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$$O(\epsilon^{-2} * log(2/\delta)) = O(\epsilon^{-2} * log(1/\delta)) \,\,\,\,\,\;\;\; for \;0 <\epsilon, \delta < 1 $$ is this true? If we apply some logarithm laws to the first Big O we get: $$O(\epsilon^{-2}*log(2) + \epsilon^{-2} * log(1/\delta))$$ where it would seem to me the equation only holds if $log(1/\delta) \ge log(2) $, but maybe some Big-O-magic makes it true for any $\delta$?
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Tue, 21 Apr 2020 18:15 GMT