Solve and assumptions too aggressive with cube root of negative numbers

[kcrisman]

#6515 did a great job helping us start to catch some assumptions when we do solving.

However, this ask.sagemath.org post catches a case where it's too aggressive, because Sage says that (-1)^(1/3) is not real.

sage: solve(x^3+1==0,x)
[x == 1/2*I*(-1)^(1/3)*sqrt(3) - 1/2*(-1)^(1/3), x == -1/2*I*(-1)^(1/3)*sqrt(3) - 1/2*(-1)^(1/3), x == (-1)^(1/3)]
sage: assume(x,'real')
sage: solve(x^3+1==0,x)
[]

What's weird about this is that the Maxima in Sage should just return x==-1.

(%i2) display2d:false;

(%o2) false
(%i3) solve(x^3+1=0,x);

(%o3) [x = -(sqrt(3)*%i-1)/2,x = (sqrt(3)*%i+1)/2,x = -1]

Not sure what's going on with that.

Upstream: Not yet reported upstream; Will do shortly.

CC: @pelegm

Component: symbolics

Work Issues: report upstream

Issue created by migration from https://trac.sagemath.org/ticket/11941

[jdemeyer]

Milestone sage-4.7.3 deleted

[mboratko]

comment:2

The fact that Maxima normally returns -1
and Sage returns (-1)^(1/3) is a bit odd,
as you mentioned. At a more basic level,
Sage doesn't seem to think that (-1)^(1/3)
is in RR:

sage: (-1)^(1/3) in RR
False
sage: (2)^(1/3) in RR
True

So if we fix that problem, then at least it would
return (-1)^(1/3). I also suspect that it would
properly simplify to -1 at that point as well,
based on the following example:

sage: solve(x^3 - 8 == 0, x)
[x == I*sqrt(3) - 1,
 x == -I*sqrt(3) - 1,
 x == 2]
sage: solve(x^3 + 8 == 0, x)
[x == I*(-1)^(1/3)*sqrt(3) - (-1)^(1/3),
 x == -I*(-1)^(1/3)*sqrt(3) - (-1)^(1/3),
 x == 2*(-1)^(1/3)]

(Note that the (1/3) exponent appears everywhere
next to the -1, as if some rule specifies that
Sage should not simplify it out.)

I am new, however, and I am not sure where next to look.

[kcrisman]

comment:3

Well, in general we do not want to do this. It's been discussed ad nauseam many times, and the sense is that:

• Similar programs don't necessarily do this
• (-1)^(1/3) is not really -1 but a primitive complex root of -1.

This ticket is about the fact that Maxima returns three solutions to the equation, but when we do the assume(x,'real') they all mysteriously vanish!

[burcin]

comment:4

Here's what maple does:

> solve( x^3+1= 0, x);
                                     1/2               1/2
                    -1, 1/2 - 1/2 I 3   , 1/2 + 1/2 I 3

I wonder why maxima is returning (-1)^(1/3). Maybe we should ask the Maxima developers.

[kcrisman]

comment:5

I wonder why maxima is returning (-1)^(1/3). Maybe we should ask the Maxima developers.

No, no! See this example from the description:

(%i2) display2d:false;

(%o2) false
(%i3) solve(x^3+1=0,x);

(%o3) [x = -(sqrt(3)*%i-1)/2,x = (sqrt(3)*%i+1)/2,x = -1]

We are somehow getting the (-1)^(1/3) by doing something nonstandard in Maxima, apparently. But their vanilla thing is just right.

[mboratko]

comment:6

It seems that sage sets domain: complex (I was made aware of this by burcin in IRC). You do get this result as follows:

{{{(%i8) domain:complex;

(%o8) complex
(%i9) solve(x^3+1=0,x);

(%o9) [x = ((-1)^{(1/3)sqrt(3)%i-(-1)}(1/3))/2,
x = -((-1)^{(1/3)sqrt(3)%i+(-1)}(1/3))/2,x = (-1)^(1/3)]
}}}

[mboratko]

comment:7

Ugh, all my comments have screwed up formatting. I wish I could edit them (can I?). I should have been:

(%i8) domain:complex;

(%o8) complex
(%i9) solve(x^3+1=0,x);

(%o9) [x = ((-1)^(1/3)*sqrt(3)*%i-(-1)^(1/3))/2,
       x = -((-1)^(1/3)*sqrt(3)*%i+(-1)^(1/3))/2,x = (-1)^(1/3)]

[kcrisman]

comment:8

Replying to @mboratko:

It seems that sage sets domain: complex (I was made aware of this by burcin in IRC). You do get this result as follows:

Yes, we do, but I didn't bother checking that. Good work.

So of course now the question becomes what the "right" thing to do is? I don't think we want to set and unset domain:real/complex in Maxima every time we use solve, because presumably this would break other things. Or? At any rate we definitely need to keep domain:complex in general, if I recall correctly other problems that occur without it.

(%i1) (-1)^(1/3);
(%o1)                                 - 1
(%i2) domain:complex;
(%o2)                               complex
(%i3) (-1)^(1/3);
                                        1/3
(%o3)                              (- 1)

Typically we would want the latter answer, e.g in

sage: a = (-1)^(1/3)
sage: a.simplify()
(-1)^(1/3)