Linear approximation and differentials: Question bank

4.1.32 · D5 · Maths › Calculus I — Limits & Derivatives › Linear approximation and differentials

Parent par wapas link: Linear approximation.

Sach ya jhooth — justify karo

L (x)

aur

f (x)

hamesha exactly

x = a

par agree karte hain.

Sach.

L (a) = f (a) + f^{'} (a) (a - a) = f (a)

(x - a)

run zero hai, isliye line aur curve touch karte hain. Approximation sirf base point par exact hoti hai.

Approximation

f (x) \approx L (x)

utni hi buri hoti jaati hai jitna

x

a

se door hota hai.

Generally sach. Bacha hua error

ε (x) (x - a)

hai; jaise

x

door jaata hai,

(x - a)

badhta hai aur

ε (x)

tiny rehna band kar deta hai, toh seedhi line curving graph se drift kar jaati hai.

Agar

f

khud ek seedhi line hai, toh

L (x) = f (x)

sirf

a

ke paas nahi balki sabhi

x

ke liye hai.

Sach. Ek line ki tangent khud wahi line hoti hai; error term

ε (x) (x - a)

har jagah exactly

0

hota hai kyunki ek line mein koi curvature nahi hoti.

d y

aur

Δ y

ek hi number hain.

Jhooth.

d y

tangent ke saath rise hai;

Δ y

curve ki actual rise hai. Ye sirf seedhe

f

ke liye coincide karte hain; warna unka gap hi curvature error hai.

Linearization error ko zero kiya ja sakta hai ek smart base point

a

choose karke.

Jhooth (genuinely curved

f

ke liye). Behtar

a

error ko shrink karta hai, lekin jab tak curve

a

aur

x

ke beech mein bend karta hai, ek seedhi line usse exactly match nahi kar sakti.

Agar

f^{'} (a) = 0

hai, toh linear approximation bekaar hai.

Jhooth. Ye

L (x) = f (a)

deta hai, ek flat estimate — bilkul valid, bas yeh kehta hai ki "

f

yahan roughly constant hai." Ye sirf crude hai, kyunki flat point ke paas leading behaviour quadratic hota hai, jo Concavity and second derivative handle karta hai.

Step

d x

ko double karne se approximation error roughly double ho jaati hai.

Jhooth. Error

(x - a)^{2}

ki tarah behave karta hai (agla Taylor term), toh step double karne se error roughly chaar guna ho jaati hai. Dekho Taylor series.

d x

positive hona zaroori hai.

Jhooth.

d x

x

mein koi bhi chosen change hai; ye negative ho sakta hai (

a

ke baayein target) ya zero. Negative

d x

bas ek negative run aur ek downhill/uphill rise deta hai accordingly.

Formula

Δ y \approx d y

sirf tab hold karta hai jab

f

increasing ho.

Jhooth. Ye kisi bhi point ke paas hold karta hai jahan

f

differentiable ho, increasing ho ya decreasing.

f^{'} (a)

ka sign sirf yeh decide karta hai ki tangent upar jaati hai ya neeche.

Kyun wale sawaal

Hum difference quotient ko

f^{'} (a) + ε (x)

ke roop mein kyun split karte hain, instead of bas yeh kehne ke ki ye

f^{'} (a)

"equals" karta hai?

Kyunki

x \neq = a

ke liye quotient sirf

f^{'} (a)

ke kareeb hota hai, equal nahi; term

ε (x)

us gap ko name karta hai, aur ye

x \to a

ke saath

\to 0

ho jaata hai limit definition ke through.

Error

(x - a)

se tezi se kyun shrink hoti hai?

Error

ε (x) \cdot (x - a)

hai — ek chhota factor

ε (x) \to 0

times ek chhota run — toh ye do shrinking cheezein ka product hai, dono mein se kisi bhi akele se tezi se vanish hota hai.

Nearest base point kyon choose karte hain jahan

f (a)

exact ho?

Nearest rakhne se run

(x - a)

tiny rehta hai, jo

(x - a)^{2}

-sized error ko shrink karta hai; exactness ensure karta hai ki

f (a)

aur

f^{'} (a)

apni khud ki errors contribute na karein.

Tangent line best straight-line approximation kyun hai, sirf ek acchi nahi?

Ye unique line hai jo

f (a)

(value) aur

f^{'} (a)

(slope) dono match karti hai; koi bhi doosri line slope mein disagree karti hai, jo

(x - a)

mein quadratically ki jagah linearly badhne wala error deti hai.

d y = f^{'} (x) d x

kyun likhte hain, instead of error propagation mein directly

Δ y

use karne ke?

Kyunki lab mein actual error nahi pata, sirf input par ek chhoti bound

d x

pata hai;

d y

us known input tolerance ko slope ke zariye output tolerance mein convert karta hai. Dekho Error propagation.

(1 + x)^{n}

a = 0

par linearize karne se itna clean formula kyun milta hai?

a = 0

set karne se run

(x - a)

sirf

x

reh jaata hai, aur

f (0) = 1, f^{'} (0) = n

hai, toh

L (x) = 1 + n x

kuch bhi carry kiye bina.

Newton's method ko repeated linear approximation kyun mana ja sakta hai?

Har step

f

ko uski tangent line se replace karta hai aur line

= 0

solve karta hai; line ki root agla guess hai, toh Newton's method linear approximation ko zero ki taraf iterate karna hai.

Under-estimate: concave-up curve apni tangent ke upar bend karti hai, toh line actual value ke neeche hoti hai. Concave-down ise flip kar deta hai — line over-estimate karti hai. Concavity and second derivative se govern hota hai.

Agar

ε (x)

x \to a

ke saath zero nahi jaata, toh kya galat ho gaya?

Tab

f^{'} (a)

actually derivative nahi tha — difference quotient uski taraf converge nahi kar raha.

ε (x) \to 0

precisely wahi statement hai ki

f^{'} (a)

define karne wali limit exist karti hai.

Kya ho agar

d x

bahut bada le liya — kya

d y

tab bhi meaningful hai?

d y = f^{'} (a) d x

hamesha defined hai (tangent par rise), lekin ye

Δ y

ko approximate karna band kar deta hai: bade

d x

ke liye curve line se kaafi door bend ho chukti hai, toh

Δ y \neq \approx d y

Connections

Derivative as a limit — $ε (x) \to 0$ statement jinpar ye traps hinge karti hain.
Tangent line — geometric object jiske baare mein har "line vs curve" sawaal hai.
Concavity and second derivative — over- vs under-estimate decide karta hai.
Taylor series — $(x - a)^{2}$ error term jo kai answers mein naam liya gaya.
Newton's method — repeated linearization.
Error propagation — kyun $d y = f^{'} (x) d x$ , $f$ directly use karne se behtar hai.

Question bank — Linear approximation and differentials

Sach ya jhooth — justify karo

Error dhundo

Kyun wale sawaal

Edge cases

Connections