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About LG (n!) Asymptotically compact supremum of

2022-06-21 22:40:00 【ash062】

$lg^{n!} = lg^{2} + lg^{3} + lg^{4} + lg^{5} + lg^{6} + lg^{7} +... + lg^{n}$

hypothesis n by 2 The power of , And $k = lg^{n}$

Reduced equation $lg^{n!}> lg^{2} + lg^{2} + lg^{4} + lg^{4} + lg^{4} + lg^{4} +... + lg^{n}$

remember $min = lg^{2} + lg^{2} + lg^{4} + lg^{4} + lg^{4} + lg^{4} +... + lg^{n} = \sum_{i = 1}^{k - 1}i\cdot 2^{i} + k$

Further treatment $min = 2min - min=(k - 1)2^{k} - \sum_{i = 1}^{k - 1}2^{i} + k=(k - 3)2^{k} + k + 2$

therefore $lg^{n!} = \Omega (nlg^{n})$

Similar to the amplification equation $lg^{n!} < lg^{2} + lg^{4} + lg^{4} + lg^{8} + lg^{8} + lg^{8} + lg^{8} + ... + lg^{n}$

remember $max = lg^{2} + lg^{4} + lg^{4} + lg^{8} + lg^{8} + lg^{8} + lg^{8} +... + lg^{n} = \sum_{i = 1}^{k}i2^{i - 1}$

Yes $max = 2max - max=k\cdot 2^{k} - \sum_{i = 1}^{k - 1}2^{i} = (k - 1)2^{k} + 2$

therefore $lg^{n!} = \bigcirc (nlg^{n})$

Sum up $lg^{n!} = \Theta (nlg^{n})$

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