Taylor's remainder theorem — error estimation: Worked examples

4.3.18 · D3 · Maths › Calculus III — Sequences & Series › Taylor's remainder theorem — error estimation

Shuru karne se pehle, sirf teen moving parts ka ek-line refresher, taaki koi symbol bina reason ke na aaye:

Recall Teen ingredients (reveal karne ke liye click karo)

$M$ = ek aisa number jo tumhe pata hai jo $∣ f^{(n + 1)} ∣$ se kam se kam utna bada hai, $a$ aur $x$ ke beech har jagah. Yeh "curve kitni wildly bend karti hai?" wala number hai.
$(n + 1)!$ = polynomial degree $n$ se ek step aage ka factorial. Factorial $= 1 \cdot 2 \cdot 3 \dots (n + 1)$ hai, ek aisa number jo explosively grow karta hai aur error ko crush kar deta hai.
$∣ x - a ∣^{n + 1}$ = centre se kitni dur tum gaye, usi $n + 1$ power tak uthaya hua. Vertical bars $∣ \cdot ∣$ ka matlab hai "distance, hamesha positive" — toh baayi taraf walk ( $x < a$ ) utni hi count hoti hai jitni seedhi taraf.

Neeche har example mein do traps baar baar aate hain. Inhe ek baar pehle hi state kar do, taaki hum inhe naam se cite kar sakein:

Scenario matrix

Taylor error ke bare mein har exam question in cells mein se ek hota hai. Neeche har worked example us cell ke saath tagged hai jo wo fill karta hai.

Cell	Kyun tricky hai	Example
C1 Chota $x > 0$ , centre $a = 0$	"easy" baseline	(1)
C2 Negative $x < 0$	kya $x$ ka sign bound ko tod deta hai?	(2)
C3 Centre $a \neq = 0$	$	x-a
C4 Bada $	x-a	>1$
C4b Boundary $	x-a	=1$ exactly
C5 Degenerate: $x = a$	zero-distance walk — error kya hai?	(5)
C6 Limiting: $n \to \infty$	kya bound $\to 0$ ho jaata hai? (convergence link)	(6)
C7 Real-world word problem with units	"X mm tak accurate" ko ek bound mein translate karo	(7)
C8 Exam twist: forgotten power / wrong factorial	Trap B, live	(8)

Worked examples

(1) Cell C1 — chota positive $x$ , baseline

Worked example

$x = 0.3$ , centre $a = 0$ par $cos x$ ke $P_{3}$ ki error bound karo. Pehle forecast karo: guess karo — kya error $1 0^{- 2}$ , $1 0^{- 4}$ , ya $1 0^{- 6}$ ke aas-paas hogi? Yeh sochke rakho.

Step 1. Polynomial likho. $cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ , toh $P_{3} (x) = 1 - \frac{x ^{2}}{2}$ (the $x^{3}$ term zero hai). Yeh step kyun? Humein $P_{n}$ chahiye taaki pata chale agla derivative kaun sa hai. Yahan $n = 3$ hai, toh remainder $f^{(4)}$ use karta hai (yeh Trap B sahi tarike se karna hai — derivative $n + 1 = 4$ hai).

Step 2. $f^{(4)}$ aur uska bound $M$ dhundho. $cos$ ke derivatives cycle karte hain $- sin, - cos, sin, cos$ , toh $f^{(4)} (t) = cos t$ . $[0, 0.3]$ par, $∣ cos t ∣ \leq 1$ , toh $M = 1$ . Yeh step kyun? $M$ ko poori walk ke worst bend ko cover karna chahiye, kisi guessed point ko nahi (Trap A). Yahan $cos t$ $[0, 0.3]$ par monotonic (decreasing) hai, toh koi interior hump nahi; safe blanket bound $∣ cos ∣ \leq 1$ vaise bhi kaam karta hai.

Step 3. Plug in karo. $n = 3$ , $a = 0$ , $x = 0.3$ ke saath: $∣ R_{3} (0.3) ∣ \leq \frac{1}{4 !} (0.3)^{4} = \frac{0.0081}{24} \approx 3.375 \times 1 0^{- 4} .$ Yeh step kyun? Agla derivative ( $M = 1$ ), agla factorial ( $4!$ ), agla power ( $(0.3)^{4}$ ) — teeno $n + 1 = 4$ carry karte hain.

