Lagrange Polynomial Approximation

From ProofWiki
Jump to navigation Jump to search


Let $f: D \to \R$ be $n + 1$ times differentiable in an interval $I \subseteq \R$.

Let $x_0, \dotsc, x_n \in I$ be pairwise distinct points.

Let $P$ be the Lagrange Interpolation Formula of degree at most $n$ such that:

$\forall i \in \set {0, \dotsc, n}: \map P {x_i} = \map f {x_i}$

Let $\map R x = \map f x - \map P x$.

Then, for every $x \in I$, there exists $\xi$ in the interval spanned by $x$ and $x_i$, $i = 0, \dotsc, n$, such that:

$\map R x = \dfrac {\paren {x - x_0} \paren {x - x_1} \dotsm \paren {x - x_n} \map {f^{\paren {n + 1} } } \xi} {\paren {n + 1}!}$


This proof is similar to the proof of Taylor's theorem with the remainder in the Lagrange form, and is also based on Rolle's Theorem.

Observe that:

$\map R {x_i} = 0$ for $i = 0, \dotsc, n$

and that:

$R^{\paren {n + 1} } = f^{\paren {n + 1} }$

Without loss of generality, assume that $x$ is different from all $x_i$ for $i = 0, \dotsc, n$.

Let the function $g$ be defined by:

$\map g t = \map R t - \dfrac {\paren {t - x_0} \paren {t - x_1} \dotsm \paren {t - x_n} \map R x} {\paren {x - x_0} \paren {x - x_1} \dotsm \paren {x - x_n} }$

Then $\map g {x_i} = 0$ for $i = 0, \dotsc, n$, and $\map g x = 0$.

Denote by $J$ the interval spanned by $x$ and $x_i$, $i = 0, \dotsc, n$.

Thus $g$ has at least $n + 2$ zeros in $J$.

The Extended Rolle's Theorem is applied in $J$ successively to $g$, $g'$, $g$ and so on until $g^{\paren {n + 1} }$, which thus has at least one zero $\xi$ located between the two known zeros of $g^{\paren n}$ in $J$.


$0 = \map {g^{\paren {n + 1} } } \xi = \map {f^{\paren {n + 1} } } \xi - \dfrac {\paren {n + 1}! \map R x} {\paren {x - x_0} \paren {x - x_1} \dotsm \paren {x - x_n} }$

and the formula for $\map R x$ follows.



This theorem gives an estimate for the error of the Lagrange polynomial approximation and is similar to the Mean Value Theorem and Taylor's Theorem with the remainder in Lagrange form.