Question bank — Dirichlet conditions for convergence

$a_n,b_n$ sab exist karte hain, to series har point par $f(x)$ ke barabar honi chahiye. False ::: Coefficients ka exist hona sirf yeh maangta hai ki integrals $\int f\cos$, $\int f\sin$ finite hon (absolute integrability). Yeh convergence se kaafi weak demand hai — "$\sim$" sign ek warning hai ki sum ko abhi bhi poore I-J-W checklist ke against test karna hoga, aur tab bhi yeh jumps par midpoint par land karta hai.

False ::: Dirichlet finitely many finite jumps allow karta hai. Har aisi jump par series sirf midpoint $\frac{f(x^{+})+f(x^{-})}{2}$ par converge karti hai $f$ ki jagah — convergence kabhi lost nahi hoti, bas re-aimed ho jaati hai.

False ::: Yeh sufficient, not necessary hain. Kuch functions jo ek condition violate karte hain phir bhi converge ho jaate hain; Dirichlet sirf ek clean fence kheenchta hai un functions ke around jahaan convergence bina extra kaam ke guaranteed ho.

False ::: Sines aur cosines symmetric ingredients hain jinhe koi left/right preference nahi, isliye sum difference split karta hai: yeh dono one-sided limits ka average deta hai, kisi ek side ka nahi.

$f$ ek period mein har jagah continuous hai, to series har point par $f(x)$ ko deti hai. True ::: Continuity ke point par $f(x^{+})=f(x^{-})=f(x)$, isliye midpoint formula collapse ho kar $f(x)$ hi ban jaata hai — provided wiggle aur integrability conditions bhi hold karein (continuity akele poora checklist nahi hai).

False ::: Overshoot jump ke paas partial sums $S_N$ ka feature hai; iska height kabhi nahi ghata, lekin $N\to \infty$ hone par yeh jump ke paas khisak jaata hai. Jump point par khud limiting value abhi bhi exactly midpoint hoti hai. Dekho Gibbs phenomenon.

$[-L,L]$ par har bounded function Dirichlet conditions satisfy karta hai. False ::: Boundedness blow-ups handle karta hai lekin wiggles nahi. $f(x)=\sin (1/x)$ $1$ se bounded hai phir bhi $x=0$ ke paas infinitely many maxima aur minima hain, isliye yeh condition 3 fail karta hai.

False ::: Finite period par, boundedness ($|f|\le M$) force karti hai ki ${\int }_{-L}^L|f|dx\le 2LM<\infty$. Absolute integrability sirf ek unbounded spike se fail ho sakti hai (jaise $1/x$ near $0$).

$f(x)=\tan x$ on $(-\pi /2,\pi /2)$ theek hai — yeh smooth hai aur koi wiggle nahi hai, isliye iska Fourier series har jagah $f$ par converge karta hai." Error hai infinite discontinuity ko ignore karna. $\tan x\to \pm \infty$ at $x=\pm \pi /2$, isliye $\int |\tan x|dx$ diverge karta hai — condition 1 (absolute integrability) aur condition 2 (no infinite discontinuities) dono fail hain.

$f(x)=x$ on $(-\pi ,\pi )$ ke liye $x=\pi$ par series $\pi$ deni chahiye, kyunki graph wahaan end hota hai." Error hai periodicity bhoolna jo wrap point par jump create karti hai: $f({\pi }^{-})=\pi$ lekin $f({\pi }^{+})$ wrap hokar $-\pi$ ban jaata hai. Isliye series deti hai $\frac{\pi +(-\pi )}{2}=0$, na ki $\pi$.

$\sin (1/x)$ Dirichlet fail karta hai kyunki yeh $x=0$ ke paas blow up hota hai." Error hai galat condition naam lena. $\sin (1/x)$ $[-1,1]$ ke andar rahta hai — yeh kabhi blow up nahi hota. Yeh fail karta hai kyunki infinitely many maxima aur minima hain (condition 3), na ki infinite discontinuity ki wajah se.

Error hai "sufficient" ko "necessary" padhna. Conditions violate karne ka matlab sirf yeh hai ki guarantee khatam — kuch aaise functions phir bhi converge karte hain; tumhe bas doosre tarike se check karna padta hai. Dekho Convergence of series (pointwise vs uniform).

Error hai ise approximation kehna. Midpoint jump par partial sums ka exact limit hai — symmetric Dirichlet kernel se derived jo dono sides ko equally sample karta hai, yeh koi fudge ya rounding nahi hai.

