Suppose I have the following functions $f(x)$ and $g(x)$, such that:

$$ f(x) = \tilde{f}(x) + O(x^{-p}) $$ $$ g(x) = \tilde{g}(x) + O(x^{q}) $$

where $p,q \in \mathbb{Z}_{++}$ and f,g are positive functions. I want to find order of the following remainder term:

$$ R_{x} = \frac{\tilde{f}(x)}{\tilde{g}(x)} - \frac{f(x)}{g(x)} $$

So do to so, I write:

$$ R_{x} = \frac{\tilde{f}(x)}{\tilde{g}(x)} - \frac{\tilde{f}(x) + O(x^{-p})}{\tilde{g}(x) + O(x^{q})} = \frac{g(x)f(x)+O(x^{q})f(x)-g(x)f(x)+g(x)Q(x^{-p})}{\tilde{g}^{2}(x) + \tilde{g}^{2}(x)O(x^{q})}$$ $$ R_{x} = \frac{O(x^{q}) \tilde{f}(x) - \tilde{g}(x)O(x^{-p})}{[\tilde{g}(x) + O(x^{q})] \tilde{g}(x)} $$ $$ R_{x} = \frac{O(x^{q}) \frac{\tilde{f}}{\tilde{g}}(x) - O(x^{-p})}{[\tilde{g}(x) + O(x^{q})] \tilde{g}(x)} $$ $$ R_{x} = \frac{O(x^{q})}{\tilde{g}(x) + O(x^{q})} = \frac{O(x^{q})}{g(x)}$$

So we have: $$ R_{x} = O(x^{q})$$

Is this correct? If not, what am I doing wrong? How can I find this remainder term?

What about the order of the following:

$$ S_{x} = \frac{\tilde{f}(x) + O(x^{-p})}{\tilde{g}(x) + O(x^{q})}$$

How could I compute this?

If we follow: Order of error of a fraction

Would the answer then be simply:

$$ S_{x} = O(x^{-p}) $$

Or something else? Would it be possible to provide a step by step solution with some explanations?


  • $\begingroup$ What are the orders of $\tilde f$ and $\tilde g$? $\endgroup$
    – Ishan Deo
    Oct 29, 2019 at 18:14
  • $\begingroup$ @IshanDeo Do you need those? In the question I linked, they managed to derive something without them. The question is -- is it correct? $\endgroup$ Oct 29, 2019 at 19:09
  • $\begingroup$ I was wondering how $O(x^{q})\frac{\tilde{f}}{\tilde{g}}(x) = O(x^{q})$ $\endgroup$
    – Ishan Deo
    Oct 29, 2019 at 19:10
  • $\begingroup$ @IshanDeo. I don't know honestly. That is why I am asking. These tilde functions are the ones I will be approximating the non-tilde functions with. So I suppose they are order O(x^-p) and O(x^q) respectively. But I really do not know if that step is correct. It probably isnt. I really dont know what I am doing here haha $\endgroup$ Oct 29, 2019 at 19:12


You must log in to answer this question.