# Big O notation for ceiling and floor functions

I want to calculate the big-O notation for ceiling and floor functions such as $$f(x)= \lfloor x^{\frac{4}{3}} \rfloor$$ and $$f(x)=x \lceil \frac{n}{2} \rceil$$
the domain of these functions is $$\mathbb{N}^*$$.

in the first function can we ignore the floor rounding and treat it as $$f(x)= x^{\frac{4}{3}}$$ then $$f(x)$$ is $$O(x^{\frac{4}{3}})$$ ?
I've never seen fractions in the power of so I'm not sure if $$O(x^{\frac{4}{3}})$$ is true or it should simply be $$O(x^2)$$ ?
I know we can ignore constants but never seen an example with a fraction as the power of

for the second function I figured we can also ignore the ceiling rounding so the function becomes $$f(x)=x \frac{x}{2}$$ and then we ignore the denominator so, $$f(x)$$ is $$O(x^2)$$

Also I want to verify if we can ignore the ceiling since big-O is used for big numbers and a ceiling/floor won't make any difference when we're dealing with really big numbers

Yes, you are right. $$f(x)$$ is $$O(x^{4/3})$$ is equivalent to the statement that there exist positive constants $$C, x_0$$ such that $$|f(x)| \le Cx^{4/3}$$ for all $$x\ge x_0$$. We can write $$f(x) = \lfloor x^{4/3} \rfloor = x^{4/3} + g(x),$$ where $$g(x)$$ is bounded in absolute value by $$1$$. Hence $$|f(x)| \le |x^{4/3}| + 1 \le 2|x^{4/3}|$$ for $$x \ge 1$$. So, $$f(x)$$ is indeed $$O(x^{4/3})$$.
$$f(x) = x\lceil \frac{x}{2}\rceil = x(\frac{x}{2} + O(1))=\frac{x^2}{2} + O(x) = O(x^2)$$. ($$O(1)$$ stands for a function that is bounded in absolute value. In this case, that function is bounded by $$1$$).