First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 3. INTERPOLATION 97
Theorem. 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.
Proof. We have already established the existence of p n without mentioning it! The Lagrange
form of the interpolating polynomial constructs p n directly:
p n (x) =
n∑
y k l k (x) =
k=0
n∑
k=0
y k
n
∏
j=0,j≠k
x − x j
x k − x j
.
Let’s prove uniqueness. Assume p n ,q n are two distinct polynomials satisfying the conclusion.
Then p n −q n is a polynomial of degree at most n such that (p n −q n )(x i )=0for i =0, 1, ..., n.
This means the non-zero polynomial (p n − q n ) of degree at most n, has (n +1) distinct roots,
which is a contradiction.
The following theorem, which we state without proof, establishes the error of polynomial
interpolation. Notice the similarities between this and Taylor’s Theorem 7.
Theorem 51. Let x 0 ,x 1 , ..., x n be distinct numbers in the interval [a, b] and f ∈ C n+1 [a, b].
Then for each x ∈ [a, b], there is a number ξ between x 0 ,x 1 , ..., x n such that
f(x) − p n (x) = f (n+1) (ξ)
(n +1)! (x − x 0)(x − x 1 ) ···(x − x n ).
The following lemma is useful in finding upper bounds for |f(x) − p n (x)| using Theorem
51, when the nodes x 0 , ..., x n are equally spaced.
Lemma 52. Consider the partition of [a, b] as x 0 = a, x 1 = a + h, ..., x n = a + nh = b. More
succinctly, x i = a + ih for i =0, 1, ..., n and h = b−a . Then for any x ∈ [a, b]
n
n∏
|x − x i |≤ 1 4 hn+1 n!
i=0
Proof. Since x ∈ [a, b], it falls into one of the subintervals: let x ∈ [x j ,x j+1 ]. Consider the
product |x − x j ||x − x j+1 |. Put s = |x − x j | and t = |x − x j+1 |. The maximum of st given
s+t = h, using Calculus, can be found to be h 2 /4, which is attained when x is the midpoint,