Power series — centre, radius of convergence, interval of convergence: Visual walkthrough

4.3.14 · D2 · Maths › Calculus III — Sequences & Series › Power series — centre, radius of convergence, interval of co

Jaate jaate humein teen prerequisite ideas ki zaroorat padegi — Geometric Series (hamara model example), Root Test aur Ratio Test (hamare measuring tools), aur Limit Superior and Limit Inferior (woh tool jo kabhi fail nahi karta). Har ek ko exactly us waqt introduce kiya jayega jab use kiya jayega.

Step 1 — Power series actually hoti kya hai

KYA humne abhi naam diya: sum ke har ingredient ko. KYun yeh matter karta hai: kyunki sirf $(x - a)^{n}$ hi $x$ par depend karta hai, "kya yeh converge karta hai?" ka poora sawaal ispar aa jaayega ki $(x - a)$ kitna bada hai — yaani uska size $∣ x - a ∣$ . PICTURE: centre $a$ ko number line par ek home base socho, aur $x$ ko kuch door ek point.

At $x = a$ distance zero hai, toh $n \geq 1$ wale har $(x - a)^{n}$ ka value zero ho jaata hai aur sum sirf $c_{0}$ reh jaata hai. Centre ek free win hai — wahan hamesha converge karta hai.

Step 2 — Sabse simple case par intuition banao: geometric series

General coefficients se pehle, sabse clean possible power series dekho, jahan har $c_{n} = 1$ aur $a = 0$ ho: $\sum_{n = 0}^{\infty} x^{n} = 1 + x + x^{2} + x^{3} + \dots$ Yeh Geometric Series hai. Yeh woh seed hai jisse poori theory ugti hai.

KYA humne kiya: ek closed form compute kiya aur uska domain note kiya. KYun yeh example: yeh pehle se hi punchline dikhata hai miniature mein — convergence $∣ x ∣ < 1$ par hoti hai ( $0$ ke aas paas ek symmetric interval) aur bahar fail ho jaati hai. Yahan magic number $R = 1$ hai. PICTURE: terms $x^{n}$ ko ek shrinking $x$ ke liye aur ek growing $x$ ke liye dekho.

Left panel (terms zero ho jaate hain) convergence hai; right panel (terms apna size rakhte hain ya badhte hain) divergence hai. Neeche jo kuch bhi hai woh isi ek picture ka general version hai.

Step 3 — Sahi measuring tool: Root Test

Kisi bhi coefficient ko handle karne ke liye humein ek aisa test chahiye jo ek series padhe aur bole "converge" ya "diverge". Hum Root Test choose karte hain.

KYA test measure karta hai: $n b_{n}$ roughly stage $n$ par "common ratio felt" hai. Agar woh ratio $1$ se neeche hai toh terms ek shrinking geometric series ki tarah behave karte hain (Step 2), toh sum converge karta hai. KYun cutoff exactly $1$ hai: $1$ par terms na decisively shrink hote hain na badhte hain — exactly Step 2 ke $∣ x ∣ < 1$ rule ki borderline. PICTURE: $n b_{n}$ ek dial ki tarah; line $1$ ke neeche = safe, upar = runaway.

Step 4 — Power series ko Root Test mein daalo

Ab $b_{n} = ∣ c_{n} (x - a)^{n} ∣ = ∣ c_{n} ∣ ∣ x - a ∣^{n}$ rakho (hum absolute values lete hain kyunki hum absolute convergence chahte hain — dekho Absolute vs Conditional Convergence). Root apply karo:

$n b_{n} = n coeff part ∣ c_{n} ∣ x part ∣ x - a ∣^{n} = depends only on n n ∣ c_{n} ∣ \cdot the n cancels! ∣ x - a ∣ .$

KYA humne kiya: term ko ek $x$ -part aur ek coefficient-part mein alag kiya. KYun yahi poora game hai: kyunki $∣ x - a ∣$ bahar aa jaata hai, limit hai $L = ∣ x - a ∣ \cdot (lim_{n \to \infty} n ∣ c_{n} ∣) .$ Convergence condition $L < 1$ ek seedhi distance condition on $x$ ban jaati hai. PICTURE: term do stacked bars mein split hoti hai, ek $x$ se fixed, ek coefficients se fixed.

Step 5 — Coefficient number $β$ , aur humein $lim sup$ kyun chahiye

Coefficient limit ko naam do $β = lim_{n \to \infty} n ∣ c_{n} ∣ .$ Ek dikkat hai: yeh ordinary limit exist nahi kar sakti (coefficients oscillate kar sakti hain). Ise theek karne ke liye hum Limit Superior and Limit Inferior use karte hain.

Toh hum $β$ ko hamesha-defined se replace karte hain $β = lim sup_{n \to \infty} n ∣ c_{n} ∣ .$

KYA humne kiya: ek possibly-missing limit ko ek aise limit se swap kiya jo hamesha exist karti hai. KYun: taaki formula har power series ke liye kaam kare, yahan tak ki jumpy coefficients waali ke liye bhi (woh parent note mein Example 5 hai). PICTURE: ek oscillating sequence — ordinary limit undefined hai, lekin jis ceiling ko woh bar bar touch karti hai woh crisp hai.

