Confused about dealing with Big $\mathcal{O}$ in functions

Question

For the following defined functions $$f(x) = \frac{1}{x}+\frac{n-1}{x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)$$ and $$y(x) = c\left(\frac{1}{x}+\frac{m-1}{x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)\right)$$ where $m > n > 0$ and $c$ is large.

I want to bound $f(x)-y(x))$.

I thought that $f(x)-y(x) \leq \frac{1}{x}+\frac{n-1}{x^2}- c\left(\frac{1}{x}+\frac{m-1}{x^2} \right)$, since we are talking about an approximation!

Is this true?

After that, I found this question where the answer states that

For every $x\geq0$, $f(x)-y(x) \leq f(x)$

Is this the tighter bound? if so, how can I get rid of the $\mathcal{O}$ in $f(x)$?

Answer 1

Regarding your first question

I thought that $ f(x) − y(x) \leq \frac{1}{x} + \frac{n−1}{x^2} −c (\frac{1}{x} + \frac{m−1}{x^2}) $ , since we are talking about an approximation!

Is this true?

No. Consider for example $ f(x) = \frac{1}{x} + \frac{n-1}{x^2} + \frac{c+1}{x^3}, g(x) = c(\frac{1}{x} + \frac{m-1}{x^2} + \frac{1}{x^3}) $. Then you get $ f(x) - g(x) = \frac{1}{x} + \frac{n-1}{x^2} - c(\frac{1}{x} + \frac{m-1}{x^2}) + \frac{1}{x^3} > \frac{1}{x} + \frac{n-1}{x^2} - c(\frac{1}{x} + \frac{m-1}{x^2}) $

Answer 2

You don't get rid of the $\mathcal{O}$, but rather the argument is, that for some reason you do not need to keep close track of it. I feel like there should be some kind of rules how to calculate with them properly. They could look like this:

$\mathcal{O}(f(x)) = c\mathcal{O}(f(x))$

$\mathcal{O}(f(x)) \pm \mathcal{O}(f(x)) = \mathcal{O}(f(x))$

and that would lead to

$f(x)-y(x) \leq \frac{1}{x}+\frac{n-1}{x^2}- c\left(\frac{1}{x}+\frac{m-1}{x^2} \right) + \mathcal{O}\left(\frac{1}{x^3}\right)$