differentiate:f'= (cn-logn)'=c-1/n. With n>n0=1/c ->f'>0->increasing function. Therefore, n>n0 ,logn<cn means logn=O(n). Do similarly, we cannot find c1 that c1n<logn when n >n0'. because logn<>omega(n)->logn<>theta(n)
On Tue, Feb 2, 2010 at 12:51 PM, Banoo <[email protected]> wrote: > hi all, > can you help me solve the following question. > > Is log n = O(n)? Is log n = omega(n)? Is log n = theta (n)? > > > Thanks > Banoo > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
