2n + 1 ∈ Θ(n) because
____ · n ≤ 2n + 1 ≤ ____ · n for all n ≥ ____
i have serveral questions like this and i have no clue as to what to do here. any help with steps of what to do would be so much help.
Here is an exerp from my book. i just dont understand it. its part of
For the sequence given by a_n=n^2−n/2,1/4n^2≤a_n≤n^2 for all n ≥ 2. Hence, a_n ∈ Θ(n^2).
Let n ≥ 2 be given. Since n is positive, n2−n/22<n^2/2, and clearly n^2/2<n^2.
This shows that a_n ≤ n^2. To show that a_n≥1/4n^2, it is much more natural to first rewrite the inequality so that it involves a comparison of a polynomial to 0. In this case the inequality n^2−n/2≥1/4n^2 is the same, after multiplying by 4 and subtracting n^2 from each side, as the inequality n^2 − 2n ≥ 0. We can now easily argue that if n ≥ 2, it follows that both n and n − 2 are positive, in which case we know that the product n(n − 2) is positive. That is, n^2 − 2n ≥ 0, as desired.