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

KYA: bas $a_n=\frac{1}{n!}$ padh lo. $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots =1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$ Yeh values kyun hain: $e^x$ ki har derivative $e^x$ hi hoti hai, aur $e^0=1$, isliye saare numerators $1$ hain. Yeh sab $x$ ke liye valid kyun hai — ratio test, concrete tarike se. Dekho ki har term apne pehle wali term se kitni badi hai. Term $n$ hai $\frac{x^n}{n!}$, term $n+1$ hai $\frac{x^{n+1}}{(n+1)!}$, toh step ratio hai $\left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right|=\frac{|x|}{n+1}.$ Koi bhi $x$ fix karo. Jaise-jaise $n$ badhta hai, $\frac{|x|}{n+1}\to 0$. Toh aakhirkar har nayi term pichli term ka ek chota sa fraction hoti hai — terms $0$ ki taraf crash karti hain chahe $x$ kitna bhi bada ho. Yahi "factorial beats power" statement hai, ab ek honest inequality ke roop mein, sirf ek slogan nahi. Neeche ki figure $e^5$ ki terms dikhati hai jo pehle badhti hain (jab $n<5$) phir collapse ho jaati hain jab $n>5$ ho jaata hai.

Answer: (B). KYun: $\sin$ ek odd function hai ($\sin (-x)=-\sin x$), isliye sirf $x$ ki odd powers aa sakti hain. Isse (C) turant khatam ho jaata hai (usmein even powers hain — yeh toh $\cos x$ hai). (A) mein odd aur even powers hain ($x^2$), aur uske denominators plain integers hain, factorials nahi — yeh toh $\ln (1+x)$ ka pattern hai ek galat sign ke saath. Toh (A) do baar galat hai. Sahi series mein factorial denominators aur alternating signs hain: $x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots$

KYA: $x=0.2$ ko $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$ mein plug karo. $1-\frac{0.04}{2}+\frac{0.0016}{24}=1-0.02+0.0000\overline{6}=0.98006\overline{6}$ Yeh kaam kyun karta hai: $x^6$ term lagbhag $\frac{(0.2{)}^6}{720}\approx 9\times 10^{-8}$ hai, jo humari zaroorat ki accuracy se bahut kam hai. True value: $\cos (0.2)=0.980066\dots$ ✓ Neeche ki figure dikhati hai ki truncated cosine polynomial kitni tezi se true curve par lock ho jaati hai jab tum $x^2$ phir $x^4$ term add karte ho.

KYA / KYun: scratch se differentiate mat karo — jaane-maane $e^x$ series mein $x\mapsto -2x$ substitute karo. Yeh legal hai kyunki $e^x$ har input ke liye converge karta hai, toh $-2x$ hamesha ek valid input hai. $e^{-2x}=1+(-2x)+\frac{(-2x{)}^2}{2!}+\frac{(-2x{)}^3}{3!}+\cdots$ $=1-2x+\frac{4x^2}{2}-\frac{8x^3}{6}+\cdots =1-2x+2x^2-\frac{4}{3}{x}^3+\cdots$

KYA: $(1+x{)}^m$ use karo jisme $m=\frac{1}{2}$ ho. $\left(\genfrac{}{}{0ex}{}{1/2}{0}\right)=1,\left(\genfrac{}{}{0ex}{}{1/2}{1}\right)=\frac{1}{2},\left(\genfrac{}{}{0ex}{}{1/2}{2}\right)=\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}=\frac{\frac{1}{2}\cdot (-\frac{1}{2})}{2}=-\frac{1}{8}$ $\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\cdots (|x|<1)$ Check $x=0.21$ par: $1+0.105-0.0055125=1.0994875\approx \sqrt{1.21}=1.1$ ✓

KYA: $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$ ko term by term differentiate karo. $\frac{d}{dx}\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \right)=1-\frac{3x^2}{3!}+\frac{5x^4}{5!}-\cdots$ Denominators kyun collapse hote hain: $\frac{3}{3!}=\frac{3}{6}=\frac{1}{2!}$ aur $\frac{5}{5!}=\frac{5}{120}=\frac{1}{4!}$ — numerator factorial ke top factor ko cancel kar deta hai. $=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots =\cos x✓$ Yahi wajah hai ki differentiation ke under $\sin \to \cos$ kyun hota hai polynomials ke level par bhi.

