Power series — centre, radius of convergence, interval of convergence

4.3.14 · Maths › Calculus III — Sequences & Series

Power series KIYA hota hai?

$a$ par centre KYU? Kyunki convergence ek khaas point se symmetrically baahir ki taraf failti hai. Wahi point hai jahan convergence guaranteed hai, isliye hum doori $∣ x - a ∣$ usi se measure karte hain.

Radius / interval of convergence KIYA hai?

Teen cases:

$R = 0$ : sirf $x = a$ par converge karta hai.
$R = \infty$ : saare real $x$ ke liye converge karta hai.
$0 < R < \infty$ : ek finite interval par converge karta hai.

Yeh hamesha interval KYU hoti hai? (First principles se derivation)

Yeh Cauchy–Hadamard idea hai. Hum absolute convergence test karte hain Root Test use karke terms $b_{n} = ∣ c_{n} (x - a)^{n} ∣ = ∣ c_{n} ∣ ∣ x - a ∣^{n}$ par.

Formula Root Test se

$R$ nikalna Root Test (apne sharp form mein) limit superior dekhta hai: $L = lim sup_{n \to \infty} n b_{n} = lim sup_{n \to \infty} n ∣ c_{n} ∣ ∣ x - a ∣^{n} = ∣ x - a ∣ \cdot β n \to \infty lim sup n ∣ c_{n} ∣ .$ Yeh step KYU? $n ∣ x - a ∣^{n} = ∣ x - a ∣$ exactly hota hai, isliye $x$ wala part bahar aa jaata hai aur sirf coefficient wala part $β = lim sup n ∣ c_{n} ∣$ bachta hai. $lim sup$ KYU, $lim$ KYU nahi? Ordinary $lim$ exist hi na kare (jaise oscillate karte coefficients), lekin $lim sup$ hamesha exist karta hai $[0, \infty]$ mein, isliye formula poori generality mein kaam karta hai.

Root Test: converge karta hai agar $L < 1$ , diverge karta hai agar $L > 1$ . Toh: $∣ x - a ∣ β < 1 ⟺ ∣ x - a ∣ < \frac{1}{β} .$ $∣ x - a ∣ < R$ se compare karne par milta hai Cauchy–Hadamard formula: $\frac{1}{R} = n \to \infty lim sup n ∣ c_{n} ∣$ (conventions ke saath: $1/0 = \infty$ aur $1/\infty = 0$ ). Jab ordinary limit $lim_{n \to \infty} n ∣ c_{n} ∣$ exist kare, toh woh $lim sup$ ke barabar hoti hai aur tum usse use kar sakte ho.

Kyunki condition $∣ x - a ∣ < R$ hai — ek distance condition — solution set automatically $a$ ke around symmetric interval ban jaati hai. Isliye yeh hamesha interval hoti hai.

Formula Practical Ratio version

Practice mein Ratio Test zyada aasaan hai (lekin yeh $R$ tabhi deta hai jab coefficient ratio kisi limit par settle down kare). Isse $b_{n} = ∣ c_{n} ∣∣ x - a ∣^{n}$ par apply karo: $lim_{n \to \infty} \frac{b _{n + 1}}{b _{n}} = lim_{n \to \infty} \frac{∣ c _{n + 1} ∣}{∣ c _{n} ∣} ∣ x - a ∣ = ∣ x - a ∣ \cdot lim_{n \to \infty} \frac{∣ c _{n + 1} ∣}{∣ c _{n} ∣} .$ Converge hota hai jab yeh $< 1$ ho, isliye — provided limit exist kare — $R = lim_{n \to \infty} \frac{c _{n}}{c _{n + 1}} .$ Yeh step KYU? Same logic: $∣ x - a ∣$ factor alag ho jaata hai; limit $< 1$ set karne par boundary $R$ milti hai. Caution (general form): agar ratio converge na kare, toh Ratio Test kuch nahi bolega; tab guaranteed-to-work tool hai $R = 1/ lim sup_{n \to \infty} n ∣ c_{n} ∣$ . (Equivalently $R = 1/ lim sup_{n \to \infty} ∣ c_{n + 1} / c_{n} ∣$ upper estimate ke roop mein.) Root/ $lim sup$ form master formula hai; Ratio limit ek convenient special case hai.

