1 1 vote 2n is the order of 3n . Is it True or False ? State with reason. Algorithms algorithms asymptotic-notations true-false + – Kapil 1.5k views answer comment Share Follow Print See 1 comment 1 1 comment reply aaggarwal commented Aug 6, 2016 reply Follow flag $2^n$ is order of $3^n$ does it mean : $2^n$=O($3^n$) or $2^n$=o($3^n$)? if it isO()..then it is false and if it is o() then true? 0 0 replyShare Please log in or register to add a comment.
3 3 votes Say f(n)=2n and g(n)=3n f(n)=o(g(n))(small-oh) //////not tightest upperbound so answer is false srestha answered Aug 5, 2016 • edited Aug 5, 2016 by Tauhin Gangwar srestha comment Share Follow See all 8 Comments 8 8 Comments reply Show 5 previous comments Arjun commented Aug 5, 2016 reply Follow flag yes, it should be. But "is the order of" is a bit ambiguous 1 1 replyShare srestha commented Aug 5, 2016 reply Follow flag 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 0 0 replyShare Arjun commented Aug 5, 2016 reply Follow flag yes. Because when we apply log we are changing the functions. i.e., $f(x)$ now becomes $g(f(x))$ where $g$ is the $\log$ function. Still this works for many functions and with some effort one can say for which all cases it won't work. 2 2 replyShare Please log in or register to add a comment.