Thread: discrete mathmatics/fill in blank to make true statement

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.

Originally Posted by bigman99

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.

To be quite honest with you, but none of that makes any sense.

Is that some sort of specialize notation that you did not bother to tell us about?

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).

PROOF

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.