3 votes 3 votes The recurrence relation $T(n) =mT(\frac{n}{2}) + tan^2$ is satisfied by O(n$^2$) O(n$^{lg \, m}$) O(n$^2 \: lg \:n)$ O(n lg n) Algorithms ugcnetcse-dec2013-paper3 algorithms asymptotic-notation recurrence-relation + – go_editor asked Jul 28, 2016 edited Jul 28, 2016 by Prashant. go_editor 2.8k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Prashant. commented Jul 28, 2016 reply Follow Share incomplete. 0 votes 0 votes Kapil commented Jul 28, 2016 reply Follow Share i think its wrong... 0 votes 0 votes Please log in or register to add a comment.
7 votes 7 votes Now according to Master Theorem a=m , b=2 and f(n)=an2 nlog2m > f(n) for all m>4 [Using case-1: Master Theorem] so solution is O(nlog2m) Ans : B Note : inplace of Sanjay Sharma answered Jul 28, 2016 Sanjay Sharma comment Share Follow See 1 comment See all 1 1 comment reply Rams B commented Jun 12, 2018 reply Follow Share Sir, I am confused as to how can we allow a variable to be taken as the value of a(=m). Also, how is m>4 and how do we compare m with 2 so as to check if it is >,< or = ? Can you please explain it 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes we can not answer ambiguous question Prateek kumar answered Nov 15, 2016 Prateek kumar comment Share Follow See all 0 reply Please log in or register to add a comment.