Lagrange Interpolation Formula

Theorem

Formulation 1

Let $\tuple {x_0, \ldots, x_n}$ and $\tuple {a_0, \ldots, a_n}$ be ordered tuples of real numbers such that $x_i \ne x_j$ for $i \ne j$.

Then there exists a unique polynomial $P \in \R \sqbrk X$ of degree at most $n$ such that:

$\map P {x_i} = a_i$ for all $i \in \set {0, 1, \ldots, n}$

Moreover $P$ is given by the formula:

$\ds \map P X = \sum_{j \mathop = 0}^n a_i \map {L_j} X$

where $\map {L_j} X$ is the $j$th Lagrange basis polynomial associated to the $x_i$.

Formulation 2

Let $f : \R \to \R$ be a real function.

Let $f$ have known values $y_i = \map f {x_i}$ for $n \in \set {0, 1, \ldots, n}$.

Let a value $y' = \map f {x'}$ be required to be estimated at some $x'$.

Then:

 $\ds y'$ $\approx$ $\ds \dfrac {y_1 \paren {x' - x_2} \paren {x' - x_3} \cdots \paren {x' - x_n} } {\paren {x_1 - x_2} \paren {x_1 - x_3} \cdots \paren {x_1 - x_n} }$ $\ds$  $\, \ds + \,$ $\ds \dfrac {y_2 \paren {x' - x_1} \paren {x' - x_3} \cdots \paren {x' - x_n} } {\paren {x_2 - x_1} \paren {x_2 - x_3} \cdots \paren {x_2 - x_n} }$ $\ds$  $\, \ds + \,$ $\ds \cdots$ $\ds$  $\, \ds + \,$ $\ds \dfrac {y_n \paren {x' - x_1} \paren {x' - x_2} \cdots \paren {x' - x_{n - 1} } } {\paren {x_n - x_1} \paren {x_n - x_3} \cdots \paren {x_n - x_{n - 1} } }$

Also known as

The Lagrange interpolation formula can also be styled as Lagrange's interpolation formula.

Source of Name

This entry was named for Joseph Louis Lagrange.