UnderDeterminedPolynomialInterpolation 2002-04-16 22:04:49 2007-03-27 14:55:02 Example interpolation \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \newcommand{\cP}{\mathcal{P}} \newcommand{\reals}{\mathbb{R}} \newcommand{\natnums}{\mathbb{N}} \newcommand{\cnums}{\mathbb{C}} \newcommand{\znums}{\mathbb{Z}} \newcommand{\lp}{\left(} \newcommand{\rp}{\right)} \newcommand{\lb}{\left[} \newcommand{\rb}{\right]} \newcommand{\supth}{^{\text{th}}} \newtheorem{proposition}{Proposition} \newtheorem{definition}[proposition]{Definition} Consider the following interpolation problem: \begin{quote} \em Given $x_1,y_1,x_2,y_2\in \reals$ with $x_1\neq x_2$ to determine all cubic polynomials $$p(x)= ax^3+bx^2+cx+d,\quad x,a,b,c,d\in \reals$$ such that $$p(x_1) = y_1,\quad p(x_2) = y_2.$$ \end{quote} This is a linear problem. Let $\cP_3$ denote the vector space of cubic polynomials. The underlying linear mapping is the multi-evaluation mapping $$E:\cP_3\rightarrow\reals^2,$$ given by $$ p\mapsto \begin{pmatrix} p(x_1)\\ p(x_2) \end{pmatrix},\quad p\in \cP_3 $$ The interpolation problem in question is represented by the equation $$E(p) = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} $$ where $p\in \cP_3$ is the unknown. One can recast the problem into the traditional form by taking standard bases of $\cP_3$ and $\reals^2$ and then seeking all possible $a,b,c,d\in\reals$ such that $$ \begin{pmatrix} \lp x_1\rp^3 & \lp x_1\rp^2 & x_1 & 1 \\ \lp x_2\rp^3 & \lp x_2\rp^2 & x_2 & 1 \\ \end{pmatrix} \begin{pmatrix} a\\b\\c\\d \end{pmatrix} = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} $$ However, it is best to treat this problem at an abstract level, rather than mucking about with row reduction. The Lagrange interpolation formula gives us a particular solution, namely the linear polynomial $$p_0(x) = \frac{x-x_1}{x_2-x_1}y_1 + \frac{x-x_2}{x_1-x_2} y_2,\quad x\in \reals$$ The general solution of our interpolation problem is therefore given as $p_0 + q$, where $q\in \cP_3$ is a solution of the homogeneous problem $$E(q)=0.$$ A basis of solutions for the latter is, evidently, $$q_1(x) = (x-x_1)(x-x_2),\quad q_2(x) = x q_1(x),\qquad x\in \reals$$ The general solution to our interpolation problem is therefore $$p(x) = \frac{x-x_1}{x_2-x_1}y_1 + \frac{x-x_2}{x_1-x_2} y_2 + (ax+b)(x-x_1)(x-x_2),\quad x\in\reals,$$ with $a,b\in \reals$ arbitrary. The general under-determined interpolation problem is treated in an entirely analogous manner.