0 votes 0 votes √logx = O(loglogx) is it true or false? and explain why? Algorithms asymptotic-notation + – suneetha asked Oct 29, 2018 suneetha 297 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Naveen Kumar 3 commented Oct 29, 2018 i edited by Naveen Kumar 3 Oct 29, 2018 reply Follow Share False. for x>0 =>$\sqrt{\log x}$ > $\log\left(\log x\right)$ 0 votes 0 votes akash.dinkar12 commented Oct 29, 2018 reply Follow Share take x = 21024 √logx = √log221024 = 2512 log2log2x = log2log2 21024 =log21024 =10 So we can say that √logx > log2log2x and hence ur given statement is false 1 votes 1 votes Hemanth_13 commented Oct 29, 2018 reply Follow Share √logx= O(loglogx) apply log on both sides 1/2* logx <= logloglogx which is false 0 votes 0 votes kumar.dilip commented Oct 29, 2018 reply Follow Share False. √logx > log2log2x 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes I hope this makes it clear. preeti0448 answered May 2, 2022 preeti0448 comment Share Follow See all 0 reply Please log in or register to add a comment.