Division of Big-Theta In an article about amortized analysis I found an evaluation$$\frac{n\cdot \Theta (1)+\Theta (n)}{n-1}=\Theta (1).$$
I don't really understand how $\Theta (n)$can be divided by $n-1$ as $n-1$ is not a Big-Theta expression, but just a number, and I didn't find such property.
Can someone explain how$$\frac{\Theta (n)}{n-1}$$evaluates to $\Theta (1)$?
 A: I'm not sure about the exact formal reasoning for this... But since clearly, $n - 1 \in \Theta(n)$, the ratio $\frac{\Theta(n)}{n-1}$ becomes $\frac{\Theta(n)}{\Theta(n)}$. It should make intuitive sense that this ratio is equivalent to $\Theta(1)$.
Why, any function in $\Theta(n)$ is also in $\mathcal{O}(n)$ and $\Omega(n)$. Thus, if $f(n) \in \Theta(n)$, it must be of the form $f(n) = a n + b$ for any non zero scalars $a$ and scalar $b$. If it was asymptotically larger (eg: $n^2, 3^n$), it wouldn't be in $\mathcal{O}(n)$, if it was any asymptotically smaller (eg: $\log(n), 1$), it wouldn't be in $\Omega(n)$. So if we have two arbitrary functions $f_1, f_2 \in \Theta(n)$, there ratio is $\frac{a_1 n + b_1}{a_2 n + b_2} = \frac{a_1  + \frac{b_1}{n}}{a_2  + \frac{b_2}{n}}$. As $n$ gets large, the $\frac{b_1}{n}$ and $\frac{b_2}{n}$ terms approach zero, so we have $\frac{a_1 + 0}{a_2 + 0} = \frac{a_1}{a_2} \in \Theta(1)$
A: $\Theta(n)$ can be thought of as representing any function with growth rate $\Theta(n).$ So when someone writes $\Theta(n)/(n-1),$ they mean "a function of the form some $\Theta(n)$ function divided by $n-1.$" But such a function always has growth rate $\Theta(1),$ and so we get the simplification.
