A First Course in Linear Algebra, 2017a

3.2. Applications of the Determinant 139
Procedure 3.48: Finding an Interpolating Polynomial
Suppose that values of x and corresponding values of y are given, such that the actual relationship
between x and y is unknown. Then, values of y can be estimated using an interpolating polynomial
p(x). Ifgivenx 1 ,...,x n and the corresponding y 1 ,...,y n , the procedure to find p(x) is as follows:
1. The desired polynomial p(x) is given by
2. p(x i )=y i for all i = 1,2,...,n so that
p(x)=r 0 + r 1 x + r 2 x 2 + ... + r n−1 x n−1
r 0 + r 1 x 1 + r 2 x 2 1 + ... + r n−1x1 n−1 = y 1
r 0 + r 1 x 2 + r 2 x 2 2 + ... + r n−1x2 n−1 = y 2
.
.
r 0 + r 1 x n + r 2 x 2 n + ... + r n−1xn
n−1 = y n
3. Set up the augmented matrix of this system of equations
⎡
1 x 1 x 2 1
··· x n−1 ⎤
1
y 1
1 x 2 x 2 2
··· x2 n−1 y 2
⎢
⎥
⎣
...
...
...
...
... ⎦
1 x n x 2 n ··· xn
n−1 y n
4. Solving this system will result in a unique solution r 0 ,r 1 ,···,r n−1 . Use these values to construct
p(x), and estimate the value of p(a) for any x = a.
This procedure motivates the following theorem.
Theorem 3.49: Polynomial Interpolation
Given n data points (x 1 ,y 1 ),(x 2 ,y 2 ),···,(x n ,y n ) with the x i distinct, there is a unique polynomial
p(x)=r 0 +r 1 x +r 2 x 2 +···+r n−1 x n−1 such that p(x i )=y i for i = 1,2,···,n. The resulting polynomial
p(x) is called the interpolating polynomial for the data points.
We conclude this section with another example.
Example 3.50: Polynomial Interpolation
Consider the data points (0,1),(1,2),(3,22),(5,66). Find an interpolating polynomial p(x) of degree
at most three, and estimate the value of p(2).
Solution. The desired polynomial p(x) is given by:
p(x)=r 0 + r 1 x + r 2 x 2 + r 3 x 3