Natural cubic spline: y(6.5)

A natural cubic spline is a piecewise cubic S(x) that passes through all data points, has continuous first and second derivatives at the interior knots, and has zero second derivative at the end points (M₀ = Mₙ = 0).

Step 1: Setup

Points: n = 4 intervals, equal spacing h = 2. Let Mᵢ = S″(xᵢ), with M₀ = M4 = 0.

For equal spacing the continuity of S′ at each interior knot gives

M(i−1) + 4Mi + M(i+1) = (6/h²)(y(i+1) − 2yi + y(i−1)),   i = 1 … 3

With 6/h² = 1.5:

4M1 + M2 = 1.5 × (9 − 2(10) + 8) = -4.5000
M1 + 4M2 + M3 = 1.5 × (12 − 2(9) + 10) = 6.0000
M2 + 4M3 = 1.5 × (5 − 2(12) + 9) = -15.0000

Solving these simultaneous equations (elimination) gives

M₀ = 0, M1 = -1.9018, M2 = 3.1071, M3 = -4.5268, M4 = 0

Step 2: Spline on the interval [5, 7]

S(x) = Mi(x(i+1) − x)³/(6h) + M(i+1)(x − xi)³/(6h)
       + (yi/h − Mi·h/6)(x(i+1) − x) + (y(i+1)/h − M(i+1)·h/6)(x − xi)

Here i = 1, x = 6.5: x2 − x = 0.5000, x − x1 = 1.5000, M1 = -1.9018, M2 = 3.1071.

S(6.5) = (-1.9018)(0.5000)³/12 + (3.1071)(1.5000)³/12
          + (5.6339)(0.5000) + (3.4643)(1.5000)
        = -0.0198 + 0.8739 + 2.8170 + 5.1964
        = 8.8675

Answer: y(6.5) ≈ 8.8675