Assume that f(x)=xⁿ+a₁x^n-1+a₂x^n-2+ … a_n=0, and that the roots of this equation x₁, x₂, x₃, … ,x_n are distinct at any (a₁,a₂, … ,a_n) considered.

If we differentiate f(x) with respect to any a_i we get ∂x/∂a_i = -x^n-i/f ‘(x), i=1, 2, 3, …, n.

It follows easily from this that a₁∂x/∂a₁ + a₂∂x/∂a₂ + …a_n∂x/∂a_n = x,

and n∂x/∂a₁+ (n-1)a₁∂x/∂a₂+ … + a_n-1∂x/∂a_n= -1.

If Δ_n² = Π_1<=i<j<=n(x_i-x_j)² and the linear operators L₁( ) and L₂( ) are defined by …

L₁(x)=Σ_1<=i<=n ia_i∂x/∂a_i and …

L₂(x)=n∂x/∂a₁+(n-1)a₁∂x/∂a₂ + … + a_n-1∂x/∂a_n then …

L₁(Δ_n²)=2Δ_n²Σ_1<=i<j<=nL₁(x_i-x_j)/(x_i-x_j)=n(n-1)Δ_n²,

and,

L₂(Δ_n²)=0.