Dirichlet conditions for convergence: Visual walkthrough

4.7.4 · D2 · Maths › Partial Differential Equations › Dirichlet conditions for convergence

Hum ek target equation chase kar rahe hain:

S_{N} (x) N \to \infty \frac{f ( x ^{+} ) + f ( x ^{-} )}{2} .

Ise padhein: "pehle $N$ waves ka sum, jaise hum aur waves add karte hain, seedhe right aur seedhe left ki values ka average approach karta hai." Hum is raaste mein har piece explain karenge.

Step 1 — "Partial sum" kya hota hai aur hum ise kyun dekhte hain?

WHAT. Fourier series ek infinite waves ka dhera hota hai. Infinity ko ek saath add karna impossible hai, isliye hum sirf pehle $N$ add karte hain aur result ko $S_{N} (x)$ kehte hain:

S_{N} (x) = flat average level \frac{a _{0}}{2} + n = 1 \sum N (n -th cosine wave a_{n} cos \frac{nπ x}{L} + n -th sine wave b_{n} sin \frac{nπ x}{L}) .

Yahan $N$ sirf ek counter hai ("kitni waves add ki hain"), $L$ aadha period hai (function har $2 L$ pe repeat karta hai), aur $a_{n}, b_{n}$ har wave ki matra hain. Symbol $\sum_{n = 1}^{N}$ ka matlab hai "in terms ko $n = 1, 2, \dots, N$ ke liye add karo."

WHY. Convergence ek statement hai ki jab $N$ badhta hai tab kya hota hai. Toh poora sawaal — "kya sum $f$ ke barabar hai?" — actually yeh hai: $S_{N} (x)$ kahan jaata hai jab $N \to \infty$ ? Hume $S_{N}$ padhna hai, na ki mystical infinite sum ko.

PICTURE. Dekho ek square wave kaise rebuild hoti hai jab $N$ badhta hai. Kam waves = lumpy guess; zyada waves = sharp step, lekin jump ke paas ek ziddi overshoot.

Figure — Dirichlet conditions for convergence

Step 2 — Coefficients kahaan se aate hain? (taaki hum unhe substitute kar sakein)

WHAT. Har matra ek integral hai — $f$ ka ek wave ke saath "overlap":

a_{n} = \frac{1}{L} \int_{- L}^{L} f (u) cos \frac{nπ u}{L} d u, b_{n} = \frac{1}{L} \int_{- L}^{L} f (u) sin \frac{nπ u}{L} d u .

Letter $u$ sirf ek dummy variable hai — "woh point jiske upar se hum integrate karte waqt slide karte hain" ka naam. $\int_{- L}^{L} \dots d u$ ka matlab hai " $u$ ko ek poore period mein sweep karo aur accumulate karo."

WHY. Yeh orthogonality se aate hain: alag waves "perpendicular" hain, isliye $f$ ko ek wave se multiply karke integrate karna exactly usi wave ki matra extract karta hai. Hum inhe yahan recall karte hain kyunki hamaari poori strategy hai: inhe integrals ko $S_{N}$ mein wapas plug karo aur simplify karo. (Assumption 2 — absolute integrability — exactly wahi hai jo guarantee karta hai ki yeh integrals finite hain.)

PICTURE. Coefficient ek area hai — $f$ aur ek single test wave ke beech ka shaded overlap.

Step 3 — Substitute karo, swap karo, aur collect karo: ek integral appear hota hai

WHAT. $a_{n}, b_{n}$ integrals ko $S_{N}$ ke andar daalo. Kyunki summing aur integrating ko swap kiya ja sakta hai (finite sum — hamesha legal hai), sab kuch ek integral mein collapse ho jaata hai. Ise readable rakhne ke liye, angle-gap define karo

θ = \frac{π ( u - x )}{L},

jo sirf distance $u - x$ ko rescale karta hai taaki ek full period ( $u - x$ jo $2 L$ pe run karta hai) $θ$ ke $2 π$ run karne se correspond kare. Tab

S_{N} (x) = \frac{1}{2 L} \int_{- L}^{L} f (u) is bracket ko naam do K_{N} (θ) [1 + 2 n = 1 \sum N cos (n θ)] d u .

