Term-by-term differentiation and integration of power series: Question bank

4.3.15 · D5 · Maths › Calculus III — Sequences & Series › Term-by-term differentiation and integration of power series

Ek hi rule hai jis par sab kuch test hota hai: open interval $(c - R, c + R)$ ke andar ek power series $\sum a_{n} (x - c)^{n}$ ek infinite polynomial ki tarah kaam karta hai — term by term differentiate/integrate karo, aur ==radius $R$ kabhi nahi badalti, lekin endpoints ka== behaviour zaroor badal sakta hai. Cauchy–Hadamard theorem aur Uniform convergence and the Weierstrass M-test ko yaad rakho — ye hi machinery hai.

True ya false — justify karo

TF1. Ek power series ko term by term differentiate karne se convergence ki radius same rehti hai.

True — factor

n

n^{1/ n} \to 1

aata hai Cauchy–Hadamard

lim sup ∣ a_{n} ∣^{1/ n}

mein, ek subgeometric nudge jo

n

-th root mein vanish ho jaata hai, isliye

R

untouched rehta hai.

TF2. Term by term differentiate karne se convergence ki interval same rehti hai.

False — radius preserve hoti hai lekin ek endpoint lost ho sakta hai; e.g. ek series jo

x = c + R

par convergent hai woh differentiate karne ke baad wahan diverge ho sakti hai.

TF3. Ek power series ko integrate karne se sirf wo set shrink ya preserve ho sakti hai jahan woh

x

ke liye converge karta hai.

False — integration coefficients ko

n + 1

se divide karta hai, jo ek endpoint par convergence gain kar sakta hai (exactly isi tarah

arctan

x = 1

milta hai).

TF4. Kyunki ek power series ek infinite polynomial ke barabar hai, tum

\frac{d}{d x}

aur

\sum

ko bina kisi condition ke hamesha swap kar sakte ho.

False — swap tabhi allowed hai kyunki derivative series

(c - R, c + R)

ke compact subintervals par uniformly converge karti hai; uniform convergence ke bina general function series ke liye swap fail ho sakta hai.

TF5. Agar

\sum a_{n} x^{n}

sirf

x = 0

par converge karta hai (to

R = 0

), term-by-term theorem ab bhi bata dega ki ise kaise differentiate karein.

False — theorem ke liye

R > 0

chahiye;

R = 0

ke saath koi open interval nahi hai differentiate karne ke liye, isliye kuch claim nahi kiya jaata.

TF6. Differentiated series

\sum_{n \geq 1} n a_{n} (x - c)^{n - 1}

aur original series hamesha ek given interior point par same value deti hai.

False — ye

f^{'}

aur

f

represent karte hain, jo generally alag functions hain; sirf convergence ki radius match karti hai, sums nahi.

TF7. Ek function jo

c

ke paas

C^{\infty}

(infinitely differentiable) hai use wahan apni Taylor series ke barabar hona chahiye.

False — smoothness se tumhe saare coefficients milte hain, lekin series kisi alag function par converge kar sakti hai (e.g.

e^{- 1/ x^{2}}

ke saare Taylor coefficients zero hain phir bhi woh nonzero hai).

TF8.

\sum a_{n} (x - c)^{n}

ka term-by-term integration same centre

c

ke saath ek genuine power series produce karta hai.

True —

C + \sum \frac{a _{n}}{n + 1} (x - c)^{n + 1}

ek power series hai

c

ke baare mein, aur iska radius original

R

ke barabar hai.

TF9. Tum ek convergent power series ko

(c - R, c + R)

ke andar jitni baar chahein term by term differentiate kar sakte ho.

True — har differentiation same radius

R > 0

ki ek power series return karta hai, to process hamesha repeat ho sakta hai; isliye power-series sums open interval par

C^{\infty}

hote hain.

TF10. Kyunki geometric series

\sum x^{n}

x = 1

par diverge karti hai, iska term-by-term derivative

\sum n x^{n - 1}

bhi

x = 1

par diverge karta hai.

True — differentiate karne se endpoint convergence kabhi create nahi hoti; terms ko

n

se multiply karne se

x = 1

aur bura ho jaata hai, aur indeed

\sum n

diverge karta hai.

