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We are not disagreeing that O(1/n) functions (in the mathematical sense) exist Obviously they do But computation is inherently discrete Something that has a lower bound, such as the running time of a program on either the von Neumann architecture or a purely abstract Turing machine cannot be O(1/n) Equivalently, something that is O(1/n) cannot have a lower. ¾iiiuuu ¡11 u vuh0h Ói @iuhdh i «. F jFûG G F¸&k 2 ©.

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Pptx Author tsuto Created Date 3/2/ 1237 PM. µYR {Y e { ¹Z¿. In a previous problem, I showed (hopefully correctly) that f(n) = O(g(n)) implies lg(f(n)) = O(lg(g(n))) with sufficient conditions (eg, lg(g(n)) >= 1, f(n) >= 1, and sufficiently large n) Now, I need to prove OR disprove that f(n) = O(g(n)) implies 2^(f(n)) = O(2^g(n)))Intuitively, this makes sense, so I figured I could prove it with help from the previous theorem.

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