How to find roots of the equation system?

**gabor7896** · September 11th 2009, 01:42 PM

$x_1+2/{x_1}=2x_2$
$x_2+2/{x_2}=2x_3$
.
.
.
$x_n+2/{x_n}=2x_1$

x is real number. How can I solve this equation system?

**running-gag** · September 11th 2009, 11:51 PM

Originally Posted by gabor7896

$x_1+2/{x_1}=2x_2$
$x_2+2/{x_2}=2x_3$
.
.
.
$x_n+2/{x_n}=2x_1$

x is real number. How can I solve this equation system?

Hi

Let $(u_n)_n$ the sequence defined by the value of $u_1$ and $u_{n+1} = \frac{u_n}{2} + \frac{1}{u_n} = f\left(u_n\right)$ where $f(x) = \frac{x}{2} + \frac{1}{x}$

I am just considering the restriction of f over $]0,+\infty[$ since it is similar over $]-\infty,0[$ . You can show that f is decreasing up to $\sqrt{2}$ and increasing after $\sqrt{2}$ and $f\left(\sqrt{2}\right) = \sqrt{2}$

Moreover $f\left(]0,\sqrt{2}]\right) \subset [\sqrt{2},+\infty[$ and $f\left([\sqrt{2},+\infty[\right) \subset [\sqrt{2},+\infty[$

This means that at least from n=2 $u_n \in [\sqrt{2},+\infty[$

You can show by induction that $\forall n ~ u_n > 0$

Then $u_{n+1}-u_n = \frac{2-u_n^2}{2u_n} < 0$

$(u_n)_n$ is decreasing (at least from n=2) and higher than $\sqrt{2}$ therefore converges towards the solution of f(x) = x which is $\sqrt{2}$

The only way to get a stationary sequence is $x_1 = x_2 = ... = x_n = \sqrt{2}$

The same for $-\sqrt{2}$

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