[Math] Show that $\frac{1}{2}n^2-3n=\Theta{(n^2)}$

Question

Show that $$\frac{1}{2}n^2-3n=\Theta{(n^2)}$$

$$$$

$\displaystyle{\frac{1}{2}n^2-3n=\Theta{(n^2)}: \\ \exists c_1, c_2 >0 , \ \ \exists n_0 \geq 1 \text{ such that } \forall n \geq n_0 \\ 0<c_1 n^2 \leq \frac{1}{2}n^2-3n \leq c_2 n^2}$

$$$$

That means that I have to find the values of $n_0, c_1, c_2$ such that
$$0<c_1 n^2 \leq \frac{1}{2}n^2-3n \leq c_2 n^2$$
$$\Rightarrow 0<c_1 \leq \frac{1}{2}-\frac{3}{n} \leq c_2$$

• $\displaystyle{0<c_1 \leq \frac{1}{2}-\frac{3}{n}} :$

It must be $\displaystyle{\frac{1}{2}-\frac{3}{n}>0 \Rightarrow \frac{3}{n} < \frac{1}{2} \Rightarrow n>6 \Rightarrow n \geq 7}$

So, $\displaystyle{n_1=7}$.

$\displaystyle{n \geq 7 \Rightarrow \frac{1}{7} \geq \frac{1}{n} \Rightarrow \frac{3}{7} \geq \frac{3}{n} \Rightarrow -\frac{3}{n} \geq -\frac{3}{7} \Rightarrow \frac{1}{2}-\frac{3}{n} \geq \frac{1}{2}-\frac{3}{7} \Rightarrow \frac{1}{2}-\frac{3}{n} \geq \frac{1}{14}}$

So $\displaystyle{c_1=\frac{1}{14}}$.

• $\displaystyle{\frac{1}{2}-\frac{3}{n} \leq c_2} :$

How can I find a $\displaystyle{n_2}$ such that $\displaystyle{\forall n \geq n_2:\frac{1}{2}-\frac{3}{n} \leq c_2}$ ??

And how can I find this $\displaystyle{c_2}$ ??

Answer 1

Hint: If $n$ is non-negative (i.e. $n \ge 0$), then $\dfrac{1}{2}n^2-3n \le \dfrac{1}{2}n^2$.

This should help you find $n_2$ and $c_2$, which is what you said you are having trouble with.