Fit exact data with a unique polynomial

Try

Given measurements (0,1),(1,2),(2,5), seek a polynomial of degree at most two passing through all three. The answer is p(x)=1+x². Instead of solving a coefficient system immediately, build three functions that select one node each. L₀=(x−1)(x−2)/2, L₁=−x(x−2), and L₂=x(x−1)/2 have values one at their own node and zero at the other two.

Predict

Predict why the degree restriction matters. Adding any multiple of x(x−1)(x−2) leaves all three data values unchanged but produces a higher-degree polynomial. Exact fitting without a degree bound is not unique. Also predict what changes if two nodes coincide with conflicting values: no ordinary function can take both values at one input. Repeated-node derivative data belong to Hermite interpolation, a different problem.

Experiment

At x=3/2, evaluate the cardinal weights.

Cardinal term	Weight	Data-weight contribution
L₀	−1/8	−1/8
L₁	3/4	3/2
L₂	3/8	15/8

The weights sum to one and the contributions sum to 13/4, matching 1+(3/2)². Change only the middle data value by 0.1 and compute the resulting change at this evaluation point.

Understand

For distinct nodes x₀,…,xₙ, define Lⱼ(x)=∏ₖ≠ⱼ(x−xₖ)/(xⱼ−xₖ). The denominators are nonzero, and direct evaluation gives Lⱼ(xᵢ)=δᵢⱼ. Therefore p(x)=ΣyⱼLⱼ(x) interpolates the data and has degree at most n. Uniqueness follows because the difference of two such polynomials has n+1 distinct roots and degree at most n. This is an algebraic existence and uniqueness theorem; it says nothing by itself about numerical conditioning, approximation between nodes or noisy observations.

Apply

For a smooth underlying function f, let ω(t)=∏(t−xⱼ). At an evaluation point x different from all nodes, define h(t)=f(t)−p(t)−Cω(t), choosing C so h(x)=0. The function has n+2 distinct zeros. Repeated Rolle’s theorem gives h^(n+1)(ξ)=0 somewhere in their enclosing interval. Since p^(n+1)=0 and ω^(n+1)=(n+1)!, the remainder is f(x)−p(x)=f^(n+1)(ξ)ω(x)/(n+1)!. Require f∈C^(n+1) there; at a node the error is zero directly. A useful bound needs control of both the derivative and the node product.

The remainder formula also explains extrapolation risk. Outside the node interval, every factor in the node product can become large, and a derivative bound must hold on the larger interval reaching the evaluation point. A bound proved only between the first and last nodes does not justify extrapolation beyond them. As an exact check, interpolate a polynomial of degree at most n: its (n+1)st derivative vanishes, so the remainder is zero everywhere. For a polynomial of degree n+1, the remainder is its leading coefficient times the node product, giving a direct test of signs and factorials.

Continue

Exact interpolation is not automatically good approximation. Data perturbations change p by ΣδyⱼLⱼ, so the sum of absolute cardinal weights measures amplification. High degree, uneven nodes and extrapolation can all be dangerous. Stable evaluation formulas and good node choices address different parts of the problem. Next, divided differences build the same polynomial incrementally and expose its coefficients in a basis adapted to the nodes. Keep exact fitting, approximation error and arithmetic error separate in every report.

Review calculus and linear algebra when a prerequisite needs practice. Continue to Build interpolation one node at a time.