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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 $x_{3}$.

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up vote 7 down vote accepted

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, $$\lim_{n \to \infty}\left( x_2^n + x_3^n \right) = \lim_{n \to \infty} 2r^n \cos(n \theta) =0 \text{ (Since $0<r < 1$)}$$

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that's what i need! Thanks! – user 1618033 Jun 24 '12 at 7:43
however, it's a bit tricky. I should have looked at this point at first. – user 1618033 Jun 24 '12 at 7:50
Your complex roots will be complex conjugates so $x_2=r e^ {i\theta} \text{ and } x_3=r e^ {-i\theta} \text { for the appropriate }r \text{ and }\theta \text {, and we have: }$ $$x_2^n + x_3^n = r^n e^ {in\theta}+r^n e^ {-in\theta} = 2r^n \cos n\theta$$ – Mark Bennet Jun 24 '12 at 7:51
@Mark Bennet: right. Marvis emphasized the same fact in his solution. Thanks. – user 1618033 Jun 24 '12 at 7:53
@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. – user 1618033 Jun 24 '12 at 8:04

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