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Answer: Option (D)

Here g(n) = 2*n$^{\sqrt{n}}$ .  

(Just Simplify it)

So, we can say that f(n) $\equiv$ g(n) .

And thus first two options are Wrong as h(n) = nn    (Just represented in different Form)

This means option D is right Only.

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Answer is D.

Solution:. Simplify the functions by taking logarithm of each function:

log(f(n)) = ( sqrt n) log (base 3) n

log(g(n)) = (sqrt n) log (base 2) n

log(h(n)) = log (n!). But according to Stirling's approximation n! = O(n^n)

So, log(h(n)) = n log n

 

Hence, we can see that log(f(n))= log(g(n)) < log(h(n)). (asymptotically)

It implies f(n) = g(n) < h(n)

The only option that satisfies this is option D which says f(n) = O(g(n)) which is true because f(n) = Theta( g(n))
Answer:

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