Reputation: 369
I couldn't prove this:
f(n) = O(g(n))
implies f(n)^k = O(g(n)^k)
where k is element of the natural, positiv numbers
I've found similar examples on the internet. But I'm not sure if it's right to implement those solutions for this example.
Upvotes: 1
Views: 1313
Reputation: 7996
Go back to the definition of big-o.
f(n) = O(g(n)) <=> \exists M \in R+,
\exists n_0 \in N,
such that:
\forall n > n_0
|f(n)| < M.|g(n)|
It is obvious that if k > 0
then |f(n)|^k < (M.|g(n)|)^k
.
If k < 0
, the relation is inversed.
Upvotes: 1