Question bank — Maclaurin series of eˣ, sin x, cos x, ln(1+x), (1+x)ⁿ — derive all

False. Yeh function ke barabar sirf apne interval of convergence ke andar hoti hai; bahar, polynomial diverge kar jaati hai jabki function bilkul theek-thaak finite hoti hai (dekho $\ln 3$ jo $\ln (1+x)$ se $x=2$ par milta hai). Figure s01 dekho — partial sums $0$ ke paas curve se chipki rehti hain aur phir alag ho jaati hain.

False. $f(x)=e^{-1/x^2}$ (jahan $f(0)=0$) ke $0$ par saare derivatives zero hain, isliye uski Maclaurin series $0$ hai, lekin function nonzero hai — series sirf analytic functions ke liye sahi hoti hai. Figure s03 dikhata hai kyun: graph axis ke khilaf itni tezi se flatten ho jaata hai ki koi bhi polynomial detect nahi kar sakti.

True. Denominator $n!$ kisi bhi fixed $x$ ke liye eventually $x^n$ se zyada badh jaata hai, isliye terms kaafi tezi se $0$ tak shrink ho jaate hain — radius of convergence infinite hai.

False. Cancel karne ko kuch nahi hota — $\cos$ ke odd derivatives $x=0$ par literally $0$ hote hain ($-\sin 0=0$, etc.), isliye woh coefficients pehle se hi zero janam lete hain.

False. Woh factorials $3!,5!,7!=6,120,5040$ hain; chota dikhne wala $3$ bas $3!=6$ likha hua hai, ek aasaan misread.

False. $k$ ko term counter maan lo, to numerator $n(n-1)\cdots (n-k+1)$ mein factor $(n-n)=0$ aa jaata hai jab $k>n$ ho kisi non-negative integer $n$ ke liye, isliye baad ke saare $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ zero ho jaate hain aur series terminate kar jaati hai — yahi ordinary binomial theorem hai.

False. Maclaurin exactly Taylor series hai centre $a=0$ ke saath; same machinery, special base point.

False. Error ka sign agle term ke sign par depend karta hai. $\ln (1+x)$ ke liye $x>0$ ke saath terms alternate karte hain, isliye partial sum oopar ya neeche ho sakta hai is baat par depend karte hue ki tum kahan ruke.

Galat. Constant term $a_0=\ln (1+0)=\ln 1=0$ hai; koi leading $1$ nahi hai. Series $x$ term se shuru hoti hai: $x-\frac{x^2}{2}+\cdots$.

$n!$ miss ho gaya. Yahan $a_n$ upar di gayi definition se $x^n$ ka weight hai. Kyunki $\frac{d^n}{d{x}^n}{x}^n=n!$, surviving term $n!a_n=f^{(n)}(0)$ hota hai, jisse $a_n=\frac{f^{(n)}(0)}{n!}$ milta hai.

Sign error. Geometric series ko ratio $r=-x$ ke saath use karne par alternating signs milte hain: $1-x+x^2-x^3+\cdots$. All-plus wala version $\frac{1}{1-x}$ hai.

Quadratic term ka sign galat hai. Coefficient $\left(\genfrac{}{}{0ex}{}{1/2}{2}\right)=\frac{\frac{1}{2}(-\frac{1}{2})}{2!}=-\frac{1}{8}$ hai, isliye yeh $1+\frac{1}{2}x-\frac{1}{8}{x}^2+\cdots$ hai — negative kyunki $n=\frac{1}{2}$ se $(n-1)$ negative ho jaata hai.

Function value exist karti hai, lekin series sirf $-1<x\le 1$ ke liye converge karti hai. $x=2$ par polynomial diverge kar jaati hai; tumhe ek alag centre ya identity chahiye hogi. Figure s02 is interval ko number line par dikhata hai.

Sign placement galat hai. $x\to -x$ substitute karo: $e^{-x}=\sum \frac{(-x{)}^n}{n!}=\sum \frac{(-1{)}^n{x}^n}{n!}$, jo alternate karta hai. Sirf odd terms ka sign badalta hai, har term uniformly nahi.

