Term-by-term differentiation and integration of power series: Worked examples

4.3.15 · D3 · Maths › Calculus III — Sequences & Series › Term-by-term differentiation and integration of power series

Kuch bhi compute karne se pehle, main char tools ke naam rakh deta hoon taaki koi bhi symbol bina wajah na aaye.

Definition Do alag endpoint tools — inhe alag rakho

Jab ek series kisi endpoint par land karti hai, do alag facts kaam karte hain, aur students inhe aksar mix kar dete hain:

Alternating series test (AST): ek convergence test hai. Agar terms alternate in sign hain aur unka size monotonically $0$ ki taraf shrink karta hai, toh series converges karti hai. Yeh batata hai ki sum koi finite number hai — lekin kaun sa number nahi batata.
Abel's theorem: ek identification tool hai. Yeh maan ke chalo ki series endpoint par pehle se converge kar rahi hai (AST yeh supply karta hai), Abel kehta hai ki uska value function ki limit ke barabar hai jab $x$ andar se endpoint ki taraf approach karta hai. Yeh batata hai ki woh number kaun sa hai.

Toh endpoint par workflow hamesha do steps ka hota hai: (1) AST — kya yeh converge karta hai? (2) Abel — agar haan, toh yeh kya equals karta hai?

Scenario matrix

Neeche har example tagged hai un cell(s) ke saath jo woh cover karta hai. Goal: koi cell khaali na rahe. Note karo two-sided rows: har "endpoint" cell check hoti hai dono $x = c + R$ aur $x = c - R$ par.

Cell	Jo scenario yeh stress-test karta hai	Covered by
A	Differentiate — sab coefficients positive	Ex 1
B	Integrate — series gains the right endpoint $x = + 1$	Ex 2
B′	Integrate — the left endpoint $x = - 1$ (opposite sign)	Ex 2, step 4
C	Integrate — series only conditionally convergent at endpoint	Ex 3
C′	Integrate — the left endpoint $x = - 1$ diverges	Ex 3, step 4
D	Differentiate — series loses an endpoint (dono sides)	Ex 4
E	Shifted centre $c \neq = 0$ (zero par centred nahi)	Ex 5
F	Degenerate / limiting: woh series jo khud ko return karti hai	Ex 6
G	Real-world word problem (physics)	Ex 7
H	Exam twist: ek unknown series ko derivative pehchan ke sum karo	Ex 8

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

Upar wali figure map hai: horizontal line $x$ -axis hai, shaded band safe zone $(c - R, c + R)$ hai, aur do dots endpoints hain — wahi jagah jahan differentiate ya integrate karna verdict flip kar sakta hai. Dono dots ko independently test karna hoga.

Cell A — ek positive series differentiate karo

Worked example Example 1 —

$\frac{1}{1 - x}$ se $\frac{1}{( 1 - x ) ^{2}}$ tak aur aage Geometric series $\frac{1}{1 - x} = n = 0 \sum \infty x^{n}$ se shuru karo, $∣ x ∣ < 1$ ke liye. $n = 1 \sum \infty n^{2} x^{n}$ ka closed form nikalo.

Forecast: padhne se pehle guess karo — ek baar differentiate karne se $\frac{1}{( 1 - x ) ^{2}}$ mila; doosri differentiation (multiply-by- $x$ trick ke saath) ek rational function deni chahiye jiske denominator mein cube ho. Apna guess likhkaro.

Ek baar differentiate karo. $\frac{d}{d x} \frac{1}{1 - x} = \frac{1}{( 1 - x ) ^{2}} = \sum_{n = 1}^{\infty} n x^{n - 1}$ . Yeh step kyun? Theorem har term $x^{n}$ ko $n x^{n - 1}$ banane deta hai; $n = 0$ term ek constant hai isliye woh vanish ho jaati hai, yahi wajah hai ki sum ab $n = 1$ se shuru hota hai.
$x$ se multiply karo taaki powers wapas line up ho jayein: $\frac{x}{( 1 - x ) ^{2}} = \sum_{n = 1}^{\infty} n x^{n}$ . Yeh step kyun? Humein eventually $\sum n^{2} x^{n}$ chahiye; exponent ko $n$ (na ki $n - 1$ ) ke barabar rakhne ka matlab hai ki अगली differentiation humein ek aur factor of $n$ cleanly de degi.
Phir se differentiate karo. $\frac{d}{d x} \frac{x}{( 1 - x ) ^{2}} = \frac{1 + x}{( 1 - x ) ^{3}} = \sum_{n = 1}^{\infty} n^{2} x^{n - 1}$ . Yeh step kyun? $n x^{n}$ differentiate karne se $n \cdot n x^{n - 1} = n^{2} x^{n - 1}$ milta hai — woh doosra factor of $n$ jo humein chahiye tha.
Ek baar phir $x$ se multiply karo. $n = 1 \sum \infty n^{2} x^{n} = \frac{x ( 1 + x )}{( 1 - x ) ^{3}}$ .

