Dirichlet conditions for convergence: Worked examples

4.7.4 · D3 · Maths › Partial Differential Equations › Dirichlet conditions for convergence

Teen conditions, ek baar clearly stated (taaki hum unhe naam de sakein)

Is poore page mein hum "condition 1/2/3" kahenge. Yahan woh hain, kisi bhi example se pehle clearly likhe gaye. Yeh periodic $f$ ke ek period par apply hone wale Dirichlet conditions hain:

Definition Dirichlet checklist

Condition 1 — absolute integrability. $∣ f ∣$ ke neeche ka total area finite hai: $\int_{- L}^{L} ∣ f (x) ∣ d x < \infty$ . (Infinite blow-ups ko ban karta hai; coefficient integrals ko exist karaata hai.)
Condition 2 — finitely many finite jumps. Har discontinuity ek jump hai jahan dono one-sided limits $f (x^{-}), f (x^{+})$ exist karti hain aur finite hain; aur aise sirf finitely many points hain. (Infinite discontinuities aur aisi points ko ban karta hai jahan koi one-sided limit na ho.)
Condition 3 — finitely many extrema. $f$ ke ek period mein sirf finitely many maxima aur minima hain. (Infinite oscillation ko ban karta hai.)

Agar teeno hold karti hain, toh series har $x$ par $S (x) = \frac{f ( x ^{+} ) + f ( x ^{-} )}{2}$ par converge karti hai. Yeh sufficient hain, necessary nahi. Mnemonic: I-J-W = Integrable, finite Jumps, finite Wiggles.

Scenario matrix

Kuch bhi work karne se pehle, har woh alag situation list karo jo Fourier convergence problem mein aa sakti hai. Neeche har worked example us ek cell ke saath tagged hai jise woh cover karta hai.

#	Case class	Kya alag hai	Example
A	Continuous interior point	$f (x^{+}) = f (x^{-})$ , series $= f (x)$	Ex 1
B	Finite jump inside the interval	$f (x^{+}) \neq = f (x^{-})$ , dono finite	Ex 2
C	Period-wrap jump at the endpoint	jump periodicity ki wajah se, $f$ ke formula ki wajah se nahi	Ex 3
D	Symmetric jump landing on zero	left aur right cancel hokar $0$ deta hai	Ex 3
E	Asymmetric jump (non-zero midpoint)	midpoint koi non-zero number hai	Ex 4
F	Removable / "hole" point (ek jagah ajeeb define kiya)	$f$ ek single point par redefine kiya gaya par limits agree karti hain	Ex 5
G	FAILS: infinite wiggles	condition 3 broken	Ex 6
H	FAILS: infinite blow-up (both sides)	condition 1 broken (not absolutely integrable)	Ex 7
K	One-sided infinite limit	ek side finite, doosri blow up karti hai	Ex 8
L	*Endpoint continuity* (wrap agrees)**	$f (π^{-}) = f (- π^{+})$ , isliye $S (π) = f (π)$	Ex 9
I	Word problem / real signal	ek physical square pulse (Gibbs overshoot ke saath)	Ex 10
J	Exam twist: series value se number series sum karna	ek clever $x$ plug karo	Ex 11

Ab hum har cell ko hit karte hain.

Warm-up: jump picture kaise padhein

Figure — Dirichlet conditions for convergence

Figure 1 — ek kaali function jisme $x_{0}$ par jump hai. Left piece open circle $f (x_{0}^{-})$ par khatam hoti hai; right piece open circle $f (x_{0}^{+})$ se shuru hoti hai; solid red dot bilkul beech mein baitha hai. Woh red dot $S (x_{0})$ hai, woh value jahan Fourier series converge karti hai. Is page ka har example bas yahi hai — "do circles dhundo, phir red dot dhundo."

Worked examples

Cell A — continuous interior point

Worked example Example 1 —

$f (x) = x$ on $(- π, π)$ , evaluate at $x = \frac{π}{2}$ Forecast: padhne se pehle series value guess karo. Kya $x = π /2$ ek jump hai ya smooth spot?