Verify: true $cos 0.3 = 0.955336$ , aur $P_{3} (0.3) = 1 - 0.045 = 0.955$ . Actual error $\approx 3.4 \times 1 0^{- 4}$ — bilkul bound ke paas (kyunki $cos$ yahan almost $1$ hai, toh $M = 1$ almost exact hai). Tumhara forecast $\sim 1 0^{- 4}$ hona chahiye tha. ✓

(2) Cell C2 — negative $x$ , kya sign tod deta hai?

Worked example

$x = - 0.4$ , centre $a = 0$ par $e^{x}$ ke $P_{2}$ ki error bound karo. Forecast: walk baayi taraf hai. Kya bound $+ 0.4$ tak seedha jaane se chota, bada, ya same hoga?

Step 1. $P_{2} (x) = 1 + x + \frac{x ^{2}}{2}$ , toh $P_{2} (- 0.4) = 1 - 0.4 + 0.08 = 0.68$ . Yeh step kyun? Standard setup; $n = 2$ ka matlab hai $f^{(3)}$ chahiye.

Step 2. $f^{(3)} (t) = e^{t}$ . $a = 0$ aur $x = - 0.4$ ke beech ke interval par (yaani $[- 0.4, 0]$ ), $e^{t}$ strictly increasing hai, isliye monotonic hai, toh uska max right end $t = 0$ par hai: $M = e^{0} = 1$ . Yeh step kyun? Kyunki $e^{t}$ yahan monotonic hai, endpoint rule safe hai — koi interior extremum exist nahi karta. $x$ ke sign ne sirf kaunsa end worst hai yeh badla.

Step 3. Bound $∣ x - a ∣^{n + 1} = ∣ - 0.4 ∣^{3} = (0.4)^{3}$ use karta hai, sirf distance: $∣ R_{2} (- 0.4) ∣ \leq \frac{1}{3 !} (0.4)^{3} = \frac{0.064}{6} \approx 1.067 \times 1 0^{- 2} .$ Yeh step kyun? Bars $∣ \cdot ∣$ minus sign mitaa dete hain — error ka size sirf distance par depend karta hai, direction par nahi. Yahi forecast ka jawab hai: $+ 0.4$ ke barabar tabhi hoga jab $M$ same ho. Yahan $M (+ 0.4) = e^{0.4} \approx 1.49$ hoga jo bada bound dega, toh baayi taraf actually tighter hai.

Verify: true $e^{- 0.4} = 0.670320$ , actual error $= 0.68 - 0.670320 = 9.68 \times 1 0^{- 3} < 1.067 \times 1 0^{- 2}$ . ✓ Bound hold karta hai aur sign ne ise nahi toda.

(3) Cell C3 — centre $a \neq = 0$

Worked example

$a = 1$ par centred $ln x$ ke $P_{2}$ se $ln (1.2)$ approximate karo. Error bound karo. Forecast: centre $1$ hai, target $1.2$ hai. $∣ x - a ∣$ kya hai? (Yahi is cell ka poora point hai.)

Step 1. $f (x) = ln x$ ke derivatives: $f^{'} = \frac{1}{x}$ , $f^{''} = - \frac{1}{x ^{2}}$ , $f^{'''} = \frac{2}{x ^{3}}$ . $a = 1$ par: $f (1) = 0, f^{'} (1) = 1, f^{''} (1) = - 1$ . Toh $P_{2} (x) = 0 + 1 \cdot (x - 1) - \frac{1}{2} (x - 1)^{2} .$ $x = 1.2$ par: $P_{2} = 0.2 - \frac{1}{2} (0.04) = 0.2 - 0.02 = 0.18$ . Yeh step kyun? Jab $a \neq = 0$ hota hai toh har term $(x - a)$ use karta hai, $x$ nahi. Yeh bhoolna classic slip hai.

