Error in polynomial interpolation: Question bank — Hinglish deep dive

4.8.11 · D5 · Maths › Numerical Methods › Error in polynomial interpolation

Intuition Is bank ko kaise use karein

Ye interpolation error topic ke conceptual landmines hain — woh wale jo logon ko tab bhi trip karte hain jab woh formula mein numbers plug kar sakte hain. Har line ek question ::: answer reveal hai. Answer cover karo, ek position pe commit karo aur ek reason bhi socho, phir check karo. Agar tumhara reason galat tha chahe answer sahi tha, toh item ne tumhe pakad liya.

Notation jo neeche har jagah use hoti hai. Hum ek closed interval $[a, b]$ pe kaam karte hain — yahi domain hai: ismein evaluation point $x$ aur saare nodes $x_{0}, x_{1}, \dots, x_{n}$ hain. Master formula hai $E (x) = f (x) - p_{n} (x) = \frac{f ^{(n + 1)} ( ξ )}{( n + 1 )!} ω (x), ω (x) = \prod_{i = 0}^{n} (x - x_{i}),$ jahan $f$ true function hai, $p_{n}$ woh degree- $\leq n$ polynomial hai jo $n + 1$ nodes se guzarti hai, $ω$ woh node polynomial hai jo har node pe zero hai, aur $ξ$ $(a, b)$ mein koi aisa point hai jo sirf exist karta hai — iska value humein kabhi nahi diya jaata. Neeche ki pictures in ideas ko visually fix karti hain traps shuru hone se pehle.

Figure — Error in polynomial interpolation

Upar ke do panels do independent knobs dikhate hain. Left pe, node polynomial $ω (x)$ : yeh har node pe zero pe pinned hai aur nodes ke beech sabse zyada swing karti hai — khaaskar equal spacing ke saath endpoints ke paas. Right pe, dikhata hai ki nodes ko ends ki taraf move karne (Chebyshev nodes) se woh swings flat kaise ho jaati hain.

Aur woh classic failure jo dimaag mein rakhni chahiye: equally spaced nodes ke saath degree badhane se true error endpoints ke paas badhti hai — Runge phenomenon — chahe function bilkul smooth ho.

True or false — justify

TRUE / FALSE — Har node

x_{i}

pe, interpolation error exactly zero hoti hai.

True. Construction ke hisaab se

p_{n} (x_{i}) = f (x_{i})

, toh

E (x_{i}) = f (x_{i}) - p_{n} (x_{i}) = 0

; equivalently

ω (x_{i}) = 0

kyunki factor

(x - x_{i})

wahan vanish ho jaata hai.

TRUE / FALSE — Do adjacent nodes ke beech error bhi kahin zero zaroor hogi.

False in general. Error nodes pe zero guaranteed hai lekin in dono ke beech ke open interval mein typically nonzero hoti hai — woh "guessing gap" hi poore topic ka reason hai.

TRUE / FALSE — Ek aur node add karne se maximum error hamesha kam hoti hai.

False. Ek node add karne se

n

n + 1

ho jaata hai, toh ab

n + 2

nodes hain aur error term ek order higher derivative use karti hai,

f^{(n + 2)} / (n + 2)!

times new

ω

(har extra node ek extra factor aur ek extra required derivative contribute karti hai). Agar

max ∣ f^{(n + 2)} ∣

factorial se tez grow kare (aur

max ∣ ω ∣

ends pe blow up ho, jaise equally spaced nodes ke liye), toh bound aur true error dono badh sakti hain — yahi Runge phenomenon hai.

TRUE / FALSE — Factor

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

guarantee karta hai ki error

n \to \infty

pe zero ho jaaye.

False. Factorial akela limit decide nahi karta; yeh

max ∣ f^{(n + 1)} ∣

se ladta hai, jo super-factorially grow kar sakta hai, aur

max ∣ ω ∣

se bhi. Convergence teeno factors milke decide karte hain, sirf factorial se nahi.

TRUE / FALSE — Degree

\leq n

ke polynomial

f

ke liye, interpolation error identically zero hai.

True. Tab

f^{(n + 1)} \equiv 0

, toh poora error term har

x

ke liye vanish karta hai; equivalently

p_{n}

f

hai kyunki

n + 1

points se guzarne wala degree

\leq n

ka interpolating polynomial unique hota hai.

TRUE / FALSE — Formula mein woh single point

ξ

har

x

ke liye same hai.

False.

ξ = ξ (x)

x

ke saath move karta hai; theorem sirf promise karti hai ki har

x

ke liye koi na koi

ξ \in (a, b)

kaam karta hai. Ise ek fixed number maanna ek classic error hai.

