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

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.

Using appropriate Newton's Interpolation Techniques, estimate y(15) and y(85) from the following data:

x1030507090
y3456452336

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

xyΔyΔ²yΔ³yΔ⁴y
103422-332224
3056-11-1146
5045-2235
702313
9036

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.

TermCoefficientDifferenceValue
y₀13434.0000
Δyp = 0.2500225.5000
Δ²yp(p−1)/2! = -0.0938-333.0938
Δ³yp(p−1)(p−2)/3! = 0.0547221.2031
Δ⁴yp(p−1)(p−2)(p−3)/4! = -0.037624-0.9023

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.

TermCoefficientDifferenceValue
yₙ13636.0000
∇yp = -0.250013-3.2500
∇²yp(p+1)/2! = -0.093835-3.2812
∇³yp(p+1)(p+2)/3! = -0.054746-2.5156
∇⁴yp(p+1)(p+2)(p+3)/4! = -0.037624-0.9023

y(85) ≈ 26.0508

Answer: y(15) ≈ 42.89, y(85) ≈ 26.05

✓ Answer checked by Claude Opus 5.5 (high effort)

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