Exercises — Taylor's remainder theorem — error estimation

Error bound (parent note se): $|R_n(x)|\le \frac{M}{(n+1)!}|x-a{|}^{n+1},M={\max }_{t\text{ between }a,x}|f^{(n+1)}(t)|.$ Ise aise padhein: "Next derivative, Next factorial, Next power" — sab kuch $n+1$ hai, polynomial degree $n$ se ek kadam aage.

• $R_n(x)=f(x)-P_n(x)$ woh exact error hai.
• $P_n$ degree-$n$ Taylor polynomial hai.
• Yeh bound Lagrange form $R_n=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a{)}^{n+1}$ se aata hai, jahan $c$ unknown hai, isliye hum $f^{(n+1)}(c)$ ko uske worst-case size $M$ se replace kar dete hain.

Symbols ka MATLAB. $n$ us polynomial ki degree hai jo aapne rakhi. Bound hamesha $n$ se ek kadam aage use karti hai.

• Derivative chahiye: $f^{(n+1)}=f^{(5)}$ — fifth derivative.
• Factorial: $(n+1)!=5!=120$.
• Power: $|x-a{|}^5$.

$n+1$ kyun aur $n$ kyun nahi. Polynomial $P_4$ pehle se hi $f$ se 4th derivative tak match karti hai; pehli cheez jo woh galat karti hai woh 5th derivative term hai, isliye wahi leftover error ko control karti hai.

$0$ par derivatives: $f(0)=\cos 0=1$, $f^{\prime }(0)=-\sin 0=0$, $f^{\prime \prime }(0)=-\cos 0=-1$. $P_2(x)=1+0\cdot x-\frac{1}{2}{x}^2=1-\frac{x^2}{2}.$ $R_2(x)=\cos x-\left(1-\frac{x^2}{2}\right)$ — true cosine aur is parabola ke beech ka exact gap.

$f(x)=e^x$, $a=0$, $n=2$, $x=0.2$.

• $f^{(3)}(t)=e^t$. $[0,0.2]$ par yeh increasing hai, $t=0.2$ par sabse bada: $e^{0.2}\approx 1.221<1.5$. $M=1.5$ lein.
• Over-estimate kyun? Hum ek safe upper bound chahte hain; $M$ thoda bada lene se guaranteed error thodi badi hogi, kabhi galat nahi. $|R_2(0.2)|\le \frac{1.5}{3!}(0.2{)}^3=\frac{1.5}{6}(0.008)=\mathrm{0.002.}$ Toh $P_2(0.2)=1.22$ $0.002$ ke andar sahi hai. (True $e^{0.2}=1.22140\dots$, actual error $\approx 0.00140$ — bound ke andar. ✓)

$f(x)=\sin x$, $a=0$, $n=3$.

• $f^{(4)}(t)=\sin t$, aur $|\sin t|\le 1$ hamesha, isliye $M=1$ (koi interval work nahi chahiye — sabse clean possible bound). $|R_3(0.5)|\le \frac{1}{4!}(0.5{)}^4=\frac{0.0625}{24}\approx 2.60\times 10^{-3}.$ $P_3(0.5)=0.5-0.020833=0.479167$; true $\sin 0.5=0.479426$, actual error $\approx 2.6\times 10^{-4}$. ✓

Derivatives: $f^{\prime }(x)=(1+x{)}^{-1}$, $f^{\prime \prime }(x)=-(1+x{)}^{-2}$, $f^{\prime \prime \prime }(x)=2(1+x{)}^{-3}$.

• Hume $M=\max |f^{\prime \prime \prime }(t)|$ chahiye $[0,0.3]$ par. Kyunki $2(1+t{)}^{-3}$ decrease karta hai jab $t$ badhta hai, uska max $t=0$ par hai: $M=2(1{)}^{-3}=2$.
• Left endpoint kyun? $(1+t{)}^{-3}$ shrink hota hai jab $t$ badhta hai, isliye worst (largest) case sabse chhote $t$ par hai. $|R_2(0.3)|\le \frac{2}{3!}(0.3{)}^3=\frac{2}{6}(0.027)=\mathrm{0.009.}$

$n=1$, toh hume $f^{(2)}(t)=-\sin t$ chahiye, jo deta hai $|f^{(2)}(t)|=|\sin t|$ $[0,\pi ]$ par.

• Max KAHAAN hai? $|\sin t|$ dono endpoints par zero hai ($\sin 0=0$, $\sin \pi =0$) aur interior point $t=\pi /2$ par $1$ tak utha hua hai. Toh maximum strictly inside hai — $M=1$ $t=\pi /2$ par, kisi endpoint par nahi. Figure dekhein: peak (red dot) beech mein baithti hai, edges par nahi.

