---
title: "Lagrange: y at x = 3.0"
subject: "Numerical Method"
chapter: 4
key: "semester_4/numerical_method:lagrange-y-at-3"
source: https://www.bctengineer.com/q/semester_4/numerical_method/lagrange-y-at-3
site: BCT Engineering
---

# Lagrange: y at x = 3.0

Numerical Method · Chapter 4: Interpolation

- Asked in: 2071 Bhadra, 8 marks
- Video: [●Lagrange Interpolation || Numerical Concept ✅️](https://www.youtube.com/watch?v=YlVOYcsyNbU) — The PrepStudio
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/lagrange-y-at-3
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-4

## Question

Use Lagrange's Interpolation formula to find the value of y when x = 3.0, from the following table.
| x | 3.2 | 2.7 | 1.0 | 4.8 | 5.6 |
|---|---|---|---|---|---|
| y | 22.0 | 17.8 | 14.2 | 38.3 | 51.7 |

## Short answer

Using Lagrange interpolation, y(3.0)=20.2672. Sum of products yᵢ·Lᵢ(3.0) with coefficients L₀=0.664773, L₁=0.361634, L₂=−0.004295, L₃=−0.030545, L₄=0.008433.

## Answer

The x values are unequally spaced and unordered, so Lagrange's formula is the right tool:

y(x) = Σ yᵢ·Lᵢ(x),   Lᵢ(x) = Π (x − xⱼ)/(xᵢ − xⱼ)  (j ≠ i)

Data: x₀ = 3.2, x₁ = 2.7, x₂ = 1.0, x₃ = 4.8, x₄ = 5.6 with y = 22.0, 17.8, 14.2, 38.3, 51.7. Required x = 3.0.

#### Lagrange coefficients at x = 3.0

For example L₀(3) = [(3−2.7)(3−1.0)(3−4.8)(3−5.6)] / [(3.2−2.7)(3.2−1.0)(3.2−4.8)(3.2−5.6)] = (0.3)(2.0)(−1.8)(−2.6) / (0.5)(2.2)(−1.6)(−2.4) = 2.808/4.224 = 0.664773.

| Term | Numerator Π(3 − xⱼ) | Denominator Π(xᵢ − xⱼ) | Value |
|---|---|---|---|
| L0(3) | (0.3) × (2.0) × (-1.8) × (-2.6) | (0.5) × (2.2) × (-1.6) × (-2.4) | 0.664773 |
| L1(3) | (-0.2) × (2.0) × (-1.8) × (-2.6) | (-0.5) × (1.7) × (-2.1) × (-2.9) | 0.361634 |
| L2(3) | (-0.2) × (0.3) × (-1.8) × (-2.6) | (-2.2) × (-1.7) × (-3.8) × (-4.6) | -0.004295 |
| L3(3) | (-0.2) × (0.3) × (2.0) × (-2.6) | (1.6) × (2.1) × (3.8) × (-0.8) | -0.030545 |
| L4(3) | (-0.2) × (0.3) × (2.0) × (-1.8) | (2.4) × (2.9) × (4.6) × (0.8) | 0.008433 |

#### Sum

| i | yᵢ | Lᵢ(3.0) | yᵢ·Lᵢ(3.0) |
|---|---|---|---|
| 0 | 22.0 | 0.664773 | 14.6250 |
| 1 | 17.8 | 0.361634 | 6.4371 |
| 2 | 14.2 | -0.004295 | -0.0610 |
| 3 | 38.3 | -0.030545 | -1.1699 |
| 4 | 51.7 | 0.008433 | 0.4360 |
|  |  | Sum | 20.2672 |

Check: ΣLᵢ = 1.000000 (must equal 1).

**Answer: y(3.0) ≈ 20.2672**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is Lagrange's formula used here?
   A. x values are unequally spaced and unordered
   B. The data are periodic
   C. x values are equally spaced
   D. Differences are constant
   Answer: A. Newton's difference formulas need equal spacing. (Key point: Why Lagrange)

2. What is Li(x)?
   A. Δ^i y / i!
   B. Product of (x - xi)/(xj - xi)
   C. Product of (x - xj)/(xi - xj) for j ≠ i
   D. Sum of (x - xj)
   Answer: C. Li equals 1 at xi and 0 at other nodes. (Key point: Formula)

3. What check holds for the Lagrange coefficients?
   A. Sum of Li(x) = 0
   B. Sum equals y
   C. Sum of Li(x) = 1
   D. Sum of Li(x) = n
   Answer: C. Interpolating a constant 1 gives ΣLi = 1. (Key point: Check)

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