Check karo ki point interior aur continuous hai. Yeh step kyun? Midpoint rule sirf jumps par $f$ se alag hota hai; baaki jagah yeh simply $f$ ke barabar hota hai. $- π$ aur $π$ ke beech formula $f (x) = x$ ek seedhi line hai — $π /2$ par koi break nahi.
Continuous hone ki wajah se, $f (x^{+}) = f (x^{-}) = f (π /2)$ . Kyun? $π /2$ ki taraf dono taraf se chalte hue, line same height par pahunchti hai.
Formula apply karo: $S (\frac{π}{2}) = \frac{f ( \frac{π}{2} ^{+} ) + f ( \frac{π}{2} ^{-} )}{2} = \frac{\frac{π}{2} + \frac{π}{2}}{2} = \frac{π}{2} .$

Verify: continuity ke point par kisi number ka apne aap se average woh number hi hota hai, isliye $S = f (π /2) = π /2 \approx 1.5708$ . Sanity check: $π /2$ strictly $(- π, π)$ ke andar hai, isliye koi wrap nahi. ✓

Cell B — finite jump inside the interval

Figure 2 — square wave. $f (0^{-}) = - 1$ aur $f (0^{+}) = + 1$ par open circles; height $0$ par red dot $S (0)$ hai.

Cells C & D — period-wrap jump landing on zero

Worked example Example 3 —

$f (x) = x$ on $(- π, π)$ , evaluate at $x = π$ Forecast: formula $f (x) = x$ bilkul continuous lagta hai. Jump kahan se aayega?

Yaad rakho ki function period $2 π$ ke saath periodic hai. Yeh step kyun? Fourier series sirf ek period jaanti hai aur phir use repeat karti hai. Isliye $x = π$ se thoda sa right mein graph neechey se restart hota hai, $- π$ par.
$x = π$ par dono limits compute karo: $f (π^{-}) = π (left piece π tak pahunchti hai), f (π^{+}) = - π (wrap - π par restart hota hai) .$ $- π$ kyun? Periodicity: $f (π^{+}) = f ((- π)^{+}) = - π$ .
Average karo: $S (π) = \frac{π + ( - π )}{2} = 0.$

Verify: sawtooth ki series $2 \sum_{n \geq 1} \frac{( - 1 ) ^{n + 1}}{n} sin (n x)$ hai; $x = π$ par har $sin (nπ) = 0$ , isliye $S (π) = 0$ . ✓ Yeh cell C hai (wrap jump) aur, kyunki do heights $\pm π$ hain, cell D bhi hai (symmetric $\Rightarrow$ midpoint zero).

Figure 3 — sawtooth $f (x) = x$ teen periods par repeat hua. $x = π$ par left circle $π$ par baitha hai, right circle (wrap) $- π$ par, aur red dot $0$ par $S (π)$ hai.

Cell E — asymmetric jump, non-zero midpoint

Worked example Example 4 — ek aisa step jo zero par average

nahi karta $f (x) = {04 - π < x < 0 0 < x < π$ , period $2 π$ . $S (0)$ aur $S (π)$ find karo.

Forecast: kaun sa jump non-zero answer deta hai, aur kaun sa zero deta hai?

$x = 0$ par: do pieces milti hain $f (0^{-}) = 0$ (left flat piece) aur $f (0^{+}) = 4$ (right flat piece). Yeh step kyun? Hum meeting point par har side ki height padhte hain. $S (0) = \frac{4 + 0}{2} = 2.$ Non-zero kyun? Dono sides zero ke baare mein symmetric nahi hain, isliye unka average ek genuinely non-zero value hai — cell E.
$x = π$ par (period wrap): just left mein top piece hai $f (π^{-}) = 4$ ; just right mein next period ki shuruat par wrap hota hai, jo bottom piece hai $f (π^{+}) = f ((- π)^{+}) = 0$ . $S (π) = \frac{4 + 0}{2} = 2.$ Yahan bhi $2$ kyun? Wrap par same do heights ( $4$ aur $0$ ) milti hain — same average.

