Taylor's remainder theorem — error estimation: Question bank

4.3.18 · D5 · Maths › Calculus III — Sequences & Series › Taylor's remainder theorem — error estimation

Core objects, taaki yahan koi mystery symbol na ho:

$f$ true function hai, $a$ centre hai, $x$ target point hai.
$P_{n} (x) = \sum_{k = 0}^{n} \frac{f ^{(k)} ( a )}{k !} (x - a)^{k}$ degree- $n$ Taylor polynomial hai (dekho Taylor & Maclaurin Series).
$R_{n} (x) = f (x) - P_{n} (x)$ exact error hai.
Lagrange form: $R_{n} (x) = \frac{f ^{(n + 1)} ( c )}{( n + 1 )!} (x - a)^{n + 1}$ kisi unknown $c$ ke liye jo strictly $a$ aur $x$ ke beech hai.
Bound: $∣ R_{n} (x) ∣ \leq \frac{M}{( n + 1 )!} ∣ x - a ∣^{n + 1}$ jahan $M \geq ∣ f^{(n + 1)} ∣$ interval par.

True or false — justify

The Lagrange remainder gives the exact error, not just a bound.

True —

R_{n} (x) = \frac{f ^{(n + 1)} ( c )}{( n + 1 )!} (x - a)^{n + 1}

ek exact equality hai; ye ek bound tabhi banta hai jab hum unknown

f^{(n + 1)} (c)

ko ek max

M

se replace karte hain.

The bound

\frac{M}{( n + 1 )!} ∣ x - a ∣^{n + 1}

always shrinks as

n

grows.

False — ye tabhi hota hai jab numerator

M ∣ x - a ∣^{n + 1}

eventually

(n + 1)!

ki race haar jaaye; bade

∣ x - a ∣

ke liye ya jin functions ke derivatives blow up ho jaate hain (jaise

\frac{1}{1 - x}

), bound shrink hone se pehle grow kar sakta hai ya kabhi shrink hi nahi karta.

R_{n} (x) \to 0

n \to \infty

, the Taylor series converges to

f (x)

at that

x

True — ye precisely definition hai: series

f

ke exactly equal hoti hai jab remainder vanish ho jaata hai, jo Radius of Convergence se analytic link hai.

A Taylor series that converges must converge to the function it came from.

False — series ka converge karna aur

R_{n} \to 0

alag statements hain; classic

e^{- 1/ x^{2}}

ka ek convergent (all-zero) Maclaurin series hai jo

0

ke barabar hai, function ke nahi, isliye

R_{n} \neq \to 0

The point

c

in the Lagrange form is the same for every

n

False —

c

depend karta hai

n

par,

x

par, aur function par; ye bas "beech mein koi point" hai jiska exist karna guaranteed hai, har remainder ke liye re-chosen hota hai.

The bound requires

f^{(n + 1)}

to exist on the whole interval between

a

and

x

True — derivation

[a, x]

par Cauchy Mean Value Theorem apply karta hai; agar

f^{(n + 1)}

andar kahin exist karna fail kare, to theorem ki hypotheses break ho jaati hain aur koi bound guaranteed nahi hai.

Making

M

larger than necessary makes your error claim invalid.

False — bada

M

ek valid lekin looser (zyada conservative) bound deta hai; claim "

∣ R_{n} ∣ \leq

bound" phir bhi hold karta hai, bas tum utni accuracy promise nahi kar rahe jitni actually hai.

For

sin x

and

cos x

, you can always take

M = 1

regardless of

n

a

, or

x

True —

sin

cos

ka har derivative

\pm sin

\pm cos

hai, sab har jagah

1

se bounded hain, isliye

M = 1

hamesha valid hai (halanki aksar loose hota hai).

Spot the error

"The error uses the

n

-th derivative and

n!

, matching the polynomial's degree."

Wrong — remainder ek step aage jaata hai:

(n + 1)

-th derivative aur

(n + 1)!

, kyunki

P_{n}

pehle se

n

-th derivative tak sab kuch match kar chuka hai, isliye pehla mismatch order

n + 1

par hota hai.

c

is between

a

and

x

, so I'll just set

c = x

to evaluate."

Wrong —

c

unknown hai; endpoint choose karne se

∣ f^{(n + 1)} ∣

under-estimate ho sakta hai aur ek bound produce ho sakta hai jo true error se chota ho. Tumhe

M = max ∣ f^{(n + 1)} ∣

interval par use karna hi hoga.

"Since

x - a = 0.3

is small,

(x - a)^{n + 1}

is negligible, so I can drop it."

Wrong — tumhe ye factor rakhna hi hoga; ye bound ke teen parts mein se ek hai. Ise drop karne se number entirely change ho jaata hai, aur agar

∣ x - a ∣ > 1

ho to wo factor actually grow karta hai aur dominate kar sakta hai.

P_{2}

has three terms, so I use

M = max ∣ f^{(2)} ∣

Wrong —

n = 2

ka matlab remainder ko

f^{(3)}

aur

3!

chahiye,

f^{(2)}

nahi; derivatives remainder se count karo, polynomial ke highest term se kabhi nahi.

e^{0.1}

is about

1.105

, so I'll take

M = 1.105

for the

e^{x}

bound."

