Power series — centre, radius of convergence, interval of convergence: Question bank

4.3.14 · D5 · Maths › Calculus III — Sequences & Series › Power series — centre, radius of convergence, interval of co

Jo tools ke baare mein aap reason karoge woh yahan hain: Ratio Test, Root Test, Limit Superior and Limit Inferior, Geometric Series, p-series, Alternating Series Test, Absolute vs Conditional Convergence, Taylor and Maclaurin Series, Term-by-term Differentiation and Integration, aur parent Power series topic note.

Vocabulary refresher (taaki har trap samajh aaye)

Definition Woh chaar words jin par yeh questions attack karte hain

Ek power series ek infinite polynomial hai jo is tarah likhi jaati hai $\sum_{n = 0}^{\infty} c_{n} (x - a)^{n} = c_{0} + c_{1} (x - a) + c_{2} (x - a)^{2} + \dots$

Centre $a$ woh special value hai $x$ ki jo har base $(x - a)$ ko zero kar deta hai; series hamesha wahaan converge karti hai (sirf $c_{0}$ bachta hai).
Coefficients $c_{n}$ woh fixed numbers hain jo har power ko multiply karte hain — sirf yahi decide karte hain ki convergence kitni door tak pahunchti hai.
Radius of convergence $R$ (ek number $[0, \infty]$ mein) woh distance hai $a$ se jiske andar series converge karti hai: yeh $∣ x - a ∣ < R$ ke liye converge karti hai, $∣ x - a ∣ > R$ ke liye diverge karti hai, aur boundary $∣ x - a ∣ = R$ par undecided hai.
Interval of convergence $x$ ka poora set hai jo kaam karta hai — open band plus jo bhi endpoints alag test mein pass ho jaate hain.

Number line ko neeche picture karo: centre $a$ beech mein hai, $R$ har direction mein reach hai, aur do dots $a \pm R$ endpoints hain jinhe hamesha haath se check karna padta hai.

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

True or false — justify karo

A power series apne centre par hamesha converge karti hai.

True.

x = a

set karna

c_{0}

ke alawa har term ko khatam kar deta hai (baaki sabmein

(x - a)^{n} = 0

ka factor hai), isliye sum sirf ek single number

c_{0}

hai — convergence wahaan automatic hai.

Agar

R = 0

hai toh series har jagah diverge karti hai.

False.

R = 0

ka matlab hai yeh sirf centre

x = a

par converge karti hai (jahaan yeh

c_{0}

deti hai); yeh har doosre point par diverge karti hai, lekin centre khud abhi bhi kaam karta hai.

Interval of convergence hamesha centre ke baare mein symmetric hota hai.

Open part symmetric hota hai —

(a - R, a + R)

— kyunki

∣ x - a ∣ < R

ek distance condition hai. Lekin closed interval behaviour mein symmetric nahi hona chahiye: ek endpoint converge kar sakta hai jabki doosra diverge kare (jaise

[1, 3)

mein).

Agar ek power series

x = 6

par converge karti hai aur uska centre

x = 2

hai, toh yeh

x = - 1

par bhi converge karni chahiye.

True.

x = 6

par convergence matlab hai yeh centre se distance

4

par converge karti hai, isliye

R \geq 4

; point

x = - 1

distance

3 < 4

par hai, safely inside, isliye wahaan converge hogi (actually absolutely mein).

Agar ek power series

x = 6

par converge karti hai (centre

2

), toh yeh

x = 6.5

par bhi converge karni chahiye.

Guaranteed nahi hai.

x = 6

endpoint

a + R

ho sakta hai (exactly

R

distance). Tab

x = 6.5

radius ke bahar hoga, jahaan series diverge karti hai. Kisi point par convergence sirf centre ke strictly closer points par convergence guarantee karta hai.

Radius of convergence

[0, \infty]

mein koi bhi number ho sakta hai.

True.

R = 0

(sirf centre par converge karta hai),

R = \infty

(har jagah converge karta hai, jaise

e^{x}

series), aur har finite positive value — sab occur karte hain.

Ek power series aur uski term-by-term derivative ke radii of convergence alag hote hain.

False. Term-by-term Differentiation and Integration

R

ko preserve karta hai — ek power series ko differentiate ya integrate karne se ek aisi series milti hai jiska same radius

R

hota hai (sirf endpoint behaviour badal sakta hai).

Ratio Test formula

R = lim ∣ c_{n} / c_{n + 1} ∣

har power series ke liye kaam karta hai.

