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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.

Elfie
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1 Answers1

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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.

fjardon
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  • considering the big-o notation, I've found another interesting problem. http://stackoverflow.com/questions/23492509/big-o-notation-algebra/23493414#23493414 I only need to know if the prove of the second question is correct? – Elfie Apr 27 '16 at 12:02