# Big O Division: Increasing a Ratio

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:

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?

• 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? Oct 28, 2019 at 20:08
• @MartianInvader Sorry Fixed. Oct 28, 2019 at 20:11
• 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. Nov 14, 2019 at 23:01