2n is the order of 3n .
Is it True or False ? State with reason.
Say f(n)=2n and g(n)=3n
f(n)=o(g(n))(small-oh) //////not tightest upperbound
so answer is false
yes @Arjun Sir
thank u :)
log is not good here
because then both value will be equal to, rather than saying 3n asymptotically larger than 2n
If we get log2 and log3 then these are constant term , So, by this we cannot say one asymptotically larger than another
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