Consider the following iteration procedure. Does it converge for any nonzero initial point? If not, why? If so, to what value? Consider the following iteration procedure. Does it converge for any nonzero initial point? If not, why? If so, to what value?
$x_{n+1} = \frac{1}{2} x_n + \frac{1}{x_n}$ Can anyone point me in the right direction on how to solve for convergence for this problem? I went to the Wiki page it was kind of vague on regards of how to actually implement the convergence analysis though. 
 A: Hint : As long as you don't get a fixed point, you can show that this type converges.  The only fixed points are $x = 0$ (not defined) and $x = \pm \sqrt 2$.
For initial value as the fixed points, the sequence is convergent. For all other initial values, if you start with a positive initial value, it converges to $\sqrt 2$, and for a negative initial value it converges to $-\sqrt 2$. 
One neat trick is to assume it converges, and then put $y = x_{n+1} = x_{n}$ and solve for y. This gives you an idea of what is happening. Then you can reason out rigorously.
A: I would go with turning it into a Newton's Method problem.
Newton's Method is used to solve for $f(x)=0$: $$x_{n+1} = x - {f(x)\over f'(x)}$$
I would look for a function $f$ such that
$$x-{f(x)\over f'(x)} = {1\over2}\,x_n + {1\over x_n}$$ or
$${f(x)\over f'(x)} = {-1\over2}\,x_n + {1\over x_n}$$
This differential equation can apparently be solved using separation, with the solution $f(x)=C(x^2-2)$ for an arbitrary constant $C$. Now look up when Newton's Method converges, and apply it to this function.
A: Clearly, the sign of all $x_n$ is the same as the sign of the initial value $x_0$, so it is sufficient to look at the positive $x_0$ only.
Consider the RHS function $f(x)=\frac{1}{2}x+\frac{1}{x}$, $x>0$.


*

*$\min_{x>0}f(x)=\sqrt{2}$ attained at $x=\sqrt{2}$ $\qquad\Rightarrow\qquad$ $x_n\ge\sqrt{2}$, $n\ge 1$.

*$f(x)\le x$ for $x\ge\sqrt{2}$ $\quad\Rightarrow\quad$ $x_{n+1}\le x_n$, $n\ge 1$.


The monotonic sequence $x_n$ has a limit $a\in[\sqrt{2},x_1]$, thus converges. To find the value, set $x_n=a$ to the recursion and solve for $a$ to get $a=\sqrt{2}$.
P.S. Similarly for negative $x_0$.
