You may work on the univariate polynamial ring in your variable of interest
over a suitable ring. A simple example :
sage: var("x, y, z")
(x, y, z)
sage: foo=x^3-x*sin(y+z)+1
sage: foo.polynomial(ring=PolynomialRing(SR,"x")).parent()
Univariate Polynomial Ring in x over Symbolic Ring
sage: foo.polynomial(ring=PolynomialRing(SR,"x")).roots()
[(-1/6*(-I*sqrt(3) + 1)*sin(y + z)/(1/6*sqrt(-4/3*sin(y + z)^3 + 9) -
1/2)^(1/3) - 1/2*(I*sqrt(3) + 1)*(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3),
1),
(-1/6*(I*sqrt(3) + 1)*sin(y + z)/(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3)
- 1/2*(-I*sqrt(3) + 1)*(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3),
1),
(1/3*sin(y + z)/(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3) +
(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3),
1)]
It might be interesting to create a more specific base ring, but only if
your expression is indeed a polynomial (e. g. no function calls as in my
oversimplified example…).
It turns out that, in this precise case, it is not necessary to explicitly
create a polynomial ; but doing so gives a two orders of magnitude speed
gain (not exactly small potatoes) :
sage: %time foo.roots()
CPU times: user 3.83 s, sys: 40.1 ms, total: 3.87 s
Wall time: 2.9 s
[(-1/6*(-I*sqrt(3) + 1)*sin(y + z)/(1/6*sqrt(-4/3*sin(y + z)^3 + 9) -
1/2)^(1/3) - 1/2*(I*sqrt(3) + 1)*(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3),
1),
(-1/6*(I*sqrt(3) + 1)*sin(y + z)/(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3)
- 1/2*(-I*sqrt(3) + 1)*(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3),
1),
(1/3*sin(y + z)/(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3) +
(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3),
1)]
sage: %time foo.polynomial(ring=PolynomialRing(SR,"x")).roots()
CPU times: user 41.2 ms, sys: 17 µs, total: 41.2 ms
Wall time: 41.1 ms
[(-1/6*(-I*sqrt(3) + 1)*sin(y + z)/(1/6*sqrt(-4/3*sin(y + z)^3 + 9) -
1/2)^(1/3) - 1/2*(I*sqrt(3) + 1)*(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3),
1),
(-1/6*(I*sqrt(3) + 1)*sin(y + z)/(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3)
- 1/2*(-I*sqrt(3) + 1)*(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3),
1),
(1/3*sin(y + z)/(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3) +
(1/6*sqrt(-4/3*sin(y + z)^3 + 9) - 1/2)^(1/3),
1)]
HTH,
Le lundi 5 juillet 2021 à 22:01:10 UTC+2, axio...@yahoo.de a écrit :
> I am trying to fix #32133, which is about translating FriCAS expressions
> containing symbolic roots of polynomials, given by their minimal
> polynomials.
>
> This is easy enough if the polynomials are univariate, so the roots are
> algebraic numbers and I can use the wonderful machinery of QQbar.
>
> However, I don't see what I can do with polynomials containing extra
> variables.
>
> Any hints appreciated!
>
> Martin
>
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