Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


share|cite|improve this question
up vote 2 down vote accepted

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.)

share|cite|improve this answer

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 $g = \frac{1}{2}(g + \frac{a}{g})$.

share|cite|improve this answer

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, $\lim(x_n)=\sqrt{a}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.