Errors of Interpolating Polynomials

Theorem

Suppose x0,x1,…,xnx_0, x_1, \ldots, x_n are distinct numbers in the interval [a,b][a, b] and f∈Cn+1[a,b]f \in C^{n+1}[a, b]. Then, for each xx in [a,b][a, b], a number ξ(x)\xi(x) (generally unknown) between x0,x1,…,xnx_0, x_1, \ldots, x_n, and hence in (a,b)(a, b), exists with:

f(x)=Pn(x)+f(n+1)(ξ(x))(x−x0)(x−x1)⋯(x−xn)(n+1)!=Pn(x)+Rn(x)f(x) = P_n(x) + \frac{f^{(n+1)}(\xi(x)) (x - x_0)(x - x_1) \cdots (x - x_n)}{(n + 1)!} = P_n(x) + R_n(x)

where Pn(x)P_n(x) is the interpolating polynomial. Rn(x)R_n(x) is the error term. The above equation can only be used for known and differentiable functions, which is usually not the case.

An alternative formula to find error is by applying a finite divided difference to approximate the (n+1)(n + 1)th derivative. Thus, we have:

Rn=f[x,xn,xn−1,…,x0](x−x0)(x−x1)⋯(x−xn)R_n = f[x, x_n, x_{n-1}, \ldots, x_0](x - x_0)(x - x_1) \cdots (x - x_n)

Since this contains the unknown f(x)f(x), it cannot be solved for the error. However, using an additional data point f(xn+1)f(x_{n+1}), the error can be estimated as follows.

Rn≈f[xn+1,xn,xn−1,…,x0](x−x0)(x−x1)⋯(x−xn)R_n \approx f[x_{n+1}, x_n, x_{n-1}, \ldots, x_0](x - x_0)(x - x_1) \cdots (x - x_n)

Wolfram Mathematica notebook

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Written by October 19, 2026 1 min read
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