Lagrange interpolation

Recall from our basics discussion that a polynomial $\phi$ of degree $d$ is a vector of $d+1$ coefficients:

\begin{align} \phi &= [\phi_0, \phi_1, \phi_2, \dots, \phi_d] \end{align}

How to compute a polynomial’s coefficients from a bunch of its evaluations

Given $n$ pairs $(x_i, y_i)_{i\in[n]}$, one can compute or interpolate a degree $\le n-1$ polynomial $\phi(X)$ such that: \(\phi(x_i)=y_i,\forall i\in[n]\)

Specifically, the Lagrange interpolation formula says that: \begin{align} \label{eq:lagrange-formula} \phi(X) &= \sum_{i\in[n]} y_i \cdot \lagr_i(X),\ \text{where}\ \lagr_i(X) = \prod_{j\in[n],j\ne i} \frac{X-x_j}{x_i-x_j} \end{align}

This formula is intimidating at first, but there’s a very simple intuition behind it. The key idea is that $\lagr_i(X)$ is defined so that it has two properties:

1. $\lagr_i(x_i) = 1,\forall i\in[n]$
2. $\lagr_i(x_j) = 0,\forall j \in [n]\setminus\{i\}$

You can actually convince yourself that $\lagr_i(X)$ has these properties by plugging in $x_i$ and $x_j$ to see what happens.

Important: The $\lagr_i(X)$ polynomials are dependent on the set of $x_i$’s only (and thus on $n$)! Specifically each $\lagr_i(X)$ has degree $n-1$ and has a root at each $x_j$ when $j\ne i$! In this sense, a better notation for them would be $\lagr_i^{[x_i, n]}(X)$ or $\lagr_i^{[n]}(X)$ to indicate this dependence.

Example: Interpolating a polynomial from three evaluations

Consider the following example with $n=3$ pairs of points. Then, by the Lagrange formula, we have:

\[\phi(X) = y_1 \lagr_1(X) + y_2 \lagr_2(X) + y_3 \lagr_3(X)\]

Next, by applying the two key properties of $\lagr_i(X)$ from above, you can easily check that $\phi(x_i) = y_i,\forall i\in[3]$: \begin{align} \phi(x_1) &= y_1 \lagr_1(x_1) + y_2 \lagr_2(x_1) + y_3 \lagr_3(x_1) = y_1 \cdot 1 + y_2 \cdot 0 + y_3 \cdot 0 = y_1\\
\phi(x_2) &= y_1 \lagr_1(x_2) + y_2 \lagr_2(x_2) + y_3 \lagr_3(x_2) = y_1 \cdot 0 + y_2 \cdot 1 + y_3 \cdot 0 = y_2\\
\phi(x_3) &= y_1 \lagr_1(x_3) + y_2 \lagr_2(x_3) + y_3 \lagr_3(x_3) = y_1 \cdot 0 + y_2 \cdot 0 + y_3 \cdot 1 = y_3 \end{align}

An important detail is that the degree of the interpolated $\phi(X)$ is $\le n-1$ and not necessarily exactly equal to $n-1$. To see this, consider interpolating the polynomial $\phi(X)$ such that $\phi(i) = i$ for all $i\in [n]$. In other words, $x_i = y_i = i$.

The inspired reader might notice that the polynomial $\phi(X) = X$ could satisfy our constraints. But is this what the Lagrange interpolation will return? After all, the interpolated $\phi(X)$ is a sum of degree $n-1$ polynomials $\lagr_i(X)$, so could it have degree 1? Well, it turns out, yes, because things cancel out. To see this, take a simple example, with $n=3$: \begin{align} \phi(X) &=\sum_{i\in [3]} i \cdot \lagr_i(X) = \sum_{i\in [3]} i \cdot \prod_{j\in[3]\setminus{i}} \frac{X - j}{i - j}\\
&= 1\cdot \frac{X-2}{1-2}\frac{X-3}{1-3} + 2\cdot \frac{X-1}{2-1}\frac{X-3}{2-3} + 3\cdot\frac{X-1}{3-1}\frac{X-2}{3-2}\\
&= \frac{X-2}{-1}\frac{X-3}{-2} + 2\cdot \frac{X-1}{1}\frac{X-3}{-1} + 3\cdot \frac{X-1}{2}\frac{X-2}{1}\\
&= \frac{1}{2}(X-2)(X-3) - 2(X-1)(X-3) + \frac{3}{2}(X-1)(X-2)\\
&= \frac{1}{2}[(X-2)(X-3) + 3(X-1)(X-2)] - 2(X-1)(X-3)\\
&= \frac{1}{2}[(X-2)(4X-6)] - 2(X-1)(X-3)\\
&= (X-2)(2X-3) - 2(X-1)(X-3)\\
&= (2X^2 - 4X - 3X + 6) - 2(X^2 - 4X +3)\\
&= (2X^2 - 7X + 6) - 2X^2 + 8X - 6\\
&= X \end{align}

Computational overhead of Lagrange interpolation

If done naively, interpolating $\phi(X)$ using the Lagrange formula in Equation \ref{eq:lagrange-formula} will take $O(n^2)$ time.

However, there are known techniques for computing $\phi(X)$ in $O(n\log^2{n})$ time. We described part of these techniques in a previous blog post, but for the full techniques please refer to the “Modern Computer Algebra” book¹.