Verify: Trick — $f = 2 + 2 g$ likho jahan $g$ Ex 2 ka $\pm 1$ square wave hai. Constant $2$ continuity ke har point aur har jump se pass-through karta hai (constants mein koi jump nahi hota). $x = 0$ par: $g (0)$ series $= 0$ , isliye $S (0) = 2 + 2 \cdot 0 = 2$ . ✓ $x = π$ par: $g$ ki series wahan $= 0$ hai, isliye $S (π) = 2$ . ✓

Cell F — removable "hole" (ek point par redefine kiya)

Pehle, woh tool jo hum use karte hain. Ek convergent Fourier series orthogonality se banaye gaye coefficients se reconstruct hoti hai:

Worked example Example 5 — ek single point ki value series nahi dekh sakti

$f (x) = x$ lo $(- π, π)$ par lekin artificially set karo $f (0) = 100$ . $S (0)$ find karo.

Forecast: kya series "dekhti" hai ki humne value spike karke $100$ kar di?

Upar ke $a_{n}, b_{n}$ $f$ ke integrals hain. Yeh step kyun? $f$ ko single point par change karne se integral zero width ki area se badalta hai, yaani kuch nahi. Isliye $a_{n}, b_{n}$ plain $f (x) = x$ jaisi hi hain — spike ka kisi bhi coefficient mein koi trace nahi raha.
One-sided limits compute karo, spike ko ignore karke. Kyun ignore karein? Limits dekhti hain ki $0$ approach karte waqt kya hota hai, at $0$ ki value nahi: $f (0^{+}) = 0, f (0^{-}) = 0.$
Average karo: $S (0) = \frac{0 + 0}{2} = 0.$

Verify: $f (x) = x$ $0$ par continuous hai dono sides se limits $0$ ke saath; artificial $100$ ek convergent Fourier series ke liye invisible hai (yeh limits par converge karti hai, assigned point value par nahi). Isliye $S (0) = 0 \neq = 100$ . ✓ Yahi wajah hai ki piecewise-smooth analysis hamesha limits ke saath kaam karti hai, isolated redefinitions ke saath kabhi nahi.

Cell G — FAILS condition 3 (infinite wiggles)

Worked example Example 6 —

$f (x) = sin (1/ x)$ for $0 < ∣ x ∣ < π$ , extended with period $2 π$ Pehle function precisely define karo. Kyun? Tum sirf ek fully specified periodic function ki Fourier series le sakte ho. $f (x) = sin (1/ x)$ set karo $0 < ∣ x ∣ < π$ par, $f (0) = 0$ assign karo (koi bhi finite value; matter nahi karega), aur period $2 π$ ke saath repeat karo. Forecast: kya Dirichlet ka theorem yahan convergence ki guarantee dega?

$x \to 0$ ke paas maxima/minima count karo. Yeh step kyun? Condition 3 per period wiggles ki sankhya cap karta hai. $sin (1/ x)$ peak hit karta hai jab $1/ x = \frac{π}{2} + 2 π k$ , yaani $x = \frac{1}{π /2 + 2 π k}$ par — infinitely many peaks $0$ ke paas bunch ho jaate hain. Condition 3 violate hoti hai.
$x = 0$ par limit bhi check karo. Yeh doosra check kyun? Kyunki $sin (1/ x)$ $x \to 0$ par hamesha $- 1$ aur $+ 1$ ke beech swing karta rehta hai, one-sided limits $f (0^{+})$ aur $f (0^{-})$ exist nahi karte — isliye condition 2 (ek finite jump ke liye dono one-sided limits ka exist karna zaroori hai) bhi fail ho jaati hai.
Do conditions fail hoti hain, isliye Dirichlet ka theorem kuch nahi kehta. Yeh kyun matter karta hai: theorem ek sufficient checklist hai; unticked boxes guarantee wapas le lete hain — woh khud se divergence prove nahi karte, bas hum bina kisi promise ke reh jaate hain.

