If a function $f(n)=\Theta(g(n))$, does it follow that $f(n/k)=\Theta(g(n))$ for a constant $k$? Suppose the following is true for some f(n) and g(n):
$f(n) = \Theta(g(n))$
Does that mean $f(n/k) = \Theta(g(n))$ for any value of $k&gt;0$?
I know that for the above to be true, there must exist positive constants $c_1$, $c_2$, $n_0$ such that for all $n \geq n_0, $ $c_1(g(n)) \leq f(n) \leq c_2(g(n))$. Thus if $f(n) = \Theta(g(n))$, then can I assume:
$\frac{1}{k}c_1(g(n)) \leq f(n/k) \leq \frac{1}{k}c_2(g(n))$ to satisfy big Theta?
My apologies if this is totally obvious or if I'm completely on the wrong track, my math background isn't the strongest. Are there any relevant properties that might describe this relationship?
Suppose $f(n) = 2^n$. Then $f(n / 2) = 2^{n / 2} = \sqrt{2^n}$ and setting e.g. $g(n) = 2^n$ as well shows $f(n) \in \Theta(g(n))$ but $f(n / 2) \notin \Theta(g(n))$, so this does not hold in general.