0 votes 0 votes F(n)=n^(sin n) G(n)=n^(cos n) Why they are non comparable. Algorithms asymptotic-notation algorithms + – Shashi Shekhar 1 asked Aug 1, 2018 retagged Jun 18, 2022 by makhdoom ghaya Shashi Shekhar 1 328 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes 1) let take n=90 ====> F(n) is grater than G(n) 2) let take n=360 ====> G(n) is grater than F(n) these are repeating right, i mean you can't tell which one grater among them after any c>0 therefore they are non-comparable. Shaik Masthan answered Aug 2, 2018 Shaik Masthan comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes When put the value of n . Then some time n^sinn is grater. Then n^cosn and some time n^cosn is grayer then n^sinn. So we can not compair them abhishekmehta4u answered Aug 2, 2018 abhishekmehta4u comment Share Follow See all 0 reply Please log in or register to add a comment.