Step 2. $n = 2 \Rightarrow$ $f^{(3)} (t) = \frac{2}{t ^{3}}$ chahiye. $[1, 1.2]$ par yeh monotonic (decreasing) hai, kyunki $2/ t^{3}$ $t$ badhne ke saath girta hai, toh uska max chote end $t = 1$ par hai: $M = \frac{2}{1 ^{3}} = 2$ . Yeh step kyun? Hum pehle monotonicity verify karte hain phir endpoint lete hain — yahi sahi order hai (Trap A avoid kiya; interior-extremum check trivially pass hua).

Step 3. Yahan $∣ x - a ∣ = ∣1.2 - 1∣ = 0.2$ hai: $∣ R_{2} (1.2) ∣ \leq \frac{2}{3 !} (0.2)^{3} = \frac{2}{6} (0.008) \approx 2.667 \times 1 0^{- 3} .$ Yeh step kyun? Distance $0.2$ hai, $1.2$ nahi — centre ne walk shift kar di. Yahi cell C3 ka poora lesson hai.

Verify: true $ln 1.2 = 0.182322$ , actual error $= 0.182322 - 0.18 = 2.32 \times 1 0^{- 3} < 2.667 \times 1 0^{- 3}$ . ✓

(4) Cell C4 — badi distance, power GROW karti hai

Worked example Tum

$e^{2}$ ko $a = 0$ par centred $P_{3}$ se approximate karne ki koshish karte ho. Error bound kya kehta hai? Forecast: $∣ x - a ∣ = 2 > 1$ ke saath, kya power $2^{n + 1}$ tumhari madad karega ya nuksan pahunchayega?

Step 1. $P_{3} (x) = 1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6}$ , toh $P_{3} (2) = 1 + 2 + 2 + \frac{8}{6} = 6.3333$ . Yeh step kyun? Baseline value; $n = 3$ ko $f^{(4)} = e^{t}$ chahiye.

Step 2. $[0, 2]$ par, $e^{t}$ monotonic increasing hai, toh max door wale end $t = 2$ par hai: $M = e^{2} \approx 7.389$ . (Hum clean safe bound ke liye $M = 8$ tak round up kar sakte hain.) Yeh step kyun? Monotonic ⇒ endpoint safe hai. Ek provable bound ke liye tum $M$ upar round kar sakte ho, lekin neeche kabhi nahi.

Step 3. Ab power dekho: $∣ R_{3} (2) ∣ \leq \frac{e ^{2}}{4 !} (2)^{4} = \frac{7.389}{24} \cdot 16 \approx 4.926.$ Yeh step kyun? $(2)^{4} = 16$ bada hai — power shrink nahi hui kyunki $∣ x - a ∣ > 1$ . Bound bahut bada hai, hume warn karta hai ki $P_{3}$ centre se door ek buri approximation hai.

Verify: true $e^{2} = 7.389056$ , actual error $= 7.389056 - 6.3333 = 1.056$ . Yeh bound $4.926$ ke andar hai (✓) lekin note karo ki dono bade hain. C4 ka lesson: door ki walks ko bahut zyada terms chahiye. Tumhara forecast "nuksan" hona chahiye tha. ✓

(4b) Cell C4b — boundary $∣ x - a ∣ = 1$ , jahan danger real hai

Worked example

$f (x) = ln (1 + x)$ ke liye $a = 0$ par, series $x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots$ hai jiska radius of convergence $1$ hai. Boundary $x = 1$ par Lagrange bound examine karo aur dikhao ki yeh obviously shrink nahi karta. Forecast: $x = 1$ par power $1$ par freeze hai. Guess karo: kya bound phir bhi $1/ (n + 1)!$ ki tarah marta hai, ya $M$ ise sabotage kar deta hai?

Step 1. General derivative dhundho. $g (x) = ln (1 + x)$ ke liye, $g^{(n + 1)} (t) = \frac{( - 1 ) ^{n} n !}{( 1 + t ) ^{n + 1}}$ , toh $g^{(n + 1)} (t) = \frac{n !}{( 1 + t ) ^{n + 1}}$ . Yeh step kyun? Bound ko $M = max ∣ g^{(n + 1)} ∣$ $[0, 1]$ par chahiye. Note karo derivative ke andar factor $n!$ — yahi sabotage hai.

