
I have to arrange the above time complexity function in increasing order of asymptotic complexity and indicate if there exist functions that belong to the same order.
So, my answer is
$[lg(n)]^2$
$lg(2^n)$ and $log(2^n)$ and $2^{lg(n)}$ these belong to the same order of O(n)
$nlog(n) + 1000 $
$n^{1.1}$
Just wish to check whether i'm right, since I'm not getting the same result as others. Thanks.
I am familiar with the question. (I am one of the "others") I believe you neglected to specify that lg is log base 2.
In increasing order of asymptotic complexity:
$lg^2(n) = O(lg^2(n))$
$lg(2^n) = n = O(n)$
$log(2^n) = n * log(2) = O(n)$
$2^{lg(n)} = n = O(n)$
$n*log(n)+1000 = O(n log(n))$
$n^{1.1} = O(n^{1.1})$