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    <p>The math being done here is as follows. When you say O(n(n + n<sup>2</sup>)), that's equivalent to saying O(n<sup>2</sup> + n<sup>3</sup>) by simply distributing the n throughout the product.</p> <p>The reason that O(n<sup>2</sup> + n<sup>3</sup>) = O(n<sup>3</sup>) follows from the formal definition of big-O notation, which is as follows:</p> <blockquote> <p>A function f(n) = O(g(n)) iff there exists constants n<sub>0</sub> and c such that for any n &ge; n<sub>0</sub>, |f(n)| &le; c|g(n)|.</p> </blockquote> <p>Informally, this says that as n gets arbitrary large, f(n) is bounded from above by a constant multiple of g(n).</p> <p>To formally prove that n<sup>2</sup> + n<sup>3</sup> is O(n<sup>3</sup>), consider any n &ge; 1. Then we have that</p> <blockquote> <p>n<sup>2</sup> + n<sup>3</sup> &le; n<sup>3</sup> + n<sup>3</sup> = 2n<sup>3</sup></p> </blockquote> <p>So we have that n<sup>2</sup> + n<sup>3</sup> = O(n<sup>3</sup>), with n<sub>0</sub> = 1 and c = 2. Consequently, we have that </p> <blockquote> <p>O(n(n + n<sup>2</sup>)) = O(n<sup>2</sup> + n<sup>3</sup>) = O(n<sup>3</sup>).</p> </blockquote> <p>To be truly formal about this, we would need to show that if f(n) = O(g(n)) and g(n) = O(h(n)), then f(n) = O(h(n)). Let's walk through a proof of this. If f(n) = O(g(n)), there are constants n<sub>0</sub> and c such that for n &ge; n<sub>0</sub>, |f(n)| &le; c|g(n)|. Similarly, since g(n) = O(h(n)), there are constants n'<sub>0</sub>, c' such that for n &ge; n'<sub>0</sub>, g(n) &le; c'|h(n)|. So this means that for any n &ge; max(c, c'), we have that</p> <blockquote> <p>|f(n)| &le; c|g(n)| &le; c|c'h(n)| = c x c' |h(n)|</p> </blockquote> <p>And so f(n) = O(h(n)).</p> <p>To be a bit more precise - in the case of the algorithm described here, the authors are saying that the runtime is &Theta;(n<sup>3</sup>), which is a stronger result than saying that the runtime is O(n<sup>3</sup>). &Theta; notation indicates a tight asymptotic bound, meaning that the runtime grows at the same rate as n<sup>3</sup>, not just that it is bounded from above by some multiple of n<sup>3</sup>. To prove this, you would also need to show that n<sup>3</sup> is O(n<sup>2</sup> + n<sup>3</sup>). I'll leave this as an exercise to the reader. :-)</p> <p>More generally, if you have <em>any</em> polynomial of order k, that polynomial is O(n<sup>k</sup>) using a similar argument. To see this, let P(n) = &sum;<sub>i=0</sub><sup>k</sup>(a<sub>i</sub>n<sup>i</sup>). Then, for any n &ge; 1, we have that</p> <p>&sum;<sub>i=0</sub><sup>k</sup>(a<sub>i</sub>n<sup>i</sup>) &le; &sum;<sub>i=0</sub><sup>k</sup>(a<sub>i</sub>n<sup>k</sup>) = (&sum;<sub>i=0</sub><sup>k</sup>(a<sub>i</sub>))n<sup>k</sup></p> <p>so P(n) = O(n<sup>k</sup>).</p> <p>Hope this helps!</p>
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    1. POHelp calculating Big O
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    1. POHelp calculating Big O
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    1. COThank you, your answer is truly mathematical. I have to go back and review a lot from my days in comp sci school in order to understand this. I'll get it eventually!
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    2. COIn your last equation, why do you assume that i is always less than k?? I mean, what if i=1000 and k=100
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      1. USIgor Ganapolsky
    3. CO@IgorG.- The variable i is a counter in the summation, so by definition it ranges from 0 to k. The idea is that a_i n^i represents one term in the polynomial. For example, in 2x^3 + x^2 - 4x - 8, the sequence a_i is 2, 1, -4, -8.
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