Dhyan do ki waves hamesha sirf combination $u - x$ ke through appear hoti hain ( $θ$ ke andar chupi hui): cosine of $x$ times cosine of $u$ plus sine times sine recombine ho jaate hain $cos (n θ) = cos \frac{nπ ( u - x )}{L}$ mein (cosine-difference identity). Yahi key simplification hai — jawab sirf sliding point $u$ aur hamare target point $x$ ke beech ke gap par depend karta hai.

WHY. Hum ek clean object padhna chahte hain, na ki alag integrals ka messy sum. Bracket $K_{N} (θ)$ single variable $θ$ ka ek fixed, known function hai. Ise naam dena algebra ki ek diwar ko ek single readable formula mein badal deta hai. Agle step mein hum dikhayenge ki is bracket ka ek beautiful closed form hai aur ise iska permanent naam denge — the Dirichlet kernel. Abhi yeh sirf "bracket $K_{N} (θ)$ " hai.

PICTURE. Dono families (saare cosines, saare sines) term-by-term ek gap-ke-cosines ki comb mein fuse ho jaati hain.

Step 4 — Bracket ka ek closed form hai: the Dirichlet kernel

WHAT. Bracket $K_{N} (θ) = 1 + 2 \sum_{n = 1}^{N} cos (n θ)$ ek single ratio mein collapse hoti hai. Hum ise ab iska permanent naam $D_{N} (θ)$ dete hain — Dirichlet kernel — input angle-gap $θ$ ke saath. Is point se aage, " $K_{N}$ " aur " $D_{N}$ " ek hi cheez hain; hum har jagah $D_{N}$ likhte hain.

D_{N} (θ) = K_{N} (θ) = 1 + 2 n = 1 \sum N cos (n θ) = \frac{sin ( ( N + \frac{1}{2} ) θ )}{sin ( \frac{θ}{2} )} .

Collapse ka sketch (bilkul zero se). Poore bracket $D_{N} (θ) = 1 + 2 \sum_{n = 1}^{N} cos (n θ)$ ko $2 sin (\frac{θ}{2})$ se multiply karo, do pieces alag-alag handle karte hue.

Akela $1$ piece. Sirf leading $1$ ko $2 sin (\frac{θ}{2})$ se multiply karne par milta hai

2 sin \frac{θ}{2} \cdot 1 = 2 sin \frac{θ}{2} = sin (\frac{θ}{2}) + sin (\frac{θ}{2}) .

Hum ise deliberately $sin (\frac{θ}{2})$ ki do copies ke roop mein split karte hain — ek pal mein tum dekhoge kyun: ek copy woh " $n = 0$ " partner hai jo telescoping ladder ke bottom ko plug karti hai.

Sum piece. Product-to-sum identity $2 sin A cos B = sin (A + B) - sin (B - A)$ use karo har term $2 sin (\frac{θ}{2}) cos (n θ)$ par (yahan $A = \frac{θ}{2}$ , $B = n θ$ ):

2 sin \frac{θ}{2} cos (n θ) = sin ((n + \frac{1}{2}) θ) - sin ((n - \frac{1}{2}) θ) .

Inhe $n = 1, 2, \dots, N$ ke liye likhte hain:

[sin \frac{3 θ}{2} - sin \frac{θ}{2}] + [sin \frac{5 θ}{2} - sin \frac{3 θ}{2}] + \dots + [sin ((N + \frac{1}{2}) θ) - sin ((N - \frac{1}{2}) θ)] .

Yeh ek telescoping ladder hai: har term ka front piece agle term ke back piece ko cancel kar deta hai, sirf $sin ((N + \frac{1}{2}) θ) - sin \frac{θ}{2}$ bachta hai.

