Term-by-term differentiation and integration of power series: Visual walkthrough

4.3.15 · D2 · Maths › Calculus III — Sequences & Series › Term-by-term differentiation and integration of power series

Step 0 — Power series kya hoti hai, pictures mein?

KYA HAI. Ek power series ek endless sum hoti hai $f (x) = \sum_{n = 0}^{\infty} a_{n} (x - c)^{n} = a_{0} + a_{1} (x - c) + a_{2} (x - c)^{2} + \dots$

Main har piece ko, bilkul wahin, naam deta hoon:

$a_{n}$ ::: the coefficient — har step $n$ ke liye ek fixed number (us term ka "weight").
$c$ ::: the centre — woh point jiske aaspaas hum build karte hain.
$(x - c)^{n}$ ::: centre se doori, ek power tak uthaya gaya — jab $x$ , $c$ se door hota hai toh yeh fast badhta hai.
$\sum_{n = 0}^{\infty}$ ::: "inhe hamesha ke liye jodते raho, $n = 0$ se shuru karke".

"Radius" ki zaroorat kyun hai. Agar $x$ , $c$ ke kareeb hai, toh powers $(x - c)^{n}$ shrink hoti hain aur sum ek number pe settle ho jaata hai. Agar $x$ door hai, toh powers explode ho jaati hain aur sum blow up ho jaata hai. "Settles" aur "explodes" ke beech ki border centre se ek distance $R$ hai, jise radius of convergence kehte hain. Us distance ke andar series ek real function hai; bahar yeh kisi kaam ki nahi.

PICTURE. Ek number line jisme centre $c$ beech mein hai aur dono taraf half-width $R$ ka ek red "safe zone" hai.

Figure — Term-by-term differentiation and integration of power series

Step 1 — Differentiate karne se har term ka kya hota hai?

KYA HAI. Ek term $a_{n} (x - c)^{n}$ pe $\frac{d}{d x}$ (slope-finder) apply karo. Power rule kehta hai: power ko neeche front mein le aao, phir power ko ek se kam karo. $\frac{d}{d x} [a_{n} (x - c)^{n}] = n a_{n} (x - c)^{n - 1}$

Term by term, kya badla:

the factor $n$ ::: naya, exponent se neeche khींcha gaya — yahi poori kahani hai.
$(x - c)^{n - 1}$ ::: power ek se gir gayi.
the $n = 0$ term ( $a_{0}$ , a constant) ::: differentiate hokar $0$ ban jaata hai aur gayab ho jaata hai, isliye sum ab $n = 1$ se shuru hota hai.

Yeh tool kyun. Hum $f$ ki slope chahte hain. Ek ordinary chhoti polynomial ke liye power rule woh machine hai jo slopes deta hai. Is chapter ka claim hai ki endless polynomial bhi usi rule ko follow karti hai — toh differentiated series hai $f^{'} (x) = \sum_{n = 1}^{\infty} n a_{n} (x - c)^{n - 1} .$

PICTURE. Coefficients ka ek tower; differentiation $n$ -th brick ko $n$ se multiply karta hai aur tower ko ek step neeche shift kar deta hai.

Step 2 — Integrate karne se har term ka kya hota hai?

KYA HAI. Ek term pe $\int \cdot d x$ (area-finder) apply karo. Rule hai: power ko ek se badhao, phir nayi power se divide karo. $\int a_{n} (x - c)^{n} d x = \frac{a _{n}}{n + 1} (x - c)^{n + 1} + C$

Term by term:

the divisor $n + 1$ ::: naya — integrate karne se coefficients chhote ho jaate hain.
$(x - c)^{n + 1}$ ::: power ek se badh gayi.
$C$ ::: poore sum ke liye ek akela constant (woh height jahan se antiderivative shuru hoti hai).

Toh integrated series hai $\int f (x) d x = C + \sum_{n = 0}^{\infty} \frac{a _{n}}{n + 1} (x - c)^{n + 1} .$

KYUN. Wahi wajah: ek chhoti polynomial ke liye yahi antiderivatives ka the rule hai, aur hum claim karte hain ki endless wali bhi saath chalti hai.

PICTURE. Wohi coefficient tower, lekin ab $n$ -th brick $n + 1$ se divide ho rahi hai aur tower ek step upar shift ho rahi hai — Step 1 ki mirror image.

