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Many of my students have asked me this simple question:

"Why does not Maple plot $y$=$x^{1/3}$ properly?"

I have examined it by myself. Also, I ploted the absolute value of $y$ and think that, it is a defect in this maths tool. Is there any suggestion? Thanks.

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closed as off-topic by Jonas Meyer, Claude Leibovici, John, Najib Idrissi, Dirk Mar 26 at 8:59

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – Jonas Meyer, Claude Leibovici, John, Najib Idrissi
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How does it plot it? Can you include an image? –  ShreevatsaR Jun 22 '11 at 14:06
General remarks, without knowing the issue here: An accurate approach to plotting relations is interval arithmetic with successive refinement, as exemplified by the excellent program GrafEq previously mentioned here, here, here. Many other programs get graphs wrong. Unfortunately GrafEq is Windows-only, and I don't know other software using this approach. –  ShreevatsaR Jun 22 '11 at 14:12
@shreevatsar: Yes! I think such this program shows some plots wrong. As Gortaur said below, Maple does not plot it for $x$<0. Thanks for your advice. –  Babak S. Jun 22 '11 at 15:46
Actually, I wouldn't say Maple is making a mistake for this particular plot; it's just using a different convention. But there are probably other plots which have subtle mistakes. :-) –  ShreevatsaR Jun 22 '11 at 19:41
+1 nice question –  Software Mar 24 '13 at 20:36

4 Answers 4

up vote 5 down vote accepted

What to you mean with a "proper" plot? As I can guess the problem is that Maple does not plot it for $x<0$. Usually $x^\alpha$ for $\alpha\notin \mathbb{N}$ is defined only for non-negative $x$.

I mean, that if we allow to calculate $(-1)^{1/3}$ then $$ -1 = (-1)^{1/3} = (-1)^{2/6} = (1)^{1/6}=1. $$

On the other hand, sometimes $x^{1/n}$ for an odd $n$ can be defined as a real root of $y^n = x$ - so this is a matter of convention.

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Thanks for explanation. Good hints. :) –  Babak S. Jun 22 '11 at 15:49

According to Maple, $x^{1/3}$ is a non-real complex number when $x<0$.
evalf((-1)^(1/3)) yields .5000000001+.8660254037*I

If you want to force the real cube root, use surd ... type ?surd for information.

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Illustrating practical example! Thanks. –  Babak S. Jun 22 '11 at 15:55
It is a Frequently Asked Question: maplesoft.com/support/faqs/detail.aspx?sid=32703&cid=358 or, see: maplesoft.com/support/help/Maple/view.aspx?path=root and other places in the Maple help system. –  acer Jun 22 '11 at 19:14

I suspect Maple, like many other computer algebra systems, decides that $a^b$ for negative real $a$ and real $b$ to be something like $|a|^b \; e^{i \pi b}$.

This makes a lot of sense as a convention (it is continuous in $b$), even if it might be unexpected for $k$ as the reciprocal of a positive integer.

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If $z < 0$, Maple is using the principal branch of the log along with the identity $$z^{1/3} = \exp((1/3)*\log(z)).$$
If you use $z = -1$, you get $$z^{1/3} = \exp((1/3)*\log(z)) = \exp(1/3*\pi) = \exp(i\pi/3) = {1 + \sqrt{3}\over 2}.$$ In a word, Maple is being "scrupulous to a fault." It's a small price to pay for the program being so complex-number savvy.

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I presume that $z^{1/)}$ should be $z^{1/3}$, not something else. –  Peter Taylor Jun 22 '11 at 16:11

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