GuideInterpolationintermediate

Lagrange interpolation

The Lagrange basis functions, the cardinal property that defines them, the polynomial as a direct combination of the data, its error and a worked example with the census data.

Cardinal basis functions

Lagrange builds the same unique polynomial in direct, symmetric form. The functions $L_i(x)$ equal 1 at their node and 0 at the others:

L_{i}(x_j)=\begin{cases}1,& i=j\\[2pt] 0,& i\ne j\end{cases}\qquad\Longrightarrow\qquad L_{i}(x)=\prod_{\substack{j=0\\ j\ne i}}^{n}\frac{x-x_j}{x_i-x_j}

The numerator has all factors $(x-x_j)$ except $(x-x_i)$ ; the denominator has all $(x_i-x_j)$ except $(x_i-x_i)$ . With these functions the polynomial is a combination of the known values:

p_n(x)=\sum_{i=0}^{n} L_{i}(x)\,f(x_i)

Why the polynomial passes through every node

To check this at a specific node $x_k$ , evaluate the polynomial there:
$p_n(x_k)=\sum_{i=0}^{n} L_i(x_k)\,f(x_i)$
The cardinal property cancels all basis functions except the one attached to the node $x_k$ :
$L_i(x_k)=\begin{cases}0,& i\ne k\\[2pt]1,& i=k\end{cases}$
Only the term $i=k$ remains, and the interpolant reproduces the datum at that node:
$p_n(x_k)=L_k(x_k)\,f(x_k)=f(x_k)$

Newton versus Lagrange

Lagrange is direct and symmetric: ideal for few nodes and for deriving rules (quadrature, differentiation).
Newton is incremental: adding a node costs one term, not a full rebuild.
Both give the same polynomial and share the same error bound.

Worked example

ExamplePopulation census (degree 4)

With the same census data (1971–2011), build the degree-4 Lagrange polynomial and estimate the 2005 population.

Each $L_i$ has the four factors of the other nodes. For example, for $x_0=1971$ :
$\begin{aligned}L_0(x)&=\frac{(x-1981)(x-1991)(x-2001)(x-2011)}{(1971-1981)(1971-1991)(1971-2001)(1971-2011)}\\&=\frac{(x-1981)(x-1991)(x-2001)(x-2011)}{240000}\end{aligned}$
The polynomial sums each $L_i$ times its value $f(x_i)$ :
$p_4(x)=\sum_{i=0}^{4} L_i(x)\,f(x_i)$

The result matches (up to rounding) the Newton one, as uniqueness requires:

p_4(2005)\approx 42.316\ \text{millones}

Derivation of the Lagrange basis

DerivationDerivation: Lagrange basis functionsView as its own page →

We want $L_i(x)$ to vanish at every node except $x_i$ . To vanish at $x_j$ ( $j\ne i$ ) it suffices to include the factor $(x-x_j)$ for each one:
$\text{numerador}=\prod_{\substack{j=0\\ j\ne i}}^{n}(x-x_j)$
That product has some value (not 1) at $x_i$ . To normalize it to 1 we divide by its value at $x_i$ , i.e. the same product evaluated there:
$L_i(x)=\prod_{\substack{j=0\\ j\ne i}}^{n}\frac{x-x_j}{x_i-x_j}$
Thus $L_i(x_i)=1$ and $L_i(x_j)=0$ . The combination $\sum_i L_i(x)\,f(x_i)$ reproduces each datum, hence interpolates.