Bill Allombert on Thu, 14 Dec 2023 23:08:30 +0100 |
[Date Prev] [Date Next] [Thread Prev] [Thread Next] [Date Index] [Thread Index]
Re: Update to qfsolve doc |
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