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When a polynomial is required to pass exactly through a set of data points $(x_0,y_0), (x_1,y_1),\dots,(x_n,y_n)$ the problem is classical interpolation. From the perspective of numerical and statistical computing we care not only that a polynomial exists and is unique (it is — there is exactly one polynomial of degree $\le n$ that interpolates $n+1$ distinct abscissas), but also how to construct it efficiently and stably, what its numerical behaviour will be, and how useful it is for modelling noisy data. Existence and uniqueness. If the $x_i$ are pairwise distinct, there exists a unique polynomial $p$ of degree at most $n$ with $p(x_i)=y_i$ for $i=0,\dots,n$. Proofs are constructive (e.g. Lagrange form) or linear-algebraic: set up the Vandermonde system $V c = y$ where $V_{ij} = x_i^{j}$ and $c$ are polynomial coefficients; $V$ is nonsingular for distinct $x_i$. Direct representations and algorithms 1. Lagrange form. The interpolant can be written as $$ p(x)=\sum_{j=0}^{n} y_j \ell_j(x),\qquad \ell_j(x)=\prod_{\substack{0\le m\le n\\ m\ne j}} \frac{x-x_m}{x_j-x_m}. $$ This formula is explicit and conceptually simple. However computing $\ell_j(x)$ directly for many $x$ is costly and can be numerically unstable for large $n$. 2. Newton divided differences. The Newton form is $$ p(x)=a_0 + a_1(x-x_0)+a_2(x-x_0)(x-x_1)+\cdots+a_n\prod_{k=0}^{n-1}(x-x_k), $$ with coefficients $a_k$ computed by the divided-difference table. Advantages: incremental (if a new data point arrives you can extend with $O(n)$ extra work), evaluation via nested multiplication (Horner-like) is $O(n)$ per $x$, and computing divided differences costs $O(n^2)$. 3. Vandermonde linear solve. Solve $V c = y$ for coefficient vector $c$. Straightforward if you want the power-basis coefficients, _______ ____ _______ ________ ________ __________ _______.
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