0 votes 0 votes Consider the two functions (logn)k and nϵ, where k>1 and ϵ>0. The solution is (logn)k= O(nϵ). My doubt is that if we take ϵ as0.000000000000000000000000000000.......000000000001 that is very small wont this result be wrong?? Can someone check? Algorithms algorithms + – sushmita asked Dec 6, 2016 edited Dec 6, 2016 by Prashant. sushmita 430 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Prashant. commented Dec 6, 2016 reply Follow Share (logn)k= (nϵ) Take log both side k. loglogn = ϵ logn so k. loglogn = O(ϵ.logn) means k. loglogn$\leqslant$ C ϵ.logn [ here c is some constant] 1 votes 1 votes sushmita commented Dec 6, 2016 reply Follow Share but ϵ can be fraction lie 0.00000000000000000000000000000000000000...........1. will it not make ϵ log n very small? 0 votes 0 votes Prashant. commented Dec 6, 2016 reply Follow Share constant C will take care of it. C can be 10000000........000000000000000000 0 votes 0 votes sushmita commented Dec 6, 2016 reply Follow Share oh yeah wow awesome. thanku so much for the explanation. 0 votes 0 votes Please log in or register to add a comment.