example of under-determined polynomial interpolation

Consider the followinginterpolation problem:

Given $x_{\mathrm{1}}\mathrm{,}{y}_{\mathrm{1}}\mathrm{,}{x}_{\mathrm{2}}\mathrm{,}{y}_{\mathrm{2}}\mathrm{\in }\mathrm{R}$ with $x_{\mathrm{1}}\mathrm{\ne }{x}_{\mathrm{2}}$ to determineall cubic polynomials

$$p(x)=a{x}^3+b{x}^2+cx+d,x,a,b,c,d\in ℝ$$

such that

$$p(x_1)=y_1,p(x_2)=y_2.$$

This is a linear problem. Let ${𝒫}_3$ denote the vector space ofcubic polynomials. The underlying linear mapping is themulti-evaluation mapping

$$E:{𝒫}_3\to {ℝ}^2,$$

given by

$$p\mapsto \left(\begin{array}{c}\hfill p(x_1)\hfill \\ \hfill p(x_2)\hfill \end{array}\right),p\in {𝒫}_3$$

The interpolation problem in question is represented by the equation

$$E(p)=\left(\begin{array}{c}\hfill y_1\hfill \\ \hfill y_2\hfill \end{array}\right)$$

where $p\in {𝒫}_3$ is the unknown. One can recast the problem into thetraditional form by taking standard bases of ${𝒫}_3$ and ${ℝ}^2$ andthen seeking all possible $a,b,c,d\in ℝ$ such that

However, it is best to treat this problem at an abstract level,rather than mucking about with row reduction. The Lagrangeinterpolation formula gives us a particular solution, namely the linearpolynomial

$$p_0(x)=\frac{x-x_1}{x_2-x_1}{y}_1+\frac{x-x_2}{x_1-x_2}{y}_2,x\in ℝ$$

The general solution of our interpolation problem is therefore givenas$p_0+q$, where $q\in {𝒫}_3$ is a solution of the homogeneousproblem

$$E(q)=0.$$

A basis of solutions for the latter is, evidently,

$$q_1(x)=(x-x_1)(x-x_2),q_2(x)=x{q}_1(x),x\in ℝ$$

The general solution to our interpolation problem is therefore

$$p(x)=\frac{x-x_1}{x_2-x_1}{y}_1+\frac{x-x_2}{x_1-x_2}{y}_2+(ax+b)(x-x_1)(x-x_2),x\in ℝ,$$

with $a,b\in ℝ$ arbitrary. The general under-determinedinterpolation problem is treated in an entirely analogous manner.