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

KYA karte hain: derivative rule lagate hain — $n$-th coefficient ko $n$ se multiply karo, power ek se kam karo, aur sum $n=1$ se shuru karo (kyunki $n=0$ wala term ek constant hai, uska derivative $0$ hota hai). Yahan $a_n=\frac{1}{n+1}$ hai, toh $n{a}_n=\frac{n}{n+1}$: $f^{\prime }(x)={\sum }_{n=1}^{\infty }\frac{n}{n+1}{x}^{n-1}.$ $R$ kyun nahi badlta: original series ka $R=1$ hai (ye geometric series jaisi lagti hai jo $\frac{1}{n+1}$ se slow ho gayi hai), aur differentiation kabhi $R$ nahi badlta. Toh answer $|x|<1$ par valid hai.

KYA karte hain: har brick ko integrate karo. Rule: $x^k$ ke coefficient ko $(k+1)$ se divide karo aur power $k+1$ kar do. Yahan power $k=2n$ hai, toh $\int x^{2n}dx=\frac{x^{2n+1}}{2n+1}$: ${\int }_0^{x}g(t)dt={\sum }_{n=0}^{\infty }\frac{(-1{)}^n}{2n+1}{x}^{2n+1}.$ YE KYA HAI: ye exactly $\arctan x$ series hai (parent mein Example 2 dekho). Constant $0$ hai kyunki definite integral $0$ se $x=0$ par $0$ hota hai.

DIFFERENTIATE KYU KARTE HAIN? Dekho $\frac{d}{dx}\frac{1}{1-x}=\frac{1}{(1-x{)}^2}$ (left side par chain rule). Toh jaani-pehchani series ko differentiate karne par anjaani series free mein mil jaati hai. $\frac{1}{(1-x{)}^2}=\frac{d}{dx}{\sum }_{n=0}^{\infty }{x}^n={\sum }_{n=1}^{\infty }n{x}^{n-1}={\sum }_{m=0}^{\infty }(m+1)x^m.$ (Re-index $m=n-1$.) Toh $h(x)=1+2x+3x^2+4x^3+\cdots$, $|x|<1$ par valid. Pehle do coefficients check karo: $h(0)=1$ ✓, aur $x$ ka coefficient $2$ hai.

INTEGRATE KYU KARTE HAIN? Kyunki $\frac{d}{dx}\ln (1-x)=\frac{-1}{1-x}$ hai. Toh $\ln (1-x)$, $-\frac{1}{1-x}=-\sum x^n$ ka antiderivative hai. $0$ se $x$ tak term by term integrate karo: $\ln (1-x)=-{\sum }_{n=0}^{\infty }\frac{x^{n+1}}{n+1}=-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots \right).$ Constant: $\ln (1-0)=\ln 1=0$, toh $C=0$. Radius: integration $R=1$ rakhta hai.

$x=1$ par: series ban jaati hai $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$. Ye ek alternating series hai jiske terms $\frac{1}{2n+1}$ $0$ ki taraf decrease karte hain, toh alternating-series test se ye converge karti hai ($\frac{\pi }{4}$ par). $x=-1$ par: plug in karo — odd powers sign flip kar dete hain, milta hai $-1+\frac{1}{3}-\frac{1}{5}+\cdots =-(1-\frac{1}{3}+\frac{1}{5}-\cdots )$, jo bhi converge karta hai ($-\frac{\pi }{4}$ par). Gain hua ya loss? Original series $\sum (-1{)}^n{x}^{2n}$ at $x=\pm 1$ hai $1-1+1-1+\cdots$, jo diverge karta hai (terms $0$ nahi jaate). Integrate karne ke baad, dono endpoints ab converge karte hain — toh integration ne endpoints par convergence gain ki. Ye parent note ke rule se match karta hai: $R$ nahi badlta, lekin integrate karne par endpoints improve ho sakte hain. Dekho Abel's theorem (endpoint behaviour) ki kyun wahan sum actually $\arctan (\pm 1)$ ke barabar hota hai.

Differentiate karo: $f^{\prime }(x)={\sum }_{n=1}^{\infty }\frac{n{x}^{n-1}}{n}={\sum }_{n=1}^{\infty }{x}^{n-1}={\sum }_{m=0}^{\infty }{x}^m=\frac{1}{1-x}$, radius abhi bhi $1$. Endpoint $x=-1$: derivative series hai $1-1+1-1+\cdots$, jo diverge karta hai (terms $0$ ki taraf nahi jaate). Conclusion: original $x=-1$ par converge karta tha, derivative nahi karta. Toh differentiation ne us endpoint par convergence kho di — L3.1 ka mirror image. Ye exactly wo "DIRT: Test endpoints again" wali warning hai.

