First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 3. INTERPOLATION 96
Summary: The interpolating polynomial p 2 (x) for the data, (−1, −6), (1, 0), (2, 6), represented
in three different basis functions is:
Monomial: p 2 (x) =−4+3x + x 2
Lagrange: p 2 (x) =− (x − 1)(x − 2) + 2(x +1)(x − 1)
Newton: p 2 (x) =− 6+3(x +1)+(x +1)(x − 1)
Similar to the monomial form, a polynomial written in Newton’s form can be evaluated
using the Horner’s method which has O(n) complexity:
p n (x) =a 0 + a 1 (x − x 0 )+a 2 (x − x 0 )(x − x 1 )+... + a n (x − x 0 )(x − x 1 ) ···(x − x n−1 )
= a 0 +(x − x 0 )(a 1 +(x − x 1 )(a 2 + ... +(x − x n−2 )(a n−1 +(x − x n−1 )(a n )) ···))
Example 50. Write p 2 (x) =−6+3(x +1)+(x +1)(x − 1) using the nested form.
Solution. −6+3(x +1)+(x +1)(x − 1) = −6+(x +1)(2+x); note that the left-hand side
has 2 multiplications, and the right-hand side has 1.
Complexity of the three forms of polynomial interpolation: The number of multiplications
required in solving the corresponding matrix equation in each polynomial
basis is:
• Monomial → O(n 3 )
• Lagrange → trivial
• Newton → O(n 2 )
Evaluating the polynomials can be done efficiently using Horner’s method for monomial
and Newton forms. A modified version of Lagrange form can also be evaluated using
Horner’s method, but we do not discuss it here.
Exercise 3.1-2: Compute, by hand, the interpolating polynomial to the data (−1, 0),
(0.5, 1), (1, 0) using the monomial, Lagrange, and Newton basis functions. Verify the three
polynomials are identical.
It’s time to discuss some theoretical results for polynomial interpolation. Let’s start with
proving Theorem 47 which we stated earlier: