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Proofing Big-O Notation: O(max(f(n), g(n)))
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[QUOTE="Iyan Makusa, post: 6224859, member: 659925"] ##h(n) + i(n) = O(max(f(n),g(n)))## if and only if there exists ##C_3 > 0, n_3 > 0## such that ##|h(n) + i(n)| \leq C_3max(f(n), g(n))~\forall{n>n_3}## I think I can start with that. \begin{align} |h(n) + i(n)| &\leq C_3max(f(n),g(n))\\ |h(n) + i(n)| &\leq C_1f(n) + C_2g(n)\\ &\leq max(C_1, C_2)(f(n) + g(n))\\ &\leq C_4max(C_1, C_2)max(f(n), g(n)) \end{align} where ##C_4max(C_1, C_2) = C_3, n > n_3##. I forgot early on the proof that ##n_3## would be ##max(n_1, n_2)## where ##h(n), n > n_1; i(n), n > n_2##. So whichever function is the "farther" one would be compared to by ##n##. Since we have found values for ##C_3, n_3## then the proof holds? ##\blacksquare{?}## I'm still not sure about this. I got more help from a slide I found. [/QUOTE]
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Proofing Big-O Notation: O(max(f(n), g(n)))
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