False. Yeh sirf tab kaam karta hai jab woh limit exist kare. Agar coefficient ratio bina settle hue oscillate kare (parent ka Example 5 dekho), toh formula kuch nahi deta aur aapko hamesha valid

lim sup

root formula use karna padega.

lim sup_{n \to \infty} n ∣ c_{n} ∣

hamesha exist karta hai (ho sakta hai

= \infty

True. Yahi toh Limit Superior and Limit Inferior use karne ka poora point hai ordinary limit ki jagah:

lim sup

har bounded-below sequence ke liye defined hai,

[0, \infty]

mein koi value lete hue, isliye Cauchy–Hadamard kabhi break nahi hota.

Spot the error

"Endpoints par ratio limit

1

ke barabar thi, aur Ratio Test kehta hai

< 1

converge karta hai, isliye endpoints converge karte hain."

Galat. Exactly

1

ki limit Ratio Test ka inconclusive case hai, na ki "

1

se kam" wala case. Aap isse endpoints ke baare mein kuch nahi seekh sakte — har ek ko substitute karo aur alag test chalao.

\sum n! x^{n}

R = \infty

hai kyunki

n!

itna fast grow karta hai."

Ulta hai. Fast-growing coefficients series ko converge karna mushkil banate hain, aasaan nahi. Ratio

(n + 1) ∣ x ∣ \to \infty

kisi bhi

x \neq = 0

ke liye, isliye

R = 0

— yeh sirf

x = 0

par converge karta hai.

\sum c_{n} (x + 3)^{n}

ka centre

a = 3

hai."

Sign error. Form

(x - a)^{n}

hai, isliye

(x + 3) = (x - (- 3))

matlab centre ==

a = - 3

== hai. Centre woh value hai

x

ki jo base ko zero banati hai.

"Maine

R = 4

find kiya aur open interval

(- 7, 1)

hai, isliye answer

(- 7, 1)

hai."

Incomplete. Aap interval tab tak state nahi kar sakte jab tak aap substitution se dono endpoints

x = - 7

aur

x = 1

ko test nahi karte; true interval mein ek, dono, ya koi bhi nahi ho sakta.

\sum (- 1)^{n} / n

converge karta hai, isliye

\sum (- 1)^{n} / n

absolutely converge karta hai."

Do ideas ko confuse kar raha hai. Yeh Alternating Series Test se conditionally converge karta hai, lekin

\sum 1/ n

(absolute version) diverge karta hai. Absolute vs Conditional Convergence dekho —

∣ x - a ∣ < R

ke andar convergence absolute hoti hai, lekin endpoint par yeh sirf conditional ho sakti hai.

"Endpoint par series hamesha ya toh dono par converge karti hai ya dono par diverge karti hai."

Symmetry ka false claim. Endpoints do alag number series hain aur independently behave karti hain —

\sum (x - 2)^{n} / n

x = 3

par diverge karta hai (harmonic) lekin

x = 1

par converge karta hai (alternating).

"Kyunki

\sum x^{n} / n!

equals

e^{x}

hai, aur

e^{x}

sab

x

ke liye finite hai, isliye series har jagah converge karni chahiye — isliye

R = \infty

hai."

Conclusion sahi hai lekin reasoning circular hai. Hum

R = \infty

pehle Ratio Test se establish karte hain (

∣ x ∣/ (n + 1) \to 0

); wahi license karta hai

e^{x}

ko sab

x

ke liye sum kehne ke liye. Aap identity assume nahi kar sakte usi convergence ko prove karne ke liye jo identity ko justify karti hai.

Why questions

Convergence set hamesha ek interval kyun hoti hai aur, say, do alag pieces kabhi nahi?

Kyunki convergence condition

∣ x - a ∣ < R

hai — ek single distance-from-a-point inequality. Iska solution set

a

ke around ek symmetric band hai, jo exactly wahi hai jo "interval" ka matlab hota hai; ek distance condition disconnected chunks nahi bana sakti.

Cauchy–Hadamard formula mein

lim

ki jagah

lim sup

kyun use karte hain?

Kyunki

lim n ∣ c_{n} ∣

exist karna fail kar sakta hai (oscillating coefficients), jabki

lim sup

hamesha

[0, \infty]

mein exist karta hai. Isse use karna formula ko har power series ke liye valid banata hai, sirf well-behaved ones ke liye nahi.

∣ x - a ∣

factor Root Test se cleanly kyun pull out hota hai?

Kyunki

n ∣ x - a ∣^{n} = ∣ x - a ∣

exactly har

n

ke liye — yeh ek fixed number hai

n

se independent. Isliye limit ko sirf coefficient part

n ∣ c_{n} ∣

deal karna hai, aur

∣ x - a ∣

untouched ride karta hai.

Endpoints ko same test se handle karne ki jagah alag se test kyun karna padta hai jisne

R

diya?

