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Interpolation polynomial in Lagrange Form

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Values $ f_k$ at $ n+1$ distinct points $ x_0,\ldots,x_n$ can be interpolated with a unique polynomial $ p$ of degree $ \le n$:

$\displaystyle p(x_k) = f_k,\quad k=0,\ldots,n \,. $

The Lagrange form of $ p$ is

$\displaystyle p(x) = \sum_{k=0}^n f_k q_k(x),\quad q_k(x) = \prod_{j\ne k} \frac{x-x_j}{x_k-x_j} \,. $

\includegraphics[width=.45\linewidth]{interpolation_Bild}

The polynomials $ q_k$ are referred to as Lagrange polynomials. They are equal to $ 1$ at $ x_k$ and vanish at all other points $ x_j$:

$\displaystyle q_{k}(x_{j})=\delta_{k,j} $

with $ \delta$ the Kronecker symbol.
The Lagrange polynomials satisfy

$\displaystyle q_k(x_j) = \delta_{j,k} \,. $

This implies

$\displaystyle p(x_j) = \sum_k f_k \delta_{j,k} = f_j \,, $

i.e., $ p$ interpolates $ f$ at $ x_j$.

To show uniqueness, we assume that $ \tilde{p}$ is another interpolating polynomial and consider the difference

$\displaystyle p - \tilde p\,. $

This polynomial has at least $ n+1$ zeros:

$\displaystyle (p - \tilde p)(x_k) = 0,\quad k=0,\ldots,n \,. $

Since the degree of $ p-\tilde p$ is at most $ n$, it follows that the difference vanishes identically, i.e., $ p=\tilde p$.

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  automatisch erstellt am 14.  6. 2016