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If we arrange the following according to increasing asymptotic complexity, which quantity comes in the second position; which one comes in the fourth position?

$(\sqrt{2}^{\log n}, n^2, 2^{\sqrt{2 \log n} }, e^n, (\log n)!, n!$
in Algorithms by Veteran (105k points) | 174 views

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+1 vote

TAKE LOG ON EACH TERMS AND EVALUATE...

by Boss (11k points)
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i think n^2 comes in second position and e^n comes in fourth [position because we know the growth of functions are like that-

2^2^n > n! >4^n >2^n >n^2 >nlogn >logn! >n >2^logn >logn^2 >underoot logn> loglogn>1
by (309 points)

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