Partfrac solves a given problem over the rational numbers,
not the (approximate) real numbers.
If you really want the (numeric, approximate) partial
fraction expansion, that can be computed, but partfrac doesn't do that.
See note below yours, for a program, fpfe, that does this.
(C8) fpfe(d1,s);
1.677050983124843 1.677050983124843
(D8) --------------------- - ---------------------
s + 2.105572809000084 s + 3.894427190999916
(C9)
.... keepfloat:true$
ratsimp(d8); gives
almost the same as what you started with.
But not exactly.
RJF
PS, as far as I know, this is not in the "share" directory. But
it could be put there. LLGPL.
Doug Stewart wrote:
>>>
> partfrac ( +3 /(s^2 +6*s +8.2),s) ;
>
> Since this has 2 real roots I would expect two terms in the answer but I
> get only 1.
> how do you make it give the two terms?
> Doug Stewart
>
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> Maxima@math.utexas.edu
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fpfe(p,x):=
block ([num, den,div, res,keepfloat:true,lroots,distinctroots,qq, L, c],
local(r,q),
p:rat(p), num:ratnumer(p), den:ratdenom(p), div: divide(num,den),
res: ratdisrep(div[1]),
num: ratdisrep(div[2]/ratcoef(den,x^hipow(den,x))),
lroots:map(rhs,allroots(den)),
r[i]:=0, for c in lroots do r[c]:r[c]+1,
distinctroots: map(first,rest(arrayinfo(r),2)),
qq:1,
for c in distinctroots do qq:qq*(x-c)^r[c],
for c in distinctroots do q[c]:qq/(x-c)^r[c],
for c in distinctroots do for i: 1 thru r[c] do
res: res+1/(r[c]-i)!
*ratsimp(subst(c,x,diff(num/q[c],x,r[c]-i)))
/(x-c)^i,
return(res));
/*
Published in Proc. ISSAC'82, "Computer Algebra and Numerical
Integration," by R. Fateman p 228 - 232 */