LECTURE 3: Polynomial interpolation and numerical differentiation

In general, an analysis of errors goes as follows: Suppose that p n is the polynomial of
least degree that interpolates f at the nodes x 0 ,x 1 ,...,x n . Then according to first theorem
on interpolating errors (Sec.3.2):
f(x)− p n (x)=
1
(n+1)! f(n+1) (ξ)w(x)
with ξ dependent on x and w(x)=(x−x 0 )...(x−x n ). Differentiating gives
f ′ (x)− p ′ n(x)=
1
(n+1)! w(x) d 1
dx f(n+1) (ξ)+
(n+1)! f(n+1) (ξ)w ′ (x)
The first observation is that w(x) vanishes at each node. The evaluation is simpler if it is
done at a node x i :
f ′ (x i )= p ′ 1
n(x i )+
(n+1)! f(n+1) (ξ)w ′ (x i )
The second observation is that it becomes simpler if x is chosen such that w ′ (x)=0. Then
f ′ (x i )= p ′ n (x i)+
1
(n+1)! w(x) d dx f(n+1) (ξ)
Example.
Derive a first-derivative formula for f ′ (x) using an interpolating polynomial p 3 (x) obtained
with four nodes x 0 , x 1 , x 2 and x 3 . Use central difference in choosing the nodes.
Solution. The interpolating polynomial written in the Newton form is
Its derivative is
p 3 (x)= f(x 0 )+ f[x 0 ,x 1 ](x−x 0 )+ f[x 0 ,x 1 ,x 2 ](x−x 0 )(x−x 1 )
+ f[x 0 ,x 1 ,x 2 ,x 3 ](x−x 0 )(x−x 1 )(x−x 2 )
p ′ 3 (x)= f[x 0,x 1 ]+ f[x 0 ,x 1 ,x 2 ](2x−x 0 − x 1 )
+ f[x 0 ,x 1 ,x 2 ,x 3 ]{(x−x 1 )(x−x 2 )+(x−x 0 )(x−x 2 )
+(x−x 0 )(x−x 1 )}
Formulas for the divided differences can be obtained
recursively:
f[x i ]= f(x i )
f[x i ,x i+1 ]= f[x i+1]− f[x i ]
x i+1 − x i
f[x i ,x i+1 ,x i+2 ] = f[x i+1,x i+2 ]− f[x i ,x i+1 ]
x i+2 − x i
We now choose the nodes using central difference:
x 0 = x−h, x 1 = x+h, x 2 = x−2h, x 3 = x+2h.