Step 2. $[0, 1]$ par, $(1 + t)^{n + 1}$ $t = 0$ par sabse chota hai, toh $∣ g^{(n + 1)} ∣$ $t = 0$ par sabse bada hai: $M = \frac{n !}{1 ^{n + 1}} = n!$ . Yeh step kyun? Monotonic ( $t$ mein decreasing) ⇒ endpoint $t = 0$ worst point hai (Trap A handle kiya). Lekin $M = n!$ $n$ ke saath grow karta hai — ek red flag.

Step 3. Bound assemble karo $∣ x - a ∣^{n + 1} = 1^{n + 1} = 1$ ke saath: $∣ R_{n} (1) ∣ \leq \frac{M}{( n + 1 )!} \cdot 1 = \frac{n !}{( n + 1 )!} = \frac{1}{n + 1} .$ Yeh step kyun? $M$ mein $n!$ $(n + 1)!$ ka zyaadatar hissa cancel kar deta hai, sirf $\frac{1}{n + 1}$ bachta hai. Yeh $0$ jaata hai, lekin sirf $1/ (n + 1)$ ki tarah — bahut dheere, aur bound muskil se convergence prove karta hai. $e^{x}$ se compare karo jahan $M$ ek constant tha aur bound $1/ (n + 1)!$ ki tarah marta tha.

Verify: $n = 0$ par, bound $= 1$ ; $n = 9$ par, bound $= 1/10 = 0.1$ . $ln 2$ ke partial sums ki true error (alternating harmonic series) sach mein $1/ (n + 1)$ ki tarah decay karti hai — tool ne wo fast factorial decay nahi di jo disk ke andar milti hai. Forecast lesson: $∣ x - a ∣ = 1$ par, ek growing $M$ bound ko gut kar sakta hai. (Compare Radius of Convergence: $x = 1$ exactly disk ki boundary hai.) ✓

(5) Cell C5 — degenerate input $x = a$

Worked example

Kisi bhi smooth $f$ ke kisi bhi $P_{n}$ ke liye $x = a$ par error bound kya hai? Forecast: tum exactly centre par khade ho. Error = ?

Step 1. Distance $∣ x - a ∣ = ∣ a - a ∣ = 0$ . Yeh step kyun? Poora bound is distance par depend karta hai; ise explicit banao.

Step 2. Plug in karo: $∣ R_{n} (a) ∣ \leq \frac{M}{( n + 1 )!} \cdot 0^{n + 1} = 0.$ Yeh step kyun? $0$ kisi bhi positive power tak uthaya $0$ hota hai, aur $n + 1 \geq 1$ hamesha. Toh bound exactly zero tak collapse ho jaata hai.

Step 3. Definition se cross-check karo. $P_{n} (a) = \sum_{k = 0}^{n} \frac{f ^{(k)} ( a )}{k !} (a - a)^{k} = f (a)$ (sirf $k = 0$ term bachta hai). Toh $R_{n} (a) = f (a) - f (a) = 0$ . Yeh step kyun? Do independent routes agree hote hain — polynomial apne khud ke centre par exact hai. Anchor point par koi approximation error nahi.

Verify: $f = e^{x}$ ke liye, $P_{2} (0) = 1 = e^{0}$ , error $= 0$ . ✓ Degenerate case ek bug nahi hai; yeh sanity anchor hai.

(6) Cell C6 — limiting behaviour $n \to \infty$

Worked example

$f (x) = sin x$ ke liye $a = 0$ par, dikhao ki error bound $n \to \infty$ ke saath har fixed $x$ ke liye $\to 0$ ho jaata hai. (Yeh convergence ka bridge hai.) Forecast: denominator mein factorial vs numerator mein power — race kaun jeetta hai?

Step 1. $sin$ ke har derivative $\pm sin$ ya $\pm cos$ hain, sab $1$ se bounded hain. Toh $M = 1$ sab $n$ ke liye kaam karta hai. Yeh step kyun? Humein ek $M$ chahiye jo har degree mein survive kare; $sin$ yeh free mein deta hai — aur kyunki $∣ sin ∣, ∣ cos ∣ \leq 1$ har jagah hai, hume interior-extremum check ki zaroorat bhi nahi padti.