Verify: $x = \frac{1}{2}$ daalo. Closed form: $\frac{\frac{1}{2} \cdot \frac{3}{2}}{( \frac{1}{2} ) ^{3}} = \frac{3/4}{1/8} = 6$ . Partial sum $\sum_{n = 1}^{6} n^{2} (\frac{1}{2})^{n} = 0.5 + 1 + 1.125 + 1 + 0.78125 + 0.5625 = 4.968 \dots$ , $6$ ki taraf badh raha hai. ✓ Radius $R = 1$ par unchanged (har differentiation ne ise rakha — radius-survival).

Cell B / B′ — integration gains an endpoint (dono sides check hoti hain)

Worked example Example 2 —

$arctan$ series aur Leibniz ka $π$ $arctan x$ ke liye series derive karo aur dikhao ki dono endpoints $x = 1$ aur $x = - 1$ par kya hota hai.

Forecast: base series $\frac{1}{1 + x ^{2}}$ $x = \pm 1$ par diverge karti hai (terms $\pm 1$ kabhi shrink nahi hote). Integrate karne ke baad, kya aap expect karte ho ki $x = \pm 1$ phir bhi fail karein, ya achanak kaam karne lagein? Guess karo.

Geometric pattern se match karo. $\frac{1}{1 + x ^{2}} = \frac{1}{1 - ( - x ^{2} )} = \sum_{n = 0}^{\infty} (- x^{2})^{n} = \sum_{n = 0}^{\infty} (- 1)^{n} x^{2 n}$ , valid jab $∣ - x^{2} ∣ < 1$ , yaani $∣ x ∣ < 1$ . Yeh step kyun? Geometric series ke " $x$ " ki jagah $- x^{2}$ substitute karna $\frac{1}{1 + x ^{2}}$ ke liye series paane ka sabse sasta tarika hai — abhi koi calculus ki zaroorat nahi.
$0$ se $x$ tak integrate karo, term by term: $arctan x = \int_{0}^{x} \frac{d t}{1 + t ^{2}} = n = 0 \sum \infty \frac{( - 1 ) ^{n}}{2 n + 1} x^{2 n + 1}$ . Yeh step kyun? $\int_{0}^{x} t^{2 n} d t = \frac{x ^{2 n + 1}}{2 n + 1}$ , aur lower limit $0$ constant ko auto-fix kar deta hai kyunki $arctan 0 = 0$ .
Right endpoint $x = 1$ — do-step. (1) AST: terms ban jaate hain $1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots$ , jo alternate karte hain aur $0$ ki taraf shrink karte hain, toh alternating series test se series converges. (2) Abel: kyunki yeh converges karta hai, Abel's theorem sum ko identify karta hai $lim_{x \to 1^{-}} arctan x = arctan 1 = \frac{π}{4}$ ke roop mein. Yeh step kyun? Dono tools ko alag rakho: AST convergence establish karta hai, Abel value pin karta hai. $2 n + 1$ se divide karne ne terms ko itna shrink kar diya — integrate karne se right endpoint gain hua.
Left endpoint $x = - 1$ — opposite sign, alag check karo. $x = - 1$ daalo: odd power $x^{2 n + 1} = - 1$ deta hai $\sum \frac{( - 1 ) ^{n} ( - 1 )}{2 n + 1} = - (1 - \frac{1}{3} + \frac{1}{5} - \dots)$ . (1) AST: same alternating, shrinking terms → converges. (2) Abel: sum $= arctan (- 1) = - \frac{π}{4}$ . Yeh step kyun? Theorem demand karta hai ki har boundary ko apne aap test kiya jaaye. Yahan $arctan$ odd hai, isliye left endpoint right ka mirror hai — lekin yeh aap tabhi jaante ho jab check karte ho, assume karke nahi.

Verify: $4 \sum_{n = 0}^{5000} \frac{( - 1 ) ^{n}}{2 n + 1} \approx 3.1414$ , $π \approx 3.14159$ ki taraf aa raha hai (dheere, oscillate karte hue). Aur $x = - 1$ par: $\sum_{n = 0}^{5000} \frac{( - 1 ) ^{n} ( - 1 )}{2 n + 1} \approx - 0.7854 \approx - \frac{π}{4}$ . ✓ Radius ab bhi $R = 1$ .

Figure dikhata hai kyun endpoints gain hote hain: base series ke terms (yellow) hamesha height $1$ par baithe hain, lekin integrate karne ke baad, naye terms (blue) $\frac{1}{2 n + 1} \to 0$ ki tarah decay karte hain — same picture $x = 1$ aur $x = - 1$ dono par.

Cell C / C′ — integration se ek conditionally convergent endpoint (dono sides)

Worked example Example 3 —

$ln (1 + x)$ aur alternating harmonic series $\frac{1}{1 + x} = \sum_{n = 0}^{\infty} (- 1)^{n} x^{n}$ se, $ln (1 + x)$ banao aur dono endpoints $x = 1$ aur $x = - 1$ par evaluate karo.

Forecast: kya endpoint sums absolutely converge karenge (freely rearrange kar sako) ya sirf conditionally (order matters)? Aur kya dono sides same behave karengi? Steps 3–4 se pehle decide karo.

Base series ko integrate karo $0 \to x$ : $ln (1 + x) = \sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n}}{n + 1} x^{n + 1} = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots$ . Yeh step kyun? $\int_{0}^{x} t^{n} d t = \frac{x ^{n + 1}}{n + 1}$ ; constant $0$ hai kyunki $ln (1 + 0) = ln 1 = 0$ .
Safe zone $∣ x ∣ < 1$ hai, toh test karne ke liye dono endpoints hain $x = 1$ aur $x = - 1$ . Yeh step kyun? Base series ka $R = 1$ hai; integrate karne se $R = 1$ rehta hai (radius-survival), isliye boundaries exactly $\pm 1$ hain.
Right endpoint $x = 1$ — do-step. (1) AST: $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$ alternate karta hai aur $0$ ki taraf shrink karta hai → converges. (2) Abel: sum $= lim_{x \to 1^{-}} ln (1 + x) = ln 2$ . Absolute version $1 + \frac{1}{2} + \frac{1}{3} + \dots$ harmonic series hai, jo diverges — isliye right endpoint sirf conditionally convergent hai. Yeh step kyun? AST (converges) ko absolute convergence (fails) se alag karna exactly wahi hai jo "conditional" ka matlab hai.
Left endpoint $x = - 1$ — alag check karo, aur yeh FAIL karta hai. $x = - 1$ daalo: $x^{n + 1} = (- 1)^{n + 1}$ deta hai $\sum \frac{( - 1 ) ^{n} ( - 1 ) ^{n + 1}}{n + 1} = \sum \frac{( - 1 ) ^{2 n + 1}}{n + 1} = - \sum_{n = 0}^{\infty} \frac{1}{n + 1}$ — negative harmonic series, jo $- \infty$ tak diverge karti hai. Yeh $ln (1 + (- 1)) = ln 0 = - \infty$ se match karta hai. Yeh step kyun? Yahi crucial asymmetry hai: right endpoint gain hua lekin left endpoint lost ho gaya. Aap yeh kabhi nahi dekhte agar sirf $x = + 1$ test karo. Hamesha dono karo.

Verify: $\sum_{n = 0}^{9999} \frac{( - 1 ) ^{n}}{n + 1} \approx 0.69310$ , aur $ln 2 \approx 0.69315$ . Left side $\sum_{n = 0}^{9999} \frac{1}{n + 1} \approx 9.79$ aur abhi bhi badh raha hai (diverges). ✓ Radius $R = 1$ .

Cell D — differentiation ek endpoint lose karta hai (dono sides)

Worked example Example 4 — jahan differentiate karna edge ko tod deta hai

Sochte hain $f (x) = n = 1 \sum \infty \frac{x ^{n}}{n ^{2}}$ . Iska radius $R = 1$ hai, aur dono $x = 1$ aur $x = - 1$ par yeh converge karta hai ( $x = 1$ par yeh $\sum \frac{1}{n ^{2}} = \frac{π ^{2}}{6}$ hai; $x = - 1$ par yeh $\sum \frac{( - 1 ) ^{n}}{n ^{2}}$ hai, absolutely convergent). Differentiate karne ke baad endpoints ka kya hoga?