Verify: peaks $x_{k} = 1/ (π /2 + 2 π k)$ par hain. $k = 0, 1, 2$ ke liye yeh hain $x_{0} = \frac{2}{π} \approx 0.6366$ , $x_{1} \approx 0.1273$ , $x_{2} \approx 0.0688$ — ek infinite shrinking sequence, isliye infinitely many extrema; aur un peaks par values $+ 1$ par pinned rehti hain jabki troughs $- 1$ par hain, isliye $0$ par koi one-sided limit exist nahi karti. ✓

Figure 4 — $sin (1/ x)$ $0$ ke paas. Pehle teen peaks (red dots) $x = 0$ ki taraf crowd karte hain; infinitely many hain, isliye graph kabhi settle nahi karta — $0$ par koi left/right limit nahi.

Cell H — FAILS condition 1 (infinite blow-up on both sides)

Worked example Example 7 —

$f (x) = \frac{1}{x ^{2}}$ on $(- π, π)$ Forecast: graph $0$ ke dono taraf $+ \infty$ ki taraf shoot karta hai. Kya coefficient integral exist bhi karta hai?

Absolute integrability test karo $\int_{- π}^{π} ∣ f ∣ d x$ . Yeh step kyun? Condition 1 ke liye $∣ f ∣$ ke neeche finite total area chahiye; uske bina $a_{n}, b_{n}$ recalled formula se defined hi nahi hote. $\int_{0}^{π} \frac{1}{x ^{2}} d x = [- \frac{1}{x}]_{0}^{π} = - \frac{1}{π} - (- \infty) = + \infty.$
Area infinite hai, isliye condition 1 fail hoti hai. Yeh step kyun? Midpoint machinery sirf tab kick in karti hai jab teeno conditions meet ho jaayein; yahan pehli hi (finite area) broken hai, isliye hum rukke aur "no guarantee" report karein na ki average karne ka dhong karein. Yeh infinite discontinuity — dono one-sided limits $= + \infty$ — bilkul wahi pathology hai jise ban karne ke liye condition 1 exist karti hai.

Verify: $\int_{ε}^{π} \frac{1}{x ^{2}} d x = \frac{1}{ε} - \frac{1}{π} \to + \infty$ as $ε \to 0^{+}$ . Not absolutely integrable. ✓ Compare karo ek finite jump se (Ex 2), jo bilkul theek hai.

Cell K — one-sided infinite limit

Worked example Example 8 — ek side finite, doosri blow up karti hai

$f (x) = ⎩ ⎨ ⎧ \frac{1}{x} 0 0 < x < π - π < x < 0$ , period $2 π$ . $x = 0$ par Dirichlet kya keh sakta hai?

Forecast: left side flat $0$ hai, right side $+ \infty$ ki taraf rocket karti hai. Kya hum phir bhi average kar sakte hain?

Har ek one-sided limit padho. Yeh step kyun? Midpoint rule ke liye dono limits finite numbers hone chahiye tabhi hum average kar sakte hain. $f (0^{-}) = 0 (finite), f (0^{+}) = lim_{x \to 0^{+}} \frac{1}{x} = + \infty (finite number ke roop mein exist nahi karta) .$
Ek side infinite hai, isliye yeh ek one-sided infinite discontinuity hai. Yeh kyun matter karta hai: condition 1 already fail ho jaati hai — $\int_{0}^{π} \frac{1}{x} d x = + \infty$ — isliye coefficients defined nahi hain, aur midpoint formula ke paas right par average karne ke liye kuch finite bhi nahi hai. Theorem koi guarantee nahi deta.
Cell B se compare karo. Yahan yeh kyun include kiya: ek finite jump (dono sides finite, Ex 2) allowed hai; jis moment ek side infinite ho jaaye, checklist toot jaati hai — doosri side ka finite hona use rescue nahi kar sakta.

