Exercises — Applications — approximation, evaluating limits

Seedha toolkit se: $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots =1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots$ Dhyan do sirf even powers aate hain, alternating $+,-,+$ ke saath. Aisa isliye hota hai kyunki $\cos$ ek even function hai ($y$-axis ke paas mirror image), isliye isme $x$ ya $x^3$ jaisi odd powers nahi aa sakti.

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$ se, $x^5$ ka term hai $+\frac{x^5}{120}$, isliye $c_5=\frac{1}{5!}=\frac{1}{120}.$

$e^x$ ka har derivative phir se $e^x$ hi hota hai. Toh $f^{(7)}(x)=e^x$, aur $x=0$ par: $f^{(7)}(0)=e^0=1.$ Kyunki $c_n=\frac{f^{(n)}(0)}{n!}=\frac{1}{n!}$ sabhi $n$ ke liye, isi wajah se exactly $e^x=\sum \frac{x^n}{n!}$ aata hai.

$\cos (0.2)\approx 1-\frac{(0.2{)}^2}{2}+\frac{(0.2{)}^4}{24}.$ Har piece compute karo:

• $\frac{(0.2{)}^2}{2}=\frac{0.04}{2}=0.02$
• $\frac{(0.2{)}^4}{24}=\frac{0.0016}{24}\approx 0.0000667$

Toh $\cos (0.2)\approx 1-0.02+0.0000667=0.9800667$. (Sahi value $0.9800666\dots$)

Yahan $\ln (1.1)=\ln (1+0.1)$, toh hum $x=0.1$ rakhte hain (safely $|x|<1$ ke andar): $\ln (1+0.1)\approx 0.1-\frac{(0.1{)}^2}{2}=0.1-0.005=\mathrm{0.095.}$ (Sahi value $0.0953102\dots$ — do terms se teen decimals tak accuracy mil gayi.)

$e^x=1+x+\frac{x^2}{2}+\cdots$ mein directly $x=-0.1$ substitute karo: $e^{-0.1}\approx 1+(-0.1)+\frac{(-0.1{)}^2}{2}=1-0.1+0.005=\mathrm{0.905.}$ (Sahi value $0.9048374\dots$) $x$ ka sign automatically carry through hota hai — koi alag rule nahi chahiye.

Value. $0.3-\frac{(0.3{)}^3}{6}=0.3-\frac{0.027}{6}=0.3-0.0045=0.2955$.

Error. $\sin$ series alternating hai aur terms size mein shrink kar rahi hain, toh Alternating Series Test se error zyada se zyada pehla omitted term hoga: $\left|R\right|\le \frac{(0.3{)}^5}{5!}=\frac{0.00243}{120}\approx 2.03\times 10^{-5}.$ Toh $\sin (0.3)=0.2955$ lagbhag $2\times 10^{-5}$ ke andar sahi hai. (Sahi value $0.2955202\dots$ ✓)

Lagrange remainder hai $R_N(x)=\frac{f^{(N+1)}(c)}{(N+1)!}{x}^{N+1}$ jahan $c\in (0,0.5)$. $f=e^x$ ke liye, $f^{(N+1)}(c)=e^c\le e^{0.5}<1.65$. $M=1.65$ lo: $|R_N(0.5)|\le \frac{1.65}{(N+1)!}(0.5{)}^{N+1}.$ Values test karo:

• $N=3$: $\frac{1.65}{24}(0.5{)}^4=\frac{1.65}{24}(0.0625)\approx 4.30\times 10^{-3}$ — bahut bada hai.
• $N=4$: $\frac{1.65}{120}(0.5{)}^5=\frac{1.65}{120}(0.03125)\approx 4.30\times 10^{-4}<10^{-3}$ ✓

Toh $N=4$ sabse chhota hai jo accuracy guarantee karta hai. Dekho kaise bound collapse hota hai jab $N$ badhta hai.

Kyun invalid hai. Series $\ln (1+x)=x-\frac{x^2}{2}+\cdots$ sirf $-1<x\le 1$ ke liye converge hoti hai — iska radius of convergence $1$ hai. $x=2$ par terms $2,-2,\frac{8}{3},\dots$ grow karti hain, isliye sum ka koi matlab nahi. Valid rewrite. Likho $\ln 3=\ln \left(\frac{1+t}{1-t}\right)$ jahan $\frac{1+t}{1-t}=3\Rightarrow t=\frac{1}{2}$ (disc ke andar). Phir use karo $\ln \frac{1+t}{1-t}=2\left(t+\frac{t^3}{3}+\frac{t^5}{5}+\cdots \right),$ jo $t=\frac{1}{2}$ ke liye quickly converge hoti hai. Isse har argument legal range ke andar rehta hai.

