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Interpolation Theory

Published: July 09, 2019

Let $x_0, x_1, … , x_n$ be distinct real or complex numbers, and let $y_0 , y_1, …, y_n$ be associated function values. We now study the problem of finding a polynomial p (x) that interpolates the given data:

By writing: $p(x) = a_0 + a_1x + … + a_mx^m$

Consider that m = n, then

It can be written that $Xa = y$, with

The matrix X is called a Vandermonde matrix.

Theorem

Given n + 1 distinct points $x_0, … , x_n$ and n + 1 ordinates $y_0, …, y_n$, there is a polynomial p(x) of $degree \le n$ that interpolates $y_i$ at $x_i$, i = 0, 1, … , n. This polynomial p(x) is unique among the set of all polynomials of degree at most n.

Below are three kinds of proofs.

Proof

• It can be shown that for the matrix X,

  $$det(X) = \prod_{0 \le j < i \le n}(x_i - x_j)$$

  This shows that $det(X) \ne 0$, since the points $x_i$ are distinct. Thus X is nonsingular and the system $Xa = y$ has a unique solution a. This proves the existence and uniqueness of an interpolating polynomial of $degree \le n$.

• The system $Xa = y$ has a unique solution iff the homogeneous system $Xb = 0$ has only the trivial solution b = 0. Therefore, assume $Xb = 0$ for some b. Using b, define

  $$p(x) = b_0 + b_1x + ... + b_nx^n$$

  From the system $Xb = 0$, we have

  $$p(x_i) = 0, \ \ \ \ i = 0, 1, ..., n$$

  The polynomial p(x) has n + 1 zeros and degree $p(x) \le n$. This is not possible unless $p(x) \equiv 0$. But then all coefficients $b = 0, i = 0, 1, … , n$, completing the proof.

• Consider a special polynomial

  $$l_i(x) = \prod_{j \ne i} (\frac{x - x_j}{x_i - x_j}), \ \ i = 0, 1, ..., n$$

  Then the polynomial can be rewritten

  $$p(x) = y_0l_0(x) + y_1l_1(x) + ... + y_nl_n(x)$$

  Since all $l_i(x)$ have degree n, degree $p(x) \le n$.

  To prove uniqueness, suppose q(x) is another polynomial of degree $\le$ n that satisfies (1). Define

  $$r(x) = p(x) - q(x)$$

  Then degree $r(x) \le n$, and

  $$r(x_i) = p(x_i) - q(x_i) = y_i - y_i = 0 \ \ i = 0, 1, ..., n$$

  Since r(x) has n + 1 zeros, we must have $r(x) \equiv 0$, which proves $p(x) \equiv q(x)$, completing the proof.

Reference

• Godfrey Harold Hardy, Edward Maitland Wright，E. W. (Edward Maitland) Wright，An Introduction to Numerical Analysis, 131-134. Clarendon Press, 1979.