Mock test

Question

Arrange the following functions in decreasing asymptotic order:

f(n)= 3^{2^n}

g(n)= n!n³

(a) f(n),g(n)

(b) g(n),f(n)

Comments

Shouldn't the answer be option (b) since factorial functions grow faster than exponential functions?

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Apply log on both functions 

log(3^{2^n}) = 2ⁿ log3 = O(2ⁿ)

log(n!n³) = log n! + 3logn = O(nlogn)  [ we know log n! = O(nlogn) ] 

O(2ⁿ) > O(nlogn) 

So, option (a) is correct.

 

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ok..got it..thanks :)