Error dhundho

SE1. "

ln (1 + x) = \sum_{n \geq 0} \frac{( - 1 ) ^{n}}{n + 1} x^{n + 1}

sabhi

x

ke liye hold karta hai, kyunki humne ek series integrate ki."

Galat — integration radius

R = 1

preserve karta hai, isliye identity sirf

(- 1, 1]

par hold karti hai;

x = - 1

par series

- \sum \frac{1}{n + 1}

hai (harmonic, divergent) aur

∣ x ∣ > 1

se pare fail ho jaati hai.

SE2. "Cauchy–Hadamard limit deta hai

\frac{1}{R ^{'}} = lim sup ∣ n a_{n} ∣^{1/ n} = lim sup (n^{1/ n} ∣ a_{n} ∣^{1/ n})

, aur kyunki

n^{1/ n} \to \infty

, radius zaroor shrink hogi."

Error hai

n^{1/ n} \to \infty

; actually

n^{1/ n} \to 1

(kyunki

\frac{l n n}{n} \to 0

), isliye factor drop out ho jaata hai aur

R^{'} = R

rehta hai.

SE3. "

\sum a_{n} x^{n}

integrate karne ke liye main likhta hoon

\sum \frac{a _{n}}{n} x^{n + 1}

Divisor

n + 1

hai,

n

nahi — power ko

n + 1

tak raise karo aur us naye exponent se divide karo;

n

se divide karne par

n = 0

term blow up ho jaata.

SE4. "

\sum_{n \geq 0} a_{n} x^{n}

differentiate karne par

\sum_{n \geq 0} n a_{n} x^{n - 1}

milta hai."

Index

n = 1

se start hona chahiye —

n = 0

ka constant term differentiate hokar

0

ban jaata hai;

n = 0

rakhne par

0 \cdot a_{0} x^{- 1}

likh jaata hai jo meaningless hai (

x^{- 1}

ek power-series term nahi hai).

SE5. "Kyunki term-by-term differentiation 'bilkul polynomials ki tarah' hai, ye kisi bhi infinite sum of functions

\sum f_{n} (x)

ke liye kaam karta hai."

Sirf special families (power series, aur more generally uniformly convergent derivative series) qualify karti hain; ek general

\sum f_{n}

converge kar sakta hai jabki

\sum f_{n}^{'}

diverge kare ya galat cheez par converge kare.

SE6. "Abel endpoint value hume bataati hai ki derivative series bhi us endpoint par converge karti hai."

Nahi — Abel's theorem us endpoint tak sum ki continuity address karta hai jahan series converge karti hai; woh differentiated series ko wahan converge karne ke liye force karne ke baare mein kuch nahi kehta.

SE7. "Kyunki

\frac{1}{1 + x ^{2}} = \sum (- 1)^{n} x^{2 n}

∣ x ∣ < 1

ke liye hai, integrate karne par

arctan x

milta hai jo sabhi real

x

ke liye valid hai kyunki

arctan

everywhere defined hai."

Function

arctan

sabhi

x

ke liye defined hai, lekin iska power-series representation sirf

∣ x ∣ \leq 1

(

R = 1

) ke liye converge karta hai; dono ko confuse nahi karna chahiye.

Why questions

WHY1. Har coefficient ko

n

se multiply karne par bhi radius kyun unchanged rehti hai, jabki har term bada ho jaata hai?

Convergence geometric hai,

∣ a_{n} ∣^{1/ n} ∣ x - c ∣

se control hoti hai; extra factor

n^{1/ n} \to 1

contribute karta hai, jo

n

-th root mein invisible hai, isliye woh sirf ek endpoint affect kar sakta hai, geometric decay rate kabhi nahi.

WHY2.

n^{1/ n} \to 1

prove karne ke liye hum logarithms kyun lete hain?

\infty^{0}

form ke ek expression ko logs se control kiya jaata hai:

ln (n^{1/ n}) = \frac{l n n}{n} \to 0

kyunki logarithm

n

ki kisi bhi positive power se haar jaata hai, aur phir

n^{1/ n} = e^{l n n / n} \to e^{0} = 1

WHY3. Uniform convergence — sirf pointwise nahi — woh property kyun hai jo

\frac{d}{d x}

aur

\sum

swap karne ki permission deti hai?

