algorithm - Compare growth rate: n·lg(n) and 0.02·n^(1.01). Which one grows faster? -


comparing n·lg(n) , 0.02·n^(1.01), 1 grows faster?

i write n^(1.01) n·n^(0.01).

doing that, question becomes then: how compare lg(n) , n^0.01.

but don't know 1 of lg(n) , n^0.01 grows faster.

how solve problem?

the logarithm grows slower positive power. assuming lg decimal logarithm, 0.02 n^(1.01) exceed n lg(n) @ n ~= 4.04192e+433 (see wolphram alpha query). if it's practical problem computational complexity though, it's reasonable n values, 0.02 n^(1.01) algorithm faster n lg(n) algorithm.


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