there is 2 functions: f(n) = n + log n and g(n) = n√n
if f(n) = O(g(n)):
n + log n <= C * n√n
else if g(n) = O(f(n)):
n√n <= C(n + log n)
stuck to prove that
n + log n <= C * n√n
n√n <= C(n + log n)
stuck to prove that
To prove
n + log(n) <= C*n*sqrt(n)
divide by n
to get
1 + log(n)/n <= C*sqrt(n)
As log(n)/n
goes towards zero when n
goes towards infinity, you get
1 <= C*sqrt(n) for n -> infinity
which is true.