I am trying to find a generator formula for the discriminant of a polynomial and believe it would be an interesting problem for the Mathematica community to know how to automate evaluation of certain terms to discover an algorithm.

I am looking at the discriminant evaluated by

parnum = 4;
parint = 1;
(m = Array[Subscript[a, ##] &, {parnum, parnum}]);
(m = m /. Subscript[a, i_, j_] :> Subscript[a, j, i] /; j < i) // 
b = Array[Subscript[be, # - 1] &, {parnum, 1}];
b = b /. {Subscript[be, 0] -> 1, 
    Subscript[be, k_] :> (-1)*Subscript[be, k] /; k > 0};
p = Transpose[b].m.b;
Collect[(1/4)*Discriminant[p, Subscript[be, parint]], Subscript[be, 2]]
(mi = Minors[m, 2]);
Minors[m, 2] // MatrixForm;

which gives me a certain discriminant \begin{eqnarray} -M_{11} - \beta_{2}^{2} M_{44} - 2 \beta_{2} M_{14} - \beta^{2}_{3} M_{55} - 2 \beta_{3} M_{15} - 2 \beta_2 \beta_3 M_{45}. \end{eqnarray} The $M_{i,j}$ expressions correspond to entries of the matrix of second minors.

Now, I am wondering if you guys could help me use FindSequenceFunction or a related function to somehow build code that allows me increase the dimensionality of the matrix denoted by parnum and the parameter I am solving for denoted by parint and also to recognize the ensuing pattern?

The above discriminant then follows a certain pattern that always involves the same generating formula, which can be written as a series expansion or rather four separate ones as outlined below.

Now similar approaches have been discussed here for a polynomial and here for integration but here I would like to have my discriminants expressed as a series of elements of the matrix of second minors.

One can quickly note that the structure of the formula is quite simple. It consists of

  1. an intercept
  2. a quadratic term in all the parameters I am not solving for
  3. an interaction term, i.e. multiplication of all parameters
  4. and linear terms for each parameter.
  • $\begingroup$ Perhaps a good start would be to just generate a sequence for a five by five matrix and then split the elements into linear, quadratic, and intercept as well as interaction terms and find separate sequences for those? $\endgroup$
    – Hirek
    Commented Mar 1, 2015 at 19:30


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