Forecast: coefficients $\frac{1}{n ^{2}}$ hain (fast decay, dono $\pm 1$ par kaam karta hai). Differentiate karne se $n$ se multiply hota hai. Kya aap predict karte ho ki $x = 1$ aur $x = - 1$ survive karenge?

Term by term differentiate karo: $f^{'} (x) = \sum_{n = 1}^{\infty} \frac{n x ^{n - 1}}{n ^{2}} = \sum_{n = 1}^{\infty} \frac{x ^{n - 1}}{n}$ . Yeh step kyun? $\frac{d}{d x} \frac{x ^{n}}{n ^{2}} = \frac{n x ^{n - 1}}{n ^{2}} = \frac{x ^{n - 1}}{n}$ : differentiate karne se aaya $n$ ka factor denominator mein $n$ ki ek power ko cancel kar gaya, ab sirf $\frac{1}{n}$ jaisa decay bacha.
Right endpoint $x = 1$ : $\sum_{n = 1}^{\infty} \frac{1}{n}$ — harmonic series, jo diverges karti hai. Right endpoint lost. Yeh step kyun? Differentiate karne se pehle $\sum \frac{1}{n ^{2}}$ converge karta tha; baad mein, $\sum \frac{1}{n}$ diverge karta hai.
Left endpoint $x = - 1$ — alag check karo. Yahan $x^{n - 1} = (- 1)^{n - 1}$ deta hai $\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n - 1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \dots$ , jo AST se converges (to $ln 2$ ). Toh left endpoint survive karta hai chahe right wala mar gaya. Yeh step kyun? Yeh sabse sharp reminder hai: dono endpoints ke opposite fate ho sakte hain. Differentiation ne $x = 1$ lose kiya lekin $x = - 1$ rakh liya. Sirf ek side test karna galat conclusion deta.

Verify: derivative closed form mein $- \frac{l n ( 1 - x )}{x}$ hai; $x = 1^{-}$ ke paas yeh $- ln (1 - x) \to + \infty$ ki tarah blow up karta hai (right endpoint gone), jabki $x = - 1$ par yeh $- \frac{l n 2}{- 1} = ln 2 \approx 0.693$ hai (finite, left endpoint kept). Numerically $x = 0.99$ par: $\sum_{n = 1}^{2000} \frac{0.9 9 ^{n - 1}}{n} \approx 4.20$ , aur $- ln (0.01) /0.99 \approx 4.65$ — badh raha hai. ✓ Radius ab bhi $R = 1$ .

Cell E — ek shifted centre $c \neq = 0$

Worked example Example 5 —

$c = 2$ par centred series Maano $g (x) = n = 0 \sum \infty \frac{( x - 2 ) ^{n}}{3 ^{n}}$ . $g^{'} (x)$ ko series mein aur closed form mein nikalo, aur iska safe zone batao.

Forecast: geometric pattern ka " $x$ " ab $\frac{x - 2}{3}$ hai. Centre $2$ hai, aur $R$ aata hai $\frac{x - 2}{3} < 1$ se. $R$ guess karo.

Geometric shape pehchano. $u = \frac{x - 2}{3}$ rakho, $g = \sum u^{n} = \frac{1}{1 - u} = \frac{1}{1 - \frac{x - 2}{3}} = \frac{3}{5 - x}$ , valid $∣ u ∣ < 1 \Rightarrow ∣ x - 2∣ < 3$ ke liye, toh $c = 2, R = 3$ . Yeh step kyun? $u$ naam dene se shift invisible ho jaata hai — yeh ab bhi sirf ek geometric series hai disguise mein.
Series differentiate karo. $g^{'} (x) = \sum_{n = 1}^{\infty} \frac{n ( x - 2 ) ^{n - 1}}{3 ^{n}}$ . Yeh step kyun? Theorem centre-agnostic hai: $\frac{d}{d x} (x - 2)^{n} = n (x - 2)^{n - 1}$ bilkul waise hi jaise $c$ zero hota.
Cross-check ke liye closed form differentiate karo. $g^{'} (x) = \frac{d}{d x} \frac{3}{5 - x} = \frac{3}{( 5 - x ) ^{2}}$ . Yeh step kyun? Do independent routes agree karne chahiye — isi se aap term-by-term kaam par trust karte ho.

