Determine whether or not the following is true: $\log((1/n) + n^2 )$ is $O(\log n)$.

I'm struggling to solve this. I'm especially confused because of the $O(\log n)$ and would appreciate some help or even a push in the right direction. I'm unconfident with logarithms in general which is why I'm struggling with this question in particular.

  • 3
    $\begingroup$ Hint: $\dfrac1n\ll n^2$. $\endgroup$ – Yves Daoust Jun 10 at 18:05
  • $\begingroup$ $\log(\frac{1}{n}+n^2) < \log(n^2+n^2) = 2 \log(2) \log(n) = O( \log n)$ $\endgroup$ – cdt Jun 10 at 18:29

$\log((1/n) + n^2 )$

If we just focus on the term inside of the logarithm.

$ let x = 1/n + n^2$

As n grows, we notice that the 1/n term effectively becomes zero and the overpowering term is $n^2$

We can now state that $\log((1/n) + n^2 )$ has growth $log(n^2)$ when n gets large.

We know from logarithms that $log(a^b) = b* log(a)$

So $log(n^2)$ can be rewritten as $2*log(n)$ for large n we notice that the constant value of 2 is insignificant and can conclude that our original expression has growth $O(log(n))$


$n^2+\frac{1}{n}$ is $O(n^2)$.

$\log(n^2)=2\log n$ is $O(\log n)$

  • $\begingroup$ The OP is clearly strugling with $O$ notations and logarithm... an explanation would be much more useful than these math lines $\endgroup$ – Thomas Lesgourgues Jun 10 at 18:28

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