$5 \log( \log n) $

$n (\log n)^2$

$\sqrt{n} \log n$


$n \log (\log n)$

$7 \sqrt{n}$

What is the ascending order of the growth function? Please give the explanation as well.


closed as off-topic by Antonio Vargas, gebruiker, Watson, Future, user228113 Jun 3 '16 at 16:32

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  • $\begingroup$ What do you mean by correct order ? $\endgroup$ – Shailesh Jun 3 '16 at 6:32
  • $\begingroup$ from most efficient to least efficient $\endgroup$ – Apples Jun 3 '16 at 6:39
  • $\begingroup$ Do you have an idea of how any of them relate? $\endgroup$ – Henrik Jun 3 '16 at 15:18
  • $\begingroup$ I know that logs are faster and fractional exponents are faster.But I don't have any depth of idea.Would you recommend something for a read? $\endgroup$ – Apples Jun 3 '16 at 16:17
  • $\begingroup$ Well, out of the 30 possible pairs, some of are clearly asymptotically bigger than others. $\endgroup$ – user228113 Jun 3 '16 at 16:31

If you take sufficiently large value of n,then ascending order according to the value will be, $5 log( log n) < 7 \sqrt{n}<\sqrt{n} log n<n log (log n)<n (log n)^2<n^{\frac{4}{3}}$

So,if we sort them from Most efficient to least efficient,it will be

$5 log( log n) > 7 \sqrt{n}>\sqrt{n} log n>n log (log n)>n (log n)^2>n^{\frac{4}{3}}$


If we want for sufficient large n, then if we name the equation from top to down as from 1 to 6 then ordering looks like 1<5<2 , 7<3<2 and 6<4.


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