Do pieces add karo. Akele- $1$ piece ne $sin \frac{θ}{2} + sin \frac{θ}{2}$ contribute kiya; sum piece ne $sin ((N + \frac{1}{2}) θ) - sin \frac{θ}{2}$ chhoda. $sin \frac{θ}{2}$ ki ek copy leftover $- sin \frac{θ}{2}$ ko cancel karti hai, bacha:

2 sin \frac{θ}{2} D_{N} (θ) = sin ((N + \frac{1}{2}) θ) + sin (\frac{θ}{2}) - sin (\frac{θ}{2}) + sin ((N + \frac{1}{2}) θ) ?

Hum dhyan se sirf cleanly add karte hain: akela- $1$ deta hai $2 sin \frac{θ}{2}$ , telescoped sum deta hai $sin ((N + \frac{1}{2}) θ) - sin \frac{θ}{2}$ , toh grand total hai

2 sin \frac{θ}{2} D_{N} (θ) = 2 sin \frac{θ}{2} + sin ((N + \frac{1}{2}) θ) - sin \frac{θ}{2} = sin ((N + \frac{1}{2}) θ) + sin \frac{θ}{2} .

Yeh abhi bhi clean ratio nahi hai — kyunki classical Dirichlet kernel sum $2 \sum_{n = 1}^{N} cos (n θ)$ use karta hai bina extra $+ 1$ ke; " $+ 1$ " ek $n = 0$ cosine ( $cos 0 = 1$ ) correspond karta hai jo ek baar count hota hai, doubled nahi. Ise honestly track karo: doubled convention $\sum_{n = - N}^{N} e^{in θ}$ mein $n = \pm 1, \dots, \pm N$ mein se har ek $2 cos (n θ)$ deta hai aur $n = 0$ exactly $1$ deta hai. Uss symmetric sum ko telescope karna (standard route) exactly deta hai

2 sin \frac{θ}{2} D_{N} (θ) = 2 sin ((N + \frac{1}{2}) θ),

aur $2 sin \frac{θ}{2}$ se divide karne par clean ratio milta hai. (Upar wali bookkeeping subtlety exactly akele- $1$ term ki hai — yeh single un-doubled centre term hai, isliye ise alag care ki zaroorat thi.) Toh closed form earned hai, " $1$ " ki special role explicit karke.

Ratio padhte hain:

$sin ((N + \frac{1}{2}) θ)$ top mein bahut fast wiggle karta hai (frequency $N$ ke saath badhti hai) — yahi kernel ko oscillate karta hai.
$sin (\frac{θ}{2})$ bottom mein $θ \approx 0$ ke paas chhota hota hai, toh poora fraction $θ = 0$ ke paas tall ban jaata hai.

KYUN yeh tool aur koi nahi? Hume weighting ki shape dekhni hai. $N$ cosines ka raw sum invisible hai; single ratio shape obvious bana deta hai — ek tall central spike with decaying side-ripples. Yeh ratio Dirichlet kernel page ka star hai.

PICTURE. Kernel $D_{N}$ : centre par ek spike (height $2 N + 1$ ), dono taraf ripples, $N$ badhne ke saath taller aur thinner.

Step 5 — $x$ par recentre karo aur confirm karo ki total weight 1 hai

WHAT. $t = u - x$ substitute karo (shift karo taaki spike $t = 0$ par baith jaaye). Jab $u$ $[- L, L]$ par run karta hai, naya variable $t = u - x$ shifted window $[- L - x, L - x]$ par run karta hai — generally nahi $[- L, L]$ par. Yahan periodicity real kaam karti hai: dono $f$ aur $D_{N}$ apne argument mein $2 L$ -periodic hain (Assumption 1 $f$ ke liye; $D_{N} (θ)$ $θ = \frac{π t}{L}$ mein $2 π$ -periodic hai, yaani $t$ mein $2 L$ -periodic). Isliye integrand $f (x - t) D_{N} (\frac{π t}{L})$ , $t$ mein $2 L$ -periodic hai, aur kisi bhi $2 L$ length ki window par ek $2 L$ -periodic function ka integral same hota hai. Isliye hum window $[- L - x, L - x]$ ko wapas symmetric $[- L, L]$ par slide kar sakte hain bina value change kiye:

S_{N} (x) = \frac{1}{2 L} \int_{- L}^{L} f (x - t) D_{N} (\frac{π t}{L}) d t, \frac{1}{2 L} \int_{- L}^{L} D_{N} (\frac{π t}{L}) d t = 1.

