Increasing Growth order

Question

Order the following functions in increasing order of their growth rate.
(i) nlogn
(ii) $n \log (\log n)$
(iii) $n \log \left(\mathrm{n}^{\mathrm{n}}\right)$
(iv) $n(\operatorname{logn})^{2}$

• A. i , ii , iii , iv
  
• B. ii , iv, i , iii
  
• C. i, iii, ii, iv
  
• D. iv, ii , i , iii

Comments

Getting increasing order as ii,i,iv,iii . pls confirm.

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nlog(log(n)) < nlog(n) <  n(log(n))^2 < n(log(n^n))

Option : 2,1,4,3

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2,1,4,3 is correct...

Answer 1

I am calculating it like below.

let n = 2^2^3

(i) nlog n = 2^8 * 8

(ii) n log(log n) = 2^8 * 3

(iii) n log(n^n) = 2^8 * 2^8 * 8

(iv) n (log n)^2 = 2^8 * 16

So the order is 3,4,1,2

Not sure if I'm doing something wrong.

Answer 2

Put $n = 2^{16}$

i) $n log  n \rightarrow 2^{16} log 2^{16} = 2^{16} . 16 = 2^{20}$

ii) $n log(log n) \rightarrow 2^{16} log (log (2^{16})) = 2^{16} . 4 = 2^{18}$

iii) $n log (n^n) \rightarrow n^2 logn = ({2^{16}})^2 log( 2^{16}) = 2^{32}.16 = 2^{36}$

iv) $n (logn)^2 \rightarrow 2^{16} (log (2^{16}))^2 = 2^{16}. 256 = 2^{16}. 2^8 = 2^{24}$

Therefore the correct increasing order is $ii < i < iv < iii.$

So, the answer is None of these.