On Mar 27, 2013 12:28 AM, "Henry Baker" <hbaker1 at pipeline.com> wrote:
> (%i7) (a*x^2+b*x+c)/(x+d)^3;
> 2
> a x + b x + c
> (%o7) --------------
> 3
> (x + d)
> (%i8) partfrac(%,x);
> 2
> a b - 2 a d a d - b d + c
> (%o8) ----- + --------- + --------------
> x + d 2 3
> (x + d) (x + d)
>
> Is there any way to tell partfrac to keep the (x+d) denominators together,
> as in %o7 ?
>
No, but it would be easy enough to combine them after the fact.
-s
> At 08:14 PM 3/26/2013, Stavros Macrakis wrote:
> >? 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.
>
>