i am trying to prove that $n^2= \theta(5n^2 + n)$ And i'm using this formula:
$c_1(5n^2 + n) ≤ n^2 ≤ c_2(5n^2 + n)$ $\forall n ≥ n_0$
Is it true? If it's true could you please help me for find the c1,c2 and n0. I found them but i'm not sure are they true or false. Thank you.
As for my attempt, in the comments it was pointed out that $n^2\leq 5n^2+n$ is already true for all positive $n$, that's why my mind confused about the $c_2$. I found the $c_1$ is $\frac{1}{5}$ and $n_0= 5$ but I'm not sure are they true or false
One among many valid final answers (and the easiest to see in my opinion) is to try to use $n_0=1$ and choose the values of $c_1,c_2$ accordingly.
We notice that $n^2\leq 1\cdot (5n^2+n)$ is true for all $n$ so setting $c_2=1$ suffices. This just follows from "if I add a positive number to another positive number then I make it bigger" as well as "if I multiply a positive number bigger than $1$ to another positive number then I make it bigger", properties that you should be well familiar with already.
Now, on the other side, we have that $5n^2+n \leq 5n^2+n^2 = 6n^2$ is true for all positive $n$ as well. This, again following from the earlier mentioned properties. By dividing both sides of this inequality by six, we arrive at $\frac{5}{6}(5n^2+n)\leq n^2$.
We have as a result, one of the many possible choices for $c_1,c_2,n_0$ as being $c_1=\frac{1}{6},c_2=1,n_0=1$ as we have that:
$$\frac{1}{6}(5n^2+n)\leq n^2\leq 1(5n^2+n)~~~~\forall n\geq 1$$
Now, as mentioned, you could have used many other choices for $c_1,c_2,n_0$. The end result for this specific problem is that you could let $c_1$ and $c_2$ be any values such that $0<c_1<\frac{1}{5}$ and $c_2\geq\frac{1}{5}$ and then let $n_0$ be however big it needed to be in order to let that work. $c_1$ being equal to $\frac{1}{5}$ is not small enough. However, none of that actually matters. All that was necessary for this problem was to find any example that works, and the one above does that just fine.