Error in polynomial interpolation

4.8.11 · Maths › Numerical Methods

HUM KYA MEASURE kar rahe hain?

Woh central result jo hum derive karenge:

YEH EXACT SHAPE KYU AATI HAI? (Derivation from scratch)

Hume dikhana hai ki $E(x)=\frac{f^{(n+1)}(\xi )}{(n+1)!}\omega (x)$.

Step 1 — Ek point fix karo aur ek helper function banao. Koi bhi aisa point $\bar{x}$ lo jo node nahi hai (agar node hai toh error $0$ hai aur kaam khatam). Hume abhi wahan error nahi pata; ek unknown constant $K$ define karo jo is tarah se defined hai: $f(\bar{x})-p_n(\bar{x})=K\omega (\bar{x}).$ Yeh step kyun? Hum error ko (constant)$\times \omega$ ki form mein force kar rahe hain. Agar hum prove kar saken ki $K=\frac{f^{(n+1)}(\xi )}{(n+1)!}$, toh formula done hai. Yeh Feynman move hai: answer ki shape assume karo, phir usse pin down karo.

Step 2 — Ek nayi variable $t$ mein ek auxiliary function define karo: $g(t)=f(t)-p_n(t)-K\omega (t).$ Yeh step kyun? Hume ek aisi function chahiye jo bahut saare points par vanish kare taaki hum Rolle repeatedly apply kar sakein.

Step 3 — $g$ ke zeros count karo.

• Har node par: $f(x_i)=p_n(x_i)$ aur $\omega (x_i)=0$, isliye $g(x_i)=0$. Yeh $n+1$ zeros hue.
• $t=\bar{x}$ par: $K$ ki definition se, $g(\bar{x})=f(\bar{x})-p_n(\bar{x})-K\omega (\bar{x})=0$. Ek aur zero.

Toh $g$ ke kam se kam $n+2$ distinct zeros hain $[a,b]$ mein.

Step 4 — Rolle's theorem baar baar apply karo. Kyun? Rolle kehta hai $g$ ke do zeros ke beech $g^{\prime }$ ka ek zero hoga.

• $g$ ke $\ge n+2$ zeros hain $\Rightarrow$ $g^{\prime }$ ke $\ge n+1$ zeros hain.
• $g^{\prime \prime }$ ke $\ge n$ zeros hain.
• … $g^{(n+1)}$ ka $\ge 1$ zero hai. Usse $\xi$ bolo.

Step 5 — $g$ ko exactly $n+1$ baar differentiate karo. $g^{(n+1)}(t)=f^{(n+1)}(t)-p_n^{(n+1)}(t)-K{\omega }^{(n+1)}(t).$

• $p_n$ ka degree $\le n$ hai $\Rightarrow p_n^{(n+1)}\equiv 0$.
• $\omega (t)=t^{n+1}+(\text{lower})$, isliye ${\omega }^{(n+1)}(t)=(n+1)!$ (ek constant hai).

Isliye $g^{(n+1)}(t)=f^{(n+1)}(t)-K(n+1)!.$

Step 6 — $\xi$ par evaluate karo. Kyunki $g^{(n+1)}(\xi )=0$: $0=f^{(n+1)}(\xi )-K(n+1)!\Rightarrow K=\frac{f^{(n+1)}(\xi )}{(n+1)!}.$

Step 7 — Wapas substitute karo. Step 1 se, $f(\bar{x})-p_n(\bar{x})=K\omega (\bar{x})=\frac{f^{(n+1)}(\xi )}{(n+1)!}\omega (\bar{x}).■$

ISKA USE KAISE KAREIN — error bound

Hume $\xi$ rarely pata hota hai, isliye hum isse bound karte hain. Maano $M_{n+1}={\max }_{[a,b]}|f^{(n+1)}(x)|$. Tab $|E(x)|\le \frac{M_{n+1}}{(n+1)!}|\omega (x)|,|\omega (x)|\le {\prod }_{i=0}^n|x-x_i|.$

Worked examples

Common mistakes (steel-manned)

Active recall

Connections

• Lagrange Interpolation — woh $p_n$ banata hai jiska yeh error hai.
• Newton Divided Differences — error term equals next divided difference $f[x_0,\dots ,x_n,x]\omega (x)$.
• Taylor's Theorem with Remainder — limiting case jab saare nodes coalesce ho jaate hain.
• Rolle's Theorem / Mean Value Theorem — derivation ka engine.
• Runge Phenomenon aur Chebyshev Nodes — node-choice ke consequences.
• Numerical Integration — Newton–Cotes error is $\omega$ factor ko inherit karta hai.

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