(%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 ?
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.