We know that log(n) = O(sqrt n ) I am wondering if is it valid to say that log(n) is theta( sqrt n ) . numerically , i proved that it is right ; yet i am not too sure about it . Would like some help
question regarding asymptotic runtime behavior
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log nis NOT inTheta(sqrt n), sincesqrt nis asymptotically greater thanlog n, meaning thatlog nisn't inOmega(sqrt n). In other words,sqrt ncannot be an asymptotic lower bound forlog n.Refer to this definition of big theta. Substitute
sqrt nforg(n)andlog nforf(n)in the definition and you will see that you can easily find ak2andn0such that the definition is satisfied (which is whylog nis inO(sqrt n)), while finding a suitablek1will prove impossible (which is whylog nis NOT inOmega(sqrt n)).