Section 3.2 #42

42. Suppose that f (x) is O(g(x)). Does it follow that 2^f (x) is O(2^g(x))?

f (x) is O(g(x)) or f(x) ≤ g(x)
Then 2^ f(x) ≤ 2^g(x)
Therefore, 2^f(x) is O(2^g(x))

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3 Responses to Section 3.2 #42

  1. Kate Poirier says:

    This is kind of a tricky question, huh? Andy, Eric is absolutely right. It looks like in your proof you’re assuming that f(x) \leq  g(x), but this doesn’t have to be true. The statement that f(x) is O(g(x)) is strictly weaker, so you can’t use the simple inequality the way that you want to.

    Eric’s link contains a nice example of functions f(x) and g(x) such that f(x) is O(g(x)) but 2^{f(x)} is not O(2^{g(x)}).

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