Multiplication of asymptotic approximation

Question

If I know that: $a = (1 - O(\frac{1}{n}))$ and $b = (1 + O(\frac{1}{n}))$, what is the asymptotic approximation of $a\cdot b$? Is answer $ab = (1 - O(\frac{1}{n^2}))$ correct or it is still $ab = (1 - O(\frac{1}{n}))$?

Are any source, when I can find some more information about action like adding, multiplication or powering this type of approximations?

Answer 1

I find it useful to think in term of example functions that satify the the big $O$. Try multiplying $1 - \frac{10}{n}$ with $1 + \frac{1}{n}$. $$ \left(1 - \frac{10}{n}\right)\left(1 + \frac{1}{n}\right) = 1-\frac{9}{n}-\frac{10}{n^2} $$ Is this $1 - O(\frac{1}{n^2})$?

Another way is to expand the multiplication using algebraic rules. Just keep in mind that the two $O(\frac{1}{n})$ do not necessarily have to be the same function.

$$ ab = \left(1 - O\left(\frac{1}{n}\right)\right)\left(1 + O\left(\frac{1}{n}\right)\right) = 1 + O\left(\frac{1}{n}\right) - O\left(\frac{1}{n}\right) - O\left(\frac{1}{n^2}\right) $$

Now it is clear that you cannot really determine the result as $1 + O(\dots)$. It can either be $1 + O(\frac1n)$, $1 - O(\frac1n)$ or $1 - O(\frac1{n^2})$. You could do it if you knew which function was dominating (or if they were equal).

I assumed that the functions in big $O$ are positive and you care about the direction you are approaching 1 from.

Answer 2

It is just $$ a b + O(\frac{1}{n}) $$ Remember what $O()$ means. For appropriate $N_0$, whenever $n > N_0$ you have: $$ a - \frac{c_a}{n} < a^* \le a \\ b \le b^* < b + \frac{c_b}{n} $$ Multiplying the inequalities gets you $$ a b - \frac{b c_a}{n} < a^* b^* < a b + \frac{a c_b}{n} $$