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 $H(f^{\prime })\subset H(f)$. 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 $f^{\prime }$. (!)

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 $f(0)\ne 0$.

Proof. Let ${\alpha }_i$ be the roots of $f$. The electric potential at any point in the plane is given by $$E(z):=\sum _i\frac{{\alpha }_i-z}{|{\alpha }_i-z{|}^2}=\sum _i\frac{1}{\overline{{\alpha }_i}-\bar{z}}.$$ Because we can shift the axes without affecting the electric field, we can choose $z=0$. So $E(z=0)=\sum _i\frac{1}{\overline{{\alpha }_i}}$. 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 $$\begin{array}{rl}{f}^{\prime }(z)& =(z-{\alpha }_2)(z-{\alpha }_3)\dots (z-{\alpha }_n)+\cdots +(z-{\alpha }_1)(z-{\alpha }_2)\dots (z-{\alpha }_{n-1})\\ & =\sum _i\prod _{j\ne i}(z-{\alpha }_j).\end{array}$$ So, $$\begin{array}{rl}{f}^{\prime }(z=0)=\sum _i\prod _{j\ne i}-{\alpha }_j& =\pm f(0)\sum _i\frac{1}{{\alpha }_i}\\ & =\pm f(0)E(0).\end{array}$$ So we have $f^{\prime }(0)=\pm f(0)E(0)$, i.e., since by hypothesis $f(0)\ne 0$, we get $f^{\prime }(0)=0⟺E(0)=0$. $◻$