$\displaystyle x_1+2/{x_1}=2x_2$

$\displaystyle x_2+2/{x_2}=2x_3$

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$\displaystyle x_n+2/{x_n}=2x_1$

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

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- Sep 11th 2009, 01:42 PMgabor7896How to find roots of the equation system?
$\displaystyle x_1+2/{x_1}=2x_2$

$\displaystyle x_2+2/{x_2}=2x_3$

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.

.

$\displaystyle x_n+2/{x_n}=2x_1$

x is real number. How can I solve this equation system? - Sep 11th 2009, 11:51 PMrunning-gag
Hi

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

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

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

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

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

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

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

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

The same for $\displaystyle -\sqrt{2}$