Existence and Uniqueness of the Interpolating Polynomial

Summary

Through $n + 1$ points with distinct abscissae there exists exactly one polynomial of degree at most $n$ that interpolates them.

Prerequisites

Polynomials and Rational Functions

Statement

Let $(x_{0}, y_{0}), \dots, (x_{n}, y_{n})$ with $x_{i} \neq x_{j}$ for $i \neq j$ . There exists a unique $P \in R [x]$ with $\deg P \leq n$ such that $P (x_{i}) = y_{i}$ for all $i$ .

Proof Sketch

Uniqueness. If $P$ and $Q$ both interpolate, then $R = P - Q$ has degree $\leq n$ and $n + 1$ roots, so $R \equiv 0$ .

Existence. The Lagrange formula

P (x) = \sum_{i = 0}^{n} y_{i} \prod_{j \neq i} \frac{x - x_{j}}{x_{i} - x_{j}}

is a polynomial of degree $\leq n$ that hits every node.

Connections

Constructions: Lagrange Polynomial, Newton Polynomial

References

This theorem is foundational numerical analysis.^[1]

NIST DLMF, §3.3 Interpolation, https://dlmf.nist.gov/3.3 ↩︎