Figure — Power series — centre, radius of convergence, interval of convergence

Interval KAISE nikaalte hain (recipe)

Ratio Test lagao $∣ a_{n} ∣$ par ( $(x - a)^{n}$ ko include karte hue). $∣ x - a ∣ \cdot k < 1$ milega.
Solve karo $∣ x - a ∣ < R$ ke liye → isse $R$ milta hai aur open interval $(a - R, a + R)$ .
Har endpoint test karo $x = a - R$ aur $x = a + R$ ko plug in karke aur koi series test use karke (yeh number series ban jaati hai).
Interval likho endpoints include/exclude karte hue.

Worked examples

Forecast-then-Verify

Recall Compute karne se pehle forecast karo

$\sum \frac{( x - 5 ) ^{n}}{3 ^{n}}$ ke liye, aage padhne se pehle $R$ predict karo. ▶ Coefficients $c_{n} = 1/ 3^{n}$ , ratio $∣ c_{n} / c_{n + 1} ∣ = 3^{n + 1} / 3^{n} = 3$ , toh $R = 3$ , $5$ par centred. Geometric: dono endpoints $\sum (\pm 1)^{n}$ dete hain jo diverge karte hain. Interval $(2, 8)$ . Kya tumhara forecast match kiya?

\sum c_{n} (x - a)^{n}

ka centre kiya hai?

Point

a

; series wahan hamesha converge karti hai (

c_{0}

milta hai).

Cauchy–Hadamard formula for

1/ R

(general, hamesha valid)?

1/ R = lim sup_{n \to \infty} n ∣ c_{n} ∣

, with

1/0 = \infty

1/\infty = 0

Ratio-test formula for

R

— yeh kab valid hai?

R = lim_{n \to \infty} ∣ c_{n} / c_{n + 1} ∣

, SIRF jab woh limit exist kare; warna

lim sup

root formula use karo.

∣ x - a ∣ = R

par kiya hota hai?

Inconclusive — har endpoint ko substitution se alag se test karna padta hai.

R = 0

ka matlab kiya hai?

Sirf centre

x = a

par converge karta hai.

R = \infty

ka matlab kiya hai?

Saare real

x

ke liye converge karta hai (jaise

e^{x}

series).

Convergence set hamesha interval KYU hoti hai?

Kyunki convergence condition ek distance condition

∣ x - a ∣ < R

hai, jo

a

ke baare mein symmetric hai.

∣ x - a ∣ < R

ke andar, convergence absolute hai ya conditional?

Absolute convergence.

\sum n! x^{n}

R

0

(saare

x \neq = 0

ke liye diverge karta hai).

\sum x^{n} / n!

R

\infty

Cauchy–Hadamard

lim

ki jagah

lim sup

KYU use karta hai?

Kyunki

lim sup

hamesha

[0, \infty]

mein exist karta hai, isliye formula

R

deta hai tab bhi jab ordinary limit fail ho jaaye.

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho point $a$ par ek campfire hai. Jitna paas khade ho, utni zyada garmi milti hai (series "kaam karti hai"). Ek magic distance $R$ hai — us ke andar tum warm ho (converge karta hai), baahir tum freeze ho rahe ho (diverge karta hai). Bilkul distance $R$ ke circle par, yeh coin-flip hai: tumhe har edge par jaana padega aur khud mehsoos karna padega ki warm hai ya nahi. Campfire centre hai, magic distance radius hai, aur warm zone (edges shayad include) interval of convergence hai.

Connections

Ratio Test aur Root Test — wo engines jo $R$ produce karte hain (Root/limsup master wala hai).
Limit Superior and Limit Inferior — KYU $lim sup$ guarantee karta hai ki $R$ hamesha exist karega.
Geometric Series — prototype: $\sum x^{n}$ , $R = 1$ .
Taylor and Maclaurin Series — power series jo function ki derivatives se bante hain.
Absolute vs Conditional Convergence — radius ke andar hamesha absolute hota hai.
p-series aur Alternating Series Test — usual endpoint tools.
Term-by-term Differentiation and Integration — $(a - R, a + R)$ ke andar valid hai, $R$ unchanged rehta hai.