Another instance of annoyance: https://github.com/sagemath/sage/pull/38564
I am fighting with the dumb synchronizer.
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Hi,
In PR https://github.com/sagemath/sage/pull/38568, we are adjusting the
style of the tabbed EXAMPLES, recently introduced. You may come by and
participate.
First, look through all screenshots in the PR.
Second, understand the candidates
in https://github.com/sagemath/sage/pull/38568#issue
This looks like RR should be fixed to match AA here.
On 28 August 2024 12:40:49 BST, Kwankyu Lee wrote:
>sage: ZZ(-1).nth_root(3)
>-1
>sage: _.parent()
>Integer Ring
>sage: QQ(-1).nth_root(3)
>-1
>sage: _.parent()
>Rational Field
>sage: RR(-1).nth_root(3)
>-1.00
>sage: _.parent()
>Rea
On 28 August 2024 12:28:54 BST, Kwankyu Lee wrote:
>
>
>On Wednesday, August 28, 2024 at 7:57:43 PM UTC+9 john.c...@gmail.com wrote:
>
>Surely the output of -1 for AA(-1)^(1/3) is correct: AA is the "Algebraic
>Real Field" and -1 has exactly one cube root in there, namely itself. On
>the ot
All of the following return the complex branch:
pari/gp, mpmath, maxima
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To
[forwarded from Dima]
This looks like RR should be fixed to match AA here.
On 28 August 2024 12:40:49 BST, Kwankyu Lee wrote:
sage: ZZ(-1).nth_root(3)
-1
sage: _.parent()
Integer Ring
sage: QQ(-1).nth_root(3)
-1
sage: _.parent()
Rational Field
sage: RR(-1).nth_root(3)
-1.00
sage: _.
[fowarded from Dima]
>
>x^(1/n) and x.nth_root(n) do not behave in the same way. All x.nth_root(n)
>gives an n-th root in the same field to which x belongs while all x^(1/n)
>gives the primitive n-th root of unity, with the exception of AA(-1)^(1/n).
the function 1/n : AA->AA is 1-1 on all AA, a
sage: ZZ(-1).nth_root(3)
-1
sage: _.parent()
Integer Ring
sage: QQ(-1).nth_root(3)
-1
sage: _.parent()
Rational Field
sage: RR(-1).nth_root(3)
-1.00
sage: _.parent()
Real Field with 53 bits of precision
sage: CC(-1).nth_root(3)
0.500 + 0.866025403784439*I
sage: _.parent()
Co
On Wednesday, August 28, 2024 at 7:57:43 PM UTC+9 john.c...@gmail.com wrote:
Surely the output of -1 for AA(-1)^(1/3) is correct: AA is the "Algebraic
Real Field" and -1 has exactly one cube root in there, namely itself. On
the other hand, QQbar(-1) has 3 cube roots and one is chosen (in som
It is puzzling, but it does not seem to be so.
If 1/3 would be a python float, (-1)**(1/3) is not defined under the usual
definition of x^y = exp(y*ln(x)) with the standard branch cut for the
logarithmic function.
If Python includes the negative axis in the domain of ln(x), and defines
ln(
Surely the output of -1 for AA(-1)^(1/3) is correct: AA is the "Algebraic
Real Field" and -1 has exactly one cube root in there, namely itself. On
the other hand, QQbar(-1) has 3 cube roots and one is chosen (in some
deterministic way).
I do not think that AA(-1)^(1/3) should return a cubroot
> >>> (-1)**(1/3)
> (0.5001+0.8660254037844386j)
Your example illustrates twice that an operation can lead to a bigger
set : 1/3 is a python float :-)
It is puzzling, but it does not seem to be so.
If 1/3 would be a python float, (-1)**(1/3) is not defined under the usual
defi
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