Exercises — Dirichlet conditions for convergence

Checklist ko ek tick ek baar mein chalao.

• Integrable? Yeh bounded hai (kabhi infinity tak nahi jaati) ek finite period par, toh $|f|$ ke neeche ka area finite hai. ✓
• Finite jumps? Exactly $4$ — ek finite number ke finite jumps. ✓
• Finite wiggles? $6$ maxima/minima finite hai. ✓

Teeno hold karte hain, toh haan — Dirichlet satisfy hoti hai aur series converge karti hai (har jump par midpoint par, baaki jagah $f$ par). Answer: HAAN.

Jab $x\to 0^{+}$, $\frac{1}{x}\to +\infty$. Yeh ek infinite discontinuity hai, finite jump nahi. Integral check karo: ${\int }_0^1\left|\frac{1}{x}\right|dx=[\ln x{]}_0^1=\infty$. Toh Condition 1 (absolutely integrable) fail ho jaati hai, aur Condition 2 (sirf finite discontinuities) bhi fail hoti hai kyunki discontinuity infinite hai. Answer: yeh absolutely integrable nahi hai / iska infinite discontinuity hai.

$x=0$ ek jump hai: left se slide karo ($x<0$) toh $f=0$, isliye $f(0^{-})=0$. Right se slide karo ($x>0$) toh $f=x\to 0$, isliye $f(0^{+})=0$. Dono sides agree karti hain, toh yeh actually value mein continuity ka point hai. $S(0)=\frac{f(0^{+})+f(0^{-})}{2}=\frac{0+0}{2}=0.$ Answer: $S(0)=0$. (Ek sabak: "piecewise" definition ka matlab automatically jump nahi hota — limits check karo.)

$\pi$ ki left se approach karo, $f=x\to \pi$, toh $f({\pi }^{-})=\pi$. $\pi$ ki right se approach karo, hum next period mein wrap ho jaate hain, jo $-\pi$ jaisi shuru hoti hai: wahan $f=0$, toh $f({\pi }^{+})=0$. $S(\pi )=\frac{\pi +0}{2}=\frac{\pi }{2}.$ Answer: $S(\pi )=\frac{\pi }{2}$.

$f(0^{-})=-3$, $f(0^{+})=+5$. $S(0)=\frac{5+(-3)}{2}=\frac{2}{2}=1.$ Answer: $S(0)=1$. Note karo ki series $1$ par land karti hai, ek aisi value jo $f$ kabhi nahi leta — sum "beech mein milta hai."

$x=0$ ke paas argument $\frac{1}{x}$ har value se race karta hai, toh $\sin (1/x)$ $+1$ se $-1$ tak infinitely baar cross karta hai jab $x\to 0$. Yeh $0$ ke aas paas kisi bhi chhoti window mein infinitely many maxima aur minima hain.

• Integrable? Haan ($|f|\le 1$, bounded). ✓
• Finite jumps? Haan (koi jumps nahi). ✓
• Finite wiggles? NAHI — infinitely many. ✗ Condition 3 fail hoti hai, toh Dirichlet's theorem apply nahi hota. Is theorem se hum convergence guarantee nahi kar sakte. Answer: FAILS (condition 3, infinite wiggles).

Ek degenerate lekin instructive case.

• Integrable? ${\int }_{-\pi }^{\pi }|c|dx=2\pi |c|<\infty$. ✓
• Jumps? Koi nahi. ✓
• Maxima/minima? Ek constant ke zero strict maxima/minima hote hain — zero finite hai. ✓ Saari conditions hold karti hain. Har jagah $f$ continuous hai, toh $S(x)=f(x)=c$. Answer: Dirichlet satisfy karta hai; $S(x)=c$ sab $x$ ke liye. (Fourier series sirf $a_0/2=c$ hai, baki sab coefficients zero hain.)

Left se, $f({\pi }^{-})={\pi }^2$. Right se wrap karo: next period $-\pi$ par shuru hoti hai, jahan $f=(-\pi {)}^2={\pi }^2$, toh $f({\pi }^{+})={\pi }^2$. Dono sides ${\pi }^2$ equal hain → koi jump nahi; $x^2$ ki periodic extension $\pi$ par continuous hai. $S(\pi )=\frac{{\pi }^2+{\pi }^2}{2}={\pi }^2.$ Answer: koi jump nahi; $S(\pi )={\pi }^2$. Isko $f(x)=x$ se compare karo (parent ka Exercise), jo $\pi$ par jump karta hai — $x^2$ ki evenness wrap ko seamless bana deti hai.

