1. ## Big Theta

Hello everyone!

No need to answer this question, I solved it.

prove or disprove: $\displaystyle 4n^3 + n^2\sqrt{n}-4n \in \Theta (n^3)$

I don't have any clue. Does anyone know how to solve this?

I think $\displaystyle 4n^3 + n^2\sqrt{n}-4n \in \Theta (n^3)$ is true, because

...

Thank you,
Rapha

2. Let $\displaystyle f(n)=4n^3+n^2\sqrt{n}-4n$. Clearly for $\displaystyle n$ large enough you have that $\displaystyle 10n^3>f(n)$ and $\displaystyle n^3<f(n)$. I'll leave it to you to show that these are true.

*note: I am sure you can come up with better bounds on the constants for $\displaystyle an^3$ for both inequalities, but it follows the same idea.