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I am a bit confused. I think there must be a mistake.

In a text I read:

The entropy is $2\ln p$, where $$p=\frac{1}{3}\left(\sqrt[3]{\frac{29+9\sqrt{31/3}}{2}}+\sqrt[3]{\frac{29-9\sqrt{31/3}}{2}}+1\right)$$ is the positive root of $x^3-x^3-1$.

But $x^3-x^3-1=-1$... I think that must be a typo? From what is $p$ the positive root?

Salamo
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1 Answers1

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There is no positive root, or any root at all, of $x^3-x^3-1=0$, since, as you noted, $x^3-x^3-1=-1$ for all $x$, and there is no $x$ that satisfies $-1=0$. This is definitely a typo.

The positive root of $$x^3-x^2-1=0$$

is exactly

$$\frac{1}{3}\left(\sqrt[3]{\frac{29+9\sqrt{31/3}}{2}}+\sqrt[3]{\frac{29-9\sqrt{31/3}}{2}}+1\right)$$

so that is almost certainly the intended polynomial.

Zubin Mukerjee
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  • Thanks. But why is it mentioned that this root is the positive one. Because there is only one and it is positive. – Salamo Jan 03 '15 at 11:01
  • @Salamo It is the positive root. It is also the only root. Authors can be redundant sometimes. – Zubin Mukerjee Jan 03 '15 at 11:02
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    Haha, okay. :-) Thanks again. – Salamo Jan 03 '15 at 11:02
  • The equation has three roots, of which just one is positive (the other two being complex). Still, it would be more natural to describe this root as the real root. Perhaps the author had in mind the need for the argument of the logarithmic function to be positive, and wanted to emphasize that it was indeed so. – John Bentin Jan 03 '15 at 11:58