Step 1 — $x=\pi /2$ par series kya equal hai? $f$ wahan continuous hai aur $f(\pi /2)=+1$, toh midpoint rule se $S(\pi /2)=1$. Step 2 — series mein $x=\pi /2$ plug karo. $\sin ((2k+1)\frac{\pi }{2})$ $k=0,1,2,\dots$ ke liye $+1,-1,+1,\dots$ cycle karta hai. $S(\frac{\pi }{2})=\frac{4}{\pi }\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots \right).$ Step 3 — dono expressions equal karo. $1=\frac{4}{\pi }\left(1-\frac{1}{3}+\frac{1}{5}-\cdots \right)\Rightarrow 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots =\frac{\pi }{4}.$ Answer: alternating sum $\frac{\pi }{4}$ ke barabar hai (Leibniz series). Is point par convergence exactly isliye valid hai kyunki Dirichlet hold karta hai aur $x=\pi /2$ ek continuity point hai.

Midpoint route: $f(0^{-})=-1,f(0^{+})=+1$, toh $S(0)=\frac{-1+1}{2}=0$. Substitution route: har term mein $\sin ((2k+1)\cdot 0)=\sin 0=0$ hota hai, toh poori series $0$ sum hoti hai. Dono $0$ par agree karte hain. Yeh reveal karta hai ki series automatically jump par midpoint $0$ produce karti hai — odd (sine-only) structure bina kisi extra kaam ke $S(0)=0$ force karta hai. Answer: dono $0$ dete hain; consistent. (Gibbs phenomenon overshoot approach ko decorate karta hai lekin limiting value exactly $0$ hai.)

Odd extension hai $\tilde{g}(x)=x$ on $(-\pi ,\pi )$ (kyunki $x$ already odd hai), periodically extend kiya gaya — classic sawtooth. Dekho Half-range expansions. (a) Checklist:

• Integrable: ${\int }_{-\pi }^{\pi }|x|dx={\pi }^2<\infty$. ✓
• Jumps: sirf wrap points $x=\pm \pi$ par (finite jumps). ✓
• Wiggles: $x$ period par monotonic hai → finitely many (zero interior) extrema. ✓ Dirichlet hold karta hai. ✓ (b) $x=0$ par: continuous, $\tilde{g}(0)=0$, toh $S(0)=0$. $x=\pi$ par: $\tilde{g}({\pi }^{-})=\pi$; wrapping se, $\tilde{g}({\pi }^{+})=\tilde{g}(-{\pi }^{+})=-\pi$. Midpoint: $S(\pi )=\frac{\pi +(-\pi )}{2}=0.$ Answer: $S(0)=0$, $S(\pi )=0$.

$f_a(0^{-})=0,f_a(0^{+})=a$, toh $S(0)=\frac{a+0}{2}=\frac{a}{2}.$ Jab $a\to 0$, jump ki height $|a-0|=|a|\to 0$ aur $S(0)=\frac{a}{2}\to 0$. Limit $a=0$ par function constant $0$ hai — koi jump nahi, aur $S(0)=0$ continuously match karta hai. Convergence kabhi khatre mein nahi padti: har $a$ ke liye function bounded hai ek single finite jump aur koi extra wiggles nahi, toh Dirichlet poore waqt hold karta hai. Jump bas smoothly kuch nahi ho jaata. Answer: $S(0)=a/2$; jab $a\to 0$ toh jump aur $S(0)$ dono gaayab ho jaate hain; convergence hamesha guaranteed hai.

Counterexample: koi bhi square wave — yeh absolutely integrable hai, phir bhi jump par series midpoint par converge karti hai, $f$ par nahi (jo $\pm 1$ hai, unke beech ki midpoint value kabhi nahi hoti). Sahi statement: absolute integrability akela sirf guarantee karta hai ki coefficients exist karte hain. Guaranteed convergence ke liye teeno Dirichlet conditions chahiye, aur tab bhi yeh $\frac{f(x^{+})+f(x^{-})}{2}$ par hoti hai — sirf continuity ke points par $f$ ke barabar. Answer: square wave; integrability ⇒ coefficients exist, convergence-to-$f$ nahi. (Yaad raho ye conditions sufficient, not necessary hain — dekho Convergence of series (pointwise vs uniform).)

• $-3$ se $+5$ tak jump par series kya value degi?
• Kya $50$ finite jumps wala bounded function Dirichlet satisfy karta hai?
• $0$ ke paas $\sin (1/x)$ Dirichlet kyun fail karta hai?
• $(-\pi ,\pi )$ par $x^2$ ke liye, kya $x=\pi$ par jump hai?
• $1-\frac{1}{3}+\frac{1}{5}-\cdots$ kya hai aur kaun sa exercise yeh prove karta hai?