For Problem #17, is f(n) = (log n)/2 okay?
It does not satisfy the requirement of the problem. You are given that f(n) grows asymptotically more slowly than log n. In the homework it states this formally by saying that f(n) = o(log n). Note that is “little-oh” NOT “big-oh”. Let me formally define little-oh for those who have not seen it. We say that f(n)=o(g(n)) if limn -> infinity [f(n)/g(n)] = 0. In other words, f(n) grows asymptotically slower than g(n). So (log n)/2 != o(log n) since the limit as n goes to infinity of 1/2 (not 0).
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