Step 3 — Woh machine jo $R$ measure karti hai

KYA HAI. Cauchy–Hadamard theorem radius seedha coefficients se deta hai: $\frac{1}{R} = lim sup_{n \to \infty} ∣ a_{n} ∣^{1/ n} .$

Har symbol decode karte hain:

$∣ a_{n} ∣^{1/ n}$ ::: $n$ -th coefficient ki size ka $n$ -th root. Yeh poochhta hai: "roughly, kaunsa constant base $b$ hai jo $∣ a_{n} ∣ \approx b^{n}$ banata hai?"
$lim sup_{n \to \infty}$ ::: un roots ka "eventual ceiling" jab $n$ badhta hai — sabse bada value jiske paas woh barabar aate rehte hain.
$\frac{1}{R}$ ::: radius ka ulta. Bada root ⇒ bada $1/ R$ ⇒ chhota safe zone. Chhota root ⇒ bada safe zone.

Yeh tool kyun aur koi kyun nahi. Geometric series bilkul tab converge karti hai jab uska base size $1$ se neeche ho. Cauchy–Hadamard kehta hai ek power series ek geometric series ki tarah behave karti hai jiski base $∣ a_{n} ∣^{1/ n} ∣ x - c ∣$ hai. Toh radius coefficients ke root behaviour se set hoti hai — aur kisi cheez se nahi. Yahi woh tool hai jo "kya yeh settle hogi?" ko ek single number mein badal deta hai.

PICTURE. $∣ a_{n} ∣^{1/ n}$ roots dots ki tarah plot kiye gaye; ek horizontal red line us ceiling ko mark karti hai jiske neeche woh hover karte hain. Woh ceiling $1/ R$ hai.

Step 4 — Woh ek limit jo poora engine chalati hai

KYA HAI. Hume yeh ek single fact chahiye: $lim_{n \to \infty} n^{1/ n} = 1.$ Shabdon mein: $n$ ka $n$ -th root $1$ ki taraf march karta hai.

Humein yeh kyun chahiye. Differentiation ne har coefficient mein ek extra factor $n$ daal diya. Cauchy–Hadamard ko $∣ a_{n} ∣^{1/ n}$ ki parwah hai. Toh extra factor root mein extra $n^{1/ n}$ contribute karta hai. Agar woh extra factor $1$ ki taraf tend karta hai, toh yeh ceiling ko $1$ se multiply karta hai — yaani kuch bhi nahi badalta. Yahi limit woh poori wajah hai ki radius survive karti hai.

Yeh sach kyun hai (logs lo). " $\infty^{0}$ " shape ki koi quantity — yahan ek badhta $n$ jo shrinking power $1/ n$ tak raised hai — ek tug-of-war hai. Logarithms power ko ek product mein badal dete hain jahan hum dekh sakte hain kaun jeetta hai: $ln (n^{1/ n}) = \frac{1}{n} ln n = \frac{l n n}{n} .$

$ln n$ ::: badhta hai, lekin dheeray (logarithm rengta hai).
$n$ ::: seedhi line mein badhta hai. Jab $n \to \infty$ , straight-line denominator crawling numerator ko crush kar deta hai, isliye $\frac{ln n}{n} \to 0$ . Log undo karo: $n^{1/ n} = e^{l n n / n} \to e^{0} = 1.$

PICTURE. Do racing curves: $ln n$ (crawling) versus $n$ (sprinting). Unka ratio, red mein, $0$ ki taraf sink karta hai — $n^{1/ n}$ ko $1$ ki taraf kheenchta hai.

Step 5 — Sab jodo: differentiated radius $R$ ke barabar hai

KYA HAI. Differentiated series ke coefficients $b_{n} = (n + 1) a_{n + 1}$ hain (re-index karo taaki dono series $n = 0$ se shuru hon). Uski radius $R^{'}$ satisfy karti hai $\frac{1}{R ^{'}} = lim sup_{n \to \infty} (n + 1) a_{n + 1}^{1/ n} .$

Root ko do factors mein split karo — yahi payoff hai: $(n + 1) a_{n + 1}^{1/ n} = \to 1 (Step 4) (n + 1)^{1/ n} \cdot same ceiling as ∣ a_{n} ∣^{1/ n} ∣ a_{n + 1} ∣^{1/ n} .$

$(n + 1)^{1/ n} \to 1$ ::: $n^{1/ n}$ waali wajah se; root ke andar ek se shift karna jo $1$ tak jaata hai, ab bhi $1$ tak jaata hai.
$∣ a_{n + 1} ∣^{1/ n}$ ::: coefficients ka apna root, lekin yahan ek subtlety chupi hai — neeche dono callouts padho.

Differentiation factors multiply karo: $\frac{1}{R ^{'}} = 1 \cdot \frac{1}{R} = \frac{1}{R}$ , isliye $R^{'} = R$ . Integration case (Subtlety B) bhi reciprocal factor se usi $R$ pe land karta hai. Safe zone dono cases mein untouched hai. ∎

PICTURE. Step 3 ka coefficient-root ceiling, jisme extra factor ek red multiplier ki tarah draw kiya gaya hai jo $1$ tak shrink ho jaata hai, ceiling ko — aur isliye $R$ ko — exactly wahin chhod deta hai.

