? partfrac
On Mar 26, 2013 11:11 PM, "Henry Baker" <hbaker1 at pipeline.com> wrote:
> Thanks, Stavros!
>
> While you're at it, is there a way to put a rational expression into
> complete 'partial fraction' form (analogous to 'Egyptian fractions' of
> rational numbers), where the numerators of the fraction terms have lower
> degree than the denominators, except for the polynomial term ?
>
> At 07:22 PM 3/26/2013, Stavros Macrakis wrote:
> >This is a variant of a frequently asked question here. The quick answer
> is that there is no general method to find the shortest or the
> fastest-to-calculate form of an expression in Maxima. It is certainly
> possible to do a combinatorial search of various kinds of rewrites of an
> expression, though some rewrites are harder to find than others. For
> example, I don't know how you'd find
> >
> > (x+y+3)^3-(x+2*y-4)^4
> >
> >starting from its expanded form.
> >
> >Some useful tools for reorganizing expressions for calculation include
> factor, horner, rat, factorsum, and optimize. Do not be deluded by the
> name 'optimize' -- it is actually very simple-minded. By the way, to
> 'factor' an expression in Maxima means to represent as a product of
> irreducible polynomial terms, not to rewrite it for efficiency or accuracy
> of calculation.
> >
> >Note also that none of these functions take into account things like
> numerical accuracy and avoiding overflows -- which can be very subtle
> problems. See George Forsythe's classic How do you solve a quadratic
> equation?
> >
> > -s
> >
> >On Mon, Mar 25, 2013 at 12:02 PM, Emmanuel Michon <
> emmanuel.michon+maxima at gmail.com> wrote:
> >Hello,
> >
> >is there a way to tell maxima to factor the most it can out of
> >
> >x^2+3*x*y+5*a^2+z^2+a^3
> >
> >which I understand as answering this
> >
> >(x+y)^2+a^2*(a+5)+x*y
> >
> >If I had to define what I want in more scientific terms, that would be
> >?the form that would comprise the less possible symbols? directly
> >related to a matter of evaluation speed.
> >
> >I found factorsum() but it's about inoperant for
> >
> >4*p[2]*p[3]^2*q[4]^2-p[2]^3*q[4]^2-p[1]^2*p[2]*q[4]^2
> > -2*p[3]^3*q[3]*q[4]+8*p[2]^2*p[3]*q[3]*q[4]
> > -2*p[1]^2*p[3]*q[3]*q[4]
> >
> +10*p[1]*p[2]*q[2]*p[3]*q[4]-4*p[2]*p[3]^2*q[3]^2
> > +p[2]^3*q[3]^2+p[1]^2*p[2]*q[3]^2
> > -10*p[1]*q[1]*p[2]*p[3]*q[3]-2*q[1]*q[2]*p[3]^3
> > -4*p[2]*q[2]^2*p[3]^2+4*q[1]^2*p[2]*p[3]^2
> >
> +8*q[1]*p[2]^2*q[2]*p[3]-2*p[1]^2*q[1]*q[2]*p[3]
> > +p[2]^3*q[2]^2+p[1]^2*p[2]*q[2]^2-q[1]^2*p[2]^3
> > -p[1]^2*q[1]^2*p[2]
> >
> >Thanks!
> >
> >***
> >
> >Sincerely yours,
> >
> >e.m.
>
>