$x=1$ par: series ban jaati hai $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$, jo alternating harmonic series hai, jo converge karti hai ($\ln 2$ par). Iske terms $0$ ki taraf shrink karte hain aur alternate karte hain — convergence ke liye itna kaafi hai. $x=-1$ par: $x=-1$ substitute karo: ${\sum }_{n=1}^{\infty }\frac{(-1{)}^{n-1}(-1{)}^n}{n}={\sum }_{n=1}^{\infty }\frac{(-1{)}^{2n-1}}{n}={\sum }_{n=1}^{\infty }\frac{-1}{n}=-(1+\frac{1}{2}+\frac{1}{3}+\cdots )$ Yeh $-\infty$ hai (harmonic series diverge karti hai). Yeh function se match kyun karta hai: $\ln (1+(-1))=\ln 0=-\infty$. Series diverge hokar singularity ko honestly report karti hai. Figure kya dikhati hai. Neeche ki figure true curve $\ln (1+x)$ (white) ko $3$, $8$ aur $20$ terms wale partial sums (blue, yellow, green) ke against plot karti hai. Do cheezein dekho: (1) $(-1,1]$ ke andar har partial sum white curve se chipka rehta hai, aur jitne zyada terms add karo utna tighter hug; (2) jaise $x\to -1^{+}$, true curve $-\infty$ ki taraf goti hai, aur partial sums us wall ke paas smoothly follow karne mein fail karte hain — finite number of terms $-\infty$ tak nahi pahunch sakti, jo red dashed line se marked singularity par series ke exactly diverge hone ka visual signature hai. Right edge $x=1$ (blue dotted line) par sums $\ln 2$ par settle ho jaate hain, jo convergence ka visual signature hai.

KYA: raw expression $x=0$ par $\frac{0}{0}$ hai — plain arithmetic se undefined. Lekin series ko $x$ se divide karo: $\frac{\sin x}{x}=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots }{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots$ Yeh legal kyun hai: $\sin x$ ki har term mein $x$ ka kam se kam ek factor hai, toh $x$ se divide karne par ek honest polynomial milta hai jisme division by zero bilkul nahi hai. Ab $x=0$ rakhne par $1$ milta hai. Toh removable hole ko $\frac{\sin x}{x}{|}_0=1$ se fill karna chahiye.

KYA: $(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots )(x-\frac{x^3}{6}+\cdots )$ multiply karo aur $x$ ki powers collect karo.

• $x^1$: $1\cdot x=x$
• $x^2$: $x\cdot x=x^2$
• $x^3$: $\frac{x^2}{2}\cdot x+1\cdot (-\frac{x^3}{6})=\frac{1}{2}{x}^3-\frac{1}{6}{x}^3=\frac{1}{3}{x}^3$ $e^x\sin x=x+x^2+\frac{x^3}{3}+\cdots$ Multiply kyun karein, product differentiate kyun nahi: do known series ke coefficients match karna product ke derivatives compute karne (product rule chains) ko poori tarah avoid karta hai.

KYA: $e^x=\sum \frac{x^n}{n!}$ mein $x\mapsto ix$ rakho, jahan $i^2=-1$. $e^{ix}={\sum }_{n=0}^{\infty }\frac{(ix{)}^n}{n!}=1+ix+\frac{(ix{)}^2}{2!}+\frac{(ix{)}^3}{3!}+\cdots$ $i$ ki powers cycle karti hain: $i^0=1,i^1=i,i^2=-1,i^3=-i,i^4=1,\dots$ Toh: $=\underset{\cos x}{\underbrace{(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots )}}+i\underset{\sin x}{\underbrace{(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots )}}$ Yeh kaam kyun karta hai: real terms (even powers) exactly $\cos x$ assemble karte hain; imaginary terms (odd powers) exactly $\sin x$ assemble karte hain. Teen series ek identity mein fuse ho jaati hain. ✓

KYA: $(1-x^2{)}^{-1/2}$. Binomial use karo $m=-\frac{1}{2}$ aur input $u=-x^2$ ke saath. $(1+u{)}^{-1/2}=1-\frac{1}{2}u+\frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}{u}^2+\cdots =1-\frac{1}{2}u+\frac{3}{8}{u}^2+\cdots$ $u=-x^2$ substitute karo: $\frac{1}{\sqrt{1-x^2}}=1+\frac{1}{2}{x}^2+\frac{3}{8}{x}^4+\cdots$ Connection kyun hai: integrate karne par milta hai $\arcsin x=x+\frac{x^3}{6}+\frac{3x^5}{40}+\cdots$, kyunki $\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}$.