Step 2. Bound ban jaata hai $∣ R_{n} (x) ∣ \leq \frac{1}{( n + 1 )!} ∣ x ∣^{n + 1} .$ Yeh step kyun? Ab yeh ek pure race hai: $∣ x ∣^{n + 1}$ (grow karta hai) vs $(n + 1)!$ (bahut zyada tezi se grow karta hai).

Step 3. Koi bhi $x$ fix karo. Figure 1 (neeche) exactly is bound, $∣ x ∣^{n + 1} / (n + 1)!$ , ko degree $n$ ke against plot karta hai fixed choice $x = 3$ ke liye. Iska pedagogical kaam yeh hai ki race ko visually dikhaye: coral curve pehle uthti hai jab $n + 1 < ∣ x ∣$ (har naya factor $\frac{∣ x ∣}{n + 1} > 1$ terms ko badhaata hai), $n \approx 2$ ke paas ek chota hump reach karta hai, aur phir, jab $n + 1$ $⌈ ∣ x ∣ ⌉ = 3$ ko cross karta hai, har naya factor $\frac{∣ x ∣}{n + 1} < 1$ se multiply hota hai, toh terms geometrically $0$ ki taraf dive karte hain. Formally, $\frac{∣ x ∣ ^{n + 1}}{( n + 1 )!} \to 0$ har fixed $x$ ke liye. Yeh step kyun? Factorial growth hamesha eventually exponential growth ko beat karta hai — isliye $sin$ ki Maclaurin series har jagah converge karti hai (link: Radius of Convergence $\infty$ hai).

Figure 1 — factorial-vs-power race.

Horizontal axis degree $n$ hai; vertical axis error bound $∣ x ∣^{n + 1} / (n + 1)!$ hai. Dekho coral curve $n \approx 2$ ke paas chote hump tak uthti hai (jahan $n + 1 \approx ∣ x ∣ = 3$ ) aur phir dive karti hai — wo dive hi convergence ki picture mein hai.

Verify: $x = 3$ par, sequence $\frac{3 ^{n + 1}}{( n + 1 )!}$ $n = 0, 1, \dots$ ke liye $3, 4.5, 4.5, 3.375, \dots$ hai phir $n = 3$ ke baad plunge karta hai. Factorial jeetta hai. ✓ (Coral curve ko Figure 1 mein dive karte dekho.)

(7) Cell C7 — real-world word problem with units

Worked example Ek lens is tarah grind kiya gaya hai ki uski surface height

$h (r) = R - R^{2} - r^{2}$ hai, radius $R = 50 mm$ ke sphere ke liye. Engineers ise parabola $P_{2} (r) = \frac{r ^{2}}{2 R}$ se approximate karte hain ( $r$ mein $n = 2$ Maclaurin polynomial). $r = 5 mm$ tak ke radii ke liye, kya approximation error $1 μ m = 1 0^{- 3} mm$ tolerance se neeche hai? Forecast: compute karne se pehle guess karo ki parabola "kaafi achha" hai ya nahi.

Step 1. $a = 0$ , $n = 2$ set karo. Exact $h$ ki Maclaurin expansion $h (r) = \frac{r ^{2}}{2 R} + \frac{r ^{4}}{8 R ^{3}} + \dots$ hai, toh wakai $P_{2} (r) = \frac{r ^{2}}{2 R}$ aur agla live term order $r^{4}$ ka hai. Humein $M = max ∣ h^{(3)} (r) ∣$ $[0, 5]$ par chahiye. Yeh step kyun? $n = 2 \Rightarrow$ remainder $h^{(3)}$ use karta hai. Units: $h$ mm mein hai aur $r$ mm mein, toh $h^{(3)}$ ke units $mm / mm^{3} = mm^{- 2}$ hain.

