Lagrange Interpolating Polynomials

The $k$ -th Lagrange basis polynomial for $n+1$ nodes is:

L_{n,k}(x) = \prod_{\substack{i=0 \\ i \neq k}}^{n} \frac{x - x_i}{x_k - x_i}

Here:

Each $L_{n,k}$ satisfies $L_{n,k}(x_k) = 1$ and $L_{n,k}(x_i) = 0$ for $i \neq k$ .

The $n$ -th Lagrange interpolating polynomial through the data $(x_k, f(x_k))$ is:

P_n(x) = \sum_{k=0}^{n} L_{n,k}(x)\, f(x_k)

Example

Interpolate through $(1, 1)$ , $(2, 4)$ , $(3, 9)$ (i.e., $f(x) = x^2$ ).

Solution Reveal Hide

Basis polynomials for $n = 2$ :

L_{2,0}(x) = \frac{(x-2)(x-3)}{(1-2)(1-3)} = \frac{(x-2)(x-3)}{2}

L_{2,1}(x) = \frac{(x-1)(x-3)}{(2-1)(2-3)} = -(x-1)(x-3)

L_{2,2}(x) = \frac{(x-1)(x-2)}{(3-1)(3-2)} = \frac{(x-1)(x-2)}{2}

P_2(x) = 1 \cdot \frac{(x-2)(x-3)}{2} - 4(x-1)(x-3) + 9 \cdot \frac{(x-1)(x-2)}{2}

At $x = 4$ : $L_{2,0}(4) = 1$ , $L_{2,1}(4) = -3$ , $L_{2,2}(4) = 3$ , so:

$P_2(4) = 1(1) + 4(-3) + 9(3) = 1 - 12 + 27 = 16 = 4^2 \checkmark$

Wolfram Mathematica notebook

The summary notebook defines every function used in this module, with documentation and no examples, as a starting point for your own problems.

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November 3, 2026 1 min read

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