I have it as a matter of homework, and I do not remember how it was studied in the classroom. Can someone point me in the right direction or have documentation on how to solve these problems?
You can prove this by applying the L'Hopitals rule to lim n-> infinity 5n / nlogn
g (n) = 5n and f (n) = nlogn
Print g (n) and f (n) so you get something like this
5 / (some things here will contain n)
5 / infinity = 0, so 5n = O (nlogn) is true
More formally ...
, f(n) = 5n, f ∈ O(n). , , k i ≥ k, f(i) ≤ ci. , c = 5 .
f(n) = 5n
f ∈ O(n)
k
i ≥ k
f(i) ≤ ci
c = 5
, , f ∈ O(n) f ∈ O(n * log n). , , k i ≥ k, f(i) ≤ ci * log i. , k , f(i) ≤ ci i ≥ 2, , ci ≤ ci * log i.
f ∈ O(n * log n)
f(i) ≤ ci * log i
i ≥ 2
ci ≤ ci * log i
QED.
O-. , 5n nlogn, . nlogn - .
, , , :
c 1 * 5 * n < c 2 * n * logn, n > c 3
c 1 c 2 - , c 3. Single c 3 in terms of c 1 and c 2and you're done.
Three years have passed since I touched the things of big O. But I think you can try to show this:
O (5n) = O (n) = O (nlogn)
O (5n) = O (n) is very easy to show, so all you have to do is show O (n) = O (nlogn), which should not be too complicated either.