Verify: $\int_{ε}^{π} \frac{1}{x} d x = ln π - ln ε \to + \infty$ as $ε \to 0^{+}$ , isliye sirf $(0, π)$ par absolute integrability fail ho jaati hai; $f (0^{+})$ infinite hai jabki $f (0^{-}) = 0$ finite hai. ✓

Cell L — endpoint continuity (wrap agrees, koi jump nahi)

Worked example Example 9 —

$f (x) = x^{2}$ on $(- π, π)$ , evaluate at $x = π$ Forecast: $f (x) = x$ ke endpoint par hum ek jump tak pahunche (Ex 3). Kya har endpoint jump deta hai? Padhne se pehle guess karo.

$x = π$ par left limit compute karo. Yeh step kyun? Hume jaanna hai ki last piece kitni height tak pahunchti hai: $f (π^{-}) = π^{2}$ .
Wrap (right) limit compute karo. Kyun? $x = π$ se thoda sa past graph ek period pehle restart hota hai, $x = - π$ par: $f (π^{+}) = f ((- π)^{+}) = (- π)^{2} = π^{2}$ . Key observation: kyunki $x^{2}$ even hai, dono ends ki same height hai — wrap agree karta hai, isliye endpoint par koi jump nahi hai.
Kyunki dono sides $π^{2}$ ke barabar hain, yeh continuity ka point hai: $S (π) = \frac{π ^{2} + π ^{2}}{2} = π^{2} .$

Verify: $f (- π^{+}) = π^{2} = f (π^{-})$ , isliye $x^{2}$ ka periodic extension har jagah continuous hai (ek smooth "scalloped" wall jaisa banta hai). Numerically $π^{2} \approx 9.8696$ , aur series wahan $f (π) = π^{2}$ ke barabar hai — koi midpoint discount nahi. ✓ Yeh endpoint picture complete karta hai: wrap ek jump create karta hai sirf tabhi jab dono ends alag hon (Ex 3); jab woh agree karein, endpoint kisi bhi interior continuous point ki tarah behave karta hai.

Figure 5 — periodic $f (x) = x^{2}$ teen periods par. $x = π$ par left height $π^{2}$ aur wrapped right height $π^{2}$ coincide karte hain (single red dot) — koi gap nahi, $S (π) = π^{2}$ .

Cell I — real-world word problem (Gibbs overshoot ke saath)

Worked example Example 10 — ek rectangular voltage pulse

Ek digital clock line har cycle ke pehle aadhe mein $0$ volts hold karti hai aur doosre aadhe mein $5$ volts: $V (t) = {05 0 < t < T /2 T /2 < t < T, period T .$ Iska Fourier reconstruction exactly switching instant $t = T /2$ par kaun sa voltage dikhata hai?

Forecast: hardware kehta hai yeh $0$ se $5$ tak clean step hai. Series us instant par kya draw karti hai?

Switch $t = T /2$ par one-sided limits padho. Yeh step kyun? $t = T /2$ ek finite jump hai — ek real physical rise time idealized to zero. $V (T / 2^{-}) = 0, V (T / 2^{+}) = 5.$
Dono finite $\Rightarrow$ midpoint applies: $S (T /2) = \frac{0 + 5}{2} = \frac{5}{2} = 2.5 volts .$

Verify: units throughout volts hain; midpoint $2.5 V$ $0$ aur $5 V$ ke beech hai jaisi kisi bhi average ki expect hai. ✓

Gibbs overshoot (Figure 6 dekho): switch ke paas finite partial sum $S_{N}$ smoothly nahi uthta — yeh jump height ka taqreeban $9%$ overshoot karta hai. Yeh kyun hota hai? Har finite partial sum ek smooth wave hai jo $5 V$ ki vertical cliff climb karne par forced hokar hamesha top se aage nikal jaati hai settle hone se pehle — baaki ripple stubborn hai. Concretely overshoot taqreeban $0.0895 \times 5 \approx 0.45 V$ hai, jo top ke upar $5.45 V$ ke paas spike deta hai aur bottom ke neeche $- 0.45 V$ ke paas matching dip. Sabse important baat yeh overshoot $N \to \infty$ ke saath shrink nahi karta (sirf narrow hota hai), phir bhi switch par value exactly $2.5 V$ rehti hai. Yeh flesh mein Gibbs phenomenon hai — overshoot jump ko decorate karta hai, par limit value phir bhi midpoint hai.

