What happens with repeated transformation of $T(x) =x -1/x$? I tested repeated application of $x = x -1/x$ on my computer and could not decide what the long term behavior is.  I iterated up to $500,000$ times and saw variation over a fairly short range from around $-4000$ to $4000$ for starting numbers up to $2,000$.  I realize that limits in the precision of the computer may be a factor. In general, what can be said about the transformation?  Is it an example of chaos?
Just a few observations

Consider the graph $\displaystyle f(x)=x-\frac{1}{x}$. This can be rewritten as $yx=x^2-1$. Since this the discriminant of this conic is positive, we conclude that this graph is a hyperbola. We can find the asymptotes of this hyperbola by considering very large $x$ or very large $y$. In both cases, the $-1$ becomes negligible, so our equation beomes $yx=x^2$, which means either $y=x$ or $x=0$. These two lines are the asymptotes of the hyperbola. For $(0,\infty)$, the graph of the hyperbola is below the line $y=x$. For $(-\infty,0)$, the graph of the hyperbola is above the line $y=x$. So, for any positive number, $x$, we have $\displaystyle x-\frac{1}{x}&lt;x$. Similarly, for any negative number, $x$, we have $\displaystyle x-\frac{1}{x}&gt;x$. The function $f(x)=x-\frac{1}{x}$ is increasing on $\mathbb{R}/\{0\}$. In other words, $0&lt;x_1&lt;x_2 \implies f(x_1)&lt;f(x_2)$ and $0&gt;x_1&gt;x_2 \implies f(x_1)&gt;f(x_2)$. 

The graph $f(f(x))$ has $4$ asymptotes: $y=x$, $x=1$, $x=-1$, $x=0$ 

The graph $f(f(f(x)))$ has $8$ asymptotes: $y=x$, $x=1$, $x=-1$, $x=0$, $x=\frac{1+\sqrt{5}}{2}$, $x=-\frac{1+\sqrt{5}}{2}$, $x=\frac{1-\sqrt{5}}{2}$, $x=\frac{\sqrt{5}-1}{2}$

In general, $f^n(x)$ has $2^n$ assymptotes. 

Let's start with a positive value $n$. We know that $f(n)&lt;n$, so if we keep applying the function, $n$ will keep decreasing, until it turns negative. Once it turns negative, if we keep applying the function, $n$ will increase until it is positive, and so on and so forth.
If this transformation converge for some starting point $x$ we would have:

$T(x) = x \rightarrow x - \frac{1}{x} = x \rightarrow \frac{1}{x} = 0$

But then $x = \pm \infty$.

Thus this transformation must diverge or start oscillating.
Easy oscillation that comes to mind is $\pm x$. Then:

$T(x) = -x \rightarrow x - \frac{1}{x} = -x \rightarrow \frac{1}{x} = 2x$

Leaving us with $x = \pm \frac{1}{2} \sqrt{2}$.

Question now remains: for what starting points will we "converge" to this 2 step oscillation and for what starting points will we diverge?
Shortly playing around with it makes it look like only way to oscillate is to start with $\frac{1}{2} \sqrt{2}$ but proof of it hasn't arrived in my head yet.

Another point one might want to address: Are there any other oscillations? So far I only handled the case $T(x) = x$ and partially $T(T(x)) = x$, but what about $T^n(x) = x$? Do those cycle exists?

Edit:
Playing around with it some more I found that numerical errors can kick you out of a cycle quite easily, and you are then put on a magical ride.
Note: $T'(x) = 1 + \frac{1}{x^2}$ is stricly larger than 1 on R so no point is a stable point, meaning that if you make some numerical error (machine precision for example), you will not converge back to the oscilation you were previously on.