TRUE / FALSE — Error formula require karta hai ki

f

n + 1

continuous derivatives hon.

True. Hypothesis hai

f \in C^{n + 1} [a, b]

; derivation helper function ko

n + 1

baar Rolle ke zariye differentiate karta hai, toh

f^{(n + 1)}

ka exist karna aur continuous hona argument ke land karne ke liye zaruri hai.

TRUE / FALSE — Better nodes choose karne se

f^{(n + 1)}

factor reduce ho sakta hai.

False. Derivative

f^{(n + 1)}

function khud ka hai aur node placement se untouched rehta hai; tum sirf

ω (x)

control kar sakte ho, yani jahan sample lete ho wahan se. Node choice sirf "distance from data" knob fix karti hai, "twistiness" knob kabhi nahi.

TRUE / FALSE — Equally spaced nodes

[a, b]

max ∣ ω (x) ∣

minimize karte hain.

False. Yeh actually endpoints ke paas

∣ ω ∣

ko blow up karte hain; woh placement jo

max ∣ ω ∣

minimize karti hai woh points ko ends ki taraf cluster karti hai — Chebyshev nodes.

TRUE / FALSE — Interpolation error aur Taylor's remainder unrelated results hain.

False. Dono mein identical

\frac{f ^{(n + 1)} ( ξ )}{( n + 1 )!}

factor hai; Taylor ek node

n + 1

baar repeat karta hai

(x - x_{0})^{n + 1}

, interpolation same degree ko distinct nodes

\prod (x - x_{i})

mein spread karta hai. Interpolation "Taylor with derivatives traded for data" hai.

Spot the error

WRONG: "

E (x) = f^{(n + 1)} (ξ) ω (x)

." Kya missing hai?

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

denominator.

n + 1

Rolle differentiations

ω

ω^{(n + 1)} = (n + 1)!

mein turn kar deti hain, aur woh constant denominator mein land karta hai — ise drop karna error ko

(n + 1)!

factor se overstate karta hai.

WRONG: "Kyunki

E (x_{i}) = 0

nodes pe, polynomial har jagah

f

ke barabar hai, toh error hamesha zero hai." Kahan galat hai?

Finitely many points pe match karna in dono ke beech agreement force nahi karta.

E

sirf

n + 1

nodes pe zero pe pinned hai; free "wiggle"

ω (x)

exactly wahi hai jo nodes se hatke disagreement measure karta hai.

WRONG: "Main

ξ

compute karunga

g^{(n + 1)} (ξ) = 0

solve karke aur exact error ke liye plug in karunga." Operationally kya galat hai?

Theorem ek existence statement hai;

ξ

generally closed form mein computable nahi hai aur

x

pe depend karta hai. Practice mein tum

f^{(n + 1)} (ξ)

M_{n + 1} = max_{t \in [a, b]} ∣ f^{(n + 1)} (t) ∣

se replace karte ho, jo domain pe

(n + 1)

-th derivative ki largest magnitude hai, taaki ek usable bound mile, exact value nahi.

WRONG: "Linear interpolation ke liye worst error ek node pe hoti hai." Ise correct karo.

n = 1

ke liye,

ω (x) = (x - x_{0}) (x - x_{1})

nodes pe roots ke saath ek downward parabola hai, toh

∣ ω ∣

midpoint pe sabse bada hai, jo

∣ ω ∣_{m a x} = h^{2} /4

deta hai aur

max ∣ E ∣ \leq \frac{h ^{2}}{8} max ∣ f^{''} ∣

. Nodes wahan hain jahan error sabse choti hai (zero), sabse badi nahi.

WRONG: "Formula ko

ξ

specifically

[x_{0}, x_{n}]

ke andar chahiye." Interval fix karo.

ξ

sirf

min {x, x_{0}, \dots, x_{n}}

aur

max {x, x_{0}, \dots, x_{n}}

ke beech guarantee hai — woh sabse chota interval jo evaluation point aur har node contain kare. Agar

x

outermost nodes ke bahar ho (extrapolation), toh

ξ

bhi

[x_{0}, x_{n}]

se aage ja sakta hai.

WRONG: "

ω^{(n + 1)} (t)

t

pe depend karta hai, toh yeh interval mein change hota hai." Ise correct karo.

ω (t) = t^{n + 1} + (lower degree)

, aur degree-

(n + 1)

polynomial ka

(n + 1)

-th derivative constant

(n + 1)!

hai. Saare lower-degree terms differentiate ho jaate hain, toh

ω^{(n + 1)}

ek pure constant hai.

