Since $P(n) = O(n^{\deg P})$, is $\log P(n) = O(P(\log n))$? Or when does this occur?
Does (or when does) $O(\log\text{poly}(n)) = O(\text{poly}(\log(n)))$?
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$\begingroup$
polynomials
algorithms
asymptotics
computational-complexity
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0Hint: is $\log^2(n)\in O(\log P(n))$ for any polynomial $P$? – 2017-02-17
1 Answers
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Hint:
$$\lim_{n \to \infty} \frac{\log P(n)}{\log(n)} = \operatorname{deg}(P) < \infty$$
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0What is that limit called and does it mean that $\log P(n) = O(\log (n))$ since $\deg P$ is a constant? – 2017-02-17
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1@FruitfulApproach I take back my last comment. The notation seems fine to me. Think of the numerator as a fixed polynomial, say $4x^2 + 4x+2$, which also means $\operatorname{deg} P$ is fixed. Oh, and note that whenever $\lim_{n \to \infty} \frac{f(x)}{g(x)} < \infty$ then we have $f(x) \in O(g(x))$ – 2017-02-17
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1@FruitfulApproach Oh, and note my comment said the limit had to be zero before I corrected it. I mixed up little-O and big-O in my head for a bit. All we need is for the limit to be finite, as we have here. This means that $\log(P(n)) = O(\log(n))$ – 2017-02-17