---
title: "Newton interpolation: y(1.6), y(4.8)"
subject: "Numerical Method"
chapter: 4
key: "semester_4/numerical_method:newton-1-6-4-8"
source: https://www.bctengineer.com/q/semester_4/numerical_method/newton-1-6-4-8
site: BCT Engineering
---

# Newton interpolation: y(1.6), y(4.8)

Numerical Method · Chapter 4: Interpolation

- Asked in: 2068 Bhadra, 8 marks
- Video: [Newton's Forward Interpolation Formula l Numerical Methods l Newton's Interpolation Formula](https://www.youtube.com/watch?v=aciJMu62Fjo) — SK Maths Tutorial
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/newton-1-6-4-8
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-4

## Question

Find the values of y at x = 1.6 and x = 4.8 from the following points using Newton's interpolation technique.
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| y | 4 | 7.5 | 4 | 8.5 | 9.6 |

## Short answer

Forward formula with x0=1, p=0.6 gives y(1.6) ≈ 8.667. Backward formula with xn=5, p=-0.2 gives y(4.8) ≈ 11.086. Both use the computed difference table.

## Answer

The points are equally spaced (h = 1). x = 1.6 is near the start of the table (Newton's forward formula) and x = 4.8 is near the end (Newton's backward formula).

#### Difference table

| x | y | Δy | Δ²y | Δ³y | Δ⁴y |
|---|---|---|---|---|---|
| 1 | 4.0 | 3.5 | -7.0 | 15.0 | -26.4 |
| 2 | 7.5 | -3.5 | 8.0 | -11.4 |  |
| 3 | 4.0 | 4.5 | -3.4 |  |  |
| 4 | 8.5 | 1.1 |  |  |  |
| 5 | 9.6 |  |  |  |  |

#### y(1.6): forward formula

x₀ = 1, p = (1.6 − 1)/1 = 0.6; top diagonal: Δy₀ = 3.5, Δ²y₀ = −7.0, Δ³y₀ = 15.0, Δ⁴y₀ = −26.4.

| Term | Coefficient | Difference | Value |
|---|---|---|---|
| y₀ | 1 | 4.0 | 4.0000 |
| Δy | p = 0.6000 | 3.5 | 2.1000 |
| Δ²y | p(p−1)/2! = -0.1200 | -7.0 | 0.8400 |
| Δ³y | p(p−1)(p−2)/3! = 0.0560 | 15.0 | 0.8400 |
| Δ⁴y | p(p−1)(p−2)(p−3)/4! = -0.0336 | -26.4 | 0.8870 |

**y(1.6) ≈ 8.6670**

#### y(4.8): backward formula

xₙ = 5, p = (4.8 − 5)/1 = -0.2; last diagonal: ∇yₙ = 1.1, ∇²yₙ = -3.4, ∇³yₙ = -11.4, ∇⁴yₙ = -26.4 (last entry of each column).

| Term | Coefficient | Difference | Value |
|---|---|---|---|
| yₙ | 1 | 9.6 | 9.6000 |
| ∇y | p = -0.2000 | 1.1 | -0.2200 |
| ∇²y | p(p+1)/2! = -0.0800 | -3.4 | 0.2720 |
| ∇³y | p(p+1)(p+2)/3! = -0.0480 | -11.4 | 0.5472 |
| ∇⁴y | p(p+1)(p+2)(p+3)/4! = -0.0336 | -26.4 | 0.8870 |

**y(4.8) ≈ 11.0862**

**Answer: y(1.6) ≈ 8.667, y(4.8) ≈ 11.086**

## Check yourself

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

1. Which formulas are used for 1.6 and 4.8?
   A. Forward for both
   B. Forward for 1.6, backward for 4.8
   C. Backward for both
   D. Backward for 1.6, forward for 4.8
   Answer: B. 1.6 near the start, 4.8 near the end. (Key point: Formula choice)

2. What is p for 4.8?
   A. 0.8
   B. -0.2
   C. -0.8
   D. 0.2
   Answer: B. p = (4.8 - 5)/1 = -0.2. (Key point: p for 4.8)

3. Which are the backward differences?
   A. 1.1, 4.5, 8.0, 3.5
   B. 9.6, 8.5, 4.0, 7.5
   C. 1.1, -3.4, -11.4, -26.4
   D. 3.5, -7.0, 15.0, -26.4
   Answer: C. Last entry of each column. (Key point: Last diagonal)

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