# Polynomial roots and discriminants

· Uncategorized
Authors

Assume that f(x)=xn+a1xn-1+a2xn-2+ … an=0, and that the roots of this equation x1, x2, x3, … ,xn are distinct at any (a1,a2, … ,an) considered.

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

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

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

If Δn2 = Π1<=i<j<=n(xi-xj)2 and the linear operators L1( ) and L2( ) are defined by …

L1(x)=Σ1<=i<=n iai∂x/∂ai and …

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

L1n2)=2Δn2Σ1<=i<j<=nL1(xi-xj)/(xi-xj)=n(n-1)Δn2,

and,

L2n2)=0.