Assume that f(x)=x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+ … a_{n}=0, and that the roots of this equation x_{1}, x_{2}, x_{3}, … ,x_{n }are distinct at any (a_{1},a_{2}, … ,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 a_{1}∂x/∂a_{1} + a_{2}∂x/∂a_{2} + …a_{n}∂x/∂a_{n} = x,

and n∂x/∂a_{1}+ (n-1)a_{1}∂x/∂a_{2}+ … + a_{n-1}∂x/∂a_{n}= -1.

If Δ_{n}^{2 }= Π_{1<=i<j<=n}(x_{i}-x_{j})^{2 }and the linear operators L_{1}( ) and L_{2}( ) _{}are defined by …

L_{1}(x)=Σ_{1<=i<=n} ia_{i}∂x/∂a_{i} and …

L_{2}(x)=n∂x/∂a_{1}+(n-1)a_{1}∂x/∂a_{2} + … + a_{n-1}∂x/∂a_{n} then …_{}

L_{1}(Δ_{n}^{2})=2Δ_{n}^{2}Σ_{1<=i<j<=n}L_{1}(x_{i}-x_{j})/(x_{i}-x_{j})=n(n-1)Δ_{n}^{2},

and,

L_{2}(Δ_{n}^{2})=0.