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    <p>To show that n<sup>k</sup> is O(2<sup>n</sup>), note that</p> <blockquote> <p>n<sup>k</sup> = (2<sup>lg n</sup>)<sup>k</sup> = 2<sup>k lg n</sup></p> </blockquote> <p>So now you want to find an n<sub>0</sub> and c such that for all n &ge; n<sub>0</sub>, </p> <blockquote> <p>2<sup>k lg n</sup> &le; c 2<sup>n</sup></p> </blockquote> <p>Now, let's let c = 1 and then consider what happens when n = 2<sup>m</sup> for some m. If we do this, we get</p> <blockquote> <p>2<sup>k lg n</sup> &le; c 2<sup>n</sup> = 2<sup>n</sup></p> <p>2<sup>k lg 2<sup>m</sup></sup> &le; 2<sup>2<sup>m</sup></sup></p> <p>2<sup>km</sup> &le; 2<sup>2<sup>m</sup></sup></p> </blockquote> <p>And, since 2<sup>n</sup> is a monotonically-increasing function, this is equivalent to</p> <blockquote> <p>km &le; 2<sup>m</sup></p> </blockquote> <p>Now, let's finish things off. Let's suppose that we let m = max{k, 4}, so k &le; m. Thus we have that</p> <blockquote> <p>km &le; m<sup>2</sup></p> </blockquote> <p>We also have that</p> <blockquote> <p>m<sup>2</sup> &le; 2<sup>m</sup></p> </blockquote> <p>Since for any m &ge; 4, m<sup>2</sup> &le; 2<sup>m</sup>, and we've ensured by our choice of m that m = max{k, 4}. Combining this, we get that</p> <blockquote> <p>km &le; 2<sup>m</sup></p> </blockquote> <p>Which is equivalent to what we wanted to show above. Consequently, if we pick any n &ge; 2<sup>m</sup> = 2<sup>max{4, k}</sup>, it will be true that n<sup>k</sup> &le; 2<sup>n</sup>. Thus by the formal definition of big-O notation, we get that n<sup>k</sup> = O(2<sup>n</sup>).</p> <p>I <em>think</em> this math is right; please let me know if I'm wrong!</p> <p>Hope this helps!</p>
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    1. POProving big O of statement
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    1. COThanks.But will this work for any k? In case of km<=m^2 - is this for any k?
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    2. CO@user629034- Yes, this works for any k. The idea is to pick the value of m (and thus of n0) based on the value of k that's picked.
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