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I was trying to draw the function $f(x)=\sqrt[3]{x^2(6-x)}$ by hand (I'm in my first year of engineering; having Calculus I; this drawing is actually an exercise given for my class) and used WolframAlpha to see if I've got it right. Well, I was expecting to miss a few things on my first try, but I didn't even consider the possibility that I would start getting its domain wrong (I thought it was all the real numbers, however, the program said it was all real numbers equal to or below $6$). So, trying to understand what I'd missed, I've concluded that Wolfram must see $\sqrt[3]{-1}$ as only as a complex number when I swore it was at least real (since the equation $x^3 = -1$ has at least a real number that solves it, i.e. $-1$). I mean, if Wolfram is right (and I'm supposing it is), then $f(x=7)$ does not return the "real number" $\sqrt[3]{-49}$ (I don't even know if this number is real anymore, and I'm getting more and more confused whilst writing this text), and my professor has put a wrong graph as an answer for the exercise above. Look, although I wanted the ability to understand what I'm not understanding so that I could be, at least, didactic, I'll try to resume my confusion with this question (and with it, I'll try to understand on my own why Wolfram didn't consider numbers like $x=7$ for $f(x)$ domain): what in heavens is this $\sqrt[3]{-1}$?

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PS: sorry if I've said anything wrong... I don't speak English fluently.

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  • $\begingroup$ Wolfram|Alpha distinguishes between the principal and real-valued root. See the output for the alternate assumption $\endgroup$
    – user170231
    Commented Feb 3, 2021 at 17:06
  • $\begingroup$ Every nonzero (complex) number has three (complex) cubic roots. If you want to talk about "the" cubic root, you need to clarify which cubic root you mean to talk about. Similarly, every nonzero number has $n$ different $n$'th roots. $\endgroup$
    – JMoravitz
    Commented Feb 3, 2021 at 17:17
  • $\begingroup$ Thank you! I didn't know about this distinguishment function. But, if you don't mind me asking, why does the program offer this option? I mean, I would treat "principal" and "real-valued" functions as synonyms at first glance. $\endgroup$
    – Nicholas Yamasaki
    Commented Feb 3, 2021 at 17:17
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    $\begingroup$ Some authors prefer the principal root to be the positive real root if it exists, else the negative real root if it exists and if neither exist then the root who when written in polar form with the angle in the principal branch has the smallest angle. Other authors merely do this as in every case treating the principal root as the root who when written in polar form with the angle in the principal branch has the smallest angle, not dealing with all of those if-then-else statements. $\endgroup$
    – JMoravitz
    Commented Feb 3, 2021 at 17:20
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    $\begingroup$ To some authors, they might have been synonymous. To other authors they very much aren't. $\endgroup$
    – JMoravitz
    Commented Feb 3, 2021 at 17:20

3 Answers 3

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When working in the real numbers, $\sqrt[3]{x}$ denotes the unique number $a$ such that $a^3=x$. Since $(-1)^3=-1$, we have $\sqrt[3]{-1}=-1$.

When working in the complex numbers, the notation $\sqrt[3]{z}$ is imprecise. Every complex number has three cube roots, and there is no natural way of defining the 'principal' cube root of a complex number. It is sensible, however, to define $\sqrt[3]{z}$ in the complex numbers if $z$ also happens to be a real number. Then, $\sqrt[3]{z}$ can be defined in the way above.

That being said, sometimes we define the 'principal' $n$-th root of a complex number in the following way. If $z=re^{i\theta}$, where $\theta$ is the principal argument of $z$*, and $r$ is the magnitude of $z$, then $$ \sqrt[n]{z} = \sqrt[n]{r} \cdot e^{i\theta/n} \, , $$ where $\sqrt[n]{r}$ denotes the principal real root of $r$. This means that $z \mapsto \sqrt[n]{z}$ is not a continuous function, but we still might choose to adopt the notation $\sqrt[n]{z}$ as a matter of convenience. This is probably the root that Wolfram Alpha was referring to: $$ \sqrt[3]{-1} = \sqrt[3]{1e^{i\pi}} = \sqrt[3]{1} \cdot e^{(i\pi)/3} = e^{(i\pi)/3} \approx 0.5 + 0.866i \, . $$

*Again, the 'principal' argument of a complex number involves a branch cut, where we require that $\theta \in (-\pi,\pi]$. Again, this does not define a continuous function.