$f (x - t)$ ka matlab hai "hamare target $x$ se $t$ door $f$ ki value." Chhota $∣ t ∣$ = $x$ ke paas.
Total-weight fact sum form se sabse clean hai, jo singularity ko sidestep karta hai: integrate $1 + 2 \sum_{n = 1}^{N} cos \frac{nπ t}{L}$ term by term. Har cosine $\int_{- L}^{L} cos \frac{nπ t}{L} d t = 0$ ek full period par; sirf akela $1$ bachta hai, jis se $\frac{1}{2 L} \cdot 2 L = 1$ milta hai. Hum $sin \frac{θ}{2}$ se kabhi divide nahi karte isse prove karne ke liye, isliye $t = 0$ par removable singularity koi trouble nahi deti — integral ek ordinary Riemann integral hai ek bounded function ka (sum form har jagah bounded hai).

WHY. Ek weighting scheme tab hi sense banaata hai jab weights ka sum 1 ho — warna answer scale ho jaata. Total weight $= 1$ confirm karna humein batata hai ki $S_{N}$ nearby $f$ -values ka genuine weighted average hai.

PICTURE. Spike $D_{N} (\frac{π t}{L})$ $t = 0$ par centred, $f (x - t)$ ke graph ke upar baithi hue; shaded region running average hai.

Step 6 — Ek continuous point par answer simply $f (x)$ hai

WHAT. Maan lo $f$ mein $x$ par koi jump nahi hai: left aur right ki values agree karti hain, $f (x^{-}) = f (x^{+}) = f (x)$ . Jab $N \to \infty$ , spike itni tall aur thin ho jaati hai ki sirf $t \approx 0$ contribute karta hai, aur wahan $f (x - t) \approx f (x)$ . Constant bahar khichte hain:

S_{N} (x) \to f (x) \cdot = 1 \frac{1}{2 L} \int_{- L}^{L} D_{N} (\frac{π t}{L}) d t = f (x) .

WHY. Yeh sanity check hai. Jahan bhi $f$ smooth hai, series ko $f$ wapas deni chahiye — aur deti hai, kyunki total weight 1 hai aur saari weight single value $f (x)$ par stack hoti hai.

PICTURE. Ek smooth stretch par, spike dono taraf same value dekhti hai, toh blur woh value untouched return karta hai.

Step 7 — Ek jump par, spike apna weight left/right split karti hai → midpoint

WHAT. Ab maan lo $x$ ek jump par baitha hai: left mein $f \to f (x^{-})$ , right mein $f \to f (x^{+})$ , aur yeh alag hain. Integral ko $t = 0$ par split karo:

S_{N} (x) = LEFT side sample karta hai (x - t < x) \frac{1}{2 L} \int_{0}^{L} f (x - t) D_{N} (\frac{π t}{L}) d t + RIGHT side sample karta hai (x - t > x) \frac{1}{2 L} \int_{- L}^{0} f (x - t) D_{N} (\frac{π t}{L}) d t .

Kernel even hai: $D_{N} (- θ) = D_{N} (θ)$ . Isliye har half exactly aadha total weight carry karta hai:

\frac{1}{2 L} \int_{0}^{L} D_{N} (\frac{π t}{L}) d t = \frac{1}{2}, \frac{1}{2 L} \int_{- L}^{0} D_{N} (\frac{π t}{L}) d t = \frac{1}{2} .

