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Newton interpolation: y(1.6), y(4.8)

Short answer

Forward formula with x0=1, p=0.6 gives y(1.6) ≈ 8.667. Backward formula with xn=5, p=-0.2 gives y(4.8) ≈ 11.086. Both use the computed difference table.

Find the values of y at x = 1.6 and x = 4.8 from the following points using Newton's interpolation technique.

x12345
y47.548.59.6

The points are equally spaced (h = 1). x = 1.6 is near the start of the table (Newton's forward formula) and x = 4.8 is near the end (Newton's backward formula).

Difference table

xyΔyΔ²yΔ³yΔ⁴y
14.03.5-7.015.0-26.4
27.5-3.58.0-11.4
34.04.5-3.4
48.51.1
59.6

y(1.6): forward formula

x₀ = 1, p = (1.6 − 1)/1 = 0.6; top diagonal: Δy₀ = 3.5, Δ²y₀ = −7.0, Δ³y₀ = 15.0, Δ⁴y₀ = −26.4.

TermCoefficientDifferenceValue
y₀14.04.0000
Δyp = 0.60003.52.1000
Δ²yp(p−1)/2! = -0.1200-7.00.8400
Δ³yp(p−1)(p−2)/3! = 0.056015.00.8400
Δ⁴yp(p−1)(p−2)(p−3)/4! = -0.0336-26.40.8870

y(1.6) ≈ 8.6670

y(4.8): backward formula

xₙ = 5, p = (4.8 − 5)/1 = -0.2; last diagonal: ∇yₙ = 1.1, ∇²yₙ = -3.4, ∇³yₙ = -11.4, ∇⁴yₙ = -26.4 (last entry of each column).

TermCoefficientDifferenceValue
yₙ19.69.6000
∇yp = -0.20001.1-0.2200
∇²yp(p+1)/2! = -0.0800-3.40.2720
∇³yp(p+1)(p+2)/3! = -0.0480-11.40.5472
∇⁴yp(p+1)(p+2)(p+3)/4! = -0.0336-26.40.8870

y(4.8) ≈ 11.0862

Answer: y(1.6) ≈ 8.667, y(4.8) ≈ 11.086

✓ Answer checked by Claude Opus 5.5 (high effort)

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