Natural cubic spline: y(0.5)

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 = 3 intervals, equal spacing h = 1. Let Mᵢ = S″(xᵢ), with M₀ = M3 = 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 … 2

With 6/h² = 6:

4M1 + M2 = 6 × (1 − 2(2.2) + 2) = -8.4000
M1 + 4M2 = 6 × (0.5 − 2(1) + 2.2) = 4.2000

Solving these simultaneous equations (elimination) gives

M₀ = 0, M1 = -2.5200, M2 = 1.6800, M3 = 0

Step 2: Spline on the interval [0, 1]

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 = 0, x = 0.5: x1 − x = 0.5000, x − x0 = 0.5000, M0 = 0.0000, M1 = -2.5200.

S(0.5) = (0.0000)(0.5000)³/6 + (-2.5200)(0.5000)³/6
          + (2.0000)(0.5000) + (2.6200)(0.5000)
        = 0.0000 − 0.0525 + 1.0000 + 1.3100
        = 2.2575

Answer: y(0.5) ≈ 2.2575