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Given $f(n)=n^2log n$ AND $g(n)=n(log n)^{10}$

Which one of the following is true?

  1. $f(n)=O(g(n))$
  2. $f(n)=\Omega(g(n))$
  3. $f(n)=\theta(g(n))$
  4. $\text{We can't compare.}$

Kindly work out this question. Stepwise solution. Thanx in advance. :)

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f(n) = n (n log n)

g(n) = (n log n) $(log n)^{9}$   

cancelling n log n on both sides we have f(n)= n, and g(n) = $(log n)^{9}$  

n < (log n)9

example:  1. n = 1 , then f(n) = 1 and g(n) = 0.

                 2. n = 4 , then f(n) = 4 and g(n) = $2^{9}$

                 3. n = 128 / $2^{7}$, then f(n) = 128 and g(n) = $7^{9}$

                 4. n= 1024 / $2^{10}$ , then f(n) = 1024 and g(n) = $10^{9}$

Therefore f(n) < c g(n) where $n_{o}$ = 4. Hence f(n) = O(g(n)).

The answer is a.
Answer:

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