Big Oh Notation Explained: What Does x = x_0 + \mathcal{O}(y) Mean?

[AxiomOfChoice]

Can someone please explain what it means to say something like
$$x = x_0 + \mathcal{O}(y)$$
?

 

[arildno]

Nothing.

You must include in your expression a "as x goes to.."

Without that, it is meaningless.

 

[g_edgar]

$\cos x = 1 + O(x^2)$ as $x \to 0$ means:
$$\frac{\cos x - 1}{x^2}$$
is bounded in some neighborhood of $0$ .