Step 2. $h^{(3)}$ explicitly derive karo, phir bound karo. $h (r) = R - (R^{2} - r^{2})^{1/2}$ differentiate karte hue: $h^{'} (r) = \frac{r}{R ^{2} - r ^{2}}, h^{''} (r) = \frac{R ^{2}}{( R ^{2} - r ^{2} ) ^{3/2}}, h^{'''} (r) = \frac{3 R ^{2} r}{( R ^{2} - r ^{2} ) ^{5/2}} .$ $[0, 5]$ par numerator $3 R^{2} r$ $r$ ke saath badhta hai aur denominator $(R^{2} - r^{2})^{5/2}$ $r$ ke saath girta hai, toh $h^{'''}$ yahan monotonic increasing hai aur uska max door wale end $r = 5$ par hai. $R = 50, r = 5$ plug karo: $h^{'''} (5) = \frac{3 \cdot 2500 \cdot 5}{( 2500 - 25 ) ^{5/2}} = \frac{37500}{247 5 ^{5/2}} \approx 1.24 \times 1 0^{- 4} mm^{- 2} .$ Safe $M = 2 \times 1 0^{- 4} mm^{- 2}$ tak round up karo. Yeh step kyun? Kyunki $h^{'''}$ $[0, 5]$ par monotonic hai (check kiya, assume nahi kiya), endpoint $r = 5$ genuinely worst hai; guaranteed bound ke liye $M$ upar round karte hain.

Step 3. Box apply karo $∣ x - a ∣ = 5 mm$ ke saath: $∣ R_{2} (5) ∣ \leq \frac{M}{3 !} (5)^{3} = \frac{2 \times 1 0 ^{- 4}}{6} \cdot 125 \approx 4.17 \times 1 0^{- 3} mm .$ Yeh step kyun? Units check: $mm^{- 2} \times mm^{3} = mm$ . Achha — answer ek length hai.

Verify: $4.17 \times 1 0^{- 3} mm = 4.17 μ m$ , jo $1 μ m$ tolerance se upar hai — toh guaranteed bound lens ko $5 mm$ tak certify nahi karta. (Direct check: true $h (5) = 50 - 2500 - 25 = 0.250627 mm$ , $P_{2} (5) = \frac{25}{100} = 0.25 mm$ , actual error $= 6.27 \times 1 0^{- 4} mm = 0.63 μ m$ .) True error (0.63 µm) actually pass karta hai, jabki bound (4.17 µm) conservative hai. Forecast lesson: bound ek guarantee deta hai; loose $M$ pessimistic ho sakta hai, toh failed bound ka matlab hamesha failed part nahi hota. ✓

(8) Cell C8 — exam twist (Trap B, live)

Worked example "Ek student

$f$ ke $P_{4}$ ki error $x = 3$ , $a = 1$ par bound karta hai, $f^{(4)} \leq M$ use karke aur likhta hai $∣ R_{4} ∣ \leq \frac{M}{4 !} ∣3 - 1 ∣^{4}$ . Do errors dhundho aur fix karo, phir $M = 15$ ke liye corrected bound evaluate karo." Forecast: aage padhne se pehle do mistakes spot karo — ek derivative order ki hai, ek factorial-and-power ki.

Step 1. Error 1 — wrong derivative order (Trap B). $P_{4}$ ki degree $n = 4$ hai, toh remainder $f^{(n + 1)} = f^{(5)}$ use karta hai, $f^{(4)}$ nahi. Student ne galat derivative bound kiya. Yeh step kyun? "Agla derivative" ka matlab $n + 1 = 5$ hai — Trap B ka pehla aadha.

Step 2. Error 2 — wrong factorial aur power (Trap B, doosra aadha). Yeh $(n + 1)! = 5!$ aur $∣ x - a ∣^{n + 1} = ∣3 - 1 ∣^{5}$ hona chahiye, $4!$ aur $4$ th power nahi. Yeh step kyun? Teeno pieces — derivative, factorial, power — sab ek hi $n + 1$ carry karen.

Step 3. Correct statement (ab $M$ $∣ f^{(5)} ∣$ ka bound hai): $∣ R_{4} (3) ∣ \leq \frac{M}{5 !} ∣3 - 1 ∣^{5} = \frac{M}{120} \cdot 32 = \frac{32 M}{120} = \frac{4 M}{15} .$ $M = 15$ ke saath: $\frac{4 \cdot 15}{15} = 4$ . Yeh step kyun? $5! = 120$ , $2^{5} = 32$ , aur $\frac{32}{120} = \frac{4}{15}$ ; clean $M = 15$ arithmetic visible bana deta hai.