  

• Yahan endpoints fail kyun hote: $t=0$ ya $t=\pi$ plug karne se $|f^{(2)}|=0$ milta hai, jo falsely zero error claim karta — clearly galat, kyunki $\sin \pi =0$ hai lekin $P_1(\pi )=\pi \ne 0$. $|R_1(\pi )|\le \frac{1}{2!}|\pi -0{|}^2=\frac{{\pi }^2}{2}\approx \mathrm{4.93.}$ Ek bada lekin honest bound: itni door linear approximation genuinely terrible hai, aur sirf interior maximum ise capture karta hai.

$f(x)=e^x$, $a=0$, $x=0.5$. $[0,0.5]$ par, $f^{(n+1)}(t)=e^t\le e^{0.5}<2$, isliye $M=2$. $|R_n(0.5)|\le \frac{2}{(n+1)!}(0.5{)}^{n+1}.$ $<10^{-5}$ ki demand karein. Values test karein:

• $n=4$: $\frac{2(0.5{)}^5}{5!}=\frac{2(0.03125)}{120}=5.2\times 10^{-4}$ — bahut bada.
• $n=5$: $\frac{2(0.5{)}^6}{6!}=\frac{2(0.015625)}{720}=4.34\times 10^{-5}$ — abhi bhi bahut bada.
• $n=6$: $\frac{2(0.5{)}^7}{7!}=\frac{2(0.0078125)}{5040}=3.10\times 10^{-6}<10^{-5}$. ✓ Answer: $n=6$ ($x^6/6!$ tak terms). Note karo ki chhota factor $(0.5{)}^{n+1}$ aapki help karta hai — parent note ke $x=1$ case se kaafi kam terms.

$f^{(5)}(t)=\sin t$ (sign tak), toh $M=1$. $|R_4(x)|\le \frac{1}{5!}|x{|}^5=\frac{|x{|}^5}{120}\le 10^{-3}.$ $|x|$ ke liye solve karein. Dono sides ko $120$ se multiply karein: $|x{|}^5\le 0.12$. Ab dono sides ki fifth root lein — yeh legal kyun hai? Function $u\mapsto u^{1/5}$ $u\ge 0$ ke liye increasing hai, isliye yeh inequality direction preserve karta hai; aur $|x|\ge 0$, toh uski fifth root well defined hai. Hence $|x|\le 0.12^{1/5}.$ Compute karein: $0.12^{1/5}=e^{(\ln 0.12)/5}=e^{-0.4241}\approx 0.6544$. Answer: $|x|\lesssim 0.654$ $10^{-3}$ accuracy guarantee karta hai. (Kyunki bound $|x|$ use karta hai, yeh range symmetric hai: $x=+0.6$ aur $x=-0.6$ dono ke liye kaam karta hai — dekho L4.3.)

Yahan $x=0.5$ aur $a=0.5$, isliye $x-a=0$. Har Taylor polynomial $P_n(0.5)=f(0.5)$ exactly deta hai. $|R_1(0.5)|\le \frac{M}{2!}|0.5-0.5{|}^2=\frac{M}{2}\cdot 0=0.$ Zero error — kyunki $|x-a{|}^{n+1}=0$. Yeh degenerate/limiting case hai: aapका centre target ke jitna close hoga, $|x-a|$ utna chhota hoga aur power $|x-a{|}^{n+1}$ error ko utni brutally crush karegi. Yahi poora reason hai ki Taylor approximations "local" hoti hain.

$a=1$, $x=1.5$, toh $x-a=0.5$, $n=3$. $\ln x$ ki derivatives: $f^{\prime }=x^{-1}$, $f^{\prime \prime }=-x^{-2}$, $f^{\prime \prime \prime }=2x^{-3}$, $f^{(4)}=-6x^{-4}$.

• $M={\max }_{[1,1.5]}|-6x^{-4}|=6x^{-4}$. Yeh $x$ mein decrease karta hai, isliye max $x=1$ par: $M=6$.
• $x=1$ kyun? $x^{-4}$ shrink hota hai jab $x$ badhta hai, isliye sabse badi value left endpoint par hai. $|R_3(1.5)|\le \frac{6}{4!}(0.5{)}^4=\frac{6}{24}(0.0625)=\mathrm{0.015625.}$ (Reference ke liye $P_3(x)=(x-1)-\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{3}$, deta hai $P_3(1.5)=0.5-0.125+0.041\overline{6}=0.41\overline{6}$; true $\ln 1.5=0.405465$, actual error $\approx 0.0112<0.0156$. ✓)

$f(x)=e^x$, $a=0$, $n=2$, target $x=-0.4$. "Between $a$ and $x$" interval ab $[-0.4,0]$ hai — yeh centre ke left ki taraf jaata hai.