Figure 6 — $0$ -to- $5 V$ pulse (black) aur ek high- $N$ partial sum. Switch ke paas red overshoot horns kabhi vanish nahi hote (Gibbs), par crossover exactly midpoint $2.5 V$ se pass hota hai.

Cell J — exam twist: series se number series sum karo

Worked example Example 11 —

$1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots$ evaluate karo Ex 2 ke square wave ki Fourier series hai $f (x) = \frac{4}{π} \sum_{k = 0}^{\infty} \frac{s i n ( ( 2 k + 1 ) x )}{2 k + 1} .$ Odd numbers ke alternating reciprocals sum karne ke liye $x = \frac{π}{2}$ plug karo.

Forecast: $x = π /2$ par left side kaun si value leti hai?

$x = π /2$ par $f$ evaluate karo. Yahan kyun? $π /2$ $(0, π)$ ke andar hai jahan $f = + 1$ hai, aur yeh continuity ka point hai (cell A), isliye $S (π /2) = f (π /2) = 1$ .
Sines evaluate karo: $sin ((2 k + 1) \frac{π}{2})$ $k = 0, 1, 2, 3$ ke liye $+ 1, - 1, + 1, - 1, \dots$ cycle karta hai. Kyun? $sin (odd \cdot \frac{π}{2}) = (- 1)^{k}$ .
Dono sides equate karo: $1 = \frac{4}{π} (1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots) ⟹ 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots = \frac{π}{4} .$

Verify: partial sum $1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} \approx 0.7440$ , (slowly) $π /4 \approx 0.7854$ ki taraf converge karta hua. ✓ Yeh Leibniz series hai — ek continuity point se janma exam favourite.

Active Recall

Recall Yeh kaun sa cell hai? (hide karke jawab do)

Ek function ek point par $- 3$ se $+ 3$ tak jump karta hai — series wahan kya deti hai? ::: Midpoint $\frac{3 + ( - 3 )}{2} = 0$ (cell D). $f (x) = x^{2}$ ek interior point par evaluate kiya — jump ya smooth? ::: Smooth (cell A); series wahan exactly $f$ ke barabar hai. $f (x) = x^{2}$ $x = π$ (endpoint) par evaluate kiya — jump ya smooth? ::: Smooth (cell L); even function same height par wrap karta hai, isliye $S (π) = π^{2}$ . Aapne $f$ ko exactly ek point par ek million redefine kiya — kya series change hoti hai? ::: Nahi (cell F); coefficients integrals hain, single points ke liye andhe hain. $1/ x^{2}$ $x = 0$ ke paas — kaun si condition fail hoti hai? ::: Condition 1 (infinite discontinuity, dono sides blow up, not absolutely integrable) — cell H. Left side $0$ par flat, right side $= 1/ x \to \infty$ — kya hum average kar sakte hain? ::: Nahi (cell K); ek infinite side checklist tod deti hai chahe doosri finite ho. Square voltage pulse switch par — series value? ::: $2.5$ volts (cell I), $0$ aur $5$ ka midpoint.

Connections

4.7.04 Dirichlet conditions for convergence (Hinglish)
Fourier Series — coefficients via orthogonality
Dirichlet kernel
Gibbs phenomenon
Piecewise smooth functions
Convergence of series (pointwise vs uniform)
Half-range expansions

Worked examples — Dirichlet conditions for convergence