0 0 votes Is sqrt(n)+ log n = omega(log n)? If yes then please explain why...? Algorithms algorithms recurrence-relation + – dhingrak 740 views answer comment Share Follow Print See 1 comment 1 1 comment reply dhingrak commented Nov 26, 2014 reply Follow flag But for larger n sqrt is greater than log n 1 1 replyShare Please log in or register to add a comment.
–1 –1 vote O(f(x) + g(x))= max {O(f(x)),O(g(x))} by putting n =15 i find log n is greater than n^1/2 so i think you are right its O(log n) Bhagirathi answered Nov 26, 2014 Bhagirathi comment Share Follow See 1 comment 1 1 comment reply Arjun commented Nov 27, 2014 reply Follow flag but questions is for omega- not big O. Also, try n = 2^100 :) 0 0 replyShare Please log in or register to add a comment.
