) = π , f ( π + ) wrap hokar − π ban jaata hai.
S ( π ) = 2 π + ( − π ) = 0.
Yeh sahi kyun hai: sawtooth ka Fourier sum odd hai aur discontinuity pe 0 se pass hona chahiye.
Worked example Example 3 — Ek function jo FAIL karta hai
f ( x ) = sin ( 1/ x ) near x = 0 ke paas infinitely many maxima/minima hain jab x → 0 . Condition 3 fail ho jaati hai, isliye Dirichlet's theorem apply nahi hota . Hum yahaan convergence guarantee nahi kar sakte.
Yeh kya sikhata hai: conditions "tame" functions ke around ek fence hain.
Common mistake "Agar coefficients exist karte hain toh series har jagah
f ke barabar hogi."
Kyun sahi lagta hai: tumne a n , b n compute kar liye, toh surely sum f hi hai . Fix: coefficients ki existence ke liye sirf integrability chahiye. Convergence (aur kis value pe ) ke liye poora Dirichlet set chahiye, aur jumps pe sum midpoint ke barabar hota hai, f ke nahi.
Common mistake "Dirichlet conditions necessary hain."
Kyun sahi lagta hai: inhe the convergence rule ki tarah bataya jaata hai. Fix: yeh sufficient, not necessary hain. Kuch functions jo inhe violate karte hain phir bhi converge karte hain; Dirichlet sirf ek clean guarantee deta hai.
Common mistake "Ek finite jump convergence khatam kar deta hai."
Kyun sahi lagta hai: jumps discontinuous aur "tute hue" lagte hain. Fix: finitely many finite jumps theek hain — series har ek pe midpoint pe converge karti hai. Sirf infinite discontinuities ya infinitely many wiggles problem create karte hain.
Common mistake "Jump pe series left (ya right) value pick karti hai."
Kyun sahi lagta hai: tumhe lagta hai yeh koi ek side choose karega. Fix: sines/cosines symmetric hote hain; sum ka koi preference nahi, toh yeh dono sides ke average pe land karta hai (Gibbs overshoot jump ko decorate karta hai lekin limit value midpoint hi hoti hai).
Recall Quick self-test (hide karke answer karo)
Teeno Dirichlet conditions batao.
Jump pe series kya value leti hai?
Kya conditions necessary hain ya sufficient?
Ek function do jo condition 3 fail karta ho.
"Ek period mein absolutely integrable" ka matlab kya hai? ∫ − L L ∣ f ( x ) ∣ d x < ∞ — ∣ f ∣ ke neeche ka area finite hai.
Dirichlet conditions kitni discontinuities allow karti hain? Finite number, aur har ek finite (jump) discontinuity honi chahiye — koi infinite discontinuities nahi.
Kitne maxima/minima allowed hain? Ek period mein sirf finitely many (koi infinite oscillation nahi).
Jump x 0 pe Fourier series kis value pe converge karti hai? Midpoint 2 f ( x 0 + ) + f ( x 0 − ) pe.
Continuity ke point pe series kis value pe converge karti hai? f ( x ) pe hi, kyunki f ( x + ) = f ( x − ) = f ( x ) .
Kya Dirichlet conditions convergence ke liye necessary hain ya sufficient? Sufficient (necessary nahi).
Partial-sum convolution ko kaunsa kernel govern karta hai? Dirichlet kernel D N ( t ) = s i n ( π t /2 L ) s i n (( N + 2 1 ) π t / L ) .
Jump pe series dono sides ko average kyun karti hai? Dirichlet kernel symmetric hota hai aur N → ∞ hone par left aur right limits ko equally sample karta hai.
"Finite maxima/minima" rule fail karne wala ek example do. x = 0 ke paas sin ( 1/ x ) .
f ( x ) = x on ( − π , π ) ke liye, x = π pe series ki value kya hai?0 , jo π aur − π ka midpoint hai.
Recall Feynman: 12-saal ke bachche ko explain karo
Socho tum ek tasveer bana rahe ho sirf smooth, wavy ribbons (sines aur cosines) use karke. Agar tasveer reasonable hai — yeh sky tak shoot nahi karti, usme hazaron tiny zig-zags nahi hain, aur sirf kuch clean "steps" hain — toh tum kar sakte ho use enough ribbons se perfectly rebuild karna. Lekin ek step pe (jahaan tasveer achanak upar jump karti hai), ribbons koi side nahi chun sakti, toh woh ek dot exactly aadhe step ke beech draw karti hain. "Dirichlet conditions" bas yeh rules hain ki "kya yeh tasveer itni reasonable hai?"
Mnemonic 3 conditions yaad karo:
"I-J-W"
I ntegrable (absolutely), finite J umps, finite W iggles. Aur jump pe: "meet in the middle."
Fourier Series — coefficients via orthogonality
Dirichlet kernel
Gibbs phenomenon (jumps ke paas overshoot, lekin limit value phir bhi midpoint hai)
Piecewise smooth functions
Convergence of series (pointwise vs uniform)
Half-range expansions
are sufficient, guarantee
spike with equal mass both sides
Single-valued and absolutely integrable
Finite number of finite jumps
Bans pathologies like sin 1 over x
Average of left and right limits