Visual walkthrough — Taylor's remainder theorem — error estimation

4.3.18 · D2 · Maths › Calculus III — Sequences & Series › Taylor's remainder theorem — error estimation

Step 1 — Approximation kya hota hai, aur "error" kya hota hai?

KYA. Hamare paas ek curvy function $f$ hai. Ek chosen point $a$ ke paas hum isse ek simpler curve se replace karte hain — ek polynomial $P_n$ — jo sirf $+,-,\times$ of $x$ se bani hoti hai (koi square roots nahi, koi $\sin$ nahi, kuch bhi exotic nahi). Dono curves $a$ par milti hain aur jaise-jaise aap door chalte ho, alag hoti jaati hain.

KYUN. Polynomials hi woh akeli functions hain jo ek calculator genuinely compute kar sakta hai. Toh hum true curve ko polynomial se trade karte hain aur ek honest sawaal poochte hain: ye dono kitni door hain?

PICTURE. Neeche, burnt-orange curve true $f$ hai; teal curve polynomial $P_n$ hai. Ye $x=a$ par ek doosre ko kiss karti hain. Target point $x$ par plum vertical gap woh cheez hai jise hum bound karna chahte hain.

Step 2 — Gap kholne wali cheez kya hai? Bending.

KYA. Do curves jo $a$ par milti hain, tabhi alag hoti hain jab ek doosri se bend karke door jaaye. Bending ki maatra derivatives se measure hoti hai — ek derivative bas "koi quantity kitni tezi se badal rahi hai" hota hai.

YE TOOL KYUN. Hum directly pooch sakte the "curves kitni door hain?", par hume $f$ ki value nahi pata (isiliye toh hum approximate kar rahe the). Jo hum often control kar sakte hain woh hai $f$ kitni curvy hai — jaise ek derivative ko bound karna: "$|f^{\prime \prime }|$ kabhi $5$ se zyada nahi hoti". Toh hum poora argument derivatives ke through route karte hain, kyunki bending woh cheez hai jise hum pin down kar sakte hain.

PICTURE. Gently bending curve (chota next derivative) apne polynomial ke paas rehti hai; violently bending curve jaldi alag ho jaati hai. Same polynomial degree, wildly different gaps.

Step 3 — Gap ko ek aisi function mein badlo jise hum slide kar sakein

KYA. Target point $x$ ko freeze karo. Ab polynomial ka centre move karo; moving centre ko $t$ kaho. Define karo: $g(t)=f(x)-{\sum }_{k=0}^n\frac{f^{(k)}(t)}{k!}(x-t{)}^k.$

KYUN. Yeh ek trick hai taaki unknown gap ek interval ke ek end par function ki value ke roop mein aaye, jisse hum baad mein ek Mean-Value tool se hit kar sakein. $g(t)$ padho as: "agar main apna sketch $a$ ki jagah $t$ par centre karta toh jo leftover error hota."

PICTURE. Centre $t$ ko $a$ se slide karke $x$ tak le jaao. Jab $t$ $x$ par pahunche, toh $x$ par centred sketch wahan perfect hota hai — gap snap karke band ho jaata hai.

Step 4 — $g$ differentiate karo: ek magic telescope

KYA. Poochho ki centre $t$ slide hone par $g$ kitni tezi se change hoti hai. Sum ke har term mein do moving pieces hain ($f^{(k)}(t)$ change hota hai, aur $(x-t{)}^k$ change hota hai), toh product rule har term se do pieces deta hai. Consecutive terms ek chain mein cancel ho jaate hain — ek telescope — aur sirf sabse aakhri bachta hai: $g^{\prime }(t)=-\frac{f^{(n+1)}(t)}{n!}(x-t{)}^n.$

KYUN. Yeh Taylor structure ka payoff hai: messy sum ka ek clean derivative hai. Saare "book-keeping" derivatives cancel ho jaate hain, sirf next derivative $f^{(n+1)}$ bachta hai — wahi bending jo humne Step 2 mein identify ki thi.

PICTURE. Dominoes ki ek row imagine karo jahan har term ka "$+$" next term ke "$-$" ko cancel karta hai. Sirf sabse door-right wala piece khada rehta hai.

