The Convex Hull of a Polynomial and its Derivatives

Recall that the convex hull of a set of points \(S\) is the smallest convex set that contains all points of \(S\) (i.e., the intersection of all convex sets which contain each point in \(S\)). Denote the convex hull of all roots of a polynomial \(f\) by \(H(f)\). Then \(H(f') \subset H(f)\). This (kind of unexpected!) result is known as the Gauss-Lucas Theorem.

The canonical proof of this theorem is really really boring perhaps not particularly insightful. Instead, I'd like to give a physical intuition (a demiproof?) for why this is true:

Consider some polynomial \(f \in \mathbb{C}[x]\) with distinct roots \(\alpha_i\), and fix an equally charged particle, say an electron, at each root. This creates an electric field everywhere in the plane which vanishes exactly at the roots of \(f'\). (!)

electrons in the plane: https://arxiv.org/pdf/2112.00110.pdf
the red points are electrons (roots of \(f\)) and the blue points are the vanishing points of the induced electric field (roots of \(f'\)) [source]

We'll prove this for a special case of \(f\), in particular, we want \(f(x) \in \mathbb{R}[x]\) to be monic, separable, and such that \(f(0) \neq 0\).

Proof. Let \(\alpha_i\) be the roots of \(f\). The electric potential at any point in the plane is given by $$ E(z) := \sum_i \frac{\alpha_i - z}{|\alpha_i - z|^2} = \sum_i \frac{1}{\overline{\alpha_i} - \overline{z}}. $$ Because we can shift the axes without affecting the electric field, we can choose \(z = 0\). So \({E(z = 0) = \sum_i \frac{1}{\bar{\alpha_i}}}\). Since \(f\) takes coefficients from \(\mathbb{R}\), we have $$ E(0) = \sum_i \frac{1}{\alpha_i}. $$ Write \(f(z) = \prod_i (z - \alpha_i)\). Then \begin{align*} f'(z) &= (z-\alpha_2)(z-\alpha_3)\dots(z-\alpha_n) + \dots + (z-\alpha_1)(z-\alpha_2)\dots(z-\alpha_{n-1})\\ &= \sum_i\prod_{j\neq i} (z-\alpha_j). \end{align*} So, \begin{align*} f'(z=0) = \sum_i\prod_{j\neq i} -\alpha_j &= \pm f(0)\sum_i\frac{1}{\alpha_i}\\ &= \pm f(0)E(0). \end{align*} So we have \(f'(0) = \pm f(0)E(0)\), i.e., since by hypothesis \(f(0) \neq 0\), we get \(f'(0) = 0 \iff E(0) = 0\). \(\Box\)

So because the electric field vanishes only at zeroes of \(f'\), and because physical intuition tells us that it doesn't make sense for the electric field to vanish outside of the convex hull of the roots of \(f\) (this is maybe worth working out formally, but not necessary to get the idea), it makes sense that the zeroes of \(f'\) occur only within the convex hull of the roots of \(f\).

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