KYA: $\cos x$ series ko numerator mein substitute karo. $\cos x-1+\frac{x^2}{2}=(1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots )-1+\frac{x^2}{2}=\frac{x^4}{24}-\frac{x^6}{720}+\cdots$ Low powers kyun cancel hote hain: $1$ aur $-\frac{x^2}{2}$ deliberately subtract kiye gaye hain, jisse pehla survivor $x^4$ par milta hai. $\frac{\text{numerator}}{x^4}=\frac{1}{24}-\frac{x^2}{720}+\cdots \stackrel{x\to 0}{\to }\frac{1}{24}$ Answer: $\frac{1}{24}$. (L'Hôpital ko chaar differentiations ki zaroorat hogi — series yeh ek hi line mein karta hai.)

KYA — radius: coefficients $\left(\genfrac{}{}{0ex}{}{1/3}{k}\right)$ hain. Ratio test apply karo — consecutive terms compare karo: $\left|\frac{a_{k+1}{x}^{k+1}}{a_k{x}^k}\right|=\left|\frac{\left(\genfrac{}{}{0ex}{}{1/3}{k+1}\right)}{\left(\genfrac{}{}{0ex}{}{1/3}{k}\right)}\right||x|=\left|\frac{\frac{1}{3}-k}{k+1}\right||x|\stackrel{k\to \infty }{\to }|x|$ Ratio $\to |x|$ kyun: $\frac{\frac{1}{3}-k}{k+1}\to -1$ magnitude mein jaise $k\to \infty$. Convergence ke liye yeh limit $<1$ chahiye, yaani $|x|<1$. Radius $=1$. Endpoint $x=1$ — converge karta hai. Bade $k$ ke liye coefficients $\left(\genfrac{}{}{0ex}{}{1/3}{k}\right)$ size mein $\frac{C}{k^{4/3}}$ jaisa behave karte hain (har naya factor $\left|\frac{\frac{1}{3}-k}{k+1}\right|=1-\frac{4/3}{k+1}+\cdots$ term ko itna shave karta hai jo accumulate hokar $k^{-4/3}$ decay deta hai), aur signs kisi point se alternate karte hain. Ek series jiske terms $k^{-4/3}$ jaisi shrink karti hain absolutely converge karti hai (kyunki $\sum k^{-4/3}$ converge karta hai, yeh $p$-series hai $p=\frac{4}{3}>1$ ke saath). Toh $x=1$ par series converge karti hai — honest value $2^{1/3}$ par. Endpoint $x=-1$ — yeh bhi converge karta hai. $x=-1$ par alternating signs hat jaate hain, toh har term ki size $\approx C{k}^{-4/3}$ hai aur sab same sign ke saath add hote hain. Lekin $\sum k^{-4/3}$ phir bhi converge karta hai (same $p$-series argument), toh series $x=-1$ par bhi converge karti hai — $0^{1/3}=0$ par, jo $(1+(-1){)}^{1/3}=0$ se match karta hai. (Function wahan finite hai, aur series theek behave karti hai.) Toh poora interval of convergence $[-1,1]$ hai. Isse compare karo $(1+x{)}^{-1/2}$ se, jahan coefficients sirf $k^{-1/2}$ jaisi decay karti hain aur endpoint $x=-1$ diverge karta hai — coefficients ki decay rate hi decide karti hai ki har endpoint par kya hoga. $|x|=1$ se thoda aage, step ratio $1$ se zyada ho jaata hai aur terms badhti hain, toh sum diverge karta hai chahe $(1+x{)}^{1/3}$ wahan perfectly finite number ho.

KYA: koi bhi $x$ fix karo aur ek integer $N>2|x|$ chuno. $n>N$ ke liye, pichli term ko multiply karne wala har naya factor $\frac{|x|}{n}<\frac{|x|}{N}<\frac{1}{2}$ hai. Yeh kyun khatam karta hai: $n=N$ ke baad, har step size ko kam se kam aadha kar deta hai: $\frac{|x{|}^n}{n!}<\frac{|x{|}^N}{N!}{\left(\frac{1}{2}\right)}^{n-N}\stackrel{n\to \infty }{\to }0$ Ek quantity jo ek shrinking geometric ratio (Geometric series) se bounded hai, woh $0$ par jaani hi chahiye. Isliye factorial hamesha power ko beat karta hai, "valid for all $x$" claim confirm karta hai — wahi halving picture jo tumne Exercise 1.1 ki figure mein dekhi thi. ✓

Kya tum memory se kar sakte ho: (1) paanchi series likho, (2) substitution se $e^{-2x}$ nikalo, (3) $\ln$ endpoint failure explain karo, (4) $e^x\sin x$ ke liye do series multiply karo, (5) $x^4$ limit karo, (6) $(1+x{)}^{1/3}$ ka radius of convergence aur dono endpoints batao? Agar koi shaky lage, toh us level ko dobara karo.