Step 5 — Ek partner function, aur Mean Value comparison

KYA. Ek simple companion introduce karo: $h(t)=(x-t{)}^{n+1},h(a)=(x-a{)}^{n+1},h(x)=0,h^{\prime }(t)=-(n+1)(x-t{)}^n.$ Ab $g$ aur $h$ ki rates ko Cauchy Mean Value Theorem se compare karo: $a$ aur $x$ ke beech ke interval par koi hidden point $c$ hai jahan: $\frac{g(a)-g(x)}{h(a)-h(x)}=\frac{g^{\prime }(c)}{h^{\prime }(c)}.$

YE TOOL KYUN, ORDINARY Mean Value Theorem NAHI. Plain MVT kehta hai ki ek single curve ka koi point hoga jiska slope average slope ke barabar hoga. Yahan hamare paas do quantities hain jo side by side race kar rahi hain aur hum unke net changes ko unke instantaneous rates at the same instant $c$ se relate karna chahte hain. "Same instant for both" exactly wahi hai jo Cauchy's version deliver karta hai — yeh MVT ka do-runner generalization hai.

PICTURE. Do runners $g$ aur $h$ same track $[a,x]$ par. Kisi time $c$ par unka speed ratio unke total-distance ratio ke barabar hoga. Woh shared instant $c$ wahan hai jahan hamara unknown rehta hai.

Step 6 — Substitute karo aur cancellation dekho

KYA. Charon anchor values daalo. Left side hai $\frac{R_n(x)-0}{(x-a{)}^{n+1}-0}$. Right side hai: $\frac{g^{\prime }(c)}{h^{\prime }(c)}=\frac{-\frac{f^{(n+1)}(c)}{n!}(x-c{)}^n}{-(n+1)(x-c{)}^n}.$

KYUN. Do minus signs cancel ho jaate hain; poora factor $(x-c{)}^n$ upar aur neeche cancel ho jaata hai; aur $n!\cdot (n+1)=(n+1)!$. Saari unknown-position clutter evaporate ho jaati hai.

PICTURE. Plum $(x-c{)}^n$ blocks numerator aur denominator se gayab ho jaate hain, ek bare stack chodke.

Step 7 — Exact se usable tak: $c$ ko ceiling $M$ se replace karo

KYA. Hum $c$ locate nahi kar sakte, par hum keh sakte hain "woh jahan bhi ho, $|f^{(n+1)}|$ interval par kabhi kisi number $M$ se zyada nahi hoti." $f^{(n+1)}(c)$ ko uski worst-case size $M$ se replace karna estimate ko sirf inflate kar sakta hai, shrink nahi: $|R_n(x)|\le \frac{M}{(n+1)!}|x-a{|}^{n+1}.$

KYUN. Ek guaranteed bound ko worst possible $c$ survive karna chahiye. Maximum $M$ use karna inequality ko har hidden $c$ ke liye airtight banata hai. (Ek single guessed $c$ choose karna — ek classic mistake — undershoot kar sakta hai aur ek false guarantee de sakta hai.)

PICTURE. True $|f^{(n+1)}|$ ek flat teal ceiling $M$ ke neeche wiggles karta hai. $c$ chahe kuch bhi ho, uska bar ceiling ke upar nahi ja sakta.

Step 8 — Edge & degenerate cases (reader ko kabhi stranded mat chhodho)

KYA / KYUN / PICTURE, teen scenarios jinhe bound survive karna chahiye:

• $x=a$ (degenerate, zero walk). Tab $|x-a{|}^{n+1}=0$, toh bound $0$ hai: sketch aur truth exactly centre par coincide karte hain. Koi error nahi, jaisa hona chahiye.
• $|x-a|>1$ (far walk). Power growti hai, shrink nahi hoti. Yahan denominator mein factorial ko growing power se aage nikalna hoga; ek fixed $x$ ke liye yeh eventually ho jaata hai (factorials kisi bhi fixed power ko beat karte hain), jo ki Radius of Convergence idea $R_n\to 0$ ka seed hai.
• Alternating series (extra help). Agar tail sign alternate kare (jaise $\sin$, $\cos$), toh consecutive errors partially cancel ho jaate hain, toh true gap often Lagrange ceiling se far below hoti hai — woh slack exactly wahi hai jo Alternating Series Estimation Theorem capture karta hai.

Ek-picture summary

Is page ki saari cheezein ek image mein compress hoti hain: true curve, polynomial, gap $R_n$, bending ceiling $M$, walk $|x-a|$, aur factorial crusher — sabh milke boxed bound banate hain. Big-O and asymptotic error language mein yeh $R_n=O\left(|x-a{|}^{n+1}\right)$ hai.

Recall

Connections

• Parent · Hinglish — woh stated result jise yeh page draw karta hai.
• Taylor & Maclaurin Series — woh polynomial $P_n$ jise bound kiya ja raha hai.
• Mean Value Theorem / Cauchy Mean Value Theorem — Steps 5–6 ka engine.
• Radius of Convergence — far-walk case aur $R_n\to 0$.
• Alternating Series Estimation Theorem — tighter bound jab tail alternate kare.
• Big-O and asymptotic error — $R_n=O(|x-a{|}^{n+1})$.