1 votes 1 votes Is ln(n!)=theta(n ln(n))? Algorithms algorithms asymptotic-notation + – nbhatt asked Nov 3, 2022 nbhatt 373 views answer comment Share Follow See 1 comment See all 1 1 comment reply nishantsharma commented Jan 11, 2023 reply Follow Share Yes, it is equal according to Stirling’s approximation. As far as GATE is concerned proof is not necessary, 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes NO! it must be $O$. because $n^n$ is upperbound of $n!$ i.e $n!$ $=$ $O(n^n)$ Pranavpurkar answered Nov 3, 2022 Pranavpurkar comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes yes it is this is from Cormen $it\ is\ \Theta \ since\ when\ an\ algorithm's\ upper\ limit\ and\ lower\ limit\ can\ be\ represented,\ we\ do\ so\ in \ \Theta \\$ $ let\ n=4$ $n!=o(n^{n})$ $ 4!<4^{4 }$ $ n!=\omega (2^{n})$ $ 4!>2^{4}$ i hope it’s clear now afroze answered Nov 3, 2022 afroze comment Share Follow See all 0 reply Please log in or register to add a comment.