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Sagemath and TI-83 giving different answers

asked 2019-11-22 09:45:44 +0100

gg gravatar image

updated 2019-11-24 12:40:53 +0100

My expression


SageMath returns a complex number:

bash sage: n((-3)^(1/3)) 0.721124785153704 + 1.24902476648341*I

while TI-83 returns -1.4422 (expected answer)

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answered 2019-11-23 02:41:11 +0100

dazedANDconfused gravatar image

updated 2019-11-25 14:11:14 +0100

I don't have knowledge of the coding decisions in SAGE but I've got some math background that helps me to understand why the SAGE answer could be considered "better" than the real answer you expected. In math terms it has to do with the fact that the nth roots of unity form a group and some of the answers are generators for the group (primitive elements). That is, knowing a generator gets you all the answers. Okay, that's a bit technical, so look at this simpler problem x^4=1. There 4 answers: 1, -1, i, -i. The root i is primitive because: i, i^2, i^3, i^4 gives i, -1, -i, 1 which is all the roots of the answer. This doesn't work with 1 since 1, 1^2, 1^3, 1^4 is always 1 and it doesn't work with -1 since -1, (-1)^2, (-1)^3, (-1)^4 gives -1,1,-1,1. SAGE has given you the primitive element needed to create all the answers. Your answer -cuberoot(3) won't generate the other 2 answers as real number times real number is always a real number. The answer SAGE has given is a primitive root that will create the other roots. It's a little bit more complicated for your example as the modulus of your number is not 1 and your number is negative but same idea is in play. Some of it is covered in precalculus when you get to DeMoivre's Theorem. There is a post at math stack exchange here relevant to your problem if you want a little more math.

EDIT: Based on Sebastien's comment below, I dusted off a book on complex number theory to find that, for nth root of a complex number the unique nth root is defined to be when k=0 in the formula for the nth roots and is called the principal nth root. So for z^3=3, r=cuberoot(3) and theta = 0 (the angle used in polar form representation of 1) so the formula for the roots is cuberoot(3)cis((0+2kpi)3) = cuberoot(3) cis(0), cuberoot(3) cis(2pi/3), cuberoot(3)cis(4pi/3). For z^3=-3, r=cuberoot(3), but now theta = pi so the formula for the roots is cuberoot(3)cis((pi+2kpi)/3)= cuberoot(3)cis(pi/3), cuberoot(3)cis(pi), cuberoot(3)cis(5pi/3). So perhaps SAGE is giving the princpal root which is then used to get the rest via the formula?

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Note that, for positive number, the positive real root is returned

sage: 3^(1/3).n()

so Sage does not always return a primitive root.

Sébastien gravatar imageSébastien ( 2019-11-24 18:52:05 +0100 )edit

Hmmm. I searched this site and found this. It sounds like nth_root() might be something to remember as long as you keep the base from looking like an integer. That is:


but not


which gives an error.

dazedANDconfused gravatar imagedazedANDconfused ( 2019-11-25 02:30:17 +0100 )edit

Another way you could easily ensure you're getting the negative real root is to just take the negative sqrt of the norm, e.g. n(-sqrt(((-3)^(1/3)).norm())) ensures the result they were looking for.

Iguananaut gravatar imageIguananaut ( 2019-11-25 12:45:47 +0100 )edit

answered 2019-11-22 11:17:10 +0100

FrédéricC gravatar image

updated 2019-11-22 11:17:45 +0100

The equation $x^3=-3$ has three roots in the complex plane.

sage: x=polygen(QQbar,'x')
sage: factor(x**3+3)
(x - 0.7211247851537042? - 1.249024766483407?*I) * (x - 0.7211247851537042? + 1.249024766483407?*I) * (x + 1.442249570307409?)
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Probably OP wants to know why taking the cube root in SageMath yields this particular root and not the real root.

rburing gravatar imagerburing ( 2019-11-22 11:37:07 +0100 )edit

@rburing yes, that's what I meant?

gg gravatar imagegg ( 2019-11-22 12:42:00 +0100 )edit

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Asked: 2019-11-22 09:45:44 +0100

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Last updated: Nov 25 '19