O(g)O(n) + O(f)O((logn)^5) = O(n^5), is the following statement true or not?

Question

Given that $f = O(n^2)$ and $g = O(n^4)$, I'm not able to conclude if the above statement is true or not, that's to say, if we can say that :

$$ O(g)O(n) + O(f)O\left(\log(n)^5\right) = O\left(n^5\right) $$

Answer 1

Do you know what the capital O notation stand for? If you do, than you should be able to verify that if $b(n)/a(n)\rightarrow 0$, then $O(a(n)+b(n))=O(a(n))$. This is one of the convenient formulas of the "ordo-calculus". It is in fact a very smart notation, and easy to use. Some other things: $O(a(n))O(b(n))+O(c(n))O(d(n))= O(a(n)b(n)+c(n)d(n))$.

Are the above hints sufficient?

Answer 2

Note that

$$\frac{O(g)O(n) + O(f)O((\log n)^5)}{n^5}=\frac{O(g)}{n^4}\frac{O(n)}n + \frac{O(f)}{n^2}\frac{O((\log n)^5)}{n^3}\to M_1\cdot M_2+M_3\cdot 0=M<\infty$$

therefore

$$O(g)O(n) + O(f)O((\log n)^5) = O(n^5)$$