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Newton interpolation: y(3), y(7)

Short answer

Using Newton's forward formula with p=0.5 and x0=2 yields y(3) ≈ 4.7125. Using Newton's backward formula with p=-2.5 and xn=12 yields y(7) ≈ 2.9969.

Compute the value of y(3) and y(7) from the following data using Newton's interpolation formula.

x24681012
y5.14.23.13.56.27.3

The points are equally spaced (h = 2), so Newton's formulas apply. Use the forward formula for x near the start of the table and the backward formula for x near the end.

Difference table

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

y(3): Newton's forward formula

y(x) = y₀ + pΔy₀ + p(p−1)/2! Δ²y₀ + p(p−1)(p−2)/3! Δ³y₀ + …, p = (x − x₀)/h

x₀ = 2, p = (3 − 2)/2 = 0.5. The differences used are the top diagonal: Δy₀ = −0.9, Δ²y₀ = −0.2, Δ³y₀ = 1.7, Δ⁴y₀ = −0.9, Δ⁵y₀ = −3.8.

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

y(3) ≈ 4.7125

y(7): Newton's backward formula

y(x) = yₙ + p∇yₙ + p(p+1)/2! ∇²yₙ + p(p+1)(p+2)/3! ∇³yₙ + …, p = (x − xₙ)/h

xₙ = 12, p = (7 − 12)/2 = -2.5. The differences used are the last diagonal: ∇yₙ = 1.1, ∇²yₙ = −1.6, ∇³yₙ = −3.9, ∇⁴yₙ = −4.7, ∇⁵yₙ = −3.8.

TermCoefficientDifferenceValue
yₙ17.37.3000
∇yp = -2.50001.1-2.7500
∇²yp(p+1)/2! = 1.8750-1.6-3.0000
∇³yp(p+1)(p+2)/3! = -0.3125-3.91.2188
∇⁴yp(p+1)(p+2)(p+3)/4! = -0.0391-4.70.1836
∇⁵yp(p+1)…(p+4)/5! = -0.0117-3.80.0445

y(7) ≈ 2.9969

All six points are used, so both formulas represent the same degree-5 polynomial; for example the forward formula with p = 2.5 also gives 2.9969.

Answer: y(3) ≈ 4.7125, y(7) ≈ 2.9969

✓ Answer checked by Claude Opus 5.5 (high effort)

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