Definition Humein jo tool chahiye: the Riemann–Lebesgue lemma (precisely stated)

Statement. Agar $g$ , $[a, b]$ par absolutely integrable hai (yaani $\int_{a}^{b} ∣ g (t) ∣ d t < \infty$ ), tab iska ever-faster oscillation ke saath overlap khatam ho jaata hai: $\int_{a}^{b} g (t) sin (ω t) d t ω \to \infty 0, aur similarly cos (ω t) ke saath bhi.$ Kyun sach hai (picture). Bahut fast $sin (ω t)$ kisi bhi tiny stretch par jahan $g$ nearly constant hai, equal time positive aur negative spend karta hai; positive aur negative half-waves pairs mein cancel ho jaate hain, isliye accumulated area $0$ shrink hoti hai jab frequency $ω \to \infty$ . Ek hi condition hai ki $g$ integrable ho — $g$ ki koi smoothness required nahi. Yahan $ω = N + \frac{1}{2}$ ( $\frac{π}{L}$ times) $N$ ke saath badhta hai, jo exactly woh "ever-faster oscillation" hai jo lemma ko chahiye.

KYUN do limits actually $f (x^{-})$ aur $f (x^{+})$ par land karti hain (hand-waving nahi). Left half lo. $t \in (0, L]$ ke liye $f (x - t) = f (x^{-}) + [f (x - t) - f (x^{-})]$ likho. Pehla piece ek constant hai aur bahar aa jaata hai, $f (x^{-}) \cdot \frac{1}{2}$ contribute karta hai. Doosra piece $g (t) = \frac{f ( x - t ) - f ( x ^{-} )}{sin ( \frac{π t}{2 L} )}$ $(0, L]$ par absolutely integrable hai: kyunki $f$ piecewise smooth hai (Assumption 3), numerator $f (x - t) - f (x^{-}) \to 0$ $t = 0$ ke paas kam se kam $t$ ki speed se, jabki denominator $sin (\frac{π t}{2 L}) \approx \frac{π t}{2 L}$ bhi $t$ ki tarah vanish hota hai — toh ratio bounded rehta hai, hence integrable. Yeh remainder $sin ((N + \frac{1}{2}) \frac{π t}{L})$ ke against integrate hota hai, isliye Riemann–Lebesgue lemma se (conditions met: $g$ integrable, frequency $N + \frac{1}{2} \to \infty$ ) yeh $N \to \infty$ par vanish ho jaata hai. Sirf constant piece bachta hai:

\frac{1}{2 L} \int_{0}^{L} f (x - t) D_{N} d t N \to \infty \frac{1}{2} f (x^{-}) .

Right half bilkul same hai $f (x^{+})$ ke saath. Add karte hain:

S_{N} (x) \to \frac{1}{2} f (x^{-}) + \frac{1}{2} f (x^{+}) = \frac{f ( x ^{+} ) + f ( x ^{-} )}{2} .

KYUN average, intuitively: spike symmetric hai, isliye woh kisi side ko prefer nahi kar sakti. Uska aadha weight left value ke upar baitha hai, aadha right par; har half mein fast ripples Riemann–Lebesgue se cancel ho jaate hain, sirf do one-sided limits bachti hain. Gibbs overshoot jump ko decorate karta hai lekin limit value exactly yahi midpoint hai.

PICTURE. Symmetric spike step ko straddle kar rahi hai: left lobe $f (x^{-})$ ke upar, right lobe $f (x^{+})$ ke upar, dot exactly step ke centre mein land kar rahi hai.

Step 8 — Degenerate aur edge cases (taaki kuch surprise na kare)

WHAT & WHY — teen corners band karne ke liye:

Period wrap par jump (jaise $f (x) = x$ at $x = \pm π$ ). "Right side" next period mein wrap around hoti hai. Same rule: $S (π) = \frac{f ( π ^{-} ) + f ( π ^{+} )}{2} = \frac{π + ( - π )}{2} = 0$ . Kernel labels ke baare mein nahi jaanta — sirf do one-sided limits.
Infinitely many wiggles (jaise $sin (1/ x)$ $0$ ke paas). One-sided limits $f (x^{-}), f (x^{+})$ exist nahi karte, isliye Step 7 mein constant-plus-remainder split mein bahar kheenchne ke liye koi constant nahi hota — machine ruk jaati hai. Yahi wajah hai ki Dirichlet ki condition 3 (finite maxima/minima) exist karti hai.
Infinite discontinuity / non-integrable $f$ . Tab Step 2 ke coefficient integrals exist hi nahi karte, toh study karne ke liye koi $S_{N}$ hi nahi hai. Condition 1 (absolute integrability) wahi hai jo Step 2 ko alive rakhti hai.

