Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am struggling still with this equations...from my class materials....

This time we deal with lower bound -> BIG OMEGA:

I know that:

$$\Omega(g(n)) = \{f(n) : \exists c, n_0 > 0\,\forall n\ge n_0\ \ cg(n) \le f(n)\}$$

1) $\frac 13 n^2 − 3n = \Omega(n^2)$

$\frac 13n^2 − 3n \ge cn^2$, if $c \le \frac 13 − \frac 3n$ which is true if $c = \frac 16$ and $n > 18$. is that right???

2) $(n + a)^b = \Omega(n^b)$ for $a, b > 0$

Here i am not sure.

$(n + a)^b \ge cn^b$, if $c <= \frac{(n+a)^b}{n^b}$ for some $a,b > 0$

HOW this works?

share|improve this question

1 Answer 1

Write $\frac{(n+a)^b}{n^b}$ as $\left(\frac{n+a}n\right)^b = \left(1 + \frac{a}{n}\right)^b$. This goes toward $1$ as $n\to\infty$, so you can use any $c<1$ if you choose a large enough $n_0$ to go with it.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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