---
title: "Newton interpolation: y(15), y(85)"
subject: "Numerical Method"
chapter: 4
key: "semester_4/numerical_method:newton-interpolation-y15-y85"
source: https://www.bctengineer.com/q/semester_4/numerical_method/newton-interpolation-y15-y85
site: BCT Engineering
---

# Newton interpolation: y(15), y(85)

Numerical Method · Chapter 4: Interpolation

- Asked in: 2072 Asoj, 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-interpolation-y15-y85
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-4

## Question

Using appropriate Newton's Interpolation Techniques, estimate y(15) and y(85) from the following data:
| x | 10 | 30 | 50 | 70 | 90 |
|---|---|---|---|---|---|
| y | 34 | 56 | 45 | 23 | 36 |

## Short answer

Use Newton's forward formula for x = 15 with x0 = 10 and p = 0.25, giving y(15) = 42.8945. Use Newton's backward formula for x = 85 with x_n = 90 and p = -0.25, giving y(85) = 26.0508.

## Answer

The data are equally spaced (h = 20). x = 15 is near the beginning of the table, so use Newton's forward formula; x = 85 is near the end, so use Newton's backward formula.

#### Difference table

| x | y | Δy | Δ²y | Δ³y | Δ⁴y |
|---|---|---|---|---|---|
| 10 | 34 | 22 | -33 | 22 | 24 |
| 30 | 56 | -11 | -11 | 46 |  |
| 50 | 45 | -22 | 35 |  |  |
| 70 | 23 | 13 |  |  |  |
| 90 | 36 |  |  |  |  |

#### y(15): forward formula

x₀ = 10, p = (15 − 10)/20 = 0.25. The top diagonal of the table gives Δy₀ = 22, Δ²y₀ = −33, Δ³y₀ = 22, Δ⁴y₀ = 24.

| Term | Coefficient | Difference | Value |
|---|---|---|---|
| y₀ | 1 | 34 | 34.0000 |
| Δy | p = 0.2500 | 22 | 5.5000 |
| Δ²y | p(p−1)/2! = -0.0938 | -33 | 3.0938 |
| Δ³y | p(p−1)(p−2)/3! = 0.0547 | 22 | 1.2031 |
| Δ⁴y | p(p−1)(p−2)(p−3)/4! = -0.0376 | 24 | -0.9023 |

**y(15) ≈ 42.8945**

#### y(85): backward formula

xₙ = 90, p = (85 − 90)/20 = -0.25. The last diagonal gives ∇yₙ = 13, ∇²yₙ = 35, ∇³yₙ = 46, ∇⁴yₙ = 24.

| Term | Coefficient | Difference | Value |
|---|---|---|---|
| yₙ | 1 | 36 | 36.0000 |
| ∇y | p = -0.2500 | 13 | -3.2500 |
| ∇²y | p(p+1)/2! = -0.0938 | 35 | -3.2812 |
| ∇³y | p(p+1)(p+2)/3! = -0.0547 | 46 | -2.5156 |
| ∇⁴y | p(p+1)(p+2)(p+3)/4! = -0.0376 | 24 | -0.9023 |

**y(85) ≈ 26.0508**

**Answer: y(15) ≈ 42.89, y(85) ≈ 26.05**

## Check yourself

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

1. Which Newton formulas suit x = 15 and x = 85?
   A. Backward for both
   B. Forward for 15, backward for 85
   C. Forward for both
   D. Backward for 15, forward for 85
   Answer: B. 15 is near the start, 85 near the end. (Key point: Choice of formula)

2. What is p for y(85) using x_n = 90, h = 20?
   A. 0.75
   B. 0.25
   C. -4
   D. -0.25
   Answer: D. p = (85 - 90)/20 = -0.25. (Key point: Value of p)

3. Which differences are used in the backward formula?
   A. The last diagonal of the table
   B. The first column only
   C. The top diagonal
   D. The middle diagonal
   Answer: A. ∇y, ∇²y, ... are the last entries of each column. (Key point: Differences used)

4. What are the final estimates?
   A. y(15) ≈ 26.05, y(85) ≈ 42.89
   B. y(15) ≈ 42.89, y(85) ≈ 26.05
   C. y(15) ≈ 45, y(85) ≈ 30
   D. y(15) ≈ 34, y(85) ≈ 36
   Answer: B. These come from summing the forward and backward series. (Key point: Results)

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