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.

This is probably a very silly question.

If $h(n)=\frac{n}{2}, \ g(n)=n$, so $$ \lim_{n \to \infty} \frac{h(n)}{g(n)} = \lim_{n \to \infty} \frac{n}{2n}=\frac{1}{2} $$ so $h(n) \leq C_1 g(n), h(n)=O(n)$

at the same time , if $g(n)=\frac{n}{10}$, this limit becomes 5, so

$h(n) \geq C_2 g(n), h(n)= \Omega(n)$

Is this logic correct enough to say that $$ \frac{n}{2}=\Theta(n) $$

share|improve this question
    
Yes (trivally). See en.wikipedia.org/wiki/… –  M.B. Jan 24 '12 at 9:39
    
In general, for any function $f(n)$ and any constant $c$, we have $c f(n) = \Theta(f(n))$. –  Srivatsan Jan 26 '12 at 16:01
add comment

1 Answer 1

up vote 3 down vote accepted

Yes but note that it also immediately follows from the definitions.

share|improve this answer
    
you mean definition of $\Theta(n)$? –  user19821 Jan 24 '12 at 10:12
    
@user19821 yes indeed, it is obvious how you have to choose the constants that it nicely fits the definition. –  Listing Jan 24 '12 at 22:31
add comment

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.