GATE | GATE-CS-2001 | Question 16

Let f(n) = n2Logn and g(n) = n (logn)10 be two positive functions of n. Which of the following statements is correct?
(A) f(n) = O(g(n)) and g(n) != O(f(n))
(B) f(n) != O(g(n)) and g(n) = O(f(n))
(C) f(n) = O(g(n)) but g(n) = O(f(n))
(D) f(n) != O(g(n)) but g(n) != O(f(n))


Answer: (B)

Explanation:  

Any constant power of Logn is asymptotically smaller than n.

Proof:



Given f(n) =n2Logn and g(n) = n (logn)10
In these type of questions, we suggest you to first cancel out the common factor in both the function.  After removing these, we are left with f(n) = n and g(n) = (logn)9. Removing a factor of nlogn from both functions.

Now n is very very large asymptotically as compared to any constant integral power of (logn) which we can verify by substituting very large value say 2100.

f(2100) = 2100 = 1030  and g(2100) = 1009 = 1018.

Always remember to substitute very large values of n in order to compare both these functions. Otherwise you will arrive at wrong conclusions because if f(n) is asymptotically larger than g(n) that means after a particular value of n, f(n) will always be greater that g(n).

 

This solution is contributed by Pranjul Ahuja.

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