2n+ 1 ∈ Θ(n) because

____ ·n≤ 2n+ 1 ≤ ____ ·nfor alln≥ ____

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.

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- January 20th 2013, 05:01 PMbigman99discrete mathmatics/fill in blank to make true statement
2

*n*+ 1 ∈ Θ(*n*) because

____ ·*n*≤ 2*n*+ 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. - January 20th 2013, 05:23 PMPlatoRe: discrete mathmatics/fill in blank to make true statement
- January 20th 2013, 05:54 PMbigman99Re: discrete mathmatics/fill in blank to make true statement
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.