Derivative series ki uniform convergence limit aur derivative operation ko exchange karne deti hai bina error ke

x

ke across unevenly accumulate hue; pointwise convergence akela us exchange ki pathological failures allow karta hai.

WHY4. Differentiate ya integrate karne ke baad hume endpoints hamesha re-test kyun karni chahiye?

Theorem sirf open-interval behaviour aur radius guarantee karta hai; boundary points

x = c \pm R

ek alag convergence question hai jise theorem deliberately open chodta hai (integration unhe gain kar sakta hai, differentiation unhe lose kar sakta hai).

WHY5. Geometric series

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

ko integrate karne par

x = 1

par convergence kyun gain hoti hai?

Integration coefficients ko

2 n + 1

se divide karta hai,

(- 1)^{n}

\frac{( - 1 ) ^{n}}{2 n + 1}

mein badal deta hai — ek alternating series jiske terms

0

tak decrease karte hain, to Leibniz's test ab convergence deta hai jahan original diverge karta tha.

WHY6. Ek power series ko apni radius ke andar infinitely many times differentiate kyun kiya ja sakta hai, jo iske sum ko

C^{\infty}

banata hai?

Har differentiation same radius

R > 0

ki ek aur power series deta hai; kyunki yeh indefinitely same open interval par repeat ho sakta hai, saare derivatives wahan exist karte hain.

WHY7. Integration ka constant centre par evaluate karke kyun fix hota hai (e.g.

x = 0

Antiderivative series ka value centre par

C

hota hai jahan saare higher terms vanish ho jaate hain, isliye target function ke known value se match karna (jaise

arctan 0 = 0

ln 1 = 0

)

C

uniquely fix kar deta hai.

Edge cases

EC1. Jab

R = \infty

ho (e.g.

e^{x}

ki series) to theorem kya kehta hai?

Term-by-term operations sabhi real

x

ke liye valid hain aur sabhi

x

ke liye valid rehti hain — ek infinite radius infinite rehti hai, koi endpoints nahi sochne ki.

EC2. Centre

x = c

par khud kya hota hai, degenerate "interval" case mein?

Har power series

x = c

par trivially converge karta hai (all terms after

a_{0}

vanish ho jaate hain), aur yeh point hamesha

(c - R, c + R)

ke interior mein hota hai jab bhi

R > 0

ho, isliye kuch special nahi chahiye.

EC3. Agar original series dono endpoints par converge karti hai, to kya integrated series dono par converge karne ki guarantee hai?

Rule ke roop mein guarantee nahi — integration endpoints ko help karta hai (

n + 1

se divide karte hue), isliye woh typically wahan convergence preserve ya gain karta hai, lekin har endpoint alag se check karna padta hai.

EC4. Agar

a_{0} \neq = 0

hai lekin tum sirf derivative chahte ho — kya constant term matter karta hai?

Nahi —

a_{0}

differentiate hokar

0

ban jaata hai aur simply drop out ho jaata hai; isliye derivative series apna index

n = 1

se shuru karta hai.

EC5. Ek aisi series consider karo jiska radius

R = 0

hai (sirf centre par converge karti hai). Kya tum ise term by term meaningfully integrate kar sakte ho?

Koi useful function nahi milta — koi open interval of convergence nahi hone par koi interval nahi hai jis par antiderivative series kuch represent kare, isliye theorem apply nahi hota.

EC6. Ek endpoint par jahan differentiated series diverge karta hai, kya original function wahan ab bhi differentiable hai?

Theorem silent hai — yeh sirf open interval par differentiability assert karta hai; boundary par

f

ka behaviour alag one-sided analysis chahta hai aur term-by-term rule se deliver nahi hota.

Recall Ek-line summary jo tumhe recite karni chahiye

Open interval ke andar tum radius preserve karte hue term by term differentiate aur integrate kar sakte ho; sirf endpoints up for grabs hain — differentiation unhe lose kar sakta hai, integration unhe gain kar sakta hai. DIRT: Differentiate/Integrate keep the Radius, Test endpoints again.

Connections

Power series and radius of convergence
Cauchy–Hadamard theorem
Uniform convergence and the Weierstrass M-test
Taylor and Maclaurin series
Geometric series
Abel's theorem (endpoint behaviour)
Solving ODEs with power series