$[0,1]$ mein har rational number par steps wala piecewise-constant staircase sirf finite jumps rakhta hai, isliye yeh Dirichlet satisfy karta hai." Error hai yeh miss karna ki "finite jumps" finitely many bhi hone chahiye. Infinitely many jumps (har rational par ek) condition 2 violate karta hai, chahe har individual jump finite ho.

average par kyun converge karti hai, kisi ek side par kyun nahi? Kyunki partial sum Dirichlet kernel $D_N(t)$ ke saath ek convolution hai, jo $t=0$ ke baare mein symmetric hai. Jab $N\to \infty$, yeh left ($t>0$) aur right ($t<0$) par equal mass wala spike ban jaata hai, isliye yeh $f(x^{-})$ aur $f(x^{+})$ dono ko equal weight $\frac{1}{2}$ se sample karta hai.

absolutely integrable kyun honi chahiye, sirf integrable kyun nahi? Absolute integrability, $\int |f|dx<\infty$, guarantee karti hai ki coefficient integrals $\int f\cos$ aur $\int f\sin$ sign cancellation se independent hokar finite hain — yeh $a_n,b_n$ ki bahut definition ko tootne se rokti hai, koi convergence question puchhe jaane se pehle.

Har smooth wave series mein kisi bhi partial-sum stage par ek fixed maximum frequency rakhti hai, isliye waves ka koi finite collection ek point ke paas infinitely rapid oscillation track nahi kar sakta — series simply keep up nahi kar paati, aur convergence guaranteed nahi hoti.

$=$" ki jagah "$\sim$" symbol kyun use hota hai? Kyunki coefficients likhna sirf yeh assert karta hai ki woh exist karte hain; yeh infinite sum ke converge hone ya kis value par hone ke baare mein kuch nahi kehta. "$\sim$" judgment reserve karta hai jab tak I-J-W checklist verify na ho, jab $=$ hold karta hai (jumps par midpoints ke saath).

$\mathbb{R}$ par nahi balki "ek period par" kyun likhi gayi hain? Ek periodic function repeat karta hai, isliye ek period par jo bhi sach hai woh automatically har jagah hold karta hai. Ek period check karna zaroori bhi hai aur kaafi bhi — aur yeh integrals ko finite rakhta hai (poore $\mathbb{R}$ par, kisi bhi nonzero periodic $f$ ke liye $\int |f|$ diverge karta).

derivative nahi, aur phir bhi converge kare — yeh kyun? Convergence ko piecewise smoothness chahiye, har jagah differentiability nahi. Corners (slope mein finite jumps) bilkul allowed hain — value continuous rehti hai, isliye series wahaan $f$ par converge karti hai; sirf $f$ mein khud ke jumps midpoint rule trigger karte hain.

continuous hai lekin sharp corner hai (jaise $|x|$ at $x=0$), wahaan series kya deti hai? $f(x)$ khud ko. Ek corner slope mein jump hai, value mein nahi, isliye $f(0^{+})=f(0^{-})=0$ aur midpoint rule $f(0)=0$ return karta hai — koi surprise nahi.

$f$ ko ek single jump point par redefine kar diya jaaye, jaise left limit ke equal, to wahaan series ki value badlegi? Nahi. Series ki value sirf one-sided limits $f(x^{\pm })$ par depend karti hai, jo exactly us point par assign ki gayi value se unaffected hain. Sum phir bhi midpoint deta hai, chahe us single point par $f$ kuch bhi defined ho.

$2L$ function ke liye endpoints $x=\pm L$ par kya hota hai? Periodicity se $f(-L)$ aur $f(L)$ circle par ek hi point hain. Agar $L$ ki taraf badhte hue left limit aur right limit ($-L$ par wrap hokar) alag hain, to woh ek jump hai, aur series unke midpoint par converge karti hai — bilkul kisi interior jump ki tarah.

$f=-1$ on $(-\pi ,0)$, $f=+1$ on $(0,\pi )$ ke liye $x=0$ par series kya value deti hai, jahan $f$ defined bhi nahi hai? $\frac{f(0^{+})+f(0^{-})}{2}=\frac{1+(-1)}{2}=0$. Series khushi se $0$ return karti hai — ek aisi value jo function kabhi actually leta hi nahi — kyunki yeh jump ke dono sides ko average karta hai.

Haan. Ek half-range expansion sirf $f$ ke odd/even periodic extension ka full Fourier series hai. Same I-J-W checklist us extension par apply hota hai, aur extension se introduced jumps (often $x=0$ ya $x=L$ par) same midpoint rule se resolve hote hain.

$f(x)=x$ ke liye $x=0$ par series ki value kya hai? $0$. Function $x=0$ par continuous hai jahan $f(0)=0$, isliye koi midpoint averaging ki zaroorat nahi — series straight $f(0)=0$ par converge karti hai. Ise $x=\pi$ se contrast karo, jahan periodic wrap $\pi$ aur $-\pi$ se midpoint $0$ force karta hai.