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Suppose I want to increase the ratio:

$$ S(x) = \frac{f(x)}{g(x)} $$

I know that

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

Where $p,q \in \mathbb{Z}_{++}$ and $f,g$ are positive real functions. Ie we have:

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

Now if I have the ratio:

$$ \hat{S(x)} = \frac{\hat{f}(x)}{\hat{g}(x)} $$

I think that:

$$ S(x) \leq \hat{S(x)} \quad \forall x $$

Ie throwing out these higher order terms increases the ratio. How can I show this?

I looked at:

Big-O notation in division General Big-O operations. Check my proof: Big O notation

But I still am confused. I tried calculating:

$$ \frac{f(x) + O(x^{-p})}{g(x) + O(x^{q})} - \frac{f(x)}{g(x)} = \frac{O(x^{q})}{g^{2}(x) + O(x^{q})} $$

But that seems to just evaluate to 1.

Any insight?

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  • $\begingroup$ I'm confused that you have $f$ and $g$ defined as functions, but seem to define $S$ as a constant. Is $S$ supposed to be some sort of limit? $\endgroup$ Oct 28, 2019 at 20:08
  • $\begingroup$ @MartianInvader Sorry Fixed. $\endgroup$ Oct 28, 2019 at 20:11
  • $\begingroup$ Then your statement is obviously false. Your only restrictions on the relationship between $S$ and $\hat{S}$ are for sufficiently large $x$, which will never allow you to make conclusions for ALL $x$. For example, let $g = \hat{g} = 1$ and let $f(x) = \hat{f}(x) + 1$ for $0 \leq x \leq 100$ and $f = \hat{f}$ elsewhere. $\endgroup$ Nov 14, 2019 at 23:01

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