---
title: "Newton interpolation: y(3), y(7)"
subject: "Numerical Method"
chapter: 4
key: "semester_4/numerical_method:newton-y3-y7"
source: https://www.bctengineer.com/q/semester_4/numerical_method/newton-y3-y7
site: BCT Engineering
---

# Newton interpolation: y(3), y(7)

Numerical Method · Chapter 4: Interpolation

- Asked in: 2070 Magh, 8 marks
- Video: [Newtons forward and backword interpolation formula|M3|vtu|suk|18mat31|21MAT31|Finite differences|](https://www.youtube.com/watch?v=HwvuCxV1iAI) — CGAT Maths
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/newton-y3-y7
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-4

## Question

Compute the value of y(3) and y(7) from the following data using Newton's interpolation formula.
| x | 2 | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|---|
| y | 5.1 | 4.2 | 3.1 | 3.5 | 6.2 | 7.3 |

## Short answer

Using Newton's forward formula with p=0.5 and x0=2 yields y(3) ≈ 4.7125. Using Newton's backward formula with p=-2.5 and xn=12 yields y(7) ≈ 2.9969.

## Answer

The points are equally spaced (h = 2), so Newton's formulas apply. Use the forward formula for x near the start of the table and the backward formula for x near the end.

#### Difference table

| x | y | Δy | Δ²y | Δ³y | Δ⁴y | Δ⁵y |
|---|---|---|---|---|---|---|
| 2 | 5.1 | -0.9 | -0.2 | 1.7 | -0.9 | -3.8 |
| 4 | 4.2 | -1.1 | 1.5 | 0.8 | -4.7 |  |
| 6 | 3.1 | 0.4 | 2.3 | -3.9 |  |  |
| 8 | 3.5 | 2.7 | -1.6 |  |  |  |
| 10 | 6.2 | 1.1 |  |  |  |  |
| 12 | 7.3 |  |  |  |  |  |

#### y(3): Newton's forward formula

y(x) = y₀ + pΔy₀ + p(p−1)/2! Δ²y₀ + p(p−1)(p−2)/3! Δ³y₀ + …,  p = (x − x₀)/h

x₀ = 2, p = (3 − 2)/2 = 0.5. The differences used are the top diagonal: Δy₀ = −0.9, Δ²y₀ = −0.2, Δ³y₀ = 1.7, Δ⁴y₀ = −0.9, Δ⁵y₀ = −3.8.

| Term | Coefficient | Difference | Value |
|---|---|---|---|
| y₀ | 1 | 5.1 | 5.1000 |
| Δy | p = 0.5000 | -0.9 | -0.4500 |
| Δ²y | p(p−1)/2! = -0.1250 | -0.2 | 0.0250 |
| Δ³y | p(p−1)(p−2)/3! = 0.0625 | 1.7 | 0.1063 |
| Δ⁴y | p(p−1)(p−2)(p−3)/4! = -0.0391 | -0.9 | 0.0352 |
| Δ⁵y | p(p−1)…(p−4)/5! = 0.0273 | -3.8 | -0.1039 |

**y(3) ≈ 4.7125**

#### y(7): Newton's backward formula

y(x) = yₙ + p∇yₙ + p(p+1)/2! ∇²yₙ + p(p+1)(p+2)/3! ∇³yₙ + …,  p = (x − xₙ)/h

xₙ = 12, p = (7 − 12)/2 = -2.5. The differences used are the last diagonal: ∇yₙ = 1.1, ∇²yₙ = −1.6, ∇³yₙ = −3.9, ∇⁴yₙ = −4.7, ∇⁵yₙ = −3.8.

| Term | Coefficient | Difference | Value |
|---|---|---|---|
| yₙ | 1 | 7.3 | 7.3000 |
| ∇y | p = -2.5000 | 1.1 | -2.7500 |
| ∇²y | p(p+1)/2! = 1.8750 | -1.6 | -3.0000 |
| ∇³y | p(p+1)(p+2)/3! = -0.3125 | -3.9 | 1.2188 |
| ∇⁴y | p(p+1)(p+2)(p+3)/4! = -0.0391 | -4.7 | 0.1836 |
| ∇⁵y | p(p+1)…(p+4)/5! = -0.0117 | -3.8 | 0.0445 |

**y(7) ≈ 2.9969**

All six points are used, so both formulas represent the same degree-5 polynomial; for example the forward formula with p = 2.5 also gives 2.9969.

**Answer: y(3) ≈ 4.7125, y(7) ≈ 2.9969**

## Check yourself

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

1. Which formula is used for y(7) with xn = 12?
   A. Lagrange with p = 2.5
   B. Backward formula with p = -2.5
   C. Stirling
   D. Forward formula with p = 0.5
   Answer: B. 7 is closer to the end; p = (7 - 12)/2 = -2.5. (Key point: Formula choice)

2. What is p for y(3) using x0 = 2, h = 2?
   A. 0.5
   B. 1
   C. 2
   D. -0.5
   Answer: A. p = (3 - 2)/2 = 0.5. (Key point: p for forward)

3. Why do forward and backward give the same value for y(7)?
   A. All six points are used, giving the same degree-5 polynomial
   B. h is 1
   C. Differences are zero
   D. Rounding
   Answer: A. Both represent the unique interpolating polynomial. (Key point: Consistency)

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