Why questions

WHY derivation auxiliary

g (t) = f (t) - p_{n} (t) - K ω (t)

invent karta hai, aur constant

K

kya hai?

K

choose kiya jaata hai taaki

g

ek extra chosen point

\overset{x}{ˉ}

pe vanish ho (woh fixed evaluation point jahan hum error chahte hain): set karo

K = \frac{f ( x ˉ ) - p _{n} ( x ˉ )}{ω ( x ˉ )}

, yani

K

woh unknown error-per-unit-

ω

hai jise hum identify karne ki koshish kar rahe hain. Us choice se

g

n + 2

known zeros hain (

n + 1

nodes plus

\overset{x}{ˉ}

), exactly wahi jo Rolle ko

n + 1

baar apply karne ke liye chahiye — ek lever jo

E

akela nahi deta.

WHY hum ek naaya symbol

\overset{x}{ˉ}

(bar-x) use karte hain instead of page ke

x

ke?

\overset{x}{ˉ}

ek frozen evaluation point hai jo fixed rehta hai jab auxiliary function

g (t)

ek free variable

t

pe run karta hai; naam alag rakhne se hum "woh jagah jahan hum error measure kar rahe hain" (

\overset{x}{ˉ}

) ko "woh running variable jisme Rolle zeros chase karta hai" (

t

) se confuse karne se bachte hain. Jab

K

pin ho jaata hai tab hum

\overset{x}{ˉ}

ko waapas

x

rename karte hain.

WHY Rolle

n + 1

baar apply karne se exactly

f^{(n + 1)}

produce hota hai koi aur derivative order nahi?

Har Rolle step guaranteed zero count ko ek se reduce karta hai, toh

n + 2

zeros se shuru karke tum exactly

n + 1

baar differentiate kar sakte ho ek guaranteed zero

ξ

pe aane se pehle; woh final order nodes-plus-one ki sankhya se match karta hai aur factor pin karta hai.

WHY

p_{n}^{(n + 1)} \equiv 0

derivation mein appear karta hai, aur yeh kya accomplish karta hai?

Degree

\leq n

ka polynomial zero

(n + 1)

-th derivative rakhta hai, toh

p_{n}

g^{(n + 1)}

se silently drop out ho jaata hai, sirf

f^{(n + 1)} (ξ) - K (n + 1)!

bacha rehta hai — jo humein

K

ke liye cleanly solve karne deta hai.

WHY "wiggly" functions kaafi nodes ke saath bhi poorly interpolate hoti hain?

Wiggliness ka matlab hai large high-order derivatives

∣ f^{(n + 1)} ∣

, aur woh factor poori error se multiply hota hai; koi bhi clever node placement use nahin laga sakta, toh roller-coaster function har polynomial fit se ladta hai.

WHY Chebyshev nodes worst-case error ke liye equally spaced nodes se better hain?

Inhe

max_{[a, b]} ∣ ω (x) ∣

minimize karne ke liye choose kiya jaata hai, node polynomial ke peaks ko flatten karte hue (jo equal spacing ke liye endpoints ke paas spike karte hain), toh woh ek factor jo tum control kar sakte ho use as small as possible banaya jaata hai.

WHY

n + 1

equally spaced nodes ke spacing

h

ke liye error

O (h^{n + 1})

hai?

Node polynomial

ω (x) = \prod (x - x_{i})

n + 1

factors ka product hai jo har

O (h)

size ke hain, toh

∣ ω ∣ = O (h^{n + 1})

; bounded derivative factor se multiply karne par, error

h^{n + 1}

ki tarah scale karta hai.

WHY hum kisi bhi accuracy tak pahunchne ke liye simply degree badhate nahi reh sakte?

Kyunki

n

badhana

max ∣ f^{(n + 1)} ∣

aur (equal spacing ke liye)

max ∣ ω ∣

dono ko

(n + 1)!

se faster inflate kar sakta hai, endpoints ke paas divergence produce karta hai — Runge ki woh warning jo piecewise/spline methods sidestep karte hain.

Sign of the error

WHY ek given

x

pe error ka sign, sirf uski size nahi, sochne layak hai?

Kyunki

E (x) = \frac{f ^{(n + 1)} ( ξ )}{( n + 1 )!} ω (x)

ek product hai, toh iska sign (sign of

f^{(n + 1)} (ξ)

)

\times

(sign of

ω (x)

) hai; yeh jaanna ki

p_{n}

f

ko over- ya under-shoot karta hai aksar magnitude utni hi importance rakhta hai.

