First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 3. INTERPOLATION 100
With this new notation, Newton’s interpolating polynomial can be written as
n∑
p n (x) =f[x 0 ]+ f[x 0 ,x 1 , ..., x k ](x − x 0 ) ···(x − x k−1 )
k=1
= 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 n ](x − x 0 )(x − x 1 ) ···(x − x n−1 )
Here is the formal definition of divided differences:
Definition 54. Given data (x i ,f(x i )),i = 0, 1, ..., n, the divided differences are defined
recursively as
where k =0, 1, ..., n.
f[x 0 ]=f(x 0 )
f[x 0 ,x 1 , ..., x k ]= f[x 1, ..., x k ] − f[x 0 , ..., x k−1 ]
x k − x 0
Theorem 55. The ordering of the data in constructing divided differences is not important,
that is, the divided difference f[x 0 , ..., x k ] is invariant under all permutations of the arguments
x 0 , ..., x k .
Proof. Consider the data (x 0 ,y 0 ), (x 1 ,y 1 ), ..., (x k ,y k ) and let p k (x) be its interpolating polynomial:
p k (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 k ](x − x 0 ) ···(x − x k−1 ).
Now let’s consider a permutation of the x i ; let’s label them as ˜x 0 , ˜x 1 , ..., ˜x k . The interpolating
polynomial for the permuted data does not change, since the data x 0 ,x 1 , ..., x k (omitting the
y-coordinates) is the same as ˜x 0 , ˜x 1 , ..., ˜x k , just in different order. Therefore
p k (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 k ](x − ˜x 0 ) ···(x − ˜x k−1 ).
The coefficient of the polynomial p k (x) for the highest degree x k is f[x 0 , ..., x k ] in the first
equation, and f[˜x 0 , ..., ˜x k ] in the second. Therefore they must be equal to each other.
Example 56. Find the interpolating polynomial for the data (−1, −6), (1, 0), (2, 6) using
Newton’s form and divided differences.