First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 3. INTERPOLATION 90
Here is an important question. How do we know that p n , a polynomial of degree at most
n passing through the data points, actually exists? Or, equivalently, how do we know the
system of equations p n (x i )=y i ,i=0, 1, ..., n, has a solution?
The answer is given by the following theorem, which we will prove later in this section.
Theorem 47. If points x 0 ,x 1 , ..., x n are distinct, then for real values y 0 ,y 1 , ..., y n , there is a
unique polynomial p n of degree at most n such that p n (x i )=y i ,i=0, 1, ..., n.
We mentioned three families of basis functions for polynomials. The choice of a family
of basis functions affects:
• The accuracy of the numerical methods to solve the system of linear equations Aa = y.
• The ease at which the resulting polynomial can be evaluated, differentiated, integrated,
etc.
Monomial form of polynomial interpolation
Given data (x i ,y i ),i =0, 1, ..., n, we know from the previous theorem that there exists a
polynomial p n (x) of degree at most n, that passes through the data points. To represent
p n (x), we will use the monomial basis functions, 1,x,x 2 , ..., x n , or written more succinctly,
φ k (x) =x k ,k =0, 1, ..., n.
The interpolating polynomial p n (x) can be written as a linear combination of these basis
functions as
p n (x) =a 0 + a 1 x + a 2 x 2 + ... + a n x n .
We will determine a i using the fact that p n is an interpolant for the data:
p n (x i )=a 0 + a 1 x i + a 2 x 2 i + ... + a n x n i
= y i
for i =0, 1, ..., n. Or, in matrix form, we want to solve
⎡
⎤ ⎡ ⎤ ⎡ ⎤
1 x 0 x 2 0 ... x n 0 a 0 y 0
1 x 1 x 2 1 x n 1
a 1
y ⎢
⎥ ⎢ ⎥ =
1
⎢ ⎥
⎣.
⎦ ⎣ . ⎦ ⎣ . ⎦
1 x n x 2 n x n n a n y n
} {{ } } {{ } } {{ }
A
a y