Step 6 — Condition solve karo → radius $R$

Ab hamare paas $L = ∣ x - a ∣ \cdot β$ hai. Root Test ka convergence rule $L < 1$ impose karo:

$∣ x - a ∣ \cdot β < 1 ⟺ ∣ x - a ∣ < \frac{1}{β} .$

Us boundary distance ko naam do $R$ : $\frac{1}{R} = n \to \infty lim sup n ∣ c_{n} ∣ (Cauchy-Hadamard) .$

$\frac{1}{R} = β$ , conventions ke saath $1/0 = \infty$ aur $1/\infty = 0$ .

PICTURE: $a$ ke aas paas half-width $R$ ka ek coloured band; andar = converge, bahar = diverge.

Step 7 — $R$ ke teen regimes (har case cover karo)

$R$ koi bhi value $[0, \infty]$ mein ho sakta hai. Teeno alag behave karte hain:

| $R$ | $β = lim sup n ∣ c_{n} ∣$ | Kahan converge karta hai | Example | |---|---|---|---| | $0$ | $\infty$ | sirf $x = a$ par | $\sum n! x^{n}$ | | $0 < R < \infty$ | finite $> 0$ | band $(a - R, a + R)$ + shayad ends | $\sum \frac{( x - 2 ) ^{n}}{n}$ | | $\infty$ | $0$ | saare real $x$ | $\sum \frac{x ^{n}}{n !} = e^{x}$ |

KYA / KYun: coefficient ceiling $β$ inversely reach control karti hai. Fast-growing coefficients ( $β = \infty$ ) series ko ek single point tak daba dete hain; fast-shrinking coefficients ( $β = 0$ ) use har jagah converge karne dete hain. PICTURE: teen number lines — ek single dot, ek finite band, poori line.

Step 8 — Endpoints: woh jagah jahan machine chup ho jaati hai

Exactly $∣ x - a ∣ = R$ par humein $L = R \cdot β = 1$ milta hai. Yahi Root/Ratio Test ki inconclusive value hai. Machine kandhe uchaati hai; humein haath se test karna padta hai.

$x = a + R$ aur $x = a - R$ par series ek plain number series ban jaati hai (ab aur koi $x$ nahi). Har endpoint independently converge ya diverge karta hai, toh interval in mein se ek hoga $(a - R, a + R), [a - R, a + R), (a - R, a + R], [a - R, a + R] .$

PICTURE: band with do hollow/solid dots jo ek ek karke decide honge.

Ek picture mein summary

Upar sab kuch ek single flow mein: term split karo, root lo, power cancel karo, band padho, ends haath se check karo.

Recall Feynman retelling — plain words mein bolo

Socho tum number line par centre $a$ par khade ho. Ek power series puchti hai: "main kitni door tak chal sakta hoon is se pehle ki infinite sum toot jaaye?" Har term ka sirf $x$ -flavoured hissa $(x - a)^{n}$ hai, ek distance jo $n$ -th power tak raised hai. Aisi cheez measure karne ke liye tum $n$ -th root lete ho — aur root power ko kha jaata hai, sirf raw distance $∣ x - a ∣$ reh jaata hai ek pure coefficient number $β$ ke saath ( $lim sup$ se liya gaya coefficient roots ka ceiling, taki woh hamesha exist kare). Series tab tak survive karti hai jab tak distance $\times β$ $1$ se neeche rahe, yaani jab tak $∣ x - a ∣ < 1/ β = R$ ho. Distance-under- $R$ rule hamesha ek symmetric band paint karta hai, toh answer hamesha ek interval hota hai. Bade coefficients band ko ek dot tak shrink kar dete hain ( $R = 0$ ); chote wale use poori line tak khol dete hain ( $R = \infty$ ). Bilkul edge par tests chup ho jaate hain, toh hum har endpoint par chalte hain, woh number daalte hain, aur us akele series ko apne dum par test karte hain.

Recall Self-check

Root test mein $x$ -dependence cleanly alag kyun ho jaati hai? ::: Kyunki $n ∣ x - a ∣^{n} = ∣ x - a ∣$ exactly — root aur power cancel ho jaate hain, ek single factor reh jaata hai. $lim sup$ kyun, $lim$ kyun nahi? ::: $n ∣ c_{n} ∣$ ki ordinary limit exist nahi kar sakti, lekin $lim sup$ $[0, \infty]$ mein hamesha exist karti hai, toh formula universal hai. Convergence set hamesha interval kyun hoti hai? ::: Condition $∣ x - a ∣ < R$ hai, ek distance-from-centre bound, jiska solution symmetric band $(a - R, a + R)$ hai. Endpoints alag kyun test karne padte hain? ::: $∣ x - a ∣ = R$ par test limit $1$ ke barabar hoti hai, jo inconclusive case hai, toh har end ko apna number-series test chahiye.

Related: Taylor and Maclaurin Series · Term-by-term Differentiation and Integration · 4.3.14 Power series — centre, radius of convergence, interval of convergence (Hinglish)