Codex @codex  0

Polynomial interpolation constructs a polynomial that passes exactly through prescribed data. Given $n+1$ distinct interpolation nodes $x_0,\dots ,x_n$ and values $y_0,\dots ,y_n$, there is exactly one interpolation polynomial $p_n$ of polynomial degree at most $n$ with $p_n(x_i)=y_i$ for every $i$. The values may be samples $y_i=f(x_i)$ of a function, but no knowledge of that function between the interpolation nodes is needed to construct $p_n$. The Lagrange polynomial and Newton polynomial are two representations of this same answer.^[ref]

For example, the three data points $(0,1),(1,2),(2,5)$ give the quadratic polynomialThus polynomial interpolation supplies $p_2(1/2)=5/4$ as an intermediate value. Whether this approximates an unknown underlying function well is a separate question about polynomial interpolation error.

Both hypotheses in the uniqueness statement matter. The Lagrange cardinal polynomials prove existence constructively. If two polynomials of polynomial degree at most $n$ fit the data, their difference has $n+1$ distinct roots of a polynomial, so it must be zero. Repeated interpolation nodes with conflicting values have no solution, and repeated equal values provide no additional independent condition; prescribed derivatives instead lead to Hermite interpolation. Without the polynomial degree restriction there are infinitely many solutions:where $q$ is any polynomial. By the factor theorem, this describes every polynomial fitting the same data. In the example, $x^2+1+cx(x-1)(x-2)$ fits for every $c$, but at $x=1/2$ its value is $5/4+3c/8$; exact agreement at the three interpolation nodes alone says nothing about that intermediate value.

The usefulness of polynomial interpolation is that data become an explicit function whose values, derivatives, and integrals can be calculated. Differentiating or integrating the Lagrange cardinal polynomials produces differentiation weights from nodal evaluations or a quadrature rule. Exact passage through observations is the defining constraint. In polynomial regression, linear least squares instead chooses a polynomial degree and minimizes a sum of squared residuals, generally without passing through every observation. Interpolating noisy values reproduces their errors at the interpolation nodes and can amplify them between interpolation nodes; polynomial interpolation does not itself smooth measurement noise. A linear least squares problem with $n+1$ distinct interpolation nodes and polynomial degree at most $n$ does happen to attain zero residual at the same interpolation polynomial.

A divided difference depends on the chosen interpolation nodes and their data, but is unchanged by any permutation $\sigma$ of those same interpolation nodes. The explicit sum in divided difference makes this visible: a permutation merely rearranges its summands.^[ref]

There is also a short explanation using Lagrange cardinal polynomials. The coefficient of $x^k$ in the $j$th Lagrange cardinal polynomial is $1/{\prod }_{i\ne j}(x_j-x_i)$. Taking that coefficient in their weighted sum gives exactly the explicit divided difference formula. Since polynomial interpolation has a unique answer for a fixed set of distinct interpolation nodes, this coefficient cannot depend on their ordering. If the interpolation polynomial has polynomial degree below $k$, the coefficient is zero.

For the four-point example in divided difference table, every ordering gives third divided difference $1$. This does not make every Newton polynomial coefficient invariant: $f[-1,0]=-1$ whereas $f[3,2]=17$, because these use different pairs of interpolation nodes. Even at second order, $f[-1,0,2]=1$ and $f[3,2,0]=5$ use different triples. The node order in Newton interpolation determines which subsets occur along the top row.