First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 3. INTERPOLATION 113
We seek a polynomial that fits the y and y ′ values, that is, we seek a polynomial H(x) such
that H(x i )=y i and H ′ (x i )=y i, ′ i =0, 1, ..., n. This makes 2n +2equations, and if we let
H(x) =a 0 + a 1 x + ... + a 2n+1 x 2n+1 ,
then there are 2n +2 unknowns, a 0 , ..., a 2n+1 , to solve for. The following theorem shows
that there is a unique solution to this system of equations; a proof can be found in Burden,
Faires, Burden [4].
Theorem 61. If f ∈ C 1 [a, b] and x 0 , ..., x n ∈ [a, b] are distinct, then there is a unique
polynomial H 2n+1 (x), of degree at most 2n +1, agreeing with f and f ′ at x 0 , ..., x n . The
polynomial can be written as:
H 2n+1 (x) =
n∑
y i h i (x)+
i=0
n∑
y i ′ ˜h i (x)
i=0
where
h i (x) =(1− 2(x − x i )l ′ i(x i )) (l i (x)) 2
˜hi (x) =(x − x i )(l i (x)) 2 .
Here l i (x) is the ith Lagrange basis function for the nodes x 0 , ..., x n , and l i(x) ′ is its derivative.
H 2n+1 (x) is called the Hermite interpolating polynomial.
The only difference between Hermite interpolation and polynomial interpolation is that
in the former, we have the derivative information, which can go a long way in capturing the
shape of the underlying function.
Example 62. We want to interpolate the following data:
x-coordinates : −1.5, 1.6, 4.7
y-coordinates :0.071, −0.029, −0.012.
The underlying function the data comes from is cos x, but we pretend we do not know this.
Figure (3.2) plots the underlying function, the data, and the polynomial interpolant for the
data. Clearly, the polynomial interpolant does not come close to giving a good approximation
to the underlying function cos x.