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    $\begingroup$ It is common also to take the principal branch cut to be instead $[0,2\pi)$ rather than $(-\pi,\pi]$. The end result is that, it all depends on what author it is and what conventions they follow. $\endgroup$
    – JMoravitz
    Commented Feb 3, 2021 at 17:40
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    $\begingroup$ @JMoravitz Absolutely. I prefer the convention $\theta \in (-\pi,\pi]$, since that means that $n$-th roots of complex numbers are in the right half-plane, but in a way all of these branch cuts are equally arbitrary. $\endgroup$
    – Joe
    Commented Feb 3, 2021 at 17:43
  • $\begingroup$ @JMoravitz So... all this sense of confusion and misunderstanding that I've felt were created because there are people who talk about the same things but they choose to do it so differently for eccentric reasons? $\endgroup$
    – Nicholas Yamasaki
    Commented Feb 3, 2021 at 17:51
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    $\begingroup$ @NicholasYamasaki for eccentric reasons? No. The different reasons are perfectly justified. Definitions are written in such a way to be useful for the problems they are meant to solve. Some problems are more easily solved with one set of definitions while other problems are more easily solved with a competing set of definitions. Recall... the problems we want to solve come first. The definitions are crafted in such a way as to be useful to us. $\endgroup$
    – JMoravitz
    Commented Feb 3, 2021 at 17:54
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    $\begingroup$ @NicholasYamasaki Definitions are for everyone, and are vital for people new to a subject. Without defining what $\sin \theta$ means, for instance, it's just a random scribble on a page. $\endgroup$
    – Joe
    Commented Feb 3, 2021 at 18:31
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Wolfram Alpha is slightly eccentric in this regard. The cube root function is a perfectly respectable function from $\Bbb R$ to $\Bbb R$, but the default behaviour of Wolfram Alpha is to evaluate cube roots of negative reals as complex numbers.

However, you can override this behaviour, by clicking on "Use the real-valued root instead". Then you can have Wolfram Alpha plot the function for any range you like.

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  • $\begingroup$ I'm still trying to understand this "evaluate cube roots of negative reals as complex numbers" function after reading some comments. Should I be concerned for not knowing what this option is about? $\endgroup$
    – Nicholas Yamasaki
    Commented Feb 3, 2021 at 17:34
  • $\begingroup$ @NicholasYamasaki Are you familiar with $n$-th roots of complex numbers in general? If not, then it will be hard to clarify things. $\endgroup$
    – Joe
    Commented Feb 3, 2021 at 17:36
  • $\begingroup$ @Joe Yes, I am. But only what I've seen at high school $\endgroup$
    – Nicholas Yamasaki
    Commented Feb 3, 2021 at 17:37
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    $\begingroup$ @NicholasYamasaki When working in the real numbers, each nonnegative number $x$ has two square roots $a$ and $-a$. This means that $\sqrt{x}$ could be $a$ or $\sqrt{x}$ could be $-a$. However, to turn $\sqrt{x}$ into a continuous function, then we could require that $\sqrt{x}$ is nonnegative. In the complex numbers, $x$ has two square roots also, but in this case, there is no way of defining $\sqrt{x}$ so that it is a continuous function. You might have seen people that $i$ is the square root of $-1$, but this is misleading: really, $i$ is a square root of $-1$. $\endgroup$
    – Joe
    Commented Feb 3, 2021 at 17:52
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    $\begingroup$ @NicholasYamasaki The only logical thing to write is $i^2 = -1$, not $i=\sqrt{-1}$. It's the same story with cube roots. Every complex number $z$ has $3$ cube roots, but there is no 'nice' way of designating a number as the cube root of $z$. $\endgroup$
    – Joe
    Commented Feb 3, 2021 at 17:54
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You are completely right. $x^3=a$, with $a<0$ has a real root, so the problem must lay in the way you are writing it in Mathematica. Try using the function CubeRoot, and see if it works. If not, I would advise you to share your code in the question.

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