--- title: "Newton forward and Stirling" subject: "Numerical Method" chapter: 4 key: "semester_4/numerical_method:newton-forward-stirling" source: https://www.bctengineer.com/q/semester_4/numerical_method/newton-forward-stirling site: BCT Engineering --- # Newton forward and Stirling Numerical Method · Chapter 4: Interpolation - Asked in: 2074 Bhadra, 8 marks - Video: [Stirling's interpolation formula](https://www.youtube.com/watch?v=UC8A9bt_-J0) — Civil learning online - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/numerical_method/newton-forward-stirling - Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-4 ## Question From the following data, compute: (a) y(3) using Newton's forward interpolation formula (b) y(6.4) using Stirling's formula | x | 2 | 4 | 6 | 8 | 10 | 12 | |---|---|---|---|---|---|---| | y | 5.1 | 4.2 | 3.1 | 3.5 | 6.2 | 7.3 | ## Short answer Use h = 2. Newton's forward formula from x0 = 2 with p = 0.5 gives y(3) = 4.7125. Stirling's central difference formula centered at x0 = 6 with p = 0.2 gives y(6.4) = 3.0214. ## Answer #### Forward difference table (h = 2) | 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 | | | | | | #### (a) y(3) by Newton's forward formula y(x₀ + ph) = y₀ + pΔy₀ + p(p−1)/2! Δ²y₀ + p(p−1)(p−2)/3! Δ³y₀ + … x₀ = 2, h = 2, so p = (3 − 2)/2 = 0.5. | 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 | Sum of the terms: y(3) = 4.7125 **Answer (a): y(3) ≈ 4.7125** #### (b) y(6.4) by Stirling's formula Stirling's formula is used near the middle of the table. Take the central value x₀ = 6, so p = (6.4 − 6)/2 = 0.2. Subscripts count steps from x₀ (y₋₁ is the value one step before x₀). Differences are read around x₀ from the table above (Δy₋₁ = −1.1, Δy₀ = 0.4; Δ²y₋₁ = 1.5; Δ³y₋₂ = 1.7, Δ³y₋₁ = 0.8; Δ⁴y₋₂ = −0.9). y(x₀ + ph) = y₀ + p·(Δy₋₁ + Δy₀)/2 + p²/2!·Δ²y₋₁ + p(p²−1)/3!·(Δ³y₋₂ + Δ³y₋₁)/2 + p²(p²−1)/4!·Δ⁴y₋₂ | Term | Coefficient | Difference | Value | |---|---|---|---| | y₀ | 1 | 3.1 | 3.1000 | | p·(Δy₋₁+Δy₀)/2 | 0.2000 | -0.35 | -0.0700 | | p²/2!·Δ²y₋₁ | 0.0200 | 1.5 | 0.0300 | | p(p²−1)/3!·(Δ³y₋₂+Δ³y₋₁)/2 | -0.0320 | 1.25 | -0.0400 | | p²(p²−1)/4!·Δ⁴y₋₂ | -0.0016 | -0.9 | 0.0014 | **Answer (b): y(6.4) ≈ 3.0214** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For y(3) with x0 = 2 and h = 2, what is p? A. 0.5 B. 1 C. 2 D. 0.25 Answer: A. p = (3 - 2)/2 = 0.5. (Key point: Value of p) 2. What is the coefficient of Δ²y0 in Newton's forward formula for p = 0.5? A. p(p+1)/2! = 0.375 B. p(p-1)/2! = -0.125 C. p²/2! = 0.125 D. p/2! = 0.25 Answer: B. 0.5 × (-0.5)/2 = -0.125. (Key point: Coefficient) 3. Why is Stirling's formula used for y(6.4)? A. It suits points near the start of the table B. It needs unequal spacing C. It suits points near the middle of the table, with x0 = 6 D. It uses no differences Answer: C. Central difference formulas like Stirling's are for x near the centre. (Key point: Choice of formula) 4. What is the first-difference term p(Δy-1 + Δy0)/2 with p = 0.2, Δy-1 = -1.1, Δy0 = 0.4? A. -0.0700 B. 0.0700 C. -0.1400 D. 0.1000 Answer: A. Average of differences = (-1.1 + 0.4)/2 = -0.35; times 0.2 = -0.07. 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