Is big O always positive? There is a function defined as follows:

$x, y, w(u), w'(u)$ are all positive real values.  $x > y \geq 2$.
Does this mean that $\Phi(x, y) > \frac{x}{\log{y}}\left(w(u)+\left(\frac{y}{2x^2}-w'(u)\right)\frac{1}{\log{y}}\right)$ ?
In other words, is the big O strictly greater than 0?
The reason I ask is that it doesn't seem to be true in the application I am using it for and I want to check if I am making a mistake.
 A: No. For $|x| < 1$
$$
\frac{1}{1-x} = 1 +x + x^2 + O(x^3)
$$
where the tail of the truncated series is negative when $x$ is negative.
A: If you “solve” the equation for the big-Oh term, $O(1/(\log y)^2)=something$, then all that is claimed is that there exists $C$ such that $|something|<\frac C{(\log y)^2}$ for $y\gg0$.
A: Here's wikipedia's definition:

One writes $f(x) = O(g(x))$ as $x\rightarrow\infty$ if the absolute value of $f(x)$ is at most a positive constant multiple of $g(x)$ for all sufficiently large values of $x$. That is, $f(x) = O(g(x))$ if there exists a positive real number $M$ and a real number $x_0$ such that $|f(x)| \le Mg(x)$ for all $x\ge x_0$

So $g(x)$ should be a positive function, but $f(x)$ is free to be negative. Big-O is a statement about its absolute value. Furthermore, big-O notation is about asymptotic behavior. $1+x$ is in $O(x)$ even though at $x=0$, $1+x$ is infinitely larger than $x$; what matters is what happens "infinitely out", not what happens for any finite value.