Verify: $x = 3$ par (andar, kyunki $∣3 - 2∣ = 1 < 3$ ): series $\sum_{n = 1}^{20} \frac{n \cdot 1 ^{n - 1}}{3 ^{n}} \approx 0.75$ ; closed form $\frac{3}{( 5 - 3 ) ^{2}} = \frac{3}{4} = 0.75$ . ✓ Centre $2$ , radius $3$ dono differentiation se unchanged. Yahan endpoints $x = - 1$ aur $x = 5$ hain; dono diverge karte hain (geometric terms $\pm 1$ shrink nahi hote), dono sides par.

Cell F — degenerate / self-returning case

Worked example Example 6 — woh series jo apni khud ki derivative ke barabar hai

Dikhao ki $E (x) = n = 0 \sum \infty \frac{x ^{n}}{n !}$ satisfy karta hai $E^{'} = E$ , aur note karo ki yeh limiting/degenerate case kyun hai jahan kuch bhi nahi badlta.

Forecast: exponential Taylor and Maclaurin series ka $R = \infty$ hai (koi endpoints nahi hote). Predict karo ki $E^{'}$ kaisi dikhegi.

Differentiate karo. $E^{'} (x) = \sum_{n = 1}^{\infty} \frac{n x ^{n - 1}}{n !} = \sum_{n = 1}^{\infty} \frac{x ^{n - 1}}{( n - 1 )!}$ . Yeh step kyun? $\frac{n}{n !} = \frac{1}{( n - 1 )!}$ — differentiate karne se aaya $n$ ka factor factorial ke top ko cancel kar deta hai.
Re-index karo $m = n - 1$ : $E^{'} (x) = \sum_{m = 0}^{\infty} \frac{x ^{m}}{m !} = E (x)$ . Yeh step kyun? Index shift karne se pata chalta hai ki differentiated series original ke identical hai.
Degeneracy note karo. Yahan $R = \infty$ hai, toh safe zone poori line hai aur gain ya lose karne ke liye koi endpoints nahi hain — na $+ \infty$ na $- \infty$ ek real boundary point hai. Yeh ek cell hai jahan "endpoint caveat" vacuously satisfied hai.

Verify: $x = 1$ par, $\sum_{n = 0}^{10} \frac{1}{n !} \approx 2.71828$ , $e$ se match karta hai. Derivative sum construction se iske barabar hai. ✓

Cell G — ek real-world word problem

Worked example Example 7 — capacitor charging, term by term (units ke saath)

Ek RC circuit mein ek capacitor charge hota hai aur uska voltage hai $V (t) = V_{0} (1 - e^{- t / τ})$ , jahan $V_{0}$ volts (V) mein hai, $t$ seconds (s) mein, aur $τ$ (time constant) bhi seconds mein. Current hai $I (t) = C V^{'} (t)$ jahan $C$ capacitance farads (F) mein hai. $V (t)$ ko $t$ mein power series ke roop mein likho aur ise differentiate karke $I (t)$ nikalo, units check karo.

Forecast: $e^{- t / τ}$ $t = 0$ ke paas roughly $1 - \frac{t}{τ}$ hai, toh $V \approx V_{0} \frac{t}{τ}$ (pehle linear ramp). $I$ ka leading term predict karo.

Exponential expand karo. $e^{- t / τ} = \sum_{n = 0}^{\infty} \frac{( - t / τ ) ^{n}}{n !}$ , toh $V (t) = V_{0} (1 - \sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n} t ^{n}}{n ! τ ^{n}}) = V_{0} \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n - 1} t ^{n}}{n ! τ ^{n}}$ . Yeh step kyun? $e^{- t / τ}$ ka $n = 0$ term $1$ hai, aur $1 - 1 = 0$ , isliye series $n = 1$ se start hoti hai: ek clean ramp bina kisi constant offset ke — $V (0) = 0$ se match karta hai.
Term by term differentiate karo. $V^{'} (t) = V_{0} \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n - 1} n t ^{n - 1}}{n ! τ ^{n}} = V_{0} \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n - 1} t ^{n - 1}}{( n - 1 )! τ ^{n}} = \frac{V _{0}}{τ} e^{- t / τ}$ . Yeh step kyun? Wahi $\frac{n}{n !}$ cancellation jaise Example 6 mein; re-summing se compact closed form milta hai.
$C$ se multiply karo. $I (t) = C V^{'} (t) = \frac{C V _{0}}{τ} e^{- t / τ}$ , aur $t = 0$ par: $I (0) = \frac{C V _{0}}{τ}$ . Yeh step kyun? Yeh peak initial current hai; term-by-term differentiation ne humein poora current curve de diya.