• Yahan $M$ kya hai. $f^{(3)}(t)=e^t$ $[-0.4,0]$ par increasing hai, isliye uska max right end $t=0$ par hai: $e^0=1$. $M=1$ lein.
• Is baar right end kyun? $e^t$ $t$ ke saath badhta hai, aur $t=0$ $[-0.4,0]$ mein sabse bada $t$ hai — L2.1 reasoning ka mirror image.
• Sign kyun disappear hoti hai. Bound $|x-a{|}^{n+1}=|-0.4-0{|}^3=|-0.4{|}^3=(0.4{)}^3$ use karta hai. Absolute value sign mita deta hai — $0.4$ units left par ek target wahi power deta hai jo $0.4$ units right par. Isliye hum kabhi worry nahi karte ki centre ke kis side par hain. $|R_2(-0.4)|\le \frac{1}{3!}(0.4{)}^3=\frac{0.064}{6}\approx 1.067\times 10^{-2}.$ Check: $P_2(-0.4)=1-0.4+0.08=0.68$; true $e^{-0.4}=0.670320$, actual error $\approx 0.00968<0.01067$. ✓

Parent note se: $[0,1]$ par, $f^{(n+1)}(t)=e^t\le e<3$, toh $M=3$ aur $|R_n(1)|\le \frac{3}{(n+1)!}<10^{-6}⟹(n+1)!>3\times 10^6.$ Kyunki $9!=362880<3\times 10^6$ lekin $10!=3628800>3\times 10^6$, hume $n+1=10$ chahiye, yaani $n=9$ ($x^9/9!$ tak terms).

Galat alternative, clearly stated. Exact Lagrange form hai $R_9(1)=\frac{e^c}{10!}\cdot 1^{10}$ kisi unknown $c\in (0,1)$ ke liye. Ek tempting shortcut yeh hai ki $c$ ki ek specific value guess karein — say $c=0$, jo $e^0=1$ aur error $\approx \frac{1}{10!}$ deta hai — aur use "the error" bulayein. Kyun galat hai: theorem sirf promise karta hai ki koi $c$ exist karta hai; woh kabhi nahi batata kaun sa. Agar true $c$ $1$ ke closer hai, toh real factor $e^c$ up to $e\approx 2.718$ ho sakta hai, aapke guess se lagbhag $3\times$ zyada — toh guessed $c$ error ko under-report kar sakta hai aur aapki guarantee tod sakta hai. Bound kyun jeetta hai: $e^c$ ko pure interval par uske safe maximum $M=3$ se replace karna ek aisi value deta hai jo valid hai chahe $c$ kahaan bhi chhupa ho. Aap thodi badi (lekin certified) number ke liye ek aisa vaada trade karte hain jo kabhi violate nahi ho sakta.

$\sin x$ ki har derivative $\pm \sin$ ya $\pm \cos$ hai, isliye $|f^{(n+1)}(t)|\le 1$ saare $t$ aur saare $n$ ke liye. Hence $M=1$ uniformly kaam karta hai: $|R_n(x)|\le \frac{1}{(n+1)!}|x{|}^{n+1}.$ YEH $\to 0$ kyun hai. Koi bhi real $x$ fix karein. Quantity $\frac{|x{|}^{n+1}}{(n+1)!}$ convergent series $\sum \frac{|x{|}^k}{k!}=e^{|x|}$ ka general term hai; ek convergent series ke terms $0$ ki taraf tend karne chahiye. Isliye $|R_n(x)|\to 0$ har $x$ ke liye. Conclusion: factorial hamesha power ko beat karta hai, chahe $|x|$ kitna bhi bada ho — series everywhere converge hoti hai (infinite radius of convergence). $■$

Lagrange (hamare formula se), $n=3$: $|R_3(1)|\le \frac{M}{4!}|1{|}^4=\frac{1}{24}\approx 0.0417$ ($M=1$ use karte hue). Alternating Series bound: Maclaurin series $\sin 1=1-\frac{1}{3!}+\frac{1}{5!}-\cdots$ alternating hai decreasing terms ke saath, isliye $x^3$ term ke baad error at most first omitted term $\frac{1}{5!}=\frac{1}{120}\approx 0.00833$ hai. Kaun tighter hai: alternating bound $0.00833$ lagbhag 5× chhota hai $0.0417$ se. Kyun. Lagrange bound blindly worst-case derivative $M=1$ use karta hai; alternating theorem $+$ aur $-$ terms ke beech cancellation exploit karta hai, jo extra information hai. Jab koi series alternate karti hai, woh theorem prefer karein; jab nahi karti, Lagrange aapka general tool hai.

Exact Lagrange form: $R_n(x)=\frac{e^c}{(n+1)!}{x}^{n+1}$ kisi $c$ ke liye $0$ aur $x$ ke beech. Jab $x\to 0$, woh $c\to 0$ bhi, toh $e^c\to 1$. Thus $R_n(x)=\frac{e^c}{(n+1)!}{x}^{n+1}=\underset{\text{ek constant ki taraf tend karta hai}}{\underbrace{\frac{1+o(1)}{(n+1)!}}}{x}^{n+1}.$ Dominant behaviour $x^{n+1}$ times ek bounded factor hai, jo exactly $R_n(x)=O(x^{n+1})$ ka meaning hai: error next power of $x$ ki tarah shrink karta hai jab $x\to 0$. Factor $\frac{1}{(n+1)!}$ woh leading constant hai jo Big-O ke andar chhupa hai.