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Where c is constant.
asked | 334 views

I am getting $\Theta$ (2^n) .

answered by Boss (7k points)
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You should use $\Theta$ though $O$ is not wrong here :)
Done.
Very nicely explained. Thanks for the answer. :)
Sir ,how to get the intution while solving a recurrence whether we should use theta or Big-O ?And why  here theta is more appropriate ?
Unless we approximate we always get $\Theta$. Theta says = while O says <=. So, Theta is more appropriate if it can be used.
But theta(n) is used when T(n) =Omega(n) as well as O(n) , so seeing the above solution how to conclude that both of them are asymptotically same .
Here, it is other way. In the proof all statements were "=". So, finally we can say it is $\Theta(2^n)$ and hence $O(2^n)$ and $\Omega(2^n)$.