0
$\begingroup$

The discriminants of polynomials $f(x)=x^n+a_1x^{n-1}+\cdots+a_n, 1\leq n\leq 4$ are

degree-n discriminant-$\Delta_n$
1 1
2 $a_1^2-4a_2$
3 $a_1^2a_2^2-27a_3^2-4a_1^3a_3-4a_2^3+18a_1a_2a_3$
4 $a_1^2a_2^2a_3^2-256a_4^3+\cdots$ (16 items)

Explicit formula for each $n$ can be obtained by the symmetric polynomial $\prod\limits_{i<j}(α_i−α_j)^2$. My question is

Is there a general term for $\Delta_n$?


I'm trying to find the pattern of these multinomials, a feasible way (I guess) is to convert them into determinants. For example

  • $\Delta_1=1$
  • $\Delta_2=\begin{array}{|cc|}a_1^2&4\\a_2&1\end{array}$.
  • $\Delta_3$ has 5 items | 3-order determinant has 6 items
  • $\Delta_4$ has 16 items | 4-order determinant has 24 items

I tried finding the determinant of $\Delta_3$ by hand but failed. Since MMA has a lot of useful functions in guessing number pattern such as FindSequenceFunction and FindLinearRecurrence, I wonder:

Is there exists an effective algorithm to convert multinomials in $\mathbb{Z}[x_1,...,x_n]$ into determinants of order k with entries in $\mathbb{Z}[x_1,...,x_n]$.

Note: Here one might require the determinant to be as "nontrivial" as possible. For instance, each entry of the determinant has the lowest degree.

$\endgroup$
5
  • 1
    $\begingroup$ The discriminant of a polynomial P is the resultant of the pair P and P' . The resultant can be calculated as a determinant from the formula here: en.wikipedia.org/wiki/Resultant#Definition So it's the determinant of a 2n-1 matrix containing coefficients from P and P' . Eg in the 2 case you get {{1, a_1, a_2 }, {1, 2* a_1 0 }, { 0 , 1 , 2 * a1}} $\endgroup$ Aug 10 at 17:15
  • $\begingroup$ An in-depth introduction to discriminants and resultants is found in "Algorithms in real algebraic geometry" by Richard M. Pollack and Saugata Basu $\endgroup$ Aug 10 at 17:25
  • $\begingroup$ Have you see the Discriminantcommand? There is Modulus option. $\endgroup$
    – yarchik
    Aug 10 at 17:49
  • $\begingroup$ You can form a matrix to take the determinant using the Sylvester or Bezout formulation. Example: In[414]:= poly[a_, x_, n_] := x^n + Array[a, n, 0] . x^Range[0, n - 1] In[417]:= ResourceFunction["SylvesterMatrix"][poly[a, x, 4], D[poly[a, x, 4], x], x] Out[417]= {{1, a[3], a[2], a[1], a[0], 0, 0}, {0, 1, a[3], a[2], a[1], a[0], 0}, {0, 0, 1, a[3], a[2], a[1], a[0]}, {4, 3 a[3], 2 a[2], a[1], 0, 0, 0}, {0, 4, 3 a[3], 2 a[2], a[1], 0, 0}, {0, 0, 4, 3 a[3], 2 a[2], a[1], 0}, {0, 0, 0, 4, 3 a[3], 2 a[2], a[1]}} $\endgroup$ Aug 11 at 1:45
  • $\begingroup$ @DanielLichtblau Thanks. This function is great! $\endgroup$
    – Rex Wang
    Aug 11 at 8:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.