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Consider the sequence $x_{k+1}=\frac{1}{2}(x_k+\frac{a}{x_k}), a\gt 0, x\in\mathbb{R}$, Assume the sequence converges, what does it converge to?

I'm having trouble seeing how to start,

Any help would be appreciated

Thanks

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3 Answers 3

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The number $x$ to which it converges should satisfy $x =\frac12\left(x+\frac a x\right)$. Solve that equation for $x$. (It may help to begin by multiplying both sides by $x$, thereby getting rid of the fraction.)

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If $x_n \to g$, then, if $g \ne 0$, we have $\frac{1}{2}(x_n + \frac{a}{x_n}) \to \frac{1}{2}(g+\frac{a}{g})$. We also have $x_{n+1} \to g$, so since $x_{n+1} = \frac{1}{2}(x_n + \frac{a}{x_n})$, we have that $g = \frac{1}{2}(g + \frac{a}{g})$.

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This sequence has a closed form. First, given $x_n$, you have $$x_{n+1}-\sqrt{a}=\frac{1}{2}\left(x+ \frac{a}{x} \right) - \sqrt{a}=\frac{1}{2}\left( \sqrt{x} - \frac{\sqrt{a}}{\sqrt{x}}\right)^2 > 0$$ Hence $x_n>\sqrt{a}$ for all $n>0$. To simplify a bit, we can safely assume that $x_0 > \sqrt{a}$ too.

You can thus write $x_n = \sqrt{a} \ \coth{t_n}$. Then $$x_{n+1} = \frac{1}{2}\left( \sqrt{a} \ \coth{t_n} + \frac{a}{\sqrt{a} \ \coth{t_n}} \right) = \frac{1}{2}\sqrt{a} \left( \coth{t_n} + \mathrm{th}\ t_n\right) = \sqrt{a} \ \coth \ 2 t_n$$ We have thus $t_{n+1}=2\ t_n$, and $t_n= 2^n t_0$, then for $x_n$ : $$x_n=\sqrt{a} \coth \left( 2^n \arg \coth \frac{x_0}{\sqrt{a}}\right)$$ And since value inside parentheses tends toward infinity, $\lim(x_n)=\sqrt{a}$.

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