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Let be $ x_{1}, x_{2}, x_{3}$ the roots of $x^3-x^2-1=0$. If $x_{1}$ is the real root of the equation, then calculate:

$$\lim_{n\to\infty} ({x_{2}}^n+{x_{3}}^n)$$

I wonder if this limit can be computed without being necessary to know the exact values of $x_{2}$ and $x_{3}$.

1 Answers 1

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We will prove that the magnitude of the complex roots is less than $1$. Let $f(x) = x^3-x^2-1$. Then $f(1) = -1 <0$ and $f(2) = 3>0$. Hence, there is a real root, $x_1$, in the interval $(1,2)$.

You can prove that this is the only real root and the other two have to be complex. This can be done by looking at the derivative in the interval $(1,2)$ which will turn out to be positive and a local maximum occurs at $x=0$ and a local minimum occurs at $x= \dfrac23$. These guarantee that the remaining two roots are complex.

The complex roots $x_2$ and $x_3$ can be written as $x_2 = r e^{i \theta}$ and $x_3 = r e^{- i \theta}$ since the complex roots occur in conjugate pairs.

But the product of the roots is $1$ i.e. $r^2 x_1 = 1 \implies r^2 = \dfrac1{x_1} < 1$.

The complex roots must have magnitude less than $1$. Hence, $$x_2^n + x_3^n = 2r^n \cos(n \theta)$$ Hence, $$\lim_{n \to \infty}\left( x_2^n + x_3^n \right) = \lim_{n \to \infty} 2r^n \cos(n \theta) =0 \text{ (Since $0

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    that's what i need! Thanks!2012-06-24
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    however, it's a bit tricky. I should have looked at this point at first.2012-06-24
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    Your complex roots will be complex conjugates so $x_2=r e^ {i\theta} \text{ and } x_3=r e^ {-i\theta} \text { for the appropriate }r \text{ and }\theta \text {, and we have: }$ $$x_2^n + x_3^n = r^n e^ {in\theta}+r^n e^ {-in\theta} = 2r^n \cos n\theta$$2012-06-24
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    @Mark Bennet: right. Marvis emphasized the same fact in his solution. Thanks.2012-06-24
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    @Marvis: for the complex roots we may also resort to the fact that $x_{1}^2+x_{2}^2+x_{3}^2=1$, i think.2012-06-24
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    @Chris Ah. nice one. I didn't see that.2012-06-24
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    How does the fact that $x_1^2 + x_2^2 + x_3^2 = 1$ help?2012-06-24
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    @RobertIsrael Chis comment was made to simplify my argument to show that the remaining two roots are complex. $x_1^2 + x^2 + x_3^2 = 1$ and I showed that $x_1 > 1$ and hence these imply $x_2^2 + x_3^2 < 0$.2012-06-24
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    You could also use Descartes' Rule of Signs to show that there's one positive real root and no negative real root.2012-06-24
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    @Robert Israel: that point was meant to be used instead of the derivative and bring the proof at an elementary level.2012-06-24