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Error in Polynomial Interpolation
Adrian Down
February 21, 2006
1 Error theorem for polynomial interpolation
Theorem. Let f (x) have n + 1 continuous derivatives in (a, b) and p(x) be the polynomial of degree ≤ n which interpolates f (x) at the n + 1 distinct points x 0 , x 1 ,... , xn in (a, b). Then
f (x) − p(x) =
(n + 1)!
f (n+1)(ζ)
∏^ n
i=
(x − xj ) (1)
where ζ ∈ (a, b).
Proof. The result is clearly true if x = xj. In this case, f (xj ) = p(xj ), since p(x) interpolates f (x) at each xj by assumption. On the right of (1), one term in the product is 0, and hence (1) reduces to 0 = 0. Now, consider x 6 = x 0 , x 1 ,... , xn. Define
w(t) =
∏^ n
i=
(t − xj )
Note. Since w(t) is comprised of n + 1 monomial factors, w(t) is of degree n + 1, and is of the form,
w(t) = tn+1^ +...
We also define
φ(t) = f (t) − p(t) − λw(t)
Note. f (t) − p(t) vanishes at each interpolated point xj. w(t) also vanishes at each interpolation point, as noted above, and hence φ(xj ) = 0.
Our goal to to arrange for φ(x) = 0 for some point x by the proper choice of λ. Evaluating this condition,
0 = φ(x) = f (x) − p(x) − λw(x)
⇒ λ =
f (x) − p(x) w(x)
With this choice of λ, φ(t) vanishes at n+2 distinct points, { x, x 0 , x 1 ,... , xn }. By the Mean Value Theorem, φ′(t) has at least n +1 zeros ∈ (a, b). Similarly, φ′′(t) has at least n zeros ∈ (a, b), and in general, φ(b)(t) has at least n + 2 − b zeros in (a, b). In particular, φ(n+1)(t) has at least 1 zero t = ζ(x) ∈ (a, b). We now evaluate this derivative explicitly.
0 = φ(n+1)(ζ) = f (n+1)(ζ) − p(n+1)(ζ) − λw(n+1)(ζ) (2)
p is an nth degree polynomial, so its n+1 derivative is 0. Recall that w(t) was a product of n + 1 monomial factors and was of the form w(t) = tn+1^ +.. .. Hence the n + 1 derivative is a constant independent of t,
w(n+1)(t) = (n + 1)!
Solving for λ in equation (2),
λ =
f (n+1)(ζ) (n + 1)!
Recall the definition of λ,
λ ≡
f (x) − p(x) w(x)
Equating these expressions for λ and inserting the definition of w(t),
f (x) − p(x) =
f (n+1)(ζ) (n + 1)!
∏^ n
i=
(x − xj )
2.1.2 Counterexample in the complex plane
Consider f (z) = (^1) z interpolated at the points
zk = e
2 πık n+1 (^0) ≤ k ≤ n
Note. • The points zk are called the nth roots of unity.
eıθ^ = cos θ + ı sin θ
This formula comes from complex analysis. It is important, and we will use it at subsequent points in this course. The interpolating polynomial is
p(ζ) = ζn
Evaluating the terms in the remainder theorem above,
p(ζ) −
ζ
eıθ
)n −
eıθ = eınθ^ − e−ıθ
= e−ı^
n− 21 ( eı^
n+1 2 θ − e−ı^
n+1 2 θ^ )
= e−ı^
n− 1 (^2) · 2 ı sin
n + 1 2
θ
The error is oscillatory, rather than convergent.
2.2 Hermite interpolation
2.2.1 Motivation
The goal of this interpolation method is to find a polynomial p(x) such that at each node xk with k = 0,... , n,
p(j)(xk) = f (j)(xk) (3)
where the number of derivatives j can be different for different interpolation points xk.
2.2.2 Connection with Taylor approximation
The simplest possible case is interpolation of a function about a single point x 0 with derivatives of order j = 0,... , l 0. Evaluating the definition (3),
p(x) = f (x 0 ) + f ′(x 0 )(x − x 0 ) +
f ′′(x 0 )(x − x 0 )^2 +... +
l 0!
f (l^0 )(x − x 0 )l^0
This is simply the l 0 order Taylor approximation of f (x) about the point x 0.
2.2.3 Standard practical case
In practice, we usually have the values of a function f and its first derivative only at n+1 points. Since there are two pieces of information associated with each point xk, there are 2n + 2 terms in the approximating polynomial. Since one of these terms is constant, the degree of the interpolating polynomial is ≤ 2 n + 1.
