Re: Update to qfsolve doc

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

On Thu, Dec 14, 2023 at 01:40:25PM -0800, Thomas D. Dean wrote:
> Here is s suggested patch to doc to reflect the 'extera' argument to
> qfsolve.
> 
> 
> Tom Dean

> *** usersch3.tex~	2023-12-09 17:05:02.737511515 -0800
> --- usersch3.tex	2023-12-14 13:35:16.753248824 -0800
> ***************
> *** 14688,14693 ****
> --- 14688,14698 ----
>   where $F$ is the factorization matrix of the absolute value of the determinant
>   of $G$.
>   
> + $G$ may be a vector [$G$,$F$] where $G$ is a symmetric matrix similar
> + to matdigonal([1,1,...,1,-$n$]) and $F$ is matrix of the factors of
> + $n$. Then, $F$ will be used to reduce the number of factorizations in
> + solving the quadratic equation.

Well the patch cuts the line above which reads
""
Given  a  square symmetric matrix G of dimension n >= 1,  solve over Q the
quadratic equation {^t}X G X = 0.   The matrix G must have
rational coefficients.  When G is integral, the argument can also be a vector
[G,F] where F is the factorization matrix of the absolute
value of the determinant of G. 
""

which is supposed to say the same thing in more generality ?

(the discriminant of a diagonal matrix is the product of its diagonal coefficients,
and if you provide F, qfsolve will not do any factorisation by itself).

Cheers,
Bill