L2.1 se kyun shuru karte hain? Hum pehle se jaante hain ${\sum }_{n=1}^{\infty }n{x}^{n-1}=\frac{1}{(1-x{)}^2}$. Hamare target mein ek extra factor of $x$ hai, toh bas dono taraf $x$ se multiply karo: $S(x)={\sum }_{n=1}^{\infty }n{x}^n=x\cdot \frac{1}{(1-x{)}^2}=\frac{x}{(1-x{)}^2}.$ Numeric evaluation: $x=\frac{1}{2}$ daalo (jo $|x|<1$ ke andar hai): ${\sum }_{n=1}^{\infty }\frac{n}{2^n}=\frac{1/2}{(1-1/2{)}^2}=\frac{1/2}{1/4}=2.$

DIFFERENTIATE KYU KARTE HAIN? Coefficient $\frac{1}{2n+1}$ signal deta hai ki koi integral hua tha — toh differentiate karne se simplify hona chahiye. Term by term differentiate karo: $T^{\prime }(x)={\sum }_{n=0}^{\infty }\frac{(2n+1)x^{2n}}{2n+1}={\sum }_{n=0}^{\infty }{x}^{2n}=\frac{1}{1-x^2}(|x|<1).$ $T$ wapas paane ke liye integrate karo. Partial-fraction split use karo $\frac{1}{1-x^2}=\frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)$: $T(x)={\int }_0^x\frac{dt}{1-t^2}=\frac{1}{2}(-\ln (1-x)+\ln (1+x))=\frac{1}{2}\ln \frac{1+x}{1-x}.$ Constant: $T(0)=0$ aur $\frac{1}{2}\ln 1=0$ ✓. Ye artanh (inverse hyperbolic tangent) series hai. $x=\frac{1}{2}$ par spot check: $T(\frac{1}{2})=\frac{1}{2}\ln 3\approx 0.5493$.

POWER SERIES KYU USE KARTE HAIN? Kyunki term-by-term differentiation ek differential equation ko coefficients par plain algebra mein badal deta hai (ye Solving ODEs with power series ka method hai). Step 1 — ansatz differentiate karo. $y^{\prime }={\sum }_{n=1}^{\infty }n{a}_n{x}^{n-1}={\sum }_{m=0}^{\infty }(m+1)a_{m+1}{x}^m$ (re-index $m=n-1$). Step 2 — $y$ se match karo. Equation $y^{\prime }=y$ ka matlab hai, $x^m$ ke coefficient by coefficient: $(m+1)a_{m+1}=a_m⟹a_{m+1}=\frac{a_m}{m+1}.$ Step 3 — initial condition use karo. $y(0)=a_0=1$. Phir $a_1=\frac{1}{1}=1$, $a_2=\frac{1}{2}$, $a_3=\frac{1}{2\cdot 3}=\frac{1}{6}$, aur generally $a_n=\frac{1}{n!}.$ Step 4 — function padhlo. $y={\sum }_{n=0}^{\infty }\frac{x^n}{n!}=e^x$, radius $R=\infty$ ke saath (ratio $\frac{a_{n+1}}{a_n}=\frac{1}{n+1}\to 0$, toh har $x$ ke liye converge karta hai).

Step 1 — integrate karo. $\frac{1}{1+t^2}=\sum (-1{)}^n{t}^{2n}$ ($|t|<1$) se, $0$ se $x$ tak integrate karo: $\arctan x={\sum }_{n=0}^{\infty }\frac{(-1{)}^n}{2n+1}{x}^{2n+1},|x|<1.$ Step 2 — endpoint convergence. $x=1$ par series hai $1-\frac{1}{3}+\frac{1}{5}-\cdots$, alternating hai aur terms $0$ ki taraf decrease karte hain, toh ye converge karti hai (L3.1 mein ye dikhaya tha). Step 3 — sum $\arctan 1$ ke barabar kyun hai. Endpoint par convergence akele kaafi nahi hai; hum Abel's theorem (endpoint behaviour) invoke karte hain: agar power series ek endpoint par converge kare, toh wahan uska sum andar se function ke limit ke barabar hota hai. Kyunki $\arctan$ continuous hai aur $\arctan 1=\frac{\pi }{4}$ hai, ${\sum }_{n=0}^{\infty }\frac{(-1{)}^n}{2n+1}=\arctan 1=\frac{\pi }{4}.$ Ye Leibniz formula hai. $■$

$\sum \frac{x^n}{n!}$ ka term-by-term derivative ::: phir se $\sum \frac{x^n}{n!}$ (ye $e^x$ hai). ${\sum }_{n\ge 1}n{x}^n$ ka closed form ::: $\frac{x}{(1-x{)}^2}$. ${\sum }_{n\ge 1}n/2^n$ ki value ::: $2$. ${\sum }_{n\ge 0}\frac{x^{2n+1}}{2n+1}$ ka closed form ::: $\frac{1}{2}\ln \frac{1+x}{1-x}$. Jo theorem $\arctan$ series ko $x=1$ tak le jaata hai ::: Abel's theorem.