LECTURE 3: Polynomial interpolation and numerical differentiation

2.10 Neville’s algorithm
Another widely used method for obtaining the interpolating polynomial for a given table
of values is given by Neville’s algorithm. It builds up the polynomial in steps, just as the
Newton algorithm does.
Let P a,b,...,s be the polynomial interpolating the given data at a sequence of nodes x a ,x b ,...,x s .
We start with constant polynomials P i (x)= f(x i ). Selecting two nodes x i and x j with i> j,
we define recursively
( ) ( )
x−x j
xi − x
P u,...,v (x)= P u,..., j−1, j+1,...,v (x)+ P u,...,i−1,i+1,...,v (x)
x i − x j x i − x j
For example, the first three constant polynomials are
P 0 (x)= f(x 0 ) P 1 (x)= f(x 1 ) P 2 (x)= f(x 2 )
Using P 0 and P 1 , we can construct P 0,1 :
( ) ( )
x−x0 x1 − x
P 0,1 (x)= P 1 + P 0
x 1 − x 0 x 1 − x 0
and similarly
P 1,2 (x)=
( ) ( )
x−x1 x2 − x
P 2 + P 1
x 2 − x 1 x 2 − x 1
Using these two polynomials of degree one, we can construct another polynomial of degree
two: ( ) ( )
x−x0 x2 − x
P 0,1,2 (x)= P 1,2 + P 0,1
x 2 − x 0 x 2 − x 0
Similarly,
P 0,1,2,3 (x)=
( ) ( )
x−x0
x3 − x
P 1,2,3 + P 0,1,2
x 3 − x 0 x 3 − x 0
Using this formula repeatedly, we can create an array of polynomials:
x 0
x 1
x 2
x 3
x 4
P 0 (x)
P 1 (x) P 0,1 (x)
P 2 (x) P 1,2 (x) P 0,1,2 (x)
P 3 (x) P 2,3 (x) P 1,2,3 (x) P 0,1,2,3 (x)
P 4 (x) P 3,4 (x) P 2,3,4 (x) P 1,2,3,4 (x) P 0,1,2,3,4 (x)
Here each successive polynomial can be determined from two adjacent polynomials in
the previous column.