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Newton forward and Stirling

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.

From the following data, compute: (a) y(3) using Newton's forward interpolation formula (b) y(6.4) using Stirling's formula

x24681012
y5.14.23.13.56.27.3

Forward difference table (h = 2)

xyΔyΔ²yΔ³yΔ⁴yΔ⁵y
25.1-0.9-0.21.7-0.9-3.8
44.2-1.11.50.8-4.7
63.10.42.3-3.9
83.52.7-1.6
106.21.1
127.3

(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.

TermCoefficientDifferenceValue
y₀15.15.1000
Δyp = 0.5000-0.9-0.4500
Δ²yp(p−1)/2! = -0.1250-0.20.0250
Δ³yp(p−1)(p−2)/3! = 0.06251.70.1063
Δ⁴yp(p−1)(p−2)(p−3)/4! = -0.0391-0.90.0352
Δ⁵yp(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

TermCoefficientDifferenceValue
y₀13.13.1000
p·(Δy₋₁+Δy₀)/20.2000-0.35-0.0700
p²/2!·Δ²y₋₁0.02001.50.0300
p(p²−1)/3!·(Δ³y₋₂+Δ³y₋₁)/2-0.03201.25-0.0400
p²(p²−1)/4!·Δ⁴y₋₂-0.0016-0.90.0014

Answer (b): y(6.4) ≈ 3.0214

✓ Answer checked by Claude Opus 5.5 (high effort)

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