On May 14, 6:40 am, Martin Albrecht <[EMAIL PROTECTED]>
wrote:
> Please excuse my ignorance, but is this definition well known or should it be
> given in the docstring? Also, how did the idea came up that SAGE needs this
> method? I am just curious and trying to learn what people use MPolynomials
On 5/14/07, Martin Albrecht <[EMAIL PROTECTED]> wrote:
>
> On Monday 14 May 2007 15:00, Michel wrote:
> > Assuming all this is correct then I can see
> >
> > > why Martin is objecting to the docstring's vagueness, but AFAIK the
> > > behavior is correct.
> >
> > Yes I now thinks so too.
>
> Pleas
On Monday 14 May 2007 15:00, Michel wrote:
> Assuming all this is correct then I can see
>
> > why Martin is objecting to the docstring's vagueness, but AFAIK the
> > behavior is correct.
>
> Yes I now thinks so too.
Please excuse my ignorance, but is this definition well known or should it be
Assuming all this is correct then I can see
> why Martin is objecting to the docstring's vagueness, but AFAIK the
> behavior is correct.
>
Yes I now thinks so too.
Michel
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To u
Using the analogy with primes, p^e||n iff p^e|n and p^{e+1} not not divide n:
y^e||f(x,y) iff y^{e}|f(x,y) and y^{e+1} does not divide f(x,y). Here I'm
implicitly assuming that y is a generator of the ring which f(x,y) belongs to.
For monomials which are not generators then I'm not sure what
"exa
>
> But y^2 isn't *exactly* divisible by y, so why is y there in the output?
>
>
So what is your definition of exactly divisible then?
Michel
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On 5/14/07, Martin Albrecht <[EMAIL PROTECTED]> wrote:
>
> Hi there,
>
> I am having trouble re-implementing MPolynomial.coeffient. The docstring
> states:
>
> If f is this polynomial, then the coefficient is the sum T/mon
> where the sum is over terms T in f that are exactly divis
