Exercises — Taylor series — derivation from power series

(a) Har power $(x-3{)}^n$ hai, toh centre se displacement $(x-3)$ hai. Iska matlab $a=3$. (b) $(x-3{)}^2$ par jo coefficient baitha hai woh $c_2=7$ hai. (c) $c_2=\frac{f^{\prime \prime }(a)}{2!}$ ko rearrange karo: $f^{\prime \prime }(a)=2!c_2=2\cdot 7=\boxed{14}$. Kyun: $2!=2$ hai, aur coefficient mein factorial pehle se divide ho chuki thi, toh raw derivative recover karne ke liye hum use wapas multiply karte hain.

Maclaurin series woh Taylor series hota hai jiska centre $a=0$ ho. Series A ke powers $x^n=(x-0{)}^n$ hain, toh $a=0$ — yahi Maclaurin wala hai. Series B mein $(x+2{)}^n=(x-(-2){)}^n$ hai, toh iska centre $a=-2$ hai. Dhyan rakho: $(x+2)$ ek minus negative chhupaata hai.

Baar baar differentiate karo; har derivative chain rule se ek factor $2$ khींchta hai: $f=e^{2x},f^{\prime }=2e^{2x},f^{\prime \prime }=4e^{2x},f^{\prime \prime \prime }=8e^{2x}.$ $a=0$ daalo (toh $e^0=1$): $f(0)=1,f^{\prime }(0)=2,f^{\prime \prime }(0)=4,f^{\prime \prime \prime }(0)=8$. $c_n=f^{(n)}(0)/n!$ apply karo: $c_0=\frac{1}{1}=1,c_1=\frac{2}{1!}=2,c_2=\frac{4}{2!}=2,c_3=\frac{8}{3!}=\frac{4}{3}.$ $\boxed{e^{2x}=1+2x+2x^2+\frac{4}{3}{x}^3+\cdots }$ Shortcut se check: $e^x=\sum x^n/n!$ mein $x\to 2x$ replace karne par $\sum (2x{)}^n/n!=\sum 2^n{x}^n/n!$ milta hai, yaani $c_n=2^n/n!$ — jo $1,2,2,\frac{4}{3}$ se match karta hai. ✓

Derivatives har chaar steps mein cycle karte hain: $\cos ,-\sin ,-\cos ,\sin ,\cos ,\dots$ $x=0$ par: $\cos 0=1,-\sin 0=0,-\cos 0=-1,\sin 0=0,\cos 0=1$. Toh $f(0),f^{\prime }(0),f^{\prime \prime }(0),f^{\prime \prime \prime }(0),f^{(4)}(0)=1,0,-1,0,1$. $n!$ se divide karo: $c_0=1,c_1=0,c_2=\frac{-1}{2!}=-\frac{1}{2},c_3=0,c_4=\frac{1}{4!}=\frac{1}{24}.$ $\boxed{\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots }$ Sirf even powers bachte hain kyunki odd-order derivatives saare $\sin 0=0$ se guzarte hain.

$f=x^{-1},f^{\prime }=-x^{-2},f^{\prime \prime }=2x^{-3}.$ $a=2$ par: $f(2)=\frac{1}{2},f^{\prime }(2)=-\frac{1}{4},f^{\prime \prime }(2)=\frac{2}{8}=\frac{1}{4}$. Coefficients: $c_0=\frac{1}{2},c_1=-\frac{1}{4},c_2=\frac{1/4}{2!}=\frac{1}{8}$. $\boxed{\frac{1}{x}=\frac{1}{2}-\frac{1}{4}(x-2)+\frac{1}{8}(x-2{)}^2-\cdots }$ $(x-2)$ ke powers kyun? Centre $a=2$ hai, toh natural building block displacement $(x-2)$ hai, $x$ nahi.

