Question

Let
x0x1xn
be n + 1 distinct points on the real line and let f(x) be a real-valued function defined on some interval I = [a, b] containing these points. Then, there exists exactly one polynomial Pn(x) of degree n, which ..... f(x) at
x0x1…xn

Answer 1

Answer on Question #61799 Programming & Computer Science

Let $x_0, x_1, \ldots, x_n$ be $n + 1$ distinct points on the real line and let $f(x)$ be a real-valued function defined on some interval $I = [a, b]$ containing these points. Then, there exists exactly one polynomial $P_n(x)$ of degree $n$ , which matches $f(x)$ at $x_0, x_1, \ldots, x_n$ . This one polynomial named an interpolation polynomial.

Solution

Let the polynomial has the form $P_{n}(x) = \sum_{k=0}^{n} a_{k} x^{k}$ and satisfying conditions of interpolation $\forall i = 0..n$ : $P_{n}(x_{i}) = f(x_{i}) = f_{i}$ . Then we have a system of $n$ linear equations with $n$ unknown coefficients $\{a_{k}\}$ :

[
    [1, x_0, x_0**2, ..., x_0**n],
    [1, x_1, x_1**2, ..., x_1**n],
    ...
    [1, x_n, x_n**2, ..., x_n**n]
]
* [a_0, a_1, ..., a_n] = [f_0, f_1, ..., f_n]

This system has exactly one solution $\{a_k\}$ , because its matrix is nonsingular Vandermonde matrix and therefore there exists exactly one polynomial $P_n(x)$ of degree $n$ , which matches $f(x)$ at $x_0, x_1, \ldots, x_n$ .

http://www.AssignmentExpert.com