True or false. If $f(n) = \Theta(n^2)$ and $g(n) = \Theta(n^2)$ then $(f-g)(n) =\Theta(n^4)$ where we define $(f-g)(n)=f(n)-g(n) \forall n$.

Question

True or false.

If $f(n) = \Theta(n^2)$ and $g(n) = \Theta(n^2)$ then $(f-g)(n) =\Theta(n^4)$ where we define $(f-g)(n)=f(n)-g(n) \forall n$.

I believe this is false.

Take $f(n) = 4n^2, g(n) = 2n^2$. Then $f(n)-g(n) = 2n^2$ and $2n^2$ is not $\Theta(n^4)$ but I cannot say for sure why.

Can anybody tell me if I am on the right track?

Answer 1

Proving that $\large{f=\Theta(n^2)}$ is the same as showing that $\large{f = O(n^2)}$ and $\large{n^2 = O(f)}$ and so clearly $\large{(f-g)(n) \neq \Theta(n^4)}$ because even though $\large{n^2 = O(n^4)}$ but $\large{n^4 \neq O(n^2)}$ meaning that you cannot find $\large{n_0}$ and $\large{c}$ such that that $\large{n^4 \leq cn^2}$ for all $\large{n \geq n_0}$