Gear2d
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I have this problem which says:
Show using rules of Theorem how to find the O notation of: (Note lg = log based 2)
5n^2*lg(n) + 20nsqrt(n) + lg^3(n) + 100/n = O(n^2 * lg(n) )
here is a link of the theorems:
http://www.augustana.ca/~hackw/csc210/exhibit/chap04/bigOhRules.html
This is what I tried to do to prove that it is O(n^2 * lg(n) ):
1) lg(n) => O(n) or O(lg(n)) [ depends if you look at Comparison Rule 2 or Log of a Power Rule]
2) 5n^2 => O(n^2)
steps 1 and 2 combined give O(n^2 * lg(n))
Steps 3, 4 and 5 I am not sure if I am right
3) 20nsqrt(n) = 20n^(3/2) => O(n^(3/2))
4) lg^3(n) => O(n) [Comparison Rule 2]
5) 100/n => O(1/n) [complete guess here]
6) Using Sum Rule and Polynomial Rule:
O(n^2 * lg(n) + n^(3/2) + n + 1/n ) = O(n^2 * lg(n))
I do not fill comfortable with steps 3 to 5 with the orders. Are all the steps correct?
Show using rules of Theorem how to find the O notation of: (Note lg = log based 2)
5n^2*lg(n) + 20nsqrt(n) + lg^3(n) + 100/n = O(n^2 * lg(n) )
here is a link of the theorems:
http://www.augustana.ca/~hackw/csc210/exhibit/chap04/bigOhRules.html
This is what I tried to do to prove that it is O(n^2 * lg(n) ):
1) lg(n) => O(n) or O(lg(n)) [ depends if you look at Comparison Rule 2 or Log of a Power Rule]
2) 5n^2 => O(n^2)
steps 1 and 2 combined give O(n^2 * lg(n))
Steps 3, 4 and 5 I am not sure if I am right
3) 20nsqrt(n) = 20n^(3/2) => O(n^(3/2))
4) lg^3(n) => O(n) [Comparison Rule 2]
5) 100/n => O(1/n) [complete guess here]
6) Using Sum Rule and Polynomial Rule:
O(n^2 * lg(n) + n^(3/2) + n + 1/n ) = O(n^2 * lg(n))
I do not fill comfortable with steps 3 to 5 with the orders. Are all the steps correct?