---
title: "Lagrange: f(2.5)"
subject: "Numerical Method"
chapter: 4
key: "semester_4/numerical_method:lagrange-f-2-5"
source: https://www.bctengineer.com/q/semester_4/numerical_method/lagrange-f-2-5
site: BCT Engineering
---

# Lagrange: f(2.5)

Numerical Method · Chapter 4: Interpolation

- Asked in: 2069 Bhadra, 8 marks
- Video: [#37 || Problem#3 || Lagrange's formula for interpolation || Numerical Methods || 18MAT21 ||](https://www.youtube.com/watch?v=uhExudBVQDA) — Math E Connect
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/lagrange-f-2-5
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-4

## Question

Use Lagrange method to find f(2.5) from the following data:
| x | 1 | 2 | 3 | 4 | [?] |
|---|---|---|---|---|---|
| f(x) | 1 | 1.414 | 1.732 | 2.00 | 2.6 |
[x row partly illegible in the scan: x values for the middle and last columns are hard to read]

## Short answer

Using cubic Lagrange interpolation on points (1,1), (2,1.414), (3,1.732), (4,2) gives f(2.5) ≈ 1.5821. The true value √2.5 is 1.5811.

## Answer

**Assumption.** In the scan only the points x = 1, 2, 3, 4 with f = 1, 1.414, 1.732, 2.000 are readable (these are the values of √x). The x value in the last column is illegible (its f value, 2.6, does not fit √x at x = 5), so the formula is applied to these four points. Required: f(2.5).

Lagrange's formula for four points (cubic):

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

```text
L₀(2.5) = (2.5−2)(2.5−3)(2.5−4) / [(1−2)(1−3)(1−4)] = (0.5)(−0.5)(−1.5)/(−6) = −0.0625
L₁(2.5) = (2.5−1)(2.5−3)(2.5−4) / [(2−1)(2−3)(2−4)] = (1.5)(−0.5)(−1.5)/(2)  =  0.5625
L₂(2.5) = (2.5−1)(2.5−2)(2.5−4) / [(3−1)(3−2)(3−4)] = (1.5)(0.5)(−1.5)/(−2)  =  0.5625
L₃(2.5) = (2.5−1)(2.5−2)(2.5−3) / [(4−1)(4−2)(4−3)] = (1.5)(0.5)(−0.5)/(6)   = −0.0625
```

| i | xᵢ | f(xᵢ) | Lᵢ(2.5) | f(xᵢ)·Lᵢ(2.5) |
|---|---|---|---|---|
| 0 | 1 | 1 | -0.0625 | -0.0625 |
| 1 | 2 | 1.414 | 0.5625 | 0.7954 |
| 2 | 3 | 1.732 | 0.5625 | 0.9742 |
| 3 | 4 | 2 | -0.0625 | -0.1250 |
|  |  | Σ | 1.0000 | 1.5821 |

f(2.5) = (−0.0625)(1) + (0.5625)(1.414) + (0.5625)(1.732) + (−0.0625)(2.000) = 1.5821

**Answer: f(2.5) ≈ 1.5821** (the true √2.5 = 1.5811).

If the fifth point (5, 2.236) is also taken, the five-point formula gives f(2.5) ≈ 1.5815, which agrees to three decimal places.

## Check yourself

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

1. What data are used since the last column is illegible?
   A. Only two points
   B. All five points
   C. Points x = 2.5
   D. Four points x = 1..4
   Answer: D. Only x = 1, 2, 3, 4 are readable. (Key point: Assumption)

2. What is L0(2.5)?
   A. -0.0625
   B. -0.5625
   C. 0.0625
   D. 0.5625
   Answer: A. (0.5)(-0.5)(-1.5)/(-6) = -0.0625. (Key point: L0)

3. True √2.5 is 1.5811. How close is the answer?
   A. 1.5821, close
   B. 2.5
   C. 3.1
   D. 1.0
   Answer: A. The estimate 1.5821 is within 0.001. (Key point: Sanity check)

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