Verify (units). $\frac{C V _{0}}{τ}$ ke units hain $\frac{F \cdot V}{s}$ . Kyunki $1 F = 1 C / V$ (coulomb per volt), $\frac{F \cdot V}{s} = \frac{C}{s} = A$ (amperes). ✓ Current amps mein — dimensionally correct. Numerically $V_{0} = 5$ V, $τ = 2$ s, $C = 1$ F, $t = 1$ s ke saath: $I = \frac{5}{2} e^{- 0.5} \approx 1.516$ A.

Cell H — exam twist

Worked example Example 8 — ek mystery series ko derivative pehchaan ke sum karo

$S = n = 1 \sum \infty \frac{n}{2 ^{n}}$ closed form mein evaluate karo.

Forecast: $n$ upar hai — yeh ek tell-tale sign hai ki koi differentiation ne ise banaya. Guess karo ki $S$ ek chota integer hai ya nahi.

Parent series dhundho. Yaad karo $\sum_{n = 1}^{\infty} n x^{n - 1} = \frac{1}{( 1 - x ) ^{2}}$ (Example 1, step 1). Yeh step kyun? $\frac{n}{2 ^{n}}$ mein $n$ ka factor chillata hai "main $\sum x^{n}$ differentiate karne se aaya hoon."
$x$ se multiply karo exponent match karne ke liye: $\sum_{n = 1}^{\infty} n x^{n} = \frac{x}{( 1 - x ) ^{2}}$ . Yeh step kyun? Humara target $x^{n}$ wala hai (yaani $x = \frac{1}{2}$ toh $x^{n} = 1/ 2^{n}$ ), $x^{n - 1}$ nahi.
$x = \frac{1}{2}$ daalo (safely $R = 1$ ke andar): $S = \sum_{n = 1}^{\infty} n (\frac{1}{2})^{n} = \frac{\frac{1}{2}}{( 1 - \frac{1}{2} ) ^{2}} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2$ . Yeh step kyun? Kyunki $∣ x ∣ = \frac{1}{2} < 1$ safe zone ke strictly andar hai, koi endpoint subtlety nahi aati — term-by-term-derived closed form valid hai, aur yeh final answer deta hai $S = 2$ .

Verify: partial sum $\sum_{n = 1}^{20} \frac{n}{2 ^{n}} \approx 1.99998$ , $2$ ki taraf converge karta hai. ✓ Final closed form $S = 2$ hai.

Recall Aapke liye kaunsa cell sabse mushkil tha?

Cell D (differentiation losing an endpoint) vs Cell B (integration gaining one) — kya aap ek sentence mein reason bata sakte ho ki har ek mein aisa kyun hota hai, aur kyun aapko dono sides check karni chahiye? Losing: differentiate karne se coefficients $n$ se multiply hote hain, decay weak ho jaati hai, isliye ek borderline endpoint toot sakta hai. Gaining: integrate karne se $n + 1$ se divide hota hai, decay strong ho jaati hai, isliye ek borderline endpoint converge karna shuru kar sakta hai. ::: Aur Ex 3/Ex 4 dikhate hain ki dono ends disagree bhi kar sakte hain — isliye $x = c + R$ aur $x = c - R$ ko alag alag, har baar test karo.

Recap

Recall Scenario matrix, memory se

Saare cells A–H ke naam batao aur woh endpoint rule jo woh enforce karte hain. A differentiate-positive; B/B′ integrate-gains-endpoint (dono sides); C/C′ integrate-conditional (right gained, left lost); D differentiate-loses-endpoint (right lost, left kept); E shifted-centre; F self-returning/degenerate; G word-problem; H exam-twist. ::: Har ek mein, radius $R$ unchanged raha — sirf endpoints ne kabhi verdict flip kiya, aur dono mein se har ek endpoint ko apne aap test karna pada.

Connections

Parent topic
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

Worked examples — Term-by-term differentiation and integration of power series