Step 5½ — Hum $\sum$ aur $\frac{d}{d x}$ swap karne ke allowed kyun hain?

KYA HAI. Ab tak sab kuch yeh assume karta raha ki hum endless sum ko ek term at a time differentiate kar sakte hain: $\frac{d}{d x} \sum_{n} a_{n} (x - c)^{n} = ? \sum_{n} \frac{d}{d x} [a_{n} (x - c)^{n}] .$ Yeh swap functions ki ek general infinite sum ke liye free nahi hai — tum aisi sums bana sakte ho jahan yeh fail ho jaaye. Toh humein power series ke liye yeh earn karna hoga.

Yeh yahan kyun hold karta hai. License hai uniform convergence. Analysis ka theorem hai: agar differentiable functions ki ek series ek point pe converge karti hai, aur unke derivatives ki series uniformly converge karti hai, toh tum term by term differentiate kar sakte ho. "Uniformly" ka matlab hai ki sum ki tail ko ek bound se chhota banaya ja sakta hai jo ek saath har $x$ ke liye kaam kare — pieces lock-step mein settle hote hain, alag $x$ ke liye alag speeds pe nahi.

Power series test kyun pass karti hai. Koi bhi closed sub-interval $∣ x - c ∣ \leq ρ$ fix karo jahan $ρ < R$ ho — strictly safe zone ke andar. Uspe, har derivative term bounded hai: $n a_{n} (x - c)^{n - 1} \leq = M_{n} (isme koi x nahi!) n ∣ a_{n} ∣ ρ^{n - 1} .$

$M_{n}$ ::: har $n$ ke liye ek pure number — sub-interval mein kahin bhi woh term sabse zyada ho sakti hai.

$\sum M_{n} < \infty$ kyun hai (geometric series se comparison). $ρ$ aur $R$ ke beech strictly ek number $σ$ chuno, yaani $ρ < σ < R$ . Kyunki $σ < R$ , differentiated series ki root ceiling (Step 5) deti hai $∣ a_{n} ∣^{1/ n} < \frac{1}{σ}$ saare bade $n$ ke liye, yaani $∣ a_{n} ∣ σ^{n} < 1$ , toh $∣ a_{n} ∣ < σ^{- n}$ . Substitute karo: $M_{n} = n ∣ a_{n} ∣ ρ^{n - 1} < \frac{n}{ρ} (\frac{ρ}{σ})^{n} .$ Ab $r := \frac{ρ}{σ} < 1$ hai, toh $\sum_{n} n r^{n}$ ek convergent series hai (ek geometric series $\sum r^{n}$ polynomial factor $n$ se nudged — aur $\sum n r^{n} = \frac{r}{( 1 - r ) ^{2}} < \infty$ for $∣ r ∣ < 1$ ; factor $n$ geometric decay se crush ho jaata hai). Isliye $\sum M_{n} < \infty$ : ek finite bound.

Weierstrass $M$-test phir kehta hai: ek single finite bound $\sum M_{n} < \infty$ derivative series ko $∣ x - c ∣ \leq ρ$ pe uniformly converge karne pe force karta hai. Yahi woh hypothesis thi jo swap-theorem ko chahiye thi. Kyunki $(c - R, c + R)$ ke strictly andar ka har point kisi na kisi $ρ$ -interval mein rehta hai, swap safe zone ke andar har jagah legal hai.

Integration swap aasaan hai: original series ka $∣ x - c ∣ \leq ρ$ pe uniform convergence (wohi $M$ -test, ab $M_{n} = ∣ a_{n} ∣ ρ^{n} < r^{n}$ , ek plain convergent geometric series ke saath) pehle se $\int$ aur $\sum$ swap karne ka license de deta hai. Toh Step 5 ka radius result ek bonus nahi hai — yahi woh cheez hai jo $M$ -test ko feed karti hai aur term-by-term calculus ko honest banati hai.

Step 6 — Edge cases: endpoints, aur extremes $R = 0$ , $R = \infty$

KYA HAI. Upar sab kuch open safe zone $∣ x - c ∣ < R$ ke baare mein tha. Teen edge situations baaki hain: do boundary points, aur do extreme radii.

Endpoints badal sakte hain. Points $x = c - R$ aur $x = c + R$ ek alag kahani hain. Wahan geometric protection gone hai (base $= 1$ ), toh leftover factor $n$ ya $\frac{1}{n + 1}$ suddenly matter karne lagta hai.

$n = 1 \sum \infty n x^{n - 1}$ at $x = 1$ is $1 + 2 + 3 + \dots$ ::: diverges (differentiation ne endpoint kho diya).
The $arctan$ build $\sum \frac{( - 1 ) ^{n}}{2 n + 1}$ at $x = 1$ converges (alternating) ::: integration ne ek endpoint gain kiya.

