Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
What is the determinant of a polynomial? –  Igor Rivin Oct 15 '12 at 2:30
1  
"discriminant" must be the intended word (up to $\pm 1$). –  Noam D. Elkies Oct 15 '12 at 3:08
add comment

2 Answers

up vote 3 down vote accepted

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.

share|improve this answer
add comment

You find a lot of information in: Dmitri Alekseevky, Andreas Kriegl, Mark Losik, Peter W. Michor: Choosing roots of polynomials smoothly, Israel J. Math 105 (1998), p. 203-233. (pdf), and in related later papers. The formula that you seek (or the inverse) is on page 7.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.