Start with variables $(a\_1, a\_2, a\_3, … a\_n)$ and transform it to the system $(x\_1, x\_2, x\_3, … x\_n)$ where the xi’s are the solutions to $x^n + a\_1x^{n-1} + a\_2x^{n-2} + a\_3x^{n-3} +…+ a\_n$. The Jacobian transformation seems to be $da\_1 da\_2 da\_3 … da\_n = J' dx\_1 dx\_2 dx\_3 … dx\_n$ where $J'$ is the square root of the negative of the determinant of $x^n + a\_1x^{n-1} + a\_2x^{n-2} + a\_3x^{n-3} +…+ a\_n$. Is this true? Is it well known?



Jacobian and determinants
Let the polynomial be $p(x).$ If you write down the Jacobian of the map which maps the $x\_i$ to the $a\_i$ (which are symmetric functions of the $x\_i$), you will see that the columns just have the coefficients of $p(x)/(x-x\_i)$ (in other words, the symmetric functions of all but the $i$-th variable). This determinant, as a polynomial in the $x\_i$ will have degree $n(n-1)/2,$ and will vanish whenever two of the $x\_i$ are equal, so it is a constant multiple of the product of $(x\_i - x\_j),$ for $i>j$ (which is a square root of the discriminant). Computing the constant is easy by induction.


