-2

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

lukalortk
  • 11
  • 1
  • 4

1 Answers1

0

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.

Support Ukraine
  • 42,271
  • 4
  • 38
  • 63