How can one show that $n \log n \subseteq O(n^{1+\epsilon})$ where $0 < \epsilon < 1$ without using limits? This question arise from a homework where I used limits to prove the relation. I'd like to know if there is another way.
Tell me more
×
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.
|
|
Let $0<\epsilon<1$ be fixed. We need only show that $\log n=O(n^\epsilon)$. Suppose $n\geq 1$ so that $\log n\geq 0$. Since $$n^\epsilon=e^{\epsilon \log n}=1+\epsilon \log n+\frac{\epsilon^2 \log^2 n}{2}\cdots \geq \epsilon \log n$$ we see that $$\log n \leq \frac{1}{\epsilon}n^\epsilon $$ which proves the desired result. |
|||||||||||
|
