0
$\begingroup$

HOMEWORK QUESTION Prove that Θ(n) + O(n^2) ⊆ O(n^2). Note that for this problem, you are proving that the set of functions on the left hand side (LHS) is a subset of the set of functions on the right hand side (RHS). The set on the LHS is the algebraic sum of two sets (not the union): an element of the LHS has the form f(n) = f1(n) + f2(n), where f1(n) ∈ Θ(n) and f2(n) ∈ O(n2).

I conceptually understand O/Theta/Omega notations fairly well I'd say, I'm just having a problem knowing how to start this proof, or how to go about showing it. I think having the '⊆' instead of '∈' is confusing me as well. Any help is appreciated.

$\endgroup$
1
$\begingroup$

Xue right? Looks like we're classmates.

Observe that since $f_1 \in \Theta (n)$, then

$f_1 (n) \le c_1 \bullet n$ $\forall n \ge N$

and also that

$c_1 \bullet n \le c_1 \bullet n^2$

so

$f_1(n) \le c_1 \bullet n \le c_1 \bullet n^2$

Thus

$f_1 \in O(n^2)$

similarly, since $f_2 \in O(n^2)$,

$f_2(n) \le c_2 \bullet n^2$ $\forall n \ge N$

Then

$f_1(n) + f_2(n) \le (c_1+c_2)n^2$ $\forall n \ge N$

And

$\Theta(n) + O(n^2) \subseteq O(n^2)$

I'm not 100% on this, but it makes sense to me so unless I can think of something better it's what I'll submit.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Lemme know if this helps or if you can think of a way to improve it. $\endgroup$ – R. Stevo Brown Jan 25 '17 at 7:51
0
$\begingroup$

Hint: You goal is to prove the statement $f(n) \in \Theta(n) + O(n^2) \implies f(n) \in O(n^2)$. This is what the statement $\Theta(n) + O(n^2) \subseteq O(n^2)$ means.

To start, use the triangle inequality on $|f(n)| = |f_1(n) + f_2(n)|$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.