Kurt Foster on Mon, 08 Dec 2008 04:24:14 +0100

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Using {fa} in nfdisc() computations

 To: Pari Users <pariusers@list.cr.yp.to>
 Subject: Using {fa} in nfdisc() computations
 From: Kurt Foster <drsardonicus@earthlink.net>
 Date: Sun, 7 Dec 2008 20:08:28 0700
 Deliverydate: Mon, 08 Dec 2008 04:24:14 +0100
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The optional {fa} parameter [factor() matrix] seems like just the
thing for speeding up the computations in a situation I'm looking at.
I've got a parametric family of polynomials. I want the exact
nfdisc()s for a whole swarm of them. I could, of course, just feed
the polynomials to nfdisc() and let it crunch away. This does work,
of course, but it occurred to me it might be faster if I could take
advantage of the fact that the poldiscs have a parametric algebraic
factorization, and the algebraic factors are generally much smaller
than the poldisc(), in the sense that
log(abs(algebraic factor))/log(abs(poldisc)
will usually be a lot less than 1. Thus, factoring the algebraic
factors should be a LOT quicker than tackling the whole poldisc.
But in order to produce the factor() matrix fa for the poldisc, I've
got to "combine" the factorization matrices of the algebraic factors.
There's probably a slick way to do this, but I'm a dunce at
programming. Would factor()ing the algebraic factors and "combining"
the results actually be likely to be faster than tackling the whole
poldisc? If so, what's an expeditious way to combine the factor()
matrices of the separate algebraic factors into the correct factor()
matrix for the poldisc?