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I've completely confused myself as to determining the domain of polynomial expressions that have rational powers, e.g. $$y = x^{2/3} \qquad \text{or} \qquad y_2 = (x^2-1)^{2/ 3}.$$

A calculus textbook I've consulted asserts that the domain of $y_2$ is all real values of $x$; however, when I plot the function using Grapher or Wolframalpha, it excludes the values for $-1 \leq x \leq 1$:

enter image description here

Similarly, when I try to use Wolfram to evaluate $(-4)^{2/ 3}$ it returns a complex number, whereas I would've expected a real value (because I thought we could think of this as $(-4)^{2/ 3} = \big((-4)^2\big)^{1/ 3} = (16)^{1/ 3} = \sqrt[3]{16}$, which I thought was the same as $\big( (-4)^{1/ 3} \big)^{2} = (-\sqrt[3]{4})^{2} = \sqrt[3]{4}^2$

I consulted this question, which clarified that the property $a^{bc} = (a^b)^c$ only applies to all $a$, if $b$ $c$ are integers (otherwise, we must assume $a > 0$), but I'm still unclear as to how I can determine the domain of these functions, given the disagreement between my textbook and what I'm finding with graphing apps.

Any clarification would be greatly appreciated!

ps. I could not determine which of the "domain" tags was appropriate, I did not see anything like "domain of a function," and most say not to use for this purpose.

Edit

I just consulted a 3rd textbook that includes the following statement:

For all real numbers $a$ for which the indicated roots exist, and for any rational number $m/n$, $$a^{m/n} = (a^{1/n})^m$$.

This prompted me to try graphing the function as $y_2 = \big((x^2-1)^{1 /3}\big)^2$; however, this yielded the same result as before (ie excluding -1 < x < 1).

On Wolfram, I tried y = (cube-root(x^2-1))^2 and this yielded the same graph as the one provided by my textbook.

I must admit, I'm a bit at a loss as to why $y_2 = \big((x^2-1)^{1 /3}\big)^2$ didn't resolve the issue. If that had worked, it would've made (some) sense to me.

Rax Adaam
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2 Answers2

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I believe the issue here is the definition of negative numbers raised to rational powers that's used by Wolfram or your graphing software. A robust definition of exponentials that's used by most software because it generalises to complex numbers is: $$a^b := \exp({b\ln a}),$$ where $\exp x=e^x$, and in the case where $a$ is complex, $\ln$ is the principal logarithm. The issue now, is when you try to evaluate something like $(-1)^{2/3}$. If we use the definition we have $$(-1)^{2/3}=\exp\left(\frac23\ln(-1)\right),$$ which clearly makes no sense in $\mathbb R$ (the $\ln$ of a negative number is undefined). So to answer your question, the reason wolfram thinks $(-1)^{2/3}$ is undefined is because of the robust definition for exponentiation, which gives $(-1)^{2/3}$ as a complex number.

YiFan Tey
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  • Thank you for taking the time to address the core of my question, and so clearly. I really appreciate it and am relieved I'll be able to explain the issue better to my students. Very grateful to you for sharing this, & for your time :) – Rax Adaam Dec 06 '18 at 07:57
  • @RaxAdaam glad it helped! – YiFan Tey Dec 06 '18 at 08:10
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When dealing with fractional powers of negatives, Wolfram Alpha prefers to take the complex branch, so that neither (-1)^(1/3) nor (-1)^(2/3) are considered real.

Of course, ((-1)^2)^(1/3) will yield the expected result, but you can also force the real branch with (cubicroot(-1))^2.

https://www.wolframalpha.com/input/?i=(cubicroot(x%5E2-1))%5E2;(x%5E2-1)%5E(2%2F3)

  • Thank you for taking the time to share the "cubicroot" command; I found an equivalent for Grapher as well. Surprising to me that a site designed for nontechnical users wouldn't have a default warning / explanation about something like that. Anyhow, thank you very much for your time and help :) – Rax Adaam Dec 06 '18 at 07:55
  • @RaxAdaam: Wolfram Alpha is for highly technical users ! Who else cares about elliptical integrals and the resolution of diophantine equations ? It is very advanced in the handling of holomorphic functions, and this must explain the choice of these branches. –  Dec 06 '18 at 08:09
  • Fair enough, but all their on-site advertising (for Wolframalpha) is targeted to students and teachers, and it was my understanding that it was an explicit goal of the site to be more accessible to nontechnical users (hence, e.g. the flexibility of input language that can be understood). But here, I just meant that the decision for this interpretation is not obvious, and seems like it deserves at least some indication / explanation / warning ... – Rax Adaam Dec 07 '18 at 20:44
  • @RaxAdaam: the first sentence on the site says "Compute expert-level answers using Wolfram's breakthrough algorithms, knowledgebase and AI technology". They are up to it. –  Dec 07 '18 at 21:19