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I wonder if anyone can solve the following problem for me:

The sequence of real numbers $x_n$ is defined inductively by $$ x_1=4 \quad\text{and}\quad x_{n+1}=\frac4{x_n}+\frac{x_n}2. $$ Show that $x_n$ converges and find its limit

Thanks to every one contribute in the solution.

LoveMath
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1 Answers1

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Look at the difference $x_{n+1}-x_n$: $$x_{n+1}-x_n=\frac{8-x_n^2}{2x_n}$$ If you can show that $$x_{n+1}-x_n\le 0$$ for all $n$ then you know that $x_n$ is monotonic. If you can further show that $x_n$ is bounded (i.e. has a lower limit), then you have proved convergence. Since $x_n$ is always positive it is sufficient to consider the difference $x_n^2-8$: $$x_n^2-8 = \left( \frac{4}{x_{n-1}} + \frac{x_{n-1}}{2} \right)^2 - 8 = \left( \frac{4}{x_{n-1}} - \frac{x_{n-1}}{2} \right)^2 \ge 0$$ This proves two things at once:

  1. $x_{n+1}-x_n \le 0,\quad$ i.e. $x_n$ is monotonically decreasing

  2. $x_n \ge \sqrt{8}=2\sqrt{2},\quad$ i.e. $x_n$ has a lower limit

Now you can simply find the limit $c=\lim\limits_{n\to\infty}x_n$ by solving

$$c = \frac{4}{c} + \frac{c}{2}$$ which gives $c=2\sqrt{2}$.

Thomas Andrews
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Matt L.
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  • Matt: Thanks very much Matt, but could you tell me how is x_n^2-8 = \left( \frac{4}{x_{n-1}} + \frac{x_{n-1}}{2} \right)^2 - 8 = \left( \frac{4}{x_{n-1}} - \frac{x_{n-1}}{2} \right)^2 and how we know that \left( \frac{4}{x_{n-1}} - \frac{x_{n-1}}{2} \right)^2 \ge 0 – LoveMath Apr 12 '13 at 10:41
  • The equality is of the form $(a+b)^2-4ab = (a^2+2ab+b^2)-4ab = (a^2-2ab+b^2) = (a-b)^2$, and since we're dealing with real numbers, the squared expression $(\cdot)^2$ must always be non-negative. – Matt L. Apr 12 '13 at 11:38
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    Rewrite this as $$x_{n+1} = x_n - \frac{x_n^2-8}{2 x_n}$$ Then we see immediately that this is the recurrence generated by Newton's method for the roots of $$x^2-8=0.$$ With initial value $x_0=4$ the convergence goes to the positive root $2\sqrt{2}.$ – Marko Riedel Apr 12 '13 at 20:54