Wrong idea, though it happens to work —

M

f^{(n + 1)} = e^{t}

ko pure

[0, 0.1]

par bound karna chahiye, aur wahan iska max

e^{0.1} \approx 1.105

hai, isliye ye us interval ke liye theek hai; galti ye sochna hai ki

M

f (x)

ko track karta hai na ki

max ∣ f^{(n + 1)} ∣

ko.

"The alternating-series bound and Taylor bound must give the same number."

Wrong — ye alag tools hain; Alternating Series Estimation Theorem pehle omitted term se bound karta hai (sirf tab jab terms alternate aur shrink karein), jabki Lagrange

\frac{M}{( n + 1 )!} ∣ x - a ∣^{n + 1}

se bound karta hai. Dono disagree kar sakte hain, aur ek dusre se sharper ho sakta hai.

Why questions

Why is the factorial

(n + 1)!

rather than

(n + 1)

2^{n + 1}

Ye directly Cauchy-MVT derivation se aata hai: Taylor structure ko differentiate karne par

\frac{f ^{(n + 1)} ( c )}{n !} (x - c)^{n}

milta hai, aur

h^{'} = - (n + 1) (x - c)^{n}

se divide karne par extra factor contribute hota hai, jisse

(n + 1)!

milta hai — ye earned hai, chosen nahi.

Why do we bound

∣ f^{(n + 1)} ∣

instead of computing

R_{n}

directly?

Agar hum

R_{n} = f (x) - P_{n} (x)

exactly compute kar sakte to hume

f (x)

pata hota aur approximation ki zarurat hi nahi hoti; next derivative ko bound karna

f (x)

jaane bina kaam chala deta hai.

Why does the Mean Value Theorem (or its Cauchy version) even appear here?

MVT woh tool hai jo "

g

R_{n}

par start hoti hai aur

0

par end" ko "koi point

c

exist karta hai jahan rate average rate ke barabar hai" mein convert karta hai, jo ek existence claim se unknown

c

ke baare mein impossible exact computation replace kar deta hai. Dekho Mean Value Theorem aur Cauchy Mean Value Theorem.

Why does the error get small so fast for a short walk (small

∣ x - a ∣

Factor

∣ x - a ∣^{n + 1}

ek chhote number ki high power hai —

0.1

ko square ya cube karna use zero ki taraf crash kar deta hai — aur

(n + 1)!

denominator aur bhi fast grow karta hai, isliye short walks mein almost perfect accuracy guaranteed hoti hai.

Why can the remainder theorem guarantee accuracy even though

c

is never found?

Hume kabhi

c

ki value nahi chahiye — bas itna ki koi valid

c

interval mein exist karta hai; pure interval par

∣ f^{(n + 1)} ∣

ko bound karna us

c

ki har possible location ko cover kar leta hai.

Why is

R_{n} = O (∣ x - a ∣^{n + 1})

near

a

a fair summary of the bound?

Jaise

x \to a

, factor

∣ x - a ∣^{n + 1}

size control karta hai jabki

\frac{M}{( n + 1 )!}

bas ek constant hai, isliye error ka leading behaviour woh power hai — Big-O and asymptotic error ki language.

Edge cases

x = a

, what is

R_{n} (x)

Exactly

0

— polynomial centre par function ke barabar hoti hai, aur bound bhi confirm karta hai kyunki

∣ x - a ∣^{n + 1} = 0

What happens to the bound when

f^{(n + 1)}

is unbounded on the interval (no finite

M

exists)?

Theorem koi usable bound nahi deta; tumhe interval shrink karna hoga taaki finite

M

exist kare, warna guarantee simply fail ho jaati hai.

Does the theorem work when

x < a

(target left of centre)?

Yes —

c

strictly

a

aur

x

ke beech hota hai kisi bhi direction mein, aur bound

∣ x - a ∣^{n + 1}

use karta hai, isliye orientation matter nahi karta.

For a polynomial

f

of degree

d

, what is

R_{n}

when

n \geq d

Exactly

0

— degree-

d

polynomial ka

(n + 1)

-th derivative

0

ho jaata hai jab

n + 1 > d

, isliye Lagrange term aur bound dono vanish ho jaate hain.

At the boundary

∣ x - a ∣ = 1

, which factor decides whether the bound shrinks with

n

Factor

∣ x - a ∣^{n + 1} = 1

inert hai, isliye race purely

M

versus

(n + 1)!

hai; factorial tab tak jeetta hai jab tak

M

(n + 1)!

se faster grow na kare.

f^{(n + 1)}

is continuous but changes sign on the interval, is the bound still valid?

Yes —

M

, absolute value

∣ f^{(n + 1)} ∣

ko bound karta hai, isliye sign changes irrelevant hain; tumhe bas sabse bada magnitude chahiye.

Connections

Taylor & Maclaurin Series — wo polynomials jinke baare mein ye traps quiz karte hain.
Mean Value Theorem / Cauchy Mean Value Theorem — kyun "koi $c$ exist karta hai".
Radius of Convergence — $R_{n} \to 0$ true/false items.
Alternating Series Estimation Theorem — "same bound?" trap.
Big-O and asymptotic error — asymptotic "why" item.