Maano $\cos x=\sum c_n{x}^n$. $x\to -x$ replace karo: left side unchanged rahta hai (even function), jabki $x^n\to (-1{)}^n{x}^n$ ho jaata hai. Match karne par: $\sum c_n{x}^n=\sum c_n(-1{)}^n{x}^n\Rightarrow c_n=(-1{)}^n{c}_n\text{ for every }n.$ Agar $n$ odd hai, toh $(-1{)}^n=-1$, jo $c_n=-c_n$ force karta hai, yaani $2c_n=0$, toh $c_n=0$. Is tarah saare odd coefficients zero ho jaate hain — sirf even powers bachte hain. Yeh L2.2 se match karta hai bina kisi calculus ke, sirf symmetry se.

$a=0$ ke baare mein: $f$ pehle se $x$ mein ek polynomial hai, toh iska Maclaurin series khud hi hai: $3x^2-x+4$. Check karo: $f(0)=4,f^{\prime }(0)=-1,f^{\prime \prime }(0)=6\Rightarrow c_2=6/2=3$. ✓ $n\ge 3$ ke liye saare coefficients $c_n=0$ hain kyunki $f^{\prime \prime \prime }\equiv 0$ hai. $a=1$ ke baare mein: $f(1)=3-1+4=6$; $f^{\prime }(x)=6x-1\Rightarrow f^{\prime }(1)=5$; $f^{\prime \prime }(x)=6\Rightarrow f^{\prime \prime }(1)=6$, toh $c_2=6/2=3$; $f^{\prime \prime \prime }=0$. $\boxed{f(x)=6+5(x-1)+3(x-1{)}^2}.$ Khaas baat: series terminate ho jaati hai — degree-$d$ polynomial ka Taylor series kisi bhi centre ke baare mein zyada se zyada $d+1$ terms wala hota hai, kyunki order $d$ se aage ke saare derivatives zero hote hain. Convergence ki koi chinta nahi; equality har jagah exact hai.

$f^{\prime }=(1-x{)}^{-2}$, $f^{\prime \prime }=2(1-x{)}^{-3}$, …, $f^{(n)}(0)=n!$. Toh $c_n=n!/n!=1$: $\frac{1}{1-x}={\sum }_{n=0}^{\infty }{x}^n=1+x+x^2+\cdots$ Yeh bilkul ek geometric series hai ratio $x$ ke saath; yeh sirf $|x|<1$ ke liye converge karti hai. $x=2$ par, sacha value $\frac{1}{1-2}=-1$ hai, lekin $1+2+4+\cdots$ diverge karta hai — coefficients "sahi" hain phir bhi, lekin series interval ke bahar $f$ ke barabar hone mein fail karti hai. Dekho Power series — radius & interval of convergence.

$e^u=1+u+\frac{u^2}{2!}+\cdots$ se shuru karo aur $u=-x^2$ substitute karo: $e^{-x^2}=1+(-x^2)+\frac{(-x^2{)}^2}{2}+\cdots =1-x^2+\frac{x^4}{2}-\cdots$ $\boxed{e^{-x^2}=1-x^2+\frac{1}{2}{x}^4-\cdots }$ Yeh legal kyun hai: convergent power series mein substitution sirf variable relabel karta hai; yeh exactly wahi reproduce karta hai jo repeated differentiation deta, kaafi sasti mein. ($e^{-x^2}$ ko haath se chaar baar differentiate karna dardnaak hai — yahi is method ka fayda hai.)

$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$ aur $\sin x=x-\frac{x^3}{6}+\cdots$ use karo. Multiply karo, $x^3$ tak ke terms rakho: $(1+x+\frac{x^2}{2}+\frac{x^3}{6})(x-\frac{x^3}{6}).$ Power ke hisaab se collect karo:

• $x^1$: $1\cdot x=x$.
• $x^2$: $x\cdot x=x^2$.
• $x^3$: $\frac{x^2}{2}\cdot x=\frac{x^3}{2}$, plus $1\cdot (-\frac{x^3}{6})=-\frac{x^3}{6}$. Sum $=\frac{3}{6}-\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$. $\boxed{e^x\sin x=x+x^2+\frac{1}{3}{x}^3+\cdots }$