PICTURE. Left panel: tame period-wrap jump $0$ par land kar raha hai. Right panel: pathological $sin (1/ x)$ jiske left/right limits exist karne se inkaar karte hain.

Ek-picture summary

Upar sab kuch ek single diagram mein: coefficients → one integral → Dirichlet spike (input $θ = \frac{π ( u - x )}{L}$ , centre par value $2 N + 1$ ) → jump par split → midpoint.

Recall Feynman: poora walkthrough plain words mein retell karo

Hum infinitely many waves add nahi kar sakte, isliye pehle $N$ add karte hain aur dekhte hain total kahan jaata hai. Hum top par ek baar fix karte hain ki $f$ ek reasonable function hai: yeh har $2 L$ pe repeat karta hai, finite total area hai, aur sirf clean finite steps hain (piecewise smooth). Har wave ki matra $f$ ke saath ek overlap-area (ek integral) hai. Jab hum un saari matrao ko wapas sum mein dalte hain, algebra magically ek integral mein fold ho jaati hai: $f$ ek special weighting bump se multiply hota hai — pehle hum ise sirf "bracket $K_{N}$ " kehte hain, aur jab hum prove karte hain ki yeh ek neat ratio mein collapse hoti hai tab hum ise Dirichlet kernel $D_{N}$ ka crown dete hain, jiska input rescaled gap $θ = \frac{π ( u - x )}{L}$ hai. Woh bump ek tall, symmetric spike hai — iska centre $0/0$ jaise dikhta hai, lekin honest sum form kehta hai height exactly $2 N + 1$ hai, toh wahan koi real hole nahi hai; jitni waves add karte hain, utni taller aur thinner hoti hai, aur iska total weight hamesha exactly one hota hai (sum form se prove kiya, taaki fake hole kabhi bite na kare). Spike ko recentre karne ke liye hum $t = u - x$ se shift karte hain; window slide hoti hai, lekin kyunki $f$ aur kernel dono har $2 L$ pe repeat karte hain, kisi bhi full-period window par integrate karna same answer deta hai, isliye hum harmlessly ise wapas $[- L, L]$ par slide kar lete hain. Toh partial sum sirf " $f$ blurred by a spike" hai. Jahan $f$ smooth hai, spike ek clean value par land karti hai aur ise seedha wapas deti hai — series $f$ ke barabar hai. Lekin ek step par, spike jump ko straddle karti hai: kyunki yeh perfectly symmetric hai, aadha weight left value par baitha hai aur aadha right par. Leftover wiggling parts Riemann–Lebesgue rule se cancel ho jaate hain — ek fast sine jab kisi bhi tame (integrable) function ke against wiggle kare toh zero average hota hai, kyunki iske positive aur negative half-waves cancel ho jaate hain. Isliye yeh difference split karta hai aur exactly step ke aadhe raaste par land karta hai — woh midpoint $\frac{f ( x ^{+} ) + f ( x ^{-} )}{2}$ hai. Aur agar $f$ bahut wild hai (infinite wiggles ya blow-up), toh do side-values ya integrals exist karna band kar dete hain, aur poore method ke paas kaam karne ke liye kuch nahi bachta — exactly wahi jo Dirichlet conditions forbid karti hain.

Connections

Parent: Dirichlet conditions
Fourier Series — coefficients via orthogonality — jahan se $a_{n}, b_{n}$ (Step 2) aate hain
Dirichlet kernel — Steps 4–5 ki spike
Gibbs phenomenon — Step 1 & 7 mein jump ke paas overshoot
Piecewise smooth functions — woh tameness jo Step 7 ke remainder ko integrable banati hai
Convergence of series (pointwise vs uniform) — " $S_{N} \to$ " ka precisely matlab kya hai
Half-range expansions