SIGN: Linear interpolation (

n = 1

) ke liye ek convex function (

f^{''} > 0

) ke liye, kya line nodes ke beech

f

ke upar ya neeche hoti hai?

Neeche. Do nodes ke beech

ω (x) = (x - x_{0}) (x - x_{1}) < 0

aur

f^{''} (ξ) > 0

, toh

E = f - p_{1} = \frac{f ^{''} ( ξ )}{2} ω (x) < 0

, matlab

f < p_{1}

: chord convex curve ke upar baithti hai, exactly ek bowl ke upar chord ki picture.

SIGN:

sign ω (x)

kaise behave karta hai jab

x

ek node cross karta hai?

Har factor

(x - x_{i})

sign flip karta hai jab

x

x_{i}

se guzarta hai, toh

ω

successive between-node intervals mein sign alternate karta hai; error isliye nodes ke across sweep karte waqt over-then-under-then-over swing karti rehti hai.

SIGN: Agar

f^{(n + 1)}

[a, b]

pe constant sign rakhta hai, toh tum keh sakte ho ki

p_{n}

f

ko kahan over- aur under-estimate karta hai?

Over/under pattern tab poori tarah

ω (x)

ke alternating sign se dictate hota hai, toh

p_{n}

ek predictable ripple mein har node ke paas

f

cross karta hai — error curve ki shape bina compute kiye guess karne ke liye useful.

Edge cases

EDGE: Kya hota hai jab evaluation point

x

ek node ke saath coincide karta hai?

Yeh exactly zero hai, kyunki

ω (x) = 0

wahan; derivation ka

K

construction explicitly is trivial case ko alag set karta hai aur sirf non-nodes handle karta hai.

EDGE: Agar do nodes merge hone diye jaayein,

x_{1} \to x_{0}

, toh formula ka kya hoga?

Distinct-node hypothesis toot jaati hai; limit mein repeated node

p_{n}

ko ek derivative bhi match karne pe force karta hai (Hermite interpolation), aur

ω

ek squared factor

(x - x_{0})^{2}

gain karta hai — interpolation picture Taylor ki taraf bend karti hai.

EDGE: Kya error formula valid hai jab

x

node interval se bahar ho (extrapolation)?

Formula

ξ

ke saath

min {x, x_{0}, \dots, x_{n}}

aur

max {x, x_{0}, \dots, x_{n}}

ke beech phir bhi hold karta hai, lekin

∣ ω (x) ∣

outermost nodes se aage bahut tez grow karta hai, toh bound explode ho jaata hai — extrapolation wahan hai jahan interpolation error least trustworthy hoti hai.

EDGE: Error term kya kehta hai jab

f

ek constant function ho?

Har derivative

f^{(n + 1)} = 0

, toh

E (x) = 0

sabhi

x

aur kisi bhi node set ke liye — ek constant even single node se perfectly fit hota hai, intuition se match karta hai.

EDGE: Sirf ek node ke saath (

n = 0

, toh

p_{0}

ek constant hai), error term kya hai?

Yeh

E (x) = f^{'} (ξ) (x - x_{0})

ban jaata hai, jo exactly Mean Value Theorem statement hai —

n = 0

interpolation error hai hi MVT, woh base case jisko poora formula generalize karta hai.

EDGE: Agar

f^{(n + 1)}

[a, b]

pe sign change kare, toh kya bound

∣ E ∣ \leq \frac{M _{n + 1}}{( n + 1 )!} ∣ ω ∣

phir bhi hold karti hai?

Haan; bound

M_{n + 1} = max_{[a, b]} ∣ f^{(n + 1)} ∣

use karta hai, sign ki parwah kiye bina largest magnitude, toh sign changes sirf true error ko bound se chota banate hain, kabhi bada nahi.

EDGE: Kya error formula numerical integration error ke connected topic pe apply hoti hai?

Haan — interpolation error term ko interval pe integrate karna exactly wahi hai jisse quadrature error formulas (jaise trapezoid ka

O (h^{2})

, Simpson ka

O (h^{4})

) derive hote hain, same

f^{(n + 1)}

-and-

ω

structure inherit karte hue.

Recall One-line survival summary

Error = (twistiness $f^{(n + 1)}$ ) $\times$ (distance-from-data $ω$ ) $\div$ (softening factorial). Tum sirf $ω$ control kar sakte ho; $ξ$ sirf exist karta hai aur $x$ ke saath move karta hai; sign $f^{(n + 1)}$ times $ω$ se aata hai; zyada nodes hamesha better nahi hote.