Ratio aur Root Tests dono

∣ x - a ∣ = R

par exactly $1$ ki limit dete hain, unka ek blind spot. Isliye har endpoint ko resolve karne ke liye ek fresh, sharper tool (p-series, Alternating Series Test, divergence test) chahiye hota hai.

Ratio version

R = lim ∣ c_{n} / c_{n + 1} ∣

kyun deta hai — coefficient ratio ka reciprocal?

Convergence ke liye

∣ x - a ∣ \cdot lim ∣ c_{n + 1} / c_{n} ∣ < 1

chahiye, yaani

∣ x - a ∣ < 1/ lim ∣ c_{n + 1} / c_{n} ∣ = lim ∣ c_{n} / c_{n + 1} ∣

. Us inequality ki boundary hi

R

hai, isliye

R

reciprocal ke roop mein nikalta hai.

Geometric Series jaise

\sum x^{n}

apna radius ek nazar mein kyun bata sakta hai?

Uske coefficients sab

1

hain, isliye

n ∣ c_{n} ∣ = 1

aur

R = 1

; equivalently yeh woh model case hai jo converge karta hai iff

∣ x ∣ < 1

aur dono endpoints

\pm 1

par diverge karta hai. Yeh woh prototype hai jiske against har power-series radius measure hoti hai.

Edge cases

Agar

c_{n}

kisi point ke baad har coefficient

0

ho, toh series kya hai aur uska radius kya hai?

Yeh ek ordinary polynomial ban jaata hai (finitely many terms). Root Test apply karo: cutoff ke baad

∣ c_{n} ∣ = 0

, isliye

n ∣ c_{n} ∣ = 0

aur

lim sup_{n \to \infty} n ∣ c_{n} ∣ = 0

, jo

R = 1/0 = \infty

deta hai — yeh sab

x

ke liye converge karta hai.

Jab

R = \infty

ho toh interval of convergence kya hota hai?

Poori real line

(- \infty, \infty)

— koi endpoints nahi hain test karne ke liye, kyunki

∣ x - a ∣ = \infty

kabhi attain nahi hota.

Agar ek power series sirf apne centre par converge karti hai, toh uska interval of convergence kya hai?

Single point

{a}

, jo

R = 0

correspond karta hai. Yeh zero length ka degenerate "interval" hai.

Kya dono endpoints diverge ho sakte hain jabki open interval converge kare?

Haan. E.g.

\sum (x - 5)^{n} / 3^{n}

R = 3

hai aur dono endpoints

\sum (\pm 1)^{n}

dete hain, jo diverge karte hain (terms

\to 0

nahi jaate), isliye interval open

(2, 8)

hai.

Kya dono endpoints converge ho sakte hain, fully closed interval dete hue?

Haan.

\sum (- 1)^{n} (x + 3)^{n} / (n^{2} 4^{n})

apne endpoints par

\sum 1/ n^{2}

\sum (- 1)^{n} / n^{2}

deta hai — dono p-series (

p = 2

) se converge karte hain, isliye interval closed

[- 7, 1]

hai.

Kya ek endpoint par series conditionally converge kar sakti hai lekin absolutely nahi?

Haan.

x = 1

par,

\sum (x - 2)^{n} / n

ban jaata hai

\sum (- 1)^{n} / n

: conditionally convergent (alternating) lekin absolutely nahi (harmonic). Radius ke andar convergence hamesha absolute hoti hai; sirf endpoints par yeh conditional tak gir sakti hai.

Jab hum ek power series ko term-by-term differentiate karte hain toh endpoints ka kya hota hai?

Radius

R

unchanged rehta hai, lekin ek endpoint jo pehle converge karta tha ab diverge karna shuru ho sakta hai (differentiation ek extra factor of

n

multiply kar deta hai, convergence ko degrade karte hue). Interval sirf ends par shrink ho sakta hai, kabhi grow nahi. Term-by-term Differentiation and Integration dekho.

Agar centre complex hai ya coefficients strange hain, toh kya

R

phir bhi well-defined hai?

Haan. Kyunki

R = 1/ lim sup n ∣ c_{n} ∣

sirf magnitudes

∣ c_{n} ∣

aur ek

lim sup

use karta hai jo hamesha exist karta hai, kisi bhi coefficient sequence ke liye ek radius defined hota hai.

Recall Traps ka ek-line summary

Trap ::: Teen aadte jo kaatte hain: endpoints test karna bhool jaana, Ratio limit par trust karna jab woh exist nahi karta, aur kisi point par convergence ko har jagah-closer convergence se confuse karna. Har baar $∣ x - a ∣ < R$ se reason karo.