Plug karo series: $\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots$, toh $1-\cos x=\frac{x^2}{2}-\frac{x^4}{24}+\cdots .$ Cancel / divide karo $x^2$ se: $\frac{1-\cos x}{x^2}=\frac{1}{2}-\frac{x^2}{24}+\cdots .$ Crawl $x\to 0$: $x^2$-term khatam ho jaata hai, bas $\boxed{\frac{1}{2}}$ bachta hai.

$e^u$ series mein $x^2$ substitute karo ($u=x^2$): $e^{x^2}=1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\cdots .$ $1+x^2$ subtract karo: $e^{x^2}-1-x^2=\frac{x^4}{2}+\frac{x^6}{6}+\cdots$. $x^4$ se divide karo: $\frac{1}{2}+\frac{x^2}{6}+\cdots \to \boxed{\frac{1}{2}}.$ Sabse chhota surviving power $x^4$ tha, jo denominator se exactly match karta hai — isliye hume series $x^4$ tak chahiye thi.

Numerator: $\ln (1+x)-x=\left(x-\frac{x^2}{2}+\cdots \right)-x=-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$ Denominator: $x\sin x=x\left(x-\frac{x^3}{6}+\cdots \right)=x^2-\frac{x^4}{6}+\cdots$ Dono $x^2$ se shuru hote hain. Top aur bottom ko $x^2$ se divide karo: $\frac{-\frac{1}{2}+\frac{x}{3}-\cdots }{1-\frac{x^2}{6}+\cdots }\stackrel{x\to 0}{\to }\frac{-\frac{1}{2}}{1}=\boxed{-\frac{1}{2}}.$

Hume $\tan x=\frac{\sin x}{\cos x}$ ko series ke roop mein chahiye. Series division use karo: $\tan x=\frac{x-\frac{x^3}{6}+\cdots }{1-\frac{x^2}{2}+\cdots }.$ Numerator ko ${\left(1-\frac{x^2}{2}\right)}^{-1}=1+\frac{x^2}{2}+\cdots$ se multiply karo: $\tan x=\left(x-\frac{x^3}{6}\right)\left(1+\frac{x^2}{2}\right)+\cdots =x+\frac{x^3}{2}-\frac{x^3}{6}+\cdots =x+\frac{x^3}{3}+\cdots$ Isliye $\tan x-x=\frac{x^3}{3}+\cdots$, aur $\frac{\tan x-x}{x^3}=\frac{1}{3}+\cdots \to \boxed{\frac{1}{3}}.$

Series. $e^{-x^2}=1-x^2+\frac{x^4}{2}-\frac{x^6}{6}+\cdots$ ($e^u$ mein $-x^2$ substitute karke). Kyunki ye $x$ mein alternating, shrinking series hai, hum term by term integrate kar sakte hain: ${\int }_0^{0.5}\left(1-x^2+\frac{x^4}{2}\right)dx={\left[x-\frac{x^3}{3}+\frac{x^5}{10}\right]}_0^{0.5}.$ $0.5$ par evaluate karo:

• $0.5$
• $-\frac{(0.5{)}^3}{3}=-\frac{0.125}{3}=-0.0416667$
• $+\frac{(0.5{)}^5}{10}=\frac{0.03125}{10}=0.003125$

Sum: $0.5-0.0416667+0.003125=0.4614583$.

Error bound. Pehla omitted integrated term aata hai $-\frac{x^6}{6}$ se: $\left|{\int }_0^{0.5}\frac{x^6}{6}dx\right|=\frac{(0.5{)}^7}{7\cdot 6}=\frac{0.0078125}{42}\approx 1.86\times 10^{-4}.$ Toh ${\int }_0^{0.5}{e}^{-x^2}dx\approx 0.46146$ error $<2\times 10^{-4}$ ke saath. (Sahi value $0.4612810\dots$ ✓)

Do expansions. $\sin x=x-\frac{x^3}{6}+\cdots ,x\cos x=x\left(1-\frac{x^2}{2}+\cdots \right)=x-\frac{x^3}{2}+\cdots .$ Subtract karo, degree ke hisaab se align karke: $\sin x-x\cos x=\left(x-\frac{x^3}{6}\right)-\left(x-\frac{x^3}{2}\right)+\cdots =-\frac{x^3}{6}+\frac{x^3}{2}+\cdots =\frac{x^3}{3}+\cdots$ Divide karo $x^3$ se: $\frac{1}{3}+\cdots \to \boxed{\frac{1}{3}}$.