I have experienced an issue while evaluating $$(x)^\frac{1}{5}-x = 0$$ The values that the program gives are 0 and 1, however, there is an additional solution, which is -1. In addition, the corrispondent graph is wrong because it excludes values of x for negative numbers. I have also noticed that when I try to evaluate: $$(x)^\frac{1}{5}=-1$$ Wolfram Alpha gives no solutions, instead the solution should be -1 (over the reals). I believe that there is a general problem in the program that comes up when trying to evaluate $$(x)^\frac{1}{n} , x<0, n = 2n+1, n = {1,2,3..}$$


closed as off-topic by Adam Hughes, Matthew Conroy, E. Joseph, Willie Wong, user91500 Nov 21 '16 at 5:30

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." – Adam Hughes, Matthew Conroy, E. Joseph, Willie Wong, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ It might be more productive to email Wolfram, we cannot change how it does its thing. $\endgroup$ – Adam Hughes Nov 20 '16 at 21:53
  • $\begingroup$ I don't see any question, here. $\endgroup$ – David Richerby Nov 21 '16 at 0:57
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    $\begingroup$ This isn't a question about mathematics. $\endgroup$ – The Great Duck Nov 21 '16 at 1:16
  • $\begingroup$ @TheGreatDuck There is a legitimate question in between the lines about wolfram-alpha, which is a valid MSE tag. To the OP: you may consider rephrasing the (mistaken) statement about "wrong answer from WA" into a question about "where is the -1 root I expected". $\endgroup$ – dxiv Nov 21 '16 at 1:35

Mind the text right under the equation:

Assuming the principal root  |  Use the real‐valued root instead

Click the real-valued root and you get the $3$ real roots $\,\{-1,0,1\}\,$, and the graph over the entire $\mathbb{R}$.

[ EDIT ]  To clarify what happens under the default Wolfram Alpha interpretation of "assuming the principal root", the equation is taken to be in complex numbers, and $x^{\frac{1}{5}}$ is assumed to be the principal value of the $5^{th}$-root complex power function. With the usual choices for the branch cut along the negative real axis and the principal root being the one with the minimum argument, the principal value of $(-1)^{\frac{1}{5}}$ is $w = \text{cis}\;\frac{\pi}{5}$. This value does not satisfy the equality $w - w^5 = 0$ and therefore WA does not return it as a root of $x^{\frac{1}{5}} - x=0$ under the "principal root" interpretation.

For related discussion and insights see for example What is the principal cubic root of −8?.

  • $\begingroup$ This answer could be improved by explaining what is the principal root, which in this case would be $\cos \frac{pi}{5} + i\sin \frac{pi}{5}$, I believe? $\endgroup$ – Pedro A Nov 20 '16 at 23:06
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    $\begingroup$ @Hamsteriffic Right. I edited some more into the answer. $\endgroup$ – dxiv Nov 20 '16 at 23:35
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    $\begingroup$ Very informative! +1 $\endgroup$ – Pedro A Nov 20 '16 at 23:43
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    $\begingroup$ @TheGreatDuck Are you serious?? There was the very obvious implied question "why doesn't WA return -1 as a root" which I replied to. A few fellow users even found the answer useful, apparently. I'd be curious to understand what you think the advantage would have been for MSE if I hadn't posted any answer. $\endgroup$ – dxiv Nov 21 '16 at 1:30
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    $\begingroup$ @TheGreatDuck The posted question is one word twist away from being a formally legitimate question. If that technicality bothers you so much, then I guess we'll just have to agree to disagree. $\endgroup$ – dxiv Nov 21 '16 at 1:37