It all starts with a simple question: given $n + 1$ points $\left\{p_0, p_1, \ldots, p_j, \ldots, p_n\right\}$ where $p_j = (x_j, y_j)$, how can we find the **unique** polnoymium of degree $n$ that traverses such points?

Let’s begin easy: “polnoymium of degree $n$” means that we are looking for a function $p(x)$ of the type:

\[p(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n\]In particular, the objective is to find the values of the coefficients $\left\{a_0, a_1, \ldots, a_n\right\}$ for which $p(x)$ traverses the $n +1$ points that were given.

Hmm…we know that the $n + 1$ points are traversed by $p(x)$, hence we can turn this information into $n + 1$ equations. This is exactly what we need to when we need to univoquely determine $n + 1$ coefficients. The problem then becomes solving the system of equations:

\[\begin{align} p(x_0) &= a_0 + a_1 x_0 + a_2 x^2_0 + \ldots + a_n x^n_0 = y_0 \\ p(x_1) &= a_0 + a_1 x_1 + a_2 x^2_1 + \ldots + a_n x^n_1 = y_1\\ &\;\;\vdots \notag \\ p(x_n) &= a_0 + a_1 x_n + a_2 x^2_n + \ldots + a_n x^n_n = y_n\\ \end{align}\]which we can write as $\mathbf{X} \cdot \mathbf{a} = \mathbf{y}$ using matrix notation, where:

\[\mathbf{X} = \begin{bmatrix} 1 & x_0 & \cdots & x^n_0 \\ 1 & x_1 & \cdots & x^n_1 \\ \vdots & \vdots & & \vdots \\ 1 & x_n & \cdots & x^n_n \\ \end{bmatrix}\quad \mathbf{a} = \begin{bmatrix} a_0 \\ a_1 \\ \vdots \\ a_n \\ \end{bmatrix}\quad \mathbf{y} = \begin{bmatrix} y_0 \\ y_1 \\ \vdots \\ y_n \\ \end{bmatrix}\]End of story? Well, not really! Finding $\mathbf{X}^{-1}$ is an $O(n^3)$ operation, so good luck with that as soon as $n$ increases. Fortunately, Lagrange came up with a very neat solution to this problem.

Lagrange proposed that $p(x)$ could be written as the sum of $n + 1$ functions, i.e.,

\[p(x) = \sum_{i=1}^n \alpha_i \phi_i(x)\]and proved that, to pass through the required $n + 1$ points, then

\[\alpha_i = y_i\] \[\phi_i(x) = \frac{\prod_{j \neq i}x - x_j}{\prod_{j \neq i}x_i - x_j}\]His idea was very clever because for any $\phi_i(x)$, it holds both that $\phi_i(x_i) = y_i$ and that $\forall j \neq i, \, \phi_i(x_j) = 0$. This means that at the x-coordinates of any of the $n + 1$ points that $p(x)$ had to cross, one of the functions composing $p(x)$ evaluated exactly to the y-coordinate of the point in question while the other $n$ where null.

A natural question that follows is whether we could not have found a more “compact” solution that the one that Lagrange proposed. For example, we could try to look for some linear dependency among the $\phi_i(x)$ functions, remove it, and come up with a solution with less than $n + 1$ functions composing $p(x)$. Well, dreaming is ok, but if that is how we come up with solutions, Lagrange clearly also dreamt, and he did it before us: his solution actually conforms a base of the $n + 1$ polynomium space. We are more used to the canonical base of this space, which is composed of functions $\left\{1, t, t^2 \ldots, t^n\right\}$. The canonical base looks simple and harmless, are we sure that the very clever formulation of Lagrange is also a base? Well, let’s prove it to convince ourselves.

We can prove that Lagrange’s solution conforms a base showing that the only way in which we the linear combination of the functions in his solution evaluate to zero for all $x$ is if and only if $\forall i, \alpha_i = 0$. If this must hold for any $x$, then in particular it must hold for all of the x-coordinates of the points that need to be traversed by $p(x)$, i.e., for the elements in $\left\{x_0, x_1, \ldots, x_n\right\}$ for which $p(x)$. Without loosing generality, taking $x_0$ as an example, we can compute:

\[p(x_0) = \alpha_0 \phi_0(x_0) + \alpha_1 \phi_1(x_0) + \ldots + \alpha_n \phi_n(x_0)\]but based on our previous analysis, we now that this ends up reducing to

\[p(x_0) = \alpha_0\]Since $p(x_0) = 0$, then $\alpha_0 = 0$. If you are still hesitating, I invite you to replicate this analysis for the remaining $n$ values. At this point, voila, we have provedd that Lagrange’s solution is indeed a base of the $n + 1$ polynomium space. His solution is simply beautiful, isn’t it?

Now that we know many things about Lagrange’s interpolation, the last point I want to address is how to implement it efficiently in code. To reason about this, let’s first write down the final expression of $p(x)$:

\[p(x) = \sum_{i=0}^n y_i \frac{\prod_{j \neq i}x - x_j}{\prod_{j \neq i}x_i - x_j}\]And I will continue writing soon…