Newton Polynomial Interpolation: Basis, Differences, and Updates

All flashcards

What is a Newton polynomial?

An interpolating polynomial expressed as a sum of Newton basis terms with divided-difference coefficients.

How is the Newton basis term $n_j(x)$ defined?

For $j>0$, $n_j(x)=\prod_{i=0}^{j-1}(x-x_i)$; $n_0(x)=1$.

How are higher-order divided differences calculated?

Subtract neighboring lower-order divided differences and divide by the distance between the outer nodes.

How can a Newton interpolant be extended to a new node?

Add a multiple of $\prod_{i=0}^{n}(x-x_i)$, which vanishes at all existing nodes.

What advantage does the Newton basis have in the interpolation system?

It produces a lower-triangular system whose coefficients can be found successively.

Quiz questions

1. For distinct nodes, what is the recursive formula for $[y_i,\ldots,y_{i+j}]$ when $j>0$?Answer: $\frac{[y_{i+1},\ldots,y_{i+j}]-[y_i,\ldots,y_{i+j-1}]}{x_{i+j}-x_i}$
2. For equally spaced nodes with $x=x_0+sh$, which expression gives the Newton forward formula?Answer: $N(x_0+sh)=\sum_{i=0}^{k}\binom{s}{i}\Delta^{(i)}y_0$
3. Where is the Newton backward formula anchored for an ordered, equally spaced set of nodes?Answer: At the final node, $x_k$.
4. If $p_n(x)$ interpolates nodes $x_0,\ldots,x_n$, what coefficient extends it to a new node $x_{n+1}$ using $w_n(x)=\prod_{i=0}^{n}(x-x_i)$?Answer: $\frac{f(x_{n+1})-p_n(x_{n+1})}{w_n(x_{n+1})}$
5. When is Newton form especially convenient compared with a fixed Lagrange representation?Answer: When data nodes are added progressively and the interpolant should be extended.