Help in understanding properties of big O notation with norms of vectors In Fast Exact Multiplication by the Hessian equation 1,
$O(\left\Vert\Delta w\right\Vert^2)$ gets taken from RHS to LHS and $\Delta w$ is substituted as $rv$ where r is small scalar and v is a vector. I understand that $O(\left\Vert rv\right\Vert^2) = O(r^2\left\Vert v\right\Vert^2)$. But what I don't get is how did the sign of $O$ not become negative when it was taken to LHS. And why did $\left\Vert v\right\Vert^2 $ disappear form $O$. Is it because as $r$ is tiny it will govern the $O$ term and $\left\Vert v\right\Vert^2$ won't matter?
 A: $O(f(x))$ refers to a quantity that's bounded (in the limit) by some multiple of $f(x)$; it properly (IMHO) represents a set. That is, to say that $g(x)\in O(f(x))$ means that there's some constant C such that for all sufficiently large (or sufficiently small, depending on the direction of the limit) $x$, $|g(x)|\lt C|f(x)|$. But this means in particular that $O()$ is 'sign-agnostic'; if $g(x)\in O(f(x))$, then $cg(x)\in O(f(x))$ for all constants $c$, positive or negative. This same fact is also what 'removes' the dependency on $\left\Vert v\right\Vert^2$; because the limit that's implicit in $O()$ is being taken with respect to $r$ and $v$ doesn't depend on $r$, the quantity $\left\Vert v\right\Vert^2$ is 'constant' and can be absorbed into the constant in $O()$.
A: When we write $x = O(z)$, where $x$ is some expression, and $z$ is some other quantity, then we mean that $x$ is a quantity such that $|x| \le Cz$ for some number $C$. If we have $x = y + O(z)$, then we really mean $x-y = O(z)$ in the sense I just described. So $x-y$ is equal to some quantity $z'$ such that $|z'| \le Cz$. So we can say
$$
y = x + O(z)
$$
too, since $y-x = -z'$, and $|\!-\!z'| = |z'| \le Cz$. So the minus sign does not need to be kept track of, and it's best to think of $O(z)$ as meaning a quantity, whose absolute value is bounded by a constant times $z$.
