I think cubic root can be extended to negative real numbers,
whereas square root cannot be extended to negative real numbers,
still giving a real numbers result, postive, zero or negative.
You would have like:
0.125^(1/3) = 0.5
(-0.125)^(1/3) = -0.5
Wolfram Alpha asks me whether I want principal root or
real-valued root. Maybe FriCAS has also such a feature somewhere?
Maybe should read the FriCAS manual.
Quick gloss over the FriCAS manual gives me, showing
the difference between principal and real-valued root:
(2) -> complexSolve(x^3 = -1/8, 0.0001)
(2)
[x = 0.25 - 0.4330127025_0231027603 %i,
x = 0.25 + 0.4330127025_0231027603_1494140625 %i, x = - 0.5]
(3) -> solve(x^3 = -1/8, 0.0001)
(3) [x = - 0.5]
But how tell integral what root to use?
Waldek Hebisch schrieb am Samstag, 21. Oktober 2023 um 21:38:28 UTC+2:
> On Sat, Oct 21, 2023 at 10:32:55AM -0700, Mild Shock wrote:
> > > Also, without ability to simplify hypergeometric answers are of
> > > little use, so we need first strong simplifier for hypergeometric
> > functions.
> >
> > I am not sure that the lack of a strong simplifier is to blame that
> > I don't see a slope in those systems that can solve the integral.
> >
> > Now I was trying to compute the slope, by using a definite
> > integral over the period instead of a indefinite integral.
> <snip>
> > The more complex problem with cos(t)^(1/3) gives me:
> >
> > (2) -> integrate(cos(t)^(1/3)*(2+cos(t))/100, t = 0..2*pi)
> >
> > (2) "potentialPole"
> >
> > How can I tell the system that ^(1/3) should be real root?
>
> (7) -> integrate(cos(t)^(1/3)*(2+cos(t))/100, t = 0..2*%pi, "noPole")
>
> (7) "failed"
> Type: Union(fail: failed,...)
>
> in 99.9999% of cases FriCAS needs indefinite integral to
> compute definite one. Since definite integral remains
> unevaluated FriCAS can not give you definite one.
>
> There is also another catch: FriCAS works with branches of
> analytic functions. But cos(t) changes sign close to 0.
> This means there are no analytic real third root close to 0.
> If you take real root for small positive t, than for small
> negative t analytic root must be complex. That must be
> taken into account in cases where FriCAS can compute
> indefinite integral.
>
> --
> Waldek Hebisch
>
--
You received this message because you are subscribed to the Google Groups
"FriCAS - computer algebra system" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
To view this discussion on the web visit
https://groups.google.com/d/msgid/fricas-devel/27d99d98-9360-4ca0-bf29-d6f0f6c22acen%40googlegroups.com.
