Dog Breeds Information and More
  Komondor - Dog Breeds Facts and Information Dog Breeds Selector A to Z dog breeds Forums

 
Dog names
Dog training
Toy dogs
Intelligence
Dog health
Dog worship
Ticks

 
Golden Retriever
Labrador Retriever
Jack Russell
 
Find a Breed
 
Dog Breeds Encyclopedia
 

Newton polynomial

(Redirected from Newtons difference method)

In the mathematical subfield of numerical analysis, a Newton polynomial, named after its inventor Isaac Newton, is the interpolation polynomial for a given set of data points in the Newton form. The Newton polynomial is sometimes called Newton's divided differences interpolation polynomial because the coefficients of the polynomial are calculated using divided differences.

As there is only one interpolation polynomial for a given set of data points it is a bit misleading to call the polynomial Newton interpolation polynomial. The more precise name is interpolation polynomial in the Newton form.

Contents

Definition

Given a set of k+1 data points

(x_0, y_0),\ldots,(x_k, y_k)

where no two xj are the same, the interpolation polynomial in the Newton form is linear combination of Newton basis polynomials

N(x) := \sum_{j=0}^{k} a_{j} n_{j}(x)

with the Newton basis polynomials defined as

n_j(x) := \prod_{i=0}^{j-1} (x - x_i)

and the coefficients defined as

a_j := [y_0,\ldots,y_j]

where

[y_0,\ldots,y_j]

is the notation for divided differences.

Thus the Newton polynomial can be written as

N(x) := [y_0] + [y_0,y_1](x-x_0) + \ldots + [y_0,\ldots,y_k](x-x_0)\ldots(x-x_{k-1})

Main idea

Solving an interpolation problems leads to a problem in linear algebra where we have to solve a matrix. Using a standard monomial basis for our interpolation polynomial we get the very complicated Vandermonde matrix. By choosing another basis, the Newton basis, we get a much simpler lower triangular matrix which can solved faster.

For k+1 data points we construct the Newton basis as

n_j(x) := \prod_{i=0}^{j-1} (x - x_i) \qquad j=0,\ldots,k

Using the these polynomials as a basis for Πk we have to solve

\begin{bmatrix}       1 &         &        &        & 0  \\       1 & x_1-x_0 &        &        &    \\       1 & x_2-x_0 & (x_2-x_0)(x_2-x_1) &        &    \\  \vdots & \vdots  &        & \ddots &    \\       1 & x_k-x_0 & \ldots & \ldots & \prod_{j=0}^{k-1}(x_k - x_j) \end{bmatrix} \begin{bmatrix}      a_0 \\      \vdots \\      a_{k}  \end{bmatrix} = \begin{bmatrix}      y_0 \\      \vdots \\      y_{k} \end{bmatrix}

to solve the polynomial interpolation problem.

This matrix can be solved recursively by solving

\sum_{i=0}^{j} a_{i} n_{i}(x_j) = y_j \qquad j = 0,...,k

Application

As can be seen from the definition of the divided differences new data points can be added to the data set to create a new interpolation polynomial without recalculation the old coefficients. And when a data point changes we usually do not have to recalculate all coefficients. Furthermore if the xi are distributed equidistantly the calculation of the divided differences becomes significantly easier. Therefore the Newton form of the interpolation polynomial is usually preferred over the Lagrange form for practical purposes.

See also

The contents of this article are licensed from Wikipedia.org under the
GNU Free Documentation License. How to see transparent copy