Lagrange Interpolation Formula

From ProofWiki
Jump to navigation Jump to search


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 $x_i$.


Recall the definition:

$\ds \map {L_j} X = \prod_{\substack {0 \mathop \le i \mathop \le n \\ i \mathop \ne j}} \frac {X - x_i} {x_j - x_i} \in \R \sqbrk X$

From this we see that:

$\map {L_j} {x_i} = \delta_{i j}$


$\ds \map P{x_i} = \sum_{j \mathop = 0}^n a_i \delta_{i j} = a_i$

Moreover, by Degree of Product of Polynomials over Integral Domain and Degree of Sum of Polynomials, the degree of $P$ as defined above is at most $n$.

It remains to show that the choice of $P$ is unique.

Aiming for a contradiction, suppose $\tilde P$ is another polynomial with the required properties.

Let $\Delta = P - \tilde P$.

By Degree of Sum of Polynomials, the degree of $\Delta$ is at most $n$.

Now we see that for $i = 0, \ldots, n$:

$\map \Delta {x_i} = \map P {x_i} - \map {\tilde P} {x_i} = a_i - a_i = 0$

Since by hypothesis the $x_i$ are distinct, $\Delta$ has $n + 1$ distinct zeros in $\R$.

But by the corollary to the Polynomial Factor Theorem this shows that:

$\ds \map \Delta X = \alpha \prod_{i \mathop = 0}^n \paren {X - x_i}$

If $\alpha \ne 0$, then this shows that the degree of $\Delta$ is $n+1$, a contradiction.


$\Delta = 0$

and so:

$P = \tilde P$

This establishes uniqueness.


Source of Name

This entry was named for Joseph Louis Lagrange.