Numerator expand karo: $e^x-1-x=(1+x+\frac{x^2}{2}+\cdots )-1-x=\frac{x^2}{2}+\frac{x^3}{6}+\cdots$ $x^2$ se divide karo: $\frac{e^x-1-x}{x^2}=\frac{1}{2}+\frac{x}{6}+\cdots \underset{x\to 0}{\to }\boxed{\frac{1}{2}}.$ Series L'Hôpital se behtar kyun hai: $\frac{0}{0}$ form ke liye L'Hôpital do baar lagana padta; series dikhata hai answer sirf numerator mein $x^2$ ka coefficient hai, ek step mein read off. Dekho Linear approximation & differentials — yeh samjhaata hai kyun leading surviving term limit control karta hai.

Differentiate karo: $f^{\prime }(x)=\frac{1}{1+x}$. Yeh ek geometric series hai (ratio $-x$): $\frac{1}{1+x}={\sum }_{k=0}^{\infty }(-x{)}^k=1-x+x^2-x^3+\cdots ,|x|<1.$ Kyunki $f(0)=\ln 1=0$ hai, $0$ se $x$ tak term by term integrate karo: $\ln (1+x)={\int }_0^x(1-t+t^2-\cdots )dt=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots ={\sum }_{n=1}^{\infty }\frac{(-1{)}^{n-1}}{n}{x}^n.$ Yeh substitution $x\to x-1$ ke saath parent Example 3 se match karta hai (yaani $a=1$ ke baare mein $\ln x$). Convergence: $|x|<1$ (aur bonus mein, $x=1$ alternating harmonic series deta hai $=\ln 2$).

$\cos x={\sum }_{k=0}^{\infty }\frac{(-1{)}^k}{(2k)!}{x}^{2k}$ se, $x^6$ ka coefficient ($2k=6\Rightarrow k=3$ lo) hai: $c_6=\frac{(-1{)}^3}{6!}=-\frac{1}{720}.$ Coefficient rule $c_6=f^{(6)}(0)/6!$ reverse karo: $f^{(6)}(0)=6!c_6=720\cdot \left(-\frac{1}{720}\right)=\boxed{-1}.$ Idea: series centre par saare derivatives ka ek encyclopedia hai — ek coefficient read off karo aur $n!$ se multiply karo, koi bhi derivative instantly milti hai.

$h(x)=\sum (b_n-d_n)(x-a{)}^n$ lo, jo $a$ ke paas saare $x$ ke liye $0$ ke barabar hai. $e_n=b_n-d_n$ rakho.

• $x=a$ daalo: $(x-a{)}^k$, $k\ge 1$ wale saare terms khatam, $e_0=0$ bachta hai.
• Ek baar differentiate karo aur $x=a$ daalo: parent note wala wahi "killer move" $1!e_1=0$ chhod'ta hai, toh $e_1=0$.
• $n$ baar differentiate karo aur $x=a$ daalo: $n!e_n=h^{(n)}(a)=0$, toh $e_n=0$. Kyunki $n!\ne 0$ hai, har $e_n=0$, isliye saare $n$ ke liye $b_n=d_n$. Yahi reason hai ki Taylor derivation THE series deti hai, kaafi mein se ek nahi — representation unique hai.

Kisi bhi series se centre read karo ($a$ nikalo)? ::: $(x-a{)}^n$ mein force karo; jo number subtract ho raha hai woh $a$ hai. Ek coefficient se raw derivative recover karo? ::: $f^{(n)}(a)=n!c_n$. Bina differentiate kiye series nikalo? ::: Known series substitute ya multiply karo (L4). Jano kab ek sahi series phir bhi $f$ ke barabar hone mein fail karti hai? ::: Radius of convergence ke bahar, ya agar $R_n\nrightarrow 0$ ho.