Koi contradiction nahi: $2!=2$, isliye $\frac{x^2}{2}$ hi $\frac{x^2}{2!}$ hai. Sine series ko differentiate karne par cosine series sahi tarike se reproduce hoti hai.

Kyunki $\frac{d^n}{d{x}^n}{x}^n=n!$; divide kiye bina, $x^n$ piece $a_n$ ki jagah $n!a_n$ contribute karta, isliye hum compensate karte hain taaki coefficients derivatives se cleanly match karein.

Har woh term jo abhi bhi $x$ contain karti hai, vanish ho jaati hai ($0^k=0$), sirf us particular derivative ka constant term bachta hai — ek clean one-equation-one-unknown extraction.

Kyunki $\frac{d}{dx}{e}^x=e^x$, isliye $0$ par har derivative $e^0=1$ ke barabar hai; saare numerators same hain, sirf $\frac{1}{n!}$ bachta hai.

$\sin$ ek odd function hai ($\sin (-x)=-\sin x$), aur sirf odd-power monomials usi symmetry ko share karte hain; even-power coefficients zero hone chahiye taaki woh symmetry preserve ho sake.

Kyunki $\frac{d}{dx}\ln (1+x)=\frac{1}{1+x}$ already ek known geometric series hai; ise term-by-term integrate karna logarithm ke higher derivatives compute karne se kaafi zyada clean hai.

$x=1$ par yeh alternating harmonic series ban jaati hai ($\ln 2$ tak converge karti hai); $x=-1$ par yeh negative harmonic series $-1-\frac{1}{2}-\frac{1}{3}-\cdots$ ban jaati hai, jo diverge karti hai — $\ln 0=-\infty$ ko reflect karte hue. Figure s02 mein closed/open endpoints dekho.

$e^x$ mein $x\to ix$ substitute karne par terms parity ke hisaab se split ho jaate hain: real even-power terms $\cos x$ rebuild karte hain aur imaginary odd-power terms $i\sin x$ rebuild karte hain, jisse $e^{ix}=\cos x+i\sin x$ milta hai.

$\sin x$ ki jagah $x-\frac{x^3}{6}+\cdots$ rakho; phir $\frac{\sin x}{x}=1-\frac{x^2}{6}+\cdots \to 1$ turant mil jaata hai, koi repeated differentiation nahi chahiye.

Woh polynomial khud hi hogi — ek finite series. Uske higher derivatives vanish ho jaate hain, isliye degree $2$ se aage ke saare coefficients $a_n$ zero hain; woh already "apni khud ki" series hai.

$\sqrt{2}$ finite hai aur series wahan $n=\frac{1}{2}$ ke liye converge karti hai, lekin endpoints par convergence delicate hoti hai aur $n$ ke sign par depend karti hai; safe interior guarantee $|x|<1$ hai (figure s02 mein shaded band).

Function constant $1$ hai; series apne pehle term $\left(\genfrac{}{}{0ex}{}{0}{0}\right)=1$ tak collapse ho jaati hai (yaad karo $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ $x^k$ ka weight hai, aur yahan sirf $k=0$ bachta hai) aur aage ke saare coefficients zero hain — terminating integer case ke saath consistent.

Haan. $x=0$ rakhne par har power term zero ho jaata hai, sirf constant $1$ bachta hai — built-in sanity check ki constant coefficient $a_0$, $f(0)$ ke barabar hota hai.

Iski $0$ par centre ki hui koi series nahi hai: $\ln x$ $x=0$ par undefined hai, isliye wahan koi derivatives exist nahi karte. Yahi wajah hai ki hum $\ln (1+x)$ expand karte hain, mushkil ko $x=-1$ par shift karte hue.

Series phir bhi converge karegi (infinite radius), lekin tumhe kaafi saare terms chahiye honge kyunki $\frac{x^n}{n!}$ sirf tabhi shrink hona shuru hota hai jab $n>x\approx 10$ ho; convergence guaranteed hai, necessarily fast nahi.