Yeh Abel's theorem (endpoint behaviour) territory hai, aur isliye parent ka mnemonic DIRT "Test endpoints again" pe khatam hota hai.

Extreme $R = 0$ survive karta hai. Agar original series sirf $x = c$ pe converge karti hai, toh $lim sup ∣ a_{n} ∣^{1/ n} = \infty$ . Differentiated ceiling hai $lim sup (n + 1)^{1/ n} ∣ a_{n + 1} ∣^{1/ n} = 1 \cdot \infty = \infty$ , toh ab bhi $R^{'} = 0$ . Ek series jo $c$ ke bahar kahin converge nahi karti, differentiate ya integrate karne ke baad bhi waisi hi rehti hai. (Integrating $n + 1$ se divide karta hai, lekin $1 \cdot \infty = \infty$ bhi hai — ek melting factor infinite ceiling cancel nahi kar sakta.)

Extreme $R = \infty$ survive karta hai. Agar original series entire hai ( $lim sup ∣ a_{n} ∣^{1/ n} = 0$ , jaise $e^{x} = \sum x^{n} / n!$ ), toh differentiated ceiling hai $lim sup (n + 1)^{1/ n} ∣ a_{n + 1} ∣^{1/ n} = 1 \cdot 0 = 0$ , toh $\frac{1}{R ^{'}} = 0$ aur $R^{'} = \infty$ : differentiated series ab bhi everywhere converge karti hai. Integration case identical hai — factor $(n + 1)^{- 1/ n} \to 1$ times ceiling $0$ phir $0$ hai, toh antiderivative bhi entire hai. Entire series dono operations ke under entire rehti hain — exactly $\frac{d}{d x} e^{x} = e^{x}$ aur $\int e^{x} d x = e^{x} + C$ ke consistent, dono har $x$ ke liye convergent.

PICTURE. Radius line finite $R$ ke liye unchanged, dono red endpoint dots apni status flip karte hue: ek ✓→✗ ho jaata hai (differentiation), doosra ✗→✓ (integration).

Ek-picture summary

Yahan poori derivation compressed hai: coefficient roots ceiling $1/ R$ set karte hain; differentiation aur integration ek aisa factor lagate hain jo $n$ -th root $1$ tak drive karta hai; ceiling — aur radius — firm rehti hai; sirf do endpoints dolta dalta karte hain.

Recall Feynman: poori kahani ek 12-saal ke bache ko batao

Socho ek endless polynomial ek "safe fence" ke andar rehta hai — apne home base ke left aur right $R$ ki doori. Fence ke andar yeh ek real function tak add up hota hai; bahar, explode ho jaata hai.

Ab slope-ing karo (differentiate karo): har brick apne step-number $n$ se multiply ho jaati hai. Yeh scary lagta hai — badi bricks! Lekin fence test karne ke liye hum har brick ka $n$ -th root lete hain, aur $n$ ka $n$ -th root rengta hua sirf $1$ tak aa jaata hai. $1$ se multiply karne se kuch nahi badalta, toh fence nahi hilti. Agar fence zero-width thi (sirf home base kaam karta tha) toh zero rehti hai; agar infinitely wide thi (har jagah kaam karti thi) toh infinite rehti hai.

Area-ing (integrate karo) ulta karta hai — bricks ko $n + 1$ se divide karta hai — aur uska $n$ -th root bhi $1$ tak jaata hai. Woh akela "+C" jo tum add karte ho woh sirf ek height hai, brick nahi, toh fence ko kabhi touch nahi karta. Fence phir unmoved.

Lekin har-piece-ko-slope-ing karna allowed kyun hai? Kyunki thodi chhoti fence ke andar, saare pieces lock-step mein saath settle ho jaate hain — ek single yardstick poori tail control karta hai (yahi uniform convergence / $M$ -test hai, jo humne shrinking geometric series se compare karke secure kiya). Sirf fence ke upar, do endpoints par, cheezein dolt karti hain. Slope-ing ek endpoint knock out kar sakta hai; area-ing ek wapas pull kar sakta hai. Toh: radius hamesha same, endpoints hamesha re-check. Yahi DIRT hai.

Recall Exactly

$n^{1/ n} \to 1$ kyun? Logs lo: $ln (n^{1/ n}) = \frac{l n n}{n}$ . Straight-line $n$ , crawling $ln n$ ko beat karta hai, toh ratio $\to 0$ hai, toh $n^{1/ n} = e^{0} = 1$ . ::: Yeh single limit woh engine hai jo radius ko fixed rakhta hai.

Connections

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

Visual walkthrough — Term-by-term differentiation and integration of power series