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asymptotic notations

0 votes
86 views
√logx  = O(loglogx) is it true or false?

and explain why?
in Algorithms 86 views
0
False. for x>0 =>$\sqrt{\log x}$ > $\log\left(\log x\right)$

edited by
1

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

0
√logx= O(loglogx)

apply log on both sides

1/2* logx <= logloglogx

which is false
0

False.

 √logx > log2log2x 

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