You are given functions f and g such that f(n)=O(g(n)). Is f(n)∗log2(f(n)c)=O(g(n)∗log2(g(n))) ? (Here c is some constant 0. You can assume that f and g are always bigger than 1.

a) true
b)false
c)depends on c
d) depends on f and g

Question

You are given functions f and g such that f(n)=O(g(n)). Is f(n)∗log2(f(n)c)=O(g(n)∗log2(g(n))) ? (Here c is some constant >0. You can assume that f and g are always bigger than 1.

a) true
b)false
c)depends on c
d) depends on f and g

anonymous · Answer

take some example of f and g
and chk if fn*logfn(n)<=gn*loggn

foolaroundmath · Answer

I think this violates the honor code of Coursera, so no replies :)
Although I can give a hint.
Use the definition of f(n) = O(g(n)) i.e.
$$f(n)  \le c*g(n) \forall n \ge n_{0}$$
try fiddling around with this, multiply same functions or same sides of an inequality.
This should be sufficient to get you to the answer.