Equations in Landau/ Big O Notation

Question

$ \underbrace{x^2 +x + O(\epsilon^2) = 0}_{\mathrm{equation \ 1}} \implies \underbrace{x^2 + x = O(\epsilon^2)}_{\mathrm{equation \ 2}} $

I'm asked why there isn't any need to write $-O(\epsilon^2)$ for the second equation. All I can think of is "Because it conveys the same information", but to answer it properly, I'd like to clarify.

I know that if $ f(\epsilon) = O(g(\epsilon))$ then f is asymptotically small to some constant multiple of g in the limit $\epsilon \to \infty$ . Also the "$=$" is abuse of notation, so I read it as "is", so "f is big Oh of g", as "$=$" isn't used as a symmetric notation.

• what is meant by $+O(\epsilon^2)$ in equation 1, because in practice I assume it to be an indicator of ignoring terms of order $\epsilon^2$ and higher, but how is it explained in terms of the equation? For equation 2 I can read it as $x^2+x $ is asymptotically small compared to $A\epsilon^2$ in the limit $\epsilon \to 0$, but my inability to "read" equation 1 means I can't sufficiently answer the question.

Answer 1

As you correctly said, $O(f(x))$ is a whole class of functions, and A = O(B) actually means A \in O(B). So actually your first equations says $-(x^2+x) \in O(\epsilon^2)$.

But this implies $x^2 + x \in O(\epsilon^2)$ and therefore $x^2 + x = O(\epsilon ^2)$.

Answer 2

like what @flawr said:

In fact $O(\epsilon^n)$ shows the error, the positive error and negative ones. So no need to write $-O(\epsilon^n)$, except in a specific condition that you know that your error is always positive (or negative), that means you ensure it doesn't change its sign. In those case you may write $O^+(\epsilon^n)$ (or $O^-(\epsilon^n)$), and in that case actually you must write $-O^+(\epsilon^n)$ (or $-O^-(\epsilon^n)$) in other side of equation. Those ones often occur when you deal with monotone functions.