Verify: corrected bound exactly $4$ deta hai. Student ka galat formula $\frac{15}{24} \cdot 16 = 10$ deta — ek alag number, toh traps genuinely answer badal dete hain. ✓

Recall Poori matrix ka one-line recap

Small/negative/far/off-centre sirf yeh badlta hai ki kaun sa end (ya interior point) $M$ deta hai aur $∣ x - a ∣^{n + 1}$ kitna bada hai; $∣ x - a ∣ = 1$ power ko $1$ par freeze kar deta hai toh sirf factorial-vs- $M$ race matter karti hai (aur ek growing $M$ , jaise $ln (1 + x)$ ke liye, bound ko $1/ (n + 1)$ tak slow kar sakta hai); $x = a$ exactly $0$ deta hai; $n \to \infty$ factorial ko jeetnee deta hai (convergence); word problems mein units add hote hain jinhe track karna zaroori hai; exam twists (Trap B) test karte hain ki derivative, factorial, aur power sab $n + 1$ hain; aur $M$ ke liye endpoint grab karne se pehle hamesha monotonicity check karo (Trap A).

Self-test:

e^{x}

ke liye

x = - 0.4

baayi taraf jaana

+ 0.4

seedha jaane se tighter bound kyun deta hai?

Kyunki

M = max e^{t}

[- 0.4, 0]

par chhota hai (max

= 1

)

[0, 0.4]

par (max

= e^{0.4}

) se; distance identical hai.

Cell C4 mein,

P_{3}

e^{2}

ki buri approximation kyun hai?

∣ x - a ∣ = 2 > 1

, toh

2^{n + 1}

grow karta hai; bound

\approx 4.9

bada hai — door ki walks ko zyada terms chahiye.

ln (1 + x)

ke liye boundary

x = 1

par, bound sirf

1/ (n + 1)

ki tarah kyun decay karta hai?

Kyunki

M = n!

n

ke saath grow karta hai aur

(n + 1)!

ka zyaadatar hissa cancel kar deta hai,

n! / (n + 1)! = 1/ (n + 1)

bachta hai.

x = a

par error bound exactly

0

kyun hota hai?

∣ x - a ∣^{n + 1} = 0^{n + 1} = 0

; polynomial apne khud ke centre par

f

ke barabar hota hai.

M

[a, x]

ke endpoint par lena kab galat hai?

Jab

f^{(n + 1)}

wahan monotonic na ho — ek interior hump true maximum ho sakta hai, toh pehle check karo.

Connections

Parent: remainder theorem — wo bound jinhe yeh examples apply karte hain.
Taylor & Maclaurin Series — $P_{n}$ polynomials jo poori jagah use hue.
Radius of Convergence — cells C6 aur C4b exactly convergence disk ki interior-vs-boundary story hai.
Alternating Series Estimation Theorem — $sin, cos$ aur $ln (1 + x)$ ke liye aksar ek aur bhi tighter alternating bound available hota hai.
Big-O and asymptotic error — har bound $O (∣ x - a ∣^{n + 1})$ hai, near- $a$ behaviour.
Cauchy Mean Value Theorem — in examples mein use hone wale Lagrange form ke peeche ka engine.

Worked examples — Taylor's remainder theorem — error estimation

Scenario matrix

Worked examples

(1) Cell C1 — chota positive $x$ , baseline

(2) Cell C2 — negative $x$ , kya sign tod deta hai?

(3) Cell C3 — centre $a \neq = 0$

(4) Cell C4 — badi distance, power GROW karti hai

(4b) Cell C4b — boundary $∣ x - a ∣ = 1$ , jahan danger real hai

(5) Cell C5 — degenerate input $x = a$

(6) Cell C6 — limiting behaviour $n \to \infty$

(7) Cell C7 — real-world word problem with units

(8) Cell C8 — exam twist (Trap B, live)

Connections

Concept Map