Join local cubics with global smoothness

Try

Interpolate (0,0),(1,1),(2,0) with a natural cubic spline. Unlike one global quadratic, the spline uses one cubic on each interval, matching values and first two derivatives at the interior node. Natural conditions set the second derivative to zero at the outer endpoints. These are modeling choices, not consequences of the measured values. They make the curve flexible while selecting a unique member of the spline family.

Predict

Predict whether natural means the same as physically correct. It does not. If the true function has nonzero endpoint curvature, imposing zero curvature introduces a boundary mismatch. Clamped splines instead prescribe endpoint slopes, while other boundary choices serve other tasks. A spline may be visually smooth and still overshoot or violate monotonicity; smoothness constraints alone do not enforce those properties.

Experiment

For equal spacings, the second-derivative equation gives 4M₁=−12.

Node	Value	Nodal second derivative
0	0	0
1	1	−3
2	0	0

The first segment is S₀(x)=3x/2−x³/2. The second is its reflection S₁(x)=S₀(2−x). Verify values, first derivatives and second derivatives at x=1, then evaluate x=1/2 without relying on a drawing.

Understand

Let hᵢ=xᵢ₊₁−xᵢ>0 and Mᵢ=S″(xᵢ). On one interval, S″ is the line joining Mᵢ and Mᵢ₊₁. Integrate it twice and choose constants to match endpoint values. The resulting cubic is Mᵢ(xᵢ₊₁−x)³/(6hᵢ)+Mᵢ₊₁(x−xᵢ)³/(6hᵢ)+(yᵢ−Mᵢhᵢ²/6)(xᵢ₊₁−x)/hᵢ+(yᵢ₊₁−Mᵢ₊₁hᵢ²/6)(x−xᵢ)/hᵢ. Matching first derivatives at interior nodes gives the tridiagonal system stated in practice. The formula already matches values and second derivatives; derivative matching supplies the remaining coupling.

Apply

For natural boundaries M₀=Mₙ=0, the interior diagonal coefficient is 2(hᵢ₋₁+hᵢ), strictly larger than the sum of adjacent off-diagonal coefficients. If a nonzero homogeneous solution existed, choose an index with largest absolute component. Its row would require the diagonal contribution to be bounded by a strictly smaller multiple of that maximum, a contradiction. Thus the system is invertible and the spline exists uniquely. Counting constraints alone would not prove this. Solving the tridiagonal system and evaluating local cubics can be done efficiently, but input ordering, duplicate nodes and boundary specifications must be checked.

To derive the interior equation concretely, differentiate the segment formula at its right endpoint and the next segment at its left endpoint. The first derivative from interval i−1 is its secant slope plus hᵢ₋₁(Mᵢ₋₁+2Mᵢ)/6. The derivative from interval i is its secant slope minus hᵢ(2Mᵢ+Mᵢ₊₁)/6. Equating these expressions yields the tridiagonal row. This derivation shows why the right-hand side is a difference of slopes, and why a reversed sign would produce the wrong curvature at a local maximum.

Continue

Splines control local degree, yet changing one datum can alter the global derivative solution. Their approximation error depends on smoothness, mesh geometry and boundary compatibility. Do not assert a fourth-order global bound without those hypotheses; natural endpoints can create lower-order boundary effects. A written implementation should test endpoint conditions, nodal values and both derivative matches independently. Next, least squares replaces exact interpolation by an explicit residual-minimization criterion, better suited to many noisy or redundant observations.

Review calculus and linear algebra when a prerequisite needs practice. Continue to Fit residuals without unnecessarily squaring conditioning.