# if $f(n)=\Theta(n^2)$ and $g(n) = \Theta(n^2)$ then $(f+g)(n)=\Theta(n^4)$ where we define $(f+g)(n)=f(n)+g(n)$ ∀$n$.

if $f(n)=\Theta(n^2)$ and $g(n) = \Theta(n^2)$ then $(f+g)(n)=Θ(n^4)$ where we define $(f+g)(n)=f(n)+g(n)$ $∀ n$.

Is the above true or false. I would say its false but honestly its a guess and i cannot work it out. Thank you for any help.

• It would be true if $\Theta(n^2)+\Theta(n^2)=\Theta(n^4)$. – Rocket Man Jun 10 '14 at 18:56

It is obviously false. Take $f(n) = g(n) = n^2$. Then $f(n)+g(n) = 2n^2$ and $2n^2$ is not $\Theta(n^4)$ because $(2n^2)/n^4 \to 0$ as $n \to \infty$.