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So I ran into this contradiction and I'd appreciate if someone could help explain what is happening: So when i perform the following integral on paper:

$$\int_{-1}^1(\sqrt[3]{x} -2)\;dx \; = [\frac 34 x^{\frac 43} \, -2x]_{-1}^1\; =$$

$$=\frac 34 1^{\frac 43} -2 -(\frac 34 (-1)^{\frac 43} +2)= \frac 34 - 2 -\frac 34 -2 = -4$$

I get -4 as an answer, and when i input the same thing into WolframAlpha, the result is a complex number (-2.875 +0.649* i), as can be seen in the following photo. Can anyone explain this discrepancy? Does it have to do with the computer using approximations?

https://i.stack.imgur.com/5bOYl.png

Harry Peter
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user232440
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3 Answers3

2

cuberoot has three possible answers. WolframAlpha (and many other software) picks the one with the smallest argument. If you tried to find the cuberoot of -1, it would give you 0.5 +0.866 i

http://www.wolframalpha.com/input/?i=%28-1%29%5E%281%2F3%29

In the future, if you want to avoid this, use sign(x)(|x|)^(1/3) (for real x)

Gappy Hilmore
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There are three complex cubic roots of $-1$. In some sense the most natural one is $e^{i\pi/3}= 1/2 + i \sqrt{3}/2$, the so-called principal one, and Wolfram Alpha uses this.

This is because one can express every non-zero complex number and thus also $-1$ as $r e^{i \theta}$ with some positive real $r$ and $\theta \in (-\pi , \pi]$, called the polar form of the complex number. To determine a cubic root, one can compute $\sqrt[3]{r}e^{i \theta/3}$. As $-1= 1 e^{i \pi}$ one gets the result above.

One gets also $-1$ in this form, as to get all the cubic roots one would consider $\sqrt[3]{r}e^{i \theta/3}$, $\sqrt[3]{r}e^{i (\theta+ 2 \pi)/3}$, $\sqrt[3]{r}e^{i (\theta+ 4 \pi)/3}$.

For $\theta = \pi$, so our case, the middle one is $-1$. But looked at it like this it is a less natural choice.

See a recent related question for further discussion What is the principal cubic root of $-8$?

Community
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quid
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0

This does not have to do with approximations, but rather the value of $\sqrt[3]{-1}$.

Wolfram Alpha is using $\sqrt[3]{-1}=e^{\pi i/3}=\frac{1}{2}+\frac{\sqrt{3}}{2}i$

Peter Woolfitt
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  • Thanks i appreciate the answer! I've just finished high-school so forgive me for the ignorant question, but would you care to elaborate on what the several values of -1 are? and why are they valid in this question, since its $(-1)^{\frac 43}$ so shouldn't the $-1^4 = 1$ then we take the cubic root of 1? – user232440 Jun 01 '15 at 23:32
  • @user232440 Well, this still works because it turns out that there are multiple cube roots of $1$ as well :P. Being the cube root of $-1$ means satisfying the equation $x^3=-1$. Notice that other than $-1$, there are two (complex) solutions to this equation. – Peter Woolfitt Jun 01 '15 at 23:40