Lagrange Interpolation Formula Numerical Problem

 

❓ Question

Given:

$$a=1,\; b=2,\; c=3$$

Using Lagrange’s Interpolation Formula:

$$y=\frac{(x-b)(x-c)}{(a-b)(a-c)}y_1 +\frac{(x-a)(x-c)}{(b-a)(b-c)}y_2 +\frac{(x-a)(x-b)}{(c-a)(c-b)}y_3$$

Find the value of $y$ at:

$$x=0$$

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🖼 Question Image

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✍️ Short Concept

Lagrange interpolation is used to:

👉 Find polynomial passing through given points
👉 Directly calculate unknown values without forming full equation.

Main idea:

$$y = \sum L_i(x)y_i$$

where $L_i(x)$ are interpolation terms.

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🔷 Step 1 — Substitute Given Values 💯

Using:

$$a=1,\; b=2,\; c=3$$ $$y_1=3,\; y_2=6,\; y_3=3$$

Formula becomes:

$$y=\frac{(x-2)(x-3)}{(1-2)(1-3)}(3)+\frac{(x-1)(x-3)}{(2-1)(2-3)}(6)+\frac{(x-1)(x-2)}{(3-1)(3-2)}(3)$$

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🔷 Step 2 — Simplify Denominators

$$y=\frac{3(x-2)(x-3)}{2} -6(x-1)(x-3) +\frac{3(x-1)(x-2)}{2}$$

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🔷 Step 3 — Put $x=0$

$$y=\frac{3(-2)(-3)}{2} -6(-1)(-3) +\frac{3(-1)(-2)}{2}$$

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🔷 Step 4 — Calculate Terms

$$y=\frac{18}{2}-18+\frac{6}{2}$$
$$y=9-18+3$$
$$y=12-18$$
$$y=-6$$

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✅ Final Answer

$$\boxed{y=-6}$$

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⭐ Golden JEE Insight

In Lagrange interpolation:

👉 No need to form full polynomial equation

Just:

$$\text{Substitute directly into interpolation formula}$$

This saves huge time in numerical questions.