3
$\begingroup$

I am trying to solve this issue but I do not know how to handle the division by $O(\sqrt{n})$.

Prove the following using limits and L’Hôpital’s Rule: That $\log n$ is $O(\sqrt{n})$.

$$ \begin{align*} \lim_{n \rightarrow \infty}\frac{\log n}{\sqrt{n}} &= \lim_{n \rightarrow \infty}\frac{\log n}{10^{1/2 \log n}}\\ \end{align*} $$

$\endgroup$
5
  • 1
    $\begingroup$ I am trying to read this issue but I can't. Also, it doesn't seem like provability is the right tag. $\endgroup$ – Lord_Farin Oct 23 '12 at 16:40
  • 1
    $\begingroup$ I have formatted your question using $\LaTeX$. Please make sure I haven't changed your intended meaning, and also consider learning some basic commands to help render your questions on this site. $\endgroup$ – Austin Mohr Oct 23 '12 at 16:50
  • 3
    $\begingroup$ In fact, $\,\log n=\cal O(n^\epsilon)\,$ , for any $\,\epsilon >0\,$ $\endgroup$ – DonAntonio Oct 23 '12 at 18:12
  • 1
    $\begingroup$ Actually $\log x < \sqrt x$ for all $x>0$. $\endgroup$ – lhf Oct 23 '12 at 18:23
  • 2
    $\begingroup$ Let $a>0$. Then $$\lim_{x\to\infty}\frac{\log x}{x^a}=0$$ $\endgroup$ – Pedro Tamaroff Oct 23 '12 at 19:10
2
$\begingroup$

Try using L'Hospital's rule:

$\log_{10} n = \frac{\ln(n)}{\ln(10)}$, and let's call $k = 1/\ln(10)$. Then, derivative of the top is $k/n$ and of the bottom is $\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$. By L'Hospital's Rule,

$\lim_{n \to \infty} \frac{\log n}{\sqrt{n}} =\lim_{n \to \infty} \frac{k/n}{\frac{1}{2\sqrt{n}}} =\lim_{n \to \infty} \frac{2k\sqrt{n}}{n} = 0$,

so indeed $\log n = O(\sqrt{n})$.

$\endgroup$
4
  • 1
    $\begingroup$ Your proof shows $\log n = o(\sqrt{n})$, which is actually stronger than $\log n = O(n)$. $\endgroup$ – Austin Mohr Oct 23 '12 at 17:57
  • $\begingroup$ Where did that $k$ come from? $\endgroup$ – Pedro Tamaroff Oct 23 '12 at 19:02
  • $\begingroup$ @AustinMohr Agree, but the last statement was only for what was asked for in the question... $\endgroup$ – gt6989b Oct 23 '12 at 20:42
  • $\begingroup$ @PeterTamaroff $k$ is defined in the answer as $1/\ln(10)$, which shows up if you interpret $\log x = \log_{10} x$, as the author of the question had done (instead of $\log x = \ln x$) $\endgroup$ – gt6989b Oct 23 '12 at 20:44
6
$\begingroup$

You need to prove that $\log x \le c \sqrt x$ for some $c>0$ and for $x$ sufficienty large.

Consider $f(x)=\sqrt x - \log x$. Then $f'(x)=\frac{1}{2\sqrt x}-\frac{1}{x}=\frac{\sqrt x -2}{2x}\ge 0$ for $x\ge4$.

Since $f(4)=2-\log 4>0$, we have $f(x)\ge 0$ for $x\ge4$.

$\endgroup$
0

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.