# Equivalence of Formulations of Lagrange Interpolation Formula

## Theorem

The following statements are equivalent:

### 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 $\tuple {x_0, \ldots, x_n}$ 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 $x_i$.