Question

Which function has asymptotically larger growth rate n^{logn} or logn^{n}

Comments

log n^n will be having larger growth

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What is approach for solving such ambiguous question..

I tried to use L-hospital but again function's are for form $\frac{\infty}{\infty}$

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(log n)^n will be larger

Answer 1

Consider $n=2^{64}$.

$n^{logn}={2^{64*64}} =$ $2^{2^{12}}$ whereas ${(logn)}^n = 64^{2^{64}}=2^{6*{2^{64}}}\ge 2^{4*{2^{64}}}$$=2^{2^{68}}$.

Considering a greater $n = 2^{256}$,we have:

$n^{logn}=2^{256*256} $$=2^{2^{16}}$ whereas $(logn)^n = {2^{8*2^{258}}}$$=2^{2^{261}}$.

You can clearly see the difference in the growth rates of the two functions. Thus asymptotic growth of $n^{logn} \le (logn)^n $

Comments

nice