I'm trying to prove (by definition) the following but to no avail:
$n^{n/2} \ne O(3^{n/2}) $
$n! \ne O(3^n)$
$(n-b)^a = \Theta(n^a)$
$a,b $ are both constants whereas $a > 0 $ and $b$ might be positive, negative or $0$.
Now , here is what I come up so far:
Lets take the first one for example (regarding the others two I'm completly lost):
Now If we asssume by contradiction that $n^{n/2} = O(3^{n/2}) $ then there exists constants $c,n_0>0$ such that $n^{n/2} < c(3^{n/2}) $ for every $n\ge n_0$.
And if ones look at the ratio:
$n^{n/2} / 3^{n/2} $ then it can be proved that its going to infinity what implies that $n^{n/2} $ is much faster than $3^{n/2} $.
But I have no idea how to show this.
regarding the other ones , can you please give me an hint?
Thanks alot!