---
title: "Difference table (3.0 to 3.8)"
subject: "Numerical Method"
chapter: 4
key: "semester_4/numerical_method:difference-table-3-0"
source: https://www.bctengineer.com/q/semester_4/numerical_method/difference-table-3-0
site: BCT Engineering
---

# Difference table (3.0 to 3.8)

Numerical Method · Chapter 4: Interpolation

- Asked in: 2071 Bhadra, 4 marks
- Video: [interpolation problem 1|| Newton's forward interpolation formula|| numerical methods](https://www.youtube.com/watch?v=TIWRyzzFUYQ) — HAMEEDA MATHTUBER
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/difference-table-3-0
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-4

## Question

Create difference table from following data.
| X | 3.0 | 3.2 | 3.4 | 3.6 | 3.8 |
|---|---|---|---|---|---|
| Y | 0.4771 | 0.5051 | 0.5315 | 0.5563 | 0.5798 |

## Short answer

With h=0.2, forward differences: Δy₀=0.0280, Δ²y₀=−0.0016, Δ³y₀=0.0000, Δ⁴y₀=0.0003. Third and fourth differences are nearly zero, indicating nearly quadratic behavior.

## Answer

The values are equally spaced with h = 0.2. Each entry of a column is the difference of the two neighbouring entries of the previous column: Δyᵢ = yᵢ₊₁ − yᵢ, Δ²yᵢ = Δyᵢ₊₁ − Δyᵢ, and so on.

#### Forward difference table

| x | y | Δy | Δ²y | Δ³y | Δ⁴y |
|---|---|---|---|---|---|
| 3 | 0.4771 | 0.0280 | -0.0016 | 0.0000 | 0.0003 |
| 3.2 | 0.5051 | 0.0264 | -0.0016 | 0.0003 |  |
| 3.4 | 0.5315 | 0.0248 | -0.0013 |  |  |
| 3.6 | 0.5563 | 0.0235 |  |  |  |
| 3.8 | 0.5798 |  |  |  |  |

Working for the first row:

```text
Δy₀  = 0.5051 − 0.4771 = 0.0280
Δ²y₀ = 0.0264 − 0.0280 = −0.0016
Δ³y₀ = −0.0016 − (−0.0016) = 0.0000
Δ⁴y₀ = 0.0003 − 0.0000 = 0.0003
```

The third and fourth differences are almost zero, which shows the data behave like a smooth, nearly quadratic function (these are log₁₀ x values). The first row (y₀, Δy₀, Δ²y₀, Δ³y₀, Δ⁴y₀) is what Newton's forward formula uses, and the last diagonal is what the backward formula uses.

## Check yourself

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

1. How is each entry of Δ²y computed?
   A. Product of Δy entries
   B. Difference of neighbouring y entries
   C. Sum of Δy entries
   D. Difference of neighbouring Δy entries
   Answer: D. Δ²yi = Δy(i+1) - Δyi. (Key point: Definition)

2. What is Δy0?
   A. 0.0280
   B. 0.5051
   C. 0.4771
   D. 0.0264
   Answer: A. 0.5051 - 0.4771 = 0.0280. (Key point: First Δy)

3. What do the near-zero third and fourth differences show?
   A. Data are smooth, nearly quadratic
   B. Data are random
   C. h is wrong
   D. The table has errors
   Answer: A. Small higher differences mean smooth behaviour. (Key point: Interpretation)

## Related in Numerical Method

- [Forward difference table](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Aforward-difference-table-1-6) — Numerical Method, chapter 4
- [Newton interpolation: steam table](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Anewton-steam-table) — Numerical Method, chapter 4
- [y'(1) and y''(1) from table](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Aderivatives-at-1-table) — Numerical Method, chapter 5
- [Angular velocity from rod data](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Avelocity-acceleration-rod-angle) — Numerical Method, chapter 5, asked 2×

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